STRUCT STEEL DESIGN
Joseph E. Bowles Professor of Ciod Engineering
McGraw-Hill Bmk Company New York St. Louis San Francisco Auckland Bog& Hambur Johannesburg London Madrid Mexico hiontreal New& Panama Paris Sao Paulo Singapore Sydney Tokyo Toron
S k;lrrLTURALSTEEL DE§IGN
CONTENTS
Cv-.:- r;ht 0 1980 by McGrdw-H111, Inc. All nghts reserved Pr: t.2 m the Unlted States of Amenca. No part of this publ~catlon n t y ~ereproduced, stored In a retneval system, or transm~tted,In any I'm or by any means, electronic, mechanical, photocopy~ng,recording, or
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Preface
Chapter 1 General Design Considerations
Thi, '..do'# was set in Times Roman by Science Typographers, Inc. Tfie ..iilurs were Julienne V. Brown and Madelaine Eichberg; thd cover was designed by Anne Canevari Green; the ?rs.,duction supenisor was Dominick Petrellese. 'The ?..:wings were done by J & R Services, Inc. K.'8. Donnelley & Sons Company was printer and binder.
'Uh.--*< of Congress Cataloging in Publication Data Bor ' s. 'oseph E rdral steel des~gn
.>graphy:
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p.
udes mdex. Bulldmg, Iron and steel. 2. Steel, Str~ 3. Structures, Theory of. I. T ~ t l e 'TA6!' "478 624'.1821 79-18155 1SBW r 37-006765-1 !
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1-1 1-2 1-3 1-4 1-5 1-6 1-7 1-8 1-9 1-10 1-1 1 1-12 1-13 1-14 1-15
Types of Structures Design Procedures Steel as a Structural Matenai Steel Products Steel Strength Temperature Effects o n Steel Structural Design Codes Building Loads Highway a n d Railroad Bridge Loads Impact Loads Earthquake Loads Fatigue Steel Structures Accuracy of Computations and Electronic Calculators Structural Engineering Computations in SI
Chapter 2 Elements of Frame, Truss, and Bridge Design 2-1 2-2 2-3 2-4 2-5 2-6
Methods of A n a l y s ~ s Beam Analysis Determinate Structures Truss Analysis h a d Frame Analys~s Bndge Analysls
2-7 2-8 2-9 2- I0 2- 1 1
The Computer Program Furnished in the Appendlx The P Matrix Load Conditions Checking Computer Output Design Examples
Chapter 3 Elastic, Plastic, and Buckling Behavior of Structural Steel Introduction Elastic versus Plastic Design Theory Safety Factors in Elastic and Plastic Design Elastic versus Plastic Design Deflections Length of Plastic Hinge Elastic versus Plastic Design Load Resistance Factor Design Local Buckling of Plates Post-Buckling Strength of Plates
Chapter 4 Design of Beams for Bending
4-8 4-9 4- 10 4- 1 1 4-12 4- 13 4-14
General Considerations Design of Beams by the Elastic Method Design of Continuous Beams Web Buckling and Crippling Shear Criteria Strong versus W e a k - h i s Bending Deflections Biaxial Bending and Bending on Unsymmetrical Sections Shear Center of Open Sections Design of Laterally Unsupported Beams Beams with Nonparallel Flanges Design of Bridge Stringers and Floor Beams Composite Beams Beam Design Using Load Resistance Factor Design (LRFD)
Chapter 5 Design of Tension Members 5- 1 5-2 5-3 5-4 5-5 5-6 5-7 5-8 5-9 5-10
Types of Tenslon Members Allowable Tension Stresses General Deslgn Charactenstlcs Stresses Due to h a 1 Load on the Net Sectlon Des~gnof AISC Tenslon Rods Net Sectlons Deslgn of Tenslon Members Design of Bndge Tenslon Members Cable Deslgn Deslgn of Tenslon Members Usmg LliFD
Axially Loaded Columns and Struts ~ntroduction The Euler Column Formula Columns a l t h End Condltlons Allowable Stresses in Steel Columns Deslgn of Bu~lt-upCompression Members Column Base Plates Lateral Brac~ngof Columns Column and Strut Design Us~ngLRFD
Beam-Column Design Introduction General Considerations of Axial Load with Bending Effective Lengths of Columns in Building Frames Developing the Beam-Column Design Formulas Determination of the Interaction Reduction Coefficient C,,, AASHTO and AREA Beam-Column Design Formulas Beam-Column Design Using Interaction Equations Stepped Columns and Columns with Intermediate Axial Load Control of Sidesway > Beam-Column Design Using LRFD
Bolted and Riveted Connections Introduction Rivets and Riveted Connections High-Strength Bolts Factors Affecting Joint Design Rivets and Bolts Subjected to Eccentric Loading Beam Framing Connections Fasteners Subjected to Tension Connections Subjected to Combined Shear and Tension Moment (Type 1) Connections Load Resistance Factor Design (LRFD) for Connections
Welded Connections General Conslderat~ons Weldmg Electrodes Types of Joints and Welds Lamella Teanng Onentation of Welds Welded Connectlons Eccentrically Loaded Welded Connections Welded Column Base Plates Welded End Plate Connections Welded Comer Connectlons Fillet Weld Design Using LRFD
Chapter 10 Plate Girders General Loads 10-3 Proportioning Flanges and Webs of Girders and Built-up Sections Partial-Length Cover Plates General Proportions of Plate Girders Plate Girder Design Theory-AISC Plate Girder Design Theory-AASHTO and AREA 10- 1 10-2
Appendix A-1 A-2
A-3
Selected Computer Programs Frame Analysis Program Load Matrix Gcnerator for AASI-IT0 Truck Loading on a Truss Bridge Load Matrix Generator for AREA Cooper's E-80 Loading on a Truss Bridge
Index
The primary purpose of this textbook is to provide the basic material for th course in structural steel design. The text contains elements of both buil and bridge design for use in the structural engineering sequence of c i d gineering programs. If the instructor wishes to emphasize building frames, text is also suitable for an introduction to structural steel design in arc programs. Approximately equal emphasis is given to fps and SI units. In the dis material both systems of units are used; the examples and homework pro are in either fps or SI. This format was arrived at throu$ discussions wi nurnber of interested faculty members and people in industry. The conse was that the text discussion should continue to use both systems of units beca transition to metric is not occurring as rapidly in the construction indust other areas'of engineering. Dual usage seems necessary to provide both and instructor with a feeling for what is a reasonable member size (num deflection, or other design parameter in both systems of units. Practical SI instruction requires use of design data, a n d since none readily available, I have assembled a set of computer-generated rolled section data tables as a supplement to the text. These tables are in ge agreement with the AISC and ASTM A-6 specifications. This bound set of also includes edited material from the AISC, AASHTO, and AREA tions. It is intended that the textbook, together with the supplemental Stnict Steel Design Data (SSDD) manual, will provide adequate material for a design course without the need for any other reference material. T b e material should be sufficient to enable students to design routine (an not-so-routine) structural members in either fps or SI units and by any the three steel design specifications which are most likely to control the deslgn
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at least in American practice. Specialized problems are not generally addressed in a classroom environment, and for these (as well as for design office practice and other nonacademic work) the reader should obtain a copy of the latest specifications from the appropriate agency. I use the digital computer as a design aid in a somewhat interactive mode (via batch processing) for the design portion of the steel design course. I have found that the use of the computer in the steel design course is one of the better academic experiences for students, because it helps them rapidly gain experience in structural behavior. This may be by acci'dent (from mispunching data on the modulus of elasticity, cross-sectional area, or moment of inertia of a member) or by iteration of a design problem in which member sizes are changed as indicated by the computer output. In either case, students readily see the effects of member section properties on structural behavior. Using the computer programs pe:;~ii'fs' thisb.with only a modest amount of work on the part of the studentno program writing. Several computer programs are listed in the Appendix to the text for those o are not already using the computer as a design aid. These programs are relatively simple, but efficient, and can easily be punched on cards for use on a local computer system. The band matrix reduction method is used so that computer cpu requirements are minimal. I can furnish these programs on tape at the cost of tape, reproduction, and mailing.for anyone who is using the text in the classroom. I have not attempted to cite, or promote the use of, desktop programmable calculators for simple tasks such as beam or column designs because of the variety of devices available (e.g., HP, TI, Sharp, Casio, etc.), each requiring a different programming method, and because of continuing rapid change in the state of the art. Listing of the multiple programs necessary for use of the various calculators would take too much text space, at the expense of more important topics. The text attempts to strike a balance between theory and "how to." The topical treatment is not so exhaustive as to obscure the fundamentals but is of sufficient depth that the reader is aware of the source of the design equations in the various specifications. A number of the equations are partially to completely derived so that the reader can be aware of the limitations. A reasonably detailed explanation is given of the basic design problems; and the illustrative examples are essentially step by step. With this format students should be able to cope with the more complex design problems on the professional level, and to obtain design solutions for the assigned home problems. Appropriate references are cited directly in the text for topics for which coverage is limited but which are sufficiently important that the reader may wish to study the subject in greater depth. The inclusion of references will generally be of more use to those in professional practice than to student users. My expcr-icnce in teaching steel design for a number of years is that most students in the ,first design course are primarily interested in learning how to design the various types of structural members they will be assigned for home or laboratory work. At this point in their professional development they are not overly
,
terested in the theoretical considerations and the extensive laboratory work researchers and theoreticians that has produced the current design equations. The complexity of semitheoretical and empirical design equations, couple with the nature of structural design and its intimate association with desim specifications and codes, makes i t necessary to take a strong '-how to" approa in teaching steel design. I t is essential to present the user with a set hypothetical (or real) data and by illustration produce a design. Students presumed to have a sufficient background in the basic engineering and sequence to appreciate what has been illustrated and are taught how to dup the steps with a similar problem to gain confidence, and, based on the illus tive problems, to extrapolate to a problem where the desim parameters considerably different, with a minimum of super-vision. Fabrication and practical considerations are introduced in the exam problems as appropriate. Fastener spacing, edge distances, erection clearanc standard gage distances, thread runout. and maintenance are considered various sections. This should give the user an appreciation of fabricati problems and other practical considerations. In conjunction with this, the te has a large number of photographs, supplemented with line drawings of structural elements and connections, which should be of particular aid to the no The reader should supplement these illustrations by observing steel frames un construction. The photographs were all taken especially for this text, to disp individual structural features as appropriate to the development of the disc sion. Plastic design is introduced briefly in Chapter 3 together with the basics o plate theory. This is done so that the design equations with origins in design or.plate theory can be efficiently referenced back to Chapter 3, there saving text space. Plastic design methods are not emphasized, for two basi reasons: there is not enough time in a first course to adequately treat the subje and elastic design seems to be preferred in professional practice. I have deviated from the current textbook, trend to reflect the fonna ibcorporated in some of the steel texts published in the 1950s. This ibcludes the use of simple illustrative examples where the design data are stated as well as more realistic design examples. These examples are anaI Chapter 2 using the computer, and selected members are subsequently designed in the later chapters. The use of simple examples gives the reader a quick grasp of the general objectives of the discussion. More detailed design examples are used to generate a sense of realism and to clearly indicate that steel d e s i g is not just a matter of manipulating numbers. The examples are accompanied with a reasonable amount of discussion of the analysis provided, Within the framework of classroom time restraints, a steel design course should be as realistic as possible. For this reason the user is encouraged to c a n y iiny structural design problems assigned in Chapter 2 through succeeding chapters, redesigning members as necessary and recycling the problem one or more times for member sizing before the connections are desiqed in Chapters 8 and 9. A false sense of security regarding the actual complexity of structural design, and even how the design loads are finally arrived at, can be developed if
fl PttZFACE $~$he>,seris simply given the loads for each design problem. Admttedly, the more reall, IC design problenls require more physical and mental effort on the part of "/ \ ,'r'lcs;udent and more grading effort on the part of the instructor. This extra , ~".ff~:t can be offset somewhat by assigning fewer total problems, but including t~i;;s In which loads are glven, to bu~ldconfidence, and some with design prc i !i31:is, to bulld des~gnsk~ll. 7 iie following text sequence might be appropriate in the semester system: i'l
L1
, dmester hour:, Rapid coverage of Chapters 1 and 3, with Chapter 2
I
assigned for reading. Reasonable coverage of Chapters 4 to 10, Probably two wceks each on Chapters 4, 7, and 10. semester hours Rapid coverage of Chapters 1 and 3. Two weeks on Chapters 2, 4, 7, and 10, followed by actual design of a building frame and highway bndge truss, or industrial building, based on the analysls in Chapter 2. One structure should be done In fps, the other in SI. A design notebook should be kept, showing computations and computer input/output. It is also suggested that this work be done in groups, each wlth no more than four students.
AC KNQWLEDGMENTS Several persons and organizations have provlded considerable encouragement arid assistance in produclng this textbook. First, I should llke to express my ,incere appreciation to Dr. Peter Z. Bulkeley, Dean of Englneenng and Technol-
a
jgy, Bradley University, who provided me with released teacbng time. I would'also like to thank Mr. Andrew Lally and Mr. Frank Stockwell, Jr., of AISC, who provided me with a prelimmary copy of the new AISC specificaand took the time to go over the major changes with me. Mr. Lally also ;..-ovided useful ~nformationon maklng the SI conversions. Mr. Robert Lorenz sf the Chicago Reg~onalOfflce, AISC, was also helpful in providing me with .ast-minute corrections to the preliminary specification changes. Both Bethlehem and US Steel corporations were most helpful ln providing copres of their new steel section profiles, nearly a year in advance of their becomlng official. This allowed work to proceed early on computer generation ,.; the Structural Steel Deslgn Data Manual tables. Particular appreciation is due to Mr. Roland Graham of US Steel, who carefully revlewed selected portions of he manuscript and the entire steel data manual and made some very useful sl:ggestions. Grateful acknowledgment is also made of the very considerable contributions of Dr. Eugene Chesson, Civil Engineering Department, University of Delaware, who carefully reviewed both the preliminary and flnal text manuscripts. Thanks are due Dr. T. V. Galambos, Civil Engineering Department, Washington Univers~ty,St. Louis, who revlewed the load resistance factor design material.
(
-1 TYPES OF STRUCTURES e structural engineer wlll be concerned wlth the design of a v a n d the follo\wng: structures including, but not necessarily I ~ m ~ t eto,
Bridges: for railroads, highways, and pedestrians. Buildings: including rigid framed, simple connected frames, load-beari Fi ,?:re 1-1 The Eads bndge across the Mississipp~&ver at St. Louis, Missouri. This rallroad and _'.way bndge completed in 1874 represents one of the first uses of steel (and high-strength F, = 50 : ? S ksl steel) in the United States for a major structure. The 192-m (630-ft) hlgh St. Louis 'It teway" arch, wlth an extenor s k ~ nof stalnless steel, can be seen In the background.
cable-stayed, and cantilevered. Numerous lateral bracing schemes, incI trussed, staggered trussed, and rigid central core, may be considered or Buildings may be further classified as to occupancy o r height as industrial, mill, high-rise, and so on. Other structures: including power transmission towers. towers for radar an installations, telephone relay towers. water supply facilities, a n d trans tion terminal facilities, including railroad, trucking. aviation, and mari In addit~on to the foregoing structures, the structural e n p e e r is engaged in the design of ships, a~rplanes,parts of vanous machines a n d mechanical equipment, automob~les,and dams and other hydraulic struc including water supply and waste d~sposal. This text will focus pnmanly on structural d e s ~ g nusmg metal, an particular standard structural shapes as produced directly by the several producers or in a few cases use of members that are built u p from steel p and shapes and fabncated either by the steel producers or in local st fabrication shops.
GENERALDESIGN COP.SIDER%TI
4,
1-2 DESIGN PROCEDURES
Structural design lnvolves application of engineering judgment to produce a structural system that will adequately satlsfy the client/owner's needs. Next, this system IS incorporated ~ n t oa mathematical model to obtain the member forces. Since the mathematical model never accurately represents the real structure, engineenng judgment is agaln required to assess the validity of the analysis so that adequate a ~ ~ o w a n can c e be made for uncertainties in both deformations and statlcs. Based on material properties, structural function, environmental considerattons and esthetics, geometrical modlf~catlonsin the analysls model are made and the solutlon process Iterated untll a solutlon is obta~nedthat produces a satisfactory balance among material selection, economics, client desires/flnancia1 ability, and various architectural cons~derat~ons. Seldom, except possibly in the most elementary structure, will a unique solution be obtained-unique in the sensc that two structural engineering firms would obtain exactly the same 82 @Qlutlon. 'r In structural englneerlng practice the designer will have available for p k, w e use numerous structural materials, including steel, concrete, wood, ' posslbly plastics and/or other metals, such as alumlnu 'occupancy/use, type of structure, location, or other design parameter " dlctait: the structural material. In thls text we will assume that the design proceeded to the point where the structural form has been decided (i.e., as trus g~rder,frame, dome, etc.) and the several possible alternative structural materia have all been eliminated in favor of using steel. We will then proceed with an .?ddA~lonalanalysis required, and make the member selection and connecti deslg~lappropriate to the topic being studied. Textbook space and classroom t ~ m elimitations will of necessity reduce the bare essentials the complexity of the design presentations. The reader shou be aware that real design 1s considerably more comp than the simplifications presented in the following chapters. Safety as a design concern takes precedence over all other design consid t q8, The "safety" of any structure, of course, de ?#&np. Since the structure is always loaded after it is built and not always i t ~ e ~ k o dore manner used in the design, the selection of design loads is pioblem in statlstlcs and probability. This part of t rather subjective and produce extremely divergent designs had not bu ,#
%)
codes been developed (and in some form or another, almost universally which place minimum required/suggested bounds f is an important factor.
1-3 STEEL A S A STRUCTURAL MATERIAL S i ~ e is l one of the most important structural matenals. Properties of particular *-ce In structural usage are h ~ g hstrength, co BP%:t:rlal, and ductil~ty.Ducabiy 1s the ability
compression before fallure Other Important cons~deratl e use of steel include widespread ava~labiiit:,~ n durability, d particularly wt odest amount of weather protection Steel 1s produced by refining iron ore and scrap metals together g agents, coke (for carbon), and oxygen in hlgh-tempe aces to produce large masses of lron called "pigs" or "pig iron." is further refined to remove excess carbon and other lm~uritiesand r metals, such as copper, nickel, chromum, man molybdenum, phosphorus, sillcon, sulfur, titanlum, columbium and vanadi to produce the desired strength, ductility, welding, and corrosion-resis characteristics The steel lngots obtained from this process are pasaed between two roU pposlte directions to produce a semiflnis called either a slab, bloom, or billet, depen nal area. From thls point the product 1s sent to other 10 ills to produce the final sect~ongeometry, Including structural shapes as plates, and pipes. The process of rolling, in addi red shape, tends to lmprove the m a t e d properties of to malleability. From these rollln:: mils the structural sh are shipped to steel fabricators or warehouses on order. The steel fabricator works from the englneerlng or architectural dra produce shop detail drawings from which the requlred dlmenslons are to shear, saw, or gas-cut the shapes to sue and to accurately locate drilling or punching. The origlnal draw~ngsalso indicate the necessary finishes to cuts. In many cases the parts are assembled in the shop to det if a proper f i t has been obtained. The pieces are marked for ease of identification and shipped rn pleces or subassemblies to the ~ o site b for erec general contractor. Some of the most important structural properties of steel are the follo 1. ~
~ of e[astlcrp, d E~ The typlcal l ~rdnge for ~ all steels (relatively l n d e ~ e n as 29 000 ksl or 200 000 MPa-
The value for deslg 2. Shear modulus, G.
E 2(1 + P ) where p = Poisson's ratlo taken as 0.3 for steel. Using ,u = 0.3 9ves G = 11 000 ksl or 77 000 MPa. as 3. coefficlenf of expnnsron, cu T h e coefficient of expansion may be G =
//
a = 11
AL
=
25 x
a(T, -
per " C
T,) L
(ft or m depending on length L )
GENERAL DESIGN CONSID
in degrees Celsius. To con eit to Celsius, use
C = ; ( F - 32)
and ultimate strength. Table 1- 1 gives the yield poin es of steel of interest to the structural designer that are mmd;z
y j o9m Q -w 0 0
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er properties o/ some interest. These properties include the mass de (1 t = 1OOO kg); or in terms o 76.975 kN/m3. The specif conversion of fps units its of kN/m and kg/m is accomplished as follows. Given: lb/ft and required to convert to:
8%3$
Note that lb mass and lb weight or force have been used interc ably in the fps system because the acceleration-producing force of gravity. This cannot be done in the SI system, since the ne derived unit that defines the force necessary to accelerate a '1 m/s2. The acceleration due to gravity is approximately 9 xample: Given: A rolled structural shape weighs 300 Required: mass/m and weight/m. Solution: mass/m = kg/m = 1.488164(300) = 446. wei&t/m = kN/m = 0.0145939(3
k 4-44
B E
V1
$i nh
M I
3
aJ Z;
E"
g3 g 7; -E 6 s 8 5
re rolled into plates of varying round, square, and rect Most of the rolling is done on hot steel, with the p steel." Sometimes the thinner plates are further rolled ' steel products. Several of following sections.
b S l RUCTURAL STEEL Ukbiut4
1 4 . 1 W Shapes X,e most commonly used structural shape is the wide-flange or W shape. This is a doubly symrnetncal (symmetrical about both the x and y axes) shape consisting i;f two rectangular-shaped flanges connected by a rectangular web plate. The flange faces are essentially parallel with the inner flange distance for most of the graiips, with a constant dimension.? There is some variation due to roll wear and other factors, but the distance is held constant within ASTM tolerances. h e shape is produced as illustrated in Fig. 1-1. Khe dssignation: W16 X 40 means a nominal overall section depth of 16 in with a weight of 40 Ib/ft. , The des~~narron: W410 X 59.5 is the same W 16 as above with a nominal depth in 1 rr~m(based on the approximate average depths of all the W16 sections and rounded to the nearest 5 mm) and with a mass of 59.5 kg/m.
&or to 1978, at least ope W section in a group designation was "exactly" the t:)$kpnal depth given (i.c., one W16 was 16.00 in deep; one W18 was 18.00 in "';,$@p). , Now the closest W16 is the W16 x 40, with a designated depth of 16.01 $0: There can be substantial deviations between the nominal and actual depth . .(<.g., the W21 ranges from 20.66 to 22.06 in). For the W14 the SI equivalent is W360, but the actual range is 349 to 570 mm (in this case the "average" was too . fr* 'from the nominal value and W360 was somewhat arbitrarily used). ' It should be noted that the rolled product will contract on cooling and at a vat~ablerate depending on the thickness at any point on the cross section. The rolls used to produce the shapes will undergo wear, and coupled with the enoinlous forces involved in the rolling process, only shapes of nominal dimension (varying from theoretical or design values) can be produced. American Soclety for Testing and Materials (ASTM) specification A-6 in Part 4 gives a:lowabie rolling tolerances, including amount of flange and web warping and deviation of web depth permitted for the section to be satisfactory. Generally, the maxipum permissible variation in depth as measured in the plane of the web is 5 in or 3 mm. Note, however, that the permissible difference in depth between two rolled beams with a theoretical depth of 16.01 can produce extreme depth:; of 15.885 to 16.135 in or a difference of 4 in or 6 mm. These variations should be' kepr rn mind, particularly when converting to SI dimensions for detailing, clearances, and mating of parts.
.
1-4.2 S Shapes These are doubly symmetric shapes produced in accordance with dimensions adopted in 1896 and were formerly called 1 beams and American Standard %
t The several sections with a constant nominal depth. Where a group consists of a large number of S~CIICJI~S, a second inner flange distance may be used.
---r --- -
i
C
-.
-
hd
[
'V shapes
S \hapes
C shape,
L shape
:;gc
Anierrian Stdndard bean1 (I-beatri)
Chdnnel
kqiral leg angle
i
to 330 MPa and refer to Figs 1-30 and 1-36. Similarly, A-44 point of 345 MPa, will have a yield strength on the order of guaranteed values converge.
UReitdllgular
"_,;id L-shdpe
7 slldpe
1 1 ~ q u dleg l dtlgle
Strut tirrel Tec i l l ! Iron1 \V.stidpc
I
C
8 Square 0Rounds Ihrt
1
Plate
k + $ i s 1-2 ktmctural shapes as drrectly produced by Be steel producen.
t--:,.j
L Shapes
i i . e , ~shapes are either equal or unequal leg angles. All angles have parallel
ed and designated A-272 (described ~n ASTM specification A-272). Sp on ASTM A-440 was wntten in 1959 for another h ~ g hstrength steel
1960 with application to weld~ng.All three of these steels have a yield at is dependent on the thickness of the metal, as shown In Table 1-1.
flal~gefaces. Angle leg dimens~onscan vary on the order of +- 1 mm in width. An L6 x 6 X IS an equal leg angle wtth nominal dimens~onsof 6 m and a 2thLE~nessof $ In. An L89 X 76 x 12.7 IS an unequal leg angle w ~ t hleg dimensions of 89 and 76 k:m, respectively, and a leg thickness of 12.7 mm (L3f x 3 x +). t
1-4.'; T Shapes St:
LC
S" WI.:1
~ r a tees l are structural members obtatned by splitting W (for WT), S (for
.. M (for MT) shapes. Generally, the spltttlng is such to produce one-half the area of the parent section, but offset splitting may be
d .: -r tze section is required. Published tables of T shapes are based slA.$~netncal splitting. No allowance is made for material loss from splitting p21 t .:t shape by sawing or flame cuttmg. 4 WT205 x 29.8 is a structural tee with a nominal depth of 205 mm m: c*c of 29.8 kg/m and is obtained by splitting the W410 x 59.5 section (f R'IG x 40). Exera1 rolled structural shapes are Illustrated in Fig 1-2.
1-3 STEEL STRENG??-I me1 design takes into consideration the yield strength of the mate yleld strength of several grades of steel available for design is given in Table 1yield strength is that minimum value guaranteed by the steel producers an All
Figure lJa Typical stress-stram structural steel.
curves for
E;lgure 1-36 Erllargement of lmti stram curve for two grades of Note that the plasbc r a g e u
.- ...---
.
Since about 1964, specifications for several other high-strength (low-alloy) steels-hzve been- incorporated into ASTM specifications as A-572 and A-588. Table 1-1 shows that the steel covered by the A-572 specification covers several $yield strengths, termed grades, such as grades 42, 45, 50, 55, 60, and 65 for the corresponding guaranteed minimum yield stress in ksl. Generally, the yield {strengths of these newer steels are also thickness-dependent, as shown in the table under the heading "plates and bar thickness." The steel producers have designated the several W shapes into five groups, depending on flange thickness (and as shown in Tables 1-1, 1-2, V-1, and V-2)t, compatible with the steel grade. The designer merely has to check these tables to see if the shape is available in the raquired/deslred yleld-stress grade. For example, in the 450-MPa grade, only shapes In group 1 qualify from flange thickness. W18 shapes are available in group 1 only from 35 to 60 Ib/ft inclusive (the five smallest sections and with a maximum part thickness of 0.695 in). Specification ASTM A-588 allows F, = 345 MPa for a high-strength low@lo$ steeltwhich may be up to 100 rnm (4 in) thick. The steel covered in this @$hPecifl,rttion is primarily for welding and is corrosion-resistant. in terms of cost/unit of mass, the A-36 steel is most economical. High stre2g:h steels have principal application where the stresses are primarily High-"irength steel beams may deflect excessively, owing to reduced rnodul:.~. The high-strength steel columns may be less economical than steel tf the slenderness ratio ( K L / r ) is large. Hybrid girders that use strength steel in the flanges, or built-up columns using high strength steels provide better solutions where member sizes are restricted. In a given cas necessary to perform an economic and availability analysis to determi suitability of using high-strength steel.
