SEAOC Seismic Design Manual Examples

Seismic Design Manual Volume I Code Application Examples

Copyright Copyright © 1999 Structural Engineers Association of California. All rights reserved. This publication or any part thereof must not be reproduced in any form without the written permission of the Structural Engineers Association of California.

Publishe Structural Engineers Association of California (SEAOC) 555 University Avenue, Suite 126 Sacramento, California 95825-6510 Telephone: (916) 427-3647; Fax: (916) 568-0677 E-mail: [email protected]; Web address: www.seaint.org The Structural Engineers Association of California (SEAOC) is a professional association of four regional member organizations (Central California, Northern California, San Diego, and Southern California). SEAOC represents the structural engineering community in California. This document is published in keeping with SEAOC’s stated mission: “to advance the structural engineering profession; to provide the public with structures of dependable performance through the application of state-of-the-art structural engineering principles; to assist the public in obtaining professional structural engineering services; to promote natural hazard mitigation; to provide continuing education and encourage research; to provide structural engineers with the most current information and tools to improve their practice; and to maintain the honor and dignity of the profession.”

Editor Gail Hynes Shea, Albany, California, [email protected]

Disclaime Practice documents produced by the Structural Engineers Association of California (SEAOC) and/or its member organizations are published as part of our association’s educational program. While the information presented in this document is believed to be correct, neither SEAOC nor its member organizations, committees, writers, editors, or individuals who have contributed to this publication make any warranty, expressed or implied, or assume any legal liability or responsibility for the use, application of, and/or reference to opinions, findings, conclusions, or recommendations included in this publication. The material presented in this publication should not be used for any specific application without competent examination and verification of its accuracy, suitability, and applicability by qualified professionals. Users of information from this publication assume all liability arising from such use.

Table of Contents

Preface .......................................................................... ................................... v Acknowledgments..................................................................... ..................................vi Introduction .......................................................................... ................................... 1 Notation .......................................................................... ................................... 3 Example 1 Earthquake Load Combinations: Strength Design ............................................... §1612.2...................... 7 Example 2 Combinations of Loads .................................... §1612.3.................... 12 Example 3 Seismic Zone 4 Near-Source Factor ................ §1629.4.2................. 17 Introduction to Vertical Irregularities ....................................... §1629.5.3................. 20 Example 4 Vertical Irregularity Type 1 ............................. §1629.5.3................. 21 Example 5 Vertical Irregularity Type 2 ............................. §1629.5.3................. 24 Example 6 Vertical Irregularity Type 3 ............................. §1629.5.3................. 26 Example 7 Vertical Irregularity Type 4 ............................. §1629.5.3................. 28 Example 8 Vertical Irregularity Type 5 ............................. §1629.5.3................. 30 Example 9 Vertical Irregularity Type 5 ............................. §1629.5.3................. 32 Introduction to Plan Irregularities ............................................. §1629.5.3................. 36 Example 10 Plan Irregularity Type 1 ................................... §1629.5.3................. 37 Example 11 Plan Irregularity Type 2 ................................... §1629.5.3................. 41 Example 12 Plan Irregularity Type 3 ................................... §1629.5.3................. 43 Example 13 Plan Irregularity Type 4 ................................... §1629.5.3................. 45 Example 14 Plan Irregularity Type 5 ................................... §1629.5.3................. 46 Example 15 Reliability/Redundancy Factor ρ ..................... §1630.1.1................. 47 Example 16 Reliability/Redundancy Factor Applications... §1630.1.1................. 52 Example 17 P∆ Effects......................................................... §1630.1.3................. 56 Example 18 Design Base Shear ........................................... §1630.2.1................. 59 Example 19 Structure Period Using Method A.................... §1630.2.2................. 61 Example 20 Simplified Design Base Shear.......................... §1630.2.3................. 65 Example 21 Combination of Structural Systems: Vertical... §1630.4.2................. 68 Example 22 Combination of Structural Systems: Along Different Axes....................................... §1630.4.3................. 71 Example 23 Combination of Structural Systems: Along the Same Axis ....................................... §1630.4.4................. 73 Example 24 Vertical Distribution of Force .......................... §1630.5.................... 74 Example 25 Horizontal Distribution of Shear...................... §1630.6.................... 76 Example 26 Horizontal Torsional Moments ........................ §1630.7.................... 81

SEAOC Seismic Design Manual

Table of Contents

Table of Contents (continued) Example 27 Example 28 Example 29 Example 30 Example 31 Example 32 Example 33 Example 34 Example 35 Example 36 Example 37 Example 38 Example 39 Example 40 Example 41 Example 42 Example 43 Example 44 Example 45 Example 46 Example 47 Example 48 Example 49 Example 50 Example 51 Example 52 Example 53 Example 54 Example 55

Elements Supporting Discontinuous Systems.. §1630.8.2.................85 Elements Supporting Discontinuous Systems.. §1630.8.2.................88 At Foundation................................................... §1630.8.3.................90 Drift .................................................................. §1630.9....................96 Story Drift Limitations ..................................... §1630.10..................98 Vertical Component ......................................... §1630.11................100 Design Response Spectrum .............................. §1631.2..................101 Dual Systems.................................................... §1631.5.7...............104 Lateral Forces for One-Story Wall Panels........ §1632.2..................107 Lateral Forces for Two-Story Wall Panel ........ §1632.2..................111 Rigid Equipment............................................... §1632.2..................116 Flexible Equipment .......................................... §1632.2..................118 Relative Motion of Equipment Attachments.... §1632.4..................121 Deformation Compatibility .............................. §1633.2.4...............123 Adjoining Rigid Elements................................ §1633.2.4.1............126 Exterior Elements: Wall Panel ......................... §1633.2.4.2............128 Exterior Elements: Precast Panel...................... §1633.2.4.2............131 Beam Horizontal Tie Force .............................. §1633.2.5...............138 Collector Elements ........................................... §1633.2.6...............139 Out-of-Plane Wall Anchorage to Flexible Diaphragm.......................................... §1633.2.8.1............142 Wall Anchorage to Flexible Diaphragms......... §1633.2.8.1............145 Determination of Diaphragm Force Fpx: Lowrise............................................................. §1633.2.9...............147 Determination of Diaphragm Force Fpx: Highrise ............................................................ §1633.2.9...............150 Building Separations ........................................ §1633.2.11.............152 Flexible Nonbuilding Structure........................ §1634.2..................154 Lateral Force on Nonbuilding Structure........... §1634.2..................157 Rigid Nonbuilding Structure ............................ §1634.3..................159 Tank With Supported Bottom .......................... §1634.4..................160 Pile Interconnections........................................ §1807.2..................161


Preface

This document is the initial volume in the three-volume SEAOC Seismic Design Manual. It has been developed by the Structural Engineers Association of Californi (SEAOC) with funding provided by SEAOC. Its purpose is to provide guidance on the interpretation and use of the seismic requirements in the 1997 Uniform Building Code (UBC), published by the International Conference of Building Official (ICBO), and SEAOC’s 1999 Recommended Lateral Force Requirements and Commentary (also called the Blue Book). The Seismic Design Manual was developed to fill a void that exists between the Commentary of the Blue Book, which explains the basis for the UBC seismic provisions, and everyday structural engineering design practice. The Seismic Design Manual illustrates how the provisions of the code are used. Volume I: Code Application Examples, provides step-by-step examples of how to use individual code provisions, such as how to compute base shear or building period. Volumes II and III: Building Design Examples, furnish examples of the seismic design of common types of buildings. In Volumes II and III, important aspects of whole buildings are designed to show, calculation-by-calculation, how the various seismic requirements of the code are implemented in a realistic design. SEAOC intends to update the Seismic Design Manual with each edition of the building code used in California.

Ronald P. Gallagher Project Manager


Acknowledgements

Authors The Seismic Design Manual was written by a group of highly qualified structural engineers. These individuals are both California registered structural engineers and SEAOC members. They were selected by a Steering Committee set up by the SEAOC Board of Directors and were chosen for their knowledge and experience with structural engineering practice and seismic design. The Consultants for Volumes I, II and III are Ronald P. Gallagher, Project Manager David A. Hutchinson Jon P. Kiland John W. Lawson Joseph R. Maffei Douglas S. Thompson Theodore C. Zsutty Volume I was written principally by Theodore C. Zsutty and Ronald P. Gallagher. Many useful ideas and helpful suggestions were offered by the other Consultants. Consultant work on Volumes II and III is currently underway.

Steering Committee Overseeing the development of the Seismic Design Manual and the work of the Consultants was the Project Steering Committee. The Steering Committee was made up of senior members of SEAOC who are both practicing structural engineers and have been active in Association leadership. Members of the Steering Committee attended meetings and took an active role in shaping and reviewing the document. The Steering Committee consisted of John G. Shipp, Chair Robert N. Chittenden Stephen K. Harris Maryann T. Phipps Scott A. Stedman


Acknowledgments

Reviewers A number of SEAOC members and other structural engineers helped check the examples in this volume. During its development, drafts of the examples were sent to these individuals. Their help was sought in both review of code interpretations as well as detailed checking of the numerical computations. The assistance of the following individuals is gratefully acknowledged Saeed R. Amirazizi Jefferson W. Asher Brent Berensen Donald A. Cushing Vincent DeVita Richard M. Drake Todd W. Erickson Daniel Fisher Kenneth Gebhar Edward R. Haninger

Thomas Hunt Mark S. Jokerst Isao M. Kawasaki John W. Lawson Ronald Lugue Robert Lyons Peter Maranian Brian McDonal Rory M. McGruer Brian Montes

Manuel Morden Farzad Naeim David A. Napoleon Josh Plummer Mehran Pourzanjani Ian Robertson John G. Shipp Donald R. Strand

Seismology Committee Close collaboration with the SEAOC Seismology Committee was maintained during the development of the document. The 1997-1998 and 1998-1999 Committees reviewed the document and provided many helpful comments and suggestions. Their assistance is gratefully acknowledged. 1998-1999

Saif M. Hussain, Chair Tom H. Hale, Past Chair Robert N. Chittenden Stephen K. Harris Douglas Hohbach Y. Henry Huang Saiful Islam Martin W. Johnson Jaiteerth B. Kinha Eric T. Lehmkuhl Simin Naaseh Hassan Sassi, Assistant to the Chair

1997-1998

Tom H. Hale, Chair Ali M. Sadre, Past Chair Robert N. Chittenden Stephen K. Harris Saif M. Hussain Saiful Islam Martin W. Johnson Eric T. Lehmkuhl Roumen V. Mladjov Simin Naaseh Carl B. Schulze Chris V. Tokas Joyce Copelan, Assistant to the Chair

Production and Art Special thanks are due Lenore Henry of R.P. Gallagher Associates, Inc. who input the entire text from handwritten copy, did all the subsequent word processing, drew al the figures, and formatted the entire document. Without her expertise, this project would never have come to fruition. SEAOC Seismic Design Manual

Suggestions for Improvement

Suggestions for Improvement In keeping with two of its Mission Statements: (1) “to advance the structura engineering profession” and (2) “to provide structural engineers with the most current information and tools to improve their practice”, SEAOC plans to update this document as seismic requirements change and new research and better understanding of building performance in earthquakes becomes available. Comments and suggestions for improvements are welcome and should be sent to the following: Structural Engineers Association of California (SEAOC) Attention: Executive Director 555 University Avenue, Suite 126 Sacramento, California 95825-6510 Telephone: (916) 427-3647; Fax: (916) 568-0677 E-mail: [email protected]; Web address: www.seaint.org

Errata Notification SEAOC has made a substantial effort to ensure that the information in this document is accurate. In the event that corrections or clarifications are needed, these will be posted on the SEAOC web site at http://www.seaint.org or on the ICBO website at http://ww.icbo.org. SEAOC, at its sole discretion, may or may not issue written errata.


Seismic Design Manual Volume I Code Application Examples

Introduction

Volume I of the SEAOC Seismic Design Manual: Code Application Examples deals with interpretation and use of the seismic provisions of the 1997 Uniform Building Code (UBC). The Seismic Design Manual is intended to help the reader understand and correctly use the UBC seismic provisions and to provide clear, concise, and graphic guidance on the application of specific provisions of the code. It primaril addresses the major seismic provisions of Chapter 16 of the UBC, with interpretation of specific provisions and examples highlighting their proper application. Volume I presents 55 examples that illustrate the application of specific seismic provisions of the UBC. Each example is a separate problem, or group of problems, and deals primarily with a single code provision. Each example begins with a description of the problem to be solved and a statement of given information. The problem is solved through the normal sequence of steps, each of which are illustrated in full. Appropriate code references for each step are identified in the right-hand margin of the page. The complete Seismic Design Manual will have three volumes. Volumes II and III will provide a series of seismic design examples for buildings illustrating the seismic design of key parts of common building types such as a large three-story wood frame building, a tilt-up warehouse, a braced steel frame building, and a concrete shear wal building. While the Seismic Design Manual is based on the 1997 UBC, there are some provision of SEAOC’s 1999 Recommended Lateral Force Provisions and Commentary (Blue Book) that are applicable. When differences between the UBC and Blue Book are significant, these are brought to the attention of the reader. The Seismic Design Manual is applicable in regions of moderate and high seismicity (e.g., Zones 3 and 4), including California, Nevada, Oregon, and Washington. It is intended for use by practicing structural engineers and structural designers, building departments, other plan review agencies, and structural engineering students.


How to Use This Document

The various code application examples of Volume I are organized in numerical order by 1997 UBC section number. To find an example for a particular provision of the code, look at the upper, outer corner of each page, or in the table of contents. Generally, the UBC notation is used throughout. Some other notation is also defined in the following pages, or in the examples. Reference to UBC sections and formulas is abbreviated. For example, “1997 UBC Section 1630.2.2” is given as §1630.2.2 with 1997 UBC being understood. “Formula (32-2)” is designated Equation (32-2) or just (32-2) in the right-hand margins. Throughout the document, reference to specific code provisions and equations (the UBC calls the latter formulas) is given in the right-hand margin under the category Code Reference. Similarly, the phrase “Table 16-O” is understood to be 1997 UBC Table 16-O. Generally, the examples are presented in the following format. First, there is a statement of the example to be solved, including given information, diagrams, and sketches. This is followed by the “Calculations and Discussion” section, which provides the solution to the example and appropriate discussion to assist the reader. Finally, many of the examples have a third section designated “Commentary.” In this latter section, comments and discussion on the example and related material are made. Commentary is intended to provide a better understanding of the example and/or to offer guidance to the reader on use of the information generated in the example. In general, the Volume I examples focus entirely on use of specific provisions of the code. No design is illustrated. Design examples are given in Volumes II and III. The Seismic Design Manual is based on the 1997 UBC, unless otherwise indicated. Occasionally, reference is made to other codes and standards (e.g., ACI 318-95 or 1997 NDS). When this is done, these documents are clearly identified.


Notation

The following notations are used in this document. These are generally consistent with that used in the UBC. However, some additional notations have also been added. AB

=

ground floor area of structure in square feet to include area covered by all overhangs and projections.

Ac

=

the combined effective area, in square feet, of the shear walls in the first story of the structure.

Ae

=

the minimum cross-sectional area in any horizontal plane in the first story, in square feet of a shear wall.

Ax

=

the torsional amplification factor at Leve x.

ap

=

numerical coefficient specified in §1632 and set forth in Table 16-O of UBC.

Ca

=

seismic coefficient, as set forth in Table 16-Q of UBC.

Ct

=

numerical coefficient given in §1630.2.2 of U BC.

Cv

=

seismic coefficient, as set forth in Table 16-R of UBC.

D

=

dead load on a structural element.

De

=

the length, in feet, of a shear wall in the first story in the direction parallel to the applied forces.

E, Eh, Em, Ev, Fi, Fn

=

earthquake loads set forth in §1630.1 of UBC.

Fx

=

design seismic force applied to Leve i, n or x, respectively.

Fp

=

design seismic force on a part of the structure.

Fpx

=

design seismic force on a diaphragm.

Ft

=

that portion of the base shear, V, considered concentrated at the top of the structure in addition to Fn.

Fa

=

axial stress.


Notation

Fy

=

specified yield strength of structural steel.

fc’

=

specified compressive strength of concrete.

fi

=

lateral force at Level i for use in Formula (30-10) of UBC.

fm’

=

specified compressive strength of masonry.

fp

=

equivalent uniform load.

fy

=

specified yield strength of reinforcing steel

g

=

acceleration due to gravity.

hi, hn,hx =

height in feet above the base to Leve i, n or x, respectively.

I

=

importance factor given in Table 16-K of UBC.

Ip

=

importance factor specified in Table 16-K of UBC.

L

=

live load on a structural element.

Level i =

level of the structure referred to by the subscript i. “i = 1” designates the first level above the base.

Level n =

that level that is uppermost in the main portion of the structure.

Level x =

that level that is under design consideration. “x = 1” designates the first level above the base.

Na

=

near-source factor used in the determination of Ca in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and slip rates as set forth in Tables 16-S and 16-U of UBC.

Nv

=

near-source factor used in the determination of Cv in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and slip rates as set forth in Tables 16-T and 16-U of UBC.

R

=

numerical coefficient representative of the inherent overstrength and global ductility capacity of lateral-forceresisting systems, as set forth in Table 16-N or 16-P of UBC.


Notation

r

=

a ratio used in determining ρ. See §1630.1 of UBC.

SA, SB, SC, SD, SE, SF

=

soil profile types as set forth in Table 16-J of UBC.

T

=

elastic fundamental period of vibration, in seconds, of the structure in the direction under consideration.

V

=

Vx

=

the total design lateral force or shear at the base given by Formula (30-5), (30-6), (30-7) or (30-11) of UBC. the design story shear in Story x.

W

=

the total seismic dead load defined in §1620.1.1 of UBC.

wi, wx =

that portion of W located at or assigned to Level i or x, respectively.

Wp

=

the weight of an element of component.

wpx

=

the weight of the diaphragm and the element tributary thereto at Level x, including applicable portions of other loads defined in §1630.1.1 of UBC.

Z

=

seismic zone factor as given in Table 16-I of UBC.

∆M

=

Maximum inelastic response displacement, which is the tota drift or total story drift that occurs when the structure is subjected to the Design Basis Ground Motion, including estimated elastic and inelastic contributions to the total deformation defined in §1630.9 of UBC.

∆S

=

Design level response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design seismic forces.

δi

=

horizontal displacement at Level i relative to the base due to applied lateral forces, f, for use in Formula (30-10) of UBC.

φ

=

capacity-reduction or strength-reduction factor.

ρ

=

Redundancy/reliability factor given by Formula (30-3) of UBC.

Ωo

=

Seismic force amplification factor, which is required to account for structural overstrength and set forth in Table 16-N of UBC.


Example 1 Earthquake Load Combinations: Strength Design

§1612.2

$%&'&

! "# #

This example demonstrates the application of the strength design load combinations that involve the seismic load E given in §1630.1.1. This will be done for the momentresisting frame structure shown below:

Zone 4 C a = 0.44 I = 1.0 ρ = 1.1 f 1 = 0.5 Snow load S = 0

A

B C

D

Beam A-B and Column C-D are elements of the special moment-resisting frame. Structural analysis has provided the following individual beam moments at A, and the column axial loads and moments at C due to dead load, office building live load, and lateral seismic forces. Dead Load D Beam Moment at A

Live Load L

Lateral Seismic Load Eh

100 kip-ft

50 kip-ft

120 kip-ft

Column C-D Axial Load

90 kips

40 kips

110 kips

Column Moment at C

40 kip-ft

20 kip-ft

160 kip-ft

Find the following:

Strength design moment at beam end A. Strength design axial load and moment at column top C.



§1612.2

Calculations and Discussion

Code Reference

Strength design moment at beam end A.

To determine strength design moments for design, the earthquake component E must be combined with the dead and live load components D and L . This process is illustrated below.

Determine earthquake load E: The earthquake load E consists of two components as shown below in Equation (30-1). E h is due to horizontal forces, and E v is due to vertical forces. E = ρE h + E v

§1630.1.1

(30-1)

The moment due to vertical earthquake forces is calculated as E v = 0.5C a ID = 0.5 (0.44 )(1.0)(100 ) = 22 k - ft

§1630.1.1

The moment due to horizontal earthquake forces is given as E h = 120 k - ft Therefore E = ρE h + E v = 1.1(120) + 22 = 154 k - ft

Apply earthquake load combinations: The basic load combinations for strength design (or LRFD) are given in §1612.2.1. For this example, the applicable equations are:

§1612.2.1

1.2 D + 1.0 E + f 1 L

(12-5)

0.9 D ± 1.0 E

(12-6)

Using Equation (12-5) and Equation (12-6), the strength design moment at A for combined dead, live, and seismic forces are determined. M A = 1.2 M D + 1.0 M E + f 1 M L = 1.2 (100) + 1.0 (154 ) + 0.5 (50) = 299 k - ft M A = 0.9 M D ± 1.0M E = 0.9 (100) ± 1.0 (154 ) = 244 k - ft or − 64 k - ft ∴ M A = 299 k - ft or − 64 k - ft



§1612.2

Specific material requirements: There are different requirements for concrete (and masonry) frames than for steel as follows. Structural Steel: Section 2210 specifies use of the load combinations of §1612.2.1 as given above without modification. Reinforced Concrete: Section 1909.2.3 specifies use of the load combinations of §1612.2.1, where Exception 2 requires the factor load combinations of Equation (12-5) and Equation (12-6) to be multiplied by 1.1 for concrete and masonry elements. ( Note: At the time of publication, April 1999, the 1.1 factor is under consideration for change to 1.0.) Therefore, for a reinforced concrete frame, the combinations are: 1.1 (1.2 D + 1.0 E + f 1 L ) = 1.32 D + 1.1E + 1.1 f 1 L

(12-5)

1.1 (0.9 D ± 1.0E ) = 0.99 D ± 1.1E

(12-6)

M A = 1.1 (299 k - ft ) = 328.9 k - ft M A = 1.1 (244 k - ft or − 64 k - ft ) = 268.4 k - ft or − 70.4 k - ft ∴ M A = 328.9 k - ft or − 70.4 k - ft for a concrete frame.

Strength design axial load and moment at column top C.

Determine earthquake load E: E = ρE h + E v

§1630.1.1 (30-1)

where E v = 0.5C a ID = 0.22 D

§1630.1.1

For axial load E = E h + E v = 1.1 (110 kips ) + 0.22 (90 kips ) = 140.8 kips For moment E = E h + E v = 1.1 (160k - ft ) + 0.22 (40k - ft ) = 184.8 k - ft



§1612.2

Apply earthquake load combinations:

§1630.1.1

1.2 D + 1.0 E + f 1 L

(12-5)

0.9 D ± 1.0 E

(12-6)

Design axial force PC at point C is calculated as PC = 1.2 D + 1.0 E + f 1 L = 1.2 (90) + 1.0 (140.8) + 0.5 (40) = 268.8 kips PC = 0.9 D ± 1.0 E = 0.9 (90 ) ± 1.0 (140.8) = 221.8 and − 59.8 kips ∴ PC = 268.8 kips compression, or 59.8 kips tension Design moment M C at point C is calculated as M C = 1.2 D + 1.0 E + f1L = 1.2 (40 k - ft ) + 1.0 (184.8 k - ft ) + 0.5 (20 k - ft ) = 242.8 k - ft M C = 0.9 D ± 1.0 E = 0.9 (40 k - ft ) ± 1.0(184.8 k - ft ) = 220.8 k - ft or − 148.8 k - ft ∴ M C = 242.8 k-ft or –148.8 k-ft Note that the column section capacity must be designed for the interaction of PC = 268.8 kips compression and M C = 242.8 k-ft (for dead, live and earthquake), and the interaction of PC = 59.8 kips tension and M C = −148.8 k-ft (for dead and earthquake).

Specific material requirements Structural Steel: Section 2210 specifies the use of the load combinations of §1612.2.1 as given above without modification.

§1630.1.1

Reinforced Concrete: The axial force PC and the moment M C must be multiplied by 1.1 per §1612.2.1.

Commentary Use of strength design requires consideration of vertical seismic load E v . When allowable stress design is used, the vertical seismic load E v is not required under §1630.1.1.



§1612.2

The incorporation of E v in the load combinations for strength design has the effect of increasing the load factor on the dead load action D. For example, consider the load combination of Equation (12-5) 1.2 D + 1.0 E + ( f 1 L + f 2 S )

(12-5)

where E = ρE h + E v and E v = 0.5C a ID this becomes 1.2 D + 1.0 (0.5C a ID + ρ E h ) + ( f 1 L + f 2 S )

(1.2 + 0.5C a I ) D + 1.0ρ E h + ( f 1 L +

f2S)

in the numerical example 0.5C a I = 0.22 Thus, the total factor on D is 1.2 + 0.22 = 1.42 For the allowable stress design load combinations of §1612.3, E v may be taken as zero. When these combinations are converted to an equivalent strength design basis, the resulting factor on dead load D is comparable to (1.2 + 0.5C a I ) in §1612.2. For example, consider the following: The basic load combinations of §1612.3.1, without increase in allowable stresses, have a 1.70 factor on D (using the procedure permitted in §1630.8.2.1 for conversion to design strength). The alternate basic load combinations of §1612.3.2 with a permitted one-third 1.70 = 1.28 factor on D. increase in allowable stress has a 1.33


Example 2 Combinations of Loads

§1612.3

&

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The code requires the use of allowable stress design for the design of wood members and their fastenings (see §2301 and §2305). Section 1612.3 permits two different combinations of load methods. These are: 1. Allowable stress design (ASD) of §1612.3.1 2. Alternate allowable stress design of §1612.3.2 This example illustrates the application of each of these methods. This is done for the plywood shear wall shown below. The wall is a bearing wall in a light wood framed building. The following information is given:

Gravity loads

Zone 4 I = 1.0 ρ = 1.0 Ca = 0.40 V E = 4.0 kips (seismic force determined from §1630.2)

VE Plywood shear wall

h = 9' Hold-down

Gravity loads: Dead w D = 0.3 klf (tributary dead load, including weight of wall) Live w L = 0 (roof load supported by other elements)

Nailing

q Pt. O 9' - 7"

L = 10'

Determine the required design loads for shear capacity q and hold-down capacity T for the following load combinations:

Basic allowable stress design. Alternate allowable stress design.



§1612.3


Code Reference

Basic allowable stress design.

§1612.3.1

The governing load combinations for basic allowable stress design are Equations (129), (12-10) and (12-11). These are used without the usua one-third stress increase except as permitted by 1809.2 for soil pressure. For wood design, however, the allowable stresses for short-time loads due to wind or earthquake may be used. D+

E 1.4

0.9 D ±

(12-9) E 1.4

(12-10)

D + 0.75L + 0.75 where

E 1.4

E = ρ Eh + Ev = (1.0) Eh + O = Eh

(12-11) (30-1)

Note that under the provisions of §1630.1.1, E v is taken as zero for ASD. Dead and live load are not involved when checking shear, and both the governing Equations (12-10) and (12-11) reduce to 1.0 E . In this example, E reduces to E h . For checking tension (hold-down capacity), Equation (12-10) governs. Whenever compression is checked, then Equations (12-9) and (12-11) must be checked.

Required unit shear capacity q. Base shear and the resulting element seismic forces determined under §1630.2 are on a strength design basis. For allowable stress design, these must be divided by 1.4 as indicated above in Equations (12-9), (12-10) and (12-11). Thus E V E 4,000 = h = e = V ASD = = 2,857 lbs 1.4 1.4 1.4 1.4 The unit shear is q=

V ASD 2,857 = = 286 plf L 10 ′

This unit shear is used to determine the plywood thickness and nailing requirements from Table 23-ΙΙ-I-1. Footnote 1 of that Table states that the allowable shear values are for short-time loads due to wind or earthquake.



§1612.3

Required hold-down capacity T. Taking moments about point O at the right edge of wall and using V E = 2,875 lbs , the value of the hold-down force TE due to horizontal seismic forces is computed 9.58T E = 9V E TE =

9V 9 ′ × 2.857 = 2.68 kips = 9.58 9.58′

Using Equation (12-10) the effect of dead load and seismic forces are combined to determine the required ASD hold-down capacity. In this example D=

1 (w D )(10 ′) = 1 (0.3)(10 ) = 1.5 kips 2 2

T = 0.9 D −

E = 0.9 D − TE = 0.9 (1.5) − 2.68 = − 1.33 kips tension 1.4

(12-10)

This value is used for the selection of the premanufactured hold-down elements. Manufacturer’s catalogs commonly list hold-down sizes with their “ 1.33 × allowable” capacity values. Here the 1.33 value represents the allowed Load Duration factor, C D , given in Table 2.3.2 of §2316.2 for resisting seismic loads. This is not considered a stress increase (although it has the same effect). Therefore, the “ 1.33 × allowable” capacity values may be used to select the appropriate hold-down element.

Alternate allowable stress design.

§1612.3.2

Under this method of load combination, the customary one-third increase in allowable stresses is allowed. However, Item 5 of §2316.2 states that the one-third increase shall not be used concurrently with the load duration factor C D . The governing load combinations, in the absence of snow load, are the following: D+L+

0.9 D ±

E 1.4

(12-13)

E 1.4

(12-16-1)

where E = ρ E h + E v = (1.0) E h + O = E h

(30-1)

Note: Equation (12-16-1) is a May 1998 errata for the first printing of the code.



§1612.3

Note that E v is taken as zero for ASD per §1630.1.1.

Required unit shear capacity q. E V E 4,000 = h = e = V ASD = = 2,857 lbs 1.4 1.4 1.4 1.4 q=

V ASD 2,856 = = 286 plf L 10

This value may be used directly to select the plywood thickness and nailing requirements from Table 23-ΙΙ-I-1. This method recognizes that Table 23-ΙΙ-I1 already includes a 1.33 allowable stress increase for seismic loading, and the one-third increase cannot be used again with the tabulated values.

Required hold-down capacity T. Taking moments about point O at the right edge of wall for only seismic forces 9.58T E = 9V E TE =

9 (2.857 kips ) = 2.68 kips 9.58

The dead load effect on the hold-down is one-half the dead load. Thus, D=

1 (w D )(10 ′) = 1 (0.3)(10 ) = 1.5 kips 2 2

The governing tension is determined from Equation (12-16-1) T = 0.9 D −

E = 0.9 D − TE = 0.9 (1.5) − 2.68 = − 1.33 kips tension 1.4

(12-10)

This value may be used directly, without modification, to select the “ 1.33 × allowable” capacity of the hold-down elements. Note that the “ 1.33 × allowable” value can be considered either as the one-third increase permitted by §1612.3.1, or the use of a load-duration factor of C D = 1.33 .



§1612.3

Commentary For wood design, the use of the load duration factor C D is not considered as an increase in allowable stress. Together with the other factors employed in establishing the allowable resistance of wood elements, it is the means of representing the extra strength of wood when subject to short duration loads and provides the allowable stress for wind or earthquake load conditions. The allowable shear values given in the Chapter 23 Tables 23-II-H, 23-II-I-1, and 23-II-1-2 are based on this use of this load duration factor. Therefore, the use of the C D factor or the aforementioned table values is permitted for the wind and earthquake load combinations of §1612.3. However, both §1622.3.1 and §2316.2, Item 5, prohibit the concurrent use of a onethird increase in the normal loading allowable stress with the load duration factor C D . It is important to note that, for other than the wind or earthquake load combinations, and for other materials such as masonry where there is no load duration factor, the equivalency of the capacity requirements for §1612.3.1 and §1612.3.2 does not apply mainly because of the prohibited use of a stress increase in §1612.3.1. In this case, the minimum required allowable stress design capacity requirements are best given by the alternate basic load combinations in §1612.3.2.


Example 3 Seismic Zone 4 Near-Source Facto

§1629.4.2

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The 1997 UBC introduced the concept of near-source factors. Structures built in close proximity to an active fault are to be designed for an increased base shear over similar structures located at greater distances. This example illustrates the determination of the near-source factors N a and N v . These are used to determine the seismic coefficients C a and C v used in §1630.2.1 to calculate design base shear.

Determine the near-source factors Na and N v for a site near Lancaster, California.


Code Reference

First locate the City of Lancaster in the book Maps of Known Active Fault NearSource Zones in California and Adjacent Portions of Nevada. This is published by the International Conference of Building Officials and is intended to be used with the 1997 Uniform Building Code. Lancaster is shown on map M-30. Locate the site on this map (see figure), and then determine the following: The shaded area on map M-30 indicates the source is a type A fault. Therefore Seismic source type: A The distance from the site to the beginning of the fault zone is 6 km. Another 2 km must be added to reach the source (discussed on page vii of the UBC Maps of Known Active Faults). Thus, the distance from the site to the source is 6 km + 2 km = 8 km. Distance from site to source: 8 km. Values of N a and N v are given in Tables 16-S and 16-T for distances of 2, 5, 10, and 15 km. For other distances, interpolation must be done. N a and N v have been plotted below. For this site, N a and N v can be determined by entering the figures at a distance 8 km. and using the source type A curves. From this N a = 1.08 N v = 1.36


Example 3 Seismic Zone 4 Near-Source Factor

§1629.4.2

Commentary The values of N a and N v given above are for the site irrespective of the type of structure to be built on the site. Had N a exceeded 1.1, it would have been possible to use a value of 1.1 when determining C a , provided that all of the conditions listed in §1629.4.2 were met.

Ref. Table 16-S

2.0

Na

Source Type A Source Type B

1.0

0.0 0

5

10

15

Distance to Source (km)

Ref. Table 16-T

2.0

Nv


1.0

0.0 0

5

10

15



Example 3 Seismic Zone 4 Near-Source Facto


Site

§1629.4.2

§1629.5.3

Introduction to Vertical Irregularities

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Vertical irregularities are identified in Table 16-L. These can be divided into two categories. The first are dynamic force distribution irregularities. These are irregularity Types 1, 2, and 3. The second category is irregularities in load path or force transfer, and these are Types 4 and 5. The five vertical irregularities are as follows: 1. Stiffness irregularity-soft story 2. Weight (mass) irregularity 3. Vertical geometric irregularity 4. In-plane discontinuity in vertical lateral-force resisting element 5. Discontinuity in capacity-weak story The first category, dynamic force distribution irregularities, requires that the distribution of lateral forces be determined by combined dynamic modes of vibration. For regular structures without abrupt changes in stiffness or mass (i.e., structures without “vertical structural irregularities”), this shape can be assumed to be linearlyvarying or a triangular shape as represented by the code force distribution pattern. However, for irregular structures, the pattern can be significantly different and must be determined by the combined mode shapes from the dynamic analysis procedure of §1631. The designer may opt to go directly to the dynamic analysis procedure and thereby bypass the checks for vertical irregularity Types 1, 2, and 3. Regular structures are assumed to have a reasonably uniform distribution of inelastic behavior in elements throughout the lateral force resisting system. When vertical irregularity Types 4 and 5 exist, there is the possibility of having localized concentrations of excessive inelastic deformations due to the irregular load path or weak story. In this case, the code prescribes additional strengthening to correct the deficiencies.


Example 4 Vertical Irregularity Type 1

§1629.5.3

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A five-story concrete special moment-resisting frame is shown below. The specified lateral forces F x from Equations (30-14) and (30-15) have been applied and the corresponding floor level displacements ∆ x at the floor center of mass have been found and are shown below. Ft + F5

∆S5 = 2.02"

10'

F4 10'

F3 10'

∆S4 = 1.75" Triangular shape

∆S3 = 1.45"

F2

∆S2 = 1.08" 10'

F1

∆S1 = 0.71" 12'

Determine if a Type 1 vertical irregularity—stiffness irregularity-soft story— exists in the first story.


Code Reference

To determine if this is a Type 1 vertical irregularity—stiffness irregularity-soft story—here are two tests:

1. The story stiffness is less than 70 percent of that of the story above. 2. The story stiffness is less than 80 percent of the average stiffness of the three stories above. If the stiffness of the story meets at least one of the above two criteria, the structure is considered to have a soft story, and a dynamic analysis is generally required under §1629.8.4 Item 2, unless the irregular structure is not more than five stories or 65-feet in height (see §1629.8.3 Item 3). The definition of soft story in the code compares values of the lateral stiffness of individual stories. Generally, it is not practical to use stiffness properties unless these can be easily determined. There are many structural configurations where the evaluation of story stiffness is complex and is often not an available output from SEAOC Seismic Design Manual


§1629.5.3

computer programs. Recognizing that the basic intent of this irregularity check is to determine if the lateral force distribution will differ significantly from the linear pattern prescribed by Equation (30-15), which assumes a triangular shape for the first dynamic mode of response, this type of irregularity can also be determined by comparing values of lateral story displacements or drift ratios due to the prescribed lateral forces. This deformation comparison may even be more effective than the stiffness comparison because the shape of the first mode shape is often closely approximated by the structure displacements due to the specified triangular load pattern. Floor level displacements and corresponding story drift ratios are directly available from computer programs. To compare displacements rather than stiffness, it is necessary to use the reciprocal of the limiting percentage ratios of 70 and 80 percent as they apply to story stiffness, or reverse their applicability to the story or stories above. The following example shows this equivalent use of the displacement properties. From the given displacements, story drifts and the story drift ratio values are determined. The story drift ratio is the story drift divided by the story height. These will be used for the required comparisons, since these better represent the changes in the slope of the mode shape when there are significant differences in interstory heights. (Note: story displacements can be used if the story heights are nearly equal.) In terms of the calculated story drift ratios, the soft story occurs when one of the following conditions exists:

1. When 70 percent of

∆ S1 h1

exceeds

∆ S 2 − ∆ S1 h2

or 2. When 80 percent of

∆ S1 h1

exceeds

1  ( ∆ S 2 − ∆ S1 ) ( ∆ S 3 − ∆ S 2 ) ( ∆ S 4 − ∆ S 3 )  + +  3  h2 h3 h4  The story drift ratios are determined as follows: ∆ S1 h1

=

(0.71 − 0) = 0.00493 144

∆ S 2 − ∆ S1 (1 .08 − 0 .71 ) = 0 .00308 = 120 h2 ∆ S3 − ∆ S2 h3

=

(1 . 45

− 1 . 08 ) = 0 . 00308 120



∆ S4 − ∆ S3 h4

=

§1629.5.3

(1 .75 − 1 .45 ) = 0 .00250 120

1 (0.00308 + 0.00308 + 0.00250 ) = 0.00289 3

Checking the 70 percent requirement: ∆ 0 .70  S 1  h1

  = 0 .70 (0.00493 ) = 0 .00345 > 0 .00308 

∴Soft story exists Checking the 80 percent requirement: ∆ 0 .80  S 1  h1

  = 0 .80 (0 .00493 ) = 0 .00394 > 0 .00289 

∴Soft story exists

Commentary Section 1630.10.1 requires that story drifts be computed using the maximum inelastic response displacements ∆ M . However, for the purpose of the story drift, or story drift ratio, comparisons needed for soft story determination, the displacement ∆ S due to the design seismic forces can be used as done in this example. In the example above, only the first story was checked for possible soft story vertical irregularity. In practice, all stories must be checked, unless a dynamic analysis is performed. It is often convenient to create a table as shown below to facilitate this exercise.

Level

Story Displacement

Story Drift

Story Drift Ratio

.7x (Story Drift Ratio)


Avg. of Story Drift Ratio of Next 3 Stories

Soft Story Status

5

2.02 in.

0.27 in.

0.00225

0.00158

0.00180

—

No

4

1.75

0.30

0.00250

0.00175

0.00200

—

No

3

1.45

0.37

0.00308

0.00216

0.00246

—

No

2

1.08

0.37

0.00308

0.00216

0.00246

0.00261

No

1

0.71

0.71

0.00493

0.00345

0.00394

0.00289

Yes



§1629.5.3

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The five-story special moment frame office building has a heavy utility equipment installation at Level 2. This results in the floor weight distribution shown below: Level

5

W5 = 90 k

4

W4 = 110 k

3

W3 = 110 k

2

W2 = 170 k

1

W1 = 100 k

Determine if there is a Type 2 vertical weight (mass) irregularity.


Code Reference

A weight, or mass, vertical irregularity is considered to exist when the effective mass of any story is more than 150 percent of the effective mass of an adjacent story. However, this requirement does not apply to the roof if the roof is lighter than the floor below. Checking the effective mass of Level 2 against the effective mass of Levels 1 and 3 At Level 1 1.5 ×W1 = 1.5 (100 k ) = 150 k At Level 3 1.5 × W3 = 1.5 (110 k ) = 165 k W2 = 170 k > 150 k ∴ Weight irregulari ty exists


Example 5

ertical Irregularity Type 2

§1629.5.3

Commentary As in the case of vertical irregularity Type 1, this type of irregularity also results in a primary mode shape that can be substantially different from the triangular shape and lateral load distribution given by Equation (30-15). Consequently, the appropriate load distribution must be determined by the dynamic analysis procedure of §1631, unless the irregular structure is not more than five stories or 65 feet in height (see §1629.8.3 Item 3).



§1629.5.3

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The lateral force-resisting system of the five-story special moment frame building shown below has a 25-foot setback at the third, fourth and fifth stories. 1

2

3

4

5

4 @ 25' = 100' Level

5

4

3

2

1

Determine if a Type 3 vertical irregularity, vertical geometric irregularity, exists.


Code Reference

A vertical geometric irregularity is considered to exist where the horizontal dimension of the lateral force-resisting system in any story is more than 130 percent of that in the adjacent story. One-story penthouses are not subject to this requirement. In this example, the setback of Level 3 must be checked. The ratios of the two levels is Width of Level 2 (100') = = 1.33 Width of Level 3 (75' ) 133 percent > 130 percent ∴ Vertical geometric irregulari ty exists



§1629.5.3

Commentary The more than 130 percent change in width of the lateral force-resisting system between adjacent stories could result in a primary mode shape that is substantially different from the triangular shape assumed for Equation (30-15). If the change is a decrease in width of the upper adjacent story (the usual situation), the mode shape difference can be mitigated by designing for an increased stiffness in the story with a reduced width. Similarly, if the width decrease is in the lower adjacent story (the unusual situation), the Type 1 soft story irregularity can be avoided by a proportional increase in the stiffness of the lower story. However, when the width decrease is in the lower story, there could be an overturning moment load transfer discontinuity that would require the application of §1630.8.2. When there is a large decrease in the width of the structure above the first story along with a corresponding large change in story stiffness that creates a flexible tower, then §1629.8.3, Item 4 and §1630.4.2, Item 2 may apply. Note that if the frame elements in the bay between lines 4 and 5 were not included as a part of the designated lateral force resisting system, then the vertical geometric irregularity would not exist. However, the effects of this adjoining frame would have to be considered under the adjoining rigid elements requirements of §1633.2.4.1.



§1629.5.3

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A concrete building has the building frame system shown below. The shear wall between Lines A and B has an in-plane offset from the shear wall between Lines C and D. B

A

C

D

3 @ 25' = 75’ Level 5 12' Shear wall 4 12'

3 25’

12'

2 50'

12'

Shear wall

1 12'

Determine if there is a Type 4 vertical irregularity, in-plane discontinuity in the vertical lateral force-resisting element.


Code Reference

A Type 4 vertical irregularity exists when there is an in-plane offset of the lateral load resisting elements greater than the length of those elements. In this example, the left side of the upper shear wall (between lines A and B) is offset 50 feet from the left side of the lower shear wall (between lines C and D). This 50-foot offset is greater than the 25-foot length of the offset wall elements. ∴ In - plane discontinu ity exists



§1629.5.3

Commentary The intent of this irregularity check is to provide correction of force transfer or load path deficiencies. It should be noted that any in-plane offset, even those less or equal to the length or bay width of the resisting element, can result in an overturning moment load transfer discontinuity that requires the application of §1630.8.2. When the offset exceeds the length of the resisting element, there is also a shear transfer discontinuity that requires application of §1633.2.6 for the strength of collector elements along the offset. In this example, the columns under wall A-B are subject to the provisions of §1630.8.2 and §1921.4.4.5, and the collector element between Lines B and C at Level 2 is subject to the provisions of §1633.2.6.



§1629.5.3

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A concrete bearing wall building has the typical transverse shear wall configuration shown below. All walls in this direction are identical, and the individual piers have the shear contribution given below. Vn is the nominal shear strength calculated in accordance with §1921.6.5, and Vm is the shear corresponding to the development of the nominal flexure strength calculated in accordance with §1921.6.6. Level

3

2

4

5

1

1

2

3

Pier

Vn

Vm

1

20 k

30 k

2

30

40

3

15

10

4

80

120

5

15

10

Determine if a Type 5 vertical irregularity, discontinuity in capacity – weak story, condition exists.


Code Reference

A Type 5 weak story discontinuity in capacity exists when the story strength is less than 80 percent of that in the story above. The story strength is considered to be the total strength of all seismic force-resisting elements sharing the story shear for the direction under consideration. Using the smaller values of Vn and Vm given for each pier, the story strengths are First story strength = 20 + 30 + 10 = 60 k Second story strength = 80 + 10 = 90 k Check if first story strength is less than 80 percent of that of the second story: 60k < 0.8 (90) = 72 k ∴ Weak story condition exists



§1629.5.3

Commentary This irregularity check is to detect any concentration of inelastic behavior in one supporting story that can lead to the loss of vertical load capacity. Elements subject to this check are the shear wall piers (where the shear contribution is the lower of either the shear at development of the flexural strength, or the shear strength), bracing members and their connections, and frame columns. Frame columns with weak column-strong beam conditions have a shear contribution equal to that developed when the top and bottom of the column are at flexural capacity. Where there is a strong column-weak beam condition, the column shear resistance contribution should be the shear corresponding to the development of the adjoining beam yield hinges and the column base connection capacity. In any case, the column shear contribution shall not exceed the column shear capacity. Because a weak story is prohibited (under §1629.9.1) for structures greater than two stories or 30 feet in height, the first story piers in this example must either be strengthened by a factor of 72/60 = 1.2, or designed for Ω o times the forces prescribed in §1630.



§1629.5.3

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A four-story building has a steel special moment resisting frame (SMRF). The frame consists of W24 beams and W14 columns with the following member strength properties (determined under 2213.4.2 and 2213.7.5): A

Beams at Levels 1 and 2: M b =ZF y = 250 kip-ft Columns on lines A, B, C, and D at both levels: M c = Z Fy − f a = 200 kip-ft at

(

)

axial loading of 1.2 PD + 0.5 PL . Column base connections at grade: M f = 100 kip-ft In addition, the columns meet the exception of §2213.7.5 such that a strong beam-weak column condition is permitted.

C

B

D

3 @ 25' Level 5 12' 4 12' 3 12' 2 12' 1 14’

Determine if a Type 5 vertical irregularity—discontinuity in capacity-weak story— condition exists in the first story:

Determine first story strength. Determine second story strength. Determine if weak story exists at first story.


Code Reference

A Type 5 weak story discontinuity in capacity exists when the story strength is less than 80 percent of that of the story above. The story strength is considered to be the total strength of all seismic force-resisting elements that share the story shear for the direction under consideration. To determine if a weak story exists in the first story, the sums of the column shears in the first and second stories—when the member moment capacities are developed by lateral loading—must be determined and compared. In this example, it is assumed that the beam moments at a beam-column joint are distributed equally to the sections of the columns directly above and below the joint. Given below is the calculations for first and second stories.



§1629.5.3

Determine first story strength.

Columns A and D must be checked for strong column-weak beam considerations. 2M c = 400 > M b = 250 ∴ strong column-weak beam condition exists. Next, the shear in each column must be determined.

V

M b 2 = 125 k - ft

V

M f = 100 k - ft

Clear height = 14 ft − 2 ft = 12 ft V A = VD =

125 + 100 = 18.75 k 12

Checking columns B and C for strong column-weak beam considerations. 2 M c = 400 < 2 M b = 500 ∴ Strong beam-weak column condition exists. Next, the shear in each column must be determined. V

Mc = 200 k-ft

Clear height = 14 ft − 2 ft = 12 ft V B = VC =

200 + 100 = 25.0 k 12

V

Mf = 100 k-ft

First story strength = V A + V B + VC + V D = 2(18.75) + 2(25.0) = 87.5 k



§1629.5.3

Determine second story strength.

Columns A and D must be checked for strong column-weak beam at Level 2. 2M c = 400 > M b = 250 ∴ strong column-weak beam condition exists.

V

M b 2 = 125 k - ft

V

M b 2 = 125 k - ft

Clear height = 12 ft − 2 ft = 10 ft VA = VD =

125 + 125 = 25.0 k 10

Checking columns B and C for strong column-weak beam considerations. 2M c = 400 < 2 M b = 500 ∴ Strong beam-weak column condition exists.

V

Mc = 200 k-ft

Clear height = 12 ft − 2 ft = 10 ft 10’

V B = VC =

200 + 200 = 40.0 k 10 V

Mc = 200 k-ft

Second story strength = V A + V B + VC + V D = 2( 25.0) + 2(40.0) = 130.0 k



§1629.5.3

Determine if weak story exists at first story.

First story strength = 87.5 k Second story strength = 130.0 k 87.5 < 0.80 (130) = 104

Table 16-L Item 5

∴ Weak story condition in first story exists


§1629.5.3

Introduction to Plan Irregularities

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Plan structural irregularities are identified in Table 16-M. There are five types of plan irregularities: 1. Torsional irregularity—to be considered when diaphragms are not flexible 2. Re-entrant corners 3. Diaphragm discontinuity 4. Out-of-plane offsets 5. Nonparallel systems These irregularities can be categorized as being either special response conditions or cases of irregular load path. Types 1, 2, 3, and 5 are special response conditions: Type 1. When the ratio of maximum drift to average drift exceeds the given limit, there is the potential for an unbalance in the inelastic deformation demands at the two extreme sides of a story. As a consequence, the equivalent stiffness of the side having maximum deformation will be reduced, and the eccentricity between the centers of mass and rigidity will be increased along with the corresponding torsions. An amplification factor Ax is to be applied to the accidental eccentricity to represent the effects of this unbalanced stiffness. Type 2. The opening and closing deformation response or flapping action of the projecting legs of the building plan adjacent to re-entrant corners can result in concentrated forces at the corner point. Elements must be provided to transfer these forces into the diaphragms. Type 3. Excessive openings in a diaphragm can result in a flexible diaphragm response along with force concentrations and load path deficiencies at the boundaries of the openings. Elements must be provided to transfer the forces into the diaphragm and the structural system. Type 4. The Type 4 plan irregularity, out-of-plane offset, represents the irregular load path category. In this case, shears and overturning moments must be transferred from the level above the offset to the level below the offset, and there is a horizontal “offset” in the load path for the shears. Type 5. The response deformations and load patterns on a system with nonparallel lateral force-resisting elements can have significant differences from that of a regular system. Further analysis of deformation and load behavior may be necessary.


Example 10 Plan Irregularity Type 1

§1629.5.3

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A three-story special moment resisting frame building has rigid floor diaphragms. Under specified seismic forces, including the effects of accidental torsion, it has the following displacements at Levels 1 and 2: δ L ,2 = 1.30"

δ R , 2 = 1.90"

δ L ,1 = 1.00"

δ R ,1 = 1.20" δR,2 Level 3

δR,1 δL,2

2 Level 2

δL,1

1 Level 1

Determine if a Type 1 torsional irregularity exists at the second story.

If it does:

Compute the torsional amplification factor Ax for Level 2.


Code Reference

A Type 1 torsional plan irregularity is considered to exist when the maximum story drift, including accidental torsion effects, at one end of the structure transverse to an axis is more than 1.2 times the average of the story drifts of the two ends of the structure.

Determine if a Type 1 torsional irregularity exists at the second story.

Table 16-M

Referring to the above figure showing the displacements δ due to the prescribed lateral forces, this irregularity check is defined in terms of story drift ∆δ X = (δ X − δ X −1 ) at ends R (right) and L (left) of the structure. Torsional irregularity exists at level x when



§1629.5.3

∆ max = ∆ R , X >

1.2(∆

R,x

+∆

2

L, x

)

(

= 1.2 ∆ avg

)

where ∆δ L, 2 = δ L , 2 − δ L ,1 ∆δ R ,2 = δ R ,2 − δ R ,1 ∆δ max = ∆δ R , X , ∆δ avg =

∆δ L, X + ∆δ R , X 2

Determining story drifts at Level 2 ∆ L, 2 = 1.30 − 1.00 = 0.30 in. ∆ R ,2 = 1.90 − 1.20 = 0.70 in. ∆ avg =

0.30 + 0.70 = 0.50 in. 2

Checking 1.2 criteria ∆ max ∆ R ,2 0.7 = = = 1.4 > 1.2 ∆ avg ∆ avg 0.5 ∴Torsional irregulari ty exists

Compute amplification factor A X for Level 2.

§1630.7

When torsional irregularity exists at a level x , the accidental eccentricity, equal to 5 percent of the building dimension, must be increased by an amplification factor Ax . This must be done for each level, and each level may have a different Ax value. In this example, Ax is computed for Level 2.  δ Ax =  max  1.2 avg

   

2

(30-16)

δ max = δ R , 2 = 1.90 in. δ avg =

δ L, 2 + δ R , 2 2

=

1.30 + 1.90 = 1.60 in. 2



§1629.5.3

2

 1.90   = 0.98 < 1.0 A2 =   1.2 (1.60)  ∴ use Ax = 1.0

Commentary In §1630.7, there is the provision that “the most severe load combination must be considered.” The interpretation of this for the case of the story drift and displacements to be used for the average values ∆δ avg and δ avg is as follows. The most severe condition is when both δ R, X and δ L, X are computed for the same accidental center of mass displacement that causes the maximum displacement δ max . For the condition shown in this example where δ R , X = δ max , the centers of mass at all levels should be displaced by the accidental eccentricity to the right side R, and both δ R, X and δ L, X should be evaluated for this load condition. While Table 16-M calls only for §1633.2.9, Item 6 (regarding diaphragm connections) to apply if this irregularity exists, there is also §1630.7, which requires the accidental torsion amplification factor Ax given by Equation (30-16). It is important to recognize that torsional irregularity is defined in terms of story drift ∆δ X while the evaluation of Ax by Equation (30-16) is in terms of displacements δ X . There can be instances where the story drift values indicate torsional irregularity and where the related displacement values produce an Ax value less than one. This result is not the intent of the provision, and the value of Ax used to determine accidental torsion should not be less than 1.0. The displacement and story drift values should be obtained by the equivalent lateral force method with the specified lateral forces. Theoretically, if the dynamic analysis procedure were to be used, the values of ∆δ max and ∆δ avg would have to be found for each dynamic mode, then combined by the appropriate SRSS or CQC procedures, and then scaled to the specified base shear. However, in view of the complexity of this determination and the judgmental nature of the 1.2 factor, it is reasoned that the equivalent static force method is sufficiently accurate to detect torsional irregularity and evaluate the Ax factor. If the dynamic analysis procedure is either elected or required, then §1631.3 requires the use of a three-dimensional model if there are any of the plan irregularities listed in Table 16-M.



§1629.5.3

For cases of large eccentricity and low torsional rigidity, the static force procedure can result in a negative displacement on one side and a positive on the other. For example, this occurs if δ L ,3 = −0.40′′ and δ R ,3 = 1.80′′ . The value of δ avg in Equation (30-16) should be calculated as the algebraic average: δ avg =

δ L ,3 + δ R ,3 2

=

(− 0.40 ) + 1.80 = 1.40 = 0.70 2

2

in.

When dynamic analysis is used, the algebraic average value δ avg should be found for each mode, and the individual modal results must be properly combined to determine the total response value for δ avg .



§1629.5.3

.#1 1 &

$%&-'0'(

The plan configuration of a ten-story special moment frame building is as shown below:

A

B

C

D

E

4 @ 25' = 100' 4

3

2

1

Determine if there is a Type 2 re-entrant corner irregularity.


Code Reference

A Type 2 re-entrant corner plan irregularity exists when the plan configuration of a structure and its lateral force-resisting system contain re-entrant corners, where both projections of the structure beyond a re-entrant corner are greater than 15 percent of the plan dimension of the structure in the direction considered. The plan configuration of this building, and its lateral force-resisting system, have identical re-entrant corner dimensions. For the sides on Lines 1 and 4, the projection beyond the re-entrant corner is 100 ft − 75 ft = 25 ft This is

25 or 25 percent of the 100 ft plan dimension. 100

For the sides on Lines A and E, the projection is 60 ft − 40 ft = 20 ft 20 or 33.3 percent of the 60 ft plan dimension. This is 60 SEAOC Seismic Design Manual

§1629.5.3


Since both projections exceed 15 percent, there is a re-entrant corner irregularity. ∴ Re - entrant corner irregulari ty exists

Commentary Whenever the Type 2 re-entrant corner plan irregularity exists, see the diaphragm requirements of §1633.2.9 Items 6 and 7.



§1629.5.3

& .#1 1 (

$%&-'0'(

A five-story concrete building has a bearing wall system located around the perimeter of the building. Lateral forces are resisted by the bearing walls acting as shear walls. The floor plan of the second floor of the building is shown below. The symmetrically placed open area in the diaphragm is for an atrium, and has dimensions of 40 ft x 75 ft. All diaphragms above the second floor are without significant openings. 1

2

3

4

125' 75' A B 40'

80'

Atrium

C D

Second floor pla

Determine if a Type 3 diaphragm discontinuity exists at the second floor level.


Code Reference

A Type 3 diaphragm discontinuity irregularity exists when diaphragms have abrupt discontinuities or variations in stiffness, including cutout or open areas greater than 50 percent of the gross enclosed area of the diaphragm, or changes in effective diaphragm stiffness of more than 50 percent from one story to the next. Gross enclosed area of the diaphragm is 80 ft × 125 ft = 10,000 sq ft Area of opening is 40'×75' = 3,000 sq ft 50 percent of gross area = 0.5 (10,000) = 5,000 sq ft 3,000 < 5,000 sq ft ∴ No diaphragm discontinu ity irregulari ty exists



§1629.5.3

Commentary The stiffness of the second floor diaphragm with its opening must be compared with the stiffness of the solid diaphragm at the third floor. If the change in stiffness exceeds 50 percent, then a diaphragm discontinuity irregularity exists for the structure. This comparison can be performed as follows: Find the simple beam mid-span deflections ∆ 2 and ∆ 3 for the diaphragms at Levels 2 and 3, respectively, due to a common distributed load w , such as 1 klf.

w = 1 klf

Level 2

∆2 Deflected shape

w = 1 klf

Level 3

∆3 Deflected shape

If ∆ 2 > 1.5∆ 3 , then there is diaphragm discontinuity.



§1629.5.3

( .#1 1 *

$%&-'0'(

A four-story building has a concrete shear wall lateral force-resisting system in a building frame system configuration. The plan configuration of the shear walls is shown below. 3 2

1

3

2 10'

1 10' Typical Floor Plan

Typical floor plan 10' A

B

C

D

E

10' 4 @ 25' = 100' 3

2

2 @ 25' = 50'

Elevation Line E

First Floor Plan

1

Ground (first) floor pla

Determine if there is a Type 4 out-of-plane offset plan irregularity between the first and second stories.


Code Reference

An out-of-plane offset plan irregularity exists when there are discontinuities in a lateral force path, for example: out-of-plane offsets of vertical resisting elements such as shear walls. The first story shear wall on Line D has 25 ft out-of-plane offset to the shear wall on Line E at the second story and above. This constitutes an out-of-plane offset irregularity, and the referenced sections in Table 16-M apply to the design. ∴Offset irregulari ty exists



§1629.5.3

* .#1 1 0

$%&-'0'(

A ten-story building has the floor plan shown below at all levels. Special moment resisting-frames are located on the perimeter of the building on Lines 1, 4, A, and F. A

B

C

D

E

F 4 @ 25' = 100'

3

2

3 @ 25' = 75'

4

1

Typical floor plan

Determine if a Type 5 nonparallel system irregularity exists.


Code Reference

A Type 5 nonparallel system irregularity is considered to exist when the vertical lateral load resisting elements are not parallel to or symmetric about the major orthogonal axes of the building’s lateral force-resisting system. The vertical lateral force-resisting frame elements located on Line F are not parallel to the major orthogonal axes of the building (i.e., Lines 4 and A). Therefore a nonparallel system irregularity exists, and the referenced section in Table 16-M applies to the design. ∴ A nonparalle l system irregulari ty exists


ρ

Example 15 Reliability/Redundancy Facto

§1630.1.1

0

5 1651 , ρ

Evaluate the reliability/redundancy factor, ρ , for the three structural systems shown below. Given information for each system includes the story shears Vi due to the design base shear V, and the corresponding element forces E h . The ρ factor is defined as ρ=2−

20

(30-3)

rmax AB

where rmax is the largest of the element-story shear ratios, ri , that occurs in any of the story levels at or below the two-thirds height level of the building; and AB is the ground floor area of the structure in square feet. Once ρ has been determined, it is to be used in Equation (30-1) to establish the earthquake load E for each element of the lateral force-resisting system. For purposes of this example, only the frame line with maximum seismic force is shown. In actual applications, all frame lines in a story require evaluation. The E h forces given include any torsional effects. Note that the story shear Vi is the total of the shears in all of the frame lines in the direction considered.


Code Reference

Braced frame structure. A

B

16'

Level

D

C

16'

16'

5 12' 4 12' 3 12' 2 12' 1 12'


Example 15 Reliability/Redundancy Factor

§1630.1.1

ρ

The following information is given: Story i

Total Story Shear Vi

Brace Force Eh

Horizontal Component Fx

1

952 kips

273 kips

218.4 kips

0.229

2

731

292

233.6

0.320

3

517

112

89.6

0.173

4

320

73.1

0.229

91.4

ri = Fx Vi

Not required above 2/3 height level (see definition of ri)

5

AB = 48 ft × 100 ft = 4,800 sq ft, where 100 ft is the building width. Horizontal component in each brace is Fx =

4 Eh 5

where E h is the maximum force in a single brace element in story i. For braced frames, the value of ri is equal to the maximum horizontal force component Fx in a single brace element divided by the total story shear Vi . rmax = 0.320 ρ=2−

20 rmax AB

=2−

20

(0.320)

4800

= 1.10

(30-3)

Moment frame structure. A

B

24'

Level

C

D

24'

24'

5.9 k

11.4 k

13.1 k

15.6 k

27.9 k

30.2 k

21.5 k

40.2 k

45.7 k

25.6 k

28.3 k

51.2 k

56.8 k

30.7

71.8 k

46.1

5

12' 7.5 k

4 12' 3

16.4 k 12'

2 12' 1

12' 38.7 k

68.6 k



ρ

§1630.1.1

AB = 72'×120' = 8,640 sq ft, where 120' is the building width Column shears are given above. E h = V A , V B , VC , V D in column lines A, B, C, D, respectively. Column Lines B and C are common to bays on opposite sides. For moment frames, ri is maximum of the sum of V A + 0.7 V B , or 0.7 ( V B + VC ), or 0.7VC + V D divided by the story shear Vi .

§1630.1.1

Section 1630.1.1 requires that special moment-resisting frames have redundancy such that the calculated value of ρ does not exceed 1.25. The story shears and ri evaluations are: Vi

1

388 kips

86.7 kips

98.3 kips

96.4 kips

0.253

2

306

64.1

75.6

70.5

0.247

3

228

49.6

60.1

57.6

0.264

151

35.1

40.7

37.5

0.270

4 5

VA + 0.7VB

0.7(VB + VC)

0.7VC + VD

ri

Story i

Not required above 2/3 height level

rmax = r4 = 0.270 ρ=2−

20

(0.270)

8640

= 1.20 < 1.25 o.k.

(30-3)

Building frame system with shear walls. B

A 10' Level

D

C 20'

20'

E 20'

5 12' 4 12' 3 12' 2 12' 1 12'


Example 15 Reliability/Redundancy Factor

§1630.1.1

ρ

AB = 70'×120' = 8,400 sq ft., where 120' is the building width E h is the wall shear V w For shear walls, ri is the maximum of

V wi Vi

 10    . The following information is given  lw 

for the walls.

Wall A-B Story i

Vi

Vwi

1

363 kips

34.1 kips

2

288

3 4 5

Wall C-D-E and C-D lwi

Vwi

lwi

10 ft

92.4 kips

40 ft

26.9

10

75.2

40

208

36.3

10

69.3

20

105

19.7

10

39.8

20

Above 2/3 height level

A-B i

VI

Vwi Vi

 10     lw 

C-D-E and C-D

Vwi Vi

 10     lw 

ri

1

363 kips

0.094

0.064

0.094

2

288

0.093

0.065

0.093

3

208

0.175

0.167

0.175

4

105

0.188

0.190

0.190

5

Not required above 2/3 height level

rmax = r4 = 0.190 ρ=2−

20

(0.190)

6000

= 0.641 < 1.0

(30-3)

∴ use ρ = 1.0



ρ

§1630.1.1

Commentary A separate value of ρ must be determined for each principal building direction. Each value of ρ is applied to the elements of the vertical lateral force-resisting system for that direction. Note that the redundancy factor does not apply to horizontal diaphragms, except in the case of transfer diaphragms. The following code provisions require the designer to provide sufficient redundancy such that ρ is less than or equal to specified values: 1. Section 1630.1.1 requires that the number of bays of special moment resisting frames be such that the value of ρ is less than or equal to 1.25. 2. Section 1629.4.2 allows that the near-source factor Na need not exceed 1.1, if along with other stated conditions, the redundancy is such that the calculated ρ value is less than or equal to 1.00.


Example 16 Reliability/Redundancy Factor Applications

§1630.1.1

% 5 1651 , 7

$%(4''

The 1997 UBC introduced the concept of the reliability/redundancy factor. The intent of this provision is to penalize those lateral force-resisting systems without adequate redundancy by requiring that they be more conservatively designed. The purpose of this example is to develop approximate relationships that will enable the engineer to estimate the number of lateral force-resisting elements required to qualify for given values of the redundancy factor ρ . These relationships are particularly useful in the conceptual design phase. Note that a redundancy factor is computed for each principal direction and that these are not applied to diaphragms, with the exception of transfer diaphragms at discontinuous vertical lateral force-resisting elements. For the following structural systems, find the approximate relation for ρ in terms of the number N of resisting elements (e.g., braces, frames, and walls).

Braced frames. Moment-resisting frames. Shear walls.


Code Reference

Before developing the approximate relationships for the three structural systems, a brief discussion of methodology is presented. For a given story level i with story shear Vi , the approximate number of lateral forceresisting elements N required a given value of ρ can be found as follows. The basic reliability/redundancy relationship given in §1630.1.1 is ρ=2−

20 rmax

(30-3)

AB

The term rmax is the maximum element-story shear ratio. This is the fraction of the total seismic shear at a given floor level that is carried by the most highly loaded element. AB is the ground floor area of the structure in square feet. The value of rmax can be approximated in terms of the story shear Vi and the number of elements N in the story. This is done for each system below to provide the approximate relationship for ρ .



§1630.1.1

Braced frames.

For a braced frame system with N braces having a maximum force component H max (this is the horizontal component of the maximum brace force), assume that the maximum component is 125 percent of the average. Thus H max = (1.25) H average = (1.25)

N braces

H max 1.25Vi 1.25 = = Vi N braces (Vi ) N braces

rmax =

ρ=2−

20 rmax

∴ρ = 2 −

Vi

AB

20 N braces 1.25 AB

, where N braces = number of braces.

Moment-resisting frames.

For a moment-resisting frame system with N bays having a maximum shear per bay of V bay,max , assume that the maximum component is 125 percent of the average component. Thus, V bay,max = (1.25)

rmax =

Vbay,max

∴ρ = 2 −

Vi

Vi N bays

=

1.25 N bays

20 N bays 1.25 AB

, where N bays = number of bays

Note that for a SMRF, ρ shall not exceed 1.25 . Thus, the number of bays of special moment-resisting frames must be increased to reduce rmax , such that ρ is less than or §1630.1.1 equal to 1.25 .



§1630.1.1

Shear walls.

Section 1630.1.1 requires that rmax be based on the number of 10-foot lengths of shear wall. For a shear wall system, let N 10 = number of 10-foot-long wall segments V  in story i, and let the maximum shear per 10-foot length be 10  w  . V w and l w  l w  max are the shear and length for a wall pier. Assuming the maximum component is 125 percent of the average. V  V 10  w  = (1.25) i N 10  l w  max

rmax

∴ρ = 2 −

V  10  w   l w  max 1.25 = = Vi N 10 20 N 10

1.25 AB

, where N 10 = number of 10-foot long segments of shear walls.

Commentary Following this page is a plot of ρ versus N for the equation ρ = 2 −

20 N

. This 1.25 AB approximate relationship can be used to estimate ρ for conceptual design. Following this 20 is a plot of ρ = 2 − . This is Equation (30-3) and can be used for final design. rmax AB



§1630.1.1

AB =

t. .f sq 00 t. ,0 .f 30 sq 00 ,0 20 ft. q. 0s ,00 10 ft. sq. 00 5,0 ft. sq. 00 2,5 . sq. ft

1,000

40

,00

,0

00

sq

ρ = 2 - 20N /(1.25A B 1/2)

A

60

80

,0

0s

q.

.f

t.

B

=1

00

00

,00

sq

0s

q.

.f

ft.

t.

ft.

1.5 1.4 1.3

ρ

1.2 1.1 1 0

1

2

3

4

5

6

7

8

9

10

11

12

N

AB = 100,0 00 sq . ft. 60,0 00 s q. ft 40 . ,00 0s q. ft. 20 ,0 00 sq .f t.

Approximate relationship of ρ for various values of N and AB

ρ = 2 - 20/(r max A B 1/2) t. .f t. ft. sq sq. f ft. t. q. 0 0 q. f sq. 0 0s 0 ,00 0 0s 00 , 0 0 0 , , 0 0 , 3 8 6 4 1

ft. sq. 00 2,0

1.5 A

1.4

ρ

. ft. 0 sq 0 0 , =1

B

1.3 1.2 1.1 1.0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

r max Reliability/redundancy factor ρ for various values of rmax and AB


Example 17 P ∆ Effects

§1630.1.3

2 ∆

$%(4''(

In highrise building design, important secondary moments and additional story drifts can be developed in the lateral force-resisting system by P∆ effects. P∆ effects are the result of the axial load P in a column being “moved” laterally by horizontal displacements, thereby causing additional “secondary” column and girder moments. The purpose of this example is to illustrate the procedure that must be used to check the overall stability of the frame system for such effects. A 15-story building has a steel special moment-resisting frame (SMRF). The following information is given:

Zone 4 R = 8.5 At the first story, ΣD = W = 8,643 kips ΣL = 3,850 kips V1 = V = 0.042W = 363.0 kips h1 = 20 ft Story drift = ∆ S1 = 0.003h1 = 0.72 in.

h1 = 20'

Determine the following:

P∆ criteria for the building. Check the first story for P∆ requirements.


Code Reference

P∆ criteria for the building.

§1630.1.3

P∆ effects must be considered whenever the ratio of secondary moments to primary moments exceed 10 percent. As discussed in Section C105.1.3 of the 1999 SEAOC Blue Book Commentary, this ratio is defined as a stability coefficient θ :

θx =

Px ∆ sx V x hx



§1630.1.3

where θ x = stability coefficient for story x Px = total vertical load (unfactored) on all columns in story x ∆ sx = story drift due to the design base shear V x = design shear in story x h x = height of story x P∆ effects must be considered when θ > 0.10 An alternative approach is to check story drift. In Seismic Zones 3 and 4, P∆ effects need not be considered for SMRF buildings whenever the story drifts satisfy the following criterion: ∆ s 0.02 0.02 ≤ = = .00235 h R 8.5

§1630.1.3

Therefore, when the story drift in a given story of an SMRF is less than or equal to .00235, P∆ effects need not be considered for that story.

Check P∆ requirements for the first story.

The first story drift ratio is ∆ S 1 0.003h1 = = 0.003 h1 h1 Check drift criteria .003 > .00235 Section 1630.1.3 requires that the total vertical load P1 at the first story be considered as the total dead (ΣD ) plus floor live (ΣL ) and snow (S ) load above the first story. These loads are unfactored for determination of P∆ effects. P1 = ΣD + ΣL + S using S = 0 for the building site P1 = 8,643 + 3,850 = 12,493 kips



§1630.1.3

θ1 =

P1 ∆ S 1 (12,493)(0.003h1 ) = = 0.103 > 0.100 V1 h1 (363.0)h1

∴P∆ effects must be considered

Commentary The 1999 SEAOC Blue Book Commentary, in Section C105.1.3, provides an acceptable P∆ analysis: for any story x where P∆ effects must be considered, the θ story shear V x must be multiplied by a factor (1 + a d ) , where a d = , and the 1− θ structure is to be re-analyzed for the seismic force effects corresponding to the augmented story shears. Also, some computer programs include the option to include P∆ effects. The user should verify that the particular method is consistent with the requirements of this §1630.1.3.


Example 18 Design Base Shea

§1630.2.1

3

$%(4'&'

# 8 "

Find the design base shear for a 5-story steel special moment-resisting frame building shown below, given the following information: Z = 0.4 Seismic source type = B Distance to seismic source = 5 km Soil profile type = SC I = 1.0 R = 8.5 W = 1,626 kips

60'

In solving this example, the following steps are followed:

Determine the structure period. Determine the seismic coefficients C a and Cv . Determine base shear.


Code Reference

Determine the structure period.

§1630.2.2

Method A to be used. Ct for steel moment-resisting frames is 0.035. T = C t (hn )

3

4

= .035 (60 )

3

4

= .75 sec .

Determine the seismic coefficients C a and Cv

(30-8) §1628

From Table 16-Q for soil profile type S C and Z = .4 C a = .40 N a From Table 16-R for soil profile type S C and Z = .4 C v = .56 N v Find N a and N v from Tables 16-S and 16-T, respectively, knowing that the seismic source type is B and the distance 5 km. N a = 1.0 SEAOC Seismic Design Manual

Example 18 Design Base Shear

§1630.2. 1

N v = 1.2 Therefore C a = .40 (1.0) = .40 C v = .56 (1.2 ) = .672

Determine base shear.

§1630.2.1

The total base shear in a given direction is determined from: V=

Cv I .672 × 1.0 × 1,626 = 171.4 kips W= 8.5 × .75 RT

(30-4)

However, the code indicates that the total design base shear need not exceed: V=

2.5C a I R

W=

2.5 × 0.4 × 1.0 × 1,626 = 191.3 kips 8.5

(30-5)

Another requirement is that total design base shear cannot be less than: V = 0.11C a IW = 0.11 × .40 × 1.0 × 1,626 = 71.5 kips

(30-6)

In Zone 4, total base shear also cannot be less than: V=

.8ZN v I R

W=

0.8 × 0.4 × 1.2 × 1.0 × 1,626 = 73.5 kips 8.5

(30-7)

In this example, design base shear is controlled by Equation 30-4. ∴V = 171.4 kips

Commentary The near source factor Na used to determine Ca need not exceed 1.1 if the conditions of §1629.4.2 are met.


Example 19 Structure Period Using Method

§1630.2.2

-

" # 9 7

$%(4'&'&

Determine the period for each of the structures shown below using Method A. Method A uses the following expression to determine period: T = C t (hn )

3

(30-8)

4

The coefficient Ct is dependent on the type of structural system used. The code also allows use of Method B for the analytical evaluation of the fundamental period. It should be noted that the computation of the fundamental period using Equation 30-10 of this method can be cumbersome and time consuming. With widespread use of personal computers and structural analysis software in practice, a computer can determine periods much more easily than through use of Equation 30-10.

Steel special moment-resisting frame (SMRF) structure. Concrete special moment-resisting frame (SMRF) structure. Steel eccentric braced frame (EBF). Masonry shear wall building. Tilt-up building.


Code Reference

Steel special moment-resisting frame (SMRF) structure.

Height of the structure above its base is 96 feet. The additional 22-foot depth of the basement is not considered in determining hn for period calculation. C t = 0.035 T = C t (hn )

§1630.2.2

96′ Superstructure

Grade

3

4

= 0.035 (96)

3

4

= 1.07 sec. 22′


Basement

Example 19 Structure Period Using M ethod A

§1630.2.2

Concrete special moment-resisting frame (SMRF) structure.

Height of tallest part of the building is 33 feet, and this is used to determine period. Roof penthouses are generally not considered in determining hn , but heights of setbacks are included. However, if the setback represents more than a 130 percent change in the lateral force system dimension, then there is a vertical geometric irregularity (Table 16-L). For taller structures, more than five stories or 65 feet in height, dynamic analysis is required for this type of irregularity.

§1630.2.2

Setback

33′ 22′

C t = 0.030 T = C t (hn )

3

4

= 0.030 (33)

3

4

= 0.41 sec .

Steel eccentric braced frame (EBF).

§1630.2.2

EBF structures use the Ct for the category “all other buildings.” C t = 0.030 T = C t (hn )

44'

3

4

= 0.030 (44 )

3

= 0.51 sec .

Masonry shear wall building.

29'

§1630.2.2

29'

10'

Typ.

60'

Front wall elevation

4

10'

Typ.

45'

Back wall elevatio


Example 19 Structure Period Using Method

§1630.2.2

For this structure, Ct may be taken as 0.020, the value for “all other buildings,” or its value may be computed from the following formula: Ct =

0.1

§1630.2.2

Ac

where  D Ac = ∑ Ae 0.2 +  e   hn

  

2

 

(30-9)

Solving for De and Ae for front and back walls, respectively, the value of Ac can be determined. Front Wall Nominal CMU wall thickness = 8” Actual CMU wall thickness = 7.625” hn = 29 ft De = 60 ft Ae = (60'−4 x10') x

7.63 = 12.7 sq ft 12

De = 2.07 hn Back Wall De = 45 ft Ae = (45'−3x10') x

7.63 = 9.5 sq ft 12

De = 1.55 hn Using Equation 30-9, the value of Ac is determined. Note that the maximum value of De /hn that can be used is 0.9.

[ (

Ac = 12.7 0.2 + 0.9 2


) ]+ [9.5 ( 0.2 + 0.9 ) ]= 22.4 sq ft 2

Example 19 Structure Period Using M ethod A

§1630.2.2

Ct =

0.1

= 0.021

22.4

T = C t (h n )

3

4

= 0.021 (29 )

3

4

= 0.26 sec .

Alternately, the period can be determined using Ct = .020 for “all other buildings” T = C t (h n )

3

4

= 0.020 (29 )

3

4

= 0.25 sec .

Under current code provisions, either period can be used to determine base shear.

Tilt-up building.

Consider a tilt-up building 150 ft x 200 ft in plan that has a panelized wood roof and the typical wall elevation shown below.

20'

15' typ.

3' typ.

8' typ.

20' typ.

150'

Typical wall elevatio

C t = 0.020 T = C t (h n )

3

4

= 0.02 (20)

3

4

= 0.19 sec .

This type of structural system has relatively rigid walls and a flexible roof diaphragm. The code formula for period does not take into consideration the fact that the real period of the building is highly dependent on the roof diaphragm construction. Thus, the period computed above is not a good estimate of the real fundamental period of this type of building. It is acceptable, however, for use in determining design base shear. It should be noted that the actual diaphragm response is approximately taken into account in the design process by increased seismic force provisions on wall anchors and by the limit of R = 4 for calculation of diaphragm loads as required under §1633.2.9.3.


Example 20 Simplified Design Base Shea

§1630.2.3

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Determine the design base shear and the design lateral forces for a three-story wood structural panel wall building using the simplified design base shear. The soil profile type for the site is unknown. The following information is known: 1

Z = 0.4 Seismic source type B Distance to seismic source = 5 km R = 5.5 W = 750k

3

2

20'

Story weight

20'

Level

150k

3 12'

300k

2 12'

300k

1 12'


Check applicability of simplified method. Determine base shear. Determine lateral forces at each level.


Code Reference

Check applicability of simplified method.

§1629.8.2

Light frame construction not more than three stories, or other buildings not more than two stories can use the simplified method. ∴ o.k.

Determine base shear.

§1630.2.3

Because soil properties for the site are not known, a default/prescribed soil profile must be used. Section 1630.2.3.2 requires that a Type S D soil profile be used in seismic Zones 3 and 4. N a = 1.0 SEAOC Seismic Design Manual

Table 16-S

Example 20 Simplified Design Base Shear

§1630.2. 3

C a = 0.44 N a = 0.44 (1.0) = 0.44 V=

Table 16-Q

3.0C a 3.0 (0.44 )750 W= = (0.24)750 = 180 k 5.5 R

(30-11)

Determine lateral forces at each level.

Fx =

§1630.2.3.3

3.0C a w x = 0.24 w x R

(30-12)

F1 = 0.24 (300) = 72 k F2 = 0.24 (300 ) = 72 k F3 = 0.24 (150) = 36 k

Commentary The following is a comparison of simplified base shear with standard design base shear. The standard method of determining the design base shear is as follows: V=

2.5C a I 2.5 (0.44 )(1.0 ) W = W = 0.2W = 0.2 (750) = 150 kips R 5.5

(30-12)

The distribution of seismic forces over the height of the structure is Fx =

(V − Ft ) w x h x n

(30-15)

∑ wi hi i =1

where V − Ft = 150 kips since Ft = 0 in this example.


Example 20 Simplified Design Base Shea

§1630.2.3

w x hx

w x hx Σw i h i

Level x

hx

wx

3

36 ft

150 kips

5,400 k-ft

0.333

50.0 kips

0.33

2

24

300

7,200

0.444

66.7

0.22

1

12

300

3,600

0.222

33.3

0.11

Σw i hi = 16,200

Fx

Fx w x

Σv = 150.0

The design base shear V and the lateral force values Fx at each level are all less than those determined by the simplified method. The principal advantage of the simplified method is that there is no need to conform to the provisions listed in §1630.2.3.4, which are otherwise applicable. Another advantage is that the value of the near-source factor N a used to determine Ca need not exceed: 1.3 if irregularities listed in §1630.2.3.2 are not present and 1.1 if the conditions of §1629.4.2 are complied with It should be noted that Section 104.8.2 of the 1999 SEAOC Blue Book has different requirements for applicability of the simplified method: Single family two stories or less Light frame up to three stories Regular buildings up to two stories Blue Book §105.2.3 allows the near source factor N a = 1.0 for evaluation of C a . The Blue Book equation V = 0.8C aW does not contain the R factor, which eliminates the sometimes difficult problem of selecting the appropriate R value for small buildings that have complex and/or mixed lateral load resisting systems.


Example 21 Combination of Structural Systems: Vertical

§1630.4.2

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In structural engineering practice, it is sometimes necessary to design buildings that have a vertical combination of different lateral force-resisting systems. For example, the bottom part of the structure may be a rigid frame and top part a braced frame or shear wall. This example illustrates use of the requirements of §1630.4.2 to determine the applicable R values for combined vertical systems. For the three systems shown below, determine the required R factor and related design base shear requirements.


Code Reference

Steel ordinary braced frame over steel SMRF.

Steel ordinary braced frame R = 5.6

Steel special moment-resisting frame R = 8.5

This combined system falls under vertical combinations of §1630.4.2. Because the rigid system is above the flexible system, Item 2 of §1630.4.2 cannot be used. Therefore, under Item 1 of §1630.4.2, the entire structure must use R = 5.6 .

Concrete bearing wall over concrete SMRF.

Concrete bearing wall R = 4.5

Concrete special momentresisting frame R = 8.5



§1630.4.2

This combined system falls under vertical combinations of §1630.4.2. Because the rigid portion is above the flexible portion, Item 2 of §1630.4.2 cannot be used. Therefore, under Item 1 of §1630.4.2, the entire structure must use R = 4.5 .

Concrete SMRF over a concrete building frame system.

Applicable criteria. This is a vertical combination of a flexible system over a more rigid system. Under §1630.4.2, Item 2, the two stage static analysis may be used, provided the structures conform to §1629.8.3, Item 4.

Concrete SMRF R = 8.5, ρ = 1.5 Avg. stiffness upper portion = 175 k/in. Tupper = 0.55 sec Tcombined = 0.56 sec

Shear walls

Concrete building frame system R = 5.5, ρ = 1.0 Stiffness = 10,000 k/in. Tlower = 0.03 sec

Check requirements of §1629.8.3, Item 4: 1. Flexible upper portion supported on rigid lower portion. o.k. 2. Average story stiffness of lower portion is at least 10 times average story stiffness of upper portion. 10,000 k/in. > 10 (175) = 1,750 k/in. o.k. 3. Period of entire structure is not greater than 1.1 times period of upper portion. 0.56 sec < 1.1 (.55) = 0.61 sec o.k. ∴ Provisions of §1630.4.2, Item 2 can be used



§1630.4.2

Design procedures for upper and lower structures.

Design upper SMRF using R = 8.5 and ρ = 1.5

Vframe

Design the lower portion of the building frame system for the combined effects of the amplified V frame force and the lateral forces due to the base shear for the lower portion of the structure (using R = 5.5 and ρ = 1.0 for the lower portion).

(

 8.5 1.5   Vframe = 1.03 Vframe Amplified Vframe =    5.5 1.0 

Vbase

)

∴V base = Amplified V frame + (V lower )


Example 22 Combination of Structural Systems: Along Different Axes

§1630.4.3

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This example illustrates determination of R values for a building that has different structural systems along different axes (i.e., directions) of the building. In this example, a 3-story building has concrete shear walls in one direction and concrete moment frames in the other. Floors are concrete slab, and the building is located in Zone 4. Determine the R value for each direction. A

B

C

D

1 Shear wall

2

3

Typical floor plan

Lines A and D are reinforced concrete bearing walls: R = 4.5 Lines 1, 2 and 3 are concrete special moment-resisting frames: R = 8.5

Determine th R value for each direction.


Code Reference

In Zones 3 and 4, the provisions of §1630.4.3 require that when a structure has bearing walls in one direction, the R value used for the orthogonal direction cannot be greater than that for the bearing wall system. ∴ Use R = 4.5 in both directions.


§1630.4.3

Example 22 Combination of Structural Systems: Along Different Axes

Commentary The reason for this orthogonal system requirement is to provide sufficient strength and stiffness to limit the amount of out-of-plane deformation of the bearing wall system. A more direct approach would be to design the orthogonal system such that the ∆ M value is below the value that would result in the loss of bearing wall capacity. The design loads for the special moment-resisting frames are calculated using R = 4.5 . However, the frame details must comply with the requirements for the R = 8.5 system.


Example 23 Combination of Structural Systems: Along Same Axis

§1630.4.4

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Occasionally, it is necessary to have different structural systems in the same direction. This example shows how the R value is determined in such a situation. A one-story steel frame structure has the roof plan shown below. The structure is located in Zone 4. Determine the R value for the N/S direction. 1

2

3

North

4

Roof plan

Lines 1 and 4 are steel ordinary moment-resisting frames: R = 4.5 . Lines 2 and 3 are steel ordinary braced frames: R = 5.6 .

Determine th R value for the N/S direction.


Code Reference

In Zones 2, 3, and 4, when a combination of structural systems is used in the same direction, §1630.4.4 requires that the value of R used be not greater than the least value of the system utilized. ∴ Use R = 4.5 for entire structure.


Example 24 Vertical Distribution of Force

§1630.5

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A 9-story building has a moment resisting steel frame for a lateral force-resisting system. Find the vertical distribution of lateral forces Fx . The following information is given: 2

1

27'

3

27'

Level

Zone 4 W = 3,762 k C v = 0.56 R = 8.5 I = 1.0 T = 1.06 sec . V = 233.8 k

Story weight

9

214k

8

405k

7

405k

6

405k

5

584k

4

422k

3

422k

2

440k

1

465k

12' 12' 12' 12' 12' 12' 12' 12'

20'


Determine Ft. Find Fx at each level.


Code Reference

Determine Ft.

§1630.5

This is the concentrated force applied at the top of the structure. It is determined as follows. First, check that the Ft is not zero. T = 1.06 sec . > 0.7 sec ∴ Ft > 0 Ft = 0.07TV = 0.07 (1.06)(233.8) = 17.3 k

(30-14)


Example 24 Vertical Distribution of Force

§1630.5

Find Fx at each level.

The vertical distribution of seismic forces is determined from Equation 30-15. Fx =

(V − Ft ) w x h x

(30-15)

n

∑ wi h i =1

where

(V − Ft ) = (233.8 − 17.3) = 216.5k Since there are nine levels above the ground, n = 9 . Therefore Fx =

216.5 wx hx 9

∑ wi hi

i =1

This equation is solved in the table below. wx

w x hx

w x hx Σw i h i

116 ft

214 kips

24,824 k-ft

0.103

8

104

405

42,120

0.174

37.7

0.093

7

92

405

37,260

0.154

33.3

0.082

6

80

405

32,400

0.134

29.0

0.072

5

68

584

39,712

0.164

35.5

0.061

4

56

422

23,632

0.098

21.2

0.050

3

44

422

18,568

0.077

16.7

0.039

2

32

440

14,080

0.058

12.6

0.028

1

20

465

9,300

0.038

8.2

0.018

Σ =3,762

241,896

Level x

hx

9

Fx w x

Fx 22.3 + 17.3 = 39.6 kips

0.185

233.8

Commentary Note that certain types of vertical irregularity can result in a dynamic response having a load distribution significantly different from that given in this section. If the structural system has any of the stiffness, weight, or geometric vertical irregularities of Type 1, 2, or 3 of Table 16-L, then Item 2 of §1629.8.4 requires that the dynamic lateral force procedure be used unless the structure is less than five stories or 65 feet in height. The configuration and final design of this structure must be checked for these irregularities. Most structural analysis programs used in practice today perform this calculation, and it is generally not necessary to manually perform the calculations shown above. However, it is recommended that these calculations be performed to check the computer analysis and to gain insight to structural behavior.


Example 25 Horizontal Distribution of Shear

§1630.6

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A single story building has a rigid roof diaphragm. Lateral forces in both directions are resisted by shear walls. The mass of the roof can be considered to be uniformly distributed, and in this example, the weight of the walls is neglected. In actual practice, particularly with concrete shear walls, the weight of the walls should be included in the determination of the Center of Mass (CM). The following information is given: Design base shear: V = 100 k Wall rigidities: R A = 300 k/in. R B = 100 k/in. RC = R D = 200 k/in. Center of mass: x m = 40 ft y m = 20 ft

Y D Shear wall below

xR

e

A

B

40'

CM Roof diaphragm

CR

yR

V = 100k

xm = 40'

ym X C 80'

Roof plan


Eccentricity and rigidity properties. Direct shear in walls A and B. Plan irregularity requirements. Torsional shear in walls A and B. Total shear in walls A and B.


Example 25 Horizontal Distribution of Shea

§1630.6


Code Reference

Eccentricity and rigidity properties.

§1630.6

The rigidity of the structure in the direction of applied force is the sum of the rigidities of walls parallel to this force. R = R A + R B = 300 + 100 = 400 k/in. The center of rigidity (CR) along the x and y axes are xR =

R B (80' ) = 20 ft. R A +R B

yR =

R D (40 ′) = 20 ft R D + RC

eccentricity e = x m − x R = 40 − 20 = 20 ft Torsional rigidity about the center of rigidity is determined as J = R A (20)2 + R B (60)2 + RC (20)2 + R D (20)2 = 300 (20 )2 + 100 (60)2 + 200 (20 )2 + 200 (20)2 = 64 × 10 4 (k/in. ) ft 2 The seismic force V applied at the CM is equivalent to having V applied at the CR together with a counter-clockwise torsion T. With the requirements for accidental eccentricity e acc , the total shear on walls A and B can be found by the addition of the direct and torsional load cases: VD,A

VD,B

D

A

VT,A

B

20'

D

VT,D

A

VT,B

B

CR

T = V (e ± eacc)

CR 20' 20'

60'

V

C

Direct shear contribution


VT,C

C

Torsional shear contributio


§1630.6

Direct shear in walls A and B.

VD, A =

RA 300 × (V ) = × 100 = 75.0 kips R A + RB 300 + 100

V D,B =

RB 100 × (V ) = × 100 = 25.0 kips R A + RB 300 + 100

Plan irregularity requirements.

The determination of torsional irregularity, Item 1 in Table 16-M, requires the evaluation of the story drifts in walls A and B. This evaluation must include accidental torsion due to an eccentricity of 5 percent of the building dimension. e acc = 0.05 (80' ) = 4.0 ft The corresponding initial most severe torsional shears V ' using e acc = 4.0 ft are: V 'T ,A =

V ( e − e acc ) ( x R ) ( R A ) 100 ( 20 − 4) ( 20) (300) = = 15.0 kips J 64 × 10 4

V ' T ,B =

V ( e + eacc ) (80 − x R ) ( R B ) 100 ( 20 + 4) (60) (100) = = 22.5 kips J 64 × 10 4

Note: these initial shears may need to be modified if torsional irregularity exists and the amplification factor Ax > 1.0 . The initial total shears are: V ' A = V ' D , A − V ' T , A = 75.0 − 15.0 = 60.0 kips (Torsional shears may be subtracted if they are due to the reduced eccentricity e − e acc ) V ' B = V ' D , B + V ' T , B = 25.0 + 22.5 = 47.5 kips The resulting displacements δ ' , which for this single story building are also the story drift values, are: δ' A =

V ' A 60.0 = = 0.20 in. RA 300


Example 25 Horizontal Distribution of Shea

δ' B =

§1630.6

V ' B 47.5 = = 0.48 in. RB 100

δ avg =

0.20 + 0.48 = 0.34 in. 2

δ max = δ ' B = 0.48 in. δ max 0.48 = = 1.41 > 1.2 δ avg 0.34 ∴ Torsional irregularity exists. Section 1630.7 requires the accidental torsion amplification factor,  δ Ax =  max  1.2δ avg 

2

2    0 . 48  =  1.2 (0.34)  = 1.38 < 3.0    

(30-16)

Torsional shears in walls A and B.

The final most severe torsional shears are determined by calculating the new accidental eccentricity and using this to determine the torsional shears e acc = Ax (4.0) = (1.38) ( 4.0) = 5.54' VT , A =

VT , B =

100 (20 − 5.54) ( 20) (300) 64 × 10 4 100 ( 20 + 5.54) (60) (100) 64 × 10 4

= 13.6 kips

= 23.9 kips

Total shear in walls A and B.

Total shear in each wall is the algebraic sum of the direct and torsional shear components. V A = V D , A − VT , A = 75.0 − 13.6 = 61.4 kips V B = V D , B + VT , B = 25.0 + 23.9 = 48.9 kips



§1630.6

Commentary Section 1630.7 requires that “the most severe load combination for each element shall be considered for design.” This load combination involves the direct and torsional shears, and the “most severe” condition is as follows: 1. For the case where the torsional shear has the same sense, and is therefore added to the direct shear, the torsional shear shall be calculated using actual eccentricity plus the accidental eccentricity so as to give the largest additive torsional shear. 2. For the case where the torsional shear has the opposite sense to that of the direct shear and is to be subtracted, the torsional shear shall be based on the actual eccentricity minus the accidental eccentricity so as to give the smallest subtractive shear.


Example 26 Horizontal Torsional Moments

§1630.7

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This example illustrates how to include the effects of accidental eccentricity in the lateral force analysis of a multi-story building. The structure is a five-story reinforced concrete building frame system. A three-dimensional rigid diaphragm model has been formulated per §1630.1.2 for the evaluation of element actions and deformations due to prescribed loading conditions. Shear walls resist lateral forces in both directions. 1

2

3

4

5

4 @ 20' = 80' A

B

C

3 @ 20' = 60'

Shear wall, typ.

xc

CMx B

A

Fx

N D

Floor plan at Level x

The lateral seismic forces Fx in the north-south direction, structure dimensions, and accidental eccentricity eacc for each level x are given below: Level x 5

Fx 110.0 kips

eacc = 0.05Lx

Lx

x cx

80.0 ft

24.2 ft

± 4.0 ft

4

82.8

80.0

25.1

± 4.0

3

65.1

80.0

27.8

± 4.0

2

42.1

80.0

30.3

± 4.0

1

23.0

80.0

31.5

± 4.0

In addition, for the given lateral seismic forces Fx a computer analysis provides the following results for the second story. Separate values are given for the application of the forces Fx at the centers of mass and the ± 0.05 Lx displacements as required by §1630.6.


Example 26 Horizontal Torsional M oments

§1630.7

Force Fx Position

x c2

x c 2 − e acc

x c 2 + eacc

Wall shear V A

185.0 k

196.0 k

174.0 k

Wall shear VB

115.0 k

104.0 k

126.0 k

Story drift

∆δ A

0.35"

0.37"

0.33"

Story drift

∆δ B

0.62"

0.56"

0.68"

Level 2 displacement

δA

0.80"

0.85"

0.75"

Level 2 displacement

δB

1.31"

1.18"

1.44"

For the second story find the following:

Maximum force in shear walls A and B. Check if torsional irregularity exists. Determine the amplification factor Ax. New accidental torsion eccentricity.


Code Reference

Maximum force in shear walls A and B.

The maximum force in each shear wall is a result of direct shear and the contribution due to accidental torsion. From the above table, it is determined that V A = 196.0 k V B = 126.0 k

Check if torsional irregularity exists.

The building is L-shaped in plan. This suggests that it may have a torsion irregularity Type 1 of Table 16-M. The following is a check of the story drifts. ∆δ max = 0.68 in. ∆δ avg =

0.68 + 0.33 = 0.51 in. 2

∆δ max 0.68 = = 1.33 > 1.2 1.2 ∆δ avg 0.51 ∴Torsional irregulari ty exists


Example 26 Horizontal Torsional Moments

§1630.7

Determine the amplification factor Ax.

Because a torsional irregularity exists, §1630.7 requires that the second story accidental eccentricity be amplified by the following factor.  δ Ax =  max  1.2δ avg 

   

2

(30-16)

where δ max = δ B = 1.44 in. The average story displacement is computed as δ avg =

1.44 + 0.75 = 1.10 in. 2 2

 1.44   = 1.19 A2 =   (1.2) (1.10) 

New accidental torsion eccentricity.

Since A2 (i.e., Ax for the second story) is greater than unity, a second analysis for torsion must be done using the new accidental eccentricity. e acc = (1.19) ( 4.0' ) = 4.76 ft

Commentary Example calculations were given for the second story. In practice, each story requires an evaluation of the most severe element actions and a check for the torsional irregularity condition. If torsional irregularity exists and Ax is greater than one at any level (or levels), then a second torsional analysis must be done using the new accidental eccentricities. However, it is not necessary to find the resulting new Ax values and repeat the process a second or third time (until the Ax iterates to a constant or reaches the limit of 3.0). The results of the first analysis with the use of Ax are sufficient for design purposes. While this example involved the case of wall shear evaluation, the same procedure applies to the determination of the most severe element actions for any other lateral force-resisting system having rigid diaphragms. When the dynamic analysis method of §1631.5 is used, rather than static force procedure of §1630.2, the following equivalent static force option may be used in lieu


§1630.7

Example 26 Horizontal Torsional M oments

of performing the two extra dynamic analyses for mass positions at x cx ± (0.05L x ) as per §1631.5.6: 1. Perform the dynamic analysis with masses at the center of mass, and reduce results to those corresponding to the required design base shear. 2. Determine the Fx forces for the required design base shear, and apply pure torsion couple loads Fx (0.05L x ) at each level x . Then add the absolute value of these couple load results to those of the reduced dynamic analysis.


Example 27 Elements Supporting Discontinuous Systems

§1630.8.2

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A reinforced concrete building has the lateral force-resisting system shown below. Shear walls at the first floor level are discontinuous between Lines A and B and Lines C and D. The following information is given: Zone 4 Concrete shear wall building frame system: R = 5.5 and Ω o = 2.8 Office building live load: f1 = 0.5 Axial loads on column C: D = 40 kips L = 20 kips E h = 100 kips A

B

C

Table 16-N §1612.4

D

Level 4 12' 3 Shear wall 12' 2 12' 1

Column C 24" x 24" f'c = 4000 psi

12'

Determine the following for column C:

Required strength. Detailing requirements.


Code Reference

This example demonstrates the loading criteria and detailing required for elements supporting discontinued or offset elements of a lateral force-resisting system.



§1630.8.2

Required strength.

§1630.8.2.1

Because of the discontinuous configuration of the shear wall at the first story, the first story columns on Lines A and D must support the wall elements above this level. Column “C” on Line D is treated in this example. Because of symmetry, the column on Line A would have identical requirements. Section 1630.8.2 requires that the column strength be equal to or greater than Pu = 1.2 D + f1 L + 1.0 E m

(12-17)

Pu = 0.9 D ± 1.0 E m

(12-18)

where E m = Ω o E h = 2.8(100) = 280 kips

(30-2)

Substituting the values of dead, live and seismic loads Pu = 1.2 (40 ) + 0.5 (20) + 280 = 338 kips compression, and Pu = 0.9(40) − 1.0(280) = − 244 kips tension

Detailing requirements.

§1630.8.2.2

The concrete column must meet the requirements of §1921.4.4.5. This section requires transverse confinement tie reinforcement over the full column height if Pu >

Ag f 'c 10

24)2 (4 ksi ) ( = = 230 kips 10

Pu = 338 > 230 kips ∴Confinemen t is required over the full height

Commentary To transfer the shears from walls A-B and C-D to the first story wall B-C, collector beams A-B and C-D are required at Level 1. These would have to be designed according to the requirements of §1633.2.6. The load and detailing requirements of §1630.8.2, Elements Supporting Discontinuous Systems, apply to the following vertical irregularities and vertical elements:



§1630.8.2

1. Discontinuous shear wall. The wall at left has a Type 4 vertical structural irregularity.

Column

2. Discontinuous column. This frame has a Type 4 vertical structural irregularity.

Transfer girder

3. Out-of-plane offset. The wall on Line A at the first story is discontinuous. This structure has a Type 4 plan structural irregularity, and §1620.8.2 applies to the supporting columns. The portion of the diaphragm transferring shear (i.e., transfer diaphragm) to the offset wall must be designed for shear wall detailing requirements, and the transfer loads must use the reliability/redundancy factor ρ for the vertical-lateral-force-resisting system.

C B A

VE

Discontinued wall

Transfer diaphragm

Offset wall

Supporting columns

It should be noted that for any of the supporting elements shown above, the load demand Em of Equation (30-2) need not exceed the maximum force that can be transferred to the element by the lateral force-resisting system.



§1630.8.2

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This example illustrates the application of the requirements of §1630.8.2 for the allowable stress design of elements that support a discontinuous lateral force-resisting system. In this example, a light-framed wood bearing wall building with plywood shear panels has a Type 4 vertical irregularity in one of its shear walls, as shown below. The following information is given: Zone 4 R = 5.5 Ω o = 2.8 f1 = 0.5

Light framed wall with plywood sheathing

Axial loads on the timber column under the discontinuous portion of the shear wall are: Timber column

Dead D = 6.0 kips Live L = 3.0 kips Seismic E h = ±7.0 kips Determine the following:

Applicable load combinations. Required column design strength.


Code Reference

Applicable load combinations.

For vertical irregularity Type 4, §1630.8.2.1 requires that the timber column have the “design strength” to resist the special seismic load combinations of §1612.4. This is required for both allowable stress design and strength design. These load combinations are:

1.2 D + f1 L + 1.0 E m

(12-17)

0.9 D ± 1.0 E m

(12-18)



§1630.8.2

Required column design strength.

In this shear wall, the timber column carries only axial loads. The appropriate dead, live and seismic loads are determined as: D = 6.0 kips L = 3.0 kips E m = Ω o E h = 2.8 (7.0 ) = 19.6 kips

(30-2)

For the required “design strength” check, both Equations (12-17) and (12-18) must be checked. P = 1.2 D + f 1 L + E m

(12-17)

P = 1.2 (6.0) + 0.5 (3.0) + 19.6 = 28.3 kips P = 0.9 D ± 1.0 E m

(12-18)

P = 0.9 (6.0) ± 1.0 (19.6 ) = 25.0 kips or − 14.2 kips

Commentary For allowable stress design, the timber column must be checked for a compression load of 28.3 kips and a tension load of 14.2 kips . In making this “design strength” check, §1630.8.2.1 permits use of an allowable stress increase of 1.7 and a resistance factor, φ , of 1.0 . The 1.7 increase is not to be combined with the one-third increase permitted by §1612.3.2, but may be combined with the duration of load increase C D = 1.33 given in Table 2.3.2 of Chapter 23, Division III. The resulting “design strength” = (1.7 )(1.0)(1.33) (allowable stress). This also applies to the mechanical hold-down element required to resist the tension load. The purpose of the “design strength” check is to check the column for higher and, hopefully, more realistic loads that it will be required to carry because of the discontinuity in the shear wall at the first floor. This is done by increasing the normal seismic load in the column, E h , by the factor Ω o = 2.8 .


Example 29 At Foundation

§1630.8.3

&7 ,

$%(4'3'(

Foundation reports usually provide soil bearing pressures on an allowable stress design basis while seismic forces in the 1997 UBC, and most concrete design, are on a strength design basis. The purpose of this example is to illustrate footing design under this situation. A spread footing supports a reinforced concrete column. The soil classification at the site is sand (SW). The following information is given: P

Zone 4 ρ = 1.0 for structural system PD = 80 k M D = 15 k - ft PL = 30 k M L = 6 k - ft PE = ± 40 k V E = 30 k Snow load S = 0

Grade

M V 4' 2'

M E = ± 210 k - ft

Find the following:

Determine the design criteria and allowable bearing pressure. Determine footing size. Check resistance to sliding. Determine soil pressures for strength design of the footing section.


Code Reference

Determine the design criteria and allowable bearing pressure.

§1630.8.3

The seismic force reactions on the footing are based on strength design. However, §1629.1 states that allowable stress design may be used for sizing the foundation using the load combinations given in §1612.3. Here it is elected to use the alternate basic load combinations of §1612.3.2. D+L+S D+L+

E 1.4

(12-13)

E 1.4 Because foundation investigation reports for buildings typically specify bearing pressures on an allowable stress design basis, criteria for determining footing size are also on this basis. 0.9 D ±

(12-12)

(12-16-1)



§1630.8.3

The earthquake loads to be resisted are specified in §1630.1.1 by Equation 30-1. E = ρE h + E v

(30-1)

Since Ev = 0 for allowable stress design, Equation 30-1 reduces to E = ρE h = (1.0 ) E h Table 18-1-A of §1805 gives the allowable foundation pressure, lateral bearing pressure, and the lateral sliding friction coefficient. These are default values to be used in lieu of site-specific recommendations given in a foundation report for the building. They will be used in this example. For the sand (SW) class of material and footing depth of 4 feet, the allowable foundation pressure p a is p a = 1.50 + (4 ft − 1 ft )(0.2 )(1.50) = 2.40 ksf

Table 18-1-A and Footnote 2

A one-third increase in pa is permitted for the load combinations that include earthquake load.

Determine footing size.

The trial design axial load and moment will be determined for load combination of Equation (12-13) and then checked for the other combinations. Pa = D + L +

P 40 E = PD + PL + E = 80 + 30 + = 138.6 kips 1.4 1.4 1.4

Ma = D + L +

M 210 E = M D + M L + E = 15 + 6 + = 171.0 k - ft 1.4 1.4 1.4

(12-13)

(12-13)

Select trial footing size. Try 9 ft x 9 ft footing size, B = L = 9 ft A = BL = 81 ft 2 ,

S=

BL2 9 3 = = 121.5 ft 3 6 6

Calculated soil pressures due to axial load and moment p=

Pa M a 138.6 171.0 + = + = 1.71 + 1.41 = 3.12 ksf A S 81 121.5

Check bearing pressure against allowable with one-third increase,



§1630.8.3

3.12 ksf < 1.33 p a = 1.33 (2.40) = 3.20 ksf , o.k. Check for the load combination of Equation (12-16-1). Pa = 0.9 D ±

P 40 E = 0.9 PD ± E = 0.9 (80) ± = 100.6 kips or 43.4 kips 1.4 1.4 1.4

(12-16-1)

M 210 E = 0.9 M D ± E = 0.9 (15) ± = 163.5 k - ft or 136.5 k - ft 1.4 1.4 1.4 (12-16-1)

M a = 0.9 D ±

Eccentricity e =

M a 163.5 k - ft 136.5 k - ft = = 1.63 ft, or = .15 ft, ∴ e = 3.15 ft governs. Pa 100.6 43.4

Check for uplift. e>

L L 9 = = 1.5 ft (where is the limit of the kern area) 6 6 6

Since e = 3.15 > 1.5, there is partial uplift, and a triangular pressure distribution is assumed to occur. Center line

4.5'

For the footing free-body: p Pa = R p = (3a )B 2 R p must be co-linear with Pa

4.5'

e

a Pa

such that the length of the triangular pressure distribution is equal to 3a . R p = Pressure resultant

p

a Rp 3a

The load combination 0.9 D −

E , with Pa = 43.4 kips and M a = 136.5 k - ft 1.4

(12-10)

governs bearing pressure a=

B − e = 4.5 − 3.15 = 1.35 ft 2



Pa =

§1630.8.3

p (3a ) B 2

or p=

  2  1  2 1 Pa   = (43.4 )   = 2.38 ksf < 1.33 p a = 3.20 ksf o.k. 3  aB  3  (1.35)(9.0)

If p had been greater than 1.33 p a , the footing size would have to be increased. Finally, check the gravity load combination (12-12) for p < p a = 3.2 ksf . Pa = D + L = PD + PL = 80 + 30 = 110 kips

(12-12)

M a = D + L = M D + M L = 15 + 6 = 21 k - ft

(12-12)

p=

Pa M a 110 21 + = + = 1.53 ksf < 3.2 ksf, o.k. 81 121.5 A S

All applicable load combinations are satisfied, therefore a 9ft x 9ft footing is adequate.

Check resistance to sliding.

Unless specified in the foundation report for the building, the friction coefficient and lateral bearing pressure for resistance to sliding can be determined from Table 18-1-A. These values are: Friction coefficient

µ = 0.25

Table 18-1-A

Lateral bearing resistance p L = 150 psf × depth below grade

Table 18-1-A

Assume the footing is 2 feet thick with its base 4 feet below grade. Average 300 + 600 = 450 psf . resistance on the 2 feet deep by 9 feet wide footing face is 2 p L = 450 psf = 0.45 ksf Load combination of Equation (12-16-1) will be used because it has the lowest value of vertical load 0.9 D = 0.9 PD ). The vertical and lateral loads to be used in the sliding resistance calculations are: P = 0.9 PD = 0.9 (80) = 72 kips


2' 300 psf 2' 600 psf Face of footing


§1630.8.3

Lateral load =

V E 30 = = 21.4 kips 1.4 1.4

The resistance due to friction is P (µ ) = 72(0.25) = 18.0 kips The resistance from lateral bearing is p L (face area) = 0.45 (2 ′ × 9 ′) = 8.1 kips The total resistance is then the sum of the resistance due to friction and the resistance due to lateral bearing pressure. Total resistance = 18.0 + 8.1 = 26.1 > 21.4 kips, o.k. ∴ No sliding occurs

Determine soil pressures for strength design of footing section.

To obtain the direct shear, punching shear, and moments for the strength design of the reinforced concrete footing section, it is necessary to compute the upward design soil pressure on the footing due to factored strength loads: 1.2 D + 1.0 E + f1 L

(12-5)

0.9 D ± 1.0 E

(12-6)

The section design must have the capacity to resist the largest moments and forces resulting from these load combinations.

Soil pressure due to load combination 1.2 D + 1.0 E + f 1 L . f1 = 0.5

§1612.2.1

Pu = 1.2 PD + 1.0 PE + 0.5PL = 1.2 (80) + 1.0 (40) + 0.5 (30) = 151 kips

(12-5)

M u = 1.2 M D + 1.0 M E + 0.5M L = 1.2 (15) + 1.0 (210) + 0.5 (6) = 231 k - ft

(12-5)



Eccentricity e =

§1630.8.3

M u 231 = = 1.53 ft Pu 151 Face of column

L 9 e > = = 1.5 ft 6 6 Therefore partial uplift occurs. a = 4.5 − e = 4.5 = 1.53 = 2.97 ft p=

p = 3.77 ksf

3a = 8.91'

  2  1  2 1  = 3.77 ksf Pu   = (151)  3  aB  3  (2.97 )(9.0) 

Soil pressure due to load combination 0.9 D ± 1.0 E : Pu = 0.9 PD ± 1.0 PE = 0.9 (80) ± 1.0 (40) = 112 kips or 32 kips

(12-6)

M u = 0.9 M D ± 1.0M E = 0.9 (15) ± 1.0 (210 ) = 223.5 k - ft or 196.5 k - ft

(12-6)

Compute pressure load due to Pu = 112 kips and M u = 223.5 k - ft Eccentricity e =

M u 223.5 = = 2.00 ft Pu 112 Face of column

e>

L = 1.5 ft 6 p = 3.32 ksf

therefore partial uplift occurs. a = 4.5 − e = 4.5 − 2.0 = 2.50 ft p=

3a = 7.50'

  2  1  2 1  = 3.32 ksf Pu   = (112 )  ( )( ) 3  aB  3 2 . 50 9 . 0  

The footing pressure is less than that for the combination of 1.2 D + 1.0 E + f 1 L . Therefore the 1.2 D + 1.0 E + f1 L combination governs. Note that the resulting direct shear, punching shear, and moments must be multiplied by 1.1 per Exception 1 of §1612.2.1. (Note: At the time of publication, the 1.1 factor is under consideration for change to 1.0). Note also that the value of p due to the strength design factored loads need not be less than 1.33 p a = 3.20 ksf, since it is used as a load for concrete section design rather than for determining footing size.


Example 30 Dri

§1630.9

(4

$%(4'A four-story special moment-resisting frame (SMRF) building has the typical floor plan as shown below. The elevation of Line D is also shown, and the following information is given: B

A

C

D

Zone 4 I = 1.0 R = 8.5 Ω o = 2.8 T = 0.60 sec

Seismic force

Typical floor plan

∆S

Deflected shape

Level 4 12' 3 12' 2 12' 1 12'

Elevation of Line D

The following are the design level response displacements ∆ S (total drift) for the frame along Line D. These values include both translational and torsional (with accidental eccentricity) effects. As permitted by §1630.10.3, ∆ S has been determined due to design forces based on the unreduced period calculated using Method B. Level

∆S

4

1.51 in

3

1.03

2

.63

1

.30

SEAOC Seismic Design M anual

Example 30 Dri

§1630.9

For the frame on Line D, determine the following:

Maximum inelastic response displacements ∆ M .

Story drift in story 3 due to ∆ M . Check story 3 for story drift limit.


Code Reference

Maximum inelastic response displacements ∆ M .

§1630.9.2

These are determined using the ∆ S values and the R-factor ∆ M = 0.7 R∆ S = 0.7 (8.5)(∆ S ) = 5.95∆ S

(30-17)

Therefore ∆S

∆M

4

1.51 in

8.98 in

3

1.03

6.12

2

0.63

3.75

1

0.30

1.79

Level

Story drift in story 3 due to ∆ M .

§1630.10

Story 3 is located between Levels 2 and 3. Thus ∆ M drift = 6.12 − 3.75 = 2.37 in.

Check story 3 for story drift limit.

§1630.10.2

For structures with a fundamental period less than 0.7 seconds, §1630.10.2 requires that the ∆ M story drift not exceed 0.025 times the story height. For story 3 Story drift using ∆ M = 2.37 in. Story drift limit = .025 (144) = 3.60 in > 2.37 in. ∴ Story drift is within limits


Example 31 Story Drift Limitations

§1630.10

(

$%(4'4

"1

For the design of new buildings, the code places limits on story drifts. The limits are based on the maximum inelastic response displacements and not the design level response displacements determined from the design base shear of §1630.2. In the example given below, a four-story steel special moment-resisting frame (SMRF) structure has the design level response displacements ∆ S shown. These have been determined according to §1630.9.1 using a static, elastic analysis. A

B

Level

Zone 4 T = 0.60 sec. R = 8.5

C

D

∆S

Deflected shape

4

2.44 in.

3

1.91

2

1.36

1

0.79

12' 12' 12'

16'


Maximum inelastic response displacements. Compare story drifts with the limit value.


0

Code Reference

Maximum inelastic response displacements.

§1630.9.1

Maximum inelastic response displacements, ∆ M , are determined from the following: ∆ M = 0.7 R∆ S

(30-17)

∴ ∆ M = 0.7 (8.5) ∆ S = 5.95∆ S

Compare story drifts

ith the limit value.

§1630.10.2

Using ∆ M story displacements, the calculated story drift cannot exceed 0.025 times the story height for structures having a period less than 0.7 seconds. Check building period. T = .60 sec < .70 sec


Example 31 Story Drift Limitations

§1630.10

Therefore, limiting story drift is 0.025 story height. Determine drift limit at each level. Levels 4, 3, and 2 ∆ M drift ≤ .025h = .025 (12 ft × 12 in./ft ) = 3.60 in.

§1630.10.2

Level 1 ∆ M drift ≤ .025h = .025 (16 ft × 12 in./ft ) = 4.80 in. For ∆ M drift = ∆ Mi − ∆ Mi−1 , check actual story drifts against limits: Level i

∆S

∆M

∆ M drift

Limit

Status

4

2.44 in.

14.52 in.

3.16 in.

3.60 in.

o.k.

3

1.91

11.36

3.27

3.60

o.k.

2

1.36

8.09

3.39

3.60

o.k.

1

0.79

4.70

4.70

4.80

o.k.

Therefore, the story drift limits of §1630.10 are satisfied.

Commentary Whenever the dynamic analysis procedure of §1631 is used, story drift should be determined as the modal combination of the story drift for each mode. Determination of story drift from the difference of the combined mode displacements may produce erroneous results because maximum displacement at a given level may not occur simultaneously with those of the level above or below. Differences in the combined mode displacements can be less than the combined mode story drift.


Example 32 Vertical Component

§1630.11

(& /

$%(4'

Find the vertical seismic forces on the non-prestressed cantilever beam shown below. The following information is given: Beam unit weight = 200 plf C a = 0.40 I = 1.0 Z = .4

A

10'

Find the following:

Upward seismic forces on beam. Beam end reactions.


Code Reference

Upward seismic forces on beam.

§1630.11

In Seismic Zones 3 and 4, the design of horizontal cantilever beams must consider a net upward seismic force. The terminology of “net upward seismic force” is intended to specify that gravity load effects cannot be considered to reduce the effects of the vertical seismic forces and that the beam must have the strength to resist the actions due to this net upward force without consideration of any dead loads. This force is computed as q E = 0.7C a IW p = 0.7 (0.40)(1.0)(200 plf ) = 56 plf

§1630.11

Beam end reactions.

qE

MA VA

V A = q E l = 56 plf (10 ft ) = 560 lbs M A = qE

l 2 56 (10) 2 = = 2,800 lb/ft 2 2

The beam must have strengths φ V n and φ M n to resist these actions.


Example 33 Design Response Spectrum

§1631.2

(( # 5 "

$%('&

Determine the elastic design response spectrum for a site in Zone 4 with the following characteristics: Soil Profile Type S D Seismic source type C Distance to nearest seismic source = 23 km

Determine design response spectrum.


Code Reference

The design response spectrum can be determined, under §1631.2, using Figure 16-3 of the code and the coefficients C a and C v . The values of C a and C v are determined from the soil profile type, seismic source type, and distance to nearest source. In Zone 4, the values of C a and C v are dependent upon the near field factors N a and N v , respectively, as given in Tables 16-Q and 16-R. Determine N a and N v

§1629.4.2

From Table 16-S with seismic source type C and distance of 23 km. N v = 1.0 From Table 16-T with seismic source type C and distance of 23 km. N v = 1.0 Determine C a and C v

§1629.4.3

From Table 16-Q with Soil Profile Type SD and Z = 0.4 C a = 0.44 N a = (0.44 )(1.0) = 0.44 From Table 16-R with Soil Profile Type SD and Z = 0.4 C v = 0.64 N v = (0.64 )(1.0) = 0.64 Once the values of C a and C v for the site are established, the response spectrum can be constructed using Figure 16-3. The peak ground acceleration (PGA) is the value of spectral acceleration at the zero period of the spectrum (T = 0). In this case it is 0.44g.



§1631.2

PGA is designated as the coefficient C a by the code. This is also called the zero period acceleration (ZPA). The peak of the response spectrum for 5 percent damping is 2.5 times C a . In this example, it is 2.5C a = (2.5)(0.44 ) = 1.1g The control periods To and T s are Ts =

0.64 Cv = = 0.58 sec 2.5C a (2.5 × .44 )

Figure 16-3

To = 0.2Ts = (0.2 )(0.58) = 0.12 sec The long period portion of the spectrum is defined as C v 0.64 = T T

Figure 16-3

From this information the elastic design response spectrum for the site can be drawn as shown below.

1.5

Sa = 1.1

1.0

0.44

To = 0.12 sec

0.5

To = 0.58 sec

Sa (g)

Sa = 0.64 / T

0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

T (sec)



§1631.2

Commentary The spectrum shown above is for 5 percent damping. If a different damping is used, the spectral accelerations of the control periods To and T s and values of C v / T must be scaled. However, the value of C a is not scaled.


Example 34 Dual Systems

§1631.5.7

(* "1

$%('0'2

This example illustrates the determination of design lateral forces for the two basic elements of a dual system. Section 1629.6.5 prescribes the following features for a dual system: 1. An essentially complete space frame for gravity loads. 2. Resistance to lateral load is provided primarily by shear walls or braced frames, but moment-resisting frames must be provided to resist at least 25 percent of the design base shear. 3. The two systems are designed to resist the total design base shear in proportion to their relative rigidities. In present practice, the frame element design loads for a dual system are usually a result of a computer analysis of the combined frame-shear wall system. In this example, a dynamic analysis using the response spectrum procedure of §1631.5 has been used to evaluate the seismic load E h at point A in the dual system of the building shown below. This is the beam moment M A . The building is classified as regular and the Eh values have been scaled to correspond to 90 percent of the design base shear determined under the requirements of §1630.2. The following information is given: Shear wall

Moment frame

Zone 4 I = 1.0 Reduced dynamic base shear V D = 0.9V = 400 kips E h = M A = 53.0 k-ft T = 0.50 sec

Eh = MA = 53.0 k-ft

Point A

VD = 400 kips

Determine the following for the moment frame system:

Design criteria.

Required design lateral seismic forces F x .

Moment at A



§1631.5.7


Code Reference

Design criteria.

Section 1629.6.5 Item 2 requires that the moment-resisting frame be designed to independently resist at least 25 percent of the design base shear, which in this case would be 0.25VD. Section 1631.5.7 allows the use of either the static force method of §1630.5 or the response spectrum analysis of §1631.5, scaled to the 0.25VD base shear. Since the independent frame, without shear wall interaction, is an idealization that never really exists, the use of the response spectrum analysis is not particularly appropriate since the true dynamic characteristics would be those of the combined frame and wall system. The purpose of a response spectrum analysis is to better define the lateral load distribution, and this would not be achieved by an analysis of the independent frame. Therefore, the use of the static force option is judged to be more consistent with the simple requirement that the frame strength should meet or exceed 0.25VD. ∴V D of frame = 0.25V D = 0.25 (400) = 100 kips

Required design lateral seismic forces F x .

Design base shear on the frame due to 0.25V D = 100 kips This base shear must be distributed over the height of the structure, and the design lateral seismic forces at each level are determined from Fx =

(V − Ft ) w x h x Σwi hi

(30-15)

where

(V − Ft ) = 0.25V D

= 100 kips

In this example, Ft = 0 because the building period of 0.50 seconds is less than 0.7 seconds. ∴ Fx =

100w x h x Σwi hi



§1631.5.7

Moment at A

Apply the F x forces to the frame structure and find the resulting seismic moments, denoted M ' A . At point A, E' h = M' A = 75.2 k-ft > M A = 53.0 k-ft The seismic moment at A must be the larger of the two values. ∴ M ' A = 75.2 k − ft In actual application, each frame element load E h due to V D in the dual system must be compared with the E'h value due to 0.25V D in the independent frame, and the element must be designed for the larger of E h or E'h .

Commentary Use of a dual system has the advantage of providing the structure with an independent vertical load carrying system capable of resisting 25 percent of the design base shear while at the same time the primary system, either shear wall or braced frame, carries its proportional share of the design base shear. For this configuration, the code permits use of a larger R value for the primary system than would be permitted without the 25 percent frame system. The dual system has been in the code for many years. The widespread use of computers in structural analysis revealed that the interaction between the frame and the shear wall (or braced frame) system produced results quite different than those obtained by the often cumbersome approximate methods used with hand calculations. For example, a shear wall system in a highrise building was found to be “loading” the frame system at the upper stories. Consequently, a dual system should be carefully analyzed as a combined system to detect critical interaction effects.


Example 35 Lateral Forces for One-Story Wall Panels

§1632.2

(0

, ;+"1 <

$%(&'&

This example illustrates the determination of the total design lateral seismic force on a tilt-up wall panel supported at its base and at the roof diaphragm level. For the tilt-up wall panel shown below, determine the out-of-plane seismic forces required for the design of the wall section. This is usually done for a representative one-foot width of the wall length, assuming a uniformly distributed out-of-plane loading. The following information is given: Top of parapet

Roof framing 4'

Zone 4 I p = 1.0

Roof

C a = 0.4 Panel thickness = 8 inches Normal weight concrete (150 pcf)

fp Tilt-up panel 20'

Assumed pin support


Ground

Out-of-plane forces for wall panel design. Shear and moment diagrams for wall panel design. Loading, shear and moment diagrams for parapet design.


Code Reference

Out-of-plane forces for wall panel design.

§1632.2

Under §1632.2, design lateral seismic forces can be determined using either: a.) Equation (32-1), or b.) Equation (32-2) with the limits of Equation (32-3). F p = 4.0C a I pW p Fp =

a p Ca I p  h  1 + 3 x Rp  hr

(32-1)   W p 

0.7C a I pW p ≤ F p ≤ 4.0C a I pW p

(32-2)

(32-3)

Generally, it is more advantageous to use Equation (32-2) with the Equation (32-3) limits, and this will be used in this example. SEAOC Seismic Design Manual


§1632.2

The wall panel is laterally supported at its base and at the roof. The value of F p to be used must represent the average of the acceleration inputs from these two attachment locations. Thus, the out-of-plane seismic forces on the wall panel are determined from the “average” of the seismic coefficients at the roof and the base. As will be shown below, the minimum force level from Equation (32-3) controls the seismic coefficient at the base. Using the coefficient method, a general expression for the force F p applied midway between the base and the top of the parapet is derived below. a p = 1.0

Table 16-O

R p = 3.0

Table 16-O

At roof level, h x = hr , and the effective seismic coefficient from Equation (32-2) is

(1.0 ) C a I p 3.0

 h 1 + 3 r  hr 

  = 1.33C a I p < 4.0C a I p  

∴ use 1.33C a I p At base level, h x = 0 , and the effective seismic coefficient from Equation (32-2) is

(1.0 ) C a I p 3.0

 0 1 + 3 hr 

  = 0.33C a I p < 0.7C a I p 

∴ use 0.7C a I p The average coefficient over the entire height of the wall may be taken as

(1.33 + 0.70 ) 2

C a I p = 1.02C a I p

The force F p is considered to be applied at the mid-height (centroid) of the panel, but this must be uniformly distributed between base and top of parapet. For the given C a = 0.4 and I p = 1.0 , the wall panel seismic force is F p = 1.02 (0.4 )(1.0)W p = 0.408W p



§1632.2

The weight of the panel between base and the top of the parapet is 8 W p =   (150) (24) = 2,400 lbs per foot of width  12  F p = 0.408 (2,400) = 979 lbs/ft The force F p is the total force on the panel. It acts at the centroid. For design of the panel for out-of-plane forces, F p must be expressed as a distributed load f p : fp =

979 lbs/ft = 40.8 plf/ft 24 ft

Shear and moment diagrams for wall panel design.

Using the uniformly distributed load f p , the loading, shear and moment diagrams are determined for a unit width of panel. The 40.8 plf/ft uniform loading is also applied to the parapet. See step 3, below, for the parapet design load. 40.8 plf/ft 4'

RR

-424

163

-326

20'

1883

9.6'

RB 392

Loading

Shear (lbs/ft)

Moment lb-ft/ft

When the uniform load is also applied to the parapet, the total force on the panel is 40.8 plf/ft (24ft ) = 979 plf RR =

979 (12) = 587 lb/ft 20

R B = 979 − 587 = 392 lb/ft The shears and moments are the E h load actions for strength design. However, the reaction at the roof, R R , is not the force used for the wall-roof anchorage design. SEAOC Seismic Design Manual


§1632.2

This anchorage force must be determined under §1633.2.8.1 when the roof is a flexible diaphragm.

Loading, shear and moment diagrams for parapet design.

Table 16-O requires a p = 2.5 , and R p = 3.0 for unbraced (cantilevered) parapets. The parapet is considered as an element with an attachment elevation at the roof level.

hx = hr The weight of the parapet is 8 W p =   (150 )(4 ) = 400 lbs per foot of width  12  The concentrated force applied at the mid-height (centroid) of the parapet is determined from Equation (32-2). Fp =

a p Ca I p  h  1 + 3 x Rp  hr

Fp =

2.5(0.4 )(1.0 )  20  1 + 3  W p 3.0 20  

  W p 

(32-2)

F p = 1.33W p = 1.33 (400) = 532 lbs/ft < 4.0C a I pW p = 1.6W p

o.k.

The equivalent uniform seismic force is fp =

532 = 133 plf/ft for parapet design 4

133 plf/ft

4'

RR 532

Shear (lbs/ft)

-1064

Moment (lb-ft/ft)

Loading


Example 36 Lateral Forces for Two-Story Wall Panel

§1632.2

(%

, +"1 <

$%(&'&

This example illustrates determination of out-of-plane seismic forces for the design of the two-story tilt-up wall panel shown below. In this example, a typical solid pane (no door or window openings) is assumed. Walls span from floor to floor to roof. The typical wall panel in this building has no pilasters and the tilt-up walls are bearing walls. The roof consists of 1½-inch, 20 gauge metal decking on open web steel joists and is considered a flexible diaphragm. The second floor consists of 1½-inch, 18 gauge composite decking with a 2½-inch lightweight concrete topping. This i considered a rigid diaphragm. The following information is given: Roof

Zone 4 I p = 1.0

2'

Wall panel 20'

C a = 0.4 Wall weight = 113 psf

nd

2 Floor

38'

Assumed pinned

16'


Out-of-plane forces for wall panel design. Out-of-plane forces for wall anchorage design.


Wall sectio

Code Reference

Out-of-plane forces for wall panel design.

§1632.2

Requirements for out-of-plane seismic forces are specified in §1632.2 for Zones 3 and 4. Either Equations (32-1) or (32-2) and (32-3) are used to determine the forces on the wall. F p = 4.0C a I p W p Fp =

a p Ca I p  h 1 + 3 × x Rp  hr

(32-1)   W p 

0.7C a I p W p ≤ F p ≤ 4.0C a I p W p R p = 3.0 and a p = 1.0 SEAOC Seismic Design Manual

(32-2)

(32-3) Table 16-O, Item 1.A.(2)


§1632.2

To determine out-of-plane forces over the height of the wall, seismic coefficients at the roof, second floor, and first floor are determined. An out-of-plane force, Fp , is determined for each story from the average of the seismic coefficients at the support points for that story. The required coefficients are evaluated as follows. Seismic coefficient at roof: a p Ca I p  36  h  1.0 (0.4 ) (1.0)  1 + 3 × x  = 1 + 3 ×  = 0.533 3.0 36  Rp  hr   4.0Ca I p = 4.0 (0.4 )(1.0) = 1.60 > 0.533 ∴ use 0.533 Seismic coefficient at second floor: a p Ca I p  16  h  1.0 (0.4 ) (1.0)  1 + 3 × x  = 1 + 3 ×  = 0.311 36  3.0 Rp  hr   Seismic coefficient at first floor: a p Ca I p  0  h  1.0 (0.4 ) (1.0)  1 + 3 × x  = 1 + 3 ×  = 0.133 36  3.0 Rp  hr   0.7Ca I p = 0.7 (0.4 ) (1.0) = 0.28 > 0.133 ∴ use 0.28 Using the average of the coefficient for the given story, the out-of-plane seismic forces are determined as follows: W p 2 = 113 (20 + 2 ) = 2,486 plf W p1 = (113) (16) = 1,808 plf Fp 2 = Fp1 =

(0.533 + 0.311)

W p 2 = 0.422W p 2 = 0.422 (2,486) = 1,049 plf

2

(0.311 + 0.280 ) 2

W p1 = 0.296W p1 = 0.296 (1,808) = 535 plf



§1632.2

F p 2 and Fp1 are the out-of-plane forces

acting on the centroids of the second and first level portions, respectively, of the tiltup wall panel. For design of the wall these forces must be uniformly distributed over their tributary height. Panel desi forces are given below.

Roof

2'

Fp2 20' nd

2 Floor 27'

Fp1

f p2 =

Fp2

(20 + 2 )

=

16'

8'

1,049 = 47.7 plf 22

Out-of-plane forces at centroids

f p1

F p1

535 = = = 33.4 plf 16 16

Alternatively, panel design forces can be determined using seismic coefficients as shown below.

22'

f p 2 = .422 (113) = 47.7 psf

fp2 = 47.7 psf

R3 = 572 plf

fp1 = 33.4 psf

16'

is not shown.

20'

R2 = 744 plf

f p1 = .296 (113) = 33.4 psf

Note that the 2-foot high parapet must be designed for seismic forces determined from Equations (32-2) and (32-3) with R p = 3.0 and a p = 2.5 . This calculation

2'

16'

R1 = 267 plf

Out-of-plane wall forces

Out-of-plane forces for wall anchorage design.

§1633.2.8.1

For design of wall anchorage, §1633.2.8.1 requires use of higher design forces than those used for panel design. Anchorage forces are determined using Equations (32-1), or (32-2) and (32-3), where W p is the weight of the panel tributary to each anchorage level. Values of R p and a p to be used at the second floor and roof are: R p = 3.0 and a p = 1.5

§1633.2.8.1, Item 1

The building of this example has a flexible diaphragm at the roof and a rigid diaphragm at the second floor. Because the code is not clear about wall anchorage requirements for buildings with both rigid and flexible diaphragms, the requirements for flexible diaphragms will be used for determination of anchorage forces at both SEAOC Seismic Design Manual


§1632.2

levels. Equation (32-3), with the limits of Equation (32-3), will be used with hx equal to the attachment height of the anchorage. Seismic anchorage force at roof:  20  W3 = (113)  + 2  = 1,356 plf  2  Fp =

a p Ca I p  h  1 + 3 x Rp  hr

F3 =

1.5 (0.4 )(1.0)  36   1 + 3  W3 = 0.8 (1,356) = 1,085 plf 3.0 36  

  W p 

(32-2)

Check limit of Equation (32-3) 4.0C a I pW p = 4.0 (0.4 )(1.0 )W3 = 1.6W3 > 0.8W3 o.k. 1,085 plf > 420 plf

(32-3) §1633.2.8.1, Item 1

∴ F3 = 1,085 plf Seismic anchorage force at second floor:  20 + 16  W2 = (113)   = 2,034 plf  2  F2 =

1.5 (0.4 )(1.0 )  16  1 + 3  W2 = .467 (2,034 ) = 950 plf 3.0 36  

(32-2)

Seismic anchorage force at first floor: At the first floor, a p = 1.0 because there is no diaphragm.  16  W1 = (113)   = 904 plf  2 F1 =

1.0 (0.4 )(1.0)  0   1 + 3  W1 = 0.133W1 3.0 36  

(32-2)



§1632.2

Check limit of Equation (32-3) 0.7C a I pW p = 0.7 (0.4 )(1.0 )W1 = 0.28W1 controls ∴ F1 = 0.28W1 = 0.28 (904 ) = 253 plf

(32-3)

F3 = 1,085 plf

22'

Note that the 420 plf minimum anchorage force of §1633.2.8.1, Item 1 does not apply at the first floor. Wall reactions for anchorage design are shown at right.

2'

20'

F2 = 950 plf

16'

16'

F1 = 253 plf

Wall anchorage forces

Commentary Anchorage forces have been determined on the basis of the weight tributary to each level using Equation (32-2), with limits of Equation (32-3) and §1633.2.8.1, Item 1. Panel forces, on the other hand, have been determined using seismic coefficients for each floor level. If reactions are determined from the uniform out-of-plane forces used for panel design, these will be different than those determined for anchorage requirements. This inconsistency is rooted in the fact that the code does not call for determination of both panel design forces and anchorage design forces from the same method. To be consistent, forces would have to first be determined at the panel centroids (between floors) and then anchorage reactions determined from statics equilibrium. In all significant California earthquakes, beginning with the 1971 San Fernando event, wall-roof anchorage for flexible diaphragms has failed repeatedly. After the 1994 Northridge earthquake, when over 200 tilt-up buildings in the city of Los Angeles experienced collapse or partial collapse of roofs and/or walls, wall-roof anchorage forces were increased significantly in the 1996 Supplement to the 1994 UBC. The 1997 UBC requirements reflect this change. It is extremely important that bearing wall tilt-up buildings maintain wall-roof (and wall-floor) connections under seismic motions. This is the principal reason that anchorage forces are 50-percent higher than those used for out-of-plane wall panel design. See §1633.2.8.1 for the special material load factors used for the design of steel and wood elements of the wall anchorage system (i.e., 1.4 for steel and 0.85 for wood).


Example 37 Rigid Equipment

§1632.2

(2 5#

$%(&'&

This example illustrates determination of the design seismic force for the attachments of rigid equipment. Attachment as used in the code means those components, including anchorage, bracing, and support mountings, that “attach” the equipment to the structure. The three-story building structure shown below has rigid electrical equipment supported on nonductile porcelain insulators that provide anchorage to the structure. Identical equipment is located at the base and at the roof of the building. Wp

Zone 4 Ca = 0.4 I p = 1.0 W p = 10 k

Nonductile attachments

Level Roof 12' 2 12' 1

Wp 12'

Find the following:

Design criteria. Design lateral seismic force at base. Design lateral seismic force at roof.


Code Reference

Design criteria.

§1632.2

The total design lateral seismic force is determined from Fp =

a p Ca I p  h  1 + 3 x Rp  hr

  W p 

(32-2)

Values of a p and R p are given in Table 16-O. Since the equipment is rigid and has nonductile attachments a p = 1.0, R p = 1.5

Table 16-O, Item 4B


Example 37 Rigid Equipmen

Design lateral seismic force at base.

§1632.2

§1632.2

hx = 0 Fp =

(1.0)(0.4 )(1.0 )  0  1 + 3  (10) = 2.67 k (1.5) 36  

Section 1632.2 has a requirement that F p be not less than 0.7C a I p W p

(32-3)

Check F p ≥ 0.7C a I p W p = 0.7 (0.4) (1.0) 10 = 2.8 k ∴ F p = 2.8 k

Design lateral seismic force at roof.

h x = h r = 36 ft Fp =

(1.0)(0.4 )(1.0 ) 1 3 36  (10) 10.7 k =   + (1.5) 36  

Section 1632.2 states that F p need not exceed 4C a I p W p

(32-3)

Check F p ≤ 4C a I pW p = 4 ( 0.4) (1.0) 10 = 16 k ∴ F p = 10.7 k

Commentary The definition of a rigid component (e.g., item of equipment) is given in §1627. Rigid equipment is equipment, including its attachments (anchorages, bracing, and support mountings), that has a period less than or equal to 0.06 seconds. The anchorage design force F p is a function of 1 R p , where R p = 1.0, 1.5, and 3.0 for nonductile, shallow, and ductile anchors, respectively. Generally, only equipment anchorage or restraints need be designed for seismic forces. This is discussed in Footnote 5 of Table 16-O. Item 3.C, also in Table 16-O states that this applies to “Any flexible equipment laterally braced or anchored to the structural frame at a point below their center of mass.” For the case where equipment, which can be either flexible or rigid, comes mounted on a supporting frame that is part of the manufactured unit, then the supporting frame must also meet the seismic design requirements of §1632.2.


Example 38 Flexible Equipment

§1632.2

(3 ,

$%(&'&

This example illustrates determination of the design seismic force for the attachments of flexible equipment. Attachment as used in the code means those components, including anchorage, bracing, and support mountings, that “attach” the equipment to the structure. The three-story building structure shown below has flexible air-handling equipment supported by a ductile anchorage system. Anchor bolts in the floor slab meet the embedment length requirements. Identical equipment is located at the base and at the roof of the building. Wp

Ductile attachments

Level Roof

Zone 4 Ca = 0.4 I p = 1.0

12' 2

W p = 10 k

12' 1

Wp

12'

Find the following:

Design criteria. Design lateral seismic force at base. Design lateral seismic force at roof.


Code Reference

Design criteria.

§1632.2

The total design lateral seismic force is determined from Fp =

a p Ca I p  h 1 + 3 x Rp  hr

  W p 

(32-2)



§1632.2

Values of a p and R p are given in Table 16-O. Since the equipment is flexible and has ductile supports a p = 2.5, R p = 3.0

Table 16-O, Item 3C

Design lateral seismic force at base.

hx = 0

Fp =

(2.5)(0.4)(1.0) 1 + 3 0  (10) = 3.33 k   (3.0 ) 36  

Section 1632.2 has a requirement that F p be not less than 0.7C a I p W p

(32-3)

Check F p ≥ 0.7C a I p W p = 0.7 (0.4) (1.0) 10 = 2.8 k ∴ F p = 3.33 k

Design lateral seismic force at roof.

h x = h r = 36 ft Fp =

(2.5)(0.4 )(1.0 ) 1 3 36  (10) 13.33 k =   + (3.0) 36  

Section 1632.2 states that F p need not exceed 4C a I p W p

(32-3)

Check F p ≤ 4C a I pW p = 4 ( 0.4) (1.0) 10 = 16 k ∴ F p = 13.33 k

Commentary The definition of flexible equipment is given in §1627. Flexible equipment is equipment, including its attachments (anchorages, bracing, and support mountings), that has a period greater than 0.06 seconds. It should be noted that the anchorage design force F p is a function of 1 R p , where R p = 1.0, 1.5, and 3.0 for nonductile, shallow, and ductile anchors, respectively. Generally, only equipment anchorage or restraints need be designed for seismic forces. This is discussed in Footnote 5 of Table 16-O. Item 3.C of that table states that this applies to “Any flexible equipment laterally braced or anchored to the SEAOC Seismic Design Manual

§1632.2


structural frame at a point below their center of mass.” For the case where the equipment, which can be either flexible or rigid, comes mounted on a supporting frame that is part of the manufactured unit, then the supporting frame must also meet the seismic design requirements of §1632.2.


Example 39 Relative M otion of Equipment Attachments

§1632.4

(5= 9 7

$%(&'*

Section 1632.4 of the UBC requires that the design of equipment attachments in buildings having occupancy categories 1 and 2 of Table 16-K, essential facilities and hazardous facilities, respectively, have the effects of the relative motion of attachment points considered in the lateral force design. This example illustrates application of this requirement. A unique control panel frame is attached to the floor framing at Levels 2 and 3 of the building shown below. The following information is given. Zone 4 Occupancy Category 1, (essential facility) Story drift: ∆ S = 0.34 in. R = 8.5 Panel frame: EI = 10 × 10 4 k/in. 2

Level

∆S

4 12' 3 Panel

12' 2


12' 1 12'

Story drift to be considered. Induced moment and shear in frame.


Deflected shape

Code Reference

Story drift to be considered.

Section 1632.4 requires that equipment attachments be designed for effects induced by ∆ M (maximum inelastic story drift). This is determined as follows: ∆ M = 0.7 R∆ S = 0.7 (8.5) 0.34 = 2.02 in.

Induced moment and shear in frame.

M=

V=

6 EI∆ M H2

=

(

)

6 10 × 10 4 (2.02)

(144 ) 2

(30-17) §1632.4

= 58.45 k - in.

58.45 M = = 0.81 k (H 2 ) 72


Example 39 Relative M otion of Equipment Attachments

§1632.4

Commentary The attachment details, including the body and anchorage of connectors, should follow the applicable requirements of §1632.2. For example, if the body of the attachment is ductile, then the induced forces can be reduced by R p = 3.0 . However, if the anchorage is provided by shallow anchor bolts, then R p = 1.5 . When anchorage is constructed of nonductile materials, R p = 1.0 . One example of a nonductile anchorage is the use of adhesive. Adhesive is a “glued” attachment (e.g., attachment of pedestal legs for a raised computer floor). It should be noted that attachment by adhesive is not the same as anchor bolts set in a drilled hole with epoxy.


Example 40 Deformation Compatibility

§1633.2.4

*4

$%(('&'*

1

A two-level concrete parking structure has the space frame shown below. The designated lateral force-resisting system consists of a two bay special momentresisting frame (SMRF) located on each side of the structure. The second level gravity load bearing system is a post-tensioned flat plate slab supported on ordinary reinforced concrete columns, A

B

C

D

E

1

2

3

4

5

Plan at second level

The following information is given: 1

Zone 4 ∆ S = 0.42 in. R = 8.5 Column section = 12 in. x 12 in. Column clear height = 12 ft Concrete E c = 3 × 10 3 ksi Find the following:

2

3

Ordinary column

4

5

SMRF

∆S

Elevation Line E

Moment in ordinary column. Detailing requirements for ordinary column.



§1633.2.4


Code Reference

Moment in ordinary column.

§1633.2.4

Section 1921.7 specifies requirements for frame members that are not part of the designated lateral force-resisting system. The ordinary columns located in the perimeter frames, and the interior flat plate/column system, fall under these requirements and must be checked for the moments induced by the maximum inelastic response displacement. For this example, the columns on Line E will be evaluated. ∆ M = 0.7 R∆ S = 0.7 (8.5) 0.42 = 2.50 in.

(30-17)

Section 1633.2.4 requires that the value of ∆ S used for this determination of ∆ M be computed by neglecting the stiffening effect of the ordinary concrete frame. The moment induced in the ordinary column due to the maximum inelastic response displacement ∆ M on Line E must be determined. For purposes of this example, a fixed-fixed condition is used for simplicity. In actual applications, column moment is usually determined from a frame analysis. M col =

6E c I c ∆ M h2

h = 12 × 12 = 144 in.

(12 )3 bd 3 = 12 = 1728 in. 4 12 12 The cracked section moment of inertia I c can be approximated as 50 percent of the gross section I g . Section 1633.2.4 requires that the stiffness of elements that are part Ig =

of the lateral force-resisting system shall not exceed one half of the gross section properties. This requirement also applies to elements that are not part of the lateral force-resisting system. Ic =

Ig 2

M col =

= 864 in. 4

(

)

6 3 × 10 3 (864 )(2.5)

(144 )2

= 1875 k − in.



§1633.2.4

Detailing requirements for ordinary column.

Section 1921.7 requires that frame members, such as the column, that are assumed not to be part of the lateral force-resisting system must be detailed according to §1921.7.2 or §1921.7.3, depending on the magnitude of the moments induced by ∆M .

Commentary In actual applications, the flat plate slab must be checked for flexure and punching shear due to gravity loads and the frame analysis actions induced by ∆ M . Section 1633.2.4 requires that the stiffening effect of those elements not part of the lateral force-resisting system shall be neglected in the structural model used for the evaluation of ∆ M . To evaluate the force induced by ∆ M in the elements not part of the lateral force-resisting system when using frame analysis, it is necessary to formulate an additional structural model that includes the stiffening effect of these elements. This model should be loaded by the same lateral forces used for the evaluation of ∆ M to obtain the corresponding element forces FM′ and displacement ∆′M . The required element forces FM induced by ∆ M can then be found by: ∆M (F ′ ) ∆′M M The values used for the displacements ∆ M and ∆′M should be those corresponding to the frame line in which the element is located. FM =

Section 1633.2.4 also requires the consideration of foundation flexibility and diaphragm deflections in the evaluation of displacement. The following criteria and procedures may be used for this consideration: 1.

Foundation Flexibility If the design strength capacity at the foundation-soil interface is less than the combined loads resulting from the special load combinations of §1612.4, then the lateral stiffness of the supported shear wall, braced frame, or column shall be reduced by a factor of .5.

2.

Diaphragm Deflection For a given diaphragm span between two lateral force-resisting elements, compare the mid-span diaphragm deflection for a given uniform load with the average of the story drifts of the two lateral force-resisting elements due to the reactions from the diaphragm load. If the diaphragm deflection exceeds 20 percent of the average story drift, then include diaphragm deflection in ∆ M .

Otherwise, for cases where the effects are critical for design, a soil-spring model of the foundation and/or a finite element model of the diaphragm may be required.


Example 41 Adjoining Rigid Elements

§1633.2.4.1

* 7># 5#

$%(('&'*'

During the 1994 Northridge earthquake in southern California, nonductile concrete and masonry elements in frame structures with ductile lateral force-resisting systems experienced failure because they lacked deformation compatibility. Deformation compatibility refers to the capacity of nonstructural elements, or structural elements not part of the lateral force system, to undergo seismic displacements without failure. It also implies that structural elements of the lateral force system will not be adversely affected by the behavior of nonstructural or nonseismic structural elements. The 1997 UBC has new requirements for deformation compatibility. These are given in §1633.2.4.1. The purpose of this example is to illustrate use of these requirements. The concrete special moment-resisting frame shown below is restrained by the partial height infill wall. The infill is solid masonry and has no provision for an expansion joint at the column faces. The maximum deflection ∆ M was computed neglecting the stiffness of the nonstructural infill wall, as required by §1633.2.4. Zone 4 ∆ M = 2.5" Column properties: f ' c = 3,000 psi

∆M

SMRF

12' 6'

E c = 3 × 10 3 ksi Ac = 144 in. 4

Infill wall

I c = 854 in. 4

Typical elevatio


Deformation compatibility criteria. Approximate column shear.


Code Reference

Deformation compatibility criteria.

§1633.2.4.1

The infill wall, which is not required by the design to be part of the lateral forceresisting system, is an adjoining rigid element. Under §1633.2.4.1, it must be shown that the adjoining rigid element, in this case the masonry infill wall, must not impair the vertical or lateral load-resisting ability of the SMRF columns. Thus, the columns must be checked for ability to withstand the ∆ M displacement of 2.5 inches while being simultaneously restrained by the 6-foot-high infill walls.


Example 41 Adjoining Rigid Elements

§1633.2.4.1

Approximate column shear.

Column shear will be determined from the frame inelastic displacement ∆ M . For purposes of the example, the expression for the fixed-fixed condition will be used for simplicity. V col =

12 E c I c ∆ M h3

=

(

12 3 × 10 3

) (854)(2.5) = 205.9 kips

(72 )3

Column clear height = 72 in Because the SMRF is the primary lateral force-resisting system, ∆ M is to be determined by neglecting the stiffness of the ordinary columns and the rigid masonry infill per §1633.2.4. Vcol = 1,447 psi . This is approximately 26 f 'c Ac and would result in column shear failure. Therefore, a gap must be provided between the column faces and the infill walls. Alternately, it would be necessary to either design the column for the induced shears and moments caused by the infill wall, or demonstrate that the wall will fail before the column is damaged. Generally, it is far easier (and more reliable) to provide a gap sufficiently wide to accommodate ∆ M . The induced column shear stress is

For this example, with the restraining wall height equal to one half the column height, ∆ the gap should be greater than or equal to M = 1.25 in . If this were provided, the 2 column clear height would be 144 inches, with resulting column shear 12 3 × 10 3 (854 )(2.5) 1 ′ = = 25.7 kips . This is of the restrained column shear V col 3 8 (144)

(

)

of 205.9 kips .


Example 42 Exterior Elements: Wall Panel

§1633.2.4.2

*& ! <

$%(('&'*'&

This example illustrates the determination of the design lateral seismic force, Fp , on an exterior element of a building, in this case an exterior wall panel. A five-story moment frame building is shown below. The cladding on the exterior of the building consists of precast reinforced concrete wall panels. The following information is known: Level 5 12' Typical exterior panel

4

Zone 4 I p = 1.0 C a = 0.4 Panel size : 11’-11” x 19’-11” Panel thickness: 6 in. Panel weight: W p = 14.4 kips

12' 3 12' 2 12' 1 12'

Find the following:

Design criteria. Design lateral seismic force on a panel at the fourth story. Design lateral seismic force on a panel at the first story.


Code Reference

Design criteria.

§1632.2

For design of exterior elements, such as the wall panels on a building, that are attached to the building at two levels, design lateral seismic forces are determined from Equation (32-2). The panels are attached at the two elevations h L and hU . The intent of the code is to provide a value of F p that represents the average of the acceleration inputs from the two attachment locations. This can be taken as the average of the two F p values at h x equal to h L and hU .



Fp =

a p Ca I p  h 1 + 3 x Rp  hr

§1633.2.4.2

  W p ≥ 0.7C a I pW p 

a p = 1.0, R p = 3.0

(32-2)

Table 16-O

Design lateral seismic force on a panel at the fourth story.

Assuming connections are 1 foot above and below the nominal 12-foot panel height hU = 47 ft h L = 37 ft h r = 60 ft F pU =

(1.0 )(0.4)(1.0)  +  47  1 3  60  W p (3.0)   

= 0.447W p

F pL =

(1.0)(0.4 )(1.0 )  +  37  1 3  60  W p (3.0)   

= 0.380W p

Fp4 =

F pU + F pL 2

=

(0.447 + 0.380) 2

Wp

F p 4 = 0.414W p = (0.414 )(14.4 ) = 5.96 kips Check: F p 4 > 0.7C a I p W p = 0.7 (0.4 )(1.0)W p = 0.2W p o.k.

(32-3)

Design lateral seismic force on a panel at the first story.

The following are known hU = 11ft hL = 0 h r = 60ft F pU =

(1.0 )(0.4)(1.0)   11  1 + 3   W p  (3.0)  60  

= 0.207W p

Check that F pU is greater than 0.7C a I pW p



§1633.2.4.2

F pU = 0.7 (0.4 )(1.0 )W p = 0.28W p

not o.k.

Also F pL < F pU < 0.28W p ∴ use F pL = F pU = 0.28W p F p1 =

F pU + F pL 2

= 0.28W p = (0.28)(14.4 ) = 4.03k

Commentary The design lateral seismic force F p is to be used for the design of the panel for outof-plane seismic forces. This can be represented by a distributed load equal to F p divided by the panel area. Note that the §163.2.4.2 Item 1 requirement to accommodate the relative movement of ∆ M is about twice the equivalent value of the previous code.


Example 43 Exterior Elements: Precast Panel

§1633.2.4.2

*( !

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This example illustrates the determination of the total design seismic lateral force for the design of the connections of an exterior wall panel to a building. Design of the body of the panel is often controlled by the non-seismic load conditions of the fabrication, transport, and erection. An exterior nonbearing panel is located at the fourth story of a five-story moment frame building. The panel support system is shown below, where the pair of upper brackets must provide resistance to out-of-plane wind and seismic forces and in-plane vertical and horizontal forces. The panel is supported vertically from these brackets. The lower pair of rod connections provide resistance to only the out-of-plane forces. 20'

Zone 4 C a = 0.4 I p = 1.0 hU = 47'

Bracket

9'

C

5' 12' 5'

hL = 37'

Height to roof h r = 60 ft Panel weight = 14.4 k ρ = 1.0 per 1632.2

9'

Wall panel

Rod

Find the following:

Strength design load combinations. Lateral seismic forces on connections and panel. Vertical seismic forces on panel. Combined dead and seismic forces on panel and connections. Design forces for the brackets. Design forces for the rods.


Code eference

Strength design load combinations.

For design of the panel connections to the building, the strength design load combinations are: 1.2 D + 1.0 E + f 1 L

(12-5)

0.9 D ± 1.0 E

(12-6)



§1633.2.4.2

where E = ρE h + E v

(30-1)

ρ = 1.0

§1632.2

E h = load due to application of Equations (32-2) and (32-3)

§1630.1.1

E v = 0.5C a I p D

§1630.1.1

Lateral seismic forces on connections and panel.

Out-of-plane panel seismic forces on the connections are determined from Equations (32-2) and (32-3) for the particular elevation of the connections. Forces at the upper level connections will be different than those at the lower level. Fp =

a p Ca I p  h  1 + 3 x Rp  hr

  W p 

(32-2)

0.7C a I pW p ≤ F p ≤ 4C a I pW p

(32-3)

a p = 1.0 and R p = 3.0

§1633.2.4.2, Item 4

W p = weight of portion of panel tributary to the connection Upper bracket connections h x = hU = 47 ft Tributary W p for the two brackets =

F pU =

14.4 = 7.2 kips 2

(1.0)(0.4 )(1.0)  +  47  1 3  60  W p (3.0)   

= 0.447W p

(32-2)

Check minimum force requirements of Equation (32-3). 0.7C a I pW p = 0.7 (0.4 )(1.0)W p = 0.28W p



§1633.2.4.2

The force on each bracket is: PB = ∴ PB =

1 × F pU 2

0.447(7.2 ) = 1.61 kips/brack et 2

Lower rod connections h x = h L = 37ft Tributary W p for the two rods =

F pL =

14.4 = 7.2 kips 2

(1.0)(0.4 )(1.0)   37  1 + 3   W p  (3.0 )  60  

= 0.38W p > 0.28W p

(32-2)

The axial force on each rod is: PR = ∴ PR =

1 × F pL 2

0.38 (7.2 ) = 1.39 kips/rod 2

Body of panel The body of the panel is also designed using a p = 1.0 and R p = 3.0 as indicated in Table 16-O, Item 1.A(2). Thus, the seismic force on the body of the panel is the sum of the forces on the upper and lower levels. Alternatively, as shown below, an equivalent coefficient for the panel body can be determined by using the average of the coefficients for the upper and the lower levels. Upper coefficient = 0.447 @ hU Lower coefficient = 0.380 @ h L Average coefficient =


(0.447 + 0.380) = 0.413 > 0.28 2

o.k.


§1633.2.4.2

The panel seismic force is the average coefficient times the weight of the entire panel:

( )

FP = 0.413 W p = 0.413 (14.4 ) = 5.95 kips This force is applied at the panel centroid C and acts horizontally in either the out-ofplane or the in-plane direction. For panel design for out-of-plane forces, this force can be made into an equivalent uniform loading: fP =

5,950 = 24.8 psf 12 × 20

Vertical seismic forces on panel.

§1630.1.1

The code requires consideration of vertical seismic forces when strength design is used. Vertical forces are determined from the equation E v = 0.5C a I p D

§1630.1.1

D = dead load effect (or weight W p of panel) E v = 0.5 (.4 )(1)W p = 0.2W p = 0.2 (14.4 ) = 2.88 kips

Combined dead and seismic forces on panel and connections.

§1630.1.1

There are two seismic load conditions to be considered: out-of-plane and in-plane. These are shown below as concentrated forces. In this example, Equation (12-5) is considered the controlling load case. Because there is no live load on the panel, the term f 1 L of this equation is zero. FpU out-of-plane seismic force at upper level

FP = in-plane seismic force at centroid

FpL out-of-plane seismic force at lower level

Out-of-plane seismic forces

± 0.2Wp = vertical seismic force at centroid

In-plane seismic forces



§1633.2.4.2

Dead load and seismic out-of-plane and vertical forces. Panel connection reactions due to dead load, out-of-plane seismic forces, and vertical seismic forces are calculated as follows: 9'

9'

5'

FpU = 3.22 k

5'

1.2Wp + 0.2Wp = 1.4Wp = 1.4 (14.4) =20.16 k

FpL = 2.78 k

Each bracket connection takes the following out-of-plane force due to lateral loads: PB =

F pU 2

=

3.22 = 1.61 kips 2

Each bracket takes the following downward in-plane force due to vertical loads: VB =

1.4W p 2

=

20.16 = 10.08 kips 2

Each rod connection takes the following out-of-plane force due to lateral loads: PR =

F pL 2

=

2.78 = 1.39 kips 2

Note that each rod, because it carries only axial forces, has no in-plane seismic loading.

Dead load and seismic in-plane and vertical forces: Panel connection reactions due to dead load, in-plane seismic forces, and vertical seismic forces are calculated as follows: 9'

9'

5'

FP = 5.95 k

C 5'

1.4Wp = 20.16 k



§1633.2.4.2

Each bracket takes the following in-plane horizontal force due to lateral seismic load: HB =

FP 5.95 = = 2.98 kips 2 2

Each bracket takes the following upward or downward force due to lateral seismic load: FB =

5 (FP ) 5 (5.95) = = ± 1.65 kips 18 18

Each bracket takes the following downward force due to vertical loads: RB =

1.4W p 2

=

20.16 = 10.08 kips 2

Under the in-plane seismic loading, each rod carries no force.

Design forces for the brackets.

Body of connection. Under §1633.2.4.2, Item 4, the body of the connection must be designed for a p = 1.0 and R p = 3.0 . These are the same values as used for the determination of F pU , F pL and FP . Therefore there is no need to change these forces. The bracket must be designed to resist the following sets of forces: PB = ±1.61 k out-of-plane together with V B = 10.08 k downward shear and

H B = ± 2.98 k horizontal shear together with FB + R B = 1.65 + 10.08 = 11.73 k downward shear

Fasteners. Under §1633.2.4.2, Item 5, fasteners must be designed for a p = 1.0 and R p = 1.0 . Thus, it is necessary to multiply the FpU , FpL and FP reactions by 3.0 since these values were based on R p = 3.0 . Fasteners must be designed to resist 3PB = 3 (1.61) = 4.83 k out-of-plane together with



§1633.2.4.2

V B = 10.08 k downward shear and 3H B = 3 (2.98) = 8.94 k horizontal shear together with 3FB + R B = 3 (1.5) + 10.08 = 15.03 k downward shear

Design forces for the rods.

Body of connection. Under §1633.2.4.2, Item 4, the body of the connection must be designed to resist PR = ±1.39 k out-of-plane

Fasteners. Under §1633.2.4, Item 5, all fasteners in the connecting system must be designed to resist a force based on R p = 1.0 : 3PR = 3 (1.39 ) = 4.17 k out-of-plane


Example 44 Beam Horizontal Tie Force

§1633.2.5

**

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8 : ,

This example illustrates use of the beam tie requirement of §1633.2.5. This requirement derives from ATC-3 and is to ensure that important parts of a structure are “tied together.” Find the minimum required tie capacity for the connection between the two simple beams shown in the example below. The following information is given: Tie

D + L = 10 k/ft

Zone 4 C a = 0.44 I = 1.0

Support, typ.

Beam

40'

40'

Determine tie force.


Code Reference

Requirements for ties and continuity are specified in §1633.2.5. For this particular example, it is required to determine the “tie force” for design of the horizontal tie interconnecting the two simply supported beams. This force is designated as E h , where E h is the horizontal earthquake load to be used in Equation (30-1). The minimum value of E h is 0.5C a I times the dead plus live load supported on the beam. Dead plus live load supported = (10 kpf )(40 ft ) = 400 kips E h = 0.5 (0.44 )(1.0)(400) = 88 kips

Commentary The tie force calculated above for 1997 UBC requirements is .22 times dead plus live load. This is on a strength design basis and is about twice the load factored value given in the 1994 UBC. The 1994 UBC value is Z 5 times dead plus live load, or .112 times dead plus live load using a 1.4 load factor.


Example 45 Collector Elements

§1633.2.6

*0

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Collectors “collect” forces and carry (i.e., drag) them to vertical shear-resisting elements. Collectors are sometimes called “drag struts.” The purpose of this example is to show the determination of the maximum seismic force for design of collector elements. In the example below, a tilt-up building with a panelized wood roof has a partial interior shear wall on Line 2. A collector is necessary to “collect” the diaphragm loads tributary to Line 2 and bring them to the shear wall. The following information is given: 1

2

3

100' A

50'

50' Tributary roof area for collector

50'

A

Collector

B 50'

Zone 4 R = 4.5 Ω o = 2.8 I = 1.0 C a = .44 Roof dead load = 15 psf Wall height = 30 ft , no parapet Wall weight = 113 psf Base shear = V = .244W

100'

Shear walls C A

Roof plan

Interior shear wall

Note: Roof framing, except collector, not shown.

50'

30'

Collector


Collector design force at tie to wall. Special seismic load of §1612.4 at tie to wall.


Elevation Section A-A

Code Reference

Collector design force at tie to wall.

§1633.2.6

The seismic force in the collector is made up of two parts: (1) the tributary out-ofplane wall forces, and (2) the tributary roof diaphragm force. Because the roof is considered flexible, the tributary roof area is taken as the 100ft by 50ft area shown on the roof plan above. Seismic forces for collector design are determined from Equation



§1633.2.6

(33-1) used for diaphragm design. This equation reduces to the following for a single story structure: F px = where

Froof Wroof

W px

F px = collector design force W px = weight tributary to collector

The term

Froof Wroof

is the base shear coefficient adjusted for the diaphragm R value of 4

required by §1633.2.9. Froof W roof

=

V W

 Rbuilding    = .244  4.5  = .275  R diaphragm   4   

F px = .275W px The tributary roof weight and out-of-plane wall weight is  30  W px = 15 psf (100)(50) + 113 psf   (100) = 75,000 + 169,500 = 244.5 kips  2  ∴ F px = .275 (244.5) = 67.2 kips

Special seismic load of §1612.4 at tie to wall.

§1633.2.6

In addition to the forces specified by Equation (33-1), collectors must resist special seismic loads specified in §1612.4. Collector load E h = 67.2 kips Required collector strength = E m = Ω o E h = 2.8 (67.2 ) = 188.2 kips

(30-2)

This load is to be resisted on a strength design basis using a resistance factor of φ = 1.0 , and 1.7 times the allowable values for allowable stress design. The connection must have the capacity to deliver this collector load to the shear wall on Line 2.



§1633.2.6

Commentary Note that the UBC in §1633.2.6 specifies that E m need not exceed the maximum force that can be delivered by the diaphragm to the collector or other elements of the lateral force-resisting system. For example, the overturning moment capacity of the shear wall can limit the required strength of the collector and its connection to the shear wall.


Example 46 Out-of-Plane Wall Anchorage to Flexible Diaphragm

§1633.2.8.1

*% ;++ < 7# , #

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For the tilt-up wall panel shown below, the seismic force required for the design of the wall anchorage to the flexible roof diaphragm will be determined. This will be done for a representative one foot width of wall. The following information is given:

4' Roof

Fanch

Zone 4 I p = 1.0 Ca = 0.4 Panel thickness = 8 in. Normal weight concrete (150 pcf)


Top of parapet

Tilt-up panel 20'

Assumed pin support

Design criteria.

Ground

Wall anchorage force.


Code Reference

Design criteria.

§1633.2

Because of the frequent failure of wall/roof ties in past earthquakes, the code requires that the force used to design wall anchorage to flexible diaphragms be greater than that used to design the panel sections. Either Equation (32-1) or Equations (32-2) and (32-3) can be used to determine anchor design forces. Normally, Equations (32-2) and (32-3) are used. Fp =

a pCa I p  3h  1 + x Rp  hr

  W p 

(32-2)

0.7C a I pW p ≤ F p ≤ 4C a I pW p

(32-3)

The wall panel is supported at its base and at the roof level. The value of F p to be used in wall/roof anchorage design is determined from Equation (32-2) using h x = hr , and W p is the tributary weight.



§1633.2.8.1

For design of elements of wall anchorage system:

R p = 3.0, a p = 1.5

§1633.2.8.1, Item 1

Also, the value of Fanch must not be less than 420 plf

§1633.2.8.1, Item 1

Wall anchorage force.

The tributary wall weight is one-half of the weight between the roof and base plus all of the weight above the roof. 8 W p = 150   (4 ′ + 10′)(1′) = 1,400 lbs/ft  12  Since

h x = h r = 20 ft R p = 3.0 a p = 1.5

The minimum force is 0.7C a I pW p = 0.7 (0.4 )(1.0 )W p = 0.28W p = 0.28 (1,400 ) = 392 plf

(32-3)

Check Equation (32-2) Fanch =

1.5 (0.4 )(1.0)  3 (20)  1 +  W p = 0.80W p 3.0 20  

(32-2)

Fanch = 0.80W p = 0.8 (1,400) = 1,120 plf > 420 plf o.k., and < 4.0C a I pW p = 1.6W p o.k. ∴ Fanch = 1,120 plf

Commentary Design of wall anchorage is crucial for successful earthquake performance of tilt-up buildings in Zones 3 and 4. Generally, it is desirable that the connections of walls to the diaphragm develop the strength of the steel. The following code sections apply to the anchorage design: 1. Sections 1605.2.3 and 1633.2.8 call for a positive direct connection. Embedded straps must be attached to, or hooked around, the wall reinforcing steel, or otherwise effectively terminated to transfer forces.



§1633.2.8.1

2. Section 1633.2.9, Item 4 states that Fanch may be carried by a subdiaphragm. 3. Section 1633.2.8.1 has the following additional anchorage requirements. Item 4: Steel elements of anchorage must be designed to take 1.4 Fanch . Item 5: Wood elements of anchorage must have strength to take 0.85Fanch , and wood elements must have minimum net thickness of 2 1 2 " (i.e., be at least 3x members). 4. Section 1633.2.8 and §1633.2.9, Item 1 require that details of anchors tolerate ∆ M of the diaphragm. 5. When allowable stress design is used, the minimum anchorage force is not 420 plf as specified in §1633.2.8.1, Item 1 but 300 plf. This is determined by substituting E = 420 plf in the load combinations of §1612.3. This gives: E 420 = = 300 plf . 1.4 1.4


Example 47 Wall Anchorage to Flexible Diaphragms

§1633.2.8.1

*2 < 7# , #

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This example illustrates use of the allowable stress design procedure for the design of steel and wood elements of the wall anchorage system in a building with a flexibl roof diaphragm. In the example below, a tilt-up wall panel is shown. It is connected near its top to a flexible roof diaphragm. The anchorage force has been calculated per §1633.2.8.1 as Fanch = 1,120 plf. The wall anchorage connections to the roof are to be provided at 4 feet on center.

Fanch

Wall panel

Subpurlin Hold-down each side

Wall-roof tie detail

Determine the strength design requirements for the following:

Design force for premanufactured steel anchorage element. Design force for wood subpurlin tie element.


Code Reference

Design force for premanufactured steel anchorage element.

The basic task is to design the steel anchorage elements (i.e., hold-downs) that connect the tilt-up wall panel to the wood subpurlins of the roof diaphragm. The anchorage consists of two hold-down elements, one on each side of the subpurlin. The manufacturer’s catalog provides allowable capacity values for earthquake loading for a given type and size of hold-down element. These include the allowabl stress increase and are typically listed under a heading that indicates a “1.33 x allowable” capacity. For the steel hold-down elements of the anchorage system, the code requires that the anchorage force PE used in strength design be 1.4 times the force otherwise required. §1633.2.8.1, Item 4 PE = 1.4 Fanch SEAOC Seismic Design Manual

Example 47 Wall Anchorage to Flexible Diaphragms

§1633.2.8.1

PE = 1.4 (1,120 plf )(4 ft ) = 6,272 lbs Since PE is determined on a strength design basis, it is the earthquake load E to be used in the design load combinations. In this example, it is elected to use the alternate basic load combinations of §1612.3.2, where the applicable combinations of E Equations (12-13), (12-16) and (12-16-1) permit to be resisted with a one-third 1.4 increase in allowable stress. The allowable stress design requirement for each pair of hold-down elements is: 6,272 P E = E = = 4,480 lbs 1.4 1.4 1.4 From the manufacturer’s catalog, select a hold-down element having a ( 1.33 × allowable) capacity of at least 4,480 = 2,240 lbs 2 Whenever hold-downs are used in pairs, as shown in the wall-roof tie detail above, the through-bolts in the subpurlin must be checked for double shear bearing. Also, the paired anchorage embedment in the wall is likely to involve an overlapping pull-out cone condition in the concrete: refer to §1923 for design requirements. When singlesided hold-downs are used, these must comply with the requirements of Item 2 of §1633.2.8.1. Generally, double hold-downs are preferred, but single-sided holddowns are often used with all eccentricities fully considered.

Design force for wood subpurlin tie element.

The strength design forces on the wood elements of the wall anchorage system can be 0.85 times the force otherwise required: PE = 0.85Fanch

§1633.2.8.1, Item 5

PE = 0.85 (1120 plf )(4 ft ) = 3,808 lbs Select the wood element such that 1.33 times the allowable capacity of the element, including dead load effects, is at least equal to PE 3,808 = = 2,720 lbs 1.4 1.4 Note that tie elements, such as the subpurlin, are required to be 3x or larger.

§1633.2.8.1, Item 5


Example 48 Determination of Diaphragm Force Fpx: Lowrise

§1633.2.9

*3

# , !

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This example illustrates determination of the diaphragm design force F px of Equation (33-1), for the design of the roof diaphragm of a single story building. A single-story tilt-up building with a panelized wood roof is shown below. This type of roof construction is generally considered to have a flexible diaphragm. 6

1 200' Normal wall

100'

A

Seismic force

Given: Zone 4 I = 1.0 C a = 0.4 R = 4.5 (bearing wall system) ρ = 1.2 Diaphragm weight = 15 psf Wall weight = 80 psf

D Normal wall

Roof plan

A

D Roof diaphragm

10' Mid-height

20'

Elevation through building

Find the following:

Diaphragm force at the roof.



§1633.2.9


Code Reference

Diaphragm force at the roof.

For buildings with tilt-up concrete walls, §1633.2.9, Item 3, requires that the flexible diaphragm design force be based on the design base shear and forces F px using an R value not exceeding 4, even though the tilt-up wall-frame system uses R = 4.5 . For a short period single story building, the diaphragm force, using R = 4 , becomes:

w px = weight of diaphragm + weight of ½ height of normal walls = 100 (15) + 2 (10)(80) = 3,100 plf F px =

2.5C a I 2.5 (0.4 )(1.0) (3,100) = 775 plf w px = 4 R

(33-1)

Note that the redundancy factor of ρ = 1.2 is not applied to the E h loads due to F px (such as chord forces and diaphragm shear loads in the diaphragm).

Commentary 1. The weight, w px , includes the weight of the diaphragm plus the tributary weight of elements normal to the diaphragm that are one-half story height below and above the diaphragm level. Walls parallel to the direction of the seismic forces are usually not considered in the determination of the tributary roof weight because these walls do not obtain support, in the direction of the force, from the roof diaphragm. 2. The single story building version of Equation (33-1) is derived as follows: n

F px =

Ft + ∑ Fi i= x

n

∑ wi

w px

(33-1)

i= x

Fi =


(30-15)

∑ w i hi i =1



§1633.2.9

For a single story building, i = 1 , x = 1 and n = 1 Ft = 0 , since T < 0.7 sec 1

∑ wi

=W

i =1

and Equation (30-15) gives F1 =

Vw1 h1 =V w1 h1

where V=

2.5C a IW R

(30-5)

Finally, for the single story building, Equation (33-1) is F p1 = F1 w p1 =


2.5C a I w p1 R

Example 49 Determination of Diaphragm Force F px : Highrise

§1633.2.9

*-

# , ! #

$%(('&'-

This example illustrates determination of the diaphragm design force F px of Equation (33-1) for a representative floor of a multi-story building. The nine-story moment frame building shown below has the tabulated design seismic forces F x . These were determined from Equations (30-14) and (30-15) and the design base shear. The following information is given: 2

1

Zone 4 W = 3,762 k C a = 0.40 C v = 0.56 R = 8.5 ρ = 1.2 I = 1.0 T = 1.06 sec V = 233.8 k Ft = 17.3 k

Level

27'

3

27'

Story Weight

9

214k

8

405k

7

405k

6

405k

5

584k

4

422k

3

422k

2

440k

1

465k

12' 12' 12' 12' 12' 12' 12' 12' 20'

Level x

h(ft)

w(k)

wh

wh Σwh

9 8

116 104

214 405

24,824 42,120

0.103 0.174

22.3 + 17.3 = 39.6 37.7

7

92

405

37,260

0.154

33.3

6

80

405

32,400

0.134

29.0

5

68

584

39,712

0.164

35.5

4

56

422

23,632

0.098

21.2

3

44

422

18,568

0.077

16.7

2

32

440

14,080

0.058

12.6

1

20

465

9,300

0.038

8.2

Σ =3,762

241,896

Fx (k)

233.8

Find the diaphragm force at Level 7.


Example 49 Determination of Diaphragm Force Fpx: Highrise


§1633.2.9

Code Reference

Diaphragm force at Level 7.

Seismic forces on floor and roof diaphragm are specified in §1633.2.9. The following expression is used to determine the diaphragm force F px at level x: n

F px =

Ft + ∑ Fi i= x

n

∑ wi

w px

(33-1)

i= x

Section 1633.2.9 also has the following limits on F px : 0.5C a Iw px ≤ F px ≤ 1.0C a Iw px For level 7, x = 7 . Fp7 =

[17.3 + (33.3 + 37.7 + 22.3)] (405) = (0.108)(405) = 43.7k (405 + 405 + 214)

Check limits: 0.5C a Iw px = 0.5 (0.40)(1.0) 405 = 81.0k 1.0C a Iw px = 1.0 (0.40)(1.0) 405 = 162.0k ∴ F p 7 = 81.0 kips Note that the redundancy factor, in this example ρ = 1.2 , is not applied to the loads E h due to F px (such as chord forces and floor-to-frame shear connections).


Example 50 Building Separation

§1633.2.11

04

$%(('&'

8# "

Building separations are necessary to prevent or reduce the possibility of two adjacent structures impacting during an earthquake. Requirements for building separations are given in §1633.2.11. In this example, the static displacements and information about each structure are given below. Separation

Structure 1 Level ∆S

Level 4

4 3

2

Structure 2 Level ∆S

1.38 in.

—

—

3

1.00

3

0.75 in

2

0.47

2

0.35

1

0

1

0

R = 8.5

R = 7.0

1

Structure 1

Structure 2

Find the required separations for the following situations:

Separation within the same building. Separation from an adjacent building on the same property. Separation from an adjacent building on another property.


Code Reference

Separation within the same building.

§1633.2.11

Expansion joints are often used to break a large building, or an irregular building, into two or more parts above the foundation level. This effectively creates separate “structures” within the same “building.” The code requires that the structures be separated by the amount ∆ MT . where ∆ MT =

(∆ M 1 )2 + (∆ M 2 )2

(33-2)

∆ M 1 = maximum inelastic displacement of Structure 1 ∆ M 2 = maximum inelastic displacement of Structure 2 The required separation is determined in the following two steps.


Example 50 Building Separations

§1633.2.11

Determine inelastic displacements of each structure. To determine the minimum separation between parts of the same “building” that are separated by an expansion joint, the maximum inelastic floor displacements under code seismic forces must be determined for each structure. These are

§1630.9.2

For Structure 1 ∆ M 1 = 0.7 R∆ s = 0.7 × 8.5 × 1.0 = 5.95 in.

(30-17)

For Structure 2 ∆ M 2 = 0.7 R∆ s = 0.7 × 7.0 × .75 = 3.68 in.

Determine the required separation. §1633.2.11 The required separation is determined from the individual maximum inelastic displacements of each structure as follows: ∆ MT =

(30-17)

(∆ ) + (∆ ) 2

M1

M2

2

=

(5.95)2 + (3.68)2

= 7.0 in.

(33-2)

Separation from an adjacent building on the same property.

If Structures 1 and 2 above were adjacent, individual buildings on the same property, the solution to this problem is the same as that shown above in Step 1. The code makes no distinction between an “internal” separation in the same building and the separation required between two adjacent buildings on the same property. ∆ MT = 7.0 in.

Separation from an adjacent building on another property.

§1633.2.11

If Structure 1 is a building under design and Structure 2 is an existing building on another property, we would generally not have information about the seismic displacements of Structure 2. Often even basic information about the structural system of Structure 2 may not be known. In this case, separation must be based only on information about Structure 1. The maximum static displacement of Structure 1 is 1.38 inches and occurs at the roof (Level 4). The inelastic displacement is calculated as: ∆ M = 0.7 R∆ s = 0.7 × 8.5 × 1.38 = 8.2 in.

(30-17)

Structure 1 must be set back 8.2 inches from the property line, unless a smaller separation is justified by a rational analysis based on maximum ground motions. Such an analysis is difficult to do, and is generally not done except in very special cases.


Example 51 Flexible Nonbuilding Structure

§1634.2

0

$%(*'&

, # " A tall cylindrical steel vessel is supported by a heavy, massive concrete foundation. The following information is given: Weight of tank and maximum normal operating contents = 150 k Occupancy Category 2 Zone 4 I = 1.25 (toxic contents per Table 16-K) C a = 0.44 C v = 0.64 N v = 1.0

Wall thickness t = 3/8"

L = 150' D = 8' Assumed base

Grade


Period of vibration. Design base shear. Vertical distribution of seismic forces. Overturning moment at base.


Code Reference

Period of vibration.

In this example, only the case with the vessel full of contents will be considered. In actual practice, other conditions may need to be considered. For calculation purposes, the base is assumed to be located at the top of the pier. The weight of the vessel is assumed to be uniformly distributed over its height. The period of the vessel must be determined by Method B. This is required by §1634.1.4. For this particular vessel, the expression for the period of a thin-walled cantilever cylinder may be used. T = 7.65 × 10 − 6 where:

2

1

 L   wD  2     D  t 

L = 150 ft , D = 8 ft t = 3 8 in. W = 150 k



w=

§1634.2

W 150,000 = = 1000 plf L 150

wD 1000 × 8 = = 256,000 ( t 0.375 / 12 ) L 150 = = 18.75 D 8 T = 7.65 × 10 −6 × 18.75 2 × 256,000 = 1.36 sec Because the period is greater than .06 seconds, the vessel is considered flexible. It should be noted that the value of the period T determined using Method B is not subject to the 30-percent limit mentioned in §1630.2.2, Item 2. This is because Method A is intended for buildings and is not applicable to structural systems that differ from typical building configurations and characteristics. Refer to Section C109.1.4 of the SEAOC Blue Book for further discussion.

Design base shear.

The design base shear for nonbuilding structures is calculated from the same expressions as for buildings. These are given in §1630.2.1. In addition, nonbuilding structures such as the vessel must also satisfy the requirements of §1634.5. V =

Cv I W RT

R = 2.9 and Ω o = 2.0 V=

(30-4) Table 16-P

0.64 (1.25) (150) = 30.4 kips 2.9 (1.36)

Under §1634.5 Item 1, design base shear must not be less than the following: V = 0.56C a IW = 0.56 (.44 )(1.25)150 = 46.2 kips

(34-2)

nor in Zone 4 less than V=

1.6 ZN v I 1.6 (.4 )(1.0)(1.25) (150) = 41.4 kips W= 2.9 R

(34-3)

∴ V = 46.2 kips



§1634.2

Vertical distribution of seismic forces.

Requirements for the vertical distribution of seismic forces are given in §1634.5 Item 2. This specifies the use of the same vertical distribution of force as for buildings, either Equation (30-13) or a dynamic analysis. The following shows use of the static procedures of Equation (30-13). n

V = Ft + ∑ Fi

(30-13)

i =1

where T = 1.36 sec > 0.7 Ft = 0.07TV ∴ Ft = 0.07 (1.36)(46.2 ) = 4.4 k < 0.25V o.k.

(30-14)

∴ F = V − Ft = 46.2 − 4.4 = 41.8 k acting at 2L/3 (centroid of triangular distribution) The vertical distribution of seismic forces on the vessel is shown below.

F = 41.8 k 2L/3 = 100'

L = 150'

Ft = 4.4 k

V = 46.2 k

Overturning moment at base.

M = 4.4 (150 ) + 41.8 (100 ) = 4,840 k − ft (at the top of the foundation)


Example 52 Lateral Force on Nonbuilding Structure

§1634.2

0&

$%(*'&

, # "

A nonbuilding structure with a concrete intermediate moment-resisting frame (IMRF) supports some rigid aggregate storage bins. Weights W1 and W2 include the maximum normal operating weights of the storage bins and contents as well as the tributary frame weight. The following information is given: Zone 4 I = 1.0 Soil Profile Type D C a = 0.44 C v = 0.64 N v = 1.0 T = 2.0 sec

W2 = 200k Level

F2

2

15'

W1 = 100k F1

1


Design base shear. Vertical distribution of seismic forces.


30'

Code Reference

Design base shear.

§1634.2

Because this is a flexible structure, the general expressions for design base shear given in §1630.2.1 must be used. Note that the Exception of §1634.2 permits use of an IMRF in Zones 3 and 4, provided the height of the structure is less than 50 feet and R does not exceed 2.8. The total base shear in a given direction is determined from V=

Cv I 0.64 (1.0 ) (200 + 100) = 0.114 (300) = 34.2 k W= RT 2.8 (2.0)

(30-4)

However, the total base shear need not exceed V≤

2.5C a I 2.5 (0.44 )(1.0) (200 + 100) = 117.9 kips W= 2.8 R

(30-5)

The total design base shear cannot be less than V ≥ 0.11C a IW = 0.11 (0.44 )(1.0)(200 + 100) = 14.5 kips


(30-6)

Example 52 Lateral Force on Nonbuilding Structure

§1634.2

In Seismic Zone 4, the total base shear also cannot be less than V≥

0.8ZN v 0.8 (0.4 )(1.0) (200 + 100) = 34.3 kips W= 2.8 R

(30-7)

In this example, design base shear is controlled by Equation (30-7). V = 34.3 kips


§1634.2

The design base shear must be distributed over the height of the structure in the same manner as that for a building structure. Fx =


∑ wi hi

=

(V − Ft )(W x h x ) (W1 h 1 + W2 h 2 )

(30-15)

i =1

Because T > 0.7 seconds, a concentrated force Ft must be applied to the top level. Ft = 0.07TV = 0.07 (2.0)(34.3) = 4.90 k F2 = 4.90 +

F1 =

(30-14)

(34.3 − 4.90)(200)(45) = 26.9 kips [200 (45) + 100 (30)]

(30-15)

(34.3 − 4.90)(100)(30) = 7.4 kips [200 (45) + 100 (30 )]

(30-15)

Commentary Section 1634.1.2 permits use of ρ = 1.0 for load combinations for nonbuilding structures using §1634.3, §1634.3 or §1634.5 for determination of seismic forces.


Example 53 Rigid Nonbuilding Structure

§1634.3

0(

$%(*'(

5# # " The code has special requirements for the determination of seismic forces for design of rigid nonbuilding structures. In this example, rigid ore crushing equipment is supported by a massive concrete pedestal and seismic design forces are to be determined. The following information is given: Zone 4 C a = 0.4 I = 1.0 T = 0.02 sec WEQUIPMENT = 100 k WSUPPORT = 200 k

CM

F2

CM

F1 30' 20'


Design base shear. Vertical distribution of seismic forces.


Grade

Code Reference

Design base shear.

§1634.3

For rigid nonbuilding structures, Equation (34-1) is used to determine design base shear. V = 0.7C a IW = 0.7 (0.4 )(1.0 )(200 + 100) = 84 kips


(34-1) §1634.3

Design base shear is distributed according to the distribution of mass F1 =

200 (84 ) = 56.0 kips 300

F2 =

100 (84 ) = 28.0 kips 300

Commentary Section 1634.1.2 permits use of ρ = 1.0 for load combinations for nonbuilding structures using §1634.3, §1634.4 or §1634.5 for determination of seismic forces. SEAOC Seismic Design Manual

Example 54 Tank With Supported Bottom

§1634.4

0*

$%(*'*

< " 8

A small liquid storage tank is supported on a concrete slab. The tank does not contain toxic or explosive substances. The following information is given:

Zone 4 C a = 0.4 I p = 1.0 Weight of tank and maximum normal operating contents = 120 kips

20'

D = 10.0' Slab Grade

Find the design base shear.


§1634.4

Code Reference

The tank is a nonbuilding structure, and seismic requirements for tanks with supported bottoms are given in §1634.4. This section requires that seismic forces be determined using the procedures of §1634.3* for rigid structures. Base shear is computed as V = 0.7Ca IW = 0.7 (0.4 )(1.0)(120) = 33.6 kips

(34-1)

The design lateral seismic force is to be applied at the center of mass of the tank and its contents. *Note: There is a typographical error on page 2-21 in some versions of the 1997 UBC in §1634.4. Section 1632 should be “Section 1634.3.”

Commentary The above procedures are intended for tanks that have relatively small diameters and where the forces generated by fluid sloshing modes are small. For large diameter tanks, the effects of sloshing must be considered. Refer to American Water Works Association Standard ANSI/AWWA D100-84 “Welded Steel Tanks for Water Storage,” or American Petroleum Institute Standard 650, “Welded Steel Tanks for Oil Storage” for more detailed guidance. Also see Section C109.5.1 of the SEAOC Blue Book for a discussion of tank anchorage methods.


Example 55 Pile Interconnections

§1807.2

00 .

$342'&

A two-story masonry bearing wall structure has a pile foundation. Piles are located around the perimeter of the building. The foundation plan of the building is shown below. The following information is given: Original grade

Zone 4 I = 1.0 (standard occupancy) Pile cap size: 3'-0" square x 2'-0" deep Grade beam: 1'-6" x 2'-0" Allowable lateral bearing = 200 psf per ft. of depth below natural grade.

1'-6" x 2'-0" Grade beam

2'-0"

Pile cap

2'-0"

Pile

Pile Cap

Dead Load

Reduced Live Load

3

46 k

16 k

14 k

0

58

16

14

0

10

Seismic N/S E/W

1

2

Section A-A: Typical pile cap

3

4

5

North 4 @ 25' = 100' A

B

2 @ 30' = 60'

A 1

2

3

4

5

A 6

7

C 8

9

10

11

12

Foundation pla


Interconnection requirements. Interconnection force between pile caps 3 and 10. Required “tie” restraint between pile caps 3 and 10.



§1807.2


Code Reference

Interconnection requirements.

§1807.2

The code requires that individual pile caps of every structure subject to seismic forces be interconnected with ties. This is specified in §1807.2. The ties must be capable of resisting in tension and compression, a minimum horizontal tie force equal to 10 percent of the larger column vertical load. The column vertical load is to be considered the dead, reduced live, and seismic loads on the pile cap. An exception to §1807.2 allows use of “equivalent restraint.”

Interconnection force between pile caps 3 and 10.

Maximum loads on each pile cap under E/W seismic forces are Pile cap 3 = 46 + 16 + 0 = 62 kips Pile cap 10 = 58 + 16 + 0 = 74 kips Minimum horizontal tie force is 10 percent of largest column vertical load P = 0.10 (74 ) = 7.40 kips

Required “tie” restraint between pile caps 3 and 10.

The choices are to add a grade beam (i.e., tie beam) connecting pile caps 3 and 10, or to try to use passive pressure restraint on the pile cap in lieu of a grade beam. The latter is considered an “equivalent restraint” under the exception to §1807.2. Check passive pressure resistance. Passive pressure =

Required length =

(400 + 800 ) ( 2

2 ft ) = 1,200 plf

7,400 lbs = 6.2 ft 1,200 plf

This is greater than 3'-0" pile cap width, but pile cap and a tributary length of N/S grade beam on either side of the pile cap may be designed to resist “tie” forces using passive pressure. This system is shown below, and if this is properly designed, no grade beam between pile caps 3 and 10 (or similar caps) is required.



§1807.2

1,200 plf 1'-6" Grade beam

B 400 psf/ft

Pile cap B 1.6'

3'

2'-0"

1.6' 800 psf/ft

6.2'

Equivalent restraint system in pla

Section B-B: Grade beam

Commentary Normally, buildings on pile foundations are required to have interconnecting ties between pile caps. This is particularly true in the case of highrise buildings and buildings with heavy vertical loads on individual pile caps. Ties are essential in tall buildings. Ties are also necessary when the site soil conditions are poor such that lateral movements, or geotechnical hazards, such as liquefaction, are possible. Also note that while §1807.2 has the wording “tension or compression,” the intent is that the ties must resist the required forces in both tension and compression. In design of relatively lightweight one- and two-story buildings, the exception to the interconnecting tie requirement of §1807.2 may permit a more economical foundation design. However, when interconnecting ties are omitted, a geotechnical engineer should confirm the appropriateness of this decision.


Seismic Design Manual Volume II Building Design Examples: Light Frame, Masonry and Tilt-up

April 2000


Publisher Structural Engineers Association of California (SEAOC) 1730 I Street, Suite 240 Sacramento, California 95814-3017 Telephone: (916) 447-1198; Fax: (916) 443-8065 E-mail: [email protected]; Web address: www.seaoc.org

The Structural Engineers Association of California (SEAOC) is a professional association of four regional member organizations (Central California, Northern California, San Diego, and Southern California). SEAOC represents the structural engineering community in California. This document is published in keeping with SEAOC’s stated mission: “to advance the structural engineering profession, to provide the public with structures of dependable performance through the application of state-of-the-art structural engineering principles; to assist the public in obtaining professional structural engineering services; to promote natural hazard mitigation; to provide continuing education and encourage research; to provide structural engineers with the most current information and tools to improve their practice; and to maintain the honor and dignity of the profession.” Editor Gail H. Shea, Albany, California Disclaimer Practice documents produced by the Structural Engineers Association of California (SEAOC) and/or its member organizations are published as part of our association’s educational program. While the information presented in this document is believed to be correct, neither SEAOC nor its member organizations, committees, writers, editors, or individuals who have contributed to this publication make any warranty, expressed or implied, or assume any legal liability or responsibility for the use, application of, and/or reference to opinions, findings, conclusions, or recommendations included in this publication. The material presented in this publication should not be used for any specific application without competent examination and verification of its accuracy, suitability, and applicability by qualified professionals. Users of information from this publication assume all liability arising from such use. ii

SEAOC Seismic Design Manual, Vol. II (1997 UBC)

Table of Contents

Table of Contents

Preface

....................................................................................................................... v

Acknowledgments .......................................................................................................... vii Suggestions for Improvement ........................................................................................ ix Introduction ....................................................................................................................... 1 How to Use This Document ............................................................................................ 2 Notation

....................................................................................................................... 3

References ..................................................................................................................... 10 Design Example 1 Wood Light Frame Residence................................................................................. 11 Design Example 2 Wood Light Frame Three-Story Structure ............................................................... 87 Design Example 3 Cold-Formed Steel Light Frame Three-Story Structure ........................................ 159 Design Example 4 Masonry Shear Wall Building ................................................................................ 213 Design Example 5 Tilt-Up Building ...................................................................................................... 247 Design Example 6 Tilt-Up Wall Panel With Openings ......................................................................... 289


iii

Table of Contents

iv


Preface

Preface

This document is the second volume of the three-volume SEAOC Seismic Design Manual. The first volume, “Code Application Examples,” was published in April 1999. These documents have been developed by the Structural Engineers Association of California (SEAOC) with funding provided by SEAOC. Their purpose is to provide guidance on the interpretation and use of the seismic requirements in the 1997 Uniform Building Code (UBC), published by the International Conference of Building Officials (ICBO), and SEAOC’s 1999 Recommended Lateral Force Requirements and Commentary (also called the Blue Book). The Seismic Design Manual was developed to fill a void that exists between the Commentary of the Blue Book, which explains the basis for the UBC seismic provisions, and everyday structural engineering design practice. While the Manual illustrates how the provisions of the code are used, the examples shown do not necessarily illustrate the only appropriate methods of seismic design, and the document is not intended to establish a minimum standard of care. Engineering judgment needs to be exercised when applying these examples to real projects. Volume I: Code Application Examples, provides step-by-step examples of how to use individual code provisions, such as how to compute base shear or building period. Volumes II and III: Design Examples, furnish examples of the seismic design of common types of buildings. In Volumes II and III, important aspects of whole buildings are designed to show, calculation-by-calculation, how the various seismic requirements of the code are implemented in a realistic design. Volume II contains six examples. These illustrate the seismic design of the following structures: (1) a two-story wood light frame residence, (2) a three-story wood light frame building, (3) a three-story cold formed light frame building, (4) a one-story masonry building with panelized wood roof, (5) a one-story tilt-up building with panelized wood roof, and (6) the design of a tilt-up wall panel with large openings. Work on the final volume, Building Design Examples, Volume III—Steel, Concrete and Cladding, is nearing completion and is scheduled for release in late Spring 2000. It is SEAOC’s present intention to update the Seismic Design Manual with each edition of the building code used in California. Work is currently underway on a 2000 International Building Code version. Ronald P. Gallagher Project Manager


v

Preface

vi


Acknowledgments

Acknowledgments

Authors The Seismic Design Manual was written by a group of highly qualified structural engineers. These individuals are both California registered civil and structural engineers and SEAOC members. They were selected by a Steering Committee set up by the SEAOC Board of Directors and were chosen for their knowledge and experience with structural engineering practice and seismic design. The Consultants for Volumes I, II and III are: Ronald P. Gallagher, Project Manager Robert Clark David A. Hutchinson Jon P. Kiland John W. Lawson Joseph R. Maffei Douglas S. Thompson Theodore C. Zsutty

Volume II was written principally by Douglas S. Thompson (Examples 1, 2, and 3), Jon P. Kiland (Example 4), Ronald P. Gallagher (Example 5), and John W. Lawson (Example 6). Many useful ideas and helpful suggestions were offered by the other Consultants. Consultant work on Volume III is currently underway. Steering Committee Overseeing the development of the Seismic Design Manual and the work of the Consultants was the Project Steering Committee. The Steering Committee was made up of senior members of SEAOC who are both practicing structural engineers and have been active in Association leadership. Members of the Steering Committee attended meetings and took an active role in shaping and reviewing the document. The Steering Committee consisted of: John G. Shipp, Chair Robert N. Chittenden Stephen K. Harris Martin W. Johnson Scott A. Stedman


vii

Acknowledgments

Reviewers A number of SEAOC members, and other structural engineers, helped check the examples in this volume. During its development, drafts of the examples were sent to these individuals. Their help was sought in both review of code interpretations as well as detailed checking of the numerical computations. The assistance of the following individuals is gratefully acknowledged: Ricardo Arevalo Gary Austin Robert Chittenden Kelly Cobeen Michael Cochran Susan Dowty Gerald Freeman Stephen K. Harris

Gary Ho John Lawson Dilip M. Khatri Harry (Hank) Martin (AISC) David McCormick Gary Mochizuki William Nelson Neil Peterson

Michael Riley George Richards Alan Robinson (for CMACN) John Rose (APA) Douglas Thompson Jerry Tucker Craig Wilcox Dennis Wish

Seismology Committee Close collaboration with the SEAOC Seismology Committee was maintained during the development of the document. The 1999-2000 Committee reviewed the document and provided many helpful comments and suggestions. Their assistance is gratefully acknowledged.

1999-2000 Martin W. Johnson, Chair Saif Hussain, Past Chair David Bonowitz Robert N. Chittenden Tom H. Hale Stephen K. Harris Douglas C. Hohbach Y. Henry Huang Saiful Islam H. John Khadivi Jaiteeerth B. Kinhal Robert Lyons Simin Naaseh Chris V. Tokas Michael Riley, Assistant to the Chair

viii




In keeping with two of its Mission Statements: (1) “to advance the structural engineering profession” and (2) “to provide structural engineers with the most current information and tools to improve their practice”, SEAOC plans to update this document as seismic requirements change and new research and better understanding of building performance in earthquakes becomes available. Comments and suggestions for improvements are welcome and should be sent to the following: Structural Engineers Association of California (SEAOC) Attention: Executive Director 1730 I Street, Suite 240 Sacramento, California 95814-3017 Telephone: (916) 447-1198 Fax: (916) 443-8065 E-mail: [email protected] Web address: http://www.seaoc.org

Errata Notification SEAOC has made a substantial effort to ensure that the information in this document is accurate. In the event that corrections or clarifications are needed, these will be posted on the SEAOC web site at http://www.seaoc.org or on the ICBO website at http://ww.icbo.org. SEAOC, at its sole discretion, may or may not issue written errata.


ix

Seismic Design Manual Volume II Building Design Examples: Light Frame, Masonry and Tilt-up

Introduction

Introduction

Seismic design of new light frame, masonry and tilt-up buildings for the requirements of the 1997 Uniform Building Code (UBC) is illustrated in this document. Six examples are shown: (1) a two-story wood frame residence, (2) a large three-story wood frame building, (3) a three-story cold formed steel light frame building, (4) a one-story masonry (concrete block) building with panelized wood roof, (5) a one-story tilt-up building with panelized wood roof, and (6) the design of a tilt-up wall panel with large openings. The buildings selected are for the most part representative of construction types found in Zones 3 and 4, particularly California and the Western States. Designs have been largely taken from real world buildings, although some simplifications were necessary for purposes of illustrating significant points and not presenting repetive or unnecessarily complicated aspects of a design. The examples are not complete building designs, or even complete seismic designs, but rather they are examples of the significant seismic design aspects of a particular type of building. In developing these examples, SEAOC has endeavored to illustrate correct use of the minimum provisions of the code. The document is intended to help the reader understand and correctly use the design provisions of UBC Chapters 16 (Design Requirements), 19 (Concrete), 21 (Masonry), 22 (Steel) and 23 (Wood). Design practices of an individual structural engineer or office, which may result in a more seismic-resistant design than required by the minimum requirements of UBC, are not given. When appropriate, however, these considerations are discussed as alternatives. In some examples, the performance characteristics of the structural system are discussed. This typically includes a brief review of the past earthquake behavior and mention of design improvements added to recent codes. SEAOC believes it is essential that structural engineers not only know how to correctly interpret and apply the provisions of the code, but that they also understand their basis. For this reason, many examples have commentary included on past earthquake performance. While the Seismic Design Manual is based on the 1997 UBC, references are made to the provisions of SEAOC’s 1999 Recommended Lateral Force Provisions and Commentary (Blue Book). When differences between the UBC and Blue Book are significant, these are brought to the attention of the reader.


1


How to Use This Document Generally, each design example is presented in the following format. First, there is an “Overview” of the example. This is a description of the building to be designed. This is followed by an “Outline” indicating the tasks or steps to be illustrated in each example. Next, “Given Information” provides the basic design information, including plans and sketches given as the starting point for the design. This is followed by “Calculations and Discussion”, which provides the solution to the example. Some examples have a subsequent section designated “Commentary” The commentary is intended to provide a better understanding of aspects of the example and/or to offer guidance to the reader on use of the information generated in the example. Finally, references and suggested reading are given under “References.” Some examples also have a “Forward” and/or section “Factors Influencing Design” that provide remarks on salient points about the design. Because the document is based on the UBC, UBC notation is used throughout. However, notation from other codes is also used. In general, reference to UBC sections and formulas is abbreviated. For example, “1997 UBC Section 1630.2.2” is given as §1630.2.2 with 1997 UBC (Volume 2) being understood. “Formula (32-2)” is designated Equation (32-2) or just (32-2) in the right-hand margins of the examples. Similarly, the phrase “Table 16-O” is understood to be 1997 UBC Table 16-O. Throughout the document, reference to specific code provisions, tables, and equations (the UBC calls the latter formulas) is given in the right-hand margin under the heading Code Reference. When the document makes reference to other codes and standards, this is generally done in abbreviated form. Generally, reference documents are identified in the right-hand margin. Some examples of abbreviated references are shown below.

Right-Hand Margin Notation

More Complete Description

23.223, Vol. 3

Section 23.223 of Volume 3, of the 1997 Uniform Building Code (UBC). Section E3.3 of the 1996 Edition of the American Iron and Steel Institute (AISI) Specification for the Design of Cold-Formed Steel Structural Members. Table 5A of the 1991 National Design Specification for Wood Construction (NDS). Table 1-A of Ninth Edition, American Institute of Steel Construction (AISC) Manual of Steel Construction, Allowable Stress Design.

96 AISI E3.3

91 NDS Table 5A Table 1-A, AISC-ASD

2


Notation

Notation

The following notations are used in this document. These are generally consistent with that used in the UBC and other codes such as ACI, AISC, AISI and NDS. Some additional notations have also been added. The reader is cautioned that the same notation may be used more than once and may carry entirely different meaning in different situations. For example, E can mean the tabulated elastic modulus under the NDS definition (wood) or it can mean the earthquake load under §1630.1 of the UBC (loads). When the same notation is used in two or more definitions, each definition is prefaced with a brief description in parentheses (e.g., wood or loads) before the definition is given.

A

=

(wood diaphragm) area of chord cross section, in square inches

A

=

(wood shear wall) area of boundary element cross section, in square inches (vertical member at shear wall boundary)

AB

=

ground floor area of structure in square feet to include area covered by all overhangs and projections.

Ac

=

the combined effective area, in square feet, of the shear walls in the first story of the structure.

Ae

=

the minimum cross-sectional area in any horizontal plane in the first story, in square feet of a shear wall.

Ap

=

the effective area (in square inches) of the projection of an assumed concrete failure surface upon the surface from which the anchor protrudes.

As

=

area of tension reinforcing steel

Ase

=

equivalent area of tension reinforcing steel

Ax

=

the torsional amplification factor at Level x.

Aconc

=

net concrete section area

a

=

depth of equivalent rectangular stress block

ap

=

numerical coefficient specified in §1632 and set forth in Table 16-O of UBC.


3

Notation

4

Btn

=

nominal tensile strength of anchor bolt in masonry, in pounds.

b

=

(concrete beam) width of compression face of member

b

=

(wood diaphragm) diaphragm width, in feet

b

=

(wood shear wall) wall width, in feet

btu

=

factored tensile force supported by anchor bolt in masonry, in pounds

Ca

=

seismic coefficient, as set forth in Table 16-Q of UBC.

Cd

=

penetration depth factor

CD

=

load duration factor

CM

=

wet service factor

Ct

=

numerical coefficient given in §1630.2.2 of UBC.

Cv

=

seismic coefficient, as set forth in Table 16-R of UBC.

c

=

distance from neutral axis to extreme fiber

D

=

(loads) dead load on a structural element.

D

=

(wood) diameter

De

=

the length, in feet, of a shear wall in the first story in the direction parallel to the applied forces.

d

=

(wood) dimension of wood member (assembly)

d

=

(concrete or masonry) distance from extreme compression fiber to centroid of tension reinforcement

d

=

(loads) distance from lateral resisting element to the center of rigidity

d

=

(wood) pennyweight of nail or spike

da

=

deflection due to anchorage details in wood shear wall (rotation and slip at tie-down bolts), in inches


Notation

E

=

(wood diaphragm) elastic modulus of chords, in psi

E

=

(wood shear wall) elastic modulus of boundary element (vertical member at shear wall boundary), in psi

Ec

=

modulus of elasticity of concrete, in psi

Em

=

modulus of elasticity of masonry, in psi

E, E ' =

(wood) tabulated and allowable modulus of elasticity, in psi

e

=

diaphragm eccentricity

en

=

nail deformation in inches (see Table 23-2-K of UBC)

E, Eh, Em, Ev,

=

(loads) earthquake loads set forth in §1630.1 of UBC.

Fb ' Fb ' =

tabulated and allowable bending design value, in psi

Fc⊥ ' Fc⊥' =

tabulated and allowable compression design value perpendicular to grain, in psi

Fv ' Fv ' =

tabulated and allowable compression shear design value parallel to grain (horizontal shear), in psi

Fx

=

design seismic force applied to Level i, n or x, respectively.

Fp

=

design seismic force on a part of the structure.

Fpx

=

design seismic force on a diaphragm.

Ft

=

(loads) that portion of the base shear, V, considered concentrated at the top of the structure in addition to Fn.

Ft

=

torsional shear force

Fv

=

direct shear force

Fy

=

specified yield strength of structural steel.

fb

=

extreme fiber bending stress

fc

=

(wood) actual compression stress parallel to grain


5

Notation

fc '

=

f c⊥

=

fi

=

lateral force at Level i for use in Formula (30-10) of UBC.

=

specified compressive strength of masonry, in psi

fp

=

equivalent uniform load.

fr

=

(masonry) modulus of rupture, in psi

fy

=

specified tension yield strength of reinforcing steel.

fv

=

(wood) actual shear stress parallel to grain

G

=

modulus of rigidity of plywood, in pounds per square inch (see Table 23-2-J of UBC)

g

=

acceleration due to gravity.

h

=

(concrete) height of wall between points of support, in inches

h

=

(wood shear wall) wall height, in feet

fm '

hi, hn, hx

6

specified compressive strength of concrete. (wood) actual compression stress perpendicular to grain

= height in feet above the base to level i, n or x, respectively

I

=

importance factor given in Table 16-K of UBC.

Icr

=

moment of inertia of cracked concrete or masonry section

Ig

=

moment of inertia of gross concrete or masonry section about centroidal axis, neglecting reinforcement

Ip

=

importance factor specified in Table 16-K of UBC.

k

=

(wood) wall stiffness

L

=

(loads) live load on a structural element, except roof live load

Lr

=

(loads) roof live load


Notation

L

=

(wood) span length of bending member

L

=

(wood diaphragm) diaphragm length, in feet

lc

=

(concrete) vertical distance between wall supports, in inches

Level i =

level of the structure referred to by the subscript i. “i = 1” designates the first level above the base.

Level n =

that level that is uppermost in the main portion of the structure.

Level x =

that level that is under design consideration. “x = 1” designates the first level above the base.

M

=

maximum bending moment

Mcr

=

nominal cracking moment strength in concrete or masonry

Mn

=

nominal moment strength

Ms

=

the maximum moment in the wall resulting from the application of the unfactored load combinations

Mu

=

factored moment at section

M.C. =

moisture content based on oven-dry weight of wood, in percent

Na

=

near-source factor used in the determination of Ca in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and slip rates as set forth in Tables 16-S and 16-U of UBC.

Nv

=

near-source factor used in the determination of Cv in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and slip rates as set forth in Tables 16-T and 16-U of UBC.

P

=

total concentrated load or total axial load

Pc

=

(concrete) design tensile strength of anchors, in pounds

Pu

=

factored axial load


7

Notation

R

=

numerical coefficient representative of the inherent overstrength and global ductility capacity of lateral-forceresisting systems, as set forth in Table 16-N or 16-P of UBC.

r

=

a ratio used in determining ρ. See §1630.1 of UBC.

SA, SB, SC, SD, SE, SF =

8

soil profile types as set forth in Table 16-J of UBC.

T

=

elastic fundamental period of vibration, in seconds, of the structure in the direction under consideration.

T

=

(loads) torsional moment

t

=

thickness

t

=

(plywood) effective thickness of plywood for shear, in inches (see Tables 23-2-H and 23-2-I of UBC)

tm

=

thickness of main member

ts

=

thickness of side member

V

=

(wood) shear force.

V

=

(loads) the total design lateral force or shear at the base given by Formula (30-5), (30-6), (30-7) or (30-11) of UBC.

Vm

=

nominal shear strength of masonry

Vn

=

(concrete or masonry) nominal shear strength

Vn

=

(wood) fastener load, in pounds

Vs

=

nominal shear strength of shear reinforcement

Vu

=

(masonry) required shear strength

Vx

=

the design story shear in Story x.

v

=

(wood diaphragm) maximum shear due to design loads in the direction under consideration, plf

v

=

(wood shear wall) maximum shear due to design loads at the top of the wall, in plf


Notation

W

=

(wood) total uniform load.

W

=

(loads) the total seismic dead load defined in §1630.1.1 of UBC.

wi, wx =

that portion of W located at or assigned to Level i or x, respectively.

Wp

=

the weight of an element or component.

wpx

=

the weight of the diaphragm and the element tributary thereto at Level x, including applicable portions of other loads defined in §1630.1.1 of UBC.

x, y

=

distance to centroid

Z

=

seismic zone factor as given in Table 16-I of UBC.

Z, Z' =

(wood) nominal and allowable lateral design value for a single fastener connection.

∆

=

(wood) the calculated deflection of wood diaphragm or shear wall, in inches.

∆M

=

maximum inelastic response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design basis ground motion, including estimated elastic and inelastic contributions to the total deformation defined in §1630.9 of UBC.

∆S

=

design level response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design seismic forces.

∆ cr

=

deflection at M cr

∆n

=

deflection at M n

∆s

=

(concrete) deflection at M s

∆u

=

deflection due to factored loads, in inches.

γ

=

load/slip modulus for a connection, in pounds per inch.

δi

=

horizontal displacement at Level i relative to the base due to applied lateral forces, f, for use in Formula (30-10) of UBC.


9

Notation

φ

=

strength-reduction factor

ρ

=

(loads) redundancy/reliability factor given by Formula (30-3) of UBC.

ρ

=

(concrete and masonry) ratio of area of flexural tensile reinforcement, As , to area bd.

ρb

=

reinforcement ratio producing balanced strain conditions.

Ωo

=

seismic force amplification factor, which is required to account for structural overstrength and set forth in Table 16-N of UBC.

∑(∆ c X ) =

sum of individual chord-splice slip values on both sides of wood diaphragm, each multiplied by its distance to the nearest support.

References The following codes and standards are referenced in this document. Other reference documents are indicated at the end of each Design Example. ACI-318, 1995, American Concrete Institute, Building Code Regulations for Reinforced Concrete, Farmington Hills, Michigan AISC, American Institute of Steel Construction, Manual of Steel Construction, Allowable Stress Design-ASD, Chicago, Illinois AISI, 1996, American Iron and Steel Institute, Specification for the Design of Cold-Formed Steel Structural Members, Washington, D.C NDS, 1991, American Forest & Paper Association, National Design Specification for Wood Construction, Washington, D.C. UBC 1997, International Conference of Building Officials, Uniform Building Code. Whittier, California. SEAOC Blue Book, 1999, Recommended Lateral Force Requirements and Commentary. Structural Engineers Association of California, Sacramento, California.

10


Design Example 1 n Wood Light Frame Residence

'HVLJQ([DPSOH :RRG/LJKW)UDPH5HVLGHQFH

Figure 1-1. Wood light frame residence

)RUHZRUG Small wood frame residences, such as the one in this example, have traditionally been designed using simplified design assumptions and procedures based largely on judgment and precedent. This example illustrates the strict, literal application of the 1997 UBC provisions. Two of the requirements shown, while required by the code, are considerably different than current California practice: 1. The use of wood diaphragms as part of the lateral force resisting system. Traditionally, light frame dwellings have been designed assuming that such diaphragms behave as infinitely flexible elements. This assumption simplifies the analysis and allows lateral forces to be distributed to the vertical elements of the lateral force resisting system by tributary area methods. The code has had a definition of a flexible diaphragm since the 1988 UBC (§1630.6 of the 1997



UBC). UBC §1630.6 permits diaphragms to be treated as flexible, only if the maximum deflection of the diaphragm under the lateral loading is equal to or greater than twice the deflection of the vertical elements supporting the diaphragm in the story below. In this example, the diaphragm has been determined not to meet these criteria, and the design is based on the rigid diaphragm assumption. However, recognizing that the diaphragms in this structure likely behave as semi-rigid elements, neither fully flexible nor fully rigid, in this example an envelope approach has been used in which two analyses are performed. The first analysis uses the traditional flexible diaphragm assumptions and the second analysis is based on rigid diaphragm assumptions. The lateral resisting elements have been designed for the most severe forces produced by either assumption. Refer to the overview portion of this design example for further discussion about using the envelope approach. Although these examples are a literal application of the 1997 UBC, the SEAOC Code and Seismology committees are of the joint opinion that the use of the more traditional design approach can provide acceptable lift-safety performance for most one- and two-family dwellings. The commentary below provides more discussion of these issues: 2.

The use of a system with limited ductility specifically cantilevered columns. In this example, the cantilevered columns are used to provide lateral resistance at the garage door openings. In conventional practice, these would be designed for forces calculated using the R value associated with that system (R= 2.2), with the balance of the structure designed with an R value with light framed shear walls (R=5.5). UBC §1630.4.4 requires that the R value used in each direction, may not be greater than the least value for any of the systems used in that same direction. Therefore, in this design example, because the R value for the cantilevered columns at the garage has an R value of 2.2, the entire structure in this direction has been designed using this R value.

Rigid versus flexible diaphragm assumptions.

Small, light frame detached one- and two-family dwellings have traditionally been designed using flexible diaphragm assumptions, or by a “hybrid” approach of treating closely spaced walls as a unit (i.e., as rigidly connected) and treating the remaining diaphragm as flexible. Also, light frame detached one- and two-family dwellings have been built with the conventional construction provisions of the code without an engineering design. These light frame structures have historically performed satisfactorily from a life-safety standpoint when subjected to strong seismic shaking. Two exceptions to light frame structures performing satisfactorily—both of which were addressed in the 1997 UBC by more stringent requirements—have been related to problems with the height-to-length ratio of shear wall panels and the use of plaster and drywall materials to resist seismic forces.



In the Commentary of the 1999 SEAOC Blue Book (§C805.3.1), it is recognized that lateral forces for many structures with wood diaphragms, mostly large buildings, may be better represented as rigid, as opposed to flexible, diaphragms. Relative to the small structure used in this example, the use of the rigid diaphragm assumptions generally will not significantly improve the seismic behavior. While the building response remains elastic, the rigid diaphragm assumptions will better reflect the initial stiffness of the building system. However, it is not practically possible to accurately calculate the stiffness of all the various elements, including the stiffness contributed by finishes and nonstructural elements and taking into account the fact that stiffness of these elements will degrade as the ground shaking intensifies. As a result, the use of the rigid diaphragm assumptions may not be significantly better than the traditional flexible diaphragm assumption for structures of this type. At the time of this publication, both the SEAOC Code and Seismology Committees agree that many one- and two-family residential structures can be safely designed using the traditional flexible diaphragm assumptions. Consequently, SEAOC recommends modification of the 1997 UBC provisions to allow use of the flexible diaphragm assumption for the design of one- and two-family dwellings. The engineer is cautioned, however, to discuss this with the building official prior to performing substantive design work.

Cantilever column elements in light frame construction.

The UBC requirement that buildings be designed using the least value R for combinations along the same axis was developed with two considerations in mind. The first is that in most structures, the building’s ability to resist seismic forces can be limited to the weakest element in the structure. The second is purely a method of discouraging the more nonductile systems. The potential for P∆ instability of cantilevered column systems limits the column’s capacity to carry large gravity loads when subjected to large building drifts. Therefore, the code has assigned a low R value to this system. However, cantilever columns used in one- and two-family dwellings are typically lightly loaded, and can not develop this P∆ instability. Further, the literal application of §1630.44 would discourage the use of ordinary moment frames and cantilever column systems in favor for the use of slender shear walls that have been known to perform poorly. Consequently, the 1999 SEAOC Blue Book §105.4.4 (page 12) recommends the following alternative approach: Exception: For light frame buildings in occupancy groups 4 and 5 and of two stories or less in height, the lateral force resisting elements are permitted to be designed using the least value of R for the different structural systems found on each independent line of resistance. The value of R used for design of



diaphragms for a given direction of loading in such structures shall not be greater than the least value used for any of the systems in that same direction. Therefore, SEAOC recommends this alternative approach. The cantilever columns (together with any shear walls along that line of force, if present) would be designed using an R = 2.2, with the shear walls located along other lines of force designed using R = 5.5. In other words, the lateral load is factored up for the line with the cantilever column elements, but the conventional R value is used on the remainder of the structure. Consult with your local building official, however, before using this recommendation.

2YHUYLHZ This design example illustrates the seismic design of a 2,800-square-foot single family residence. The structure, shown in Figures 1-1, 1-2, 1-3, 1-4 and 1-5, is of wood light frame construction with wood structural panel shear walls, roof, and floor diaphragms. Roofing is clay tile. Due to the high h/w (height/width) ratios of the walls next to the garage doors, cantilevered column elements are used to provide lateral support. As shown in Figure 1-3, there is an out-of plane offset from the cantilevered column elements on Line E to the glulam beams (GLBs) supporting the shear walls above Line D. The wood structural panel shear walls over the GLBs in the garage do not meet the required h/w ratios without the addition of straps and blocking above and below the window. The residence cannot be built using conventional construction methods for reasons shown in Part 8 of this design example. The following steps illustrate a detailed analysis for some of the important seismic requirements of the 1997 UBC that pertain to design of wood light frame buildings. As stated in the introduction of this manual, these design examples, including this one, are not complete building designs. Many aspects of building design are not included, and only selected parts of the seismic design are illustrated. As is common for Type V construction (see UBC §606), a complete wind design is also necessary, but is not given in this design example. Although the code criteria only recognize two diaphragm categories, flexible and rigid, the diaphragms in this design example are judged to be semi-rigid. Consequently, the analysis in this design example will use the envelope method, which considers the worst loading condition from both the flexible and rigid diaphragm analyses for vertical resisting elements. It should be noted that the envelope method, although not explicitly required by the code, will produce a more predictable performance than will use of only flexible or rigid diaphragm assumptions.



This design example will first determine the shear wall nailing and tiedown requirements obtained using the flexible diaphragm assumption to determine shear wall rigidities for the rigid diaphragm analysis. The method of determining shear wall rigidities used in this design example is by far more rigorous than normal practice, but is not the only method available to determine shear wall rigidities. The Commentary at the end of this design example illustrates two other simplified approaches that would also be appropriate.

2XWOLQH This example will illustrate the following parts of the design process:

Design base shear and vertical distributions of seismic forces.

Lateral forces on shear walls and shear wall nailing assuming flexible diaphragms.

Rigidities of shear walls and cantilever columns at garage.

Centers of mass and rigidity of diaphragms.

Distribution of lateral forces to the shear walls with rigid diaphragms.

Reliability/redundancy factor ρ.

Diaphragm deflections and whether diaphragms are flexible or rigid.

Does residence meet requirements for conventional construction provisions?

Design shear wall frame over garage on line D.

Diaphragm shears at the low roof over garage.

Detail the wall frame over the GLB on line D.

Detail the anchorage of wall frame to the GLB on line D.

Detail the continuous load path at the low roof above the garage doors.



*LYHQ,QIRUPDWLRQ Roof weights (slope 5:12): Tile roofing ½-in. sheathing Roof framing Insulation Miscellaneous Gyp ceiling D (along slope) =

Floor weights: Flooring 5/8" sheathing Floor framing Miscellaneous Gyp ceiling

10.0 psf 1.5 4.0 1.0 0.2 2.8 19.5 psf

1.0 psf 1.8 4.0 0.4 2.8 10.0 psf

D = dead load D = (horiz. proj.) = 19.5 (13/12) = 21.1 psf (the roof and ceilings are assumed to be on a 5:12 slope, vaulted) Weights of respective diaphragm levels, including exterior and interior walls: W roof = 64,000 lb (roof and tributary walls) W floor = 39,000 lb (floor and tributary walls above and below) W

= 103,000 lb

Weights of diaphragms are typically determined by adding the tributary weights of the walls to the diaphragm, e.g., add one-half the height of walls at the second floor to the roof and one-half the height of second floor walls plus one-half the height of first floor walls to second floor diaphragm. It is acceptable practice to ignore the weight of shear walls parallel to the direction of seismic forces to the upper level and add 100 percent of the parallel shear wall weight to the level below, instead of splitting the weight between floor levels. Weights of bearing partitions (not shear walls) should still be split between floors. Unlike commercial construction, the code minimum of 20 psf (vertical load) and 10 psf (lateral load) is often exceeded in residential construction. Framing lumber is Douglas Fir-Larch grade stamped No. 1S-Dry. APA-rated wood structural panels for shear walls will be 15/32-inch thick Structural I, 32/16 span rating, 5-ply with Exposure I glue, however, 4-ply is also acceptable. Three-ply 15/32-inch sheathing has lower allowable shears and the inner ply voids can cause nailing problems. The roof is 15/32-inch thick APA-rated sheathing (equivalent to C-D sheathing in Table 23-II-4), 32/16 span rating with Exposure I glue. The floor is 19/32-inch thick APA-rated Sturd-I-floor 16 inches o.c. rating (or APA-rated sheathing, 42/20 span rating) with Exposure I glue.



Boundary members for the shear walls are 4x posts. Common wire nails are to be used for diaphragms, shear walls, and straps. Sinker nails are to be used for design of the shear wall sill plate nailing at the second floor. (Note: many nailing guns use the smaller diameter box and sinker nails instead of common nails. Closer nail spacing may be required for smaller diameter nails). Seismic and site data: Z = 0.4 (Zone 4) I = 1.0 (standard occupancy) Seismic source type = B Distance to seismic source = 12 km Soil profile type = S C

Table 16-I Table 16-K

S C has been determined by geotechnical investigation. Without a geotechnical investigation, S D can be used as a default value.

Figure 1-2. Foundation plan (ground floor)



Figure 1-3. Second floor framing plan and low roof framing plan

Figure 1-4. Roof framing plan



Figures 1-2 through 1-4 depict the shear walls as dark solid lines. This has been done for clarity in this example. Actual drawings commonly use other graphic depictions. Practice varies on how framing plans are actually shown and on which level the shear walls are indicated. Actual drawings commonly do not call out shear wall lengths. However, building designers should be aware that some building departments now require shear wall lengths to be called out on plans.

)DFWRUV7KDW,QIOXHQFH'HVLJQ Prior to starting the seismic design of the residence, three important related aspects of the design bear discussion. These are the effect of moisture content on lumber, the level of engineering design required to meet code requirements in present-day California practice, and effects of box nails on wood structural panel shear walls. Moisture content in lumber connections.

91 NDS Table 7.3.3

This design example is based on dry lumber. Project specifications typically call for lumber to be grade stamped S-Dry (Surfaced Dry). Dry lumber has a moisture content (MC) less than or equal to 19 percent. Partially Seasoned or Green lumber grade-stamped S-GRN (surfaced green) has a MC between 19 percent and 30 percent. Wet lumber has a MC greater than 30 percent. Construction of structures using lumber with moisture contents greater than 19 percent can produce shrinkage problems in the structures. Also, many engineers and building officials are not aware of the reduction requirements, or wet service factors, related to installation of nails, screws, and bolts (fasteners) into lumber with moisture contents greater than 19 percent at time of installation. For fasteners in lumber with moisture contents greater than 19 percent at the time of installation, the wet service factor, C M = 0.75 for nails and C M = 0.67 for bolts, lags and screws (91 NDS Table 7.3.3). In other words, in lumber whose moisture content exceeds 19 percent, there is a 25 percent to 33 percent reduction in the strength of connections, diaphragms, and shear walls that is permanent. Drying of the lumber after installation of the connectors does not improve the connector capacity. The engineer should exercise good engineering judgment in determining whether it is prudent to base the structural design on dry or green lumber. Other areas of concern are geographical area and time of year the structure will be built. It is possible for green lumber (or dry lumber that has been exposed to rain) to dry out to a moisture content below 19 percent. For 2x framing, this generally takes about two to 3 weeks of exposure to dry air. Thicker lumber takes even longer. Moisture contents can easily be verified by a hand-held “moisture meter.”



Level and type of engineering design required for California residences.

The residence structure in this design example was chosen because it contains many of the structural problem areas that are commonly present in residential construction. These include: 1.

The discontinuous shear wall at the north end of the line 5. (Although this is not a code violation per se, selection of a shear wall location that is continuous to the foundation would improve performance).

2.

Lack of a lateral resisting element along line 4. (Although this is not a code violation per se, the addition of a shear wall at this location would improve performance).

3.

The reduced scope of many structural engineering service contracts, such as “calculation and sketch” projects where the structural engineer provides a set of calculations and sketches of important structural details and the architect produces the actual plans and specifications. This often leads to poorly coordinated drawings and missing structural information. This method also makes structural observation requirements of the building code less effective when the engineer responsible for the design is not performing the site observation. Refer to the Commentary at the end of this design example for further discussion on this subject.

An important factor in the design of California residences, and residences in other high seismic zones, is the level of sophistication and rigor required by the designer. In this design example, a complete, rigorous analysis has been performed. In some jurisdictions, this may not be required by the building official or may not be warranted given the specifics of the design and the overall strength of the lateral force resisting system. The designer must chose between use of the more rigorous approach of considering a rigid diaphragm with torsional resistance characteristics with the more common approach of considering flexible diaphragms with tributary mass. The former may not be necessary in some situations, while at the same time recognizing that the laws of physics must be obeyed. In all cases, the completed structure must have a continuous lateral load path to resist lateral forces. Complete detailing is necessary, even for simple structures. Effects of box nails on wood structural panel shear walls.

This design example uses common nails for fastening wood structural panels. Based on cyclic testing of shear walls and performance in past earthquakes, the use of common nails is preferred. UBC Table 23-II-I-1 lists allowable shears for wood structural panel shear walls for “common or galvanized box nails.” Footnote number five of Table 23-II-I-1, states that the galvanized nails shall be “hot-dipped or tumbled” (these nails are not gun nails). Most contractors use gun nails for diaphragm and shear wall installations. The UBC does not have a table for allowable shears for wood structural panel shear walls or diaphragms using box nails.



Box nails have a smaller diameter shank and a smaller head size. Using 10d box nails would result in a 19 percent reduction in allowable load for diaphragms and shear walls as compared to 10d common nails. Using 8d box nails would result in a 22 percent reduction in allowable load for diaphragms and shear walls as compared to 8d common nails. This is based on comparing allowable shear values listed in Tables 12.3A and 12.3B in the 1997 NDS for one-half-inch side member thickness (t s ) and Douglas Fir-Larch framing. In addition to the reduction of the shear wall and diaphragm capacities, when box nails are used, the walls will also drift more than when common nails are used. A contributor to the problem is that when contractors buy large quantities of nails (for nail guns), the word “box” or “common” does not appear on the carton label. Nail length and diameters are the most common listing on the labels. This is why it is extremely important to list the required nail lengths and diameters on the structural drawings for all diaphragms and shear walls. Another problem is that contractors prefer box nails because their use reduces splitting, eases driving, and they cost less. Just to illustrate a point, if an engineer designs for “dry” lumber (as discussed above) and “common” nails, and subsequently “green” lumber and “box” nails are used in the construction, the result is a compounding of the reductions. For example, for 10d nails installed into green lumber, the reduction would be 0.81 times 0.75 or a 40 percent reduction in capacity.

&DOFXODWLRQVDQG'LVFXVVLRQ

&RGH5HIHUHQFH

Design base shear and vertical distribution of seismic forces.

§1630.2.2

This example uses the total building weight W applied to each respective direction. The results shown will be slightly conservative since W includes the wall weights for the direction of load, which can be subtracted out. This approach is simpler than using a separated building weight W for each axis under consideration.

D

Design base shear.

Period using Method A (see Figure 1-5 for section through structure): T = Ct (hn )3 / 4 = .020(23)3 / 4 = .21sec.

(30-8)

where: hn is the center of gravity (average height) of diaphragm above the first floor.



With seismic source type B and distance to source = 12 km N a = 1.0

Table 16-S

N v = 1.0

Table 16-T

For soil profile type S C and Z = 0.4 C a = 0.40 N a = 0.40(1.0 ) = 0.40

Table 16-Q

C v = 0.56 N v = 0.56(1.0 ) = 0.56

Table 16-R

North-south direction: For light framed walls with wood structural panels that are both shear walls and bearing walls: R = 5.5

Table 16-N

Design base shear is: CV I 0.56(1.0 ) W = W = 0.485W RT 5.5(.21) (Note that design base shear in the 1997 UBC is now on a strength design basis) V =

(30-4)

but need not exceed: V =

2.5C a I 2.5(.40 )(1.0 ) W = W = 0.182W R 5.5

(30-5)

A check of Equations 30-6 and 30-7 indicates these do not control: ∴ V N − S = 0.182W

Comparison of the above result with the simplified static method permitted under §1630.2.3 shows that it is more advantageous to use the standard method of determining the design base shear. V=

3.0C a 3.0(.40) W= W = 0.218W > 0.182W R 5.5

(30-11)

All of the tables in the UBC for wood diaphragms and shear walls are based on allowable loads.



It is desirable to keep the strength level forces throughout the design of the structure for two reasons: 1.

Errors in calculations can occur and confusion on which load is being used— strength or allowable stress design. This design example will use the following format: Vbase shear

=

strength

F px

=

strength

Fx v

= =

force to wall (strength) wall shear at element level (ASD)

=

ASD

v=

2.

Fx 1.4b

This design example will not be applicable in the future, when the code will be all strength design. E = ρE h + E v = 1.0 E h + 0 = 1.0 E h

(30-1)

where: E v is allowed to be assumed as zero for allowable stress design, and ρ is assumed to be 1.0. This is the case for most of Type V residential construction structures. Since the maximum element story shear is not yet known, the value for ρ will have to be verified. This is done later in Part 6.

The basic load combination for allowable stress design is: D+

E E E = 0+ = 1.4 1.4 1.4

(12-9)

VN − S = 0.182W ∴V N −S = 0.182(103,000 lb ) = 18,750 lb

§1612.3.1

East-west direction: Since there are different types of lateral resisting elements in this direction, determine the controlling R value. For light framed walls with wood structural panels that are both shear walls and bearing walls: R = 5 .5



For cantilevered column elements: R = 2 .2

Table 16-N

For combinations along the same axis, the UBC requires the use of least value for any of the systems utilized in that same direction, therefore the value for the cantilevered column elements must be used for the entire east-west direction. This provision for combinations along the same axis first appeared in the 1994 UBC. R = 2 .2

§1630.4.4

Design base shear is: V =

CV I 0.56(1.0 ) W = W = 1.21W RT 2.2(.21)

(30-4)


2.5C a I 2.5(.40)(1.0 ) W = W = 0.454W R 2.2

(30-5)

A check of Equations 30-6 and 30-7 indicates that these do not control: ∴ V E −W = 0.454W

This is less than that obtained with the simplified static method: V =

3.0C a 3.0(.40 ) W = W = 0.545W > 0.454W R 2.2

(30-11)

V E −W = 0.454W

VE −W = 0.454(103,000 lb ) = 46,750 lb

§1612.3.1

Discussion of R factors.

The UBC places a severe penalty on the use of cantilevered column elements. The design base shear for the east-west direction is two and a half times that for the north-south direction. Some engineers use the greater R factor for light framed walls (e.g., R = 5.5) , determine the design base shear, and then factor up the force for the respective frame element by using the ratio of the R for the shear walls over the R for the frame element (e.g., 5.5 2.2 = 2.5) . However, under a strict interpretation of the UBC, the factoring up approach does not appear to meet the intent of the UBC requirements. Another approach could be to design the residence



using a rigid diaphragm assumption with the wood shear walls taking 100 percent of the lateral force using R = 5.5 . Then design the cantilever columns using R = 2.2 and a flexible diaphragm. Usually in residential construction, cantilevered column elements are preferred over moment frames by engineers and builders because of the elimination of field welding. The 1999 Blue Book has added an exception for light frame buildings in Occupancy Groups 4 and 5 and of two stories or fewer in height. The local building department should be consulted on whether or not they will accept this exception. A higher force level could be counter productive in terms of splitting caused by added close nailing. An ordinary moment-resisting frame could be used with an R value equal to 4.5. This would produce design base shear values only 22 percent higher than in the north-south direction. Additionally, the architecture could be modified to provide shear wall lengths that meet the h/w ratio limit of 2:1. With the plate height at 9’-0", the minimum wall length needed would be 4’-6". Another solution would be to increase the concrete curb height at the base of the wall such that the h/w ratio limit of 2:1 is not exceeded. For illustrative purposes, this design example uses the cantilevered column elements with the higher design base shear for the entire east-west direction. This conforms to the 1997 UBC. Pre-manufactured proprietary trussed wall systems and factory-built wood shear wall systems are also available. Special design considerations should be given when using these systems as outlined below: 1.

Building system R values are to be based on officially adopted evaluation reports, such as ICBO reports.

2.

Pre-manufactured systems should not be used in the same line as field-built shear walls because of deformation compatibility uncertainties.

3.

Pre-manufactured systems should be limited to the first floor level only (of multi-story wood frame buildings) until testing is completed for these systems that sit on wood framing and are not rigidly attached to a concrete foundation.

4.

Many of the these “systems” exceed not only the new aspect ratio limit of 2:1, but also exceed the old aspect ratio limit of 3½: 1. Some are as narrow as 16 inches wide, leaving unanswered the question of whether this is a shear wall or a cantilever column (by comparison, if the “system” were a steel channel with the same width, it would be considered a cantilever column).

5.

Many building officials are requesting that the same aspect ( 2:1) ratio limit for wood structural panel shear walls be adhered to for the pre-manufactured systems.



E


The vertical distribution of seismic forces is determined from Equation 30-15.

F px =

(V − Ft )wx hx (30-15)

n

∑ wi hi i =1

where: h x is the average height at level i of the sheathed diaphragm in feet above the base.

Since T = 0.21 seconds < 0.7 seconds, Ft = 0 Determination of F px is shown in Table 1-1.

Figure 1-5. Cross-section through residence



Table 1-1. Vertical distribution of seismic forces w x hx

∑wi hi

Fpx N −S

Fpx N −S

Fpx E −W

Fpx E −W

(lb)

wx

(lb)

wx

14,800

0.231

36,950

0.577

21

3,950

0.101

9,800

0.251

100

18,750

0.182

46,750

0.454

wx

hx

w x hx

(lb)

(ft)

(lb-ft)

Roof

64,000

23.0

1,472,000

79

Floor

39,000

10.0

390,000

Σ

103,000

—

1,862,000

Level

(%)

Lateral forces on shear walls and shear wall nailing assuming flexible diaphragms.

Determine the forces on shear walls. As has been customary practice in the past, this portion of the example assumes flexible diaphragms. The UBC does not require torsional effects to be considered for flexible diaphragms. The effects of torsion and wall rigidities will be considered later in Part 5 of this design example. The selected method of determining loads to shear walls is based on tributary areas with simple spans between supports. Another method of determining loads to shear walls can assume a continuous beam. A continuous beam approach may not be accurate because of shear deformations in the diaphragm. The tributary area approach works with reasonable accuracy for a continuous beam with 100 percent shear deflection and zero bending deflection. This design example uses the exact tributary area to the shear walls, an approach that is fairly comprehensive. An easier and more common method would be to use a uniform load equal to the widest portion of the diaphragm, which results in conservative loads to the shear walls.

D

Forces on east-west shear walls.

Roof diaphragm: Roof area = 2,164 sq ft f p roof = w1 =

36,950 lb = 17.07 psf 2 ,164 sf

(17.07 psf ) (43.0 ft ) = 734 plf



w2 =

(17.07 psf )(37.0 ft ) = 632 plf

w3 =

(17.07 psf )(32.0 ft ) = 546 plf

Figure 1-6. Roof diaphragm loading for east-west forces

Check sum of forces: 1,092 + 4,106 + 4,256 + 4,788 + 5,080 + 8,074 + 8,074 + 1,468 = 36,938 lb

V Roof = 36,938 lb ≈ 36,950 lb

o.k.

Note that Figures 1-6, 1-7, 1-8 and 1-9 are depicted as a continuous beam. From a technical standpoint, “nodes” should be shown at the interior supports. In actuality, with the tributary area approach, these are considered as separate simple span beams between the shear wall “supports” (Figure 1-6 has three separate single span beams).

Floor diaphragm: Second floor area = 1,542 sf

f p floor

w4

=

9,800 lb 1,542 sf

=

6.36 psf

=

(6.36 psf )(16.0 ft )

=

102 plf



w5

=

(6.36 psf )(20.0 ft )

=

127 plf

w6

=

(6.36 psf )(33.0 ft )

=

210 plf

w7

=

(6.36 psf )(28.0 ft )

=

178 plf

w8

=

(6.36 psf )(32.0 ft )

=

204 plf

PD

=

(1,092 lb + 4,106 lb)

=

5,198 lb

Figure 1-7. Second floor diaphragm loading for east-west forces

Check sum of forces: 408 + 5,655 + 3,640 + 1,470 + 1,470 + 1,233 + 1,136= 15,012 lb Subtract PD from the sum of forces: 15,012 − 5,198 = 9,814 lb

V floor = 9,814 lb ≈ 9,800 lb o.k.



E

Required edge nailing for east-west shear walls using 10d common nails.

Table 23-II-I-1

Table 1-2. East-west shear walls at roof level (second floor to roof)1, 2, 3, 4, 5, 6, 7, 8, 9 Wall (grid line) A B C D

Σ Notes: 1.

2.

3.

4. 5. 6. 7.

8.

9.

∑ Fabove ∑ Fx (lb) 0 0 0 0 0

(lb) 9,542 13,154 9,044 5,198 36,938

Ftot (lb)

b (ft)

v=

Ftot (b )1.4 (plf)

9,542 13,154 9,044 5,198 36,938

10.0 14.0 8.5 6.0 38.5

681(6) 671(6) 760(6) 619(6)

Sheathing(5) 1 or 2 sides One Two Two Two(8)

Allowable Shear (plf) 870 1330 1330 1740

Edge Nail Spacing (in.) 2(2) (4) 3(4) 3(4) 2(2) (4)

Minimum framing thickness. The 1994 and earlier editions of the UBC required 3x nominal thickness stud framing and blocking at abutting panel edges when 10d common nails were spaced 2 inches on center or when sheathing is installed on both sides of the studs without staggered panel joints. The 1997 UBC (Table 23-II-I-1 footnotes) requires 3x nominal thickness stud framing at abutting panel edges and at foundation sill plates when the allowable stress design shear values exceed 350 pounds per foot or if the sheathing is installed on both sides of the studs without staggered panel joints. Sill bolt washers. Section 1806.6.1 requires a minimum of 2-inch-square by 3/16-inch-thick plate washers to be used for each foundation sill bolt (regardless of allowable shear values in the wall). These changes were a result of the splitting of framing studs and sill plates observed in the Northridge earthquake and in cyclic testing of shear walls. The plate washers are intended to help resist uplift forces on shear walls. Because of vertical displacements of holdowns, these plate washers are required even if the wall has holdowns designed to take uplift forces at the wall boundaries. The washer edges shall be parallel/perpendicular to the sill plate. Errata to the First Printing of the 1997 UBC (Table 23-II-I-1 footnotes) added an exception to the 3x foundation sill plates by allowing 2x foundation sill plates when the allowable shear values are less than 600 pounds per foot, provided that sill bolts are designed for 50 percent of allowable values. Refer to Design Example 2 for discussions about fasteners for pressure—preservative treated wood and the gap at bottom of sheathing. APA Structural I rated wood structural panels may be either plywood or oriented strand board (OSB). Note forces are strength level and shear in wall is divided by 1.4 to convert to allowable stress design. It should be noted that having to use a nail spacing of 2 inches is an indication that more shear wall length should be considered. However, in this example, the close nail spacing is a direct result of R = 2.2 for the cantilever column elements. Some jurisdictions, and many engineers, as a matter of judgment, put a limit of 1,500 plf on wood shear walls. A minimum of 3-inch nail spacing with sheathing on only one side is required to satisfy shear requirements. In this design example, sheathing has been provided on both sides with closer nail spacing in order to increase the stiffness of this short wall. The 1999 Blue Book recommends special inspection when the nail spacing is closer than 4-inch on center.



Table 1-3. East-west shear walls at floor level (first floor to second floor) Wall (grid line) A B C D E

∑ Fabove ∑ Fx (lb)

(lb)

9,542 13,154 9,044 5,198 0 36,938

1,136 2,703 5,110 0 6,063 15,012

Σ Notes: See notes for Table 1-2.

F

Ftot (lb) 10,678 15,857 14,154 0 6,063 46,752

v=

b (ft)

Ftot

(b)1.4

Sheathing 1 or 2 sides

Allowable Shear (plf)

Edge Nail Spacing (in)

One Two Two(3)

870 1330 1330

2 3 3

(plf) 10.0 14.0 19.0 0 Frame 43.0

763(2) 809(2) 532(2) 0 Frame

Forces on north-south shear walls.

Roof diaphragm:

f p roof

=

14 ,800 lb 2 ,164 sq ft

=

6.84 psf

w1

=

(6.84 psf ) (55.0 ft )

=

376 plf

w2

=

(6.84 psf ) (40.0 ft )

=

274 plf

w3

=

(6.84 psf ) (34.0 ft )

=

233 plf

Figure 1-8. Roof diaphragm loading for north-south forces



Check sum of forces: 466 + 713 + 767 + 726 + 848 + 5,264 + 5,264 + 752 = 14,800 lb

Vroof = 14,800 lb ≈ 14,800 lb

o.k.

Floor diaphragm:

f p floor

=

3,950 lb 1,542 sq ft

=

2.56 psf

w4

=

(2.56 psf ) (9.0 ft )

=

23.0 plf

w5

=

(2.56 psf ) (60.0 ft )

=

154 plf

w6

=

(2.56 psf ) (43.0 ft )

=

110 plf

w7

=

(2.56 psf ) (38.0 ft )

=

97.2 plf

w8

=

(2.56 psf ) (23.0 ft )

=

58.9 plf

w9

=

(2.56 psf ) (14.0 ft )

=

35.8 plf

Figure 1-9. Second floor diaphragm loading for north-south forces

Check sum of forces: 99 + 126 + 1,653 + 2,028 + 46 = 3,952 lb

V floor = 3,952 lb ≈ 3,950 lb

o.k.



G

Required edge nailing for north-south shear walls using 10d common nails.

Table 23-II-I-1

Table 1-4. North-south shear walls at roof level (second floor to roof) Wall 1 2 3 5

Σ

∑ Fabove

∑ Fx

(lb) 0 0 0 0 0

Ftot

b

(lb)

(lb)

(ft)

1,179 1,493 6,112 6,016 14,800

1,179 1,493 6,112 6,016 14,800

18.0 10.0 15.0 26.0 69.0

v =

Ftot

(b )1.4



Edge Nail Spacing (in.)

(plf) 47 107 291 165

One One One One

510 510 510 510

4 4 4 4



Edge Nail Spacing (in)

One One One

510 510 510

4 4 4

Table1-5. North-south shear walls at floor level (first floor to second floor) Wall 2 3 5

Σ

∑ Fabove

∑ Fx

(lb) 1,493 6,112 6,016 13,621

Ftot

(lb)

(lb)

b (ft)

99 1,779 2,074 3,952

1,592 7,891 8,090 17,573

10.0 22.0 14.0 46.0

v=

Ftot (b )1.4 (plf)

114 256 413

Rigidities of shear walls and cantilever columns at garage.

D

Estimation of wood shear wall rigidities.

Determination of the rigidities of wood shear walls is often difficult and inexact, even for design loads. In addition, when walls are loaded substantially beyond their design limits, as occur under strong earthquake motions, rigidity determination becomes even more difficult. It is complicated by a number of factors that make any exact determination, in a general sense, virtually impossible short of full-scale testing. There is the well-known expression for shear wall deflection found in UBC Standard 23-2. This expression, shown below, is used to estimate deflections of shear walls with fixed bases and free tops for design level forces. =

8vh 3 vh h + + 0.75hen + d a EAb Gt b


§23.223, Vol. 3


The expression above was developed from static tests of solid wood shear walls, many typically 8-foot x 8-foot in size. Until recently, there was very little cyclic testing of wood shear walls (to simulate actual earthquake behavior) or testing of walls with narrow aspect ratios. In modern wood frame building construction, shear walls take many forms and sizes, and these are often penetrated by ducts, windows, and door openings. Also, many walls in residences are not designed as shear walls, yet have stiffness from their finish materials (gypsum board, stucco, etc.). In multi-story structures, walls are stacked on the walls of lower floors, producing indeterminate structural systems. In general, it is difficult to calculate wall rigidities with the UBC equation alone. As will be shown in subsequent paragraphs, things like shrinkage can significantly effect deflection and subsequent stiffness calculations. Further, in strong earthquake motions, shear walls may see forces and displacements several times larger than those used in design, and cyclic degradation effects can occur that significantly change the relative stiffness of shear walls at the same level. It can be argued that wall rotation of the supporting wall below needs to be considered when considering shear wall rigidities. However, considering rotation of the supporting wall below would be similar to measuring the shear wall as the cumulative height, as opposed to the accepted floor-to-floor clear height. Not considering rotation of the supporting wall below is appropriate for determining relative wall rigidities. At the present time, there are number of ways to estimate shear wall rigidities, particularly when only relative rigidities are desired (see Blue Book §C805.3). These include: 1.

Rigidity based on estimated nail slip.

2.

Rigidity calculated from UBC Standard 23-2 (the four term equation given above).

3.

Rigidity incorporating both UBC Standard 23-2 and shrinkage.

4.

Several other procedures.

Only one of these approaches is given in this design example. By using this one approach, SEAOC does not intend to establish a standard procedure or indicate a standard of care for calculation of wood shear wall rigidities. It is merely one of the present-day methods. At present, CUREe (California Universities for Research in Earthquake Engineering) is conducting a large testing program to study earthquake effects on wood structures, including research on shear walls and diaphragms. It is expected that in the years ahead, new approaches will be developed and/or existing approaches reaffirmed or refined. Until then, the practicing structural engineer must use judgment in the method selected to determine wood shear wall rigidities.



It is recommended that the local building official be contacted for determination of what is acceptable in a particular jurisdiction.

E

Discussion of rigidity calculation using the UBC deflection equation.

Since the rigidity, k , of a shear wall or cantilever column is based on its displacement, ∆ , the displacements will first be computed using the Ftot forces already determined above in Tables 1-2 and 1-3. Compute values for k : F = k∆

or k = F ∆ The basic equation to determine the deflection of a shear wall is the four-term equation shown below. =


§23.223, Vol. 3

The above equation is based on a uniformly nailed, cantilever shear wall with a horizontal point load at the top, panel edges blocked, and reflects tests conducted by the American Plywood Association. The deflection is estimated from the contributions of four distinct parts. The first part of the equation accounts for cantilever beam action using the moment of inertia of the boundary elements. The second term accounts for shear deformation of the sheathing. The third term accounts for nail slippage/bending, and the fourth term accounts for tiedown assembly displacement (this also should include bolt/nail slip and shrinkage). End stud elongation due to compression or tension is not considered, nor the end rotations of the base support. The UBC references this in §2315.1. Testing on wood shear walls has indicated that the above formula is reasonably accurate for aspect ratios (h w) lower than or equal to 2:1. For higher aspect ratios, the wall drift increases significantly, and testing showed that displacements were not adequately predicted. Use of the new aspect ratio requirement of 2:1 (1997 UBC) makes this formula more accurate for determining shear wall deflection/ stiffness than it was in previous editions of the UBC, subject to the limitations mentioned above. Recent testing on wood shear walls has shown that sill plate crushing under the boundary element can increase the deflection of the shear wall by as much as 20 percent to 30 percent. For a calculation of this crushing effect, see the deflection of wall frame at line D later in Part 3c.



Fastener slip/nail deformation values (en).

Volume 3 of the UBC has Table 23-2-K for obtaining values for en . However, its use is somewhat time-consuming, since interpolation and adjustments are necessary. Footnote 1 to Table 23-2-K requires the values for en to be decreased 50 percent for seasoned lumber. This means that the table is based on nails being driven into green lumber and the engineer must use one-half of these values for nails driven in dry lumber. The values in Table 23-2-K are based on tests conducted by the APA. The 50 percent reduction for dry lumber is a conservative factor. The actual tested slip values with dry lumber were less than 50 percent of the green lumber values. It is recommended that values for en be computed based on fastener slip equations from Table B-4 of APA Research Report 138. Note that this Research Report is the basis for the formulas and tables in the UBC. Both the Research Report and the UBC will produce the same values. Using the fastener slip equations from Table B-4 of Research Report 138 will save time, and also enable computations to be made by a computer. For 10d common nails there are two basic equations: When the nails are driven into green lumber: en = (Vn / 977 )1.894

APA Table B-4

When the nails are driven into dry lumber: en = (Vn / 769 )3.276

APA Table B-4

where: Vn is the fastener load in pounds per fastener.

These values are based on Structural I sheathing and must be increased by 20 percent when the sheathing is not Structural I. The language in footnote a in UBC Table B-4 states “Fabricated green/tested dry (seasoned)…” is very misleading. The values in the table are actually green values, since the lumber is fabricated when green. Don’t be misled by the word “seasoned.” It is uncertain whether or not the d a factor is intended to include wood shrinkage and crushing due to shear wall rotation, because the code is not specific. This design example includes both shrinkage and crushing these in the d a factor. Many engineers have a concern that if the contractor installs the nails at a different spacing (too many or too few), then the rigidities will be different than those calculated. However, nominal changing of the nail spacing in a given wall does not significantly change the stiffness.



Determination of the design level displacement ∆s.

§1630.9.1

For both strength and allowable stress design, the 1997 UBC requires building drifts to be determined by the load combinations of §1612.2, which covers load combinations using strength design or load and resistance factor design. Errata for the second and third printing of the UBC unexplainably referenced §1612.3 for allowable stress design. The reference to §1612.3 is incorrect and will be changed back to reference §1612.2 in the fourth and later printings. Wood design using the 1997 UBC now means that the engineer must use both strength-level forces and allowable stress forces. This can create some confusion, since the code requires drift checks to be strength-level forces. However, all of the design equations and tables in Chapter 23 are based on allowable stress design. Drift and shear wall forces will be based on strength-level forces. Remember that the structural system factor R is based on using strength-level forces.

F

Estimation of roof level rigidities.

Roof design level displacements.

To determine roof level wall rigidities, roof level displacements must first be determined. Given below are a series of calculations, done in table form, to estimate the roof level displacements ∆s in each shear wall connecting to the roof (Table 1-7). Because there is a wall with openings supported by a GLB on line D, the ∆s for this wall must also be determined. Finally, roof level wall rigidities are summarized in Table 1-8 and a drift check is given in Table 1-9. Table 1-6. Determine tiedown assembly displacements for roof level shear walls1 ASD Wall A1 A2 B C 1 2 3 5a 5b

Uplift/1.4(2) (lb) 5,915 5,915 5,975 7,430 0 0 830 0 0

Strength Design

Tiedown(3) Device

Uplift (lb)

Bolted Bolted Bolted Bolted Not required Not required Strap Not required Not required

8,280 8,280 8,365 10,400 0 0 1,160 0 0

Tiedown(4) Elongation (in.) 0.13 0.13 0.13 0.17 0 0 0.004 0 0

Tiedown Assembly Displacement Shrink(5) Crush(6) Slip(7) 0.19 0.02 0.04 0.19 0.02 0.04 0.19 0.02 0.04 0.19 0.04 0.05 0.02 0.02 0 0.19 0.02 0 0.19 0.02 0.002 0.19 0.02 0 0.19 0.02 0

da(8) (in.) 0.38 0.38 0.38 0.45 0.04 0.21 0.21 0.21 0.21

Notes: 1. 2.

Tiedown assembly displacement is calculated at the second floor level. Uplift force is determined by using the net overturning force (M OT − M OR ) divided by the distance between the centroid of the tiedown to the end of the shear wall. With 4x members at the ends of the wall, this equates to the length of the wall minus 1¾ inches for straps, or the length of wall minus 5½ inches when using a bolted holdown with 2-inch offset from post to anchor bolt. Using allowable stress design, tiedown devices



3. 4.

5.

need only be sized by using the ASD uplift force. The strength design uplift force is used to determine tiedown assembly displacement in order to determine strength-level displacements. Continuous tie rod holdown systems can also be used. See Design Example 2 for method of calculating tiedown assembly displacement. Tiedown elongation is based on actual uplift force divided by tiedown capacity times tiedown elongation at capacity (from manufacturer’s catalog). Example for tiedown elongation at A1: tiedown selected has a 15,000 lb allowable load for a 5½-inch-thick (net) member. From the manufacturer’s ICBO Evaluation Report, the tiedown deflection at the highest allowable design load (15,000 lb) is 0.12 inches. Since there are two tiedown devices (one above and one below the floor), the total elongation is twice the tiedown deflection of one device. Therefore the total tiedown elongation is (8,280 15,000)0.12 × 2 = 0.13 inches. Wood shrinkage based on a change from 19 percent moisture content (MC) to 13 percent MC with 19 percent MC being assumed for S-Dry lumber per project specifications. The MC of 13 percent is the assumed final MC at equilibrium with ambient humidity for the project location. The final equilibrium value can be higher in coastal areas and lower in inland or desert areas. This equates to (0.002 ) (d ) (19-13), where d is the dimension of the lumber (see Figure 1-10). Shrinkage: 2 × DBL Top Plate + 2 × sill plate = (0.002 ) (3 × 1.5 in.) (19 − 13) = 0.05

2 × 12 Floor Joist

6.

7.

=

(0.002)(11.25) (19 − 13)

=

0.14

= 0.19 in. The use of pre-manufactured, dimensionally stable, wood I joists are considered not to shrink, and would thereby reduce the shrinkage to 0.05 inches. Per 91 NDS 4.2.6, when compression perpendicular to grain ( f c⊥ ) is less than

0.73F ’c⊥ crushing will be approximately 0.02 inches. When f c⊥ = F ’c⊥ crushing is approximately 0.04 inches. The effect of sill plate crushing is the downward effect with uplift force at the opposite end of the wall and has the same rotational effect as the tiedown displacement. Short walls that have no uplift forces will still have a wood crushing effect and contribute to rotation of the wall. Per 91 NDS 7.3.6γ = (270,000)(1)1.5 = 270,000 lb/in. plus 1/16" oversized hole for bolts. For nails, values for en can be used. Example for slip at tiedown at A1 (tiedown has five 1-inch diameter bolts to post): Load/bolt = 8,280 / 5 = 1,656 lb/bolt

= (270,000) (1)1.5 = 270,000 lb/in. slip = (1,656 270,000) = 0.006 in. Since there are two tiedown devices (one above and one below the floor), the total slip is twice the bolt slip. Good detailing practice should specify the tiedown bolts to be re-tightened just prior to closing in. This can accomplish two things: it takes the slack out of the oversized bolt hole and compensates for some wood shrinkage. This design example will assume that about one-half of the bolt hole slack is taken out.  1 1 Therefore, total slip equals (0.006 × 2) +   = 0.04 inches.  16  2 8.

d a is the total tiedown assembly displacement. This also could include mis-cuts (short-studs) and lack of square cut ends.



Figure 1-10. Second floor diaphragm connection to shear wall

Table 1-7. Deflections of the shear walls at the roof level1,2,6,10 Wall

ASD v (plf)

A1 A2 B(8) C(8) 1 2 3 5a(9) 5b(9)

681 681 336 380 47 107 291 194 120

Strength v (plf) 953 953 470 532 66 150 407 271 168

h (ft)

A(3) (sq in.)

9.0 9.0 10.0 10.0 15.25 9.0 9.0 9.0 9.0

19.25 19.25 19.25 19.25 19.25 12.25 12.25 12.25 12.25

E (psi) 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6

b (ft)

G(4) (psi)

t (in.)

Vn

5.0 5.0 14.0 8.5 18.0 10.0 15.0 16.0 10.0

90,000 90,000 90,000 90,000 90,000 90,000 90,000 90,000 90,000

0.535 0.535 0.535 0.535 0.535 0.535 0.535 0.535 0.535

159 159 118 133 22 50 136 90 56

(lb)

en (5) (in.) 0.0057 0.0057 0.0022 0.0032 8.8E-6 0.0001 0.0034 0.0009 0.0002

da

∆ S (7)

(in.)

(in.)

0.38 0.38 0.38 0.45 0.04 0.21 0.21 0.21 0.21

0.93 0.93 0.39 0.68 0.06 0.22 0.23 0.18 0.23

Notes: 1. 2. 3. 4.

5. 6. 7.

8vh 3 vh h + + 0.75he n + d a §23.223, Vol. 3 EAb Gt b h values are from the bottom of the sill plate to the bottom of the framing at diaphragm level (top plates). A values are for 4 × 6 posts for walls A1, A2, B, C, and wall 1. A values are for 4 × 4 posts for walls 2, 3, 5a, and 5b. G values are for Structural I sheathing. Testing of shear walls has indicated that the G values are slightly higher for oriented strand board (OSB) than plywood, but not enough to warrant the use of different values. ∆S =

en values for Structural I sheathing with dry lumber = (V n 769 )3.276

The use of a computer spreadsheet is recommended. This will not only save time, but also eliminate possible arithmetic errors with these repetitive calculations. Deflection of walls (∆ S ) is based on strength level forces. The shear wall deflections must be determined using the strength design forces. The calculated deflection of a shear wall is linear up to about two times



9.

the allowable stress design values. Since there are tiedown assembly displacements, and dead loads that resist overturning, the factoring up approach of ASD forces is not appropriate. When sheathing is applied to both sides of the wall, the deflection of the shear wall is determined by using one-half the values from Table 1-2. In-plane shears to walls 5a and 5b are proportioned based on relative lengths (not per §23.223, Volume 3).

10.

Example for wall at line 5a: R = 162 162 + 102 = 72 percent, which is appropriate for two walls in a line, but not necessarily for three or more walls in line. Attempting to equate deflections is desirable. However, the calculations are iterative and indeterminate, and the results are very similar. For deflection of shear wall at line D, see the following Part 3c.

8.

(

)

Determine deflection of wall frame at line D (with force transfer around openings).

The deflection for the shear wall can be approximated by using an analysis similar to computing the stiffness for a concrete wall with an opening in it. The deflection for the solid wall is computed, then a deflection for a horizontal window strip is subtracted, and the deflection for the wall piers added back in. Engineering judgment may be used to simplify this approximation. However, the method shown below is one way to approximate the deflection.

Figure 1-11. Elevation of wall frame on line D



First, determine deflection of the entire wall, without an opening: Deflection of solid wall: =


§23.223 Vol. 3

Sheathing is on both sides of wall with 10d common nails @ 2 inches o. c. Wall has 2 × 6 studs with 4 × 6 at ends. V = 5,198 lb v=

5,198 lb = 260 plf (2 )10.0 ft

With edge nailing at 2 inches on center: Vn = load per nail = 260(2 12 ) = 43 lb/nail e n = (43 769 )3.276 = 0.0001 inch

With a tiedown elongation of 0.05 in., wood shrinkage of 0.13 in., and wood crushing of .02, it gives a tiedown assembly displacement of 0.20 in. For crushing: from Part 9e, the strength level overturning moment M OT = 52,452 ft-lb. Dividing by the distance L = 9.7 ft computes the seismic downward component of the 4 × 6 post: P = 52,452 9.7 = 5,407 lb fc = P A

f c = 5,407 (3.5 × 5.5) = 281 psi < 0.73(625) = 456 psi ∴ crush = 0.02 in

For shrinkage of GLB fabricated to AITC specifications at 17 percent MC: 0.002(17-13) 16.5= 0.13 in.



PL + strap nail slip = 0.05 in. AE

For strap:

d a = 0.05 + 0.13 + 0.02 = 0.20 in.

∆=

8(260 )9.03 260(9.0 ) 9.0 (0.20 ) + + 0.75(9.0) 0.0001 + = 0.23 in. 1.7 E6(19.25) 10.0 (90,000) 0.535 10.0

Second, determine deflection of window strip: V = 5,198 lb (strength)

With sheathing on both sides: v=

5,198 lb = 260 plf (2 )10.0 ft

Vn = load per nail = 260(2 12 ) = 43 lb/nail e n = (43 769 )3.276 = 0.0001 in

Since the boundary elements are connected to continuous posts that extend above and below the opening, the value of d a equals the sheathing nail deformation value calculated above (boundary element “chord” elongation is neglected): d a = 0.0001 in.

−∆=

8 (260) 4.0 3 260 (4.0) 4.0 (0.0001) + + 0.75 (4.0) 0.0001 + = 0.02 in. 10.0 1.7Ε6 (19.25)10.0 (90,000) 0.535

Note that this deflection is negative because it is subtracted from the sum of the deflections, as shown later.



Third, determine deflection of wall piers:

V=

5,198 lb = 2,599 lb 2

v =

2,599 lb = 433 plf (2 )3.0 ft

Vn = load per nail = 433(2 12 ) = 72 lb/nail e n = (72 769 )3.276 = 0.0004 in.

Since the boundary elements are connected to continuous posts that extend above and below the opening, the value of d a equals the sheathing nail deformation value calculated for the wall piers. d a = 0.0004 in.

∆=

8 (433) 4.0 3 433 (4.0) 4.0 (0.0004 ) + + 0.75 (4.0 ) 0.0004 + = 0.04 in. 1.7 Ε (19.25) 3.0 (90,000) 0.535 3.0

Last, determine the sum of the deflections: ∆ = 0.23 − 0.02 + 0.04 = 0.25 in

Thus the stiffness of the wall is (0.23 0.25) , or 92 percent of that of the solid wall.

Determine deflection of wall due to deflection of GLB (see Figure 1-12).

∆h = Shear wall deflection due to deflection of the support beam

tan θ =

∴ ∆h =

∆V ∆h = b h

h (∆V ) b



ROT =

Vh b

ROT =

5,198 lb(9.0 ft ) = 4,678 lb (strength) 10.0 ft

For 5.125 × 16.5 GLB 24 FV 4 : E = 1,800,000 psi I = 1,918 in.4

∆V =

ROT a 2 b 2 3EIL

∆V =

4 ,678(8.0 × 12 )2 (10.0 × 12 )2 = 0.278 in. 3(1.8E6 )1,918(18.0 × 12 )

∆h =

h (∆ V ) b

∆h =

(9.0 × 12 )(0.278 ) = 0.25 in. (10 .0 × 12 )

Total deflection of shear wall including GLB rotation and tiedown assembly displacement: ∆h = 0.25 + 0.25 = 0.50 in.



Figure 1-12. Wall elevation at line D

Table 1-8. Wall rigidities at roof level1(walls from second floor to roof) Wall

∆s(2) (in.)

A1 A2 B C D 1 2 3 5a 5b

0.93 0.93 0.39 0.68 0.50 0.06 0.22 0.23 0.18 0.23

Ftot (lb) 4,771 4,771 13,154 9,044 5,198 1,179 1,493 6,112 4,332 1,684

k=

Ftot (k/in.) ∆s 5.130 5.130

k=

Ftot (k/in.) ∆s 10.26 33.73 13.30 10.40 19.65 6.79 26.57

24.07 7.32

31.39

Notes: 1. 2.

Deflections and forces are based on strength force levels.

∆S

is the design level displacement from Table 1-7 and calculations of wall frame.



Determination of ∆M.

§1630.9.2

Before checking drift, the maximum inelastic response displacement ∆ M must be computed. This is done as follows: ∆ M = 0.7 R∆ S

R

= 5.5 for the north-south direction

R

= 2.2 for the east-west direction

∆ M = 0.7(5.5)∆ S = 3.85∆ S for the north-south direction ∆ M = 0.7(2.2 )∆ S = 1.54∆ S for the east-west direction

Determination of maximum drift.

§1630.10.2

The calculated story drift using ∆ M shall not exceed the maximum ∆ M which is 0.025 times the story height for structures that have a fundamental period less than 0.7 seconds. The building period for this design example was calculated to be 0.21 seconds, which is less than 0.7 seconds, therefore the 0.025 drift limitation applies.

North-South

East-West

Table 1-9. Drift check at roof level

Wall

∆S (in.)

h (ft)

∆M (in.)

Max. ∆M (1) (in.)

Status

A1 A2 B C D 1 2 3 5a 5b

0.93 0.93 0.39 0.68 0.50 0.06 0.22 0.23 0.18 0.23

9.0 9.0 10.0 10.0 9.0 15.25 9.0 9.0 9.0 9.0

1.43 1.43 0.60 1.05 0.77 0.23 0.85 0.88 0.69 0.88

2.70 2.70 3.00 3.00 2.70 4.57 2.70 2.70 2.70 2.70

ok ok ok ok ok ok ok ok ok ok



G

Estimation of second floor level rigidities.

First floor level design displacements.

First floor level rigidities are determined by first calculating tiedown displacements (Table 1-10) and then deflections of shear walls at the second floor level (Table 1-11). The drift check, discussed in Part 3c, is given in Table 1-12, and wall rigidities calculated in Table 1-13. Table 1-10. Tiedown assembly displacements for first floor level walls1 ASD Wall A1 A2 B C1 C2 2 3 5

Uplift/1.4 (lb) 13,450 13,450 12,675 11,335 3,890 0 825 400

(2)

LRFD Tiedown Device Bolted Bolted Bolted Bolted Bolted Not req’d Strap Strap

Uplift (lb) 18,830 18,830 17,745 15,870 5,445 0 1,155 560

Tiedown(3) Elongation (in.) 0.15 0.15 0.14 0.13 0.04 0 0.05 0.03

Tiedown Assembly Displacement Shrink(4) Crush(5) Slip(6) 0.01 0.04 0.05 0.01 0.04 0.05 0.01 0.04 0.05 0.01 0.04 0.04 0.01 0.02 0.04 0.01 0.02 0 0.01 0.02 0.002 0.01 0.02 0.002

da (in.) 0.25 0.25 0.24 0.22 0.11 0.03 0.08 0.06

Notes: 1. 2.

3.

4.

5.

Tiedown assembly displacement is calculated at the foundation. Uplift force is determined by using the net overturning force (M OT − M R ) , divided by the distance to the centroids of the boundary elements assuming 4x members at the ends of the shear wall. This equates to the length of the wall minus 3½ inches for straps, or the length of wall minus 7¼ inches when using a bolted holdown, which includes a 2-inch offset from post to tiedown bolt. Tiedown elongation is based on actual uplift force divided by tiedown capacity multiplied by the tiedown elongation at capacity from manufacturer’s catalog. Example of tiedown elongation at A1: Tiedown selected has a 15,000 lb allowable load for a 5½-inch member. From the manufacturer’s ICBO approval, the tiedown deflection at the highest allowable design load (15,000 lb) is 0.12 inches, giving a tiedown elongation of (18,830 15,000 0)0.12 = 0.15 inches. Since the tiedown device has an average ultimate strength of 55,000 lb, the displacement can be assumed to be linear and therefore extrapolated. Wood shrinkage is based on a change from 15 percent MC to 13 percent MC. This equates to 0.002 × d × (15-13). Where d is 2.5 inches for a 3 × sill plate. Pressure-treated lumber has a moisture content of less than 15 percent at completion of treatment. Per 91 NDS 4.2.6, when compression perpendicular to grain ( f c ⊥ ) is less than 0.73F ’c ⊥ crushing will be approximately 0.02 inches, when f c ⊥ = F ’c ⊥ crushing is approximately 0.04 inches.