G E N E R U DESIGN CONSIDERATIO
"C x 100
14 Effect of elevated temperatures on either y e l d or ult~matetensile strength ucpresse f strength at room temperature of approximately 70°F
/
TEMPERATURE EFFECTS ON STEEL
1-6.1 High-Temperature Effects
Steel is not a flammable material; however, the strength is heavily temperaturedependent, as illustrated in Fig. 1-4. Both the yield and tensile strength at 1000°F is about 60 to 70 percent of that at room (about 70°F) temperature. The drop in strength 1s rather marked at higher temperatures, as shown on the figure, where the strength at 1600°F is only about 15 percent of that at room temperature. Steel frames enclos~ngrnater~alsthat are flammable will require fire protection to control the ternperature of the metal for a sufficient time for the occupants to seek safely or for the fire to either consume the flammables or be extir~guishedbefore the building collapses. In many cases the building does not
t See footnote a of Table 1-1 In J. E. Bowles, Structura/ Steel Desrgn Data Manual, McGraw-Hd, New York, 1980
1-6.2 IAV-TemperatureEffects Brittle fracture is a failure often associated with low temperatures. Essentially, brittle fracture is failure that takes place without material yielding. The stress-strain curves of Fig. 1-3 indicate that m the usual failure of a tensile specimen, considerable elongation takes place. As a matter of fact, a minimum percent elongation is specified for steel in the ASTM standard tensile test. Inlplicit in steel design is the resultant deformation (yieldixig) of the material
Fireproofing materials Cinder concrete Gypsum board plaster, cclnetit and sand Expanded shale concrete Vertnicul~te Perlite
I
Unit w e ~ g h t Pcf l 10 30-40 100
I I Usually use 35* nim o f flreprotection for 2-h fire rating; obtain specific thickness values from either tests o r from the producers o f gypsuln, perlite, etc.
k~/rn~ 17.3 4.7-6.3 15.7
,.,
rigure 1-5 Methads of producing fireproofing of structural steel members. ( a ) Sprayed fiber. (6) Lath and plaster. (c) Lightweight concrete (formed). ( d ) Gypsum board-use boards to build 'hick;us. ( e ) Corner detail of sprayed on fireproofing. Thickness is built in several sprayings. (fl Bear. dzwl of sprayed fireproofing.
16
STRUCTURAL STEEL DESIGN
GENERXL. DESIGN CONSIDERATI
any abrupt change in cross-sectional area-to tension situation.
inhibit lateral contraction
brittle fracture. This may initiate as a crack that propagates to a member f
occurring is of little aid in settling the resulting damage claims that are sure follow. Brittle fracture can be controlled in several ways: 1. Detail niembers and their connections to minimize stress concentrations. 2. Specify the fabrication and assembly sequence to minimize residual ten stresses. 3. Use steels that are especially alloyed for low-temperature environments
tures are encountered.
.
Code, published by, and available from, the iation, 85 John Street, New York, N.Y. 10038. Building Code by International Conference of Building Workman Mill Road, Whittier, California 90601. lding Code, Building Officials and Code Administrators Inte East 60th Street, Chicago, Illinois 60637 (formerly BOCA). itute (ANSI), Minimum Design ngs and Other Strucrures, ANSI 58-1, 1430 Broadway, New Yo
.
5 , If possible, machine (or grind) the notch into a sm
6. P-educe the rate of tensile strain application.
1-7 STRUCTURAL DESIGN CODES
tructural design. On the one hand, it sometimes takes ials and methods; on the other hand, th fast." If the local .building code is care minimum design requirements met, or exceeded, and a catastro of exists that good engineering practice has been followed. ng codes are supposed to reflect that part of the structural unique for that locale, such as temperatures, earthquakes, ntity, frost depth, and average wind velocities. list gives several design codes and/or specifications may have occasion to use:
.
Lo.~,d.lbuilding departments almost always require structural de
titGte of Steel Construction (Specifications), Steel th ed. (1979), 101 Park Avenue, New Y0rk;N.Y. 1001 lding Society (AWS), Structural Welding Code, 2501 iami, Florida 33125. and Steel Institute (AISI), 1000 Sixteenth Street, . Publishes various specifications for using iron and ste Association of State Highway and Transportation Of l.ITO), Specifications for Highway Bridges, 341 National Press Bu
The various state departments of transportation generally use the specifications put forth by AASHTO, several railroads generally use specifications put forth by The structural designer doing highway or railroad work closely the design specifications of these publications, partic government is involved with any of the financing.
the owner/client may require a more stringent design than the building co c,piteria. Only in rare cases can the designer get a variance from the local governing body to deviate in a less conservative manner from the code. Vari-
Railway Engineering Association (AREA), Speci/ica ay Bridges, 59 East Van Buren Street, Chicago, Illinois 6060 national and city codes use specification standards as applic zations, such as AWS, AISC, and AIS cies for the other construction materials. nd AISC, as well as the steel producer tables of structural shape design data as well as data on other steel , wire, and bolts. Certain of these data together AISC, AASHTO, arid AREA, have been pretnrctzlrnl Steel Design Data (SSDD) man by McGraw-Hill Book Company and used as a supplement with manual was developed so that both fps and SI units would
18 STRUCTURAL ST
carry the dead 1
GENLRAL DESIGN CONS1
fps: R SI:
=
0.0008 x area
R = 0.0086 x area
(when area (area
>
150 ft2)
> 11.2 m2)
include: Ceiling materials, including duct work for environmental co supplies. -xterior walls supported by the frame, including windows, doors, and ba Interior walls Qat are permanently placed. 'iechanical eqmpment (heating, air conditioning, ventilati ' (such as elevators, including cage, cables, motors). r .reproofing. Beams, girders, and columns, including the footings making frame. From this list it is evident that any part of the building w nstalled contributes to the total dead load. Dead loads can
(some codes limit R
< 0.40 for horizontal membe
here R= reduction factor used as ( 1 - R ) x L,,,,, D = dead load, psf or kPa (kllonewtons/m) L = live load, psf or kPa, but L 1s limlted to not over 100 psf or generally, values larger than this are not reduced lic assembly (such as auditoriums), garages, and roofs.
I
Example 1-1 A port~onof an office (multistory) floor plan is shown El-1. The floor is 4-111concrete on a metal deck over steel bar joists. W the reduced live load for the floor beams and for an exterior column floors down from the top floor?
prescribed by building codes based on occupancy and 1 anow, and earthquake loads are considered. In addition to !oads include: People, as in auditoriums, assembly halls, and classrooms. Movable room partitions. Office equipment and production machines if they are m Warehouse products. Furniture. Building code values of live loads tend to be based on SOLUTIONEstimate the dead load on the contributory (centered ber) floor area as: Concrete floor and finish: 4 x 144/ 12 = 48 psf computational convenience and because the actual buildin :lot known.
Ceiling, metal deck, steel bar joists
f As p e n m the several natlonal b u l l l n g codes c ~ t e dearher
= 12 psf
GENERAL DESIGN COKSID
From Table IV-4 of SSDD, the live load = 5.00 kPa. The r r for a grder based on a contnbutory area as shown is El-1, 18 X 22 (area ABCD): R = 0.0008(18 X 22) = 0.32
< 0.60
-
R = ----- = 6o 8o - 0.40 < 0.60 4.33 L 4.33(80) Use the smaller value of R computed, 0.32. The reduced live loa L' = (1 - 0.32)80 = 54.4 say 55 psf Compute R for the column; the contributory area is centered on the column of 9 X 22 (AEFD); but for the accumulation of three stories, we have R = 0.0008(3 x 9 x 22) = 0.475 < 0.60 O.K. +
+
R = ------60 80 - 0.40 < 0.60 4.33(80) Using the smaller value of R, 0.40, the reduced live load on L' = (1 - 0.4) x 80 = 48 psf We note that O.$0 is the maximum R for the column and is uppermost floor level. +
R
R
+ 8)/2 X 91 = 0.774 > 0.40 (and also 0. D +L = ----- = 3.703 + 5.00 = 0.402 < 0.60
=
0.0086[(12 4.33 L
4.33(5.00)
Since the problem statement limlts live-load reduction to not mor 0.40, the reduced l~veload is L ' = ( 1 - 0.4) x 500 = 300kPa
d loads have been extensively studied in recent years, particularly for larger -rise structures. Generally, for tall structures ~wnd-tunnelstudies should bz or smaller regular-shapsd 30 m, the wind pressure ory to use. The Nation
Example 1-2 A meeting/banquet room in a hotel has 22 X 27 m. The floor is 125 mm of concrete with a tile surf
15
0.75
d this pressure times
SOLUTION Note that a public room is not the same as ,whece the loading is pnmarily seating in fixed or movable definition we may use a live-load reduction factor. First, estimate the dead load using Table IV-3 of S weight of concrete = 23.5 kN/m3. Weight of concrete: 0.125 x 23.5
o allowance is commonly made for the shielding effect of adjacent structure r from ground cover. ge ground level at th The wind pressure is commonly computed between floor levels and prorate he adjacent floors using simple beam theory if the vertical distance compare
The sever31 wind values are shown in Fig. El-36. The data display is convenient for computer programming for frame stresses using the com~uter-gr,ogramdiscussed in Chap. 2.
///
Wind pressures can be approximately computed as
where V is mi/h or km/h. This equation is readily derived as q = f mu2, where the mass density of air is approximately 0.00238 lb s2/ft4. Since wind is a transient load, the building codes usually allow a one-third increase in the allowable design stresses with wind included as a part of the load condition as long as the required section is not less than required in the load condition of dead + live loads alone. For example, if a stress of 20 ksi is allowed, then with the wind load condition a stress of 20 X 1.33 = 26.6 ksi could be used.
-
1-8.2 Snow Loads Snow loads are live loads acting on roofs. Snow and any other live loads are taken with respect to the horizontal projection of the roof, as illustrated in Fig. 1-8. Figure 1-9 is a map illustrating snow loads that may be used in the absence of spzcific load building code requirements. Even in areas where snow loads are minir~~al, a minimum roof live load should be.used. The NBC stipulates the larger of the snow load or 20 psf or 1.0 kPa. Since 10 in of snow approximates 1 in of water, a 20-psf snow load corresponds to a roof snow depth of nearly 40 in -easily obtained where snow drifting occurs. When rain later falls on snow, however, the saturated snow weighs considerably more and the unit weight can approach that of water.
Snow and other live loads
lilllllllllllllllII1111111111'
Figure 1-8 Snow and other roof live and dead loads.
In addition to the types of pressure or area loads noted, building codes may stipulate checking for a concentrated load of some magnitude which may be placed anywhere on the floor or roof. Where roofs are used as recreational areas or sun decks, the live loads must be adjusted to values based on occupancy in addition to considering snow and/or wind. Po~idingis a special roof load that may require investigation. Ponding is a condition where water collects on a flat roof which has deflected locally (possibly due to an overload, poor construction, foundation settlement, or plugged roof drain), causing a concentration of water which in turn increases the load and deflection, causing a further concentration of water. Noting that a water depth of 1 in results in a live-load pressure of 5.2 psf, loads are readily eveloped which can locally fail roof members. Through progressive failure, the roof may collapse. Ponding design is considered in some detail by Marho July 1966 AISC Engineering Journal.
Erection loads are not directly considered in biiildiiig c o d ~ s 'rhese . loads may control the design of certain members, particularly very high rise buildings, cantilevered bridges, or cable-supported structures. The engineei responsible for esy phase of the erection may be held legally responsible for damages or loss of life resulting frorn a structural failure during erection. Most structural failures (at least that are reported) tend to occur during erection rather than later. Erection methods andequipment tend to vary from project to project; thus ii is not practical in textb~oksto do more than point out this very important design area. The engineer musf+d$termine what equipment will be used, where it is placed, loads to be lifted; quantities of material, and the storage locations,so tLat the affected individual steel structural members can be checked for adequacy using princip chapters on design.
'
'
1-9 HIGHWAY AND RAILROAD BRIDGE LOADS 'The American Association of Highway and Transportation Officials (AASHTO stacdard truck loadi General lucarion of rndxrrnurn momsnr
is oLtained with eith ti,c i l S truck and span lengths, shears are as follows: -. -
,
),
,,, ,*,.
,. . .
31 kips IJJ kN 24 k ~ p s108 k N I6 kips 7 3 kS
Bridge span
---
33.8 to 145.6 ft
M,.,
= :[(0.9~
+ 4.206)(0.5L + 2.33) -
1l 2 L ] It - kips
26 ?a :27.5 ft
2.53 to 38.86 m Over 127.5 ft Over 38.86 m a W = 40, 30, and 20 kips or equivalent in kN (and is the basic truck load, not the total).
&re 1-10 Standard AASHTO truck loadings for bridges.
GENERAL DESIGN COWIDEIW
Moments, shears, and floor-beam reactions for Cooper's E-81) loa
al axle lodd nd of train:
sfS'RUCTURAL STEEL DESIGN ~ a b 1-2 ~ d (Continued)
. !
,. 3
.-.' 1
- ..
23 712.0~ 35 118.0b 48 800.0~
.
-
.
,
17 990.0 27 154.0 38 246.0
479.6
282.0
128.1
522.0 626.4 729.3
306.8 367.3 426.4
197.9
1225.3
.U1 values shown are for one rail (one-half track load). Axle loads shown in diagram. Obtain for other E loads by proportion. ,, .4t center of span; other moment values are usually close to center of span, so one may obtain !utal moment as the sum of w ~ ' / 8 for dead load + live load value shown in the table.
. . :s
,;
I.
,
f ::-cd
on the locomotive weight, the Cooper load is designated as E-40, E-50, E.50, E-75, E-80, or E-110 and is directly proportional (i.e., E-60 = $XE-80). The cl:rrent AREA design criteria are based on the E-80 (sometimes E-110) loading s.:dwn in Fig. 1-11. Table 1-2 can be used to obtain the bending moments and s.+c.;;rs at selected locations for girder bridges, with values given for a single rail fbading (based on one-half the axle load shown in Fig. 1-1I). Where multitracks or road lanes are carried by the bridge, the live load is as ici!ows:
total uniform loading, however, must not be less than the following:
2. AREA wind requirements: Pressure, force/area Unloaded span fps, psf
SI, kPa
Loaded span fps. psf
SI, kPa
Percent of live load Lane or track
AASHTO
AREA
2
100 90 75 75
100 2X100+1X50 2~100+1X50+1X25 As specified by designer
3
4 XI.xe than 4
Other bridge loadings that must be considered include impact, wind, and longitudinal forces. Impact and longitudinal forces allow for dynamic effects from rolling equipment going across as well as for starts and stops made on the bridge. Impact will be considered in the next section. The wind force is >elf-explanatory and in the case,of a loaded railroad bridge, the wind against the train may be'a substantial load.
AREA: 0.15
X
live load (without impact).
Other loadings that may require consideration include differential temperatures between top and bottom flanges or chords, ice and snow loads, possibIe overloads, and for continuous bridges, support (pier) settlements.
.-
<..*,,.,
6. ..4
- -..'i
.- - . llr,
....,-.., -
G E N E W . DESIGN CONSIDERATIONS
~ L O I U I Y
nt; must be divided by 100 to use in de
Railroad bridges make a distinction between those bridges which consi
een centers of single or groups of longitudinal s which frame into transverse floor beams or girders, or between trusses or girders, ft or m = length between transverse floor beams or between supports as app cable, ft or m beneath the ballast.
e 90 percent of I, computed above for ballasted-deck bridges.
Example 1-4 Given a highway truss bridge with HS 20 loading. The truss anels are 7.5 m, and the distance between reactions is 37.5 m. T h e distance etween trusses (width) is 14.1 m. What is the impact factor, I!?
1-10 IMPACT LOADS
OLUTION The impact factor will vary for the floor beams, stringers, and russ, depending on their lengths. For the stringers the impact factor is l5 = 0.330 > 0.30 therefore. use 1/ = 0.30 //
impact load as
Item Elevator loads Macbnery and other moving loads
1.00
ilroad bridge consists in two trusses spa sses are made up of seven panels at 27.60 ft/ at is the impact factor?
> 0.25
r,unoN Since L > 80 ft, L = 7(27.60) = 193.2 = 25.6 percent 600 193.2 - 30 The AASHTO impact requirement is SI
f PS
I,
=.
-
2 < 0.30 I,+ 125 -
I,
-l5
i0.30
L+38-
where L is the length of span or portion of span that is loaded, in ft or m. The AREA impact specifications depend on the rolling equipment. For diesel and electric locomotives and tenders:
Example 1-6 What is the impact for the floor beams of the AREA truss of Example 1-5? Floor beams are transverse members connecting the two trusses at panel points. SOLUTION S = 27.6 ft, L = 17 f t
< 80.
3(17)* + 40 - = 43.1 percent 1600
HQUAKE LOADS general trends. One is to attempt to model tfie asses and springs and use a digital computer to s assumed earthquake accelerations- The the earthquake accelerations based on earthcitation based on building geometry, and apply
GENERAL DESIGN
roof
Elevation
Figure 1-12 Earthquake zone map for the Umted States. (After Unrform Bu~ldngCode, 19
Example 1-7 A 10-story apartment building with basement is as shown Fig. El-7a. The exterior is insulated curtain walls and thennopane windo with an estimated weight of 15 psf. The interior walls are generally stud partitions plastered on both sides with insulation between apartments. Use 4-in concrete floors (tiled or carpeted) on corrugated metal pan supports carried by open web steel bar joists. The building site is in Memphis, Tennessee. Estimate the earthquake force and corresponding story load.
SOLUTIONEstimate the roof and floor dead loads as follows:
...
Alloaf....
Wood sheathing
= 3 psf
5-ply felt roofing
= 7 psf
Ceiling and bar joists
= I1 psf
.
,,.,.'.b..*
"
..
-.
.,.
G E N E W DESIGN CONSID 9
L
Any floor:
Partitions in 40 X 30 apartment at 8-ft height and cross-walls at 20 psf gives: (40 x 2
.
+ 30 x 2)(8)(20)/(40 x 30)
,I
.4
/
4 Floor: -(144) (concrete) 12 Ceiling (estimated) Bar joists and metal pan Exterior wall at 10-ft height (2 x 40 + 2 x 90)(15)(10)/(40
x 90)
Total = 93.5 psf Total floor weight = 0.0935(40 x 90) = 336.6 kips These weights are illustrated in Fig. El-7a. For easi weights of 76 and 337 kips, respectively, for remaining wo Fig 1-12, the Z factor is 0.75. Take I = 1.00; take K = 0. Table 1-3. The total building weight = 76 + lO(337) = 3446 The earthquake force in the E-W direction is computed D=40ft T = : 0'05(100) = 0,7906
Figure El-76
Since the frame is of steel, the alternative computation for period is T = 0.1
x
number of stories
=
0.1(10) = 1.0 s
The author will average the two values of T to obtain T = 0.895 s. 1 1 C = ----= = 0.0705 1 5 r ~ 15Substitution of this accumulation of factors/weight into Eq. (1-4) F = 0.75(1.0)(0.80)(0.0705)(1.5)(3446) = 218.6 kips The roof value is F,,, = 0.07TF = 0.07(0.895)(218.6) = 13.7 kips ( T > 0.7 s) The story loads are found using hn = distance grou follows: Z W,hn = 337(90) + 337(80) + 337(60) + . . 151 650 f t . kips
F
=
(F- F
lop
)-=Wn A, 2 W,hn
(218.6 - 13.7)
For tenth floor
F,, = 0.001351(337
X
90) = 40.98 kips
F, = 0.001351(337 X 80) = 36.42 kips '
F8 = 0.00135 l(337 X 70) = 3 1.87 kips
F,
=
0.001351(337
X
10) = 4.55 kips
(shown in Fig. El-7b)
Summing the horizontal floor loads and including the top value of 13.7 218.59 versus 218.6 kips as a check. These lateral floor loads further prorated to the several bays in the E-W direction for the fr analysis load condition(s), which includes earthquake forces.
12 FATIGUE failures which have been attributed t ue. Fatigue failure is a material fracture caused by a sufficiently Iarg ulsating stresses, or stress reversals. Th ture of the material at a location where oscopic in size) exists. A crack form nding on stress level, rapidly or slowly (sometimes so slowly that the re) progresses to failure of the part Most metals tested under repeated or cyclic loadings display stress r s qualitatively illustrated in Fig. Irly, these curves were commonly displayed as stress level versus cycIes. t, the stress range is used as the parameter of interest. The stress range
GENERAL DESIGN CONSIDERATIONS
4(8 STRUCTURAL STEEL DESIGN
to
Base metal with rolled surfaces
41
100 to 500 500 to 2OCO Over 2
!@I
T or R e p 60 (415)
36 (250)
24 (165)
24 (165)
T o r Rev
32 (221)
19 (131)
13 (90)
lob(
T or Rev
2 1 (145)
12.5 (86)
8 (55)
T o r Rev
45 (310)
27.5 (190)
18 (124)
16
TorRev
27(186)
16(110)
10(69)
7(48
T or Rev
45 (310)
27.5 (1%)
18 (124)
16 (110
T or Rev
45 (3 10)
27.5 (190)
18 (124)
16 (110
T or Rev
21 (145)
12.5 (86)
8 (55)
plates and shapes with full or
of girder webs or flanges adjacent to welded transverse stiff-
Figure 1-13 Qualitative plot of stress range F,, versus number of cycles to failure. . .-",., ... ',"....-,. -
Base metal at end pr partial
3.
anically fastened connections
can be defined as
The AISC, AASHTO, and AREA specifications are very nearly .identical (where rnembcr failure is not catastrophic) in specifying the stress range and number of stress cycles. These specifications are based on a large number of fatigue tests performed (see Fisher, "Fatigue Strength of Steel Members with Welded Details," AISC Engineering Journal, No. 4, 1977). Typical stress range values which may be used for all three specifications are given in Table 1-4. The reader should consult the Structural Welding Code, Sec. 9-14, which is the ori of the material used in the three specifications or the appropriate des specification for a more complete presentation of fatigue cases and F,. The largest value in each cycle category in Table 1-4 is generally applicable to buildings. Lesser values than shown are necessary for reduced sections, certain types of joints, type of joining material, and for certain members in industrial buildings. The AISC (Appendix B) manual, AWS Structural Welding Code, the AASHTO specifications (Sec. 1-7.2) or AREA should be consulted for those situations where fatigue must be considered. Note that fatigue is not usually considered with wind or earthquake loadings on buildings. Fatigue is usually not considered for routine building design, since 10 load cycles/day over
_ . . _.
. . - " .,.....
N = 10 x 365 x 2 0 = 73 OM ,, This is seldom enough cycles of whatever thk;,itress.rang6 to require a reduction
se 12 ksi or 83 MPa for girder webs.
___-________
._ _. ---
-.--.--------_I-
.A STRUCTURAL STEEL DESIGN
Example 1-8 A rolled section will undergo an expected 1 X 106 cycles of stress over the design period of the structure. The stress analysis gives f,, = P,,,/A = 16 ksi; f, = - P,,/A = - 12 ksi. The basic stress'for this member is Fa = 22 ksi (tension) and 16.5 ksi (compression). 'The structural configuration limits F, = 24 ksi in the base metal. Is the section satisfactory? S l u l r ~ p l sb ~ bent y
SOLUTION f,,=16-(-12)=28ksi>24ksi
N.G.
The section is inadequate for this number of cycles; increase the section so that f,, < 24 ksi. Note that the section is adequate for "allowable" static stresses.
Tal)crcd CUILIIIIII .lnJ r ~ k1r i n a r i g ~ di r ~ n hcnt ~ r
1-13 STEEL STRUCTURES Slructures of steel include bridges, buildings, trans sign supports, and even art objects. The primary focus of this asid buildings, since these are the most common projects invol Buildings are commonly classified according :,ti?ry buildings. Little use is made at present of s .: zpt in multistory apartments. .
'L-13.1 Industrial Buildings
T ~ L I ~ S -soli~rnn on
1-14 Several bents used in steel building frames.
Iildustrial. buildirigs are commonly one- or two-story structures fnr industrial (such as manufacturing, storage, or retail/wholesal and institutional (including schools, hospitals, Other structures may include gymnasiums, aren tr: rlsportation terminals (land, sea, and air). Thes frame, as illustrated in Fig. 1-14, or have a roo resting on load-bearing walls (see Fig. 2-4). The rnay"8e~"?lgidor pinned; may be a two or th t;uss-on-column system. The truss may be rig frames under construction are illustrated in Fig. 4-1. A building frame is a three-dimensional skeleton b rigid in only one plane. Some buildings are rigid in both t but this: type of frame will not be considered resulting' from considering only the principal frame termed a bent and may be one or more stories in iiiusirate; terms defined here and later). The ter vlhether'rigid, truss-on-column, rafters-on-colu to span between columns in the principal pla the third dimension is the bay spacing. Span 1-15 Additional terms used to identify structural members in industriai buildings.
GENERAL DESIGN CONSIDEIUTIONS
h type, where traffic passes between the trusses. The deck-type trus preferred if clearance beneath the truss is not a factor, because pr Many truss bridges combine both types (see Fig. 1-19) o of a truss for the longer spansand girders for the short s a common practice. This latter scheme is illustrated I with bridge trusses are shown in Fig. 1-20 (see also Fi n bridge design is to use girder structures, which req sses. In all cases as much welding is used as lions either welded or fabricated using high-strength
ACCURACY leng:;. uf the diagonal members.
1-13.2 Bridges
reduced to reduce member Iengors.
OF COMPUTATIONS AND ELECTROMC
.
the 10-in slide rule was the principal computational tool in the structur er's office, the computations rarely exceeded three significant dipits. This satisfactory, for the reasons presented earlier in this y and as implied in the example computations. Presently, the electronic calculator and/or the digital computer are almost nivetsally used for structural computations because of both the greater corn lexity of structural configurations and the greater speed of performing calcul ally setting the decimal. Now it is almost mandato ethical and economical reasons) to provide several iterations on a he design. This step almost always requires use of These calculating devices can give a rather large number of digits to any blem is how to treat this increased computing capacity. ot any better than the input, but with a large number of arently significant digits it looks very impressive. In nearly all design offices, design computations are checked by a second person as a design precaution, ner carries a large number of string calculations on an ronic calculator, the results will differ from those .obtained where the ker truncates intermediate steps, then reenters these values and continues computations. Where the discrepancies are not large, the question arises of ether the problem has been "checked" or whether one (or both) of the persons has made a design omission. For these reasons it is suggested that regardless of the initial input data accuracy, computations should be camed to as many decimal places as is aps) to obtain good checking convergence. The extra alculation effort is minimal. Any intermediate steps should be written to the me precision as they are used in subsequent calculations (e.g., do not write 06.1 and then use 106.153 in the following computations).
50
STRUCTURAL S T B ~ LDESIGN
The reader should note that the several intermkdiate produce the end results 62.4 and 9.807 are not shown. Several other useful conversion factors are as follows:
To convert
to
kilogram (kg) :mund (lb) pound ';ips (1000 Ib) ib/ft : ips/ft !b/ft psi psi ksi
kilonewton (kN) kg kN kN kN/m kN/m kg/m MPa kPa MPa kPa kPa kPa kN . m kN m
.ASI.
psf isf Lip . ft (moment) kip . in
What are the wind forces at points 1 through 9 of the roof shown in Fig. Pi-7? Answer: ~ ( 5 = ) - 18.07 kips.
Multiply by
.
0.009806650 0.4535924 , 0.004448222 4.448222 0.014593727 14.593727 1.488 16404 0.006894757 6.8947577 6.8947577 6894.7577 0.04788026 47.88026 1.35584 0.1 129862
cago Building Code requires 1.25 kPa? What is the maximum shear and bending moment in a bridge span of 92 f t for an HS 15 at is the impact factor? Answer: V = 48.5 kips, I = 0.23.
PROBLEMS 1-1 What is E, for a steel with F, = 40 ksi? 1-2 What is E, for a steel with F, = 365 MPa? Answer: 0.00182.
1-3 What is the increase in weight/ft of a W16 x 40 beam with 2 i 1-5) fireproofing as in Fig. PI-3? Note that the method of applica exact geometric section; therefore, the coinputations should not be , Answer: Aw 30 Ib/ft.
-
1- rep roofing
Figure PI-3 .
.,,.
c,
.,.
,.,,., ".,..,..
1.1What is the increase in weight of a W410 X 46.1 rolled section with 55 mm of vermiculite fi:.eproofing as in Fig. P1-31 (See the comment in Prob. 1-3 before starting this problem.) i-5 What is the R factor for the interior column at the top floor and three floors down of Ex 1-I? 1-6 What is the R factor for beam B1 and column B2 of Example 1-2?
Answer: M = 7865 kN . m; I
=
40.1 percent.
-12 What are the story shears in the N-S direction of the building of Example 1-7?