6.

( )

Per 91 NDS 7.3.6 γ = load/slip modulus = (270,000) D1.5 plus 1/16" oversized hole for bolts. For nails, values for en can be used. Example for slip at tiedown at A1 (Tiedown has five 1-inch diameter bolts to post). Load/bolt = 18,830 5 = 3,766 lb/bolt



γ = (270,000) (1)1.5 = 270,000 lb/in. Slip = (3,766 270,000) = 0.014 in. Good detailing should specify the tiedown bolts to be re-tightened just prior to closing in. This can accomplish two things: it takes the slack out of the oversized bolt hole, and compensates for some wood shrinkage. This design example assumes that about one-half of the bolt hole slack is taken out.

 1 1 Therefore the total slip = (0.014 ) +   = 0.05 in.  16  2

Table 1-11. Deflections of the shear walls at the second floor level 1,2,3,4

(§23.222 Vol. 3)

Wall

ASD v (plf)

Strength v (plf)

h (ft)

A (sq in.)

E (psi)

b (ft)

G (psi)

t (in.)

Vn

A1 A2 B C1(5) C2(5) 2 3 5

763 763 404 279 251 114 256 413

1,067 1,067 566 391 351 159 359 578

9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0

19.25 19.25 19.25 19.25 19.25 12.25 12.25 12.25

1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6

5.0 5.0 14.0 10.0 9.0 10.0 22.0 14.0

90,000 90,000 90,000 90,000 90,000 90,000 90,000 90,000

0.535 0.535 0.535 0.535 0.535 0.535 0.535 0.535

178 178 141 98 88 53 120 192

(lb)

en (in.) 0.0083 0.0083 0.0039 0.0012 0.0008 0.0002 0.0023 0.0106

da

∆S

(in.)

(in.)

0.25 0.25 0.24 0.22 0.11 0.03 0.08 0.06

0.74 0.74 0.29 0.29 0.19 0.06 0.12 0.23

Notes: 1. h values are from bottom of sill plate to bottom of framing at diaphragm level (top plates). 2. 3. 4. 5.

8vh 3 vh h + + 0.75hen + d a §23.223, Vol. 3 EAb Gt b G values are for Structural I sheathing. Testing of shear walls has indicated that the G values are slightly higher for OSB than plywood, but not enough to warrant different values. ∆S =

en values for Structural I sheathing with dry lumber = (Vn 769 )3.276 Shear distributed to walls C1 and C2 are proportioned based on relative lengths. Attempting to equate deflections is desirable, however the calculations are iterative and indeterminate, and the results are very similar. The average ∆ for walls A, B, and C at the second floor level is 0.42 inches. For deformation compatibility, it has been decided to size the cantilever column elements at line E for the deflections nearest shear wall at C, where the average is ∆ = 0.24 inches. Another approach would be to use a weighted average that includes the force in the wall. For example, if 99 percent of the load is carried by a stiff wall with ∆ = 0.10 inches and 1 percent is carried by wall with ∆ = 1.00 inches, then the weighted average approach is appropriate. ∆ = 0.10 × 0.99 + 1.0 × 0.01 = 0.11 inches, this assumes no rotation and a rigid diaphragm. If the diaphragm is flexible, then deflection compatibility is not an issue. The engineer should exercise good engineering judgment in determining deformation compatibility.



NorthSouth

East-West

Table 1-12. Drift check at second floor level Wall

∆S (in.)

h (ft)

∆M (in.)

Max. ∆M (1) (in.)

Status

A1 A2 B C1 C2 2 3 5

0.74 0.74 0.29 0.29 0.19 0.06 0.12 0.23

9.0 9.0 9.0 9.0 9.0 9.0 9.0 9.0

1.14 1.14 0.44 0.44 0.29 0.23 0.46 0.88

2.70 2.70 2.70 2.70 2.70 2.70 2.70 2.70

ok ok ok ok ok ok ok ok

Drift for cantilever columns at line E.

The cantilever column is assumed to be fixed at the base. This can be accomplished by setting the column on a footing and then casting the grade beam around the column. With this type of connection, the stresses in the flange of the column caused by concrete bearing at the top of the grade beam should be checked. Another approach is to provide a specially detailed base plate with anchor bolts that are bolted to the top of the grade beam. The bolts and base plate will allow for some rotation, which should be considered in computing the column deflections. The grade beam should have a stiffness of at least 10 times greater than that of the column for the column to be considered fixed at the base. It is common for columns of this type to have drift control the size of the column rather than bending. ∆=

PL3 3EI

It should be noted that if the steel columns were not needed to resist lateral forces (gravity columns only), and all lateral forces were resisted by the wood shear walls, then only relative rigidities of the wood shear walls would need to be calculated. From Figure 1-7 at line E, the force to each of the three cantilever columns: P = (5,655 lb+408 lb)/ 3 = 2 ,021 lb/column

I req ’d

2 ,022 (9 × 12 )3 = = 122 in.4 6 3 29 × 10 0.24

(

)

Use TS10 × 5 × 3 8



I x = 128 in. 4  122  ∆ TS =  0.24 = 0.23 in.  128  M = (2,021 / 1.4 ) = 12 ,992 ft - lb (allowable stress design) fb =

M 12 ,992 × 12 = = 6,115 psi < 0.66(46,000 ) S 25.5

o.k.

Table 1-13. Wall rigidities at second floor level (walls from first to second floor)(1) Wall A1 A2 B C1 C2 E 2 3 5

∆S

(2) (in.)

0.74 0.74 0.29 0.29 0.19 0.23 0.06 0.12 0.23

Ftot (lb) 5,339 5,339 15,857 7,820 6,334 6,063 1,592 7,891 8,090

k=

Ftot (k/in.) ∆s 7.215 7.215

k=

Ftot (k/in.) ∆s

14.43 54.68

26.96 33.34

60.30 26.36 26.53 65.76 35.17

Notes: 1.

Deflections and forces are based on strength force levels.

2. ∆S is the design level displacement from Table 1-11.

Determine centers of mass and rigidity of diaphragms.

It has been a common practice for practicing engineers to assume flexible diaphragms and distribute loads to shear walls based on tributary areas. This has been done for many years and is a well-established conventional design assumption. In this design example, the rigid diaphragm assumption will be used. This is not intended to imply that seismic design of residential construction in the past should have been necessarily performed in this manner. However, recent earthquakes and testing of wood panel shear walls have indicated that expected drifts are considerably higher than what was known or assumed in the past. This knowledge of the increased drifts of short wood panel shear walls has increased the need for the engineer to consider the relative rigidities of shear walls. This, and the fact that diaphragms tend to be much more rigid than the shear walls, has necessitated consideration of diaphragm rigidities. In this Part, the diaphragms are assumed to be rigid. See Part 7 for later confirmation of this assumption.


Design Example 1

4a. 4a

!

Wood Light Frame Residence

For roof diaphragm.

Figure 1-13. Roof diaphragm centers of rigidity and mass

Determine center of mass of roof diaphragm from wall loads.

Using diaphragm loading from flexible diaphragm analysis for east-west direction (Figure 1-6) and summing forces about line D:

734 plf (34.0 ft )

=

632 plf (6.0 ft )

=

546 plf (15.0 ft )

=

  34 24,956 lb ×  + 6 + (15 − 2 ) ft   2  6 3,792 lb ×  + (15 − 2) ft  2   15 8,190 lb ×  − 2   2 36,938 lb


=

898,416 ft - lb

=

60,672 ft - lb

=

45,045 ft - lb 1,004,133 ft - lb

51

Design Example 1

!


∴ ym =

∑ wx = 1,004,133 ft − lb = 27.2 ft @ roof 36,938 lb ∑w

Using diaphragm loading from flexible diaphragm analysis for north-south direction (Figure 1-8) and summing forces about line 1:

376 plf (32.0 ft ) = 12,032 lb × 25.0 ft

=

300,800 ft - lb

274 plf (5.0 ft )

=

1,370 lb × 6.5 ft

=

8,905 ft - lb

233 plf (6.0 ft )

=

1,398 lb × 1.0 ft

= 1,398 ft - lb

10,800 lb ∴ xm =

311,103 ft - lb

∑ wy = 311,103 ft − lb = 21.0 ft @ roof 14,800 lb ∑w

Determine center of rigidity for roof diaphragm.

Using the rigidity values R from Table 1-8 and the distance y from line D to the shear wall:

y=

∑ (k xx y ) ∑ k xx

or y ∑ k xx = ∑ k xx y

y (10.40 + 13.30 + 33.73 + 10.26) = 10.40(0) + 13.30(15.0) + 33.73(29.0 ) + 10.26(51.0) 1700.9 ∴ yr = = 25.1 ft @ roof 67.69 x=

∑ (k yy x ) ∑ k yy

or x ∑ k yy = ∑ k yy x

x (19.65 + 6.79 + 26.57 + 31.39 ) = 19.65(0) + 6.79(6.0 ) + 26.57(11.0) + 31.39(39.0) 1557.2 ∴ xr = = 18.5 ft @ roof 84.40

52


Design Example 1

4b. 4b

!


For second floor diaphragm.

Figure 1-14. Second floor diaphragm centers of rigidity and mass

Determine center of mass of floor diaphragm from wall loads.

Using diaphragm loading from flexible diaphragm analysis for east-west direction (Figure 1-7) and summing forces about line E:

102 plf (17.0 ft )

=

1,734 lb × 49.5 ft

=

85,833 ft - lb

127 plf (5.0 ft )

=

635 lb × 38.5 ft

=

24,448 ft - lb

210 plf (14.0 ft )

=

2,940 lb × 29.0 ft

=

85,260 ft - lb

178 plf (15.0 ft )

=

2,670 lb × 14.5 ft

=

38,715 ft - lb

204 plf (9.0 ft )

=

1,836 lb × 2.5 ft

=

4,590 ft - lb

9,815 lb ∴ ym =

238,846 ft - lb

238,846 ft − lb = 24.3 ft @ second floor 9,815 lb


53

Design Example 1

!


Using diaphragm loading from flexible diaphragm analysis for north-south direction (Figure 1-9) and summing forces about line 2:

23 plf (2.0 ft )

=

46 lb × 34.0 ft

=

5,564 ft - lb

154 plf (16.0 ft )

=

2,464 lb × 25.0 ft

=

61,600 ft - lb

110 plf (4.0 ft )

=

440 lb × 15.0 ft

=

6,600 ft - lb

97.2 plf (8.0 ft )

=

778 lb × 9.0 ft

=

7,002 ft - lb

58.9 plf (2.0 ft )

=

118 lb × 4.0 ft

=

471ft - lb

35.8 plf (3.0 ft )

=

107 lb × 1.5 ft

=

160 ft - lb 77,397 ft − lb

3,953 lb

∴ xm =

77,397 ft − lb = 19.6 ft @ second floor 3,953 lb

Determine center of rigidity for floor diaphragm.

Using the rigidity values k from Table 1-13 and the distance y from line E to the shear wall: y r (26.36 + 60.30 + 54.68 + 14.43) = 26.36(0 ) + 60.30(22.0) + 54.68(36.0) + 14.43(58.0)

∴ yr =

4132.0 = 26.5 ft @ second floor 155.77

Using the rigidity values k from Table 1-13 and the distance x from line 2 to the shear wall: x r (26.53 + 65.76 + 35.17 ) = 26.53(0.0) + 65.76(5.0) + 35.17(33.0)

∴ xr =

54

1489.4 = 11.7 ft @ second floor 127.5


Design Example 1

5.

!


Distribution of lateral forces to the shear walls with rigid diaphragms.

§1630.6

Using the rigid diaphragm assumption, the base shear was distributed to the two levels in Part 1. In this Part, the story forces are distributed to the shear walls that support each level. The code requires that the story force at the center of mass to be displaced from the calculated center of mass a distance of 5 percent of the building dimension at that level perpendicular to the direction of force. This is to account for accidental torsion. The code requires the most severe load combination to be considered and also permits the negative torsional shear to be subtracted from the direct load shear. However, lateral forces must be considered to act in each direction of the two principal axis. This design example does not consider eccentricities between the center of masses between levels. In this example, these eccentricities are small and are therefore considered insignificant. The engineer must exercise good engineering judgment in determining when these effects need to be considered.

5a.

For the roof diaphragm (Figure 1-13).

Forces in the east-west (x) direction: Distance to the calculated CM : y m

=

27.2 ft

Displaced e y = (0.05 × 55 ft )

=

2.7 ft

New y to displace CM

=

27.2 ft ± 2.7 ft

Distance to the calculated CR : y r

=

25.1 ft

e y = 29.9 − 25.1 = 4.8 ft or e y

=

25.1 − 24.5

=

29.9 ft or 24.5 ft

=

0.6

Note that displacing the center of mass by 5 percent can result in the CM being on either side of the CR and can produce added torsional shears to all walls.

Tx = Fx e y = 36,950 lb (4.8 ft ) = 177,360 ft − lb or

Tx = Fx e y = 36,950 lb (0.6 ft ) = 22,170 ft − lb


55

Design Example 1

!


Forces in the north-south (y) direction: Distance to the calculated CM : x m

=

21.0 ft

Displaced e x = (0.05 × 43 ft )

=

2.2 ft

New x to displace CM

=

21.0 ft ± 2.2 ft

Distance to the calculated CR : x r

=

18.5 ft

e x = 23.2 − 18.5 = 4.7 ft or e x

=

18.8 − 18.5

T y = Fy e x = 14,800 lb (4.7 ft )

=

69,560 ft − lb

=

4,440 ft − lb

=

23.2 ft or 18.8 ft

=

0.3

or

T y = Fy e x = 14,800 lb (0.3 ft )

56

Fe-w

=

36,950 lb (Table 1 - 1)

Fn-s

=

14,800 lb (Table 1 - 1)

Tx

=

177,360 ft - lb for walls A and B

Tx

=

22,170 ft - lb for walls C and D

Ty

=

69,560 ft - lb for wall 5

Ty

=

4,440 ft - lb for walls 1, 2, and 3


Design Example 1

5b.

!


For roof wall forces.

The direct shear force Fv is determined from:

R ∑R

Fv = F

and the torsional shear force Ft is determined from:

Ft = T

Rd J

where:

J = ∑ Rd x2 + ∑ Rd y2 R = rigidity of lateral resisting element d = distance from lateral resisting element to the center of rigidity T = Fe

North-South

East-West

Table 1-14. Distribution of forces to shear walls below the roof level Wall

Rx

A B C D

10.26 33.73 13.30 10.40 67.69

Σ 1 2 3 5 Σ

Ry

19.65 6.79 26.57 31.39 84.40

dx

-18.5 -12.5 -7.5 20.5

Σ

dy

Rd

Rd 2

25.9 3.9 10.1 25.1

265.7 131.5 134.3 261.0

6,883 513 1,357 6,552 15,305

-363.5 -84.8 -199.3 643.5

6,725 1,061 1,495 13,192 22,473

Direct Force

Torsional Force Ft

Total Force

5601 18,412 7,260 5,677 36,950

1247 617 79 151

6,848 19,029 7,339 5,828

3,446 1,191 4,659 5,504 14,800

-42 -10 -24 1,185

3,404 1,181 4,635 6,689

Fv

Fv + Ft

37,778

For simplicity, many engineers will add 5 percent or 10 percent of the direct force shears to account for torsional effects. The average torsional force added to the shears walls in this design example is 11 percent of the direct force. Adding only 5 percent of the wall shears can be unconservative. Torsional forces are subtracted from direct forces for this design example as now allowed by code. This only occurs when both of the displaced center of mass is on


57

Design Example 1

!


the same side of the center of rigidity for a given direction. When the center of rigidity occurs between the two displaced centers of mass, then torsional forces can not be subtracted (which occurs at the roof in the east-west direction). Many engineers still neglect these negative forces.

5c.

For the floor diaphragm (Figure 1-14).

Forces in the east-west (x) direction:

=

24.3 ft

=

3.0 ft

New y to displace CM

=

24.3 ft ± 3.0 ft

Distance to the calculated CR : y r

=

26.5 ft

Tx = Fx e y = 46,750 lb (0.8 ft )

=

37,400 ft − lb

or Tx = Fx e y = 46,750 lb (5.2 ft )

=

243,100 ft − lb

=

19.6 ft

= = = = =

1.7 ft 19.6 ft ± 1.7 ft 11.7 ft 17.9 − 11.7 180,000 ft − lb

Distance to the calculated CM : y m

Displaced e y = (0.05 × 60 ft )

=

27.3 ft or 21.3 ft

=

21.3 ft or 17.9 ft

=

6.2 ft

e y = 27.3 − 26.5 = 0.8 ft or e y = 26.5 − 21.3 = 5.2 ft

Forces in the north-south (y) direction: Distance to the calculated CM : x m Displaced e x = (0.05 × 35 ft ) New x to displace CM Distance to the calculated CR : x r

e x = 21.3 − 11.7 = 9.6 ft or e x T y = Fy e x = 18,750 lb (9.6 ft ) or

= T y = Fy e x = 18,750 lb (6.2 ft ) 116,250 ft − lb = (36,950 + 9,800) = 46,750 lb (adding forces from roof and floor from Table 1-1) Fe-w Tx 37,400 ft - lb for walls A and B = Tx 243,100 ft - lb for walls C and E = = (14,800 + 3,950) = 18,750 lb (adding forces from roof and floor from Table 1-1) Fn-s Ty 116,250 ft - lb for walls 2 and 3 = Ty 180,000 ft - lb for wall 5 =

58


Design Example 1

!


Table 1-15. Distribution of forces to shear walls below the second floor level

East-West

Wall A B C E

North-South

Σ

Ry

Rx

dx

14.43 54.68 60.30 26.36 155.77

Direct Force Fv

Torsional Force Ft

Total Force

dy

Rd

Rd 2

31.5 9.5 4.5 26.5

454.5 519.5 271.3 698.5

14,318 4,935 1,221 18,511 38,985

4,331 16,410 18,097 7,910 46,750

276 316 1,072 2,760

4,607 16,726 19,169 10,670

Fv + Ft

2

26.53

-11.7

-310

3,632

3,903

-585

3,318

3 5

65.76 35.17 127.46

-6.7 21.3

-440 749

2,952 15,956 22,540

9,674 5,173 18,750

-831 2,191

8,843 7,364

Σ Σ

61,525

Table 1-16.Comparison of loads on shear walls using flexible versus rigid diaphragm analysis and recheck of nailing in walls Rigid/ Flexible Ratio

b (ft)

6,848 19,029 7,339 5,828 3,404 1,181 4,635 6,689

0.72 1.44 0.81 1.12 2.89 0.79 0.76 1.11

10.0 14.0 8.5 6.0 18.0 10.0 15.0 26.0

4,607 16,726 19,169 10,670 3,318 8,843 7,364

0.43 1.05 1.35 1.76 2.08 1.12 0.91

10.0 14.0 19.0 — 10.0 22.0 14.0

F flexible

Frigid

(lb)

(lb)

A B C D 1 2 3 5

9,542 13,154 9,044 5,198 1,179 1,493 6,112 6,016

A B C E 2 3 5

10,678 15,857 14,154 6,063 1,592 7,891 8,090

Wall


v=

Fmax (b )1.4




(plf) Roof Level 682 970 760 693 135 107 292 184 Floor Level 762 853 721 — 237 287 413

One Two Two Two One One One One

870 1330 1330 1740 510 510 510 510

2 3 3 2 4 4 4 4

One Two Two — One One One

870 1330 1330 — 510 510 510

2 3 3 — 4 4 4

59

Design Example 1

!


Shear walls with shears that exceed 350 pounds per lineal foot will require 3x framing at abutting panel edges with staggered nails. See also notes at bottom of Table 1-2. Where rigid diaphragm analysis shows seismic forces to the shear walls are higher than from flexible diaphragm analysis, the wall stability and anchorage must be reevaluated. Engineering judgment should be used to determine if a rigid diaphragm analysis should be repeated due to changes in wall rigidity. If rigid diaphragm loads are used, the diaphragm shears should be rechecked for total load divided by diaphragm length along the individual wall lines.

6.


§1630.1.1

The reliability/redundancy factor penalizes lateral force resisting systems without adequate redundancy. In this example (in Part 1), the reliability/redundancy factor was assumed to be ρ = 1.0 . This will now be checked. ρ = 2−

20

(30-3)

rmax AB

where: rmax = the maximum element-story shear ratio. For shear walls, the ratio for the wall with the largest shear per foot at or below two-thirds the height of the building is calculated. Or in the case of a three-story building, the lower two levels. The value of rmax is computed from the total lateral load in the wall multiplied by 10 l w and divided by the story shear. l w = length of wall in feet AB = the ground floor area of the structure in square feet. ri =

Vmax (10 l w ) F

AB = 1,542 sq ft

60


Design Example 1

!


For east-west direction: Using strength-level forces for wall C: rmax =

16,726(10 14.0 ) = 0.26 46,750

ρ = 2−

20 0.26 1,542

= 0.04 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, there is no increase in base shear required due to lack of reliability/redundancy. The SEAOC Seismology Committee added the sentence “The value of the ratio 10 l w need not be taken as greater than 1.0” in the 1999 SEAOC Blue Book—which will not penalize longer walls, but in this design example has no effect. Note that the cantilevered column elements are not considered to be a moment frame and are not subject to the ri and ρ requirements of §1630.1.

For north-south direction: Using strength-level forces for wall 5: rmax =

8,090(10 14.0 ) = 0.31 18,750

ρ = 2−

20 0.31 1,542

= 0.36 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, for both directions there is no increase in base shear required due to lack of reliability/redundancy.

7.

Diaphragm deflections and whether diaphragms are flexible or rigid.

This step is shown only as a reference for how to calculate horizontal diaphragm deflections. Since the shear wall forces were determined using both flexible and rigid diaphragm assumptions, there is no requirement to verify that the diaphragm is actually rigid or flexible.


61

Design Example 1

!


The design seismic force in the roof and floor diaphragms using Equation 33-1 must first be found. The design seismic force is then divided by the diaphragm area to determine the horizontal loading in pounds per square foot (refer to Figures 1-13 and 1-14 ). The design seismic force shall not be less than 0.5C a Iw px nor greater than 1.0C a Iw px . The basic equation to determine seismic forces on a diaphragm is shown below. The following will compute the seismic forces in the north-south direction. n

F px =

Ft + ∑ Fi i= x

n

∑ wi

w px

(33-1)

i= x

Ft = 0 in this example because T < 0.7 seconds. Note that the forces in the east-west direction are higher. Fp

roof

=

(36,950 × 64,000) = 36,950 lb

Fp

roof

=

36,950 lb = 17.07 psf 2,164 sq ft

64,000

For the uppermost level, the above calculation will always produce the same force as computed in Eq. (30-15). Fp

floor

Fp

min

Fp

floor

=

(36,950 × 9,800)× 39,000 = 17,701 lb (governs) (39,000 + 64,000)

= 0.5C a Iw px = 0.5(0.40 )(1.0 )w px = 0.20(39,000 ) = 7,800 lb =

§1633.2.9

17 ,701 lb = 11.48 psf 1,542 sq ft

In this example, the roof and floor diaphragms spanning between line A and line B will be used to illustrate the method. The basic equation to determine the deflection of a diaphragm is shown below. ∆=

62

5vL3 vL + + 0.188 Len + 8 EAb 4Gt

∑ (∆ c X ) 2b

§23.222 Vol. 3


Design Example 1

!


The above equation is based on a uniformly loaded, uniformly nailed, simple span diaphragm with blocked panel edges and is based on monotonic tests conducted by the American Plywood Association (APA). The equation has four separate parts. The first part of the equation accounts for beam bending, the second accounts for shear deformation, the third accounts for nail slippage/bending, and the last part accounts for chord slippage. The UBC references this in §2315.1. For the purpose of this calculation, assume the diaphragm is a simple span supported at A and B (refer to Figures 1-13 and 1-14). In reality, with continuity at B, the actual deflection will be less.

7a.

Roof diaphragm.

Check diaphragm shear. Based on the F p roof = 17.07 psf as computed above, find roof shear to line A for the east-west direction. 1.

Area of roof including over hangs is 22' x 43'.

2.

Wall length is 39 ft.

3.

Diaphragm shears are converted to allowable stress design by dividing by 1.4. v=

(17.07 )43.0 (22.0) = 148 plf < 190 plf allowable 1.4 (39.0 )2

From Table 23-II-H, the allowable shear of 190 plf is based on 15/32-inch APArated wood structural panels with unblocked edges and 10d nails spaced at 6 inches on center at boundaries and panel edges. APA-rated wood structural panels may be either plywood or oriented strand board (OSB). Check diaphragm deflection. The UBC specifies that the deflection be calculated on a unit load basis. In other words, the diaphragm deflection should be based on the same load as the load used for the lateral resisting elements, not F px total force at the level considered. Since the code now requires building drifts to be determined by the load combinations of §1612.2 (see Part 3b for additional comments), determine strength loads on building diaphragm. fp

roof

=

36 ,950 lb = 17.07 psf 2,164 sq ft


63

Design Example 1

!


v=

(17.07 psf ) 43.0 ft (22.0 ft ) = 207 plf 2(39.0 ft )

With nails at 6 inches on center, the load per nail is 207(6 / 12 ) = 104 lb/nail = V n = 22.0 ft L = 39.0 ft b = 90,000 psi G Table 23-2-J Vol. 3 = 1,700,000 psi E A 2×4 chords = 5.25 sq in. × 2 = 10.50 sq in.

Sum of individual chord-splice slip.

Note that the area for 2 − 2 × 4 top plates (chord) has been used. All top plates are connected with metal straps. If a metal strap is not used, then use of the area for one top plate is recommended. Also note that the top plates at line 1 are 2 − 2 × 6 . The deflection calculation will conservatively use the chord area of the 2 2x4s at line 5. Fastener slip/nail deformation values (en). en = 1.20(104 769 )3.276 = 0.0017 t = 0.298 in. (for CDX or Standard Grade)

Table 23-2-H

The chord-splice of the diaphragm will be spliced with a 12 gauge metal strap using 10d nails. Assume a chord splice of the diaphragm at mid-span. The slippage for both the diaphragm chords is to be included. The nail slip value from APA Research Report 138 can be used: e n = (V n 769 )3.276 = (120 769 )3.276 = 0.002 in. where: The allowable load is 120 pound per nail (from NDS Table 12.3F for a 10d nail in a 12-gauge strap). Vn = 120 lb/nail in the strap. The elongation of the metal strap is assumed to be 0.03 inches. Therefore, the chord slip is: ∆ c = 0.002 + 0.03 ∆ c = 0.032 in.

64


Design Example 1

!


∑ (∆ c X ) = (0.032 )11.0 ft (2 ) = 0.70 in. - ft Where the distance to the nearest support is 11'–0" and to get the sum for both chords you multiply by 2. ∆=

5(207 )22 .0 3 207 (22 .0 ) 0.70 + + 0.188 (22 .0 )0.0017 + = 0.06 in. 2 (39 .0 ) 8(1.7 E 6 )10 .50 (39 .0 ) 4 (90 ,000 )0 .298

This deflection is based on a blocked diaphragm. The UBC does not have a formula for an unblocked diaphragm. The APA is currently working on a simplified formula for unblocked diaphragms. Based on diaphragm deflection test results performed by the APA, an unblocked diaphragm will deflect between 2 to 2½ times more than that of a blocked diaphragm or can be proportioned to allowable shears. The roof diaphragm is also sloped at 5:12, which is believed to increase the deflection (but has not been confirmed with tests). This design example has unblocked panel edges for the floor and roof diaphragms, so a conversion factor is necessary. It is assumed that the unblocked diaphragm will deflect: ∆ = 0.06(2.5) = 0.15 in. Note that at gable ended roofs, when the chord is in the plane of the roof (pitched), the chord connection at the ridge should be carefully detailed to accommodate the uplift component of the chord.

7b.

Floor diaphragm.

Check diaphragm shear. Based on the F p

floor

= 11.48 psf as computed in Part 7 above, find floor shear to

line A for the east-west direction (area of floor is 22 × 16 ). Diaphragm shears are converted to allowable stress design by dividing by 1.4 where: v=

(11.48 psf )16.0' (22.0') = 90 plf < 190 plf (1.4)2(16.0)

Table 23-II-H

Allowable shear of 190 plf is based on 15/32-inch APA-rated sheathing with unblocked edges and 10d nails spaced at 6 inches on center at boundaries and panel edges supported on framing. APA-rated wood structural panels may be either plywood or oriented strand board (OSB).


65

Design Example 1

!


Check diaphragm deflection: =

9,800 = 6.36 psf 1,542

fp

floor

v=

(6.36 psf )16.0 ft (22.0 ft ) = 70 psf 2(16.0 ft )

With nails at 6 inches on center the load per nail is 70(6 12 ) = 35 lb/nail = Vn L = 22.0 ft b = 16.0 ft

G = 90,000 psi

Table 23-2-J, Vol. 3

E = 1,700,000 psi A2×4 chords = 5.25 sq in. × 2 = 10.50 sq in. e = 1.2(35 769 )3.276 = 4.8 E-05 n t = 0.319 in

Table 23-2-H

Using an assumed single chord-splice slip of 0.032-inch at the mid-span of the diaphragm: Σ∆ c X = (0.032 )11.0 ft (2 ) = 0.70 in. 5 (70) 22.0 3 70 (22.0) 0.70 ∆= + + 0.188 (22.0 ) 4.8E − 05 + = 0.04 in. 8 (1.7 E6 )10.50 (16.0) 4 (90,000 )0.319 2(16.0) Converting to an unblocked diaphragm: ∆ = 0.04(2.5) = 0.10 in.

7c. 7c

Flexible versus rigid diaphragms.

§1630.6

The maximum diaphragm deflection is 0.15 inches, assuming a simple span for the diaphragm. The average story drift is on the order of 0.62 inches (see Part 4, Tables 1-9 and 1-12 for the computed deflections of the shear walls). For the diaphragms to be considered flexible, the maximum diaphragm deflection will have to be more than two times the average story drift, or 1.25 inches. This would be eight times the computed “simple span” deflections of the diaphragms. As defined by the UBC, 66


Design Example 1

!


the diaphragms are considered rigid. Since some amount of diaphragm deformation will occur, the analysis is highly complex and beyond the scope of what is normally done for this type of construction. Diaphragm deflection analysis and testing to date has been performed on level/flat diaphragms. There has not been any testing of sloped (e.g. roof) and complicated diaphragms as found in the typical wood-framed single-family residence. Consequently, some engineers perform their design based on the roof diaphragm being flexible and the floor diaphragm being rigid. In this procedure, the engineer should exercise good engineering judgment in determining if the higher load of the two methodologies is actually required. In other words, if the load to two walls by rigidity analysis is found to be 5 percent to line A, 95 percent to line B, but by flexible analysis it is found to be 50 percent to line A and 50 percent to line B, the engineer should probably design for the larger of the two loads for the individual walls. Note that the same definition of a flexible diaphragm has been in the UBC since the 1988 edition. However, it generally has not been enforced by building officials for Type V construction. The draft of the IBC 2000 has repeated this same definition in Chapter 23 (wood) definitions. For further discussion, see the Commentary at end of this example.

8.

Does residence meet requirements of conventional construction provisions.

§2320

The UBC has had prescriptive provisions for Type V (light frame) construction for many years. It used to be quite common for building officials to allow developers, architects, building designers, and homeowners to build structures under these provisions without any engineering design. The size and style of current singlefamily residences now being constructed—with vaulted ceilings and large floor openings, tile roofs, and larger window sizes—require an engineering design be done. Due to misuse of the conventional construction requirements, more stringent limitations on the usage of these provisions were placed in the 1994 UBC. Following is an analysis of the construction of the residence proposed in this design example compared with conventional construction requirements and an explanation of why an engineering design is required for both vertical and lateral loads. As engineered design code changes continue to get more restrictive, the “gap” between the double standard (i.e. conventional construction vs. engineered design) continues to widen. The structure must be checked against the individual requirements of §2320.1. Additionally, because this structure is in Seismic Zone 4, it must also be checked against §2320.5. Results of these checks are shown below.

Roof total loads.

Dead load of roof exceeds the 15 psf limit


§2320.1

67

Design Example 1

!


Unusually shaped buildings.

Exterior braced wall panels at line D over the garage are horizontally offset from the bracing systems at the floor below and therefore not in one vertical plane.

§2320.5.4.1

Floor opening exceeds 12 feet and 50 percent of the least floor dimension at line A.

§2320.5.4.4

Floor is not laterally supported by braced wall lines on all edges.

§2320.5.4.2

Cantilever column bracing at the garage door does not conform to prescribed methods.

§2320.11.3

Stud height exceeds 10'–0" without lateral support at line 1.

§2320.11.1

Braced wall lines.

Spacing between braced wall lines 3 and 5 exceeds 25 feet maximum. Minimum individual panel length is less than 4'-0" at second floor at line D. ∴

9.

§2320.5.1 §2320.11.3

The residence cannot be designed using the conventional construction provisions of the code.

Design shear wall over garage on line D.

V = 5,828 lb (from Table 1-16) Converting to allowable stress design for the wall frame: V = 5,828 1.4 = 4,159 lb (refer to Figures 1-11 and 1-15) Determine h w aspect ratios for the shear walls: h w = 9.0 3.0 = 3.0 Maximum h w = 2.0 for Seismic Zones 3 and 4

Table 23-II-G

Therefore, the wall piers need to be designed to transfer forces around opening.

Figure 23-II-1

New h w ratio = 4.0 3.0 = 1.33 < 2.0

68

o.k.


Design Example 1

9a. 9a

!


Design of wall frame (perforated shear wall with force transfer around opening).

It is possible to get the misleading impression from Table 23-II-1 that all a designer needs to do is add some blocking and straps in order to reduce the h/w ratio. This design example has a structure with 9'-0" plate heights, which makes using a wall frame feasible. However, when the plate height is 8'-0", which is a more common plate height, there are chord development and panel nailing capacity problems. Most often, the wall shears above and below the opening will be higher than in the wall piers. This design example analyzes the wall frame and neglects gravity loads, although from a technically correct standpoint, some engineers will argue that vertical loads need to be considered when determining wall shears. The standard practice of neglecting gravity loads when considering wall shears is considered appropriate. Gravity loads are considered for anchorage of the wall in Part 9b. Using statistics, determine the shears and forces in each free body panel. This is a two-step procedure as follows: First: Find forces acting on upper left corner of wall frame (Figure 1-15). Second: Break up wall frame into free-body panel sections and balance forces for each panel starting with upper left corner forces already determined (Figure 1-16).

Figure 1-15. Wall frame elevation at line D


69

Design Example 1

!


Figure 1-16. Free-body individual panels of wall on line D

70


Design Example 1

!


Many engineers will arbitrarily add tiedowns at the window jamb members (Figure 1-18). However, with this type of design, the tiedowns at these locations are not necessary, but shear stresses above and below the window may become higher. Adding tiedowns at the window jambs would increase the wall frame performance and help prevent sill plate uplift at the window jambs, which occurs (to some degree) when they are not provided.

9b. 9b

Design horizontal tie straps above and below windows (Figure 1-18).

Determine the tie force for the horizontal strap (from Figure 1-16). Tie force is maximum at header beam. Ftie = 1,546 lb Consult ICBO Evaluation Reports for the allowable load capacity of premanufactured straps. Check penetration depth factor: C d : for 10d nail thru-strap and ½" sheathing penetration = 3.0 − 0.060 − 0.5 = 2.4" Required penetration for full value = 12 D = 12 × 0.148 = 1.8 < 2.4"

o.k.

Allowable load per 10d common nail with 16 ga metal side plate = 113 lb Number of 10d nails required each end =

91 NDS Table 12.3F

1,546 lb = 10.3 nails 113 lb/nail × 1.33

(nailing does not control) Use a continuous 16 gauge x 1¼-inch strap across the opening head and sill to blocking. Allowable strap load is (1.25)0.06(0.6 × 33)1.33 = 1,975 lb > 1,546 lb


o.k.

71

Design Example 1

9c. 9c

!


Load combinations using allowable stress design.

§1612.3

The basic load combinations of §1612.3.1 do not permit stress increases. However, the alternate basic load combinations of §1612.3.2 do permit stress increases. The Errata to the first printing of the UBC added 0.9 D ±

E to the alternate basic 1.4

load combinations as Eq. (12-16-1). Since this exact same load combination is listed in the basic load combinations, the UBC is in contradiction and is confusing (to say the least). This design example uses the alternate basic load combinations with the one-third stress increase.

9d. 9d

Check shear panel nailing in wall frame.

From Figure 1-16: Maximum panel shear = 773 plf 2-inch edge nailing with sheathing both sides v allowable = 2 × 870 = 1,740 plf

o.k.

Table 23-II-I-1

Note that sheathing on both sides of this wall does not appear to be required by the code. To eliminate sheathing on one side, a complete design would recheck the force distribution with the reduced wall rigidity. An inspection of Figure 1-13 would indicate that the center of rigidity would shift to the north and hence add more torsional force to the wall.

9e. 9e

Determine anchorage of wall to the supporting GLB.

The former UBC provision of using 85 percent of the dead loads for consideration of uplift effects has now been replaced with the basic load combinations in UBC §1612.3.1 or §1612.3.2 From Figure 1-17:

wDL = 100 plf (triangle loading from hip roof) PDL = 700 lb Wall DL = 1,100 lb E = V = 5,828 lb h M OT = 5,828 lb (9.0 ft) = 52,452 ft - lb (strength level)

72


Design Example 1

!


Figure 1-17. Wall frame elevation at D at second floor

Determine anchorage at A: M R = 100 plf (10.0 ft 2 )(10.0 × 2 3) + 1,100 (10.0 ft 2 ) + 700 lb (10.0 ft ) = 8,833 ft − lb 3.5 in. = 9.7 ft With a 4 × 6 post at each end wall L = 10.0 − 12 E The critical loading condition is: 0.9 D ± (12-10) 1.4 (52,452 1.4) − (8,833 × 0.9) = 3,043 lb Uplift at A = 9.7 ft Determine anchorage at B: M R = 100 plf (10.0 ft 2 )(10.0 3) + 1,100 (10.0 ft 2 ) + 700 lb (10.0 ft ) = 14,167 ft − lb Uplift at B =

(52,452 1.4 ) − (14,167 × 0.9 ) = 2,548 lb 9.7 ft

Elements supporting discontinuous systems

§1630.8.2

Since location A does not continue to the foundation, check special seismic load combination for elements supporting discontinuous systems. 1.2 D + f1 L + 1.0 E m

(12-17)

0.9 D ± 1.0 E m

(12-18)


73

Design Example 1

!


where: f1 = 0.0 for roof live loads (non-snow)

§1612.4

f1 = 0.5 for live loads

§1612.4

Em = Ωo Eh

(30-2)

Determine the seismic force overstrength factor Ωo

§1630.3.1

Ω o = 2.8 for wood structural panel wall

Table 16-N

Ω o = 2.0 for cantilevered column building systems

Table 16-N

For east-west axis of structure R = 2.2 for cantilevered building systems Therefore, Ω o = 2.0 Determine anchorage force at A for special seismic load combination: E m = Ω o E h = 2.0(5,828 lb ) = 11,656 lb M OT = 11,656 lb(9.0 ft ) = 104,904 ft - lb Therefore, uplift =

(104,904 1.4) − (8,833 × 0.9) = 6,905 lb (10.0 ft − 0.3 ft )

Consult ICBO Evaluation Reports for the allowable load capacity of premanufactured straps. Allowable load per 10d nail common with 14 ga metal side plate = 115 lb

91 NDS Table 12.3F

From Part 9b, with 3-inch nails penetration factor C d = 1.0 . For allowable stress design, the allowable stress increase factor is 1.7 for steel. Number of 10d common nails required =

§1630.8.2.1

6,905 lb = 26.5 nails 115 lb/nail (1.7 )(1.33)

Use a continuous 14 gauge x 3-inch strap bent around GLB.

74


Design Example 1

!


Note that §1630.8.2.1 allows the combination of allowable stress increase of 1.7 with the duration of load increase in Chapter 23. Note that the adequacy of the GLB to resist the overturning of the wall must be checked using the special seismic load combinations. As permitted in §1612.4 and §1630.8.2.1, an allowable stress increase of 1.7 can be used in addition to the duration of load increase of 1.33 for C D . Also, the boundary post at the wall corner must be checked for orthogonal effects with shear wall 5 (and on other locations in the structure with common corners).

10. 10

§1633.1

Diaphragm shears at the low roof over garage (Figure 1-20).

From Table 16-M, this has plan irregularity type 4. The diaphragm between lateral resisting elements C and E is required to transfer the design seismic force from shear wall D due to the offset between D and E. UBC §1633.2.9 requires the diaphragm force used in UBC Equation (33-1) to be used. UBC §1630.8.2 references special seismic load combinations of §1612.4 and does not allow the one-third increase permitted under §1612.3.2 From Part 7 in this design example: fp

floor

= 11.48 psf

From Table 16-P: Ω o for cantilever column type structures is 2.0. f p Ω o = 11.48 × 2.0 = 22.96 psf For simplification of analysis, assume the diaphragm over the garage is a simple span between lateral resisting elements at lines C and E. Load from wall D above = 5,828 lb VE = 22.96 (28.0 ft )(22.0 2 ) + 5,828 lb (15.0 ft 22.0 ft ) = 11,045 lb v E = 11,0451 lb 1.4 (28.0) = 281 plf > 215 plf (for unblocked)

n.g.

Table 23-II-H

Therefore, panel edges need to be blocked. Since the allowable shear values in Table 23-II-H already include a increase for short-term loading, (C D ), the duration of load increase (§1612.3.1 and §1612.3.2) cannot be used concurrently with the 1.7 increase, as prohibited in §2316.2, Item 5.


75

Design Example 1

!


From Table 23-II-H, the allowable diaphragm shear for 19/32-inch APA sheathing, with 10d common wire nails spaced at 6-inch centers, with blocked edges, is 320 plf. 320 plf>281 plf

o.k.

∴ Use 10d @6 inches o.c. with blocked edges on 19/32-inch sheathing.

11. 11

Detail the wall frame over the GLB.

Wall frame details must be shown on the drawings. Depending on the variations, when multiple wall frames are on a project, it is necessary at times to have individual details for each condition. While the detail shown in Figure 1-18 is somewhat generic, it should be noted that a separate anchorage detail (keynote 10) may be necessary where the end of the GLB is connected to the supporting post.

Figure 1-18. Details of wall frame on line D at second floor

76


Design Example 1

12. 12

!


Detail the anchorage of wall frame to the GLB.

Cross-grain shrinkage of the GLB may be a problem when using a connection of the type shown in Figure 1-19. Also, nails above the neutral axis of the GLB should be left out from the design to avoid cross-grain tension. In other words, only the nails below the neutral axis are considered effective for uplift forces. To avoid confusion in the field, all nail holes are to be filled. It should be noted that a separate anchorage detail may be necessary where the end of the GLB is connected to the supporting post (intersection of grids D and 5).

Figure 1-19. Detail of anchorage at point A (see also Figure 1-18)


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13. 13

!


Detail the continuous load path at the low roof above the garage doors.

The low roof above the garage is an important part of the continuous load path. Historically, this type of detail has been mis-detailed and mis-constructed. This detail has two load paths: the loads from the roof can either go through the pitched roof, or down the wall to the GLB and across the horizontal diaphragm to the exterior wall. Figure 1-20 shows one way that the shear transfer can be made. Also note that the chord/drag tie of the top plates will be interrupted by the GLB-to-post connection and will require detailing at grids D3 and D5.

Figure 1-20. Detail of load path for low roof over garage

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Commentary Following are some issues and topics related to the seismic design of wood frame residences that can be used to improve design practices and/or understanding of important aspects of design.

“Calc and sketch” philosophy.

In wood frame construction, particularly for single-family residences, it has been a common design practice to have an engineer provide only calculations and sketches for the architect to include on the architectural drawings.” This is done to provide a cost savings to the owner. This approach has some significant problems based on reviews of how residential framing is actually being constructed, the “calc and sketch only” service is a practice which should be discontinued, with a few exceptions. Architects and building officials need to be encouraged to adopt the following standards: 1.

Any new building (or remodel requiring the existing building to be brought into conformance with the current building code) that cannot be clearly shown to conform with building code conventional construction framing requirements should require submittal of structural drawings and calculations signed for by a licensed civil or structural engineer.

2.

Structural framing plans and details should be separate from the architectural drawings. Most new wood residential building designs are complex and beyond the scope and intent of the prescriptive conventional construction requirements of the UBC. Misuse of these conventional requirements has led to structures with incomplete lateral force systems, resulting in poor performance in earthquakes. Since the engineer generally is not asked to review the architect’s final drawings, the use of calculations and sketches lends itself to poorly coordinated drawings and missing structural information. The common practice of referring to details on architectural drawings as “similar” leads to further confusion as to the design intent. The structural observation requirements of the code, when enforced (many jurisdiction do not require structural observation for single-family residences), are even less effective, since the architect did not design the structural system and often can not identify what is missing or incorrect.

Rigid versus flexible diaphragm.

This design example illustrates seismic design using both flexible and rigid diaphragms. It also illustrates that most one- and two-family dwellings have rigid


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diaphragms as defined by code. This being the case, a design based on flexible diaphragm assumption would not be required if the design is based on the rigid diaphragm assumption. Using the common approach of basing wall rigidities on deflections of shear walls and other vertical elements, the engineer first needs to know or assume how the shear walls will be constructed (e.g., nail size and spacing). Without performing a preliminary analysis, the procedure of just doing a design based on rigid diaphragms may be subject to a trial and error process. One method (as used in this design example) to avoid this process is to first perform an analysis based on flexible diaphragms, then use the construction required from the flexible diaphragms for determining the wall rigidities. Part 2 of this design example uses flexible diaphragms to determine shear wall construction. Parts 3, 4, and 5 of this design example use rigid diaphragms per UBC requirements. The shear wall deflections used in this design example use UBC equations. This needs to be viewed as one possible approach that is substantiated by the code. However, other approaches can also be used. Two of these are given below: 1.

The rigidities of the shear walls can be based on the length of the wall times the allowable shear capacity. This method can be appropriate provided the tiedown assembly displacements are kept to a minimum. This may involve using specific types of tiedown devices that limit displacements to less than 1/8".

2.

Shear wall rigidities can be based on graphs of the four-term shear wall code deflection equation (see Part 3b). As shown in Figure 1-21, a chart of these is included in this section and is also considered appropriate in determining wall rigidities.

Tiedown location.

When designing shear walls, the engineer needs to consider where the tiedown posts will actually be located. The tiedown posts occur where shear walls stack from floor to floor. The lower level wall requires tiedown devices on each side of the tiedown post. However, the upper shear wall only requires a tiedown device on one side of the tiedown post. Since the posts must align between story levels, the upper level tiedown post will need to be offset inward in order to line up with the post below. Based on actual tiedown post locations, the upper level shear wall design may have to be rechecked once the lower level shear wall design is complete. The use of tiedown devices on each side of the post will improve the shear wall performance, since eccentricity in the connection, as occurs when there is only a single-sided tiedown, is avoided. Double-sided tiedowns are generally preferred over singlesided.

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Design comments.

This design example illustrates a detailed analysis for some of the important seismic requirements of the 1997 UBC. To complete this design, the engineer will have to check all the major structural elements along the various lateral load paths of the residence, including the foundations. The seismic calculations and details for this example residence are approximately 50 percent complete. Normal engineering design of this type of structure may omit many of the calculations shown in this example and rely on good engineering judgment. This design example illustrates a very comprehensive approach to the engineering calculations. This design example fills a void in the available engineering literature on the subject—many engineers have stated that there simply are not sufficient reference documents available on this subject. In the so called “big one,” it is expected that actual peak earthquake forces may be 2 to 3 times greater than the equivalent static forces required by the UBC and used in this example. The use of good detailing practices with ductile elements to absorb energy, clear construction documents with adequate detailing, structural site observation, and special inspection are considered every bit as important as a comprehensive set of structural calculations.

80.0 K = stiffness = F/d = (Vb )/d 70.0

d = deflection =(8vh 3)/(EAb ) +(vh )/(Gt ) + 0.75he n + d a [A1] h = 8 ft

Stiffness K (kips/in.)

60.0 Where: E = modulus of elasticity = 1.8x106 psi G = shear modulus = 90x103 psi h = wall height (ft) b = wall depth (ft) t = plywood thickness = 15/32 in. A = area of end post = 12.25 in.2 v = shear/foot d a = slip at hold down = 1/8 in. e n = nail deformation slip (in.) F = applied force = Vb (kips)

50.0

40.0

30.0

[A2] h = 10

[B1] h = 8 ft [B2] h = 10 ft [C1] h = 8 ft

[C2] h = 10 ft [D1] h = 8 ft [D2] h = 10

20.0 [A] [B] [C] [D]

10.0

edge nail spacing at 2” o.c. edge nail spacing at 3” o.c. edge nail spacing at 4” o.c. edge nail spacing at 6” o.c.

(v=870 plf, (v=665 plf, (v=510 plf, (v=340 plf,

e n =0.024) e n =0.033) e n =0.033) e n =0.033)

0.0 0

5

10

15

20

25

30

35

40

Wall Depth b (ft)

Figure 1-21. Stiffness of one-story ½-inch Structural-I plywood shear walls


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References American Forest and Paper Association, 1996, Wood Construction Manual. American Forest and Paper Association, Washington D.C. American Plywood Association, 1997, Design/ Construction Guide – Diaphragms and Shear Walls. Report 105, Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1997, Diaphragms and Shear Walls. Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1993, revised, Wood Structural Panel Shear Walls. Report 154, Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1994, Northridge, California Earthquake. Report T94-5. Engineered Wood Association, Tacoma, Washington. American Plywood Association, Performance Standards and Policies for Structural– Use Panels [Sheathing Standard, Sec. 2.3.3]. Standard PRP–108. Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1988, Plywood Diaphragms, Research Report 138. American Plywood Association, Tacoma, Washington. Applied Technology Council, 1995, Cyclic Testing of Narrow Plywood Shear Walls ATC R-1. Applied Technology Council, Redwood City, California. Applied Technology Council, 1981, Guidelines for Design of Horizontal Wood Diaphragms, ATC-7. Applied Technology Council, Redwood City, California. Applied Technology Council, 1980, Proceedings of a Workshop on Design of Horizontal Wood Diaphragms, ATC-7-1. Applied Technology Council, Redwood City, California. Building Seismic Safety Council, 1997, National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings. Building Seismic Safety Council, Washington D.C. Bugni, David A., 1999, “A Linear Elastic Dynamic Analysis of a Timber Framed Structure.” Building Standards, International Conference of Building Officials, Whittier, California

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Cobeen, K.E., 1996, “Performance Based Design of Wood Structures.” Proceeding: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Coil, J., 1999, “Seismic Retrofit of an Existing Multi-Story Wood Frame Structure,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Commins, A. and Gregg, R., 1996, Effect of Hold Downs and Stud-Frame Systems on the Cyclic Behavior of Wood Shear Walls, Simpson Strong-Tie Co., Pleasanton, California. Countryman, D., and Col Benson, 1954, 1954 Horizontal Plywood Diaphragm Tests. Laboratory Report 63, Douglas Fir Plywood Association, Tacoma Washington. CUREe, 1999, Proceedings of the Workshop on Seismic Testing, Analysis, and Design of Wood Frame Construction. California Universities for Research in Earthquake Engineering. Dolan, J.D., 1996, Experimental Results from Cyclic Racking Tests of Wood Shear Walls with Openings. Timber Engineering Report No. TE- 1996-001. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J. D. and Heine, C.P., 1997a, Monotonic Tests of Wood Frame Shear Walls with Various Openings and Base Restraint Configurations. Timber Engineering Report No. TE-1997-001, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D. and Heine, C.P., 1997b, Sequential Phased Displacement Cyclic Tests of Wood Frame Shear Walls with Various Openings and Base Restrain Configurations. Timber Engineering Report No. TE-1997-002, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D., and Heine, C.P., 1997c, Sequential Phased Displacement Test of Wood Frame Shear Walls with Corners. Timber Engineering Report No. TE-1997003, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Earthquake Engineering Research Institute, 1996, “Reconnaissance Report: Northridge Earthquake of January 17, 1994,” Earthquake Spectra. Vol. 11, Supplement C. Earthquake Engineering Research Institute, Oakland, California. Faherty, Keith F., and Williamson, Thomas G., 1995, Wood Engineering Construction Handbook. McGraw Hill, Washington D.C.


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Federal Emergency Management Agency, 1998, National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings. Federal Emergency Management Agency, Washington D.C. Ficcadenti, S.K., T.A. Castle, D.A. Sandercock, and R.K. Kazanjy, 1996, “Laboratory Testing to Investigate Pneumatically Driven Box Nails for the Edge Nailing of 3/8" Plywood Shear Walls,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Foliente, Greg C., 1994, Analysis, Design and Testing of Timber Structures Under Seismic Loads. University of California Forest Products Laboratory, Richmond, California. Foliente, Greg C., 1997, Earthquake Performance and Safety of Timber Structures. Forest Products Society, Madison Wisconsin. Forest Products Laboratory, 1999, Wood Handbook Publication FPL – GTR- 113. Madison, Wisconsin. Goers R. and Associates, 1976, A Methodology for Seismic Design and Construction of Single-Family Dwellings. Applied Technology Council, Redwood City, California. International Code Council, 1999, International Building Code – Final Draft, 2000. International Code Council, Birmingham, Alabama. Ju, S. and Lin, M. ,1999, “Comparison of Building Analysis Assuming Rigid or Flexible Floors,” Journal of Structural Engineering. American Society of Civil Engineers, Washington, D.C. Mendes, S., 1987, “Rigid versus Flexible: Inappropriate Assumptions Can Cause Shear Wall Failures!” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Mendes, S., 1995, “Lessons Learned From Four Earthquake Damaged Multi-Story Type V Structures,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. NFPA, 1991a, National Design Specification for Wood Construction. National Forest Products Association, Washington D.C. NFPA, 1997b, National Design Specification for Wood Construction. Natural Forest Products Association, Washington D.C.

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Rose, J. D., 1998, Preliminary Testing of Wood Structural Panel Shear Walls Under Cyclic (Reversed) Loading. Research Report 158, APA – Engineered Wood Association, Tacoma, Washington. Rose, J.D., and E.L. Keith, 1996, Wood Structural Panel Shear Walls with Gypsum Wallboard and Window [ Sheathing Standard, Sec. 2.3.3 ]. Research Report 158. APA - The Engineered Wood Association, Tacoma Washington. SEAOC, 1997, Seismic Detailing Examples for Engineered Light Frame Timber Construction. Structural Engineers Association of California, Sacramento, California. SEAOC, 1999, Guidelines for Diaphragms and Shear Walls. Structural Engineers Association of California, Sacramento, California. SEAOC, 1999, Plan Review – Codes and Practice. Structural Engineers Association of California, Sacramento, California. Shipp, J., 1992, Timber Design. Volumes IV and V. Professional Engineering Development Publications, Inc., Huntington Beach, California. Steinbrugge, J., 1994, “Standard of Care in Structural Engineering Wood Frame Multiple Housing,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California.


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Wood Light Frame Three-Story Structure

Design Example 2 Wood Light Frame Three-Story Structure

Figure 2-1. Wood light frame three-story structure elevation

Foreword After careful consideration and extensive discussion, SEAOC is recommending that large wood frame structures, such as the three-story building in this design example, be designed for seismic forces considering both rigid and flexible diaphragm assumptions. This method represents a significant change from current practice. At present, California practice has almost exclusively used the flexible diaphragm assumption for determining distribution of story shears to shear walls. There are two principal reasons for considering both rigid and flexible diaphragms. First, since adoption of the 1988 UBC, there has been a definition of diaphragm flexibility in the code (§1630.6 of the 1997 UBC). Arguably, when introduced in 1988, this definition may not have been intended to apply to wood framed diaphragms. After considerable discussion and re-evaluation, it is now the joint opinion of the SEAOC Code and Seismology Committees that this definition should be considered in wood framed diaphragms. The application of this definition in wood construction often requires the use of the rigid diaphragm assumption, and subsequent calculation of shear wall rigidities, for distribution of story shears to shear walls. In fact, this definition results in many, if not most, diaphragms in wood frame construction being considered rigid. Many engineers feel that exclusive use of the flexible diaphragm assumption results in underestimation of forces on some shear walls. For example, a rigid SEAOC Seismic Design Manual, Vol. II (1997 UBC)

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diaphragm analysis is judged more appropriate when the shear walls are more flexible compared to the diaphragm, particularly where one or more lines of shear walls (or other vertical resisting elements) are more flexible than the others are. Second, in some instances, the use of flexible diaphragm assumptions can actually force the engineer to provide a more favorable lateral force resisting system than would occur by only using rigid diaphragm assumptions. Flexible diaphragm assumptions encourage the placement of shear walls around the perimeter of the floor and roof area, therefore minimizing the need to have wood diaphragms to resist torsional forces. In this design example, the floor diaphragms are constructed using screw shank nails, sheathing is glued to the framing members (to reduce floor squeaks), and lightweight concrete fill is placed over the floor sheathing (for sound insulation). Additionally, gyp board is applied to the framing underside for ceiling finish. These materials in combination provide significantly stiffer diaphragms than those represented by the diaphragm deflection equation of UBC standard 23-2. For the part of the analysis that assumes a rigid diaphragm, the engineer must also select a method to estimate shear wall rigidities (and rigidities of other vertical resisting elements). This also requires use of judgment because at the present time there is no consensus method for estimating rigidities. In the commentary of Design Example 1, several alternatives are discussed. Prior to starting design of a wood light frame structure, users of this document should check with the local jurisdiction regarding both the level of analysis required and acceptable methodologies.

Overview This design example illustrates the seismic design of a three-story 30-unit hotel structure. The light frame structure, shown in Figures 2-1, 2-2, 2-3, and 2-4, has wood structural panel shear walls, and roof and floor diaphragms. The roofs have composite shingles and are framed with plated trusses. The floors have a 1½-inch lightweight concrete topping framed with engineered I joists. The primary tiedowns for the shear walls use a continuous tiedown system. This structure cannot be built using conventional construction methods for reasons shown in Part 6 of this design example. The following sections illustrate a detailed analysis for some of the important seismic requirements of the 1997 UBC. This design example is not a complete building design, and many aspects of a complete design, including wind design (see UBC §606 ), are not included. Only selected items of the seismic design are illustrated. In general, the UBC recognizes only two diaphragm categories: flexible and rigid. However, the diaphragms in this design example are considered to be semi-rigid. 88


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Hence, the analysis will use the envelope method, which considers the worst loading condition from the flexible and rigid diaphragm analyses for each vertical shear resisting element. It should be noted that the envelope method, although not explicitly required by code, is deemed necessary and good engineering practice for this design example. Initially, the shear wall nailing and tiedown requirements are determined using the flexible diaphragm assumption. Secondly, use these shear wall forces to determine shear wall rigidities for the rigid diaphragm analysis. Finally, further iterations may be required with significant stiffness redistributions. The method of determining shear wall rigidities used in this design example is by far more rigorous than normal practice but is not the only method available to determine shear wall rigidities. The commentary following Design Example 1 illustrates two other simplified approaches that would also be appropriate for this design example.

Outline This example will illustrate the following parts of the design process: 1.


2.

Lateral forces on the shear walls and required nailing assuming flexible diaphragms.

3.

Rigidities of shear walls.

4.

Distribution of lateral forces to the shear walls.

5.


6.

Does structure meet requirements of conventional construction provisions?

7.

Diaphragm deflections to determine if the diaphragm is flexible or rigid.

8.

Tiedown forces for shear wall on line C.

9.

Tiedown connection at the third floor for the shear wall on line C.

10. 10

Tiedown connection at the second floor for the shear wall on line C.

11. 11

Anchor bolt spacing and tiedown anchor embedment for shear wall on line C.


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12. 12

Detail of tiedown connection at the third floor for shear wall on line C (Figure 2-9).

13. 13

Detail of tiedown connection at the second floor for shear wall at line C. (Figure 2-10).

14. 14

Detail of wall intersection at exterior shear walls (Figure 2-11).

15. 15

Detail of tiedown connection at foundation (Figure 2-12).

16. 16

Detail of shear transfer at interior shear wall at roof (Figure 2-13).

17. 17

Detail of shear transfer at interior shear walls at floors (Figure 2-14).

18. 18

Detail of shear transfer at interior shear walls at foundation (Figure 2-15).

19. 19

Detail of sill plate at foundation edge (Figure 2-16).

20. 20

Detail of shear transfer at exterior wall at roof (Figure 2-17).

21. 21

Detail of shear transfer at exterior wall at floor (Figure 2-18).

Given Information Roof weights (slope 6:12): Roofing ½" sheathing Trusses Insulation Miscellaneous Gyp ceiling DL (along slope)

3.5 psf 1.5 3.5 1.5 0.7 2.8 13.5 psf

Floor weights: Flooring Lt. wt. concrete 5/8" sheathing Floor framing Miscellaneous Gyp ceiling

1.0 psf 14.0 1.8 5.0 0.4 2.8 25.0 psf

DL (horiz. proj.) = 13.5 (13.41/12) = 15.1 psf Stair landings do not have lightweight concrete fill Area of floor plan is 5,288 sq ft Weights of respective diaphragm levels, including tributary exterior and interior walls: Wroof W3rd floor W2nd floor W 90

= 135,000 lb = 2300,000 lb = 230,000 lb = 595,000 lb


Design Example 2

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Weights of diaphragms are typically determined by taking one-half height of walls at the third floor to the roof and (with equal story heights) full height of walls for the third and second floor diaphragms. Framing lumber is Douglas Fir-Larch (DF-L) grade stamped No. 1 S-Dry. (Note: The designer must recognize the increased potential for shrinkage problems when green lumber is used. The shrinkage of lumber can effect the architectural and mechanical systems as well as the structural system. The potential for wood shrinkage problems proportionally increases with the number of stories in the structure.) Foundation sill plates are pressure-treated Hem-Fir. APA-rated wood structural panels for shear walls will be 15/32-inch-thick Structural I, 32/16 panel index span rating, 5-ply with Exposure I glue is specified. However, 4-ply is also acceptable. Three-ply 15/32-inch sheathing has lower allowable shears and the inner ply voids can cause nailing problems. The roof is 15/32-inch-thick APA-rated sheathing (equivalent to C-D in Table 23-II-4), 32/16 span rating with Exposure I glue. The floor is 19/32-inch-thick APA-rated Sturd-I-Floor 24" o/c rating (or APA-rated sheathing, 48/24 span rating) with Exposure I glue. Common wire nails are used for diaphragms, shear walls, and straps. Sinker nails will be used for design of the shear wall sill plate nailing at the second and third floor. (Note: Many nailing guns use the smaller diameter box and sinker nails instead of common nails. Closer nail spacing may be required if the smaller diameter nails are used). Seismic and site data: (Zone 4) I = 1.0 (standard occupancy) Seismic source Type = B Distance to seismic source = 12 km Soil profile type = S C




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Figure 2-2. Foundation plan (ground floor) 92


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Figure 2-3. Floor framing plan (second and third floors)

Note: Shear walls on lines 2 and 3 do not extend from the third floor to the roof.


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Factors That Influence Design Before starting the example, four important related aspects of the design will be discussed. These are the effect of moisture content on lumber, the use of pre-manufactured roof trusses, proper detailing of shear walls at building pop-outs, and effects of box nails on wood structural panel shear walls. Moisture content in lumber connections.

This design example is based on dry lumber. Project specifications typically call for lumber to be grade-stamped S-Dry (Surfaced Dry). Dry lumber has a moisture content (MC) less than or equal to 19 percent. Partially seasoned or green lumber grade stamped S-GRN (surfaced green) has a MC between 19 percent and 30 percent. Wet lumber has a MC greater than 30 percent. Construction of structures using lumber with moisture contents greater than 19 percent can produce shrinkage problems. Note that UBC §2304.7 requires consideration of lumber shrinkage. Also, many engineers and building officials are not aware of the reduction requirements or wet service factors related to installation of nails, screws, and bolts (fasteners) into lumber with moisture content greater than 19 percent. For fasteners installed in lumber with moisture content greater than 19 percent, the wet service factor C M = 0.75 for nails and C M = 0.67 for bolts, lags and screws (91 NDS Table 7.3.3) are used. For construction using lumber of MC greater than 19 percent, there is a 25 percent to 33 percent reduction in the strength of connections, diaphragms, and shear walls that is permanent. The engineer needs to exercise good engineering judgment in determining whether it is prudent to base the structural design on dry or green lumber. Other areas of concern are the geographical area and the time of year the structure is built. It is possible for green lumber (or dry lumber that has been exposed to rain) to dry out to a moisture content below 19 percent on the construction site. For 2 × framing, this generally takes about 2 to 3 weeks of exposure to dry air, 4 × lumber takes even longer. Drying occurs when the surfaces are exposed to air on all sides, not while stacked on pallets (unless shimmed with stickers). Moisture content can easily be verified by a hand held “moisture meter.”

Use of pre-manufactured roof trusses to transfer lateral forces.

The structural design in this design example uses the pre-manufactured wood roof trusses. Under seismic forces, these must transfer the lateral forces from the roof diaphragm to the tops of the interior shear walls. To accomplish this, special considerations must be made in the design and detailed on the plans. In particular, any trusses that are to be used as collectors or lateral drag struts should be clearly indicated on the structural framing plan. The magnitude of the forces, the means by which the forces are applied to the trusses and transferred from the trusses to the shear walls must be shown on the plans. In addition, if the roof sheathing at the hip SEAOC Seismic Design Manual, Vol. II (1997 UBC)

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ends breaks above the joint between the end jack trusses and the supporting girder truss, the lateral forces to be resisted by the end jacks should be specified so that an appropriate connection can be provided to resist these forces. The drawings also must specify the load combinations and whether or not a stress increase is permitted. If ridge vents are being used, special detailing for shear transfers must be included because normal diaphragm continuity is disrupted. Proper detailing of shear walls at building pop-outs.

The structure for this design example has doubled-framed walls for party walls and exterior “planted-on” box columns (pop-outs). The designer should not consider these walls as shear walls unless special detailing and analysis is provided to substantiate that there is a viable lateral force path to that wall and the wall is adequately braced. Effects of box nails on wood structural panel shear walls.

This design example uses common nails for fastening wood structural panels. Based on cyclic testing of shear walls and performance in past earthquakes, the use of common nails is preferred. UBC Table 23-II-I-1 lists allowable shears for wood structural panel shear walls for “common or galvanized box nails.” Footnote number five of Table 23-II-I-1, states that the galvanized nails shall be “hot-dipped or tumbled” (these nails are not gun nails). Most contractors use gun nails for diaphragm and shear wall installations. The UBC does not have a table for allowable shears for wood structural panel shear walls or diaphragms using box nails. Box nails have a smaller diameter shank and a smaller head size. Using 10d box nails would result in a 19 percent reduction in allowable load for diaphragms and shear walls as compared to 10d common nails. Using 8d box nails would result in a 22 percent reduction in allowable load for diaphragms and shear walls as compared to 8d common nails. This is based on comparing allowable shear values listed in Tables 12.3A and 12.3B in the 1997 NDS for one-half-inch side member thickness (t s ) and Douglas Fir-Larch framing. In addition to the reduction of the shear wall and diaphragm capacities, when box nails are used, the walls will also drift more than when common nails are used. A contributor to the problem is that when contractors buy large quantities of nails (for nail guns), the word “box” or “common” does not appear on the carton label. Nail length and diameters are the most common listing on the labels. This is why it is extremely important to list the required nail lengths and diameters on the structural drawings for all diaphragms and shear walls. Another problem is that contractors prefer box nails because their use reduces splitting, eases driving, and they cost less. Just to illustrate a point, if an engineer designs for “dry” lumber (as discussed above) and “common” nails, and subsequently “green” lumber and “box” nails are used in the construction, the result is a compounding of the reductions. For 96


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example, for 10d nails installed into green lumber, the reduction would be 0.81 times 0.75 or a 40 percent reduction in capacity.


1.


1a. 1a

Design base shear.

Code Reference

§1630.2.2

Determine period using Method A (see Figure 2-5 for section through structure): T = Ct (hn )3 / 4 = .020(33.63)3 / 4 = 0.28 sec

(30-8)

Figure 2-5. Typical cross-section through building


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With seismic source type B and distance to source = 12 km N a = 1.0

Table 16-S

N v = 1.0

Table 16-T


Table 16-Q

C v = 0.56 N v = 0.56(1.0 ) = 0.56

Table 16-R

Because the stud walls are both wood structural panel shear walls and bearing walls

Table 16-N

R = 5.5 Design base shear is: V=

Cv I 0.56 (1.0 ) W = 0.364W W= 5.5 (0.28) RT

(30-4)

Note: design base shear is now on a strength design basis. but need not exceed: V=

2.5C a I 2.5 (0.40 )(1.0) W= W = 0.182W R 5.5

(30-5)

V = 0.11C a IW = 0.11 (0.40)(1.0 )W = 0.044W < 0.182W

Check Equation 30-7: V=

0.8ZN v I 0.8 × 0.4 × 1.0 × 1.0 W= W R 5.5

V = 0.058W < 0.182W All of the tables in the UBC for wood diaphragms and shear walls are based on allowable loads.

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It is desirable to use the strength level forces throughout the design of the structure for two reasons: 1.

2.

Errors in calculations can occur and which load is being used—strength design or allowable stress design—may be confused. This design example will use the following format: Vbase shear =

strength

F px

=

strength

Fx v

= =

force-to-wall (strength) wall shear at element level (ASD)

v=

Fx = 1.4b

ASD

Future editions of the code will use only strength design. E = ρE h + E v = 1.0 E h + 0 = 1.0 E h

(30-1)

where: E v is permitted to be taken as zero for allowable stress design, and ρ will be assumed to be 1.0 (under most cases is 1.0 for Type V construction with interior shear walls). Since the maximum element story shear is not yet known, the assumed value for ρ will have to be verified. (This will be shown in Part 5.) The basic load combination for allowable stress design for horizontal forces is: D+

E E E = 0+ = 1.4 1.4 1.4

(12-9)

For vertical downward loads: D+

E E  or D + 0.75 L + (Lr or S ) + 1.4 1.4  

(12-10,12-11)

For vertical uplift: 0.9 D ±

E 1.4

(12-10)

V = 0.182W §1612.3.1 ∴V = 0.182(595,000 lb ) = 108,290 lb


99

Design Example 2

1b. 1b

!


Vertical distributions of forces.

The base shear must be distributed to each level. This is done as follows: F px =

(V − Ft )wx hx

(30-15)

n

∑ wi hi i =1

Where h x is the average height at level i of the sheathed diaphragm in feet above the base. Since T = 0.28 second < 0.7 second, Ft = 0 Determination of F px is shown in Table 2-1.

§1630.5

Note: Although not shown here, designers must also check wind loading. In this example, wind loading may control the design in the east-west direction.

Table 2-1. Vertical distribution of seismic forces Level Roof 3rd Floor 2nd Floor

Σ

2.

w x (k)

h x (ft)

135.0 230.0 230.0 595.0

33.6 18.9 9.4

w x h x (k-ft) 4,536 4,347 2,162 11,045

w x hx (%) ∑ wi hi 41.1 39.4 19.5

F px (k) 44.5 42.7 21.1 108.3

F px wx 0.330 0.186 0.092

Ftot (k) 44.5 87.2 108.3

Lateral forces on the shear walls and required nailing assuming flexible diaphragms.

In this step, forces on shear walls due seismic forces will be determined. As has been customary practice in the past, this portion of the example assumes flexible diaphragms. The UBC does not require torsional effects to be considered for flexible diaphragms. The effects of torsion and wall rigidities will be considered in Part 4 of this design example. Under the flexible diaphragm assumptions, loads to shear walls are determined based on tributary areas with simple spans between supports. Another method of determining loads to shear walls can assume a continuous beam. This design example uses the total building weight W applied to each respective direction. The results shown will be slightly conservative, since the building weight W includes the wall weights for the direction of load, which can be subtracted out. This example converts the story forces into seismic forces per square foot of floor or 100


Design Example 2

!


roof area. This may result in loosing a certain amount of precision, but in turn results in much simpler calculations. This approach is generally considered acceptable unless there is seen to be a concentration of dead load in a particular area (e.g., a mechanical penthouse). A detailed analysis will include the derivation of these tributary weights, which includes the tributary exterior and interior wall weights. Using forces from Table 2-1 and the area of the floor plan = 5,288 sf, calculate tributary weights. For roof diaphragm: Roof area = 5,288 sq ft f p roof =

44.5 × 1,000 = 8.415 psf 5,288

For third floor diaphragm: Floor area = 5,288 sq ft f p 3rd =

42.7 × 1,000 = 8.075 psf 5,288

For second floor diaphragm: Floor area = 5,288 sq ft f p 2 nd =

21.1 × 1,000 = 3.990 psf 5,288


101

Design Example 2

!


1, 2, 3

Table 2-2. Forces to walls and required panel nailing for east-west direction Wall

A B C E F G H

Σ A B C E F G H

Σ A B C E F G H

Σ Notes: 1.

2.

3.

Trib Area (sq ft)

170 746 1,344 1,344 960 554 170 5,288 170 746 1,344 1,344 960 554 170 5,288 170 746 1,344 1,344 960 554 170 5,288

∑ FAbove (lb)

∑ Fx

(lb)

Ftot (lb)

b (4) (ft)

v=

Ftot (5) Sheathed (1.4)b 1 or 2 sides

(plf) 0 0 0 0 0 0 0 0

1,430 6,280 11,310 11,310 8,080 4,660 1,430 44,500 2,805 12,305 22,160 22,160 15,830 9,135 2,805 87,200

Allowable Shear(6) (plf)


1,430 6,280 11,310 11,310 8,080 4,660 1,430 44,500

Shear Walls at Roof Level (7) 1,430 12.5 6,280 22.0 11,310 43.0 11,310 43.0 8,080 43.0 4,660 22.0 1,430 12.5 44,500 198

85 205 190 190 135 155 85

1 1 1 1 1 1 1

340 340 340 340 340 340 340

6 6 6 6 6 6 6

1,375 6,025 10,850 10,850 7,750 4,475 1,375 42,700

Shear Walls at Third Floor Level 2,805 12.5 12,305 22.0 22,160 43.0 22,160 43.0 15,830 43.0 9,135 22.0 2,805 12.5 87,200 198

160 400 370 370 265 300 160

1 1 1 1 1 1 1

340 510 510 510 510 510 340

6 4 4 4 4 4 6

Shear Walls at Second Floor Level 3,485 12.5 200 15,280 22.0 500 27,525 43.0 460 27,525 43.0 460 19,660 43.0 330 11,345 22.0 370 3,485 12.5 200 108,300 198

1 1 1 1 1 1 1

340 665 665 665 665 665 340

6 3 3 3 3 3 6

680 2,975 5,365 5,365 3,830 2,210 680 21,100

Minimum framing thickness: The 1994 and earlier editions of the UBC required 3 × nominal thickness stud framing at abutting panel edges when 10d common nails were spaced 3 inches on center or closer (2" on center for 8d) or if sheathing is installed on both sides of the studs without staggered panel joints. The 1997 UBC (Table 23-II-I-1 Footnote 2 and 3) requires 3 × nominal thickness stud framing at abutting panels and at foundation sill plates when the allowable shear values exceed 350 pounds per foot or if the sheathing is installed on both sides of the studs without staggered panel joints. Sill bolt washers: Section 1806.6.1 of the 1997 UBC requires that a minimum of 2-inch-square by 3/16-inch-thick plate washers be used for each foundation sill bolt (regardless of allowable shear values in the wall). These changes were a result of splitting of framing studs and sill plates observed in the Northridge earthquake and in cyclic testing of shear walls. The plate washers are intended to help resist uplift forces on shear walls. Because of observed vertical displacements of tiedowns, these plate washers are required even if the wall has tiedowns designed to take uplift forces at the wall boundaries. The washer edges shall be parallel/perpendicular to the sill plate. Errata to the First Printing of the 1997 UBC (Table 23-II-I-1 Footnote 3) added an exception to the 3 × foundation sill plates by allowing 2 × foundation sill plates when the allowable shear values are less than 600 pounds per foot, provided that sill bolts are designed for 50 percent of allowable values. The 1999 SEAOC Blue Book recommends special inspection when the nail spacing is closer than 4" on center.

102


Design Example 2

4. 5. 6. 7.

!


The shear wall length used for wall shears is the “out-to-out” wall length. Note that forces are strength level and that shear in wall is divided by 1.4 to convert to allowable stress design. APA Structural I rated wood structural panels may be either plywood or oriented strand board (OSB). Allowable shear from UBC Table 23-II-I-1. Shear walls at lines C, E, and F extend to the bottom of the prefabricated wood trusses at the roof level. Shear transfer is obtained by framing clips from the bottom chord of the trusses to the top plates of the shear walls. Project plans call for trusses at these lines to be designed for these horizontal forces (see also comments in Part 8). Roof shear forces are also transferred to lines A, B, G, and H.

1, 2, 3

Table 2-3. Forces to walls and required panel nailing for north-south direction Wall

Trib. Area (sq ft)

∑ FAbove

∑ Fx

2,644 0 0 2,644 5,288

0 0 0 0 0

22,250 0 0 22,250 44,500

1,202 1,442 1,442 1,202 5,288

22,250 0 0 22,250 44,500

9,705 11,645 11,645 9,705 42,700

1,202 1,442 1,442 1,202 5,288

31,955 11,645 11,645 31,955 87,200

4,795 5,755 5,755 4,795 21,100

(lb)

(lb)

Ftot (lb)

b (4) (ft)

v=

Ftot ( 4) (plf) (1.4)b

Sheathed 1 or 2 sides



Shear Walls at Roof Level (5) 1 2 3 4

Σ 1 2 3 4

Σ 1 2 3 4

Σ Notes:

22,250 0 0 22,250 44,500

64.5 0 0 64.5 129.0

250 0 0 250

1

340

6

1

340

6

355 140 140 355

1 1 1 1

510 340 340 510

4 6 6 4

Shear Walls at Second Floor Level 36,750 64.5 410 17,400 60.0 210 17,400 60.0 210 36,750 64.5 410 108,300 249.0

1 1 1 1

510 340 340 510

4 6 6 4

Shear Walls at Third Floor Level 31,955 64.5 11,645 60.0 11,645 60.0 31,955 64.5 87,200 249.0

1. Minimum framing thickness: The 1994 and earlier editions of the UBC required 3 × nominal thickness stud framing at abutting panel edges when 10d common nails were spaced 3 inches on center or closer (2" on center for 8d) or if sheathing is installed on both sides of the studs without staggered panel joints. The 1997 UBC (Table 23-II-I-1 Footnote 2 and 3) requires 3 × nominal thickness stud framing at abutting panels and at foundation sill plates when the allowable shear values exceed 350 pounds per foot or if the sheathing is installed on both sides of the studs without staggered panel joints. 2. Sill bolt washers: Section 1806.6.1 of the 1997 UBC requires that a minimum of 2-inch-square by 3/16-inch-thick plate washers be used for each foundation sill bolt (regardless of allowable shear values in the wall). These changes were a result of splitting of framing studs and sill plates observed in the Northridge earthquake and in cyclic testing of shear walls. The plate washers are intended to help resist uplift forces on shear walls. Because of observed vertical displacements of tiedowns, these plate washers are required even if the wall has tiedowns designed to take uplift forces at the wall boundaries. The washer edges shall be parallel/perpendicular to the sill plate. Errata to the First Printing of the 1997 UBC (Table 23-II-I-1 Footnote 3) added an exception to the 3 × foundation sill plates by allowing 2 × foundation sill plates when the allowable shear values are less than 600 pounds per foot, provided that sill bolts are designed for 50 percent of allowable values. 3. The 1999 SEAOC Blue Book recommends special inspection when the nail spacing is closer than 4" on center. 4. Note that forces are strength level and that shear in wall is divided by 1.4 to convert to allowable stress design. SEAOC Seismic Design Manual, Vol. II (1997 UBC)

103

Design Example 2

!


5. The interior shear walls at lines 2 and 3 were not used to brace the roof diaphragm. This is because installing wall sheathing (blocking panels) perpendicular to plated trusses is labor intensive. Often it is not installed correctly, and occasionally it is not even installed due to contractor error. This approach will increase the third floor diaphragm transfer (redistribution) forces. With rigid diaphragms, you must carefully follow the load paths.

3.


3a. 3a

Rigidity calculation using the UBC deflection equation.

Determination of wood shear wall rigidities is not a simple task. In practice, approximate methods are often used. The method illustrated in this example is by far the most rigorous method used in practice. There are other methods that are more simplified, and use of these other more simplified methods is often appropriate. The alternate methods are briefly discussed in the Commentary to Design Example 1. It must be emphasized, that at the present time every method is approximate, particularly for multistory structures such as in this example. Until more definite general procedures are established through further testing and research, the designer must exercise judgment in selecting an appropriate method to be used for a given structure. When in doubt, consult with the local building official regarding methods acceptable to the jurisdiction. At the time of this publication, the type of seismic design required for a project of this type varies greatly from jurisdiction to jurisdiction. Wall rigidities are approximate. The initial rigidity R of the structure can be significantly higher due to stucco, drywall, stiffening effects of walls not considered, and areas over doors and windows. During an earthquake, some low-stressed walls may maintain their stiffness and others degrade in stiffness. Some walls and their collectors may attract significantly more lateral load than anticipated in flexible or rigid diaphragm analysis. It must be understood that the method of analyzing a structure using rigid diaphragms takes significantly more engineering effort. However, use of the rigid diaphragm method indicates that some lateral resisting elements can attract significantly higher seismic demands than from tributary area (i.e., flexible diaphragm) analysis methods. In this example, shear wall rigidities (k) are computed using the basic stiffness equation: F = k∆

or: k=

104

F ∆


Design Example 2

!


The basic equation to determine the shear wall deflections is shown below. This should be viewed as one possible approach that can be substantiated with code equations. There are other approaches that can also be used. ∆=


§23.223 Vol. 3

where: v = shear in the wall in pounds per lineal foot h = height from the bottom of the sill plate to the underside of the framing at diaphragm level above (top plates)

A = area of the boundary element in square inches At the third floor, the boundary elements consist of 2-2x4s (see Figure 2-9) At the second floor, the boundary elements consist of 3-2x4s (see Figure 2-10) At the ground floor, the boundary elements consist of 3-3x4s: b = is the shear wall length in feet

G = shear modulus values from Table 23-2-J, in pounds per square inch t = equivalent thickness values from Table 23-2-I, in inches Vn = load per fastener (nail) in pounds en = nail slip values are for Structural I sheathing with dry lumber = (Vn 769 )3.276 d a = displacement of the tiedown due to anchorage details in inches The above equation is based on tests conducted by the American Plywood Association and on a uniformly nailed, cantilever shear wall with fixed base and free top, a horizontal point load at top, and panel edges blocked, and deflection is estimated from the contributions of four distinct parts. The first part of the equation accounts for cantilever beam action using the moment of inertia of the boundary elements. The second term accounts for shear deformation of the sheathing. The third term accounts for nail slippage/bending, and the fourth term accounts for tiedown assembly displacement (this also should include bolt/nail slip and shrinkage). The UBC references this equation in §2315.1.


105

Design Example 2

!


The engineer should be cautioned to use the units as listed in §23.223 (and as listed above). Do not attempt to change the units. Testing on wood shear walls has indicated that the above deflection formula is reasonably accurate for wall aspect ratios (h w) lower than or equal to 2:1. For higher aspect ratios, the wall drift increases significantly, and displacements were not be adequately predicted by the formula. Using the new aspect ratio requirement of 2:1 (UBC 1997) makes this formula more accurate for determining shear wall deflection/stiffness than it was in previous editions of the UBC, subject to the limitations mentioned above. Recent testing on wood shear walls has shown that sill plate crushing under the boundary element can increase the shear wall deflection by as much as 20 to 30 percent. For a calculation of this crushing effect, see the deflection of wall frame at line D later in this same Part 11c.

Faster slip/nail deformation values (en).

Volume 3 of the UBC has Table 23-2-K for obtaining values for en . However, its use is somewhat time-consuming since interpolation and adjustments are necessary. Footnote 1 to Table 23-2-K requires the values for en to be decreased 50 percent for seasoned lumber. This means that the table is based on nails being driven into green lumber and the engineer must use one-half of these values for nails driven in dry lumber. The values in Table 23-2-K are based on tests conducted by the APA. The 50 percent reduction for dry lumber is a conservative factor. The actual tested slip values with dry lumber were less than 50 percent of the green lumber values. It is recommended that values for en be computed based on fastener slip equations from Table B-4 of APA Research Report 138. This research report is the basis for the formulas and tables in the UBC. Both the research report and the UBC will produce the same values. However, using the fastener slip equations from Table B-4 of Research Report 138 will save time and also enable computations to be made by a computer. For 10d common nails used in this example, there are two basic equations: When nails are driven into green lumber: en = (Vn 977 )1.894

APA Table B-4

When nails are drive into dry lumber: en = (Vn 769 )3.276

APA Table B-4

where: Vn = fastener load in pounds per fastener

106


Design Example 2

!


These values from the above formulas are based on Structural I sheathing and must be increased by 20 percent when the sheathing is not Structural I. The language in Footnote A in Research Report 138, Table B-4, which states “Fabricated green/tested dry (seasoned)…” is potentially misleading. The values in the table are actually green values, since the assembly is fabricated when green. Don’t be misled by the word “seasoned.” It is uncertain whether or not the d a factor is intended to include wood shrinkage and crushing due to shear wall rotation, because the code is not specific. This design example includes shrinkage and crushing in the d a factor. Many engineers are concerned that if the contractor installs the nails at a different spacing (too many or too few), then the rigidities will be different than those calculated. However, nominal changing of the nail spacing in a given wall does not significantly change the stiffness.

3b. 3b

Calculation of shear wall rigidities.

In this example, shear wall rigidities are calculated using the four-term code deflection equation in §23.223 of Volume 3. These calculations are facilitated by the use of a spreadsheet program, which eliminates possible arithmetic errors from the many repetitive computations that must be made. The first step is to calculate the displacement (i.e., vertical elongation) of the tiedown assemblies and the crushing effect of the boundary element. This is the term d a . The force considered to act on the tiedown assembly is the net uplift force determined from the flexible diaphragm analyses of Part 2. These forces are summarized in Tables 2-4, 2-9, and 2-13 for the roof at the third floor and second floor, respectively. After the tiedown assembly displacements are determined, the four-term deflection equation is used to determine the deflection ∆ S of each shear wall. These are summarized in Tables 2-5 and 2-6 for the roof level, and in Tables 2-10 and 2-11 for the third floor level, and in Table 2-14 and 2-15 for the second floor level. Finally, the rigidities of the shear walls are summarized in Tables 2-7, 2-12, and 2-16 for the roof, third floor, and second floor, respectively. For both strength and allowable stress design, the 1997 UBC now requires building drifts to be determined by the load combinations of §1612.2, which covers load combinations using strength design or load and resistance factor design. Errata for the second and third printing of the UBC unexplainably referenced §1612.3 for allowable stress design. The reference to §1612.3 is incorrect and will be changed back to reference §1612.2 in the fourth and later printings.


107

Design Example 2

!


Using strength level forces for wood design using the 1997 UBC now means that the engineer will use both strength-level forces and allowable stress forces. This can create some confusion, since the code requires drift checks to be strength-level forces. However, all of the design equations and tables in Chapter 23 are based on allowable stress design. Drift and shear wall rigidities should be calculated from the strength-level forces. Remember that the structural system factor R is based on using strength-level forces.

3c. 3c

Estimation of roof level rigidities.

To determine roof level wall rigidities, roof level displacements must first be determined. Given below are a series of calculations, done in table form, to estimate the roof level displacements ∆s in each shear wall. First, the shear wall tiedown assembly displacements are determined (Table 2-4). These, and the parameters given in Table 2-5, are used to arrive at the displacements ∆s for each shear wall at the roof level (Table 2-5 and 2-6). Rigidities are estimated in Table 2-7 for walls in both directions. Once the ∆s displacements are known, a drift check is performed. This is summarized in Table 2-8. 1

Table 2-4. Determine tiedown assembly displacements at roof level ASD Strength Design TIedown Assembly Displacement Wall Tiedown(3) Tiedown Uplift/1.4 (2) Uplift (lb) (lb) Device Elongation (in.) Shrink(4) Crush(5) Slip(6) A B1 B2 C1 C2 E1 E2 F1 F2 G1 G2 H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f

0 840 840 100 100 100 100 0 0 500 500 0 120 0 0 0 0 120

Not required Strap Strap Strap Strap Strap Strap Not required Not required Strap Strap Not required Strap Not required Not required Not required Not required Strap

0 1,175 1,175 140 140 140 140 0 0 700 700 0 170 0 0 0 0 170

0 0.04 0.04 0.02 0.02 0.02 0.02 0 0 0.02 0.02 0 0.02 0 0 0 0 0.02

0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05

0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02 0.02

0 0.002 0.002 0.002 0.002 0.002 0.002 0 0 0.002 0.002 0 0.002 0 0 0 0 0.002

d a (7) (in.) 0.07 0.11 0.11 0.09 0.09 0.09 0.09 0.07 0.07 0.09 0.09 0.07 0.09 0.07 0.07 0.07 0.07 0.09

Notes: 1. 2.

108

Tiedown assembly displacements for the roof level are calculated for the tiedowns at the third floor level. Uplift force is determined by using the net overturning moment (M OT − M OR ) divided by the distance between the centroids of the boundary elements with 4x members at the ends of the shear wall. This equates to the length of the wall minus 3½ inches for straps or the SEAOC Seismic Design Manual, Vol. II (1997 UBC)

Design Example 2

3.

4.

5.

!


length of wall minus 7¼ inches when using a bolted tiedown with 2-inch offset from post to anchor bolt. Using allowable stress design, tiedown devices need only be sized by using the ASD uplift force. The strength design uplift force is used to determine tiedown assembly displacement in order to determine strength-level displacements. The continuous tiedown (rod) system selected for this structure will have a “shrinkage compensating” system. Most of these systems have shrinkage compensation by either pre-tensioning of cables or a “self-ratcheting” hardware connector and are proprietary. The device selected in this design example has adjusting grooves at 1/10-inch increments, meaning the most the “system” will have not compensated for in shrinkage and crushing will be 1/10-inch. If the selected device does not have a shrinkage compensating device then, shrinkage of floor framing, sill plates, compression bridges, crushing of bridge support studs, and collector studs will need to be considered. See Design Example 1, Part 3c for an example calculation for a bolted connection. The tiedown rod at line B will elongate as follows: PL for 5 8" rod: ∆ = = 6,090 lb(4.5)(12 ) 0.31(29 E6 ) = 0.04 in AE Note that the rod length is 4.5 feet (Figure 2-12). The elongation for the portion of the rod at the level below will be considered at the level below. For level below (Table 2-13) rod length is 9.44 feet (Figure 2-12): PL for 5 8" rod: ∆ = = 12,040 lb(9.44)(12 ) 0.31(29 E6 ) = 0.15 in . AE Wood shrinkage is based on a change in moisture content (MC) from 19 percent to 15 percent, with 19 percent MC being assumed for S-Dry lumber per project specifications. The MC of 15 percent is the assumed final MC at equilibrium with ambient humidity for the project location. The final equilibrium value can be higher in coastal areas and lower in inland or desert areas. This equates to (0.002 )(d )(19 − 15) , where d is the dimension of the lumber (see Figure 2-11). Pressure-treated lumber has moisture content of less than 16 percent at treatment completion. Shrinkage of 2 × DBL Top Plate + 2 × DBL sill plate = (0.002 )(4 × 1.5 in )(19 − 15) = 0.05 in .

Per 91 NDS 4.2.6, when compression perpendicular to grain ( f c⊥ ) is less than 0.73F ' c⊥

crushing will be approximately 0.02 inches. When f c⊥ = F 'c⊥ crushing is approximately 0.04 inches. The effect of sill plate crushing is the downward effect at the opposite end of the wall with uplift force and has the same rotational effect as the tiedown displacement. Short walls that have no uplift forces will still have a crushing effect and contributes to rotation of the wall. 6.

( )

Per 91 NDS 7.3.6 load/slip modulus γ = (270,000) D1.5 , plus an additional 1/16" for the

oversized hole for bolts. For nails, values for en can be used. 7.

d a is the total tiedown assembly displacement. This also could include mis-cuts (short studs) and lack of square cut ends.


109

Design Example 2

!


Table 2-5. Deflections of shear walls at the roof level in east-west direction Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H

ASD v (plf)

Strength v (plf)

h (ft)

A

(in.2)

E (psi)

b (ft) 12.5 11.0 11.0 22.0 21.5 21.5 43.0 21.5 21.5 43.0 21.5 21.5 43.0 11.0 11.0 22.0 12.5

85 205 205

119 287 287

8.21 8.21 8.21

10.5 10.5 10.5

1.7E6 1.7E6 1.7E6

190 190

266 266

8.21 8.21

10.5 10.5

1.7E6 1.7E6

190 190

266 266

8.21 8.21

10.5 10.5

1.7E6 1.7E6

135 135

189 189

8.21 8.21

10.5 10.5

1.7E6 1.7E6

155 155

217 217

8.21 8.21

10.5 10.5

1.7E6 1.7E6

85

119

8.21

10.5

1.7E6

Nail Spacing (in.)

Vn

en

da

∆S

(lb)

(in.)

(in.)

(in.)

0.535 0.535 0.535

6 6 6

60 144 144

0.0002 0.0041 0.0041

0.07 0.11 0.11

0.07 0.16 0.16

90,000 90,000

0.535 0.535

6 6

133 133

0.0032 0.0032

0.09 0.09

0.10 0.10

90,000 90,000

0.535 0.535

6 6

133 133

0.0032 0.0032

0.09 0.09

0.10 0.10

90,000 90,000

0.535 0.535

6 6

95 95

0.0011 0.0011

0.07 0.07

0.07 0.07

90,000 90,000

0.535 0.535

6 6

109 109

0.0017 0.0017

0.09 0.09

0.12 0.12

90,000

0.535

6

60

0.0002

0.07

0.07

G

t

(psi)

(in.)

90,000 90,000 90,000

Table 2-6. Deflections of shear walls at the roof level in north-south direction Wall 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4

110

ASD v (plf) 250 250 250 250 250 250

Strength v (plf) 350 350 350 350 350 350

h (ft) 8.21 8.21 8.21 8.21 8.21 8.21

A (2)

(in.) 10.5 10.5 10.5 10.5 10.5 10.5

E (psi)

b (ft)

1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6

8.0 14.0 11.5 11.5 11.5 8.0 64.5

G (psi) 90,000 90,000 90,000 90,000 90,000 90,000

t (in.) 0.535 0.535 0.535 0.535 0.535 0.535

Nail Spacing (in.)

Vn

en

da

∆S

(lb)

(in.)

(in.)

(in.)

6 6 6 6 6 6

175 175 175 175 175 175

0.0078 0.0078 0.0078 0.0078 0.0078 0.0078

0.09 0.07 0.07 0.07 0.07 0.09

0.21 0.16 0.17 0.17 0.17 0.21


Design Example 2

Table 2-7. Shear wall rigidities at roof level Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4

∆ S (2) (in.) 0.07 0.16 0.16 0.10 0.10 0.10 0.10 0.07 0.07 0.12 0.12 0.07 0.21 0.16 0.17 0.17 0.17 0.21

F (lb) 1,430 3,140 3,140 6,280 5,655 5,655 11,310 5,655 5,655 11,310 4,040 4,040 8,080 2,330 2,330 4,660 1,430 2,760 4,830 3,965 3,970 3,965 2,760 22,250

ki =

!


1

F (k/in.) ∆s

20.43 19.62 19.62 39.24 56.55 56.55 113.1 56.55 56.55 113.1 57.71 57.71 115.4 19.42 19.42 38.84 20.42 13.14 30.19 23.32 23.35 23.32 13.14 126.5

k total (k/in.) 20.43

39.24

113.1

113.1

115.4

38.84 20.42

126.5

Notes: 1. 2.

Deflections and forces are based on strength force levels. ∆ S are the design level displacements from Tables 2-5 and 2-6.


111

Design Example 2

3d. 3d

!


Drift check at roof level.

§1630.10.2

To determine drift, the maximum inelastic response displacement ∆ M must be determined. This is defined in §1630.9.2 and computed as follows: ∆ M = 0.7 R∆R S

(30-17)

R = 5.5

Table 16-N

∆ M = 0.7(5.5)∆ S Under §1630.10.2, the calculated story drift using ∆ M shall not exceed 0.025 times the story height for structures having a fundamental period less than 0.7 seconds. The building period for this design example was calculated to be 0.28 seconds, which is less than 0.7 seconds, therefore the 0.025 drift limitation applies. The drift check is summarized in Table 2-8.

Table 2-8. Drift check at roof level

North-South

East-West

Wall

112

A B C E F G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f

∆ S (in.)

Height (ft.)

∆ M (in.)

Max. ∆ M (in.)

Status

0.07 0.16 0.10 0.10 0.07 0.12 0.07 0.21 0.16 0.17 0.17 0.17 0.21

8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21 8.21

0.27 0.62 0.38 0.38 0.27 0.46 0.27 0.81 0.62 0.65 0.65 0.65 0.81

2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46 2.46

ok ok ok ok ok ok ok ok ok ok ok ok ok


Design Example 2

3e.

!


Estimation of third floor level rigidities.

Shear wall rigidities at the third floor are estimated in the same manner as those a the roof. The calculations are summarized in Tables 2-9, 2-10, 2-11, and 2-12. A drift check is not shown.

1

Table 2-9. Tiedown assembly displacements at third floor level ASD Strength Design Tiedown Tiedown AssemblyDisplacement Wall Tiedown Uplift Uplift/1.4(2) Elongation (3) (lb) Device (lb) Crush(5) Slip(6) Shrink(4) (in.) A 135 Strap 190 0.02 0.05 0.02 0.002 B1 4,350 Rod 6,090 0.04 0 0 0.10 B2 4,350 Rod 6,090 0.04 0 0 0.10 C1 2,000 Strap 2,800 0.02 0.05 0.02 0.002 C2 2,000 Strap 2,800 0.02 0.05 0.02 0.002 E1 2,000 Strap 2,800 0.02 0.05 0.02 0.002 E2 2,000 Strap 2,800 0.02 0.05 0.02 0.002 F1 550 Strap 770 0.02 0.05 0.02 0.002 F2 550 Strap 770 0.02 0.05 0.02 0.002 G1 2,800 Rod 3,920 0.02 0 0 0.10 G2 2,800 Rod 3,920 0.02 0 0 0.10 H 135 Strap 190 0.02 0.05 0.02 0.002 1a, 4a 2,275 Strap 3,185 0.02 0.05 0.02 0.002 1b, 4b 0 Not req’d 0 0 0.05 0.02 0 1c, 4c 0 Not req’d 0 0 0.05 0.02 0 1d, 4d 0 Not req’d 0 0 0.05 0.02 0 1e, 4e 0 Not req’d 0 0 0.05 0.02 0 1f, 4f 2,275 Strap 3,185 0.02 0.05 0.02 0.002 2a, 3a 0 Not req’d 0 0 0.05 0.02 0 2b, 3b 0 Not req’d 0 0 0.05 0.02 0 2c, 3c 0 Not req’d 0 0 0.05 0.02 0

da

(7)

(in.) 0.09 0.14 0.14 0.09 0.09 0.09 0.09 0.09 0.09 0.12 0.12 0.09 0.09 0.07 0.07 0.07 0.07 0.09 0.07 0.07 0.07

Notes: 1. 2.

Tiedown assembly displacements for the third floor level are calculated for the tiedowns at the second floor level. Footnotes 2-6, see Table 2-4.


113

Design Example 2

!


Table 2-10. Deflections of shear walls at third floor level in east-west direction Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H

ASD v (plf)

Strength v (plf)

A

(ft)

(in.2)

E (psi)

h

160 400 400

224 560 560

9.43 9.43 9.43

15.7 15.7 15.7

1.7E6 1.7E6 1.7E6

370 370

518 518

9.43 9.43

15.7 15.7

1.7E6 1.7E6

370 370

518 518

9.43 9.43

15.7 15.7

1.7E6 1.7E6

265 265

371 371

9.43 9.43

15.7 15.7

1.7E6 1.7E6

300 300

420 420

9.43 9.43

15.7 15.7

1.7E6 1.7E6

160

224

9.43

15.7

1.7E6

da

∆S

(in.)

(in.)

12.5 11.0 11.0 22.0 21.5 21.5 43.0 21.5 21.5 43.0 21.5 21.5 43.0 11.0 11.0 22.0 12.5

0.0018 0.0097 0.0097

0.09 0.14 0.14

0.13 0.31 0.31

173 173

0.0075 0.0075

0.09 0.09

0.20 0.20

4 4

173 173

0.0075 0.0075

0.09 0.09

0.20 0.20

0.535 0.535

4 4

124 124

0.0025 0.0025

0.09 0.09

0.13 0.13

90,000 90,000

0.535 0.535

4 4

140 140

0.0038 0.0038

0.12 0.12

0.22 0.22

90,000

0.535

6

112

0.0018

0.09

0.13

Space (in)

Vn

G (psi)

t (in.)

90,000 90,000 90,000

0.535 0.535 0.535

6 4 4

112 187 187

90,000 90,000

0.535 0.535

4 4

90,000 90,000

0.535 0.535

90,000 90,000

b (ft)

(lb)

en (in.)

Table 2-11. Deflections of shear walls at the third floor level in north-south direction Wall 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

114

ASD v (plf)

Strength (v) (plf)

h (ft)

(in.2)

E (psi)

b (ft)

G (psi)

355 355 355 355 355 355

497 497 497 497 497 497

9.43 9.43 9.43 9.43 9.43 9.43

15.7 15.7 15.7 15.7 15.7 15.7

1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6

140 140 140

196 196 196

9.43 9.43 9.43

15.7 15.7 15.7

1.7E6 1.7E6 1.7E6

8.0 14.0 11.5 11.5 11.5 8.0 64.5 18.0 24.0 18.0 60.0

A

(in.)

Space (in.

90,000 90,000 90,000 90,000 90,000 90,000

0.535 0.535 0.535 0.535 0.535 0.535

4 4 4 4 4 4

166 166 166 166 166 166

90,000 90,000 90,000

0.535 0.535 0.535

6 6 6

98 98 98

t

Vn (lb)

da

∆S

(in.)

(in.)

0.0066 0.0066 0.0066 0.0066 0.0066 0.0066

0.09 0.07 0.07 0.07 0.07 0.09

0.27 0.20 0.21 0.21 0.21 0.27

0.0012 0.0012 0.0012

0.07 0.07 0.07

0.09 0.08 0.09

en (in)


Design Example 2

Table 2-12. Shear wall rigidities at third floor Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

∆ S (in.) (2) 0.13 0.31 0.31 0.20 0.20 0.20 0.20 0.13 0.13 0.22 0.22 0.13 0.27 0.20 0.21 0.21 0.21 0.27 0.09 0.08 0.09

F (lb) 2,805 6,152 6,153 12,305 11,080 11,080 22,160 11,080 11,080 22,160 7,915 7,915 15,830 4,568 4,567 9,135 2,805 3,965 6,936 5,696 5,696 5,696 3,966 31,955 3,494 4,657 3,494 11,645

ki =

!


1

F (k/in.) ∆s

21.58 19.84 19.84 39.68 55.40 55.40 110.80 55.40 55.40 110.80 60.88 60.88 121.70 20.76 20.76 41.52 21.58 14.68 34.68 27.12 27.12 27.12 14.68 145.40 38.82 58.21 38.82 135.80


39.68

110.80

110.80

121.70

41.52 21.58

145.40

135.80

Notes: 1. 2.

Deflections and forces are based on strength levels. ∆s are the design level displacements form Tables 2-10 and 2-11.


115

Design Example 2

3f. 3f

!


Estimation of second floor level rigidities.

Shear wall rigidities at the second floor level are estimated in the same manner as those for the roof and third floor. The calculations are summarized in Tables 2-13, 2-14, 2-15, and 2-16. A drift check is not shown.

1

Table 2-13. Tiedown assembly displacements at second floor level ASD Strength Design Tiedown Tiedown Assembly Displacement Wall Tiedown Uplift/1.4(2) Uplift (lb) Elongation(3) (lb) Device Crush(5) Slip(6) Shrink(4) (in.) A 1,090 Strap 1,525 0.02 0.01 0.02 0.002 B1 8,600 Rod 12,040 0.15 0 0 0.10 B2 8,600 Rod 12,040 0.15 0 0 0.10 C1 4,380 Rod 6,130 0.08 0 0 0.10 C2 4,380 Rod 6,130 0.08 0 0 0.10 E1 4,380 Rod 6,130 0.08 0 0 0.10 E2 4,380 Rod 6,130 0.08 0 0 0.10 F1 1,565 Rod 2,200 0.03 0 0 0.10 F2 1,565 Rod 2,200 0.03 0 0 0.10 G1 5,700 Rod 7,980 0.10 0 0 0.10 G2 5,700 Rod 7,980 0.10 0 0 0.10 H 1,090 Strap 1,525 0.02 0.01 0.02 0.002 1a, 4a 5,240 Rod 7,340 0.10 0 0 0.10 1b, 4b 0 Not req’d 0 0 0.01 0.02 0 1c, 4c 1,000 Strap 1,400 0.02 0.01 0.02 0.002 1d, 4d 1,000 Strap 1,400 0.02 0.01 0.02 0.002 1e, 4e 1,000 Strap 1,400 0.02 0.01 0.02 0.002 1f, 4f 5,240 Rod 7,340 0.10 0 0 0.10 2a, 3a 0 Not req’d 0 0 0.01 0.02 0 2b, 3b 0 Not req’d 0 0 0.01 0.02 0 2c, 3c 0 Not req’d 0 0 0.01 0.02 0

d a (7) (in.) 0.05 0.25 0.25 0.18 0.18 0.18 0.18 0.13 0.13 0.20 0.20 0.05 0.20 0.03 0.05 0.05 0.05 0.20 0.03 0.03 0.03

Notes: 1. 2.

116

Tiedown assembly displacements for the second floor level are calculated for the tiedowns at the first floor level. See Table 2-4 for footnotes 2-6.


Design Example 2

!


Table 2-14. Deflections of shear walls at the second floor level in east-west direction Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H

ASD v (plf)

Strength v (plf)

A

h (ft)

(in.2)

E (psi)

b (ft) 12.5 11.0 11.0 22.0 21.5 21.5 43.0 21.5 21.5 43.0 21.5 21.5 43.0 11.0 11.0 22.0 12.5

200 500 500

280 700 700

9.43 9.43 9.43

26.2 26.2 26.2

1.7E6 1.7E6 1.7E6

460 460

644 644

9.43 9.43

26.2 26.2

1.7E6 1.7E6

460 460

644 644

9.43 9.43

26.2 26.2

1.7E6 1.7E6

330 330

462 462

9.43 9.43

26.2 26.2

1.7E6 1.7E6

370 370

518 518

9.43 9.43

26.2 26.2

1.7E6 1.7E6

200

280

9.43

26.2

1.7E6

G (psi)

t (in.)

Space (in.)

Vn

en (in.)

(lb)

da

∆S

(in.)

(in.)

90,000 90,000 90,000

0.535 0.535 0.535

6 3 3

140 175 175

0.0038 0.0078 0.0078

0.05 0.25 0.25

0.12 0.42 0.42

90,000 90,000

0.535 0.535

3 3

161 161

0.0060 0.0060

0.18 0.18

0.25 0.25

90,000 90,000

0.535 0.535

3 3

161 161

0.0060 0.0060

0.18 0.18

0.25 0.25

90,000 90,000

0.535 0.535

3 3

115 115

0.0020 0.0020

0.13 0.13

0.16 0.16

90,000 90,000

0.535 0.535

3 3

130 130

0.0030 0.0030

0.20 0.20

0.30 0.30

90,000

0.535

6

140

0.0038

0.05

0.12

Table 2-15. Deflections of shear walls at the second floor level in north-south direction Wall 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

ASD v (plf)

Strength v (plf)

A

Space (in.)

Vn

en

da

∆S

h (ft)

(in.2)

E (psi)

b (ft)

G (psi)

t (in.)

(in.)

(in.)

(in.)

8.0 14.0 11.5 11.5 11.5 8.0 64.5 18.0 24.0 18.0 60.0

90,000 90,000 90,000 90,000 90,000 90,000

0.535 0.535 0.535 0.535 0.535 0.535

4 4 4 4 4 4

191 191 191 191 191 191

0.0104 0.0104 0.0104 0.0104 0.0104 0.0104

0.20 0.03 0.05 0.05 0.05 0.20

0.43 0.21 0.23 0.23 0.23 0.43

90,000 90,000 90,000

0.535 0.535 0.535

6 6 6

147 147 147

0.0044 0.0044 0.0044

0.03 0.03 0.03

0.10 0.10 0.10

410 410 410 410 410 410

574 574 574 574 574 574

9.43 9.43 9.43 9.43 9.43 9.43

26.2 26.2 26.2 26.2 26.2 26.2

1.7E6 1.7E6 1.7E6 1.7E6 1.7E6 1.7E6

210 210 210

294 294 294

9.43 9.43 9.43

26.2 26.2 26.2

1.7E6 1.7E6 1.7E6


(lb)

117

Design Example 2

!


Table 2-16. Wall rigidities at second floor Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

∆ S (2) (in.) 0.12 0.42 0.42 0.25 0.25 0.25 0.25 0.16 0.16 0.30 0.30 0.12 0.43 0.21 0.23 0.23 0.23 0.43 0.10 0.10 0.10

F (lb) 3,485 7,640 7,640 15,280 13,762 13,763 27,525 13,762 13,763 27,525 9,830 9,830 19,660 5,672 5,673 11,345 3,485 4,558 7,978 6,552 6,552 6,552 4,558 36,750 5,221 6,958 5,221 17,400

ki =

1

F (k/in.) ∆S

29.04 18.19 18.19 36.38 55.05 55.05 110.1 55.05 55.05 110.1 61.44 61.44 122.8 18.91 18.91 37.82 29.04 10.60 37.99 28.48 28.48 28.48 10.60 144.6 52.21 69.58 52.21 174.0


36.38

110.1

110.1

122.8

37.82 29.04

144.6

174.0

Notes: 1. 2.

4.

Deflections and forces are based on strength force levels. ∆s are the design level displacements from Tables 2-14 and 2-15.


§1630.6

The base shear was distributed to the three levels in Part 2. In this step, the story forces are distributed to the shear walls supporting each level using the rigid diaphragm assumption. See Part 7 for a later confirmation of this assumption. It has been a common engineering practice to assume flexible diaphragms and distribute loads to shear walls based on tributary areas. This has been done for many years and is a well-established conventional design assumption. In this design example, the rigid diaphragm assumption will be used. This is not intended to imply that seismic design of wood light frame construction in the past should 118


Design Example 2

!


have been necessarily performed in this manner. However, recent earthquakes and testing of wood panel shear walls have indicated that drifts can be considerably higher than what was known or assumed in the past. This knowledge of the increased drifts of short wood panel shear walls and the fact that the diaphragms tend to be much more rigid than the shear walls has increased the need for the engineer to consider the relative rigidities of shear walls. The code requires that the story force at the center of mass to be displaced from the calculated center of mass (CM) a distance of 5 percent of the building dimension at that level perpendicular to the direction of force. This is to account for accidental torsion. The code requires the most severe load combination to be considered and also permits the negative torsional shear to be subtracted from the direct load shear. The net effect of this is to add 5 percent accidental eccentricity to the calculated eccentricity. However, lateral forces must be considered to act in each direction of the two principal axis. This design example does not consider eccentricities between the centers of mass between levels. In this design example, these eccentricities are small and are therefore considered insignificant. The engineer must exercise good engineering judgment in determining when those effects need to be considered. The direct shear force Fv is determined from: R ∑R and the torsional shear force Ft is determined from: Fv = F

Ft = T

Rd J

where: J = ΣRd x 2 + ΣRd y 2 R = shear wall rigidity d = distance from the lateral resisting element (e.g., shear wall) to the center of rigidity (CR)

T = Fe F = 44,500 lb (for roof diaphragm) e = eccentricity


119

Design Example 2

4a. 4a

!


Determine center of rigidity, center of mass, eccentricities for roof diaphragm.

Forces in the east-west (x) direction:

yr =

∑ k xx y ∑ k xx

or y r ∑ k xx = ∑ k xx y

Using the rigidity values k from Table 2-7 and the distance y from line H to the shear wall: y r (20.43 + 39.24 + 113.1 + 113.1 + 115.4 + 38.84 + 20.42 ) = 20.43(116 ) + 39.24(106) + 113.1(82.0) + 113.1(50.0 ) + 115.4(26.0) + 38.84(10.0 ) + 20.42(0) Distance to calculated CR y r =

24,847.3 = 53.9 ft 460.53

The building is symmetrical about the x-axis (Figure 2-6) and the center of mass is determined as: ym =

116.0 = 58.0 ft 2

The minimum 5 percent accidental eccentricity for east-west forces, e y , is computed from the length of the structure perpendicular to the applied story force. e y = (0.05 × 116 ft ) = ±5.8 ft The new y m to the displaced CM = 58.0 ft ± 5.8 ft = 63.8 ft or 52.2 ft The total eccentricity is the distance between the displaced center of mass and the center of rigidity y r = 53.9 ft ∴ e y = 63.8 − 53.9 = 9.9 ft or 52.2 − 53.9 = − 1.7 f t Note that displacing the center of mass 5 percent can result in the CM being on either side of the CR and can produce added torsional shears to all walls.

120


Design Example 2

!


Note that the 5 percent may not be conservative. The contents-to-structure weight ratio can be higher in wood framing than in heavier types of construction. Also, the location of the calculated center of rigidity is less reliable than in other structural systems. Use engineering judgment when selecting the eccentricity e .

Forces in the north-south (y) direction: The building is symmetrical about the y-axis (Figure 2-6). Therefore, the distance to the CM and CR is: xm =

48.0 = 24.0 ft 2

e' x = (0.05)(48 ft ) = ± 2.4 ft Because, the CM and CR locations coincide, e x = e' x ∴ e x = 2.4 ft or − 2.4 ft


121

Design Example 2

!


Figure 2-6. Center of rigidity and location of displaced centers of mass for second and third floor diaphragms

122


Design Example 2

4b. 4b

!


Determine total shears on walls at roof level.

The total shears on the walls at the roof level are the direct shears Fv and the shears due to torsion (combined actual torsion and accidental torsion), Ft . Torsion on the roof diaphragm is computed as follows: Tx = Fe y = 44,500 lb(9.9 ft ) = 440,550 ft - lb for walls A, B, and C or Tx = 44,500 lb(1.7 ft ) = 75,650 ft - lb for walls E, F, G, and H T y = Fe x = 44,500 lb(2.4 ft ) = 106,800 ft - lb Since the building is symmetrical for forces in the north-south direction, the torsional forces can be subtracted for those walls located on the opposite side from the displaced center of mass. The critical force will then be used for the design of these walls. Table 2-17 summarizes the spreadsheet for determining combined forces on the roof level walls.

4c. 4c

Determine the center of rigidity, center of mass, and eccentricities for the third and second floor diaphragms.

Since the walls stack with uniform nailing, it can be assumed that the center of rigidity for the third floor and the second floor diaphragms will coincide with the center of rigidity of the roof diaphragm. Torsion on the third floor diaphragms F = (44,500 + 42,700) = 87,200 lb Tx = Fe y = 87,200 lb(9.9 ft ) = 863,280 ft - lb for walls A, B, and C or 87,200 lb(1.7 ft ) = 148,240 ft - lb for walls E, F, G, and H T y = Fe x = 87,200 lb(2.4 ft ) = 209,280 ft - lb Results for the third floor are summarized in Table 2-18.


123

Design Example 2

!


Torsion on the second floor diaphragms: F = (44,500 + 42,700 + 21,100) = 108,300 lb Tx = Fe y = 108,300 lb(9.9 ft ) = 1,072,170 ft - lb for walls A, B, and C 108,300 lb(1.7 ft ) = 184,110 ft - lb for walls E, F, G, and H

or

T y = Fe x = 108,300 lb(2.4 ft ) = 259,920 ft - lb Results for the second floor are summarized in Table 2-19.

4d. 4d

Comparison of flexible vs. rigid diaphragm results.

Table 2-20 summarizes wall forces determined under the separate flexible and rigid diaphragm analysis. Since nailing requirements were established in the flexible diaphragm analysis of Part 2, they must be checked for results of the rigid diaphragm analysis and adjusted if necessary (also given in Table 2-20).

Table 2-17. Distribution of forces to shear walls below the roof level

East-West

Wall A B C E F G H

North-South

Σ

124

1 4

Σ Σ

Rx

Ry

dx

20.43 39.24 113.10 113.10 115.40 38.84 20.42 460.53 126.5 126.5 253.0

24.0 -24.0

dy

Rd

62.1 52.1 28.1 3.9 27.9 43.9 53.9

1,269 2,044 3,178 441 3,220 1,705 1,101 3,036 -3,036

Direct Force

Torsional Force

Fv

Ft

78,786 106,513 89,305 1,720 89,829 74,853 59,324 500,330

1,970 3,791 10,932 10,932 11,153 3,752 1,970 44,500

+865 +1394 +2167 +52 +377 +200 +129

2,835 5,185 13,099 10,984 11,530 3,952 2,099

72,864 72,864 145,728

22,250 22,250 44,500

+502 -502

22,752 21,748

Rd 2

Total Force

Fv + Ft

646,058


Design Example 2

!


East-West

Table 2-18. Distribution of forces to shear walls below the third floor level

North-South

Total Force

Rd 2

21.58 39.68

62.1 52.1

1,340 2,067

83,221 107,708

4,024 7,399

1,685 2,559

5,709 9,998

110.8 110.8 121.7 41.52 21.58 467.66

28.1 3.9 27.9 43.9 53.9

3,113 432 3,395 1,823 1,163

87,489 1,685 94,732 80,018 62,694 517,547

20,660 20,660 22,692 7,741 4,024 87,200

3,914 93 733 393 251

24,574 20,693 23,425 8,134 4,275

3,490 340 -340 -3,490

83,750 849 849 83,750 169,198

22,544 21,056 21,056 22,544 87,200

1,064 259 -259 -1,064

23,608 21,315 20,797 21,480

A B C E F G H

Σ

Torsional Force

Rd

Rx

1 2 3 4

Ry

Direct Force Fv

dy

Wall

145.4 135.8 135.8 145.4 562.4

Σ Σ

dx

24.0 2.5 -2.5 -24.0

Ft

Fv + Ft

686,745

East-West

Table 2-19. Distribution of forces to shear walls below second floor level Wall

Rx

A B C E F G H

29.04 36.38 110.1 110.1 122.8 37.82 29.04 475.28

North-South

Σ 1 2 3 4

Σ Σ

Ry

144.6 174.0 174.0 144.6 637.2

Direct Force Fv

Torsional Force

Total Force

Ft

Fv + Ft

dy

Rd

Rd 2

62.1 52.1 28.1 3.9 27.9 43.9 53.9

1,803 1,911 3,094 429 3,426 1,660 1,565

111,990 98,750 86,936 1,675 95,589 72,887 84,367 552,194

6,617 8,290 25,088 25,088 27,982 8,618 6,617 108,300

2,682 2,843 4,602 109 875 424 400

9,299 11,133 29,690 25,197 28,857 9,042 7,017

3470 435 -435 -3470

83,290 1,088 1,088 83,290 168,756

24,576 29,574 29,574 24,576 108,300

1,251 157 -157 -1,251

25,827 29,731 29,417 23,325

dx

24.0 2.5 -2.5 -24.0


720,950

125

Design Example 2

!


Table 2-20. Comparison of loads on shear walls using flexible versus rigid diaphragm analysis and recheck of nailing in walls Wall

F flexible

Frigid

(lb)

(lb)

Rigid/ Flexible ratio

A B C E F G H 1 4

1,430 6,280 11,310 11,310 8,080 4,660 1,430 22,250 22,250

2,835 5,185 13,099 10,984 11,530 3,952 2,099 22,752 22,752(3)

+98% -17% +15% -3% +43% -15% +46% +2% +2%

A B C E F G H 1 2 3

2,805 12,305 22,160 22,160 15,830 9,135 2,805 31,955 11,645 11,645

5,709 9,998 24,574 20,693 23,425 8,134 4,275 23,608 21,315 21,315(3)

+103% -18% +11% -7% +48% -11% +52% -26% +83% +83%

4

31,955

23,608(3)

-26%

A B C E F G H 1 2 3 4

3,485 15,280 27,525 27,525 19,660 11,345 3,485 36,750 17,400 17,400 36,750

9,299 11,133 29,690 25,197 28,857 9,042 7,017 25,827 29,731 29,731(3) 25,827(3)

+167% -27% +7% -9% +47% -20% +100% -30% +70% +70% -30%

b (ft) Roof Level 12.5 22.0 43.0 43.0 43.0 22.0 12.5 64.5 64.5 Third Floor 12.5 22.0 43.0 43.0 43.0 22.0 12.5 64.5 60.0 60.0 64.5 Second Floor 12.5 22.0 43.0 43.0 43.0 22.0 12.5 64.5 60.0 60.0 64.5

Fmax (b )1.4

Plywood 1 or 2 sides

Allowable Shear (plf) (1)(2)


165 205 220 190 195 155 120 255 255

1 1 1 1 1 1 1 1 1

340 340 340 340 340 340 340 340 340

6 6 6 6 6 6 6 6 6

330 400 415 370 390 300 245 355 255 255

1 1 1 1 1 1 1 1 1 1

340 510 510 510 510 510 340 510 340 340

6 4(2) 4 4 4 4 6 4 6 6

355

1

510

4

535 500 495 460 480 370 400 410 355 355 410

1 1 1 1 1 1 1 1 1 1 1

510 665 665 665 665 665 510 510 340 340 510

4 3 3 3 3 3 4 4 6(5) 6(5) 4

v=

(plf)

Notes: 1. 2. 3. 4. 5.

Allowable shears from UBC Table 23-II-I-1 Shear walls with shears that exceeds 350 pounds per lineal foot will require 3 × framing at abutting panel edges with staggered nails. See also notes at bottom of Table 1-3. Designates the force used was the higher force for the same wall at the opposite side of the structure. The shear of 535 plf exceeds allowable of 510 plf therefore the nail spacing will need to be decreased to 3 inch spacing. A redesign will not be necessary. The shear of 355 plf exceeds allowable of 340 plf, therefore the nail spacing will need to be decreased to 4-inch spacing. A redesign will not be necessary.

126


Design Example 2

!


Where forces from rigid diaphragm analysis are higher than those from the flexible diaphragm analysis, wall stability and anchorage must be re-evaluated. However, engineering judgment may be used to determine if a complete rigid diaphragm analysis should be repeated due to changes in wall rigidity. If rigid diaphragm loads are used, the diaphragm shears should be rechecked for total load divided by diaphragm length along the individual wall lines.

5.

Determine reliability/redundancy factor ρ.

§1630.1.1

The reliability/redundancy factor penalizes lateral force resisting systems that do not have adequate redundancy. In Part 1 of this example, the reliability/redundancy factor was previously assumed to be ρ = 1.0 . This will now be checked. ρ = 2−

20 rmax AB

(30-3)

where: rmax = the maximum element-story shear ratio. For shear walls, the ratio for the wall with the largest shear per foot at or below two-thirds the height of the building is calculated. Or in the case of a three-story building, the ground level and the second level are calculated (see the SEAOC Blue Book Commentary §C105.1.1.1). The total lateral load in the wall is multiplied by 10 l w and divided by the story shear. l w = length of wall in feet AB = the ground floor area of the structure in square feet ri =

Vmax (10 l w ) F

AB = 5,288 sq ft


127

Design Example 2

!


For ground level.

For east-west direction: Using strength-level forces for wall A: rmax =

(9,299)(10 12.5)

ρ = 2−

108,300 20 0.068 5,288

= 0.068

= −2.0 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, there is no increase in base shear due to lack of reliability/redundancy.

For north-south direction: Using strength-level forces for walls 1 and 4: Load to wall: 36,750 × 11.5 64.5 = 6,552 lb ri =

(6,552 )(10 11.5) 108,300

= 0.053

Note that this is the same as using the whole wall. rmax =

(36,750)(10 64.5)

ρ = 2−

108,300 20 0.053 5,288

= 0.053

= −3.2 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, for both directions there is no increase in base shear required due to lack of reliability/redundancy.

128


Design Example 2

!


For second level.

For east-west direction: Using strength-level forces for wall B: rmax =

(24,574 × 5)(10 21.5) 87,200

ρ = 2−

20 0.065 5,288

= 0.065

= −2.2 < 1.0 minimum o.k.


For north-south direction: Using strength-level forces for walls 1 and 4: rmax =

(31,955)(10 64.5)

ρ = 2−

87,200 20 0.057 5,288

= 0.057

= −2.8 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, there is no increase in base shear due to lack of reliability/redundancy. The SEAOC Seismology Committee added the sentence “The value of the ratio of 10/lw need not be taken as greater than 1.0” in the 1999 SEAOC Blue Book— which will not penalize longer walls, but in this design example has no effect.


129

Design Example 2

6.

!


Determine if structure meets requirements of conventional construction provisions.

While SEAOC is not encouraging the use of conventional construction methods, this step is included because conventional construction is allowed by the UBC (however, it is often misused) and can lead to poor performing structures. The structure must be checked against the individual requirements of §2320, and because it is in Seismic Zone 4, it must also be checked against §2320.5.2. Results of these checks are shown below.

6a. 6a

Floor total loads.

§1230.5.2

The dead load weight of the floor exceeds the limit of 20 psf limit, and therefore the structure requires an engineering design for vertical and lateral forces.

6b. 6b

Braced wall lines.

§2320.5.2

The spacing of braced wall lines exceeds 25 feet on center, and therefore the entire lateral system requires an engineering design. Therefore, the hotel structure requires an engineering design for both vertical and lateral loads. If all walls were drywall and the floor weight was less than 20 psf, then use of conventional construction provisions would be permitted by the UBC. However, conventional construction is not recommended for this type of structure.

7.

Diaphragm deflections to determine if the diaphragm is flexible or rigid.

This step is shown only as a reference for how to calculate horizontal diaphragm deflections. Since the shear wall forces were determined using both flexible and rigid diaphragm assumptions, there is no requirement to verify that the diaphragm is actually rigid or flexible. The roof diaphragm has been selected to illustrate the methodology. The design seismic force in the roof diaphragm using Eq (33-1) must first be determined. The design seismic force is then divided by the diaphragm area to determine the horizontal loading in pounds per square foot. These values are used for determining diaphragm shears (and also collector forces). The design seismic force shall not be less than 0.5C a IW px nor greater than 1.0C a IW px .

130


Design Example 2

7a. 7a

!


Roof diaphragm check.

The roof diaphragm will be checked in two steps. First, the shear in the diaphragm will be determined and compared to allowables. Next, the diaphragm deflection will be calculated. In Part 7b, the diaphragm deflection is used to determine whether the diaphragm is flexible or rigid. Check diaphragm shear: The roof diaphragm consists of 15/32"-thick sheathing with 10d @ 6" o/c and panel edges are unblocked. Loading on the segment between C and E, where: v=

(8.41)48.0' (32.0') = 96 plf 1.4(48.0')2

Diaphragm span = 32.0 ft Diaphragm depth = 48.0 ft Diaphragm shears are converted to allowable stress design by dividing by 1.4 From Table 23-II-H, the allowable shear of 190 plf is based on 15/32-inch APA-rated wood structural panels with unblocked edges and 10d nails spaced at 6 inches on center at boundaries and supported panel edges. APA-rated wood structural panels may be either plywood or oriented strand board (OSB).

Check diaphragm deflection: The code specifies that the deflection is calculated on a unit load basis. In other words, the diaphragm deflection should be based on the same load as the load used for the lateral resisting elements, not F px total force at the level considered. Since the UBC now requires building drifts to be determined by the load combinations of §1612.2 (see Step 4 for additional comments), strength loads on building diaphragm must be determined. The basic equation to determine seismic forces on a diaphragm is shown below. n

F px =

Ft + ∑ Ft i= x

n

∑ wi

w px

(33-1)

i=x


131

Design Example 2

!


where Ft = 0 in this example because T < 0.7 seconds f p roof =

(44.5 × 135.0) = 44.5 k 135.0

For the uppermost level, the above calculation will always produce the same force as computed in Eq (30-15). Then divide by the area of the diaphragm to find the equivalent uniform force. f p roof =

44.5 × 1,000 = 8.41 psf 5,288

In this example, the roof and floor diaphragms spanning between C and E will be used to illustrate the method. The basic code equation to determine the deflection of a diaphragm is shown below. ∆=

5vL3 vL + + 0.188 Le n + 8 EAb 4Gt

∑ (∆ C X )

§23.222, Vol. 3

2b

The above equation is based on a uniformly nailed, simple span diaphragm with panel edges blocked and is based on monotonic tests conducted by the American Plywood Association (APA). The equation has four parts. The first part accounts for beam bending, the second accounts for shear deformation, the third accounts for nail slippage/bending, and the last part accounts for chord slippage. The UBC references this in §2315.1. For the purpose of this design example, the diaphragm is assumed to be a simple span supported at C and E (refer to Figure 2-4). In reality, with continuity, the actual deflection will be less. With nails at 6 inches on center the strength load per nail is 96 × 1.4(6 12 ) = 67 lb/nail = Vn . Other terms in the deflection equation are: L = 32.0 ft b = 48.0 ft G = 50,000 psi

Table 23-2-J Vol. 3

E = 1,700,000 psi A2×4 chords = 5.25 sq in × 2 = 10.50 sq in.

132


Design Example 2

!


Fastener slip/nail deformation values (en ) are obtained as follows: Volume 3 of the UBC uses Table 23-2-K for obtaining nail slip values en , however, its use is somewhat time-consuming, since interpolation and adjustments are necessary. Footnote 1 in Table 23-2-K requires the nail slip values en be decreased 50 percent for seasoned lumber. This means that the table is based on nails being driven into green lumber and the engineer must use half of these values for nails driven in dry (seasoned) lumber. The values in Table 23-2-K are based on tests conducted by the APA. The 50 percent nail slip reduction for dry lumber is a conservative factor. The actual tested slips with dry lumber were less than 50 percent of the green lumber slips. Values for en can be computed based on fastener slip equations from Table B-4 of APA Research Report 138. This will save time, be more accurate, and also enable computations to be made by a computer. Using the values of en from Volume 3 of UBC requires interpolation and is very time-consuming. For 10d common nails, there are 2 basic equations: When the nails are driven into green lumber: en = (Vn 977 )1.894

APA Table B-4

When the nails are driven into dry lumber: en = (Vn 769 )3.276

APA Table B-4

where: Vn is the fastener load in pounds per fastener These values are based on Structural I sheathing and must be increased by 20 percent when the sheathing is not Structural I. Footnote a in UBC Table B-4 states “Fabricated green/tested dry (seasoned)…” is very misleading. The values in the table are actually green values, since the lumber is fabricated when green. Again, don’t be misled by the word “seasoned.” en = 1.20(67 769)3.276 = 0.0004 t = 0.298 in. (for CDX or Standard Grade)

Table 23-2-H

Assume chord-splice at the mid-span of the diaphragm that will be nailed. The allowable loads for fasteners are based on limit state design. In other words, the deformation is set at a limit rather than the strength of the fastener. The deformation limit is 0.05 diameters of the fastener. For a 16d nail, a conservative slippage of 0.01 inch will be used.


133

Design Example 2

!


Using strength level diaphragm shear:

∑ (∆ C X ) = (0.01)16.0 ft (2 ) = 0.32 in. - ft ∆=

5 (96 × 1.4 )32.0 3 96 × 1.4 (32.0 ) 0.32 + + 0.188 (32.0 )0.0004 + = 0.08 in. 2 (48.0) 8(1.7 E 6)10.50 (48.0) 4 (50,000 )0.298

This deflection is based on a blocked diaphragm. The UBC does not have a formula for an unblocked diaphragm. The APA is currently working on a simplified formula for unblocked diaphragms. Based on diaphragm deflection test results (performed by the APA), an unblocked diaphragm will deflect between 2 to 2½ times that of a blocked diaphragm or can be proportioned to the allowable shears of a blocked diaphragm divided by the unblocked diaphragm. The roof diaphragm is also sloped at 6:12, which is believed to increase the deflection (but this has not been confirmed with tests). This design example has unblocked panel edges for the floor and roof diaphragms, so a conversion factor is necessary. This conversion is for the roof diaphragm. The floors will similarly neglect the stiffening effects of lightweight concrete fill and gluing of sheathing. It is assumed that the unblocked diaphragm will deflect: ∴ ∆ = 0.08(2.5) = 0.20 in.

7b. 7b

Flexible versus rigid diaphragms.

§1630.6

In this example, the maximum diaphragm deflection was estimated as 0.20 inches. This assumes a simple span for the diaphragm, and the actual deflection would probably be less. The average story drift is on the order of 0.10 inches at the roof (see Step 3c for the computed deflections of the shear walls). For the diaphragms to be considered flexible, the maximum diaphragm deflection will have to be more than two times the average story drift. This is right at the limit of a definition of a flexible diaphragm. The other diaphragm spans would easily qualify as “rigid” diaphragms. As defined by the code, the diaphragms in this design example are considered rigid. In reality, some amount of diaphragm deformation will occur, and the true analysis is highly complex and beyond the scope of what is normally done for this type of construction. Diaphragm deflection analysis and testing has been performed on level/flat diaphragms. There has not been any testing of sloped and complicated diaphragms, as found in the typical wood framed structure. Therefore, some engineers perform their design based on the roof diaphragm as flexible and the floor diaphragms as rigid. In using this procedure, the engineer should exercise good engineering judgment in determining if the higher load of the two methodologies is actually required. For example, if the load to two walls by rigidity analysis is found to be 5 percent to line 134


Design Example 2

!