ELEMENTS OF FRAME, TRUSS, AND BRIDGE DESIGN
'
. of
ben Ib )
Figure 11.1 Two of world's tallest buildings using structural steel frameworks. ( (442-4 Sears Tower building in Chicago, Illinois-c-ently the tallest buil
337-111John Hancock Center building, also in Chicago, Illinois. View from the sears
ANA I
.*
requires dev eloping frame member forces (axial, based on the dead and cntical live-load combina-
The method of analysis depends oh. the complexlty of the structure and whether it is rigid (indeterminate) or simply framed. An analysis may consider the structure as either two- or three-dimensional. Simple framed structures are generally determinate; that is, the three equations of statics (2FA,2 F,, Cbf = are sufficient to obtain the internal member forces. In any case, with simpEe ming the ends of the members are assumed to have no moment resistance ransfer to adjacent members. The internal member forces of determinate tructures are readily obtained by hand calculations and with considerabIe efficiency using pocket calculators. Rigid,framed structures are generally indeterminate since the member ends transfer shear forces and moments to the adjacent members. Indeterminate ructures require d e f o m o n compatibility to supplement the equations of atics to determine3e internal member forces. DigitaEomputers are used to obtain solutions for all but the simplest indeterminate structures. Continuous beams and certain simple rigid frame structures have solutions that can be
ELEMENTSOF FXAME,TRUSS, M .D
obtained from handbooks (or readily derived using mechanics-of-m methods). A few beam solutions are presented in Part I manual, and in most engineering handbooks. A fundamental part of structural design is to es .f.rai,c,.is to be-rigid or simple. A rigid frame gener moments but does not necessarily result in a more economical design: This because: 1. Practical considerations of quality control/design o connections. 2. Tendency to use the same depth beams across a bay even though certain spans are shorter or carry less load. 3. The use of a constant column size through at least two and' often three or more floors to reduce splicing. Since the colum it is over designed in the upper floor(s). ? . The stiffness (formerly called slope deflection) method of analysis is most commonly used for.indeterminate structural analysis u This method is particularly adapted for'computer use,. as sparse and symmetrical. Advantage can be taken of inversion effort so that rapid solutions of very large economically. . All indeterminate structure solutions are iterative a in that the output depends on*the input; or, stated di have member properties (area, A ; moment o displacement compatibility. Since digital comp rapidly, one may initialize a problem using relative va this output to select preliminary members and iterate as required: Wh input is considerable, with frames containing lar be better to use some approximate methods of member sizes. Approximate methods include simply columns and beams (e.g., A = constant, Ib = 1.21, Moment distribution (but limited to not more than used together with relative values of A and I for pr The-portal and cantilever methods of approxim for building frames of one or more story heights. primarily to obtain the effects of lateral (wind) forces on a frame. The portal method (refer to Fig. 2-1) makes the following assumptions:
'
,q = 4 + 6 + 3 = I3
'
'
1. Point of contraflexure occurs at midheight of all columns. 2. Sum of wind (lateral) load is distributed as shears in proportion to ba 3. Beam or girder moment is zero at midspan. 4. Interior columns carry.. na axial loads. ~--. ... .
--
With these four assumptions and application of statics, joint moments can be obtained at any location.
~ I ,
21 portal
of approximate frame andysis [br a typical three-ba~bent in
The cantilever method (Fig. 2-2) is an alternative method of appro
beams and columns.
. Axial loa_d_s_isin~-al columns and are pro~ortionalto the neutral axis of the bent treated as a vertical cantilever beam. The location the neutral axis is based on column areas (since these are not g known, a value of A = 1 unit area is usually used). The equivalentmo inertia of the vertical cantilever beam is computed as I =C
A
The column loads are computed as '
Mc v.=I
~
~
:'
STRUCTURAL STEEL DESIGN
ELEMENTS OF FRAVE, TRUSS, AND BRJDGE DESIG
The concentrated loads produce a moment of M
:
!
= -3-P- -L= -
2 2
PL 4
PL 2
'The error is L(0.5 - 0.375)/0.5](100) = 25 percent (too small). The error ra decreases and reverses sign as the number of con (i.~~ting that the situation is one where the total beam load is a illcreasing with the number of loads). For example, if the tota ." 1" and placed at five equal spaces, the M, value is still 3 p L / 9 .';'L/25. Beams may be simply supported, overhanging, cantilevered, fixed, or con. t;iious, as illustrated in Fig. 2-3. The designer must always m ~ ~ o i involving is point loads as well as how realistic the fixed, cantilever, continuous-beam models will, compare to the actual member geometry. It seldom possible and never desirable to have actual point application l;nds/reactions, although this is a common assumption made in all analy ,rfie.:.ods. 'The proportioning of beams can be done as soon as the s di.1 :rams can be obtained. The general differential equation for a beam is EZylV=-w 2.1.:
one in which much of the roof load is carried by the perime
+,C ,
- wx2 EIy" = M = moment = -+ c 1 x 2
+ c2
- wx3 c1x2 EZy' = slope = -+ 6 2
+ C2x + C,
;
TERMINATE STRUCTURES
'.
successively: EIy"' = V = shear = - wx
Load-brar~rigwail -,
- wx4 EIy = deflection = 24 Tile general equation for the elastic curve for a beam produ~cesfour integration. Constants C2 and C4 = 0 for simple beams, and the w for beams loaded with concentrated loads. The general beam equation is sometimes useful in that the appro i n f ~ i n a t i o nmay be approximately obtained by replacing a ser ce~tratedloads with the equivalent uniform load value. ,@,earnde.fl,c.ctionsare often limited for both buildings and bridges. A va of L/360 to as low as L/1000 can be fo occupancy and/or interior finish. Where' deflection approximate deflections should be computed before a de The reader should note that highly refined deflectio possible because of the uncertainties in loading. It should also b deflection is heavily depende,nCton the moment of inertia_an constant for steel. Therefore, where a rigid deflection criterion i use of high-strength steel ma'y not be economical, since the sectlo fix?? by the moment of inertia rather than by the bending stress.
ical sense almost all beam-column connections can, and do, transmi
n
=
2j
--
R
n = number.of bar members
e-rea&Iy- _ optimized for least -- - weigh~,-since--hes_geometry. Indeterminate trusses are not SO readily .
60 snivcl.rfRA~STEEL DESIGN
ELEMENTS OF F W I E , TRUSS, Ah?) BRIDGE D
optimized, since the bar forces will depend on both s-ndeterminate trusses are commonly used wh some members, unsymmetrical loading conditions extra members are used to reduce the L / r ratio members. ~ontrolling.'theL / r ratio is likely 'to dictate the geometry trusses. Where the L / r ratio controls the design, op very difficult. The reader should bear in mind, however, that the (or any frame) js that which"ba1ances weight, fabrication, and against the total client cost. Minimum weight alone does not sati of "optimization." -ate trusses are readily solved by hand using the m equilibrium or the alternative of a cut section. Both determinate and indeterminat using one of the large number of computer progr computer program in the Appendix can be used t indeterminate trusses. This program uses the stiffness method of matrix analysis as outlined below. The user is expected to have taken or be taking a course in advanced structural analysis, so only a brief outline of the stiffness method as used in the computer program is given. Referring to Fig. 2-5, the coding of a typical truss element is as shown, where the P-X code refers to nodal effects in a truss system of which only ,P, through P, are shown as affecting the ith member. The internal member force' of
V Z ) The ~ . nodal equilibrium in matrix notation is
P=AF.,
member forces can be written as F = S e = SATX
combining Eqs. (c) and (a), we obtain
',
'
n ( d ) can be inverted to obtain
x = ( A S A ~ ) -P'
mpute member forces: e general A-matrix' entries for any truss member are the direction cosines: - COS a
p4
- x4
rforming matrix multiplication, we can obtain the S A
as
Truss member connecting 2 nodes
e product of A x SA
Truss member
PI + F I cos cx = 0 ZFv = O
gives the element stiffness matrix as follows:
66
STRUCTURAL STEEL DESIGN
determine the number of P-X entries (NP) and those definin (NP + 1 = NPPI), where the X values (displacements) are zero. members (NM) is also determined from this step. The ho distances from one end to the other of each member is obtained from t building geometry. The program computes the actual member length, using' and V from this step as input data. If the corrert member cross-sectional area A and moment of inertia I values are input, the displacements (X matrix) are actual values. If relative (or incorrect) values are used, the displacements are not the true values (also the member ,;orces may be incorrect in many cases). The member forces of a determinate .,ass can be obtained from the use of any area (use area = 1.0 so program does i;ot divide by zero), but again the displacements will only be correct if the correct member area is used. The moment of inertia is not used in determinate or indeterminate pinned truss computations. The member forces of inde rerminate truss configurations are dependent on the member cross-sectiona nreas. The member forces in rigid frames depend on b area and moment of inertia; thus this is always an iterative use some estimated intial values, obtain output, revise required, and make additional solutions as required. If i t is desired to obtain floor beam deflections (or deflections along highway bridge), it is only necessary to add nodes at those locations illustrated in Fig. 2-8. While adding nodes increases the size of the matrix to inverted, the routine (a banded reduction method) used in th gram here only uses a part of the global matrix, so that very la be very efficiently solved: To take full advantage of this reduction meth coding should be such that the difference between the P, on the near end (for trusses) or P, at the far end is as small as possible, sin [NBAND = NPE(4 or 6) - NPE(1) + I] sets the size of the reduced. Note that NBAND is only 4 for the beam coding This means that a very large number of beam segments alternatively, a continuous bridge with five or six spans with span can be easily solved using a very small amount of computer core, since value of STIFF(1) is NBAND X NP = 4 X NP; with NP = 100, there are o 400 STIFF(1) entries versus 100 X 100 = 10,000 entries for inversion. The solution time is down to about that of the 100 X 100 mat ..
,
,
,," ..,.,.....,.
~
2-8 THE P MATRIX 'The P matrix is developed from the structure loads c element a s either a truss element that has just node forces team element based on using statics and the external for member, such as wind, roof loads, and wheel loads. It wheel loads between truss panels are prorated to adjacent Seam analysis, as if that truss member is a beam with the re
ELEMENTS OF FWLIE,
TRUSS, AND BRIDGE DES
elements making up a beam are considered as a seri s; thus the P-matrix entries will include the fixed-end mome ed for the several beam elements meeting at a joint in additio the shears as illustrated in Fig. 2-2. In both beam and truss analyses, the node forces (equal and opposite t those acting on the connecting elements) are resolved into components par to the translation Pi's. Where beam and beam-columns (axial force and bending) are analyze is convenient to have the computer program compute the FEMs and shears the several beam elements using the beam loading, and make the summ process as each element contribution to a node is found and then buiId tho matrix. The program should also be able to read in selected additional P-ma entries, which can be used and/or added to the value(s) already in the P mat The computer program in the Appendix allows this procedure. The user should note in using the computer program to develop th P-matrix entries that all dead and live loads are applied along the length of beam element in the direction of gravity using a (+) sign. This alIows program to correctly compute the P-matrix entries (and signs) for sloping be Wind loads applied to sloping, flat, and vertical beam members are aIwa applied along and normal to the member axis. If the slope is 0 _< 0 _< 30°, t sign will usually be ( - ) because of the aerodynamic (uplift) effect. This the program to develop the P matrix for the wind NLC. Since gravity an loads along a member cannot be combined because of a horizontal compone of wind on a sloping member, it is convenient to store the D + L (or D o nd treat the wind case totally separately, then combine wind with L analysis to obtain the design case for D + L + I.V. Specificatio the allowable steel stresses to be increased one-third in afTy stress co ing wind (as long as the resulting member is not smaller than rately). This stress increase is equivalent to reducing the membe including wind (axial forces and moments), by 25 percent. The comput m also does this, so that the designer merely obtains the maximum ax^ r moment from any of the NLC for design of that element. T o antage of the program ability to do this, it is necessary to order the win g NLW) after the D + L load NLC. This procedure is illustrate xamples displaying computer output..The reader should parti lrly inspect the input/output for Example 2-7, which uses wind pressure on loping beam element. The coding of NP and noting NP + 1 = NPPl for specifying zero displac nts (including but not limited to) for reactions excludes those global matr ries from consideration. Those P-matrix entries developed by the comput re not used in the analysis. The reader'should note that thes directly into the reaction and therefore do not cause intern mber forces (see the output check in Fig. E2-4d). One of the very early decisions the structural designer must make is wheth he superstructure (columns in particular) is pinned (allowing rotation) or rigdl
P S X E N T S OP PPAUE, TRUSS, AND BIUDGE
p6
I
VP = 6 1VlII = 2 NPPI = 7
I
SOLUTIONIt will be necessary to develop the necessary equations for for sloping members. For this refer to Fig. E2-2b and note that loads no to member cause FEM. Now, returning to the initial problem, fo sloping member 0 = 45" : FEM =
12(0.707 11) 6 (-)' 0.707 11 12
For the horizontal member:
RL=--4(72) 6
- 48.0
,= 50.91
kN - m
70 STRUCTURAL
STEEL DESIGN
Now consider the following sketch for nodes 1 , 2, and 3: P, = 0 p2 = 64 - 50.91 = 13.09 (note +)
I
P4 =
- (50.91 + 48.0) = - 98.91 kN
e several load conditions for a bridge would include Dead Dead
+ live, including impact + live + wind
i
.",,.,,
4,
.,. ,,.,,.... ,"..-"
7 mPP1)
-(does
not matter)
several load conditions. Computer programs are included in the Appendix to deveIop trix via punched cards for analysis of a highway or railroad truss.
0 CHECKING COMPUTER OUTPUT
2-9 LOAD CONDITIONS of two basic steps:
condition (NLC in computer program), The load conditions !;)ads making.up the P matrix in the equation
X
=
(ASA~)-'{P}
night include:
NLC
b a d combination
--
~ l e a d+ live Dead + live in alternate bays Dead + live snow Dead live f snow Dead + live + snow one side Dead live wind
+
+
..
1 2 3
+ +
+
the end of the coding sequence is checked and both satisfy statics, problem has been correctly solved for that irlpltt data. Problem checks can be performed in several ways, as illustrated in
4
+ wind
5
6
of computational effort-often
by inspection. Summing forces above
72. STRUCTURAL STEEL DESIGN
a story are also often convenient, as the column axial forces are direct1 Advantage can often be taken of symmetry of structure and lo two office building examples that follow (without wind). Sometimes t matrix can be used to advantage, as illustrated in the column check of th building. Truss checking is similar to that for rigid frames. Always take truss geometry by checking members in which the internal force is zero. If the computer output does not give zero and the load there is something surely wrong with the input data, such as mispunching an NP or a mismatch of H and V (either length or signs). Perform any additional statics checks near each end of the truss at joints where a minimum of bars connect. If these joints do not satisfy statics, the more either and the problem needs to be reprogrammed.
2-11 DESIGN EXAMPLES The Sallawmg several design examples will further illustrate frame coding and computer input/output for the analysis computer program in computer output from these examples will be used in many examples in later chapters toillustrate the problems in structural ( a manner somewhat like that which would take place in actual design. interest of saving text space and maintaining reader interest, the examp considerably edited from actual design problems.
L
7'-6"
1 Z
2 1'1" 7 3
(2)
(5)
6
(3) 16)
(9)
110)
7
(12)
(13)
10 ( 1 6 )
I I
(19)
(20)
14123)
I j
(75)
(26)
117)
124)
,.s.v 4 (;
17 45.47.47
---.\'P = 4 b .YPPI = 47
18 47
r.2.i Elevation o f typical interior bent
,
5 @ 18'-6"
Example 2-3 A small three-story office building is as shown in plan elevation on Fig. E2-3a. Wind bracing will be used in the Ebay and locations inarked wb) together 'with a simple fra rigid frame (with resulting member end moments) will be used in the N-S direction. A brick veneer exterior will be carried by lintel beams to the exterior columns and at each floor. The framing method allows a corridor between the interior columns with a clear space for the office areas. Rental space will be in blocks: a, b, c, and so on. This space will contain miscellaneous office furniture, partitions, and so on, based and desires. With three stories, a flight of stairs at each elevator on the west end for large equipment will be the to allowance. Air conditioning and heating will utilize a h auxiliary shed (not shown). The basement will contain the remainder of the environmental equipment and provide. extra storage. Use the NBC for general design. We will use a flat roof which may be used for worker exercising during the day. For this additional activity, the roof will be designed for a live load of 80 psf (as opposed to 20 to 30 psf for usual live and/or snow). The reader "should note that building codes and material specifications stipulate minimum requirements-the designer can always use larger values.
-
-
- -
Plan view
~-
STRUCTURAL STEEL DESIGN
SOLUTION The general computer coding is partially shown .in Fig. E2-3a and also on Fig. E2-3d (computer output sheet). This coding gives the following computer program control data information:
NP = 46 NBAND = 11 The next step is to develop the beam and column loadings for this typical interior bent (refer to Fig. E2-3b, which displays loads after comcputations). Roof: Dead load: estimate 5 in of concrete on a metal deck supported by steel bar joists. NM
=
Dead load
=
0.080(18.5)
Live load office
26(members)
=
Live load corridor
0.080(0.70)(18.5)= 1.036 kips/ft =
O.lO(l8.5)
= 1.85 kips/ft
These loads are shown in Fig. E2-36.
.02(6)18.5 = ? . 2 ? k
Concrete: (5/12)144 Metal deck and joists (estimated)
,
Ceiling, ductwork, electrical, etc. Total
=
5 psf
-
= 5 psf
=70 psf
Live load: check reduction in 21.3 x 18.5 spa live load is not reduced; however, we are not using a stand loading. R = 21.3(18.5)(0.08)
=
31.5 percent
"
+
(100) = 43.3 percent R = - D+L(lOO)=4.33 L 4.33(80)
or
Note that the L = 80 psf value is from Table IV-4. Use a 30 perce reduction. Live load = 8OC1.00- 0.30) = 56 psf In a 7.5-ft span, R = 7.5(18.5) = 138.8 ft2 < 150 (no reduction). The equivalent beam loads for the roof are: ..
. ... ..,..,.... . .
Dead load = 0.070(18.5) = 1.295 kips/ft Live load in 18.5-ft span = 0.056(18.5) = 1.036 kips/ft Live load in 7.5-ft span = 0.080(18.5) = L.48 kips/ft
Other floors:
Estimate dead load with increase for floor surfacing and ceiling finishing
The exterior brick veneer wall will contribute column loads for' t
Corridors (Table IV-4 and conservative use) Office space
The office space will have a 30 percent reduction in 1
terior wall finish, fixtures, etc., of 5 psf gives a total ='45 psf.. .." Column load
=
12(18.5)(0.045) = 9.99 kips
STRUCTURAL STEEL DESIGN
The longitudinal walls defining the corridor and lo Ute interior column loads: L
Tile wall at 8 in = 35 psf, Beam at 50 lb/ft (est.) = 50 plf Column shear = 12(18.5)(0.035) + 18.5(0.050) = 8.695 kip , For roof = 0.050(18.5) i=0.925 kip (interior columns)
i
L'
i
~ & l e c any t beam weight e tributing shear to the roof line of exteri columns, since the values are tw small to be either reliable or to affect t design. Now make some preliminary member size estimates with the followin practical considerations:
r rO T rr 0C0 0 0 0.0. .. -0 e 0 0CC 0 *r a ~0O 0 oo - O ~ .O O* o C c ~ - ~ O
M,
1
I u
"I
;f the moment is M = lue is somewhere - - - in --A
.a
U
J
~
=
0.925
+
A
-
-9-0000---0000-*-000O_t_O rr* * D * 0-0
0 0 O O O 0 C c 0 O C C 0 0 0 3 0 0 0 0 0 0 9 0 0 0 l 0 0 0 ~ 0 C 3 0 0 1 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 C 0 0 0 0 0 C 0 C 0 C 0 0 0 0 0 00 00 00 00 00 00 00 00 C0 00 00 00 00 00 00 00 0
--ldddd222dddd222;ddd22fdd 0000000000000000000000000 Q O C O C o o ~ O 9 O O O O I O e O O O O * o - ~
;d;;;;;;d;;;;;;dJ;;:;n.dddO * ~ * * . * * n D r * * * i
= 142.7'ft , kips
h
N----N
* r O r * * ~ * ~ w ~ ~ ~ N----N *----.+ L--
6 -.
I
1 I.OU
in-
1, = 3 I 6.u in'
I
This computation utions + the uniform beam loads from roof to the first flo 'c01umn~ Z
,
C D D o C o O O O O o O 0 E O O O O o c O o o o o O
c
For columns: Estimate a length coefficient (Chap. 7) K = 1.2 -+KL = 12 x 14.4 ft. Estimate F, = 16 ksi (som thing less than 22 ksi).
Papp,,
I
........................
12!0(12) = 60 in3 24
n =
,
O O o C O O O O O O O O O o O O O O O O O O o o
-
dl,d;dd;ddddld;,hdb:dD'O'd:: ----*.-----*<-----*.-wm~=*
*rw----nrn----ccn----.oQ*--
000000000000000000000000 000000000000000000000000 00000000000000000000O0O0
dddddddddddddddddddddddd 000000000000000000000000 O O O O o O O O o O O O O O O O o O O O O O O O
000000000000000000000000
dddddddddddddddddddddddd 0 0 0 0 ~ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ~ 0 0 0000000000000000000000~0 000000000000000000000000
1
.=.
O O O o 0 O O O O O o O O C C O O o o o O o o o
=
-u.
-- ,--a
8
................._...._._ - - - *---=* - --
The required section modulus based on an allowable 24 ksi (Chap. 7) is
ox
-,8
,
0 0 0 - - - - C 3 0 - - - - C 0 0 - - - - ~ 0 0
8
Sx =
-
~ * r 0 0 0 0 = - C o O O C O - ~ 0 0 O O O O e o * C r 0 0 ' 0 0 . ~ . 0 0 0 0 * - * 0 0 0 0 * C * 0
8 (142.7)- == 95.1 ft . kips 12 Ma, = 118.9, say 120 ft . kips M,
8
8
0 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 0 0 C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 C 0 0 0 c 0 0 0 0 0 0 ~ 0 0 C C 0 0 0 0 ~ 0
+ 1.48)(21'3)2
=
-
. L
1. Use continuous column (no splices) for full 36 ft of hei interior columns (maybe) at fi st floor level.] 2. Use constant-depth beams acr ss the bent. For beams: If the roof beam is simply su fully fixed, the moment is M = between, so taking an average: -
. c . . 0-0
x
-
.-.
;l;dddd:;:dddd-.;:dddd2;*
*
dddddddddddddddddddddddd 000000000000000000000000 000000000000000000000000 000000000000000000000000
dddddddddddddddddddddddd
00000000000000000000000000 00000000000000000000000000
'
ddd;;;;ddd;;;;ddd;;;:ddddd
.
:
'"I.
' Z
4 W .
'4-
\ l I T F I I I tN"IE U h/D
YO YP ND N> ND NO NP YO yo NO Yo LIP qo YP YD q P NP LIP ND
=
1 7
-
3
=
4
;
r
5
=
4
-
P
=
10
= = = = = = =
YP = " I = VP = NO = No = NO
I
VP = YP =
YO = LIP = ND = No = YO
=
.,D YO NP NP
= = = =
NO
=
yo = NO = NP = NP = YD T NO =
= No = NO
OANJ h I O T H =
-
13
AND U NX = NX = NX = NX = WX = NX = VX = NX NX = NX = NX = NX = MX = NX NX= YX = NX = NX = NX = NX = NX = YX = NX = NX = NX = NX = NX = NX = NX = NX = NX = NX = NX = NX = NX = FIX = NX = NX = NX = NX = NX = NX = YX = YX = NX = NX =
-
7
=
= = =
690
9
11 I 2
-
114 15 16 17 18 19 20 21 22 23 24 25 26 27 2R 20 30 31 32 33 34
35 36 37 38 30 40 41 42 4' 44 45 46
1 2 4 5
6 7
8 0
10 11 12 13 14 19 16 17 18 l a 20 71 22 23 24 25 76 77 ZP 29 10 31 32 31 34 3< 36 77
38 39 40 41 42 43 44 4T 46
0.00394 O.OO4Rl -0.09541 -0.00169 0.00033 -0.19343 0.00169 -0.00033 -0.19343 -0.00394 -0.00481 -0.09941 0.00272 -0.00081 -0.08302 -0.00121 -0.00006 -0.17721 0.00121 0.00006 -0.17721 -0.00272 0.00082 -0.OR302 0,00275 0.00061 -0.05121 -0.00125 0.00007 -0.14018 0.00125 -0.00007 -0.14018 -0.00275 -0.00061 -0.05121 0.00353 -0.00136 0.00023 -0.08232 0.00136 -0.00023 -0.Oq232 -0.00353 0.00068 -0.00068
Sketch lo ~denttfyP - X codlng
XAL F3RCE. K
3 4 5
6 (I
9
10 11 12 13 14 15 16 17
-6.07 -22.77 -38.21 -38.21
-82.42 34.19 -20.14 20.1+
38.11 3R.70 -22.49 22-49
1.01 0.*6 1.01 -58.49 -81.26 -81.21 -58.64 -0.72 -0.54 -0.12
-64.52 -49.19 -87.27 30.45 -18.14 19.14 -30.45 -63.13 -50.34 -81.36
87-21 49.19 64.52 30.1.3 -11.44 11.94 -30.33 81-36 50.34 63.13
DESIGN EN0 *O*tNTS CORRECTED F n R F E I A N 0 WIN0 lNE.49 EN0 FIRSTI. -26.18-
64-31
I-F1
STRUCTURAL STEEL DESIGN ELWlE?4TS OF FRAME,TRUSS, AND BRIDGE DESIG
check if O.K.: K L / r , = 14.4(12)/3.53 = 49. The allowable col ess from Table 11-5 .= 18.5 ksi. We will try this value for the first design iteration, since there will moments on the columns of an unknown amount that will effective increase P. We will use a smaller section for exterior columns. Use WS
A = 9.13 in2
For the two basement columns, use W8
Check CF, for t o p story.
110 in"
I,
=
X
48:
C/',l,,,8d,, = ? . 3 3 1 ( 2 1 . 3 ) ( 2 ) + 2 . 5 1 6 ( 7 . 5 ) + 0.925(2) = 120.02 kips
A = 14.10 in2
From computer o u t p u t for values shown which have been reduced 25 percent I'or wind 5o.that o u t p u t is directly comparable t o D + L o u t p u t :
T F , = 120.01 kips O.K. . .-.
7.
..l
1,
... ...,...*.,_
101.55 Actual F = ------ = ]35,4k 0.75
..
R3= 12.09~
'Mom have been reduced for wlnd
V1= 8 965
C h e ~ k l n gcomputer o u t p u t for column (member 25) uslng computer o u t p u t and the statics o f node 1 4
Figure E23f Output checking for NLC = 2 (with wind).
Use W8
X
40:
A = 11.70 in2
Ix = 146 in4
I, = 184.0 inJ
These data are used to make up a set of member data cards, b loading cards (interspersed as appropriate in member data), and a P-m data set. Note the P-matrix entries read (PR(i, j)) are zero for NLC = 1 ake the values shown in Fig. E2-36 for NLC = 2. The remainder o trix is built by the computer program using beam loading data and th umn shears separately computed and also shown on Fig. E2-36. Th ut is written back as a part of the output for designer checking (Fig. 3c). The remainder of the output is shown on Figs. E2-3d and ether with some output checks. Figure E2-3f further illustrates eck of NLC = 2. e reader should note that when wind is considered, the allowable stre increased by one-third. So that the design is all on the same b gram multiplies all member forces by 0.75 for NLC that have wind forces as ut. With this adjustment, the designer merely has to scan all the output for maximum moment or axial force. The largest value in any load condition i hen used with the allowable stress to determine if the member is adequate. Not in checking the output (as illustrated in Fig. E2-lj-) the fact that this 0 or has been used must be considered so that statics is satisfied.
ample 2-4 We will design the small office building of Example 2units. Note that the dimensions are slightly different than when oft conversion. The same general design considerations/parameters pply as in that example except that the brick veneer wall will be take 00 mm (approximately 2 x thickness of Example 2-3). Figure E2-4a lays the general building layout and SI dimensions. LUTION The computer program will solve either SI or fps probIems. I cessary to include a "UNITS" card with fps or SI identification for use in the FORMAT statements to identify the units of the output. Since'the same building design criteria are being used, we may pi% directly to coding the joints (refer back to Fig. E2-3d since frame is s ~ i l x ) eveloping the frame loads.
STRUCTURAL STEEL DESIGN ELEhlENTS OF F I U M E , TRUSS, A N D BRI
oof loads: Dead load: estimate 130 mm of concrete on a metal deck steel bar joists.
Concrete: 0.13(23.5 k ~ / m ' )
= 3.055
kPa
Metal deck and joists (estimated)
=0.263 kPa
Ceiling, finishing, electrical, etc. Total
= 0.263
kPa 3.581 kPa
Live load: a live-load reduction can be used in the 6.5 x 5.9 m area Basic live load = 4.0 kPa (Table IV-4).
Use a 30 percent live-load reduction. The reduced live load = 4.q1.0 0.30) = 2.8 kPa. D o not use a live-load reduction across the 2.3-m span The roof beam loads are: =
21.1 k N / m
Live load in 6.5-m span: 2.8(5.9)
=
16.5 k N / m
Live load in 2.3-m span: 4(5.9)
=
23.6 kN/m
Dead load Elevat~ono f t y p ~ c a lInterlor bent
59
,! 1
59
59 1
L 1
5.9
, 1
59
N
=
3.581(5.9)
Frame loads for other floors: Use 30 percent reduction for office area live loads but none in co
Plan vrew
Figure E2-4a
3.77 kPa (brick)
+ f~nish:use 4.0 kPa
58100'0
I
-
11590'0 ',5100'0ZLZOO'O 010b9'1OhlCh', . . 01200'0fbO1L.kEUOUb'f EVIOU-0 0kEII.EbV8bb.b ~0100'0U0015'19b9L0.4 bLkOO.0 ObULP'ZO99Z5'8 8?Z00'0siSOE'ZEZEIS.8 E~IOO'O bLLL0.bIEIZS'Y blIOO'0SZBfb'ZVEZSS'8 Zli00'0
A
l'd
01100'0 LLEOL'I91LOO'O OLlOO'OOCZOO'O IlbO9'1ZfU20.0ULZOO'Ob b 9OZOO.OSETOO'O ElZIZ'EZIb00'0 5E100'0bObO9'1LZOEO'O bLZOO.0 ULEUS'ZILPZO'O 18200'01Ulbl'tbuZOO'O LE100'0 bZ1bl-295000'0EiIOO'OOEEYS'ZIECZO-00UZ00'0
12 09 bk dC ~r
=
VZ
= XN = XN XN = Xh - Xh ‘ XN Xh xh XN Xh = Xh XN Xh
52
=
?C
;
VL >< ZE rL
ZL le Or bZ
VZ LZ
iZ 22 1Z UZ 01 11 LI VI 51 ?I LI
Ol.bCb511 sb'51Z00'0 OI'bZbS110u'Z~ObLl UL'LIZ-
KN
= XN = XN
-
-
= = = =
OI'bZb$ll 56'5120 0 '0 Ol'bZb'YI1OU'ZLOI,CI Ok'iIZ-
Of br Hk
LC
9 LC
6ZZEl'O ZZfOO'O
Z 1
XN XN XN XN
XN = XN
OU'ZlObLI5b.GIZ00'0 01-bzb511 11'91Z00'0 01'501911-
1 1 - 1 00-0 01-SO19I1
11'01Z00'0 01'G01011
so3
NIS
1
I
v
A
?I 01
(11
1:
lfr ;I
6 B L
P \ P
PSI;.
19L.b
XI
S6SXOItM
1 f ; I
-
131113W Z
-
= XN = XN
-.. .
..
OZ'OI OE'EBEZEI
00-0 0€'r@EZ~1 MY-W
~ H L
--
d
- o h
= = =
dN oh dh
4
AllJIlSVl3 OOY Od3Z-YON ON 92 = S138M3Y
=
dh
OOb 3J
=
ON
-
0
U L 9 5
b 2 I
= cN = dh oh ON dN ON
= = = = = =
dN
= dk
S3181N3
tIl3jllS
030151 3 3 l ~ l iL N 31N U N i l Oi dlh G V 9C - dN dN
dh
= dk
'XIdllW-d
P ~ H L ~ +N d ~E ~ NZ ~ NI ~ .N3 8 h z r
Z
21 I1 01
NW
ONV
000'58 OOZ'LZ. OOZ'LZ 000-0 000'0
O.OOOOOL 11 31I1JM E = llONJ3 O V J l 30 ON
62 HC
= XN
OOZ.LZ OOZ'LZ 000'0 000-0 OOZ'LZ 002-LZ 000'58 000'58 000'0 000'0 OOZ'LZ OOZ'LZ 000.5u OOO'b8 000.0 000'0 OOZ'LZ OOZ'LZ 000-0 000'0 000-0 000'0 009-t OO*..
CC Zi II C 0k
= -
= XN = XN = XN = XN
= H l O l H ONVB
1~33
dN
ah dk ON dN
5k
21
= XN
S N V i O V M MO Yvl ' X I d l V W - X
2 ~ 3 3
= =
00'0
Ob'ZlOhCI56*$1Z00'0 O I ' ~ L ~ ~ I 1I'VIZ00'0 01~b01011-
XI%
. -. .
OCZSZ'OI OESOO'O
= dk = dk
01
00'0
OZ'ZZIOZ'ZZI00-0 00'0 O L ~ E B E 2 E I -O C - E B E Z E I 00'uLI00.8~100'0 00'0 OO'uL9211 00'8L9ZII 00'8L100'8LI00-0 00'0 OO'LfL9ZII- 00'8L9ZII~Aluo)
= dN = dh = dN
3H1
.
w
ELEMENTS OF FRXLIE, TRUSS, AND BRIDGE DESIGN
89
e reactions at the basement wall are computed:
+ axial load,, + R i + axial load,, + axial load?, + axial load, 387.29 + 402.51 + 626.26 + 665.79 = 111.95 + 119.28 + u,..-/>
= RI
-2081.85
0.75 Now consider C F,,
=
53.64
- 3007.03 (vs. 3006.36 kN)
O.K.
0: 17.60
14.92
J1
LJ 47.18
Member forces
!
. .
.'
m ~ c * ~m c~c * ~c o ,~ r c ~o c u o m ~c o ~ m o ~- o -~* D ~O
dd:dd'::;;;'dd;';:;;;;f;;r:
0 0 f f " m N * O * C N " ~
I
4
l
-
l
"IOtCN"
l
-
m
l
l
m m m R - C I O n O m O m N n o m C ~ m O * C C O O o
m F C E m N N m ~ O C * m n * 0 ~ 0 0 3 e 0 3 0 0 0
d;:;;;;;d'd';;;d:;:d;;d;;d ~IDP~N-C*QONV I
l
l
I
I , , ,
m n n o m *
I
o
w
o
m
~
~
a
e
c
c
~
~
-
. .. . .
f
-
*l,-"...a~
n m r p t
""ti 'i"
'(Mem.
22)
'(24)
Forces on two basement reactions, including effects from horizontal beam members 22 and 24: CI;;t(applied externally)
= 10.2
, ,,,=
CFh
=
+
15.3(2) = 40.8 k N
-22.52+ 15.15 + 3 7 . 3 3 + 11.80+4.32-5.11 40.93 kN
O.K.
Example 2-5 This is a partial design of an industrial building (fo manufacturing process) with a general elevation view shown in Fig. ,I$together with a reduced size plan view. With the large unsupporied' cl span and height required, roof beams are not deemed pr~ctical,so roo trusses with purlins will be used. These trusses will both reduce roof deflections and increase the overall frame rigidity, particularly by making e columns continuous to the top of the main roof truss. Making the lumn continuous will reduce column rotation and translation at the roof
ELEMENTS OF F R h X E ,
line. The purlins (beams across bays to support th spaced at about 6-ft centers; thus one will fall bet will produce bending in the top chord members in addition to the axial from the truss analysis. It will be necessary to make an initial estimate of member siz program these data, inspect the computer output. and iterate as required. first iteration (not shown) used W14 x 68 columns. The most severe Ioadi condition (with wind) produced lateral displacements on the order of 36 which would never be acceptable. The columns were resized using W27 178 for the second iteration shown here. The building plan shows 22 bays at 25-ft sp carry wind bracing (not shown here or designed in be longitudinal beams between columns at app spacings to carry the exterior walls. These memb load to the column and also provide lateral braci the weak axis. The wall loads are not considered in The wall contribution can be readily accounted for by shears directly to the axial force on the computer output when it determined what the exterior siding will be (sheet metal, metal and woo etc.). General design parameters for a typical interior bent (or frame) inclu Roof
+ purlins, estimated at
Miscellaneous additional details Total
= 5 psf
-
=20 psi
Load/ft of roof = 25(0.02) = 0.50 kip/ft
SOLUTIONThe dead load of the truss is obtained after a n approximate analysis (not indeterminate truss) has been made to estimate possibIe forces. This analysis is very approximate and not shown here, as there very crude assumptions and there are several which make to obtain similar results. From this initial estimate the following member sizes are selected: T o p and bottom chords of shed and main roof tr wt/ft
=
20.6 Ib/ft
A = 6.06 in2
Vertical and diagonal members: two L4 wt = 11.6 Ib/ft
x
3 x
i:
.A = 3.38 in2
The approximate truss weight is computed as follows: Length of top
+ bottom chord (approx.) main truss = 2 x
Verticals: av. height x 1 1
=
(15
+ 27)(11)/2
Diagonals: av. length x 10 = (27.3
+ 19.2)(10)/2
120
=2
92
STRUCTURAL STEEL DESIGN
The average truss weight: W(120) = 240(0.0206)
w=-
120
+ 0.0116(231 + 232) = 10.31 kips
= 0.09 kip/ft
use 0.10 kip/ft
Truss weight for side sheds is as follows:
+ bottom chord (approx.) = 2 X 72 Verticals: (3 + 15)5/2 Diagonals: (12.4 + 19.2)(5)/2
Length of top
= 144 ft
= 45 ft = 80ft
The average truss weight: 144(0.0206) + 0.01 16(80 + 45) = 4.42 kips 4.42 w=-use 0.07 kip/ft - 0.06 kip/ft 72
W(72) I
=
t
The weights of the trusses have been rounded up to account for connections. The truss will be analvzed for two load conditions.
NLC = 1: dead load + snow NLC = 2: wind + snow + dead on left shed + wind (suction) main truss + 1.5 snow + dead on right shed
+ dead on
The truss ,will be symmetrically designed, since wind can blow either direction. The rationale for this combination of loads is as follo 1. Wind from the right will not blow snow off the left shed. Also, the vertical wall will create some stagnation, so the direction of the wind is normal downward. 2. On the main truss, the slope is such that aerodynamic action will result (in addition to blowing snow off) in a suction. 3. The right shed should be protected from the wind, but the snow from the roof can accumulate; to account for this, the snow is increased to
p
B
We can now begin to compute the node forces. For dead load: Apply truss weight (computed as kips/ft of span) to the top node along with the roof dead load since the small total weight causes too much additional work for the increase in computation precision. Shed dead loads: note that both shed and main roof truss have 12-ft panels.
+
Interior nodes: (0.07 0.50)(12) Exterior nodes: 6.84/2
= 6.84 kips =
3.42 kips
% STRUC'IVRAL STCCI. DESIGPJ
Main truss: Interior nodes: (0.10 0.50)(12) = 7.20 kips .,... Exterior nodes: 7.20/2 = 3.60 kips Snow load at 25 psf of horizontal (span) projection:
+
d
Interior nodes: 0.025(25)(12) Exterior nodes: 7.50/2
4812
=
= 7.50 kips =
3.75 kips
5 74 12 1 1 31' 60 sln 0 = 0 19612 LOS 0 = 0 98058 12 L==1224' cos 0
0 = tdn
-=
Main roof truss
7.50' 8 = tan-' '1
Py
18 72
- = 14 04'
3 87'
12
- = 12 37'
Los e = 0 025(25)(12 3 7 ) = 7 73'
r':, =
7.73 cos B = 7.50'
Ph = 7.73 sin 0 = 1 .88'
Figure E t S c
Shed trusc (left side)
ELEMENTS OF FRA!!IE,
% STRUCTURAL STEEL DESIGN
~ 1 1 0 1 9c ~ rhotTlo*
? N O I T I O N NO
YQ
STIFFIII
THF P - Y A T R I X ,
ENTRIES
Y
AND
=
IN-K
954
IL FORCE. K &XI
D E S I G N EN0 VOlENlS CO99EClE0 GO. F E N bNC Y l Y O INEL9 EN0 F I R S T I . FT-X
----
8.15
----
--.. -..-
YO r qp = '$0 = YP = YO = YP YO = YD = YP = NO = N O = NO z NP =
.
q
10 11 12 13
14 15
16 17
1R
=
14 20 21 22 23
=
24
L
25
= =
26
= YD NP NO ND No NP NO ND N'P YP NO NP NP YO NP NP NO ND Yo YD YP NP l o rlD
R
5
; .
=
27 28
24
t . PO'-
= =
31 32
= = =
79 34 35 96 37
r
= = 5
=
38 30
; .
40 41 47 41
=
44
=
45
r
=
;:
$2
.NO
yb NP ND NP ND NP yo YO NP NO
.
=
= r
49 50 51
= = = =
52 53
= = =
56
54 55 57 5R
Figure E2-5e
TRUSS, A N D BRIDGE DESIGN
-----.. ----------------------------
----
---.
-----.----
----------------------
0.00
*---
. . .
----
-------
----
--.-
-------------
...-
----
----
rjrsFs
NO
rrlu FO~CE.I
-
2
OESIG* EN0 .C*E*TS COPPECIEO FOP U W # * E L M ENO F I S S ~ I . FT
ELEMENTS OF FRXME, TRUSS, A N D BRIDGE D a 1
chapters to design (or redesign the members). Note that the maxim lateral displacement is now on the order of 4.7 in at the base of the sh truss and only 4.0 in at the top of the main truss, indicating some additio bending in the interior columns. Note that this displacement is occunin the wind NLC, as one would reasonably expect.
Example 2-6 Deslgn the ~ n d u s t n a lwarehouse bullding In SI units with the general dimens~onsshown in Fig E2-6a Use the same assumptions and problem parameters as used for Example 2-5 SOLUTION The initial member sizes are selcc~ed as follows (and bein considerably guided by the computer output of Example 2-5, which professional terminology is using "experience"): Top and bottom chord members of both main and shed truss: two L152 x 102 x 7.9 mm:
I,
=
not needed for truss-type members
Intermediate truss members (verticals and diagonals): two L102 6.3 mm: wt = 0.169 k N / m A = 2.18 x mZ Shed columns (member numbers 1 and 93): W610 X 241.1: wt = 0.72 k N / m
A
=
30.77 x lo-' m'
I,
=
2151.9
X
X
76 X
low6m
Two interior columns (members 24, 25, 27, 28, 67, 68. 69, 70): W690
The computation of frame loads now follows. Roof? Miscellaneous, including purllns Total
= 0.72
kPa
= 0.24
kPa = 0.96 kPa
Load/m on bent = 0.96(7.6) = 7.3 k N / m Dead load of truss: note that this is computed directly as a horiz projection. Main truss: =72 m Length of top and bottom chord = 2(36) =70.7 m Approx. length of diagonals = lO(5.84 + 8.3)/2 = 70.4 m Approx. length of verticals = 1 l(4.6 + 8.2)/2
t
Take as horizontal pro~ect~on, as truss slope
IS
very small and values are estimates.
'1@ .. .KUCTURAL
STEEL DESIGN
I
Note (-1 s~gnfor P m a t r ~ x
~u-. J I K U L ~UKAL
SIEEL DESIGN
KN &NO
=
YP
1
2
ND ; N = VD = VP = LiO
3 4 5 6 7
;
NP
=
qu
;
NO
,
NO
=
vu = rrr = =
8 9 10 11 12
17
UO
;
'40
:
\ Y
=
16
NP
r
17
5
19
.
Nu NP
Y O : '40
,do
NO L(P
.. 2
= YP =
Ye =
v; NU
*iP 40
7
=
14 15
1R 70 71
77 73 24
7r, 26 27
%.
2R 2s
=
30
31
U I
= = -
YO
=
34 35
NP
YP
= UP = ND = UD
NO
;
S' N %,
-
.1R
= = =
'40
=
VP
= = No = LID = YO
NO
37
3, 36 37 38 39 40 41 42 43 44 45
16 47
Vo
=
"R
yo
;
4Q
'40
:
50
No
-
U'
7
NP
=
V?
hi"
= =
:Z 53 54 55
YO
=
56
YO
.
YP
:
= NP = NO YO
=
','o i LIP =
No = YP = Yo = ND = NV ; NO = hi0 ; NO
-
.,D
= =
VP
:
VO
YO = =
Urn ND
=
ND
=
= LID = YO
ii,.,..
57 58 59 60 61
62 61 64 55 66 67 be 69
70 71
'2 73 74 75 Ih 77 78 79 80
K-MR
!X,
MU 01 R A D I A N S
CONOI'IOH
YO
AXIAL FORCE, K N
I
106 STRUCTURAL STEEL DESIGN
present maximum horizontal deflections are: X, = 101.7 mm X,, = 105.1 mm X6 = 106.4 mm X,, = 90.4 mm XI, = 106.4 mm X,, = 88.5 mm The difference X,, - X,, = 90.4 - 85.8 = 4.6 mrn
gles with long legs back to back for both top and bottom chords, find the an
tension and compression members and connections.
PROBLEMS The following problems are of two 'types. The first four problems are for the student to obtain familiarity with using the computer program given in the Appendix to solve structural problems (other computer programs, such as STRESS or STRUDL, may also be used, but the program in the Appendix is likely to be considerably faster to run). Use E = 200 000 MPa or 29 000 ksi. Obtain the solution in either fps or SI units, as assigned by the instructor. One or more of the last three problems should be used for the design projects to be carried along with other later chapter problems. 2-1 Given the following beam. Obtain a W section that limits the deflection at point A to 1 in or 25 mm. Note that all NPE(2) can be made NPPI. Answer: MA = 171.6 ft kips or 236.2 kN . m.
w=
4 kipsift or 60 k N / m
*
4 p a n t i s '0 20' = 80' ( 4 paneis '.i 7 rn = 78 rn)
sok 2.4
200 k N
Wind
=
NBC value (take height as 9 rn to points B and E for wind in eith
Vertical load from a hoist at point D = 20 kips or 90 kN Bay spacing = 25 ft or 8 m Try to limit iterations to five.
Figwe PZ1 2-2 Given the following academic beam/frame. Find a W section that limits the'deflection at point A to 0.5 in or 12.5 rnm.
Note. Joints B. E. and Fare "iixed." Joints .-I. C. a n d D arc "pinned."
110
STRUCTURAL STEEL DESIGN
ELEMENTS OF FRAME, TRUSS,
BRlDGE DESX
dunensions for your analysis: Dunensions Part a' b c d e
30 rt 3.0 29 t 0.5 13.75 r 0.5 7.5 r 0.5 20.0 t 2.0 3.25 6.0 -C 1.5
f g h
9.1 r 1.0 8.8 r 0.20 4.20 r 0.15 2.3 r 0.2 6.1 r 0.6 1 2.0 +- 0.5
for the analysls program for the wheel loads at selected pos~tlonsalong the span. It is sugg use wheel d~stance~ncrementsof 5 f t or 1.8 m.
Make t h s dunension consistent w t h d~mensione. Notes
'
-
-
1. The knee brace is plnned to the column, but the column is contmuous to the truss. 2. The lateral crane lmpact load wdl be applied to the column at the attachment of the runway guder to the column, as shown m the figure. 3. An optional vehcal member is used to reduce the K L / r of the bottom chord. If you use h s member, it should be pinned to the bottom chord, but the bottom chord is continuous across this connecuon (thus wll have bendng). 4. Sketch where you would place w n d bracmg, but do not design. 2-7 Given the hghway bndge truss shown in Fig. P2-7, make a computer analysis for the s truck loadmg ass~gnedby the mstructor. Use the followmg dimensions and mtial data: Dimensions/section,
fps: 25 r 3 ft
3.
Members Bottom chord (1,4, . . ,21,25) Top chord (6, 10, 15, 19, 23) Vertxxils (3, 7, . . . , 20, 24)
.
W8 x 48 2 C12 x 30 W8 X 40
SI: 7.5
+Im
ELASTIC, PLASTIC, AND BUCKLING BEHAVIO OF STRUCTURAL STEE
m-1The Chicago "Picasso". A massive sculpture using corrosion-resistant steel designed
NTRODUCTION
-.
9
.
.
.
,.
.
... ,,",,>, .... ,,,,..,,.,.. ,. * .
.
tendency of unsupported structural elements to buckle under
gned using equations that have been developed as a combination
l i e STRUCW
STEEL BESIGN
ELASTIC, PLASTIC, AND BUCKLING \
From Eqs. (c) and (b), el
=
h we obtaiq 60P3 = 60%
e, =
BEHAVIOR OF SlRUC?UIWL SEEL
S o ~ u n oWe ~ must apply a factored load of sufficient magnitude t
0.8 P3(48) - P3(60) 0.64(29 000) 1.0(29 000)
velop f,
which checks displacements = constant
=
Fy in the three bars. At this time, the bar forces are simply P I = A, Fy P2 = A2F, P3 = A3F, P I + P2 + P, = P,,,,,,, P,,= 36(0.64) + 36(0.75)
. :jlso, f r o b Eq. (c):
or simply P,
+ 36(1.0) = 84.04 kips
Since the actual load is only 30 kips, the load factor is P , / P : 86 04 L F = -= 2.87 30 Several comments are in order: ' ~ wsubstituting into Eq. (a), we obtain 0.8 P,
1. Plastic analysis is much simpler. 2. The rigid bar ABC will rotate under the applied load P, = 84.04 ki which was not the case in the elastic analysis. Why? 3. The elastic analysis (Example 3-1) indicates bar 2 yields first. Wh When F, = 36 ksi in bar 2, it carries no additional load but mer elongates with any additional load being carried by adjacent bars un they reach F, in turn. When Fy is reached in bar 2, the load in the bar i P2 = 0.75(36) = 27 kips.
+ 1.25P3 + P, = 30 kips P, = 30/3.05
= 9.836 kips
P2 = 1.25(9.836)= 12.295 &ips PI = 0.8(9.836) = Total :!ack substitution for e in each of Eq. ( c ) gives e r ~ a d e should r verify).
$ 0.02035 in (which the
By proportion from Example 3-1 the load at this point is
///
p = - 27'0 (30) = 65.88 kips
12.295
l 'ow let us reconsider Example 3-1 using "plastic analysis" in the following exnrple.
amble 3-2 For
the sketch shown in Big. E3-2 (same as Example 3-I),
n hat are the bar forces when all three bars have yielded?
Beam behavior based on a plastic analysis is similar to the bar problem. Consider the beam shown in Fig. 3-1. If we apply a bending moment to the ection, the moment-rotation (M-+) curve is linear? to !tfy. From the point at ch the most stressed beam fiber is at F, (producing the yield moment itl,) to point at which all of the beam fibers are at F, (either tension or compression epending on which side of the neutral axis we are inspecting) and producing plastic moment M,, the curve is nonlinear. When iCfp is reached, the beam ply rotates at this point, with no further increase in moment capacity (or tress) and we say that a plastic "hinge" has formed. There is some small itional increase in moment capacity when some of the beam fibers most ant from the neutral axis reach strains into the strain-hardening region. This ect depends on beam cross-sectional geometry of the flanges and web and, of he beam span and boundary conditions.. If the beam is loaded with a greater than My (but not M,) and unloaded, the curve branch BE is with a permanent amount of residual beam rotation OE.
t
Within material homogeneity and rolling tolerances, as well as practical measuring limitations.
120
ELASTIC, PLASTIC, kWD BUCKLING BEHAVIOR OF STRUCKJR.U
STRUCTURAL STEEL DESIGN
The plastic section modulus is obtained as the statical moment about the neutral axis, which div~desthe area equally. Note that necessary in order to satisfy statlcs on the section of CF, = 0.
proportions are such that the section can become fullv ~ l a s t i cbefore the nnce strain hardening (i.e., .depth/web thickness and flange width/flange' thicknes not too large). moment in t W e ' will now investigate - .- --- --- ---- rnnrent --*--vy. - in detail the Plastic following several paragraphs. Referring to Fig. 3-1, the moment at initial yield i A
A..
.LLW
My = S,F, where S, is the section modulus, I / c . The moment of inertia I and the distanc from the neutral axis to the extreme fiber c are as in any mechanics-of-material. ~ ~----C F C P Fshnwn textbook. The plastic moment M...Y , bv ins~ectionof - - the ---- S---..".. nn t*. ~ e cross section in Fig. 3-1 with a fully plastic section, M = Mp and noting that the neutral axis at this point divides the area in two- narts avo3 --- -- with ---- rliqtnnre /17 t* nv u.vu centroid from neutral axis, is
..
- --
The value of A7 is called the plastic modulus, Z , so that we may rewrite the moment as 3 ,
Mp
--
= Lr,,
'lle,.ratia.of Z / S is termed the shape factor, f.
Example 3-3 What is the section modulus S,, plastic modulus Z , and th shape factor f for the rectangular shape shown in Fig. E3-3?
The shape factor is computed as - 2 8
'
..
* a
The plastic modulus and shape factor for a W shape can be computed manner similar to the rectangular shape of Example 3-3. Here convenient use made of the tables for T shapes, as illi~stratedin the following ex,mple.
Example 3-4 Compute the plastic section modulus and shape factor for W610 x 241.1 rolled shape. SOLUTION The value of A/2 is readily obtained from the WT table CVT30 X 120.5), since this T shape is made from splitting a W610 shape. The value in the table also locates the center of the area of the T but is with respect to,the flange. From Table V-18 of SSDD, obtain
I I ,
I'
I c-
E
0 0 xr I1 iY
+
X
From Table V-3, the depth of a W610 x 241.1 is 635 mm. The total area = 30.77 X lo-' m2.
&,
= d - 2& = 635 - 2(68.6) = 497.8
rnrn = 8.498 m
AJ = 15.39 x lo-' x 0.498 = 7.664 x Z =-
Figure E3-3
I
SOLUTIONThe elastic section modulus is computed us materials equations:
J
ll
S
=I - =bh3= . - bh2 c 12(h/2) 6
m3 2 m3 and The value given in' Table V-3 for 2, = 7.659 x discrepancy is due to the extra digits used by the computer in computing directly as opposed to rounding for Table V-18 and the use of 0.497 0.498 above. From Table V-3, the section modulus of a W6 10 x 241.1 is S,
=
6.78 x
m3
and the shape factor f can be directly computed as S =
200(0.4)~
6
= 5.333
x
m3
,
8
122 STRUCTLTRAL STEEL DESIGN
ELASTIC,
PLASTIC, AND BUCKLING
BEHAVIOR OF STRUCTURAL
portional to strain as defined b
-
Z bli2/6 /=I50
Z=
h( b/12 - b,/iZ 1
/ ' = L.SI1
( bh2 - -
bill,?
(bli'
b,/i:
Z = b/iZ/3 f = 2.00
esign method. The current elastic design procedures as found in the several esign specifications is based on the linear stress-strain response tb the elastic mit, but there is implicit recognition of the steel behavior beyond the elastic limit. Elastic design as commonly used places the limiting steel stress as the yield that used the limiting stress of F, in the elastic design safety factor F = 1. A,.safety factor of 1 is unacceptble, as it allows for no future changes in structural use/occupancy, or for material properties (flaws, under dimensions of sections, and minor mztallurgj-
,
value of F > 1 is required. Ideally, every element of a steel structure should have the same factor of fety. In practice, this is not the case. Flexural member response tends to be the ost reliable to predict and those members have a minimum value of F.
z = 4r3/3
z
/=I70 J
= + ( r i -- r j )
ri
32 --=ri + r j
z
= from tables
f = Z / S = 1.10 t o 1.18 Average = 1.1 4
Modal value = 1.12
results in a structure collapse, rightfully have the largest values of F. The basic safety factor for the steel members in building construction is obtained as follows. Let the computed strength of the member be defined as the d be defined as R. The safety factor can
Figure 3-3 Plastic section modulus and shape factor for selected cross sections.
computed service load The shape factor is a measure of the increase in plastic moment capacity over the value of yield moment My, and we have M, = ZF,
= f(SF,)
For F = 1, this new ratio becomes The plastic section modulus and resulting shape factor for several cross .sectisn.s.isshown in Fig. 3-3.
Now if we take A S / S
=
AR/R
=
0.25 and noting that S / R
=
F, we obtain
3-3 SAFETY FACTORS IN ELASTIC AND PLASTIC DESIGN Solving for F, we obtain I
-
Steel design may be based on the yield strength (termed plastic, or limit states, design) or on an elastic design. In limit states design the analysis proceeds based on assumed plastic behavior-the member continues to strain from E, to e,, (see Fig. 1-3b) with no increase in load. Elastic design allows for this unique behavior of steel but limits the working stresses to the elastic region of the
This value of F is taken as the basic value of F for use in the elastic design ethod for structural steel for structures other than bridges for highways and railroads. Railroad and highway bridges are generally subjected to a more
ELASTIC, PLASTIC, AND BUCKLING BEHAVIOR
124 STRUCTURAL STEEL DESIGN
overloading, so the un hostile environment and a greater possibility factor A S = AR is taken as 0.29, which gives F = 1/0.55 = 1.82. The value of F = 1/0.6 ,is modified to 1/0.66 when the cross-section geometry is such that a plastic hinge can fully develop at the most high stressed point. Rolled shapes whose section geometry is such that the plast hinge can fully develop so that the basic value of F = 1/0.66 can be used are termed compact shapes. The geometry criteria for these shapes will be considered in Chap. 4. For A-36 steel the basic allowable stresses using the previously defined safety factor becomes Fa
=
s obtained based on section properties from these ultimate loa red to the yield stress Fy and adjustments made until computedf;,
EFLECI-IONS uld a plastic hinge develop at a point along a beam or column, a very ection would result. This deflection would, however, have no meaning ould result in a structure collapse. No structure is
0.6< = 0.6(36) = 21.6 ksi
(the AISC specification allows use of 22 ksi for this case) Fa = 0.6(250) = 150 MPa (in SI units)
,
OF STRUCTURAL
r v
For the AREA and AASHTO specifications, we have
under actual working load conditions will be elastic values. Since plastic deflections result in a structure co ignificance, so for this reason deflections stic analytical procedures for both general metho
Fa = 0.55(36) = 19.8 ksi (these specifications allow use of 20 ksi for this single case) Fa = 0.55(250)
=
137.5 MPa
We note that the optional rounding of 21.6 ksi to 22 ksi (as should be done using desk calculations) can create a slight computational discrepancy if a digital computer is used, unless a rounding procedure for this grade of steel is set in the computer program. The author suggests that the rounding to 22 ksi be done, since it is allowed and A-36 steel is the most common grade used. It is not recommended (at this time) to round 137.5 MPa to 140 MPa, since some rounding up has already taken place to obtain 250 MPa from 36 ksi.