A and 95 percent to line B, but by flexible analysis it is found to be 50 percent to line A and 50 percent to line B, the engineer should probably design for the larger of the two loads for the individual walls. Note that though the same definition of a flexible diaphragm has been in the UBC since the 1988 edition, it has not been enforced by building officials for Type V construction. The draft of the IBC 2000 has repeated this same definition into Chapter 23 (wood) definitions.

8.

Tiedown forces for the shear wall on line C.

Tiedowns are required to resist the uplift tendency on shear walls caused by overturning moments. In this step, tiedown forces for the three-story shear wall on line C are determined. The design chosen uses continuous tiedowns below the third floor. At the third floor, conventional premanufactured straps are used. Not included in this design example, but it should be noted: the code has two new provisions for one-hour wall assemblies—Footnotes 17 and 18 of Table 7-B in Volume 1. Footnote 17 requires longer fasteners for gypsum sheathing when the sheathing is applied over wood structural panels. Footnote 18 requires values for F ' c to be reduced to 78 percent of allowable in one-hour walls.

8a. 8a

Discussion on continuous tiedown systems.

The continuous tiedown system is a relatively new method for resisting shear wall overturning. Similar to the many metal connectors used for wood framing connections, most are proprietary and have ICBO approvals. All of the systems have some type of rod and hardware connector system that goes from the foundation to the top of the structure. A common misconception that engineers have with these types of systems is that the elongation of the rod will produce large displacements in the shear walls. Contrary to that perception, these systems are in many instances superior to the one-sided bolted tiedowns. Investigations after the Northridge earthquake as well as independent testing of the conventional one-sided bolted tiedowns, have concluded that there can be large displacements associated with this type of connection. The large displacements are a result of eccentricity with the boundary element, deflection of the tiedown, wood shrinkage, wood crushing, and oversized holes for the through-bolts. Some of the proprietary systems compensate for shrinkage either by pre-tensioning of the rod or by a self-ratcheting connector device. Shrinkage-compensating devices are desirable in multi-level wood frame construction. These devices will also compensate for other slack in the tiedown system caused by crushing of plates, seating of posts, studs, etc.


135

Design Example 2

8b. 8b

!


Determine strength shear wall forces.

The shear wall on line C is shown on Figure 2-7. Forces at each story are determined as follows (from Table 2-20): Froof = 13,099 2 = 6,550 lb Fthird = (24,574 − 13,099 ) 2 = 5,738 lb Fsecond = (29,690 − 24,574 ) 2 = 2,558 lb

Figure 2-7. Shear wall C elevation

136


Design Example 2

!


The distance between the centroid of the boundary forces that represent the overturning moment at each level must be estimated. This is shown below. e

=

the distance to the center of tiedown rod and boundary studs or collectors studs (Figure 2-12)

e

=

2 × 2.5 in. + (13 2 ) = 11.5 in. = 0.958 in.

Use

e

=

1.0 ft

d=

the distance between centroids of the tiedown and the boundary studs, in feet. (Note that it is also considered acceptable to use the distance from the end of the shear wall to the centroid of the tiedown.)

d

=

21.5 ft − 2(1.0 ft ) = 19.5 ft at second floor for third level (Figure 2-12)

d

=

21.5 ft − (2 × 0.125) = 21.25 ft at third floor for roof level (Figure 2-11)

The resisting moment M R is determined from the following loads: Wroof = 13.5 psf (2.0 ft ) = 27.0 plf W floor = 25.0 psf (2.0 ft ) = 50.0 plf Wwall = 10.0 plf

Table 2-21. Tiedown forces for shear wall C Level Roof Third Second

M OT (ft-lb)

MR (ft-lb)

53,775 169,774 309,920

25,216 58,590 91,965

M R × 0.9 (1) (ft-lb) 22,694 52,731 82,769

(M OT

Uplift 1.4 ) − 0.9 M R (lb) d 740 3,515 7,110

Differential Load (2) (lb) 740 2,775 3,595

Notes: 1. 2.

The UBC no longer has the 0.85 DL provision for stability, this has been replaced with the basic load combinations of §1612.3.1. The differential is the load difference between the uplift force at level x and the level above.


137

Design Example 2

9.

!


Design tiedown connection at the third floor for the shear wall on line C.

Figure 2-11 illustrates the typical tiedown connection for the shear wall on line C at the third floor. This is the conventional pre-manufactured strap and is fastened to the framing with nails. The total uplift force at this level is 740 lb. P 1= 740 lb The tiedowns will be designed using allowable stress design.

§1612.3

The basic load combinations of §1612.2.1 do not permit stress increases. The alternate basic load combinations of §1612.2.2, however, do permit stress increases. E The Errata to the first printing of the code added 09 D ± , Eq. (12-16-1), to the 1.4 alternate basic load combinations. This exact same load combination is listed in the basic load combinations. This is confusing to many engineers on this topic, because the basic load combinations are based on duration factors (see 1999 SEAOC Blue Book Commentary, §C101.7.3 for further explanation). This design example will use the one-third stress increase of the alternate basic load combination method. With a 16-gauge × 1.25 -in strap and 10d common nails. Allowable load per nail is ZC D = 113(1.33) = 150 lb/nail

NDS Table 12.3F

Number of nails required = 740 150 = 4.9 ∴use 5 With nails at 1.5 inches on center the length of strap required is 2(0.75 in. + 5 × 1.5 in.) + 6 in. = 22.5 in. ∴use 24-inch-strap

10. 10

Design tiedown connection at the second floor for the shear wall on line C.

As previously mentioned, the second floor tiedown will be part of the continuous tiedown system used below the third level. Refer to Figure 2-12 for illustration of this system and the location of forces P1 , P2 , and P3 . The total uplift force at the second floor is 3,515 lb (Table 2-21). P1 = P2 = total uplift force from above = 740 lb

138


Design Example 2

!


P3 = uplift force for the collector studs = differential load/2 = 2,775 lb/2 = 1,388 lb Since the strap from above is only connected to one pair of collector studs, the total uplift force for the outside set of collectors is equal to the uplift force plus the uplift force on the second floor shear wall from the third floor. Taking a free-body diagram of the system, the tension in the tiedown rod is increased due to cantilever action between the centroids of the forces. A downward component is actually applied to the interior-most support stud (Figure 2-8):

Figure 2-8. Free-body force diagram of compression bridge

Next, the tension in the tiedown rod between the second floor and the compression bridge is the differential load plus the tension load, as computed above. This will produce the total force P2 on support stud (Figure 2-9):

Figure 2-9. Free-body force diagram of compression bridge


139

Design Example 2

!


Determine spacing for the flat nailing: Pmax = 2,028 lb The allowable lateral load for a 16d common nail in a 1½-inch side member is: ZC D = 141(1.33) = 187 lb

NDS Table 12.3B

With 2 rows of 16d nails, the number of nails per row is 2,028 lb 2 × 187 = 5.4 nails ∴use 6 nails Maximum spacing = 48 in (6 + 1) = 6.8 in. ∴Use 6-inch o.c. for the flat nailing Check compression perpendicular to grain for the bridge support studs to compression bridge: Critical at P2 f c max = 2,028 lb (1.5 × 3.5) = 386 psi < Fc ⊥ = 625 psi o.k.

NDS Supp. Table 4A

Check the bearing perpendicular to grain on bearing plate: F = T1 = 4,255 lb f c ⊥ = 4,255 lb 3.25 × 5.0 = 262 psi < Fc ⊥ = 625 psi

o.k.

Check bearing perpendicular to grain on the top plate from the collector studs from below: First floor is framed with 3 × 4 studs Force at P3 = 1,388 lb f c ⊥ = P A = 1,388 lb (2.5 × 3.5) = 160 psi < Fc ⊥ = 625 psi

o.k.

Check shear on 4 × 8 compression bridge (assume tiedown is at center of wall and not at party wall, see Figure 2-12): T1 = 4,255 lb

140


Design Example 2

!


Assuming compression bridge to take all shear: V =

T1 4,255 = = 2,130 lb 2 2

fV =

2,130 × 1.5 = 126 psi 3.5 × 7.25

For Douglas Fir-Larch No. 1: FV ' = FV C D = 95 × 1.33 = 126 psi o.k. Check bending on 4 × 8 compression bridge: T1 = 4,255 lb T1 × L 4,255 × (10 + 1.5) = = 12,235 in. - lb 4 4 S x for 4 × 8 with hole for 5 8" rod = (3.5 − 0.69)7.25 2 6 = 24.6 in 3 M=

fb =

M 12,235 = = 497 psi S 24.6

For Douglas Fir Larch No. 1: Fb ' = Fb C D C F = 1,000(1.33)(1.3) = 1,729 psi o.k. Check shear on plates at floor: Tiedown connector reaction is the differential load, which is 3,595 lb. T = 3,595 lb Assuming 2 sill plates and 2 top plates to take all shear: V =

T 3,595 = = 1,800 lb 2 2

fV =

1,800 × 1.5 = 130 psi 4(1.5 × 3.5)


141

Design Example 2

!


Since plate have no spits C H = 2.0 (plates rarely check on the edges) FV ' = FV C H C D = 95(2.0 )(1.33) = 252 psi o.k. Therefore, the tiedown connection shown on Figure 2-12 meets the requirements of code.

11. 11

Design tiedown connection and anchor bolt spacing for shear wall on line C.

11a. 11a

Design anchor bolt spacing of sill plate on Line C.

See discussion about fasteners for pressure-preservative treated wood and in Step 19. From Table 2-20: V = 29,690 lb v=

V 29,690 lb = = 690 lb/ft L 43 ft

The 1997 UBC references the 1991 NDS, which specifies in §8.2.3 that the allowable bolt design value, Z , is equal to t m = Z ts = twice the thickness of wood member. The problem is, there aren’t any tables for 6x to 6x members, leaving only the Z formulas. In lieu of using the complex Z formulas, an easier method would be to use the new tables in the 1997 NDS, which are specifically for ledgers and sill plates. For a side member, thickness = 2.5 inch in Hem-Fir wood (note that designing for Hem-Fir will require a tighter nail and bolt spacing): Z11 = 1,350 lb/bolt Table 8.2E 97 NDS Z C (1,350)(1.33)(1.4 ) = 3.6 ft = 43 in. Required spacing = 11 D = v 690 where 1.4 is the strength conversion factor ∴Use ¾" diameter bolts at 32 inches on center.

142


Design Example 2

11b. 11b

!


Determine tiedown anchor embedment.

In this calculation, the tiedown anchor will be assumed to occur at the center of the exterior wall. This will produce a lower capacity than if the rod were located at the double-framed wall shown in Figure 2-13. From Table 2-21: T = 7,110 lb T y = 7,110 × 1.4 × 1.3 = 13,000 lb where 1.4 is the strength conversion factor and 1.3 is for special inspection per §1923.2. Neglecting the area of bolt head bearing surface, the effective area A p

( )

of the projected (Figure 2-10), assumed concrete failure surface is:

(A p ) = πl2e

2

+ 1.75(l e )2

For l e = 15 in. A p = 406 in. 2 ΦPC = Φλ 4 A p

f ' c = 0.65 × 1.0 × 4 × 406 3,000 = 57.8 k

PSS = 0.9 × 0.307 × 60,000 = 16,580 lb > 13,000 lb (critical) Provide an oversized hole for the tiedown rod in the foundation sill plate. The rod has no nut or washer to the sill plate, therefore, assume V = 0 lb in the rod. Tiedown bolts resist vertical loads only, anchor bolts are designed to resist the lateral loads.

11c. 11c

Check the bearing perpendicular to grain on sill plates.

Assuming all compressive force for overturning will be resisted by end boundary elements, the critical load combination is:  E  D+ L+   1.4 


(12-13)

143

Design Example 2

!


From Table 2-21, the strength level overturning moment is: M OT = 309,920 ft - lb The seismic compressive force is obtained by dividing by the distance d. Conversion to allowable stress design is obtained by dividing by 1.4. Pseismic =

[

M OT 309,920 = = 11,350 lb d (1.4) 19.5(1.4 )

(

)

]

PDL = Wroof + W floor + Wwall (27 ft )

 16"+8"  PDL = [27.0 + 2 (50.0 ) + 10.0(27' )]  = 795 lb  12"   16"+8"  PLL = (40 psf × 2'×2 )  = 320 lb  12" 

∑ P = 11,350 + 795 + 320 = 12,465 lb with full width bearing studs bearing on both sill plates (Figure 2-13), the bearing area is equal to six 3x4 studs. fc max =

12,465 = 240 psi < Fc'⊥ = 626 psi 6(8.75)

o.k.

where the area of a 3 × 4 is 8.75 square inches. Note that if a Hem-Fir sill plate is used the allowable compression perpendicular to grain Fc'⊥ = 405 psi . f c < 0.73Fc'⊥ = 0.73(405) = 295 psi

NDS Supp. Table 4A

Therefore, the assumed crushing effect of 0.02 inches (Table 2-13) is correct. This crushing will be compensated by the ratcheting effect of the continuous tiedown system as discussed in the notes for Table 2-4.

144


Design Example 2

!


Figure 2-10. Tiedown bolt


145

Design Example 2

12. 12

!


Detail of tiedown connection at the third floor for shear wall on line C.

Note that since the boundary element is a double stud and the wall panel edge nailing is nailed to the end stud, the 16d at 12 inches o.c. internailing of the two tiedown studs should have the capacity to transfer one-half the force to the interior stud (Figure 2-11). These nails may be installed from either side (normally nailed from the outside). See Figure 2-16 for the location of the top plates and commentary about plate locations.

Figure 2-11. Tiedown connection at the third floor for shear wall C.

146


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13. 13

!


Detail of tiedown connection at the second floor for shear wall on line C.

This tiedown rod system (Figure 2-12) may also be extended to the third floor instead of using the conventional metal strap shown in Figure 2-11. See Figure 2-16 for the location of the top plates and commentary about plate locations.

Figure 2-12. Tiedown connection at second floor for shear wall C SEAOC Seismic Design Manual, Vol. II (1997 UBC)

147

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14. 14

!


Detail of wall intersection at exterior walls.

The detail shows full-width studs at tiedown (Figure 2-13). This is desirable when sheathing is applied to both stud walls. It is also desirable for bearing perpendicular to grain because the bearing area is doubled. When full-width studs are used for bearing, both sill plates will need to be 3x thickness (not as shown in Figure 2-17). Tiedowns may be located at the center of the stud wall that is also sheathed. It is good practice to tie the wall together. In this case, there is no design requirement or minimum shear wall to shear wall connection requirement other than that required by the UBC standard nailing schedule.

Figure 2-13. Wall intersection at shear wall (plan view)

148


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15. 15

!


Detail of tiedown connection at foundation.

The manufacturer of the tiedown system usually requires the engineer of record to specify the tiedown forces at each level of the structure. This can easily be done in a schedule (Figure 2-14).

Figure 2-14. Tiedown connection at foundation


149

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16. 16

!


Detail of shear transfer at interior shear wall at roof.

Note: Edge nailing from roof sheathing to collector truss may need to be closer than the roof sheathing edge nailing due to shears being collected from each side of the truss. It is also common to use a double collector truss at these locations. The 2 × 4 braces at the top of the shear wall need to be designed for compression or provide tension bracing on each side of the wall (Figure 2-15).

Figure 2-15. Shear transfer at interior shear wall at roof

150


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17. 17

!


Detail of shear transfer at interior shear wall at floors.

This detail uses the double top plates at the underside of the floor sheathing (Figure 2-16). This is advantageous for shear transfer. Another detail that is often used is to bear the floor joists directly on the top plates. However, when the floor joist is on top of the top plates, shear transfer is required through the glue joint in the webs and heavy nailing from the joist chord to the top plate.

Figure 2-16. Shear transfer at interior shear wall at floor

Note: The nailers for the drywall ceiling need to be installed after the wall sheathing and wall drywall have been installed.


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18. 18

!


Detail of shear transfer at interior shear wall at foundation.

Figure 2-17. Shear transfer at foundation

19. 19

Detail of sill plate at foundation edge.

Fasteners for pressure- or preservative-treated wood.

Sections 2304.3 and 1811.3 of the 1997 UBC added a new requirement for corrosion-resistant fasteners. Although it does not appear to be the intent of the provision, a literal interpretation of the section would require hot-dipped zinc-coated galvanized nails and anchor bolts. The code change was proposed by the wood industry, and §2304.3 is from a report in the wood handbook by the Forest Products Lab, where fasteners were found to react with the preservative treatment when “… in the presence of moisture….” However, it is uncertain whether a sill plate in a finished “dried-in” building is “in the presence of moisture.” This can create a construction problem because hot-dipped zinc coated nails have to be hand-driven, requiring the framer to put down his nail gun and change nailing procedures.

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An additional caution for sill plates is the type of wood used. The most common species used on the west coast for pressure treatment is Hem-Fir, which has lower fastener values for nails and bolts than for Douglas-Fir-Larch. A tighter nail spacing to the sill plate is necessary, or a double stagger row can be used. Figure 2-18 shows two rows of edge nailing to the sill plate as a method of compensating for a Hem-Fir sill plate.

Gap at bottom of sheathing.

Investigations into wood-framed construction have found that plywood or oriented strand board sheathing that bear on concrete at perimeter exterior edges can “wick” moisture up from the concrete and cause corrosion of the fasteners and rotting in the sheathing. To help prevent this problem, the sheathing can be placed with a gap above the concrete surface. A ¼-inch gap is recommended for a 3x sill plate and an 1/8-inch gap is recommended for a 2x sill plate (Figure 2-18).

Figure 2-18. Sill plate at foundation edge

Note: The UBC only requires a minimum edge distance of 3/8-inch for nails in sheathing. Tests have shown that sheathing with greater edge distances have performed better.


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Detail of shear transfer at exterior wall at roof.

Figure 2-19. Shear transfer at exterior wall at roof

Note: The roof truss directly above the exterior wall is also a “collector” truss. Roof edge nailing to this truss and the 16d nails to the blocking need to be checked for the “collector” load. Double top plates are also a chord and collector.

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21. 21

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Detail of shear transfer at exterior wall at floor.

Figure 2-20. Shear transfer at exterior wall at floor

Note: This detail uses double top plates at the underside of the floor sheathing. Another detail that is often used is bearing the floor joists on the double top plates. See Figure 2-16 for additional commentary.


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References American Forest and Paper Association, 1996, Wood Construction Manual. American Forest and Paper Association, Washington D.C. American Plywood Association, 1997, Design/ Construction Guide – Diaphragms and Shear Walls. Report 105, Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1997, Diaphragms and Shear Walls. Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1993, revised, Wood Structural Panel Shear Walls. Report 154, Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1994, Northridge, California Earthquake. Report T-94-5. Engineered Wood Association, Tacoma, Washington. American Plywood Association, Performance Standards and Policies for Structural– Use Panels [Sheathing Standard, Sec. 2.3.3]. Standard PRP–108. Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1988, Plywood Diaphragms, Research Report 138. American Plywood Association, Tacoma, Washington. Applied Technology Council, 1995, Cyclic Testing of Narrow Plywood Shear Walls ATC R-1. Applied Technology Council, Redwood City, California. Applied Technology Council, 1981, Guidelines for Design of Horizontal Wood Diaphragms, ATC-7. Applied Technology Council, Redwood City, California. Applied Technology Council, 1980, Proceedings of a Workshop on Design of Horizontal Wood Diaphragms, ATC-7-1. Applied Technology Council, Redwood City, California. Building Seismic Safety Council, 1997, National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings. Building Seismic Safety Council, Washington D.C. Bugni, David A., 1999, “A Linear Elastic Dynamic Analysis of a Timber Framed Structure.” Building Standards, International Conference of Building Officials, Whittier, California

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Cobeen, K.E., 1996, “Performance Based Design of Wood Structures.” Proceeding: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Coil, J., 1999, “Seismic Retrofit of an Existing Multi-Story Wood Frame Structure,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Commins, A. and Gregg, R., 1996, Effect of Hold Downs and Stud-Frame Systems on the Cyclic Behavior of Wood Shear Walls, Simpson Strong-Tie Co., Pleasanton, California. Countryman, D., and Col Benson, 1954, 1954 Horizontal Plywood Diaphragm Tests. Laboratory Report 63, Douglas Fir Plywood Association, Tacoma Washington. CUREe, 1999, Proceedings of the Workshop on Seismic Testing, Analysis, and Design of Wood Frame Construction. California University for Research in Earthquake Engineering. Dolan, J.D., 1996, Experimental Results from Cyclic Racking Tests of Wood Shear Walls with Openings. Timber Engineering Report No. TE- 1996-001. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J. D. and Heine , C.P., 1997a, Monotonic Tests of Wood Frame Shear Walls with Various Openings and Base Restraint Configurations. Timber Engineering Report No. TE-1997-001, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D. and Heine, C.P., 1997b, Sequential Phased Displacement Cyclic Tests of Wood Frame Shear Walls with Various Openings and Base Restrain Configurations. Timber Engineering Report No. TE-1997-002, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D., and Heine, C.P., 1997c, Sequential Phased Displacement Test of Wood Frame Shear Walls with Corners. Timber Engineering Report No. TE-1997-003, Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Earthquake Engineering Research Institute, 1996, “Northridge Earthquake of January 17, 1994,” Reconnaissance Report, Earthquake Spectra. Vol. 11, Supplement C. Earthquake Engineering Research Institute, Oakland, California. Faherty, Keith F., and Williamson, Thomas G., 1995, Wood Engineering Construction Handbook. McGraw Hill, Washington D.C.


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Federal Emergency Management Agency, 1998, National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings. Federal Emergency Management Agency, Washington D.C. Ficcadenti, S.K., T.A. Castle, D.A. Sandercock, and R.K. Kazanjy, 1996, “Laboratory Testing to Investigate Pneumatically Driven Box Nails for the Edge Nailing of 3/8" Plywood Shear Walls,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Foliente, Greg C., 1994, Analysis, Design and Testing of Timber Structures Under Seismic Loads. University of California Forest Products Laboratory, Richmond, California. Foliente, Greg C., 1997, Earthquake Performance and Safety of Timber Structures. Forest Products Society, Madison Wisconsin. Forest Products Laboratory, 1999, Wood Handbook Publication FPL – GTR- 113. Madison, Wisconsin. Goers R. and Associates, 1976, A Methodology for Seismic Design and Construction of Single-Family Dwellings. Applied Technology Council, Redwood City, California. International Code Council, 1999, International Building Code – Final Draft, 2000. International Code Council, Birmingham, Alabama. Ju, S. and Lin, M. ,1999, “Comparison of Building Analysis Assuming Rigid or Flexible Floors,” Journal of Structural Engineering. American Society of Civil Engineers, Washington, D.C. Mendes, S., 1987, “Rigid versus Flexible: Inappropriate Assumptions Can Cause Shear Wall Failures!” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Mendes, S., 1995, “Lessons Learned From Four Earthquake Damaged Multi-Story Type V Structures,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. NFPA, 1991a, National Design Specification for Wood Construction. National Forest Products Association, Washington D.C. NFPA, 1997b, National Design Specification for Wood Construction. Natural Forest Products Association, Washington D.C.

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Rose, J. D., 1998, Preliminary Testing of Wood Structural Panel Shear Walls Under Cyclic (Reversed) Loading. Research Report 158, APA – Engineered Wood Association, Tacoma, Washington. Rose, J. .D., and E.L. Keith, P. E., 1996, Wood Structural Panel Shear Walls with Gypsum Wallboard and Window [ Sheathing Standard, Sec. 2.3.3 ]. Research Report 158. APA - The Engineered Wood Association, Tacoma Washington. SEAOC, 1997, Seismic Detailing Examples for Engineered Light Frame Timber Construction. Structural Engineers Association of California, Sacramento, California. SEAOC, 1999, Guidelines for Diaphragms and Shear Walls. Structural Engineers Association of California, Sacramento, California. SEAOC, 1999, Plan Review – Codes and Practice. Structural Engineers Association of California, Sacramento, California. Shipp, J., 1992, Timber Design. Volumes IV and V. Professional Engineering Development Publications, Inc., Huntington Beach, California. Steinbrugge, J., 1994, “Standard of Care in Structural Engineering Wood Frame Multiple Housing,” Proceedings: Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California.


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Design Example 3

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Cold-Formed Steel Light Frame Three-Story Structure

Design Example 3 Cold-Formed Steel Light Frame Three-Story Structure

Figure 3-1. Cold-formed light frame three-story structure elevation

Foreword The building in this example has cold-formed light-gauge steel framing, and shear walls and diaphragms that are sheathed with wood structural panels. This example presents a new approach to the seismic design of this type of building. This is because the past and present California design practice in seismic design of light framed structures has almost exclusively considered flexible diaphragms assumptions when determining shear distribution to shear walls. However, since the 1988 UBC, there has been a definition in the code (§1630.6 of the 1997 UBC) that defines diaphragm flexibility. The application of this definition often requires the use of the rigid diaphragm assumption, and calculation of shear wall rigidities for distribution to shear walls. While the latter is rigorous and complies with the letter of the code, it does not reflect present-day practice. In actual practice, for reasons of simplicity and precedence, many structural engineers routinely use the flexible diaphragm assumption.


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A rigid diaphragm analysis is recommended where the shear walls can be judged by observation to be flexible compared to the diaphragm, and particularly where one or more lines of either shear walls, moment frames, or cantilever columns are more flexible than the rest of the shear walls. This design example has floor diaphragms with lightweight concrete fill over the floor sheathing (for sound insulation), making the diaphragms significantly stiffer than that determined using the standard UBC diaphragm deflection equations. Before beginning design, users of this Manual should check with the local jurisdiction regarding the level of analysis required for cold-formed light framed structures.

Overview This design example illustrates the seismic design of a three-story cold-formed (i.e., light-gauge) steel structure. The structure is shown in Figures 3-2, 3-3 and 3-4. The building in this example is the same as in Design Example 2, with the exception that light-gauge metal framing is used in lieu of wood. The structure has wood structural panel shear walls, and roof and floor diaphragms. The roofs have composite shingles over the wood panel sheathing that is supported by light-gauge metal trusses. The floors have 1½ inches of lightweight concrete fill and are framed with metal joists. The following steps illustrate a detailed analysis of some of the important seismic requirements of the 1997 UBC. As stated in the introduction of the manual, this example is not a complete building design. Many aspects have not been included, and only selected steps of the seismic design have been illustrated. As is common for Type V construction (see UBC §606), a complete wind design is also necessary, but is not given here. Although code requirements recognize only two diaphragm categories, flexible and rigid, the diaphragms in this example are judged to be semi-rigid due to the fact that the diaphragms do deflect. The code also requires only one type of analysis, flexible or rigid. The analysis in this design example will use the envelope method. The envelope method considers the worst loading condition from both flexible and rigid diaphragm analyses to determine the design load on each shear-resisting element. It should be noted that the envelope method is not a code requirement, but is deemed appropriate for this design example, because neither flexible nor rigid diaphragm analysis may accurately model the structure.

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Outline This example will illustrate the following parts of the design process. 1.


2.


3.


4.


5.

Tiedown forces for shear wall on line C.

6.

Allowable shear and nominal strength of No. 10 screws.

7.

Tiedown connection at third floor for wall on line C.

8.

Tiedown connection at the second floor for shear wall on line C.

9.

Boundary studs for first floor wall on line C.

10. 10

Shear transfer at second floor on line C.

11. 11

Shear transfer at foundation for walls on line C.

12. 12

Shear transfer at roof at line C.

Given Information Roof weights ( slope 6:12 ): Roofing 3.5 psf ½" sheathing 1.5 Trusses 3.5 Insulation 1.5 Miscellaneous 0.7 Gyp ceiling 2.8 DL (along slope ) 13.5 psf

Floor weights: Flooring Lt. wt. concrete 5/8" sheathing Floor Framing Miscellaneous Gyp ceiling

1.0 psf 14.0 1.8 5.0 0.4 2.8 25.0 psf

DL (horiz. proj.) = 13.5 (3.41/12) = 15.1 psf Stair landings do not have lightweight concrete fill Area of floor plan is 5,288 sq ft


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Weights of respective diaphragm levels, including tributary exterior and interior walls: Wroof = 135,000 lb W3rd floor = 230,000 lb W2 nd floor = 230,000 lb = 595,000 lb The same roof, floor, and wall weights used in Design Example 2 are also used in this example. This has been done to better illustrate a side-by-side comparison of cold-formed light-gauge steel construction with the more traditional wood frame construction used in Design Example 2. This side-by-side comparison has been done so that the engineer can have a better “feel” for the similarities and differences between structures with wood studs and structures with cold-formed metal studs. It should be noted that roof, floor, and wall weights for light-gauge steel framed structures are typically lighter than similar structures constructed of wood framing. Because of light-gauge steel framed structures being lighter, a more accurate estimate of building weight for this structure would be about 560 kips instead of the 595 kips used in this example. Consequently, wall shears and overturning forces would be reduced accordingly. Weights of diaphragms are typically determined by taking one-half height of walls at the third floor to the roof and full height of walls for the third and second floors diaphragms. Wall framing is ASTM A653, grade 33'-4" × 18-gauge metal studs at 16 inches on center. These have a 1-5/8-inch flange with a 3/8-inch return lip. The ratio of tensile strength to yield point is at least 1.08. Studs are painted with primer. ASTM A653 steel is one of three ASTM steel specifications used in light frame steel construction. The others are A792 and A875. The difference between the specifications are primarily the coatings which are galvanized, 55 percent aluminum-zinc (A792), and zinc-5 percent aluminum (A875) respectively. The recommended minimum coating classifications are G60, AZ50 and GF60 respectively. It should be noted that the studs do not require painting with primer. It should be noted that the changing stud sizes or thickness of studs at various story heights is common (as is done in wood construction). The thickness of studs and tracks should be identified by visible means such as coloring or metal stamping of gauges/sizes on studs and tracks. APA-rated wood structural panels for shear walls will be 15/32-inch-thick Structural I, 32/16-span rating, 5-ply with Exposure I glue is specified, however 4-ply is also acceptable. 164


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Framing screws are No. 8 by 5/8-inch wafer head self-drilling with a minimum head diameter of 0.292-inch, as required by footnote 2 of Table 22-VIII-C of the UBC. The roof is 15/32-inch thick APA-rated sheathing, 32/16-span rating with Exposure I glue. The floor is 19/32-inch thick APA-rated Sturd-I-Floor 24" o/c rating (or APA-rated sheathing, 48/24-span rating) with Exposure I glue. Seismic and site data: Z = 0.4 (Zone 4) I = 1.0 (standard occupancy) Seismic source type = B Distance to seismic source = 12 km Soil profile type = S C




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Figure 3-2. Foundation plan (ground floor)

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Note: Shear walls on lines 2 and 3 do not extend from the third floor to the roof.

Figure 3-3. Floor framing plan (second and third floors)


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Factors That Influence Design Requirements for seismic design of cold-formed steel stud wall systems are specified in Division VIII of the UBC. Division VIII is a new addition to the UBC and it contains information previously found in §2211.11 of the 1994 UBC relating to seismic design. Division VIII has provisions for both wind and seismic forces for shear walls with wood structural panels framed with cold-formed steel studs. The tables for shear walls (Tables 22-VIII-A, 22-VIII-B and 22-VIII-C) are primarily based on static and cyclic tests conducted by the Light-gauge Steel Research Group at the Santa Clara University Engineering Center for the American Iron and Steel Institute (AISI). Before starting the example, several important aspects of cold-formed construction will be discussed. These are: Stud thickness Screw type Material strength Use of pre-manufactured roof trusses to transfer lateral forces Proper detailing of shear walls at building “pop-outs” AISI Specification for design of cold-formed steel

Stud thickness.

Section 2220.3 of Division VIII states that the uncoated base metal thickness for the studs used with wood structural panels shall not be greater than 0.043-inch. Since an 18-gauge stud has 0.0451-inch thickness, this implies that the heaviest gauge studs that can be used are 20-gauge studs, which can not support a significant bearing or out-of-plane loading. At the time the code change proposal by AISI was submitted to ICBO for inclusion in the 1997 UBC, testing had been performed on only 33 mil (0.033-inch) studs. The SEAOC Seismology Committee felt, and AISI agreed, that there should be a cap on the maximum thickness permitted until testing could be performed on thicker studs. It was felt at the time that limiting the system to 20- and 18-gauge studs would be acceptable for attaching sheathing with #8 screws. Since the UBC is no longer referencing gauge, the 0.043-inch thickness was intended to be a nominal thickness. Subsequent to the code change proposal, AISI has modified this limitation by taking the average thickness between the old 18- and 16-gauges and placed a limitation of 0.043-inch in the AISI code. The 0.043-inch thickness represents 95 percent of the design thickness and is the minimum acceptable thickness delivered to the job site for 18-gauge material based on Section A3.4 of the 1996 AISI Code. Thus, 18-gauge studs can be used, and are used in this example (Table 3-1). The industry has gone away from the use of the gauge designation and is, for the purposes of framing applications, switching to a mil (thousandths of an inch)


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designation. In the future, studs, joists, and track will have their thickness expressed in mils.

Table 3-1. Stud thicknesses Mils Min. Delivered Thickness 33 0.033 inch 43 0.043 inch 54 0.054 inch

Min. Design Thickness 0.0346 inch 0.0451 inch 0.0566 inch

Gauge Reference 20 18 16

The reason for the limitation on maximum stud thickness of 43 mils (18-gauge) is for ductility. At the time of this publication (March 2000), the cyclic tests to date of wood structural panels fastened to 16- and 14-gauge studs with screws have shown nonductile (brittle) failures with the screws shearing off at the face of the stud flange. Cyclic tests for the 20- and 18-gauge studs resulted in ductile behavior with the screw fasteners rocking (tilting) about the plane of the stud flange. Tests are still being conducted by AISI and other organizations on wall systems using the thicker 16- and 14-gauge studs in an attempt to come up with a fastening system that will be ductile. The failure mode of the tests with 33-mil studs for screw spacings of 3 inches and 2 inches on center was end stud compression failure. Subsequent to the code change proposal included in the 1997 UBC, the assemblies have been retested using 43 mil end studs, and higher capacities have been proposed for such assemblies. The values in Table 22-VIII-C are for seismic forces and are nominal shear values. Values are to be modified for both allowable stress design (ASD) and load and resistance factor design (LRFD or strength design). For ASD, the allowable shear values are determined by dividing the nominal shear values by a factor of safety (Ω) of 2.5. For LRFD the design shear values are determined by multiplying the nominal shear values by a resistance factor (φ) of 0.55. Comparing the difference to the two designs: 2.5(0.55)=1.375. In other words, design shears for LRFD (or strength design) are 1.375 times higher than shears for ASD or working stress design. This is consistent with the ASD conversion factor of 1.4 in §1612.3. The values in Table 22-VIII-C for 15/32-inch Structural I sheathing using No. 8 screws are almost identical to the values for the same sheathing applied to Douglas Fir with 8d common nails at the same spacing.

Screw type.

Footnote 2 of UBC Table 22-VIII-C requires the framing screws to be self-drilling. The reason for the self-drilling screws (or drill point screws) is to be able to penetrate 43-mil steel and thicker steel. Self-piercing screws can also be used in 170


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33-mil steel, but with some difficulty. Both self-drilling and self-piercing screws have performed equally well in the shear tests. There is a significant concern in screw installation when there is a gap between the stud flange and the sheathing after installation (e.g., jacking). When jacking occurs, the stiffness of the shear wall is significantly reduced. The drill point alone will not prevent jacking. Jacking occurs when the drill point spins for a rotation or two before the drill point pierces the metal. Only a blank shaft (i.e. smooth with no threads) for the depth of the sheathing will remove the jacking created by the drill point spin prior to piercing. A detailed drawing or explicit specifications should be included in the design drawings and should specify that the distance from the screw head to the beginning of the thread portion be equal to or less than the thickness of the plywood or OSB (oriented strand board). The “unused portion” of the screw protruding from the connection of sheathing and metal stud can be used as a simple inspection gauge to see if jacking has occurred.

Material strength.

Common practice is for material 16-gauge and heavier to have a yield strength of 50,000 psi; for 18-gauge and lighter, 33,000 psi. This practice holds true for studs and track, but not for manufactured hardware (straps, clips and tiedown devices).

Use of pre-manufactured roof trusses to transfer lateral forces.

The structural design in this design example utilizes pre-manufactured roof trusses to transfer the lateral forces from the roof diaphragm to the tops of the interior shear walls. Special considerations need to be included in the design and detailed on the plans for this including: 1.

Provision that any trusses used as collectors (i.e., drag struts) should be clearly indicated on the structural framing plan.

2.

The magnitude of the forces, the means by which the forces are applied to the trusses, and how the forces are transferred from the trusses to the shear walls should be shown.

3.

If the roof sheathing at the hip ends breaks above the joint between the end jack trusses and the supporting girder truss, the lateral forces to be resisted by the end jacks should be specified so that an appropriate connection can be provided to resist these forces.

4.

The drawings should also specify the load combinations and whether or not a stress increase is permitted.

5.

If ridge vents are being used, special detailing for shear transfers need to be indicated in the details.


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Proper detailing of shear walls at building “pop-outs. ”

The structure for this design example has double framed walls for party walls, exterior “planted-on” box columns (pop-outs). The designer should not consider these walls as shear walls unless special detailing and analysis is provided to substantiate that there is a viable lateral force path to that wall and the wall is adequately braced.

AISI Specification for design of cold-formed steel.

The code uses the 1986 version of AISC Specification for Design of Cold-Formed Steel Structural Members as an adopted “Standard” by reference (UBC §2217). Section 2218 amends the 1986 manual. These for the most part are from the 1996 version of the manual. Some sections of the 1996 sections have been used for the solution of this design example.


Code Reference

1.


1a. 1a

Design base shear.

§1630.2.2

Period using Method A (See Figure 3-5 for section through structure): T = Ct (hn )3 / 4 = .020(33.63)3 / 4 = 0.28 sec

(30-8)

With seismic source type B and distance to source = 12 km

172

N a = 1.0

Table 16-S

N v = 1.0

Table 16-T


Design Example 3

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Figure 3-5. Typical cross section through building


Table 16-Q

C v = 0.56 N v = 0.56(1.0 ) = 0.56

Table 16-R

Since the stud walls are both wood structural panel shear walls and bearing walls: R = 5.5

Table 16-N

Design base shear is: V =

Cv I 0.56(1.0 ) W = 0.364W W = 5.5(0.28) RT


(30-4)

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Note that design base shear is now on a strength design basis, but need not exceed: V=

2.5C a I 2.5 (0.40 )(1.0) W= W = 0.182W R 5.5

V = 0.11C a IW = 0.11 (0.40 )(1.0)W = 0.044W < 0.182W

(30-5)

Check Equation 30-7: V=

0.8ZN v I 0.8 × 0.4 × 1.0 × 1.0 W= W = 0.058W < 0.182W R 5.5

V = 0.182W

§1612.3.1

∴V = 0.182 (595,000 lbs ) = 108,290 lb In this Design Example, the designer may choose either allowable stress design or strength design. In Design Example 2, however, allowable strength design must be used. It is desirable to use the strength level forces throughout the design of the structure for two reasons: 1. Errors in calculations can occur and confusion on which load is being used, strength or allowable stress design. This Design Example uses the following format: Vbase shear = strength F px = strength Fx = force-to-wall strength v = wall shear at element level - ASD F v = x = ASD 1.4b 2. This design example is not paving the way for the future, when the code will be all strength design. E = ρE h + E v = 1.0 E h + 0 = 1.0 E h

174

(30-1)


Design Example 3

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where: E v is permitted to be taken as zero for allowable stress design and initially ρ will be assumed to be 1.0, and in most cases ρ = 1.0 for Type V construction with interior shear walls. Since the maximum element story shear is not yet known, the assumed value for ρ will have to be verified. This is done later in Part 4. The basic load combination for allowable stress design for horizontal forces is: D+

E E E = 0+ = 1.4 1.4 1.4

(12-9)

For vertical downward: D+

E E  or D + 0.75 L + (Lr or S ) + 1.4 1.4  

(12-10, 12-11)

For vertical uplift: 0.9 D ±

1b. 1b

E 1.4

(12-10)

Vertical distribution of forces.

The design base shear must be distributed to each level, as follows: F px =

(V − Ft )wx hx n

(30-15)

∑ wi hi i =1

Where h x is the average height at level i of the sheathed diaphragm in feet above the base. Since T = 0.28 seconds < 0.7 seconds, Ft = 0 Determination of F px is shown in Table 3-2.


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Table 3-2. Vertical distribution of seismic forces Level Roof 3rd Floor 2nd Floor Σ

w x (k) 135 230 230 595

h x (ft) 33.6 18.9 9.4

wx hx (k-ft) 4,536 4,347 2,162 11,045

w x hx (%) ∑ wi hi 41.1 39.4 19.5

F px (k) 44.5 42.7 21.1 108.3

F px

Ftot

wx

(k)

0.330 0.186 0.092

44.5 87.2 108.3

Note: Although not shown here, designers must also check wind loading. In this example, wind load may control the design in the east-west direction.

2.


2a. 2a

Deflection of panel assemblies with metal studs.

At the time of this publication, there is not a UBC formula, nor any accepted guideline, for determining the deflection for a diaphragm or shear wall framed with metal studs and structural wood panels. This does not mean that the deflections, drifts, and shear wall rigidities need not be considered (though some engineers may argue otherwise). The formula in UBC Standard §23.223, Vol. 3, can be used with somewhat reasonable results. Given below is a comparison of results from shear panel tests conducted by the Light-gauge Steel Research Group and those determined using the UBC formula. For an 8 ft × 8 ft test panel with 15/32-inch APA-rated sheathing and #8 screw fasteners at 6-inch spacing to 3½-inch x 20-gauge studs and 485 pounds per foot shear, the measured deflection was 0.5 inch. In this Design Example 3, 4-inch × 18-gauge studs are used. Tests have indicated that measured deflections are partially dependent on the stiffness of the studs used. The shear panel test results should not be compared to the nominal shear values from UBC Table 22-VIII-C. Using this table would give an allowable shear of 780 2.5 = 312 plf . This panel test is used only to show the relationship of the measured deflection with results using the UBC formula. Deflection using the formula of UBC standard §23.223, Vol. 3 is shown below: ∆=

176

8vh 3 vh h + + 0.75hen + d a = 0.40 in. ≈ 0.50 in. as tested EAb Gt b

§23.223, Vol. 3


Design Example 3

!


where: v = 485 plf h = 8 ft E = 29 × 10 6 psi A = 0.250 in.2 for 3 1 2 - inch × 20 gauge stud G = 90,000 psi t = 0 . 298 in.

Table 23-2-J, Vol. 3 Table 23-2-H, Vol. 3

Vn = load per screw = (485)6 12 = 242 lb/screw en = 1.2(242 769 )3.276 = 0.0272 in. b = 8 ft d a = 0.0625 in (assumed at 1 16 in.)

2b. 2b

Calculation of shear wall rigidities.

In this Design Example 3, shear wall rigidities (k) are computed using the basic stiffness equation. F = k∆

or k=

F ∆

To simplify the calculations compared to the more rigorous approach used in Design Example 2, this example uses wall rigidities based on the chart in Figure 3-6. This chart is based on the shear wall deflection equation given in UBC Standard §23.223. It should be noted that Design Example 2 considered wood shrinkage and tiedown displacements. With metal framing, shrinkage is zero. This Design Example also assumes a fixed base and pinned top for all shear walls. The chart in Figure 3-6 uses a tiedown displacement (e.g., elongation) of 1/8 inch, which is based on judgment and considered appropriate for this structure. Actual determinations of shear wall rigidities at the roof, third floor, and second floor are shown in Figures 3-3, 3-4, and 3-5, respectively.


177

Design Example 3

!


80.0 K = stiffness = F/d = (Vb )/d 70.0

d = deflection =(8vh 3)/(EAb ) +(vh )/(Gt ) + 0.75he n + d a [A1] h = 8 ft

Stiffness K (kips/in.)

60.0 Where: E = modulus of elasticity = 1.8x106 psi G = shear modulus = 90x103 psi h = wall height (ft) b = wall depth (ft) t = plywood thickness = 15/32 in. A = area of end post = 12.25 in.2 v = shear/foot d a = slip at hold down = 1/8 in. e n = nail deformation slip (in.) F = applied force = Vb (kips)

50.0

40.0

30.0

[A2] h = 10

[B1] h = 8 ft [B2] h = 10 ft [C1] h = 8 ft

[C2] h = 10 ft [D1] h = 8 ft [D2] h = 10

20.0 [A] [B] [C] [D]

10.0

edge nail spacing at 2” o.c. edge nail spacing at 3” o.c. edge nail spacing at 4” o.c. edge nail spacing at 6” o.c.

(v=870 plf, (v=665 plf, (v=510 plf, (v=340 plf,

e n =0.024) e n =0.033) e n =0.033) e n =0.033)

0.0 0

5

10

15

20

25

30

35

Wall Depth b (ft)

Figure 3-6. Stiffness of one-story Structural-I 15/32-inch plywood shear walls

178


40

Design Example 3

!


Table 3-3. Shear wall rigidities at roof level Wall Depth Edge Fastener k (From Fig. 3— Wall Spacing (in.) b (ft) 6) (k/in.) A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4

12.5 11.0 11.0 — 21.5 21.5 — 21.5 21.5 — 21.5 21.5 — 11.0 11.0 — 12.5 8.0 14.0 11.5 11.5 11.5 8.0 —

6 6 6 — 6 6 — 6 6 — 6 6 — 6 6 — 6 6 6 6 6 6 6 —


8.0 7.5 7.5 15.0 15.0 15.0 30.0 15.0 15.0 30.0 15.0 15.0 30.0 7.5 7.5 15.0 8.0 6.0 10.0 8.0 8.0 8.0 6.0 46.0

k total (k/in.) 8.0 — — 15.0 — — 30.0 — — 30.0 — — 30.0 — — 15.0 8.0 — — — — — — 46.0

179

Design Example 3

!


Table 3-4. Shear wall rigidities at third floor

180

Wall

Wall Depth

A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

12.5 11.0 11.0 — 21.5 21.5 — 21.5 21.5 — 21.5 21.5 — 11.0 11.0 — 12.5 8.0 14.0 11.5 11.5 11.5 8.0 — 18.0 24.0 18.0 —

b (ft)

Edge Fastener Spacing (in.) 6 4 4 — 4 4 — 4 4 — 4 4 — 4 4 — 6 4 4 4 4 4 4 — 6 6 6 —

k (From Fig. 3-6) (k/in.) 8.0 10.0 10.0 20.0 19.0 19.0 38.0 19.0 19.0 38.0 19.0 19.0 38.0 10.0 10.0 20.0 8.0 7.0 12.0 10.0 10.0 10.0 7.0 56.0 12.0 15.0 12.0 39.0

k total (k/in.) 8.0 — — 20.0 — — 38.0 — — 38.0 — — 38.0 — — 20.0 8.0 — — — — — — 56.0 — — — 39.0


Design Example 3

!


Table 3-5. Shear wall rigidities at second floor Wall A B1 B2 B C1 C2 C E1 E2 E F1 F2 F G1 G2 G H 1a, 4a 1b, 4b 1c, 4c 1d, 4d 1e, 4e 1f, 4f 1, 4 2a, 3a 2b, 3b 2c, 3c 2, 3

2c. 2c

Wall Depth (ft) 12.5 11.0 11.0 — 21.5 21.5 — 21.5 21.5 — 21.5 21.5 — 11.0 11.0 — 12.5 8.0 14.0 11.5 11.5 11.5 8.0 — 18.0 24.0 18.0 —

b

Edge Fastener Spacing (in.) 6 3 3 — 3 3 — 3 3 — 3 3 — 3 3 — 6 4 4 4 4 4 4 — 6 6 6 —

k (From Fig. 3-6) (k/in.) 8.0 11.0 11.0 22.0 22.5 22.5 45.0 22.5 22.5 45.0 22.5 22.5 45.0 11.0 11.0 22.0 8.0 7.0 12.0 10.0 10.0 10.0 7.0 56.0 12.0 15.0 12.0 39.0

k total (k/in.) 8.0 — — 22.0 — — 45.0 — — 45.0 — — 45.0 — — 22.0 8.0 — — — — — — 56.0 — — — 39.0

Determination of the design level displacement ∆s.

§1630.9.1

For both strength and allowable stress design, the UBC now requires building drifts to be determined by the load combinations of §1612.2, these being the load combinations that use strength design, or LRFD. An errata for the second and third printing of the UBC unexplainably referenced §1612.3 for allowable stress design. The reference to §1612.3 (Allowable Stress Design) is incorrect and will be changed back to reference §1612.2 (Strength Design) in the fourth and later printings. Shear wall displacements for a structures of this type (generally) are well below the maximum allowed by code and the computation of these displacements is


181

Design Example 3

!


considered not necessary. Refer to Design Example 2 for an illustration of this procedure.

3.


§1630.6

In this part, story shears are distributed to shear walls with the diaphragms assumed to be rigid. (Refer to Design Example 2 for a code confirmation of the applicability of this assumption). It has been common practice for engineers to assume flexible diaphragms and distribute loads to shear walls based upon tributary areas. The procedures used in this Design Example 3 are not intended to imply that seismic design of light frame construction in the past should have been performed in this manner. Recent earthquakes and testing of wood panel shear walls have indicated that drifts can be considerably higher than what was known or assumed in the past. Knowledge of the increased drifts of short wood panel shear walls has increased the need for the engineer to consider relative rigidities of shear walls. Section 1630.6 requires the center of mass (CM) to be displaced from the calculated center of mass a distance of 5 percent of the building dimension at that level perpendicular to the direction of force. Section 1630.7 requires the most severe load combination to be considered and also permits the negative torsional shear to be subtracted from the direct load shear. The net effect of this is to add 5 percent accidental eccentricity to the actual eccentricity. The direct shear force Fvi in wall i is determined from: Fvi = F

R ∑R

and the torsional shear force Fti in wall i is determined from: Fti = T

Ri d i J

where: i = wall number J = ΣRd x 2 + ΣRd y 2 R = shear wall rigidity d = distance from the lateral resisting element (e.g., shear wall) to the center of rigidity (CR). T = Fe F = story shear e = eccentricity 182


Design Example 3

3a. 3a

!


Determine center of rigidity, center of mass, eccentricities for roof diaphragm.

Forces in the east-west (x) direction: yr =

∑ k xx y ∑ k xx

or

y r = ∑ k xx = ∑ k xx y

Using the rigidity values k from Table 3-3 and the distance y from line H to the shear wall: y (8.0 + 15.0 + 30.0 + 30.0 + 30.0 + 15.0 + 8.0 ) = 8.0 (116) + 15.0 (106) + 30.0 (82.0) + 30.0 (50.0) + 30.0 (26.0 ) + 15.0 (10.0 ) + 8.0 (0 ) ∴ yr =

7,408 = 54.5 ft 136.0

The building is symmetrical about the x-axis and the center of mass is determined as: ym =

116.0 = 58.0 ft 2

The minimum 5 percent accidental eccentricity for east-west forces, e' y , is computed from the length of the structure perpendicular to the applied story force. e' y = (0.05)(116 ft ) = ±5.8 ft The y m to the displaced CM = 58.0 ft ± 5.8 ft = 63.8 ft or 52.2 ft The total eccentricity is the distance between the displaced center of mass and the center of rigidity y r = 54.5 ft ∴ e y = 63.8 − 54.5 = 9.3 ft

or

52.2 − 54.5 = − 2.3 ft

Note that the distance is slightly different than in Design Example 2. Note that in this Design Example, displacing the center of mass 5 percent can result in the CM being on either side of the CR and can produce added torsional shears to all walls.


183

Design Example 3

!


Note that the 5 percent may not be conservative. The contents-to-structure weight ratio can be higher in light framed structures than in heavier types of construction. Also, the location of the calculated center of rigidity is less reliable for light framed structures than for other structural systems. Use engineering judgment when selecting the eccentricity e .

Forces in the north-south (y) direction: The building is symmetrical about the y-axis. Therefore, the distance to the CM and CR is xm =

48.0 = 24.0 ft 2

min. e' x = (0.05)(48 ft ) = ±2.4 ft Because the CM and CR locations coincide, e x = e ′x ∴ e x = 2.4 ft

184

or

− 2.4 ft


Design Example 3

!


Figure 3-7. Center of rigidity and location of displaced centers of mass for diaphragms


185

Design Example 3

3b. 3b

!


Determine total shears on walls at roof level.

The total shears on the walls at the roof level are the direct shears Fv and the shears due to torsion (combined actual and accidental torsion) Fti . Torsion on the roof diaphragm is computed as follows: T x = Fe y = 44,500 lb (9.3 ft ) = 413,850 ft - lb for walls A, B, and C or T x = 44,500 lb (2.3 ft ) = 102,350 ft - lb for walls E, F, G, and H T y = Fe x = 44,500 lb (2.4 ft ) = 106,800 ft - lb Since the building is symmetrical for forces in the north-south direction, the torsional forces can be subtracted for those walls located on the opposite side from the displaced center of mass. However, when the forces are reversed then the torsional forces will be additive. As required by the UBC, the larger values are used in this Design Example. The critical force is then used for the design of these walls. Table 3-6 summarizes the spreadsheet for determining combined forces on the roof level walls.

Table 3-6. Distribution of forces to shear walls below the roof level Direct Force 2 Ry dy Rx dx Wall Rd Rd

East-West

Fv

A B C E F G H

North-South

Σ

186

1 4

Σ Σ

8.0 15.0 30.0 30.0 30.0 15.0 8.0 136.0

61.5 51.5 27.5 4.5 28.5 44.5 54.5 46.0 46.0 92.0

24.0 -24.0

Torsional Force Ft

Total Force

Fv + Ft

492.0 772.5 825.0 135.0 855.0 667.5 436.0

30,258 39,784 22,688 608 24,368 29,704 23,762 171,172

2,617 4,910 9,815 9,815 9,815 4,910 2,618 44,500

+908 +1,426 +1,523 +62 +390 +305 +199

3,525 6,336 11,338 9,877 10,205 5,215 2,817

1,104 -1,104

26,496 26,496 52,992

22,250 22,250 44,500

+526 -526

22,776 21,724

224,164


Design Example 3

3c. 3c

!


Determine center of rigidity, center of mass, and eccentricities for the third floor diaphragm.

Since the walls stack with uniform fasteners, it can be assumed that the center of rigidity for the third floor and the second floor diaphragms will coincide with the center of rigidity of the roof diaphragm. Torsion on the third floor diaphragm is: F = (44,500 + 42,700) = 87,200 lb T x = Fe y = 87,200 lb (9.3 ft ) = 810,960 ft - lb for walls A, B, and C or 87,200 lb (2.3 ft ) = 200,560 ft - lb for walls E, F, G, and H T y = Fe x = 87,200 lb (2.4 ft ) = 209,280 ft - lb Results for the third floor are summarized in Table 3-7.

East-West

Table 3-7. Distribution of forces to shear walls below the third floor level Direct Ry dy Rx dx Wall Rd Rd 2 Force F v A B C E F G H

North-South

Σ 1 2 3 4

Σ Σ

8.0 20.0 38.0 38.0 38.0 20.0 8.0 170.0

61.5 51.5 27.5 4.5 28.5 44.5 54.5 56.0 39.0 39.0 56.0 190.0

24 2.5 -2.5 -24


Torsional Force Ft

Total Force

Fv + Ft

492 1030 1045 171 1083 890 436

30,258 53,045 28,738 770 30,865 39,605 23,762 207,043

4,104 10,258 19,492 19,492 19,492 10,258 4,104 87,200

+1,467 +3,071 +3,116 +126 +798 +656 +329

5,571 13,329 22,608 19,618 20,290 10,914 4,433

1,344 97.5 -97.5 -1,344

32,256 244 244 32,256 65,000

25,700 17,900 17,900 25,700 87,200

+1,034 +76 -76 -1,034

26,734 17,976 17,824 24,666

272,043

187

Design Example 3

3d. 3d

!


Determine center of rigidity, center of mass, and eccentricities for the second floor diaphragm.

Torsion on the second floor diaphragm is. F = (44,500 + 42,700 + 21,100) = 108,300 lb T x = Fe y = 108,300 lb (9.3 ft ) = 1,007,190 ft - lb for walls A, B, and C or 108,300 lb (2.3 ft ) = 249,090 ft - lb for walls E, F, G, and H T y = Fe x = 108,300 lb (2.4 ft ) = 259,920 ft - lb Results for the second floor are summarized in Table 3-8.

North-South

East-West

Table 3-8. Distribution of forces to shear walls below second floor level Torsional Total Force Direct Ry dy dx Wall R x Rd Rd 2 Force Fv Force Ft Fv + Ft A 8 61.5 492 30,258 4,444 +1,695 6,139 B 22 51.5 1,133 58,350 12,218 +3,901 16,119 C 45 27.5 1,238 34,031 24,992 +4,263 29,255 E 45 4.5 203 911 24,992 +172 25,164 F 45 28.5 1,283 36,551 24,992 +1,093 26,085 G 22 44.5 979 43,565 12,218 +834 13,052 H 8 54.5 436 23,762 4,444 +371 4,815 195 227,428 108,300 Σ 1 56.0 +24.0 1,344 32,256 31,920 +1,195 33,115 2 39.0 +2.5 97.5 244 22,230 +87 22,317 3 39.0 -2.5 -97.5 244 22,230 -87 22,143 4 56.0 -24.0 -1,344 32,256 31,920 -1,195 30,725 190.0 65,000 108,300 Σ 292,428 Σ

188


Design Example 3

3e. 3e

!


Comparison of flexible vs. rigid diaphragm results.

Table 3-9 summarizes wall forces determined under the separate flexible and rigid diaphragm analysis. Fastener requirements were established in Part 2 in Design Example 2. These determinations should be checked for results of the rigid diaphragm analysis and adjusted if necessary (also shown in Table 3-9). Table 3-9. Comparison of loads on shear walls using flexible versus rigid diaphragm results and recheck of wall fastening Wall

F flexible (lbs)

(3)

Frigid (lbs)

Rigid/ Flexible ratio

A B C E F G H 1 4

1,430 6,280 11,310 11,310 8,080 4,660 1,430 22,250 22,250

3,525 6,336 11,338 9,877 10,205 5,215 2,817 22,776 22,776(4)

+147% +1% 0% -13% +26% +11% +97% +2% -2%

A B C E F G H 1 2 3 4

2,805 12,305 22,160 22,160 15,830 9,135 2,805 31,955 11,645 11,645 31,955

5,571 13,329 22,608 19,618 20,290 10,914 4,433 26,734 17,976 17,976(4) 26,734(4)

+99% +8% +2% -11% +28% +19% +58% -16% +54% +54% -16%

A B C E F G H 1 2 3 4

3,485 15,280 27,525 27,525 19,660 11,345 3,485 36,750 17,400 17,400 36,750

6,139 16,119 29,255 25,164 26,085 13,052 4,815 33,115 22,317 22,317(4) 33,115(4)

+77% +5% +6% -9% +33% +15% +38% -10% +28% +28% -10%

b (ft)

v=

Fmax (b )1.4

(plf) Roof Level 12.5 205 22.0 205 43.0 190 43.0 190 43.0 170 22.0 170 12.5 165 64.5 255 64.5 255 Third Floor 12.5 320 22.0 435 43.0 375 43.0 370 43.0 340 22.0 355 12.5 255 64.5 355 60.0 215 60.0 215 64.5 355 Second Floor 12.5 350 22.0 525 43.0 485 43.0 460 43.0 435 22.0 425 12.5 275 64.5 410 60.0 265 60.0 265 64.5 410

Plywood 1 or 2 sides

Allowable Shear(1) (plf)


1 1 1 1 1 1 1 1 1

310 310 310 310 310 310 310 310 310

6 6 6 6 6 6 6 6 6

1 1 1 1 1 1 1 1 1 1 1

310 400 400 400 400 400 310 400 310 310 400

6(2) 4(2) 4 4 4 4 6 4 6 6 4

1 1 1 1 1 1 1 1 1 1 1

310 585 585 585 585 585 310 400 310 310 400

6(2) 3 3 3 3 3 6 4(2) 6 6 4(2)

Notes:


189

Design Example 3

1.

!


Allowable shears are determined from UBC Table 22-VIII-C for 15/32-inch Structural I sheathing using nominal shear values divided by factor of safety (Ω ) of 2.5. Sheathing may by either plywood or oriented stand board (OSB). 2. Screw spacing needs to be decreased from that required for Design Example 2 forces. See also discussion about building weight for the two example problems. 3. Forces taken from Design Example 2. 4. Designates the force used was the higher force for the same wall at the opposite side of the structure.

Comment: Wall rigidities used in this analysis are approximate. The initial rigidity R can be significantly higher than estimated due to the stiffening effects of stucco, drywall walls not considered, and areas over doors and windows. During an earthquake, some low stressed walls may maintain their stiffness and others may degrade in stiffness. Some walls and their collectors may attract significantly more lateral load than anticipated in either a flexible or rigid diaphragm analysis. It must be understood that the method of analyzing a structure using rigid diaphragms takes significantly more engineering effort. This rigid diaphragm analysis method indicates that some lateral resisting elements can attract significantly higher seismic demands than those determined under tributary area analysis methods.

4.

Reliability/redundancy factor ρ. The reliability/redundancy factor penalizes lateral force resisting systems without adequate redundancy. In this Design Example, Part 1, the reliability/redundancy factor was previously assumed to be ρ = 1.0. This will now be checked: ρ = 2−

20

(30-3)

rmax AB

where: rmax = the maximum element-story shear ratio. For shear walls, the wall with the largest shear per foot at or below two-thirds the height of the building; or in the case of a three-story building, the ground level and the second level. See the SEAOC Blue Book Commentary §C105.1.1.1. The total lateral load in the wall is multiplied by 10 l w and divided by the story shear. l w = length of wall in feet AB is the ground floor area of the structure. ri =

Vmax (10 l w ) F

AB = 5,288 sq ft 190


Design Example 3

!


For ground level.

For east-west direction: Using strength level forces for wall B: Vmax = 16,119 lb applied to 2 walls. ri =

(16,119 × 0.5)(10 11.0) 108,300

ρ = 2−

20 0.068 5,288

= 0.068

= −2.0 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, there is no increase in base shear due to lack of reliability/redundancy. For north-south direction: Using strength level forces for walls 1 and 4: Load to wall: 36,750 × 11.5 64.5 = 6,550 lbs ri =

(6,550)(10 11.5) 108,300

= 0.053

Note that this is the same as using the whole wall. ri =

(36,750)(10 64.5)

ρ = 2−

108,300 20 0.053 5,288

= 0.053

= −3.2 < 1.0 minimum o.k.

∴ ρ = 1.0


191

Design Example 3

!


For second level.

For east-west direction: Using strength-level forces for wall B: rmax =

(13,329 )(10 11.0) 87,200

ρ = 2−

20 0.069 5,288

= 0.069

= −1.9 < 1.0 minimum o.k.


For north-south direction: Using strength-level forces for walls 1 and 4: rmax =

(31,955)(10 64.5)

ρ = 2−

87,200 20 0.057 5,288

= 0.057

= −2.8 < 1.0 minimum

o.k.

∴ ρ = 1.0 Therefore, for both directions, there is no increase in base shear required due to lack of reliability/redundancy. The SEAOC Seismology Committee added the sentence “The value of the ratio of 10 l w need not be taken as greater than 1.0” in the 1999 Blue Book—which will not penalize longer walls, but in this Design Example has no effect.

192


Design Example 3

5.

Tiedown forces for the shear wall at line C.

5a. 5a

Determination of tiedown forces.

!


§2220.2

Tiedowns are required to resist the uplift tendency of shear walls caused by overturning moments. In this step, tiedown forces for the three-story shear wall on line C (Figure 3-8) are determined. Since there are two identical shear walls on line C, forces from Table 3-7 must be divided by two. Computation of story forces for one of the two walls is shown below. Note that forces are on strength design basis. Froof = 11,338 2 = 5,669 lb/wall (two walls on line C) Fthird = (22,608 − 11,338) 2 = 5,635 lb Fsec ond = (29,255 − 22,608) 2 = 3,324 lb Ω o = 2.8 bearing wall system

Table 16-N

Figure 3-8. Typical shear wall C elevation


193

Design Example 3

!


The distance between the centroid of the boundary forces that represent the overturning moment at each level must be estimated. This is shown below. e = the distance to the center of tiedown and boundary studs or collectors studs (Figure 3-10) e = 3 in. = 0.25 ft d = the distance between centroids of the tiedowns and the boundary studs. Note that it is also considered acceptable to use the distance from the end of the shear wall to the centroid of the tiedown.

d = 21.5 ft − 2(0.25 ft ) = 21.0 ft The resisting moment M R is determined from the following dead loads: wroof = 13.5 psf (1.33 ft ) = 18.0 plf w floor = 25.0 psf (1.33 ft ) = 33.0 plf wwall = 10.0 plf Overturning resisting moments are determined from simple statics. Calculations are facilitated by use of a spreadsheet. Table 3-10 summarizes the tiedown (i.e., uplift) forces for the shear walls on line C.

Table 3-10. Tiedown forces for shear wall C

Level

M OT (ft-lb)

Ω o M OT M (ft-lb) R (ft-lb)

Strength Uplift

ASD Uplift

d

Ω o M OT − 0.85M R d (1.4 )

0.85M R (1) Ω o M OT − 0.85M R (ft-lb)

(lbs) Roof Third Second

46,545 153,255 291,340

130,330 429,115 815,755

23,135 52,580 82,025

19,665 44,695 69,720

5,275 18,315 35,525

(lbs) 3,770 13,080 25,375

Notes: 1.

194

The 0.85 dead load factor of §2213.5.1 is different from the 0.9 factor of §1612.4.


Design Example 3

5b. 5b

!


Load combinations using allowable stress design.

The UBC has two special sections for shear walls with light framing in Seismic Zones 3 and 4. For metal framing, §2220 is used, and for wood framing, Section 2315.5.1. Section 2220.2 specifies requirements for steel stud wall boundary members and anchorage and refers to §2213.5.1 for load combinations. Section 2315.5.1 deals with wood stud walls and does not have any such special requirements. In the case of identical building types (as in Design Example 2 and Design Example 3 of this manual) this would give an apparent advantage to wood framing. The basic load combinations of §1612.3.1 do not permit stress increases. The alternate basic load combinations if §1612.3.2 permit stress increases. Errata to the First Printing added Equation (12-16-1): 0.9 D ±

E to the alternate basic load combinations 1.4

(12-16-1)

Since this exact same load combination is listed in the basic load combinations the code is in contradiction and confusing (to say at least). This Design Example will use one-third stress increase of §1612.3.2.

6.

Allowable shear and nominal shear strength of No. 10 screws.

Tiedown connections for the line C shear wall will utilized 12-gauge straps at the third floor. This part shows determination of the shear strength of the No. 10 screws that will be used to connect the tiedown straps to the 18-gauge boundary studs. There are two basic ways of determining the shear strength of the screws. The first is to use the values established in an ICBO Evaluation Report with appropriate conversion to strength design. The second is to compute the shear strength of a screw using the ’96 AISI specification. Both methods are shown below.


195

Design Example 3

6a. 6a

!


Nominal shear strength determined from ICBO Evaluation Report.

The Metal Stud Manufacturers’ Association provides ICBO ER No. 4943. Shear values on an ASD basis are provided for various gauge studs having a minimum yield strength of 33 ksi and a minimum ultimate strength of 45 ksi. For No. 10 screws in an 18-gauge stud, the allowable shear is given as 258 lbs per screw. This must be increased as shown below to convert to the strength design basis used in this example. Pns = ΩPas where: Pns = nominal shear strength per screw Pas = allowable shear strength per screw Ω = 3.0

96AISI E4

Pns = 3.0(258 lb ) = 774 lb per screw Note that ER No. 4943 also specifies a minimum edge distance and a minimum on center spacing of 9/16 inch for No. 10 screws.

6b. 6b

Calculation of nominal shear strength using strength design.

The nominal shear strength is the screw capacity without the appropriate reduction factors for allowable stress design (Ω) or load and resistance factor design (φ). d = 0.190 in. Fu1 = 45,000 psi Note: some connector straps and hardware have an Fu = 65,000 psi , which will give higher screw capacities. Fu 2 = 45,000 psi

196


Design Example 3

!


Case I: Strap applied to stud flange (Figure 3-9). Assume 12-gauge galvanized strap: t1 = 0.1017 in. With 18-gauge studs: t 2 = 0.0451in. t 2 t1 = 0.0451 0.1017 = 0.44 < 1.0

(

Pns = 4.2 t 2 3 d

)

12

Fu 2 = 789 lb

96 AISI (E4.3.1-1)

Pns = 2.7t1dFu1 = 2,348 lb

96 AISI (E4.3.1-2)

Pns = 2.7t 2 dFu 2 = 1,041 lb

96 AISI (E4.3.1-3)

Using the smallest value of Pns : Pns = 789 lb per screw Note how this value is almost equal to the 774 lb determined from Part 6a, above.

Case 2: Strap applied to double stud webs (Figure 3-10). Assume 10-gauge galvanized strap: t1 = 0.138 in. With 18-gauge studs: Since there are two stud webs, thickness t 2 is doubled. t 2 = 0.0451 × 2 = 0.0902 in. t 2 t1 = 0.0902 0.138 = 0.65 < 1.0

(

Pns = 4.2 t 2 3 d

)

12

Fu 2 = 2,232 lb

96 AISI (E4.3.1-1)

Pns = 2.7t1 dFu1 = 3,186 lb

96 AISI (E4.3.1-2)

Pns = 2.7t 2 dFu 2 = 2,082 lb

96 AISI (E4.3.1-3)


197

Design Example 3

!


Using the smallest value of Pns : Pns = 2,082 lb

6c. 6c

Calculation of allowable shear using ASD.

Case I: Strap applied to stud flange (Figure 3-9): From Part 6b, above: Pns = 789 lb Pas = Pns Ω = 789 3.0 = 263 lb per screw

Case II: Strap applied to double stud webs (Figure 3-10): From Part 6b, above: Pns = 2,082 lb Pas = Pns Ω = 2,082 3.0 = 694 lb per screw

7.

Tiedown connection at third floor for wall on line C.

Shown below is the strength design of the tiedown strap to be used for the shear walls on line C at the third floor. The configuration at the tiedown is shown on Figure 3-9. Uplift = 3,770 lb Try a 12-gauge × 3 inch strap and No. 10 screws: Pns = 789 lb per screw LRFD design strength = ϕPns

198

96 AISI (3.1)


Design Example 3

!


where: ϕ = 0.50 ϕPns = 0.5(789) = 395 lb Number of screws required: 3,770 395 = 9.5 ∴ Use 12 minimum With 2 rows of #10 screws @ 3½ inches on center the length of strap required: Strap is pre-manufactured, use half spacing for end distance or 1¾ inch. Net spacing is screws is 1.75 inches on center. Need to add in thickness of 1½ inch lightweight concrete and ¾-inch sheathing, plus the 12-inch depth for the floor joist:

(1.75 + (1 + 12 )1.75 + 1.75)2 + (1.5 + 0.75 + 12) = 65.0 ∴Use 72-inch-long strap Check capacity of strap for tension: Strap to be used will be a pre-manufactured strap for which there is an ICBO Evaluation Report. The rated capacity, including 33 percent increase for wind or seismic loading, is given as 9,640 lb. 9,640 lb > 3,770 lb

o.k.

If the strap does not have an ICBO rated capacity, the manufacturer should be contacted to determine the strength of the steel used. It is probable that the steel used in the strap will have strengths that differ from the steel used in the studs. Generally, strengths differ from manufacturer to manufacturer. Checking capacity of strap: Tn = An F y

96 AISI (C2-1)

b = 3.0 in. (strap width) t = 0.1046 in. (strap thickness) d = 0.171in. (diameter of holes)


199

Design Example 3

!


An = 0.2987 in.2 (net area of strap) F y = 45,000 psi (yield strength of particular manufacturer) Tn = 0.2987(45,000 ) = 13,443 lb (nominal strength of strap)

For ASD: Tie force = 3,770 lb (Table 3-10) Tn 13,443 = 8,050 lb > 3,770 lb = allowable tension = 1.67 Ωt

o.k.

For LRFD: Tie force = 5,275 lb (Table 3-10) ϕTn = tension strength = 0.95(13,443) = 12,770 lb => 5,275 lb

o.k.

Use 12-gauge × 3 in. × 72 in. strap with 12 #10 screws @ 3½ inches o.c. each end.

Figure 3-9. Typical tiedown connection at the third floor on line C. 200


Design Example 3

8.

!


Tiedown connection at second floor for wall on line C.

Design of the pre-manufactured tiedowns for the second floor shear walls on line C is shown below. Figure 3-10 shows the configuration of the tiedown. Uplift = 13,080 lb from Table 3-10 The connector is an ICBO approved, pre-manufactured holdown device. The rated capacity including the 33 percent increase for wind or seismic loading is 9,900 lb. Using two holdowns, one on each boundary stud, the capacity is: 2 × 9,900 = 19,800 lb > 13,080 lb

o.k.

In general, when using pre-manufactured tiedowns, consult with ICBO Evaluation Service or the manufacturer for the necessary approvals for hardware selection.


201

Design Example 3

9.

!


Boundary studs for first floor wall on line C.

The studs at each end of the shear walls on line C must be designed to resist overturning forces. In this example, double studs as shown in Figure 3-10 will be used at each end. The critical aspect of design is checking the studs for axial compression. This is shown below.

Figure 3-10. Typical tiedown connection at the second floor on line C

Note that §2220.2 of Division VII (Lateral Resistance of Steel Stud Wall Systems) requires use of the requirements of §2315.5.1. This includes use of the seismic force amplification factor Ω o to account for structural overstrength. This requirement does not apply for boundary elements of wood stud shear walls.

202


Design Example 3

!


For axial compression, the load combination to be used is: 1.0 PDL + 0.7 PLL + Ω o PE

§2213.5.1

PLL = [40 psf + (16 12 )(16 12 ) ]2 = 145 lb PDL = [13.5 psf + (25.0)2](16 12 ) + [10 psf (8 ft )(16 12 )(3) ] = 405 lb From Table 3-10: Ω o M OT = 815,696 ft - lb Ω o PE = 815,755 ft - lb (21.0 ft )(1.4 ) = 27,745 lb Thus, the design load to boundary studs using the equation of §2213.5.1 is: 1.0 (405) + 0.7 (145) + 27,745 = 28,250 lb With a computer program using 1996 AISI Specifications, the allowable axial load for a 4"× 18-gauge stud with 2-inch flanges is 4,042 lb with the flanges braced at mid-height. No. of studs required =

28,250 = 4.1 4,042 × 1.7

where: 1.7 is the allowable stress increase Therefore, use 5 studs at ends of wall as follows: Use two back-to-back studs, plus two back-to-back studs with additional stud (Figure 3-10).


203

Design Example 3

10. 10

!


Shear transfer at second floor on line C.

Shear forces in the second floor diaphragm are transferred to the shear walls below as shown in Figure 3-11. From Table 3-9, the ASD shears in the wall are: v = 485 lb/ft Try using #8 screws, 18-gauge metal side plates and Douglas Fir plywood: Z = 119 lb/screws

91NDS Table II.3B

C D = 1.33 Maximum spacing =

§1612.3.2 ZC D 119(1.33)12 = = 3.9 in. v 485

∴Use # 8 screws at 3 inches on center. Capacity of the #8 screws in the 18-gauge tracks and runner channels are O.K. by inspection.

Figure 3-11. Typical detail for shear transfer through floor on line C

204


Design Example 3

11. 11

!


Shear transfer at foundation for walls on line C.

Shown below is the design of the connection to transfer the shear force in the walls on line C to the foundation. This detail is shown in Figure 3-12. From Table 3-9: v = 485 lb/ft Allowable load based on bolt bearing on track:

96 AISI (E3.3)

For 5/8" bolts and 18-gauge track: Pn = 2.22 Fu d

96 AISI, Table E3.3-2

where: Pn = nominal resistance Fu = 45 ksi (minimum value) d = 0.625 in. t = 0.0451in. Pn = 2.22 (45)(0.625)(0.0451) = 282


k bolt

205

Design Example 3

!


Figure 3-12. Detail for shear transfer at foundation on line C.

Allowable service load on embedded bolts in concrete is determined as follows. For 5/8" bolts and 3000 psi concrete: Allowable shear = 2,750

lb bolt

Table 19-D

Therefore the bolt in concrete governs the required spacing: Maximum spacing

2,750 = 5.67 ft o.c. 485

∴Use 5/8" diameter bolts at 4'−0" o.c. spacing

206


Design Example 3

12. 12

!


Shear transfer at roof on line C.

Shear forces in the roof diaphragm are transferred to the shear walls below as shown in Figure 3-13. From Table 3-9 are the ASD shears in the wall. v = 190 lb/ft From manufacturer’s catalog, allowable load for the 6-3/8-inch-long framing clip is 915 pounds. With framing clips at 4.0 ft centers, the design ASD force is:

(190)(4) = 760 lb < 915 lb

o.k.

Figure 3-13. Shear transfer at roof at line C

Note that double studs are used for sound control, but that only one stud is considered in shear wall calculations.


207

Design Example 3

!


Commentary The code does not have conventional construction provisions for cold-formed steel similar to the conventional light frame construction provisions for wood. The 2000 International Residential Code (IRC) has included prescriptive provisions for cold-formed steel for one- and two-family dwellings. It should be noted that the structure shown in example could not use the IRC prescriptive provisions. Inasmuch as there is no one standard for the manufacturing of the studs, the process to design gravity load members is a tedious method and should not be done by prescriptive means. The AISI Specification for Design of Cold-Formed Steel Structural Members has complex equations and is considered by most engineers too difficult to be readily used in design. Due to the complex nature of the equations, in the AISI code it is recommended that engineers designing in cold-formed steel utilize computer software for design.

References American Iron and Steel Institute, 1996. Cold Formed Steel Design Manual, 1996 Edition. American Iron and Steel Institute, Washington, D.C. American Iron and Steel Institute, 1986. Cold Formed Steel Design Manual, 1986 Edition. American Iron and Steel Institute, Washington, D.C. American Plywood Association, 1997. Design/Construction Guide – Diaphragms and Shear Walls. Report 105, Engineered Wood Association, Tacoma, Washington. American Plywood Association, 1993. Revised. Wood Structural Panel Shear Walls. Report 154, Engineered Wood Association, Tacoma, Washington. Applied Technology Council, 1995. Cyclic Testing of Narrow Plywood Shear Walls, ATC R-1. Applied Technology Council, Redwood City, California. BSSC, 1997. National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings, 1997. Building Seismic Safety Council, Washington, D.C.

208


Design Example 3

!


Building Seismic Safety Council, 1997. National Earthquake Hazard Reduction Program, Recommended Provisions for Seismic Regulations for New Buildings. Building Seismic Safety Council, Washington D.C. Cobeen, K.E., 1996. “Performance Based Design of Wood Structures,” Proceedings, Annual SEAOC Convention. Structural Engineers Association of California, Sacramento, California. Countryman, D., and Col Benson, 1954. 1954 Horizontal Plywood Diaphragm Tests. Laboratory Report 63, Douglas Fir Plywood Association, Tacoma, Washington. Dolan, J.D., 1996. Experimental Results from Cyclic Racking Tests of Wood Shear Walls with Openings. Timber Engineering Report No. TE- 1996-001. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J. D. and Heine , C.P., 1997a. Monotonic Tests of Wood-frame Shear Walls with Various Openings and Base Restraint Configurations. Timber Engineering Report No. TE-1997-001. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D. and Heine, C.P., 1997b. Sequential Phased Displacement Cyclic Tests of Wood frame Shear Walls with Various Openings and Base Restrain Configurations. Timber Engineering Report No. TE-1997-001. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Dolan, J.D., and Heine, C.P., 1997c, Sequential Phased Displacement Test of Wood-frame Shear Walls with Corners. Timber Engineering Report No. TE-1997-003. Virginia Polytechnic Institute and State University, Blacksburg, Virginia. Earthquake Engineering Research Institute, 1996. “Northridge Earthquake of January 17, 1994, Reconnaissance Report,” Earthquake Spectra. Vol. 11, Supplement C. Earthquake Engineering Research Institute, Oakland, California. Foliente, Greg C., 1994. Analysis, Design and Testing of Timber Structures Under Seismic Loads. University of California Forest Products Laboratory, Richmond, California. Foliente, Greg C., 1997. Earthquake Performance and Safety of Timber Structures. Forest Products Society, Madison Wisconsin. Forest Products Lab, 1999. Wood Handbook Publication FPL – GTR – 113. Madison, Wisconsin.


209

Design Example 3

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Goers R. and Associates, 1976. A Methodology for Seismic Design and Construction of Single-Family Dwellings. Applied Technology Council, Redwood City, California. International Code Council, International Building Code – Final Draft, 2000. International Code Council, Birmingham, Alabama. Ju, S., and Lin M., 1999. “Comparison of Building Analysis Assuming Rigid or Flexible Floors,” Journal of Structural Engineering. American Society of Civil Engineers, Washington D.C. Light-gauge Steel Engineers Association, Tech Note 558b-1. Lateral Load Resisting Elements: Diaphragm Design Values. Light-gauge Steel Engineers Association., 2400 Crestmoor Road, Nashville, Tennessee 37215. Light-gauge Steel Engineers Association, Tech Note 556a-6, Vertical Lateral Force Resisting System Boundary Elements. Light-gauge Steel Engineers Association, 2400 Crestmoor Road, Nashville, Tennessee 37215. Light-gauge Steel Engineers Association, Tech Note 556a-4. Shear Transfer at Top Plate: Drag Strut Design. Light-gauge Steel Engineers Association, 2400 Crestmoor Road, Nashville, Tennessee 37215. Light-gauge Steel Engineers Association, Tech Note 565c. Screw Fastener Selection for Light-gauge Steel Frame Construction. Light-gauge Steel Engineers Association. 2400 Crestmoor Road, Nashville, Tennessee 37215, February 1997. Metal Stud Manufacturer’s Association, 1993. ICBO Evaluation Report No. 4943. Metal Stud Manufacturer’s Association, P.O. Box 1211, Corvallis, Oregon97339, revised December 1993. National Forest Products Association, 1991. National Design Specification for Wood Construction. National Forest Products Association, Washington D.C. Performance Standards and Policies for Structural-Use Panels [Sheathing Standard, Sec. 2.3.3]. APA Standard PRP-108. APA-The Engineered Wood Association. Tacoma, Washington 98411. Rose, J.D., 1998, Preliminary Testing of Wood Structural Panel Shear Walls Under Cyclic (Reversed) Loading. Research Report 158. APA – The Engineered Wood Association, Tacoma, Washington. Rose, J.D., and E.L. Keith, 1996. Wood Structural Panel Shear Walls with Gypsum Wallboard and Window [ Sheathing Standard, Sec. 2.3.3].Research Report 158. APA - The Engineered Association, Tacoma Washington.

210


Design Example 3

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Structural Engineers Association of California, 1999. “Recommended Lateral Force Requirements and Commentary,” Structural Engineers Association of California, Sacramento, California. Serrette, R. 1996. Final Report: Shear Wall Values for Lightweight Steel Framing. Santa Clara University Engineering Center, Santa Clara, California 95053. Yu, Wei-wen, 1991. Cold-Formed Steel Design, John Wiley & Sons, Inc., New York, New York.


211

Design Example 3

212

!



Design Example 4

!

Masonry Shear Wall Building

Design Example 4 Masonry Shear Wall Building

Figure 4-1. Schematic CMU building elevation

Overview Reinforced concrete block masonry is frequently used in one-story and lowrise construction, particularly for residential, retail, light commercial, and institutional buildings. This type of construction has generally had a good earthquake performance record. However, during the 1994 Northridge earthquake, some one-story buildings with concrete masonry unit (CMU) walls and panelized wood roofs experienced wall-roof separations similar to that experienced by many tilt-up buildings. This building in this Design Example 4 is typical of one-story masonry buildings with wood framed roofs. The building is characterized as a heavy wall and flexible roof diaphragm “box building.” The masonry building for this example is shown schematically in Figure 4-1. Floor and roof plans are given in Figure 4-2 and 4-3, respectively. The building is a one-story bearing wall building with CMU shear walls. Roof construction consists of a plywood diaphragm over wood framing. An SEAOC Seismic Design Manual, Vol. II (1997 UBC)

213

Design Example 4

!


elevation of the building on line A is shown in Figure 4-4. A CMU wall section is shown in Figure 4-5, and a plan view of an 8'-0" CMU wall/pier is shown in Figure 4-6. The design example illustrates the strength design approach to CMU wall design for both in-plane and out-of-plane seismic forces.

Outline This example will illustrate the following parts of the design process.

214

1.

Design base shear coefficient.

2.

Base shear in the transverse direction.

3.

Shear in wall on line A.

4.

Design 8'-0" shear wall on line A for out-of-plane seismic forces.

5.

Design 8'-0" shear wall on line A for in-plane seismic forces.

6.

Design 8'-0" shear wall on line A for axial and in-plane bending forces.

7.

Deflection of shear wall on line A.

8.

Requirements for shear wall boundary elements.

9.

Wall-roof out-of-plane anchorage for lines 1 and 3.

10. 10

Chord design.


Design Example 4

!


Given Information Roof weights: Roofing+ one re-roof ½" plywood Roof framing Mech./elec.

7.5 psf 1.5 4.5 1.5

Insulation Total dead load Roof live load

1.5 17.0 psf 20.0 psf

Exterior 8-inch CMU walls: 75 psf (fully grouted, light-weight masonry) f ' m = 2,500 psi f y = 60,000 psi

Seismic and site data: Z = 0.4 (Seismic Zone 4) I = 1.0 (standard occupancy) Seismic source type = A Distance to seismic source = 5 km Soil profile type = S D


Figure 4-2. Floor plan SEAOC Seismic Design Manual, Vol. II (1997 UBC)

215

Design Example 4

!


Figure 4-3. Roof plan

Figure 4-4. Elevation of wall on line A

216


Design Example 4

!


Figure 4-5. Section through CMU wall along lines 1 and 3

Figure 4-6. Reinforcement in 8’-0” CMU shear walls on lines A and D


217

Design Example 4

!


Calculations and Discussion Discussion

1.

Code Reference


§1630.2.2

Period using Method A (see Figure 4-5 for section through structure): T = C t (h n )3 / 4 = .020 (16 ft )3 / 4 = 0.16 sec

(30-8)

Near source factors for seismic source type A and distance to source = 5 km N a = 1.2

Table 16-S

N v = 1.6

Table 16-T

Seismic coefficients for Zone 4 and soil profile type S D are: C a = 0.44 N a = 0.53

Table 16-Q

C v = 0.6 4 N v = 1.02

Table 16-R

The R coefficient for a masonry bearing wall building with masonry shear walls is: R = 4.5

Table 16-N

Calculation of design base shear: V=

Cv I 1.02 (1.0 ) W = 1.417W W= RT 4.5 (0.16 )

(30-4)

but need not exceed: V=

2.5C a I 2.5 (0.53)(1.0 ) W= W = 0.294W R 4.5

(30-5)

The total design shear shall not be less than: V = 0.11C a IW = 0.11 (0.53)(1.0 )W = 0.058

218

(30-6)


Design Example 4

!


In addition, for Seismic Zone 4, the total base shear shall also not be less than: V=

0.8ZN v I 0.8 (0.4 )(1.60)(1.0 ) W= W = 0.114W R 4.5

(30-7)

Therefore, Equation (30-5) controls the base shear calculation and the seismic coefficient is thus: V = 0.294W

2.

Base shear in transverse direction.

This building has a flexible roof diaphragm and heavy CMU walls (see Figure 4-3). The diaphragm spans as a simple beam between resisting perimeter walls in both directions and will transfer 50 percent of the diaphragm shear to each resisting wall. However, in a building that is not symmetric or does not have symmetric wall layouts, the wall lines could have slightly different wall shears on opposing wall lines 1 and 3 and also on A and D. The building weight (mass) calculation is separated into three portions: the roof, longitudinal walls, and transverse walls for ease of application at a later stage in the calculations. The reason to separate the CMU wall masses is because masonry walls that resist ground motions parallel to their in-plane directions resist their own seismic inertia without transferring seismic forces into the roof diaphragm. This concept will be demonstrated in this example for the transverse (north-south) direction. For the transverse direction, the roof diaphragm resists seismic inertia forces originating from the roof diaphragm and the longitudinal masonry walls (out-of-plane walls oriented east-west) on lines 1 and 3, which are oriented perpendicular to the direction of seismic ground motion. The roof diaphragm then transfers its seismic forces to the transverse masonry walls (in-plane walls oriented north-south) located on lines A and D. The transverse walls resist seismic forces transferred from the roof diaphragm and seismic forces generated from their own weight. Thus, seismic forces are generated from three sources: the roof diaphragm; in-plane walls at lines 1 and 3; and out-of-plane walls at lines A and D. The design in the orthogonal direction is similar and the base shear is the same. However, the proportion of diaphragm and in-plane seismic forces is different. The orthogonal analysis is similar in concept, and thus is not shown in this example. Roof weight: Wroof = 17 psf (5,400 sf ) = 92 kips


219

Design Example 4

!


For longitudinal wall weight (out-of-plane walls), note that the upper half of the wall weight is tributary to the roof diaphragm. This example neglects openings in the top half of the walls.

(19 ft )2  19 ft   1   = 75 psf (180 ft ) W walls, long = 75 psf (2 walls)(90 ft )(19 ) = 152 kips   2(16 ft )  2   16 ft  For forces in the transverse direction, seismic inertial forces from the transverse walls (lines A and D) do not transfer through the roof diaphragm. Therefore, the effective diaphragm weight in the north-south direction is: Wtrans. diaph = Wroof + W walls, long = 92 k + 152 k = 244 kips The transverse seismic inertial force (shear force), which is generated in the roof diaphragm is calculated as follows: Vtrans. diaph = 0.294Wtrans. diaph = 0.294(244 kips ) = 72 kips The seismic inertial force (shear force) generated in the transverse walls (in-plane walls) is calculated using the full weight (and height) of the walls (with openings ignored for simplicity). Vtrans. walls = 0.294 (75 psf )(19 ft )(60 ft )(2 walls) = 50 kips The design base shear in the transverse direction is the sum of the shears from the roof diaphragm shear and the masonry walls in-plane shear forces. ∴Vtrans. = Vtrans. diaph + Vtrans. walls = 72 k + 50 k = 122 kips

3.

Shear wall on line A.

The seismic shear tributary to the wall on line A comes from the roof diaphragm (transferred at the top of the wall) and the in-plane wall inertia force: VA =

4.

Vtrans. diaph 2

+

Vtrans. walls 2

=

72 kips 50 kips + = 61 kips 2 2

Design 8'-0" shear wall on line A for out-of-plane seismic forces.

In this part, the 8'-0" shear wall on line A (Figure 4-4) will be designed for out-of-plane seismic forces. This wall is a bearing wall and must support gravity loads. It must be capable of supporting both gravity and out-of-plane seismic forces, and gravity plus in-plane seismic forces at different instants in time 220


Design Example 4

!


depending on the direction of seismic ground motion. In this Part, the first of these two analyses will be performed. The analysis will be done using the “slender wall” design provisions of §2108.2.4. The analysis incorporates static plus P∆ deflections caused by combined gravity loads and out-of-plane seismic forces and calculates an axial plus bending capacity for the wall under the defined loading.

4a. 4a

Vertical loads.

Gravity loads from roof framing tributary to the 8'-0" shear wall at line A:  60 ft   30 ft  PDL = (17 psf )   = 7,650 lb  2  2  Live load reduction for gravity loads: R = r ( A − 150) ≤ 40 percent

§1607.5

A = (30 ft )(15 ft ) = 450 sq ft R = 0.8 (450 sq ft − 150 sq ft ) = 24 percent DL    17  Rmax = 23.11 +  = 23.11 +  = 42.7 percent LL    20  ∴ R = 24 percent The reduced live load is:  60 ft   30 ft  PRLL = (20 psf )  (100 percent − 24 percent ) = 6,840 lb  2  2  Under §2106.2.7, the glulam beam reaction load may be supported by the bearing width plus four times the nominal wall thickness. Assuming a 12-inch bearing width from a beam hanger, the vertical load is assumed to be carried by a width of wall 12 in. + 4 (8 in.) = 44 in. PbeamD +L =

PbeamD =

(7,650 lb + 6,840 lb) = 3,952 plf (44 in. 12 in.) 7,650 lb = 2,086 plf (44 in. 12 in.)


221

Design Example 4

!


Wall load on 8-foot wall (at wall mid-height):  16 ft  Pwall DL = (75 psf )(8 ft ) + 3 ft  = 6,600 lb  2  6,600 lb w wall DL = = 825 plf 8 ft Dead load from wall lintels:  20 ft  PL intel D = (75 psf )(9 ft )  = 6,750 lb  2  l = (96 in. − 44 in.) 2 = 26 in. w L int elD =

6,750 lb = 3,115 plf 26 in. 12 in.

Since the lintel loads are heavier than the beam load, and since dead load combinations will control, the loads over the wall/pier length will be averaged. The gravity loads on the 8'-0" wall from the weight of the wall, the roof beam, and two lintels are:

∑ PDL = (6,600 lb + 7 ,650 lb + 6,750 lb + 6,750 lb) = 27,750 lb ∑ PRLL = 6,840 lb 4b. 4b

Seismic forces.

Out-of-plane seismic forces are calculated as the average of the wall element seismic coefficients at the base of the wall and the top of the wall. The coefficients are determined under the provisions of §1632.2 using Equation (32-2) and the limits of Equation (32-3). Fp =

a p Ca I p  h 1 + 3 x Rp  hr

 W p 

0.7C a I pW p ≤ F p ≤ 4.0C a I pW p

222

(32-2)

(32-3)


Design Example 4

!


At the base of the wall: Fp =

=

a p Ca I p  h 1 + 3 x Rp  hr

(1.0)C a I p Rp

 W p 

 0 ft  1 + 3 W p 16 ft  

= 0.133C a I pW p ≤ 0.7C a I pW p ∴Use 0.7C a I pW p

F p = 0.7 (.53)(1.0 )W p = 0.37W p = 0.37(75 psf ) = 27.8 psf At roof: Fp =

=

a p Ca I p  h 1 + 3 x Rp  hr

 W p 

(1.0)C a I p  Rp

0'  1 + 3 W p 16'  

= 1.33C a I pW p ≤ 4.0C a I pW p ∴Use 1.33C a I pW p

F p = 1.33(.53)(1.0 )W p = 0.37W p

= 0.70(75 psf ) = 52.5 psf

Thus, use the average value of F p = (1 2 )(27.8 psf + 52.5 psf ) = 40.2 psf Calculation of wall moments due to out-of-plane forces is done using the standard beam formula for a propped cantilever. See Figure 4-7 for wall out-of-plane loading diagram and Figure 4-8 for tributary widths of wall used to determine the loading diagram.


223

Design Example 4

!


3'

R2 W1

16'

10'

W2 R1

Figure 4-7. Propped cantilever loading diagram

Tributary width of wall considered 18'-0"

W1 9'-0"

Lintel beam resists out-of-plane forces

W2

10'-0"

8'-0"

10'-0"

“h”

10'-0"

Figure 4-8. Tributary width of wall for out-of-plane seismic inertial force calculations

224


Design Example 4

!


W1 = (10 ft + 8 ft + 10 ft )(40.2 psf ) = 1,125 plf W2 = 8 ft (40.2 psf ) = 322 plf Using simple beam theory to calculate moment M oop for out-of-plane forces, the location of maximum moment is at h = 9.8 feet: M oop = 15,530 lb - ft = 186,360 lb - in. Comparison of seismic out-of-plane forces with wind (approximately 25 psf) indicate that seismic forces control the design.

4c. 4c

Design for out-of-plane forces.

§1612.2.1

The wall section shown in Figure 4-6 will be designed. The controlling load combinations for masonry are: 1.2 D + 1.6 Lr

(12-3)

1.1(1.2 D + 1.0 E ) = 1.32 D + 1.1( E h + E v )

(12-5)

1.1E v = 1.1(0.5)C a ID = 0.55(0.53)(1.0) D = 0.30 D Note: Exception 2 of §1612.2.1 requires that a 1.1 factor be applied to the load combinations for strength design of masonry elements including seismic forces. The SEAOC Seismology Committee has recommended that this factor be deleted. However; this example shows use of the factor because it is a present requirement of the code, thus: PD + RLL = 1.2 (27 ,750 lb ) + 1.6 (6840 lb ) = 44 ,244 lb

(12-3)

Pu = PD + L + E = PD + 1.1E v = 1.32(27 ,750 lb ) + (0.30 )(27 ,750 lb ) = 44 ,955 lb

(12-5)

The controlling load case by examination is Equation (12-5) for gravity plus seismic out-of-plane forces. Slender wall design of masonry walls with an axial load of 0.04 f m' or less are designed under the requirements of §2108.2.4.4.


225

Design Example 4

!


Check axial load vs. 0.04 f m' using unfactored loads: Pw + Pf Ag

≤ 0.04 f m

27,750 lb = 38 psi ≤ 0.04 (2500 psi ) = 100 psi (7.625 ft )(8 ft )(12 in.) ∴ o.k.

Calculate equivalent steel area Ase : Ase =

As f y + Pu fy

(0.31 in. )(6 bars)(60,000 psi) + 44,955 lb = 2.61 in. =

(8-24)

2

2

60,000 psi

Calculate I cr :

(Pu + As f y )

a=

.85 f ' m b

=

(

c=

a = 0.86 in. .85

n=

E s 29 ,000 ,000 psi = = 15.46 1,875,000 psi Em

I cr =

=

)

44 ,955 lb + 1.86 in. 2 (60 ,000 psi ) = 0.77 in. .85 (2500 psi )(96 in.)

(8-25)

§2106.2.12.1

bc3 + nAse (d − c )2 3

96 in.(0.90 in.)3 + (15.46) 2.62 in. 2 (3.81 in. − 0.90 in )2 = 365.0 in. 4 3

(

)

Calculate M cr using the value for f r from §2108.2.4.6, Equation (8-31):  96 in.(7.625 in.)2 M cr = S g f r =   6 

226

  (4.0 )(2,500 )1 2 = 186,050 lb - in.  

(8-30)


Design Example 4

!


Calculate I g : Ig =

(96 in.) (7.625 in.) 3 = 3546.6 in. 4 12

Calculate M u based on Equation (8-20) of §2108.2.4.4: First iteration for moment and deflection (note that eccentric moment at mid-height of wall is one-half of the maximum moment): M u = M out −of − plane + M eccentric = 1.1E + 1.1(1.2 D ) + 1.1(1.6)(L = 0 ) M u = M out −of − plane + M eccentric = 1.1 (186,360 lb - in.)

+ 1.32 (7,650 lb )(6 in.) 2 = 235,960 lb - in. ∆u =

∆u =

+

5M cr h 2 5 (M u − M cr )h 2 + 48 E m I g 48 E m I cr 5 (186,050 lb - in.)(192 in.)2

(

48 (1,875,000 psi ) 3,546.6 in. 4

(8-28)

)

5 (235,290 lb - in. − 186,050 lb - in.)(192 in.)2

(

(8-20)

48 (1,875,000 psi ) 365.0 in. 4

)

= 0.11 in. + 0.28 in. = 0.38 in.

Note: The deflection equation used is for uniform lateral loading, maximum moment at mid-height, and pinned-pinned boundary conditions. For other support and fixity conditions, moments and deflections should be calculated using established principals of mechanics. Beam deflection equations can be found in the AITC or AISC manuals or accurate methods can be derived. Second iteration for moment and deflection: M u = 235,290 lb - in. + 44,955 lb (0.38 in.) = 252,540 lb - in. ∆ u = 0.11 in. +

5(252,540 lb - in. − 186,050 lb - in.)(192 in.)2

(

48 (1,875,000 psi ) 365.0 in. 4

)

= 0.11 in. + 0.37 in. = 0.48 in.


227

Design Example 4

!


Third iteration for moment and deflection: M u = 235,290 lb - in. + 44,955 lb - in.(0.48 in.) = 256,891 lb - in. ∆ u = 0.11 in. +

5 (256,891 lb - in. − 186,050 lb - in.)(192 in.)2

(

48 (1,875,000 psi ) 365.0 in. 4

)

= 0.11 in. + 0.40 in. = 0.51 in. Final moment (successive iterations are producing moments within 3 percent, therefore convergence can be determined): M u = 235,290 lb - in. + 44,955 lb (0.51 in.) = 258,217 lb - in. Calculation of wall out-of-plane strength: a  φM u = φAse f y  d −  2 

(

)

0.73 in.   = 0.80 2.47 in. 2 (60,000 psi ) 3.81 in. −  2   = 408,439 lb - in. ≥ 258,217 lb - in. Since the wall strength is greater than the demand, the wall section shown in Figure 4-4 is okay. Note that out-of-plane deflections need to be checked using same iteration process, but with service loads per §2108.2.4.6, (i.e., PD = 27,750 lbs). Since ultimate deflections are within allowable, there is no need to check service deflections in this example. The limiting deflection is 0.007 h per §2108.2.4.6 is 0.007(16'×12") = 1.34". The deflection from this analysis is 0.50 inches. Thus the deflection is within allowable limits. Check that the wall reinforcement is less than 50 percent of balanced reinforcement per §2108.2.4.2: ρb =

ρ=

.85β1 f ' m 87,000 + = 0.0178 fy 87,000 + f y

(6)(0.31 in. 2 ) = 0.0051 ≤ 0.0089 (3.81 in.)(96 in.)

∴ o.k.

Check the unbraced parapet moment: 228


Design Example 4

!


a p = 2.5

Table 16-0

R p = 3.0 Fp =

a pCa I p  h 1 + 3 x Rp  hr

 (2.5)(.53)(1.0)   16 ft    W p =   W p 1 3 +  ( ) 16 ft 3 . 0    

= 1.76W p = 1.76 (75 psf ) = 132.5 psf M u (132.5 psf )(3 ft )2 8 = 596 lb - ft = 7,155 lb - in. ≤ 408,439 lb - in. ∴ Wall section is okay at parapet.

5.

Design 8'-0" shear wall on line A for in-plane seismic forces.

5a. 5a

Shear force distribution.

The shear force on line A must be distributed to three shear wall piers (6', 8', and 6' in width, respectively) in proportion to their relative rigidities. This can be accomplished by assuming that the walls are fixed at the tops by the 9-foot-deep lintel. Reference deflection equations are given below for CMU or concrete walls with boundary conditions fixed top or pinned top. For this Design Example, the fixed/fixed equations are used because the deep lintel at the wall/pier tops will act to fix the tops of wall piers. ∆i =

Vi h 3 1.2Vi h + for walls/piers fixed top and bottom AG 12 E m I

∆i =

Vi h 3 1.2Vi h + for walls/piers pinned top and fixed at bottom AG 3E m I

G = 0.4 E m for concrete masonry under §2106.2.12.13

(6-6)

1 where ∆ is the deflection under load Vi . Using the ∆ fixed/fixed equation, the percentage shears to each wall are shown in Table 4-1. Relative rigidity is thus


229

Design Example 4

!


Table 4-1. Distribution of line A shear to three shear walls. Wall Moment Shear Deflection Total Deflection Distribution to Rigidity (1/in.) Wall Shear (k) Length (ft) Deflection (in.) (in.) (in.) Piers (%) 6 1.17E-05 3.50E-07 1.20E-05 83,28 26.6% 16.2 8

6.56E-06

2.62E-07

6.82E-06

146,635

46.8%

28.6

6

1.17E-05

3.50E-07

1.20E-05

83,28

26.6%

16.2

Totals

313,200

100%

61.0

The seismic shear force E h to the 8-foot pier is (0.468)61 k = 28.6 k. Calculation of reliability/redundancy factor ρ is shown below. For shear walls the maximum element story shear ratio ri is determined as:

§1630.1.1

ri8 = (28.6 k )(10 ) / 8 ft / 122 k = 0.29 for 8 ft segment ri 6 = (16.2 k )(10) / 6 ft / 122 k = 0.22 for 6 ft segment ∴ rmax = 0.29 ρ=2−

20 rmax AB

=2−

20

(0.29 )

5,400 ft

(30-3)

2

∴ ρ = 1.06 The strength design shear for the 8'-0" wall is: ∴V8' wall = 1.06(28.6 k ) = 30.3 k

5b. 5b

Determination of shear strength.

The in-plane shear strength of the wall must be determined and compared to demand. The strength of the wall is determined as follows. Vertical reinforcement is #5@16 inches o.c. Try #4@16 inches o.c. horizontally. Note that concrete masonry cells are spaced at 8-inch centers, thus reinforcement arrangements must have spacings in increments of 8 inches (such as 8 inches, 16 inches, 24 inches, 32 inches, 40 inches, and 48 inches). Typical reinforcement spacings are 16 inches and 24 inches for horizontal and vertical reinforcement.

230


Design Example 4

!


Calculate M Vd : 151.5 k - ft M = = 0.625 V d (30.3 k )(8 ft ) From Table 21-K and by iteration, the nominal shear strength coefficient C d = 1.8 Vn = Vm + V s Vm = C d Amv

(8-36) f m = (1.80)(7.625 in.)(96 in.) 2,500 psi = 65.9 k

Vs = Amv ρ n f y

(8-37) (8-38)

for φ = 0.80, with #4 @16" o.c. horizontally:

(

)

(

)

  0.20 in. 2 φV s = φAmv ρ n f y = (0.80)(7.625 in.)(96 in.)   (60,000 psi ) = 57.6 k  (7.625 in.)(16 in.) for φ = 0.60, with #4 @16" o.c. horizontally:   0.20 in. 2 φV s = φAmv ρ n f y = (0.60)(7.625 in.)(96 in.)   (60,000 psi ) = 43.2 k  (7.625 in.)(16 in.) Thus, conservatively, using φ = 0.60 φV n = 0.6 (65.9 k ) + 43.2 k = 82.7 k The designer should check the failure mode. If failure mode is in bending, φ = 0.80. If failure mode is in shear, φ = 0.60. For this example, we will conservatively use φ = 0.60. The method of checking the failure mode is to check how much moment M u is generated when the shear force is equal to shear strength Vn with φ = 1.0. Then that moment is compared with the wall Pn and M n with a φ = 1.0. If there is reserve moment capacity, there will be a shear failure. If not, there will be a bending failure. Later in the example this will be checked. The reason the failure mode should be checked is to understand whether a brittle shear failure will occur or a ductile bending failure. Since the bending failure is more desirable and safer, the φ factor is allowed to be higher. Vu = 1.1(30.3 k ) = 33.3 k ≤ φVn = 82.7 k , for 0.60,∴ o.k. ∴ Use #4 @16" horizontal reinforcement in the wall/pier.


231

Design Example 4

6.

!


Design 8'-0" shear wall on line A for combined axial and in-plane bending actions.

Part 5 illustrated the design of the wall for shear strength. This Part illustrates design for wall overturning moments combined with gravity loads. A free body diagram of the wall/pier is needed to understand the imposed forces on the wall. The load combinations to be considered are specified in §1612.2.1. These are as follows (with the 1.1 factor of Exception 2 applied): 1.1(1.2 D + 0.5 L + 1.0 E ) (floor live load, L = 0)

(12-5)

1.1 (0.9 D − 1.0 E )

(12-6)

E = ρE h + E v

(30-1)

E v = 0.5C a ID = 0.5 (0.53)(1.0) D = 0.27 D

§1630.1.1

The resulting Equation (12-5) is: 1.1(1.2 D + 0.27 D + 1.0 E h ) = 1.61D − 1.1E h The resulting Equation (12-6) is: 1.1 (0.9 D + 0.27 D + 1.0 E h ) = 0.63D − 1.1E h E h = V8' −0" wall = 1.1 (30.3 k ) = 33.3 k Axial loads Pu are calculated as Pu1 and Pu 2 for load combinations of Equations (12-5) and (12-6): Pu1 = 1.61(27,750 lb ) = 44.7 kips Pu 2 = 0.63(27,750 lb ) = 17.5 kips

232


Design Example 4

!


By performing a sum of moments about the bottom corner at point A (Figure 4-9):

Pu

Mu, top

Vu

10'-0"

A Vu

Mu,bottom

8'-0"

Figure 4-9. Free body diagram of 8’-0” shear wall

∑ M A = 0 = 2 M u − Vu (10 ft ) M u , top ≈ M u , bottom =

(33.3 k )(10 ft ) 2

= 166.5 k - ft

The reader is referred to an excellent book for the strength design of masonry Design of Reinforced Masonry Structures, by Brandow, Hart, Verdee, published by Concrete Masonry Association of California and Nevada, Sacramento, CA, Second Edition, 1997. This book describes the calculation of masonry wall/pier strength design in detail. The axial load vs. bending moment capacity (P-M) diagram for the wall must be calculated. For this, the designer must understand the controlling strain levels that define yielding and ultimate strength. At yield moment, the steel strain is the yielding strain (0.00207 in./in. strain) and the masonry strain must be below 0.002 in./in. (for under-reinforced sections). At ultimate strength, the masonry has reached maximum permissible strain (0.003 in./in.) and the steel strain is considered to have gone beyond yield strain level (see§2108.2.1.2 for a list of design assumptions). See Figure 4-10 for concrete masonry stress-strain behavior. A representation of these strain states is shown in Figures 4-11 and 4-12 (the pier width is defined as h ).


233

Design Example 4

!


Compressive stress

fm

f 'm

(psi)

0.5 f ' m Figure 4-10. Assumed masonry compressive stress versus strain curve

0.002 Strain,

0.003

em

Figure 4-11. Strain diagram at yield moment; steel strain =0.00207 in./in.; masonry strain is less than yield for under-reinforced sections

ε m ≤ 0.002

ε s2

ε s3

ε s1 = 0.00207

Figure 4-12. Strain diagram at ultimate moment; masonry strain =0.003 in./in.; steel strain has exceeded 0.00207 in./in.; the Whitney stress block analysis procedure can ε s1 ≥ 0.00207 be used to simplify calculations 234

c

ε m ≥ 0.003

ε s2

ε s3 c


Design Example 4

!


Note that masonry strain may continue to increase with a decrease in stress beyond strains of 0.002 in./in. at which time stresses are at f ' m . At strains of 0.003, masonry stresses are 0.5 f ' m . With boundary element confinement, masonry strains can be as large as 0.006 in./in. By performing a summation of axial forces F , the axial load in the pier is calculated as:

∑ F = P = C1 = T1 = T2 = T3 The corresponding yield moment is calculated as follows: h h h h c    M y = T1  d 1 −  + T2  d 2 −  + T3  d 3 −  + C  −  2 2 2  2 3    The ultimate moment is calculated as: h h h h a    M u = T1  d 1 −  + T2  d 2 −  + T3  d 3 −  + C  −  2 2 2 2 2    Strength reduction factors, φ , for in-plane flexure are determined by Equation (8-1) of §2108.1.4.1.1 φ = 0.80 −

Pu , 0.6 ≤ φ ≤ 0.8 (Ae f ' m )

(8-1)

Strength reduction factors for axial load, φ = 0.65. For axial loads, φPn , less than 0.10 f ' m Ae , the value of φ may be increased linearly to 0.85 as axial load, φPn , decreases to zero. The balanced axial load, Pb , is determined by Equations (8-2) and (8-3). Pb = 0.85 f ' m ba b    em a b = 0.85d  f  em + y  Es 

(8-2)       


(8-3)

235

Design Example 4

!


Pb = 0.85 (2,500)(7.625 in.)(0.85)(92 in.)(0.003 0.00507 ) = 750 kips φPb = 0.65(750 kips ) = 487 kips A P-M diagram can thus be developed. The P-M diagrams were calculated and plotted using a spreadsheet program. By observation, the design values Pu and M u (Pu = 43 k, M u = 167 k - ft ) are within the nominal strength limits of φPn , φM n values shown in Figure 4-13. Plots for Pn vs. M n can be seen in Figure 4-13 and for φPn vs. φM n in Figure 4-14.

2,000 1,800 1,600

Pn (kips)

1,400 1,200 1,000 800 600 400 200 0 0

200

400

600

800

1,000

1,200

1,400

Mn (k-ft)

Figure 4-13. The Pn-Mn nominal strength curve with masonry strain at 0.003 in./in.

236


Design Example 4

!


2,000 1,800 1,600

φPn (kips)

1,400 1,200 1,000 800 600 400 200 0 0

200

400

600

800

1,000

1,200

1,400

φMn (k-ft)

Figure 4-14. The φPn-φMn design strength curve with masonry strain at 0.003 in./in.

Check for type of wall failure by calculating wall moment at shear Vn :  82.7 k   (10') Vn (10')  0.60  Mu = = 689 k - ft = 2 2 Pu = 43.7 k By looking at the Pn − M n curve, this Pu − M u load is just outside the Pn , M n curve. The shear wall failure will likely be a bending failure. However, the designer might still consider a φ = 0.60 for shear design to be conservative.

7.

Deflection of shear wall on line A.

§1630.10

In this part, the deflection of the shear wall on line A will be determined. This is done to check actual deflections against the drift limits of §1630.10. Deflections based on gross properties are computed as: ∆s =

Vi h 3 1.2Vi h + for wall/piers fixed top and bottom AG 12 E m I


237

Design Example 4

!

∆s


(28.6 k )(120 in.)3 1.2 (28.6 k )(120 in.) = + = 0.011 in. 3 12 (1,875 ksi ) (562,176 in ) (732 in 2 ) (750 ksi )

Assume cracked section properties and I cr = 0.3I g (approximately): ∆s =

(28.6 k )(120 in.)3 1.2 (28.6 k )(120 in.) + = 0.021 in. 3 12 (1,875 ksi ) (168,652 in ) ( 732 in 2 ) (750 ksi )

∆ m = 0.7 R∆ s = 0.7 (4.5)(0.021 in.) = 0.066 in.

(30-17)

Thus, deflections are less than 0.025h = 3.0 in. ∴ o.k.

8.

Requirements for shear wall boundary elements.

§2108.2.5.6

Section §2108.2.5.6 requires boundary elements for CMU shear walls with strains exceeding 0.0015 in./in. from a wall analysis with R = 1.5 . The intent of masonry boundary elements is to help the masonry achieve greater compressive strains (up to 0.006 in./in.) without experiencing a crushing failure. The axial load and moment associated with this case is: Pu = 44.7 kips Mu =

4.5 (166.5 k - ft ) = = 619 k - ft 1.1 1.1

This P-M point is not within the P-M curve using a limiting masonry strain of 0.0015 in./in. (see Figure 4-15). From an analysis it can be determined that the maximum c distance to the neutral axis is approximately 22 inches. For this example, boundary ties are required. Note that narrow shear wall performance is greatly increased with the use of boundary ties. The code requires boundary elements to have a minimum dimension of 3 × wall thickness, which is 24 inches due to yield moments. After yield moment capacity is exceeded, the c distance is reduced. Thus, if boundary element ties are provided at each end of the wall/pier extending 24 inches inward, the regions experiencing strain greater than 0.0015 in./in. are confined. Space boundary ties at 8-inch centers. The purpose of masonry boundary ties is not to confine the masonry for compression, but to support the reinforcement in compression to prevent buckling. Tests have been performed to show that masonry walls can achieve 0.006 in./in. compressive strains when boundary ties are present. 238


Design Example 4

!


1,100 1,000 900 800

P (kips)

700 600 500 400 300 200 100 0 0

500

1,000

1,500

2,000

M (k-ft)

Figure 4-15. P-M curve for boundary element requirements; masonry strain is limited to 0.0015 in./in.

The P-M curve shown in Figure 4-15 is derived by setting masonry strain at the compression edge at 0.0015 in./in. and by increasing the steel tension strain at the opposite wall reinforcement bars. Moments are calculated about the center of the wall pier and axial forces are calculated about the cross-section. P-M points located at the outside of the denoted P-M boundary element curve will have masonry strains exceeding the allowable, and thus will require boundary element reinforcement or devices. It can be seen that boundary reinforcement is required for the point (Pu = 45 k, M u = 619 k ) . Boundary element confinement ties may consist of #3 or #4 closed reinforcement in 10-inch and 12-inch CMU walls. At 8-inch CMU walls pre-fabricated products such as the “masonry comb” are the best choice for boundary reinforcement because these walls are too narrow for reinforcement ties (even #3 and #4 bars). The boundary reinforcement should extend around three vertical #4 bars at the ends of the wall.


239

Design Example 4

9.

!


Wall-roof out-of-plane anchorage for lines 1 and 3.

CMU walls should be adequately connected to the roof diaphragm around the perimeter of the building. In earthquakes, including the 1994 Northridge event, a common failure mode has been separation of heavy walls and roofs leading to partial collapse of roofs. A recommended spacing is 8’-0" maximum. However, 6'-0" or 4'-0" might be more appropriate and should be considered for many buildings. This anchorage should also be provided on lines A and D, which will require similar but different details at the roof framing perpendicular to wall tie condition. UBC §1633.2.9 requires that diaphragm struts or ties crossing the building from chord to chord be provided that transfer the out-of-plane anchorage forces through the roof diaphragm. Diaphragm design is presented in Design Example 5, and is not presented in this example. Per §1633.2.8.1, elements of the wall out-of-plane anchorage system shall be designed for the forces specified in §1632 where R p = 3.0 and a p = 1.5 . Fp =

a p Ca I p  h 1 + 3 x Rp  hr

 W p 

Fp =

16'  1.5(.53)(1.0 )  1 + 3 × W p = 1.06W p 16'  3.0 

(32-2)

or: f p = 1.06w p , where w p is the panel weight of 75 psf (see Figure 4-16) loading.

a

qroof

fp h

Figure 4-16. Wall-roof connection loading diagram

240


Design Example 4

!


Calculation of the reaction at the roof level is: q roof =

w p (h + a )2 2h

=

(1.06)(75 psf )(16 ft + 3 ft )2 2 (16 ft )

= 897 plf

Section 1633.2.8.1 requires a minimum wall-roof anchorage of q roof = 420 plf q roof = 897 plf ≥ 420 plf ∴ use q roof = 897 plf The design anchorage reaction at different anchor spacings is thus: at 4'-0" centers, q roof = 3,588 lb at 6'-0" centers, q roof = 5,382 lb at 8'-0" centers, q roof = 7,175 lb Therefore, choose wall-roof anchors that will develop the required force at the chosen spacing. The roof diaphragm must also be designed to resist the required force with the use of subdiaphragms (or other means). The subject of diaphragm design is discussed in Design Example 5. For this example, a double holdown connection spaced at 8'-0" centers will be used (see Figure 4-19). This type of connection must be secured into a solid roof framing member capable of developing the anchorage force. First check anchor capacity in concrete block of Tables 21-E-1 and 21-E-2 of Chapter 21. Alternately, the strength provisions of §2108.1.5.2 can be used. The required tension, T, for bolt embedment is T = E 1.4 = 7,175 lbs 1.4 = 5,125 lb . For ¾-inch diameter bolts embedded 6 inches, T = 2,830 lb per Table 21-E-1 and 3,180 lb per Table 21-E-2. These values are for use with allowable stress design (ASD).


241

Design Example 4

!


Figure 4-17. Intersection of anchor bolt tension failure cones

The anchor bolts are spaced at 6-5/8 inches center to center (considering purlin and hardware dimensions) and have 12-inch diameter pull-out failure cones. Thus, the failure surfaces will overlap (Figure 4-17). In accordance with §2108.1.5.2, the maximum tension of this bolt group may be determined as follows: Calculate Bt n per bolt using the strength provisions of Equation (8-5): Btn = 1.04 A p

(

)

f ' m = 1.04 113 in. 2 (50 psi ) = 5,876 lb

(8-5)

Calculate one-half the area of intersection of failure surfaces from two circles with radius 6 inches and centers (2-1/16" + 2½" + 2-1/16") 6 5/8" apart. A p = 37.8 in.2 from Equations (8-7) and (8-8). Thus the bolt group tension can be calculated as:

(1.0) (2 × 113 in. 2 − 2 × 37.8 in. 2

)

2 (50 psi ) = 9,410 lb

φ Btn ≥ Btu ∴ 0.8(9,410 lb ) = 7,528 lb ≥ 7,175 lb ∴ o.k.

By choosing a pair of pre-fabricated holdown brackets with adequate capacity for a double shear connection into a 2½-inch glued-laminated framing member, the brackets are good for 2 × 3,685 lb = 7 ,307 lb (ASD) > 7,175 lb × 1.4 steel element factor/1.4 ASD factor = 7,175 lb. Thus, the brackets are okay.

242


Design Example 4

!


Also check bolt adequacy in the double shear holdown connection with metal side plates (2½-inch main member, 7/8-inch bolts) per NDS Table 8.3B. T = 2 × 3,060 lb × 1.33 = 8,140 lb > 7 ,175 lb, if the failure is yielding of bolt (Mode IIIs or IV failure). If the failure is in crushing of wood (Mode I m failure), the required force is 0.85 × 5,125 lb = 4 ,356 lb. Therefore, the double shear bolts and pre-fabricated holdown brackets can be used. Thus, use two holdown brackets on each side of a solid framing member connecting the masonry wall to the framing member with connections spaced at 8'-0" centers. Verify that the CMU wall can span laterally 8'-0" between anchors. Assume a beam width of 6'-0" (3' high parapet plus an additional three feet of wall below roof) spanning horizontally between wall-roof ties. w = q roof = 897 plf 2 wl 2 (897 plf )(8 ft ) = = 7,176 lb - ft 8 8

Mu =

The wall typically has #4@16-inch horizontal reinforcement, therefore a minimum 4-#4 bars in 6'-0" wall section.

a=

As f y .85 f ' m b

=

(

)

4 .20 in. 2 (60,000 psi ) = 0.314 in. .85 (2,500 psi )(72 in.)

a  φM n = φAs f y  d −  2 

(

)

.314 in.   1    = 11,689 lb - ft ≤ 7,176 lb - ft φM n = 0.8 (4 ) .20 in. 2 (60,000 psi ) 3.81 in. −  2   12 in.   ∴ o.k.

Per §1633.2.8.1, item 5, the wall-roof connections must be made with 2½-inch minimum net width roof framing members (2½-inch GLB members or similar) and developed into the roof diaphragm with diaphragm nailing and subdiaphragm design.


243

Design Example 4

!


Anchor bolt embedment and edge distances are controlled by §2106.2.14.1 and §2106.2.14.2. Section 2106.2.14.1 requires that the shell of the masonry unit wall next to the wood ledger have a hole cored or drilled that allows for 1-inch grout all around the anchor bolt. Thus, for a 7/8-inch diameter anchor bolt, the core hole is 2-7/8-inch in diameter at the inside face masonry unit wall. Section 2106.2.14.2 requires that the anchor bolt end must have 1½ inches clearance to the outside face of masonry. The face shell thickness for 8-inch masonry is 1¼ inches, thus the anchor bolt end distance to the inside face of the exterior shell is 7-5/8"-1¼"-6" = 3/8". It is recommended that the minimum clear dimension is ¼-inch if fine grout is used and ½-inch if coarse pea gravel grout is used (Figure 4-18).

Figure 4-18. Embedment of anchor bolts in CMU walls (MIA, 1998)

244


Design Example 4

10. 10

!


Chord design.

Analysis of transverse roof diaphragm chords is determined by calculation of the diaphragm simple span moment wl 2 8 divided by the diaphragm depth.

(

wdiaph, trans. =

)

(72 k + 50 k ) = 1,356 plf 90'

Modify w for R = 4.0 by factor (4.5/4.0) = 1.125

§1633.2.9, Item 3

M diaph. = wl 2 8 = 1.125 (1,356 plf )(90 ft )2 8 = 1,545 k - ft Tu = C u = 1,545 k - ft 60 ft = 25.7 kips Using reinforcement in the CMU wall for chord forces: As , required =

Tu 25.7 k = = 0.54 in. 2 φf y (0.80 )(60 ksi )

(

)

Thus 2-#5 chord bars As = 0.62 in.2 are adequate to resist the chord forces. Place chord bars close to the roof diaphragm level. Since roof framing often is sloped to drainage, the chord placement is a matter of judgment.

Figure 4-19. CMU wall section at wall-roof ties


245

Design Example 4

!


References ACI 530-99 / ASCE 6-99 / TMS 402-99, 1999, Building Code Requirements for Masonry Structures. American Concrete Institute, Farmington Hills, Michigan, American Society of Civil Engineers, Reston, Virginia, The Masonry Society, Boulder, Colorado. Amrhein, J.E., 1996, Reinforced Masonry Engineering Handbook, 5th Edition. Masonry Institute of America, Los Angeles, California. Brandow, G E., Hart, G., and Virdee, A., 1997, Design of Reinforced Masonry Structures. Concrete Masonry Association of California and Nevada (CMACH), Sacramento, California. MIA, 1998, Reinforced Concrete Masonry Construction Inspector’s Handbook: Conforming to the 1997 UBC. Masonry Institute of America, 2550 Beverly Boulevard, Los Angeles, California 90057. Paulay, T. and Priestly, M.J.N., 1992, Seismic Design of Reinforced Concrete and Masonry Buildings. John Wiley & Sons, Inc., New York. Robinson, A. and Uzarski, J., 1999, CMD97, Concrete Masonry Design to the 1997 UBC. Computer Aided Design of Reinforced Concrete and Clay Masonry Elements in Accordance with the 1997 Uniform Building Code, Concrete Masonry Association of California and Nevada (CMACH), Sacramento, California.

246


Design Example 5

!

Tilt-Up Building

Design Example 5 Tilt-Up Building

Figure 5-1. Tilt-up building of Design Example 5

Overview In this example, the seismic design of major components of a tilt-up building are presented. Many tilt-up buildings have suffered severe structural damage in earthquakes, particularly during the 1971 San Fernando and 1994 Northridge events. The most common problem is wall-roof separation, with subsequent partial collapse of the roof. In the 1997 UBC, substantial improvements, including higher wall-roof anchorage forces, have been added to help prevent the problems that appeared in tilt-up buildings built to codes as recent as the 1994 UBC. The example building is the warehouse shown in Figure 5-1. This building has tilt-up concrete walls and a panelized plywood roof system. The building’s roof framing plan is shown in Figure 5-2, and a typical section through the building is given in Figure 5-3. The emphasis in this Design Example 5 is the seismic design of the roof diaphragm, wall-roof anchorage, and a major collector.


247

Design Example 5

!

Tilt-Up Building

Outline This example will illustrate the following parts of the design process: 1.


2.

Design the roof diaphragm.

3.

Design typical north-south subdiaphragm.

4.

Design wall-roof ties for north-south subdiaphragm.

5.

Design continuity ties for north-south direction.

6.

Design of collector along line 3 between lines B and C.

7.

Required diaphragm chord for east-west seismic forces.

8.

Required wall panel reinforcing for out-of-plane forces.

9.

Deflection of east-west diaphragm.

10. 10

Design shear force for east-west panel on line 1.

Given Information The following information is given: Roof: dead load = 14.0 psf Walls: thickness = 7.25" height = 23' normal weight concrete = 150 pcf f ' c = 4,000 psi A615, Grade 60 rebar f y = 60 ksi

(

)

Seismic and site data: Z = 0.4 (Zone 4) I = 1.0 (Standard occupancy) seismic source type = B distance to seismic source = 13km soil profile type = S D ρ N/S = 1.0 ρ E/W = 1.5 (due to short wall on line 3)

Roof sheathing: Structural I plywood

248


Design Example 5

!

Tilt-Up Building

Figure 5-2. Roof framing plan of tilt-up building

Figure 5-3. Typical cross-section


249

Design Example 5

!

Tilt-Up Building


1.

Code Reference


§1630.2.2

Using Method A, the period is calculated as: T = Ct (hn ) 4 = .020(21) 3

3

4

= .20 sec

(30-8)

Comment: The building’s lateral force-resisting system has relatively rigid walls and a flexible roof diaphragm. The code formula for period does not take into consideration that the real period of the building is highly dependent on the roof diaphragm construction. Consequently, the period computed above using Equation (30-8) is not a good estimate of the real fundamental period of the building, however it is acceptable for determining design base shear. With seismic source type B and distance to source = 13 km N a = 1.0

Table 16-S

N v = 1.0

Table 16-T

For soil profile type SD and Z = .4 C a = .44 N a = .44(1.0 ) = .44

Table 16-Q

C v = .64 N v = .64(1.0 ) = .64

Table 16-R

Since tilt-up concrete walls are both shear walls and bearing walls: R = 4.5

Table 16-N

Design base shear is calculated from: V =

Cv I .64(1.0 ) W = W = .677W RT 4.5(.21)

(30-4)

but base shear need not exceed: V =

250

2.5C a I 2.5(.44 )(1.0 ) W = .244W W = 4.5 R

(30-5)


Design Example 5

!

Tilt-Up Building

A check of Equations (30-6) and (30-7) indicate these do not control, therefore the base shear in both directions is V = .244W Note that the base shear is greater than that required under the 1994 UBC. The principal reason for this is that base shear under the 1997 UBC is determined on a strength design basis. If allowable stress design (ASD) is used, the base shear is divided by 1.4 according to §1612.3.

2.

Design the roof diaphragm.

2a. 2a

Roof diaphragm weight.

Seismic forces for the roof are computed from the weight of the roof and the tributary weights of the walls oriented perpendicular to the direction of the seismic forces. This calculation is shown below: roof area = 110 ft (64 ft ) + 140.67 ft (224 ft ) = 38,550 sq ft roof weight = 38,550 sq ft (14 psf ) = 539.7 kips wall weight =

7.25 × 150 = 90.6 psf 12

north-south walls = 90.6 psf (2 ft + 10.5 ft )(140.67 ft )(2 ) = 318.6 kips east-west walls = 90.6 psf (2 ft + 10.5 ft )(288 ft )(2 ) = 652.3 kips In this example, the effect of any wall openings has been neglected. This is considered an acceptable simplification because the openings usually occur in the bottom half of the wall.

2b. 2b

Roof diaphragm shear.

The roof diaphragm must be designed to resist seismic forces in both directions. The following formula is used to determine the total seismic force, Fpx , on the diaphragm at a given level of a building. In general, separate forces are computed for each direction.


251

Design Example 5

!

Tilt-Up Building

n

F px =

Ft + ∑ Fi i= x

n

∑ Wi

W px

(33-1)

i= x

Base shear for this building is V = .244W . This was determined using R = 4.5 as shown in Part 1 above. For diaphragm design, however, §1633.2.9 requires that R not exceed 4. Since this is a one-story building with Ft = 0 , and using R = 4 , Equation (33-1) becomes the following: F px =

4.5  V  4.5 (.244) W px = 0.275W px   W px = 4 W  4

Fpx need not exceed 1.0C a IW px = 1.0 (.44 )(1.0 )W px = .44W px

§1633.2.9

but cannot be less than 0.5C a IW px = 0.5 (.44 )(1.0 )W px = .22W px

§1633.2.9

Therefore, for diaphragm design use F px = .275W px Note: The reliability/redundancy factor ρ is not applied to horizontal diaphragms, except transfer diaphragms. (Refer to Examples 15 and 16 in Volume I of the Seismic Design Manual for a discussion of the ρ factor.) North-south direction: W px = 539.7 k + 318.6 k = 858.3 kips F px = .275 (858.3) = 236.0 kips The equivalent uniform load on the diaphragm can be computed as: w=

236.0 kips = 1,678 plf 140.67'

In this calculation, an approximation has been made that the uniform load between lines A and B is the same as that between B and E. The actual load on the A-B segment is less, and the load on the B-E segment is slightly greater than that shown. This has been done to simplify the computations. Because the panelized wood roof diaphragm in this building is considered flexible (see §1630.6 for definition of flexible diaphragm), lines A, B and E are considered lines of resistance for the north-south seismic forces. A collector is needed along line B to drag the tributary north-south diaphragm forces into the shear wall on line B. The shear diagram is shown below. 252


Design Example 5

!

Tilt-Up Building

25.7 k

A 30'-8" 25.7 k

B

92.3 k

w = 1,678 plf 110'-0"

92.3 k

E

Shear

Loading

Figure 5-4. Seismic loading and shear diagram for north-south diaphragm

3

1

Diaphragm shear at line A and on the east side of line B is:

10 224'-0"

64'-0"

25,700lbs = 115 plf 224' Diaphragm shear at the west side of line B and at line E is: 92,300 lb = 320 plf 288 ft

w = 1,138 plf Loading

East-west direction:

127.5 k

Diaphragm forces for the east-west direction are computed using the same procedure and assumptions as the north-south direction. The actual load on segment 1-3 is less than that shown, and the load on 3-10 slightly greater.

36.4 k

36.4 k

W px = 539.7 k + 652.3 k = 1,192. 0 kips F px = .275 (1,192.0 k ) = 327.8 kips

127.5 k Shear

Equiv. w =

327.8 k = 1,138 plf 288 ft


Figure 5-5. Seismic loading and shear diagram for east-west diaphragm

253

Design Example 5

!

Tilt-Up Building

Diaphragm shear at line 1 and the north side of line 3 is: 36,400 lb = 331 plf 110 ft Diaphragm shear at the south side of line 3 and at line 10 is: 127,500 lb = 906 plf 140.67 ft

2c. 2c

Design of east-west diaphragm.

The east-west diaphragm has been selected to illustrate the design of a plywood roof diaphragm. Allowable stress design (ASD) will be used. The basic earthquake loading combination is given by Equation (30-1). When ASD is used, vertical effects need not be considered, and in this example of the diaphragm design, they would not come into use even if strength design was being used. As discussed earlier, the reliability/redundancy factor does not apply to the diaphragm, and ρ=1 in Equation (30-1). E = ρE h + E v = 1.0 E h + 0 = 1.0 E h

(30-1)

For ASD, the basic load combination to be used to combine earthquake and dead load is Equation (12-9). This simplifies to the following: D+ E = 0+ E = E 1.4 1.4 1.4

(12-9)

Assume the diaphragm is to be constructed with ½-inch Structural I plywood with all edges supported. Refer to use UBC Table 23-II-H for nailing requirements. Sheathing arrangement (shown in Figure 5-2) for east-west seismic forces is Case 4. Diaphragm shear forces must be divided by 1.4 to convert to ASD. Because open web truss purlins with double 2x4 chords are used in this direction, the framing width in the east-west direction is 3½ inches. However, in the north-south direction, the framing consists of 2 × subpurlins, and strength is therefore limited by the 2-inch nominal width. Required nailing for panel edges for various zones of the roof (for east-west seismic only) is given in Table 5-1 below. Minimum field nailing is 10d @ 12 inches. A similar calculation (not shown) must be done for north-south seismic forces.

254


Design Example 5

Table 5-1. Diaphragm nailing for east-west seismic forces Boundary and East-West Edge North-South Zone Edge Nailing (2) Allowable Shear Nailing (1) A 10d @ 2½" 4" 640 plf

!