LENGTH OF PLASTIC HINGE of the plastic hinge for a W shape ca d-end beam (refer to Fig. 3-4) as foll e horizontal at midspan, the offset to t yo = IM, - ( - 1tfp)
=
2hlP a
L
The equation of a parabola with the origin of s as shown is
3-3.1 Factor of Safety for Current Plastic Design The factor of safety used (called load factor) for plastic design according to t present AISC procedure is obtained by using the average shape factor f defin in Sec. 3-2 and illustrated in the computations for a typical rectangular shape Example 3-3. In elastic design with compact sections, the value of F = 1/0.66 = 1.52. The value of plastic moment Mp = fMy, where the shape factor = 1.12 as the modal value for all the rolled W shapes. Now using the same working stress fb for either design method, we have MY- = -= 1.52s
MP F,S
l l l i l l l l l l l l l l l l l l l l
fMY
F,S
Canceling the section modulus S, we obtain F, = 1,.52f = 1.52(1.12) = 1.70
(as used in Part 2 of the AISC specifi
This value of F is used in plastic (or limit states) design as a load factor which the working or design loads are multiplied to obtain the "ultimate'
beam.
ELASTIC, PLASTIC,AND BUCKLING BEHAVIOR OF STRUCIIIRAL NOWwe need to find x such that M, have
=
My = S,F,. Since y = Mp -
Y = F,(Z, - S,) = F,(fS, - S,)
Substituting for y and Mp in Eq. (3-l), we obtain 1
8(fF, s,)($)I = f ~ ( 1- 7)s, "Caricdi'i and F, and rearranging, we obtain for the hinge length defined by the length x (which is half the rnidspan hinge length), Figure 3-5 Plastic hinge formation for several beams loaded as shown. (a) Simply s u p p hinge for failure. (b) Propped cantilever two hinges for failure. ( c ) Fixed-cnd, three hinges fo
When f
=,
1.12, the length 2 x of the plastic hinge in the center of the b
beams with concentrated loads, and load. combinations, can be obtained in when the collapse mechanism has been determined. These two
3-6 ELASTIC VERSUS PLASTIC DESIGN There are several advantages in using plastic design fo small one- or two-story structures: s to produce a collapse mechanism fop 1. The rapidity of obtaining design moments.
sary for the beams in the following e Offsetting these advantages are several disadvantages: 1. Widespread availability of computer programs, which can rapidly sol simple and complicated structures using elastic methods. 2. Most designers have more familiarity with elastic design methods. 3. Difficulty of obtaining the collapse mode 'if the structure is reasonably complicated. 4. There is little savings in column design (and sometimes for other members depending on the fabrication methods). 5. Difficult to design for fatigue.
I
In plastic design it is necessary to determine the location of the plastic hinges that form at locations where M, develops. It is necessary that enough
Example 3-5 Derive an expression for hfp for the fixed-end beam s and select a W shape with adequate Z for P,,, = 120 kN. P, = 170 k N
SIRUCIWRAL STEEL DESIGN
SOLUTION Three hinges are necessary to produce a collapse mechanism Note that the beam is indeterminate to second degree (no horizontal load so Z F, = 0 has no significance). From symmetry the three hinges necess to form the mechanism must be as shown in Fin. - E3-5. The effect of fixed-end moments is to reduce the simple beam moment diagram as shown byi,the dashed lines. For hinges to form it is necessary that the moment val'ue be Mp, and it is evident that Mp will form first at the fixed-end locations, since the elastic moment is largest at those points. Further increases in moment increases the elastic moment into the plastic range. It is also evident that the only other possible location for M, is under the concentrated load, since the moment at this point will be the next location where the elastic moment is large enough that increases in Pw to P, will orce the moment into the plastic range. When this hinge forms, the . !ructure cbllapses (theoretically) and no further increase in load is possible. With this consideration, we have (again referring to Fig. E3-5)
The total Z, required is Z, = 0.612
+ 0.0058 = 0.6178 < 0.6566 x
m3 furnished
O.K.
Use a W360 x 38.7 beam. It is still necessary to check bracing requirements. For an elastic design using Fa = 0.6Fy (commonly used allowable stress), the beam would be M = - =PL 8- = 9 0120(6) k 8N . m The required section modulus S is
Use a W410 x 38.7 section. from which O.K. Sx(reqd, = 0.60 + 0.0076 = 0.6076 < 0.629 X l o u 3 furnished By coincidence we have found a section that has exactly the same mass per meter; in most cases sections obtained by plastic design methodswe somewhat lighter than those obtained using elastic design, at least when the /// beam is indeterminate.
!-'or A-36 steel, Fy = 250 MPa.
Pu = Pw X load factor
= 120(1.7) = 204 kN
,,q
The required plastic section modulus is
From Table VI-2 of SSDD, select
Zx = 0.6566 x m3 The beam must cany its own weight, so for self-weight the simple beam moment is M = w ~ ~ /For 8 .plastic analysis use the same concept as for the concentrated load, which gives W360 X 38.7
For the W360 x 38.7, the weight/m
=
0.38 kN/m (Table V-3).
Example 3-6 Given the propped cantilever beam shown in Fig. U-6, it is required to obtain a general expression for itl, and design the beam if ww = 5 kN/m and F, = 250 MPa. Also derive a general expression for the location of M, in the span.
LJU STRUCTURAL STEEL DESIGN
ELASTIC,
PLASTIC, AND BUCKLING
BEHAVIOR OF STRUCTLRU STEEL
SOLUTIONThe collapse mechanism will consist of two hinges located as shown. From statics the moment Mp at B is a maxjmum. From mechanics of materials, V = 0, where M = maximum. This gives Rc = wux Also, using statics Z M B = 0 for segment BC, which gives wux2
2- R,x
0 Taking moments of beam segment AC about A ( Z M , = 0) gives M~ +
Mp
=
+ R,L-;=Ow L2
2 Substituting Eq. ( a ) for Rc in Eq. (c), then substituting Eq. (c) into Eq. (b) for Mp, we obtain
x 2 + 2 x L - L* = 0 Solve this by completing the square, we obtain
Figure E3-7
Right span:
x = 1.414L - L = 0.414L Now a general expression for Mp can be obtained from Eqs. ( a ) and (b): Mp
=
0.08579~~~~
Using Eq. (3-4) with the given beam length and loading, the value of Mp is
M, = 0.08579(45
X
1 . 7 ) ( 6 )= ~ 236.26 kN . m
z,=--236'26 - 0.9451 x 250
The maximum value of M, from either span is used for design (since beam runs across both spans using a constant section).
Mp = 15(1.7)(18) = 459 f t . kips low3m3 M~=20(1.7)(15)=510ft.kips
usethis
beam weight is
AZ,
=
0.5 1 45
-(0.945 1 )
=
Zx.,,,d, = 0.9451 + 0.0107
0.0107 =
0.9558
!< 1.0861 x
m3
O.K.
bracing requirements).
From Table 11-2, select a W24 x 68 w ~ t hZ,= 176.4 in3. Check tde beam weight effect as approx~mately (if a borderline case is found, one may be justified in the additional work for an exact analys~s):
AM' w, L~ AM; + 2 2 = -8-
Example 3-7 Given the two-span continuous beam shown in Fig. E3-7, select an economical W section using plastic design and A-36 steel. I
.
Sofiyno~Two hinges are necessary to collapse at least one span. The values of Mp to accomplish this are: Left Lett span:
and the required AZ, is
AZ, = 170.0 Total Zx(reqd)
=
2.17(12) 36
-= 0.72 in3
+ 0.72 = 170.72 < 176.4 furnished
Use a W24 x 68 section with Z, = 176.4 in3.
O.K.
?32 STRUCTURAL STEEL DESIGN
3-7 LOAD RESISTANCE FACTOR DESIGN
occupancy; other values are also used (e.g., 1.5 for maximum sn
Load resistance factor design (or LRFD) is a recent proposal which undergoing some development as an alternative approach to the cumently used ~ ~ ~ fully accepted by elastic design method. It is expected that L R F D W become the AISC within the useful life of this textbook. This forecast is based on the facts that this procedure (at least the essentials, called limit states design) is .,!;eady accepted in Canada and several other countries outside the United States. The current AASHTO bridge specifications (12th edition) provide an .,hernative design method in steel termed load factor design for simple and ;3~tin~,o.!!s*beams and girders of moderate length which use compact sections. .--!li the LRFD designs are very similar to each other and to the strengt ,jrocedure used in reinforced concrete design. In LRFD, as in rei concrete, +,factors are used to reflect uncertainties in the material (in t h s t-h% specified steel strength, F,). These factors are under current study with the .&rent suggestions as in Table 3- 1. LRFD uses an equation of the general form
case
+R = $(FdD
+R 2 l . l ( l . 1 0
+ 1 . 4 1 + 1.6PVm,,)
+R
l.l(l.10
+
l.l(l.10
+ 1.4L)
+R
>
1.5Sm,,)
The general objective with LRFD is to assess each item that influences design of a structure rather than "lumping" several effects together, as, example, simply adding the dead and live loads to obtain the composite lo Larger factors are used with those items that carry more uncertainty, such snow and wind loads (live-load factors of 1.5 and 1.6 versus the dead-load fa
+ FLL)
where $= analysis factor (also termed importance factor); value cu suggested, 1.1 Fd = uncertainty factor for dead load with a value of 1.1 suggeste FL = uncertainty factor for live load with a value of 1.4 sugges
2%ble 3-1 Current recommendations for
equations for several loadings including wind and snow mi&t re
LOCAL BUCKLING OF PLATES
factors
Stress condition
Suggesteda
Canada
AASHTO
~ & s i o nmembers Yielding (4) Fracture (F,)
0.88 0.74
0.90 0.90
1.O
sending Rolled sections and plate girders
0.86
0.90
1.o
0.90
1.o
1.0
Columnsb TJ 2 0.16 0.16 < 7 1.0 TJ > 1.0 :!hear Webs of beams and girders
0.86
0.90
1.o
Cvnnections BoltsC 'Welds
0.70- 1.00 0.80
0.90 0.90
-
0.86 0.90-0.25~~ 0.65
a1 stresses, perfectly plane, homogeneous, and isotropic that is subjecte o m compressive load along opposite edges. Under this stress the ~ l a t press uniformly until the buckling stress is reached, When the buc stress is reached, the plate will deflect in a single wave or a series of wav depending on the edge (boundary) conditions and length to width ( a / b ) ra with a resultant redistribution of the compressive stresses until, with the additio of load, the entire plate is buckled. From the theory of plates as proposed by several authorities,j the critic elastic buckling stress F,, is
" See Journal of Structural Division, ASCE ST9, September 1978 (contains eight Fapers on LRFD). 7 = ( K L / ~ ~ ) V F (~K/ =E length factor as given in Chap. 6). See Sec. 8-10.
Metal Structures (New York: McGraw-Hill Book Company). or Johnston, Guide ro Srabiliry Deri2n Criteria for Metal Structures, 3rd ed. (New York: John Wiley 8i Sons. Inc.).
, 7
136
STRUCTURAL STEEL DESIGN
where
ELASTIC, PLASTIC,
AND BUCKLING BEIMVIOR OF STRUCTURAL STEEL
F,,=
steel stress at the proportional limit (a value of F, 0.70 to 0.75FY may be used) F, = yield stress of steel Fcr= critical buckling stress of Eq. (3-6).
+,
rC
If we attempt to solve Eq. (3-6) for the critical buckling stress, s problems develop, particularly if A < 1. First, we must determine kc. Wh general expression for kc has been given, it is necessary to adjust this f rious boundary conditions that are possible. This has been done by several 7lthorities, but as a convenience the author has further combined the effect of ?/[12(1 - p2)] = 0.9038 to give values shown in Table 3-2 for k; [i.e., a(0.9038) = 3.6151. If the X term is less than 1, it is necessary to iterate to Fcr. B' is is illustrated as follows. Rewrite Eq. (3-6) in terms of k:, to obtain
:p 9
u7
h'[
i--- h ---I i =O;Y
k c = I 1 5 ,re111 FIdnze ,dnle >I.
it
,h.111=
k; = 0 63 l l ~ n g e \ k: = 4 5 for web
a
,
J,
$,
Figure 3-7 Compress~oncharactens~csfor rolled shapes shown. Note that 111s generafly nec to mvesbgate the cntical b / t ratlo, which may be as shown for a W shapc w t h a cover plate welded or bolted.
2
8
Fcr = X E ~ ; ( ; )
13
.
dew divide through by 2 Fcr A -~k:(i) Ye will temporanly hold this equation. Now using Eq. (3-7) for X 9 t h F,, = 0.755, we can, with some rearranging, obtain
Fcr
-=
0.1875
c2
4 - Fcr
Compression charactenstics (kc and wldth b in compress~on)of three rolled shapes are g~venin Fig. 3-7. The kc values shown have been found to agree reasonably with tests. Adjustments In kc are necessary because very few plates are free of imperfections and residual stresses If we use the value of k: = 0.63 shown in Flg 3-7 for the flange of a W section and a SF = 2.00 and Fc, = F,, we obtain
Since Fc, is on both sides of Eq. (3-9), we must solve for Fcr/X by trial. Once the value is obtained, this can be used in Eq. (3-8a) for the ratio of b / t , which is ~sually,the item of interest. From Eq. (3-8a) the b / t ratio is
Lt '9
For kc
=
= 11 3
=pC
65 say -
vz
Fcr / A 3.615, X = 1, and Fc, = 0.754, the limiting b / t ratio for A-36 steel is
",/@=Em t 0.75 x 36
= 62.3
The current AISC specification allows a b / t (uses b,/21/) ratio of 6 5 / 6 . Note also that if we consider the web, k: = 4.9, we obtaln 190 say -
I t is often useful in using Eq. (3-9) to set up a table of A vs. F, with values from FCr = 0.754 to F,. For A-36 steel, typical values are as follows: /
i;,,ksl /F
27., 28.0;
2ii 0 33.0 36
-
A
E,, ksl
1.00 0.922 0.741 0.407 0.0
29 M)O 26 738 21 489 11 803 0
say 11
6
which is also in AISC. Values of k, = k, are also shown in Table 3-2 for the cr~ticalstress to produce shear buckling. The critical buckling stress for shear can be derived in a similar manner to that for compression, with the substitution of an appropriate buckling coefficient k, to obta~n,from Eq. (3-6), F C ~=S
k,n2XE 12(1 - p 2 ) ( b / l ) 2
I
.
138 s d u m
I
ELASTIC, PLASTIC, AND BUCKLING
STEEL DESIGN
BEHAVIOR OF STRUCTURAL STEEL
139
C
I is usual to assume that the four plate edges are simply supported in shear and
i
t e sha'ar stress F,, = \/5. This value must be combined with the safety factor of 1/0.6 so that the design shear stress becomes . . ." C r7 C crs l y - 'y Fs = SFX* 1.67xfl 2.89 Mo~t~practlcal steel design problems consider either buckling in compression 6r b'uckling in shear. w h e r e both stresses act simultaneously, the reader should consult books such as those by Bleich, Johnston, and Timoshenko and Goodier cited in an earlier footnote. a
where all terms have been previously identified except be, which is shown in Fig. 3-8. For a very long, thin, simply supported plate, it appears that the theoretical value of k: is 3.615 for this equation. The use of post-buckling strength of plates is not often directly evaluated. It is used more often in a more indirect manner; for example, AISC allows use indirectly via Appendix C3, which states: "When the width-thickness ratio of a uniformly compressed stiffened element exceeds the applicable limit given in Sec. 1-9.2.2, a reduced effective width, b,, shall be used in computing the . . . ."
PROBLEMS
3-9 POST-BUCKLING STRENGTH OF PLATES Experimental evidence shows that a buckled plate does not result in immediate failure. Rather, there is a considerable strength reserve attributed to the effect of the adjacent plate material, which restrains the buckling and allows transfer of any post-buckling load increase to the unbuckled zones. This situation is idealized in Fig. 3-8, which illustrates the central buckled zone in the loaded width b. On either side are strips that confine the buckling and are loaded to a lesser effective load fe. The concept of effective width be is applied as the sum of the two strip widths on each side of the buckled zone. When the effective stress on these two edge strips (stress on width of be) reaches a value such that deformation is constant with no further increase in load, the full load capacity of the plate has been reached. The difference between the initial buckling load and this new value is the post-buckling strength of the plate. The stress fe may be evaluated using Eq. (3-8), to obtain
3.1 What is the allowable b,/2t, for any W section using a steel with F;. = 50 ksi? Answer: 9.2. 3.2 What is the allowable b,/2!, for any W section with F, = 250 MPa?
3-3 What is the plastic moment capacity (kN . m) of a W 9 2 0 x 200.9 section for Fy Answer: 2865.2 kN . m. 34 Verify 2, in Table V - 3 for a W 6 1 0 x 241.1 rolled shape.
=
345 MPa?
3-5 What is Z, for the geometrical shape shown in Fig. P3-5 if it is used for a beam? *
3-6 Select the lightest W section to satisfy F, = 36 ksi and plastic design.
bending for the span and loading shown in Fig. P3-6. Use
4 kipa't't
3-7 Select the lightest W section to sorisfy bending for the span and loading shown in Fig. P3-7. Use Fy = 250 MPa and plastic design. Answer: W 6 1 0 X 101.2. 340 k;\'
50 kN1m
Ngum 3-8 Effective width for post-buckling plate capacity.
+7 -in
8 IT]--1
FIgwp P3.7
3-8 Select the lightest W section for bending for the beam shown in Fig. P3-8. Use F, = 345 MPa and plastic design.
3-9 What is the uncertainty factor on both S and R to produce F,
reasonable alternate nonequal values to produce the same effect? Answer: S = R = $ .
-
0.5Fy? What are two
i
>
144 STRUCTURAL STEEL. DESIGN .
of our attention will.be' directed to cases where . ,
1. Has no end moments (simple beam), or 2. Has moments at the ends of each span (continuous or I .
Beam loadings will consist of both dead and live loads. in 'terms of whether a series of concentr roduced depends on the general framing pl rhe beam dead load is its own weight. Wher self-weight may be a significant part of the total beam load. where th small and/or the external loads are also sma small, and here the strength/weight ratio of any case, the section should always be chec loads, including the beam weight. Beams may be classified as: 1. Girders: main load-carrying members into which floor beams
as shown in Fig. 4-1. Chapter 10 consider for bridges. 2. Joists: members used to carry roofing and floors of buildings. 3. Linfels: beam members used to carry wall loads over wall openings. 3. Spandre/s:exterior beams at the floor level in building construction used to
compressive stresses in the masonry. 5. Stringers: members used in bridges parall slab and commonly frames into transverse ( 6. Floor beams: secondary members of a floor members in bridge construction into which stringers f 4- 14).
----
.-ii--ii--
#!fI
In most cases, particularly for maximizing economy, a loaded so that the bending is about the strong a section property tables in SSDD). Occasionally, the bending :he weak (Y- Y) axis and in some instances there is simultane both the X and Y axes. In nearly all of the appli bending, the load is considered to be applied th or S shapes. The shear center for these shapes ' .,ad position produces simple bending about ei When the load does not pass through the channels, angles, and some built-up sections, u a torsional moment is produced along with the taken into account to avoid overstressing the member. The design of beams requires an analytical iterative a n the shear and moment diagrams based on the se ms simple framed to columns. Wind is resisted in both directions using cross bracing at fou ( e ) Girder (left to right) and floor beam system in an office building. Cloxup k$s.*
(n
'framing of floor beams and joists.
( n ) Roof beams and spandrel for f l a t roof buildin?.
(hi
- .-.
"-...,.--
I...
U l L L b
ULOlUIY
Y
--------
-
inoment together with the loads can be used to back-compute the critical shear.
on mechanics of materials: For bending:
For extreme fiber stresses:
For shear:
N OF BEAMS BY THE ELASTIC METHOD The W shapes will usually be used for beams. On occasion, S shapes, M shapz channels are used, depending on location, mode of connection to
desired (see Fig. 4-2) c = distance from neutral axis to extreme fiber
I = moment of inertia of cross section S= I / c = section modulus of section (both I and S are tabulated in
section property tables such as Tables 1-4 and V-4 of SSDD) Q = statlcal moment of area above point where shear stress is being determined = A7 (refer also to Flg. 4-2) t = thickness of beam at point of shear stress investigation V= critical shear force from shear diagram or direct computations
e same time one with an adequate S.. ISC) have the sections commonly used r beams ranked in descending order of S and wth respect to the X-X (strong) weight, so that considering a group as defined by extra llne spacings, we have
The section is checked to ke sure that it is adequate to carry its own weight, and finally the working load deflestions are checked. Sometimes deflections
2. The hghtest section in the group is at the top and the heavlest is at the
1. The largest S In any group
is
at the top of that group
radang allows rapid detemnatlon of the most economical roIIed
.
"4.
.-
I
STRUCTLTRAL STEEL DESIGN
/DESIGN OF B
I
SOLUTIONDraw shear and moment diagrams as shown in Fig. E4-1 Next, obtain M,,, = 276 ft - kips. With a uniform loading on the beam, the top flange is in contact with something; thus we will assume full lateral support, so that
Chechng the beam weight, we obtain
ST,,, = 138 + 3.24 = 141.2 < 143 in3
Lb = 0 < LC rind use Fb = 0.665 = f x 36 = 24 ksi. Rearranging Eq. (4-l), the required section modulus with strodg axis bending is
We must find a section with a somewhat greater value of S so that the beam. , weight can be carried. From Table 11-1, select
furnished
Use a W14 x 90 beam. Example 4-3 Given the beam span and loading shown in Fig. E4-3 approximately the beam system of Example 4-1. Most of the SI will be approximately the same as the fps problems (but not conversion of units), so the reader may be asslsted in developing a feeling for the SI units and selection of section sizes. Use F, = 250 MPa.
<-
Checking the beam weight, the additional S required is
S,,,, = 138
+ 2.45 = 140.45 > 140
furnished
If we use this beam, the actual fb= 24(140.45)/140 = 24.077 ksi, or this beam will be overstressed 77 psi based on the given load conditions. We may: 1. Use engineering judgment that this small overstress is acceptable, or 2. Use the next lightest section, which in this case is a W24
x 68: Sx = 154.0 > 140.45
With Sx so much larger and beam weight the same as in the earlier section, it is not necessary to again check the beam weight. The reader should observe that from a deflection stand~ointthe W24 is the better selection. ///. Example 4-2 What size of beam should be used in Example 4-1 if the beam depth is limited to 16 in? Sreqd is approximately 145 in3, since the weight will have to be SOLUTION larger than 68 lb/ft of the most economical section selected in the example Always use the maximum depth possible when the beam is laterally sup :;ported. Therefore, by inspection of the W16's in Table 1-3, select ,
SOLUTION First, draw shear and moment diagrams. Second, obtain M,,, = 436 kN . m. Assuming full lateral support, w obtain Fb = 0.67 X 250 = 167 MPa The required section modulus is
,
Try
W16 X 89:
S = 155.0 in3 d = 16.75 in
W 14 x 90:
S = 143.0 in3 d = 14.02 in
W12 x 106: S
=
145.0 in3 d
=
12.89 in
> 16
N.G.
(next try)
Note that 167 MPa x lo3 = 167 x l d kPa, which is consistent with 436 kN . m. Check Table VI-I for Sx somewhat larger. Find:
154
STRUCTURAL STEEL DESIGN b
Check AS, for beam weight = 1.11 kN/m:
Col.
-r
?
S,,,
= S,
+ AS,
= 2.611
+ 0.0532 = 2.6642 < 2.8841
Col >,-
:!.,.
..
cur .,,.-,
..
I
O.K.
m3
Use a W610 x 113.1 beam.
major difference Continuous beams are designed similar to simple beams. when using AISC specifications is that if the section is compact and is not a " , . .-.. cantilever beam, the section may be designed on the basis of using either: 1. 0.9 X largest negative moment in span, or 2. Positive moment based on maximum positive moment from moment diagram 0.1 x average of the negative span moments,
+
+ +
chever is larger. These moments are based on gravity (D + L, D L S, etc., and not wind) load moments. Where the beam or girder is rigidly framed into a column, the design moment value for the column at this point can also be reduced 10 percent. This procedure is based on recognition of the method of plastic hinge formation and resultant transfer of moment from the neg+tive zone to the pos~tivezone until a hinge finally forms at that point. This procedure will now be illustrated. Example 4-4 Use the computer output of Example 2-2 and check and/or redesign members 22, 23, and 24 (refer to Fig. E4-4 for the gravity load moment (which inspection of Fig. E2-4d indicates are the critical values for design). Use F, = 250 MPa. *
+ M for design is + M = 1i1.3 + 57.86 +2 140.6 (0.1) = 121.2 < 126.5kN - rn
Since the largest moment is - 126.5 kN . m, this is used to select the sectionNote that the compression flange is part on the top and part on the bottom, with inflection points as shown on the moment diagram of Fig. E4-4. Tentatively assume that LC will be larger than the values for the bottom flange; if so, no lateral bracing will be required and we can use Fb = 0.665,
SOLUTION From the computer, output obtain beam end moment values of
+
.
- 57.86 + 140.64 - 86.86 86.88 - 140.64 +57.86 kN m These values are needed to complete the shear and moment diagrams shown in Fig. E4-4. Since we are using the same section for all spans, the largest design ( k m o m e n t is in span 1 or 3, by inspection. Note the moment in span 2 is all t n ~ a t i v e ,requiring possible lateral bracing on the bottom (compression) .?/hange if & = 2.3 > LC.We must check this possibility when the section is selected.
r,,
,The
From Table VI- I , select a W410 x 46.1 : S = 0.7735 x lo-' m3 LC = 1.78 m > 1.20 O.K. for no bracing L, = 2.16 m The point of inflection in either outside span produces 0.87 m of unsupported compression flange on bottom of beam. This will be deemed adequate compression flange bracing, since the remainder of the compression flange (on the top) is braced by the floor. We will have to use a
156 W U C l V R A L
1
STEEL DESIGN
,
midspan brace to the bottom flange for the center span, to produce
& = -2'3 - - 1.13 rn < 1.78
2 &Check the beam for self-weight (weight = 0.45 kN/m). By proportion ( s i ~ c ethe beam loading is uniform) t
S,,,, = S
DESIGN OF B W L I S FOR B M N G
"I
+ A S = 0.7575 + 0.0086 = 0.7611 x
l o v 3m3 < 0.7735 furnished
Use a W410 x 46.1 beam. The reader should verify that the method used to obtain A S is both correct and the most practical means available. gince a W410 x 59.5 beam was used in the initial computer analysis, it appears that the problem will have to be reprogrammed after the column design/revision has been made '.i?a later chapter. ///
4-4 WEB BUCKLING AND CRIPPLING Web buckling is an out-of-plane web distortion resulting from a combination of iarge d/tw ratio and bending stress. The unbraced length of compression flange ,nay also contribute to web buckling. Web buckling is controlled by limiting dtfier.the.d/tw ratio or the stress that can be used for the given d/tw ratio. This 1s allowed for in the several specifications. Web buckling is illustrated in Fig. 4-5b. Web crippling can occur if the web in-plane compressive stresses are 5ufficiently large. This can occur if reaction distances or load-bearing plates used to deliver column loads to the beam flange are too narrow. Web crippling can also occur if a uniform load on the flange is too large for the web thickness. Web crippling control will be obtained by determining the required reaction
distance or column base plate width in the following way. The needed reactio distance is obtained by considering an area in web compression defined by the reaction length + an additional distance using a 1 : 1 (45") slope through the k distance of the section. The section property tables tabulate k for the several rolled sections. The k distance is measured from the outer flange face to the top of the fillet transitioning',the web-to-flange interface. At this location the resulting web area in compression is nearly (if not exactly) a minimum. At a reaction the area in web compression is A, = (A' + k ) t , The allowable stress at this location is taken by AISC (see SSDD Sec. 1-10.10.1 to be F, At a reaction with j
=
=
0.75Fy
R I A , , we obtain
At a concentrated load in the span, the distance k can develop on both sides of the load as illustrated in Fig. 4-6. For thls condition, we obta~n
where
N = reaction width; a basic value of 3 f in or 89 mm (width of standard brick) is often assumed; iV = width of column or load-deliverin element for interior loads R =.reaction or other concentrated load, kips or kN
7
Figure 4-5 Web failures to avoid in design. (a) Web crippling. (6) Web buckling. Figure 4-6 Bearing length for concentrated loads on beams accord~ngto AISC spcnf~cahons.
DESIGN OF BE4bf.5 FOR B
-5 SHEAR CRITERZA The shear stress distribution across any section subjected to bending c computed using the equation presented earlier:
When large uniform loads are carried through the flange to the we be necessary to check the compression stress& and limit the valuedto
&
VQ
5 0.75Fy
f, = It
Example 4-5 What is the allowable reaction for a W16 x 40 using the basic value of N = 3; in, with A-36 steel? What column load can be transmitted using a W 8 X 31?
A plot of shear stress using this equation is illustrated in Fig. 4-7. We note that the average shear stress based on
v
f" = dlw
SOLUTIONFrom Table 1-3, obtain for a W16 x 40: k t, mr.
= =
1.03 in 0.305 in
R = (N
+ k)tW(0.75<) = (3.5 + 1.03)(0.305)(0.75 X 36) = 37.3 kips.
in = N.
differs somewhat from the maximum value shown in Fig. 4-7 (in this case about 23 percent) but is considerably easier to compute. AISC allows use of Eq. (4.T) for either rolled or fabricated (plate girders) sections. The USHTO and AREA specifications simply allow computation of f, based on the area of the "gross" section. This can be interpreted as in the AISC specifications (i.e., f, = V/dt,). The allowable shear stress F, for rolled sections is computed as:
+ 2k)t(0.75F,) = [8.W + 2(1.03)](0.305)(0.75 X 36) = 82.8 kips
P = (N
F, = 0.405
~ x a m ~ 4-6 l e What is the allowable reactidn for a W460 x 74.4 .section using the basic value of N = 89 mm and F, = 345 MPa? What column load can be transmitted using a W200 x 46.1?
F,
=
(AISC)
(AASHTO)
0.335
F, = 0.35Fy
(AREA)
Shear stresses seldom govern in building construction unless the section both very short and heavily loaded, as illustrated in the following example-
k = 27.8 mrn tw = 9.0 mm 50
Note that MPa X mm2 x = kN (low3not shown). The column lo (assuming base plate same size as column depth) is: for a ~ 2 0 dx 46.1, the depth, d = N = 203 mm.
P = [203 + 2(27.8)](0.009)(0.75 X 345) = 602.2 kN
\ X: 9 ~
C114h0 X 1 0 5 . 7
///
Bridge reactions are generally supplied by special fabricated bearings, and -'~A;4'SHfO-~pecifications require that web stiffeners will be used when the web shear at the bearing is f, > 0.75F0. Column loads are usually not carried by
100 \I PJ
1 ' 1 1 ' 1 ' 1
1
1
~
~
I:,< T r ~ b i ero gcr T. 1 i=
107 4
Y rzc
\!I'd
-1.7
/
71.6
4-7 Theoretical and average shear stress distribution on section for conditions shown-
160
STRUCTURAL STEEL DESIGN
4-7 What is the span length and load/ft for a beam of A shown Fig. E4-7 such that either F, or Fb will control? m a t length required for the reaction? II
4-6 STRONG- VERSUS WEAK-AXIS BENDING
k~p\/ft
. -- ,L-
Figure E4-7
SOLUTION Data for a W24 x 94: dz24.31in tw = 0.515 in
Depth limitations or other considerations may require onentation so bending is produced about the Y-Y m s . The allowable bending stress in AISC specification for compact rolled sections bent about the Y- Y axis and solid round, square, or rectangular bars may be taken as
k=1.53ln S,
=
222.0 in3 where B= 0.005 for fps units = 0.0019 for SI units
wL f,=-- 0.44 2twd W L ~
f -8Sx = 0.665 From Eq. (a),
0.4(36)(2)(24.31)(0.515) L Substltutlng this w into Eq. (b), we obtain =
W
360.6 L /
=-
4-7 DEFLECTIONS
-360'6L - 0.66(36) L
=
8% 24(8) (222) 360.6 x 12
=
Back substltutlon gives w = 360.6/9.85 tain fb =
=
= 2(24.3 1)(0.515)
=
(N .
1
indows to fail to function properly. Cracks can be produced in
*
36.61 hps/ft. Checking, we ob-
14.4 ksi
The reaction length N based on R 0.75Fy = 27 ksi is 1
Deflect~onestimates under working load are often required to ensure that floo
O.K.
36.61(9.85)2(12) = 24.0 ka 8(222.0) 36.61(9.85)
fv
9.85 ft
N = - -R
about both the X and Y axis takes place.
0.K.
= WL/~=
180.4 kips and F, =
+ k)twF, = R
a failure of the roof drainage system by pluggng from accumulat flections for simple beams can be computed using superposition of
k
t w Fa
-
180.3 - 1.53 = 11.44 in 0.515(27)
can be computed using the general differentlal equation EIylV = - w
L STEEL DESION
For L and
=.
6.5 m, we obtain C, = - 173.98 (computer output = 0.0043 rad
It appears that Ax occurs at + M = M,,, at x = 0.87 + 2.37 = 3.24 m from the left end. For this value of x in the preceding equation we compute
=
12.6 mm (approximately
in)
//
-.-8 BIAXIAL BENDING AND BENDING ON UNSYMMETRICAL SECTIONS
use superposition to flnd the stresses at cntical locations, using
.,
r! .
and for shear
Equation (4-9) is commonly used for spandrel beams and roof purli where there are both vertical and horizontal force components. When the load is applied to the top flange of the member (common for purlins on sloping roofs) and is separated into components perpend~cularand pafallel to the X and Y axes, the force parallel to the Y axis does not pass through the origin of This is a very complex stress state that involves biaxial bending and torsion. approximate solution is obtained as
I6 1.
:
The preceding sections have considered bending about either the X or the Y axi of W, M, or S shapes. In all these cases the moment produced by the load zpplied perpendicular to the X or Y axis with a line of action through the o of axes. This produced the type of bending stress that can be computed usi Eq. (4-1). The W, M, and S sections are symmetrical with respect to ,lane containing the X and Y axes and thus produce principal axes. xres are such that
that is, using one-half the section modulus of the section for the tangential bending component and the (+) sign used to obtain the stress. An alternative form of Eq. (4-9) is widely used for biaxial bending and similar form for combining bending and axial stresses. To obtain this form, setf, = Fb and then divide both sides of the equation by the allowable stress (noting that this is done even if Fbx# Fb), delete the (-), and reorder to ob 1%
FbXSX
+ -=
MY
1.0
F b s,
When this is done, a section with any combination of moments and s modulus producing 1.0, or less, is satisfactory; one should try to o b t a u v as nearly 1.0 as possible. Any value larger than 1.0 represents an overstress. The latter equation for Ixy is a product of the inertia term produced when at least one of the X and Y axes is not an axis of symmetry. Principal axes a produced as mutually perpendicular axes through the centroid of area such th the moments of inertia are a maximum with r e s ~ e c tto one of the axes and minimum with respect to the other. The principal axes are axes of symmetry symmetrical sections but can be found for unsymmetrical sections. If product of inertia Ixy is not zero, the axes of interest are not principal axes.
4-8.1 Symmetrical Sections with Biaxial Bending ,the X and Y axes are principal axes but th to the principal axes (but the resultant ikin ipf axes), biaxial bending is produced. For this .PQ '" * cessary to determine the components of load perpend I
1
Example 4-10 Deslgn a roof purlin using a channel section for the si sheds and for the most cntical loadlng cond~tionof Example 2-5. Space the purlins 6 ft horizontally as shown in Fig. E4-10 and use sag rods at the midbay spacing, giving an unsupported length of 12.5 ft for bendin moments about the Y axis of the purlin. SOLUTION Additional data from Example 2-5: Dead load Live load
= 20.0
=
snow load
=
25 cos 14.04
=
25 cos B re
= 24.25
'Total .
.?
Vertical load/ft
s
psi
9
= 44.25
psi psf
= 6.1 8(0.04425) = 0.273
kip/ft
>? .!&
15
DESIGN OF B M M S FOR BENmING
STRUCTURAL STEEL DESIGN
1
P
i. {
values, we obtain
-*hk tJ
I
fb = - 2046x - 1220y
(if set to 0, gives location of neutqal
t point A , x = - 0.0419, y = + 0.1357 m. fb = - 2046( - 0.0419) - 1220(0.1357) = - 79.8
b t point B, x = - 0.0419, y fb
MPa (compression)
0.0673. = 167.8 MPa (
veneer to crack if it actually occurred. Eng~neeringjudgment would h be applied to decide if this angle is satisfactory with a deflection that coul be as large as 14 mm but has a good probability of being much less than this.
4-9 SHEAR CENTER OF OPEN SECTIONS
= -
+ = tension)
The shear center locates the point with respect to a cross section to apply flexural load so that no twisting (or torsion) occurs when shear stresses due t bending act on the plane through the point. Thus. if the loading passes through the shear center, the section may be analyzed for simple bending and shear using Eqs. (4-9) and (4-9a). I f the beam loading does not pass through the she center, a torsion moment is developed that produces torsional s stresses of Ve' t
where Ve'= shear and shear eccentricity with respect to the shear center t = thickness of element where shear stress is desired J = torsional constant of section, for a thin rectangle bt J= b/t 4 10 (section webs and some flanges) 3
At point C, x
=
0.1 10, y fb =
=
- 0.0673.
- 142.9 MPa ( -
=
compression)
If we use fb = M c / I : At point A :
14'21(0'1357) (lo3) = 57.3 MPa (compression) 33.63 At points B and C: fb =
14'21(0.0673) (lo3) = 28.4 MPa (tension) 33.63 These values are considerably different'from those computed using metrical bending. Check the deflections approximately using Table IV-5 for a supported beam: fb =
bt J = - - 0.21t4 b/r 4 (stubby flanges as for channeis) 3 The computation of the shear center is complicated for all but the sim shapes. Fortunately, most sections have the shear center at a convenient Iocation (see Fig. 4-10), for example:
1. If a section contains an axis of symmetry, the shear center is on the axis. 2. From (1) it follows that the shear center of all sections with symmetry abo both axes is on the intersection of the two axes (all W, M, and S shapes). 3. For all sections consisting of two intersecting plate elements (angles, tee etc.), the shear center is at the plate intersection. The shear center E, (using symbols as in Table 1-6) for channels is rea derived to be bl E, = 2b,$ + h1t,/3 b, = b, - t,/2 where h'= d - t/ = average section depth 9, t,= flange and web thickness, respectively
rr
Using these values for a C10 x 30 with d = 10.00 in. b, = 3.033 in, t f = 0.436 in, and t, = 0.673 in, we obtain E, = 0.705 in, as in AISC and Table 1-6 This deflection is computed at 14.2 mm (over
f in), which could cause the
/$
1':l STRUCTURAL STEEL DESIGN
DESIGN OF BEAAIS FOR BENDCjG
At midspan: L
x=--
f,
Shear enter l o ~ a t ~ o n
E,
- 4-10 Shear center location for several rolled shapes.
=
v = o
82 20.7
-= 3.96 ksi << 0.6FV(22ksi)
At x ,= L / 4
=
3.13 ft:
M,
=
w(L - x2) - 0.35(12.5 - 3.13')(12) 2 2
=
j.7 in kips
Qe reader should note that there are several equations used in various
dp,;g-~ tables which give Eo values for channels slightly different from those of
d ,,!-14), depending on the assumptions made for shear flow in the flanges. The computation for shear center of more complicated shapes is beyond the s c c ~ ~ofe this text, and the reader is referred to any text on advanced mechanics of ?,~terials. 1' I-.: bending stress for conditions where torsion is developed by the applied lo22 * ,t passing through the shear center is approximately
whe .e M c / I = usual bending stress computation V = shear at a distance x along beam from origin by= width of flange being stressed x = distance from origin of axis to V and where fb is desired If= moment of "inertia of flange being stressed; approximately Iy/2 for a channel
Example 4-12 What is the approximate bending and torsion stress in a C10 X 30 channel loaded as shown in Fig. E4-12? .,", 4
"
_-. ,,
= 0.3
+ 31.6 = 31.9 ksi >> 0.65,
Note that if the channel beam were designed without considering torsio and shear center location, the apparent bending stress at rnidspan indicates that the section is considerably overdesigned. When considering the sh center effects, the section is not adequate unless it can be assumed that analysis is too approximate and that the end connections are such as to restrain rotation so that the larger bending stresses do not develop. The author would assume that the section is considerably underdesigned and use a different section, as there is too much design risk involved for the small amount of savings obtainable from using a lighter section. The torsion shear stress can be evaluated as Ve' t,
L=,
J = contribution of web 4 two flanges
A
U
B r ~ i kv e n e e r
=
I+-
Figure E4-12
I 2 4-5'
SOLUTIONFrom Table 1-6, the section properties of a C10 X 30 are: S, = 20.70 in3 I, = 3.94
= 21,
Eo = 0.705 in b, = 3.033 in
0.927
+ 0.152 = 1 08 1n4(vs
1 22 In SSDD)
The discrepancy between J as computed above and In the SSDD tables is due to using a more refined computatlon in the tables, which accounts for the increased stiffness at the flange-to-web intersect~on.We w11 use the table value of J = 1.22. e' = Eo + 7 = 0.75 1 + 0.649 = 1.40 in f, = 1'092(1'40)(0.436) 1.22
=
0.546 ksi which appears satisfactory
///
174 STRUCTURAL STEEL DESIGN .
4-10 DESIGN OF LATERALLY UNSUPPORTED B
DESIGN OF B E A M FOR BENDING
= shear modulus =
1
= I 1 150ksifory=0.3
I,.;: moment of inertia about the Y axis
s,= section modulus about the X axis L= unsupported length floor or roof. Cases where the floor system consists of metal decking placed onto the bea 5 ay not provide adequate lateral flange support if the resistance is $$4 ~olely4 . f s i c t i . n . In these situations, carefuily placed welds to take, say; 2 to . 3 ~erc$$., ::%he flange force (0.02 to 0.03A, f,) will provide adequate lateral bracing."'&>,v6ry large lateral restraint is not required, as evidenced by the . 'i applicatidrrq~only hand force to restrain reasonably sized beams against lateral f buckling in.laboratory tests. The lateral bending and warping of laterally unsupported beams was illustrated.,in Figs. 4-2 and 4-3. There are several situations, particularly involving vertical flexural members and beams used to c a m column loads across large '"Pen area4 crane runway girders, and continuous beams in large spans, where the compression flange is not in contact with decking and others, where it is not practical to provide lateral bracing except at the ends and/or at only a few interior span points. For conditions where the compression flange is laterally unsupported for some distance, the resulting column-type action may result in flange buckling, a warping (partial-to-full section rotation), and lateral bending. Just the superposition effect of vertical compressive bending stresses adding to the compressive stresses due to lateral bending as the section deflects out of plane may produce yield or buckling stresses in one side of the compression flange. In any case, where this situation is possible, the allowable bending stresses are reduced. This reduction is critical when the section is light and deep, since with lateral bending the warping resistance of the flanges and web is small. If lfo,true for long unsupported lengths, since, similar to buckling of a long ~'~?~&h~f~t'rakes a smaller load (or stress) to cause the compression flange to laterally out of plane and producing a tendency for warping. 'A 'theoretical combination of torsion and lateral bending resistance can be made to obtain the critical compression flange buckling stress as ,
ll",,. ,
'
hen*
or, alternatively,
where
J=c,t:orsion constant previously defined and given in the table of ib :'section properties in SSDD C, = warping constant =$ t,bf3(d - t,)2
~h~ AISC specifications can now be developed using the preceding several ations and some additional simplification and rounding of numbsrs. Since ut 1946, the AISC specifications have used the following formula based On g a safety factor of 1.67 and a factor Cb to
.
where rT= radius of gyration of the compression flange + 1/3 of the sion web area taken about the Y axis; these values are also given in
a c - r ~ n tfor moment gradient (the latest modification) into Eq. (4-21), to obtain
Fb = Fb =
I
12 OOOC, Ld/A j 82 700Cb MIAj
stead of 0.41d, to obtain
[AISC Eq. (1.5-7)]
2 0.6.
(SI units)
wher: ,I,= area of compression flange = bjtf d=depth of section; note that the ratio d/A, is computed and tabulated in tables of section properties Cb = 1.75 -I- 1.05(Ml/M2) + 0.3(M,/ M,)' 5 2.3. In this equation M, is always the smaller and M, the larger moment at the end of the' '*: unbraced length. M l / M 2 = (+) when moments are of same sign (producing reversed curvature) and Ml/M2 = (-) when of opposite sign. Use Cb = 1.0 when the moment in the interior of the unbraced length is larger than either of the end moments and regardless of sign.
The minimum L/rT ratio for using Eq. (4-26) is found by equating F, .6Fy to obtain (with slight rounding):
T!I-: Column Research Council proposed as a n alternative to Eq. (4-20) the basic ,. Iumn formula
T a k i ~:SF = 1.67, this equation becomes SC Eq. 1.5-7) when ( L / r , is less thdn that obtained from the equation abo e minimum L / r , ratio at which Eq. (4-27) applies is found by equating E (4-26) and (4-27) to obtain (again slightly rounded): where the term Cc= slenderness ratio for which residual stresses cause inelastic buckliilg or the transition L / r from inelastic to Euler (or elastic) buckling:
-
Substituting this value and using 0.667 instead of 0.6 for the first term (1.0) and approximately 0.75 for the second term in Eq. (4-24), we obtain
I
F,
[AISC Eq. (1.5-6a)]
(4-26)
1/U STRUCTURAL STEEL DESIGN
m*y be used according to AISC, which means that we may use
+ [Eq. (1.5-7)12 2
Eq. (1.5-6b)
but will be neglected. SOLUTIONUsing the given load geometry, complete Fig. E4-13 by drawing the shear and moment diagrams also shown. Next, from Table 11-1 of SSDD for a W36 x 300, obtain LC = 17.6 f t and L, = 35.3 ft. Since L, > I, (46 > 35.3), we immediately note that Fb < 0.6Fk.
acEording to Eq. (4.22) and as illustrated in Fig. 4-1 1. The design of laterally unsupported beams may be summarized as follows: *
M = 204P
.
1. Initially assume that Fb = 0.6Fy and make a tentative section selection using a table such as Table 11-1 of SSDD, which also gives LCand L,. If the actual unbraced length Lb I LC or L,, a direct solution can be obtained since the unbraced length will not be a factor. 2. If the tentative section indicates L, < Lb, the unbraced length.may be a design factor. Using the tentative section or one somewhat larger, compute Fb using Eq. (4-23). If this equation gives a value of Fb that satisfies bending, a so:ution (but possibly not the best) is obtained. 3. If Eq. (4-23) does not supply a (satisfactory) solution, the designer must use either Eq. (4-26) or (4-27), depending on the L / r , ratio. Use the largest F, from either Eq. (4-23) or from the controlling equation (426) or (4-27) as determineckby L/r,.
+ 952.2 in . kips
From Table 1-3, obtain the section properties for a W36 x 300:
Alternative computation of d / A,: b, d = 36.74 in
=
16.655 in
r,
=
1.680 in
Alternative computation of r,:
A table of unbraced lengths versus allowable bending moments such as Table 11-3 or VI-3 may be used to obtain a direct design or to give an indication of sections that may possibly prove to be adequate. The following examples illustrate the use of the equations for laterally unsupported beams.
Example 4-13 Given a girder using a W36 x 300 supporting two columns as shown in Fig. E4-13, what is the maximum column load using the AISC specifications and A-36 steel? Assume that the girder is restrained against rotation only at the ends. The columns may provide some lateral restraint
The slight discrepancy is due to the somewhat approximate computations used; however, this method is satisfactory where r, must be computedCb = 1.0, since the moment diagram shows that the end moments are 0 and the interior span moment is larger.
Check to see if AISC Eq. 1.5-66 [Eq. (4-27)] applies:
Use AISC Eq. 1.5-66 [Eq. (4-27)]:
-m
"
rli
I&.; :TRUCTLTRALSTEEL DESIGN
Estimate the beam weight as approximately 0.1 x load:
Also use AISC Eq. 1.5-7 [Eq. (4-23)]: 12 oOo Fb = 12 oOo(1) Ld/A, 46(12)(1.3 1)
=
(0.03 x 35 + 30)(0.1) 35
16.6 ksi
=
0.09 kips/ft
Use , f b = 16.6 ksi (largest value): t.
M S 952.2
204P
+
M due to beam
Fb + M = SFb
f b C - =
=
=
85.6 ki
'"he preceding example was easy to check, since the beam size has been s e l e c ~ ~and d it is only necessary to determine the allowable bending stress. most design situations the problem is more of an iterative process, in that wh th? loads.~w,igiven,we do not know what section wi1l:be 011c;:casion: one may use charts such as those in AISC; which give the allbwa mom:nt fbr several unbraced lengths or, alternatipely, computer-genera ta&s such as Table 11-3 or VI-3 of SSDD, which giye'the a1 se!ccted shapes for several unbraced lengths Lb. We\should also observe that Eq. (4-23) controls the design, the use of A-36 steel is the 'most 'economic so!:~!ion (the reader should verify why this is true). Example 4-14 Given the laterally unsupported girder for a crane runway in an industrial warehouse. Select t that also limits deflection to L/360. Use A-36 steel. trolley travels on a 90-lb railroad rail fastened to the top of the flange (90 lb = 90 lb/yd = 30 lb/ft).
-=
= 237 ft
M,,,,,
16.6(1110)
P = l7 473'8 204
=
. kips ,. ..
.-
.'.
With a large unsupp somehow estimate the bea selected sections with moments givkn up to ft is only slightly more, use this table as a guide and select the fol tentative sections: ~
1 x 296: 14 x 90:
W16
x
S,
131.0
L,
=
39.9
S, = 143.0
L,,
=
34
L,
=
28
100: S,
W18 x 97:
=
175.0
=
S, = 188.0
(Table 11-4 for W12)
L, = 24.1
ry a W14 x 90 and check AISC Eq. (1.5-7). Take Cb = 1.0. 12000 Fb = (35)(12)(1.36) =21ksi M , = F bS
.r
21(143) =-------=250ft,kips>237 12
O.K.
Check the deflections: I, = 999.0 in4
From Fig. 4-8, obtain the needed deflections at forward load: ~ . .,
,,
,,,
.., .,..,..,-.
P
= 3ok
Figure E4-14
S ~ L U T IFind ~ N the maximum moment. Write an equation for M in terms of and take d M / d x = 0. [15x + 15(x + 6)][L - ( x f 6 ) ] M = L dM - 0 = -60x - 270 + 30L x = 13 ft from left end dx M = (30 x 13 + 90)(35 - 13 - 6 ) = 219.4 f t . kips 35
x
Aload
=
15(19)'(1612 3 EIL 1
= -(47.80
3 IL -=--
+
15(13)(16) ( L Z6EIL
1-31 -
162)
+ 43.03)(1728) = 1.5 in
'('2)-l.2
DESIGN OF BEAMS FOR B .
where & = 10 ft or 3.05 m W = roadway width (including any shoulder allowance) The number of lanes N is an integer; thus a lane fraction should be rounded to the next integer. A reduction in load intensity is allowed when the number lanes N > 2 as follows: N
Percent of live load
2 3 4 or more
100 90 75
The distribution of wheel loads to stringers (or longitudinal beams) as as transverse floor beams is based on a paper by Newmark ("Design of I-B Bridges," Transactions, ASCE, Vol. 114, 1949) and is given in terms of span divided by a coefficient as given in Table 4- 1. The computation of shear and bending moments for deck stringers is bas on a'simple beam analysis for the critical load (either truck or lane). After thes values are computed and adjusted for impact, the distribution factors from Table 4-1 are used to obtain the design effect on any of the interior stringers.
Table 4-1 Wheel load distribution coefficients for AASHTO bridge design used as S/coefficient where S = stringer (beam) spacing 1 lane
Bending moment and shear (lateral distribution on steel interio? beams) For:
concrete floor steel grid deck steel grid deck
2
+ lanes
( ) = SI value
5.5 (1.676) 4.0 (1219) 5.0 (I 524)
7.0 (2.134) 4.5 ( 1.372) 6.0 (1.828)
<'4 in (102 mm) > 4 in (102 mm)
Note: For exterior stringers use statics to o b m n effectwe wheel loads causmg bending and shear (refer to Fig. 4- 15). Bending moment in transverse floor beams w~thoutstnngers: With concrete deck With steel grid < 4 ~n (102 mm) > 4 in (102 mm)
6.0 (1.828) 4.5 (1.372) 6.0 (1.828)
Bending in transverse floor beams w t h stnngers: see Fig. 4-15. Bending moments in concrete deck slabs (mam relnforcement transverse): S = effective span length = clear flange edge-(+flange edge dlstance + b,/2 P, = rear wheel load of H 20, H 15, H 10, etc. (H 20 orSHS + 0.61 20 = 32/2 = IQkips = 72 !d)
.P, LN M = - -S + 2 P , f t . ~ p ~ / f t M=----9.74 r, 32
, am* I
.
,
9
m
*..*
""
a
Exterior stnngers to be at least as large as mtekor.stnngers. to allow for future bndge w d z n i n g
488
STRUCTURU STEEL QESIGN
DESIGN OF B%W
Step 3. Find the beam section. The required sectlon modulus is based on F, pression flange is laterally supported.
Ballast plate: include 10 percent for corrosion protection 0.015(77 k~/m~)(0.76)(1.10) Estimate cross-beam weight
S =-=-= F, 137.5 591
Miscellaneous maintenance, storage of ties, material Total =9.01 kN/m
the dead-load moment is
- 9.01(5.79)2 Y*
8 The dead-load shear is
4'44/2(5.79 2
= 37.8 + 4.8 = 42.6 kN
.m
1.15PwD s 1.15(110/80)(80 x 4.448)(0.76) = 280.6 kN 1.524 with P / 2 placed on each rail. The live-load shear is =
Pw -_ -280'6 = 2
2
140.5 kN
The impact factor is 3 L2 I = -30'5 + 40 - S 150
where S = beam spacing, L
3(5 79)2 30.5 = 79.5 percent + 40 0.76 150 The design live-load shear is 140.5 x 1.795 = 252.2 kN The design live-load moment is =-
ML
=
280.6(2.177 x 1.795) 2
The total design moment is M,,,,, = M, + ML = 42.6 &vt
=
548 kN
.
c,, 1'
+ 548 = 591 kN . m
The total design shear is Vdeslgn
=
'd
+
= 28.3 a
.
"
&
'L
+ 252.2 = 280.5 kN
4.30~10-~m'
m3 d
=
t , = 13.8
> 8.50 mm
758 mm
We note that the weight is 0.08 kN/m larger than assumed, but the sec modulus is more than adequate for this small difference. Check the sh f
Vd = 9.01(5.79) + 4.44 - 28.3 kN 2 2 Step 2. Find the live load on the beam. (D = 0.76 m, s = 5 ft8=1.524 m) P
x
t, = 19.3 mm
+
0.555, since th
ould use a W760 X 147.3 section; however, we will arbitrarily go ost economical section, W760 x 160.7/ 1.58: Sx = 4.8997
WL2 + track Md = 8
=
FO
"
- V dtw
280.5 0.758(13.8)
=
26.8 << 0.35 F,
O.K.
(The end connections to the plate grder wlll be designed in Example 8-9 Use a W760 x 160.7/ 1.58 rolled section.
4-13 COMPOSITE BEAMS A composite beam is one whose strength depends upon the mechanical interac tion between two or more matenals. Reinforced concrete beams are actualIy composite members but are not generally designated as such. Most often the term "composite beam" in building and bndge construction is applied to a ste section on which a concrete floor or bndge deck has been cast. The concrete securely bonded to the steel section via carefully designed shear connectors so that the concrete and steel act together as a tee beam. Figures 4-16 and 4-17 illustrate shear connectors in position to bond the concrete to the beams so composite action is obtained. The shear connectors shown are called shears and are welded to the beam (and through the deck pan in Fig. 4-16). Other of shear connectors can be used, but shear studs are most common. When there is no particular effort to bond the steel beam and concrete flo eck, relative slip occurs at the interface of the two materials and the result 1 omposite section. In some cases we may design for a specified amoun site action. Actually, there wlll always be some small slip due to nequal deformations in the shear studs, concrete, and steel ractical purposes it can be neglected In composite design. The brief introd composite design made here will only consider full composite action. e method of construction and code specifications are significant desl meters in design of composite sections. Figure 4-18 illustrates the effe h of concrete to use in a transformed section b' using the modular n = E,/E, as in several mechan~cs-of-matenalstexts. The modular ratio depends on the 28-day concrete design strength 1, shown in Table 4-2.