ASD Shear

Tilt-Up Building

Status

906/1.4 = 647 plf

say o.k.

B

10d @ 4"

6"

425 plf

583/1.4 = 416 plf

o.k.

C

10d @ 6"

6"

320 plf

331/1.4 = 236 plf

o.k.

Notes: 1. 2.

The east-west running sheet edges are the “continuous panel edges parallel to load” mentioned in Table 23-II-H. The north-south sheet edges are the “other panel edges” in Table 23-II-H. Note that the nailing for north-south running diaphragm boundaries is 10d @ 2½ inches.

The demarcation between nailing zones A and B is determined as follows. It was decided to use 10d at 2½-inch spacing in A and 4-inch spacing in B. The limiting shear for 10d at 4 inches (from Table 23-II-H) is 425 plf. Shear reduces from a maximum of 906 plf at lines 3 and 10 to 595 plf (i.e., 425 plf × 1.4 = 595 plf) at 38.4 feet from lines 3 and 10. Rounding to the nearest 8-foot increment because purlins are spaced at 8 feet o.c., zone A extends a distance of 40 feet from lines 3 and 10 as shown below.

1

3

10 40'-0"

40'-0" 64'-0"

144'-0"

A

B

A

C

Figure 5-6. Nailing zones for east-west roof diaphragm

The above illustrates design of the east-west diaphragm for shear. Design of the chord for the east-west diaphragm is shown in Part 7 of this example. Design of ledger bolts, required to transfer the diaphragm shear to the wall panels, is not shown.


255

Design Example 5

3.

!

Tilt-Up Building

Design typical north-south subdiaphragm.

Subdiaphragms are used to transfer out-of-plane seismic forces from the tilt-up wall panels to the main diaphragm. Consequently, subdiaphragms are considered to be part of the wall anchorage system as defined in §1627. In the example below, design of a typical subdiaphragm for north-south seismic forces is shown. Design of subdiaphragm for east-west seismic forces is similar but not shown.

3a. 3a

Check subdiaphragm aspect ratio.

Maximum allowable subdiaphragm ratio is 2.5 to 1 From Figure 5-2, the maximum north-south subdiaphragm span =

Minimum subdiaphragm depth =

§1633.2.9 110 ft = 36.67 ft 3

36.67 ft = 14.67 ft 2.5

Typical roof purlin spacing = 8 ′ − 0 ′′ Minimum subdiaphragm depth = 16 ′ − 0 ′′ ∴ Must use subdiaphragm at least = 16 ′ − 0 ′′ deep

3b. 3b

Forces on subdiaphragm.

Because subdiaphragms are part of the out-of-plane wall anchorage system, they are designed under the requirements of §1633.2.8.1. Seismic forces on a typical north-south subdiaphragm are determined from Equation (32-2) with R p = 3.0 and a p = 1.5. w p = 90.6 psf Fp =

a p Ca I p  h 1 + 3 x Rp  hr

 W p 

(32-2)

The value of F p to be used in wall-roof anchorage design is determined from Equation (32-2) using h x = hr , and W p is the tributary weight. The tributary wall weight is one-half of the weight between the roof and base plus all of the weight above the roof.

256


Design Example 5

!

Tilt-Up Building

W p = 90.6 psf (2 ft + 10.5 ft )(1 ft ) = 1,133 lb/ft Fp =

1.5 (.44 )1.0  21  1 + 3 ×  W p = .88W p 3.0 21  

Solving for the uniform force per foot, q , at the roof level q = .88W p = .88 (1,133)= 997 plf

2'-0" q 10'-6"

.88Wp

10'-6"

Figure 5-7. Loading diagram for wall-roof anchorage design

Check minimum wall-roof anchorage force 997 plf > 420 plf

o.k.

§1633.2.8.1(1)

∴ q = 997 plf

3c. 3c

Check subdiaphragm shear.

Assume a 32-foot deep subdiaphragm as shown below. This is done for two reasons. First, the GLB along Line 9 can be used as a chord. Second, the deeper than required subdiaphragm depth (32 feet vs. 16 feet) makes the subdiaphragm displacement more compatible with that of the main north-south diaphragm.


257

Design Example 5

!

Tilt-Up Building

Figure 5-8. Typical north-south subdiaphragm

Shear reaction to glulam beams along lines C and D: R=

997 plf (36.67 ft ) = 18,280 lb 2

Maximum shear =

18,280 lb = 571 plf 32

From Table 5-1, the minimum nailing in Zone A (Figure 5-6) is 10d @ 4 in. along north-south edges, except at boundaries. Load on an ASD basis with the 0.85 load factor of §1633.2.8.1(5) applied is 0.85

(571 plf ) = 347 plf 1.4

Check 10d @ 4 in. for Case 2, capacity = 640 plf > 347 plf o.k.

Table 23-II-H

∴Use of Zone A nailing for subdiaphragm okay

258


Design Example 5

3d. 3d

!

Tilt-Up Building

Check GLB as subdiaphragm chord.

Glulam beams (GLB) along lines 2 and 9, and the continuous horizontal reinforcement in panels along lines 1 and 10, act as chords for the subdiaphragms. Check to see if the GLB can carry additional seismic force within incremental one-third allowable tension increase using ASD. Note that 0.85 load factor of §1633.2.8.1(5) is applied to the chord force when checking the tension stress in the GLB. Chord force =

997 plf (36.67 )2 = 5,237 lb 8(32 )

Assume GLB 6 3 4 × 24 with 24F-V4 DF/DF A = 162 in.2 Ft = 1,150 psi ft =

Table 5A, 91 NDS

0.85 (5,237 lb) 1.4 × 162 in.

2

= 20 psi <

1 × 1,150 psi = 383 psi o.k. 3

Comment: In reality, the GLB along line 9 may not act in tension as a subdiaphragm chord as shown above. It will be loaded in tension only when compressive wall anchorage forces act on the diaphragm. Under this loading, the seismic forces probably do not follow only the subdiaphragm path shown above but are also transmitted through the wood framing to other parts of the diaphragm. Even if subdiaphragm action does occur, the subdiaphragm may effectively be much deeper than shown. However, because it is necessary to demonstrate that there is a system to resist the out-of-plane forces on the diaphragm edge, the subdiaphragm system shown above is provided.

3e. 3e

Determine minimum chord reinforcement at exterior concrete walls.

This Design Example 5 assumes that there is continuous horizontal reinforcement in the walls at the roof level that acts as a chord for both the main diaphragm and the subdiaphragms. The 1.4 load factor of §1633.2.8.1(4) must be applied to the reinforcement. Subdiaphragm chord force = P = 5,237 lb As =

1.4 (5,237 ) P = = 0.14 in.2 φf y 0.9 (60,000 )


259

Design Example 5

!

Tilt-Up Building

This is a relatively small amount of reinforcement. Generally, the main diaphragm chord reinforcement exceeds this amount. In present California practice, the subdiaphragm chord steel requirement is not added to the chord steel requirement for the main diaphragm. Determination of the main chord reinforcement is shown in Part 7.

4.

Design wall-roof ties for north-south subdiaphragm.

§1633.2.8.1

The key elements in the wall anchorage system, defined in §1627, are the wall-roof ties. Wall-roof ties are used to transfer out-of-plane seismic forces on the tilt-up wall panels to the subdiaphragms. Requirements for connection of out-of-plane wall anchorages to flexible diaphragms are specified in §1633.2.8.1.

4a. 4a

Seismic force on wall-roof tie.

Seismic forces are determined using Equations (32-1) or (32-2). Values of R p and a p are: R p = 3.0 a p = 1.5

§1633.2.8.1(1)

Forces on the anchorage were computed above in Part 3, using the same values of R p and a p , and are q = 997 plf .

4b. 4b

Design typical wall-roof tie.

Minimum required thickness of a subpurlin used as wall-roof tie = 2½ inches

§1633.2.8.1(5)

Try ties at 8 ft-0 in. spacing, and determine F p F p = 8 ft × 997 plf = 7,976 lb Comment: When tie spacing exceeds 4 feet, the SEAOC Blue Book (§108.2.6) recommends that walls be designed to resist bending between anchors. Try prefabricated metal holdowns with two ¾-inch bolts in subpurlin and two ¾-inch bolts connecting the subpurlin to the wall panel. This connection (Figure 5-9) is designed to take both tension and compression as recommended by the SEAOSC/COLA Northridge Tilt-up Building Task Force and the SEAOC Blue Book (§C108.2.8.1). Design of the holdown hardware not shown. Consult ICBO Evaluation Reports for allowable load capacity of pre-manufactured holdowns. Note that if a one-sided holdown is used, eccentricities in the subpurlin must be considered, as specified in §1633.2.8.1(2). Generally, one-sided wall-roof anchorage is not recommended.

260


Design Example 5

!

Tilt-Up Building

precast wall panel

plywood sheathing

6"

¾"anchor bolt (2 total)

holdown each side of subpurlin w/ 2-¾" M.B.

3x subpurlin 5"

7¼"

nut and washer, typical ledger

Figure 5-9. Typical subpurlin wall-roof tie

Check capacity of the two ¾-inch bolts in DF-L subpurlin using ASD:

(2,630)(2 bolts)(1.33) = 6,996 lb > 0.85 (7 ,976 lb) = 4,843 lb 1.4

Table 8.3B, 91 NDS

o.k.

Note that the .85 load factor of §1633.2.8.1(5) is used to reduce the seismic force. This applies to forces on nails and bolts connecting brackets or strips to the wood framing because these are considered “wood elements” under the code (see SEAOC Blue Book §C108.2.8.1). Comment: The Blue Book (§C108.2.8.1) makes a recommendation for the minimum length to diameter ratio of the through-bolts connecting the holdowns to the subpurlin. In this case, the l/d ratio is 2.5/0.75 = 3.3. The minimum recommended value is 4.5. This ratio is necessary to maintain a ductile failure mode (e.g., bending of the through-bolts). To satisfy the Blue Book recommendation, a 4x subpurlin would be required in this situation. Minimum required end distance = 7 D = 7 (.75) = 5.25 in.

Table 8.5.4, 91 NDS

A distance of 6 inches from the through-bolt in the holdown to the ledger will be used. Often, there is a gap of 1/8-inch or more between the end of the subpurlin and the side of the ledger due to panelized roof erection methods, and the use of a 6-inch edge distance will ensure compliance with the 7D requirement. A larger


261

Design Example 5

!

Tilt-Up Building

distance can be used to ensure that through-bolt tear out does not occur in the 3 × subpurlin. Check tension capacity of two ¾-inch A307 anchor bolts using ASD: Ft = 20.0 ksi

Table 1-A, AISC-ASD

P = Ft AB (2 bolts)(1.33)

(

P = (20.0 ksi ) 0.4418 in. 2

) (2 bolts)(1.33) = 23.5 k > 1.4 (17.,4976) lb = 8.0 kips

o.k.

As specified in §1633.2.8.1(4), the 1.4 steel factor has been used to increase the seismic force. Check compression capacity of two ¾-inch A307 anchor bolts using ASD: Radius of gyration of ¾-inch rod = 0.75-inch/4 = 0.1875-inch Assume L = 4½-inch L 4.5" = = 24, r 0.1875"

Fa = 20.35 ksi

Table C-36, AISC-ASD

P = Fa AB (2 rods )(1.33)

(

P = (20.35 ksi ) 0.4418 in. 2

) (2 rods)(1.33) = 23.9 k > 8.0 kips

o.k.

Check tension capacity of anchor bolts in wall panel for concrete strength: The tilt-up panels are exterior wall elements, but the requirements of §1633.2.4.2 do not apply. This is because the tilt-up panels are both bearing walls and shear walls. The requirements of §1633.2.8 are the appropriate design rules in this situation. This section requires that wall anchorage using straps be attached or hooked so as to transfer the forces to the reinforcing steel. In this case, we are using cast-in-place bolts instead of straps, and the bolts are not required to be “hooked” around the wall reinforcement. In fact, headed anchor bolts have been shown to be more effective than L-bolts in resisting pull-out forces [Shipp and Haninger, 1982]. Try anchor bolts with a 5-inch embedment. Although this embedment is considered shallow anchorage under §1632.2, Rp is 3.0 regardless of whether the anchorage has shallow embedment because §1633.2.8.1 is applicable. The material specific load factors of §1633.2.8.1 (1.4 for steel and 0.85 for wood) are intended to provide the nominal overstrength necessary to resist brittle failure of the wall anchorage system when subjected to the maximum anticipated roof accelerations of flexible diaphragms. Section 1633.2.8.1 is intended as a stand-alone section, and 262


Design Example 5

!

Tilt-Up Building

the more restrictive requirements on R p of §1632.2 do not apply (see Blue Book §C108.2.8.1). F p = 7,976 lb Actual bolt spacing is: 2½ in. (width of 3 × subpurlin) +4¼ in. (2 times bolt edge distance of holdown flange) 6¾ in. From Table 19-D, required spacing for full capacity is 9 inches. Minimum spacing is 50 percent of this, or 4½ in. Interpolation for 6¾ in. spacing is shown below with f 'c = 4,000 psi and assuming Special Inspection. Alternately, using strength design, the requirements of §1923.2 could be used with computation for overlapping pull-out cones. If §1923.2 is used, a load multiplier of 1.3 and a strength reduction factor of 0.65 would be used: Tension Capacity (w/Special Inspection) 6,400 lb/bolt 4,800 3,200 Fp 2

=

Bolt Spacing 9 in. 6¾ in. 4½ in.

7 ,976 = 3,988 lb/bolt 2

Allowable = 4,800 lb (1.33) = 6 ,384 lb >

3,988 = 2,849 1.4

o.k.

Comment: The code in §1633.2.8.1 requires that material-specific load factors be applied in the design of elements of the wall anchorage system. These factors are 1.4 for steel, 1.0 for concrete, and 0.85 for wood. They are applied to the anchorage force determined from Equation (32-2). A background discussion on this is given in the Blue Book Commentary §C108.2.8.1, where the load factors are shown to provide a connection having nominal overstrength of approximately 2.0. This is required to meet the maximum expected roof acceleration of four times the peak ground acceleration. The latter is also discussed in §C108.2.8.1 and is shown to be equivalent to doubling the design anchorage force F p . Thus, an anchorage connection designed under §1633.2.8.1 should have the overstrength that just meets the maximum expected demand of 2 F p . This overstrength approach was selected, in lieu of a ductility approach, after wall anchorage failures were observed in steel strap connectors with limited yield and deformation range. Because anchor bolt pull-out is a critical and brittle failure mode, it must be prevented by having sufficient embedment strength. The nominal factor of two SEAOC Seismic Design Manual, Vol. II (1997 UBC)

263

Design Example 5

!

Tilt-Up Building

overstrength for concrete anchorage just meets the expected maximum demand. This is based on dividing the 1.3 load factor by a φ-factor of 0.65 as discussed in §C108.2.8.1 of the Blue Book. Shown below is the calculation of the strength of the anchorage shown in Figure 5-9 using the method of §1923.3.2 (an alternate method is given in Cook, 1999). In this calculation, a φ-factor of 0.65 is used to provide an additional margin of safety beyond the code minimum. If the overstrength desired was only 2.0, then φ=1.0 would be used. Note that the capacity φPc is greater than 2 F p . φPc = φλ 4 A p

f c'

§1923.3.2

For ¾ in. bolts with hex heads, the width across the flats is 1 18 in. , and A p is computed as follows. A p = 0.785 (10 + 1.125)2 + 6.75(10 + 1.125) = 172 in. 2 < 2(0.785)(11.125)2 = 194 in. 2 φ = 0.65

§1923.3.2

λ =1.0

§1923.3.2

(

φPc = (0.65)(1.0)4 172 in.2

) 4,000

= 28.3 kips > 2 F p = 16.0 kips

o.k.

Therefore, the anchorage in Figure 5-9 is strong enough to resist the expected pull-out forces for code-level ground motions. In general, it is recommended that the concrete pull-out strength exceed the bolt yield strength. If this is not possible, it is recommended that the concrete pull-out strength exceed the code minimum by a substantial margin (as shown above). An alternate wall-roof tie connection is in Figure 5-10. However, this connection, which utilizes a heavy-gauge strap, does not offer the same compression resistance as the bolt scheme (Figure 5-9). Compression forces in the subpurlin generally must be carried by the strap and/or plywood sheathing because subpurlins are typically not installed snugly against the ledgers. Often there is a 1/8-inch to 1/4-inch gap at each end. Providing both tension and compression capability in wall-roof ties protects the diaphragm edge nailing under the reversible seismic forces. In this case, the strap is hooked around a reinforcing bar to meet the requirements of §1633.2.8. The code requires that different loads be applied to the various materials involved in the wall anchorage system. However, most hardware manufacturer’s catalogs provide only a single allowable stress capacity for the component, which often includes concrete, steel, and wood elements. To properly apply code requirements, the design engineer must compute the capacity of each element separately.

264


Design Example 5

!

Tilt-Up Building

pre-manufactured 12-gauge strap

plywood sheathing

#5 bar

3x subpurlin 4"

ledger

7¼"

Figure 5-10. Alternate wall-roof tie

4c. 4c

Design connection to transfer seismic force across first roof truss purlin.

Under §1627, continuity ties in the subdiaphragms are considered part of the wall anchorage system. Consequently, the forces used to design the wall-roof ties must also be used to design the continuity ties within the subdiaphragm. F p = wall-roof tie load = 7,976 lb If the subdiaphragm is 32-foot deep and roof truss purlins are spaced at 8 feet, then the connection at the first roof truss purlin must carry three-quarters of the wall-roof tie force. Comment: Some engineers use the full, unreduced force, but this is not required by rational analysis.

(32 − 8) × F 32

p

=

3 × 7 ,976 = 5,982 lb 4

At the second and third roof truss purlins, the force to be transferred is one-half and one-fourth, respectively, of the wall-roof tie force. 1 × 7 ,976 = 3,988 lb 2


265

Design Example 5

!

Tilt-Up Building

1 × 7 ,976 = 1,994 lb 4 Try 12-gauge metal strap with 10d common nails. Design of strap not shown. Consult ICBO Evaluation Reports for allowable load capacity of pre-manufactured straps. Note that the 1.4 load factor of §1633.2.8.1(4) applies to the strap design and that the 0.85 load factor of §1633.2.8.1(5) applies to the nails. Tension on the gross and net areas of the strap must be checked separately. The tensile capacity of the strap, which is generally not indicated in the ICBO Evaluation Report, is usually controlled by the nails. Consult with the strap manufacturer for appropriate values of F y and Fu . The following calculation shows determination of the number of 10d common nails required at the first connection: 0.85 (5,982 lb ) = 22.8 120 lb (1.4 )(1.33)

Table 23-III-C-2 Table 12-3F, 91 ND

∴ Use 12-gauge metal strap with 24-10d nails each side

12-gauge strap with 24-10d nails each side of roof purlin

subpurlin open web roof truss purlin

Figure 5-11. Subpurlin continuity tie at first purlin

266


Design Example 5

!

Tilt-Up Building

Note that both subpurlins in Figure 5-11 would be 3 × members because of the heavy strap nailing. Design of the second and third connections is similar to that shown above.

5.

Design continuity ties for north-south direction

In a tilt-up building, continuity ties have two functions. The first is to transmit the subdiaphragm reactions (from out-of-plane seismic forces on the wall panels) and distribute these into the main roof diaphragm. The second function is that of “tying” the interior portions of the roof together. In this example, the continuity ties on lines C and D will be designed.

5a. 5a

Seismic forces on continuity ties on lines C and D.

Force in the continuity tie at line 10 is the wall-roof tie force: P10 = (997 plf )(8 ft ) = 7 ,976 lb Force in continuity tie at the glulam beam splice north of line 9 is the sum of both subdiaphragm reactions. P9 =

997 plf (36.67 ft ) (2 subdiaph.) = 36,560 lb 2

The splice near line 9 must also be checked for the minimum horizontal tie force of §1633.2.5. Assume the splice is at fifth point of span as shown on the roof plan of Figure 5-2. This requirement imposes a minimum tie force on the GLB connections and is based only on the dead and live loads carried by the beams. F p = 0.5C a IWD + L

§1633.2.5

W DL = 14 psf , W LL = 12 psf 32 ft 32 ft   W D + L = (14 psf + 12 psf )(36.67 ft ) 32 ft − −  = 18,306 lb 5 5   F p = 0.5 (.44)(1.0 )(18,306 lb ) = 4 ,027 lb < 36,560 lb ∴ Subdiaphragm reaction controls


267

Design Example 5

5b. 5b

!

Tilt-Up Building

Design glulam beam (continuity tie) connection to wall panel.

In this example, walls are bearing walls, and pilasters are not used to vertically support the GLBs. Consequently, the kind of detail shown in Figure 5-12 must be used. This detail provides both vertical support for the GLB and the necessary wall-roof tie force capacity. The tie force is the same as that for wall-roof tie of Part 5a (P10 = 7,976 lb ) . The detail has the capacity to take both tension and compression forces. Details of the design are not given. The horizontal force design is similar to that shown in Part 4.

ledger plywood sheathing

P Stud (typical)

bracket

GLB

5"

7¼"

Figure 5-12. Bracket for wall-roof anchorage at GLB

It should be noted that the alternate wall-roof tie of Figure 5-10 is not acceptable in this situation because the strap cannot resist compression. Comment: Although not required by code, some designers design the wall-GLB tie to take all of the tributary wall-roof forces (assuming the subpurlin wall ties carry none) and carry this force all across the building as the design force in the continuity ties. In this example, this force is P9 = 36,560 lb . This provides for a much stronger “tie” between the wall and the GLB for buildings without pilasters (the usual practice today) to help prevent loss of support for the GLB and subsequent local collapse of the roof under severe seismic motions.

268


Design Example 5

5c. 5c

!

Tilt-Up Building

Design continuity tie across glulam splice.

P9 = 36,560 lb at splice near line 9 The ASD design force for the continuity tie is computed below. Note that the 0.85 wood load factor of §1633.2.8.1(5) is used for bolts in wood (see discussion in Blue Book §C108.2.8.1). P=

0.85 (36 ,560 lb ) = 22,197 lb 1.4

6-3/4” glulam beam

22.2 k

22.2 k

vertical slotted holes

hinge connector

Figure 5-13. Typical continuity tie splice

Try four 7/8-inch bolts in vertical slotted holes at center of hinge connector. Design of hinge connector hardware not shown. Consult ICBO Evaluation Reports for allowable load capacity of pre-manufactured hinge connectors. Note that the bolt capacity is based on the species of the inner laminations (in this case DF-L). 4 (4 ,260 lb )(1.33) = 22 ,663 lb > 22 ,197 lb o.k.

5d. 5d

Table 8.3D, 91 NDS

Check GLB for continuity tie force.

The glulam beams along lines C and D must be checked for the continuity tie axial force. See Part 6 for an example of this calculation. Note that use of the amplified force check of §1633.2.6 is not required for continuity ties that are not collectors.


269

Design Example 5

6.

!

Tilt-Up Building

Design collector along line 3 between lines B and C.

The collector and shear wall ledger along line 3 carry one-half of the east-west roof diaphragm seismic force. The force in the collector is “collected” from the tributary area between lines B and E and transmitted to the shear wall on line 3.

6a. 6a

Determine seismic forces on collector.

From diaphragm shear diagram for east-west seismic forces, the maximum collector load on at line 3 is:  110.0 ft  127.5 k = 136.1 kips tension or compresion R = 36.4 k +   140.67 ft  Uniform axial load in collector can be approximated as the total collector load on line 3 divided by the length of the collector (110'-0") in this direction. q=

6b. 6b

R 136,100 lb = = 1,237 plf L 110.00 ft

Determine the collector force in GLB between lines B and C.

Assume the collector is a GLB 6 3 4 × 21 with 24F-V4 DF/DF and it is adequate to support dead and live loads. A = 141.8 in.2 , S = 496 in.2 , and w = 34.5 plf. Calculate seismic force at mid-span. Tributary length for collecting axial forces is l = 110.00 ft −

36.67 ft = 91.67 ft 2

P = ql = 1.237 klf (91.67 ft ) = 113.4 kips tension or compression in beam

6c. 6c

Check GLB for combined dead and seismic load as required by §1612.3.2.

D+L+S +

E 1.4

(12-16)

w DL = 8 ft (14 psf ) + 34.5 plf = 146.5 plf M DL = 270

0.147 k / ft (36.67 ft )2 = 24.7 kip −ft 8 SEAOC Seismic Design Manual, Vol. II (1997 UBC)

Design Example 5

Fb * = 2,088 psi fb =

P=

24.7 k −ft (12,000) 496 in.3

!

Tilt-Up Building

Table 5A, 91 NDS = 598 psi

113.4 = 81.0 kips tension or compression on ASD basis 1.4

Ft = 1,150 psi

Table 5A, 91 NDS

Fc = 1,650 psi

Table 5A, 91 NDS

ft = fc =

81,000 lb = 571 psi 141.8

Because there is a re-entrant corner at the intersection of lines B and 3, a check for Type 2 plan irregularity must be made. Requirements for irregular structures are given in §1629.5.3. North-south direction check: .15 × (288) = 43.2 ft < 64 ′ − 0 ′′

Table 16-M

East-west direction check: .15 × (110.0 + 30.67 ) = 21.1' < 30'−8"

Table 16-M

Since both projections are greater than 15 percent of the plan dimension in the direction considered, a Type 2 plan irregularity exists. The requirements of Item 6 of §1633.2.9 apply, and the one-third allowable stress increase cannot be used. Checking combined bending and axial tension using Equation (3.9-1) of NDS: fb f + t ≤ 1.00 Fb * Ft '

3.9.1, 91 NDS

598 571 + = 0.29 + 0.50 = 0.79 < 1.00 o.k. 2,088 1,150 Equation (3.9-2) of NDS o.k. by inspection.


271

Design Example 5

!

Tilt-Up Building

Checking combined bending and axial compression using Equation (3.9-3) of NDS and considering the weak axis of the GLB laterally braced by the roof: 2

fb  fc   ′ +  f  Fc  Fb′ 1 − c FcE 

  

≤ 1.0

3.92, 91 NDS

Find Fc ' by first calculating the column stability factor C p . l e = k e l = 1.0 (36.67) = 36.67 ft FcE =

K cE E'

(le /d )

2

=

0.418(1,600,000 )

(36.67 × 12 / 21)2

3.7.1.2, 91 NDS = 1,523 psi

3.7.1.5, 91 NDS

Fc * = Fc = 1650 psi

Table 5A, 91 NDS Supplement

1 + (FcE /Fc* ) F /F * 1 + (FcE /Fc* ) − cE c −   2c 2c c   2

Cp =

1 + (1,523 / 1,650 ) − Cp = 2(0.9 )

Eq. 3.7-1, 91 NDS

1 + (1,523 / 1,650 ) 1,523 / 1,650 −   2(0.9 ) 0.9   2

C p = 0.73

( )

Fc′ = Fc C p = 1,650 (0.73) = 1,205 psi

Table 2.3.1, 91 NDS

2

598  571  = 0.22 + 0.46 = 0.68 < 1.0 1,205  + 571     2,0881 −   1,523 

6d. 6d

o.k .

Check GLB collector for amplified force requirements.

The GLB must also be checked for the special collector requirements of §1633.2.6. Using ASD, an allowable stress increase of 1.7 may be used for this check. The relevant equations are:

272

1.2 D + f1 L + 1.0 Em

(12-17)

0.9 D ± 1.0 E m

(12-18)

Em = Ω o Eh

(30-2) SEAOC Seismic Design Manual, Vol. II (1997 UBC)

Design Example 5

!

Tilt-Up Building

Em is an estimate of the maximum force transmitted by the collector elements in the seismic event. Unless a more refined analysis is done and the maximum force that the diaphragm, or the shear wall, can transmit to the collector determined, the seismic force Eh is scaled by the amplification factor Ω o for estimating Em . Ω o = 2.8

Table 16-N

E h = 113.4 kips from Part 6b, above E m = 2.8 (113.4 ) = 317.5 kips tension or compression in beam Comment: The axial force E m = 317.5 kips in the above calculations is 1.4 times greater than that which would be obtained using the 3R w / 8 factor applied to collector forces obtained under the 1994 UBC provisions. This is because forces in the 1997 UBC are strength based and were established to be 1.4 times greater than those of the 1994 UBC. Unfortunately, the 1997 UBC does not first reduce the forces by the 1.4 ASD factor when increasing the axial force by the Ω o = 2.8 factor. This appears to result in an unnecessarily conservative design for elements like the GLB collector in this example. Under both §1612.2.1 and §1612.4, roof live load is not included in the seismic design load combinations. Generally, Equation (12-17) controls over Equation (12-18). Because the 6 3 4 × 21 GLB will not work, a 6 3 4 × 27 beam will be tried. A = 182 in. 2 , S = 820 in. 3 , and w = 44.3 plf . Dead load bending stress at mid-span is (neglecting small increase in beam weight): M DL = 24.7 kip −ft fb =

24,700 lb −ft (12) = 361 psi 820

Fb = 0.85 (2,400 psi ) = 2,040 psi ft = fc =

Table 5A, 91 NDS

317,500 lb = 1,745 psi 182

Check combined dead plus tension and compression seismic stresses using Equation (12-17). The load factors are 1.2 on dead load and 1.0 on seismic forces, and the allowable stress increase is 1.7. Check tension using NDS Equation (3.9-1):


273

Design Example 5

!

Tilt-Up Building

 1.2 f b   1.0 f t    +   ≤ 1.0  1.7 Fb*   1.7 Ft'   1.2 (361)   1,745    +   = 0.12 + 0.89 = 1.01 ≈ 1.0  1.7 (2 ,040 )   1.7 (1,150) 

say o.k.

NDS Equation (3.9-2) is o.k. by inspection. Check compression using NDS Equation (3.9-3) as modified below: 2

 1.0 f c  +  ' 1.7 Fc 

FcE =

1.2 f b  1.0 f c 1.7 Fb' 1 − FcE 

K cE E'

(le /d )

2

=

  

≤1.0

0.418 (1,600 ,000 )

(36.67 × 12 / 27 )2

= 2,518 psi

3.7.1.5, 91 NDS

C p = 0.88

( )

Fc' = Fc C p = 1,650 (0.88) = 1,452 psi

Table 2.3.1, 91 NDS

1.2 (361) = 0.50 + 0.41 = 0.91 < 1.0  1,745  1.7 (2 ,040)1 −   2 ,518  ∴Use GLB 6 3 4 × 27 2

 1,745    + 1.7 (1,452 )

o.k.

Note that the special collector requirement of §1633.2.6 has necessitated that the size of the GLB be increased from 6 3 4 × 21 to 6 3 4 × 27 .

6e. 6e

Collector connection to shear wall.

The design of the connection of the GLB to the shear wall on line 3 is not given. This is an important connection because it transfers the large “collected” seismic force into the shear wall. The connection must be designed to carry the same seismic forces as the beam, including the amplified collector force of §1633.2.6. Because there is also a collector along line B, there is similarly an important connection of the GLB between lines 3 and 4 to the shear wall on line B. Having to carry two large tension (or compression) forces through the intersection of lines B and 3 (but not simultaneously) requires careful design consideration.

274


Design Example 5

7.

!

Tilt-Up Building

Required diaphragm chord reinforcement for east-west seismic forces.

Chords are required to carry the tension and compression forces developed by the moments in the diaphragm. In this building, the chords are continuous reinforcement located in the wall panels at the roof level as shown in Figure 5-14. (These must be properly spliced between panels.)

A E

precast wall panel

plywood sheathing

ledger

chord reinforcement

Figure 5-14. Diaphragm chord

The east-west diaphragm spans between lines 1 and 3 and lines 3 and 10. The plywood diaphragm is considered flexible, and the moments in segments 1-3 and 3-10 can be computed independently assuming a simple span for each segment. In this example, the chord reinforcement between lines 3 and 10 will be determined. This reinforcement is for the panels on lines A and E. Equiv. w = 1,138 plf from Part 2 wl 2 1.14 klf (224 ) M= = = 7,150 kip −ft 8 8 2


275

Design Example 5

!

Tilt-Up Building

The chord forces are computed from T =C=

7,150 k −ft = 50.8 kips 140.67 ft

The chord will be designed using strength design with Grade 60 reinforcement. The load factor of Equations (12-5) and (12-6) is 1.0 for seismic forces. As =

50.8 k T = = 0.94 in.2 φ f y 0.9 (60 ksi )

∴Use minimum 2-#7 bars, As = 1.20 in 2 > 0.94 o.k. Comment: The chord shown above consists of two #7 bars. These must be spliced at the joint between adjacent panels, typically using details that are highly dependent on the accuracy in placing the bars and the quality of the field welding. Alternately, chords can also be combined with the ledger such as when steel channels or bent steel plates are used, and good quality splices can be easier to make.

8.

Required wall panel reinforcing for out-of-plane forces.

In this part, design of a typical solid panel (no door or window openings) is shown. The panel selected is for lines 1 and 10, and includes the reaction from a large GLB. The wall spans from floor to roof, and has no pilaster under the GLB. There are no recesses or reveals in the wall.

8a. 8a

Out-of-plane seismic forces.

Requirements for out-of-plane seismic forces are specified in §1632.2. Equation (32-2) is used to determine forces on the wall. Fp =

a p Ca I p  h 1 + 3 x Rp  hr

  W p 

(32-2)

F p min = 0.7C a I pW p

(32-3)

F p max = 4.0C a I pW p

(32-3)

R p = 3.0 and a p = 1.0

Table 16-0, Item 1.A.(2)

C a = 0.44 276


Design Example 5

!

Tilt-Up Building

F p can be determined by calculating the equivalent seismic coefficient at the ground and roof levels. The average of the two values is used to determine the uniform out-of-plane seismic force applied over the height of the wall. At the ground level, hx = 0, and the effective seismic coefficient from Equation (32-2) is: aa Ca I p  h 1 + 3 x Rp  hr

 1.0 (.44)(1.0 )  0  = 1 + 3  = 0.147 3.0 21   

Check minimum value from Equation (32-3): 0.7C a I p = 0.7 (.44 )(1.0 ) = 0.308 > 0.147 ∴ Use 0.308

At the roof level, hx = hr , and the effective seismic coefficient from Equation (32-2) is: aa Ca I p  h 1 + 3 x Rp  hr

 1.0 (.44)(1.0 )  21   = 1 + 3  = 0.587 3.0 21   

Check maximum value from Equation (32-3): 4.0C a I p = 4.0 (.44 )(1.0) = 1.76 > 0.587 ∴ Use 0.587

The average force over the height of the wall is: F p = 12 (0.308 + 0.587 )W p = 0.448W p Design of the wall for moments from out-of-plane seismic forces is done by assuming the force Fp to be uniformly distributed over the height of the wall as shown in Figure 5-15. Solving for the uniform force per foot f p : f p = .448 (90.6 psf ) = 40.6 psf


277

Design Example 5

!

Tilt-Up Building

2'-0"

q

21'-0"

fp = 40.6 psf

Figure 5-15. Loading diagram for out-of-plane wall design

8b. 8b

Check applicability of alternate slender wall design criteria.

The panel to be designed is shown in Figure 5-16. The section at mid-height carries the maximum moment from out-of-plane seismic forces. At the same time, this section also carries axial load, from the weight of the panel and the GLB, as well as bending moments due to the eccentricity of the GLB reaction on the wall and P∆ effects. The tributary width of wall for support of the vertical loads of the GLB was determined as follows. The GLB is supported on the wall as shown in Figure 5-12. The vertical reaction on the wall is assumed to be at the bottom of the GLB, and the wall is assumed to span from finished floor to roof in resisting out-of-plane forces. These are conservative assumptions made for the convenience of the analysis. Other assumptions can be made. For example, the center of the stud group (see Figure 5-12) can be assumed to be the location of the GLB reaction on the wall. This assumption would result in a wider effective width of wall to carry vertical loads. The mid-depth of the beam could be assumed to be the point to which the wall spans for out-of-plane forces. This assumption would result in a lower moment in the wall due to the out-of-plane forces. Assume 6 3 4 × 25 1 2 GLB bearing on wall  6.75   21.0   25.5  Tributary width = t GLB + H 2 − d GLB =  + −  = 8.94 ft  12   2   12 

278

§1914.8.2(4)


2.13 ft

Design Example 5

!

Tilt-Up Building

GLB

8.37 ft

H/2

roof

2

A

H = 21.0 ft

1

H/2

B

fin. floor 8.94 ft

Figure 5-16. Typical panel supporting a GLB; line A-B denotes the tributary width of wall to be checked for the vertical load of the GLB and the moment due to out-of-plane seismic forces

Generally, it is advantageous to use the alternate design slender wall criteria of §1914.8. This will be shown below. As a first step, check the limitations on the use of this criteria. These are indicated in §1914.8.2. 1.

Check that vertical service load is less than 0.04 f c ' Ag : Proof =

§1914.8.2(1)

14 psf (36.7 )(32 / 2 ) = 0.92 kip / ft 8.94

 21.0  Pwall = 90.6 psf  + 2.0  = 1.13 kip / ft  2  P = Proof + Pwall = 0.92 + 1.13 = 2.05 kip / ft 0.04 f ' Ag = 0.04 (4 ,000 )(12 )(7.25) = 13.9 kip / ft > 2.05 kip / ft

∴ Vertical service load is less than 0.04 f c ' Ag


279

Design Example 5

2.

!

Tilt-Up Building

Check that the reinforcement does not exceed 0.6ρ b .

§1914.8.2(2)

Assume vertical #4 @ 12 inches o.c. in center panel: ρ=

As 0.20 = = 0.00459 bd (12 )(3.63)    

ρb =

0.85β1 f c '  87,000  87,000 + f y fy 

ρb =

0.85 (0.85)(4 ,000)  87 ,000    = 0.0285 60 ,000  87 ,000 + 60 ,000 

(8-1)

0.6ρ b = 0.6(0.0285) = 0.0171 > 0.00459 o.k.

∴ Reinforcement does not exceed 0.6ρ b 3.

Check that φM n > M cr :

§1914.8.2(3)

Before φM n is calculated, φ must be determined. Calculate φ based on requirements of §1909.3.2.2. The axial load considered to determine M n is the factored vertical load, and this is also used in determining φ . Because strength design is being used, the load effect of vertical motion, E v , must be added to the vertical load. E v = 0.5 C a ID = 0.5 (0.44 )(1.0 ) D = 0.22 D

§1630.1.1

Ev has the effect of increasing the dead load by 0.22 D to a total of 1.42 D. The load factors of Equation (12-5) must be multiplied by 1.1 for concrete as required by Exception 2 of §1612.2.1. The net effect of this is shown below. Pu = 1.1 (1.2 D + 0.22 D ) = 1.56 D

(

§1612.2.1

)

Pu = 1.56 Proof + Pwall = 1.56 (2.05) = 3.20 kip / ft Section 1909.3.2.2 states that φ may be increased up to 0.9 as φPn decreases from the smaller of φPb and 0.1 f c ' Ag to zero. Calculate φPb and 0.1 f c ' Ag : Pb =

b

f y bd − As f y = (0.0285)(60 )(12 )(3.63) − (0.20)(60 ) = 62.4 kip / ft

§1910.3.2

φPb = (0.7 )62.4 = 43.7 kip / ft 280


Design Example 5

!

Tilt-Up Building

0.1 f ' c Ag = 0.1 (4.0 )(12 )(7.25) = 34.8 kip / ft < φPb

∴ Use 0.1 f c ' Ag in calculating φ φ = 0.9 −

0.2 ( Pu ) 0.2 (3.20) = 0.9 − = 0.882 34.8 0.1 f 'c Ag

§1909.3.2.2

Calculate M n for the given axial load of 2.05 kip / ft . Note that values of Ase and a are taken from Part 8c below. 0.372  a   M n = Ase f y  d −  = 0 .253 (60) 3.63 −  = 52.3 kip − in. 2 2    φM n = 0.882 (52.3) = 46.1 kip − in. Calculate the cracking moment M cr . bh 3 12 (7.25) Ig = = = 381in.4 12 12 3

M cr =

( )

5 f c′ I g yt

=

5 4000 (381) = 33.2 kip − in. 3.63

§1914.0

M cr < φM n o.k. 4.

A 2:1 slope may be used for the distribution of the concentrated load throughout the height of the panel (Figure 5-16).

§1914.8.2(4)

∴ Slender wall criteria may be used


281

Design Example 5

8c. 8c

!

Tilt-Up Building

Check wall strength.

Combine factored moment due to out-of-plane seismic forces with moment due to roof vertical load eccentricity and the moment due to P∆ effects. Calculate P∆ moment using the maximum potential deflection, ∆ n . E c = 57,000 f c ' = 3,605 ksi

§1908.5.1

E s = 29,000 ksi

§1908.5.2

n=

E s 29,000 = = 8.04 Ec 3,605

Ase =

a=

c=

Pu + As f y Fy Ase f y

0.85 f ' c b

=

=

3.20 + 0.20 (60 ) = 0.253 in. 60

§1914.8.4

0.253 (60 ) = 0.372 in. 0.85 (4.0)12

a 0.372 = = 0.438 in. 0.85 β1

d = 3.63 in. I cr = n Ase (d − c )2 +

bc 3 3

= 8.04 (0.253)(3.63 − 0.438)2

12 (0.438)3 + = 21.1 in. 4 3

§1914.8.4

Maximum potential deflection is: ∆n =

5 M n l c2 5 (52.3)(21 × 12)2 = = 4.55 in. 48 E c I cr 48 (3,605)(21.1)

Assuming the GLB reaction is 2 in. from the face of wall e = 2.0 +

t wall 7.25 = 2.0 + = 5.63 in. 2 2

Pu, roof = 1.56 Proof = 1.56 (0.92 ) = 1.44 kip / ft

282


Design Example 5

!

Tilt-Up Building

Pu, wall = 1.56 Pwall = 1.56 (1.13) = 1.76 kip / ft Required factored moment at mid-height of the wall is: Mu =

Mu =

f p l c2 8

+

Pu,roof (e ) 2

+ Pu ∆ n

40.6 (21.0)2 1.44 (5.63) 3.20 (4.55) + + 8 (1,000) 12 (2)(12)

M u = 2.24 + 0.34 + 1.21 M u = 3.79 kip - ft = 45.5 kip − in. φM n = 46.1 kip − in. > M u

o.k.

§1914.8.3

Required factored shear is: Vu ≈

f p lc 2

+

Pu,roof (e ) h

φV c = 0.85 (2.0)

(

=

40.6 (21.0 ) 1.44 (5.63) + = 0.458 kip / ft 2 (1,000) 12 (21.0 )

)

(

)

f c' bd = 0.85 (2.0) 4 ,000 (12 )(3.63) = 4.68kips/ft. >> Vu

o.k.

∴ Wall strength is o.k.

8d. 8d

Check service load deflection.

§1914.8.4

The mid-height deflection under service lateral and vertical loads cannot exceed the following: ∆s =

lc 21.0 (12) = = 1.68 in. 150 150

(14-3)

The service level moment M s is determined as follows: Proof (e )

40.6 (21.0 )2 0.92 (5.63) 2.05 (1.68) Ms = + + P∆ s = + + (1.4)(8)(1,000) 2 (12) (1.4)8 2 12 M s = 2.10 kip − ft = 25.2 kip − in. f p l c2

Note M s < M cr


283

Design Example 5

!

Tilt-Up Building

∆s =

5M s l c2 5 (25.2 )(21 × 12 )2 = = 0.12 in. < 1.68 in. 48 E c I g 48 (3,605)(381)

∴ Use #4 @ 12 in. o.c. vertical reinforcing in wall.

8e. 8e

Additional comments.

1. The parapet must be checked as a separate structural element for seismic forces determined from Equation (32-2) with R p = 3.0 and a p = 2.5. This check is not shown. 2. Attention must be given to the location of panel joints and wall openings. These can change the tributary width of wall available to resist combined axial loads and moments. 3. An iterative approach to the calculation of M u and M s may allow for a less conservative analysis. 4. The effective depth of the wall must be modified for architectural reveals, if these are used.

9.

Deflection of east-west diaphragm.

Diaphragm deflections are estimated primarily to determine the displacements imposed on attached structural and nonstructural elements. Columns and walls connected to the diaphragm must satisfy the deformation compatibility requirements of §1633.2.4. An acceptable method of determining the horizontal deflection of a plywood diaphragm under lateral forces is given in §23.222 of 1997 UBC Standard 23-2. The following equation is used: ∆=

Σ(∆ c X ) 5vL3 vL + + 0.188 Len + 8 EAb 4Gt 2b

The deflection of the diaphragm spanning between lines 3 and 10 will be computed. Values for each of the parameters in the above equation are given below: v=

wl (1,138 plf )(224 ft ) = = 906 plf 2b 2 (140.67 ft )

L = 224'−0"

284


Design Example 5

!

Tilt-Up Building

E = 29 × 10 6 psi A = 2 #7 bars = 2 × .60 = 1.20 in.2 b = 140.67 ft G = 90,000 psi

Table 23-2-J

t = 0.54

Table 23-2-I

en = see Table 5-2, below.

Table 23-2-K

∆ c = 0 (Assume no slip in steel chord.)

Table 5-2. Determination of en Zone

L

Nails

s

Shear per nail

en

A

80'-0"

10d

2½"

906(2.5/12) = 189 lb

.042

B

144'-0"

10d

4"

583(4.0/12) = 194 lb

.044

Substituting the above parameters into the deflection equation, the deflection (in inches) at mid-span of the diaphragm is determined. ∆=

5 (906)(224 )3

(906)(224 ) + 0.188 (80)(0.042 ) + 0.188 (144 )(0.044 ) + 0 8 (29 × 10 6 ) (1.20 )(140.67 ) 4 (90,000)(0.54 ) +

∆ = 1.30 + 1.04 + 0.63 + 1.19 = 4.16 in. Under §1633.2.4, all structural framing elements and their connections that are part of the lateral force-resisting system and are connected to the roof must be capable of resisting the “expected” horizontal displacements. The “expected” displacements are amplified displacements taken as the greater of ∆ M or a story drift of 0.0025 times the story height. In this example, the “expected” displacement is: ∆ M = 0.7 R∆ S = 0.7 (4 )(4.16 in ) = 11.6 in.

(30-17)

Note that the R value used above is R = 4 . This is the R value used to determine the shear in the diaphragm in Part 2b under the requirements of §1633.2.9(3).


285

Design Example 5

!

Tilt-Up Building

Comment: The diaphragm deflection calculation shown above is based on strength design seismic forces. Under the 1994 UBC, seismic forces are based on ASD loads, and a smaller deflection would be calculated.

10. 10

Design shear force for east-west panel on line 1.

In this part, determination of the in-plane shear force on a typical wall panel on line 1 is shown. There are a total of five panels on line 1 (Figure 5-1). The panel with the large opening is assumed not effective in resisting in-plane forces, and four panels are assumed to carry the total shear. From Part 2, the total shear on line 1 is 36.4 kips . This force is on a strength basis and was determined using R = 4 for the diaphragm. Except for the diaphragm, the building is designed for R = 4.5 , and an adjustment should be made to determine in-plane wall forces. Earthquake loads on the shear walls must also be modified by the reliability/redundancy factor ρ . This factor varies between a minimum of 1.0 and a maximum of 1.5. Because the shear wall on line 3 (not shown) has large openings for a truck dock, the maximum element-story shear ratio, rmax of §1630.1.1, is large and the resulting reliability/redundancy factor for the east-west direction is the maximum value of 1.5. This requires that shear forces in individual east-west panels, determined from the analysis shown in Part 2, be increased by a factor of 1.5 as shown below. Finally, seismic forces due to panel weight must also be included. These are determined using the base shear coefficient (.244) from Part 1. The panel seismic force is determined as follows: Panel weight: width =

110 ft = 22 ft 5

 7.25  W p = 0.15   (23 ft )(22 ft ) = 45.9 kips  12  Seismic force due to panel weight: F p = 0.244W p = 0.244 (45.9 k ) = 11.2 kips

286


Design Example 5

!

Tilt-Up Building

The total seismic force on the panel, E , is the horizontal shear transferred from the diaphragm and the horizontal seismic force due to the panel weight, both adjusted for the reliability/redundancy factor. This calculation is shown below: E = ρE h + E v

(30-1)

1 4  Eh =   (36.4 k ) + (11.2 k ) = 19.3 kips 4  4.5  Ev = 0 ∴ V panel = ρE h + E v = 1.5 (19.3) + (0 ) = 29.0 kips per panel Comment: The 1997 UBC introduced the concept of the reliability/redundancy factor. The intent of this provision is to penalize those lateral force resisting systems without adequate redundancy by requiring that they be more conservatively designed. A redundancy factor is computed for each principal direction. In general, they are not applied to diaphragms, except transfer diaphragms.

References ACI, 1996. Practitioner’s Guide to Tilt-Up Construction. American Concrete Institute, P.O. Box 9094, Farmington Hills, Michigan 48333. Breyer, D.E., Fridley, K.J., and Cobeen, K.E., 1999. Design of Wood Structures Allowable Stress Design, Fourth edition. McGraw Hill, Inc., New York. Brooks, Hugh, 1997. The Tilt-up Design and Construction Manual, Fourth edition. HBA Publications, 2027 Vista Caudal, Newport Beach, California 92660. City of Los Angeles Division 91. Earthquake Hazard Reduction in Existing Tilt-up Concrete Wall Buildings, Los Angeles Dept. of Building and Safety, 200 N. Spring Street, Los Angeles, California 90012. Cook, R.A., 1999. Strength Design of Anchorage to Concrete, Portland Cement Association, Skokie, Illinois. Hamburger, R., and McCormick, D., 1994. “Implications of the January 17, 1994 Northridge Earthquake on Tilt-up and Masonry Buildings with Wood Roofs,”


287

Design Example 5

!

Tilt-Up Building

1994 Fall Seminar Notes. Structural Engineers Association of Northern California (SEAONC), 74 New Montgomery Street, Suite 230, San Francisco, California 94105-3411. SEAOSC/COLA, 1994. 1994 Northridge Earthquake (Structural Engineers Association of Southern California/City of Los Angeles) Special Investigation Task Force, Tilt-up Subcommittee. Final report dated September 25, 1994. Shipp, J.G., and Haninger, E.R., 1982. “Design of Headed Anchor Bolts,” Proceedings of 51st Annual Convention, Structural Engineers Association of California, September 30-October 2, 1982.

288


Design Example 6

!

Tilt-Up Wall Panel with Openings

Design Example 6 Tilt-Up Wall Panel With Openings

32'-0"

28'-0" 12' × 14' opening

3'-0"

12'-0"

3' × 7' door

4'-0"

3'-0"

Figure 6-1. Wall elevation and section

Overview Walls designed under the alternative slender wall method of UBC §1914.8, are typically tilt-up concrete panels that are site-cast, cured, and tilted into place. They are designed to withstand out-of-plane forces and carry vertical loads at the same time. These slender walls differ from concrete walls designed under the empirical design method (UBC §1914.5) in that there are greater restrictions on axial loads and reinforcement ratios. In addition, secondary effects of eccentricities and p-delta moments play an important role in analysis and design of these slender tilt-up panels.

In this example, the out-of-plane lateral design forces for a one-story tilt-up concrete slender wall panel with openings are determined, and the adequacy of a proposed reinforced concrete section is checked. The example is a single-story tilt-up concrete wall panel with two openings, site-cast, and tilted up into place. The pier between the two openings is analyzed using the slender SEAOC Seismic Design Manual, Vol. II (1997 UBC)

289

Design Example 6

!


wall design method (UBC §1914.8). Analysis of the wall panel for lifting stresses or other erection loads is not a part of this example. Outline This example will illustrate the following parts of the design process: 1.

Out-of-plane lateral design forces.

2.

Basic moment from the out-of-plane forces.

3.

Vertical design forces acting on the pier.

4.

Nominal moment strength φMn.

5.

Factored moment including eccentricity and p-delta effects.

6.

Service load out-of-plane deflection.

7.

Special horizontal reinforcing.

Given Information Wall material: f’c = 3000 psi normal weight concrete Reinforcing steel material: fy = 60,000 psi Wall thickness = 9¼ inches with periodic ¾-inch narrow reveals. Reinforcing steel area = 7 #5 each face at wall section between openings. Reinforcing depth based on 1-inch minimum cover per UBC §1907.7.1 item 4. Loading data: Roof loading to wall = uniform loading; 40-foot span of 12 psf dead load; no snow load. Roof loading eccentricity = 4 inches from interior face of panel. Seismic Zone = Zone 4 Near-source influence = more than 10 km to any significant seismic source (Na = 1). Soil profile = SD Seismic importance factor = 1.0 Wind does not govern this wall panel design. 290


Design Example 6


!


291

Design Example 6

!



1.

Code Reference

Out-of-plane lateral design forces.

The wall panel is subdivided into a design strip. Typically, a solid panel is subdivided into one-foot-wide design strips for out-of-plane design. However, where wall openings are involved, the entire pier width between openings is generally used as the design strip for simplicity. The distributed loading accounts for the strip’s self-weight, as well as the tributary loading from above each opening.

W3

W1

W2 tributary load area

W3

parapet

W2 W1 roof

design strip

3' × 7' door

12' × 14' opening

4'-0"

floor

4'-0"

Figure 6-2. Design strip and distributed out-of-plane loading

292


Design Example 6

1a.

!


Seismic coefficient of wall element.

The wall panel is considered an element of a structure, thus §1632.2 applies in determining the lateral seismic force. UBC Equations 32-2 and 32-3 are used to determine forces for design. Fp =

ap Ca Ip  hx  1 + 3 Wp Rp  hr 

(32-2)

Except : FP is limited by 0.7CaIpWp ≤ Fp ≤ 4CaIpWp

(32-3)

ap = 1.0

Table 16-O

Rp = 3.0

Table 16-O

Ca = 0.44

Table 16-Q

Ip = 1.0

Table 16-K

Therefore, the limits on Fp are: 0.308Wp ≤ Fp ≤ 1.76Wp hx is defined as the attachment height above grade level. Since the wall panel is connected at two different heights, an equivalent lateral force will be obtained using the average of the roof Fp and the at-grade Fp [ref. 1999 SEAOC Blue Book Commentary §C107.2.3]. Fp roof =

(1.0)(0.44)(1.0)  hr  1 + 3 Wp = 0.587Wp 3.0 hr  

(1.0)(0.44)(1.0)  0 1 + 3 Wp = 0.147Wp , 3.0 hr   but Fp min = 0.308Wp governs. Fp grade =

Fp wall =

Fp grade + Fp roof 2

=

0.587 + 0.308 = 0.448Wp 2

Note: The seismic coefficient 0.448 is virtually the same as the 1994 UBC coefficient 0.30 when adjusted for strength design and the different seismic zone coefficient Ca defaults: Fp (1994 UBC equivalent) =


0.448Wp  0.40    = 0.291 ≈ 0.30 1.4  0.44 

293

Design Example 6

1b. 1b

!


Load combinations for strength design.

For this example, the use of load combination (12-5) of §1612.2.1 is applicable, and governs for concrete strength design under seismic loading. 1.2D + 1.0E + (f1L + f2S)

(12-5)

where: D = self weight of wall and dead load of roof L = 0 (floor live load) S = 0 (snow load) E = ρEh + Ev where ρ = 1.0 (§1632.2) and Ev = 0.5CaID Load combination (12-5) reduces to: (1.2 + 0.5CaI)D + 1.0Eh or (1.2 + 0.22)D + 1.0Eh or 1.42D + 1.0Eh

(30-1)

Note: Exception 2 under §1612.2.1, which multiplies strength design load combinations by 1.1, has been determined to be inappropriate by SEAOC and others, and has not been included in the 1999 SEAOC Blue Book, Recommended Lateral Force Requirements and Commentary. For the purposes of this example, the 1.1 multiplier has been included in order to conform to the 1997 UBC as originally published. For additional information, see “Design of Reinforced Concrete Buildings under the 1997 UBC,” by S.K. Ghosh, published in Building Standards, May-June 1998, ICBO. Load combination (12-5) increases to: 1.1(1.42D + 1.0Eh) = 1.56D + 1.1Eh

1c. 1c

Lateral out-of-plane wall forces.

The lateral wall forces Eh are determined by multiplying the wall’s tributary weight by the lateral force coefficient. Three different distributed loads are determined due to the presence of two door openings of differing heights. See Figure 6-2. Wall weight =

(

9.25 150 pcf = 116 lb/ft2 12

)

Fp wall = 0.448 116 lb/ft 2 = 52lb/ft 2 W1 = 52 lbs/ft2 x 4 ft = 208 plf W2 = 52 lbs/ft2 x 3/2 ft = 78 plf W3 = 52 lbs/ft2 x 12/2 ft = 312 plf 294


Design Example 6

2.

!


Basic moment from out-of-plane forces.

W3 W2 W1

7,212 lbs

14'

W 3=312 plf

x maximum moment

W 2=78 plf

7'

W 1=208 plf

7'

4,618 lbs

Loading

Shear

Moment

Figure 6-3. Corresponding loading, shear, and moment diagrams

Locate the point of zero shear for maximum moment. Ignore the parapet’s negative moment benefits in reducing the positive moment for simplicity of analysis. If the designer decides to use the parapet’s negative moment to reduce the positive moment, special care should be taken to use the shortest occurring parapet height. For this analysis, the seismic coefficient for the parapet shall be the same as that for the wall below (ap = 1.0, not 2.5). The parapet should be checked separately later, but is not a part of this example. This example conservatively assumes the maximum moment occurs at a critical section width of 4'-0". In cases where the maximum moment occurs well above the doors, a more comprehensive analysis could consider several critical design sections, which would account for a wider design section at the location of maximum moment and for a narrower design section with reduced moments near the top of the doors.


295

Design Example 6

2a. 2a

!


Determine the shear reactions at each support.

Rgrade = shear reaction at grade level for design strip Rroof = shear reaction at roof level for design strip  28)2 21)2 14 )2  1 ( ( ( + 78 + 312 Rgrade = 208 = 4,618 lbs  2 2 2  28  Rroof = [208( 28) + 78( 21) + 312(14)] − 4618 = 7,212 lbs Determine the distance of the maximum moment from the roof elevation downward (Figure 6-3): X =

2b. 2b

7212 =12.1 feet to point of zero shear (maximum moment) (208 + 78 + 312)

Determine Mu basic

This is the primary strength design moment, excluding p-delta effects and vertical load eccentricity effects, but including the 1.1 load factor (see the earlier discussion of this load factor in Step 1b, above):  12.1)2  ( Mu basic = 1.17212(12.1) − (208 + 78 + 312 )  = 47,837 lb-ft 2   Mu basic = 47.8 k-ft

3.

Vertical design forces acting on the center pier.

The pier’s vertical loads are comprised of a roof component Proof and a wall component Pwall. The applicable portion of the wall component is the top portion Pwall top above the design section. Proof = gravity loads from the roof acting on the design strip The appropriate load combinations using strength or allowable stress design do not include roof live load in combination with seismic loads. However, strength designs considering wind loads must include a portion of roof live loads per §1612.2.1.

296


Design Example 6

!


Proof = (roof dead load) x (tributary width of pier) x (tributary width of roof) 3 12  40  Proof = (12 psf ) 4 + +  = 2,760 lb 2 2 2  Note: When concentrated gravity loads, such as from a girder, are applied to slender walls, the loads are assumed to be distributed over an increasing width at a slope of 2 vertical to 1 horizontal down to the flexural design section height (§1914.8.2.4). Pwall top = the portion of the wall’s self weight above the flexural design section. It is acceptable to assume the design section is located midway between the floor and roof levels 3 12   28   Pwall top = (116 psf ) 4 + +   + 4  = 24,012 lbs 2 2  2   Ptotal = Proof + Pwall top = 2760 + 24012 = 26,772 lbs Check the vertical service load stress for applicability of the slender wall design method (UBC §1914.8.2 item 1). Use the net concrete section considering the reveal depth: stress =

Ptotal 26772 = = 66 psi < 0.04 f c′ = 0.04(3000) = 120 psi Aconc 48 (9.25 − 0.75)

o.k.

The compressive stress is low enough to use the alternative slender wall method; otherwise a different method, such as the empirical design method (§1914.5), would be required along with its restrictions on wall height.

4.

Nominal moment strength φMn.

The nominal moment strength φMn is given by the following equation: a  φMn = φAsefy d −  2  where: 0.2 (1.56) ( 26772) 0.2 Pu φ = 0.9 − = 0.9 − = 0.83 0.10 ( 3000) ( 48) (9.25 − 0.75) 0.10 f c′ Aconc Ase =

a=

§1909.3.2.2

Pu + Asfy 1.56 ( 26772) + 7 (0.31) (60000) = = 2.87 in. 2 fy 60000

Pu + Asfy 1.56 ( 26772) + 7 (0.31) (60000) = = 1.40 in. 0.85 f c′b 0.85 (3000) ( 48)


297

Design Example 6

!


Reinforcing depth is based on new tilt-up cover provision §1907.7.1 item 4. d = thickness − reveal − cover − tie diameter − 1 2 bar diameter d = 9 1 4 − 3 4 − 1 − 3 8 − (1 2 )(5 8 ) = 6.8 in.

3/4" reveal

#3 ties

9 1/4" thick

d = depth

Figure 6-4. Design section

Thus: 1.40   Mn = 2.87 (60000)  6.8 −  = 1050 k − in = 87.5 k − ft 2   φMn = 0.83 (87.5) = 72.6 k − ft

Verify that Mcr < φMn to determine the applicability of the slender wall design method (UBC §1914.8.2 item 3). Mcr is defined uniquely for slender walls in UBC §1914.0.

M cr = 5 f c′

Ig yt

=

5 3000 ( 48) 9.25 2

(9.25) 3 12 = 187,458 lb − in. = 15.6 k − ft

M cr = 15.6 k − ft < φ M n = 72.6 k − ft

§1914.0

o.k.

Sufficient reinforcing is provided to use the alternative slender wall method, otherwise the empirical design method of UBC §1914.5 would be necessary. Note: For the purposes of §1914.8.2 item 3, Ig and yt are conservatively based on the gross thickness without consideration for reveal depth. This approach creates a worst-case comparison of Mcr to φMn. In addition, the exclusion of the reveal depth in the Mcr calculation produces more accurate deflection values when reveals are narrow. 298


Design Example 6

!


Verify the reinforcement ratio ρ ≤ 0 .6ρ b to determine the applicability of the slender wall design method (§1914.8.2 item 2): 0.6ρ b = 0.6

ρ=

0.85β1 f c′ 87000 0.85(0.85)3000 87000 = 0.6 = 0.0128 fy 87000 + f y 60000 (87000 + 60000)

(8-1)

As 7(0.31) = = 0.0066 < 0.0128 o.k. bd 48(6.8)

Therefore, the slender wall method is applicable.

5.

Factored moment, including eccentricity and p-delta effects.

Determine the design moment including the effects from the vertical load eccentricity and p-delta (P∆): Mu = Mu basic + Mu eccentricity + Mu P∆ Use the figures below to determine Mu eccentricity and Mu P∆:

eccentricity "e"

eccentricity "e"

H

H Proof

Proof Pwall top

lc 2

Pwall top

lc deflected shape

Pwall

bottom

M

2 ∆n 3

∆n 3

Figure 6-6. Freebody of upper half 2∆n 3

∆n 3

Figure 6-5. Vertical loading


299

Design Example 6

5a. 5a

!


Determine force component H from statics (moment about base of wall).

From Figure 6-5, assuming a parabolic deflected shape:

H=

( Pwall top + Pwall bottom )

2∆ n − Proof e 3

lc

Since the panel’s openings are not positioned symmetrically with the panel’s mid-height, Pwall bottom will be less than Pwall top. For ease of calculation, conservatively assume Pwall bottom = Pwall top, as is similar to panels without openings. H =

5b. 5b

4 Pwall top ∆ n 3l c

−

Proof e lc

Determine moment component M from statics using Figure 6-6 to account for eccentricity and P∆ ∆ effects:

M = Proof (∆ n + e) + Pwall top M = Proof

5c. 5c

∆n l +H c 3 2

e + ( Pwall top + Proof )∆ n 2

Determine the wall's deflection at full moment capacity ∆n.

5M n l c2 ∆n = 48E c I cr

§1914.8.4

where: Mn is from Step 4. E c = 57 f c′ = 3122 ksi 1.40 a = = 1.65 in. 0.85 0.85 29,000 48(1.65) 3 = 2.87(6.8 − 1.65) 2 + = 779 in. 4 3,122 3

I cr = nAse ( d − c ) 2 + I cr

300

bc 3 ; 3

§1908.5.1 where c =


Design Example 6

∆n =

!


5 (87.5) ( 28) 2 (12) 3 = 5.1 in. 48 (3122) (779)

Section 1914.8.3 requires the maximum potential deflection ∆n be assumed in the calculation of the P∆ moment, unless a more comprehensive analysis is used. An iterative approach or use of a moment magnifier are examples of acceptable “more comprehensive” analyses, but are beyond the scope of this example.

5d. 5d

Determine and check the total design moment Mu.

Mu = Mu basic + Mu eccentricity + Mu P∆ e Mu = 47.8 + Pu roof + Pu wall top + Pu roof ∆ n 2 1 9.25 − 0.75  1 1 Mu = 47.8 + 1.56( 2.76)  4 +  + 1.56( 24.0 + 2.76)(5.1) 2 2 12  12 Mu = 47.8 + 1.5 + 17.7

(

Mu =

)

67.0k − ft < φM n = 72.6k − ft o.k.

(14-2)

Therefore, the design section’s strength is acceptable.

6.

Service load out-of-plane deflection.

6a. 6a

Determine if the wall’s cross-section is cracked.

The service load moment Ms is determined with the following formula where the denominators are load factors to convert from load combination (12-5) to load combination (12-13): Ms =

M u basic M u eccentricity + + M s P∆ 1.1 (1.4) 1.56

Assume the service load deflection is the maximum allowed ∆s

Maximum

(

=

lc : 150

lc 28 (12) = = 2.24 in. 150 150

(14-3)

)

M s P∆ = Pwall + Proof ∆ s = (24.0 + 2.76 ) 2.24 = 59.9 k − in. = 5.00 k − ft Ms =

M u basic M u eccentricity + + M s P∆ 1.1(1.4) 1.56


301

Design Example 6

!


Ms =

47.8 1.5 + + 5.00 = 37.0 k − ft 1.1(1.4) 1.56

M cr = 15.6k − ft < M s Therefore, section is cracked and Equation (14-4) is applicable for determining ∆s. If the section is uncracked, Equation (14-5) is applicable.

6b. 6b

Determine the deflection at initiation of cracking ∆cr.

∆ cr =

5M cr l c 2 5 (15.6) ( 28) 2 (12) 3 = = 0.22 in. 48E c I g ( 48) (9.25) 3 48 (3122) 12

§1914.8.4

Ig is based on gross thickness, without consideration for the architectural reveal depth, since this produces more accurate results when the reveals are narrow.

6c. 6c

Determine and check the service load deflection ∆s.

 M − M cr ∆ s = ∆ cr +  s  M n − M cr

 (∆ n − ∆ cr ) 

(14-4)

 37.0 − 15.6  ∆ s = 0.22 +   (5.1 − 0.22 ) = 1.67 in.  87.5 − 15.6  ∆ s = 1.67 in. <

lc = 2.24 in. o.k. 150

§1914.8.4

Therefore, the proposed slender wall section is acceptable using the alternative slender wall method.

302


Design Example 6

7. 7a. 7a

!


Special horizontal reinforcing.

Determine the horizontal reinforcing required above the largest wall opening for out-of-plane loads.

The portion of wall above the twelve-foot-wide door opening spans horizontally to the vertical design strips on each side of the opening. This wall portion will be designed as a one-foot unit horizontal design strip and subject to the out-of-plane loads computed in this example earlier. Fp wall = 0.448(116 lbs/ft2) = 52 lb/ft2 The moment is based on a simply supported horizontal beam with the 1.1 multiplier per Exception 2 under §1612.2.1:  (opening width )2 M u = 1.1 F p  8  = 1030 lb − ft = 1.03 k − ft

2    = 1.1 52 12   8  

   

Try using #5 bars at 18-inch spacing to match the same bar size as being used vertically at the maximum allowed spacing for wall reinforcing. a  φMn = φAsfy  d −  2  where:  12  φ = 0.9 and As = 0.31   = 0.21 in. 2  18  a=

(0.21) ( 60000) Asfy = = 0.41 in. 0.85 f c′b 0.85 (3000) (12)

Assume the reinforcing above the opening is a single curtain with the vertical steel located at the center of the wall’s net section. The horizontal reinforcing in concrete tilt-up construction is typically place over the vertical reinforcing when assembled on the ground. d=

1

d=

1

2

(thickness − reveal ) − bar diameter

2

(9 1 4 − 3 4 ) − 5 8 = 3.63 in.


303

Design Example 6

!


0.41   φMn = 0.9 (.21) (60)  3.63 −  = 38.8 k − in. 2   = 3.24 k − ft ≥ 1.03 k − ft φMn ≥ M u

o.k.

o.k.

Therefore, the horizontal reinforcing is acceptable.

7b. 7b

Typical reinforcing around openings.

Two #5 bars are required around all window and door openings per §1914.3.7. The vertical reinforcing on each face between the openings provides two bars along each jamb of the openings, and thus satisfies this requirement along vertical edges. Horizontally, two bars above and below the openings are required to be provided. In addition, it is common to add diagonal bars at the opening corners to assist in limiting the cracking that often occurs due to shrinkage stresses (Figure 6-7).

7c. 7c

Required horizontal (transverse) reinforcing between the wall openings.

The style and quantity of horizontal (transverse) reinforcing between the wall openings is dependent on several factors relating to the in-plane shear wall design of §1921.6. Sections conforming to “wall piers,” as defined in §1921.1, shall be reinforced per §1921.6.13. Wall pier reinforcing has special spacing limitations and is often provided in the form of closed ties. In narrow piers, these ties are often preferred so as to assist in supporting both layers of reinforcing during construction, even if not required by the special wall pier analysis (Figure 6-7). Configurations not defined as wall piers, but which have high in-plane shears, also have special transverse reinforcing requirements per §1921.6.2.2. In these situations, the transverse reinforcing is required to be terminated with a hook or “U” stirrup.

304


Design Example 6

!


typical horizontal reinforcing #5 at 18" o.c. vertical reinforcing (7) #5 each face reinforcing around openings (2) #5

transverse reinforcing

design section (see Figure 6-4)

Figure 6-7. Typical wall reinforcing

Commentary The UBC section on the alternative slender wall method made its debut in the 1988 edition. It is largely based on the equations, concepts, and full-scale testing developed by the Structural Engineers Association of Southern California and published in the Report of the Task Committee on Slender Walls in 1982. The American Concrete Institute (ACI) has incorporated similar provisions for slender wall design in their publication ACI 318-99. Tilt-up wall construction has become very popular due to its versatility and its erection speed. However, wall anchorage failures at the roofline have occurred during past earthquakes. In response to these failures, the 1997 UBC anchorage design forces and detailing requirements are significantly more stringent than they have been under past codes (see Design Example 5).


305

Design Example 6

!


References Recommended Tilt-up Wall Design, Structural Engineers Association of Southern California, 1979. 5360 Workman Mill Road, Whittier, CA 90601 (562) 908-6131. Report of the Task Committee on Slender Walls, Southern California Chapter American Concrete Institute and the Structural Engineers Association of Southern California, 1982. Brooks, Hugh, The Tilt-up Design and Construction Manual, HBA Publications, Fourth Edition, 1990. 2027 Vista Caudal, Newport Beach, CA 92660. Tilt-Up Concrete Structures Reported by ACI Committee 551, American Concrete Institute, 1997. P.O. Box 9094, Farmington Hills, MI 48333 (248) 848-3700. “Design of Reinforced Concrete Buildings under the 1997 UBC,” Building Standards. S.K. Ghosh, ICBO, May-June 1998. 5360 Workman Mill Road, Whittier, CA 90601

306


Design Example 6


!


307

Vertical Irregularity Type 1 Example 1

!

§1629.4.2

Example 1 Seismic Zone 4 Near-Source Factor

§1629.4.2

The 1997 UBC introduced the concept of near-source factors. Structures built within close proximity to an active fault are to be designed for an increased base shear over similar structures located at greater distances. This example illustrates the determination of the near-source factors N a and N v . These are used to determine the seismic coefficients Ca and Cv used in §1630.2.1 to calculate design base shear. 1.

Determine the near-source factors N a and N v for a site near Lancaster, California.

Calculations and Discussion 1.

Code Reference

Determine N a and N v .

First locate the City of Lancaster in the book Maps of Known Active Fault NearSource Zones in California and Adjacent Portions of Nevada. This is published by the International Conference of Building Officials and is intended to be used with the 1997 Uniform Building Code. Lancaster is shown on map M-30. Locate the site on this map (see figure), and then determine the following: The shaded area on map M-30 indicates the source is a type A fault. Therefore Seismic source type: A The distance from the site to the beginning of the fault zone is 6 km. Another 2 km must be added to reach the source (this is discussed on page vii of the UBC fault book). Thus, the distance from the site to the source is 6 km + 2 km = 8 km. Distance from site to fault zone: 8 km. Values of N a and N v are given in Tables 16-S and 16-T for distances of 2, 5, 10, and 15 km. For other distances, interpolation must be done. N a and N v have been plotted below. For this site, N a and N v can be determined by entering the figures at a distance 8 km. and using the source type A curves. From this N a = 1.08 N v = 1.36


1

§1629.4.2

!

Example 1

Vertical Irregularity Type 1

Commentary The values of N a and N v given above are for the site irrespective of the type of structure to be built on the site. Had N a exceeded 1.1, it would have been possible to use a value of 1.1 when determining Ca , provided that all of the conditions listed in §1629.4.2 were met.

2



!

§1629.4.2

R

2.0

Source Type A a 1.0

Source Type B

0.0 0

5

10

15


R

2.0


N 1.0

0.0 0

5

10

15



3

§1629.4.2

4

!

Example 1



Introduction to Vertical Irregularities §1629.5.3

Introduction to Vertical Irregularities Vertical irregularities are identified in Table 16-L. These can be divided into two categories. The first are dynamic force distribution irregularities. These are irregularity Types 1, 2, and 3. The second category are irregularities in load path or force transfer, and these are Types 4 and 5. The five vertical irregularities are as follows: 1. Stiffness irregularity—soft story 2. Weight (mass) irregularity 3. Vertical geometric irregularity 4. In-plane discontinuity in vertical lateral-force resisting element 5. Discontinuity in capacity—weak story The first category, dynamic force distribution irregularities, requires that the distribution of lateral forces be determined by combined dynamic modes of vibration. For regular structures without abrupt changes in stiffness or mass (i.e., structures without “vertical structural irregularities”), this shape can be assumed to be linearly-varying or a triangular shape as represented by the code force distribution pattern. However, for irregular structures, the pattern can be significantly different and must be determined by the combined mode shapes from the dynamic analysis procedure of §1631. The designer may opt to go directly to the dynamic analysis procedure and thereby bypass the checks for vertical irregularity Types 1, 2, and 3. Regular structures are assumed to have a reasonably uniform distribution of inelastic behavior in elements throughout the lateral force resisting system. When vertical irregularity Types 4 and 5 exist, there is the possibility of having localized concentrations of excessive inelastic deformations due to the irregular load path or weak story. In this case, the code prescribes additional strengthening to correct the deficiencies.


5

§1629.5.3

!

Example 2



§1629.5.3

For example: A five-story concrete special moment-resisting frame is shown below. The specified lateral forces Fx from Equations (30-14) and (30-15) have been applied and the corresponding floor level displacements ∆x at the floor center of mass have been found and are shown below. F t + F5

∆S5 = 2.02"

10' F4 10' F3 10' 10'

F2

∆S4 = 1.75" Triangula r shape

∆S3 = 1.45" ∆S2 = 1.08"

F1 ∆S1 = 0.71"

12'

1.

Determine if a Type 1 vertical irregularity—stiffness irregularity-soft story—exists in the first story.

Calculations and Discussion 1.

Code Code Reference

To determine if this is a Type 1 vertical irregularity, stiffness irregularity—soft story, there are two tests:

1. The story stiffness is less than 70-percent of that of the story above. 2. The story stiffness is less than 80-percent of the average stiffness of the three stories above. If the stiffness of the story meets at least one of these two criteria, the structure is considered to have a soft story, and a dynamic analysis is generally required under §1629.8.4 Item 2, unless the irregular structure is not more than five stories or 65feet in height (see §1629.8.3 Item 3). The definition of soft story in the code compares values of the lateral stiffness of individual stories. Generally, it is not practical to use stiffness properties unless these can be easily determined. There are many structural configurations where the evaluation of story stiffness is complex and is often not an available output from computer programs. Recognizing that the basic intent of this irregularity check is to determine if the lateral force distribution will differ significantly from the linear pattern prescribed by Equation (30-15), which assumes a triangular shape for the 6



!

§1629.5.3

first dynamic mode of response, this type of irregularity can also be determined by comparing values of lateral story displacements or drift ratios due to the prescribed lateral forces. This deformation comparison may even be more effective than the stiffness comparison because the shape of the first mode shape is often closely approximated by the structure displacements due to the specified triangular load pattern. Floor level displacements and corresponding story drift ratios are directly available from the computer programs. To compare displacements rather than stiffness, it is necessary to use the reciprocal of the limiting percentage ratios of 70 and 80 percent as they apply to story stiffness or reverse their applicability to the story or stories above. The following example shows this equivalent use of the displacement properties. From the given displacements, story drifts and the story drift ratio values are determined. The story drift ratio is the story drift divided by the story height. These will be used for the required comparisons since these better represent the changes in the slope of the mode shape when there are significant differences in inter-story heights. (Note: story displacements can be used if the story heights are nearly equal.) In terms of the calculated story drift ratios, the soft story occurs when one of the following conditions exists:


∆S 1 ∆ − ∆S 1 exceeds S 2 ., or h1 h2


∆S 1 exceeds h1

1  (∆S 2 − ∆S 1 ) ( ∆S 3 − ∆S 2 ) (∆S 4 − ∆S 3 )  + +   3 h2 h3 h4  The story drift ratios are determined as follows:

∆S1 (0.71 − 0) = = 0.00493 h1 144 ∆S 2 − ∆S1 (1 .08 − 0 .71 ) = 0 .00308 = 120 h2

(1 . 45 − 1 . 08 ) = 0 . 00308 ∆S 3 − ∆S 2 = 120 h3 ∆S 4 − ∆S 3 (1 .75 − 1 .45 ) = 0 .00250 = 120 h4


7

§1629.5.3

!

Example 2


1 (0.00308 + 0.00308 + 0.00250 ) = 0.00289 3

Checking the 70 percent requirement: ∆  0.70  S 1  = 0.70 (0.00493 ) = 0.00345 > 0.00308  h1 

∴ Soft story exists Checking the 80 percent requirement: ∆  0 .80  S 1  = 0 .80 (0 .00493 ) = 0 .00394 > 0 .00289  h1 

∴ Soft story exists

Commentary Section §1630.10.1 requires that story drifts be computed using the maximum inelastic response displacements ∆M . However, for the purpose of the story drift, or story drift ratio, comparisons needed for soft story determination, the displacements ∆S due to the design seismic forces can be used as done in this example. In the example above, only the first story was checked for possible soft story vertical irregularity. In practice, all stories must be checked, unless a dynamic analysis is performed. It is often convenient to create a table as shown below to facilitate this exercise.

8

Level

Story Displacement

Story Drift

Story Drift Ratio



Avg. of Story Drift Ratio of Next 3 Stories

Soft Story Status

5 4 3 2 1

2.02 in. 1.75 1.45 1.08 0.71

0.27 in. 0.30 0.37 0.37 0.71

0.00225 0.00250 0.00308 0.00308 0.00493

0.00158 0.00175 0.00216 0.00216 0.00345

0.00180 0.00200 0.00246 0.00246 0.00394

---------0.00261 0.00289

No No No No Yes



!

1629.5.3

Example 3 Vertical Irregularities Type 2

§1629.5.3

The five-story special moment frame office building has a heavy utility equipment installation at Level 2. This results in the floor weight distribution shown below: 5

W 5 = 90k

4

W 4 = 110k

3

W 3 = 110k

2

W 2 = 170k

1

W 1 = 100k


Code Reference

A weight, or mass, vertical irregularity is considered to exist when the effective mass of any story is more than 150 percent of the effective mass of an adjacent story. However, this requirement does not apply to the roof if the roof is lighter than the floor below. Checking the effective mass of Level 2 against the effective mass of Levels 1 and 3 At Level 1 1.5 × W1 = 1.5(100k ) = 150k At Level 3 1.5 × W3 = 1.5(110k ) = 165k W2 = 170k > 150k ∴ Weight irregularity exists

Commentary As in the case of irregularity Type 1, this type of irregularity also results in a primary mode shape that can be substantially different from the triangular shape and lateral load distribution given by Equation (30-15). Consequently, the appropriate load distribution must be determined by the dynamic analysis


9

§1629.5.3

!

Example 3


procedure of §1631, unless the irregular structure is not more than five stories or 65-feet in height (see §1629.8.3 Item 3)

10



Example 4 Vertical Irregularity Type 3 1.

!

§1629.5.3

§1629.5.3

Determine if there is a Type 2 vertical weight (mass) irregularity.


Code Reference

A weight, or mass, vertical irregularity is considered to exist when the effective mass of any story is more than 150 percent of the effective mass of an adjacent story. However, this requirement does not apply to the roof if the roof is lighter than the floor below. Checking the effective mass of Level 2 against the effective mass of Levels 1 and 3 At Level 1 1.5 × W1 = 1.5(100k ) = 150k At Level 3 1.5 × W3 = 1.5(110k ) = 165k W2 = 170k > 150k ∴ Weight irregularity exists

Commentary As in the case of irregularity Type 1, this type of irregularity also results in a primary mode shape that can be substantially different from the triangular shape and lateral load distribution given by Equation (30-15). Consequently, the appropriate load distribution must be determined by the dynamic analysis procedure of §1631, unless the irregular structure is not more than five stories or 65-feet in height (see §1629.8.3 Item 3). The lateral force-resisting system of the five-story special moment frame building shown below has a 25 foot setback at the third, fourth and fifth stories.


11

§1629.5.3

1

2

3

4

!

Example 4


5

4 @ 25' = 5

4

3

2

1

1.

Determine if a Type 3 vertical irregularity, vertical geometric irregularity, exists.


Code Reference

A vertical geometric irregularity is considered to exist where the horizontal dimension of the lateral force-resisting system in any story is more than 130 percent of that in the adjacent story. One-story penthouses are not subject to this requirement. In this example, the set-back of Level 3 must be checked. The ratios of the two levels is Width of Level 2 (100') = = 1.33 Width of Level 3 (75') 133 percent > 130 percent ∴ Vertical geometric irregularity exists

Commentary The more than 130-percent change in width of the lateral force-resisting system between adjacent stories could result in a primary mode shape that is substantially different from the triangular shape assumed for Equation (30-15). If the change is a decrease in width of the upper adjacent story (the usual situation), the mode shape

12



!

§1629.5.3

difference can be mitigated by designing for an increased stiffness in the story with a reduced width. Similarly, if the width decrease is in the lower adjacent story (the unusual situation), the Type 1 soft story irregularity can be avoided by a proportional increase in the stiffness of the lower story. However, when the width decrease is in the lower story, there could be an overturning moment load transfer discontinuity that would require the application of §1630.8.2. When there is a large decrease in the width of the structure above the first story along with a corresponding large change in story stiffness that creates a flexible tower, then §1629.8.3, Item 4 and §1630.4.2, Item 2 may apply. Note that if the frame elements in the bay between lines 4 and 5 were not included as a part of the designated lateral force resisting system, then the vertical geometric irregularity would not exist. However, the effects of this adjoining frame would have to be considered under the adjoining rigid elements requirements of §1633.2.4.1.


13

§1629.5.3 ! Example 5 ! Vertical Irregularity Type 4


§1629.5.3

A concrete building has the building frame system shown below. The shear wall between Lines A and B has an in-plane offset from the shear wall between Lines C and D. 1.

Determine if there is a Type 4 vertical irregularity, in-plane discontinuity in the vertical lateral force-resisting element. B

A

C

D

3 @ 25' = 75’ 5 12' Shear wall 4 12' 3 12'

25'

2 12'

50'

Shear wall

1 12'


Code Reference

A Type 4 vertical irregularity exists when there is an in-plane offset of the lateral load resisting elements greater than the length of those elements. In this example, the left side of the upper shear wall (between lines A and B) is offset 50-feet from the left side of the lower shear wall (between lines C and D). This 50-foot offset is greater than the 25-foot length of the offset wall elements. ∴ In-plane discontinuity exists

Commentary The intent of this irregularity check is to provide correction of force transfer or load path deficiencies. It should be noted that any in-plane offset, even those less or equal to the length or bay width of the resisting element, can result in an overturning moment load transfer discontinuity that requires the application of §1630.8.2. When the offset exceeds the length of the resisting element, there is also a shear transfer discontinuity that requires application of §1633.2.6 for the strength

14



!

§1629.5.3

of collector elements along the offset. In this example, the columns under wall A-B are subject to the provisions of §1630.8.2 and §1921.4.4.5, and the collector element between Lines B and C at Level 2 is subject to the provisions of §1633.2.6.


15

Seismic Design Manual Volume III Building Design Examples: Steel, Concrete and Cladding

November 2000


Publisher Structural Engineers Association of California (SEAOC) 1730 I Street, Suite 240 Sacramento, California 95814-3017 Telephone: (916) 447-1198; Fax: (916) 443-8065 E-mail: [email protected]; Web address: http://www.seaoc.org/ The Structural Engineers Association of California (SEAOC) is a professional association of four regional member organizations (Central California, Northern California, San Diego, and Southern California). SEAOC represents the structural engineering community in California. This document is published in keeping with SEAOC’s stated mission: “to advance the structural engineering profession, to provide the public with structures of dependable performance through the application of state-of-the-art structural engineering principles; to assist the public in obtaining professional structural engineering services; to promote natural hazard mitigation; to provide continuing education and encourage research; to provide structural engineers with the most current information and tools to improve their practice; and to maintain the honor and dignity of the profession.” Editor: Gail Hynes Shea, Albany, California. Cover photos, clockwise from upper right: 900 E. Hamilton Ave. Office Complex, Campbell, Calif.—Joe Maffei, Rutherford & Chekene; Clark Pacific; SCBF connection—Buehler & Buehler; UBC; RBS “Dog Bone” connection—Buehler & Buehler.

Disclaimer Practice documents produced by the Structural Engineers Association of California (SEAOC), and all narrative texts, drawings, calculations, and other information herein, are published as part of SEAOC’s educational program. The material presented in this publication is intended for educational purposes only; it should not be used or relied on for any specific application without the competent examination and verification of its accuracy, suitability, and applicability to a specific project by a qualified structural engineer. While the information presented in this publication is believed to be correct, neither SEAOC nor its member organizations, committees, writers, editors, individuals, or entities which have in any way contributed to it make any warranty, express or implied, or assume any legal liability or responsibility for the use, application of and/or reference to the text, drawings, calculations, samples, references, opinions, findings, conclusions, or recommendations included in this publication. Users of this publication and its contents assume all liability arising from such use.

Table of Contents

Table of Contents

Preface ............................................................................................................................... v Acknowledgements .............................................................................................................vi Introduction ......................................................................................................................... 1 How to Use This Document ................................................................................................ 3 Notation ............................................................................................................................... 4 Design Example 1 1A Special Concentric Braced Frame ....................................................................... 19 1B Ordinary Concentric Braced Frame ..................................................................... 67 1C Chevron Braced Frame........................................................................................ 77 Design Example 2 Eccentric Braced Frame ............................................................................................. 89 Design Example 3 3A Steel Special Moment Resisting Frame............................................................. 143 3B Steel Ordinary Moment Resisting Frame........................................................... 189 Design Example 4 Reinforced Concrete Wall......................................................................................... 209 Design Example 5 Reinforced Concrete Wall with Coupling Beams...................................................... 237 Design Example 6 Concrete Special Moment Resisting Frame............................................................. 271 Design Example 7 Precast Concrete Cladding....................................................................................... 313

SEAOC Seismic Design Manual, Vol. III (1997 UBC)

iii

iv


Preface

Preface This document is the third volume of the three-volume SEAOC Seismic Design Manual. The first volume, Code Application Examples, was published in April 1999. The second volume, Building Design Examples: Light Frame, Masonry and Tilt-up was published in April 2000. These documents have been developed by the Structural Engineers Association of California (SEAOC) with funding provided by SEAOC. Their purpose is to provide guidance on the interpretation and use of the seismic requirements in the 1997 Uniform Building Code (UBC), published by the International Conference of Building Officials (ICBO), and in SEAOC’s 1999 Recommended Lateral Force Requirements and Commentary (also called the Blue Book). The Seismic Design Manual was developed to fill a void that exists between the Commentary of the Blue Book, which explains the basis for the UBC seismic provisions, and everyday structural engineering design practice. While the Seismic Design Manual illustrates how the provisions of the code are used, the examples shown do not necessarily illustrate the only appropriate methods of seismic design, and the document is not intended to establish a minimum standard of care. Engineering judgment must be exercised when applying these Design Examples to real projects. Volume I: Code Application Examples, provides step-by-step examples of how to use individual code provisions, such as how to compute base shear or building period. Volumes II and III: Design Examples furnish examples of the seismic design of common types of buildings. In Volumes II and III, important aspects of whole buildings are designed to show, calculation-by-calculation, how the various seismic requirements of the code are implemented in a realistic design. Volume III contains ten examples. These illustrate the seismic design of the following structures: 1. 2. 3. 4. 5. 6. 7.

Three steel braced frames (special, ordinary, and chevron) Eccentric braced frame Two steel moment-resisting frames (special and ordinary) Concrete shear wall Concrete shear wall with coupling beams Concrete special moment-resisting frame Precast concrete cladding

It is SEAOC’s present intention to update the Seismic Design Manual with each edition of the building code used in California. Work is presently underway on an 2000 International Building Code version. Ronald P. Gallagher Project Manager

SEAOC Seismic Design Manual, Vol. III (1997UBC)

v

Acknowledgments

Acknowledgments

Authors The Seismic Design Manual was written by a group of highly qualified structural engineers. These individuals are California registered civil and structural engineers and SEAOC members. They were selected by a Steering Committee set up by the SEAOC Board of Directors and were chosen for their knowledge and experience with structural engineering practice and seismic design. The Consultants for Volumes I, II and III are: Ronald P. Gallagher, Project Manager Robert Clark David A. Hutchinson Jon P. Kiland John W. Lawson Joseph R. Maffei Douglas S. Thompson Theodore C. Zsutty Volume III was written principally by David A. Hutchinson (Design Examples 1A, 1B and 1C, and 3A and 3B), Jon P. Kiland (Design Examples 2 and 6), Joseph R. Maffei (Design Examples 4 and 5), and Robert Clark (Design Example 7). Many useful ideas and helpful suggestions were offered by the other consultants.

Steering Committee Overseeing the development of the Seismic Design Manual and the work of the Consultants was the Project Steering Committee. The Steering Committee was made up of senior members of SEAOC who are both practicing structural engineers and have been active in Association leadership. Members of the Steering Committee attended meetings and took an active role in shaping and reviewing the document. The Steering Committee consisted of: John G. Shipp, Chair Robert N. Chittenden Stephen K. Harris Martin W. Johnson Scott A. Stedman vi


Reviewers

Reviewers A number of SEAOC members, and other structural engineers, helped check the examples in Volume III. During its development, drafts of the examples were sent to these individuals. Their help was sought in both review of code interpretations as well as detailed checking of the numerical computations. The assistance of the following individuals is gratefully acknowledged: Vin Balachandran Raymond Bligh Dirk Bondy David Bonowitz Robert Chittenden Michael Cochran Anthony Court Juan Carlos Esquival Brent Forslin

S. K. Ghosh Jeff Guh Ronald Hamburger Douglas Hohbach Dominic Kelly Edward Knowles Kenneth Lutrell Robert Lyons Peter Maranian

Harry (Hank) Martin, (AISC) James O’Donnell Richard Phillips Paul Pina Mehran Pourzanjani Rafael Sabelli C. Mark Saunders David Sheppard Constantine Shuhaibar

Seismology Committee Close collaboration with the SEAOC Seismology Committee was maintained during the development of the document. The 1999-2000 Committee reviewed the document and provided many helpful comments and suggestions. Their assistance is gratefully acknowledged. 1999-2000

Martin W. Johnson, Chair Saif Hussain, Past Chair David Bonowitz Robert N. Chittenden Tom H. Hale Stephen K. Harris Douglas C. Hohbach Y. Henry Huang Saiful Islam


H. John Khadivi Jaiteeerth B. Kinhal Robert Lyons Simin Naaseh Chris V. Tokas Michael Riley, Assistant to the Chair

vii


Suggestions for Improvement In keeping with two of its Mission Statements: (1) “to advance the structural engineering profession” and (2) “to provide structural engineers with the most current information and tools to improve their practice”, SEAOC plans to update this document as seismic requirements change and new research and better understanding of building performance in earthquakes becomes available. Comments and suggestions for improvements are welcome and should be sent to the following: Structural Engineers Association of California (SEAOC) Attention: Executive Director 1730 I Street, Suite 240 Sacramento, California 95814-3017 Telephone: (916) 447-1198; Fax: (916) 443-8065 E-mail: [email protected]; Web address: http://www.seaoc.org

Errata Notification SEAOC has made a substantial effort to ensure that the information in this document is accurate. In the event that corrections or clarifications are needed, these will be posted on the SEAOC web site at http://www.seaoc.org or on the ICBO website at http://ww.icbo.org. SEAOC, at its sole discretion, may or may not issue written errata.

viii


Seismic Design Manual Volume III Building Design Examples: Steel, Concrete and Cladding

Introduction

Introduction Seismic design of new steel and concrete buildings, and precast cladding, for the requirements of the 1997 Uniform Building Code (UBC) is illustrated in this document. Ten examples are shown: 1A 1B 1C 2 3A 3B 4 5 6 7

Steel special concentric braced frame Steel ordinary concentric braced frame Steel chevron braced frame Eccentric braced frame Steel special moment-resisting frame Steel ordinary moment-resisting frame Concrete shear wall Concrete shear wall with coupling beams Concrete special moment-resisting frame Precast concrete cladding

The buildings selected are for the most part representative of construction types found in Zones 3 and 4, particularly California and the western states. Designs have been largely taken from real world buildings, although some simplifications were necessary for purposes of illustrating significant points and not presenting repetitive or unnecessarily complicated aspects of a design. The Design Examples are not complete building designs, or even complete seismic designs, but rather they are examples of the significant seismic design aspects of a particular type of building. In developing these Design Examples, SEAOC has endeavored to illustrate correct use of the minimum provisions of the code. The document is intended to help the reader understand and correctly use the design provisions of UBC Chapter 16 (Design Requirements), Chapter 19 (Concrete), and Chapter 22 (Steel). Design practices of an individual structural engineer or office, which may result in a more seismic-resistant design than required by the minimum requirements of UBC, are not given. When appropriate, however, these considerations are discussed as alternatives. In some examples, the performance characteristics of the structural system are discussed. This typically includes a brief review of the past earthquake behavior and mention of design improvements added to recent codes. SEAOC believes it is essential that structural engineers not only know how to correctly interpret and


1

Introduction

apply the provisions of the code, but that they also understand their basis. For this reason, many examples have commentary included on past earthquake performance. While the Seismic Design Manual is based on the 1997 UBC, references are made to the provisions of SEAOC’s 1999 Recommended Lateral Force Provisions and Commentary (Blue Book). When differences between the UBC and Blue Book are significant, these are brought to the attention of the reader.

2



How to Use This Document Each Design Example is presented in the following format. First, there is an “Overview” of the example. This is a description of the building and the seismic aspects to be designed. This is followed by an “Outline” indicating the tasks or steps to be illustrated in each example. Next, “Given Information” provides the basic design information, including plans and sketches given as the starting point for the design. This is followed by “Calculations and Discussion,” which provides the solution to the example. Some Design Examples have a subsequent section designated “Commentary.” The commentary is intended to provide a better understanding of aspects of the example and/or to offer guidance to the reader on use of the information generated. Finally, references and suggested reading are given at the end of the example. Some Design Examples have a section entitled “Factors that Influence Design” that provides remarks on salient design points. Because the document is based on the UBC, UBC notation is used throughout. However, notation from other codes is also used. In general, reference to UBC sections and formulas is abbreviated. For example, “1997 UBC Section 1630.2.2” is given as §1630.2.2 with 1997 UBC (Volume 2) being understood. “Formula (32-2)” is designated Equation (32-2) or just (32-2) in the right-hand margins of the Design Examples. Similarly, the phrase “Table 16-O” is understood to be 1997 UBC Table 16-O. Throughout the document, reference to specific code provisions, tables, and equations (the UBC calls the latter formulas) is given in the right-hand margin under the heading Code Reference. When the document makes reference to other codes and standards, this is generally done in abbreviated form. Generally, reference documents are identified in the right-hand margin. Some examples of abbreviated references are shown below. Right-Hand Margin Notation

More Complete Description

Table 1-A AISC-ASD

Table 1-A of Ninth Edition, American Institute of Steel Construction (AISC) Manual of Steel Construction, Allowable Stress Design, 1989.

AISC-Seismic §15.3b

Section 15.3b of the American Institute of Steel Construction, Seismic Provisions for Structural Steel Buildings, Chicago, Illinois, 1997.

SEAOC C402.8

Section C402.8 of Commentary of SEAOC Recommended Lateral Force Requirements and Commentary (Blue Book), 1999.


3

Notation

Notation The following notations are used in this document. These are generally consistent with that used in the UBC and other codes such as ACI and AISC. Some additional notations have also been added. The reader is cautioned that the same notation may be used more than once and may carry entirely different meaning in different situations. For example, E can mean the tabulated elastic modulus under the AISC definition (steel) or it can mean the earthquake load under §1630.1 of the UBC (loads). When the same notation is used in two or more definitions, each definition is prefaced with a brief description in parentheses (e.g., steel or loads) before the definition is given.

4

AB

=

ground floor area of structure in square feet to include area covered by all overhangs and projections

ABM

=

cross-sectional area of the base material

Ab

=

area of anchor, in square inches

Ac

=

the combined effective area, in square feet, of the shear walls in the first story of the structure

Ach

=

cross-sectional area of a structural member measured out-toout of transverse reinforcement

Acv

=

net area of concrete section bounded by web thickness and length of section in the direction of shear force considered

Ae

=

the minimum cross-sectional area in any horizontal plane in the first story of a shear wall, in square feet

Af

=

flange area

Ag

=

gross area of section

Ap

=

the effective area of the projection of an assumed concrete failure surface upon the surface from which the anchor protrudes, in square inches

As

=

area of nonprestressed tension reinforcement


Notation

Ash

=

total cross-sectional area of transverse reinforcement (including crossties) within spacing s and perpendicular to dimension hc

Ask

=

area of skin reinforcement per unit height on one side face

As,min

=

minimum amount of flexural reinforcement

Ast

=

area of link stiffener

Av

=

area of shear reinforcement within a distance s, or area of shear reinforcement perpendicular to flexural tension reinforcement within a distance s for deep flexural members

Avd

=

total area of reinforcement in each group of diagonal bars in a diagonally reinforced coupling beam

Avf

=

area of shear-friction reinforcement

Aw

=

(web) link web area

Aw

=

(weld) effective cross-sectional area of the weld

Ax

=

the torsional amplification factor at Level x

a

=

(concrete) depth of equivalent rectangular stress block

a

=

(concrete spandrel) shear span, distance between concentrated load and face of supports

ac

=

coefficient defining the relative contribution of concrete strength to wall strength

ap

=

in-structure component amplification factor, given in §1632 and Table 16-O of UBC

b

=

(concrete) width of compression face of member

bf

=

flange width

bw

=

web width

b/t

=

member width-thickness ratio

Ca

=

seismic coefficient, as set forth in Table 16-Q of UBC


5

Notation

Ce

=

combined height, exposure, and gust factor coefficient as given in Table 16-G of UBC

Cq

=

pressure coefficient for the structure or portion of structure under consideration as given in Table 16-H

Ct

=

numerical coefficient as given in §1630.2.2

Cv

=

seismic coefficient as set forth in Table 16-R

Cm

=

coefficient defined in Section H1 of AISC-ASD

c

=

distance from extreme compression fiber to neutral axis

D

=

dead load on a structural element

De

=

length, in feet, of a shear wall in the first story in the direction parallel to the applied forces

d

=

effective depth of section (distance from extreme compression fiber to centroid of tension reinforcement)

db

=

(anchor bolt) anchor shank diameter

db

=

(concrete) bar diameter

dz

=

column panel zone depth

E

=

(steel) modulus of elasticity

EI

=

flexural stiffness of compression member

E, Eh, Em, Ev, Fi, Fn=

6

(loads) earthquake loads set forth in §1630.1

Ec

=

modulus of elasticity of concrete, in psi

Es

=

(concrete) modulus of elasticity of reinforcement

e

=

EBF link length

Fa

=

axial compressive stress that would be permitted if axial force alone existed

Fb

=

bending stress that would be permitted if bending moment alone existed

FBM

=

nominal strength of the base material to be welded SEAOC Seismic Design Manual, Vol. III (1997 UBC)

Notation

FEXX

=

classification number of weld metal (minimum specified strength)

Fp

=

design seismic force on a part of the structure

Fu

=

specified minimum tensile strength, ksi

Fw

=

(steel LRFD) nominal strength of the weld electrode material

Fw

=

(steel ASD) allowable weld stress

Fy

=

specified yield strength of structural steel

Fyb

=

Fy of a beam

Fyc

=

Fy of a column

Fye

=

expected yield strength of steel to be used

Fyf

=

Fy of column flange

Fyh

=

(steel) specified minimum yield strength of transverse reinforcement

Fyw

=

Fy of the panel-zone steel

fa

=

computed axial stress

fb

=

bending stress in frame member

f c'

=

specified compressive strength of concrete

fct

=

average splitting tensile strength of lightweight aggregate concrete

fut

=

minimum specified tensile strength of the anchor

F' e

=

23(Kλb / rb )2

fi

=

lateral force at Level i for use in Formula (30-10)

fm'

=

specified compressive strength of masonry

fp

=

equivalent uniform load

12 π 2 E


7

Notation

fr

=

modulus of rupture of concrete

Ftt

=

through-thickness weld stresses at the beam-column interface

fy

=

(concrete) specified yield strength of reinforcing steel

f x, f y, f r = g

=

acceleration due to gravity

h

=

overall dimensions of member in direction of action considered

hc

=

(concrete) cross-sectional dimension of column core, or shear wall boundary zone, measured center to center of confining reinforcement

hc

=

(steel) assumed web depth for stability

he

=

assumed web depth for stability

hi, hn,hx =

8

(steel) weld stresses at connection interface

height in feet above the base to Level i, n, or x, respectively

hr

=

height in feet of the roof above the base

hw

=

height of entire wall or of the segment of wall considered

I

=

(loads) importance factor given in Table 16-K

I

=

(concrete) moment of inertia of section resisting externally applied factored loads

Icr

=

moment of inertia of cracked section transformed to concrete

Ig

=

(concrete, neglecting reinforcement) moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement

Ig

=

(concrete, transformed section) moment of inertia of cracked section transformed to concrete.

Ip

=

importance factor specified in Table 16-K

Ise

=

moment of inertia of reinforcement about centroidal axis of member cross section


Notation

It

=

moment of inertia of structural steel shape, pipe or tubing about centroidal axis of composite member cross section

Iw

=

importance factor as set forth in Table 16-K of UBC

K

=

(steel) effective length factor for prismatic member

k

=

effective length factor for compression member

L

=

(loads) live load due to occupancy and moveable equipment, or related internal moments and forces

L

=

(steel) unbraced beam length for determining allowable bending stress

Lp

=

limiting laterally unbraced length for full plastic flexural strength, uniform moment case

lc

=

(steel RBS) length of radius cut in beam flange for reduced beam section (RBS) design

lc

=

length of a compression member in a frame, measured from center to center of the joints in the frame

lh

=

distance from column centerline to centerline of hinge for reduced bending strength (RBS) connection design

ln

=

clear span measured face to face of supports

lu

=

unsupported length of compression member

lw

=

length of entire wall, or of segment of wall considered, in direction of shear force.

Level i =

level of the structure referred to by the subscript i; i = 1 designates the first level above the base

Level n =

the level that is uppermost in the main portion of the structure

Level x =

the level that is under design consideration; x = 1 designates the first level above the base

M

=

(steel) maximum factored moment

Mc

=

factored moment to be used for design of compression member


9

Notation

Mcl

=

moment at centerline of column

Mcr

=

moment causing flexural cracking at section due to externally applied loads (see §1911.4.2.1)

MDL, MLL, Mseis = unfactored moment in frame member

10

Mf

=

moment at face of column

Mm

=

(concrete) modified moment

Mm

=

(steel) maximum moment that can be resisted by the member in the absence of axial load

Mn

=

nominal moment strength at section

Mp

=

(concrete) required plastic moment strength of shearhead cross-section

Mp

=

(steel) nominal plastic flexural strength, Fy Z

Mpa

=

nominal plastic flexural strength modified by axial load

Mpe

=

nominal plastic flexural strength using expected yield strength of steel

Mpr

=

(concrete) probable moment strength determined using a tensile strength in the longitudinal bars of at least 1.25 fy and a strength reduction factor φ of 1.0

Mpr

=

(steel RBS) probable plastic moment at the reduced beam section (RBS)

Ms

=

(concrete) moment due to loads causing appreciable sway

Ms

=

(steel) flexural strength; member bending strength at plastic capacity ZFy

Mu

=

(concrete) factored moment at section

Mu

=

(steel) required flexural strength on a member or joint

My

=

moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution


Notation

M1

=

smaller factored end moment on a compression member, positive if member is bent in single curvature, negative if bent in double curvature

M2

=

larger factored end moment on compression member, always positive

Na

=

near-source factor used in the determination of Ca in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and slip rates as set forth in Tables 16-S and 16-U

Nv

=

near-source factor used in the determination of Cv in Seismic Zone 4 related to both the proximity of the building or structure to known faults with magnitudes and to slip rates as set forth in Tables 16-T and 16-U

P

=

(steel) factored axial load

P

=

(wind) design wind pressure

PDL, PLL, Pseis

= unfactored axial load in frame member

Pb

=

nominal axial load strength at balanced strain conditions (see §1910.3.2)

Pbf

=

connection force for design of column continuity plates

Pc

=

(concrete) critical load

Pc

=

(concrete anchorage) design tensile strength

Pe

=

(23/12)F'e A, where F'e is as defined in Section H1 of AISC-ASD

Pn

=

nominal axial load strength at given eccentricity, or nominal axial strength of a column

Po

=

nominal axial load strength at zero eccentricity

Psc

=

1.7 Fa A

Psc,Pst =

Psi

=

strength level axial number force for connection design or axial strength check (see §2213.5) Fy A


11

Notation

12

Pu

=

(concrete) factored axial load, or factored axial load at given eccentricity

Pu

=

(steel) nominal axial strength of a column, or required axial strength on a column or a link

Pu

=

(concrete anchorage) required tensile strength from loads

Py

=

nominal axial yield strength of a member, which is equal to Fy Ag

PDL

=

axial dead load

PE

=

axial load on member due to earthquake

PLL

=

axial live load

qs

=

wind stagnation pressure at the standard height of 33 feet, as set forth in Table 16-F

R

=

numerical coefficient representative of the inherent overstrength and global ductility capacity of lateral force resisting systems, as set forth in Table 16-N or 16-P

Rn

=

nominal strength

Rnw

=

nominal weld strength

Rp

=

component response modification factor, given in §1632.2 and Table 16-0

Ru

=

required strength

Ry

=

ratio of expected yield strength Fye to the minimum specified yield strength Fy

r

=

(loads) a ratio used in determining ρ (see §1630.1)

r

=

(steel) radius of gyration of cross section of a compression member

ry

=

radius of gyration about y axis

s

=

spacing of shear or torsion reinforcement in direction parallel to longitudinal reinforcement, or spacing of transverse reinforcement measured along the longitudinal axis


Notation

SA, SB, SC, SD, SE, S F = soil profile types as set forth in Table 16-J SRBS

=

section modulus at the reduced beam section (RBS)

T

=

elastic fundamental period of vibration, in seconds, of the structure in the direction under consideration

tf

=

thickness of flange

tw

=

thickness of web

tz

=

column panel zone thickness

U

=

required strength to resist factored loads or related internal moments and forces

V

=

the total design lateral force or shear at the base given by Formula (30-5), (30-6), (30-7) or (30-11)

Vc

=

(concrete) nominal shear strength provided by concrete

Vc

=

(concrete anchorage) design shear strength

VDL, VLL, Vseis = unfactored shear in frame member Vn

=

(concrete) nominal shear strength at section

Vn

=

(steel) nominal shear strength of a member

Vp

=

(steel) shear strength of an active link

Vpa

=

nominal shear strength of an active link modified by the axial load magnitude

Vs

=

(concrete) nominal shear strength provided by shear reinforcement

Vs

=

(steel) shear strength of member, 0.55 Fy dt

Vu

=

(concrete anchorage) required shear strength from factored loads

Vu

=

(concrete) factored shear force at section, including shear magnification factors for overstrength and inelastic dynamic effects

Vu

=

(loads) factored horizontal shear in a story


13

Notation

Vu

=

(steel) required shear strength on a member

Vu *

=

factored shear force at section, including shear magnification factors for overstrength and inelastic dynamic effects

Vx

=

the design story shear in story x

W

=

(seismic) the total seismic dead load defined in §1620.1.1

W

=

(wind) load due to wind pressure

Wp

=

the weight of an element of component

wc

=

weights of concrete, in pcf

wi, wx =

14

that portion of W located at or assigned to Level i or x, respectively

wpx

=

the weight of the diaphragm and the element tributary thereto at Level x, including applicable portions of other loads defined in §1630.1.1

wz

=

column panel zone width

Z

=

(loads) seismic zone factor as given in Table 16-I

Z

=

(steel) plastic section modulus

ZRBS

=

plastic section modulus at the reduced beam section (RBS)

∆

=

design story drift

∆M

=

maximum inelastic response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design basis ground motion, including estimated elastic and inelastic contributions to the total deformation, as defined in §1630.9

∆O

=

relative lateral deflection between the top and bottom of a story due to Vu, computed using a first-order elastic frame analysis and stiffness values satisfying §1910.11.1

∆S

=

design level response displacement, which is the total drift or total story drift that occurs when the structure is subjected to the design seismic forces


Notation

δί

=

horizontal displacement at Level i relative to the base due to applied lateral forces, f, for use in Formula (30-10)

φ

=

(concrete) capacity reduction or strength reduction factor (see §1909.3)

φb

=

(steel) resistance factor for flexure

φc

=

(steel) resistance factor for compression

φv

=

resistance factor for shear strength of panel zone of beam-tocolumn connections

∝

=

(concrete) angle between the diagonal reinforcement and the longitudinal axis of a diagonally reinforced coupling beam

∝, β

=

(steel) centroid locations of gusset connection for braced frame diagonal

∝c

=

coefficient defining the relative contribution of concrete strength to wall strength

βc

=

ratio of long side to short side of concentrated load or reaction area

β1

=

factor defined in §1910.2.7.3

ρ

=

(loads) redundancy/reliability factor given by Formula (30-3)

ρ

=

(concrete) ratio of nonprestressed tension reinforcement (As/bd)

ρb

=

reinforcement ratio producing balanced strain conditions (see §1910.3.2)

ρn

=

ratio of area of distributed reinforcement parallel to the plane of Acv to gross concrete area perpendicular to that reinforcement.

ρs

=

ratio of volume of spiral reinforcement to total volume of core (outto-out of spirals) of a spirally reinforced compression member

ρv

=

ratio of area of distributed reinforcement perpendicular to the plane of Acv to gross concrete area Acv

λ

=

lightweight aggregate concrete factor; 1.0 for normal weight concrete, 0.75 for “all lightweight” concrete, and 0.85 for “sandlightweight” concrete


15

Notation

16

λp

=

limiting slenderness parameter for compact element

la

=

length of radius cut in beam flange for reduced beam section (RBS) connection design

lh

=

distance from column centerline to centerline of hinge for RBS connection design

ln

=

clear span measured face to face of supports

lu

=

unsupported length of compression member

lw

=

length of entire wall or of segment of wall considered in direction of shear force

Ωo

=

(loads) seismic force amplification factor, which is required to account for structural overstrength and set forth in Table 16-N

Ωo

=

(steel) horizontal seismic overstrength factor

µ

=

coefficient of friction


References

References ACI-318, 1995. American Concrete Institute, Building Code Regulations for Reinforced Concrete, Farmington Hills, Michigan. AISC-ASD, 1989. American Institute of Steel Construction, Manual of Steel Construction, Allowable Stress Design, Chicago, Illinois, 9th Edition. AISC-LRFD, 1994. American Institute of Steel Construction, Manual of Steel Construction, Load and Resistance Factor Design, Chicago, Illinois, 2nd Edition. AISC-Seismic. Seismic Provisions for Structural Steel Buildings, American Institute of Steel Construction, Chicago, Illinois, April 15, 1997 and Supplement No. 1, February 15, 1999. SEAOC Blue Book, 1999. Recommended Lateral Force Requirements and Commentary, Structural Engineers Association of California, Sacramento, California. UBC, 1997. International Conference of Building Officials, Uniform Building Code, Whittier, California.


17

18


Design Example 1A

!

Special Concentric Braced Frame

Design Example 1A Special Concentric Braced Frame

Figure 1A-1. Four-story steel frame office building with special concentric braced frames (SCBF)

Foreword Design Examples 1A, 1B and 1C show the seismic design of essentially the same four-story steel frame building using three different concentric bracing systems.

"

Design Example 1A illustrates a special concentric braced frame (SCBF).

"

Design Example 1B illustrates an ordinary concentric braced frame (OCBF).

"

Design Example 1C illustrates a chevron braced frame design.

These Design Examples have been selected to aid the reader in understanding design of different types of concentric braced frame systems. Design of eccentric braced frames (EBFs) is illustrated in Design Example 2.


19

Design Example 1A

!


Overview The 4-story steel frame office structure shown in Figure 1A-1 is to have special concentric bracing as its lateral force resisting system. The typical floor plan is shown on Figure 1A-2, and a building section is shown in Figure 1A-3. Figure 1A-4 depicts a two-story x-brace configuration and elevations. Design of the major lateral force resisting structural steel elements and connections uses AISC Allowable Stress Design (ASD). The 1997 UBC design provisions for special concentric braced frames (SCBFs) are attributed to research performed at the University of Michigan. The basis for SCBF bracing is the proportioning of members such that the compression diagonals buckle in a well behaved manner, without local buckling or kinking that would result in a permanent plastic deformation of the brace. Research performed has demonstrated that systems with this ductile buckling behavior perform well under cyclic loading. Several references are listed at the end of this Design Example.

Figure 1A-2. Typical floor framing plan

20


Design Example 1A

!


Figure 1A-3. Typical building section

Elevation A

Elevation B

Figure 1A-4. Braced frame elevations


21

Design Example 1A

!


Outline This Design Example illustrates the following parts of the design process: 1.

Design base shear.

2.

Distribution of lateral forces.

3.

Interstory drifts.

4.

Typical diaphragm design.

5.

Braced frame member design.

6.

Bracing connection design.

Given Information Roof weights: Roofing Insulation Concrete fill on metal deck Ceiling Mechanical/electrical Steel framing

Live load:

4.0 psf 3.0 44.0 3.0 5.0 7.0 66.0 psf 20.0 psf

Exterior wall system weight: steel studs, gypsum board, metal panels Structural materials: Wide flange shapes Tube sections Weld electrodes Bolts Shear Plates Gusset plates

22

Floor weights: Flooring Concrete fill on metal deck Ceiling Mechanical/electrical Steel framing Partitions

Live load:

1.0 psf 44.0 3.0 5.0 9.0 10.0 72.0 psf 80.0 psf

15 psf

ASTM A36 (Fy = 36 ksi) ASTM A500 grade B (Fy = 46 ksi) E70XX ASTM A490 SC ASTM A572 grade 50 (Fy = 50 ksi) ASTM A36 (Fy = 36 ksi)


Design Example 1A

Site seismic and geotechnical data: Occupancy category: Standard Occupancy Structure Seismic Importance Factor: I=1.0 Soil Profile Type “Stiff Soil”: Type S D (default profile) Seismic zone: Zone 4, Z = 0.4 Seismic Zone 4 near-source factors: Seismic source type: Type B Distance to seismic source: 8 km Near source factors: N a = 1.0, N v = 1.08

!


§1629.2 Table 16-K §1629.3, Table 16-J §1629.4.1, Table 16-I §1629.4.2 Table 16-U Tables 16-S, 16-T

The geotechnical report for the project site should include the seismologic criteria noted above. If no geotechnical report is forthcoming, ICBO has published Maps of Known Active Fault Near-Source Zones in California and Adjacent Portions of Nevada [ICBO, 1998]. These maps (prepared by the California Department of Conservation Division of Mines and Geology, in cooperation with the Structural Engineers Association of California) provide a means for easily determining the seismic source type and distance to the seismic source.

Factors that Influence Design Requirements for design of steel braced frames are given in the 1997 UBC. These cover special concentric braced frames (SCBF), ordinary concentric braced frames (OCBF), and chevron (or V) braced frames. After the adoption of the 1997 UBC provisions by ICBO, the 1997 AISC Seismic Provisions for Structural Steel Buildings (AISC-Seimsic) became available. Although not adopted into the code, these represent the state-of-the-art and are recommended by SEAOC, particularly for design of SCBF connections. The following paragraphs discuss some important aspects of braced frame design. This discussion is based on SEAONC seminar notes prepared by Michael Cochran, SE. Permissible types of concentric braced frames.

Shown in Figure 1A-5 are various types of concentric braced frames permitted by the code. Each of these can be design as either an ordinary concentric braced frame (OCBF) or a special concentric braced frame (SCBF). It should be noted that the only difference between an SCBF and an OCBF is the connection detailing and some prescriptive code requirements.


23

Design Example 1A

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a. Zipper

b. 2-story-X

c. X-bracing

d. Inverted V (or chevron)

e. V-bracing

Figure 1A-5. Permissible types of braced frames

All of the frames shown in Figure 1A-5 are essentialy variations on the chevron brace, except for the one-story X-brace (Figure 1A-5c). Single diagonal braced frames are also permissible by the code, but these are heavily penalized since they must take 100 percent of the force in compression unless multiple single diagonal braces are provided along the same brace frame line. Grades of steel used in SCBFs.

SCBF members are typical wide flange sections (ASTM A36, Fy = 36 ksi, or A572, grade 50, Fy = 50 ksi), tube sections (ASTM A500, grade B, Fy = 46 ksi), or pipes (ASTM A53, grade B, Fy = 35 ksi). When designing brace connections, the actual yield strength of the steel needs to be considered. The AISC-Seismic provisions address this overstrength issue using the R y factor, which is not addressed by the UBC or considered in this Design Example. The gusset plate material used in SCBF connections should be of equal yield strength to the brace member. Since the actual expected yield strength of most structural sections used as brace members is in excess of 50 ksi, the strength of the gusset plate material should be at least 50 ksi. High strength steel is required in order to keep the gusset plate thickness and dimensions to a minimum. Use of A36 material (as shown in this Design Example) will generally result in larger connections. Brace behavior.

Concentric braced frames are classified by the UBC as either ordinary or special. The title “special” is given to braced frames meeting certain detailing and design parameters that enable them to respond to seismic forces with greater ductility. The Blue Book Commentary is an excellent reference for comparison and discussion of these two systems. 24


Design Example 1A

!


Both inverted V-frames and V-frames have shown poor performance during past earthquakes due to buckling of the brace and flexure of the beam at the midspan connection instead of truss action, therefore the zipper, 2-story-X and X-bracing schemes are the preferred configurations.

V

Figure 1A-6. Chevron brace post-buckling stage and potential hinging of columns

The SEAOC Blue Book (in Section C704) has gone as far to recommend that chevron bracing should not be used unless it is in the Zipper or 2 story x configuration in high seismic zones. The reader is referred to the SEAOC Blue Book for a further discussion on chevron braces. Generally, the preferred behavior of bracing is in-plane buckling when fixity is developed at the end connections and three hinges are required to form prior to failure of the brace. The problem is that it is difficult to develop this type of fixity when you are using gusset plate connections which tend to lend themselves to outof-plane buckling of the brace and behave more like a pin connection.

There are limited structural shapes availble that can be oriented such that the brace will buckle in-plane. The following is a list of such shapes: 1. Hollow structural sections about their weak axis, for example, a TS 6x3x1/2 arranged as shown in Figure 1A-7a (Note: there can be a problem with shear lag in HSS sections). 2. Double angles with short legs back to back (Figure 1A-7b). 3. Wide flange shapes buckling about their weak axis (Figure 1A-7c).


25

Design Example 1A

!


y

x

y

x

y a. Flat tube (HSS)

x

x

x

y

y b. Double angles (SLV)

c. Wide flange (weak axis)

Figure 1A-7. Various brace shapes oriented for in-plane buckling

When designing a brace to buckle in-plane, it is recommended that the ratio of rx ry not exceed 0.65 to ensure that the brace will buckle in-plane.

Two architectural restrictions typically occur that inhibit in-plane buckling. First, the architect may not want to reduce the floor space by putting the brace in the flat position, and second, often there are infill steel studs above and below the brace, which may prevent the brace from buckling in-plane and force it to buckle out-of-plane. Both AISC and UBC steel provisions provide an exception that when met, allow for the brace to buckle out-of-plane. With the predominate use of gusset plates, this exception is probably used 95 percent of the time in brace design. The brace connection using a vertical gusset plate has a tendancy to buckle outof-plane due to the lack of stiffness in this direction. As can be seen in the Figure 1A-8, the gusset plate has significantly less stiffness in the out-of-plane direction. If the brace is symmetrical, you have a 50-50 chance as to whether it will buckle in-plane or out-of-plane, and the end connections then have a great influence as to how the brace will actually buckle. Since there is significantly less stiffness in the out-of-plane direction, the brace will buckle outof-plane. When a brace buckles out-of-plane relative to the gusset plate, it attempts to form a hinge line in the gusset plate. In order for the brace to rotate and yield about this hinge line (act as a pin connection), the yield lines at each end of the brace must be parallel. This is illustrated in Figure 1A-9 and Figure 1A-10.

26


Design Example 1A

!


buckling perpendicular to gusset plate (least resistance) y x yield line (hinge) x y gusset plate x x

Figure 1A-8. In-plane vs out-of-plane buckling of braces; gusset plate stiffness can influence brace buckling direction

Plan view

force

yield line

C

T

Isometric view

Figure 1A-9. Out-of-plane buckling of the brace; gusset plates resist axial loads without buckling, but can rotate about the yield line to accommodate the brace buckling


27

Design Example 1A

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yield line 90 degrees to slope of brace

Figure 1A-10. Yield line in gusset plate must be perpendicular to the brace axis

To ensure that rotation can occur at each end of the brace without creating restraint, the axis of the yield line must be perpendicular to the axis of the brace. Another requirement to allow for rotation about the yield line to occur, is a minimum offset from the end of the brace to the yield line, as shown in Figure 1A-11. If this distance is too short, there physically is insufficent distance to accomodate yielding of the gusset plate without fracture. Figure 1A-11 depicts the minimum offset requirement of the building codes. Due to erection tolerances and other variables, it is recommended that this design offset not be less than three times the gusset plate thickness (3t).

2t (min) 4t (max offset

plastic hinge forms at yield line

brace

gusset plate (t) yield line 90 degrees to slope of brace

Beam

Figure 1A-11. Yield line offset requirements; in practice 3t is often used to allow for erection tolerances

28


Design Example 1A

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There has been a misconception in some previous interpetations of the yield line offset, that all that was necessary was shape the end of the brace relative to the yield line so that they both were parallel to each other. Inherently, what happens is that the yield lines at the opposite ends of the brace are not parallel (see Figure 1A-10 for parallel yield line illustration) to each other and restraint builds up in the gusset plate as it attempts to buckle out-of-plane. The only way to relieve the stress is for the gusset plate to tear at one end of the brace, until the yield lines at each end of the brace are again parallel to each other.

possible yield line 90 degrees to axis of brace 2t offset (from brace tip)

brace

detailed 2t offset from yield line

gusset plate theoretical curved yield line as gusset attempts to bend around tip

2t offset (clamp force)

Beam

Note: This detail is not recommended.

Figure 1A-12. Shaping end of brace creates restraint

Figure 1A-12 (not recommended) depicts what happens when you try to shape the end of the brace to match the yield line slope. Due to the offset in the end of the brace, the yield line will attempt to bend around corner of the brace. This creates several problems, in that it is impossible to bend the plate about a longer curved line, since the curve creates more stiffness than a shorter straight line between two points that wants to be the hinge. The end tip of the brace along the upper edge is generally not stiff enough to cause a straight yield line to bend perpendicular to the brace axis about the tip end of the brace since there is only one side wall at this location to apply force to the gusset plate.


29

Design Example 1A

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Detailing considerations.

Floor slabs, typically metal deck and concrete topping slab in steel frame buildings, can cause additional restraint to buckling out-of-plane and must be taken into account during design. If the yield line crosses the edge of the gusset plate below the concrete surface, more restraint occurs, the gusset plate will likely tear along the top of the concrete surface. The SCBF connections design details in Design Example 1A have been simplified, but need to consider the potential restraint that occurs due to the floor deck since it will impact the gusset plate design. To keep the gusset plate size as small as possible, the gusset plate should be isolated from the concrete slab so the yield line can extend below the concrete surface. Figure 1A-13 shows how the gusset plate could be isolated from restraint caused by the slab. Note that the entire gusset plate does not have to be isolated, just that area where the yield line occurs. The compressible material which can be used would be a fire caulk that has the same required fire rating as the floor system. compressible material gusset plate 1" ±

2t (min) 4t (max) offset Plan

brace gusset plate yield line 90 degrees to slope of brace

compressible material each side of gusset plate

concrete slab

.

2" min

Beam

Figure 1A-13. For the yield line to develop in the gusset plate, the gusset plate must be isolated from the slab

30


Design Example 1A

!


A recent development in the design of gusset plate connections is the need to consider the length of the unstiffened edge of the gusset plate and the possibility of a premature buckling. For additional information about this subject, as well as additonal gusset plate design and sizing criteria such as the “Critical Angle Concept” and other practical design information, the reader is referred to the recent SEAONC (May, 2000) and SEAOSC (November, 1999) seminar notes on the design and detailing of SCBF steel connections. Field inspection of SCBFs.

Because of the critical importance of the connections, the actual field erection of SCBFs must be carefully inspected. Shop drawings often show erection aids such as clip angles and erection bolts. These are used to properly center the brace on the gusset plate. In the case of tube bracing, it is very common to have an erection bolt hole placed at each end of the brace. Occasionally, erector crews ignore these erection aids while placing the bracing over the gusset plates and making the weldments without verifying that the required 2t to 4t offset from the yield line has been maintained. The design engineer needs to remember that structural steel is erected using the shop drawings and that the structural drawings are often not checked, even though it is common practice to provide some form of general note that states “shop drawings are an erection aid, and structural drawings shall take precedent over the shop drawings…”. The following is a list of items that should be included in the checklist given to the Special Inspector: 1. Verify that the 2t minimum, 4t maximum offset from the yield line to brace end is maintained at each end of the brace. 2. Verify that the 1-inch minimum offset from the brace to the edge of the gusset plate is maintained and that the gusset plate edge slopes are the same slopes as shown on shop drawings and structural drawings. 3. Verify that the gusset plate yield line has been isolated from the concrete slab and that is is away from an edge stiffener plates.


31

Design Example 1A

!


Calculations Calculations and Discussion

1. 1a.

Code Reference

Design base shear.

§1630.1

Check configuration requirements.

§1629.5, Table 16-L

The structure is L-shaped in plan and must be checked for vertical and horizontal irregularities. Vertical irregularities. Review Table 16-L. By observation, the structure has no vertical irregularities; the bracing is consistent in all stories with no discontinuities or offsets, and the mass is similar at all floor levels. Plan irregularities. Review Table 16-M.

§1633.2.9, Table 16-M, Items 6 & 7

The building plan has a re-entrant corner with both projections exceeding 15 percent of the plan dimension, and therefore is designated as having Plan Irregularity Type 2. Given the shape of the floor plan, the structure is likely to have Torsional Irregularity Type 1. This condition will be investigated with the computer model used for structural analysis later in this Design Example. Plan Irregularity Type 2 triggers special consideration for diaphragm and collector design, as delineated in §1633.2.9, Items 6 and 7.

1b.

Classify structural system and determine seismic factors.

§1629.6

The structure is a building frame system with lateral resistance provided by special concentrically braced frames (SCBFs) (System Type 2.5.a per Table 16-N). The seismic factors are: R = 6.4 Ω o = 2.2 hmax = 240 ft

32

§1630.3, Table 16-N


Design Example 1A

1c.

!


Select lateral force procedure.

§1629.8

The static lateral force procedure is permitted for irregular structures not more than five stories or 65 feet in height (§1629.8.3). Although the structure has a plan irregularity, it is less than 65 feet in height. A dynamic analysis is not required, so static lateral procedures will be used.

1d.

Determine seismic response coefficients Ca and Cv.

§1629.4.3

For Zone 4 and Soil Profile Type S D :

1e.

C a = 0.44(N a ) = 0.44(1.0 ) = 0.44

Table 16-Q

C v = 0.64(N v ) = 0.64(1.08) = 0.69

Table 16-R

Evaluate structure period T.

§1630.2.2

Per Method A: T A = C t (hn ) 4 C t = 0.020 3

T A = 0.02(62 )

3

4

(30-8)

= 0.44 sec

Per Method B: From three-dimensional computer model, the periods are: North-south direction: TB = 0.66 sec East-west direction: TB = 0.66 sec Maximum value for TB = 1.3 T A = 1.3(0.44) = 0.57 sec Therefore, upper bound on period governs use T = 0.57 sec


§1630.2.2

33

Design Example 1A

1f.

!


Determine design base shear.

The total design base shear for a given direction is determined from Equation (30-4). Since the period is the same for both directions, the design base shear for either direction is: V =

Cv I 0.69(1.0 ) W = W = 0.189W RT 6.4(0.57 )

(30-4)

Base shear need not exceed: V =

2.5Ca I 2.5(0.44)(1.0 ) W = = 0.172W R 6.4

(30-5)

For Zone 4, base shear shall not be less than: V =

0.8ZN v I 0.8(0.4 )(1.08)(1.0) W = = 0.054W R 6.4

(30-7)

Equation (30-5) governs base shear. ∴ V = 0.172W

1g.

Determine earthquake load combinations.

§1630.1

Section 1630.1.1 specifies earthquake loads. These are E and E m as set forth in Equations (30-1) and (30-2). E = ρE H + E v

(30-1)

Em = Ω o E H

(30-2)

The normal earthquake design load is E . The load E m is the estimated maximum earthquake force that can be developed in the structure. It is used only when specifically required, as will be shown later in this Design Example. Before determining the earthquake forces for design, the reliability/redundancy factor must be determined. Reliability/redundancy factor ρ = 2 −

34

20 rmax Ab

(30-3)


Design Example 1A

!


Ab = (180)2 + 180(132 + 192 ) = 90,720 ft 2 To estimate an initial value for ρ , for purposes of preliminary design, an assumption for the value of rmax is made. For rmax , assume that the highest force in any brace member is 10 percent greater than average for the 18 total braces. ∴ rmax =

1.10 = 0.061 18

§1630.1.1

and: ρ = 2−

20 0.061(90,720 )1 / 2

= 0.91

and: 1.0 ≤ ρ ≤ 1.5 ∴ Use ρ = 1.0 The value for ρ should be confirmed upon completion of the computer analysis for the brace forces. For load combinations of §1612, E and E m are as follows: E = ρE h + E v = 1.0(V )

(30-1)

( E v = 0 since allowable stress design is used in this Design Example) Em = Ω o Eh = 2.2(V )

(30-2)

Note that seismic forces may be assumed to act non-concurrently in each principal direction of the structure, except as per §1633.1.

2. 2a.


Calculate building weights and mass distribution.

Calculated building weights and centers of gravity at each level are given in Table 1A-1. Included is an additional 450 kips (5.0 psf) at the roof level for mechanical equipment. Building mass properties are summarized in Table 1A-2. Braced frame locations are noted in Figure 1A-14 below.


35

Design Example 1A

!


A denotes two braced bays B denotes one braced bay

Figure 1A-14. Braced frame location plan

Table 1A-1. Building weight Roof Weight (1) Mark2 I II III Walls Totals ∴

w DL (psf) 71 71 71 15

Area (sf) 23,760 32,400 34,560 16,416

Wi (kips) 1,687 2,300 2,454 246 6,687

Ycg

W X cg

(ft) 90 90 276 168

(ft) 66 222 222 175

(lbs) 151,826 207,036 677,238 41,368 1,077,468

I II III Walls Totals ∴

( )

W Ycg

(lbs) 111,339 510,689 544,735 43,092 1,209,855

X cg = 1,077,468 6,687 = 161.1 ; Ycg = 1,209,885 6,687 = 180.9

4th, 3rd, & 2nd Floor Weights (2) Mark2

( )

X cg

w DL (psf) 72 72 72 15

Area (sf) 23,760 32,400 34,560 20,520

Wi (kips) 1,711 2,333 2,488 308 6,840

X cg

Ycg

(ft) 90 90 276 168

(ft) 66 222 222 175

( )

W X cg (lbs) 153,965 209,952 686,776 51,710 1,102,404

( )

W Ycg

(lbs) 112,908 517,882 552,407 53,865 1,237,061

X cg = 1,102,404 6,840 = 161.1 ; Ycg = 1,237,061 6,840 = 180.9

Note: 1. Roof weight: wDL = 66.0 + 5.0add'l mech = 71.0 psf ; exterior walls: wwall = 15 psf ; wall area = (7.5 + 4.5)(1,368 ft ) = 16,416 ft 2 2. wDL = 72.0 psf ; exterior walls: wwall = 15 psf ; wall area = (15)(1,368 ft ) = 20,520 ft 2

36


Design Example 1A

Table 1A-2. Mass properties summary Level Roof 4th 3rd 2nd Total

WDL (kips) 6,687 6,840 6,840 6,840 27,207

!