~i
190
a*",: STRUCTURAL STEEL DESIGN
Figure 4-16 Composite building construction. (a) Shear studs in construction and metal deck for form. ( b ) View of inetal deck encased by concrete floor. (c) Bottom vlew of metal deck resting on floor beams and girders. Metal deck is welded to floor beams wlth shear studs.
DESIGN OF B U \ S FOR B
4-17 Composite bridge deck (two lanes interstate hlghway uslng six stringers. (a site bridge deck nearly ready for concrete wit reinforcing and shear studs in place. Side vertical is for parapet. ( b ) Closeup wew of shear studs (c) Undernew showng stringers and unshored work for deck.
The two basic construction methods for producing composite beams are: 1. Shored construction. The steel beams are put in place and the formwork for the concrete slab is added. This assembly is then shored (braced or propped) so that no (or relative small amounts of) deflection can occur, and the concrete is poured. After the concrete has hardened for'about 7 days (about 70 to 75 percent off,' obtained), the shoring is removed. At this point the stresses in the composite beam are due to the dead weight of the steel beam plus a proportionate share of the concrete deck. 2. Unshored construction. The steel beams are placed and formwork (metal decking may be the necessary formwork as in Fig. 4-16) supplied for the concrete deck (refer to Fig. 4-17). The concrete is poured and at this time the steel beam carries the dead load of steel, formwork (as used), and the concrete. After the concrete hardens, any formwork is r at this stage of the construction that the steel beam h stressed with the weight of the steel beam plqs a proporti weight of the concrete deck.
II
9
*
.0
(,'I
i =
*pJl#
se srn.~llestvalue ol b b = LI-I
b=b,+i
AASHTO ~ = I s + ~ I 2, I h=b,+ 16i
-\-\SIITO h = I:[
4-18 Effectwe flange wdth of composite sections by both AlSC and M H T O s w c a t i o n s ny differences ldentrfled. ( a ) Edge. (b) Intenor.
w
Y
192 STRUCTURAL STEEL DESIGN
use an overst~e~C@torof 135 percent (1.35 to account for &e te capacity), tKk" stress on the bottom flange of a steel composite
'"
2000 2500 3000 3500 4000 5000 6000
13.8 17.2 20.7 24.1 27.6 34.5 41.4
2550 @SI) 2850 3100 3400
3600 4000
4400
The actual bending stresses in a composite beam are limited by the materi
ccvnstruchon methods is illustrated in Fig. 4-19. The ultimate load of a shored composite beam has been found to be on th
S, = section modulus of steel beam referred to tension flange S,
=
section modulus of composite beam referred to tension flang
Shored
!-----
b -----4
igure 4-20 Stress dstnbutlon at ultmate load m compoate beam
DESIGN OF BE4!!S FOR
Use is
$
connectors with a capaclty of 8 hps/connector; the nu = 264.6
fq I = -2 4
= 33.07 8
use 34 connectors
Use two connectors at each section at a spacing w = 4d = center to center. The longitudinal connector spacing is S
30(12) = 10.6 in > 6 0 (minimum) = -----2(34)/2
< 81 (maximum) Step 7. Check the shear with both dead and live loads actingV = 30.96 klps
V j--=-= dt,
30.96 = 4.9 ksi << 0 . 4 5 1g(0.355)
O.K.
Step 8. Check the deflection under live load. Note that th deflection has been built out of the floor via flow of the wet con placing.
A
~ =~ W = L! 5(0.13X8)(3@)(12~) = 0.32 in 384EI 384(29 000)(2056.1)
L <360
Example 4-17 Design a composite bridge section for a two-Ian 12-ft lanes + two 8-ft shoulders and pedestrian walkways) as E4-17a. Other data: HS 20 loading; /; = 27.6 MPa (class A ksi);f, = 415 MPa (reinforcing steel); A-36 steel for the stnngen; specifications; 2 X lo6 loading cycles. SOLUTIONThe two parapets and rallmgs,.ulll bg pl ed after the placed. We will assume that this load~n$ is &%ed by the two stringers, which are not deslgned here(ddjsd.t& interior s t k g e r checke see if it is adequate as an extenor stnnger, since the exterior stringer mur at least as large as the interior stnngers). We wlll also assume that a fu 40 mm of wearing surface will be added as the initial surface deteriora Step 1. Design the concrete deck slab. The steel beam spacing is selected at 2033 rnm. as shown in Fig. Estimate the stringer flange width b, = 420 mm (a W920 se compute the effective stnnger spaclng (AASHTO Sec. 1-3.2) of the interior beams as s = sac< - b,
b/ = 2033 - 210 = +-
1823 mm 2 The slab dead load, including future wearing surface based on D = 230
m
STRUCTURAL STEEL DESIGN
roadway, the live-load moment is
The design moment for the deck slab using strength design is M,,,
=
M,
~ M =L 2.33 + -----5(18.7) = 33.5 kN - m +3 3
Take d = D - 40 (allowing 25 rnm clear cover for bottom reinforcement bridge decks AASHTO Art. 1.6.16): .
I
•
40 rnm haunch
.
I
A: - 0.02148AS = - 1.0140 x loT5 Solve for A, by completing the square: I
I
I
I
I
v Cross frames
. . . ..
.'I..,
77.200 4 (? 5550
The maximum ratio of A,/A, = 0.0214, based on f: and f, and taking 75 percent of the value to ensure initial steel yield. The actual steeI percentage is 0'0004829 = 0.0025 << 0.0214 O.K. = 0.19(1) The minimum percentage is
of concrete, is (0.230 + 0.040)(23.6 kN/m3) = 6.37 kPa Increase 10 percent for roadway debris, snow, etc.: 6.37 X 1.10 = 7.0 kPa The dead-load moment is computed as
The impact factor for the slab is (using L = s = 2.033 m)
z2= l5 = 0.375 > 0.30
+
use 1 = 0.30
L 38 Since the s / L = 2.033/22.4 = 0.091 is so small, one-way principal slab reinforcement perpendicular to the traffic flow will be used in the slab design. For this case and five spans providing slab continuity across the
JY
Use A, = 0.0033(0.19 x 1.0) = 0.000627 m2/m. Step 2. Design the steel stringers. Note it was necessary to design the slab so that the dead load carried by the stringers could be computed. Assume unshored construction-the stringers must carry the dead weight of the deck slab until the concrete hardens. Dead load of slab: (0.230 + 0.040)(23.6)(2.032) = 13.00 kN/m Haunch: 0.040(0.42)(23.6)
= 0.40
kN/m
Rolled beam assumed
=3.60 kN/m
I
2~iscellaneous, including debris, formwork, etc. Total (I I!
= 1.00 kN/m = 18.00 kN/m
r
A4J.Z STRUCTURAL STeEL DESIGN
DESIGN OF BWki.5 FOR BENDWG' 21)3
The maximum dead-load moment at the center of span is
The maximum live-load moment, including impact, is based on the AASHT3 truck and the stringer spacing. I
=
conservative, so that we may check stresses at the center of span under and add the composite effect of M L (unshored construction sequznce). Step 3. Find the properties of the composite section (refer to E4- 17b). Assume that 15 mm of concrete is not usable in composite because of wear and surface deterioration. Neglect the area of con the haunch.
15 = 0.25 22.4 38
+
From Sec. 1-9, the live-load moment due to total truck load on span of this length is b = 0.42 + 1.613 = 2.033 rn b = 0.42 + 161 = 3.86 m Use b = 2.033 m; for f,' = 27.6 MPa, n = 8.
The distribution factor for live load to a stringer of a bridge with two or more lanes is obtained from Table 4-1 as Factor
=
-= 2.033 = 1.213 wheels
1.676 1.676 Slnce 1.213 wheels = 1.213/2 = 0.6065 axle, the adjusted (for deck1 [slab action) stringer bending moment is
It is usually sufficient to select a steel beam one or more sizes Iess than that required for noncomposite action. The section required for noncomposite action is approximately (and flange laterally supported) s x = - = MT 0.555 Tentatively, try a W920
The maximum shear occurs when one of the rearmost truck wheel's [144-kN (32-kip) axle] load is at the beam end:
1129 + 1090.5 0.55(250)
X
=
16.14 x 10-3
m3
.*
364.6/3.58:
Ix=6701.3x10-6m4
S,=14.67X10-3m3
I
t, = 34.3 mm
This value must also be adjusted for plate action and impact:
b, = 419 mrn A = 46.52 x
m2
Check the shear: We note that M, is at the center of the stringer span, whereas ML is slightly off center. The error of summing MD + ML for design is negligible and = 26.1 MPa
F
<< 2 3
O.K.
Check the dead-load deflect~on:
= 0.04402 m (44.02 mm)
This much deflection in a 22.4-m span would tend to aggravate during placing the concrete to a level top surface (requires an additional 44 mm of
:
$>
'
I*.
i..)
sgF-
b
i k c r e t e at the center of the span. This will require a temporary or, preferably, have the stringers cambered at the mill with a (and being sure to place the cambered side up on the job d'faster would occur). The remainder of the beam properties ek.) are within the assumptions, so we can continue. Compute I of the composite section (refer to Fig. E4-17c).
@
,
6'
s
'
snnL DnsmN
,
AY = ZM*., A = 46.52 b'd' = 46.52 + 0.254(215) = 101.13 x
Final:
I top '%=-!!??!-=26.1Mpa Stop
41.77
26.1 8
3.26 MPa << 0.41
f,=-=
O.K.
+
m2
The section stress profiles are shown in Fig. E4-17d. Check 100 percen overload stresses against 150 percent of allowable stresses:
f, = ----2(1091) -
f,,,,,,, =
106.2. MPa 20.54 77.0 + 106.2 = 183.2 MPa
<
1 5(0 55F,)
also O.K.
Figure E4-17c
= 6701.3
+ 46.52(0.327)~x
Id
+
Dedd load
Lnc. l o ~ d
Figure E4-17d
= 6701.3
+ 4974.3 + 210.4 + 4235.7
= 16 121.7
x
m4
The section modulus values are
I
""P
= I31
SbOt
=
l6 121a7 + 40 215 = 41.77 x 10-3
+
l6 121'7 = 20.54 x 10-3 458 327
+
m3
The stresses are:
Initial:
=
f;=f,=-= M"
S,
.
.
m3
Step 4. Design the shear connectors. Use three 20-mm studs, as shown in Fig. E4-17e, with L = 150 mm Thus the L / d = 150/20 > 4 is 0.K: The studs must resist the smaller of 0.85f:bdf vh~= -7 (neglect haunch)
-I 129
14.67
- 77.0 MPa
5130 kN
A F v , = x = 2
controls (after computing V,) 46.52(250) 2
=
5815 kN
@6
STRUCTURAL STEEL DESIGN
DESIGN OF B W M S FOR B
1.
-.v
95
1. Ka
R .A
Figure W17f
At X = 7 m from supports, the live-load shear plus impact is Figure W 1 7 e
I'
[
22.4Ra
=
4078.8
R,
=
182 kN
The ultimate strength of the shear stud is S,,
=
0 . 4 d 2 m
V = 182(0.6065)(1.25) = 138 k N
E=4 7 4 0 e
p
= 0.4(0.020)~(27.6 x 24 900); (w/SI adjustments
N, =
5130 = 38.7 132.6
4 use 39 (multiples of 3) I
With studs in groups of three, there will be 13 groups. Now set,the spacing based on fatigue. . The maximum range of live-load shear is 0 to 214.6 kN. Zr = c d 2 = 0.0541(20)~= 21.6 kN
At X = 3.5 m from supports, the live-load shear plus impact is (refer to E4- 17f)
=
0.5022 mm
,
0- .---K~
Using the spacings above, the first 3.44 m requires 1 1 groups of studs, next 3.43 m requires 9 groups, and the next 4.33 m requires 9 zrou~s.for total of 29 groups at 3 studs each, for a total of 87 studs vs. th for ultimate shear considerations. Step 5. Design the diaphragms. Lateral load: Wind at 24 kPa X exposed surface area and, referring to Fig E4-17a: H = 4 6 0 + 225 + 270+ 916mm = 1.871 m Curb load at 500 lb/ft for side opposite Wind
nz. = ---301.6) = 0+323 Pitchp = S, 200.6
0.323(%)
< 0.6 1 max. s ~ a c i "n zallowed
= 132.6 kN
. The minimum number of studs required from the end to the middbn tance of L = 22.4/2 = 11.2 m) is
=
=
24(1.871) Total
Note: there is no wind load from the truck because the deck acts as diaphragm. With diaphragms spaced at 5.55 m, Pdiaph = 52.2(5.55)
=
289.7 kN
(interior diaphragm)
- 289.7
- 144.9 kN (exterior diaphragms) 2 Use all same-size diaphragms; also, the diaphragms must be at Ieast one third and preferably one-half of the grder depth. Arbitrarily select nr<
W530 $rk@yproportion since Q / I = constant,
X
65.5/0.64:
d = 525 mm 1, =
8.9 rnrn
A = 8.39 x r, = 32 mm
Y to > --j-
-
> 8.0 mm
O.K.
O.K.
m' L = 2.033 m
STRUC'IURAL STEEL DESIGN S
A check (not shown) as a column indicates P = 994 kN >> 289.7 k ~ so, section is amply adequate. Use the same section for both ends ahd inten points. Step 6. Check the live-load deflection. Convert the live-load moment to an equivalent uniform load as a first approximation.
I
1-
Flgum FX-18
SOLUTION Since the girder is symmetrical on the floor plan and are closely spaced, assume that the loads will give a uniform girder I
Taking the maximum allowable deflection as L 22400 -_ - 22.4 mm > 17.6 mm O.K. 1000 1000 If the designer assumes that the actual moment diagram resulting from thre wheel loads is sufficiently close to a uniform load diagram, the deflection a computed is adequate; otherwise, a more exact analvsis should be m a d e
bv, =
+(FdD
+ FLL)
Using load factors glven in Sec. 3-7 (and Table 3-1) and noting th no live load reduction for floor area since the load factors are the statistical terms, we obtain
3 -
I
I
I
I
--I
w,
=
1.1 [ 1.1(22 x 0.075) + 1.4(22 x 0.080)]
=
l.l(l.815
+ 2.464) = 4.71 kips/ft
The design moment is
4-14 BEAM DESIGN USING LOAD RESISTANCE FACTOR -,-.,-.--=-
,w
---.
wUL' 4.7 1 (25)2 Mu=-= 367.97 f t - kips 8 8 (4 = 0.86 from Table 3-1) Mu = OFyZ
I
The design of a beam for bending using LRFD is relatively straightforward. It necessary to have separate values for dead and live loads. The allowable bendin stress is taken as Fb = +F,
+ = 0.86
Mu = +F,Z
vu=
'
+( v3 )
dt,"
and Mu is computed using factored dead and live loads. This is illustrated in the following example.
Example 4-18 Given the floor system shown in Fig. E4-18 for an off and live 80 psf; use LRFD design and A-36 steel. Make a preliminary design selection of a W shape with depth no factor.
From Table 11-2 of SSDD, obtain a W21 x 62 with Z = 144.4 20.99 in, and t, = 0.0400 in. Check the beam weight: Awu
=
1.1(1.1)(0.062) = 0.075 kip/ft
By proportion, Aw Z 0.075(142.6) A Z = U -= 2.27 in3 4.7 1 Wu
O.K.
Check the shear:
: building using simple framing. Take the loads as dead = 75 psf
=
+- F~
Use a W 2 1
v3
X
dtw = 0.86- 36 (20.99)(0.40) = 180.3 >> 58.9 kips
62 section.
v3
O.K.
, ,
,
212
STRUCTURAL STEEL DESIGN
.$r"
-be
7'-
i
DESIGN OF'BEAMSFO
gn the floor stringers, noting full lateral support for the compressi = 250 MPa. Use the AASHTO specifications.
interior floor stringers of Fig. P4-19 using composite daigo ction under live load to L/800. Use j; = 4 ksi.
no twisting occurs due to the eccentricity of the brick. Use A-36 steel.
Design the interior floor stringers of Fig. P4-19 using composite desip and unshored cons Limit live-load deflections to L/800. Use f; = 28 MPa. ~ollowingare miscellaneous beam problems for laterally unsupported spans and considerations. 4-a A column load of 160 kips is carried across an open work area as in Fig. P4-23. The unsupported length is 39.5 ft. Select the lightest W shape with the deflection limited to L any grade of steel if A-36 is not adequate. Note that you must assume your end conditio Answer: W36 X 300. (Simply supported.)
Answer: W410 x 59.8 for Al.
Answer: W410 x 46.1. 4-14 Design the floor beams spanning between columns assuming supports and = 345 ~ p ~ Answer: W410 x 46.1. 4-15 Do Prob. 4-8 using unshored composite construction. U s e x = 3000 psi. 4-16 Do h o b . 4-12 using composite unshored construction but use F, = 250 MPa steel (instead of
F,
.
,7 +(I
= 345 MPa steel; why?). U s e x = 21 MPa.
4-17 Redo Example 4-16 using shored construction instead of the "unshored" construction of he example. 46 and see if a workable solution can be obtained. For Probs. 4-19 to 4-22, refer to the cross section shown in Fig. PC$-19. The bridge span will be assigned by the instructor (36 to 46 ft or 11 to 14 m). If not assigned, use 40 ft or 12.5 m for the span. For exterior beams the maximum possible load due to truck is one-half of the truck load. The dead
4-18 Redo Example 4-16 using a W18
an ,HS 20 truck and A-36 or&
X
y6
P=
& T *
[II
30 5 ' i .; :. 111
i
t>OZ
- 'lo
L\
--
B J . ~ pi~r:,
il
t
'1
i /i iI
250 MPa steel.
I ._.I' ,
steel. Use the AASHTO specifications. A m e r : W33 x 130.
24 ~~d~ h o b . 4-23 with p = 790 kN and the metric dunensions shown
g. 4-6 is required for the beam selected in h o b . 4-23? Anrwer: N = 8.50 in. 4% ~f the beam selected in Prob. 4-24 supports the column load of 790 kN carried 200 x 86.3 column section, what length N of Fig. 4-6 is required? Answer: N = 222 m.
w section is to be used in a span of 10 f t to carry a midspan concentrated load of 650 lect the section, check for shear stresses, and compute the values for both the reaction and ncentrated load, using A-36 steel and AISC specifications. 28 A W section is to be used in a span of 3.5 m to cany a midspan concentrated load of 2ICO lect the section, check for shear stresses, and compute the N values for both the reaction and ncentrated load using Fy = 250 MPa steel and the AISC specifications. Answer: W840 x 299.1, N = 260 mm for reaction. 27 A
11000 - 14000 rnm
Figme P4-19
4-29 Redo Example 4-18 if @ = 0.90 and FL = 1.5. 4-30 Redo Example 4-18 if Fy = 50 ksl. Annuer: W21 X 50. 4-31 ~~d~ Example 4-18 if D = 3.75 kpa, L = 4.0 k ~ a L, = 7.75 m (bay). and the joist l e W h = = 250 MPa and LRFD with factors suggested in k c . 3-7. 6.75 m. Use Answer: W530 X 92.3.
I
,~
' ,
,
',
, ,
214 STRUCTURAL STEEL DESIGN 4-32 Redo Example 4-10 for the lightest W or M section. Answer: W10 X 22 or MI0 X 22.9. 4-33 Design the roof purlins for Prob. 5-20 (of the next chapter) using the lightest rolled W or C section and with a sag rod at midspan. Use the AISC specifications and A-36 steel. Answer: W14 X 43. 434 Obtain the lightest W section for a beam span of 45 ft with two 20-kip concentrated loads at 15 ft from each end. Lateral support is available only at the ends and concentrated loads. Use A-36 steel and the AISC specifications. Answer: W27 X 84. 4-35 Obtain the lightest W section for a 30-kips/ft uniform load on a 15-ft span. Also find the reaction distance N. Use A-36 steel and the AISC specifications.
218
STRUCTURAL STEEL DESIGN Round bar
I I Squdre bar J
I
Fldt bar o r plate %;
I ,
n all these uses the tensile' strength of the steel is used. In 'thi uration plate buckling or warping is not a consideration. In some s, however, specifications will require a minimum amount. of ss for esthetic and safety reasons. enerally, tension members may be categorized as rods and bars, tural shapes, built-up members, and wires or cables. Several of bers are illustrated in Figs. 5-1 and following.
ALLOWABLE T E N S I O N S T R E S S E S e AISC allowable tension stress of members, except eyebars, is limited to F,
1
W shape
=
0.6 F,
(gross section area)
F, = 0.5FU
(net section area)
(5- 1)
e AASHTO and AREA allowable tension stress is somewhat more conservaF,
=
0.555
(5-2) cification further limits this basic stress to the lesser value of
ut the net section is used for both these equations. For steel with not over 80 i the basic tensile steel stress is governed by Eq. (5-2) for AASHTO design. On the net section across the pin hole of an eyebar (see Fig. 5-3), th owable AISC stress is
Figure 5-1 Tenston members. See Fig. 5-2 for cables used as tension members. (a) Structural shapes used for tens!?? members. (b) Upset bar. ( c ) Threaded bar and use of a turnbuckle to adjust bar lei&. Applicable for square and round bars.
Allowable tensile stresses for several steel grades are shown in Table 5-1, ere the reader should note that established practice allows rounding of the ues for A-36 steel to the values shown for both AISC and AASHTO/AREA In all cases, except eyebars, the tension stresses must be computed based o both the gross and net cross-sectional area when using AISC specifications. Ody the net area is required for AASHTO and AREA specifications. The net area is the gross (total) area where welded connections are used. The net area is the least effective cross-sectional area for all other cases as where bolt or rivet holes are used for mechanical fasteners at the ends or where holes and/or reductions occur along the member. The effective net area at approximately the root of the thread of thread ension members using the AISC specification is
$+'7 1
where D = nominal outslde diameter of threads n= number of threads/ln (or the SI equivalent of n/25.4)
DESIGN OF TENSIOS & E M B E R S
UP
le 5-1 Allowable tensile stresses for specifications and equations shown
Anchor rod
On gross section not at pinholes. On gross section at pinholes. For A-572 steel.
(strand or rope)
-3 GENERAL DESIGN CONSIDERATIONS
principal factor is how to affect a connection of the tension member to the tension member.
,
5-3. Next in simplicity would be some kind of threaded bar or cable. Here 1
2.'- STRUClVRAL
STEEL DESIGN
,$4:; ii. .
r-
t
.
1
.
I,.
= A*
1%.
'> ., .
4.
5 81
I
!
,.. .,...,.
1.33Ab S A , 5 1.5
r 2 0.5" (12.5 rnm)
4 i-
..
1.;
2
t
I
I
I
/
I
I
I
5-3 AISC eyebar dimensions.
several problems develop, including the area in tension for the and fitting the member into the structure. This problem is usually solved b of a turnbuckle or by having an extra threaded distance on one end to ta the slack. Bars and cables used as tension members are generally giLen a initial tension when installed to eliminate any tendency to s rattling when the structure vibrates under service loads. The . sometimes useful in "tightening" up the remainder of the structure., Cables may be strands of 'wire rope, with the terms "bridge stra "bridge rope" being used to specify the structural quality of the cables. wires are not used in structural applications; rather, strands that are a n~ectsof 7 to 61 or more single wires are wrapped around a cen produce a symmetrical section. Wire rope is produced by laying several helically around a wire core. Commonly wire rope.consists of 6 to 37 T.:bie 5-2 gives design data for selected sizes of both bridge r0y.e. Cable connections commonly take the form of one of the configurations i1lus::ated in Fig. 5-2. In the connections shown, the end of the cable is carefully c!:~ned and then fed through the opening in the connector. The cable end is then broomed (strands separated somewhat), carefully cleaned, and molten zinc at about 850°F is poured into the wire matrix. After the zinc cools, and the connection is again cleaned and assembled, the member is ready to be installed. Tiiis method of attaching the end connector produces a joint at least as strong as the cable., Occasionally (but not shown), the cable can be inserted into a longer c~.:r?ectorwhich is squeezed (termed swaging) to produce a friction connection.
using r, of element to compute L ' / r between stitch bolts.
d stretching it so that the component parts are fitted together. The modulus of elasticity of bridge strand and rope may be taken
67 (and larger)
23 000
158
CCO
DESIGN OF TENSION ~ ( E . M B ~
Table 5-2 Selected cable design data
b
rl
!
Welght
4
in
Ib/ft
mm
kN/m
Area in2
pu
m2 ( X lo-')
bps
kN
project and design uncertainties.
I
B dge Strand (single strand with multiple wires) I
13 16 18
0.52 0.82 0.99
0.008 0.012 0.014
f
19 22
1.18 1.61
0.017 0.023
; ; f
: 1
10 13 16 119 "22 25
.
0.15 0.234
0.0968 0.1510
29.0 46.6
129 207
0.284 0.338 0.459
0.1832 0.2181 0.2961
56.2 66.0 89.2
250 294 397
Bndge Rope [6 X 7 (6 strands of 7 wues/strand)] 0.24 0.004 0.065 0.0419 042 0.006 0.119 0.0768 0.65 0.009 0.182 0.1 174 0.95 0.014 0.268 0.1729 1.28 0.019 0.361 0.2329 1.67 :0.024 0.471 0.3039
13.0
58
23.0 36.0 52.0 70.0 91.4
102 160 231 31 1 407
286.0 372.0 576.0
1272 1655 2562
Bridge rope [6 X 37 (6 strands of 37 wires/strand)] 15.1 0.220 4.25 2.7419 824.0 21.0 0.306 5.83 3.7613 1110.0 27.0 0.394 7.56 4.8774 1460.0
3665 4938
,
-3.2 Shear Flow and Maximum Effective Cross-sectional.Areas
3. For all other shapes, including built-up shapes, with at least three fasteners a line (Fig. 5-5 has five fasteners in line), A, = 0.85A,. 4. Any tension members with only two fasteners in a line, A, = 0.75A,.
Bndge Rope (6 X 19 (6 strands of 19 wires/strand)]
1; 2 2f
44 50 64
3 3f 4
75 90 1,00
--
5.24 6.85 10.60
0.076 0.100 0.155
1.47 1.92 2.97
0.9484 1.2387 1.9161
At point I : Uniform >tress d ~ s t r i b i i t ~ u~i is s u r n e dI I I a n s l y s ~ i 2 . S o ~ r i eload traiisferred to gusset p b t e s I e ~ v l n pP' Load transfer is at flanges w h ~ c hresults In qurllitati~estresses across sectlon sliown. Interior web stresscs w1II be lower than P1.4, due to shear lag. Thus. In long j o ~ n t sthe web may tear d u e t o large diifercntial s t r ~ i n sresult~ngIn I! progressive tension failure at [tie tbrwdrd end ot'joint 3: Additional load trsnsferred to gusset plates I e ~ v i n gY . 4: A11 load transferred t o gusset plates and tension member stresses are zzro. '
6494
. -
,
. -
228 S p U C l U R A L STEEL DESIGN ..',...--,.. A,
,-.,,a,
..4
(0)
T ~ p i i . l ilr u , ,
\.ur<
Figure 5-8 Lacing and other means of producing a built-up member with access to interior. fabrication shop practice gives preference to use of batten or perforated cover plates which are welded. (a) Single lacing. (b) Double lacing. (c) Batten plates. (d) Perforated cover plates.
(f) Figure 5-7 Cross sections of several built-up sections. General cross-section configuration is limited only by designer's need and ingenuity and may include W and S shapes with'lacing and/or plates. Where additional area is needed, more plates may be added to any of the above sections. ( a ) Four ;, angles with lacing. ( 6 ) Four angles with both plates and lacing. (c) Two angles and one or two . ( d ) Two channels with lacing. ( e ) Two channels with flanges reversed from (d) and both plate and lacing. (A Four plates welded to form box section. $,
'.i
4.
$
not occur. It is also necessary that the member(s) be constructed so that painting of the complete member can be affected. This requirement generally precludes use of fully enclosed box or other built-up sections with .enclosed cavities. Instead, the built-up sections are open on one or more sides, with continuity being obtained on those sides by use of lacing bars or by use of perforated cover plates. Either of these configurations allows maintenance of the interior as well as the exterior of the member, as shown in Fig. 5.8.
iccrlon
i b ~ l i i t n gIS prescntiy used I r i i i i i l o i rivering.
,
parts, making the connection sufficiently strong that no relative movement can take place across the hole. Tension tests using plastic and photoelastic techniques indicate stress concentrations at the edges of the bolt hole which have a maximum value on the order of two to three times the average s
f, =
5,it is reasonable to assume a uniform stress distribution across
indeterminate, but should the working load be inaccurately estimated so h a t redistributed until the load is ber of stress reversals or to range to the member. The id a fatigue failure if the er of stress cycles and/or the stress range is large. is usual to assume that fatigue can be neglected in normal building load cycles/day, the total es over a period of 20 years is only N,,,,,, = 20(365)(20) = 146 000
232 STRUC-IVRAL STEEL DESIGN
DESIGN OF TENSIOK
?