(1)

X cg

Ycg

(ft)

(ft)

161.1 161.1 161.1 161.1

180.9 180.9 180.9 180.9

M (2)

MMI (3)

17.3 17.7 17.7 17.7 70.4

316,931 324,183 324,183 324,183

Notes: 1. Mass (M) and mass moment of inertia (MMI) are used in analysis for determination of fundamental period (T). 2. M = (W 3.86.4 )(kip ⋅ sec in.)

(

)(

3. MMI = (M A) I x + I y kip ⋅ sec 2 ⋅ in

2b.

)


As noted above, Equation (30-5) governs, and design base shear is: V = 0.172W = 0.172(27207) = 4,680 kips

2c.

Determine vertical distribution of force.

§1630.5

For the static lateral force procedure, vertical distribution of force to each level is applied as follows: V = Ft + ∑ Fi

(30-13)

where: Ft = 0.07T (V ) Except Ft = 0 where T ≤ 0.7 sec

(30-14)

For this structure Ft = 0 , and the force at each level is Fx =

(V − Ft )W x hx ∑ Wi hi

 W h = V  x x  ∑ Wi hi


   

(30-15)

37

Design Example 1A

!


The vertical distribution of force to each level is given in Table 1A-3 below. Table 1A-3. Distribution of base shear

2d.

Level

wx (kips)

hx (ft)

Roof 4th 3rd 2nd Total

6,687 6,840 6,840 6,840 27,207

62 47 32 17

w x hx (k-ft) 414,594 321,480 218,880 116,280 1,071,234

w x hx Σw x hx 0.39 0.30 0.20 0.11 1.00

Fx (kips) 1,811.3 1,404.5 956.2 508.0 4,680.0

ΣV

(kips) 1,811.3 3,215.8 4,172.0 4,680.0

Determine horizontal distribution of force.

§1630.6

Structures with concrete fill floor decks are generally assumed to have rigid diaphragms. Forces are distributed to the braced frames per their relative rigidities. In this Design Example, a three-dimensional computer model is used to determine the distribution of seismic forces to each frame. For rigid diaphragms, an accidental torsion must be applied (in addition to any natural torsional moment), as specified in §1630.6. The accidental torsion is equal to that caused by displacing the center of mass 5 percent of the building dimension perpendicular to the direction of the applied lateral force. For our structural computer model, this can be achieved by combining the direct seismic force applied at the center of mass at each level with the accidental torsional moment (M z ) at that level. North-south seismic: M t = 0.05(372 ft )Fx = (18.6)Fx East-west seismic: M t = 0.05(312 ft )Fx = (15.6 )Fx Using the direct seismic forces and accidental torsional moments given in Table 1A-4, the distribution of forces to the frames is generated by computer analysis. (For the computer model, member sizes are initially proportioned by preliminary hand calculations and then optimized by subsequent iterations.)

38


Design Example 1A

!


Table 1A-4. Accidental torsional moments Level

Fx (kips)

N-S M t (k-ft)

E-W M t (k-ft)

Roof 4th 3rd 2nd

1,811.3 1,404.5 956.2 508.0

33,690 26,124 17,785 9,449

28,256 21,910 14,917 7,925

From the computer analysis, forces in each bracing member are totaled to obtain the seismic force resisted by each frame. The frame forces are then summed and compare to the seismic base shear for a global equilibrium check. Forces at the base of each frame are summarized in Table 1A-5 below:

North-South Direction

East-West Direction

Table 1A-5. Distribution of forces to frames Frame

Direct Seismic (kips)

Torsional Force (kips)

Direct + Torsion (kips)

A1 A2 A3 A4 B1 Total

1,023 1,067 1,063 1,018 509 4,680

61 65 26 87 12

1,084 1,132 1,089 1,105 521 4,931

977

77

1,054

937 1,005 1,280 481 4,680

76 13 134 6

1,013 1,018 1,414 487 4,986

A5 A6 A7 A8 B2 Total

Note that the torsional seismic component is always additive to the direct seismic force. Sections 1630.6 and 1630.7 require that the 5 percent center-of-mass displacement be taken from the calculated center-of-mass, and that the most severe combination be used for design.

2e.

Determine horizontal torsional moments.

§1630.7

As shown above, the accidental torsional moment has been accounted for as required by §1630.6. However, we must check for a torsional irregularity (per Table 16-M, Type 1) to determine if a torsional amplification factor (Ax ) is required under the provisions of §1630.7.


39

Design Example 1A

!


Torsional irregularity exists when the drift at one end of the structure exceeds 1.2 times the average drifts at both ends, considering both direct seismic forces plus accidental torsion. For this evaluation, total seismic displacements at the roof level are compared. The displacements in Table 1A-6 below are taken from the computer model for points at the extreme corners of the structure.

Table 1A-6. Roof displacements North-South Direction East-West Direction

@ Line A 0.95 in @ Line 1 1.05 in

@ Line N 1.3 in @ Line 11 1.22 in

Average 1.125 Average 1.135

Ratio (max/avg) 1.16 o.k. Ratio (max/avg) 1.07 o.k.

Because the maximum drift is less than 1.2 times the average drift, no torsional irregularity exists. The relative displacements at the 2nd, 3rd, and 4th floors are similar to those at the roof; no torsional irregularities were found to exist at those levels.

3. 3a.

Interstory drift. Determine ∆s and ∆m.

§1630.9

The design level response displacement (∆ S ) is obtained from a static elastic analysis using the seismic forces derived from the design base shear. When determining displacements, §1630.10.3 eliminates the upper limit on TB , allowing for a reduction in seismic forces calculated using Equation (30-4). For this example, the base shear could be reduced about 5 percent using TB with Equation (30-4), with a proportional reduction in calculated drifts. The maximum inelastic response displacement (∆ M ) includes both elastic and estimated inelastic drifts resulting from the design basis ground motion: ∆ M = 0.7(R )∆ S = 0.7(6.4 )∆ S = 4.48∆ S

(30-17)

The greatest calculated values for ∆ S and ∆ M are to be used, including torsional effects. For determination of ∆ M , P∆ effects must be included. Story drift ratios are calculated from lateral displacements at each level for both the north-south and east-west directions (as generated by the computer analysis), and are presented in the Table 1A-7.

40


Design Example 1A

!


East-West Displacements

North-South Displacements

Table 1A-7. Story displacements and drift ratios Story

Height (in.)

∆S (in.)

∆M (in.)

Drift Ratio (1) (2)

4th

180

(1.30-1.04) = 0.26

1.16

0.0064

3rd

180

(1.04-0.70) = 0.34

1.52

0.0084

2nd

180

(0.70-0.34) = 0.36

1.61

0.0089

1st

204

(0.34-0.0) = 0.34

1.52

0.0075

4th

180

(1.22-0.98) = 0.24

1.08

0.0060

3rd

180

(0.98-0.67) = 0.31

1.39

0.0077

2nd

180

(0.67-0.34) = 0.33

1.48

0.0082

1st

204

(0.34-0.0) = 0.34

1.52

0.0075

Notes: 1. Interstory drift ratio = ∆ M /story height. 2. Maximum drift occurs at Line N for north-south direction and Line 11 for east-west direction.

3b.

Determine story drift limitation.

§1630.10

Story drift limits are based on the maximum inelastic response displacements, ∆ M . For structures with T < 0.7 the maximum allowable drift is 0.025 times the story height. A review of drift ratios tabulated in Table 1A-7 shows that all interstory drift ratios are less than 0.025 using the period of Equation (30.4). (Note: Using the full value for TB would result in a lower base shear and smaller story displacement.)

4.


The building has rigid diaphragms at all levels, including the roof. In this Part, seismic forces on each diaphragm will be determined, and the roof level diaphragm designed. The roof was selected because it is the most heavily loaded diaphragm.

4a.

Determine diaphragm load distribution.

§1633.2.9

In multistory buildings, diaphragm forces are determined by the following formula: F px =

Ft + ∑ Fi

∑ wi

(w px )

(33-1)

where: 0.5C a IW px < F px ≤ 1.0C a IW px


§1633.2.9 Item 2

41

Design Example 1A

!


The diaphragm forces at each level, with the upper and lower limits, are calculated as shown in Table 1A-8. Table 1A-8. Diaphragm forces (kips) Level

Fi

ΣFi

Roof 4th 3rd 2nd

1,811.3 1,404.5 956.2 508.0

1,811.3 3,215.8 4,172.0 4,680.0

wx 6,687 6,840 6,840 6,840

Σw i

Fpx

0.5Ca Iw px

1.0Ca Iw px

6,687 13,527 20,367 27,207

1,811.3 1,626.1 1,401.1 1,176.6

1,471.1 1,504.8 1,504.8 1,504.8

2,942.3 3,009.6 3,009.6 3,009.6

Note: C a = 0.44 and I = 1.0 .

4b.

Determine diaphragm shear.

The maximum diaphragm design force occurs at the roof level. To facilitate diaphragm and collector design, this force is divided by the plan area to obtain an average horizontal seismic force distribution, q roof . q roof =

1,811 = 0.020 kips/ft 2 90,720

The maximum diaphragm span occurs between Lines A and N, so the north-south direction will control. Both loading and shear for the roof diaphragm under northsouth seismic forces are shown in Figure 1A-15.

Figure 1A-15. Roof diaphragm north-south seismic load and shear

42


Design Example 1A

!


The computer model assumes rigid diaphragms or load distribution to the frames. In lieu of an exact analysis, which considers the relative stiffness of the diaphragm and braced frames, we envelop the solution by next considering the diaphragms flexible. Shears at each line of resistance are derived assuming the diaphragms span as simple beam elements under a uniform load. w1 = q roof (312 ft ) = 0.020(312) = 6.24 kips/ft

w2 = q roof (180 ft ) = 0.020(180 ) = 3.6 kips/ft

Diaphragm shears:  180  V A = VGA = 6.24   = 562 k  2   192  VGN = V N = 3.6   = 346 k  2  To fully envelop the solution, we compare the flexible diaphragm shear at Line N with the force resisted by Frame A8 (Figure 1A-14) assuming a rigid diaphragm. From the computer model, we find at Frame A8: Froof = 440 k . The force from the rigid analysis (440 k) is greater than the force from the flexible analysis (346 k), so the greater force is used to obtain the maximum diaphragm shear at Line N: q N = 440 180 = 2.44 k/ft at Line N

§1612.3.2

Using allowable stress design and the alternate load combinations of §1612.3.2, the (12-13) basic load combination is:  E     1.4 

(12-13)

Maximum design shear:  2.44  qN =   = 1.74 kips/ft  1.4  With 3-1/4 inch lightweight concrete over 3"×20 gauge deck, using 4 puddle welds per sheet, the allowable deck shear per the manufacturer’s ICBO evaluation report is: Vallow = 1.75 > 1.74 kips/ft


o.k.

43

Design Example 1A

!


Other deck welds (e.g., parallel supports, seam welds) must also be designed for this loading. At seismic collectors, it is good practice to place additional welded studs in every low flute of the deck for shear transfer.

4c.

Determine collector and chord forces.

Using a flexible analysis and assuming diaphragm zone III acts as a simple beam between Lines G and N (Figure 1A-16), for north-south seismic loads the maximum chord force on lines 1 and 7 is: wl 2 3.6(192) 2 CF = = = 92.2 kips 8d 8(180)

§1633.2.9 Items 6 and 7

Note that this value must be compared to the collector force at Lines 1 and 7, and the largest value used for design.

Figure 1A-16. Roof diaphragm zones

For structures with plan irregularity type 2, the code disallows the one-third stress increase for allowable stress design for collector design (§1633.2.9, Item 6). This code section also requires chords and collectors be designed considering “independent movement of the projecting wings,” for motion of the wings in both the same and opposing directions. There are two ways to achieve this:

44


Design Example 1A

!


1.

Use a three dimensional computer model with membrane or thin-shell diaphragm elements to capture the relative stiffness between the floor and braces.

2.

Make a simplifying assumption that gives reasonable values for collector forces at the re-entrant corner.

For this example, the second option is chosen. If each wing is assumed to be flexible relative to the central diaphragm (Zone II), the wings can be considered as “fixed-pinned” beams. The maximum moment at Line G is: M fixed

w2l 2 3.6(192)2 = = = 16,589 kips-ft 8 8

The maximum tie force (TG ) along Lines 1 and 7 at the intersections with Line G is: TG = 16,589 180 = 92.2 kips With allowable diaphragm shear of 75 k/ft, this tie force must be developed back into diaphragm zone II over a length of at least: 92.2 kips = 37.6 ft (1.4)1.75 kips/ft Next, the collector forces for east-west seismic loads are determined. For Zone III between Lines 1 and 7, the equivalent uniform lateral load is: w3 = q (depth ) = 0.020(372 ) = 7.44 k/ft The collector force at Line 1 is: R1 = 7.44(180 2) = 670 kips From the computer model, at the roof level the frames on Line 1 (Frames A1 and A2) resist loads of 405 kips and 425 kips, respectively. R1 = 405 A1 + 425 A2 = 830 kips > 670 kips


45

Design Example 1A

!


Therefore, the “rigid diaphragm” analysis governs, and the shear flow along Line 1 (q1 ) , is: q1 = 830 372 = 2.23 kips/ft As shown in Figure 1A-17, collector forces at points a, b, c, and d are: Fa = 2.23(30 ) = 67 kips Fb = 2.23(90 ) + 405 = 204 kips Fc = 2.23(244 ) + 405 = 140 kips Fd = 2.23(64 ) = 143 kips The maximum collector force as shown in Figure 1A-17 is T = 204 kips .

Figure 1A-17. Collector force diaphragm at Line 1

The collector forces for east-west seismic loads exceed the chord forces calculated for north-south seismic, and therefore govern the collector design at Line 1. Use maximum T1 = 204 kips and minimum T1 = 140 kips . The collector element can be implemented using either the wide flange spandrel beams and connections or by adding supplemental slab reinforcing. In this example, supplemental slab reinforcing is used. Under §1633.2.6, using the strength design method, collectors must be designed for the special seismic load combinations of §1612.4. E m = Tm = Ω oT = (2.2)T

46

§1633.2.6


Design Example 1A

!


Using the factored loads of §1612.4: Tmu = (1.0 )E m = (1.0 )(2.2 )T

§1612.4

Maximum Tmu = 2.2(204) = 449 kips

(30-2)

Minimum Tmu = 2.2(140) = 308 kips Maximum As = Tmu φ f y = 449 0.9 (60) = 8.3 in.2

(

∴ Use 11-#8 As = 8.69 in.2

(12-18)

)

Minimum As = 308 0.9 (60 ) = 5.7 in. 2 ∴ Use 8-#8 (As = 6.32 in.2) On Line 1, place 8-#8 bars continuous from Lines A to N, and additional 3-#8 (for a total of 11) along frame A1 to Line G. With slab reinforcing, the collected load must be transferred from the slab to the frame. This can be done with ¾" diameter headed studs, again using the special seismic load combination of §1612.4. At Frame A1:  Ω F  1.0 (2.2 ) 405 vu = 1.0  o A1  = = 14.9 kips/ft 60  LA1  The shear strength of ¾" diameter headed studs as governed in this case by the concrete strength ( f ' c = 3,000 psi ) is derived from §1923.3.3: φVc = φ800 Abλ f 'c = 0.65 (800)(0.44 )(0.75) 3,000 1,000 = 9.4 kips/stud

§1923.3.3

The required number of studs per foot (n ) is: n=

14.9 kips/ft = 1.59 studs/ft 9.4 kips/stud

∴ Use 2-3/4" diameter studs at 12-inch cc over length of Frame A1.


47

Design Example 1A

5.

!



§2212

In this part, the design of a typical bay of bracing is demonstrated. The design bay, taken from Elevation A, Figure 1A-4, is shown in Figure 1A-18. Member axial forces and moments are given for dead, live, and seismic loads as output from the computer model. All steel framing will be designed per Chapter 22, Division V, Allowable Stress Design. Requirements for special concentrically braced frames are given in §2213.9 of Chapter 22.

Figure 1A-18. Typical braced bay

TS brace @ 3rd story: PDL = 24 kips PLL = 11 kips Pseis = 348 kips PE = ρ (Pseis ) = 1.0(348 ) = 348 kips

WF beam @ 3rd floor: M DL = 1,600 kip-in. M LL = 1,193 kip-in. V DL = 14.1 kips V LL = 10.3 kips 48


Design Example 1A

!


Pseis = 72 kips PE = ρ (Pseis ) = 1.0(72 ) = 72 kips

WF column @ 3rd story: PDL = 67 kips PLL = 30 kips Pseis = 114 kips M seis ≈ 0 PE = 1.0 (Pseis ) = 1.0(114 ) = 114 kips

5a.

rd

Diagonal brace design at the 3 story.

§1612.3.1

The basic ASD load combinations of §1612.3.1 with no one-third increase are used. D+

348 E : Ρ1 = 24 + = 273 k (compression) 1.4 1.4

0.9 D ±

348 E : Ρ2 = 0.9(24 ) − = −227 kips (tension) 1.4 1.4

 348    E  D + 0.75 L +  = 219 kips (compression)  : Ρ3 = 24 + 0.7511 + 1.4   1.4   

(12-9)

(12-10)

(12-11)

The compressive axial load of Equation (12-9) controls. The clear unbraced length (l ) of the TS brace is 18.5 feet, measured from the face of the beam or column. Assuming k = 1.0 for pinned end, kl = 1.0(18.5) = 18.5 ft Maximum slenderness ratio:


§2213.9.2.1 kl 1,000 ≤ r Fy

49

Design Example 1A

!


For a tube section, F y = 46 ksi ∴

Minimum r =

1,000 46

= 147.4

12(18.5) kl = = 1.51in. 147.4 147.4

§2213.9.2.4

 b  110 = 16.2 Maximum width-thickness ratio   ≤ Fy t Try TS 8 × 8 × 5 8 : r = 2.96 > 1.51 in.

o.k

b 8 = = 12.8 < 16.2 t 0.625

o.k.

For kl = 19 ft, Pallow = 324 kips > 273 kips

o.k.

AISC-ASD, pp. 3-41

∴ Use TS8 × 8 × 5 8

5b.

rd

Girder design at the 3 floor.

The girder will be designed using the basic load combinations of §1612.3.1 as noted above. The loads are: D + L : M D +L = 1,600 + 1,193 = 2,793 kip-in. D±

E : 1.4

Pseis =

(12-8)

72 = 51.4 kips 1.4

(12-9)

M DL = 1,600 kip-in.   E   72  D + 0.75 L +   : Pseis = 0.75   = 38.6 kips  1.4   1.4  

(12-11)

M D + L+ seis = 1,600 + 0.75(1,193) = 2,495 kip-in.

50


Design Example 1A

!


For the girder, use ASTM A36 steel with F y = 36 ksi . Assume that the bottom beam flange is braced at third points ∴ly =

30 = 10.0 ft 3

As a starting point for design, assume a beam with a cross-section area of area of 20 in.2 Find the required beam section modulus. fa =

51.4 = 2.6 ksi , and maximum Fa = 0.6(36 ) = 21.6 ksi then, 20

fa 2.6 = = 0.12 Fa 21.6 For an allowable bending stress, use: f b = (1 − 0.12 )(0.60)(36 ) = 19.0 ksi ∴ S req'd

2,793 = 147 in.3 19.0

Try W 24 × 68 beam S = 154 in.3 A = 20.1 in.2 rx = 9.55 in. ry = 1.87 in. 12(30 )  kl  = 37.7   = 9.55  r x 12 (10.0 )  kl  = 64.2   = 1.87  r y Fa = 17.02 ksi (compression governs)


AISC-ASD, pp. 3-16

51

Design Example 1A

!


Maximum f a =

51.4 = 2.55 ksi 20.1

fa 2.55 = = 0.149 < 0.15 Fa 17.02

o.k.

For combined stresses, use AISC Equation H1-3.

AISC-ASD Part 5, Ch. H

Check load combination of Equation (12-8). fb 2,793 = = 0.84 < 1.0 Fb 154(21.6 )

o.k.

Check load combination of Equation (12-9). fa f 2.55 1,600 + b = + = 0.15 + 0.48 = 0.63 < 1.0 Fa Fb 17.02 154(21.6 )

o.k.

Check load combination of Equation (12-11). fa f 38.6 2,495 + b = + = 0.11 + 0.75 = 0.86 < 1.0 Fa Fb 20.1(17.02 ) 154(21.6 )

o.k.

∴ Use W 24 × 68 girder Note that §2213.9.1 requires the girders to be continuous through brace connections between adjacent columns. For chevron bracing configurations, several additional requirements are placed on the girder design. Those requirements are addressed in Design Example 1C. The X-bracing configuration shown in this Example ensures the desired post-buckling capacity of the braced frame without inducing the large unbalanced seismic loading on the girder that occurs in a chevron brace configuration.

5c.

rd

Column design at the 3 floor.

The frame columns will also be designed using the basic load combinations of §1612.3.1 with no one-third increase. D + L : P0 = 67 + 30 = 97 kips (compression) D+

52

E 114 = 148.4 kips (compression) : P1 = 67 + 1.4 1.4

(12-8) (12-9)


Design Example 1A

0.9 D ±

!


E 114 : P2 = 0.9(67 ) − = 21.1 kips (tension) 1.4 1.4

(12-10)

 114   E   D + 0.75 L +  = 150.6 kips (compression)  : P3 = 67 + 0.75 30 + 1.4    1.4  

(12-11)

Per the requirements of §2213.9.5, the columns must have the strength to resist the special column strength requirements of §2213.5.1: ΡDL + 0.7 ΡLL + Ω o ΡE : Pcomp = 67 + 0.7(30 ) + 2.2(114 ) = 339 kips (compression)

§2213.9.5, Item 1

0.85ΡDL ± Ω o ΡE : Ρtens. = 0.85(67 ) − 2.2(114 ) = −194 kips (tension)

§2213.5.1, Item 2

For the columns, ASTM A36 steel with F y = 36 ksi will be used. The unbraced column height (floor height less ½ beam depth) is: h = 15 − 1 = 14 ft Try a W 10 × 49 column with kl = 14 ft

Pallow = 242 kips > 150.6 kips

o.k.

AISC-ASD, pp. 3-30

Check the column for the special column strength requirements of §2213.5 using member strength per §2213.4.2: Psc = 1.7 Pallow Psc = 1.7(242 ) = 411 > 339 kips (compression) Pst = F y A = 36(14.4) = 518.4 > 194 kips (tension)

o.k. o.k.

§2213.4.2

Note that §2213.5.2 places special requirements on column splices. To ensure the column splice can meet the ductility demand from the maximum earthquake force (E m ) , full-penetration welds at splices are recommended. The splice must occur within the middle one-third of the column clear height, not less than 4 feet above the beam flange. Finally, §2213.9.5 requires that the columns meet the width-thickness ratio limits of §2213.7.3: SEAOC Seismic Design Manual, Vol. III (1997 UBC)

53

Design Example 1A

bf 2t f

!


≤ 8.5 for F y = 36 ksi

For a W 10 × 49

bf 2t f

=

§2213.7.3

10 (0.56) = 8.9 > 8.5 2

no good

Division III, §2251N7

Try a W 10 × 54 bf 2t f

= 8.1 < 8.5

o.k.

AISC-ASD, pp. 5-96

Thus, the column design is governed by the local buckling compactness criterion. ∴ Use W10 x 54

6.


In this part, the connection of the TS8 × 8 brace to the W 10 column and W 24 girder will be designed. Connection of the braces to the mid-span of the girder is similar, and is shown in Example 1C.

6a.

Determine connection design forces.

§2213.9.2

Section 2213.9.3.1 requires that bracing connections have the strength to resist the lesser of: 3. The strength of the brace in axial tension, Pst . 4. Ω o times the design seismic forces, plus gravity loads. 5. The maximum force that can be transferred to the brace by the system. For the TS8 × 8 × 5 8 brace used in the design bay, the connection force is taken as the lesser of: Pst = Fy A = 46(17.4 ) = 800.4 kips

54

controls


Design Example 1A

!


or: Pm = PD + PL + Ω o PE = (24 + 11) + 2.2 (348) = 800.6 kips ∴ Use 800.4 kips for design

6b.

Design procedure using the uniform force method.

Based on research by AISC [Thornton, 1991], the Uniform Force Method (UFM) has been presented as an efficient, reliable procedure for design of bracing connections. The basis for the UFM is to configure the gusset dimensions so that there are no moments at the connection interfaces: gusset-to-beam; gusset-tocolumn; and beam-to-column. [For more information on the UFM, refer to AISC 1994 LRFD, Volume II, Connections.] Figure 1A-19 illustrates the gusset configuration and connection interface forces for the UFM. Note that the distances to the centroids of the gusset connection, ∝ and β , are coincident with the brace centerline. To achieve the condition of no moments at the interfaces, the following relationship must be satisfied: ∝ − β tan θ = eb tan θ − ec The connection forces are then given by these equations: r=

(α + ec )2 + (β + eb )2

α H b =  Ρ r e  Vb =  b  Ρ  r  β Vc =   Ρ r e  H c =  c Ρ  r  If the connection centroids do not occur at ∝ and β , moments are induced on the connection interface. The UFM can also be applied to this condition (see the LRFD Connections manual for the Special Case No. 2 example). In some cases, it may be beneficial to first select proportions for the gusset, then design the welds using unbalanced moments computed per the UFM Special Case No. 2. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

55

Design Example 1A

6c.

!


Gusset plate configuration and forces.

Application of the UFM essentially involves selecting of gusset dimensions, then analyzing plate and connection stresses and capacities at the interfaces. It is inherently a trial and error solution, and can readily be formatted for a spreadsheet solution. For this example, welded connections are used from gusset-to-beam and gusset-to-column. The beam-to-column connection will be made with highstrength bolts. A suggested starting point for determining the length of weld between gusset and column (2 β ) is to assume half the total length of weld to the brace. Note that per the AISC reference, these welds should be designed for the larger of the peak stress or 140 percent of the average stress. The 40 percent increase is intended to enhance ductility in the weld group, where gusset plates are welded directly to the beam or column. For this example brace connection, these parameters are fixed: θ = 45° ec =

10.0 = 5.0" (W 10 × 54) 2

eb =

23.7 = 11.9" (W 24 × 68) 2

α − β tan θ = eb tan θ − ec α − β(1.0) = 11.9(1.0 ) − 5.0 ∴ α = 6.9 + β After a few trials, the following are selected: α = 15.9" and β = 9.0" Using the axial strength of the brace, Pst = 800.4 kips , the connection interface forces are as follows: r=

(15.9 + 5)2 + (9.0 + 11.9)2

= 29.56"

Gusset-to-beam:  15.9   11.9  H b = 800.4  = 431 kips , Vb = 800.4  = 322 kips  29.56   29.56 

56


Design Example 1A

!


Gusset-to-column:  9.0   5.0  Vc = 800.4  = 244 kips , H c = 800.4  = 135 kips  29.56   29.56  From review of the computer output for the braced frame at the third floor, the collector force (Ab ) to the beam connection is: Ab = 41 kips

6d.

Brace-to-gusset design.

Bracing connections must have the strength to develop brace member forces per §2213.9.3.1. The capacities of the connection plates, welds and bolts are determined under §2213.4.2. Determine TS brace weld-to-gusset. For 5/8-in. tube, minimum fillet weld is ¼-in. Try ½-in. fillet weld using E70 electrodes. Per inch, weld capacity = 1.7(8)(0.928) = 12.62 kips-in. lreq =

AISC-ASD Table J2.5

800.4 = 15.9" @ 4 locations 12.62 ( 2)(2)

∴ Use 18-inches of ½-in. fillet each side, each face Check minimum gusset thickness for block shear:

[

RBS = (1.7 ) 0.30 Av Fu + 0.50 A t Fu

]

Fu = 58 ksi (A36 plate) where: Av = net shear area At = net tension area For TS 8 × 8 with Lweld = 18 in. Av = 2(18)t , At = (8)t SEAOC Seismic Design Manual, Vol. III (1997 UBC)

57

Design Example 1A

!


RBS = (1.7 )[0.3(36 ) + 0.5(8)](58)(tmin ) = 1,361kips tmin = 0.93 in. ∴ Use 1-in. plate gusset minimum, ASTM A36, F y = 36 ksi Check gusset plate compression capacity.

§2213.9.3.3

Section 2213.9.3.3 requires the gusset plate to have flexural strength exceeding that of the brace, unless the out-of-plane buckling strength is less than the in-plane buckling strength and a setback of 2t is provided as shown in Figure 1A-19. The gusset plate must also be designed to provide the required compressive capacity without buckling. The 2t setback is a minimum requirement. A setback of 3t provides for construction tolerance for brace fit-up, and should be considered during design. From Figure 1A-19, the gusset plate provides much greater in-plane fixity for the tube. The effective length factor (k ) for out-of-plane buckling is by observation greater than the in-plane factor (k ) , so the out-of-plane buckling strength will be less than the in-plane buckling strength. The setback of 2t promotes enhanced post-buckling behavior of the brace by allowing for hinging in the gusset instead of the brace. The gusset plate must be designed to carry the compressive strength of the brace without buckling. Using the Whitmore’s Method (see AISC LRFD Manual Vol. II), the effective plate width at Line A-A of Figure 1A-19a is: b = tube width + 2 (λ w ) tan 30° = 8 + 2 (18) tan 30° = 28.8 in. The unsupported plate length Lu is taken as the centerline length from the end of the brace to the edge of beam or column. From Figure 1-19a, this length measures 20 in. As recommended by Astaneh-Asl [1998], a value of k = 1.2 will be used. Maximum l u = 20 in. r=

t 1.0 = = 0.289 in. 12 3.464

kl 1.2 (20 ) = = 83.0 ∴ for F y = 36 ksi, r 0.289

AISC-ASD, Table C-36

Fa = 15.0 ksi

Gusset capacity: Pplate = 1.7(1.0)(28.8)(15.0 ) = 734 kips 58

§2213.4.2 SEAOC Seismic Design Manual, Vol. III (1997 UBC)

Design Example 1A

!


TS 8 × 8 brace compression capacity: Pbrace = 1.7(324 ) = 551 < 734 kips

o.k.

Comment: Where tube sections are slotted for gusset plates, as shown in Figure 1A-19, recent testing has shown that over-cut slots are of concern. Net section fracture at the end of the slot should be checked considering shear lag at the connection. If required, it is recommended that the tube section be reinforced with a cover plate at the end of the slot.

a. Symbols for connection design

c. Force diagram at column

b. Force diagram at gusset plate

d. Force diagram at beam

Figure 1A-19. Connection design using the uniform force method (UFM)


59

Design Example 1A

6e.

!


Gusset-to-beam design.

In this section, the connection of the 1-inch-thick plate gusset to the W24 beam will be designed. The weld length from gusset to beam is the plate length less the 1-inch clear distance between the beam and column. l w = 2(15.9 − 1.0 clr ) = 29.8" Per inch of effective throat area, weld stresses are: fx =

Hb 431 = = 7.23 ksi (x-component) 2(l w ) 2(29.8)

fy =

Vb 322 = = 5.40 ksi (y-component) 2(l w ) 2(29.8)

(7.23)2 + (5.40)

fr =

2

= 9.0 ksi (resultant)

For E70 electrodes, the allowable weld strength is:

§2213.4.2

Fw = 1.7(0.3)(70 ksi ) = 35.7 ksi The required weld size is: t weld =

9.0 = 0.36 in. 35.7(0.707 )

Under AISC specifications (Table J2.4), the minimum weld for a 1-inch gusset plate is 5/16-in., but as noted in Part 6c, we increase the weld size by a factor of 1.4 for ductility. t weld = 0.36(1.4 ) = 0.50 in. use ½-in. fillet weld Comparing the double-sided fillet to the allowable plate shear stress, the minimum plate thickness is: t pl =

2 (0.707 )(21)(0.50 ) = 1.0 in. 0.4 (36.0 )

∴ 1-inch plate

60

o.k.


Design Example 1A

!


Check compressive stress in web toe of W 24 × 68 beam: t w = 0.415 in. k = 1.375 in.

N = lw = 29.8 in. R = Vb = 322 kips R ≤ 1.33(0.66 )F y t w (N + 2.5 k ) 322 kips = 23.3 ksi ≤ 1.33 (0.66 )(36 ksi ) = 31.6 ksi (0.415)(29.8 + 2.5 (1.375))

6f.

AISC-ASD, K1.3

o.k.

Gusset-to-column design.

The gusset plate connection to the column is designed using the same procedure as the gusset-beam connection. The weld length to the column is: lw = 2(9 ) = 18 in. Per inch of effective throat area, weld stresses are: fx =

Hc 135 = = 3.75 ksi (x-component) 2(l w ) 2(18)

fy =

Vc 244 = = 6.77 ksi (y-component) 2(l w ) 2(18)

fr =

(3.75)2 + (6.77 )2


Determine the required weld size, with the 1.4 factor to enhance ductility of the weld.  7.75 ksi  t weld = 1.4   = 0.42 in.  35.7(0.707 )

∴ ½-in. fillet weld

o.k.


61

Design Example 1A

!


Check compressive stress in the web toe of the W 10 × 54 column: R 135 = = 17.3 ksi t (N + 2.5k ) (0.37 )(18 + 2.5(1.25)) 1.33(0.66 )(36 ksi ) = 31.6 ksi > 17.3 ksi

6g.

AISC-ASD K1.3

o.k.

Beam-to-column connection.

The connection of the W 24 beam to the W 10 column must carry the dead and live loads on the beam as well as the vertical and horizontal components of the brace force transferred from the gusset plates to the top and bottom of the beam. From the diagonal brace above the beam (see Figure 1A-19d), the connection forces to the beam are: Ab + H c = 41 + 135 = 176 kips Rb = V DL + V LL = 14.1 + 10.3 = 24.4 kips Rb + Vb = 24.4 + 322 = 346 kips The diagonal brace below the beam also contributes to the beam-to-column connection forces. The horizontal component from the brace below (H c ) acts opposite to the brace above, while the vertical component (Vb ) adds to that from the brace above. The connection forces above are based on the tensile capacity of the brace, so it is reasonable to use the compressive strength of the brace below. Assuming a TS8 × 8 × 5 8 -in. brace below: Psc = 1.7(324 ) = 551 kips ∴ Vb = 322(551 800 ) = 222 kips H c = 135(551 800) = 93 kips The net beam-to-column connection forces (as shown in Figure 1A-19b) are: Ab + H c = 176 − 93 = 83 kips Rb + Vb = 346 + 222 = 568 kips

62


Design Example 1A

!


Using an eccentricity of ± 3 inches: M ecc = (3)(568) = 1,704 kip-in. Try a single shear plate (A572 grade 50) with 2 rows of 7-1¼-inch diameter A490 SC bolts (14 bolts total) and a complete penetration weld from the shear tab to the column. Slip critical bolts are required for connections subject to load reversal per AISC. Check the plate and weld stresses with capacities per §2213.4.2. Assuming a plate thickness of 1-inch, stresses are: Shear tab length = 6(3") + 3" = 21 in. fx =

83 = 3.95 ksi (x-component) (21)(1)

fy =

568 = 27.0 ksi (y-component) (21)(1)

Z plastic =

(21)2 4

= 110.3

f x⋅x =

1,704 = 15.4 ksi (rotation) 110.3

fr =

(27.0)2 + (3.95 + 15.4)2


(

)

Required minimum plate thickness F y = 50 ksi : t PL =

f r (1) 33.2 = = 0.66 in. Fy 50

Try ¾-in. shear tab with complete penetration weld to column.

§2213.4.2

Check shear capacity of plate. Vs = 0.55 F y dt = 0.55 (50 )(21)(0.75) = 433 kips < 568 kips

no good

Try 1-inch plate.  1.0  Allowable Vc = 433   = 577 kips > 568 kips  0.75 

o.k.

∴Use 1-in. plate shear tab SEAOC Seismic Design Manual, Vol. III (1997 UBC)

63

Design Example 1A

!


Check shear plate net area for tension.

§2213.9.3.2 §2213.8.3.2

Ae 1.2αF * ≥ Ag Fu

(13-6)

where: F* =

83 = 3.95 ksi (1.0)(21)

1.2αF * 1.2(1.0 )3.95 = = 0.073 Fu 65 Ae = 21(1.0) − 7 (1.375)(1.0) = 11.38 in. Ae 11.38 = = 0.54 > 0.073 Ag 21.0

o.k.

Check bolt capacity for combined shear and tension. Per bolt: Fx =

83 = 5.9 kips 14

Fy =

568 = 40.6 kips 14

FR =

(5.9)2 + (40.6)2

= 41.0 kips

For 1-1/4-in. diameter A490-SC bolts, the allowable shear bolt is: Vbolt = 1.7(25.8) = 43.9 kips > 41.0 kips

o.k.

∴Use 1¼-inch A90-SC bolts

64


Design Example 1A

!


Commentary As shown on the frame elevations (Figure 1A-4), a horizontal steel strut has been provided between the columns at the foundation. Welded shear studs are installed on this strut with the capacity to transfer the horizontal seismic force resisted by the frame onto the foundations, through grade beams or the slab-on-grade. This technique provides redundancy in the transfer of seismic shear to the base, and is recommended as an alternate to transferring the frame shear force solely through the anchor bolts.

References AISC-ASD, 1989. Manual of Steel Construction, Allowable Stress Design. American Institute of Steel Construction, Chicago, Illinois. 9th Edition. AISC/LRFD, 1994. Manual of Steel Construction, Load and Resistance Factor Design. Volumes I and II. American Institute of Steel Construction, Chicago, Illinois. 2nd Edition. Astaneh-Asl, A., 1998. “Seismic Behavior and Design of Gusset Plates,” SteelTips. Structural Steel Educational Council. Cochran, Michael, 2000. “Design and Detailing of Steel SCBF Connections,” SEAONC Seminar Series. Structural Engineers Association of California, Sacramento, California. Hassan, O. and Goel, S., 1991. Seismic Behavior and Design of Concentrically Braced Steel Structures. Dept. of Civil Engineering, University of Michigan. ICBO, 1998. Maps of Known Active Fault Near-Source Zones in California and Adjacent Portions of Nevada. International Conference of Building Officials, Whittier, California. Lee, S. and Goel, S., 1987. Seismic Behavior of Hollow and Concrete Filled Square Tubular Bracing Members. Dept. of Civil Engineering, University of Michigan. Sabelli, R., and Hohbach, D., 1999. “Design of Cross-Braced Frames for Predictable Buckling Behavior,” Journal of Structural Engineering. American Society of Civil Engineers, Vol.125, no.2, February 1999. Thornton, W., 1991. “On the Analysis and Design of Bracing Connections,” National Steel Conference Proceedings. American Institute of Steel Construction, pp. 26.1-26.33 Chicago, Illinois. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

65

Design Example 1A

66

!



Design Example 1B

!

Ordinary Concentric Braced Frame

Design Example 1B Ordinary Concentric Braced Frame

Figure 1B-1. Four-story steel frame office building with ordinary concentric braced frames (OCBF)

Overview This Design Example illustrates the differences in design requirements for an ordinary concentric braced frame (OCBF) and a special concentric braced frame (SCBF) (illustrated in Design Example 1A). The same four-story steel frame structure from Example 1A is used in this Design Example (Figure 1B-1). Building weights, dimensions, and site seismicity are the same as Example 1A. Coefficients for seismic base shear are revised as required for the OCBF. The “typical design bay” is revised for the OCBF, and the results compared to those for the SCBF structure. It is recommended that the reader first review Design Example 1A before reading this Design Example. Refer to Example 1A for plans and elevations of the structure (Figures 1A-1 through 1A-4).


67

Design Example 1B

!


In the Blue Book Commentary (C704.12), OCBFs are not recommended for areas of high seismicity or for essential facilities and special occupancy structures. SCBFs are preferred for those types of structures, since SCBFs are expected to perform better in a large earthquake due to their ductile design and detailing. OCBFs are considered more appropriate for use in one-story light-framed construction, non-building structures and in areas of low seismicity.


Design base shear.

2.


3.

Interstory drifts.

4.


5.



1. 1a.

Code Reference

Design base shear.

Classify the structural system.

§1629.6

The structure is a building frame system with lateral resistance provided by ordinary braced frames (System Type 2.4.a of Table 16-N). The seismic factors are: R = 5.6 Ω = 2.2 hmax = 160 ft

68

Table 16-N


Design Example 1B

1b.

!



§1629.8.3

The static lateral force procedure will be used, as permitted for irregular structures not more than five stories or 65 feet in height.

1c.

Determine seismic response coefficients.

§1629.4.3

For Zone 4 and Soil Profile Type SD:

1d.

C a = 0.44(N a ) = 0.44(1.0 ) = 0.44

Table 16-Q

C v = 0.64(N v ) = 0.64(1.08) = 0.69

Table 16-R


From Design Example 1A: TB = 0.57 sec

1e.


V =

Cv I 0.69(1.0) W = W = 0.216W RT 5.6(0.57)

§1630.2.2

§1630.2.1

(30-4)

Base shear need not exceed:

V =

2 .5C a I 2 . 5 ( 0 . 44 ) (1 . 0 ) W = = 0 . 196 W R 5 .6

(30-5)

For Zone 4, base shear shall not be less than: V =

0.8ZN v I 0.8(0.4)(1.08)(1.0) W = = 0.062W R 5.6

(30-7)

Equation 30-5 governs base shear. ∴ V = 0.196W


69

Design Example 1B

1f.

!



§1630.1

20

Reliability/redundancy factor ρ = 2 −

(30-3)

rmax Ab

From Design Example 1A, use ρ = 1.0 . For the load combinations of §1630.1:

2. 2a.

E = ρE h + E v = 1.0(V )

(30-1)

E m = ΩE h = 2.2(V )

(30-2)


Building weights and mass distribution.

The weight and mass distribution for the building is shown in Table 1B-1. These values are taken from Design Example 1A. Table 1B-1. Mass properties summary

2b.

Level

W (kips)

Roof 4th 3rd 2nd

6,687 6,840 6,840 6,840

Total

27,207

X cg

Ycg

(ft) 161.1 161.1 161.1 161.1

MMI

(ft)

M (kip⋅ sec2/in.)

(kip ⋅ sec2 ⋅ in.)

1,80.9 1,80.9 1,80.9 1,80.9

17.3 17.7 17.7 17.7

316,931 324,183 324,183 324,183

70.4

Determine total base shear.

As noted above, Equation (30.5) governs, and V = 0.196W = 0.196(27207) = 5,333 kips

70

(30-5)


Design Example 1B

2c.

!



§1630.5

For the Static lateral force procedure, vertical distribution of force to each level is applied as follows: Fx =

 W h (V − Ft )W x h x = V  x x ∑ Wi hi  ∑ Wi hi

   

(30-15)

Table 1B-2. Distribution of base shear

3. 3a.

Level

wx (kips)

hx (ft)

w x hx (ft)

w x hx Σw x hx

Fx (kips)

ΣV (kips)

Roof 4th 3rd 2nd

6,687 6,840 6,840 6,840

62 47 32 17

414,594 321,480 218,880 116,280

0.39 0.30 0.20 0.11

2,064 1,600 1,090 579

2,064 3,665 4,754

Total

27,207

1,071,234

1.00

5,333

5,333

Calculate interstory drift.

Determine ∆M.

The maximum inelastic response displacement, ∆ M , is determined per §1630.9.2 as: ∆ M = 0.7(R )∆ S = 0.7(5.6 )∆ S = 3.92∆ S

3b.

(30-17)

Check story drift.

The maximum interstory drift (obtained from a computer analysis and summarized in Table 1A-7 of Design Example 1A) occurs in the north-south direction at the second story, and is 0.36 inches with R = 5.6 . This value must be adjusted for the R = 6.2 used for OCBF systems.  6.2  ∆ S drift =   (0.36") = 0.40 in.  5.6  ∆ M drift = 0.40(3.92 ) = 1.57 in. Drift ratio =

1.57 = 0.009 < 0.025 180


o.k.

1630.10.2 71

Design Example 1B

!


Comment: The elastic story displacement is greater for the SCBF than the OCBF, but the maximum inelastic displacement (∆ M ) is equivalent to the SCBF. Drift limitations rarely, if ever, govern braced frame designs. And, as a design consideration, there is essentially no difference in the calculated maximum drifts for OCBFs and SCBFs.

4.


Braced frame member design will be done using the same typical design bay as shown in Example 1A. SCBF member seismic forces are increased proportionally for the OCBF using a ratio of the R values. Member axial forces and moments for dead load and seismic loads are shown below (Figure 1B-2). All steel framing is designed per Chapter 22, Division V, Allowable Stress Design. Requirements for braced frames, except SCBF and EBF, are given in §2213.8.

Figure 1B-2. Typical braced bay

TS brace @ 3rd story:

Ρ DL = 24 kips ΡLL = 11 kips ΡE = 400 kips

72


Design Example 1B

!


WF beam @ 3rd floor: M DL = 1600 kip-in. M LL = 1193 kip-in. V DL = 14.1 kips V LL = 10.3 kips ΡE = 83 kips WF column @ 3rd story: ΡDL = 67 kips ΡLL = 30 kips ΡE = 130 kips ME ≈ 0

4a.

rd

Diagonal brace design at the 3 story.

The basic ASD load combinations of §1612.3.1 with no one-third increase will be used. D+

400 E : P1 = 24 + = 310 kips (compression) 1.4 1.4

0.9 D ±

400 E : P2 = 0.9(24 ) − = −264 kips (tension) 1.4 1.4

 400    E  D + 0.75 L +  = 246 kips (compression)  : P3 = 24 + 0.7511 + 1.4   1.4   

(12-9)

(12-10)

(12-11)

The compressive axial load of Equation (12-9) controls. The unbraced length, lw, of the TS brace is 18.5 feet. The effective length kl = 1.0(18.5) = 18.5 feet .


73

Design Example 1B


!

Maximum slenderness ratio: kl 720 ≤ r Fy

§2213.8.2.1

For a tube section: Fy = 46 ksi ∴

720 = 106 46

Minimum r =

12(18.5) kl = = 2.09 in. 106 106

 b  110 = 16.2 Maximum width-thickness ratio   ≤ Fy t

§2213.8.2.5

Try TS 10 × 10 × 5 8 . r = 3.78 > 2.09"

o.k.

b 10 = = 16.0 < 16.2 t 0.625

o.k.

For an OCBF, the capacity of bracing members in compression must be reduced by the stress reduction factor “B” per §2213.8.2: Fas = BFa

(13-4)

B = 1 /{1 + [(K l r ) / (2C c )]}

(13-5)

where: Cc =

2π 2 E Fy

( Kl ) / r =

B=

74

AISC-ASD §E2

1.0(12)(18.5) = 58.7 3.78

1 = 0.79 1 + [58.7 2 (111.6 )]


Design Example 1B

!


For kl = 18.5 ft Pallow = 482 kips

AISC-ASD, pp. 3-41

Pas = (0.79)(482) = 380 > 310 kips

o.k.

∴ Use TS 10 × 10 × 5 8

4b.

rd

Girder design at the 3 story.

From a review of Design Example 1A, the vertical load moment governs the girder design. With only a nominal increase in axial force from seismic loading, the girder is okay by inspection.

4c.

rd

Column design at the 3 story.

The columns will be designed using the basic ASD load combinations with no onethird increase. D + L : Ρ1 = 67 + 30 = 97 kips (compression) D+

(12-8)

130 E : Ρ1 = 67 + = 160 kips (compression) 1.4 1.4

0.9 D ±

(12-9)

130 E : Ρ2 = 0.9(67 ) − = 33 kips (tension) 1.4 1.4

(12-10)

 130   E   D + 0.75 L +  = 159 kips (compression)  : Ρ3 = 67 + 0.7530 + 1.4   1.4   

(12-11)

For the columns, ASTM A36 steel with F y = 36 ksi . The unbraced column height is: h = 15 − 1 = 14 ft Per AISC-ASD manual, p. 3-30, select a W 10 × 49 column with kl = 14 ft . Pallow = 242 > 160 kips

o.k.

AISC-ASD pp. 3-30

∴ Use W 10 × 49 column


75

Design Example 1B

!


Note that without the local buckling compactness requirement of §2213.9.2.4, the W 10 × 49 works in the OCBF, where a W 10 × 54 is required for the SCBF of Example 1A. Also note that the special column strength requirements of §2213.5.1 do not apply to the OCBF. The relaxation of ductility requirements for the OCBF reflects lesser inelastic displacement capacity than the SCBF, hence the greater seismic design forces for the OCBF.

5.

Braced connection design.

§2213.8.3

The design provisions for OCBF connections are nearly identical to those for SCBF connections, with one significant difference. The SCBF requirements for gusset plates do not apply to OCBF connections. Therefore, the minimum “2t” setback, as shown in Figure 1A-19(a) of Design Example 1A for the SCBF, may be eliminated. This allows the end of the tube brace to extend closer to the beamcolumn intersection, thereby reducing the size of the gusset plate. Under the requirements of §2213.8.3.1, the OCBF connections must be designed for the lesser of: 1.

PST = F y A = 46( 22.4) = 1030 kips

§2213.8.3.1

2. PM = PD + PL + Ω M PE = ( 24 + 11) + 2.2( 400) = 915 kips 3. Maximum force that can be transferred to brace by the system.

The remainder of the connection design follows the same procedure as for Design Example 1A, with all components designed for the 915 kip force derived above.

76


Design Example 1C

!

Chevron Braced Frame

Design Example 1C Chevron Braced Frame

Figure 1C-1. Four-story steel frame office building with chevron braced frames

Overview This Example illustrates the additional design requirements for chevron bracing designed as either an Ordinary Concentric Braced Frame (OCBF) or a Special Concentric Braced Frame (SCBF). The typical design bay from Design Example 1A is modified for use in this example. For comparison, the member forces are assumed to be the same as for Design Examples 1A and 1B. It is recommended that the reader first review Design Examples 1A and 1B before reading this example. Refer to Design Example 1A for plans and elevations of the structure (Figures 1A-1 through 1A-4).


77

Design Example 1C

!



Bracing configuration.

2.

Chevron bracing design under OCBF requirements.

3.

Chevron bracing design under SCBF requirements.

4.

Brace to beam connection design.


1.

Code Reference

Bracing configuration.

§2213.2, 2213.8

Section 2213.2 defines chevron bracing as “…that form of bracing where a pair of braces located either above or below a beam terminates at a single point within the clear beam span.” It also defines V-bracing and inverted V-bracing as chevron bracing occurring above or below the beam (Figure 1C-2).

Chevron V-bracing

Chevron inverted V-bracing

Figure 1C-2. Chevron bracing elevations

As discussed in the Blue Book Commentary §C704.9, the seismic performance of chevron braces can degrade under large cyclic displacements if the diagonals have poor post-buckling behavior. For this reason, the design force for chevron bracing in OCBF systems is increased so that the bracing members remain elastic during moderate earthquakes. Chevron bracing in SCBF systems has demonstrated enhanced post-buckling behavior, due to the additional design parameters placed 78


Design Example 1C

!


on SCBF members and connections. Chevron braces designed to SCBF requirements are therefore not subject to the load amplification factor (§2213.8.4.1, Item 1) imposed on chevron braces in OCBF systems. Recognizing that the buckling capacity of the compression diagonals is critical to all forms of braced frame performance, §2213.8.2.3 requires that no more than 70 percent of the diagonals act in compression along any line of bracing. By providing some balance in the distribution of tension and compression diagonals, ultimate inelastic story drifts are compatible for both directions. The typical design bay from Design Example 1A is re-configured for chevron inverted V-bracing, as shown below in Figure 1C-3.

Figure 1C-3. Typical chevron braced bay under OCBF requirements

2.

Chevron bracing design under OCBF requirements.

For comparison, assume the forces to the diagonal bracing members are the same as for Example 1B: TS brace @ 3rd story: PDL = 24 kips PLL = 11 kips PE = 400 kips


79

Design Example 1C

!


For OCBF chevron bracing, §2213.8.4.1 requires that the seismic force be increased by a factor of 1.5: PE = 1.5(400 ) = 600 kips

§2213.8.4

Also note that the same section requires the beam to be continuous between columns, and that the beam be capable of supporting gravity loads without support from the diagonal braces. From Design Example 1A, the W 24 × 68 girder satisfies these conditions. For the diagonal brace at the third story, we have the following basic ASD load combinations with no one-third increase: D+

600 E : P1 = 24 + = 453 kips (compression) 1.4 1.4

0.9 D ±

600 E : P2 = 0.9(24 ) − = −407 kips (tension) 1.4 1.4

 600   E   D + 0.75 L +  = 354 kips (compression)  : P3 = 24 + 0.75 11 + 1.4   1.4   

(12-9)

(12-10)

(12-11)

The compressive axial load of Equation (12-9) controls. From Design Example 1B, the capacity of a TS 10 × 10 × 5 8 tube section, adjusted by the stress reduction factor (B ) of §2213.8.2.2 is: Pas = 342 kips < 453 kips

n.g.

§2213.8.2.5

The TS 10 × 10 × 5 8 is the largest section that satisfies the width-thickness ratio for tubes as required by §2213.8.2.5. A wide flange section using A572 grade. 50 steel F y = 50 ksi will be required in lieu of a tube section.

(

)

Effective length @ centerline: kl = 1.0 (18.5) = 18.5 ft Maximum slenderness ratio:

For F y = 50 ksi;

80

720 50

§2213.8.2.1

kl 720 ≤ r Fy

= 102


Design Example 1C

∴ Minimum r =

!


12(18.5) kl = = 2.18 in. 102 102

 bf Maximum width-thickness ratio   2t

 65 ≤ = 9.2  Fy 

AISC-ASD, Table B5.1

Try W 12 × 120 brace: ry = 3.13 > 2.18 in. bf 2t

= 5.6 < 9.2

o.k.

o.k.

Stress reduction factor:

§2213.8.2.2

Pas = BPa

(13-4)

B = 1 /{1 + [(kl / r ) / 2C c ]}

(13-5)

kl / ry = B=

1.0(12)(18.5) = 70.9 3.13

1 = 0.75 1 + [70.9 / 2(107)]

For kl = 18.5 Pa = 733 kips Pas = (0.75)(733) = 550 > 453 kips

AISC-ASD, pp. 3-27 o.k.

∴Use W 12 × 120 brace member

3.

Chevron bracing design under SCBF requirements.

§2213.9.4.1

For SCBF chevron bracing, §2213.9.4.1 does not require the seismic force to be increased by a factor of 1.5 as is required for OCBF chevron braces. This provision is waived for SCBF chevron bracing due to an additional requirement for beam design. As for OCBF braces, §2213.9.4.1 also requires the beam to be continuous between columns, and that the beam be capable of supporting gravity loads without support from the diagonal braces. Additionally, for special chevron bracing, the beam intersected by chevron braces is to have sufficient strength to resist gravity loads combined with unbalanced brace forces. This requirement SEAOC Seismic Design Manual, Vol. III (1997 UBC)

81

Design Example 1C

!


provides for overall frame stability, and enhanced post-buckling behavior, with reduced contribution from the buckled compression bracing members. For comparison, assume the member forces remain the same as for Design Example 1A. TS brace @ 3rd story: PDL = 24 kips PLL = 11 kips PE = 348 kips WF beam @ 3rd story: M DL = 1,600 kip-in. M LL = 1,193 kip-in. V DL = 14.1 kips V LL = 72 kips PE = 72 kips

3a.

Diagonal brace design.

The diagonal brace design for the SCBF chevron brace remains the same as that of the two-story X-brace presented in Design Example 1A.

∴ Use TS 8 × 8 × 5 8 brace member

3b.

rd

Beam design at the 3 floor.

As demonstrated in Design Example 1A, the W 24 × 68 beam satisfies the basic load combinations of §1612.3.1. However, the unbalanced brace force specified in §2213.9.4.1 imposes a severe mid-span point load to the beam. Using a TS 8 × 8 × 5 8 section, the brace forces are as follows:

( )

Pst = A F y = 17.4(46 ) = 800.4 kips Psc = 1.7 Pallow = 1.7(324 ) = 551 kips 82


Design Example 1C

!


The maximum unbalanced brace force Pb is taken as the net difference of the vertical components of Pst and 0.3Psc as show in Figure 1C-4.

P st

§2213.9.4.1

0.3P sc

Figure 1C-4. Unbalanced chevron brace forces

Pb = 0.707[800.4 − 0.3(551)] = 449 kips M b = Pb L 4 = 449(12)(30 ) 4 = 40,410 kip-in. The beam must have the strength to resist load combinations similar to the Special Seismic Combinations of §1612.4: 1.2 D + 0.5L + Pb

§2213.9.4.1, Item 3

M max = 1.2(1,600 ) + 0.5(,1193) + 40,410 = 42,927 kip-in. 0.9 D − Pb M min = 0.9(1600) − 40410 = −38970 kip − in. Neglecting consideration of composite beam action, and using the flexural strength, the minimum required plastic modulus Z is solved below (using A572 grade 50 steel).

( )

M s = Z F y > M max ∴ Z reqd ≥ 42927 50 = 859 in.3 Try W 36 × 232 Z = 936 in.3 > 859 in.3

o.k.

∴Use W 36 × 232 beam


83

Design Example 1C

!


To complete the beam design, the beam-to-column connection should be checked for the reaction from vertical load plus (Pb 2) . Comment: From the foregoing examples in Parts 2 and 3, it is apparent that compared to X-bracing, chevron bracing will require a substantial increase in member sizes. For the OCBF chevron-braced system, the brace size will increase, possibly resulting in larger demands at the connections. For the SCBF chevron bracing, the beam size increases to provide the capacity to meet the strength demand imposed by the unbalanced, post-buckling brace forces. Given their superior cyclic performance, it is recommended that SCBF chevron bracing be used in regions of moderate to high seismicity.

4.

Brace to beam connection design.

§2213.9.3.1

The brace to beam connection is shown in Figure 1C-5 below. This Example uses the SCBF bracing and forces. The design for the OCBF connection is similar, without the 2t setback between the end of the brace and the line of restraint for the gusset plate, as required for SCBF systems.

Figure 1C-5. Chevron brace-to-beam connection

84


Design Example 1C

4a.

!


Gusset plate design.

From Design Example 1A, the TS 8 × 8 × 5 8 brace strength is used for connection design. The brace-to-gusset design is as given in Part 6d of Design Example 1A: Connection force:

( )

Pst = A F y = 800.4 kips Brace weld to gusset: 18" of

1

2"

fillet weld each side each face

Gusset plate thickness: 1" plate gusset minimum The gusset plate is also checked for shear and bending at the interface with the beam. From Figure 1C-5 we determine the plate length to be 86 inches. Check plate shear stress: V Plate =

fv =

2(800.4) 2

= 1,132 kips

1,132 kips = 13.1 ksi 1.0(86 in.)

Allow Fv = 0.55 F y = 0.55(36 ) = 19.8 ksi

o.k.

§2213.4.2

Check plate bending stress. From Figure 1-4, use an assumed moment couple length as distance between intersections of brace centerlines with beam flange. M plate =

Z=

2(18)(800.4 ) 2

= 20,375 kip-in.

1.0(86 )2 = 1,849 in.4 4

fb =

20,375 = 11.0 ksi 1,849


85

Design Example 1C

!


The allowable compressive bending stress is governed by the unsupported plate length perpendicular to the beam. From Figure 1C-5: l 2 = 10" and assume k = 1.2 kl 1.2(10 ) = = 41.4 r 0.29(1.0 )

AISC-ASD, Table C-36

∴ Fa = 19.08 ksi Allowable Fsc = 1.7(Fa ) = 1.7(19.08) = 32.4 ksi > 11.0 ksi

o.k.

∴Use 1-inch plate gusset

4b.

Gusset to beam design.

Length of weld to beam is l w = 86 inches. Minimum fillet weld for 1-inch plate is 5/16-inch. Per inch of effective throat area, weld stresses are: fx =

V 1,132 = = 6.58 ksi (x-axis) 2(l w ) 2(86)

fy =

20,375(6 ) M = = 8.26 ksi (y-axis) Sw 2(86 )2

fr =

(6.58)2 + (8.26)2


Allow Fw = 1.7(0.3)70 = 35.7 ksi Required weld size: t w =

§2213.4.2

10.56 = 0.41in. 0.707(35.7 )

∴ Use 1/2-inch fillet weld each side plate

86


Design Example 1C

!


Commentary The Blue Book Commentary warns that even with the strong-beam SCBF chevron, configurations may be susceptible to large inelastic displacements and P-delta effects. To mitigate these effects, chevron configurations that use two-story X-bracing or zipper columns are recommended. These bracing configurations are presented in the section Factors That Influence Design at the beginning of Design Example 1A.


87

Design Example 1C

88

!



Design Example 2

!

Eccentric Braced Frame

Design Example 2 Eccentric Braced Frame

Figure 2-1. Eccentric braced frame (EBF) building

Overview Use of eccentric braced frames (EBFs) in steel frame buildings in high seismic regions is a fairly recent development. This system was introduced in the 1988 UBC. While the concept has been thoroughly tested in laboratories, it has not yet been extensively tested in actual earthquakes. Many structural engineers, however, feel that it offers superior earthquake resistance. Following the problems with steel moment frame connections in the 1994 Northridge earthquake, many buildings that previously would have been designed as SMRF structures are now being designed with EBF systems. Eccentric braced frames may be configured with several geometric patterns, including centrally located links (as chosen in this problem) or with links located adjacent to columns. When links are located adjacent to columns, a seismic SMRF connection is required at the link beam/column intersection. Several papers and many practitioners recommend that configurations using centrally located links be chosen to avoid the use of link beam/column SMRF connections, which increase the risk of brittle failure. Braces may be oriented to slope up to central link beams (inverted “V” braces) or down (“V” braces) to central link beams. Also, a twostory frame section can be designed with upper and lower braces meeting at a common link beam located between the two levels. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

89

Design Example 2

!


It is also desirable to prevent single-story yield mechanisms. Some options for this include using inverted braces at two levels with common link beams, which ensures two story yield mechanisms, or zipper columns at either side of link beams, extending from the second level to the roof, which ensures multi-story mechanisms. In this Design Example, the five-story steel frame building shown schematically in Figure 2-1 is to have eccentric braced frames for its lateral force resisting system. The floor and roof diaphragms consist of lightweight concrete fill over steel decking. A typical floor/roof plan for the building is shown in Figure 2-2. A typical EBF frame elevation is shown in Figure 2-3. The typical frame is designed in both allowable stress design (ASD) and load and resistance factor design (LRFD) because the code allows a designer the choice of either design method. The LRFD method is from the 1997 AISC-Seismic, which is considered by SEAOC to be the most current EBF design method. The ASD method has been in the UBC for several cycles and is considered to be older, not updated, code methodology.

Outline This Design Example illustrates the following parts of the design process. 1.


2.

Reliability/redundancy factor.

3.

Design base shear and vertical distribution of shear.

4.

Horizontal distribution of shear.

5.

EBF member design using allowable stress design (ASD).

6.

EBF member design using load and resistance factor design (LRFD).

7.

Typical EBF details.

Note: Many calculations in this Design Example were performed using a spreadsheet program. Spreadsheet programs carry numbers and calculations to ten significant figures of accuracy, and thus will have round off errors when compared to hand calculations with two or three significant figures. The round off errors are usually within a percent or two. The reader should keep this in mind when comparing tables and calculations performed by hand.

90


Design Example 2

!


Given Information Roof weights: Roofing Insulation Steel deck and fill Roof framing Partition walls (10 psf) Ceiling Mechanical/electrical Total

6.0 psf 3.0 47.0 8.0 5.0 seismic 3.0 2.0 74.0 psf

Floor weights: Floor covering Steel deck and fill Framing (beams and columns) Partition walls Ceiling Mechanical/electrical Total

1.0 psf 47.0 13.0 10.0 3.0 2.0 76.0 psf

Live load:

50.0 psf

Live load:

20.0 psf

Exterior curtain wall, steel studs, gypsum board, EIFS skin, weight:

20.0 psf

Structural materials: Wide flange shapes and plates

ASTM A572, Grade 50 F y = 50 ksi

Weld electrodes Light weight concrete fill

(

E70XX f c ' = 3,000 psi

Seismic and site data: Z = 0.4 (Seismic Zone 4) I = 1.0 (standard occupancy) Seismic Source Type = A Distance to seismic source = 5 km Soil profile type = S D


)


91

Design Example 2

!


Figure 2-2. Typical floor and roof framing plan

Figure 2-3. Typical frame elevation at frame EBF4

92


Design Example 2


1.

!


Code Reference


The static force procedure will be used and the building period is calculated using Method A. T = Ct (hn )3 4 = .030(62 )3 4 = .66 sec

§1630.2

§1630.2.2 (30-8)

Near source factors for seismic source type A and distance to source of 5 km are: N a = 1.2

Table 16-S

N v = 1.6

Table 16-T

Seismic coefficients for Zone 4 and soil profile type S D are: C a = 0.44 N a = 0.44(1.2 ) = 0.53

Table 16-Q

Cv = 0.64 N v = 0.64(1.6 ) = 1.02

Table 16-R

R coefficient for a steel frame building with eccentric braced frames: R = 7.0 , height limit is 240 feet Calculation of design base shear: V =

Cv I 1.02(1.0 ) W = W = 0.22W RT 7(0.66 )

Table 16-N §1630.2.1 (30-4)


2.5C a I 2.5(.53)(1.0 ) W = W = 0.189W R 7

(30-5)

The total design shear shall not be less than: V = 0.11C a IW = 0.11(.53)(1.0)W = 0.058W


(30-6)

93

Design Example 2

!


In addition, for Seismic Zone 4, the total base shear shall also not be less than: V =

0.8ZN v I 0.8(0.4)(1.6)(1.0) W = W = 0.073W R 7

(30-7)

Therefore, Equation (30-5) controls the base shear calculation. ∴ V = 0.189W

2.

Reliability/redundancy factor.

§1630.1.1

The reliability/redundancy factor ∆ must be estimated. The factor was added to the code to penalize non-redundant systems. It varies from a minimum of 1.0 to a maximum of 1.5. It is determined for each principal direction. Since the building in this Design Example has four frames in the east-west direction, ∆ is determined based on eight braces (two per frame) and a maximum torsional contribution of 2 percent (thus 1.02). The assumption is that all frames will be identical and that the horizontal component carried by each brace is equal. This assumption can be checked after final analysis. However, in this analysis it is determined without a structural analysis. ρ = 2−

20

(30-3)

rmax AB

AB = 212′ ×15′ = 32,224 ft 2 rmax =

1 = 0.128 (8 braces, 2 percent from torsion) 8(1.02)

1.0 ≤ ρ ≤ 1.5 ρ = 2−

20 .128 32,224

§1630.1.1 = 1.13

(30-3)

∴ ρ = 1.13 for east-west direction ρ = 1.0 for north-south direction

94


Design Example 2

3.

!


Design base shear and the vertical distribution of shear.

§1630.5

The floor area at each level is 32,224 square feet. The perimeter of the exterior curtain wall is 728 feet. The roof parapet height is 4 feet. Assume that the curtain wall weights distribute to each floor by tributary height. The building mass calculation is shown in Table 2-1.

Table 2-1. Building mass calculation

3a.

Level

Floor area (sf)

wi (psf)

Wr f (kips)

Length exterior walls (ft)

h Walls (ft)

Roof 5 4 3 2

32,224 32,224 32,224 32,224 32,224

74 76 76 76 76

2,385 2,449 2,449 2,449 2,449

728 728 728 728 728

10 12 12 12 13

Totals

161,120

w i Walls W Walls (kips) (psf) 20 20 20 20 20

12,181

Wi (kips)

146 175 175 175 11

2,530 2,624 2,624 2,624 2,660

871

13,062

Design base shear.

§1630.2.1

Using the design base shear coefficient from Part 1, the base shear for the east-west direction is V = 1.13 × 0.189W = 1.13 × 0.189(12900 ) = 2,789 k

3b.

Vertical distribution of shear.

§1630.5

The total lateral force (i.e., design base shear) is distributed over the height of the building in accordance with §1630.5. The following equations apply: n

V = Ft + ∑ Fi

(30-13)

Ft = 0.07TV ≤ 0.25V

(30-14)

i =1

Ft = 0 for T ≤ 0.7 sec , T = 0.66 sec for this Design Example Fx =

(V − Ft )w x hx ∑ wx hx


(30-15)

95

Design Example 2

!


Using the building mass tabulated in Table 2-1 above, the vertical distribution of shear is determined as shown in Table 2-2 below.

Table 2-2. Vertical distribution of shear Level R 5 4 3 2 Totals

4.

wx (k)

w (k)

2,530 2,530 2,624 5,154 2,624 7,778 2,624 10,401 2,660 13,062 13,062

hx (ft)

h (ft)

w x hx (k-ft)

w x hx Σw i h i (%)

Fx (k)

ΣVI (k)

62 50 38 26 14

12 12 12 12 14

156,871 131,187 99,702 68,217 37,242 493,220

32 27 20 14 7 100

887 742 564 386 211 2,789

887 1,629 2,193 2,598 ,789 2

Horizontal distribution of shear.

§1630.6

Although the centers of mass and rigidity coincide, §1630.6 requires designing for an additional torsional eccentricity, e , equal to 5 percent of the building dimension perpendicular to the direction of force regardless of the relative location of the centers of mass and rigidity. eew = (0.05)(150 ) = 7.5 ft for east-west direction ens = (0.05)(210) = 10.5 ft for north-south direction Assume that all frames have the same rigidity, since all are similar EBFs. This assumption can be refined in a subsequent analysis, after members have been sized and an elastic deflection analysis has been completed. Many designers estimate the torsional contribution for a symmetric building by adding 5 percent to 10 percent to the element forces. However, in this Design Example the numerical application of the code provisions will be shown. Assume R1 = R2 = ...R14 = 1.0 , where Ri is the rigidity of each EBF frame. The calculation of direct shear plus torsion for a given frame is based on the following formula:  V ec  V  i Vi = Ri  i  ± Ri  2  ∑R  ∑ R xy c

96

   


Design Example 2

!


Table 2-3 gives the distribution of direct shear and torsional shear components as percentages of shear force (based on geometry).

Table 2-3. Calculation of direct shear plus torsion as percentage of story shear Frame X(ft) (1) Y(ft) (1) ID Longitudinal 1 2 3 4 Transverse 5 -110 6 -110 7 10 8 10 9 100 10 100

75 75 75 75

X 2Ri

Ri XRi YRi 1 1 1 1

-75 -75 75 75

1 -110 1 -110 1 10 1 10 1 100 1 100

Y 2Ri

J= ΣRd

2

5,625 5,625 5,625 5,625

Sum Vi Vy Sum Ty (%) (3) V Vi / Vy (2) Tx (%) (3) V (%) (2) i I (%) 25% 25% 25% 25%

12,100 12,100 100 100 10,000 10,000

-0.84% -0.84% 0.84% 0.84%

25.00% 25.00% 25.84% 25.84%

-1.23% -1.23% 0.11% 0.11% 1.12% 1.12%

Totals

-1.18% -1.18% 1.18% 1.18% 16.7% 16.7% 16.7% 16.7% 16.7% 16.7%

-1.73% -1.73% 0.16% 0.16% 1.57% 1.57%

16.7% 16.7% 16.9% 16.9% 18.3% 18.3%

0% 100% 0% Notes: 1. X and Y are distances from the center of mass (i.e., the center of the building) to frames in the X and Y directions, respectively. 2. Vx and Vy are direct shears on frames in the X and Y directions, respectively. 3. Tx and Ty are shear forces on frames that resist torsional moments on the building. These shear forces are either in the X or Y directions and can be additive or subtractive with direct shear forces.

4.

66,900(4) 100%

2

∑ Rd 2 = ∑ x Ri + ∑ y 2 Ri

Based on the direct and torsional shear values tabulated in Table 2-3, and on the vertical distribution of shear tabulated in Table 2-2, the story forces to be used for design of the typical eccentric braced frame (EBF4) are as follows:

Table 2-4. Story shear forces for design of frame EBF4 Frame ID

Level

Story V x (kips)

Story Tx (ft-kips)

Frame V4 (kips)

Frame T4 (kips)

(kips)

Story Fx ,4 (kips)

4 4 4 4 4

R 5 4 3 2

887 1,629 2,193 2,578 2,789

6,653 12,217 16,445 19,338 20,918

222 407 548 645 697

7 14 18 22 23

229 421 567 666 721

229 192 146 99 55


Vi ,6

97

Design Example 2

5.

!


EBF member design using (ASD).

In the 1997 UBC, a designer has a choice of whether to design using allowable stress design (ASD) methods or whether to use load and resistance factor design (LRFD) methods. In part 5, the ASD method is illustrated. In part 6, the LRFD method is illustrated. The results are slightly different, depending on the method chosen.

5a.

Seismic forces for initial member design.

§2213.10

Seismic forces on a typical EBF, in this case EBF4 on line 6, will be determined. The forces E , applied to EBF4 are calculated first by determining the seismic load along line 6. The unit shear load along line 6, vi 6 , is thus Vi 6 210 feet. Frame EBF4 has a tributary collector length of 210 feet / 2 = 105 feet, and tributary lengths on the west side of the frame of 60 feet and on the east side of the frame of 45 feet. The frame forces are thus F4iL = vi 6 (60 feet) and F4iR = vi 6 (45 feet). The compression force in the link is equal to half the story shear tributary to the frame, minus the frame force at the right side (F4iL + F4iR ) 2 − F4iR . Table 2-5 summarizes the forces at each level of frame EBF4.

Table 2-5. Axial forces through shear links on frame EBF4 Level

Frame Fx ,4

R 5 4 3 2

5b.

(kips)

Line 6 Vi 4 (klf)

229 192 146 100 54

2.18 1.83 1.39 0.95 0.51

FxiL (west) (kips)

FxiR (east) (kips)

C =T (link) (kips)

131.0 109.5 83.2 57.0 31.1

98.2 82.1 62.4 42.7 23.3

16.4 13.7 10.4 7.1 3.9

Link length.

The inelastic behavior of a link is influenced by its length, e . The shorter the link length, the greater the influence of shear forces on the inelastic performance. Shear yielding tends to occur uniformly along the link length. Shear yielding of short links is very ductile with an inelastic capacity in excess of that predicted by calculations. The following is a summary of link behavior as a function of the link length e . MS is the flexural strength of the link and VS is the shear strength. Both are defined in §2213.4.2. 98


Design Example 2

1.0

M MS ≤ e ≤ 1.3 S VS VS

!


Ensures shear behavior and is the recommended upper limit for shear links. Links less than 1.0 M s Vs the link may not yield as expected.

e ≤ 1.6

MS VS

Elastic behavior is controlled by shear behavior, however, region is transition between shear governed behavior and bending governed behavior.

e > 2.0

MS VS

Link behavior is theoretically balanced between shear and flexural yielding.

e ≥ 3.0

MS VS

Elastic deformation is controlled by flexural yielding.

The shorter the link length, the stiffer the EBF frame will be; however, the greater the link rotation. The code sets limits on link plastic rotation of 0.090 radians (ASD) and 0.080 radians (LRFD) due to ∆ m deflections. For most designs, link lengths of 1.0 to 1.3 M s Vs work well.

5c.

Preliminary EBF frame member sizes.

Preliminary sizes of the EBF frame beams are determined by calculating the required shear area (dt w ) due to the story forces and frame geometry. The load combinations for allowable stress design procedures are given in Equations (12-7) through (12-11) or (12-12) through (12-16) in §1612.3. These load combinations use load values of E 1.4 to account for allowable stress design.

ΣV3 = 666kips/1.4/2 = 237.8 kips

ΣV2 = 721k/1.4/2 = 257.4 kips

15' 15'

Level 3

Level 2 V2, link

V3, link 12'

14'

Link analysis at Level 2

Link analysis at Level 3 (levels 4, 5, R, similar)

Figure 2-4. Preliminary link analysis


99

Design Example 2

!


For initial sizing, shear forces in the links may be approximated as follows: Vi ,link =

ΣVi ( h) ΣVi / 2( h) = l l/2

V2,link

 721 kips  (14' )    = 240.2 kips =  1.4   30'    

V3,link

 666 kips  (12' )    = 190.4 kips =  1.4   30'    

V4,link

 567 kips  (12' )    = 161.9 kips =  1.4   30'    

V5,link

 421 kips  (12' )    = 120.3 kips =  1.4   30'    

V R ,link

 229 kips  (12' )    = 65.5 kips =  1.4   30'    

The values for dt w , VS and M S are calculated as follows: Minimum dt w =

V i ,link 0.80 × 0.55 F y

§2213.10.5

V s = .55 F y dt w M s= Zx F y Preliminary beam sizes are determined as shown in Table 2-6 (forces are E 1.4 ).

100


Design Example 2

!


Table 2-6. Preliminary link analysis and sizing for frame EBF4 Vi ,link min. Fi Story Vi req. dtw Level h 2 2 (ft) (kips) (kips) link (in.2) shear R

12

5 4

81.9 81.9

Link Beam Size

d (in.)

1.3

tw (in.)

dtw (in.2)

Zx Ms Vs M s Vs (in.3) (k-in.) (kips) (in.)

1.6 M s Vs

47.2

Ω

(in.)

Link Lg. (in.)

58.1

24

3.16

65.5

2.98

W16x77

16.52 0.46

7.52

150.0

7500 207

12

150.3 68.5 120.3

5.47

W18X86

18.39 0.48

8.83

186.0

9300 243

49.8

61.3

34

2.02

12

202.4 52.0 161.9

7.36

W18X97

18.59 0.54

9.95

211.0 10550 274

50.1

61.7

36

1.69

3

12

238.0 35.6 190.4

8.65

W18X97

18.59 0.54

9.95

211.0 10550 274

50.1

61.7

36

1.44

2

14

257.4 19.4 240.2 10.92 W21X132 21.83 0.65 14.19

333.0 16650 390

55.5

68.3

46

1.62

The most efficient link sections usually: 1. Optimize the required shear area, i.e., minimize dt w . 2. Are the deepest section possible while complying with the compact web criteria , i.e., maximize dt w . 3. Have compact flanges with sufficient bending capacity to ensure shear failure of the section under ultimate load. 4. The frames must meet the deflection and link rotation limitations and thus be sized for stiffness. The recommended [Engelhardt and Popov, 1989] link length is emax = 1.3

MS VS

A computer model has been created for EBF4. The results of the computer analysis, including forces and displacements, have been determined. The computer model was analyzed with moment resisting connections, which more closely estimates the real behavior of the frame with end moments much less than M p . For the first story, the EBF member design will be based on use of a W 21× 132 link beam at Level 2.

5d.

Link rotation.

The frame displacement at the second level, ∆ S 2 , was determined from a separate computer analysis (not shown) using the design base shear (not divided by 1.4) and not increased by ∆ because frame distortion limits are based on calculations using applied strength level seismic forces not increased by the redundancy factor. ∆ S 2 = 0.48 in.


101

Design Example 2

!


The corresponding maximum inelastic response displacement at the second level, ∆ M 2 is estimated as follows: ∆ M = .7 R∆ s = .7 (7 )0.48" = 2.40 in.

(30-17)

The link rotation is computed as a function of the frame story drift and frame geometry. For a frame of story height h , bay width l , link length e , and (l − e) , the link rotation may be calculated by the following dimensions a = 2 formula [Becker and Ishler, 1996]. Link rotations, θ , must be limited to 0.090 radians per §2213.10.4. θ=

∆M  2 a  1.37"  2(157 ")  l + = l +  = 0.060 radians ≤ 0.090 e  180"  h  46" 

∴ o.k. Note that the frame height, h , in the first story is 180 inches, or 15 ft-0 in. because the base plate is anchored 12 inches below the slab.

5e.

Link shear strength.

§2213.10.5

The purpose of EBF design is to ensure that any inelastic behavior in the structure under seismic motions occurs in the links. To achieve this, all elements other than the links are designed to have strengths greater than the forces that will be induced in them when the links experience yielding. Therefore, if the links have excess capacity, all other elements in the frame (braces, columns, link beams outside the link lengths) will also have corresponding excess capacity. Section 2213.10.5 requires than the link shear does not exceed 0.8Vs under design seismic forces. Thus links have a minimum overstrength factor Ω min = (1.0 0.8) = 1.25 which provides a safety factor on shear capacity. Depending on the actual link beam chosen for design, the link overstrength factor, Ω , may be greater than 1.25. Thus, for the W 21× 132 link beam with applied shear Vi ,link = 240.2 kips (see Table 2-6): V s = .55 F y dt w = .55 (50 ksi )(21.83")(.650") = 390.2 kips Ω=

390.2 k Vs = = 1.62 ≥ 1.25 V i,link 240.2 k

∴ o.k .

§2213.4.2

The link beam in this Design Example is sized for stiffness to thus limit deflections and link rotations under code loads. It therefore has greater strength than required 102


Design Example 2

5f.

!


Beam compact flange.

§2213.10.2

Check to assure that the beam flanges are compact to prevent flange buckling. 12.44" 52 bf = = 6.0 ≤ = 2t f 2 (1.035") Fy

52 = 7.36 50 ksi

∴ o.k.

5g.

Link length.

The length of the link will determine whether the link yields in shear or in bending. To ensure shear yielding behavior, the link beams have been limited to lengths less than 3 M s Vs . V s = .55 F y dtw = 390.2 kips e ≈ 1.3 M s = 55.5 in. Vs

§2213.4.2 [Popov, Englehardt, and Ricles]

3 M s = Z x F y = (333 in. ) (50 ksi) = 16,650 kip-in.

§2213.4.2

For frame stiffness, drift, and rotation control purposes at the second level, use e = 46 in. Thus: eVs (46" ) (390.2 kips) = = 1.08 Ms 16,650 kip − in. ∴ o.k.

5h.

Beam and link axial loads.

The summation of story forces down to level 3, ΣFi = V3 in Table 2-4, (the sum of level shears from the roof to level 3) is 666k (476k on an ASD basis). The ASD frame forces in level 2 at the left connection and right connection are F2 L = 31.1 k 1.4 = 22.2 k and F2 L = 23.3 k 1.4 = 16.7 k . The link beam outside the link must be checked for combined bending, plus axial loads. The link must be checked for bending plus axial loads using the flanges only (because the web is assumed to have yielded in shear and not capable of carrying axial load).


103

Design Example 2

!


Therefore, the axial force in the link on an ASD basis is: C link = T link =

(31.1 k − 23.3 k ) = 2.8 k 2 × 1.4

The axial force can be factored up to account for actual link design overstrength, Ω . For this link, Ω = 1.62 and the link axial force can be factored to be 4.5 kips.

5i.

Beam compact web.

The maximum d/tw ratio permitted for compact beam sections is dependent on the axial load in the beam. Wide flange sections listed in the AISC W shapes tables (AISC-ASD) have compact webs for all combinations of axial stress when the yield strength is less than the tabulated values of F y . If a beam section is chosen that does not have a compact web for all axial loads, the section should be checked using allowable stress design of UBC Chapter 22, Division V, Table B5.1 of (AISC-ASD). The web should be compact along the full length of the beam. The UBC does not allow doubler plates to reduce d/tw requirements for a link beam (see §2213.10.5). For the W 21× 132 beam at the second level of EBF4: dt w = 33.6 A = 38.8 in.2 Maximum axial force in link beam outside the link:  V 3   666 kips + 31.1 kips   + F 2L   2  2 =  = 260 kips P 2L = 1.4 1.4 fa= fa Fy

104

=

P 2L = 260 k = 6.7 ksi A 38.8 in.2 6.7 ksi = 0.13 ≤ 0.16 50 ksi



Design Example 2

!


For f a ≤ 0.16 F y , the allowable d/tw to prevent local buckling is determined from the equation below. d tw

=

f  640  l − 3.74 a  =  F y  Fy

6.7 ksi  640   l − 3.74  = 45.1in.3 50 ksi  50 ksi 


∴ d tw = 33.6 in.2 ≤ 45.1in.3 ∴ o.k.

5j.

Combined link loads.

This calculation is made to check the combined bending plus axial strength of the link (using loads anticipated to yield the link with the link design overstrength factor, Ω = 1.62 ). P2,link = 2.8 k (1.62) = 4.5 kips M 2,link = VS , 2

(46") = 8,975 kip-in. e = 390.2 k 2 2

A f = b f t f = (12.440") × (1.035") = 12.875 in.2

(

)(

(

)

)

Z f = d − t f b f t f = (21.83"−1.035") 12.875 in.2 = 267.7 in.3 P2,link 2 Af

+

M 2,link Zf

=

4.5 k 8,975 ksi + = 33.7 ksi ≤ 50 ksi 2 2 12.875 in. 267.7 in.3

(

)

∴ Link combined axial plus bending capacity is o.k.

5k.

Verification of link shear strength.

§2213.10.3

The strength of the link is used to establish the minimum strength required of elements outside the link. The link shear strength Vs was determined using the web area d/tw, of the beam. When a beam has reached flexural capacity, shear in the link may be less than the shear strength of the section. If this is the case, the flexural capacity of the section will limit the shear capacity of the link. Section 2213.10.3 requires that the flexural capacity of the section, reduced for axial stress, be considered as a possible upper limit of the link capacity. This will be checked below.


105

Design Example 2

!


Vs = 390.2 kips

(

M rs = Z x f y − f a fa =

P2,link 2 Af

=

)

§2213.10.3

4.5 k = 0.17 ksi 2 × 12.875 in.2

Z x = 333 in.4

AISC-ASD, pp. 1-21

M rs = 333 in.3 (50.0 ksi − 0.17 ksi ) = 16,593 kip-in. Vrs =

2 M rs 2 (16,593 kip − in.) = = 721 kips e (46")

The controlling shear capacity is the least of Vs or Vrs . In this case, Vs = 390 kips and Vrs = 721 kips . Therefore the controlling shear capacity is 390 kips. Thus, the controlling mode of yielding is shear in the link, because the shear required to yield the beam in bending will not be developed.

5l.

Required beam brace spacing.

§2313.10.18

Section 2213.10.18 requires lateral braces for the top and bottom flanges at the ends of the link beams. The maximum interval l u ,max is determined below. l u ,max = 76

bf Fy

= 76

(12.87") = 138.4" ≅ 11'−6" 50 ksi

§2313.10.18

Therefore the beam bracing at 10 ft 0 in. is adequate. (Note: the composite steel deck and lightweight concrete fill is not considered effective in bracing the top flange.)

5m.

Beam analysis (outside of link).

§2213.10.13

The beam outside the link is required to resist 130 percent of the bending, plus axial forces generated in the link beam. The combined beam bending plus axial interaction equations are referenced from AISC-ASD, Section N. Note that the ASD version of capacity design is being used because the beam is being checked under forces generated with a yielding link element in shear. Forces are from a hand evaluation of EBF frame behavior and from computer model analysis: 106


Design Example 2

!


Axial force in beam outside link: PE = 260 kips From computer model: PD = 11 kips Increased axial load on beam outside the link: P = 1.3ΩP2,link + 1.3PDL = (1.3 × 1.62 × 260 k ) + (1.3 × 11 kips ) = 564 kips From EBF frame analysis: M E = 8,974 k-in. From computer analysis: M D = 188.4 k-in. Increased moment on beam outside the link: M = 1.3

V link e + 1.3M DL = 1.3 (8,974 k-in.) + 1.3 (188.4 k-in.) = 11,912 k-in. 2

Beam slenderness parameters, assuming k = 1.0 : =

(1.0)(120") = 41.0

=

(1.0)(150") = 16.4

kl ry kl rx

2.93"

9.12"

Allowable axial stress based on beam slenderness and bracing:

Fay

 ( kl / ry ) 2   ( 41.0) 2  F 1 −  50 ksi y 1 − 2 2C c 2    2(107)   = = = 25.7 ksi 3 5 3 ( 41.0) (41.0) 3 5 3 ( kl / ry ) (kl / ry ) + − + − 3 8 (107) 8 (107) 3 3 8C c 8C c 3


AISC-ASD §E2

107

Design Example 2

!


Euler buckling stress multiplied by a safety factor: Fey' =

12π 2 E

(

23 kl / ry

)2

=

12 (3.14 )2 (29,000,000 psi ) 23 (41.0 )2

= 88,834 psi = 88.8 ksi

AISC-ASD §E2

Beam slenderness parameter: Cc =

12 (3.14 )2 (29,000 ksi ) 12π 2 E = = 107 Fy (60 ksi )

AISC-ASD §E2

ASD axial capacity: Pcr = 1.7 Fa A = 1.7 (25.69 ksi )(38.8 sq in.) = 1,695 kips

AISC-ASD §N4,

Euler buckling capacity:

(

)

AISC-ASD §N4

Py = F y A = (50 ksi ) 38.8 in.2 = 1,940 kips

AISC-ASD §N4

 23   23  Pe =   Fe' A =   (88.8 ksi ) 38.8 in.2 = 6,603 kips  12   12  ASD axial yielding load:

(

)

Maximum moment that can be resisted by the member in the absence of axial load:

(

)

M m = M p = F y Z x = (50 ksi ) 333 in.3 = 16,650 k-in.

AISC-ASD §N4

Coefficient for sidesway: C m = 0.85 Check AISC Equations (N4-2) and (N4-3): P Pcr

+

564 kips 0.85 (11,912 k − in.) Cm M + = 1,695 kips   Pbu  564 kips   1  M m  1 16,650 k − in. Pe    6,603 kips 

AISC-ASD (N4-2)

= 0.33 + 0.67 = 1.0 ∴ Say

108

o.k.


Design Example 2

P Py

+

564 kips 11,912 k − in. M = + 1.18 M p 1,940 kips 1.18 (16,650 k − in.)

!


AISC-ASD (N4-3)

= 0.29 + 0.61 = 0.90 ≤ 1.0 ∴ o.k.

5n.

Beam stiffeners.

§2213.10.7

There are two types of stiffeners required in links: link stiffeners at ends at brace connections and intermediate stiffeners (Figures 2-7 and 2-11). Link end stiffeners.

Full depth web stiffeners are required on both sides of the link beam at the brace connections. The stiffeners are used to prevent web buckling and to ensure ductile shear yielding of the web. The stiffeners shall have a combined width not less than bf - 2tw and a thickness not less than 0.75t w or 3/8 inch. For the W 21× 132 beam B f − 2t w = 12.440"−2 (.650") = 11.14" use 2 × 5.625"

§2213.10.10

The minimum thickness of the stiffener is t stiff ≥ 0.75t w = 0.75 × .650" = 0.49" use ½ in. stiffeners. Therefore, use 55/8 in. × ½ in. link beam stiffeners at link ends at each side of web (total 4). Intermediate link stiffeners.

Section 2310.10.8 requires intermediate full depth web stiffeners (see Part 7, Figure 2-7) for either of the following conditions: 1.

Where link beam strength is controlled by Vs .

2.

Where link beam strength is controlled by flexure and the shear determined by applying the reduced flexural strength, M rs exceeds 0.45F y dt w .

Therefore, intermediate web stiffeners are required for this Design Example.


109

Design Example 2

!


The spacing limits are a function of the link rotation per §2310.10.9. For a link rotation 0.09 radians, the maximum allowed, the spacing shall not exceed 38t w − d w 5 . For link rotation of 0.03 radians, the spacing shall not exceed 56t w − d w 5 . Linear interpolation may be used between link rotations of 0.03 and 0.09 radians. Thus, 38t w −

dw  21.83"  = 38 (.650") −   = 20.33 in. 5  5 

§2213.10.9

56tw −

dw  21.83"  = 56 (.650") −   = 32.03 in. 5  5 

§2213.10.9

Since the link rotation is 0.088 radians for the beam, interpolation must be used to determine the maximum spacing of intermediate stiffeners. This is shown below.  0.090 rad − 0.088 rad    (32.03"−20.33") + 20.33" = 20.72 in.  0.090 rad − 0.030 rad  Since the link length is 46 inches, use three equal spacings of 46/3 =15.33 inches. The web stiffener location is determined in accordance with §2313.10.10. Since the link beam is a W21, one sided stiffeners are required of thickness 3/8-inch. The width shall not be less than:

(b f 2)− tw + (12.44" 2) − .650"+5.57 in. Therefore, use 5-5/8 in. × 3/8 in. intermediate (one-sided) stiffener plates (2 total).

Web stiffener welds.

Fillet welds connecting the web stiffener to the web shall develop a stiffener force of: Ast F y = (5.625"× .375")(50 ksi ) = 105.5 kips The minimum size of fillet weld, per AISC Table J2.4, is ¼-inch to the link web and 5/16 in. to the link flange. Using E70XX electrodes and 5/16-inch fillet welds each side, the weld capacity is 1.7 allowable. The required weld length is 1required =

105.5 kips (70 ksi )(1.7 )(2 × 5 16")(.707 ) = 6.7 in. .3

Therefore, 5/16 in. fillet welds, both sides of the stiffener, at the flanges and the web are adequate.

110


Design Example 2

!


Fillet welds connecting the web stiffener to the flanges shall develop a stiffener force of Ast F y / 4 = (5.625"×.375")(50 ksi ) / 4 = 26.4 kips 1, required =

26.4 kips (70 ksi )(1.7 )(2 × 5 16")(.707 ) = 1.7 in. .3

Therefore, 5/16-inch fillet welds, both sides of the stiffener, at the flanges are adequate.

5o.

Link beam design.

Tables 2-7a through 2-7g presents tabular calculations that show the results from procedures from Parts 5a through 5s applied to all beams in the frame EBF4. The link beam design for all levels is as shown below in tabular form following the equations given above (each link beam at each level of the frame has a row calculation which extends through the full table):

Table 2-7a. Link beam section properties Fy

Level

Link

A (in.2)

Zx (in.3)

br (in.)

tr (in.)

d (in.)

tw (in.)

e (in.)

a (in.)

h (in.)

Af (in.2)

Zf (in.3)

(ksi)

R 5 4 3 2

W16x77 W18X86 W18X97 W18X97 W21X132

22.60 25.30 28.50 28.50 38.80

150.0 186.0 211.0 211.0 333.0

10.30 11.09 11.15 11.15 12.44

0.76 0.77 0.87 0.87 1.04

16.52 18.39 18.59 18.59 21.83

0.46 0.48 0.54 0.54 0.65

24 34 36 36 46

168 163 162 162 157

144 144 144 144 168

15.6 17.1 19.4 19.4 25.8

123.3 150.5 171.8 171.8 267.7

50 50 50 50 50

Table 2-7b. Compact flange, compact web Level

Vs

Ω

Ms

bf 2tf

R 5 4 3 2

206.7 242.7 273.5 273.5 390.2

3.16 2.02 1.69 1.44 1.62

7,500 9,300 10,550 10,550 16,650

6.77 7.20 6.41 6.41 6.01


Compact 1.3 Flange Compact Flange Limit M s Vs Results bf 2tf 7.35 7.35 7.35 7.35 7.35

o.k. o.k. o.k. o.k. o.k.

47.2 49.8 50.1 50.1 55.5

fa

fa Fy

dtw

4.14 6.33 7.36 8.53 6.71

0.08 0.13 0.15 0.17 0.13

36.3 38.3 34.7 34.7 33.6

Compact Compact web Web Limit dtw Results 62.5 47.7 40.7 36.3 45.1


111

Design Example 2

!


Table 2-7c. Combined link stresses, controlling shear, unsupported length Shear Levels Level Above (kips) R 5 4 3 2

0 229.2 420.9 566.6 666.3

Level at level (kips)

Pmax Link Beam (kips)

fa (ksi)

Plink Diaph. M link (kips) Factor (k-in.)

Comb. fa Vs Link (psi) Stress (kips) (psi)

M rs (k-in.)

Vrs (kips)

Vmin (kips)

lu Max (in.)

131.0 109.5 83.3 57.0 31.1

94 160 210 243 260

4.14 6.33 7.36 8.53 6.71

16.4 13.7 10.4 7.1 3.9

0.75 0.64 0.50 0.44 0.29

7.388 9.181 10.444 10.456 16.553

615.7 540.0 580.2 580.9 719.7

206.7 242.7 273.5 273.5 390.2

110.7 119.2 119.8 119.8 133.7

1.00 1.12 1.31 1.69 2.70

785.9 2.044.5 2.914.1 3.426.8 5.525.6

6.7 13.9 17.2 20.2 20.8

206.7 242.7 273.5 273.5 390.2

Table 2-7d. Calculation of design forces, beam outside the link Level

P (kips)

M = Vs e 2 (k-in.)

Link Ω

Beam Overstress Factor

Pcomp

Mcomp

DL (kips)

R 5 4 3 2

94 160 210 243 260

2,480 4,127 4,923 4,923 8,975

3.16 2.02 1.69 1.44 1.62

1.3 1.3 1.3 1.3 1.3

10 8.73 11.2 10 11

"DL" (k-in.)

Beam Overstress Factor

Pbu Design (kips)

208.8 226.8 213.6 200.4 188.4

1.3 1.3 1.3 1.3 1.3

397 431 475 467 564

M bu Design (kips) 3,496 5,660 6,678 6,661 11,912

Table 2-7e. Beam properties Level

Section

A (in.2)

Z (in.3)

R 5 4 3 2

W16x77 W18X86 W18X97 W18X97 W21X132

22.6 25.3 28.5 28.5 38.8

150 186 211 211 333

Fy

ry

(ksi)

Lu (ft)

rx (in.)

kl ry

Cc (ksi)

kl r y C c

(in.)

50 50 50 50 50

10 10 10 10 10

5.89 7.77 7.82 7.82 9.12

1.92 2.63 2.65 2.65 2.93

62.5 45.6 45.3 45.3 41.0

107.0 107.0 107.0 107.0 107.0

0.58 0.43 0.42 0.42 0.38

Cm

P Design (k)

M Design (k-in.)

AISCASD (N4-2)

AISC ASD (N4-3)

Results

0.85 0.85 0.85 0.85 0.85

397 431 475 467 564

3,496 5,660 6,678 6,661 11,912

0.98 0.99 1.00 0.99 1.00

0.75 0.86 0.87 0.86 0.90


Table 2-7f. AISC-ASD equations (N4-1) and (N4-2) Level

Fa (ksi)

F'e (ksi)

Pcr (k)

Pe (k)

Py

M m ,M p

(k)

(k-in.)

R 5 4 3 2

22.3 25.0 25.1 25.1 25.7

38.2 71.7 72.8 72.8 89.0

856 1,076 1,214 1,214 1,695

1,655 3,478 3,978 3,978 6,620

1,130 1,265 1,425 1,425 1,940

7500 9300 10550 10550 16650

Table 2-7g. Link rotations at each level

112

Level

Delta S Deflection (in.)

Delta M Drift (in.)

Rotation (rad)

Results

R 5 4 3 2

1.01 0.87 0.69 0.46 0.24

0.69 0.88 1.13 1.08 1.18

0.0715 0.0649 0.0783 0.0749 0.0548

o.k. o.k. o.k. o.k. o.k. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

Design Example 2

5p.

!


Brace design.

§2213.10.13

The braces are required to be designed for 1.3Ω times the earthquake forces in the braces, plus 1.3 times the gravity loads. There is a misprint in 97 UBC §2213.10.13, where the brace and beam overstrength factor is both 1.5 and 1.3. However, the factor 1.5 was from the 1994 UBC and should have been deleted. The factor 1.3 should be used. PE = 1.3Ω Pcomputer due to

E loads 1.4

M E = 1.3Ω M computer due to

E loads 1.4

Using plastic design procedures outlined in AISC Section N, obtaining forces from a computer analysis, and showing calculations in tabular form. Design forces for braces ( P and M ) are calculated as 1.3φ times seismic forces plus 1.3 times gravity forces. Column shear forces are not a controlling factor and are not shown for brevity. Tables 2-8a through 2-8c show tabular design of braces for EBF4 at all levels.

Table 2-8a. Brace forces Level

PE E/1.4

ME E/1.4

Ω

Brace Overstress Factor

PD D

MD D

Brace Overstress Factor

P Design

M Design

5 4 3 2 1

106 194 262 302 372

10.2 11.7 23.4 26.7 38.5

3.16 2.02 1.69 1.44 1.62

1.5 1.5 1.5 1.5 1.5

11.8 14.6 14.7 14.4 13.9

5.1 4.4 4.3 4.3 3.4

1.5 1.5 1.5 1.5 1.5

519.5 609.3 686.0 672.4 927.2

55.9 42.0 65.7 64.0 98.9

kl ry

Cc (ksi)

kl / ry / Cc

74.4 73.1 72.5 71.0 69.4

107.0 107.0 107.0 107.0 107.0

Table 2-8b. Brace section properties Level

Brace Section

A (in.2)

Z (in.3)

5 4 3 2 1

W12X87 W12x87 W12x87 W12X106 W12X120

25.60 25.60 25.60 31.20 35.30

132.0 132.0 132.0 164.0 186.0


Fy

ry

(ksi)

L (ft)

rx (in.)

(in.)

50 50 50 50 50

20.5 20.2 20.2 20.2 19.9

5.34 5.38 5.43 5.57 5.66

3.31 3.32 3.34 3.41 3.44

0.70 0.68 0.68 0.66 0.65

113

Design Example 2

!


Table 2-8c. Brace, axial plus bending interaction calculations Level 5 4 3 2 1

5q.

Fa

F' e

Pcr

Pe

Py

M m ,M p

(ksi)

(ksi)

(k)

(k)

(k)

(k-in.)

20.1 20.3 20.5 20.7 21.0

262.1 262.1 262.1 262.1 262.1

875.2 885.6 890.8 1100.5 1262.8

12,860 12,860 12,860 15,673 17,732

1280 1280 1280 1560 1765

6600 6600 6600 8200 9300

Cm 0.85 0.85 0.85 0.85 0.85

P Design

M Design

(k)

(k-in.)

450.2 528.0 594.5 582.8 803.5

659.0 493.7 778.7 757.5 1178.6

AISC

AISC

(N4-2)

(N4-3)

0.60 0.66 0.77 0.61 0.75

0.35 0.41 0.46 0.37 0.46

Column design.

Results o.k. o.k. o.k. o.k. o.k.

§2213.10.14

The columns are required to resist 1.25 times the strength developed in the links to assure that the yielding mechanism is the link beams (Section 2213.10.14). Design forces ( P and M ) are calculated as 1.25Ω times (frame analysis) seismic forces plus 1.25 times gravity forces. Column shear forces are not a controlling factor and are not shown for brevity. Tables 2-9a through 2-9c show tabular design of columns for EBF4 at all levels

Table 2-9a. Design column forces Level

PE E/1.4

ME E/1.4

km5 4 3 2 1

106 3.16 2.02 1.69 372

10.2 1.25 1.25 1.25 38.5

Ω

11.8 14.6 14.7 1.62

Brace Overstress Factor 5.1 4.4 4.3 1.25

PD D

MD D

1.25 1.25 1.25 13.9

4.3 3.4

Brace Overstress factor

1.25 1.25

P Design

M Design

432.9 507.7 571.7 560.3 772.6

46.6 35.0 54.8 53.3 82.4

Table 2-9b. Column section properties

114

Level

Column Section

A (in.2)

Z (in.3)

5 4 3 2 1

W12X65 W12X65 W12X65 W12X87 W12X87

19.10 19.10 19.10 25.60 25.60

96.8 96.8 96.8 132.0 132.0

Fy

ry

(ksi)

L (ft)

rx (in.)

(in.)

50 50 50 50 50

12 12 12 12 14

5.28 5.28 5.28 5.38 5.38

5.67 5.67 5.67 5.72 5.72

kl ry

Cc (ksi)

kl / ry / Cc

2.48 2.48 25.4 25.2 29.4

107.0 107.0 107.0 107.0 107.0

0.02 0.02 0.24 0.24 0.27


Design Example 2

!


Table 2-9c. Column axial plus bending interaction calculations Level 5 4 3 2 1

5r.

Fa

F' e

Pcr

Pe

Py

M m ,M p

(ksi)

(ksi)

(k)

(k)

(k)

(k-in.)

29.8 29.8 27.7 27.7 27.2

262.1 262.1 262.1 262.1 262.1

968 968 899 1,206 1,185

955 955 955 1280 1280

4840 4840 4840 6600 6600

9,594 9,594 9,594 12,860 12,860

Cm 0.85 0.85 0.85 0.85 0.85

P Design

M Design

(k)

(k-in.)

432.9 507.7 571.7 560.3 772.6

559.4 420.2 657.5 639.9 989.0

AISC

AISC

(N4-2)

(N4-3)

0.55 0.60 0.76 0.55 0.79

0.45 0.53 0.60 0.44 0.60

Results

Foundation design considerations.

In EBF design, special consideration should be given to the foundation design. The basis for design of the EBF is that the yielding occurs in the EBF links. Thus, all other elements should have the strength to develop the link beam yielding strengths. The code does not require the foundation design to be capable of developing the link beam strengths. However, if only a minimum code foundation design is performed, the foundation will generally not develop the EBF link beam strengths, and yielding will occur in the foundation. This is not consistent with the design philosophy for EBF frames. The SEAOC Blue Book recommends that the foundation be designed to develop the strength of the EBF frame. The intention is to have adequate foundation strength and stability to ensure the development of link beam yield mechanisms to achieve the energy dissipation anticipated in the eccentric braced frames. A static pushover analysis of an EBF frame can give a good indication of the foundation adequacy.

5s.

Final frame member sizes (ASD).

Table 2-10. Final frame member sizes for EBF4 (ASD) Level

Beams

Link Lengths

Roof 5 4 3 2 1

W16X77 W18X86 W18X97 W18X97 W21X132

24" 34" 36" 36" 46"


Columns

Braces

W12X65 W12X65 W12X65 W12X87 W12X87

W12X87 W12X87 W12X87 W12X106 W12X120

115


Design Example 2

!


Figure 2-5. EBF4 frame member sizes (ASD)

6.

EBF member design using (LRFD).

In the 1997 UBC, a designer has a choice of whether to design using allowable stress design (ASD) methods or whether to use load and resistance factor design (LRFD) methods. In part 5, the ASD method is illustrated. In part 6, the LRFD method is illustrated. The results are slightly different, depending on the method chosen. In this part, the frame EBF4 that was designed to ASD requirements in Part 5 is now designed to LRFD requirements of AISC-Seismic. LRFD design provisions for EBF frames are contained in Section 15 of the AISC document, “Seismic Provisions for Structural Steel Buildings,” published in 1997. This document is commonly known as AISC-Seismic. Note that the Seismic Provisions for Structural Steel Buildings, 1992 edition, is included in the AISCLRFD Manual, Part 6, which is adopted by reference in the code in Chapter 22, Division II, §2206. However, the 1997 AISC-Seismic provisions have been updated and are recommended in the SEAOC Blue Book, Section 702.

116


Design Example 2

6a.

!



The link shear strength Vn can be found from the minimum values of V p or 2 M p e . The values for V p are calculated as follows:

(d − 2t f )t w ≥ 0.90V(0i,link .60)

AISC-Seismic §15.2d

Fy

V p = 0.60 F y t w (d − 2 t f )


M p = ZxFy Preliminary beam sizes are determined as shown in Table 2-11. Note that seismic forces for LRFD procedures use both E h and E v . The E v seismic force is additive to dead load D and is included in the load combination of Equation (12-5). 1.2 D + f1l + 1.0 E

(12-5)

E = ∆E h + E v Ev = 0.5Ca ID = 0.5(0.53)(1.0 )D = 0.265 D Substituting for E h , E v , and f1 in Equation (12-5) 1.2 D + 0.5l + 1.0(∆E h + E v ) = 1.2 D + 0.5l + 1.0(1.13E h + 0.265D ) = 1.465D + 0.5l + 1.13E h Tables 2-11a through 2-11c show preliminary link analysis and sizing (LRFD). Table 2-11a. Design seismic forces at EBF frame Level R 5 4 3 2

Story Forces 229 192 146 100 54

Frame Forces, E h

Frame Forces, E h

Left

Right

C, T link

Fil

Fir

Vi

131.0 109.5 83.2 57.0 31.1

98.2 82.1 62.4 42.7 23.3

16.4 13.7 10.4 7.1 3.9

131.0 109.5 83.2 57.0 31.1

98.2 82.1 62.4 42.7 23.3

229.2 420.9 566.6 666.2 720.6


Fi

Vi 2

229.2 191.7 145.7 99.7 54.4

114.6 210.4 283.3 333.1 360.3

117

Design Example 2

!


Table 2-11b. Preliminary link beam sizes and properties Level

Story Height

R 5 4 3 2

12 12 12 12 15

Fi 2 114.6 95.8 72.8 49.8 27.2

Fi 2 114.6 95.8 72.8 49.8 27.2

Vli 105.8 194.3 261.5 279.8 415.8

(d − 2t f )t w

Size

d

tw

tf

W14X38 W16X89 W21X111 W21X122 W27X178

14.10 16.75 21.51 21.68 27.81

0.31 0.53 0.55 0.60 0.73

0.52 0.88 0.88 0.96 1.19

min. 3.92 7.19 9.68 10.36 15.40

Table 2-11c. Preliminary link beam results Level

(d − 2t f )t w

Results

Zx

R 5 4 3 2

4.05 7.88 10.87 11.86 18.44


61.50 175.00 279.00 307.00 567.00

Mp

φVp

1.3 M p Vp

1.6 M p Vp

Link e

Ratio M p Vp

Ω

CDR

3,075 8,750 13,950 15,350 28,350

109 213 293 320 498

36.5 53.5 61.8 62.3 74.0

45.0 65.8 76.1 76.7 91.1

32 48 56 56 66

1.26 1.30 1.31 1.30 1.29

1.15 1.22 1.25 1.27 1.33

1.03 1.09 1.12 1.14 1.20

For the first (ground level) story, the EBF link beam design will be based on use of a W 27 × 178 link beam at Level 2. Note that §15.2 of AISC-Seismic limits the yield strength of the link beam to F y = 50 ksi .

6b.

Link rotation.

The frame displacement at the second level, ∆ S 2 , was determined from a separate computer analysis (not shown) using the design base shear without ∆ . ∆ S 2 = 0.28 in. The corresponding inelastic displacement, ∆ M 2 may be estimated from a static analysis by the following formula: ∆ M = .7 R∆ s = .7(7 )0.28" = 1.37 in.

118

(30-17)


Design Example 2

!


The link rotation is computed as a function of the frame story drift and frame geometry. For a frame of story height h , bay width l , link length e, and dimensions a = l − e 2 , the link rotation may be calculated by the following formula. Link rotations, θ , must be limited to 0.080 radians per AISC-Seismic §15.2g. θ=

∆M  2a  1.37"  2 (147 ")  1 + = 1 +  = 0.042 radians ≤ 0.080 e  180 "  H  66 

∴

o.k.

Comment: The above formula makes the assumption that all deformation occurs within the link rotation at a particular level. It has been observed that there is significant contribution to deformations from column and brace elongation and shortening. A more accurate analysis of link rotation can be made looking at joint displacements and calculating rotations based on relative joint displacements. Another simple method is to perform an analysis using very strong column and brace section properties in the model and force all deformations into the link beam for purposes of evaluating the link rotations.

6c.



The nominal shear strength of the link, Vn , is equal to the lesser of V p or 2 M p e . Solving for the design strength φVn . φVn ≤ Vi ,link at any given level

(

)

φVn = 0.9 (0.60 )F y t w d − 2t f = 0.9 (0.6 )(50 ksi )(.73")[27.8"−2 (1.19")] = 498 kips φ2 M p e

=

0.9 (2 )M p e

=

0.9 (2) F y Z x e

=

(

)

0.9 (2 )(50 ksi ) 567.0 in.3 = 773 kips 66"

φVn = 498 kips φVn =

498 kips = 553 kips 0.9

The design overstrength factor for this link beam Ω is calculated as follows: Ω=

Vn 553 kips = = 1.33 Vi ,link 416 kips


119

Design Example 2

!


The minimum link design shear overstrength ratio is controlled by the φ factor. Thus, the minimum Ω is Ω min = 1.0 φ = 1.0 0.9 = 1.11 . The significance of the overstrength ratio is that the link will not yield until seismic forces overcome the link yield point. The overstrength factor Ω is a relationship between code forces and design overstrength forces which will likely yield the link. Note that the Ω factor does not include the R y factor for expected yield stress of the steel. The link beam in this Design Example has been sized for strength and stiffness. In beams above the level under discussion, it was found necessary to add cover plates for the beams outside the links (for increased beam capacity outside the link). The attempt was made to balance the design between good ratios of Mp /Vp of approximately 1.3 and the requirement for cover plates outside the link. It was decided to use cover plates to meet strength requirements for EBF beams outside the link to maintain desired ratios of Mp /Vp. The trade-off is to lessen the ratio of Mp /Vp and not require cover plates. It is believed that the performance of the link is more important than the cover plate requirement, and thus it was not possible to size beams to meet requirements outside the link without beam cover plates for this configuration of EBF frame.

6d.

Beam compact flange.

Check the W 27 × 178 beam to ensure that the flanges are compact to prevent flange buckling. 14.09" 52 bf = = 5.92 ≤ = 2 t f 2(1.19") Fy ∴

6e.

52 = 7.35 50 ksi

o.k.

AISC-Seismic, Table I-9-1

Link length.

The length of the link will determine whether the link yields in shear or in bending deformations. To ensure the desired shear yielding behavior (see discussion in Part 5b), the link beams have been limited to lengths less than 1.3Mp /Vp. From part 6c, Vp and Mp are calculated: V p = 553 kips

(

)

3 M p = Z x F y = 567 in. (50 ksi ) = 28,350 kip-in.

120


Design Example 2

!


Check that the 1.3 M p V p criteria is not exceeded. eV p Mp ∴

=

(66")(553 kips ) = 1.29 ≤ 1.3 28,350 k − in.

o.k.

Second floor link length of 66 inches is o.k.

6f.

Verification of link shear strength.

The strength of the link is used to establish the minimum strength required of elements outside the link. The link shear strength Vp was determined using the web area (d-2tf) of the beam. When the beam has reached flexural capacity, shear in the link may be less than the shear strength of the section. If this is the case, the flexural capacity of the section will limit the shear capacity of the link. AISCSeismic §15.2f requires that the shear strength of the section be the minimum of shear yielding strength or shear required for plastic moment yielding behavior. V p = 553 kips 2M p e

=

(

)

2 (50 ksi ) 576 in.3 = 872 kips 66"

The controlling nominal shear capacity Vn is the minimum of V p or 2V p e . From Part 6c, Vn = 553 kips . By selecting the W27x178 section as the link beam, the controlling mode of yielding is shear yielding in the link and therefore bending yielding will not be developed.

6g.

Required beam brace spacing.

§2313.10.18

The limiting unbraced length for full plastic bending capacity, L p , is determined as follows. Lateral beam braces for the top and bottom flanges at the ends of the link beams are still required. Lp =

300ry F yf

=

300(3.26") 50 ksi

= 138.3" ≅ 11'−6"

AISC-LRFD (F1-4)

Therefore, the beam bracing at 10 ft.-0 in. is adequate. (Note: the composite steel deck and lightweight concrete fill is not considered effective in bracing the top flange.)


121

Design Example 2

6h.

!


Beam and link axial loads.

The summation of story forces down to level 3, ΣFi = V3 in Table 2-4 (the sum of level shears from the roof to Level 3) is 666 k. The frame forces in Level 2 at the left connection and right connection are F2 L = 31.1 k and F2 R = 23.3 k . If the required axial strength of the link Pu is equal to or less than 0.15 Py , the effect of axial force on the link design shear strength need not be considered. Therefore, the axial force in the link is: C link = T link =

(31.1 k - 23.3 k ) = 3.90 k 2

The maximum axial stress in the link must be checked for the requirements of §15.2e of AISC-Seismic: fa =

Ω(3.9 kips ) 1.33(3.9 kips ) = = 0.10 ≤ 0.15 F y Ag 52.30 in.2

Therefore, the effect of axial force on the link design shear strength need not be considered.

6i.

Beam compact web.

AISC-Seismic §9.4

The maximum hc t w ratio permitted for compact beam sections is dependent on the axial load in the beam. Sections noted Fy′′′ in the AISC-LRFD (2nd Edition) have compact webs for all combinations of axial stress when the yield strength is less than the tabulated values. If a beam section is chosen that does not have a compact web for all axial loads, the section should be checked using Table I-9-1, of AISC-Seismic. The web should be compact along the full length of the beam. Both the UBC and AISC-Seismic do not allow the use of doubler plates for a link beam. For a W 27 × 178 beam. A = 52.30 in.2 hc d − 2k 27.81"−2(1.875") = = = 32.9 tw tw 0.73"

122


Design Example 2

!


Maximum axial force in link beam outside the link: V 3   666 kips  + 31.1 kips  = 484 kips P 2L = Ω  + F 2L  = 1.33  2 2    Pu 484kips = = 0.21 ≥ 0.125 φ b Py 0.90 (50 ksi ) 52.30 in.2

(

)


For Pu φ b Py ≥ 0.125 , allowable d t w to prevent local buckling is determined from the equation below.  hc  191  2.75 Pu  = 2.33 − φ b Py  tw  F y 

 =  

 (364 kips )  = 58.8 ≥ 253 = 5.06  2.33 − 0.9 (2,615 kips )  Fy 50 ksi  191

∴ hc / t w = 32.9 ≤ 58.8 ∴ o.k.

6j.


Combined link loads.

The combined bending plus axial strength of the link must be checked and compared with the yield stress. In the link, axial and bending stresses are resisted entirely by flanges. Pu = 3.9 kips (Ω ) = 3.9 kips(1.33) = 5.2 kips Pu 364 kips = = 0.14 ≤ 0.15 Py (50 ksi ) 52.30 in.2

(

)

AISC-Seismic §15.2f

Moment from yielding link shear: Mu = Vp

(66") = 18,249 kip − in. e = 553 k 2 2

A f = b f t f = (14.09")× (1.19") = 16.77 in.2


123

Design Example 2

!


(

)(

(

)

)

Z f = d − t f b f t f = (27.81"−1.19") 16.77 in. 2 = 446.2 in.3 Pu M 10.5 kips 18,249 k − in. + u = + = 40.9 ksi ≤ 50 ksi 2 Zf 2Af 2 16.77 in. 446.2 in.3

(

)

∴ Link combined axial plus bending capacity is o.k.

6k.

Beam analysis (outside of link).

AISC-Seismic, AISC §15.6b

Link beams have difficulty resisting the link beam moments increased by 1.1 and Ry when using a lower bound strength not including Ry. Although AISC-Seismic allows the LRFD design strength to be increased by Ry, it is not very clear how AISC-Seismic had intended it to be performed. In conversation with representatives of AISC-Seismic, it was conveyed to the author of this Design Example that the intention was simply to increase LRFD design strengths (Pn, Mn) by an Ry factor. It was not the intention of the AISC-Seismic subcommittee to increase Fy by Ry and carry those values through all the LRFD design equations. The solution in this Design Example has the beam outside the link resisting the entirety of the link beam moment. A more refined analysis can be performed where the brace contributes to the resistance of moment, which would reduce the moment on the beam outside the link. The analysis in this Design Example includes the use of flange cover plates to increase the bending capacity of the beam outside the link. The beam outside the link is required to resist 110 percent of the bending and axial forces corresponding to the link beam yield, using its nominal strength Ry. The combined beam bending plus axial interaction equations are referenced from AISC-LRFD Section H. Axial load analysis is referenced from AISC-LRFD Section E and bending analysis is referenced from AISC-LRFD Section F. The steps below yield forces from the hand evaluation of EBF frame behavior and from the computer model (not shown). Axial force in beam outside link is: PE = 364 kips From computer model, the load combination of Equation (12-5), including E v = 0.265 D , is: 1.2 D + 0.265 D + 0.5l + 1.0 E h 1.465D + 0.5l ; PD +L = 18 kips

124


Design Example 2

!


From EBF4 frame analysis: M E = 18,249 kip-in. Pu = (1.1)(1.33)(1.3)(364 k ) + (1.15)(18 kips ) = 712 kips

AISC-Seismic §15.6a

Pu = 1.1ΩR y PE + 1.1PD + L From computer analysis, load combination Equation (12-5): 1.2 D + 0.265 D + 0.5l + 1.0 Eh M D + L = 307 kip-in. Mu=

=

1.1R y V p e 2

+ 1.1M D + L

1.1(1.3)(553 kips )(66") + 1.1(307 kip-in.) 2

= 26,443 kip-in. Beam section properties.

Combined section properties are given in Table 2-12, the reader should understand how to convert typical beam section properties to those with cover plates: The beam at Level 2 does not require cover plates. The beams at Levels 3-Roof all require cover plates and thus have transformed section properties for use in the following equations. For W 27 × 170 beam without cover plates: A = 52.3 in.2 Z x = 567 in.3 Z f = 446 in.3 I x = 6,990 in.4 S x = 503 in.3 ry = 3.26 in. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

125

Design Example 2


!

I y = 555 in.4 J = 19.5 in.4 C w = 98,300 in.6 X 1 = 2,543 X 2 = 0.00375 Beam slenderness parameters: kl ry

=

(1.0)(120") = 36.8 3.26"

Slenderness parameter for beam-column lc is calculated as follows: lc =

kl Fy 36.8 (50 ksi ) = 0.487 = rπ E 3.1416 29,000 ksi

AISC-LRFD (E2-4)

The critical axial stress Fcr is calculated: For lc ≤ 1.5 :

(

2 2 Fcr =  0.658lc  Fy = 0.658.487  

) (50 ksi) = 45.3 ksi

AISC-LRFD (E2-2)

φ c = 0.85 Nominal axial strength is calculated as follows:

(

)

Pn = Ag Fcr = 52.3 in.2 (45.3 ksi ) = 2,368 kips R y Pn = 1.3(2,368 kips ) = 3,078 kips

AISC-LRFD (E2-1) AISC-Seismic §15.6b

Bending capacity calculations are calculated: φ b = 0.90

AISC-LRFD§F1.1

M n = M p for a limit state if flexural yielding

(

AISC-LRFD (F1-1)

)

M p = Z x Fy = 567 in.3 (50 ksi ) = 28,350 k-in. 126


Design Example 2

!


Check lateral torsional buckling stability and allowable strength:   Lb − L p   ≤ M p M n = C b  M p − ( M p − M r )   L L −  p   r 

AISC-LRFD (F1-2)

C b = 1.0 Unbraced length: Lb = 120 in. Limiting laterally unbraced length for full plastic yielding: Lp =

300ry F yf

=

300(3.26) 50 ksi

= 138 in.

AISC-LRFD (F1-2)

Limiting laterally unbraced length for inelastic lateral torsional buckling: Lr =

ry X 1 FL

1 + 1 + X 2 FL 2

AISC-LRFD (F1-6)

Limiting buckling moment: M r = FL S

AISC-LRFD (F1-7)

Beam buckling factors, X 1 and X 2 : X1 =

π Sx

X2 = 4

EGJA 2

Cw  S x    I y  GJ 

AISC-LRFD (F1-8) 2

AISC-LRFD (F1-9)

FL is the smaller of the yield stress in the flange minus compressive residual stresses (10 ksi for rolled shapes) or web yield stress. AISC-LRFD §F1.2a FL = (50 ksi − 10 ksi ) = 40 ksi Lr =

ry X 1 FL

1 + 1 + X 2 FL 2 =


(3.26)(2,543) (40 ksi )

1 + 1 + (0.00375)(40 ksi )2 = 396

127

Design Example 2

!


(

)

M r = FL S x = (40 ksi ) 503 in.3 = 20,108 k-in.  M n = Cb  M p − M p − M r 

(





) LLb −− LLp  

r

p



  120"−138"  = 1.0 28,350 − (28,350 − 20,108)   396"−138"   = 28,933 k-in. ≥ M p = 28,350 k-in.

∴ M n = 28,350 k-in. R y M n = 1.3 (28,350 k-in.) = 36,855 k-in. Comparison of lateral torsional buckling moment with plastic yield moment indicates that plastic yield moment is the controlling yield behavior. AISC-LRFD Section H, combined axial plus bending interaction equations are as follows: For the case: Pu ≥ 0.2 φ c R y Pn

AISC-LRFD (H1-1a)

Pu 8 M ux + ≤ 1.0 φ c R y Pn 9 φ b R y M nx For the case: Pu < 0.2 φ c R y Pn

AISC-LRFD (H1-1b)

Pu M ux + ≤ 1.0 2φ c R y Pn φ b R y M nx Thus, for this Design Example: Pu 712 kips = = 0.27 ≥ 0.2 φ c R y Pn 0.85(3,078 kips )

128


Design Example 2

!


Pu 8 M ux 712 kips  8  26,443 k − in. = 0.98 ≤ 1.0 + = +  φc R y Pn 9 φb R y M nx (0.85)(3,078 kips )  9  0.90 (36,855 k − in.) ∴ o.k. Therefore, W 27 × 178 beam outside the link is okay. The EBF beams above Level 2 require cover plates and thus utilize combined section properties in the above equations.

6l.

Beam stiffeners.

AISC-Seismic §15.3

There are two types of stiffeners required in links: 1.) link stiffeners at ends at brace connections; and 2.) intermediate stiffeners. These are shown in Figure 2-7. Link end stiffeners.

Full depth web stiffeners are required on both sides of the link beam at the brace connections. The stiffeners are used to prevent web buckling and to ensure ductile shear yielding of the web. The stiffeners shall have a combined width not less than bf - 2tw and a thickness not less than 0.75t w or 3/8 inch, whichever is larger. For the W 27 × 178 beam: b f − 2t w = 14.09"−2(0.73") = 12.63" use 2 × 6.375"

AISC-Seismic §15.3a

The minimum thickness of the stiffeners is: 0.75t w = 0.75(0.73") = 0.548" use 5/8" stiffeners ∴ Use 6 3/8 in. × 5/8 in. stiffeners each side of beam (total 4) Intermediate stiffeners.

AISC-Seismic §15.3b requires intermediate full depth web stiffeners (Figure 2-7) where link lengths are 5 V p M p or less. Where link lengths are 1.6 V p M p or less, the spacing shall not exceed 30t w − d w 5 for link rotation of 0.08 radians and 52t w − d w 5 for link rotations of 0.02 radians. Linear interpolation may be used between link rotations of 0.02 and 0.08 radians. Thus,


129

Design Example 2

!


30tw −

d  27.81"  = 30(0.73") −   = 16.33 in. 5  5 


52t w −

d  27.81"  = 52(0.73") −   = 32.43 in. 5  5 


Since the link rotation is 0.040 radians for the beam, interpolation must be used to determine the maximum spacing of intermediate stiffeners. This is shown below.  0.080 rad − 0.040 rad    (32.43"−16.33") + 16.33" = 27.0 in.  0.080 rad − 0.020 rad  Since the link length is 72 inches, therefore use three equal spacings of 24 inches. Since the link beam is a W 27 , stiffener depth is 27.81 in. – 2 (1.19 in.) = 25.4 in. Under §15.3b, Item 5, AISC-Seismic, intermediate stiffeners of depth greater than 25 inches are required to be placed on both sides of the beam. One-sided stiffeners are required for depths less than 25 inches. The width shall not be less than b f 2 − t w = 12.44" 2 − .650" = 5.57 in. Therefore use 6 3/8 in. × 5/8 in. stiffeners on both sides of the beam. Web stiffener welds.

The web stiffener welds are required to develop a stiffener force of Ast F y = (6.375")(0.625")(50 ksi ) = 199 kips

AISC-Seismic §15.3c

The minimum size of fillet weld, per AISC-LRFD Table J2.4, is ¼-inch to the link web and 5/16-inch to the link flange. Using E70XX electrodes and 5/16-inch fillet welds each side, the weld capacity is 0.6FEXX. The required weld length on the beam web is: 1required =

199 kips = 10.72 in. 0.60(70 ksi )(2 × 5 16")(.707 )

Therefore, use 5/16-inch fillet welds, both sides of the stiffener, at flanges and web. Note: One-fourth of the above required weld is required at the flanges.

130


Design Example 2

6m.

!


Tabulated link beam design.

Tables 2-12a through 2-12h present tabular calculations that show the results from procedures in Parts 6a through 6l applied to all beams in the frame EBF4. The link beam design for all levels is as shown below in tabular form following the equations given above (each row/level is a continuation of the table above).

Table 2-12a. Link beam section properties Aweb

Level

Link Beam

e

a

h

Fy

A

bf

tf

d

tw

t w (d − 2t r )

Size

R 5 4 3 2

W14X38 W16X89 W21X111 W21X122 W27X178

32 48 56 56 66

164 156 152 152 147

144 144 144 144 168

50 50 50 50 50

11.20 26.20 32.70 35.90 52.30

6.77 10.37 12.34 12.39 14.09

0.52 0.88 0.88 0.96 1.19

14.10 16.75 21.51 21.68 27.81

0.31 0.53 0.55 0.60 0.73

4.05 7.88 10.87 11.86 18.44


Table 2-12a. Link beam section properties (continued) Level

Link Beam

Af

Zf

Zx

lx

Sx

rx

ly

Sy

ry

J

Cw

R 5 4 3 2

W14X38 W16X89 W21X111 W21X122 W27X178

7.0 18.1 21.6 23.8 33.5

47.4 144.0 222.8 246.5 446.2

61.5 175.0 279.0 307.0 567.0

385.0 1,300.0 2,670.0 2,960.0 6,990.0

54.6 155.0 249.0 273.0 502.0

5.86 7.04 9.04 9.08 11.56

26.7 163.0 274.0 305.0 555.0

7.89 31.45 44.41 49.23 78.81

1.54 2.49 2.89 2.91 3.26

0.80 5.45 6.83 8.98 19.50

1,230 10,200 29,200 32,700 98,300

Table 2-12b. Combined section properties (beams plus cover plates) Level

Link Beam

Plate b

t

At

Zx

Zf

lx

Sx

ry

ly

J

Cw

X1

X2

R 5 4 3 2

W14X38 W16X89 W21X111 W21X122 W27X178

6 6 6 6 0

0.375 0.250 0.250 0.250 0.000

16 29 36 39 52

94 201 312 340 567

80 169 255 279 446

621 1517 3025 3321 6990

84 176 275 299 503

1.60 2.43 2.82 2.84 3.26

40 172 283 314 555

0.8 5.5 6.8 9.0 19.5

1,230 10,200 29,200 32,700 98,300

1,697 2,872 2,274 2,499 2,543

0.01065 0.00197 0.00533 0.00369 0.00375

Table 2-12c. Compact flange, web Level φVp R 5 4 3 2

109.4 212.6 293.4 320.1 497.8

Ω

Mp

φVp Mp

1.15 3,075 1.26 1.22 8,750 1.30 1.25 13,950 1.31 1.27 15,350 1.30 1.33 28,350 1.29

b b f 2t f 6.57 5.92 7.05 6.45 5.92


Comp. Comp. 1.1R y Flange Flange Limit Results M p Vp 7.35 7.35 7.35 7.35 7.35


34.0 49.8 57.5 58.0 68.9

Pu

Py

Pu φPn

tw

130.97 224.13 293.69 340.24 364.21

560.00 1,310.00 1,635.00 1,795.00 2,615.00

0.28 0.20 0.21 0.22 0.16

42.2 28.6 35.9 32.9 35.1

h

Comp. Comp. Web Web Limits Results 55.5 57.5 57.2 56.9 58.5


131

Design Example 2

!


Table 2-12d. Combined link stresses, unsupported length Shear Level Above Level R 5 4 3 2

Fi

0 131.0 229.2 109.5 420.9 83.2 566.6 57.0 666.2 31.1

Pmax Link Beam

Plink

131.0 224.1 293.7 340.2 364.2

16.4 1,944.8 15.3 5,670.0 13.7 9,129.1 12.0 9,959.0 10.5 18,252.4

M link

Comb Allow. Link Link Link Result Loads Stress 42.2 39.8 41.3 40.7 41.1

50 50 50 50 50


2φM pa 2

Vu

Value

Lu max.

2,780 8,558 13,504 14,680 28,794

156.4 320.9 434.1 471.8 785.3

106.4 209.5 288.7 314.3 492.9

1.14 1.17 1.18 1.17 1.16

72.8 111.4 132.6 133.2 151.4

Ry

Pbu (kips)

1.3 1.3 1.3 1.3 1.3

217 403 542 638 712

Vpa

M pa

106.4 209.5 288.7 314.3 492.9

Table 2-12e. Beam outside link, design forces Level Pu ,1.0Eh R 5 4 3 2

M u ,SEISMIC

131 224 294 340 364

Ω

Vp e 2 1,945 5,670 9,129 9,959 18,252

1.15 1.22 1.25 1.27 1.33

Beam Pu ,D +L M u ,D +L Beam Overstr. 1.465D+ 1.465D+ Overstr. Factor Factor 0.5L 0.5L 1.10 1.10 1.10 1.10 1.10

2.0 12.0 17.0 17.4 17.6

342.0 359.0 303.0 318.0 307.0

1.10 1.10 1.10 1.10 1.10

Pbu (k-in.) 3,157 8,503 13,388 14,591 26,439

Table 2-12f. Axial compression parameters Level

Section

Lu (ft)

kl r y

λc

Fcr (ksi)

φc

Pn (kips)

R y Pn

R 5 4 3 2

W14X38 W16X89 W21X111 W21X122 W27X178

10 10 10 10 10

75.0 49.4 42.6 42.2 36.8

0.991 0.653 0.563 0.558 0.487

33.14 41.82 43.78 43.89 45.28

0.85 0.85 0.85 0.85 0.85

520 1,221 1,563 1,707 2,368

676 1,587 2,032 2,219 3,078

(kips)

Table 2-12g. Flexural strength parameters and combined axial plus bending results (LTB=lateral torsional buckling yield mode) Level φ b R 5 4 3 2

132

0.9 0.9 0.9 0.9 0.9

Mn = M p (k-in.) 4,703 10,025 15,582 16,994 28,350

C b Lb L p LTB (in.) (in.) 1.0 1.0 1.0 1.0 1.0

120 120 120 120 120

68 103 119 121 138

X1

X2

1,697 2,872 2,274 2,499 2,543

0.01065 0.00197 0.00533 0.00369 0.00375

40 40 40 40 40

Mn LTB (k-in.)

Mn (k-in.)

Ry Mn (k-in.)

Pu φR y Pn

156 3,344 3,895 3,895 304 7,034 9,771 9,771 324 10,995 15,569 15,570 338 11,977 17,007 16,995 396 20,108 28,933 28,350

5,064 12,703 20,241 22,093 36,855

0.38 0.30 0.31 0.34 0.27

FL Lr (ksi) (in.)

Mr (k-in.)

AISC- AISCLRFD LRFD H1-1a H1-1b 0.99 0.96 0.97 0.99 0.98


NA NA NA NA NA

Design Example 2

!


Table 2-12h. Link rotations

6n.

Level

∆S

Story Drift ∆S

Story ∆M (i.n)

h (in.)

a (in.)

e (in.)

Rot θ (rad)

R 5 4 3 2

1.21 1.01 0.78 0.53 0.28

0.20 0.23 0.25 0.25 0.28

0.98 1.13 1.23 1.23 1.37

144 144 144 144 180

164 156 152 152 147

32 48 56 56 66

0.0766 0.0587 0.0547 0.0547 0.0416

AISC-Seismic brace design.

§15.6, AISC-Seismic

The braces are required to be designed for 1.25R yV p times the yielding link strength plus 1.25 times gravity load combinations. PE = 1.25ΩR y Pcomputer due to E h loads. M E = 1.25 R yV p e / 2 Using strength design procedures outlined in AISC-LRFD Section H, obtaining forces from a computer analysis, and showing calculations in tabular form (Tables 2-13a through 2-13e), the design forces for braces ( P and M ) are calculated. Column shear forces are not a controlling factor and are not shown for the sake of brevity.

Table 2-13a. Brace section properties ry

Level

Section

A (in.2)

Zx (in.3)

Sx (in.3)

L (ft)

rx (in.)

(in.)

5 4 3 2 1

W12X87 W12X152 W12X210 W12X230 W12X252

26 45 62 68 74

132 243 348 386 428

118 209 292 321 353

18 18 18 18 19

5.38 5.66 5.88 5.98 6.06

3.07 3.19 3.28 3.31 3.34


kl r y 71.4 68.7 66.8 65.5 68.6

Fy (ksi) 50 50 50 50 50

133

Design Example 2

!


Table 2-13b. Brace design loads Level

Section

PE

ME

Ω

Overstr. Factor

R 5 4 3 2

W12X87 W12X72 W12X79 W12X106 W12X120

150 276 378 446 565

1,512 3,036 3,744 4,200 3,996

1.15 1.22 1.25 1.27 1.33

1.25 1.25 1.25 1.25 1.25

Pgravity

M gravity

1.465D+ 0.5L

1.465D+ 0.5L

18.0 24.0 24.0 25.0 25.0

276.0 247.0 180.0 181.0 105.0

Overstr. Factor

Ry

Pbu design

M bu design

1.25 1.25 1.25 1.25 1.25

1.3 1.3 1.3 1.3 1.3

303 575 796 953 1,253

3,168 6,309 7,811 8,903 8,770

Table 2-13c. Brace axial design parameters Level

Section

Lu (ft)

kl r y

λc

Fcr (ksi)

φc

Pn (kips)

R 5 4 3 2

W12X87 W12X72 W12X79 W12X106 W12X120

18 18 18 18 19

71.4 68.7 66.8 65.5 68.6

0.943 0.908 0.883 0.865 0.906

34.45 35.40 36.08 36.55 35.46

0.85 0.85 0.85 0.85 0.85

881.9 1,582.4 2,229.5 2,474.5 2,627.3

X1

X2

FL (k-in.)

Lr

Mr (k-in.)

Mn (k-in.)

Table 2-13d. Brace bending design parameters Level

5 4 3 2

φb (ksi)

Mn = M p

0.9 0.9 0.9 0.9 0.9

Lp

(k-in.)

Cb (kips)

Lb (in.)

(in.)

6,600.0

1.0

219

130

3,869

0.0006

40

459

4,720

6,092.5

12,150.0 17,400.0 19,300.0 21,400.0

1.0 1.0 1.0 1.0

219 219 217 229

135 139 140 142

3,225 3,524 4,650 5,231

0.0012 0.0008 0.0003 0.0002

40 40 40 40

423 460 572 639

8,360 11,680 12,840 14,120

11,045.1 15,974.5 18,158.5 20,120.9

Table 2-13e. Brace, combined axial plus bending results

134

Level

Section

Pu φPn

AISC LRFD H1-1a

AISC LRFD H1-1b

5 4 3 2 1

W12X87 W12X152 W12X210 W12X230 W12X252

0.40 0.43 0.42 0.45 0.56

0.92 0.99 0.90 0.94 0.99

NA NA NA NA NA


Design Example 2

6o.

!


AISC-Seismic column design.

AISC-Seismic §15.8

The design of the columns for frame EBF4 for the requirements of AISC-Seismic is shown in Tables 2-14a through 2-14e. The columns are required to resist an axial force corresponding to 1.1RyVn, which is the shear strength of the links to ensure that the yielding mechanism is within the link beams. Design forces (P and P) are calculated as 1.1ΩRy times seismic forces plus 1.1 times factored gravity load combinations. Column shear forces are not a controlling factor and are not shown for the sake of brevity.

Table 2-14a. Column, section properties ry

Level

Section

A (in.2)

Zx (in.3)

Sx (in.3)

L (ft)

rx (in.)

(in.)

5 4 3 2 1

W12X87 W12X87 W12X87 W12X170 W12X170

26 26 26 50 50

132 132 132 275 275

118 118 118 235 235

18 18 18 18 19

5.38 5.38 5.38 5.74 5.74

3.07 3.07 3.07 3.22 3.22

kl r y 71.4 71.4 71.4 67.4 71.3

Fy (ksi) 50 50 50 50 50

Table 2-14b. Column, design loads Level

Section

R 5 4 3 2

W12X87 W12X87 W12X87 W12X170 W12X170

PE

ME

Ω

Overstr. Factor

0 84 238 458 683

276 432 504 552 972

1.15 1.22 1.25 1.27 1.33

1.10 1.10 1.10 1.10 1.10

Pgravity

M gravity

1.465D+ 0.5L

1.465D+ 0.5L

4.0 22.0 44.0 67.0 87.0

168.0 180.0 144.0 120.0 60.0

Overstr. Factor

Ry

Pbu design

1.10 1.10 1.10 1.10 1.10

1.3 1.3 1.3 1.3 1.3

4 170 473 906 1,395

M bu design 638 949 1,057 1,136 1,915

Table 2-14c. Column, axial design parameters Level

Section

Lu (ft)

kl r y

λc

Fcr (ksi)

φc

Pn (kips)

R 5 4 3 2

W12X87 W12X87 W12X87 W12X170 W12X170

12 12 12 12 15

46.9 46.9 46.9 44.8 56.0

0.620 0.620 0.620 0.592 0.740

42.56 42.56 42.56 43.18 39.76

0.85 0.85 0.85 0.85 0.85

1,089.6 1,089.6 1,089.6 2,159.0 1,988.1


135

Design Example 2

!


Table 2-14d. Column, bending design parameters Level

φb (ksi)

Mn = M p (k-in.)

Cb (kips)

Lb (in.)

(in.)

R 5 4 3 2

0.9 0.9 0.9 0.9 0.9

6,600.0 6,600.0 6,600.0 13,750.0 13,750.0

1.0 1.0 1.0 1.0 1.0

144 144 144 144 180

130 130 130 136 136

Lp

X1

X2

FL (k-in.)

Lr

Mr (k-in.)

Mn (k-in.)

3,869 3,869 3,869 7,173 7,173

0.0006 0.0006 0.0006 0.0001 0.0001

40 40 40 40 40

459 459 459 824 824

4,720 6,521.0 4,720 6,521.0 4,720 6,521.0 9,400 13,702.1 9,400 13,474.4

Mn (k-in.) 6,521 6,521 6,521 13,702 13,474

Table 2-14e. Column, combined axial plus bending results

6p.

Level

Section

Pu φPn

AISC LRFD H1-1a

AISC LRFD H1-1b

5 4 3 2 1

W12X87 W12X87 W12X87 W12X170 W12X170

0.00 0.18 0.51 0.49 0.83

NA NA 0.67 0.58 0.97

0.11 0.25 NA NA NA

Final frame member sizes (LRFD).

Table 2-15. Final frame member sizes for EBF4 (LRFD) Level

Beams

Links (in.)

Beam Cover Plate (in.) (1)

Roof 5 4 3 2 1

W14x38 W16x89 W21x111 W21x122 W27x178

32 48 56 56 66

6x¼ 6x¼ 6x¼ 6x¼ Not req’d

Columns

Braces

W12X65 W12X65 W12X65 W12X87 W12X87

W12X87 W12X87 W12X87 W12X106 W12X120

Note: 1. Top and bottom flanges outside link.

136


Design Example 2

!


Figure 2-6. EBF4 Frame member sizes (LRFD)

7.

Typical EBF details.

Figures 2-7 through 2-14 are examples of typical EBF connection details. These are shown for both wide-flange and tube section braces.

Figure 2-7. EBF brace-beam connection at link using wide flange brace


137

Design Example 2

!


Figure 2-8. EBF brace-column connection using wide flange brace

Figure 2-9. EBF beam-brace connection at link using TS brace

138


Design Example 2

!


Figure 2-10. Brace-beam connection with TS brace

Figure 2-11. EBF stiffeners at links


139

Design Example 2

!


Figure 2-12. EBF beam stability bracing

Figure 2-13. Partial plan of EBF beam stability bracing

140


Design Example 2

!


Figure 2-14. Link beam cover plates (beam outside the link)

Commentary EBF frames are considered a quality seismic system because of their ability to yield with a known behavior at controllable locations and to demonstrate very good hysteretic behavior during cyclical loading. The possibility exists of discrete postearthquake repairs in local areas if yielding of a frame occurs in an earthquake. The construction of these frames is not difficult, and the cost is only slightly greater than the cost of special concentric braced frame systems. As can be seen, the LRFD design in accordance with AISC-Seismic yields more conservative results. However, the provisions of AISC-Seismic are considered state-of-the-art and more likely to yield an EBF frame with the superior performance that is expected of EBF systems. It was found that by designing an EBF link beam that meets all of the most desirable attributes of EBF design, that the beam outside the link might require cover plates to achieve the required strength. The designer will struggle with optimization of the link design and the requirement for cover plates outside the link. It is believed that optimization of the link is the most important element in the system and if cover plates are required outside the link, that is a cost worth paying. In the ASD example, the link lengths (to 1.3Vs/Ms), were not optimized and thus did not need cover plates. However, from a performance standpoint, the ASD frame may not be as good a design as the LRFD frame because its link lengths are much shorter.


141

Design Example 2

!


References Becker and Ishler, “Seismic Design Practice for Eccentrically Braced Frames, Based on the 1994 UBC,” Steel Tips. Structural Steel Educational Council, Moraga, California, December 1996. Popov, Kasai, and Engelhardt, 1987. “Advances in Design of Eccentrically Braced Frames,” Earthquake Spectra. Earthquake Engineering Research Institute, Oakland, California, Vol.3, no.1. Engelhardt and Popov, 1989. “On Design of Eccentrically Braced Frames,” Earthquake Spectra. Earthquake Engineering Research Institute, Oakland, California, Vol. 5, No. 3. Popov, Engelhardt, and Ricles, 1989. “Eccentrically Braced Frames: US Practice,” Engineering Journal. American Society of Civil Engineers, Reston, Virginia, AISC, 2nd quarter. Kasai and Popov, 1986. “General Behavior of WF Steel Shear Link Beams,” Journal of Structural Engineering. American Society of Civil Engineers, Reston, Virginia, Vol. 112, no. 2.

142


Design Example 3A

!

Steel Special Moment Resisting Frame

Design Example 3A Steel Special Moment Resisting Frame

Figure 3A-1. Four-story steel office building with steel special moment resisting frames (SMRF)

Foreword This Design Example illustrates use of the 1997 UBC provisions for design of a steel special moment resisting frame (SMRF). During the course of the development of this Volume III, an intensive steel moment frame research program, including considerable full-scale testing, was conducted by the SAC project. As a result of this effort, new SAC guidelines have been developed. However, these came after the finalization of this Design Example. Consequently, the SMRF example given in this document shows only 1997 UBC and FEMA-267/267A methodology. With the help of member of the SAC team, comments have been added to this Design Example indicating where the anticipated new SAC guidelines will be different than the methodology shown in this Design Example.


143

Design Example 3A

!


Overview Since the 1994 Northridge earthquake, the prior design procedures for steel moment resisting frames have been subject to criticism, re-evaluation, and intensive reseach. Given the observed earthquake damage attributed to brittle connection fractures in the 1994 Northridge earthquake, it was determined that the 1994 UBC requirements for moment resisting joint design were inadequate and should not continue to be used in new construction. In September 1994, the International Conference of Building Officials (ICBO) issued an emergency code amendment that eliminated the prescriptive code design procedures for special moment resisting frame (SMRF) beam-column connections. Those procedures were replaced with code language requiring qualification of SMRF connection design through prototype testing or calculation. A SMRF conection is now required to demonstrate by testing or calculation the capacity to meet both the strength and inelastic rotation performance as specified by 1997 UBC §2213.7.1. To address the research needs precipitated by the SMRF connection concerns, the SAC Joint Venture was formed by SEAOC, the Applied Technology Council (ATC), and the California Universities for Research in Earthquake Engneering (CUREe). SAC was charged with developing interim recommendations for professional practice, including design guidelines for use in new SMRF connections. To this end, FEMA-267, Interim Guidelines: Evaluation, Repair, Modification and Design of Welded Steel Moment Frame Structures was published in August, 1995. This was followed by FEMA-267A, Interim Guidelines; Advisory No. 1, published in March, 1997. As a prelude to possible future code requirements, FEMA-267A offers design procedures and calculation methodologies for certain SMRF connection configurations. While these procedures are subject to further refinement, they represent the current state of practice for SMRF connection design. This Design Example follows the procedures as presented in FEMA-267A, with the reduced beam section (RBS) the selected joint configuration. Test results for the RBS joint configuration indicate that it provides the requisite inelastic rotation capacity, and is one of the more cost-effective of the current SMRF connection options. Following publication of the FEMA-267 series, the SAC Joint Venture entered into a supplemental contract with FEMA to perform additional research and develop final design guidelines. That work, recently completed, culminated with the publication of FEMA-350, Recommended Seismic Design Criteria for New Moment Resisting Steel Frame Structures. FEMA-350 will present design details and criteria for ten different types of connections that are prequalified for use within certain limits. The FEMA-350 criteria are similar, but not identical, to those illustrated here.

144


Design Example 3A

!


The 4-story steel office structure shown in Figure 3A-1 is to have special moment resisting frames as its lateral force resisting system. The typical floor plan is shown on Figure 3A-2 and the moment frame elevation is provided in Figure 3A-3 at the end of this Design Example.


Design base shear.

2.


3.

Interstory drifts.

4.


5.

SMRF member design.

6.

SMRF beam-column connection design.

Given Information The following information is given: Roof weights: Roofing Insulation Concrete fill on metal deck Ceiling Mechanical/electrical Steel framing Total

4.0 psf 2.0 44.0 3.0 4.0 6.0 63.0 psf

Floor weights: Floor covering Concrete fill on metal deck Ceiling Mechanical/electrical Steel framing Partitions (seismic DL) Total

1.0 psf 44.0 3.0 5.0 9.0 10.0 76.0 psf

Live load:

20.0 psf

Live load:

80.0 psf

Exterior wall system weight: steel studs, gypsum board, fascia panels

20.0 psf


145

Design Example 3A

!


Site seismic and geotechnical data: Occupancy category: Seismic importance factor: Soil profile type: Seismic Zone: Seismic source type: Distance to seismic source: Near source factors:

Structural materials: Wide flange shapes Plates Weld electrodes

Standard Occupancy Structure I = 1.0 Type S D (default profile) Zone 4, Z = 0.40 Type C

§1629.2 Table 16-K §1629.3, Table 16-J §1629.4.1, Table 16-I §1629.4.2

10 km

Table 16-S

N a = 1.0 N v = 1.0

Table 16-T Table 16-U

(

ASTM A572, Grade 50 f y = 50 ksi

)

ASTM A572, Grade 50 E70XX

Figure 3A-2. Typical floor framing plan

146


Design Example 3A

!


Figure 3A-3. Frame elevation at Line A


1. 1a.

Code Reference

Design base shear.

Check configuration requirements.

§1629.5

Check the structure for vertical and horizontal irregularities. Vertical irregularities—review Table 16-L.

Table 16-L

By observation, the structure has no vertical irregularities. The moment frames have no discontinuities or offsets, and the mass is similar at all levels. Plan irregularities—review Table 16-M.

Table 16-M

The floor plan has no re-entrant corners exceeding 15 percent of the plan dimension, nor are there any diaphragm discontinuities. Therefore, the structure has no plan irregularities.


147

Design Example 3A

1b.

!



§1629.6

The structure is a moment-resisting frame system with lateral resistance provided by steel special moment resisting frames (SMRF) (system type 3.1, Table 16-N). The seismic factors are: R = 8.5

Table 16-N

Ω = 2.8 hmax = no limit

1c.


§1629.8.3

The static lateral force procedure will be used. This is permitted for regular structures not more than 240 feet in height.

1d.


§1629.4.3

For Zone 4 and Soil Profile Type S D :

1e.

C a = 0.44(N a ) = 0.44(1.0 ) = 0.44

Table 16-Q

C v = 0.64(N v ) = 0.64(1.0 ) = 0.64

Table 16-R


§1630.2.2

Per Method A:

(30-8)

T = Ct (hn )3 4 C t = 0.035 T A = 0.03(55.5)3 4 = 0.71 sec Per Method B: Using a computer model, in lieu of Eq. (30-10), with assumed member sizes and estimated building weights, the period is determined:

148


Design Example 3A

!


North-south ( y ) : TBy = 1.30 sec East-west (x ):

§1630.2.2 Para. #2

TBx = 1.16 sec For Seismic Zone 4, the value for Method B cannot exceed 130 percent of the Method A period. Consequently, Maximum value for TB = 1.3T A = 1.3(0.71) = 0.92 sec

1f.


§1630.2.1

The total design base shear for a given direction is: V =

Cv I 0.64(1.0 ) W = W = 0.082W RT 8.5(0.92 )

(30-4)

The base shear need not exceed: V =

2.5Ca I 2.5(0.44 )(1.0 ) W = = 0.129W R 8.5

(30-5)

But the base shear shall not be less than: V = 0.11C a IW = 0.11(0.44)(1.0)W = 0.048W

(30-6)

And for Zone 4, base shear shall not be less than: V =

0.8ZN v I 0.8(0.4)(1.0)(1.0) W = = 0.038W R 8.5

(30-7)



(30-4)

149

Design Example 3A

!


Note that if the period from Method A (T = 0.71sec) was used, the base shear would be V = 0.106W . Method A is based on empirical relationships and is not considered as accurate as Method B. To avoid unconservative use of Method B, the code limits the period for Method B to not more than 1.3 times the Method A period.

1g.


§1630.1

Section 1630.1.1 specifies earthquake loads. These are E and E m as set forth in Equations (30-1) and (30-2). E = ρE h + E v

(30-1)

Em = Ω o Eh

(30-2)

The normal earthquake design load is E . The load E m is the estimated maximum earthquake force that can be developed in the structure. It is used only when specifically required, as will be shown later in this Design Example. Before determining the earthquake forces for design, the reliability/redundancy factor must be determined. Reliability/redundancy factor: ρ = 2 −

20

(30-3)

rmax Ab

Ab is the ground floor area of the structure. Note that per the exception in §1630.1, Ab may be taken as the average floor area in the upper setback portion in buildings with a larger ground floor area and a smaller upper floor area. Ab = (140 × 240) − 8(8.5)2 / 2 = 33,311 ft 2 The element story shear ratio ri is the ratio of the story shear in the most heavily loaded single element over the total story shear at a given level i . The value for rmax is the greatest value for ri occurring in any story in the lower two-thirds of the structure. In structures with setbacks or discontinuous frames, the value of ri should be checked at each level. For this Design Example, the frames are uniform at all levels and will resist approximately the same relative lateral force at each story. For moment frames, ri is taken as the maximum of the sum of the shears in any two adjacent columns in a moment frame bay, divided by the story shear. The exception is that for interior columns in multi-bay frames, 70 percent of the shear may be used in the column shear summation.

150


Design Example 3A

!


By observation, the moment frame with the highest total shear per bay will govern the value for rmax . For this Design Example, the design base shear is equal for both north-south and east-west directions. Referring to the floor framing plan (Figure 3A-2), the east-west direction has 16 moment frame columns, while the north-south direction has 12 moment frame columns; so the north-south rmax will be greatest. Although a different value of ρ may be used for each direction, the larger rmax will be used for both directions in this Design Example to be conservative. Assume that the frames at Lines A and H each take half the story shear. Using the portal method for the frame at Line A (Figure 3A-4), the four interior columns take approximately 80 percent of the frame shear, and the two exterior columns 20 percent of the frame shear.

ΣF=50%

0.05V

0.1V

0.1V

0.1V

0.1V

0.05V

Figure 3A-4. Frame at Line A

At the exterior bay: ri = 0.05 + 0.7 (0.1) 1.0 = 0.12 At the interior bays: ri = 0.7 (0.1 + 0.1) 1.0 = 0.14 The interior bay governs with the larger value of ri . Per the SEAOC Blue Book Commentary (§C105.1.1.1), ri is to include the effects of torsion, so a 5 percent increase will be assumed. rmax = 1.05(0.14) = 0.147 ∴ρ = 2−

20 0.147(33,311)1 / 2

= 1.25

o.k.

(30-3)

Note that ρ cannot be less than 1.0, and that for SMRFs, ρ cannot exceed 1.25 per §1630.1.1. If necessary, moment frame bays must be added until this requirement is met.


151

Design Example 3A

!


For the load combinations per §1612, and anticipating using allowable stress design (ASD) in the frame design: E = ρE h + E v = 1.25(V ) ( E v = 0 for allowable stress design)

(30-1)

E m = ΩE h = 2.8(V )

(30-2)

Note that seismic forces may be assumed to act nonconcurrently in each principal direction of the structure, except as per §1633.1. Although for this Design Example the same value of ρ is used in either direction, a different value of ρ may be used for each of the principal directions.

2. 2a.


Building weights and mass distribution.

Calculate the building weight and center of gravity at each level. Include an additional 90 kips (3.0 psf) at the roof level for estimated weight of mechanical equipment. Distribute the exterior curtain wall to each level by tributary height.

Figure 3A-5. Typical floor

152


Design Example 3A

!


Table 3A-1. Building mass properties Roof Level Mass Properties

Roof: w DL = 63.0 + 3.0add'lmech = 66.0 psf ; Exterior Walls: w wall = 20 psf ; Wall Area = (6.5 + 4.0)(696 ft ) = 7,308 ft 2 Mark Floor Walls Totals

w DL (psf) 66.0 20.0

Wi (kips) 1,920 146 2,066

Area (sf) 29,090 7,308

X cg

Ycg

(ft) 100 100

(ft) 70 70

( )

( )

W X cg

W Ycg

191,994 14,616 206,610

134,396 10,231 144,627

X cg = 206 ,610 2,066 = 100.0ft ; Ycg = 144,627 2,066 = 70.0ft 4th, 3rd, & 2nd Level Mass Properties

Floor: w DL = 72.0 psf ; Exterior Walls: w wall = 15 psf ; Wall Area = (13.5)(696 ft ) = 9,396 ft 2 Mark Floor Walls Totals

w DL (psf) 72.0 15.0

Wi (kips) 2,094 141 2,235

Area (sf) 29,090 9,396

X cg

Ycg

(ft) 100 100

(ft) 70 70

( )

( )

W X cg

W Ycg

209,448 14,094 223,542

146,614 9,866 156,479

X cg = 223 ,542 2,235 = 100.0ft ; Ycg = 156,479 2,235 = 70.0ft

Table 3A-2. Mass properties summary Level

W (kips)

X cg

Ycg

(ft)

(ft)

Roof

2,066

100

70

(1)

M (2)

MMI (3)

5.3

26,556

4th

2,235

100

70

5.8

28,728

3rd

2,235

100

70

5.8

28,728

2nd

2,235

100

70

5.8

28,728

Total 8,771 22.7 Notes: 1. Mass (M) and mass moment of inertia (MMI) are used in analysis for determination of fundamental period (T). 2. M = (W/386.4) (kips-sec2/in.) 3. MMI = M/A (lx + ly) (kips-sec2-in.)


153

Design Example 3A

2b.

!



As noted above, Equation (30-4) governs, and: V = 0.082W = 0.082(8,771) = 720 kips

2c.


§1630.5

For the static lateral force procedure, vertical distribution of force to each level is applied as follows: V = Ft + ∑ Fi where Ft = 0.07T (V ) ≤ 0.25(V )

(30-13)

Except Ft = 0 where T ≤ 0.7 sec

(30-14)

For this structure: T = 0.92 sec ∴ Ft = 0.07(0.92)(720) = 46.4 kips The concentrated force Ft is applied at the roof, in addition to that portion of the balance of the base shear distributed to each level per §1630.5: Fx =

(V − Ft )Wx hx = (673.6 )  

W x hx  Wi hi

∑ Wi hi

∑

   

(30-15)

Table 3A-3. Vertical distribution of shear Level

wx (kips)

hx (ft)

w x hx (k-ft)

w x hx Σwx

Fx (kips)

Roof

2,066

55.5

114,663

0.375

299.0

4th

2,235

42.0

93,870

0.307

206.8

ΣV (kips) 299.0

3rd

2,235

28.5

63,698

0.208

140.3

505.8

2nd

2,235

15.0

33,525

0.110

73.9

646.1

Total

8,771

305,756

1.000

720.0

720.0

Note: Froof = 0.38 (673.6) + 46.4 = 299.0 kips

154


Design Example 3A

2d.

!


Determine horizontal distribution of shear.

§1630.6

Structures with concrete fill floor decks are typically assumed to have rigid diaphragms. Seismic forces are distributed to the moment frames according to their relative rigidities. For structures with assumed rigid diaphragms, an accidental torsion must be applied (in addition to any actual torsional moment) equal to that caused by displacing the center of mass 5 percent of the building dimension perpendicular to the direction of the applied lateral force. For the structural computer model of this Design Example, this can be achieved by combining the direct seismic force applied at the center of mass at each level with a torsional moment at each level: North-south: M t = 0.05(204 ft )FX = (10.2 )FX East-west: M t = 0.05(144 ft )FX = (7.2)FX

Table 3A-4. Horizontal distribution of shear Level

Fx (kips)

N/S M t (k-ft)

E/W M t (k-ft)

Roof

299.0

3,050

2,153

4th

206.8

2,109

1,489

3rd

140.3

1,431

1,010

2nd

73.9

754

532

Note: Mt = horizontal torsional moment

Using the direct seismic forces and torsional moments noted above, the force distribution to the frames is generated by computer analysis. The torsional seismic component is always additive to the direct seismic force. For the computer model, member sizes are initially proportioned by extrapolation from the tested configurations for SMRF reduced beam section joints, as discussed in Part 6 below. From the preliminary computer analysis, the shear force at the ground level is determined for each frame column. As shown in Figure 3A-5, there are a total of six rigid frames: A1, A2, B1, B2, B3, and B4. Frames A1 and A2 are identical. Frames B1, B2, B3, and B4 are also identical. Recognizing that the building is symmetrical, the frame forces are the same for Frames A1 and A2, as well as for Frames B1 through B4. Frame forces at the base of each frame type, A1 and B1 are summarized in Tables 3A-5 and 3A-6.


155

Design Example 3A

!


Table 3A-5. North-south direction, frame type A1 Column Shears (kips)

Line A/1.2 (kips)

Line A/2 (kips)

Line A/3 (kips)

Line A/4 (kips)

Line A/5 (kips)

Line A/5.8 (kips)

Total (kips)

Direct Seismic

41.8

75.2

69.7

69.7

75.2

41.8

373.4

Torsion Force

2.6

4.6

4.3

4.3

4.6

2.6

23.0

Direct + Torsion

44.4

79.8

74.0

74.0

79.8

44.4

396.4

Line 1/B (kips)

Line 1/C (kips)

Line 1/C.8 (kips)

Total (kips)

Table 3A-6. East-west direction, frame type B1 Column Shears (kips)

Line 1/A.2 (kips)

Direct Seismic

33.4

59.9

59.9

33.4

186.6

Torsion Force

1.3

2.3

2.3

1.3

7.2

Direct + Torsion

34.7

62.2

62.2

34.7

193.8

As a check on the computer output, compare the total column shears with the direct seismic base shear of 720 kips: North-south: ΣFtype A = 2(373.4 ) = 746.8 > 720 kips

o.k.

East-west: ΣFtype B = 4(186.6 ) = 746.4 > 720 kips

o.k.

The summation of the column shears is about 3 percent greater than the design base shear input to the computer model. This is mostly due to the inclusion of P∆ effects in the computer analysis. As required by §1630.1.3, P∆ effects are to be considered when the ratio of secondary (i.e., moment due to P∆ effects) to primary moments exceeds 10 percent. Next, to refine the initial approximation for rmax and ρ , the actual column shears for Frame A1 from Table 3A-5 above will be used. rmax = 0.7(79.8 + 74.0 ) / 747 = 0.144 ∴ ρ=2−

156

20 0.144(33,311)1 / 2

= 1.24 ≈ 1.25

o.k.


Design Example 3A

3. 3a.

!


Interstory drift.

Determine ∆S and ∆M.

The design level response displacement ∆ S is the story displacement at the center of mass. It is obtained from a static-elastic analysis using the design seismic forces derived above. For purposes of displacement determination, however, §1630.10.3 eliminates the upper limit on TB , used to determine base shear under Equation (30-4). The maximum inelastic response displacement ∆ M includes both elastic and estimated inelastic drifts resulting from the design basis ground motion. It is computed as follows: ∆ M = 0.7(R )∆ S = 0.7(8.5)∆ S = 5.95∆ S

(30-17)

The maximum values for ∆ S and ∆ M are determined, including torsional effects (and including P∆ effects for ∆ M ). Without the 1.3T A limit on TB , the design base shear per Equation (30-4) is: North-south: TBy = 1.30 sec Vn / s =

Cv I 0.64(1.0 ) W = W = 0.058W = 509 kips 8.5(1.30) RT

(30-4)

East-west: TBx = 1.16 sec Ve / w =

Cv I 0.64(1.0 ) W = W = 0.064W = 561 kips 8.5(1.16 ) RT

Note that §1630.9.1 and §1630.1.1 require use of the unfactored base shear V, with ρ = 1 . Using these modified design base shears, the accidental torsion and force distribution to each level are adjusted for input to the computer model. The structure displacements and drift ratios are derived as shown below in Table 3A-7.


157

Design Example 3A

!


Table 3A-7. Interstory displacements North-South Interstory Displacements Story

Height h (in.)

∆ S Drift (in.)

∆ M Drift (in.)

Drift Ratio (∆ M h )

4th 3rd 2nd 1st

162 162 162 180

(1.36 - 1.16)= 0.20 (1.16 - 0.85)= 0.31 (0.85 - 0.47)= 0.38 (0.47 - 0.0) = 0.47

1.19 1.84 2.26 2.80

0.0073 0.0114 0.0140 0.0156

East-West Interstory Displacements Story

Height h (in.)

∆ S Drift (in.)

∆ M Drift (in.)


4th 3rd 2nd 1st

162 162 162 180

(1.17 - 1.01)= 0.16 (1.01 - 0.73)= 0.27 (0.73 - 0.40)= 0.33 (0.40 - 0.0) = 0.40

0.95 1.61 1.96 2.38

0.0059 0.0099 0.0121 0.0132

Note: Interstory drift ratio = ∆M/story height.

3b.

Determine the story drift limitation.

§1630.10

For structures with T > 0.7 , the allowable story drift is: ∆ M = 0.020 (story height). A review of drift ratios tabulated in Table 3A-7 shows that all interstory drift ratios are less than 0.020, using seismic forces corresponding to the actual period TB in base shear Equation (30-4). Also, note that all drift ratios are less than (0.95)(0.020 ) = 0.019 . This 5 percent reduction in the drift limit is required for reduced beam section joint designs under FEMA-267A. Looking ahead to the SMRF member design, §2213.7.10 imposes certain conditions on moment frame drift calculations, including bending and shear contributions from clear beam-column spans, column axial deformation, and panel zone distortion. These conditions are met by most general purpose structural analysis programs used in building design, except for the contribution to frame drift from panel zone distortion. The code provides an exception whereby a centerline analysis may be used if the column panel zone strength can develop 80 percent (0.8ΣM s ) of the strength of the girders framing into the joint. As will be seen from the SMRF beam-column joint design, this condition will always be met under the current performance criteria. Moreover, the FEMA-267A provisions produce stronger, stiffer column panel zone designs than previously permitted by the UBC. Therefore, panel zone distortion will generally not contribute significantly to overall frame drift.

158


Design Example 3A

!


To gain a feel for the influence of beam-column joint stiffness on overall frame drift, two conditions are modeled for east-west seismic forces, with the lateral displacements at the roof derived as follows: Centerline analysis: 1.37 inches 50 percent rigid joint analysis: 1.17 inches The centerline analysis produces a displacement 17 percent greater than the 50 percent rigid joint analysis. Most engineers feel that the centerline analysis over-estimates, and the 100 percent rigid joint underestimates, the actual frame drift. The 50 percent rigid joint analysis is an accepted standard of practice for providing reasonable design solutions for frame displacements.

4. 4a.


§1633.2.9

Determine diaphragm load distribution.

In multi-story buildings, diaphragm forces F px are determined by the formula: F px =

Ft + ∑ Fi

∑ wi

(w px )

and 0.5C a IW px < F px ≤ 1.0C a IW px

(33-1)

The diaphragm forces at each level, with the upper and lower limits, are calculated as shown in Table 3A-8 below. Note that the 0.5C a IW px minimum controls for this building.

Table 3A-8. Diaphragm load distribution Level

Fi (1) (kips)

ΣFi (kips)

wx (kips) (1)

Σw i (kips)

FPx (kips) (2)

0.5Ca Iw Px (kips) (3)

1.0Ca Iw Px (kips) (3)

Roof 4th 3rd 2nd

299.0 206.8 140.3 73.9

299.0 505.8 646.1 720.0

2,066 2,235 2,235 2,235

2,066 4,301 6,536 8,771

299.0 262.8 220.9 183.5

454.5 491.7 491.7 491.7

909.0 983.4 983.4 983.4

Notes: 1. See Table 3A-3. 2. Ft = 46.4 kips (see Part 2c) 3. Ca = 0.44 kips (see Part 1d)


159

Design Example 3A

4b.

!


Determine diaphragm shear.

(

The diaphragm design is governed by the minimum seismic force 0.5C a I pW p

)

and the 491.7 kip force at the floor levels is used for design. This value is not factored up by ρ per §1630.1.1. The reliability/redundancy factor ρ is only applied to transfer diaphragms (see Blue Book §105.1.1). ∴ E floor = FP = 491.7 kips

(30-1)

The maximum diaphragm span occurs between Lines A and H, so the north-south direction will control. Although the computer model assumes rigid diaphragms for load distribution to the frames, we now consider the diaphragm as a horizontal beam. Shears at each line of resistance are derived assuming the diaphragm spans as simple beams under a uniform load.

w1 = E floor / (200′) = 491.7 / (200) = 2.46 k/ft Diaphragm shear:  200  VA = VH = 2.46   = 246 kips  2 

Figure 3A-6. Diaphragm shear

160


Design Example 3A

!


Using the alternate basic load combination of Equation (12-13) for allowable stress design, the factored diaphragm design shear at Line A is (E/1.4): qA =

(V ) =

1.4

246 = 1.25 k-ft 1.4(140')

Using 3¼-inch light weight concrete over 3"× 20 gauge deck, with 4 welds per sheet at end laps and button punch at 12 in. side laps, the allowable deck shear per the manufacturer’s ICBO Evaluation Report is: Vallow = 1.75 > 1.25 k-ft

4c.

o.k.

Determine collector and chord forces.

Assuming the diaphragm acts as a simple beam between Lines A and H (and this is the usual assumption), the maximum chord force at Lines 1.2 and 5.8 for northsouth seismic is: CF =

2.46(200)2 = 100.0 k 8(123)

Because the beam framing is continuous on Lines 1.2 and 5.8, these lines are chosen to resist the chord force. [Lines 1 and 6 have indentations in the floor plan (Figure 3A-2).] The chord force must be compared to the collector force at these lines, and the greatest value used for design. For east-west seismic loads, the factored shear flow at Line 1.2 is approximately: q1.2 =

491.7 = 1.23 k-ft (2)(200')

Figure 3A-7 shows the collector force diaphragm for Line 1.2.

Figure 3A-7. Collector force diagram at Line 1.2


161

Design Example 3A

!


The maximum collector force is: Fa = Fd = 1.23(8.5) = 10.5 kips Fb = Fc = 1.23(75.5) − 123 = 30.1 kips Per §1633.2.6, seismic collectors must be designed for the special seismic load combinations of §1612.4. Note that the value for E M does not include the ρ factor. E M = Ω o (FP ) = 2.8(30.1) = 84.3 kips

(30-2)

The seismic drag tie or chord can be implemented using supplemental slab reinforcing. With the strength design method for concrete per §1612, including Exception 2, the factored collector and chord forces are: Factored chord force: Tu = 1.1(E ) = 1.1(100.0 ) = 110.0 kips

§1612.4

Factored collector force: Tu = 1.0(E M ) = 1.0(84.3) = 92.7 kips

(12-17)

The factored chord forces for north-south seismic loads govern the design at Line 1.2. The required slab chord reinforcing is calculated as: Required As = Tu φf y = 110.0 0.9(60) = 2.0 in.2 ∴ Use 4-#7, As = 2.4 in.2

5.

SMRF member design.

In this Part, representative beam and column members of Frame A1 are designed under the provisions of §2213.7. Certain provisions of §2213.7 pertaining to joint design have been modified by the recommendations of FEMA-267A. These provisions, including the strong column-weak beam and panel zone requirements, are discussed with the RBS joint design in Part 6 of this Design Example. From past experience, steel moment frame designs have typically been drift controlled. Frame members were chosen with sufficient stiffness to meet the drift limits, and then checked for the SMRF design requirements. However, to meet the intent of §2213.7.1, the design process begins by selecting beam-column combinations extrapolated from tested RBS joint assemblies. The rationale for selection of the member sizes is also presented in Part 6, with a W 30 × 108 beam and W 14 × 283 column chosen for this Design Example.

162


Design Example 3A

5a.

!


rd

Design typical beam at 3 floor.

The typical beam selected to illustrate beam design is a third-floor beam in Frame A1. This is shown in Figure 3A-8 below.

Figure 3A-8. Typical beam at third floor of Frame A1.

From a review of the computer output prepared separately for this Design Example, the moments and shears at the right end of the beam are greatest. The moments and shears at the face of the column at Line 5 are: M DL = 1,042 kip-in. M LL = 924 kip-in. M seis = ± 3,590 kip-in. M E = ρM seis = 1.25(3,590 ) = ± 4,487 kip-in. V DL = 16.4 kips V LL = 13.3 kips Vseis = ± 22.3 kips VE = ρVseis = 1.25(22.3) = ± 27.9 kips


§1630.1.1

163

Design Example 3A

!


The basic load combinations of §1612.3.1 (ASD) are used, with no one-third increase. (These were selected to illustrate their usage, although generally it is more advantageous to use the alternate basic load combinations of §1612.3.2.) D + L : M D +L = 1,042 + 924 = 1,966 kip-in.

(12-8)

V D +L = 16.4 + 13.3 = 29.7 kips D+

E 4,487 : M D +E = 1,042 + = 4,247 kip-in. 1.4 1.4 VD + E = 16.4 +

(12-9)

27.9 = 36.3 kips 1.4

   E   4,487   D + 0.75 L +   : M D + L + E = 1,042 + 0.75924 +    = 4,139 kip-in.  1.4   1.4    

(12-11)

  27.9  VD + L + E = 16.4 + 0.7513.3 +   = 41.3 kips  1.4  

Try W 30 × 108 , ASTM A572, Grade 50 beam. Check flange and web width-thickness ratios per §2213.7.3 (flange and web compactness criteria to mitigate premature formation of local buckling): bf 2t f

≤

52 50

= 7.35

For W 30 × 108 : and

bf 2t f

and

d 640 ≤ = 90.5 tw 50

= 6.9 < 7.35

d 29.83 = = 54.7 < 90.5 t w 0.545

o.k. o.k.

Check the beam bracing requirements of §2213.7.8: Maximum brace spacing = 96ry = 96 (2.15) 12 = 17.2 ft Place minimum bracing at one-third points: L = 96 28.0 3 = 9.33 ft on center

164


Design Example 3A

!


Check allowable moment capacity: From AISC-ASD (p. 2-10) for W 30 × 108 :

( )

Lu = 9.8 > 9.33 ∴ Fb = 0.60 F y = 30.0 ksi Allowable M a = 299(30.0 ) = 8,970 kip-in. > 4,247 kip-in.

o.k.

Check allowable shear capacity: For W 30 × 108 :

( )

∴ Fv = 0.4 F y

h 29.83 − 2(0.76 ) 380 = = 51.9 < = 53.7 tw 0.545 50 = 0.4(50 ) = 20.0 ksi

Allowable Va = 20.0(0.545)(29.83) = 325 kips > 41.3 kips

o.k.

∴ Use W 30 × 108 beam Note: The W 30 × 108 beam is much larger than required by allowable stress considerations. The reason for this is that this shape has been part of the beamcolumn assemblies tested with RBS configurations.

5b.

nd

Design typical column at 2

story.

The column to be designed is the second-story column of Frame A1 shown in Figure 3A-9. For the second-story column at Line 5, the maximum column forces generated by the frame analysis (not shown) are: M DL = 236 kip-in. M LL = 201 kip-in. M seis = 3,970 kip-in. M E = 1.25(3,970) = 4,963 kip-in. V DL = 3.1 kips V LL = 2.7 kips Vseis = 56.8 kips


165

Design Example 3A

!


V E = 1.25(56.8) = 71 kips PDL = 113 kips PLL = 75 kips Pseis = 28 kips PE = 1.25(28) = 35 kips The maximum strong axis moments occur at the bottom of the column, and are taken at the top flange of the second-floor beam.

Figure 3A-9. Typical second story column at Frame A1

Using the basic load combinations of §1612.3.1: D + L: M D + L = 236 + 201 = 437 kip-in.

(12-8)

PD + L = 113 + 75 = 188 kips VD + L = 3.1 + 2.7 = 5.8 kips

166


Design Example 3A

D+

!


4,963 E : M D + E = 236 + = 3,781 kip-in. 1.4 1.4

(12-9)

35 = 138 kips 1.4 71.0 = 3.1 + = 53.8 kips 1.4

PD + E = 113 + VD + E 0.9 D −

E 35 : PD − E = 0.9(113) − = 76.7 kips compression 1.4 1.4

   4,963   E  D + 0.75 L +   = 3,046 kip-in.  : M D + L + E = 236 + 0.75201 +   1.4   1.4   

(12-10)

(12-11)

71.0   = 43 kips VD + L + E = 3.1 + 0.75 2.7 + 1.4     35  PD + L + E = 113 + 0.75 75 +   = 188 kips  1.4   Under the requirements of §2213.5.1, columns must have the strength to resist the following axial load combinations (neglecting flexure): PDL + 0.7 PLL + ΩPseis : Pcomp = 113 + 0.7(75) + 2.8(28) = 244 kips compression 0.85PDL − ΩPseis :

Ptens = 0.85(113) − 2.8(28) = −18 kips compression

The intent of these supplemental load combinations is to ensure that the columns have adequate axial strength to preclude buckling when subjected to the maximum seismic force that can be developed in the structure. Try W 14 × 283 , ASTM A572, Grade 50 column. Unbraced column height (taken from top of framing at bottom to mid-depth of beam at top): h = 13.5 − (2.5 2 ) = 12.25 ft


167

Design Example 3A

!


Under §2213.5.3, the factor k can be taken as unity if the column is continuous, drift ratios are met per §1630.8, and f a ≤ 0.4 F y . The example column is

( )

continuous, complies with the drift ratios, and: Maximum f a = 188 / 83.3 = 2.26 ksi < 0.4(50) = 20.0 ksi ∴ k = 1.0 12(12.25)  kl  = 21.6   = 6.79  r x 12(12.25)  kl  = 35.3   = 4.17  r y ∴ Fa = 26.5 ksi Maximum

f a 2.26 = = 0.085 < 0.15 Fa 26.5

Therefore, AISC-ASD Equation H1-3 is used for combined stresses. From AISC-ASD manual (p. 3-21) for W 14 × 283 , Grade 50: Lc = 14.4 > 12.5

( )

∴ Fb = 0.66 Fy = 33.0 ksi Check combined stresses for the critical load combinations. D+

f f E 138 3,781 : a + bx = + = 0.063 + 0.250 = 0.313 < 1.0 o.k. 1.4 Fa Fb 83.3(26.5) 459(33.0 )

3,046 E  f a f bx  + = 0.085 + = 0.286 < 1.0 D + 0.75 L + :  459(33.0 ) 1.4  Fa Fb 

o.k.

(12-9)

(12-11)

Check column shear capacity: Allowable Va = 0.4(50)(16.74)(1.29) = 432 kips > 53.8 kips

168

o.k.


Design Example 3A

!


Next, check required axial strength per §2213.5. Compression: Psc = 1.7 Pallow = 1.7 (83.3)(26.5) = 3,753 kips > 244 kips

o.k.

Tension: Pst = F y A = 50(83.3) = 4,165 kips

o.k.

The column width-thickness ratio limit of §2213.7.3 is to meet the requirement of AISC-ASD, Chapter N, Plastic Design, Section N7. Columns meeting this criterion are expected to achieve full plastic capacity prior to local flange buckling. bf 2t f

≤ 7.0 for F y = 50 ksi

For W 14 × 283 :

bf 2t f

= 3.89 < 7.0

o.k.

∴ Use W 14 × 283 column Note: The W 14 × 283 column is much larger than required by allowable stress considerations. The beam-column assemblies selected for this Design Example have been tested with the RBS configuration.

6.

SMRF beam-column connection design.

As discussed in FEMA-267 (Sections 7.3 and 7.5), SMRF joint designs may be acceptable without testing of a particular beam-column combination only with the following qualifications: 1. Joint design calculations are based on comparisons with tested assemblies. 2. The joint configuration considered closely mirror the tested detail. 3. Calculated member sizes are extrapolated from tested combinations. 4. A qualified third party peer review is performed.

Where such calculations are determined to be acceptable, the design provisions of FEMA-267A may be applied to member sizes extrapolated or interpolated from tested configurations. Use of calculations alone, without testing to form a basis for reasonable extrapolation, is not recommended.


169

Design Example 3A

!


This Design Example utilizes tests conducted at the University of Texas Ferguson Laboratory [Engelhardt et al., 1996]. Testing of additional RBS joint combinations was performed as part of the SAC Phase II program. Results of these tests will be published by SAC when available; updates may be found at SAC’s web site: http://quiver.eerc.berkeley.edu:8080/design/conndbase/index.html. Using the circular cut reduced beam section, the following beam-column joint assemblies were successfully tested at the University of Texas:

Table 3A-9. Tested RBS beam-column joint assemblies Specimen

Column

Beam

DB2 DB3 DB4 DB5

W14x426 W14x426 W14x426 W14x257

W36x150 W36x170 W36x194 W30x148

Each of these specimens achieved plastic chord rotation capacity exceeding 0.03 radians, the recommended acceptance criterion per FEMA-267A (Section 7.2.4). The parameters for extrapolation or interpolation of beam-column test results are difficult to determine. When extrapolating, it should be done only with a basic understanding of the behavior of the tested assembly. The California Division of the State Architect (DSA), in the commentary to its Interpretation of Regulations 27-8 (DSA IR 27-8), has established guidelines for extrapolation of joint tests. Until further testing is completed, DSA recommends that members sizes taken from tested configurations be extrapolated, by weight or flange thickness, no more than 15 percent upward or no more than 35 percent downward. Using the DSA criteria for extrapolation with the lightest column section (DB5) of the tested sizes noted above, the following possible beam-column size combinations are possible: W 14 × 257 column: Max. weight = 296 lbs.

Max. t f = 2.17 in.

Min. weight = 167 lbs.

Min. t f = 1.22 in.

∴ Use W 14 × 176 to W 14 × 283

170


Design Example 3A

!


W 30 × 148 beam: Max. weight = 170 lbs. Min. weight = 96 lbs.

Max. t f = 1.36 in. Min. t f = 0.77 in.

∴ Use W 30 × 108 to W 30 × 173 For compatibility with this test configuration, beam-column pairs are selected from the ranges noted above. After evaluating several combinations for weak beam/strong column and panel zone strength criteria, the combination of a W 30 × 108 beam and W 14 × 283 column is selected for use in this Design Example. Note that this combines the lightest beam with the heaviest column in the available range. The W 30 × 108 beam was selected after confirming that with this combination, the overall frame drifts per the computer analysis are within the code limits (as shown in Part 3b above). The W 14 × 283 column was chosen to eliminate the requirement for doubler plates. When given the option, steel fabricators have elected to use heavier columns in lieu of doubler plates for economy. Also, tests have shown that the weld of the doubler plate to the column fillet (k) region may be detrimental to joint performance. As shown in Figure 3A-10, the W 14 × 283 columns are to be full-height, one length. Full-height columns without splices were found to be the least-cost option. Column splices in SMRFs must comply with §2213.5.2. However, it is suggested that column splices be made with complete penetration welds located near midheight. Note: Where referenced, the FEMA-267/267A sections are noted with a preceding “FEMA” in the remainder of this Design Example (e.g. FEMA §7.2.2.1). The reduced beam section (RBS) joint configuration used in this Design Example is shown in Figure 3A-10.

6a.

Determine member and material strengths.

When determining the strength of a frame element, FEMA §7.2.2 defaults back to §2213.4.2. Material strength properties are stipulated in FEMA §7.5.1, Table 7.5.1-1. FEMA-267A modified the allowable through-thickness stress to 0.9 (Fy) in recognition of improved joint performance for configurations locating the plastic hinge away from the face of the column. For this Design Example, material strengths are taken as:


171

Design Example 3A

!


W 30 × 108 beam, Shape Group 2, A572 Grade 50: F y = 50 ksi F ym = 58 ksi Fu = 65 ksi Through-thickness FTT = 0.9(50 ) = 45 ksi W 14 × 283 column, Shape Group 4, A572 Grade 50: F y = 50 ksi F ym = 57 ksi Fu = 65 ksi Through-thickness FTT = 45 ksi

6b.

Establish plastic hinge configuration and location.

The fundamental design intent espoused in FEMA-267 is to move the plastic hinge away from the column face. The RBS design achieves that goal in providing a well-defined, relatively predictable plastic hinge region. Of the various RBS options, the circular curved configuration is chosen due to its combination of tested performance and economy of fabrication. The distance c from the face of the column (see Figure 3A-10) to the beginning of the circular cut, and the length of the cut l c , are based on prior RBS tests. It is desirable to minimize c to reduce the amplification of M f at the face of the column. FEMA-267A recommends that c = d b / 4 , while Englehardt [1998] recommends 0.5b f ≤ c ≤ 0.75b f . As the member sizes for this Design Example are extrapolated from testing by Englehardt, c ≅ 0.6b f is selected. Both FEMA-267A and Englehardt recommend l c ≅ 0.75d .

172


Design Example 3A

!


Figure 3A-10. RBS (“dog bone”) geometry

W 30 × 108 : 0.5b f = 0.5(10.5) = 5.25 in. 0.75b f = 0.75(10.5) = 7.88 in. ∴ Use c = 6.0 in. lc = 0.75d = 0.75(29.83) = 22.37 in. ∴ Use lc = 24.0 in. The depth of the cut n should be made such that 40 percent to 50 percent of the flange is removed. This will limit the projection of moments at the face of the column to within 90 percent to 100 percent of the plastic capacity of the full beam section. With a 45 percent reduction in the flange area:  bf n = 0.45  2

 0.45(10.5)  = = 2.36 in. 2 

Use 2 ¼-in. cut 4n 2 + l c2 4(2.25)2 + 24 2 ∴R = = = 33.1 in. radius 8n 8(2.25) The plastic hinge may be assumed to occur at the center of the curved cut per FEMA §7.5.3.1, so that: lh = (16.74 / 2) + 6.0 + (24 / 2) = 26.37 in.


173

Design Example 3A

!


and: L = 28.0 ft. ∴ L' = 28 − 2(26.37 / 12 ) = 23.6 ft The length between the plastic hinges L ' (see Figure 3A-11) is used to determine forces at the critical sections for joint analysis.

Figure 3A-11. Plastic hinges

The circular curved cut provides for a gradual transition in beam flange area. This configuration also satisfies the intent of §2213.7.9.

6c.

Determine probable plastic moment and shear at the reduced beam section.

The plastic section modulus at the center of the reduced beam section is calculated per FEMA §7.5.3.2 as:

[ (

Z RBS = Z x − br t f d − t f

)]

FEMA-267A, Eqn. (7.5.3.2-1)

where b r is the total width of material cut from the beam flange. br = 2(2.25) = 4.5 in. and Z RBS = 346 − [4.5(0.76 )(29.83 − 0.76)] = 247 in.3

174


Design Example 3A

!


Next, the probable plastic moment at the reduced beam section Mpr is calculated as: M pr = Z RBS β(FY )

FEMA-267A, Eqn. (7.5.3.2-2)

The factor β accounts for both variations in the beam steel average yield stress and strain hardening at the plastic hinge. Per FEMA §7.5.2.2, for ASTM A572 steel, β = 1 .2 . Therefore: M pr = 247(1.2)(50 ) = 14,820 kip-in. As illustrated in FEMA §7.5.2.3, the shear at the plastic hinge is derived by statics, considering both the plastic moment at the hinge and gravity loads. For simplicity, the beam shear from the frame analysis for dead and live loads at the hinge is used. To be consistent with this strength design procedure, the special seismic load combinations of §1612.4 are used:

L’

Mpr

Mpr VE

VE

Figure 3A-12. Beam equilibrium under the probable plastic moment Mpr

VE =

2 M pr L'

=

2(14,820 ) = 104.7 kips 12(23.6)

and: V P = 1.2(V D ) + 0.5(V L ) + 1.0(V E ) ∴ V P = 1.2(16.4 ) + 0.5(13.3) + 1.0(104.7 ) = 131 kips

6d.

Calculate strength demands at the critical sections of beam-column joint.

There are two critical sections for the joint evaluation. The first section is at the interface of the beam section and the face of the column flange. The strength demand at this section is used to check the capacity of the beam flange weld to the column, the through-thickness stress on the column flange (at the area joined to the beam flange), and the column panel zone shear strength. The second critical section occurs at the column centerline. The moment demand at this location is SEAOC Seismic Design Manual, Vol. III (1997 UBC)

175

Design Example 3A

!


used to check the strong column-weak beam requirement per FEMA §7.5.2.5 (UBC §2213.7.5).

a. Column face

b. Column centerline

Figure 3A-13. Critical sections at beam-column joint

The moment at the face of the column is: M f = M pr + V P (x ) = 14,820 + 131(6 + 12) = 17,178 kip-in.

FEMA §7.5.2.4

The moment at the centerline of the column is: M cl = M pr + VP (lh ) = 14,820 + 131 (26.37 ) = 18,274 kip-in.

6e.

Evaluate the RBS joint strength capacity.

Section 7.5.3.2 of FEMA-267A lists four criteria for the evaluation of RBS joint capacity: 1. At the reduced section, the beam must have the capacity to meet all code required forces (i.e. dead, live & seismic per §1612).

176

2.

Code required drift limits must be met considering effects of the RBS.

3.

The beam-to-column flange weld must have adequate strength.

4.

The through-thickness stress on the face of the column at the beam flange must be within the allowable values listed in FEMA §7.5.1. (Note: In SEAOC Seismic Design Manual, Vol. III (1997 UBC)

Design Example 3A

!


subsequent studies conducted by the SAC project, typical rolled column shapes were found insensitive to through thickness stress. In FEMA-350, the requirement to check this parameter has been eliminated, and the connection is designed to produce near-yield conditions at the beam flange to column joint.) Check reduced section for code design forces.

At the reduced section, the section modulus S RBS is: S RBS =

[4,470 − 2(4.5)(0.76)(14.92 − 0.78) ] = 203in. 2

3

14.92

The allowable moment M a , with Fb = 33.0 ksi (see Part 5a), is M a = 203(33.0 ) = 6,700 kip-in. > 4,247 kip-in. Thus, the reduced W 30 × 108 section is adequate for the moments derived for the load combinations of §1612.3.1. Check frame stiffness for code drift limits.

As discussed in FEMA §7.5.3, the RBS will reduce overall frame stiffness approximately 5 percent, thereby increasing calculated frame displacements about 5 percent proportionally. To account for this increase, the allowable drift limits are reduced 5 percent for comparison to calculated frame lateral deflections from the computer analysis. As shown in Part 3b, the structure drift ratios are found to be within the reduced code limits. Check beam-to-column welded connection.

The W 30 × 108 beam and W 14 × 283 column are extrapolated from specimen sizes tested in an RBS configuration at the University of Texas. In the tested configuration, the beam webs have complete-penetration welds to the column flange. Under FEMA §7.8.2, the web connection should be consistent with the tested assemblies—this weld is shown in Figure 3A-17. Note: In FEMA-350, RBS and other connections have been prequalified for application within ranges of member and frame sizes. As long as framing falls within prequalified limits, reference to specific test data is not required. Using the cross-sectional area of the beam flange and web weldments at the face of the column (Figure 3A-14), the elastic section modulus S c of the beam is calculated from the information in Table 3A-10.


177

Design Example 3A

!


Figure 3A-14. Built-up section at column face

Table 3A-10. Built-up section properties Area (in.2)

Mk 1 2 3 4 5

0.545(26.73)=14.58 0.76(10.48)=7.96 0.76(10.48)=7.96 0.31(10.48)=3.28 0.31(10.48)=3.28

Y (in.) 0.00 14.54 14.54 15.07 15.07

Sum

A(y)2

Io (in.4)

0 1,682 1,682 745 745

869 0 0 0 0

4,854

869

The calculated section properties are: I c = 4,854 + 869 = 5,723 in.4 ∴ S c = 5,723 15.23 = 376 in.3 As given in FEMA §7.2.2.1, for complete penetration welds, the weld strength is taken at the beam yield stress of 50 ksi. The maximum weld stress is calculated using Mf (see Figure 3A-11). The moment demand on the weld at the face of the column: f weld = 17,178 / 376 = 45.7 ksi < 50 ksi

o.k.

With the beam web welded to the column, the plastic shear demand should be checked against the beam shear strength. The plastic shear demand is calculated in Part 6b above.

178


Design Example 3A

!


VS = 0.55 (50.0)(0.545)(29.83) = 447 kips > V p = 131 kips

o.k.

FEMA §7.8.2

In this Design Example, the shear tab shown in Figure 3A-17 is present only for steel erection. For beam web connections using shear tabs, the shear tab and bolts are to be designed to resist the plastic beam shear Vp. The bolts must be slipcritical, and the shear tab may require a complete penetration weld to the column. However, in September 1994, ICBO issued an emergency code change to the 1994 UBC, which deleted the prior requirement for supplemental welds from the shear tab to the beam web. An example beam-column shear tab connection design is given in Design Example 1A, Part 6g. Check the through-thickness stress at the column.

Under FEMA §7.5.3.2, the through-thickness stresses at the interface of the beam flange with the column face is determined as f t −t = M f Sc

FEMA §7.5.3.2

where M f and S c are as determined above. ∴ f t −t = 17,178 / 376 = 45.7 ksi ≈ 0.9(50) = 45.0 ksi

o.k.

Although the through-thickness stress is at the upper limit of the recommended allowable stress, RBS joints have been successfully tested with calculated stresses as high as 58 ksi [Englehardt, et al., 1996]. The success of these tests is attributed to locating the plastic hinge away from the column face and into the beam span.

6f.

Verify the strong column-weak beam condition.

The strong column/weak beam requirement given in FEMA §7.5.2.5 is similar to §2213.7.5. Per FEMA §7.5.2.5 the beam moments are derived from M pr (see Part 6c above), whereas the UBC sums moments at the column centerline. The column moments ΣM c are taken at the top and bottom of the column panel zone as shown in Figure 3A-15.


179

Design Example 3A

!


Figure 3A-15. Joint forces and moments

(

ΣZ C F yc − f a

)

ΣM C

≥ 1.0

FEMA Eqn. (7.5.2.5-1)

where:

(

M Ct = VC ht ;

)

M Cb = VC + V f hb

and: ΣM C = M Ct + M Cb V f is the incremental seismic shear to the column at the 3rd floor. From the computer analysis (not shown): V f = 16.4 kips Summing moments at the bottom of the lower column: VC =

[

( )]

2 M pr + lh V p − V f (hb + d P / 2)

(hb + d P + ht )

ht = hb = ∴ VC =

180

(13.5)(12) − 29.83 = 66.1 in. 2

2

2[14,820 + 26.4(131)] − 16.4(66.1 + 29.83 / 2 ) = 217.4 kips [2(66.1) + 29.83]


Design Example 3A

!


The column moments, taken at the top and bottom of the panel zone are: M Ct = 217.4(66.1) = 14,370 kip-in. M Cb = (217.4 + 16.4 )(66.1) = 15,454 kip-in. ∴ M C = 14,370 + 15,454 = 29,834 kip-in. From Part 5b above, the maximum column axial stress is f a = 2.26 ksi . For the W 14 × 283 column, Z x = 542 in.3 :

(

ΣZ C Fyc − f a ΣM C

) = 2[542(50 − 2.260)] = 1.74 > 1.0 29,824

o.k.

FEMA Eqn. (7.5.2.5-1)

Therefore, the columns are stronger than the beam moments 2 M pr , and the strong column-weak beam criteria is satisfied.

6g.

Check column panel zone strength.

Column panel zone strength is evaluated per FEMA §7.5.2.6. FEMA-267A modifies the panel zone provisions of UBC §2213.7.2. The provision (in the 1994 UBC) allowing panel zone strength to be proportioned for “…. gravity loads plus 1.85 times the prescribed seismic forces …” has been eliminated. This modification produces stiffer/stronger panel zones than previously permitted under the UBC. Heavier columns are often preferable to use of doubler plates. Thus, panel zone strength may well dictate the selection of column sizes. (Note: In FEMA-350, this criteria has changed again to produce balanced yielding between the beam and panel zone, such that yielding initiates in the panel zone simultaneously—or slightly after—yielding in the RBS. This is compatible with, but not identical to, the FEMA-267 procedures.) Per FEMA §7.5.2.6, the panel zone (Figure 3A-16) is to be capable of resisting the shear required to develop 0.8ΣM f of the girders framing into the joint (where Mf

(

)

is the moment at the face of the column). The panel zone shear strength is derived as follows:


181

Design Example 3A

!


Figure 3A-16. Panel zone forces

H = 2(66.1) + 29.83 = 162 in. d p = 29.83 − 0.76 / 2 = 29.45 in. M f = 17,178 kip-in. (see Part 6d) VC =

Ff =

[ ( )] = 2(0.8)(17,178) = 170 kips

2 0.8 M f H

2(0.8)ΣM f dp

162

=

2(0.8)(17,178) = 933 kips 29.45

VZ = F f − VC = 933 − 170 = 763 kips The panel zone shear strength is determined from §2213.7.2.1.  3bc t cf2  V = 0.55 F y d c t 1 +   d b d c t 

182

(13-1)


Design Example 3A

!


where: bc =

width of the column flange

db =

depth of the beam

dc =

column depth

t cf =

thickness of the column flange

t=

total thickness of the panel zone, including doubler plates

For the W 14 × 283 column, the panel zone shear strength is:  3 (16.11)(2.07 )2  V = 0.55(50 )(16.74)(1.29 ) 1 +  = 785 > 763 kips  (29.83)(16.74 )(1.29 )

o.k.

(13-1)

The W 14 × 283 column panel zone strength is just adequate when matched with the W 30 × 108 beam without doubler plates. Again, this configuration is selected in lieu of a lighter column with doubler plates as the most economical design. Note that if the design does include doubler plates, then compliance with §2213.7.2.3 is required. The minimum panel zone thickness t z is also checked per §2213.7.2.2: t z ≥ (d z + w z ) / 90 where: d z = panel zone depth between continuity plates wz = panel zone width between column flanges t z = 1.29" for W 14 × 283 t z = 1.29" ≥ [(29.73 − 0.76) + (16.74 − 2.07 )/ 90] = 0.48 in.


o.k.

(13-2)

183

Design Example 3A

6h.

!


Check column continuity plates.

Subject to further research, FEMA-267 §7.8.3 recommends that continuity plates always be provided. The plate thickness should match the beam flange thickness. Complete penetration welds from the continuity plate to the column flanges are recommended, and fillet welds to the column web are acceptable. (Note: In FEMA-350, this criteria has been relaxed, permitting omission of continuity plates for columns with heavy flanges.) The minimum continuity plate area is validated for conformance with §2213.7.4 using AISC-ASD Section K1.8, Equation K1-9. UBC §2213.7.4 stipulates that for this equation the value for Pbf is to be taken as: 1.8bt f F y .

(

)

For W 30 × 108 : Pbf = 1.8(0.76)(10.48)(50) = 717 kips

§2213.7.4

AISC-ASD Eq. (K1-9) yields: Ast =

Pbf − F yc t wc (t b + 5k ) F yst

=

717 − 50(1.29 )[0.76 + 5(2.75)] = −4.38 50

As the area calculated is negative, stiffeners are not required per Equation K1-9 of AISC-ASD, and continuity plates with a thickness matching the beam flange are adequate. With complete penetration welds to the column flanges, the continuity plate corners should be clipped to avoid the column k-area. This leaves a fillet weld length to the column web of: lw = d c − 2(k ) = 16.74 − 2(2.75) = 11.2 in. The fillet weld to the column web is designed for the tensile strength of the continuity plate. Using a 3 4 "× 7" plate on each side of the web (top and bottom), the weld size is determined. Plate strength: Pst = 0.75 (7.0 ) 50.0 = 263 kips

184


Design Example 3A

!


Weld size (16ths): n=

Pst ( 263) = = 7.4 2lw (1.7 )(0.928) 2(11.2 )(1.7 )(0.928)

where weld strength per 1/16th inch with E70XX electrodes is 0.3(70 ksi) (1/16) (.707) = 0.928 kip-in. per AISC-ASD Table J2.5. ∴ Use a ½" fillet top and bottom of continuity plate to column web.

6i.

Evaluate beam-to-column joint restraint.

§2213.7.7

To preclude SMRF column members from out-of-plane or lateral torsional buckling, §2213.7.7 specifies requirements for beam-column joint restraint. The W 14 × 283 frame column has a perpendicular beam framing into it at each level, providing both column lateral support and joint restraint. The column flanges need to be laterally supported only at the beam top flange if the column remains elastic. By satisfying one of the four conditions listed in §2213.7.7.1, a column may be considered elastic for purposes of determining lateral bracing. Check condition #1: Strong column-weak beam strength ratio > 1.25 From a review of Part 6f above: (strength ratio) = 1.74 > 1.25

o.k.

The column flanges therefore need lateral bracing only at the beam top flange. The bracing force is taken at 1 percent of the beam flange capacity, perpendicular to the plane of the frame. By observation, the bolted connection from the beam framing perpendicular to the column is adequate.

6j.

Provide beam lateral bracing at RBS flange cut.

FEMA §7.5.3.5

Lateral bracing is next considered for the beam flanges adjacent to the RBS cut. As stated in FEMA §7.5.3.5, lateral braces for the top and bottom beam flanges are to be placed within d/2 of the reduced section. (Note: This requirement is dropped in FEMA-350 when a composite concrete slab is present. ) Lateral support of the top flange is ordinarily provided by shear studs to the concrete fill over metal deck. Either diagonal angle bracing or perpendicular beams can provide bottom flange lateral bracing. Generally, bracing elements may be designed for about 2 percent of the compressive capacity of the member being braced. Figure 3A-17 shows an example for angle bracing of the bottom flange.


185

Design Example 3A

6k.

!


Detailing considerations.

As noted in FEMA-267A, the reduced beam section SMRF design entails a few unique considerations:

6l.

"

At the cut edge of the reduced section, the beam flange should be ground parallel to the flange to a mirror finish (surface roughness < 1000 per ANSI B46.1).

"

Shear studs should be omitted over the length of the cut in the beam top flange, to minimize any slab influence on beam hinging.

"

A 1-inch-wide gap should be placed all around the column so as to the slab to reduce the slab interaction with the column connection. (Note: FEMA-350 has relaxed this requirement.)

Welding specifications.

To ensure that the SMRF joint welded connections are of the highest possible quality, the design engineer must prepare and issue project-specific welding specifications as part of the construction documents. The guidelines presented in FEMA-267, Section 8.2 provide a comprehensive discussion of welding specifications. For an itemized list of welding requirements, see California Division of the State Architect (DSA), Interpretation of Regulations #27-8, Section K – Welding. A few of these requirements are noted below:

186

"

The steel fabricator is to prepare and submit a project Welding Procedure Specification (WPS) per AWS D1.1, Chapter 5 for review by the inspector and Engineer of Record.

"

Weld filler materials are to have a rated toughness, recommended at 20ft-lbs. absorbed energy at –20o F per Charpy V-notch test.

"

Pre-heat and interpass temperatures are to be strictly observed per AWS D1.1, Chapter 4.2, and verified by the project inspector.

"

Weld dams are prohibited, and back-up bars (if used) should be removed, the weld back-gouged, and a reinforced with a fillet weld.

"

All complete penetration welds shall be examined with ultrasonic testing/inspection for their full length.


Design Example 3A

6m.

!


Tests and inspections.

Quality control is presented in Chapter 9 of FEMA-267. Guidelines are presented for inspector qualifications, as well as suggested scope of duties for the inspector, engineer and contractor. The extent of testing is discussed, with a recommendation that the contract documents clearly identify the required testing. An example Quality Assurance Program is given in FEMA §9.2.7. It is recommended that the structural engineer incorporate similar requirements into the project specifications.

Figure 3A-17. Reduced beam section joint detail


187

Design Example 3A

!


References AISC, 1997, 1999. Seismic Provisions for Structural Steel Buildings. American Institute of Steel Construction, April 1997 with Supplement No. 1, February 1999, DSA IR 27-8, 1998. Interpretation of Regulations 27-8. California Division of the State Architect, Sacramento, California. Englehardt, M., 1998. Design Recommendations for Radius Cut Reduced Beam Section Moment Connections. University of Texas, Austin. Englehardt, M., et al., 1996. “The Dogbone Connection, Part II,” Modern Steel Construction. American Institute of Steel Construction. FEMA-267, 1995. Interim Guidelines: Evaluation, Repair, Modification, and Design of Welded Steel Moment Frame Structures. SAC Joint Venture, funded by the Federal Emergency Management Agency, Washington, D.C. FEMA-267A, 1997. Interim Guidelines Advisory No. 1, Supplement to FEMA-267, Federal Emergency Management Agency, Washington, D.C. FEMA-267B, 1999. Interim Guidelines Advisory No. 2, Supplement to FEMA-267, Federal Emergency Management Agency, Washington, D.C. Steel Tips, 1999. “Design of Reduced Beam Section (RBS/Moment Frame Connections,” Steel Tips. Structural Steel Educational Council, Moraga, California.

188


Design Example 3B

!

Steel Ordinary Moment Resisting Frame

Design Example 3B Steel Ordinary Moment Resisting Frame

Figure 3B-1. Four story steel office building with steel ordinary moment resisting frames (OMRF)

Foreword Steel ordinary moment resisting frames (OMRF) differ from special moment resisting frames (SMRF) in several important ways. The most significant differences lie in the details of the beam-column joints and in the consideration of strong column-weak beam effects in member selection. Because of these and other factors, the SMRF structure has a higher R-factor (8.5) and no height limit, while OMRF structures have a low R-factor (4.5) and are limited to 160 feet in height. In general, SMRF structures are expected to perform much better in earthquakes than OMRF structures. This Design Example uses the same 4-story structure used in Design Example 3A to illustrate design of a steel OMRF. The choice of this structure was based on both convenience and the fact that the differences between OMRFs and SMRFs could be easily shown.


189

Design Example 3B

!


It should be noted, however, that SEAOC does not recommend use of steel OMRFs in buildings over two stories. In fact, SEAOC recommends use of SMRFs in all steel moment frame structures of any height, particularly mid-rise and taller structures, in high seismic regions. Typical uses of OMRF systems in high seismic regions include structures such as one-story open front retail buildings, two-story residential structures with open lower levels, penthouses and small buildings.

Overview Steel ordinary moment resisting frames are required to meet the provisions of §2213.6. The OMRF requirements are essentially the same as stipulated in prior UBC editions, and were not addressed in the emergency code amendment for SMRF design issued in the 1996 Supplement to the 1994 UBC. However, both the SEAOC Blue Book and FEMA-267 recommend against the use of OMRFs in areas of high seismicity. The OMRF provisions are retained in the code for use in light on- or two-story buildings, and structures in low seismic hazard zones. The UBC requires OMRFs to be designed for about twice the lateral seismic force that would be required for a SMRF in the same structure. As such, the plastic rotation demand for OMRF connections should be roughly half that of the SMRF. The connection ductility requirements for OMRFs are therefore less stringent than for SMRFs. Notwithstanding code provisions, OMRF connections should receive similar attention to joint detailing as for SMRFs. In particular, lessons learned from the Northridge earthquake concerning weld procedures and filler materials should also be applied to OMRFs. As suggested in FEMA-267 (see p.7-2), OMRFs in areas of high seismicity may be acceptable if the connections are designed to remain elastic for the design level earthquake, while the beam and column members are designed per UBC OMRF requirements. This can be achieved by applying an R factor of 1 in deriving design base shear and confirming that the connection stresses do not exceed yield. This enhanced OMRF design approach is also illustrated in this Design Example. This Design Example uses the 4-story steel office structure from Design Example 3A to illustrate OMRF design. The same building weights, frame elevations and site seismicity are used as for Design Example 3A. Although this Design Example is for a 4-story structure, the design procedure is applicable to all OMRFs, including such uses as one-story, single bent frames at garage door openings. It is recommended that the reader first review Design Example 3A before reading this Design Example. Refer to Example 3A for plans and elevations of the structure.

190


Design Example 3B

!



Design base shear.

2.


3.

Interstory drift.

4.

OMRF member design.

5.

OMRF beam-column joint design.


1. 1a.

Code Reference

Design base shear.


§1629.6

The structure is a building frame system with lateral resistance provided by steel ordinary moment resisting frames (system type 3.4.a of Table 16-N). The seismic factors are: R = 4.5

Table 16-N

Ω = 2.8 hmax = 160 ft

1b.


§1629.4.3

For Zone 4 and Soil Profile Type S D : C a = 0.44(N a ) = 0.44(1.0 ) = 0.44

Table 16-Q

C v = 0.64(N v ) = 0.64(1.0 ) = 0.64

Table 16-R


191

Design Example 3B

1c.


!


§1630.2.2

Per Method A:

(30-8)

T = Ct (hn )3 4 C t = 0.035 T A = 0.03(55.5)3 4 = 0.71 sec Per Method B: From Design Example 3A, assuming we retain the same beam and column sizes: North-south:

(y ) : TBy

= 1.30 sec

§1630.2.2

= 1.16 sec

Para. #2

East-west:

(x ) : TBx

For Seismic Zone 4, the value for Method B cannot exceed 130 percent of the Method A period. Consequently, Maximum value for TB = 1.3T A = 1.3(0.71) = 0.92 sec

1d.


The total design base shear for a given direction is: V =

Cv I 0.64(1.0 ) W = W = 0.155W RT 4.5(0.92 )

(30-4)

The base shear need not exceed: V =

192

2.5C a I 2.5(0.44)(1.0 ) W = W = 0.244W R 4.5

(30-5)


Design Example 3B

!


But the base shear shall not be less than: V = 0.11C a IW = 0.11(0.44)(1.0)W = 0.048W

(30-6)

And for Zone 4, base shear shall not be less than: V =

0.8ZN v I 0.8(0.4)(1.0)(1.0) W = = 0.071W R 4.5

(30-7)


1e.

(30-4)


Reliability/redundancy factor: ρ = 2 −

20 rmax Ab

§1630.1

(30-3)

From Design Example 3A, use ρ = 1.25 . For the load combinations §1612, and anticipating using allowable stress design (ASD) for the frame design: E = ρEh + Ev = 1.25(V )

(30-1)

( E v = 0 for allowable stress design) E m = ΩE h = 2.8(V )

(30-2)

Note that seismic forces may be assumed to act nonconcurrently in each principal direction of the structure, except as per §1633.1.


193

Design Example 3B

2. 2a.

!



Building weights and mass distribution (from Design Example 3A).

Table 3B-1. Mass properties summary

2b.

Level

W (kips)

Roof 4th 3rd 2nd

2,066 2,235 2,235 2,235

Total

8,771

X cg

Ycg

(ft) 100 100 100 100

(ft) 70 70 70 70

M (k-sec 2 / in.)

MMI (k-sec 2-in.)

5.3 5.8 5.8 5.8

26,556 28,728 28,728 28,728

22.7


As noted above, Equation (30-4) governs, and: V = 0.155W = 0.155(8,771) = 1,360 kips

2c.

(30-4)


For the static lateral force procedure, vertical distribution of force to each level is applied as follows: V = Ft + ∑ Fi

(30-13)

where: Ft = 0.07T (V ) ≤ 0.25(V ) Except Ft = 0 where: T ≤ 0.7 sec

194


Design Example 3B

!


For this example structure: T = 0.92 sec ∴ Ft = 0.07 (0.92)(1,360 ) = 87.6 kips The concentrated force Ft is applied at the roof, in addition to that portion of the balance of the base shear distributed to each level per §1630.5: Fx =

(V − Ft )Wx hx = (1,360 − 87.6)  Wx hx   ∑W h  ∑Wi hi i i  

(30-15)

Table 3B-2. Vertical distribution of shear Level

wx (kips)

hx (ft)

w x hx (k-ft)

w x hx Σwx

Fx (kips)

ΣV (kips)

Roof 4th 3rd 2nd

2,066 2,235 2,235 2,235

55.5 42.0 28.5 15.0

114,663 93,870 63,698 33,525

0.375 0.307 0.208 0.110

564.8 390.6 265.1 139.5

564.8 955.4 1,220.5

1.000

1,360.0

1,360.0

Total 8,771 305,756 Note: Froof = 0.375 (1,272.4) + 87.6 = 564.8 kips

2d.

Determine horizontal distribution of shear.

As in Design Example 3A, the direct seismic force, Fx , applied at the center of mass is combined with an accidental torsional moment, M t , using a 5 percent eccentricity, at each level. This is shown in Table 3B-3. North-south: M t = 0.05(204′)Fx = (10.2)Fx East-west: M t = 0.05(144′)Fx = (7.2 )Fx


195

Design Example 3B

!


Table 3B-3. Horizontal distribution of shear Level

Fx (kips)

N-S M t (k-ft)

E-W M t (k-ft)

Roof 4th 3rd 2nd

564.8 390.6 265.1 139.5

5,761 3,984 2,704 1,423

4,067 2,812 1,909 1,004

Note: Mt = horizontal torsional moment

With the direct seismic forces and torsional moments given in Table 3B-3 above, the force distribution to the frames is generated by computer analysis (not shown). For this Design Example, the beam and column sizes from Design Example 3A are used in the computer model. From the computer analysis, the shear force at the ground level is determined for each frame column. Frame forces at the base of frame types A1 and B1 are summarized in Tables 3B-4 and 3B-5.

Table 3B-4. North-south direction, frame type A1 Column Shears (kips)

Line A/1.2 (kips)

Direct Seismic Torsion Force

Line A/2 (kips)

Line A/3 (kips)

Line A/4 (kips)

Line A/5 (kips)

79.4

143.1

132.6

132.6

143.1

4.9

8.8

8.2

8.2

8.8

4.9

43.8

151.9

84.3

754.0

84.3

Direct + Torsion

151.9

140.8

140.8

Line A/5.8 (kips) 79.4

Total (kips) 710.2

Table 3B-5. East-west direction, frame type B1 Column Shears (kips)

Line 1/A.2 (kips)

3a.

Line 1/C (kips)

113.1

113.1

Line 1/C.8 (kips)

Total (kips)

Direct Seismic

63.1

Torsion Force

2.4

4.3

4.3

2.4

13.4

65.5

117.4

117.4

65.5

365.8

Direct + Torsion

3.

Line 1/B (kips)

63.1

352.4

Interstory drift.

Determine ∆S and ∆M.

§1630.9

The design level response displacement ∆S is obtained from a static-elastic analysis using the design seismic forces derived above. For purposes of displacement determination, however, §1630.10.3 eliminates the upper limit on TB, used to determine design base shear under Equation (30-4). The maximum inelastic response displacement ∆ M includes both elastic and estimated inelastic drifts resulting from the design basis ground motion. It is computed as follows: 196


Design Example 3B

!


∆ M = 0.7(R )∆ S = 0.7(4.5)∆ S = 3.15∆ S

(30-17)

The maximum values for ∆ S and ∆ M are determined, including torsional effects (and including P∆ effects for ∆ M ). Without the 1.3T A limit on TB , the design base shear per Equation (30-4) is: North-south: TBy = 1.30 sec Vn − s =

Cv I 0.64(1.0) W = W = 0.109W = 956 kips RT 4.5(1.30 )

(30-4)

East-west: TBx = 1.16 sec Ve − w =

Cv I 0.64(1.0 ) W= W = 0.123W = 1,079 kips RT 4.5(1.16 )

§1630.1.1

Note that §1630.1.1 stipulates use of the unfactored base shear (V ) , with ρ = 1 . Using these modified design base shears, the accidental torsion and force distribution to each level are adjusted for input to the computer model. The structure displacements and drift ratios are derived below in Table 3B-6.

Table 3B-6. Interstory displacements North-South Interstory Displacements Story

Height h (in.)

∆ S drift (in.)

∆ M drift (in.)


4th

162

(2.41 -2.06)= 0.35

1.10

0.0068

3rd

162

(2.06 -1.52)= 0.54

1.70

0.0105

2nd

162

(1.52 -0.82)= 0.70

2.21

0.0136

1st

180

(0.82 -0.0) = 0.82

2.58

0.0143

East-West Interstory Displacements Story

Height h (in.)

∆ S drift (in.)

∆ M drift (in.)


4th

162

(2.24 -1.92)= 0.32

1.01

0.0062

3rd

162

(1.92 -1.41)= 0.51

1.61

0.0099

2nd

162

(1.41 -0.77)= 0.64

2.01

0.0124

1st

180

(0.77 -0.0) = 0.77

2.43

0.0135

Note: Interstory drift ratio = ∆M/story height


197

Design Example 3B

3b.

!


Determine the story drift limitation.

§1630.10

For structures with T > 0.7 seconds, the maximum allowable drift is: ∆ M = 0.020 (story height) per §1630.10.2. A review of the drift ratios tabulated above in Table 3B-6 shows that all interstory drift ratios are less than 0.020, using the actual period TB in base shear Equation (30-4). The maximum drift ratio of 0.0143 occurs at the first story in the north-south direction, and is a little more than 70 percent of the 0.020 allowable. As expected, the maximum ∆ M displacements for the OMRF are very close to the values for the SMRF from Design Example 3A. At this point in the design process, the beam and column sizes could be reduced to make the displacements closer to the code limit. However, using more conservative ∆ M drift ratios produces stiffer frame designs, which mitigates possible deformation compatibility issues in other elements such as cladding and non-frame (P∆ ) column design. The same beam and column sizes previously selected will be retained. The next step will be to check member stress levels.

4.

OMRF member design.

§2213.6

Using the W 30 × 108 beam and W 14 × 283 column from Design Example 3A (see Figure 3A-3 for frame on Line A) for preliminary sizes, the OMRF frame members are designed per §2213.6.

4a.

rd

Design typical beam at 3 floor.

The typical beam designed is the third floor beam shown in Figure 3B-2.

Figure 3B-2. Typical beam at third floor of Frame A1

198


Design Example 3B

!


From a review of the computer output (not shown), the moments and shears at the right end of the beam are greatest. Note that the seismic moment and shear are about twice that for the SMRF example. The moments and shears, at the face of the column at Line 5 are: M DL = 1,042 kip-in. M LL = 924 kip-in. M seis = 6,780 kip-in. M E = ρM seis = 1.25(6,780) = ±8,475 kip-in. V DL = 16.4 kips V LL = 13.3 kips Vseis = 42.2 kips V E = ρVseis = 1.25(42.2) = ±52.7 kips

§1630.1.1

Using the basic load combinations of §1612.3.1 (ASD), with no one-third increase. D + L : M D +L = 1,042 + 924 = 1,966 kip-in.

(12-8)

V D +L = 16.4 + 13.3 = 29.7 kips D+

E 8,475 : M D +E = 1,042 + = 7,096 kip-in. 1.4 1.4 V D +E = 16.4 +

(12-9)

52.7 = 54.0 kips 1.4

   8,475   E  D + 0.75 L +   = 6,275 kip-in.  : M D + L + E = 1,042 + 0.75924 +   1.4   1.4   

(12-11)

  52.7  V D + L+ E = 16.4 + 0.7513.3 +   = 54.6 kips  1.4   Try W 30 × 108 , ASTM A572, Grade 50 beam.


199

Design Example 3B

!


Check flange width-thickness ratios per AISC-ASD, Table B5.1 (Note: AISCASD is adopted, with amendments, in Division III of the code): bf 2t f

≤

65 50

= 9.19

and: d 640 ≤ = 90.5 tw 50 For W 30 × 108 :

bf 2t f

= 6.9 < 9.19

o.k.

And: d 29.83 = = 54.7 < 90.5 t w 0.545

o.k.

As in Design Example 3A, provide beam bracing at one-third points. The maximum unbraced length is: L = 28.0 3 = 9.33 ft Check allowable moment capacity. From AISC-ASD, p. 2-10; for W 30 × 108 : Lu = 9.8 > 9.33

( )

∴ Fb = 0.60 Fy = 30.0 ks Allowable M a = 299(30.0) = 8,970 kip-in. > 7,096 kip-in.

o.k.

Check allowable shear capacity.

200


Design Example 3B

!


For W 30 × 108 : h 29.83 − 2(0.76 ) 380 = = 51.9 < = 53.7 tw 0.545 50

( )

∴ Fv = 0.4 F y = 0.4(50 ) = 20.0 ksi Allowable Va = 20.0(0.545)(29.83) = 325 kips > 54.6 kips

o.k.

∴ Use W 30 × 108 beam

4b.

nd

Design typical column at 2

story.

The column to be designed is the second-story column of Frame A1 shown in Figure 3B-3. For the 2nd story column at Line 5, the maximum column forces generated by the OMRF frame analysis (not shown) are: M DL = 236 kip-in. M LL = 201 kip-in. M seis = 7,501 kip-in. M E = 1.25(7,501) = 9,376 kip-in. V DL = 3.1 kips V LL = 2.7 kips Vseis = 107 kips V E = 1.25(107 ) = 134 kips PDL = 113 kips PLL = 75 kips Pseis = 53 kips PE = 1.25(53) = 66 kips The maximum strong axis moments occur at the bottom of the column, and are taken at the top flange of the beam.


201

Design Example 3B

!


Figure 3B-3. Typical second-story column of Frame A1

Using the basic load combinations of §1612.3.1: D + L : M D + L = 236 + 201 = 437 kip-in.

(12-8)

PD + L = 113 + 75 = 188 kips VD + L = 3.1 + 2.7 = 5.8 kips D+

9,376 E : M D + E = 236 + = 6,933 kip-in. 1.4 1.4 PD + E = 113 + VD + E = 3.1 +

0.9 D −

202

(12-9)

66 = 160 kips 1.4

134 = 99 kips 1.4

66 E : PD − E = 0.9(113) − = 54.5 kips compression 1.4 1.4

(12-10)


Design Example 3B

!


   9,376   E  D + 0.75 L +   = 5,410 kip-in.  : M D + L + E = 236 + 0.75201 +   1.4   1.4   

(12-11)

134   = 77 kips VD + L + E = 3.1 + 0.752.7 + 1.4     66  PD + L + E = 113 + 0.7575 +   = 205 kips  1.4   Under the requirements of §2213.5.1, columns must have the strength to resist the following axial load combinations (neglecting flexure): PDL + 0.7 PLL + ΩPseis : Pcomp = 113 + 0.7(75) + 2.8(53) = 314 kips compression 0.85PDL − ΩPseis :

Ptens = 0.85(113) − 2.8(53) = −52 kips tension

Try W 14 × 283 , ASTM A572, Grade 50 column: Unbraced column height: h = 13.5 − (2.5 2 ) = 12.25 ft Maximum f a = 205 / 83.3 = 2.46 ksi 1.0(12 )(12.25)  kλ  = 35.3   = 4.17  r y ∴ Fa = 26.5 ksi Maximum

f a 2.46 = = 0.092 < 0.15 Fa 26.5

Therefore, AISC-ASD Equation H1-3 is used for combined stresses. From AISC-ASD manual (p. 3-21) for W 14 × 283 , Grade 50: Lc = 14.4 > 12.5

( )

∴ Fb = 0.66 Fy = 33.0 ksi


203

Design Example 3B

!


Check combined stresses for the critical load combinations. D+

E f f 160 6,933 : a + bx = + = 0.073 + 0.458 = 0.530 < 1.0 1.4 Fa Fb 83.3(26.5) 459(33.0 )

5,410 E  f a f bx  + = 0.092 + = 0.449 < 1.0 : D + 0.75 L +  459(33.0 ) 1.4  Fa Fb 

o.k.

o.k. (12-9)

(12-11)

Check column shear capacity. Allowable Va = 0.4(50)(16.74 )(1.29 ) = 432 kips > 99 kips

o.k.

Next, check required axial strength per §2213.5. Compression: Psc = 1.7 Pallow = 1.7(83.3)(26.5) = 3,753 kips > 314 kips

o.k.

Tension: Pst = F y A = 50(83.3) = 4,165 kips >> −52 kips

o.k.

∴Use W 14 × 283 column

5.

OMRF beam-column joint design.

§2213.6

As shown above, the W 30 × 108 beam and W 14 × 283 column taken from the SMRF of Design Example 3A have the capacity to meet the load combinations for an OMRF per §1612.3. Section 2213.6 requires that OMRF beam-to-column connections are to either meet the SMRF connection criteria (see §2213.7.1), or be designed for gravity loads plus Ω times the calculated seismic forces. As discussed in FEMA-267 (Section 7.1), OMRF joints may be considered acceptable if designed to remain elastic, with an R of unity (1.0). Using an R factor of 1 is marginally more stringent than multiplying the seismic forces by Ω o . With R = 1 , it is appropriate to use the full calculated period (TBx = 1.30) to determine the base shear for joint design. Therefore, the north-south base shear is taken as: Vn / s =

204

Cv I 0.64(1.0 ) W = W = 0.492W = 4,315 kips 1.0(1.30 ) RT


Design Example 3B

!


For an OMRF (with Ω = 2.8 ), the UBC base shear for connection design is: Vn / s = 2.8(0.155)W = 0.434W = 3,807 kips The ratio of base shears is: FEMA/UBC = 4,315 / 3,807 = 1.13 Thus, there is a 13 percent increase with R = 1 as recommended in FEMA-267. Using the unreduced seismic base shear, the beam-column joint stresses are checked to remain elastic. For this, §1612.4, Special Seismic Load Combinations, is used with a resistance factor φ of one.

5a.

Determine beam forces with R=1.

The beam end moment and shear are scaled up to the unreduced seismic force level by the ratio of the base shears, as follows:  0.492  VE ' =   Vseis = 3.17(42.2 ) = 138 kips  0.155   0.492  M E' =   M seis = 3.17(6,780 ) = 21,493 kip-in.  0.155  The special seismic load combination from §1612.4 is: 1.2 D + 0.5 L + 1.0 E M

(12-17)

M D + L + E = 1.2(1,042) + 0.5(924) + 1.0(21,493) = 23,205 kip-in. VD + L + E = 1.2(16.4) + 0.5(13.3) + 1.0(138) = 164 kips


205

Design Example 3B

5b.

!


Check beam-to-column weld.

As was done in Design Example 3A, the beam webs are to have complete-penetration welds to the column flange. (Note that this weld is shown in Figure 12-4). Note also that the flanges are reinforced with 5/16" fillet welds. Using the cross-sectional area of the beam flange and web weldments at the face of the column, the elastic section modulus S c of the beam is calculated from information in Table 3B-7.

Figure 3B-4. Built-up section at column face.

Table 3B-7. Built-up section properties Mk

Area (in.2)

1 2 3 4 5

0.545(26.73)=14.58 0.76(10.48)=7.96 0.76(10.48)=7.96 0.31(10.48)=3.28 0.31(10.48)=3.28

Sum

206

Y (in.)

A(y)2

Io (in.4)

0.00 14.54 14.54 15.07 15.07

0 1,682 1,682 745 745

869 0 0 0 0

4,854

869


Design Example 3B

!


The calculated section properties are: I c = 4,854 + 869 = 5,723 in.4 ∴ S c = 5,723 15.23 = 376 in.3 Per FEMA §7.2.2.1 for complete penetration welds, the weld strength is taken as the beam yield stress of 50 ksi. The maximum weld stress is calculated using the maximum moment (M D + L+ E ) at the face of the column: f weld = 23,205 / 376 = 61.7 ksi > φF y = 1.0(50 ) = 50 ksi

n.g.

The W 30 × 108 connection (weld) stresses to the column are not within the elastic limit. At this point, we can choose to either add cover plates, or make the beam larger. With similar weld patterns, a W 33 × 152 is required to obtain an adequate connection section modulus S c = 575 in.3 :

(

)

f weld = 23,205 / 575 = 40.4 ksi < 50 ksi

o.k.

If we choose to instead add cover plates, we would need 10"× 3 / 4" plates at the top and bottom flanges. With complete penetration welds at the cover plates to the column, the increased moment of inertia and section modulus are: I c = 5,723 + 2(7.5)(15.3)2 = 9,234 in.3 S c = 9,234 15.98 = 578 in.3 and: f weld = 23,205 / 578 = 40.1 ksi < 50 ksi

o.k.

The cover plates should be about half the beam depth in length, with fillet welds to the beam flange as required to develop the tensile capacity of the plate. The minimum size for ¾" plate is a 5/16" fillet weld. Cover plate capacity: TPl = 0.75(10 )(50.0 ) = 375 kips


207

Design Example 3B

!


5/16" fillet capacity: q = 1.7(0.707 )(0.313)(21.0) = 7.9 kip-in. Required weld length:

lw = 375/ 7.9 = 47" Use a 20-inch long plate, which will provide for a total weld length of: 2(20) + 10 = 50" > 47"

o.k.

As noted above, the beam web is to have a complete penetration weld to the column face. The allowable beam shear of 325 kips from Part 4a above exceeds the unreduced seismic shear demand of 164 kips. For beam-to-column connections with bolted shear plates in lieu of welded webs, the connection plate and bolts must be designed for this maximum shear force. See Design Example 3A, Part 6g for a beam-to-column shear plate connection design.

5c.

Additional considerations.

Although the UBC does not explicitly require any further OMRF connection analysis, it is good practice to check the strong column-weak beam criteria and the column panel zone shear strength. The column panel zone shear strength should be reviewed for capacity to resist the maximum beam moment from the unreduced seismic force. The strong column-weak beam analysis would be similar to that of the SMRF Design Example 3A, Part 6f. The OMRF joint should also include continuity plates, and expanded welding procedures as for the SMRF. OMRFs designed to comply with the foregoing parameters can be expected to provide a high level of seismic performance. The objective of maintaining connection stresses within the elastic range is shown to be reasonable even for the unreduced seismic demand. The resulting frame design produces a structure that may respond to the design level ground motion without damage (i.e., plastic deformations). Moreover, OMRF designs will likely produce nominally heavier members, thereby reducing overall building drift and decreasing the potential for damage to nonstructural components.

208


Design Example 4

!

Reinforced Concrete Wall

Design Example 4 Reinforced Concrete Wall

Figure 4-1. Eight-story reinforced concrete parking garage (partial view)

Overview The structure in this Design Example is an 8-story parking garage with loadbearing reinforced concrete walls (shear walls) as its lateral force resisting system, as shown in Figure 4-1. This Design Example focuses on the design and detailing of one of the 30'-6" long walls running in the transverse building direction. The purpose of this Design Example is twofold: 1.

Demonstrate the design of a solid reinforced concrete walls for flexure and shear, including bar cut-offs and lap splices.

2.

Demonstrate the design and detailing of wall boundary zones.


209

Design Example 4

!


The Design Example assumes that design lateral forces have already been determined for the structure, and that the forces have been distributed to the walls of the structure by a hand or computer analysis. This analysis has provided the lateral displacements corresponding to the design lateral forces.


Load combinations for design.

2.

Preliminary sizing of wall.

3.

Moment strength of wall.

4.

Lap splices and curtailment of vertical bars.

5.

Design for shear.

6.

Sliding shear (shear friction).

7.

Boundary zone detailing.

Given Information The following information is given: Seismic zone = 4 Soil profile type = S D Near field = 5 km from seismic source type A Reliability/redundancy factor, ρ = 1.0 Importance factor, I = 1.0 Concrete strength, f ' c = 5,000 psi Steel yield strength, f y = 60 ksi Figure 4-2 shows the typical floor plan of the structure. Figure 4-3 shows the wall elevation and shear and moment diagrams. The wall carries axial forces PD (resulting from dead load including self-weight of the wall) and PL (resulting from live load) as shown in Table 4-1. Live loads have already been reduced according to §1607.5. The shear V E and moment M E resulting from the design lateral earthquake forces are also shown in Table 4-1.

210


Design Example 4

!


Figure 4-2. Floor plan

Figure 4-3. Wall elevation, shear, and moment diagram


211

Design Example 4

!


Table 4-1. Design loads and lateral forces Level R 8 7 6 5 4 3 2 1

PD (k)

PL (k)

VE (k)

ME (k-ft)

216 436 643 851 1060 1270 1470 1730

41 81 122 162 203 244 284 325

96 262 438 625 821 1030 1270 1470 0

0 960 3760 8530 15400 24400 35600 49600 75500

For this Design Example, it is assumed that the foundation system is rigid and the wall can be considered to have a fixed base. The fixed-base assumption is made here primarily to simplify the example. In an actual structure, the effect of foundation flexibility and its consequences on structural deformations and strains should be considered. Using the fixed base assumption and effective section properties, the horizontal displacement at the top of the wall, corresponding to the design lateral forces, is 2.32 inches. This displacement is needed for the detailing of boundary zones according to the UBC strain calculation procedure of §1921.6.6, which is illustrated in Part 7 of this Design Example. The design and analysis of the structure is based on an R factor of 4.5 (UBC Table 16-N) for a bearing wall system with concrete shear walls. Concrete wall structures can also be designed using an R factor of 5.5, if an independent space frame is provided to support gravity loads. Such a frame is not used in this Design Example.


1.

Code Reference


Load combinations for the seismic design of concrete are given in §1612.2.1. (This is indicated in §1909.2.3, and in the definition of “Design Load Combinations” in §1921.1.) Equations (12-5) and (12-6) of UBC Chapter 16 are the seismic design load combinations to be used for concrete. Exception 2 of §1612.2.1 states “Factored load combinations of this section multiplied by 1.1 for concrete and masonry where load combinations include seismic forces.” Thus, the load combinations for Equations (12-5) and (12-6) for the seismic design of concrete can be written:

212


Design Example 4

!


1.32 D + 1.1E + 1.1( f1 L + f 2 S ) 0.99 D ± 1.1E The factors f1 and f 2 are defined in §1612.2.1. The additional 1.1 factor is eliminated in the SEAOC Blue Book and in the 2000 International Building Code, for the reasons given in Blue Book §101.7.1, and as presented in the section below on SEAOC-recommended revisions to load combinations. Load combinations for nonseismic loads for reinforced concrete are given in §1909.2. Equations (12-1) through (12-4) of §1612.2.1 are not used for concrete. The allowable stress design load combinations of §1612.3 are also not used for concrete design. Horizontal and vertical components of earthquake force E.

§1630.1.1

The term E in the load combinations includes horizontal and vertical components according to Equation (30-1): E = ρE h + E v

(30-1)

Equation (30-1) represents a vector sum, and E v is defined as an addition to the dead load effect, D . Substituting into Equation (30-1): E = ρEh ± 0.5Ca ID Substituting this into the seismic load combinations for concrete gives:

(1.32 + 0.55C a I )D + 1.1ρE h + 1.1( f1 L +

f2S )

(0.99 − 0.55C a I )D ± 1.1ρE h SEAOC-recommended revisions to load combinations.

Blue Book §101.7.2.1

SEAOC recommends revisions to the load combinations of §1612, as indicated in Blue Book §101.7.2.1. As shown in Blue Book Section C403.1, the SEAOC recommended load combinations for the seismic design of reinforced concrete omit the 1.1 multiplier, and can be written:

(1.2 + 0.5C a I )D + ρE h + ( f1 L +

f2S )

0.9 D ± ρE h


213

Design Example 4

!


Load combinations used in this Design Example.

For this Design Example, it is assumed that the local building department has indicated approval of the SEAOC recommended revisions to the UBC load combinations. For examples using the UBC load combinations instead of the SEAOC recommendations, see Seismic Design Manual Volume II. Since the given structure is a parking garage, f1 = 1.0 , per §1612.2.1, and since there is no snow load, S = 0 . For Soil Profile Type S D , Seismic Zone 4, the factor C a is calculated as 0.44 N a , according to Table 16-Q. From Table 16-S, the factor N a is given as 1.2 (5km from Seismic Source Type A). However, the structure meets all of the conditions of §1629.4.2 and therefore the value of N a need not exceed 1.1. Thus, C a = 0.44(1.1) = 0.484 . With I = 1.0 and ρ = 1.0 , the governing load combinations for this Design Example are:

[1.2 + 0.5(0.484)]D + E h + L = 1.44 D + E h + L 0.9 D ± E h Actions at base of wall.

For the example wall, the dead and live loads cause axial load only, and the earthquake forces produce shear and moment only. The second of the above combinations gives the lower bound axial load. For a wall with axial loads below the balance point, the lower bound axial load governs the design for moment strength. (Typically, axial loads in concrete walls are well below the balance point, as is the case in this Design Example, as shown in Figure 4-8). The governing axial load at the base of the wall is thus: Pu = 0.9 PD = 0.9 (1,730 k ) = 1,560 k The governing moment and shear at the base of the wall is: M u = M E = 75,500 k - ft Vu = V E = 1,470 k - ft

214


Design Example 4

2. 2a.

!


Preliminary sizing of wall.

Shear stress and reinforcement ratio rules of thumb.

The dimensions and required number of walls in a building can be selected by limiting the average shear stress in the walls, corresponding to factored lateral forces, to between 3 f ' c and 5 f ' c . Limiting the average shear stress to between 3 f ' c and 5 f ' c helps prevent sliding shear failure of the walls. Walls with higher levels of shear stress are permitted by the UBC. For the example wall, the maximum factored shear force equals 1470 k. Conservatively using a 3 f ' c criterion, for a wall length of 30'-6", the wall thickness equals: 1,470,000# = 19.0 in. 366′ 3 5,000 psi

(

)

Say b = 20 in.

2b.

Minimum wall thickness to prevent wall buckling.

§1921.6.6(1.1)

For structures with tall story heights, the designer should check that the wall thickness exceeds l u 16 , where l u is the clear height between floors that brace the wall out-of-plane. This is based on §1921.6.6.6, paragraph 1.1, applicable to walls that require boundary confinement. The SEAOC Blue Book Commentary (C407.5.6, page 178) recommends “that the wall boundary thickness limit of l u 16 be applied at all potential plastic hinge locations, regardless of whether boundary zone confinement is required.” For the example wall, the clear height at the first story is 17 feet. Minimum thickness = l u 16 = 17 (12) 16 = 12.8" < 20"


o.k.

215

Design Example 4

2c.

!


Layout of vertical reinforcement.

Based on brief calculations and the preliminary sizing considerations discussed here, the wall section and reinforcement layout shown in Figure 4-4 is proposed for the base of the wall:

Figure 4- 4. Layout of vertical reinforcement at wall base

The reinforcement layout considers the following issues:

"

Vertical bars are spaced longitudinally at 9 inches on center. This spacing exceeds 6db of the largest bars used #11: 6db = 6(1.41) = 8.46 in. This offers the best conditions for lap splicing of reinforcement, as indicated in the CRSI rebar detailing chart [CRSI, 1996]. A closer spacing of vertical bars might typically be used in the boundary regions of the wall, but such a spacing could require longer lap splice lengths.

"

The maximum center-to-center spacing of vertical bars is 12 inches in boundary regions of walls where confinement is needed, according to §1921.6.6.6 Paragraph 2.4. This means that at the ends of the 20-inch-thick wall, three bars are used as shown in Figure 4-4.

Section 1921.6.2.1 specifies a minimum reinforcement ratio of 0.0025 for both vertical and horizontal reinforcement of shear walls. For the proposed layout, at the center portion of the wall’s length: ρv = As bs = 1.58 in.2 (9"× 20") = 0.0056 > 0.0025

216

o.k.


Design Example 4

3.

!


Moment strength of wall.

As recommended in the SEAOC Blue Book Commentary (§C407.5.5) the vertical reinforcement in the web of the wall and axial load contributions to the moment strength of wall sections should not be neglected. The 1991 and earlier editions of the UBC required wall boundaries to carry all moment and gravity forces. This practice results in higher moment strengths in walls, which can lead to poor earthquake performance because it makes shear failure more likely to occur. This design practice is no longer accepted by the code. Wall moment strength can be computed by hand calculations, spreadsheet calculations, or a computer program such as PCACOL. All three calculation approaches are demonstrated below. All of the calculation methods are based on an assumed strain distribution and an iterative calculation procedure.

3a.

Assumed reinforcement strain.

As indicated in the SEAOC Blue Book Commentary (§C407.4.4), for cyclic loading all vertical reinforcement along the wall can be assumed to yield in either tension or compression. This assumption simplifies the hand calculation of moment capacity and is used in the hand calculations shown below. Alternatively the reinforcement strain can be assumed to be directly proportional to distance from the neutral axis, as discussed in §1910.2. This assumption is used in the spreadsheet calculations demonstrated here and is also used by the PCACOL computer program. The assumption of all reinforcement yielding results in a slightly greater moment strength compared to the strain assumption of §1910.2, but the difference is not significant. The two possible assumed strain distributions are illustrated in Figure 4-5 below. The assumption of all reinforcement yielding is typically closer to the actual strain distribution in a wall section under cyclic displacements than is the strain assumption of §1910.2, which is derived from monotonic loading.


217

Design Example 4

!


c

Pn Mn CS2

CS1

TS2

TS1

CC fy Steel stress, cyclic loading

-fy Steel stress, monotonic loading

Figure 4-5. Steel stress and neutral axis depth

In calculating moment strength, it is necessary to determine the neutral axis depth, c , as shown in Figure 4-5. A typical calculation of moment strength is based on the following steps:

218

1.

An initial estimate of c . c = 0.15lw can be used as an initial estimate.

2.

Calculation of the steel reinforcement tension and compression forces.

3.

Balancing the forces to calculate the concrete compressive force,

4.

Calculation of the stress block length a , which corresponds to C c .

5.

Calculation of c equal to a β1 , and a reiteration of Steps 1 through 4 if necessary.

Cc = (Pn + ΣTs − ΣCs ) .


Design Example 4

3b.

!


Hand calculation.

The calculation of moment strength is based on the free-body diagram shown in Figure 4-6. The force reduction factor, φ , is calculated as a function of axial load according to §1909.3.2.2, as follows. 0.10 f ' c Ag = 0.10(5.0ksi )(20′′)(366′′) = 3,660 kips Pu = 1,560 kips (see Part 1) φ = 0.9 − 0.2(1,560 3,660 ) = 0.815 Pn = Pu φ = 1,560 0.815 = 1,910 kips

0

2

4

8 ft

x PN

CS1

TS3

CS2

TS2

TS1

CC

Figure 4-6. Free body diagram for moment strength

The iterative calculation of neutral axis depth and moment strength is shown in Tables 4-2 and 4-3 below.

§1909.2

Table 4-2. First iteration for c and Mn Force CS1 CS2 TS3 TS2 TS1 Pn Cc

Reinforcement Bars

As in.2

Asfy kips

3-#11 12-#11 54-#8 12-#11 3-#11

4.68 18.7 42.7 18.7 4.68

-281 -1122 2562 1122 281 1910 -4472 0


x in.

Asfy*x kip-in.

3 34.5 183 332 363 183 26.3

-842 -38,700 469,000 373,000 102,000 350,000 -126,000 1,130,000 =Mn (kip-in.)

219

Design Example 4

!


Table 4-3. Second iteration for c and Mn Force CS1 CS2 TS3 TS2 TS1 Pn Cc

Reinforcement Bars

As in.2

3-#11 12-#11 52-#8 12-#11 3-#11

4.68 18.7 41.1 18.7 4.68

Asfy kips

x in.

-281 -1122 2465 1122 281 1910 -4375 0

Asfy*x kip-in.

3 34.5 187 332 363 183 25.7

-842 -38,709 460,918 372,504 101,930 349,530 -123,369 1,121,961=Mn (kip-in.) 93,497=Mn (kip-ft)

First iteration, assume c = 60 in.

Therefore, 15-#11 bars yield in compression, 54-#8 bars (all web vertical bars) plus 15-#11 bars yield in tension. (Assume all reinforcement yields in either tension or compression.) Solve for C c to balance forces, C c = 4,470 kips

(

Calculate a corresponding to C c , a = C c .85 f ' c b

)

a = 52.6 c = 65.8

Calculate c = a β1 = 52.6 0.80 Second iteration, assume c = 65 in.

Therefore, 15-#11 bars yield in compression, 52-#8 bars plus 15-#11 bars yield in tension. Neglect force in 2-#8 located at x = 67 inches. Therefore, centroid of 52-#8 bars is at x = 187 in. Assume all other reinforcement yields. Solve for C c to balance forces, C c = 4,375 kips

(

Calculate a corresponding to C c , a = C c .85 f ' c b Calculate c = a β1

)

a = 51.5 c = 64.3

solution converged

This results in M n = 93,500 k - ft φM n = 0.815(93,500 k - ft ) = 77,200 k - ft > M u 75,500 k - ft

220

o.k.


Design Example 4

3c.

!


Calculation using a general spreadsheet.

The approach used above to calculate flexural strength can be done on a spreadsheet or by hand. A more generally applicable spreadsheet to calculate wall flexural strength can also be created. Such a spreadsheet is shown in Figure 4-7. This spreadsheet is set up so that each individual layer of reinforcement is represented by a spreadsheet row. The input variables are at the top of the spreadsheet. The user adjusts the input value of the neutral axis depth, c , on the spreadsheet until the tension and compression forces on the section are balanced, as indicated by the added notes on the section. The spreadsheet gives a design moment capacity, φM n , of the selected section equal to 76,150 k-ft, nearly identical to that calculated by hand in the previous section.

3d.

Calculation by PCACOL.

The computer program PCACOL can also be used to design wall sections for flexure and axial load. The example wall section was run on PCACOL and the moment strength obtained was the same as that calculated by the hand and spreadsheet methods. The printed screen output of the PCACOL run is shown in Figure 4-8.


221

Design Example 4

!


Figure 4-7. General spreadsheet to calculate flexural strength

222


Design Example 4

!


Figure 4-8. Analysis of wall section by PCACOL

4. 4a.

Lap splices and curtailment of vertical bars.

Bar cut-offs.

§1912.10.3

Section 1912.10.3 addresses the development of flexural reinforcement and states “Reinforcement shall extend beyond the point at which it is no longer required to resist flexure for a distance equal to the effective depth of the member or 12d b , whichever is greater.” For a wall, the effective depth may be considered equal to 0.8l w , according to §1911.10.4. Section 402.7 of the SEAOC Blue Book clarifies this requirement and recommends that the requirement be applied to concrete walls. Applying the bar cut-off requirement to the example wall, the moment strength is reduced in two steps over the height of the wall: above Level 5 and above Level 7. The dimensions of the wall section and the number of vertical bars are unchanged at these transitions—only the size of the reinforcement is reduced. The selection of vertical reinforcement sizes and cut-offs is shown in the wall elevation of Figure 4-10. A summary of flexural reinforcement and moment strength over the wall height is given in Table 4-4, below.


223

Design Example 4

!


Table 4-4. Boundary and vertical web reinforcement Location

Vertical Bars Each Boundary

Web Vertical Bars

Level 1 – Level 5 Level 5 – Level 7 Level 7 – Level 9

15-#11 15-#10 15-#8

54-#8 54-#7 54-#6

Axial Load Pu=0.9PD 1560 k 766 k 392 k

Design Moment Strength, ΦMn 76,200 k-ft 59,200 k-ft 40,400 k-ft

The moment strengths for each reinforcement arrangement were calculated using the spreadsheet procedure described in Part 3c, above. The moment strength above Level 5 is checked by the calculation below. For simplicity, the moment diagram is assumed to be linear over the building height. This also addresses higher mode effects according to the recommendations of Paulay and Priestley [1992]. Height of reinforcement cut-off above base Height after subtracting 0.8l w bar extension Moment demand M u at the base of the wall Overall wall height, hw Moment demand at h = 29.8' based on linear moment diagram

= 51'-0" + 3'-2" lap splice = 54.2' – 0.8(30.5')

= 54.2' = 29.8' = 75,500 k-ft = 95.3'

= (75,500)(95.3 – 29.8)/95.3

= 51,900 k-ft. < 59,200 o.k.

Similarly, the moment strength above Level 7 is checked by the following calculation: Height of reinforcement cut-off above base Height after subtracting 0.8l w bar extension Moment demand at h = 51.5' based on linear moment diagram

= 73'-2" + 2'-9" lap splice = 75.9 – 0.8(30.5)

= 75.9' = 51.5'

= (75,500)(95.3 – 51.5)/95.3

= 34,700 k-ft. < 40,400 o.k.

The calculations for bar cut-off locations are illustrated in Figure 4-9.

224


Design Example 4

Amount of vertical reinforcement

!


Moment demand assuming linear variation

Figure 4-9. Calculation of required moment strength at bar cut-off locations

4b.

Splices of reinforcement.

The lap splices of the vertical reinforcement are shown in the wall elevation of Figure 4-11. Lap splice lengths are taken from the CRSI rebar detailing chart [CRSI, 1996]. Lap splices are not used over the first two stories of the wall, because this is the anticipated plastic hinge region. Although not specifically required by the code, lap splices of flexural reinforcement should be avoided in plastic hinge regions of walls. As indicated in 1999 Blue Book Sections C402.7 and C404.3 (and in the commentary to Section 21.3.2 of ACI 318 [1999], applicable to flexural members of frames), lap splices in plastic hinge regions are likely to slip unless they are surrounded by confining ties. Even well-confined lap splices (§C402.7) that do not slip are undesirable in plastichinge regions because they prevent an even distribution of yielding along the length of the flexural reinforcement. Paulay and Priestley [1992] note that splices in plastic hinge zones tend to progressively unzip and that attempting to mitigate the problem by making lap splices longer than required is unlikely to ensure satisfactory performance.


225

Design Example 4

!


Welded splices or mechanical couplers.

§1921.2.6

Properly designed welded splices or mechanical connection splices are preferable to lap splices in plastic hinge regions. Ideally, the welded or mechanical splices should be able to develop the breaking strength of the bar. As a minimum, mechanical splices must be Type 2 splices according to §1921.2.6. If used in plastic hinge regions, SEAOC recommends that welded or mechanical splices be staggered so that no more than one-half of the reinforcement is spliced at one section, and the stagger is not less than 2 feet. Staggering of the splices is not required by the UBC. Plastic hinge length and zone in which to exclude lap splices.

§1921.6.6.5

Section 1921.6.6.5 specifies that the equivalent plastic hinge length, l p , of a wall section “shall be established on the basis of substantiated test data or may be alternatively taken as 0.5l w .” Based on the work of Paulay and Priestley [1993] and FEMA-306 [1999], l p for walls can be taken as 0.2lw + 0.07 M V , where M V is the moment to shear ratio at the plastic hinge location. For the example wall, l p is calculated by both methods as shown below: l p = 0.5l w

= 0.5(30.5') = 15.2'

l p = 0.2l w + 0.07 M V

= 0.2(30.5') + 0.07(68,600 k-ft / 1340 k) = 6.1' + 3.6' = 9.7'

For this Design Example, we will take 9.7 ft as l p , based on the substantiated test data reviewed by Paulay and Priestley [1993]. Equivalent plastic hinge lengths, as calculated above, are used to relate plastic curvatures to plastic rotations and displacements (for example in §1921.6.6.5). The actual zone of yielding and nonlinear behavior typically extends beyond the equivalent plastic hinge length. For flexural members of frames, §1921.3.2.3 indicates that flexural yielding may be possible “within a distance of twice the member depth from the face of the joint.” This distance is conservatively defined to be larger, by a factor of two or more, than the equivalent plastic hinge length, lp. Thus, for this Design Example wall, the expected zone of yielding should be taken as equal to at least 2l p (19.4 ft), and lap splices should be avoided over this height. In the Design Example, lap splices are excluded over the first two stories, i.e., over a height of 28.8 ft, as shown in the wall elevation of Figure 4-10. Because of potential construction difficulties in using continuous vertical bars from the 226


Design Example 4

!


foundation through Level 3, an option to use welded or mechanical connection splices can be specified as shown in Figure 4-10. Such splices require an ICBO Evaluation Report or acceptance by the local building official.

Figure 4-10. Wall elevation

5.

Design for shear.

The SEAOC Blue Book Section 402.8.1 requires that “the design shear strength φVn shall not be less than the shear associated with the development of the nominal moment strength of the wall.” A design for shear forces based on code requirements will not necessarily achieve this objective. Thus, the code provisions covered in Part 5(a) should be considered as minimum requirements for the shear design of walls. Designing for amplified shear forces as recommended in the Blue Book is covered in Part 5(b) below.


227

Design Example 4

5a.

!


UBC requirements. Shear demand.

If designing to the minimum requirements of the UBC, the shear demand is taken directly from the design forces, factored by the load combinations discussed in Part 1 of this Design Example. At the base of the wall: Vu = V E = 1,470 k Shear capacity.

Section 1911.10 gives shear provisions for walls designed for nonseismic lateral forces such as wind or earth pressure. Section 1921.6.5 gives shear strength provisions for walls designed for seismic forces. Since the subject wall has a ratio of hw l w greater than 2.0, Equation (21-6) governs wall shear strength: Vn = Acv  2 f ' c + ρ n f y    As prescribed in §1909.3.4.1, the shear strength reduction factor, φ , shall be 0.6 for the design of walls if their nominal shear strength is less than the shear corresponding to development of their nominal flexural strength. φVn = 0.6(20")(366") [2 + ρ n (60,000 psi )] = 621 k + 263,000 ρ n At each level, the amount of horizontal reinforcement provided for shear strength is given in Table 4-5. Note that for all levels above Level 2, the minimum reinforcement ratio of 0.0025 governs the amount of horizontal reinforcement. (§1921.6.2.1)

Table 4-5. Horizontal reinforcement for UBC shear strength requirements Level R 8 7 6 5 4 3 2

228

VE kips 95 262 438 625 821 1030 1260 1470

Horizontal Reinforcement

ρn

φVn kips

#5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #6@12” E.F.

0.00258 0.00258 0.00258 0.00258 0.00258 0.00258 0.00258 0.00367

1300 1300 1300 1300 1300 1300 1300 1585


Design Example 4

!

UBC §1921.6.5.6 requires that Vn shall not be taken greater than 8 Acv 8 Acv

5b.

f ' c = 8 (20")(366") 5,000 = 4,140 kips > 1,585 kips


f 'c .

o.k.

Blue Book recommendations. Shear demand.

To comply with the Blue Book requirement of providing shear strength in excess of the shear corresponding to wall flexural strength, an amplified shear demand is considered. Section C402.8 of the Blue Book commentary gives the following equation for the shear amplification factor, ωv , that accounts for inelastic dynamic effects. For application to designs according to the UBC, the amplification factor recommended by Paulay and Priestley [1992] can be reduced by a factor of 0.85, because the Paulay and Priestley recommendations use a different strength reduction factor, φ , than does the UBC. ωv

= 0.85(1.3 + n/30), for buildings over 6 stories, where n = number of stories

ωv

= 0.85(1.3 + 8/30) = 1.33

As indicated in the Blue Book, the ωv factor is derived for analysis using inverted triangular distributions of lateral forces. If a response spectrum analysis is carried out, a slightly lower ωv factor can be justified in some cases. For this Design Example, the shear demand is taken at the nominal strength. For further conservatism, one could base the shear demand on the upper bound of flexural strength, which can be taken as the “probable flexural strength,” Mpr, defined in §1921.0. M n is calculated using a strength reduction factor, φ , of 1.0, and taking the upper bound of axial load from the load combinations of UBC §1921.0. The probable and nominal moment strengths for the higher axial load are as shown in Table 4-6. The nominal moment strength previously calculated is shown for comparison.

Table 4-6. Moment strength comparison Quantity Probable strength Nominal strength Nominal strength

Axial Load Considered Pu = 1.44PD + PL Pu = 1.44PD + PL Pu = 0.9PD


=2820 k =2820 k =1560 k

Reinforcement Strength 1.25 fy = 75 ksi fy = 60 ksi fy = 60 ksi

Moment Strength Mpr = 125,000 k-ft Mpr = 111,000 k-ft Mn = 93,500 k-ft

229

Design Example 4

!


At the base of the wall, the magnified shear demand Vu * is calculated as follows: Vu * = ωv (M n M u )(VE )= 1.33(111,000 k − ft 75,500 k − ft )(1,470 k ) = 2,870 k Shear capacity.

Since this Design Example uses nominal shear strength to exceed the shear corresponding to flexural strength, a strength reduction factor, φ , of 0.85 can be used. As before, Equation (21-6) is used to calculated shear capacity:

[

]

φVn = 0.85 (20")(366") 2 + 5,000 + ρ n (60,000 psi ) = 880 k + 373,000ρ n

§1921.6.5

For the shear demand of 2870 k, the required amount of horizontal reinforcement is calculated: ρ n = (2,870 k − 880 k ) 373,000 = 0.00535 Try #8 @ 12" o.c. each face

(

ρ n = 2 0.79 in.2

) (12"× 20") = 0.00658 > 0.00535

o.k.

This amount of shear reinforcement is provided over the bottom two stories of the wall. For the other stories, the recommended amount of horizontal reinforcement, based on the magnified shear demand Vu*, is calculated as shown in Table 4-7.

Table 4-7. Horizontal reinforcement based on Blue Book shear design recommendations Level R 8 7 6 5 4 3 2

VE (k) 95 262 438 625 821 1030 1260 1470

Vu* (k) 186 512 856 1220 1610 2010 2460 2870


ρn

#5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #5@12” E.F. #7@12” E.F. #8@12” E.F. #8@12” E.F.

0.00258 0.00258 0.00258 0.00258 0.00258 0.00500 0.00658 0.00658

φVn (k) 1841 1841 1841 1841 1841 2742 3331 3331

Paulay and Priestley [1992] recommend equations for shear strength that are somewhat different than Equation (21-6), and in which the shear strength at plastic hinge zones is taken to be less than that at other wall locations. For the wall design in this Design Example, the Paulay and Priestley shear strength equations result in nearly identical amounts of horizontal reinforcement as does Equation (21-6).

230


Design Example 4

5c.

Discussion of UBC and Blue Book results for shear reinforcement.

!


Blue Book §C407.2.5

A comparison of Tables 4-6 and 4-7 shows that the Blue Book recommendation (§C407.2.5) of providing shear strength that exceeds flexural strength results in more horizontal reinforcement in the bottom three stories of the wall than that required by the code. The Blue Book approach is recommended by SEAOC, as it leads to more ductile wall behavior. In the upper five stories of the wall, the code minimum amount of horizontal steel (ρ n = 0.0025) is adequate to meet both the UBC requirements and the Blue Book recommendations. Overall, the additional cost of heavier bars in the first three stories, as determined under the Blue Book requirements, should not be significant. The wall elevation of Figure 4-10 shows the horizontal reinforcement per the Blue Book recommendation.

6.

Sliding shear (shear friction).

§1911.7

At construction joints and flexural plastic hinge zones, walls can be vulnerable to sliding shear. Typically lowrise walls are more vulnerable. If construction joint surfaces are properly prepared according to §1911.7.9, taller walls should not be susceptible to sliding shear failure. Sliding shear can be checked using the shear friction provisions of §1911.7. Shear strength is computed by Equation (11-25): Vn = Avf f y µ µ is the coefficient of friction, which is taken as 1.0λ , where λ = 1.0 for normal weight concrete. Avf is the amount of shear-transfer reinforcement that crosses the potential sliding plane. For the wall in this Design Example, all vertical bars in the section are effective as shear-transfer reinforcement [ACI-318 Commentary §R11.7.7]. At the base of the wall:

(

) (

)

Avf = 30 1.56 in. 2 + 54 0.79 in.2 = 89.5 in.2


231

Design Example 4

!


Section 1911.7.7 indicates that “permanent net compression” can be taken as additive to the force Avf f y , thus the lower bound axial load, 0.9 PD , can be included in Equation (11-25):

(

)

Vn = Avf f y + 0.9 PD µ

[(

]

)

= 89.5 in.2 (60 ksi ) + 1,560 k (1.0 ) = 6,930 k Section 1911.7.5 requires that the shear friction strength not be taken greater than 0.2 f ' c or 800 psi times the concrete area. For the example wall with f ' c = 5,000 psi , the 800 psi criterion governs: Vn ≤ (800 psi )(20"× 366") = 5,860 k > Vu * = 2,870 k

o.k.

By inspection, the sliding shear capacity at higher story levels of the building is also okay.

7.

Boundary zone detailing.

The code gives two alternatives for determining whether or not boundary zone detailing needs to be provided: a simplified procedure, §1921.6.6.4, and a strain calculation procedure, §1921.6.6.5.

7a.

UBC simplified procedure.

§1921.6.6.4

Under §1921.6.6.4, boundary zone detailing need not be provided if: Pu ≤ 0.10 Ag f ' c

( Pu ≤ 0.05 Ag f ' c for nonsymmetrical wall sections)

and either: M u (Vu l w ) ≤ 1.0 or: Vu ≤ 3 Acv

232

f 'c


Design Example 4

!


Use of this procedure for the wall in this Design Example is shown below: Pu = 1.44 PD + PL = 2,820 k 0.10 Ag f ' c = 0.10(20"× 366")(5.0 ksi ) = 3,660 k > 2,820 k M u (Vu l w ) = 75,000 k − ft [(1,470 k )(30.5')] = 1.68 > 1.0 3 Acv f 'c = 3(20"×366") 5,000 psi = 1,550,000# = 1,550 k > Vu = 1,470 k Therefore, boundary zone detailing as defined in §1921.6.6.6 is not required.

7b.

UBC strain calculation procedure.

§1921.6.6.5

Section 1921.6.6.5 requires the calculation of total curvature, φ t , at the plastic hinge region of the wall. The procedure applies only when the plastic hinge is located at the base of the wall, which is the case for the example wall. Total curvature is calculated by the following equation: φt =

∆i hw − l p 2 l p

(

)

§1630.9.2

where ∆ i = ∆ t − ∆ y and ∆ t = ∆ m ,when the analysis has used effective stiffness (cracked section) properties ∆ m is defined in Equation (30-17) of §1630.9.2 as ∆ m = 0.7 R∆ s ∆ s is the design level response displacement. For the example wall at the top, it is the displacement ∆ s = 2.32 inches, taken from the analysis. ∆ m = 0.7 R∆ s = 0.7(4.5)(2.32") = 7.32"

(

)

∆ y is the yield displacement of the wall, taken as M ' n M E ∆ E . For the example

wall, ∆ E , the displacement corresponding to M E , is equal to ∆ s (= 2.32"), the displacement taken from the analysis.


233

Design Example 4

!


Calculation of Mń requires a re-calculation of the moment strength at the base of the wall, this time using the axial load Pú = 1.2 PD + 0.5PL . The results of the calculation, including the neutral axis depth, are shown in Table 4-8, below.

Table 4-8. Summary of Mń calculation Quantity

Axial Load Considered

M'n

P'u = 1.2PD + 0.5PL

(

= 2240 k

Reinforcement Strength fy = 60 ksi

Moment Strength M'n = 103,000 k-ft

Neutral Axis Depth c'u = 78.0”

)

∆ y = M ' n M E ∆ E = (103,000 k − ft 75,500 k − ft )(2.32") = 2.54" ∆ i = ∆ t − ∆ y = 7.31"−2.54" = 4.15" The height of the wall, hw , equals 95.3 ft (1140 in.), and the plastic hinge length, l p will be taken as 0.5l w (183 in). The yield curvature φ y , can be estimated as 0.003 / l w . Substituting these values into Equation (21-9): φt =

4.15" + 0.003 366" = 29.8(10 )−6 in.−1 (1,140"−183" 2)183"

The compressive strain at the extreme fiber of the section equals the total curvature times the neutral axis depth:

(

)

ε c = φ t c ' u = 29.8(10 )−6 in.−1 (78") = 0.00233 < 0.003 ∴ Boundary confinement not required. Note that assuming a smaller plastic hinge length, l p = 9.7 ft = 116" , as defined in Part 4b above, results in a strain of 0.00321, which would require that boundary confinement be provided.

7c.

Blue Book recommendations.

Blue Book §402.11.1

Section 402.11 of the Blue Book modifies the UBC, including a revised formula for ∆ t that gives a more realistic estimate of inelastic seismic displacements and corrects a tendency for the UBC strain calculation procedure to give unconservative results. Section 402.11.1 of the 1999 Blue Book replaces the definition of ∆ t to give: ∆ t = R∆ s

234


Design Example 4

!


For the example wall in this Design Example, this gives: ∆ t = R∆ s = 4.5(2.32") = 10.4" ∆ i = ∆ t − ∆ y = 10.4"−3.17" = 7.28" Plugging this value of ∆ i into Equation 21-9 gives: φt =

7.28" + 0.003 366" = 46.1(10)−6 (1,140"−183" 2)183"

The compressive strain at the extreme fiber of the section equals the product of the total curvature and the neutral axis depth:

(

)

εc = φt c 'u = 46.1(10)−6 in.−1 (78") = 0.00360 > 0.003 ∴ Boundary confinement is required. Assuming a smaller plastic hinge length, l p = 9.7 ft = 116 in., as defined in Part 4b above, results in a strain of 0.00515, further indicating the prudence of adding boundary confinement to the subject wall. Section 402.12 of the SEAOC Blue Book requires that all wall edges in potential plastic hinge regions have ties spaced at 6d b or 6 inches maximum, to restrain the buckling of bars. For the wall in this Design Example, #4 tie sets at 6 inches on center, with a tie leg located at each of the #11 bars, as shown in Figure 4-11, and on the wall elevation of Figure 4-10, should be provided as a minimum.

Figure 4-11. Boundary reinforcement at wall base


235

Design Example 4

!


References ATC-43, 1999. Evaluation of Earthquake Damaged Concrete and Masonry Wall Buildings, prepared by the Applied Technology Council (ATC-43 project) for the Partnership for Response and Recovery. Federal Emergency Management Agency, Report No. FEMA-306, Washington, D.C. CRSI, 1996. Rebar Design and Detailing Data – ACI. Concrete Reinforcing Steel Institute, Schaumberg, Illinois. Maffei, Joe, 1996. “Reinforced Concrete Structural Walls — Beyond the Code,” SEAONC Fall Seminar Proceedings. Structural Engineers Association of Northern California, San Francisco, California, November. Paulay, T., and M.J.N. Priestley, 1992. Reinforced Concrete and Masonry Buildings, Design for Seismic Resistance. John Wiley & Sons, Inc., New York. (Chapter 5 covers seismic behavior and design of reinforced concrete walls, including examples. The book is not based on the ACI or UBC codes, but explains the principles that underlie several code provisions.) Paulay, T., and M.J.N. Priestley, 1993. Stability of Ductile Structural Walls. ACI Structural Journal, Vol. 90, No. 4, July-August 1993. PCA, 1999. “PCACOL: Design and Investigation of Reinforced Concrete Column Sections,” Portland Cement Association, Skokie, Illinois.

236


Design Example 5

!

Reinforced Concrete Wall with Coupling Beams

Design Example 5 Reinforced Concrete Wall with Coupling Beams

Figure 5-1. Six-story concrete office building (partial view)

Overview The structure in this Design Example is a 6-story office building with reinforced concrete walls (shear walls) as its lateral force resisting system. The example focuses on the design and detailing of one of the reinforced concrete walls. This is a coupled wall running in the transverse building direction and is shown in Figure 5-1. The example assumes that design lateral forces have already been determined for the building, and that the seismic moments, shears, and axial loads on each of the wall components, from the computer analysis, are given.


237

Design Example 5

!


The purpose of this Design Example is to illustrate the design of coupling beams and other aspects of reinforced concrete walls that have openings. Research on the behavior of coupling beams for concrete walls has been carried out in New Zealand, the United States, and elsewhere since the late 1960s. The code provisions of the UBC derive from this research.



2.

Preliminary sizing of shear wall.

3.

Coupling beam design.

4.

Design of wall piers for flexure.

5.

Plastic analysis of flexural mechanism in walls.

6.

Design of wall piers for shear.

7.

Boundary zone detailing of wall piers.

8.

Detailing of coupling beams.

Given Information The following information is given: Seismic zone = 4 Soil profile type = S D Near-field = 5 km from seismic source type A Redundancy/reliability factor, ρ = 1.0 Importance factor, I = 1.0 Concrete strength, f 'c = 4000 psi Steel yield strength, f y = 60 ksi The wall to be designed, designated Wall 3, is one of several shear walls in the building. The wall elevation, a plan section, and the design forces are shown in Figure 5-2. An elastic analysis of the wall for lateral forces, using a computer program, gives the results shown in Figure 5-3, which shows the moments and shear for each coupling beam (i.e., wall spandrel), and the moments, shear and axial forces for each vertical wall segment (i.e., wall pier). 238


Design Example 5

!


Lateral story displacements, corresponding to gross section properties, are also shown on the figure. Where displacements are used in design they should correspond to effective section properties rather than gross section properties, as indicated in §1633.2.4. Typical practice is to use a percentage of the gross stiffness, e.g., 50 percent, for the effective stiffness. In such a case, the displacements from the gross section model can be uniformly factored up. The displacements for a linear elastic model using 50 percent of Ig will be two times the displacements using the gross section properties. In this Design Example, the displacement output is not used. In an actual building design, the displacements would need to be considered for: 1.) design of elements not part of the lateral-force-resisting system, 2.) building separations, 3.) boundary design by the strain calculation procedure, and 4.) P∆ analysis. Other recommendations for member stiffness assumptions are given in Section 5.3 of Paulay and Priestley [1992]. Gravity loads are not included in the computer model. Gravity effects are added separately by hand calculations.

2

Plan

Elevation

Figure 5-2. Wall elevation, plan section, and design forces of Wall 3


239

Design Example 5

!


Units: P=kips beam moment at edge of wall piers V=kips pier moments at floor levels M=kips-inch

Figure 5-3. Results of ETABS computer analysis for Wall 3

240


Design Example 5

!



1.

Code Reference


Load combinations for reinforced concrete are discussed in detail in Part 1 of Design Example 4. As in that example, we assume here that the presiding building department has indicated approval of the SEAOC recommended revisions to the UBC load combinations. Thus the governing load combinations become:

(1.2 ± 0.5Ca I )D ± ρEh + ( f1L +

f2S )

Blue Book §101.7.2.1

0.9 D ± ρEh

Blue Book §C403.11

Since the given structure is an office building, f1 = 0.5 . And since there is no snow load, S = 0 . The same seismic zone, soil profile, near-field, redundancy, and importance factors are assumed as for Design Example 4, thus C a = 0.484 . With I = 1.0 and ρ = 1.0 , the governing load combinations for this Design Example are: 0.9 D ± Eh

[1.2 ± 0.5(0.484)]D ± Eh + L

{

= 1.44 D ± Eh + 0.5 L = 0.958D ± Eh + 0.5L

does not govern

The forces shown in Figure 5-3 correspond to Eh .

2.

Preliminary sizing of shear wall.

For walls with diagonally reinforced coupling beams, the required wall thickness is often dictated by the layering of the reinforcement in the coupling beam. Typically, a wall thickness of 15 inches or larger is required for diagonally reinforced coupling beams conforming to the 1997 UBC. For the subject wall, a wall thickness, bw , of 16 inches will be tried.


241

Design Example 5

3. 3a.

!


Coupling beam design.

Requirement for diagonal reinforcement.

Code requirements for the diagonal reinforcement of coupling beams (§1921.6.10.2) are based on the clear-length to depth ratio for the coupling beam, l n d , and on the level of shear stress in the coupling beam. For the wall in this Design Example, it will be assumed that d equals 0.8 times the overall depth, so that l n d = 72" (0.8 × 72") = 1.25 for the typical coupling beam, and l n d = 72" (0.8 × 120") = 0.75 for the coupling beams at the second floor. As shown in Table 5-1 (6th column), for five of the nine coupling beams the shear exceeds 4 f 'c bw d . For these coupling beams, diagonal reinforcement is required. For the four coupling beams that have lower shear stress, diagonal reinforcement is not required by the UBC. Designing these 4 coupling beams without diagonal reinforcement, using horizontal reinforcement to resist flexure and vertical stirrups to resist shear, might lead to cost savings in the labor to place the reinforcing steel. In this Design Example, however, diagonal reinforcement is used in all of the coupling beams of the wall because: 1.) it can simplify design and construction to have all coupling beams detailed similarly, and 2.) research results show that diagonal reinforcement improves coupling beam performance, even at lower shear stress levels, as discussed in §C407.7 of the SEAOC Blue Book.

Table 5-1. Coupling beam forces and diagonal reinforcement Grid Line Level C-D C-D C-D C-D C-D C-D D-E D-E D-E

Roof 6th 5th 4th 3rd 2nd 4th 3rd 2nd

Vu (kips) 151 325 447 211 180 285 319 454 406

h (in.) 72 72 72 72 72 120 72 72 120

d (in.) 57.6 57.6 57.6 57.6 57.6 96.0 57.6 57.6 96.0

Vu bw d f 'c

(1)

Diagonal Bars

Ad (in.2)

α (degrees)

φVn (kips)

φVn Vu

4-#8 4-#10 6-#10 4-#9 4-#9 4-#9 6-#9 6-#10 4-#10

3.16 5.08 7.62 4.00 4.00 4.00 6.00 7.62 5.08

37.9 37.9 36.0 37.9 37.9 53.1 36.0 36.0 53.1

198 318 456 251 251 326 359 456 414

1.31 0.98 1.02 1.19 1.39 1.14 1.13 1.00 1.02

2.6 5.6 7.7 3.6 3.1 2.9 5.5 7.8 4.2

Note: Diagonal bars are required when this ratio exceeds 4.

242


Design Example 5

3b.

!


Design of diagonal reinforcement.

Diagonal reinforcement is provided in the coupling beams according to Equation (21-1) of §1921.6.10.2: φVn = 2φf y sin αAvd

(21-1)

Each group of diagonal bars must consist of at least 4 bars (§1921.6.10.2). The calculation of the required diagonal reinforcement is shown in Table 5-1. For coupling beams with higher shear stresses, 6 bars are needed in each group, as shown in Table 5-1. The angle α of the diagonal bars is calculated based on the geometry of the reinforcement layout, as shown in Figure 5-4. The value of α depends somewhat on overall dimension of the diagonal bar group and on the clearance between the diagonal bar group and the corner of the wall opening. This affects the dimension x shown in Figure 5-4 and results in a slightly different value of α for a group of 6 bars compared to that for a group of 4 bars, as shown in Table 5-1. The provided diagonal bars are shown in Figure 5-5.

Figure 5-4. Geometry of coupling beam diagonal bars


243

Design Example 5

!


Figure 5-5. Diagonal bars provided in coupling beams

4.

Design of wall piers for flexure.

The design of the vertical wall segments for flexure is carried out following the procedures and recommendations given for conventional “solid” walls. This is shown in Part 3 of Design Example 4. From Figure 5-3, the critical wall segments (i.e., those with the highest moments or earthquake axial forces) include the wall pier at the 4th floor on Line D, and the wall piers at the base on Lines C and E. The 20-foot long wall pier on Line D at the base is also checked.

244


Design Example 5

4a.

!


Critical moments and axial forces.

As can be seen from Figure 5-2, the gravity loads on each wall pier are not concentric with the wall pier centroid. Therefore, gravity load moments must be considered in the design of flexural reinforcement. The dead and live loads (except wall self-weight shown in Table 5-2) in Figure 5-2 act at the column grid lines, and have an eccentricity, eDF , with respect to the section centroid, as given in Table 5-3 (Note: The calculation of weights, section centroids, eDF, and eDW is not shown). The wall self-weight provides additional dead load at each level, equal to the values given in Table 5-2.

Table 5-2. Dead load from wall self-weight Level Above 6th Above 5th Above 4th Above 3rd Above 2nd At base

Line C Sum of Wall Eccentricity, Weight (kips) eDW (ft) (1) 26 53 79 106 132 166

2.06 2.06 2.06 2.06 2.06 2.03

Line D Sum of Wall Eccentricity, Weight (kips) eDW (ft) (1) 26 53 79 132 185 252

-2.06 -2.06 -2.06 -3.71 -2.65 -1.94

Line E Sum of Wall Eccentricity, Weight (kips) eDW (ft) (1) 0 0 0 26 53 86

-2.06 -2.06 -2.00

Note: 1. eDW = distance between centroid of weight and centroid of wall section.

The calculation of the factored forces on the critical wall piers is shown in Table 5-3. In this table, gravity moments are calculated about the section centroid, using the gravity loads acting at the column centerline, PDF and PL , plus the dead load from wall self-weight, PDW. Earthquake moments, ME, are taken from Figure 5-3. Loads are factored according to the combinations discussed in Part 1 of this Design Example, giving two cases for each wall pier: minimum axial load and maximum axial load. The minimum axial load case is based on the combination of Eh with 0.9 D , and the maximum axial load case is based on the combination of Eh with 1.44 D + 0.5 L . Considering that larger axial compression generally increases moment strength, potentially governing combinations are shown as shaded areas in Table 5-3.


245

Design Example 5

!


Table 5-3. Calculation of factored axial forces and moments on critical wall piers Level Line 4th 4th 1st 1st 1st 1st 1st

D D C C E E D

PDF eDR (kips) (ft) 428 428 874 874 874 874 874

PDW e DW PL Direction PE (kips) (ft) (kips) of force (kips)

-4.13 79 -4.13 79 4.13 166 4.13 166 -4.13 86 -4.13 86 0 252

-2.06 -2.06 2.03 2.03 -2.00 -2.00 -1.94

44 44 100 100 100 100 100

west east west east west east west

-923 923 1,600 -1,600 -1,179 1,179 -421

ME (k-ft) -6,070 6,070 -4,105 4,105 -4,191 4,191 -13,250

MD M L Minimum Axial Maximum Axial MU PU MU (k-ft) (k-ft) PU 1,603 1,603 -3,268 -3,268 3,433 3,433 -489

182 182 -413 -413 413 413 0

-467 -4,628 1,379 7,512 2,536 -7,047 -664 1,164 -315 -1,101 2,043 7,281 592 -13,690

-171 -3,671 1,675 8,469 3,148 -9,018 -52 -807 253 959 2,611 9,341 1,250 -13,954

Notes: PDF = dead load distributed over floor area, which acts at the column line. e DF = distance between PDF and centroid of wall section. PDW = dead load from wall self-weight. e DW = distance between PDW and centroid of wall section.

4b.

Vertical reinforcement.

The program PCACOL [PCA, 1999] is used to design the reinforcement in each wall pier. Figure 5-6 shows a wall section with the typical layout of vertical reinforcement. Typical reinforcement in the “column” portion of the wall piers is 8-#9 and typical vertical reinforcement in the wall web is #7@12. The PCACOL results of Figure 5-7a, 5-7b, and 5-7c show that this reinforcement is adequate in all locations except Line D at the 4th floor where 8-#10 are required instead of 8-#9. Figure 5-7d shows that the typical reinforcement provides adequate moment strength to the 20-foot long wall pier on Line D. Figure 5-8 shows the vertical reinforcement provided in the wall piers to satisfy moment strength requirements. Note that the vertical reinforcement in the column portion of the 4th floor piers is increased to 8-#11 (from 8-#9 used at the lower levels), and that at the 5th and 6th floors is increased to 8-#10. The reasons for this will be discussed in Part 5 of this Design Example.

Figure 5-6. Section through wall pier in vicinity of Line C 246


Design Example 5

!


a.

b.

c.

d.

Figure 5-7. PCACOL results for design of vertical reinforcement


247

Design Example 5

!


Figure 5-8. Elevation of vertical wall reinforcement

4c.

Lap splice locations.

In general, lap splices should be avoided in potential plastic hinge regions of concrete structures. This is discussed in Part 4b of Design Example 4 and in Blue Book §C404.3. For this example wall, plastic hinging is expected (and desired) at the base of each wall pier and in the coupling beams. Plastic hinging may also be possible above the wall setback, in the 4th floor wall piers. (This will be investigated in more detail in Part 5 of this Design Example.) Lap splices of the vertical wall reinforcement are located to avoid the potential plastic hinge regions in first floor and fourth floor wall piers, as shown in Figures 5-10 and 5-11 and in Tables 5-5 and 5-6 in Part 5B, below.

248


Design Example 5

5.

Plastic analysis of flexural mechanism in walls.

!


Blue Book §C402.8, C407.5.5.2

This part of the Design Example presents a plastic analysis methodology that is not a code requirement. It is included to assist the reader in understanding the postelastic behavior of coupled shear walls and how they can be analyzed for seismic forces when elements of the wall are yielding. Plastic analyses are not required by the UBC, but they are recommended in the SEAOC Blue Book: 1.) to establish shear demand corresponding to flexural strength, and 2.) to identify potential plastic hinge regions where special boundary and splicing requirements may be necessary. With the trend toward nonlinear static analysis (pushover) procedures, as called for in performance-based structural engineering guidelines [FEMA-273, 1997 and ATC-40, 1996], the ability to use plastic analyses will become increasingly important. The first three chapters of the textbook Plastic Design in Steel [ASCE, 1971] summarize the basic principles and methods of plastic design, and these are recommended reading for the interested reader. Given below is an illustration of plastic analysis for the reinforced concrete walls and coupling beams of this Design Example.

5a.

Probable moment strength.

The “probable flexural strength,” Mpr, will be determined in calculating shear demands, according to the Blue Book recommendations. As defined in §1921.0, Mpr is calculated assuming a tensile stress in the longitudinal bars of 1.25 f y , and a strength reduction factor, φ , of 1.0. For the purposes of this plastic analysis, we will neglect earthquake axial forces Ev in calculating Mpr for each wall pier and assume an axial load of 1.2 PD + 0.5 PL . In reality, the wall pier with earthquake axial tension will have a decreased moment strength, while the wall pier with earthquake axial compression will have an increased moment strength. These effects tend to cancel out so that our plastic analysis will give a good estimate of 1.) the governing mechanism of response, and 2.) the shear corresponding to the development of a mechanism at probable flexural strength. Table 5-4 shows Mpr values for the critical wall piers, based on the PCACOL results shown in Figure 5-9.


249

Design Example 5

!


a.

b.

c.

d.

(

Figure 5-9. PCACOL calculation of probable moment strength M pr fy = 75 ksi, φ = 1.0 250

)


Design Example 5

!


Table 5-4. Approximate probable moment strengths of wall piers for plastic analysis

5b.

Level

Grid Line

Reinforcement of Column Portion

Axial Load Considered 1.2PD + 0.5PL (kips)

4th 4th 1st 1st 1st 4th 4th

C D C D E C D

8-#9 8-#10 8-#9 8-#9 8-#9 8-#11 8-#11

630 630 1,300 1,400 1,200 630 630

M pr (k-ft) 10,500 7,500 12,500 28,000 10,000 13,000 8,000

Mechanism with plastic hinging at the base.

The preferred behavior of the wall occurs when plastic hinges occur at the base of the wall piers and in the coupling beams. This produces the desirable situation of flexural yielding, energy dissipation, and avoidance of shear failures. Table 5-5 shows calculations of the shear strength of the preferred plastic mechanism, which has plastic hinges forming at the base of each wall pier and in each coupling beam. The equivalent plastic hinge length at the pier base, lp, is taken equal to 5 feet. The plastic hinge length is used in the calculation of external work shown in Table 5-5. The calculation is not sensitive to the value of lp assumed, since lp /2 is subtracted from hi, the height above the base. In this case, the value of 5 feet is taken as one-half the wall length of the external wall piers. Although the central pier is longer, it is assigned the same plastic hinge length. Note that in the strain calculation procedure for wall boundary design, the value used for lp has a significant effect on the results. This is discussed in Part 7 of Design Example 4. Plastic lateral story displacements, ∆ i , increase linearly with height above the midpoint of the base plastic hinges. ∆ i is arbitrarily set equal to 1.00 feet at the roof. The external work equals the sum of each lateral story force, fxi, times ∆ i .


251

Design Example 5

!


The plastic rotation angle of the wall piers, θ , equals the roof displacement divided by the roof height above the midpoint of the plastic hinge. Thus, θ = 1.00 85.5 . The plastic rotation angle and internal work of the coupling beams can be calculated as follows: θ cb = θ

lc ln

where: l n = clear length of the coupling beam lc = distance between centroids of wall pier sections Internal work

(

)

= Σ θcb × M pr for each end of each coupling beam = Σ(θcb × 1.25Vn ln 2 ) = Σ(θ ×1.25Vn lc 2 ) = Σ(θ × 1.25Vn lc ) for each coupling beam (sum of 2 ends)

The internal work of the base plastic hinges equals the sum of Mpr times θ for each of the three base plastic hinges. The summation of the internal work is shown in Table 5-5. Equating internal work with external work gives the solution of V = 2,420 kips .

252


Design Example 5

!


Table 5-5. Plastic mechanism calculations assuming plastic hinging (1) at base and in all coupling beams External Work Level R 6th 5th 4th 3rd 2nd Sum

hi (ft)

hi − l p 2 (ft)

∆i (ft)

fxi V

Work / V (ft)

88 74 60 46 32 18

85.5 71.5 57.5 43.5 29.5 15.5

1.000 0.836 0.673 0.509 0.345 0.181

0.254 0.240 0.195 0.149 0.104 0.058 1.000

0.254 0.201 0.131 0.076 0.036 0.011 0.708

Work (k-ft)

Internal Work, Coupling Beams Grid Line

Level

1.25Vn (k)

lc (ft)

C-D C-D C-D C-D C-D C-D D-E D-E D-E

R 6th 5th 4th 3rd 2nd 4th 3rd 2nd

291 468 671 368 368 480 528 671 609

21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5 21.5

θ = 1.00/85.5

Internal Work, Wall Piers Grid Line

Level

C D E

base base base

73 118 169 93 93 121 133 169 153 1,120

M pr (k-ft)

Work (k-ft)

12500 28000 10000

146 327 117 591

V= (1120 + 591)/0.708 = 2,420 kips

Note: 1. See Figure 5-10 for illustration of hinge locations.


253

Design Example 5

!


Figure 5-10. Mechanism with plastic hinges at base of wall

5c.

th

Mechanism with plastic hinging at the 4 floor.

Table 5-6 shows calculations of the shear strength of another possible plastic mechanism, which has plastic hinges forming at the 4th floor wall piers and only in the coupling beams at the 5th, 6th, and roof levels. This plastic mechanism is less desirable than a mechanism with hinging at the base, because energy dissipation is concentrated in fewer yielding locations, and because plastic rotations in the wall piers would need to be much greater to achieve the same roof displacement. As in the previous calculation, plastic lateral story displacements, ∆ i , increase linearly with height above the midpoint of the base plastic hinges, and ∆ i is set equal to 1.00 feet at the roof. For this mechanism, the plastic rotation angle of the wall piers, θ , equals 1.00/39.5. The plastic analysis solution, based on equating internal and external work, gives V = 2,300 kips . Since this is less than 2,420 kips, the mechanism having plastic hinging at the 4th floor governs (i.e., is more likely to form than the preferred base mechanism shown in Figure 5-10). To help prevent plastic hinging in the 4th floor piers, their flexural strength can be increased. Reinforcement of the column portions of these wall piers is increased to 8-#11. Table 5-6 shows revised internal work calculations. The solution gives V = 2,460 kips . Since this is greater than 2420 kips, the preferred mechanism now governs. Note that the calculation of the governing plastic limit load, V, depends on the assumed vertical distribution of lateral forces, which in actual seismic response can vary significantly from the inverted triangular pattern assumed. Thus the difference between V = 2,420 kips and 2,460 kips does not absolutely ensure against plastic hinging in the 4th floor wall piers.

254


Design Example 5

!


Inelastic dynamic time-history analyses by computer generally show less predictability of yield locations than plastic analyses imply. For the wall of this Design Example, a time-history analysis might show some wall pier yielding both at the base and at the 4th floor. Interaction of the wall with other walls in the structure and with gravity framing can also influence the mechanism of yielding. Plastic analyses are simpler to carry out and understand than most other analysis methods, particularly inelastic time-history analyses, and they offer valuable insight into the seismic performance of a structure. For this Design Example, the plastic analyses indicate that strengthening the 4th floor piers will protect the upper stories above the setback against high ductility demands, and make it more likely that the preferred mechanism will form.

Table 5-6. Plastic mechanism calculations assuming plastic hinging th (1) at 4 floor piers External Work Level R 6th 5th 4th 3rd 2nd Sum

hi (ft)

hi − l p 2 (ft)

∆i (ft)

fxi V

Work / V (ft)

42 28 14

39.5 25.5 11.5

1.000 0.646 0.291 0.000 0.000 0.000

0.254 0.240 0.195 0.149 0.104 0.058 1.000

0.254 0.155 0.057 0.000 0.000 0.000 0.466

Internal Work, Coupling Beams Grid Line

Level

1.25Vn (k)

lc (ft)

Work (k-ft)

C-D C-D C-D Sum

R 6th 5th

291 468 671

17 17 17

125 201 289 615 θ = 1.00/39.5


Level

C D Sum

4th 4th

M pr (k-ft)

Work (k-ft)

10500 7500

266 190 456

V= (615 + 456)/0.466 = 2,300 kips

Note: 1. See Figure 5-11 for illustration of hinge locations.


255

Design Example 5

!


Table 5-7. Plastic mechanism calculations assuming plastic hinging th th at 4 floor piers—revised for stronger piers at 4 floor θ = 1.00/39.5


Level

C D Sum

4th 4th

M pr (k-ft) 13000 8000

Work (k-ft) 329 203 532

V= (615 + 532)/0.466 = 2,460 kips

th

Figure 5-11. Mechanism with plastic hinges at 4 floor wall piers

6.

Design of wall piers for shear.

In this part, the wall piers will be designed for shear. Both the UBC and Blue Book approaches will be illustrated. Design for the minimum UBC requirements is given in Part 6a below. As discussed in Part 5 of Design Example 4, the SEAOC Blue Book contains more restrictive requirements than does the UBC for the shear design of reinforced concrete walls. The SEAOC approach, in Part 6b of this Design Example, is recommended for the reasons given in Design Example 4.

256


Design Example 5

6a.

!


Design under UBC requirements. Shear demand.

If designing to the minimum requirements of the UBC, the shear demand is taken directly from the design forces, factored by the load combinations discussed in Part 1. For the example wall, all of the significant shear on the wall piers results from earthquake forces, thus Vu = VE , where the values VE are those shown in Figure 5-3. The highest shears are at the 4th floor, Line D, with VE = 544 kips in an 11-foot-long wall pier (48.5 k/ft), and at the 1st floor, Line D, with VE = 731kips in a 20-foot long wall pier (36.6 k/ft). Shear capacity.

§1921.6.5

UBC §1911.10 gives shear provisions for walls designed for nonseismic lateral forces such as wind or earth pressure. Section 1921.6.5 gives shear strength provisions for walls designed for seismic forces. In Equation (21-7), wall shear strength depends on α c , which depends on the ratio hw l w .

(

Vn = Acv α c

f 'c + ρ n f y

)

(21-7)

Per §1921.6.5.4 the ratio hw l w is taken as the larger of that for the individual wall pier and for the entire wall. Overall wall

hw l w = 88' 54'

= 1.63

11' long by 8' clear-height pier

hw l w = 8' 11'

= 0.73

20' long by 8' clear-height pier

hw l w = 8' 20'

= 0.40

Thus the value hw l w = 1.63 governs for all wall piers. The coefficient α c varies linearly from 3.0 for hw l w = 1.5 to 2.0 for hw l w = 2.0 . α c = 3.0 − 1.0(1.63 − 1.5) (2.0 − 1.5) = 2.74


257

Design Example 5

!


As prescribed in §1909.3.4.1, the shear strength reduction factor, φ , shall §1921.6.5.3 be 0.6 for the design of walls if their nominal shear strength is less than the shear corresponding to development of their nominal flexural strength. For the 11-foot long wall piers:

[

]

φVn = 0.6(16") lw 2.74 4,000 + ρ n (60,000 psi ) = lw (1.66 k − in. + 576 k − in. ρn ) For the wall sections with highest shear, the amount of horizontal shear reinforcement is given in Table 5-8.

Table 5-8. Design for shear by the UBC Level

Grid Line

lw (in.)

VE (kips)


ρn

φVn (kips)

4th 4th 1st 1st 1st

C D C D E

132 132 132 240 132

371 544 283 731 316

#4@10” E.F. #6@10” E.F. #4@10” E.F. #4@10” E.F. #4@10” E.F.

0.00250 0.00550 0.00250 0.00250 0.00250

409 637 409 744 409

Vu φAcv f 'c

(1)

4.63 6.79 3.53 5.02 3.95

Note: 1. Under §1921.6.5.6, the value of Vu φAcv for an entire wall section.

f 'c shall not exceed 10 for any wall pier, or 8

As shown above, for all wall pier locations except the 4th floor at Line D, the minimum reinforcement ratio of 0.0025 (required under §1921.6.2.1) is sufficient to meet UBC shear strength requirements.

6b.

Design using Blue Book recommendations. Shear demand.

SEAOC 402.8, C402.8

To comply with the Blue Book requirement of providing shear strength in excess of the shear corresponding to wall flexural strength, an amplified shear demand must be considered. For this Design Example, shear strength in excess of that corresponding to the development of probable flexural strength will be provided. This has been calculated by the plastic analysis in Part 5 of this Design Example as V = 2,420 kips at the base of the wall. Section C402.8 of the Blue Book Commentary gives the following equation for the shear amplification factor, ωv , that accounts for inelastic dynamic effects. For application to designs according to the UBC, the amplification factor recommended by Paulay and Priestley [1992] can be reduced by a factor of 0.85, because the Paulay and Priestley recommendations use a different strength reduction factor, φ , than does the UBC. 258


Design Example 5

ωv

!


= 0.85(0.9 + n / 10), for buildings up to 6 stories, where n = number of stories = 0.85(0.9 + 6 / 10) = 1.28

As indicated in the Blue Book, the ωv factor is derived for analysis using inverted triangular distributions of lateral forces. If a response spectrum analysis is carried out, a slightly lower ωv factor can be justified in some cases. At the base of the wall, the magnified shear demand Vu * is calculated as follows:

(

)

Vu * = ωv M pr M u (VE ) = (ωv 2,420 kips ) = 1.28(2,420 ) = 3,100 kips In the plastic analysis, the amplification effect considered by ωv can instead be considered by using a different vertical distribution of the lateral forces, fxi. Rather than using the inverted triangular distribution, a vertical distribution with a resultant located lower in the building, such as a uniform distribution pattern, could be used in the plastic analysis to give shear forces. Shear capacity.

Since we are designing for the nominal shear strength to exceed the shear corresponding to flexural strength, a strength reduction factor, φ, of 0.85 can be used. As before, UBC Equation (21-6) is used to calculate shear capacity:

(

Vn = Acv α c

f 'c + ρ n f y

)

(21-7)

[

]

φVn = 0.85(16") lw 2.74 4,000 + ρ n (60,000 psi ) = lw (2.36 k − in. + 816 k − in. ρn ) For the shear demand of 3100 k over the net wall length of 42 feet (504 inches) at the first floor, the required amount of horizontal reinforcement is calculated: φVn = 504(2.36 + 816ρ n )= 1,190 + 411,000ρ n ≥ 3,100 ρ n = (3,100 k − 1,190 k ) 411,000 = 0.00464 Try #6 @ 12" o.c. each face

(

)

ρ n = 2 0.44 in.2 (12"×16") = 0.00458

o.k.

For the other stories of the building, the shear demands are magnified from the analysis results by the same proportion as for the first floor. The recommended amount of horizontal reinforcement can be calculated as shown in the Table 5-9.


259

Design Example 5

!


Table 5-9. Design for shear by the Blue Book recommendations Level 6th 5th 4th 3rd 2nd 1st

VE (kips) 338 656 915 1,150 1,250 1,310

Vu * (kips) (1)

l w net (in.)


ρn

φVn (kips)

264 264 264 504 504 504

#5@12” E.F. #6@12” E.F. #6@8” E.F. #6@12” E.F. #6@12” E.F. #6@12” E.F.

0.00323 0.00458 0.00688 0.00458 0.00458 0.00458

1,320 1,610 2,100 3,070 3,070 3,070

788 1,530 2,130 2,680 2,920 3,100

Note: 1. Vu * = magnified shear demand.

At the 4th floor wall piers, the vertical reinforcement must be increased from #7@12" to #8@12" to provide ρ v ≥ ρ n , per §1921.6.55.5. The Blue Book deletes this requirement for the reasons given in Blue Book §C402.9. However, in this case, the increase in flexural strength of the 4th floor wall piers is desirable, as discussed in Part 5C, above.

6c.

Recommended shear reinforcement.

A comparison of the Tables 5-8 and 5-9 shows that the Blue Book recommendations for ensuring that shear strength exceeds flexural capacity results in increased horizontal reinforcement compared to that required by the UBC. The Blue Book approach is recommended, as it leads to more ductile wall behavior.

7.

Boundary zone detailing of wall piers.

The UBC gives two alternatives for determining whether or not boundary zone detailing needs to be provided: a simplified procedure (§1921.6.6.4), and a strain calculation procedure (§1921.6.6.5). For this Design Example, the simplified procedure will be used, and for comparison the Blue Book recommendations for the strain calculation procedure will be checked. For an illustration of the UBC strain calculation procedure, see Design Example 4.

260


Design Example 5

7a.

!


UBC simplified procedure.

§1921.6.6.4

Under the requirement of §1921.6.6.4, boundary zone detailing need not be provided in the example wall if the following conditions are met: Pu ≤ 0.10 Ag f 'c

( Pu ≤ 0.05 Ag f 'c for unsymmetrical wall sections)

and either M

u

(Vu l w ) ≤ 1.0

or Vu ≤ 3 Acv

f 'c

For the critical piers of the example wall, Pu /Agf′c calculated as shown in Table 5-10. All of the piers are geometrically unsymmetrical, except for those on Line D at the 1st, 2nd, and 3rd stories. Of the unsymmetrical piers, only those at the 6th floor have Pu Ag f ' c ≤ 0.005 and Vu ≤ 3 Acv f ' c . All three of the symmetrical piers have Pu / Ag f c′ ≤ 0.01 and Vu ≤ 3 Acv

f ' c . Therefore all piers require th

boundary confinement except those at the 6 floor, and those on Line D at the 1st, 2nd, and 3rd floors. The required boundary zone length is calculated as a function of Pu / Ag f c′ per §1921.6.6.4. The code requires that shear walls and portions of shear walls not meeting the conditions of §1921.6.6.4 and having Pu < 0.35Po shall have boundary zones at each end over a distance that varies linearly from 0.25l w to 0.15l w as Pu varies from 0.35Po to 0.15Po . The boundary zone shall have a minimum length of 0.15l w and shall be detailed in accordance with §1921.6.6.6. The results of this determination are shown in Table 5-10.

Table 5-10. Boundary zone strength requirement by the UBC simplified procedure Level

Line

6th 4th 1st 1st 1st

C,D D C E D

Pu

(1.44PD + 0.5PL + PE ) (kips) 388 1,675 3,148 2,611 1,250


(in.2)

Pu Ag f 'c

(Required Boundary Length) ÷ lw

Required Boundary Length (in.)

2,300 2,300 2,300 2,300 4,030

0.042 0.182 0.342 0.284 0.078

not required 0.166 0.246 0.217 not required

not required 21.9 32.5 28.6 not required

Ag

261

Design Example 5

!


At the column end of each wall pier, confining the 8 column bars plus two wallweb bars gives a boundary zone length of 34 inches. At the inside (doorway) end of each wall pier, confining 8 bars give a boundary zone length of 39 inches. The confinement details are shown in Figure 5-12. The required area of boundary ties is calculated according to Equation (21-10): Ash = 0.09 shc f ' c f y

(21-10)

Figure 5-12. Boundary ties required by the UBC simplified procedure

Calculations of Ash are given in Table 5-11, corresponding to section cuts A, B, C, D, and E through the boundary zones as shown in Figure 5-10.

Table 5-11. Required boundary zone ties by the UBC simplified procedure Section Cut

hc (in.)

s (in.)

Ash Required (in.2)

Tie legs

Ash Provided (in.2)

A B C D E

20.5 12.5 32 12.5 37.5

6 6 6 4 4

0.74 0.45 1.12 0.45 0.90

3-#5 2-#5 4-#5 2-#5 4-#5

0.93 0.62 1.24 0.62 1.24

Note: 1. See Figure 5-12.

262


Design Example 5

7b.

!


Blue Book recommendations.

SEAOC §402.11

Section 402.11 of the Blue Book contains significant revisions to the UBC provisions for wall boundary confinement. Sections 402.11.1 and 402.11.2 revise definitions used in the strain calculation procedure of §1921.6.6.5. Blue Book §402.11.3 adds the following two exceptions to the UBC procedure: Exception 1: Boundary zone details need not be provided where the neutral axis depth c'u is less than 0.15l w . Exception 2: The length of wall section at the compression boundary over which boundary zone detailing is to be provided may be taken as cc , where cc is the larger of c'u = 0.1lw or c'u 2 . In applying these recommendations to the example wall, the wall piers with the largest neutral axis depth-to-length ratio, c′u /lw, govern the design. The largest neutral axis depth at the column end of a wall pier occurs at the 1st floor at Line C, where a large downward earthquake axial force occurs: P’u = (1.2PD + 0.5PL) + PE = 1,300 kips + 1,600 kips = 2,900 kips The neutral axis depth, c’u, for this case is calculated by PCACOL to be 48 inches. c'u l w = 48" 132" = 0.36 ≥ 0.15 therefore boundary zone detailing is required cc = c'u −0.1l w = 48"−0.1(132") = 35 in. cc = c'u 2 = 48" 2 = 24 in.

governs

does not govern

The calculation of cc = 35 inches can be compared to the required UBC boundary length of 32.5 inches shown in the Table 5-10. The largest neutral axis depth at the inside (doorway) end of a wall pier occurs at the 1st floor Line E. Compression at this end of the wall pier corresponds to the loading direction that has earthquake axial force acting upward: P'u = (1.2 PD + 0.5PL ) + PE = 1,200 kips − 1,180 kips = 20 kips The neutral axis depth, c’u, for this case is calculated by PCACOL to be 20 inches. c'u l w = 20" 132" = 0.15 ≥ 0.15


263

Design Example 5

!


Thus, the requirement for boundary confinement at the inside (doorway) ends of the wall piers is marginal. cc = c'u −0.1lw = 20"−0.1(132") = 7 in. cc = c'u 2 = 20" 2 = 10 in.

does not govern

governs

The calculation of cc = 10" can be compared to the required boundary length of 28.6 inches shown in the Table 5-10. Figure 5-6 shows the ties resulting from the Blue Book recommendation, which can be compared to those required by the UBC simplified procedure, shown in Figure 5-12.

8.

Detailing of coupling beams.

The detailing of coupling beams may require a number of preliminary design iterations to determine required bar sizes and the lateral dimensions of the diagonal bar group. Preliminary design iterations are not shown in this Design Example.

8a.

Layering of reinforcement.

For this Design Example, the recommended layering of reinforcement in the coupling beams is shown in Figure 5-13. The proposed layering corresponds to a clear cover of 1 inch in the coupling beam and 1 3/8 inches in the wall pier. Section 1921.6.10.3 requires transverse reinforcement around each group of diagonal bars of the coupling beam. Figure 5-13 assumes that these ties are No. 4 in size and extend over the portion of the diagonal bars within the coupling beam length, as shown in Figure 5-14. Thus the diagonal bars, but not the ties around them, must pass between the reinforcement curtains of the wall pier. The layering shown in Figure 5-13 results in a diagonal bar cage with lateral “core” dimensions of 9.0 inches by 14.8 inches, measured outside-to-outside of the ties. These dimensions conform to the requirement of §1921.6.10.2 that the lateral core dimensions be “not less than bw 2 or 4 inches.”

264


Design Example 5

!


Figure 5-13. Section through coupling beam showing layering of reinforcement

8b.

Ties around diagonal bars.

§1921.4.4

Under the requirements of §1921.6.10.3, the required transverse reinforcement around diagonal bars must conform to §1921.4.4.1 through §1921.4.4.3. Section 1921.4.4.2 requires a maximum tie spacing of 4 inches or one-quarter of the minimum member dimension. Equations (21-3) and (21-4) must be checked in each direction.

(

Ash = 0.3 shc f 'c f y

)([ Ag

) ]

Ach − 1

Ash = 0.09 shc f ' c f y


(21-3)

(21-4)

265

Design Example 5

!


The quantity Ag is calculated assuming the minimum cover per §1907.7 around each diagonal bar core. For walls with No. 11 bars and smaller, without exposure to weather, this minimum cover equals ¾ inch. Thus: Ag = [9.0 + 2 (0.75)] × [14.8 + 2 (0.75)] = 10.5 × 16.3 = 171 in. and Ach = 9.0 ×14.8 = 133 in. Although Ach is based on outside-to-outside of tie dimensions, hc is based on center-to-center of tie dimensions. Assuming No. 4 ties, hc = 9.0 – 0.5 = 8.5 inches in the horizontal direction, and hc = 14.8 – 0.5 = 14.3 inches in the other lateral dimension. For hc = 8.5:

(

Ash = 0.3 shc f 'c f yh

)[(Ag

) ]

Ach − 1

(21-3)

= 0.3[(4")(8.5")(4 ksi ) 60 ksi ] (171 133 − 1) = 0.194 in .2 Ash = 0.09 shc f 'c f yh = 0.09 (4")(8.5")(4 ksi ) (60 ksi ) = 0.204 in .2

governs

(21-4)

For hc = 14.3 :

(

Ash = 0.3 shc f 'c f yh

)[(Ag / Ach )− 1]

(21-3)

= 0.3 [(4")(14.3")(4 ksi ) 60 ksi ] (171 133 − 1) = 0.327 in .2 Ash = 0.09 shc f 'c f yh = 0.09(4")(14.3")(4 ksi ) (60 ksi ) = 0.343 in .2

governs

(21-4)

A single #4 tie around the six diagonal bars provides two tie legs in each direction and Ash = 0.40 in .2 A #3 perimeter tie with a #3 crosstie would provide Ash = 0.22 in .2 across the shorter core direction and Ash = 0.33 in .2 across the longer core direction, which would not quite meet the Ash requirement of 0.343 in.2 Per §1921.4.4.3, crossties shall not be spaced more than 14 inches on center. For the heaviest diagonal reinforcement of 6-#10 bars, the center-to-center dimension of the #10 bars is given as 12 inches in Figure 5-14. The center-to-center hoop dimension in this direction thus equals 12 inches plus one diameter of a #10 bar plus one diameter of a #4 tie, equal to 12.0 + 1.27 + 0.5 = 13.8 inches. Since this is less than 14 inches, a crosstie is not needed.

266


Design Example 5

!


The diagonal bars must be developed for tension into the wall piers. Following the recommendation of Paulay and Priestley [1992], the bars are extended a distance of 1.5l d beyond the face of the supporting wall pier, as shown in Figure 5-14, where l d is the development length of a straight bar as determined under §1912.2. Crossties are added at the intersection of the diagonal bars at the center of the coupling beam, and along their development into the wall piers, as shown in Figure 5-14. The crossties are also added in locations where ties around the diagonal bars are not used.

Figure 5-14. Elevation showing detailing of a coupling beam

8c.

Reinforcement “parallel and transverse.”

§1921.6.10.4

Section 1921.6.10.4 requires reinforcement parallel and transverse to the longitudinal axis of the coupling beam, conforming to §1910.5, §1911.8.9, and §1911.8.10. The Blue Book contains less restrictive requirements (in §402.13) for this reinforcement, and the Blue Book Commentary notes that the UBC requirements referenced should not be applied because the diagonal bars, not the parallel and transverse bars, act as the principal flexural and shear reinforcement.


267

Design Example 5

!


UBC requirements.

By §1911.8.9, for #4@6 transverse (vertical) bars: Av ≥ 0.0015bw s = 0.0015 (16")(6") = 0.144 in .2 ≤ 0.40 in .2

o.k.

By §1911.8.10, for 14-#4 longitudinal (horizontal) bars: Avh ≥ 0.0025bw s2 = 0.0025 (16")(72" 7 ) = 0.41in .2 ≅ 0.40 in .2

o.k.

By §1910.5.1: As , min = 200 bwd f y = 200 (16")(0.8 × 72") 60,000 psi = 3.07 in .2

(10-3)

This requires 7-#6 longitudinal bars (As = 7(0.44 in.2) = 3.08 in.2 ) both top and bottom of the coupling beam, or 14-#6 longitudinal bars total. Per the discussion below, these are not recommended by SEAOC to be used, and are not shown in Figure 5-14. Blue Book recommendations.

Blue Book Commentary §C402.13 cautions against providing excess longitudinal reinforcement in the coupling beam, as required by the application of UBC §1910.5.1. The 1999 ACI code eliminates the requirement of UBC §1910.5.1. The Blue Book recommends using less longitudinal reinforcement. This can be justified on the basis of UBC §1910.5.3, which states that the requirements of §1910.5.1 need not be applied if the reinforcement provided is “at least one-third greater than that required by analysis.” Since the diagonal bars resist the entire flexural tension forces, it could be interpreted that no additional longitudinal reinforcement is required by analysis. In §402.13 of the Blue Book requires the reinforcement parallel to the longitudinal axis of the beam to be at least No. 3 in size, spaced at not more than 12 inches on center. The reinforcement transverse to the longitudinal axis of the beam must be at least No. 3 in size, spaced at not more than 6 inches on center. Figure 5-14 shows the recommended parallel and transverse reinforcement: 14-#4 bars longitudinally and #4 ties @ 6" transversely. Per the Blue Book recommendations of §402.13, the longitudinal reinforcement is extended 6 inches into the wall pier, as shown in Figure 5-14, but is not developed for tension.

268


Design Example 5

!


References ASCE, 1971, Plastic Design in Steel, A Guide and Commentary, American Society of Civil Engineers, New York. ATC-40, 1996. Seismic Evaluation and Retrofit of Concrete Buildings, Applied Technology Council, Redwood City, California. FEMA 273, 1997. NEHRP Guidelines for the Seismic Rehabilitation of Buildings, Federal Emergency Management Agency, Washington, D.C. FEMA 306/307, 1998. Evaluation of Earthquake Damaged Concrete and Masonry Wall Buildings, Federal Emergency Management Agency, Washington, D.C. Ghosh, S. K., 1998. “Design of Reinforced Concrete Buildings under the 1997 UBC,” Building Standards, May-June, pp. 20-24. International Conference of Building Officials, Washington, D.C. Maffei, J., 1996. “Reinforced Concrete Structural Walls: Beyond the Code” SEAONC 1996 Fall Seminar Notes, Structural Engineers of Northern California, San Francisco, California. Paulay, T., and Priestley, M.J.N., 1992. Reinforced Concrete and Masonry Buildings, Design for Seismic Resistance, John Wiley & Sons, New York, N.Y. PCA, 1999. “PCACOL: Design and Investigation of Reinforced Concrete Column Sections,” Portland Cement Association, Skokie, Illinois.


269

Design Example 5

270

!



Design Example 6

!

Concrete Special Moment Resisting Frame

Design Example 6 Concrete Special Moment Resisting Frame

Figure 6-1. Seven-story concrete special moment resisting frame (SMRF) building

Overview Concrete frame buildings, especially older, nonductile frames, have frequently experienced significant structural damage in earthquakes and a number have collapsed. Following the 1971 San Fernando earthquake, special requirements for ductile concrete frames were introduced in the code. Today these ductile frames are designated as special moment resisting frames (SMRF). All reinforced concrete frame structures built in Seismic Zones 3 and 4 must be SMRF, as required by §1633.2.7. Ordinary moment resisting frames (OMRF) and intermediate moment resisting frames (IMRF) are prohibited in Zones 3 and 4, except that IMRF are permitted for some nonbuilding structures under §1634.2.


271

Design Example 6

!


In this Design Example, the seismic design of a seven-story concrete SMRF is illustrated. A conceptual elevation of the building is shown in Figure 6-1. The structure is a reinforced concrete office building with the typical floor plan shown in Figure 6-2. The building is seven stories and has a SMRF on each perimeter wall. A typical building elevation is shown in Figure 6-3.


Design base shear coefficient and reliability/redundancy factor.

2.

Vertical and horizontal distribution of shear.

3.

Frame nodal and member forces.

4.

Analysis and evaluation of frame drifts.

5.

Beam design.

6.

Column design.

7.

Joint shear analysis.

8.

Detailing of beams and columns.

9.

Foundation considerations.

Given Information The building has a floor system that consists of post-tensioned slabs and girders. Vertical loads are carried by a frame system. Use of perimeter SMRF frames and interior frames is designed to allow freedom for tenant improvements. Seismic and site data: Z = 0.4 (Seismic Zone 4) I = 1.0 (standard occupancy) Seismic source type = A Distance to seismic source = 10 km Soil profile type = S D

272



Design Example 6

!


Average story weights (for seismic design) Roof weights: Roofing Concrete slab (8 in.) Girders Columns Partitions Curtain wall Mechanical/electrical Miscellaneous Total

9.0 psf 100.0 27.0 4.0 5.0 5.0 5.0 3.0 158.0 psf

Typical floor weights: Covering Concrete slab (8 in.) Girders Columns Partitions* Curtain wall Mechanical/electrical Miscellaneous Total

(3rd-7th floors) 2.0 psf 100.0 48.0 8.0 10.0 10.0 5.0 3.0 186.0 psf

(2nd floor) 2.0 psf 100.0 48.0 10.0 10.0 10.0 5.0 3.0 188.0 psf

*Partitions are 2 psf for gravity calculations and 10 psf for seismic calculations. Structural materials: Concrete f c ' = 4,000 psi (regular weight) Reinforcing A706, Grade 60 f y = 60 ksi

(


)

273

Design Example 6

!


Figure 6-2. Typical floor plan

Figure 6-3. Typical frame elevation, Line A

274


Design Example 6

!



1.

Code Reference

Design base shear coefficient and reliability/redundancy factor.

Two key design parameters, the design base shear coefficient and the reliability/redundancy factor ρ , are determined in this part. The 1997 UBC significantly revised the determination of base shear and introduced the concept of the reliability/redundancy factor to penalize lateral force resisting systems that have little redundancy. Base shear is now determined on a strength basis, whereas base shear in the 1994 UBC was determined on an allowable stress basis, with forces subsequently increased by load factors for concrete strength design. The 1997 UBC also introduced design for vertical components of ground motion E v . Period using Method A. T = Ct (hn )3 / 4 = .030(86 )3 / 4 = .85 sec

(30-8)

Near source factors for seismic source type A and distance to source = 10 km N a = 1.0

Table 16-S

N v = 1.2

Table 16-T

Seismic coefficients for Seismic Zone 4 (0.4) and soil profile type S D : C a = 0.44 N a = 0.44(1.0 ) = 0.44

Table 16-Q

C v = 0.64 N v = 0.64(1.2 ) = 0.77

Table 16-R

The R coefficient for a reinforced concrete building with an SMRF system is: R = 8.5

Table 16-N

Note that Table 16-N puts no limitation on building height when a SMRF system is used.


275

Design Example 6

1a.

!


Calculation of design base shear coefficient.

§1630.2.2

The four equations for design base shear are as follows: V=

Cv I 0.77(1.0 ) W= W = 0.107W RT 8.5(0.85)

(30-4)

but the design base shear need not exceed: V =

2.5C a I 2.5(.44 )(1.0) W = W = 0.129W R 8.5

(30-5)

The total design base shear shall not be less than: V = 0.11C a IW = 0.11(.44 )(1.0 )W = 0.048W

(30-6)

In addition, for Seismic Zone 4, the total base shear shall also not be less than: V =

0.8ZN v I 0.8(.4)(1.20 )(1.0 ) W = W = 0.045W R 8.5

(30-7)

Therefore, Equation (30-4) controls the base shear calculation. ∴ V = 0.107W

1b.

Calculation of reliability/redundancy factor.

§1630.1

The reliability/redundancy factor is determined in accordance with §1630.1 by comparing the shear in the highest loaded moment frame bay with the base shear at that level. This calculation is completed using an iterative process with knowledge of results from the frame analysis presented later in this Design Example. The two columns with the largest base shears are used to define the highest loaded bay. If the columns are part of adjacent bays, 70 percent of their shear values are used in this computation.

276


Design Example 6

!


Column base shear reactions from computer model of the building are shown below (Figure 6-4). These base shear reactions are based on a computer analysis of the frame as described later, including an accidental torsion moment.

116 k

176 k

168 k

176 k

116 k

Figure 6-4. Column shears at frame base (from computer analysis with 1.0Eh )

The maximum element story-shear ratio rmax is defined as the largest individual element story-shear ratios at or below the two-thirds height of the building. For this building rmax is calculated as shown below. Calculation of r at interior SMRF bay: r=

§1630.1.1

0.70(176 k + 168 k ) = 0.16 1,475 k

Calculation of r at exterior SMRF bay: r=

116 kips + 0.70(176 kips ) = 0.16 1475 k

Note that r should be evaluated at all moment frame bays and for the bottom twothirds levels of the building. Since no other r values control, other calculations are not shown. Equation (30-3) is used to calculate ρ as shown below. AB = (120')(90') = 10,800 ft 2 ρ = 2−

20 rmax AB

= 2−

20 .16 10800

= 0.82 ≤ 1.0

(30-3)

∴ ρ = 1.0


277

Design Example 6

!


For moment resisting SMRF frames, ρ must be less than 1.25. If ρ is greater than 1.25, additional bays must be added such that ρ is less than or equal to 1.25.

1c.

Vertical component of earthquake ground motion.

§1630.1.1

Because the design of the concrete frames will use strength design, the vertical component E v must be considered in the load combination of Equation (30-1). Determination of E v is shown below. E v = 0.5C a ID = 0.5(0.44 )(1.0 )W = 0.22W The effect of E v is added to the gravity loads that are used in combination with horizontal seismic loads. Thus, the following earthquake load is used in the earthquake load combinations: E = ρE h + E v

2.

(30-1)

Vertical and horizontal distribution of shear.

§1628.4 and §1628.5

In this part, the seismic forces on the concrete frame are determined.

2a.

Story masses (weights).

Table 6-1. Calculation of building and story weights

278

Level

Area (sf)

wi (psf)

WI (kips)

R 7 6 5 4 3 2

10,800 10,800 10,800 10,800 10,800 10,800 10,800

158.0 186.0 186.0 186.0 186.0 186.0 188.0

1,706 2,009 2,009 2,009 2,009 2,009 2,030

Total

75,600

13,781


Design Example 6

2b.

!


Base shear and vertical distribution of shear.

Using the results of Part 1a, the base shear is V = .107W = .107(13,781 k ) = 1,475 kips The building period is 0.85 seconds using Method A. Therefore, the concentrated force at the top is determined from §1630.5 as follows Ft = 0.07TV = 0.07(0.85)(1,475 k ) = 87 kips

(30-14)

The vertical distribution of shear is determined from Equation (30-13) n

V = Ft + ∑ Fi

(30-13)

i =1

The calculation of story forces and story shears is shown in Table 6-2 below.

Table 6-2. Vertical distribution of shear Wi (k)

ΣW i (k)

hi (ft)

Story H (ft)

Wi h i (k-ft)

Wi h i ΣW i h i (%)

Ft = R 7 6 5 4 3 2

1,706 2,009 2,009 2,009 2,009 2,009 2,030

1,706 3,715 5,724 7,733 9,742 11,750 13,781

86 74 62 50 38 26 14

12 12 12 12 12 12 14

146,750 148,651 124,546 100,440 76,334 52,229 28,426

Totals

13,781

677,376

Level


Fi (k)

ΣFi (k)

22% 22% 18% 15% 11% 8% 4%

87 301 30 255 206 156 107 58

388 304 255 206 544 651 709

100%

1,475

279

Design Example 6

!


Figure 6-5. Computer model of the frame on Line A

3.

Frame nodal and member forces.

The longitudinal frame along Line A is designed in this part. First, dead and live loads on the beams are determined using a tributary width of 15 feet. The gravity loads applied to the beams in the frame analysis are summarized below in Table 6-3.

Table 6-3. Beam gravity loads for analysis Framing Level

Dead Load (plf)

Live Load (plf)

Roof Floor 6th Floor 5th Floor 4th Floor 3rd Floor 2nd Floor

2,250 2,886 2,886 2,886 2,886 2,886 2,879

300 750 750 750 750 750 750

7th

A torsional analysis of the building using a 5 percent accidental torsion (using an eccentricity equivalent to 5 percent of the perpendicular building dimension) gives results such that all frames on the four faces of the building resist torsional shears of approximately 2 percent of the base shear. Thus the seismic forces in the frame analysis were increased by 2 percent to account for accidental torsion (per

280


Design Example 6

!


§1630.7). Each of the perimeter frames on Lines A, D, 1 and 5, will be designed to resist a base shear of 52 percent of the total building design base shear, V . A two dimensional frame analysis is performed for the frame along Line A. The frame forces are determined from story forces above. Forces are distributed to frame nodes in proportion to their location along Line A. Thus, at longitudinal frames (Lines A and D), 12.5 percent of the story force is applied to end column nodes and 25 percent of the story force is applied to the interior column nodes. The force distribution at transverse frames (Lines 1 and 5) is 16.7 percent to exterior column nodes and 33 percent to interior column nodes. The frame nodal loads for longitudinal and transverse frames are summarized below in Table 6-4. Frame joint and member numbers are shown in Figure 6-5.

Table 6-4. Column nodal forces for analysis Level

Story Forces (kips)

R 7 6 5 4 3 2

388 304 255 206 156 107 58

Total

Long. Frame End Column Node Forces (kips) 24.7 19.4 16.3 13.1 10.0 6.8 3.7

Long. Frame Interior Col. Node Forces (kips) 49.5 38.8 32.5 26.2 19.9 13.6 7.4

Trans. Frame End Column Node Forces (kips) 33.0 25.9 21.7 17.5 13.3 9.1 4.9

Trans. Frame Interior Col. Node Forces (kips) 66.0 51.7 43.4 35.0 26.6 18.2 9.9

1,475

The loads shown in Table 6-4 add to 50 percent of the design base shear. To account for torsion, a load factor of 1.02 was used in the frame analysis program. This problem was solved on a two dimensional frame program. Any elastic finite element analysis program could be used, including those with three dimensional capability.

4.

Analysis and evaluation of frame drifts.

Under §1630.10.2, story drifts are limited to 0.020 times story heights for drifts corresponding to the maximum inelastic response displacement ∆ m for structures with periods 0.7 seconds or greater. Under §1630.10.2 ∆ m = 0.7 R∆ s or: ∆ m = 0.7(8.5)∆ s = 5.95∆ s SEAOC Seismic Design Manual, Vol. III (1997 UBC)

281

Design Example 6

!


Table 6-5 summarizes the calculation of the allowable frame drifts.

Table 6-5. Allowable story deformations and displacements Story

Total Height (ft)

Story Height (ft)

Allowable ∆s (in.)

Sum Σ∆ s (in.)

Allowable ∆M (in.)

Sum Σ∆ M (in.)

R 7 6 5 4 3 2

86 74 62 50 38 26 14

12 12 12 12 12 12 14

0.484 0.484 0.484 0.484 0.484 0.484 0.565

3.469 2.985 2.501 2.017 1.533 1.049 0.565

2.88 2.88 2.88 2.88 2.88 2.88 3.36

20.64 17.76 14.88 12.00 9.12 6.24 3.36

The frame analysis is thus performed using a standard frame analysis program. Columns, beams, and grade beams were sized to meet allowable drift limits. Member section properties were chosen to represent the cracked structure. In accordance with §1910.11.1, 70 percent of the gross section properties are used for columns and 35 percent of gross section properties are used for beams to estimate the contribution of cracked sections on frame behavior. Selected sections were 42 × 42 corner columns, 36 × 44 interior columns, 30x48 beams and 60 × 48 foundation grade beams. The designer must size a frame which meets drift limitations and also meets strength criteria. For the design of this frame, the controlling parameters are frame stiffness and strength of beams. Using the member sizes chosen, frame analysis gives the lateral story displacements, given below in Table 6-6. Note that the frame analysis gives ∆ s deflections, thus the comparison is made using ∆ s deflections and that the ρ factor is not used in the deflection analysis.

Table 6-6. Displacements determined from analysis

282

Story

Total Height (ft)

Story Height (ft)

R 7 6 5 4 3 2

86 74 62 50 38 26 14

12 12 12 12 12 12 14

From Analysis ∆s Story Drifts (in.) 0.38 0.48 0.48 0.48 0.47 0.44

Maximum Allowable ∆s Story Drifts (in.)

From Analysis Σ∆ s (in.)

Maximum Allowable Σ∆ s (in.)

0.48 0.48 0.48 0.48 0.48 0.48 0.56

3.18 2.80 2.34 1.82 1.34 0.87 0.43

3.47 2.98 2.50 2.02 1.53 1.05 0.57


Design Example 6

!


As shown in Table 6-6, story drifts are determined to be within allowable limits. The iteration between frame stiffness and member strengths has resulted in a frame design with conservative drifts. The designer must iterate between frame analysis and member section design.

5. 5a.

Beam design.

Load combinations.

The next procedure is frame member design. Frame beams are designed to support gravity loads and resist seismic forces. Beams are sized to limit frame drift and to resist the corresponding moment with a nominal strength φM n . The φ factor for bending analysis is 0.90. The controlling load combinations are given in §1612.2.1 and are summarized below. Note that Exception 2 of §1612.2.1 requires the load combinations to be multiplied by 1.1 as shown below. 1.1(1.2 D + 0.5 L + 1.0 E + 0.22 D ) = 1.58 D + 0.55 L + 1.1E

(12-5)

1.1(0.9 D − 0.22 D − 1.1E ) = 0.75D − 1.1E

(12-6)

Note: The SEAOC Seismology Committee does not support the 1.1 factor for concrete and masonry elements under seismic loads and the 1.1 factor is not included in the 1999 SEAOC Blue Book. However, until ICBO makes a different ruling, it is part of the 1997 UBC and is thus included in this Design Example.

5b.

Design requirements for frame beams.

The nominal beam strength is calculated using the following formulas and ignoring compression steel for simplicity: a  φ M n = φ As f y  d −  ≥ M u 2  Note that historic practice has been to consider the frame beam to have a rectangular section without consideration of the contribution of the adjacent slab for both compression and tension stresses. That is still true for design under the 1997 UBC. The ACI-318-99 has included new provisions requiring that the adjacent slab be included in consideration of the frame beam analysis. These provisions will be required in the adoption of future codes. The probable flexural strength, Mpr, is calculated per §1921.5.1.1 using 1.25 f y for the reinforcing steel stress. Recalculating the beam strength using φ = 1.0 , thus: SEAOC Seismic Design Manual, Vol. III (1997 UBC)

283

Design Example 6

!


a pr    M pr = 1.25 As f y  d −  2   The shear strength of the beam must be designed to be greater than required in order to resist Mpr, at both ends of the beam. L is the distance from column face to column face. For this Design Example the distance is L = 30 ft – 48 in. (columns) = 26 ft – 0 in. The φ factor for shear analysis is 0.85 per §1909.3.2.3. Thus, the ultimate shear load is calculated as: Vu=

+M

pr

− (− M L

pr

)

+

w FACTORED , GRAVITY L 2

≤ φV n

φV n = φ V c + φ V s φ V c = 0;

φVs = .85 Av f

y

d s

Under §1921.3.4.2, the shear contribution from concrete Vc is considered zero when both of the following conditions occur: 1.) the earthquake-induced shear force represents more than one-half of the total shear force; and 2.) factored axial compressive force is less than Ag F ' c 20 per §1921.3.4.2. In the region of plastic hinges, transverse ties are required to resist shear forces. Maximum spacing of ties cannot exceed any of the following: 1.

d 4.

2.

8 times the diameter of the smallest longitudinal reinforcement.

3.

24 times the diameter of the hoop bars.

4.

12 inches.

§1921.3.3.2

An example beam design for Beam 36 (Figure 6-5) is shown. The controlling load combinations, including seismic forces, are Equations (12-5) and (12-6). Depending on the direction of seismic inertial force, seismic moments add with gravity moments at one beam end and subtract at the other end. Beyond regions of potential plastic hinges, stirrups with seismic ties are required at a maximum spacing of d 2 throughout the length of the beam under §1921.3.3.4. Diagrammatic shear and moment diagrams are shown below in Figure 6-6.

284


Design Example 6

!


Gravity loading Gravity moment

Gravity shear

Seismic moment

Seismic shear

Gravity + seismic moment

Gravity + seismic shear

Figure 6-6. Moment and shear diagrams for beams

A review of the moment and shear diagrams for gravity loads and seismic loads (Figure 6-6) will help the designer realize that seismic moment and negative gravity moment at beam ends will be additive for top reinforcement design and subtractive for bottom reinforcement design. Since seismic moment is usually considerably greater than gravity moment, the reinforcement design will be controlled by load combinations including seismic loads. However, greater amounts of top reinforcement will be required than bottom reinforcement. Since the frame behavior produces beam moments as depicted in Figure 6-6, load combination Equation (12-5) will maximize negative moments for top reinforcement design and load combination Equation (12-6) will maximize positive moments for bottom reinforcement design.


285

Design Example 6

!


An example calculation for Beam 36 is as follows: From the frame analysis, Equation (12-5), negative moment is –1,422 k-ft. For a beam with b = 30 in. and h = 48 in., d = 45 in. Try 5-#11 top bars, As = 7.80 in.2 Per §1921.3.2.1: As,min =

a=

200bwd 200(30")(45") = = 4.5 in.2 ≤ 7.80 in.2 ∴ fy 60,000 psi

o.k.

(7.80 in. ) (60,000 psi) = 4.59 in. 2

0.85(4.000 psi )(30")

(

)

4.59"   1   1 kip    φM n = (0.90 ) 7.80 in.2 (60,000 psi ) 45"−   2   12"   1,000 lbs   = 1,498 k-ft ≥ 1,422 k-ft ∴ o.k.

From the frame analysis, Equation (12-6), positive moment is 905 k-ft. Try 5-#9 bottom bars, As = 5.0 in.2 a=

(5.0 in. )(60,000 psi) = 2.94 in. 2

0.85(4000 psi )(30")

(

)

2.94"   1   1 kip    φM n = (0.90 ) 5.0 in.2 (60,000 psi ) 45"−   2   12"   1,000 lbs   = 979 k-ft ≥ 905 k-ft ∴

o.k.

Thus, the Beam 36 design will have 5-#11 top bars and 5-#9 bottom bars. Note that §1921.3.2.2 requires that positive moment strength (bottom reinforcement) be a minimum 50 percent of negative moment strength at the joints and that neither the positive nor negative moment strength along the beam be less than one-quarter of the strength at either joint (end).

286


Design Example 6

5c.

!


Beam skin reinforcement.

If the effective depth of a beam exceeds 36 inches, longitudinal skin reinforcement shall be distributed along both side faces of a beam for a distance d 2 nearest the flexural tension reinforcement per §1910.6.7. The skin reinforcement shall be spaced a maximum of the lesser of d 6 or 12 inches. Thus, for a 48-inch deep beam with d = 45 inches, d 6 is 7 ½ inches. The beam will have flexural tension regions at the top and bottom of the beam, thus four quantities of Ask are required at the top and bottom of each side. Ask = 0.012(2 − 30")(d / 12") = 0.012(45"−30")(45" / 12") = 0.675 in.2

(

)

Ask = 2 0.675 in. 2 = 1.35 in. 2 ∴ Use 5-#5 bars, Ask = 1.55 in. 2 each side of beam spaced 7½ inches apart ∴

5d.

o.k.

Beam shear design.

As noted above, the beam will also have 5-#5 side bars on each side of the beam. For this Design Example, the assumption is made that 3-#5 side bars each side contribute to the plastic moment. For shear design, the designer allows for plastic hinge formation that will produce shear forces greater than those from frame analysis. Vu= +a =

+M

pr

− (−M

pr

)

L

+

wGRAVITY L 2

(1.25)(7.80 + 1.86)(60,000 psi ) = 7.10 in. 0.85(4,000 psi )(30")

(

)

(

)

  7.10"  7.10"   1    = 2,328 k - ft + M pr = (1.25) 7.80 in.2 (60,000 psi ) 45"−   + (1.25) 1.86 in.2 (60,000 psi )  30"− 2  2   12,000    

− a pr =

(1.25)(5.0 + 1.86)(60,000 psi ) = 5.04 in. 0.85(4,000 psi )(30")

(

)

(

)

 5.04"  5.04"   1     = 1,647 k - ft − M pr = (1.25) 5.0 in.2 (60,000 psi )  45"−    + (1.25) 1.86 in.2 (60,000 psi ) 30"− 2 2   12,000     


287

Design Example 6

!


Shear from dead load is calculated from the load combination of Equation (12-5):  26'  V gravity = [(1.58)(2,879 plf ) + (0.55)(750plf )]  = 65 kips  2  ∴ Vu =

(2,328 k - ft + 1,647 k - ft ) + 65 kips = 246 kips 22'

The design shear Vu is thus the sum of the shear from the plastic end moments plus the gravity shear. Seismic stirrups at the plastic hinge regions are calculated as shown below. Note that the plastic hinge region is a distance of 2h from the column face. Try #4 ties with four vertical legs at 6-inch spacing over the 2h length (86 inches). φVn = φVc + φV s φVc = 0 φVs =

φAv f y d s

φVn = 0 +

(

)

0.85(4 ) 0.20 in.2 (60,000 psi )(45") = 306 kips ≥ 246 kips 6"

∴ o.k. Therefore, use 4 legs, #4 stirrup ties at 6-inch spacing at plastic hinge regions at beam ends. Seismic stirrups in the beam between plastic hinge regions are calculated as follows. Try #4 ties at 8-inch spacing:  13'−3"−2 × 45"  Vu = 181 kips + 65 kips   = 209 kips 13'−3"  

(

)

φVs = .85 .80 in.2 (60,000 psi )(45") 8" = 229 kips ≥ 209 kips ∴ o.k.

288


Design Example 6

!


Therefore, the final design for Beam 36 is a 30-inch wide by 48-inch deep beam with 5-#11 top bars, 5-#9 bottom bars, 5-#5 side bars, and 4 legs - #4 stirrup ties at 6-inch spacing each end with 4 legs - #4 stirrup ties at 8 feet between.

5e.

Design of all Frame A beams.

Following these same procedures and using the forces from the frame analysis, the Frame A beam designs for flexural strength are shown in Table 6-7.

Table 6-7. Beam member longitudinal steel design Member

M u ,i

M u, j

b d h (Eq. 12-6) (Eq. 12-5) (in.) (in.) (in.)

36

-1,405 905

37

-1,389 858

38

-1,392 856

39

-1,422 876

Level 3 40

-1,568 1,093

41

-1,569 1,036

42

-1,564 1,036

43

-1,637 1,036

Level 4 44

-1,281 781

45

-1,304 772

46

-1,304 772

47

-1,334 781

Bar No. Location bars

Bar Size

Bar Area (in.2)

As (in.2)

a (in.)

φM n Bending DCR (1) (k-ft) Results

30 30 30 30 30 30 30 30

48 48 48 48 48 48 48 48

45 45 45 45 45 45 45 45

Top Bottom Top Bottom Top Bottom Top Bottom

5 5 5 5 5 5 5 5

#11 #9 #11 #9 #11 #9 #11 #9

1.56 1.00 1.56 1.00 1.56 1.00 1.56 1.00

7.80 5.00 7.80 5.00 7.80 5.00 7.80 5.00

4.59 2.94 4.59 2.94 4.59 2.94 4.59 2.94

1,499 979 1,499 979 1,499 979 1,499 979

o.k. o.k. o.k. o.k. o.k. o.k. o.k. o.k.

0.94 0.92 0.93 0.88 0.93 0.87 0.95 0.89

30 30 30 30 30 30 30 30

52 52 52 52 52 52 52 52

49 49 49 49 49 49 49 49


5 5 5 5 5 5 5 5

#11 #9 #11 #9 #11 #9 #11 #9

1.56 1.00 1.56 1.00 1.56 1.00 1.56 1.00

7.80 5.00 7.80 5.00 7.80 5.00 7.80 5.00

4.59 2.94 4.59 2.94 4.59 2.94 4.59 2.94

1,639 1,069 1,639 1,069 1,639 1,069 1,639 1,069


0.96 1.02 0.96 0.97 0.95 0.97 1.00 0.97

30 30 30 30 30 30 30 30

44 44 44 44 44 44 44 44

41 41 41 41 41 41 41 41


5 5 5 5 5 5 5 5

#11 #9 #11 #9 #11 #9 #11 #9

1.56 1.00 1.56 1.00 1.56 1.00 1.56 1.00

7.80 5.00 7.80 5.00 7.80 5.00 7.80 5.00

4.59 2.94 4.59 2.94 4.59 2.94 4.59 2.94

1,359 889 1,359 889 1,359 889 1,359 889


0.94 0.88 0.96 0.87 0.96 0.87 0.98 0.88


289

Design Example 6

!


Table 6-7 (continued) Level 5 48

-1,273 783

49

-1,298 766

50

-1,297 766

51

-1,343 780

Level 6 52

-854 337

53

-878 346

54

-878 346

55

-887 346

Level 7 56

-775 257

57

-799 267

58

-799 266

59

-806 266

Roof 40

-593 206

41

-603 198

42

-599 196

43

-610 199

30 30 30 30 30 30 30 30

44 44 44 44 44 44 44 44

41 41 41 41 41 41 41 41


5 5 5 5 5 5 5 5

#11 #9 #11 #9 #11 #9 #11 #9

1.56 1.00 1.56 1.00 1.56 1.00 1.56 1.00

7.80 5.00 7.80 5.00 7.80 5.00 7.80 5.00

4.59 2.94 4.59 2.94 4.59 2.94 4.59 2.94

1,359 889 1,359 889 1,359 889 1,359 889


0.94 0.88 0.96 0.86 0.95 0.86 0.99 0.88

24 24 24 24 24 24 24 24

36 36 36 36 36 36 36 36

33 33 33 33 33 33 33 33


4 4 4 4 4 4 4 4

#11 #8 #11 #8 #11 #8 #11 #8

1.56 0.79 1.56 0.79 1.56 0.79 1.56 0.79

6.24 3.16 6.24 3.16 6.24 3.16 6.24 3.16

4.59 2.32 4.59 2.32 4.59 2.32 4.59 2.32

862 453 862 453 862 453 862 453


0.99 0.74 1.00 0.76 1.00 0.76 1.00 0.76

24 24 24 24 24 24 24 24

36 36 36 36 36 36 36 36

33 33 33 33 33 33 33 33

Top Top Top Top Top Top Top Top

4 4 4 4 4 4 4 4

#11 #8 #11 #8 #11 #8 #11 #8

1.56 0.79 1.56 0.79 1.56 0.79 1.56 0.79

6.24 3.16 6.24 3.16 6.24 3.16 6.24 3.16

4.59 2.32 4.59 2.32 4.59 2.32 4.59 2.32

862 453 862 453 862 453 862 453


0.90 0.57 0.93 0.59 0.93 0.59 0.93 0.59

24 24 24 24 24 24 24 24

36 36 36 36 36 36 36 36

33 33 33 33 33 33 33 33


4 4 4 4 4 4 4 4

#10 #8 #10 #8 #10 #8 #10 #8

1.27 0.79 1.27 0.79 1.27 0.79 1.27 0.79

5.08 3.16 5.08 3.16 5.08 3.16 5.08 3.16

3.74 2.32 3.74 2.32 3.74 2.32 3.74 2.32

712 453 712 453 712 453 712 453


0.83 0.46 0.85 0.44 0.84 0.43 0.86 0.44

Note: 1. DCR=demand to capacity ratio

With longitudinal beam reinforcement proportioned as indicated in Table 6-7 above, the plastic moment Mpr and shear design is as follows. Note that Mpr is calculated including contribution of perimeter reinforcement. VU , gravity is calculated as the factored combination of D + L loads : VU , gravity = 1.58D + 0.55L .

290


Design Example 6

!


Table 6-8. Beam member shear reinforcement design Mem ID Level 2 36 37 38 39 Level 3 40 41 42 43 Level 4 44 45 46 47 Level 5 48 49 50 51 Level 6 52 53 54 55

(k-ft)

Vpr V u , GR Vu φV c Ties Avs (kips) (kips) (kips) (kips) # legs (in.2)

φV s φVn s Result DCR (1) (in.) (kips) (kips)

7.10 5.04 7.10 5.04 7.10 5.04 7.10 5.04

2,389 1,708 2,389 1,708 2,389 1,708 2,389 1,708

215 154 215 154 215 154 215 154

70

285

0.0

4

0.80

6.0

306

306

o.k.

0.93

70

285

0.0

4

0.80

6.0

306

306

o.k.

0.93

70

285

0.0

4

0.80

6.0

306

306

o.k.

0.93

70

285

0.0

4

0.80

6.0

306

306

o.k.

0.93

2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64

7.68 5.62 7.68 5.62 7.68 5.62 7.68 5.62

2,769 2,028 2,769 2,028 2,769 2,028 2,769 2,028

253 185 253 185 253 185 253 185

70

323

0.0

4

0.80

6.0

333

333

o.k.

0.97

70

323

0.0

4

0.80

6.0

333

333

o.k.

0.97

70

323

0.0

4

0.80

6.0

333

333

o.k.

0.97

70

323

0.0

4

0.80

6.0

333

333

o.k.

0.97

5-#11 5-#9 5-#11 5-#9 5-#11 5-#9 5-#11 5-#9

1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20

6.62 4.56 6.62 4.56 6.62 4.56 6.62 4.56

2,055 1,435 2,055 1,435 2,055 1,435 2,055 1,435

182 127 182 127 182 127 182 127

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90


5-#11 5-#9 5-#11 5-#9 5-#11 5-#9 5-#11 5-#9

1.20 1.20 1.20 1.20 1.20 1.20 1.20 1.20

6.62 4.56 6.62 4.56 6.62 4.56 6.62 4.56

2,055 1,435 2,055 1,435 2,055 1,435 2,055 1,435

182 127 182 127 182 127 182 127

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90

70

252

0.0

4

0.80

6.0

279

279

o.k.

0.90


4-#11 4-#8 4-#11 4-#8 4-#11 4-#8 4-#11 4-#8

5.74 2.90 5.74 2.90 5.74 2.90 5.74 2.90

1,175 623 1,175 623 1,175 623 1,175 623

101 54 101 54 101 54 101 54

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

Bar Loc.

As T&B

As side

a (in.2)


5-#11 5-#9 5-#11 5-#9 5-#11 5-#9 5-#11 5-#9

1.86 1.86 1.86 1.86 1.86 1.86 1.86 1.86


5-#11 5-#9 5-#11 5-#9 5-#11 5-#9 5-#11 5-#9


M pr


291

Design Example 6

!


Table 6-8 (continued) Level 7 56 57 58 59 Roof 40 41 42 43


4-#11 4-#8 4-#11 4-#8 4-#11 4-#8 4-#11 4-#8

5.74 2.90 5.74 2.90 5.74 2.90 5.74 2.90

1,175 623 1,175 623 1,175 623 1,175 623

101 54 101 54 101 54 101 54

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76

70

171

0.0

4

0.80

6.0

224

224

o.k.

0.76


4-#10 4-#8 4-#10 4-#8 4-#10 4-#8 4-#10 4-#8

4.67 2.90 4.67 2.90 4.67 2.90 4.67 2.90

974 623 974 623 974 623 974 623

84 54 84 54 84 54 84 54

48

132

0.0

4

0.80

6.0

224

224

o.k.

0.59

48

132

0.0

4

0.80

6.0

224

224

o.k.

0.59

48

132

0.0

4

0.80

6.0

224

224

o.k.

0.59

48

132

0.0

4

0.80

6.0

224

224

o.k.

0.59

Note: 1. DCR=demand to capacity ratio.

Check longitudinal skin reinforcement per §1910.6.7. The code requires skin reinforcement for beams with d greater than 36 inches. This reinforcement is calculated as Ask = .012(d − 30 ) per foot depth on each side face. This reinforcement is required on the tension half of the section, and thus is required both top and bottom since seismic loads could cause tension stresses on the bottom half of the section. For a 48-inch deep beam, d = 45 inches: Ask = 0.012(45"−30")(48" / 12") = 0.72 in.2 This skin reinforcement is required on each side of the beam and in each tension region a distance d 2 from the tension reinforcement. Thus, four quantities of this reinforcement are required. The reinforcement may be spaced a maximum distance apart of the lesser of 12 inches or d 6 .

(

)

Therefore, use 5-#5 bars Ask = 1.55 in.2 / 1.44 in.2 each side spaced d 6 = 45 in. / 6 = 7.5 in. along the side face of the beam. Having satisfied both the design for bending and shear, the final beam designs are thus chosen as shown in Table 6-9. See Figure 6-7 for a beam cross-section showing dimensions and reinforcement.

292


Design Example 6

!


Table 6-9. Final beam designs Level

Width (in.)

Depth (in.)

Long. Reinf. Top

Long. Reinf. Bottom

Skin Reinf.

Roof

24

36

4-#10

4-#6

None

7

24

36

4-#11

4-#7

None

6

24

36

4-#11

4-#7

None

5

30

42

5-#11

5-#9

4

30

42

5-#11

5-#9

3

30

52

5-#11

5-#9

2

30

48

5-#11

5-#9

5 - #4 ea. face 5 - #5 ea. face 5 - #6 ea. face 5 - #5 ea. face

Shear Reinf. In Hinge Regions

Shear Reinf. Between Hinge Regions

4 legs #4 ties@ 6" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6"

4 legs #4 ties@ 12” 4 legs #4 ties@ 9" 4 legs #4 ties@ 9" 4 legs #4 ties@ 8" 4 legs #4 ties@ 8" 4 legs #4 ties@ 6" 4 legs #4 ties@ 6”

Figure 6-7. 30 x 48 beam at Level 2


293

Design Example 6

6.

!


Column design.

Columns should be designed to ensure that the plastic hinges are located in the beams (i.e., strong column-weak beam behavior) and to resist column shears. To ensure strong column-weak beam behavior, columns must be designed to have nominal bending strengths 120 percent stronger than beams per §1921.4.2.2. This is achieved by summing the M e of columns above and below a joint and comparing that with the sum of M g for beams on both sides of a joint. ∑ M e ≥ (6 / 5)∑ M g

(21-1)

The controlling girder location occurs at Level 3. The girder is a 30 in. by 52 in. with 5-#11s top, 5-#9s bottom, and 5-#6s shin reinforcement each side. The assumed two skin bars are effective in calculation of M g , or alternatively a computer program can be used for more accurate results. Calculation of − M g (negative, at beam tops).

a=

[5 (1.56 in. ) + 4 (0.44 in. )](60,000 psi) = 5.62 in. 2

2

0.85(4,000 psi )(30")

(

)

(

)

5.62"  5.62"    2 − M g = (0.90 ) 7.80 in.2 (60,000 psi ) 49"−  + (0.90 ) 1.76 in. (60,000 psi ) 37.5"−  2  2    − M g = 22,752 kip-in. = 1,896 kip-ft Calculation of M g (positive, at beam bottoms).

a=

[5 (1.00 in. )+ 4 (0.44 in. )](60,000 psi) = 3.98 in. 2

2

0.85(4,000 psi )(30")

(

)

(

)

3.98"  3.98"    2 M g = (0.90 ) 5.00 in.2 (60,000 psi ) 49"−  + (0.90 ) 1.76 in. (60,000 psi ) 37.5"−  2  2    M g = 16,067 kip-in. = 1,339 kip-ft

294


Design Example 6

!


Therefore, at interior columns: 6 6 M g = (1,896 kip-ft + 1,339 kip-ft) = 3,882 kip-ft ∑ 5 5 Therefore, at end columns: 6 6 M g = (1,896 kip-ft) = 2,275 kip-ft ∑ 5 5 The girder moments are resisted by two column sections, the column above the joint and the column below the joint. The required column strengths, M e , for interior and end columns are given below. Me =

1 (3,882 kip-ft) = 1,941 kip-ft 2

Me =

1 (2,275 kip-ft) = 1,138 kip-ft 2

or:

6a.

Forces on columns due to factored load combinations.

For column design, the load combinations of Equations (12-5) and (12-6) are used. Also, because strength design is used, the effect of the vertical seismic component Ev must be included. Equations (12-5) and (12-6) are given below. Tables 6-10 and 6-11 provide axial forces and moments on the columns of Frame A for Equations (12-5) and (12-6), respectively. 1.1(1.2 D + 0.5 L + 1.0 E + 0.22 D ) = 1.58 D + 0.55 L + 1.0 E h

(12-5)

1.1(0.9 D − 1.0 E ) = 1.1(0.9 D − 0.22 D − 1.0 E h ) = 0.75 D − 1.1E h

(12-6)


295

Design Example 6

!


Table 6-10. Column loads for Equation (12-5) Member 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

296

Pu (kips) 145 141 148 136 123 81 34 1,001 850 700 553 405 258 111 1,002 853 705 557 408 260 112 990 843 698 552 406 259 111 868 724 566 428 290 181 78

Vu (kips) 114 71 62 51 39 23 -21 192 196 180 158 128 88 62 185 196 181 160 130 93 61 195 195 185 162 132 94 61 140 137 127 115 103 90 55

M u bottom (k-ft) 1,604 476 505 323 276 -69 -305 2,227 1,212 1,255 942 874 326 102 2,142 1,214 1,262 954 886 346 86 2,259 1,193 1,289 963 901 346 80 1,719 902 894 709 668 318 45

M u top (k-ft) -226 -374 -241 -287 -190 -347 -54 -842 -1142 -903 -957 -665 -733 -642 -822 -1133 -913 -969 -670 -770 -647 -868 -1141 -926 -983 -680 -783 -651 -520 -744 -625 -675 -570 -762 -610


Design Example 6

!


Table 6-11. Column loads for Equation (12-6) Member 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Pu (kips)

Vu (kips)

-140 -102 -53 -23 7 8 4 441 374 307 243 178 114 51 438 373 309 244 180 115 51 430 367 305 242 179 115 50 583 481 365 270 174 108 47

122 90 80 69 57 42 0 193 196 181 159 129 90 62 185 196 181 160 130 93 61 194 195 183 161 131 92 61 133 118 108 97 85 71 34


M u bottom (k-ft) 1636 597 615 432 386 40 -206 2236 1206 1264 948 882 332 96 2142 1214 1262 954 886 346 86 2250 1198 1279 957 894 340 86 1686 782 784 600 557 209 -53

M u top (k-ft) -309 -478 -350 -397 -297 -464 -207 -849 -1142 -910 -964 -669 -747 -645 -822 -1133 -913 -969 -670 -770 -647 -860 -1141 -919 -976 -675 -769 -648 -437 -639 -517 -566 -462 -644 -458

297

Design Example 6

6b.

!


Design of column for bending strength.

Section 1921.4.3 requires the longitudinal reinforcement ratio of columns to be between 1 and 6 percent. Design of columns is usually performed by calculating a column axial force-moment capacity (P − M ) interaction diagram. The major points used to construct such a diagram are φPn for compression, (φPb , φM b ) at the balance point, φM n for pure moment, and φTn for pure tension. The φ factor for column calculations is 0.70 for tied columns and 0.75 for spiral tied columns meeting requirements on §1910.9.3. In accordance with §1909.3.2.2, the φ factor may be increased linearly to 0.9 for columns or other axial load carrying members as φPn decreases from 0.10 f ' c Ag (or φPb whichever is less) to zero. The equation for φPn is given in §1910.3.5.

[

(

)

φPn = 0.85φ 0.85 f 'c Ag − Ast + f y Ast

]

(10-1)

Note that φ = 0.70 for members with axial compression and flexure (not with spiral shear reinforcement) per §1909.3.2.2. Calculation of the balance point is determined by using 0.002 strain for reinforcing steel at yield and 0.003 for concrete strain at crushing (§1910.3.2.). By summing forces and moments, the balanced axial load and moment (φPb , φM b ) can be determined. The nominal moment strength is determined by using 0.002 strain for steel yielding and by calculating tension forces and compression forces such that they add up to 0. The resulting moment is thus φM n , where φ = 0.90 . The equation for tension members is: φTn = φf y Ast Note that φ = 0.90 for members with axial tension and axial tension with flexure per §1909.3.2.2. The designer may use a commercial program such as PCACOL developed by the Portland Cement Association to develop a P − M diagram for the column axial load-moment interaction, including effects for slenderness of columns. From the frame analysis for Frame A, the controlling load cases are summarized in Table 6-12.

298


Design Example 6

!


Table 6-12. Critical column loads for Frame A Column

Level

Location

Size (in.)

Load Comb. Equation

Pu (kips)

Vu (kips)

Mu (k-ft)

22 1

1 1

interior end

36x44 42x42

12-5 12-6

990 -140

195 121

2,258 1,636

Note: See Figure 6-5 for locations of columns.

Column 22 represents the controlling load combination for a column in compression and Column 1 represents the controlling load combination for Column 22 in tension. Using the PCACOL program, check 36 × 44 interior column with 18 #10 bars around perimeter. The resulting P − M diagram is shown in Figure 6-8.

φ Pn (kips)

P-M diagram

column 22 point

φ Mn (kip-ft)

Figure 6-8. Column P-M diagram for 36 x 44-inch interior Column 22


299

Design Example 6

!


Check 42 × 42 corner Column 1 with 20-#10 bars around perimeter. The resulting P − M diagram is shown in Figure 6-9.

φ Pn (kips)

P-M diagram

column 1 point

φ Mn (kip-ft)

Figure 6-9. Column P-M diagram for 42-inch square Column 1

By comparing the design loads against the column P − M diagrams of Figures 6-8 and 6-9, it can be seen that both columns have adequate strength. Both column sections achieve 120 percent of beam moment strength, and thus have adequate strength to develop the plastic moments of beams. φM n for interior columns is approximately 2,550 kip-ft and for end columns is approximately 2,450 kip-ft at the axial load of approximately 1,000 kips. 6

∑Me = 5 ∑M ∑ M e,interior ∴

300

2(2,550 kip - ft ) = 7,284 kip-ft ≥ 3,882 kip-ft 0.7

o.k.

M e,end = ∴

=

2(2,450 kip - ft ) = 7,000 kip-ft ≥ 2,275 kip-ft 0.7

o.k.


Design Example 6

!


It is assumed by the code that the design of columns to be 120 percent greater in flexural strength than girders will ensure plastic hinge formation in the beams, and this is probably true in most cases. Since that is what is required in the 1997 UBC, that is what is shown in this Design Example. Some engineers believe that they should design the columns to develop the strength of the beam plastic moments Mpr. While this is not explicitly required by the 1997 UBC, it is probably a good idea. The reasoning is that the yielding elements in the frame are the beam plastic moments located at beam ends followed by column plastic moments at column bases. When all nonyielding aspects of the frame are designed to be stronger than the yielding elements, the anticipated frame yield behavior is ensured. Thus, the shear design of beams, columns, and joints, column flexural strengths, and foundation elements are all designed to have adequate strengths to resist the anticipated flexural yield mechanism of the frame.

Table 6-13. Column axial and flexural design strengths

6c.

Column

Size (in.)

φP n at M = 0 (kips)

φP b (kips)

φM b (k-ft)

φM n at P = 0 (k-ft)

Interior End

36x44 42x42

3,750 4,100

1,600 1,900

2,750 2,850

1,950 2,100

Design of columns for shear strength.

Columns must be designed for shear strength Ve required by §1921.4.5.1 and for the special transverse reinforcement required by §1921.4.4.1. The design shear force Ve shall be determined from the consideration of the maximum forces that can be generated at the faces of the beam/column joints at the ends of beams framing into the joint. These joint forces are determined in one of three methods: 1.

Using the maximum probable moment strengths, M pr , of the column at the top and bottom between joints along with the associated factored axial loads on the column.

2.

The column shear Ve need not exceed that determined based on the probable moment strength, M pr , of the beams framing into the joint.

3.

Ve shall not be less than the factored shear determined from analysis.

It is likely that the second method described above will control the shear design of the column, since strong column behavior of the frame will force plastic hinges to form in the beams. At the columns in the first story, the controlling case is from SEAOC Seismic Design Manual, Vol. III (1997 UBC)

301

Design Example 6

!


column top moments based on M pr of beams and column bottom moments based on M pr of the column calculated with associated axial loads. For the interior column, 36 × 44 , at stories one and two, the maximum shear need be determined from maximum shear that can be transferred from beam strength, M pr , as shown below. Interior column at first story.

Clear height of column = 14 ft-0 in. – 4 ft-0 in. = 10 ft-0 in. M pr of beams framing into top of column is based on negative moment from one beam and positive moment from the other beam.

∑ M pr

= 2,389 kip-ft + 1,708 kip-ft = 4,097 kip-ft

Distribution of beam moments to columns is in proportion of 4 EI L of columns below and above the joint. Since columns are continuous, 4 EI is constant, and moments are distributed based on 1 L of columns. The lower column has height 14 ft-0 in. and the upper column has height 12ft-0 in. The lower column will have a moment determined as follows at its top: 1      12'  M = 4,097 kip-ft  14'  = 4,097 kip-ft   = 1,890 kip-ft 1   1 26'   +    14' 12'  The lower column could develop a maximum of M pr at its base. The moment M pr for the column is determined with the PCA column program using a reinforcement strength of 1.25 F y or 75 ksi. M pr determined with the PCA column for an axial load of 1,000 kips is approximately 4,000 kip-ft. The shear Ve is determined as follows based on clear column height Ve =

(4,000 kip − ft + 1,890 kip − ft ) = 589 kips 10' 0"

This value is compared with frame analysis Vu = 176 kips, thus Ve controls.

302


Design Example 6

!


Interior column at second story.

Clear height of column = 12 ft-0 in. – 4 ft-2 in. = 9 ft-10 in. M pr of beams framing into top and bottom of column is based on negative moment from one beam and positive moment from the other beam.

∑ M pr ,above = 2,769 kip-ft + 2,028 kip-ft = 4,797 ∑ M pr ,below = 2,389

kip-ft

kip-ft + 1,708 kip-ft = 4,097 kip-ft

The second story column will have moments of:  12'  M top = 4,797 kip-ft   = 2,399 kip-ft  24'   14'  M bottom = 4,097 kip-ft   = 2,206 kip-ft  26' 

∑ M col

= 2,399 kip-ft + 2,206 kip-ft = 4,605 kip-ft

thus column shear Ve is determined as follows based on clear column height Ve =

4,605 kip − ft = 588 kips 7'10"

This value is compared with frame analysis Vu = 195 kips , thus Ve controls. The tabulated calculation of column shears is shown in Table 6-14 below.


303

Design Example 6


!

Table 6-14. Calculation of column shear forces, Ve Col. at Grid Lines 1, 5

2,3,4

Level/ Story 1 2 3 4 5 6 7 1 2 3 4 5 6 7

−M pr +M pr Col. ΣM pr Clear (joint (joint at Height above) above) Joint (ft) (kip-ft) (kip-ft) 10 7.83 8.5 8.5 9 9 9 10 7.83 8.5 8.5 9 9 9

2,389 2,769 2,055 2,055 1,175 1,175 974 2,389 2,769 2,055 2,055 1,175 1,175 974

Dist ΣM pr

to col.

−M pr +M pr M ΣM pr at Col. (joint (joint at Top below) below) Joint (kip-ft) (kip-ft) (kip-ft)

to col.

M Ve at Col. ΣM Bot. (kip-ft) at Col. (kips ) (kip-ft) 4,000 1,285 1,385 1,028 1,028 588 588 4,000 2,204 2,399 1,745 1,745 899

1,104 2,670 2,412 2,055 1,615 1,175 1,562 1,893 4,603 4,144 3,490 2,644 1,798

510 341 284 242 179 131 174 589 588 487 411 294 200

899

2,496

277

Dist. ΣM pr

0 0 0 0 0 0 0 1,708 2,028 1,435 1,435 623 623

2,389 2,769 2,055 2,055 1,175 1,175 974 4,097 4,797 3,490 3,490 1,798 1,798

0.462 0.5 0.5 0.5 0.5 0.5 1 0.462 0.5 0.5 0.5 0.5 0.5

1,104 1,385 1,028 1,028 588 588 974 1,893 2,399 1,745 1,745 899 899

0 2,389 2,769 2,055 2,055 1,175 1,175 0 2,389 2,769 2,055 2,055 1,175

0 0 0 0 0 0 0 0 1,708 2,028 1,435 1,435 623

0 2,389 2,769 2,055 2,055 1,175 1,175 0 4,097 4,797 3,490 3,490 1,798

0.462 0.538 0.5 0.5 0.5 0.5 0.5 0.462 0.538 0.5 0.5 0.5 0.5

623

1,597

1

1,597

1,175

623

1,798

0.5

Special transverse reinforcement per §1921.4.4.

The total cross-section area of rectangular hoop reinforcement shall not be less than that required by Equations (21-3) and (21-4). Ash = 0.3 shc f ' c / f yh

)[(Ag

(

)

(

Ash = 0.09 shc f ' c / f yh

) ]

(21-3)

Ach − 1

(21-4)

Transverse reinforcement shall be spaced at distances not exceeding 1.) one-quarter minimum member dimension and 2.) 4 inches. The transverse reinforcement should extend beyond any joint face a distance l o equal to the larger of: 1.) one column member depth; 2.) 1/6 of the clear column span; or 3.) 18 inches. Spacing between transverse reinforcement should not exceed 6 bar diameters of the longitudinal steel or 6 inches. Table 6-15 below shows calculations for special transverse reinforcement.

304


Design Example 6

!


Table 6-15. Special transverse reinforcement in columns Col. Size

Eq.

36x44 (21-3) (21-4) (21-3) (21-4) 42x42 (21-3) (21-4) (21-3) (21-4)

b

d

36 36 36 36 42 42 42 42

44 44 44 44 42 42 42 42

hc Trans

hc Long 32 32

40 40 38 38 38 38

f 'c

fy

Ag

Ach

s

Ash

4,000 4,000 4,000 4,000 4,000 4,000 4,000 4,000

60,000 60,000 60,000 60,000 60,000 60,000 60,000 60,000

1,584

1,390

4 4 4 4 4 4 4 4

0.357 0.768 0.446 0.96 0.397 0.912 0.397 0.912

1,584 1,764 1,764

1,390 1,560 1,560

No. Legs

Size Bars

5

#4

6

#4

6

#4

6

#4

Calculations for the required shear steel are shown in Table 6-16. The final column design at the first level is summarized in Table 6-17. The column design may be used for the full height columns or the reinforcement can be reduced slightly at the upper portion of the frame. Since the longitudinal reinforcement is only 1.44 percent, the longitudinal reinforcement cannot be reduced below 1 percent in any portion of the columns.

Table 6-16. Shear strength fy

Col.

Shear Vu (kips)

Shear Ve (kips)

b (in.)

d (in.)

f 'c (psi)

(psi)

36x44 42x42

195 140

510 589

36 42

44 42

4,000 4,000

60,000 60,000

φVc Av (kips) (sq. in.) 159 176

1.2 1.2

s (in.)

φVs (kips)

φVn (kips)

DCR

4 4

627 597

786 773

0.65 0.76

Table 6-17. Final column design at first level Column

Longitudinal Reinforcement

Long. Stirrups Within Yielding Zones, l o

Long. Stirrups Beyond Yielding Zones, l o

Trans. Stirrups wWthin Yielding Zones, l o

Trans. Stirrups Beyond Yielding Zones, l o

36x44 42x42

18-#10 20-#10

6-#4@4" 6-#4@4"

6-#4@6" 6-#4@6"

5-#4@4" 6-#4@4"

5-#4@6" 6-#4@6"

Figures 6-10 and 6-11 show the column cross-section with dimensions and reinforcement indicated. Note: Crossties can have 90 degree and 135 degree bends at opposite ends. 90 degree bends should be alternated with 135 degree bends at each successive tie set and at adjacent bars.


305

Design Example 6

!


Figure 6-10. 36 x 44 column

Figure 6-11. 42 x 42 column

306


Design Example 6

6d.

!


Orthogonal effects for columns.

§1633.1

The code requires that columns that are part of two or more intersection lateral force resisting systems be analyzed for orthogonal effects. However, the code excepts columns where the axial force caused by seismic forces from systems in any direction is less than 20 percent of the column capacity (per §1633.1). In this Design Example, the corner columns are required to be part of both the longitudinal and transverse seismic frames. An analysis would indicate that these columns fall below the 20 percent threshold and thus do not require an orthogonal analysis.

7.

Joint shear analysis.

Beam-column joints of frames must be analyzed for joint shear in accordance with §1921.5. The shear forces from analysis and the joint strength are calculated in Table 6-18.

Table 6-18. Joint shear analysis Element

8. 8a.

Location

Shear from Analysis (kips)

Vpr , Plastic Shear (kips)

Nominal Shear Stress

Aj (in.2)

Joint Strength (kips)

Result s

Interior Beam

Level 3

155

253

φ15 f'c A j

1,320

1,064

o.k.

End Beam

Level 3

157

253

φ12 f'c A j

1,260

813

o.k.

Interior Column

Level 2

195

588

φ15 f'c A j

1,320

1,064

o.k.

End Column

Level 2

133

341

φ12 f'c A j

1,260

813

o.k.

Detailing of beams and columns.

Beam reinforcement.

Beams should be detailed with top, bottom and side reinforcement as shown in Figure 6-7. In accordance with §1921.3.3, beam shear reinforcement, which meets the spacing requirements of §1921.3.3.2, should be provided over a distance 2d from the faces of columns. The tie spacing shall not exceed: 1.) d 4 ; 2.) 8d b of minimum beam longitudinal bar diameters; 3.) 24d b of stirrup bars; and 4.) 12 inches. These requirements result in a 9-inch maximum tie spacing. However, from analysis, ties required are #5 ties spaced at 6-inch centers. For ties between beam hinge regions, ties are required at d 2 spacing. However, based on analysis # 5 SEAOC Seismic Design Manual, Vol. III (1997 UBC)

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ties at 9-inch spacing are adequate across the remaining length of the beam (outside the hinge areas at each end). Longitudinal beam bars should be spliced away from the beam-column joints and a minimum distance of 2h from the face of the columns, per §1921.3.2.3. At the Level 2 beams for this Design Example, the beam clear spans are approximately 26 ft and 2h is 2(46") = 7 ft-8 in. The designer might consider splicing beam longitudinal reinforcement at the quarter-, third-, or half-span locations. In this case, the quarter-span locations would not be away from hinge regions. However, the one-third, or mid-span, locations would also be okay. Increased shear reinforcement is required at the lap splice locations per §1921.3.2.3. The maximum spacing of ties in these regions shall not exceed d 4 or 4 inches. In this case, the beam mid-point is the best place to locate the lap splices, which for the #11 top bars at Class B splices would have a splice length of 110 inches or 7 ft-2 in. The lap splice length for #9 bottom bars at a Class B lap splice is 69 inches or 5 ft-9 in. Longitudinal reinforcement can be shipped in 60 ft-0 in. lengths on trucks, thus two locations of longitudinal beam lap splices would be required in the frame along Line A, conceivably on the two interior spans.

8b.

Column reinforcement.

Column splices should occur at column mid-story heights (or within the center half of the column heights) per §1921.4.3.2. Special transverse reinforcement is required per §1921.4.4 over a length l o above and below beams at spacing not greater than: 1.) the column depth; 2.) one-sixth the column clear span; or 3.) a maximum of 18 inches. For this Design Example the column depth would control which is either 42 inches or 44 inches depending on the column. For column sections between the locations where special transverse reinforcement is required, the spacing requirements of §1907.10.5.2 apply where ties should be spaced a maximum of 16 longitudinal bar diameters, 48 tie bar diameters or the least dimension of the column. This would require ties at 20 inches; however for this Design Example, it is recommended not to space column tie bars greater than 6 inches per §1921.4.4.6 and 4 inches at lap splices.

9.

Foundation considerations.

The foundation system should be capable of resisting column base moments sufficient to cause plastic hinges to be located in the beams and column bases. If the plastic hinge location is forced into the columns, the foundation elements need not be designed for yielding or ductility. The foundation should also be adequate to keep soil pressures within allowable values and adequate for frame overturning stability. For this analysis, a 60-inch wide by 48-inch deep grade beam was used and cracked beam properties were used in the computer analysis (Figures 6-12 through 6-16). Note that ASD combinations of loads are used for calculation of soil pressures. The actual design of foundation elements is not performed in this Design Example. 308


Design Example 6

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Figure 6-12. Beam-column joint

Figure 6-13. Beam reinforcement lap splice


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Figure 6-14. Beam-column joint reinforcement at exterior span

Figure 6-15. Beam reinforcement at interior spans

Figure 6-16. Beam column corner joint at roof

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Commentary Deformation compatibility should be checked at interior columns due to seismic drifts ∆ M . This will lead to a conservative design for punching shear at slab/column joints. These joints may require drop panels or shear head reinforcement in the slab over interior columns. The building period in this Design Example was calculated using Method A. Method B could be used as long as the resulting period was not more than 130 percent of the Method A period (in Seismic Zone 4) or 140 percent of the Method A period (in Seismic Zones 1, 2, and 3). If Method B is used to determine the period, the designer should keep in mind that nonseismic elements can cause stiffness in the building and thus cause a decrease to the Method B period determination. Thus, interior nonseismic columns or other important stiffening elements should be included in Method B period calculations to ensure conservative period calculation results. Reinforced concrete SMRF frames can provide very ductile seismic systems for buildings with highly desirable performance characteristics. The yielding mechanisms can be predicted and the seismic performance will be ductile and not brittle. Care should be taken to ensure adequate shear strength at beams, columns, and joints, so that ductile flexural yielding will occur as anticipated. Care should also be taken with lap splices and detailing of reinforcement and with specified couplers. Reinforcement should be ASTM-A706, which has more ductile performance characteristics that ASTM-A615 reinforcement.


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References A. H. Nilson and G. Winter, 1986. Design of Concrete Structures, Tenth Edition. McGraw-Hill, New York. ATC-40, 1996. Seismic Evaluation and Retrofit of Concrete Buildings. Applied Technology Council, Redwood City, California. J. C. McCormac, Design of Reinforced Concrete. Third Edition. Harper-Collins, New York. J. G. MacGregor, 1992. Reinforced Concrete Mechanics and Design, Second Edition. Prentice Hall, New Jersey. R. Park and T. Paulay, 1975. Reinforced Concrete Structures. John Wiley and Sons, New York. Paulay, T. and Priestley, N.J., 1992. Seismic Design of Reinforced Concrete and Masonry Buildings. John Wiley and Sons, New York.

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Design Example 7

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Precast Concrete Cladding

Design Example 7 Precast Concrete Cladding

front elevation

typical wall section

Figure 7-1. Typical precast concrete panel elevation

Overview This Design Example illustrates the seismic design of precast concrete cladding Panels A and B shown in the partial wall elevation of Figure 7-1. This cladding example is for a 4-story steel moment frame structure located in Seismic Zone 4. The architect has chosen precast concrete panels for the façade. Current standard practice is to specify that the fabricator perform the design for the panel and connections. The structural Engineer of Record for the building typically reviews the fabricator’s design for compliance with the project design SEAOC Seismic Design Manual, Vol. III (1997 UBC)

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specifications, and for compatibility with the structural framing. It is important that the structural Engineer of Record understand that panel loads are concentrated at discrete points to the structure. These points of attachment will usually require additional support steel to reach the panel connection hardware. These supports will typically induce eccentric loads into the beams and columns that must be accounted for in design of the structure. Wind loads will also be considered in this example, since some elements of the connection and panel reinforcing may be controlled by wind, while seismic forces may control other parts. Earthquake-damaged cladding can become a severe falling hazard, particularly damaged cladding on highrise buildings in congested urban areas. Cladding is typically connected at a few discrete points, which limit the redundancy of the system. For this reason, code seismic requirements for the “attachment” of cladding require a more conservative design than other building components. Building cladding is also required to resist realistic story drifts without failure through flexible connections and adequate panel joints. These requirements are detailed in §1633.2.4.2 and will be illustrated in this Design Example. This Design Example provides an overview of the design procedure for precast concrete cladding panels and their connections to the structure.

Outline This Design Example illustrates the following parts of the design process:

314

1.

Governing loading conditions and forces.

2.

Selection of panel thickness.

3.

Selection of the panel connection scheme.

4.

Panel reinforcing design.

5.

Connection forces.

6.

Typical connection design.

7.

Panel joint widths to accommodate drift.

8.

Typical connection details.


Design Example 7

!


Given Information Exterior wall system weight: Precast concrete panel (5″ thickness) = 62.5 psf Window system = 10 psf Metal stud and gypsum board, 5 psf Wind design data: Basic wind speed = 70 mph Wind exposure = C Importance factor, Iw = 1.0 Seismic design data: Occupancy category: standard occupancy structure Seismic importance factor, Ip = 1.0 Soil profile type: stiff soil type SD (default profile) Seismic zone 4, Z=0.4 Near-source factors: Seismic source type A Distance to seismic source, 7 km Maximum inelastic response displacement, ∆M = 3.2 in. Building design data: Building mean roof height = 64 ft Top of parapet = elevation 66 ft 6 in. Building plan dimensions = 150 ft x 70 ft Material specifications: Concrete: Compressive strength fć = 4,000 psi, ASTM C39 Aggregates, ASTM C33 Portland Cement, ASTM C150 Admixtures, ASTM C494 Unit weight 150 pcf, ASTM C138 Steel: Structural shapes, plates and bars Fy = 36 ksi, ASTM A36 Hollow structural section: round Fy = 33 ksi, ASTM A53, Grade B Hollow structural section: rectangular Fy = 46 ksi, ASTM A500, Grade B Welded Reinforcing steel fy = 60 ksi, ASTM A706 Non-welded reinforcing steel fy= 60 ksi, ASTM A615, Grade 60 Coil rods, ASTM A108 Weld electrodes: Shielded metal arc welding FEXX = 70 ksi, AWS A5.1 E70XX Flux-cored arc welding FEXX = 70 ksi, AWS A5.20 E7XT


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1.

Code Reference

Governing loading conditions and forces.

§1605.1

Cladding panels must be designed to resist both vertical loads and lateral forces. Typically the vertical loads consist of the panel weight and the weight of any windows or other miscellaneous architectural items attached to the panel. Normally, two bearings points are provided and the panel is treated as a simply supported beam for vertical loads. The lateral forces consist of both wind and seismic effects. Wind forces are included in this Design Example because they are an integral part of the design process for cladding and to illustrate the application of load combinations for all the loading cases. Where structural effects of creep, shrinkage, and temperature change may be significant in the design, they shall be included in the load combinations.

1a.

Design wind pressures.

§1909.2.7

Chapter 16, Division III

Wind pressures are determined from Equation (20-1) using the 70 mph basic wind speed. This process is shown below. (20-1)

P = Ce Cq qs Iw qs = 12.6 psf

Table 16-F

h = mean roof height = 64 ft Interpolation is used to determine the combined height and exposure factor Ce.

Table 16-G

Interpolation for h = 64 ft (mean roof height). Ce = (1.53 − 1.43)

(64 − 60) + 1.43 = 1.45 (80 − 60)

Interpolation for h = 66.5 ft (top of parapet). Ce = (1.53 − 1.43)

(66.5 − 60) + 1.43 = 1.46 (80 − 60)

The pressure coefficients for the exterior elements are given in Table 16-H. The resulting pressures are summarized in Table 7-1 below.

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Table 7-1. Design wind pressures Element

Direction

Cq

p (psf)

In

1.2

21.92

Typical panel & connection

Out

1.2

21.92

Corner panel & connection

Out

1.5

27.41

Corner panel & connection

In

1.2

21.92

Parapet panel & connection

In/out

1.3

23.91

Typical panel and connection

Note: The inward pressure may be calculated for the actual height of that element; however, the outward pressure is based on the mean roof height and is considered to be constant along the height of the building. For simplicity the inward pressure is calculated using the mean roof height. The outward corner pressure may be reduced based on the actual tributary area being considered . Since seismic forces will usually govern the connection design for precast panels, this reduction has not been applied in Table 7-1.

1b.

Design seismic forces.

§1632

Seismic forces for elements of structures, such as the precast panels of this example, are specified in §1632. These are summarized below.

panel

in-plane

out-of-plane

Figure 7-2. In-plane and out-of-plane force on panel

The basic equation is: Fp = 4.0 Ca Ip Wp

(32-1)

This represents an upper bound of element force levels and is seldom used. The alternate equation, more frequently used is: Fp =

a p Ca I p  h  1 + 3 x W p Rp  hr 


(32-2)

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Limits are set on Equation (32-2) such that Fp shall not be less than 0.7 CaIpWp and need not be more than 4 CaIpWp.

(32-3)

Typically the alternate Equation (32-2) is used since the results for panel and body loads will be more in line with the previous code force levels. Pertinent values for ap and Rp , taken from Table 16-O, are given below in Table 7-2.

Table 7- 2, Horizontal Force Factors, ap and Rp Wall Elements of Structure

ap

Rp

Unbraced (cantilevered) parapets

2.5

3.0

Exterior walls at or above the ground level

1.0

3.0

All interior bearing and nonbearing walls

1.0

3.0

The structural Engineer of Record must specify the near-source factor and distance to the fault zone. In many cases the seismic coefficient Ca is specified, but for this example we will start with Na and the fault distance. The seismic coefficient Ca is found from Table 16-Q. For seismic zone 4 and soil profile type Sd Ca = 0.44 Na

Table 16-Q

Since the distance to the source is 7 km and the source is type A, Na is found by interpolation as permitted by Table 16-S. Na =

(1.2 − 1) (10 − 7 ) + 1.0 = 1.12 (10 − 5)

Ca = 0.44 (1.12) = 0.493

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The maximum, minimum, and the value at height hx of Fp are: Maximum

Fp = 4.0 (0.493) (1.0) Wp = 1.97 Wp

Minimum

Fp = 0.7 (0.493) (1.0) Wp = 0.345 Wp

At hx : Fp =

(1.0)(0.493)(1.0) 1 + 3 hx  = 0.164 1 +  

3.0

 64 

 

hx   21.33 

Additional requirements for exterior elements are given in §1633.2.4.2. These apply to the “attachments” of the panel to the structure. For the body of the connection system: ap = 1.0

Rp = 3.0

For the fasteners of the connection system: ap = 1.0

Rp = 1.0

Table 7-3 below summarizes the seismic coefficients, which multiplied by the tributary weight Wp, are used to determine the design lateral force Fp. Note that the seismic coefficients for the fasteners are substantially higher than those for the panel or the body of the connection. Use of these is illustrated later in this example.

Table 7-3. Seismic coefficients Level

hx/hr

Fp (panel)

Fp (body)

Fp (fastener)

0

0.00

0.345

0.345

0.493

1

0.25

0.345

0.345

0.862

2

0.50

0.411

0.411

1.232

3

0.75

0.534

0.534

1.602

4

1.00

0.657

0.657

1.970

Parapet

1.00

1.643

1.643

1.970

Note: When the difference in elevation of connections becomes significant, the current interpretation of the code requires a calculation of Fp at each level of connections for the area of panel tributary to those connections. Examples are full story wall panels where the bottom connections are made to one floor while the top connections are made to the floor above.

In this Design Example, the floor elevation where the upper connections are attached was used to calculate Fp. For out-of-plane forces, this is conservative since the other connections are below this point. For in-plane forces this would follow the current interpretation since all primary reactions occur at this level. SEAOC Seismic Design Manual, Vol. III (1997 UBC)

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Design Example 7

2.

!


Selection of panel thickness.

In general the final precast design begins with the panel thickness as a fixed dimension and the connection system is developed from that point forward. The panel thickness is a decision that must be made early in the design process by the architect. Consultation with a precast manufacturer is recommended to help with shipping and handling considerations. Any changes to the panel thickness after the project has proceeded can have significant impact on other portions of work. There are many factors to consider when deciding on a panel thickness. Some of these are listed below: Architectural considerations: Fire resistance Thermal insulation Sound insulation Weather resistance Structural considerations: Total weight of exterior elements Weight supported by exterior beams and columns Deflection and cracking Fabrication and installation: Minimum weight for handling, shipping and erection Adequate thickness for efficient handling Adequate stiffness for an efficient connection scheme For this project, the panels are specified to be 5 inches thick. This thickness provides adequate anchorage depth for the connection hardware and also allows the panel to be handled easily. Another consideration is the warping and bowing that may occur during curing. Thin long panels will bow or warp more than thick short panels.

3.

Selection of panel connection scheme.

The primary goal in developing a connection system is to minimize the number of connections and provide connections that have adequate tolerance with the structural frame. For this example we will try 4 connections first as shown in Figure 7-3. Because of the moment frame structural system, the bearing connection must either be located off of the column or on the beam away from any potential hinge location. In this case we will assume a support is provided off of the column so that the bearing connections will be close to the end of the panel.

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▼ resists forces in all directions ● resists out-of-plane forces only 1.5 ft, typ

27 ft

Figure 7-3. Initial connection scheme for Panel A

Compare wind loading versus seismic loading.

The tributary height is 16 ft because the precast panels transfer wind load on both the glazing and panels to the structure. Total uniform wind loading on panel. pw = 21.92 psf (16 ft) = 351 plf Assume the panel under consideration is located on Level 3. The working level load for the seismic forces is: ps =

Fp 1.4

wp =

0.534 [(62.5 psf )(7 ft ) + (10 psf )(9 ft )] = 201.1 plf 1.4

Therefore, wind controls for panel design. This is typical for a spandrel panel. Check panel moment at mid-span. 2

2

M = 0.351 klf (27 /8 – 1.5 /2) = 31.5 k-ft 2

Sy = (84 in.)(5 in.) / 6 = 350 in.

3

fby = My / Sy = 1.08 ksi The modulus of rupture for concrete is f r = 7.5 f c′ = 7.5 4000 = 474 psi

(9-9)

This panel stress is well above the modulus of rupture and the panel will not satisfy the deflection criteria because of the reduced moment of inertia from cracking (§1909.5.2.3).


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Although the code does not specifically address out-of -plane deflection of cladding panels, some guidance can be found in Table 16A-W of the 1998 California Building Code. Typically, the deflection is limited to L/240 because of the other elements that are attached. Also, in order to satisfy the crack control criteria of the code (See §1910.6.4), large amounts of reinforcing may be required. Consequently, connections will be provided at mid-span to reduce the panel stresses and deflections.

4.

Panel design.

Wind controls the panel design and bending moments are determined using the load combination of Equation (12-6). Note that the 1.1 multiplier of Exception 2 of §1612.2.1 is not applied for wind. Wind: M f = pw

l2 27' = 13.5' = 8.0 k-ft where l = 8 2

Mc Mu 1

M c = pw

2

a = 0.39 k-ft where a = 1.5' 2

Mu 2

centerline

1   M u1 = 1.3 M f − M c  = 10.14 k-ft, moment over middle support 2    1M  M u 2 ≈ 1.3 M f −  u1 + M c  = 5.07 k-ft, approx. moment between supports 2  1.3   Determine reinforcing required for strength. Consider a one-foot width: Mu = 10.14 k-ft / 7 ft = 1.45 k-ft b = 12"

d = 2"

a  φ M n = φAs f y  d −  2  a=

As f y 0.85 f c′b

Solving directly for As leads to 0.17 in.2/ft. 322


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Minimum reinforcement required for walls.

§1914.3.3

For deformed bars not larger than #5 with fy ≥ 60,000 As / bh = 0.0020 As ,min = 0.002(12")(5") = 0.12 in.2 /ft < As , as required Flexural minimum steel requirements: As ,min =

3 f c′ fy

bw d =

3 4000 (12")(2") = 0.076 in.2/ft 60,000

§1910.5 (10-3)

But not less than: 200bw d 200 (12")(2") = = 0.08 in.2 /ft fy 60,000 The ratio of reinforcement ρ provided shall not exceed 0.75 of the ratio ρ b that would produce balanced strain conditions for the section.  0.85β1 f c′   87,000  ρb =    87,000 + f y  f y  

  0.85 (0.85)(4 )   87,000  =  = 0.0285    60   87,000 + 60,000  

(8-1)

As, max = 0.75ρ b bw d = 0.75 (0.0285)(12")(2") = 0.51 in.2/ft Use #4 at 12 inch o.c., As = 0.20 in.2 /ft Check crack control requirements.

§1910.6.4

Consider a one-foot-wide strip at the bottom of the panel. bh 2 12 (5) Sy = = = 50.0 in.3 6 6 2

M wind =

7.8 k − ft = 1.1 k-ft/ft 7′

f by =

M wind 1.11(12") = = 0.267 ksi Sy 50.0

Sx =

hb 2 5(84 )2 = = 5,880 in.3 6 6


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Neglect small cantilever at the ends. M DL = f bx =

0.528 klf (27')2 = 48.1 k-ft 8

M DL 48.1(12") = = 0.098 ksi Sx 5,880

f tot = f by + f bx = 0.36 ksi < f r = 0.474 ksi Therefore, there is no cracking under service loads, and the crack control requirements of §1910.6.4 are not applicable. The maximum deflection under service wind loading is: ∆ = 0.03" <

13.5' (12 ) L = = 0.675" 240 240

o.k.

Deflection is o.k.

5.

Connection forces.

In this part, connection forces will be determined. Seismic forces are determined for a 1g loading. These will then be appropriately scaled in Part 6. The distribution factors used to determine reactions at the various supports were determined from a generic moment distribution. For brevity, that analysis is not shown here. Element weights:

324

Panel A

Wpa = 62.5 psf (7ft) (30 ft)

= 13.13 k

x = 15 ft

z = .208 ft

Column cover B

Wpb = 62.5 psf (4 ft/2) (9 ft)

= 1.125 k

x = 1 ft

z = .208 ft

Column cover B

Wpb = 62.5 psf (4 ft/2) (9 ft)

= 1.125 k

x = 29 ft

z = .208 ft

Window

Ww =10.0 pst (9 ft) (26 ft)

= 2.34 k

x = 15 ft

z = .10 ft

Total

Wtot =

= 17.72 k

x = 15 ft

z = .19 ft


Design Example 7


!

Gravity.

For gravity loads, the panel is treated as a simply supported beam using two bearing connections to support the vertical load. Since the vertical support reaction does not line up with the center of gravity in the z-direction, additional reactions are necessary in the z-direction to maintain equilibrium, as shown in Figure 7-4. ez = 0.33 ft (distance from the back of the panel to the center of the bearing bolt) R1y = 17.72/2 = 8.86 k R1z = 8.86 (0.19 + 0.33)/5.25 = .88 k R3z = -R1z y ez y

R1y R1z x

R3z z

Figure 7-4. Gravity load reactions

Seismic out-of-plane ( 1g ).

Connection distribution factors for a uniform load applied to a symmetric two span continuous beam with cantilevers at the ends are shown below and are used to distribute the uniform panel weight applied transverse to the panel. These can be found by moment distribution or other suitable means of continuous beam analysis. DFe = 0.223 (fraction of total load resisted by outside support) DFm = 0.554 (fraction of total load resisted by mid-span support) Connection distribution factors for a uniform load applied to a symmetric two span continuous beam without cantilevers at the end are given below. These will be used to distribute the uniform window load to the connections. DFe = 0.1875 (fraction of total load resisted by exterior support) DFm = 0.625 (fraction of total load resisted by middle support)


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Design Example 7

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The total reactions (Figure 7-5) are as follows: R1z = [(0.223)(13.13 kips ) + (0.1875)(2.34 kips ) + 1.125](2.5' 5.25') = 2.19 kips R3 z = [(0.223)(13.13 kips ) + (0.1875)(2.34 kips ) + 1.125](2.75' 5.25') = 2.35 kips R5 z = [(0.554 )(13.13 kips ) + (0.625)(2.34 kips )](2.5' 5.25') = 4.16 kips R6 z = [(0.554 )(13.13 kips ) + (0.625)(2.34 kips )](2.75' 5.25') = 4.58 kips

y

y

2.75 ft R5z

R1z

2.50 ft x

R3z

R6z

z

Figure 7-5. Seismic out-of-plane reactions

Seismic in-plane (1g)

In-plane seismic forces (Figure 7-6) are typically resisted by connections at the level the panel is supported. Overturning forces are resisted by vertical reactions at the supports. el = 0.50 ft R1x = 17.72/3 = 5.91 k R1y = 17.72 k (2.75'/27') = 1.80 k R1z = 1.80 k (.19 +.33)/5.25' + 17.72 k (.19 + .5)/27' = 0.63 k

y el

y R2x R1x

R5x

R1z R1y R3z

x R2y

z

Figure 7- 6. Seismic in-plane reactions 326


Design Example 7

!


Wind loading.

The distribution of total load is similar as was done for seismic out-of-plane forces (Figure 7-5). Pw = 21.92 psf (16')(30') = 10.52 k R1z = [(0.223)(10.52 k)](2.5'/5.25') = 1.12 k R3z = [(0.223)(10.52 k)](2.75'/5.25') = 1.23 k R5z = [(0.554)(10.52 k)](2.5'/5.25') = 2.78 k R6z = [(0.554)(10.52 k)](2.75'/5.25') = 3.05 k

6.

Connection design.

Design of the bearing connection will be done using strength design for both concrete and steel elements of the connection. This is illustrated in the parts below.

6a.

Application of load factors.

The basic load combinations are defined in §1612.2.1. Normally there are no floor live loads, roof live loads, or snow loads on cladding panels. The load combinations of Equations (12-1) through (12-6) reduce to the following. Parts of the load combinations not used have a strike line through them. 1.4D

(12-1)

1.2D + 1.6L + 0.5(Lr or S)

(12-2)

1.2D + 1.6 (Lr or S) + (f1L or 0.8W)

(12-3)

1.2D + 1.3W + f1L + 0.5 (Lr or S)

(12-4)

1.2D + 1.0E + (f1L +f2S)

(12-5)

0.9D + (1.0E or 1.3W)

(12-6)

For concrete design, Exception 1 to §1612.2.1 applies for combinations that include seismic forces. For all other combinations, Exception 2 refers to §1909.2. These equations reduce to the following:


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1.4D + 1.7 L

(9-1)

0.75 (1.4D + 1.7 L + 1.7 W)

(9-2)

0.9D + 1.3 W

(9-3)

1.1(1.2D + 1.0E)

(12-5)

1.1 (0.9D ± 1.0E)

(12-6)

For concrete anchors, additional load factors can be found in §1923.2. A load factor of 1.3 is normally applied for panel anchorage when special inspection is provided. When special inspection is not provided, a factor of 2 is applied. In addition, when anchors are embedded in the tension zone of a member, an anchor factor of 2 is required for special inspection and an anchor factor of 3 is required for no special inspection. These factors are not considered applicable to cladding panels, since the connector load is already raised significantly for nonductile portions of the connector. It should be noted that §1632.2 requires the design of shallow anchors to be based on forces using a response modification factor, Rp , of 1.5. Most embedded anchors in panels fall within the shallow anchor criteria. Since the fastener force is based on an Rp equal to 1.0, the shallow anchor requirement is superceded by the more stringent fastener force requirement. The total seismic force is defined as follows, where Fp is used for Eh and Ev is defined in §1630.1.1: E = ρ Eh + Ev

(30-1)

Ev = 0.5 Ca I D

§1630.1.1

Under §1632.2, the reliability/redundancy factor, ρ, may be taken equal to 1.0 for component design. The 1997 UBC load factors do not distinguish between members of the lateral force-resisting system and components, as the 1994 UBC did. Therefore, wording in the 1997 code is such that Ev should be considered for strength design of components similar to the requirements for the structure design. Ev was added to the code to make the load factors consistent with the load combination 1.4 (D + L + E), which applied to lateral force-resisting systems. For component design, the normal ACI and AISC load factors were appropriate, and hence no inconsistency was created. The addition of Ev for component design creates a higher load factor on dead load when compared to the 1994 UBC requirements.

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Design Example 7

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Application of load factors for typical bearing angle design. Ca = 0.493 Ip = 1.0 Ev = 0.5 Ca Ip D = 0.25 D For steel design the equivalent load factor for dead load is 1.2 + 0.25 = 1.45. For concrete design the equivalent load factor for dead load is 1.1(1.2 + 0.25) = 1.60. Assuming this panel is located at Level 3, Fp (body) = 0.534(Wp); Fp (fastener) = 1.602(Wp)

Table 7-3. Connection 1: bearing support Loading (1) D Et Et Wo Wi

X-Direction

Body Force Y-Direction

Z-Direction

0.00 0.00 3.17 0.00 0.00

8.86 0.00 0.96 0.00 0.00

0.88 1.17 0.34 1.12 1.12

Concrete Load Combinations 1.4D 1.60D + 1.1Et 0.99D - 1.1Et 1.60D + 1.1El 0.99D – 1.1El 1.05D + 1.275Wo 0.9D – 1.3 Wi Steel Load Combinations 1.4D 0.00 1.45D + 1.0Et 0.00 0.9D - 1.0Et 0.00 1.45D + 1.0El 3.16 0.9D – 1.0El -3.16 1.2D + 1.3Wo 0.00 0.9D – 1.3Wi 0.00

Fastener Force X-Direction Y-Direction Z-Direction 0.00 0.00 9.48 0.00 0.00 0.00 0.00 0.00 13.56 -13.56 0.00 0.00

12.40 12.82 7.97 13.78 7.01 10.63 7.97

1.23 2.44 -0.38 1.61 0.45 2.51 -0.66

0.00 0.00 0.00 9.48 -9.48 0.00 0.00

8.86 0.00 2.88 0.00 0.00 Anchor Factor = 16.13 18.33 11.40 22.45 7.28 12.09 10.37 12.40 12.82 7.97 15.70 5.09 10.63 7.97

0.88 3.51 1.02 1.12 1.12 1.3 (2) 1.60 6.84 -3.89 3.28 - 0.33 3.06 -0.83 1.23 4.78 -2.72 2.29 - 0.23 2.51 -0.66

Notes: 1. D = dead load Et = seismic out-of-plane El = seismic in-plane Wo = wind out Wi = wind in 2. From §1923.3, assuming special inspection.


329

Design Example 7

6b.

!


Typical connection design.

A typical bearing support is illustrated below and is used in this example to outline the design procedure for a panel connection. Most cladding panels use a threaded bolt to support the gravity loads. The bolt can be turned to adjust the panel into its final position. The embed is usually an angle with a threaded hole oriented as shown is Figure 7-7. This provides a low profile that can be hidden within the interior finishes.

Figure 7-7. Typical bearing connection

Determine angle size using LRFD.

§2206

Make preliminary selection of angle thickness. Note the critical section occurs at the root of the fillet or a distance k from the heel of the angle. M u = Rvu (e − k ) = 13.78 (4 − 1.5) = 34.45 k-in.

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Design Example 7

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Let width of angle b = 8 in. t=

4 Mu = 0.73" Φ fy b

∴ Use t = 1.0" Try L6 × 6 × 1 × 0' – 8" The body of connection forces for the load combination of 1.45 D + 1.0 El are shown below. Note that the moment is determined at the k-distance (see p. 1-58 of AISC–LRFD Manual). Muy = Ry1 (e1 – k) = 13.78 (4–1.5) = 34.45 k-in. Mux = Rx1 (e2 – k) = 3.16 (6–1.5) = 14.22 k-in. Pu = 1.61 k φM ny

0.9(36 ksi )(8")(1)2 = φ fyZy = = 64.8 k-in. 4

φM nx = φ f y Z x =

0.9 (36 ksi )(1")(8)2 = 518.4 k-in. 4

φPnt = φ f y A t = 0.9 (36 ksi )(8")(1") = 259.2 k M uy M ux 1 Pu + + = 0.56 < 1.0 2 φ Pnt φ M nx φ M ny

∴

o.k.

Use L6 × 6 × 1 × 0'−8"

6c.

Anchorage to concrete.

§1923

The concrete anchors consist of flat bar metal straps bent in a U-shape and welded to the back of the angle, as shown in Figures 7-8 and 7-9. Reinforcing bars are then placed in the inside corners of the bends to effectively transfer the anchor forces into the concrete. By doing this, the strength reduction factor, φ , may be taken as 0.85 instead of 0.65 per §1923.3.2.


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Design Example 7

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Headed studs are also used to transfer the forces to the concrete. The pull-out calculation for design is similar to the procedure for bent straps.

le

w

le

le bs

le

le

bs le

Figure 7-8. Single strap Ap

Figure 7-9. Double strap Ap

Single strap pull-out capacity.

§1923.3.2

Find the pull-out capacity of one leg of a 2 in. x 5/16 in. flat bar using the projected area of the shear cone. bs = 2 in. ts = 0.3125 in. le = l – ts = 4 – 0.3125 = 3.69 in.

(

)

A p = 2l e bs + πl e 2 = 57.54 in.2 φPnc1 = φλ 4 A p

f c′ = 0.85 (1.0)(4 )(57.54) 4,000 / 1,000 = 12.37 k

Double leg strap pull-out capacity.

Find the pull-out capacity of both legs of the 2-inch x5/16-inch flat bar using the projected area of the shear cone. Width w = 8 – 2 (1/2) – 0.3125 = 6.69 in.

(

)

A p 2 = 2l e bs + πl e 2 + 2l e w = 106.9 in. φPnc 2 = φλ 4 A p

f c′ = 0.85 (1.0 )(4 )(106.9 ) 4,000 / 1,000 = 22.9 k < 2φPc1 = 24.75

∴ φPc 2 controls

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Design Example 7

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Check pullout of bottom straps (double leg).

dy

el

Cu

a/2

Tu1

a/2 Rtu Cu d

ez Rvu Tu2 Rxu

Figure 7-10. Forces on straps

a=

φPnc 22.9 = = 0.84 in. 0.85 f c′b 0.85 (4 )(8)

Tu 2 =

(Rvu )(e) (22.45)(4) = = 19.61 k < φPnc 2 (d − a / 2) (5 − 0.84 / 2)

o.k.

Check pullout of top straps (single leg).

Tu1 =

(R xu )(e1 ) R zu + (d − a / 2) 2

−

Cu min (13.56)(6) + (− 0.33) − 1 × (7.28)(4) = 8.56 k < φP ∴ o.k. = c1 (7.34 − 0.5) 2 2 2 (5 − 0.4 )

Use reinforcing steel to resist vertical and horizontal shear forces. Computations of required reinforcement is shown below. Asv =

R yu φf y

=

22.45 = 0.32 in.2 (1.3)(0.9)(60 ksi )

Use 2-#4 vertical bars welded to angle. Ash =

Rxu 13.56 = = 0.19 in.2 φf y (1.3)(0.9 )(60 ksi )

Use 1-#5 horizontal bar welded to angle.


333

Design Example 7

6d.

!


Weld design: plate to support.

Rzu/2 2 in.

Rzu/2

Figure 7-11. Typical weld Out-of-plane forces.

Vertical load is supported by bearing (i.e. leveling) bolt. Rzu = 4.78 k (factored steel load , fastener level) Try a fillet weld 3 inches long. fv =

1 R zu 1 4.78 = = 0.80 k/in. shear component 2 lw 2 3

ft =

1 Rzu e 1 4.78(2 / 2 ) = = 1.59 k/in. tension component 2 Sw 2 32 / 6

fr =

f x 2 + ft 2 = 1.78 k/in. resultant

The weld capacity can be found in Table J2.5 [AISC-LRFD]. φRnw = φt eff 0.6 FEXX = 0.75(0.707 )(0.25)(0.6 )70 ksi = 5.57 k/in. > fr

o.k.

Use ¼-inch fillet weld by 3 inches long on each plate. Since the plate is designed for body loads, a plate of the same length and thickness will work. Use plate 5/16 x 3 x 0 ft-5 in.

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Design Example 7

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In-plane forces.

Rxu = 9.48 k (factored steel load, fastener level) Try a fillet weld 4 inches long. fv =

R xu 9.48 = = 2.37 k/in. lw 4

ft =

1 R xu e 1 9.48(2) = = 3.55 k/in. 2 Sw 2 42 / 6

fr =

f v 2 + f t 2 = 4.27 k/in.

φRnw = φt eff 0.6 FEXX = 0.75(0.707 )(0.25)(0.6 )70 ksi = 5.57 k/in. > fr

o.k.

Use ¼-inch fillet weld by 4 inches long.

7.

Drift analysis.

§1633.2.4.2(1)

One of the most important aspects of cladding design is to ensure that the panel connections and joints allow for the interstory drift that occurs as a result of lateral deflection of the frame from wind, seismic loads, temperature, and shrinkage forces. For most structures in Seismic Zones 3 and 4, seismic drift will control. For seismic drift, all cladding elements must accommodate the maximum inelastic story drift (∆M) that is expected for the design basis earthquake forces. The 1994 UBC estimated the inelastic drift as 3/8(Rw) times the calculated elastic story drift caused by design seismic forces. Now the inelastic drift is computed as 0.7 R∆S per §1630.9.2 or by a more detailed analysis. A comparison of the two values is shown below:


335

Design Example 7

!


1994 UBC

∆M =

R  ∆ M = 0.7 R∆ s ≈ 0.7 w 1.4∆ ≈ 0.7 Rw ∆  1.4 

3 Rw∆ 8

If T < 0.7 sec , ∆ ≤ ∆m =

1997 UBC

0.04 h Rw

3 0.04h Rw ≤ 0.015 h Rw 8

If T ≥ 0.7 sec , ∆ ≤ ∆M =

0.03 h Rw

3 0.03h Rw ≤ 0.01125h 8 Rw

If T < 0.7 sec ∆ M ≤ 0.025h If T ≥ 0.7 sec ∆ M ≤ 0.020h

The maximum inelastic drift can be as much as 78 percent higher under the provisions of the 1997 UBC compared to that calculated under the 1994 UBC. This can have a major impact on the cladding elements and must be considered early in the planning process. Fortunately, the majority of structures have drift less than the maximum. It is also important to coordinate the mechanism by which this drift is accommodated with other elements and components of the cladding system, such as the window system. Drift requirements are: 1.

2(∆wind )

2.

∆M = 3.2 in.

3.

∆ min = 0.5 in.

§1633.2.4.2 (1)

Infill panels, such as the column cover (Panel B), require special review when it comes to movement. Typical these panels are attached to other elements and see the full story drift, but the height over which this movement occurs is much less than the story height. Therefore, the rotation that the panel undergoes can be more than two times the rotation of the column.

336


Design Example 7

!


Figure 7-12. Cladding interaction with frame displacements

Consider the column cover in this case: hs = typical story height (ft) φ = ∆ M / hc = 3.2"/(9')(12 in./ft) = 0.0296 radians δv = φ (wc – a) = 0.0296 (48" – 12") = 1.06 in. Since this is an estimate of the maximum movement, round the joint size to the nearest ¼-inch. tj = 1.25 in. δv

φ hc

a

pivot point

wc

Figure 7-13. Rocker panel


337

Design Example 7

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As the beam hinge location moves toward the interior, the spandrel panel can also experience up and down movement at each support point. θ = ∆ M / hs = 3.2"/(16')(12') = 0.01667 radians δv = θxb = 0.01667 rad (18") = 0.30 in. Differential displacements out-of-plane of the panel should also be considered.

8.

Typical details.

Figures 7-14 and 7-15 illustrate typical connection details.

Figure 7-14. Tieback connection at bottom of cladding

338


Design Example 7

!


1" return welds

top of floor slab

Figure 7-15. Bearing and shear connection at top of cladding

References Iverson, James K. and Hawkins, Neil M., 1994. Performance of Precast/Prestressed Concrete Building Structures During Northridge Earthquake, PCI Journal, Vol. 39. McCann, R.A., 1991. Architectural Precast Concrete Cladding Connections, Implementation and Performance of Structural Details, 1991 Fall Seminar, Session 2, Structural Engineers Association of Northern California. PCI, 1999. PCI Design Handbook – Precast and Prestressed Concrete, 5th edition. Precast/Prestressed Concrete Institute, Chicago Illinois. Sheppard, D. A. and Phillips, W. R., 1989. Plant Cast Precast and Prestressed Concrete: A Design Guide,3rd edition. McGraw-Hill, New York, New York.


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Design Example 7

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