L:
'
0.4 195 kipsift
0.066 kiplft
+
p" .y;,
a
,,
it.
'
,
t,i
!, .". ,),<.. ,;;*,.,i >\ , ,
,?,$
)
.
"
:.
,
',
14 P= 1
.03~
Vol. 89, 1922, pp. 847-848) and is almost universally used. The meth Figure E5-1b
Notes:
';y;;,~'~~,,
.
.
, , .v
:
.
..., (
'1 .";Tie rod size could be reduced in the lower-half group of purlins for a rod force of 6 X 1.03 = 6.18 kips. 2. A rafter, substantial girt, or other member spanning the 25-ft bays will be required to carry the concentrated rnidspan load of 12.18 kips without excessive outward deflection. The roofing will undoubtedly reduce this effect considerably, but the designer will have to decide how much. 3. Computations neglect the benefits of the roof in contact with the top flange of the purlins, and may reduce the sag rod force 50 percent or more. \
one hole width for each hole encountered. For each change in direction from one hole to the next hole, add ba
,
;
,
me possible path as just cited, which vior along irregular failure lines. The
'
5-6 NET SECTIONS When the cross section contains a row of holes, the critical section will occur through one of the holes, as shown in Fig. 5-9. When there is more than one row of holes, the designer must determine a failure section that will yield the minimumiarea. The path across the section producing the minimum area is t critic&l:ine;tsection. -4 li$@ig. 5-9b, which is a portion of a repeating bolt or rivet pattern @mi& be btained in a large plate ass ~ .d t # & ~ r e )the , critical section is ; :.. .- .\: ;. ... .......... .
where s= pitch, or longitudinal distance between adjacent holes g= gage distance between adjacent holes across the-width
"hole" on the diagonal, thus canceling that hole reduction on the net The effective hole diameter (except for slotted holes) is taken as AISC specifications AASHTO and AREA specifications
The hole is always larger than the fastener by at least
$ in (or
1.5
,t,
""
, I ,
for possible metal damage around the hol iameter as D + in or'D + 3 rnm.
\[-I/
Example 5-2 What is the critical net section of the hole pattern shown Fig. E5-2? Take holes for i-in-diameter bolts. Note that this pattern is rno academic than practical, but is used to illustrate the method of using s2/4 Use the AISC specifications for the hole diameter.
r
0
0
Total width w = 2(1.5)
fy,
(a)
Mgure 5-9 Critical net section for tension members with holes. (a) One or more rows of holes w the net section is obvious. (b) Hole pattern where net section must be found by trial.
D
=:+$=
+ 3 + 2(2)
I.OOin
t width along path ABDF': w = 10 - 2(1.W) = 8.00 in
DESIGN OF TEXSION Sf33
Net width along path ABCDFG: rv, = 13.25 - 4(0.875) Net width along path ABDEFG:
+ 0.333 + 0.333 =
10.42 in
>vn = 13.25 - 4(0.875) + 0.222 + 0.400 = 10.37 in Check the 85 percent requirement: ~ v ,= 13.25(0.85) = 1 1.26 in
Based on the critical net width of 10.37 in (note that bo connected, so a shear lag reduction is not required) and c = 0.7 A, = 10.37(0.75)
Pallow= A,F, P
5-10 Net section of angle used as tension member with staggered holes in both legs.
,
=
A,F,
=
7.78 in2
=
13.25(0.75)(0.65)
=
7.78(0.5Fu)
whichever is smaller.
J
* 2d"
5-7 DESIGN OF AISC TENSION MEiMBERS Example 5-4 What is the critical section of the 8 x 6
X
angle with
s computations to determine the tional computations to determine the critical net section where for mechanical fasteners. Member selection proceeds based o n ncluding the 85 percent maxim are in the member, and any reduction in net area from be met in all the specificatio and AREA) have minimum mum connection requirement is
F4W-e E5-4
SOLUTION
.
'
D = 0.75 Net width along path ABDFG:
+ 0.125 = 0.875 in
,&,w = 8 + 6 - 0.75 - 3(0.875) = 10.625 in Net width til&gpath ABCDEFG:
= 13.25 -' 4.375 = 10.83 in
+ 0.667 + 0.667 + 0.222 + 0.400
Example 5-5 Select the lightest single angle section for the vertical of the side shed truss of the industrial building of Example desirable to use the same size angle section for all the verticals, section for all the diagonals, and similarly, top and bottom chords d s s i p e using constant section sizes. Only the vertical web members will be d e s i g e in this example. Use F, = 250 MPa and the AISC specifications.
1
9 3 STRUCTURAL STEEL DESIGN
!'
which S o ~ u n o NFrom inspection of the computer output (Pa* shown in Example 2-6), the following values are obtained from the tw conditions used: Left side
Right side Member 4 8 12 16' 20
LC-1
@pq
LC-2 (kpq
0.0 31.18 62.36 93.54 124.72
Member 74 78 82 86 90
0.0 48.62 97.26 145.88 194.50
0
LC-1 @W 124.73 93.56 62.37 31.18 0.0
157.57 118.18 78.80 39.40 0.0
I n the table above,
d
Since the signs are (+), all the vertical members have only tension Iorces Ior either load condition. Note that the computer program automatically output with wind by the factor 0.75 so that all loads are on the same design basis. Member 20 has the largest axial tension force with wind from the left; wind from the right would produce this design value in member 74. Member 20 is the longest vertical of the side shed members and has L = 4.5 m. If One of the other members had been longer, that member length the ~ / , . - i ~this case both the controlling length and the maximum force are in the same member. Design P = 194.5 kN. L/,. = 240 (per AISC, Set. 1-8.4, and assuming that this is a main member with such a large axial force). The minimum radius of gyration is
I
I
1 1 b*
3i
I
4.5(1000) = 18.75 mm 240 A preliminaly side computation indicates that the bolt Pattern !&Own to ~ 5 - 5can probably be used, as L / r rather than stress is m i s pattern may be able to use standard gage distances and economical. For two bolts at the section and using 22-mm A-325 D = 22 + 3.0 = 25.0 mm me effectiveangle area using AISC criteria for shear 1% (see Set. 5-3.2 A, = 0.90An
L
~~~~don F, = 0.6FY,the gross angle section must be at least
'mm
I
i:
't
i1
faa I" i
I
t
m2
must be Based on F, = 0 . 5 ~ "(use Table 5-1 for F,), the effective net area 194 5 A, = = 0.9725 x 10-3 m2 200 m i s effective net area A, must be obtained from a gross section of least
-
=
0.9725 0.85(0.90)
=
1.2712 x
2Dt
=
2(25.0)(0.0063) = 0.315 x 10-3 m2
An,, = A, - Aholes2
0.9725 = 1.081 x 0.90
m'
Here the radius of gyrauon controls the selectlon of he section than the area requirements. Tentatively use: L 152 x 89 6.3 (as ae lightest). We must select a section that can be used in hejoint as shownCheck if it is possible to get two 22-mm bolts side by side in the long leg of the angle. Use Table V-13 and obtain: 'Onsidered
-
Standard gage distances: g, = 57.2 mm g, = 63.5 AISC specifications: center to center of hole = 2,670 ( 3 ~ specifications: center of hole to edge of angle leg ~ 1.16.4 SSDD) = 28 rnm (or 1; in)
=
A, = ---194.5 = 1.2967 x lo-' 0.6(250)
A,
=
Ahole,
> 2.670 0 . ~ . Actual hole diameter = 22 + 1.5 = 23.5 mm Edge distance = 152 - 57.2 - 63.j = 3 1.3 > 28 g2 = 63.5
Use an L152
X
89 x 6.3 section.
~
b
O.K.
designed for both tension and compression (and p o ~ ~ l b lfor y fatigue). F~~~ the 'Omputer output, members 39 and 47 will be crltica] for axial forms-
LC-1 OrN)
LC-2 GN)
L, m
39
- 63.65
47
+70.18
+ 20.39 - 37.60
6.76 8 2 (longest of verticals)
Member
m2 < 1.2967
P,, = 70.18
- (-37.60) = 107 78 L N
l
~
' 2&
1
STRUCTURAL STEEL DESIGN
Member 47 is critical for tension design. Minimum r * '
=
8.2(1000) = 34.17 mm 240
+
Use 25-mm bolts: D = 25 3.0 = 28.0 mm. Use two L sections with a 12-mm gusset plate as in Fig. E5-6. Assume two lines of holes, since P is only 70.18 kN, which gives A, = 0.90An.
DESIGN OF TEU'SION LIMBERS
Example 5-7 A portion of a highway bridge truss is shown in Fig. E5-7. required to select the lightest W12 section for member 9 using A-36 and the AASHTO specifications. Assume 2 X lo6 load cycles for the scrwce life of the structure. I,:
,
SOLUTION From the computer output, we obtaln the following (includes impact in live loads): =
+ 80.3 kips
Maximum live load =
+ 60.2 kips
Dead load For A, = 0.75An, the gross angle area is at least
Minimum live load
= - 22.8
kips
From which Use long legs of unequal leg angles back to back. (Why?). Again set up a table of double angles using Tables V-10 and V-11 of SSDD [A, = smaller of (A, or 0.85Ag) X 0.901:
1,
mm
6.3 7.9
'
Al 0.353 0.442
m2 Anet, m2 Ac, m2 Section 2.307 3.468
2.03 2.991
2L127 X 8 9 X 6.3 2L127X 127 X7.9
r,
mm ,A ,
37.5 39.9
(X
P,, P,,,
= 80.3
+ 60.2 = 140.5 kips
= 80.3 - 22.8 = 57.5 kips
The force range (analogous to stress) is P,, = 140.5 - 57.5
2.660 3.910
"
p:' ,
A,
83 klps
We will assume four holes in the flange at any net section, as shown in the insert of Fig. E5-7. Use i-in-diameter bolts, so that the effective hole diameter D = + = 1.0 in. Aholcs
There is nothing lighter, than these two double angles. Note that minimum %v L / r ' controls the design. Now checking stress range and using A,, we have f = -psr =
=
mZ
107.78 = 54.0 MPa 2.03
This value of 63.6 is much smaller than any value of F,, in Table 1-4 up to 2000 X lo3 stress cycles for "base metal" at mechanically fastened joints. Use two L127 X 89 X 6.3 sections. ///
5-8 DESIGN OF BRIDGE TENSION MEMBERS
$
The design of bridge tension members is similar to that using the AISC specifications except that fatigue will have to be considered, as outlined in Secs. 5-4 and 1-9. This will be illustrated by the following example.
= 4(l)tj
High-strength bolts will be used to connect the member to the joint, producing stress range conditions for "base metal at friction fasteners." The data given in Sec. 1-12 are based on an allowable stress range when a catastrophic failure does not immediately occur when the member fails. In this case it is not likely the truss would collapse if member 9 fails. With this consideration we obtain the allowable stress range:
F,,
=
16 ksi
(first column and checking footnote)
We will use lateral and sway bracing across the top and spanning between the two trusses to satisfy stability, but this wdl not reduce the unbraced length of the several web members with respect to the Y axis. (AASHTO requires the effective depth of this bracing to be at least 5 ft or 1.8 m.) Joint fabrication requires orienting the X axis in the plane of the truss for rolled sections. Note that commonly the transverse floor beams are
242
STRUCTURAL STEEL DESIGN
{ t
,* ;" r
somewhat below the point of intersection of the tmss web me bottom chord, but this should not affect the unbraced length L Y. The AASHTO limitation for L/r = 200 for main tension rers and no stress reversals (Sec. 1-7.5), and for a P = 140.5 kips, this would seem deem ttn qualify. The net area, assuming that Fa = 0.55F,, is -- 140e5 = 140e5 Anet = Fa 20
,
7.025 7.025 in2 in2
25(1.414)(12) = 2.12 in (for r,,) 200 We will arbitrarily use the AISC 85 percent reqmrement, so A, is at least rmn rmn =
A >-- 025 - 8.26 in2 - 0.85 By tnal set up the following table, where Ahole.= 4(1.0)$; AWqd= - - reqa
- Aholcs
I
* $4
Figure 5-11 Cable geometry ior derelopmg dei~pnequauobs. N@tc Bat m%rs conf~gurauonthe T at :' the o r i p is larger than the T at the other end iif $e cable.
a
Y d ,
Section -
+, Ln
Aholcs,
W12x53 W12~50
0.575 0.640
2.30 2.56
ln2
Arcqd,
13.30 12.14
ln2
Arm, m2
15.60 > 8.26 14.70
2.48 O.K.
In general, the cable sag y at any point IS 4h x y = - ~ ( u- L ) L Differentiat~ng,we obtain
The W12 x 53 sectlon is the lightest W12 that is satisfactory for both area and L / r requirements. This section 1s also selected so that the connection can be more easily fabricated when the vert~calmember (No. 7) is designed in Chap. 6. Check the stress range for the W12 x 53 section: Psr A , = - -PSr = - =-
A,
83.0 13.30
- 6.24 ksi << 16 ksi
O.K.
+
x
tan8
+ =8 h x- 4-Lh + t a n ~ .
-
(6)
~2
Since
dr = [ I
+ ($)2]1'2
H and
T.= cos 0 -
we obtain a general equation for the tension force in the cable at any point as
Although fatigue is not a controlling parameter for this member, all the truss the truss members should be checked similarly. Use a W12 x 53 section.
///
5-9 'CABLE DESIGN * ' H moment summation about a convenient location with respect to the parabolic "x-bie geometry as shown in Fig. 5-1 1 gives
Noting that all the terms under the square root are relatively imi,onif icant except the first, third, and fourth, we may simplify to obtain
b t :
H z -wL2 8h xhere = mdspan sag, as shown in Fig. 5-1 1 w = uniform cable loading/unit of length; be . . . -- there will always load caused by the cable weight L = span length (the cable length is always somewhat longer)
43
I
-J-
-
When the ends of the cable are at greatly dlffenng elevations, Eq. ( 5 - 7 ) rather than Eq. (5-8) should be used, because the tension at the upper end of h e cable will be considerably different (it is "caqing" the weight of the cable + any additional cable loading). For horizontal cables T has the same value at b o b
STRUCTURAL
1
.,
,
DESIGN OF TMSION bEEJ
STEEL DESIGN
ends and is directly computed as
L . The . , , .. cable ..,..... ,.. ... length
T=~
[ + l16($)]
1/2
is approximately given by S I T ~ I2nd I > I u * r r L ~ b l pruduic c ,riiii.ning 6ileir r o ,~s[e111 Id)
Cables have been used to support large-span roof structures as well as bridges and guys for towers. In buildings the cable roof is constructed by stringing cables across the open space at sufficiently close spacing and pretension force T to produce the desired sag based on Eq. (5-9). The sag in buildings is on the order of to $ (bridges may be as large as In round to structures the cables may be attached to a large compression ring at the building perimeter and terminate in the center in a tension ring. This configuration is the most desirable, since the cable tension must be carried by some kind of anchorage. The compression ring is the most desirable, since large compressive stresses can be used if the nng is made of steel. The use of prestressed concrete roofing "plank" directly on the cables produces the necessary roof and at the same time tends to reduce the vibrations, since the concrete planking is rather heavy and develops a large system mass. Where the use of concrete planking is not sufficient to restrict damping, other means must be employed. Damping tends to modify the natural amplitude of a vibrating system and if reliably introduced, eliminates resonance amplitudes of vibration. The natural frequency of a cable system (the same as any other type of natural frequency computation) is
i).
,
whe?e rn = w / g = mass of the entire cable (including any attachments, such as a roof) n = any integer, such as 1, 2, or 3, used to obtain the fundamental modes; the value at n = 1 is of primary interest, but we may need values for n = 2, 3, and perhaps 4 T= cable tension; can be written as T AT, so that it is evident that a change in T produces a new fn
+
A convenient way to dampen a cable to control vibration is to attach it to a second cable, as illustrated in Fig. 5-12, which has a different natural frequency. This can be accomplished by use of Eq. (5-1 l), which indicates that changes in T different f , values. It will then be necessary to install the cables using ruts to a tension sufficient to satisfy design forces and at T values such values at any of the fundamental modes does not match, which would uce resonance (very large vibration amplitudes that would likely cause a tructural collapse).
C.ible JrTJnpelllcnI m.1) 'cc r ~ d i . i l i n g f r o m rov.
Figure 5-12 Several conflguratlons of cables In building construction (u) Cabies spaced across span for a rectangular buildlng plan ( 6 ) S~nglecable using a central tenalon nng/;md for clrculr b u i g (c) Double-cable system for increased stiffness and mbration control For round buridmg p h - ( 6 ) Cables used In roof support system If cables rad~are.the towers may be esamnally self-supprtug.
I
STRUCTURAL STEEL DESIGN
The safety factors when using cables have been suggeste k i n g from 2.5 to 3.0; compare the factored ultimate tension tables such as Table 5-2. The factored load may be
U = 1.5 dead + 3 live U = 2.5(D + L)
With SF
U=2(D + L + WorE) U = 2.0 X erection loads
=
3, the anchorage resistance would be 3 x 154.7 = 464
xample 5-8 Design the guy cables at The horizontal forces for design are shown in Fig. E5-8. used, but each is to be designed for the force shown. 20-ft centers and will be in compression. The design of the considered in Chap. 6.
-200
m+-
120 k N
-100
m--L
I10 k N
Figure E5-8
S ~ L U T For ~ ~ the N 100-m level, using Eq. (5-6), we obtain -wL2 =
8h
8(110) -w ---h lOd
llOkN
Roofing dead load, including weight of cables (horizontal projection) Live load at 40 psf:
40 x 4
- 0.088
Also, the cable tension is (Eq. (5-7):
[
(:r
T = H I + 16 ,
shown, and this tension will be [combine Eqs. (5-6) and (5-8) to obtain th
+tan20-8
obtained with slight simplification of terms. By Ulal and using SF = 2.5,
19 = 45" +tan O = 1.0: Trial
h = w/0.088
T
T"
1 2
0 (assumed) 0.272 m
155.6 154.7
389 386
Taking 1/2 the roof load as w, 0.024
Close enough to initial
DESIGN OF TENSION hD31BEXS
.M8 S - I U C W R A L STEEL DESIGN
The bottom cable will carry the uniform load, producing the 23.3-ft sag of the upper cable plus the total roof load:
249
e used. Once this is done and the ultimate axial tension load P, is obtained, k
1 1 Q V V
p,, The required cable areas using F = 3.0: Top cable: P,,= 3(74.8) = 224.4 kips -+ use I f-in-diameter bridge strand From Table 5-2, w = 4.73 Ib/ft. Bottom cable: P, = 3(224.4) = 673.2 -+ use 2f-in-diameter bridge strand From Table 5-2, w = 12.8 Ib/ft. The struts wlll carry a compression load based on q producing an h value in the upper cable of 23.3 ft. This gives an equivalent uniform strut "or diaphragm" load of 0.220/2 = 0.1 10 kip/ft. At a 20-ft spacing, the strut load is 20(0.110) = 2.2 kips. Check the natural frequency of the top and bottom cables: First find the cable length:
=
A,+Fy
(+ = 0.68, Table 3-1)
Example 5-10 Given a roof truss member with a length of 11.5 ft in a building frame with a dead load of 18.52 kips and a live load of 22.54 kips (snow load). Use the LRFD method, A-36 steel, and i-in-diameter A-325 high-strength bolts. Design the member using the lightest C shape possible. SOLUTION We will use a bolt pattern as shown In F-lg. Ej-10 so that the AISC shear lag reduction factor will be 0.85 (at leasr three fasteners in the line of stress).
= 354.13 ft With dead and live loads in contact with the top cable and using 350 ft as the contribution span for these loads, the mass is
With two holes out at the critical section, the net area is Anet= A, - 2(;
For n = 1, 2, and 3, we obtain f , = 0.96, 1.92, and 2.88 Hz. For the bottom cable, only the bottom cable weight will be used, since only the struts make continuity with the top cable:
+ ;)lw
From Sec. 3-7, obtain ( 9 factors from Table 3-1) Pu = l . l ( l . 1 0 4- 1.55') =
1.1[1.1(18.52) t 1.5(22.54)] = 59.6 hps
P U 59.6 - 1.88 in' A >-..-=--.----- OFy 0.88(36)
For n = 1, 2, and 3, we obtainf, = 6.78, 13.6, and20.3 Hz. Slnce the natural frequencies f , of the two cables are considerably different, no resonance 1s likely to occur. When one cable is at resonance, the other is at a different frequency, which acts to dampen the resonance vibrations so that the total vibration amplitude is kept small. ///
A >
"
59.6 - = ----p u
- dF
1.39 in?
0.74(58)
Using the largest A,, the gross section using the AISC efficiency factor and the shear lag factor gives the gross area as at least
5-10 DESIGN OF TENSION MEMBERS USING LRFD The design of tension members using LRFD is relatively straightforward. Again the dead and live loads must be identified so that the appropriate load factor
L
Try a C7 * x 9.80: A, = 2.87 in2
3'
ry = 0.625 in
> 0.575
O.K.
250
STRUCTURAL STEEL DESIGN
DESIGN OF m & I O N L
NOWcheck if we can get two bolts in the web, as illustrated in
What is the bolt pitch in Fig. P5-3 so that the critical net section is at [east 205 Answer: 62 mm.
*"5-10. g, = 2.50 in ih
4,:+ ,
rl
:
d
eel is used and the plate
2g, = 2(2.50) = 5.00 in
This leaves a bolt hole spacing of 7.00 - 5.00 = 2.00
,':, ,
A > ,
and
< 2.67(7/8)
Answer: 97.9 kips. 5-8 What is the allowable plate capacity of Fig. P5-3 ~f F, = 250 &(Pa steel is used and the thickness is 12 mm? Answer: 423 kN. 5-9 What is the allowable tensile load for the plate shoan in Fig. P5-9 using F, = 345 20-mm-diameter bolts, and the AISC specifications?
N.G.
Try a C8 x 1 1.50: A = 3.38 in2
g , = 2.50 in
r, = 0.625 in > 0.575
O.K.
t, = 0.220 in
This leaves a center-to-center bolt spacing of 8-5=3in>3D Check the net area.: with two holes out: A, = 3.38 - 2
Check the shear lag:
O.K.
(f+ ;)
- 0.22
A, = 2.940(0.85) = 2.499 Joint efficiency and shear lag: X
2.940 in2
> 1.88 in2
A , = 3.38(0.85)(0.85) = 2.44
Use a C8
=
>
1.88
Flgum PS-9
O.K.
5-10 What pitch is necessary in Prob. 5-9 so that only three bolt holes are deducted from to produce the net section? Answer: 32 mm. 5-11 Select the lightest single angle for a tension load of 50 kips. The length is 6 ft and I bolts wiU be used as shown in the pattern on Fig. P5- 11. Use A-36 steel and the AISC s Answer: L6 X 4 X f .
O.K.
11.50 member.
PROBLEMS .For all problems, assume adequate fastener strength so that only the net/gross section requirement$, control.
*
5-1 Design an eyebar to cany a tension load of 40 kips using a I-in-diameter pin. Use A-36 steel and the AISC specifications. Use t 2 in and w in multiples of in. , .
5-2 Design an eyebar to carry a tension load of 200 kN using a 25-mm-diameter pin. Use MPa and the AISC specifications. Use t 2 15 mm and w in multiples of 3 mm. 5-3 What is the net width of the plate shown in Fig. P5-3 using the given fps units? Answer: 6.75 in.
v, + v, rl
75
3 0 40
50
2"
1,s"
= 250
Figure P5-1 1 210 kN. The length is 1-9 m a d shown in Fig. P5-11. Use 5 =. E O 5-13 Select the lightest pair of angles back to back to c a m a tensile load of 433 kN. Ux 22---diameter bolts, a 12-mm gusset plate, and F, = 3 4 5 &{Pa steel. The member lenglh is 3.2 EL Use the bolt pattern of Fig. P5-13 and the AISC specifications. Answer: L102 X 76 X 6.3.
3"
40
F,
Flsurep5-3
5-4 What is the net width of the plate shown in Fig. P5-3 using the given SI units? Answer: 176.3 mm. 5-5 What is the bolt pitch in Fig. P5-3 so that the critical net section is at least 8 in? Flgure F5-13
' ~ 2STRUCTURAL : STEEL DESIGN 5-14 Select the lightest pair of angles back to back to cany a tensile load of 92.5 kips. i-in-diameter bolts, a f-in gusset plate, and F, = 50 ksi steel. The member is 8.375 ft long. bolt pattern of Fig. P5-13 and the AISC specifications..' Answer: P = 154 kips.
5-15 Select the lightest single angle for a tension load of 68 kips, assuming one the critical section. The member is 7.5 ft long. Use A-36 steel, the bolt pattern and both the AISC and AASHTO specifications. Assume no stress reversals fo Answer: By AASHTO, L7 x 4 x
A.
F!
..,. .
4
,.,
.,,..,.
..
P
Figure P5-15
5-16 Select the lightest single angle for a tension load of 220 kN assuming one 20-mm-diameter bolt at the critical section. The member is 4.3 m long. Use F, = 345 MPa steel, the bolt pattern shown in Fig. P5-15, and both the AISC and AASHTO specifications. Assume no stress reversals for AASHTO. 5-17 Design the bottom chord members to satisfy tension for the side shed truss of Example 2-5 using WT (structural tee), i-in-diameter bolts, A-36 steel, and the AISC specifications. Assume two bolts at the critical section is the web of the tee. The tee is continuous across the critical joint. Answer: WT9 X 27.5. 5-18 Design member 5 of Fig. E5-7 using a W12 section if the computer output (including impact) is Dead load = 160.5 1 kips Live load = 77.17 kips (maximum) Live load = - 5.8 1 kips (minimum) Use the AASHTO specifications and A-36 steel. A m e r : W12 x 53.
5-19 Design the guy cable for the 200-111 level of the TV antenna of Example 5-8. 5-20 Design sag rods for the purlins shown in Fig. P5-20. The purlin span is 28 ft and spaced on 8-ft centers. The sag rod is at the midspan of the purlin. The roof slope is as shown. Dead load = 25 psf of ioof surface
5.23 Do Prob. 5-21 using the AASHTO specifications. Do not consider fatigue. 5-24 Do Prob. 5-22 using the AREA specifications. Do not consider fatigue. Answer: 747.4 kN.
4
Live load = 45 psf (horizontal project) Use A-36 steel and the AISC specifications. Answer: Diameter = 1$ in.
Answer: L3;
Figure P5.20
,,
8
,
3
X
7/16.
5-26 Redo Example 5-10 if the truss member length is 14.5 f t instead of 11.5 ft. 5-27 Given the bottom chord of a truss using a pair of C2CO X 17.1 1 back to back with a 15-mnr gusset plate between them. Using two 25-mm-diameter A-325 bolts at the critical section and for a dead-load bar force of 120 kN, what is the maximum live-load bar force that is d o w e d = k g FL = 1.67? Use F, = 250 MPa and a panel length of 5.1 rn. Assume at least thrrt fastrnrrs i n , ~ e line of stress and the AISC specifications as applicable. Answer: LL = 365 kN. 5-28 Design the bottom chord member (No. 12) of the highway bridge truss oi E w p l e 5-7 (refer to Fig. E5-7) using a built-up section. Use Examples 6-7 and 8-3 as a guide in selecting the rolled sections to make up the cross section. Loads: dead = 336.9 hps (tension); live mimimum == 13.8 kips (tension); live minimum = 0.0 kip. Use the AASHTO splcifications, i-in-diameter highstrength bolts, and A-36 steel. Panel length = 25 ft, as shown in Fig. Ej-7.
45 lb/ft2
..&&+ha.%
X
4,%W '
AXIALLY LOADED COLUMNS AND STRUTS
6-1 INTRODUCTION The vertical compression members in a structure are commonly identified as columns (sometimes stanchions in fore~gnliterature). Sometimes verti pression members are called posts. The diagonal compression members ing the top chord of bridge approaches are end posts. The diagonals of a truss members used in wind bracing may be called stnlts. Short compression m at the junction of columns and roof trusses or beams may be called knee In all cases, however, the member under consideration is carrying a compress1 load. A structural member carrying a compression load is termed a column if length is sufficiently great. For lesser lengths the member may be called compression block. The length which divides these two classifications is such it affects the maximum compressive stress which can be developed under load. The length is seldom used alone in describing column behavior. Rather, a n offshoot of the Euler column fonnula developed in the next section, the r of column length to radius of gyration ( L l r ) is used.
construction. Joints are field-fabricated using high-strength bolts.
Material
Limiting L/r (approximately)
Steel Aluminum Wood
60 30 10
S T R U C STEEL ~ DESIGN
AXIALLY
LOAD COLUSD~S
reasons, which include the following: 1. The difficulty 'of determining the exact point of demarcation between co pression blocks and columns. 2. Columns, although appearing straight and homogeneous, may have sm imperfections and always have residual stresses from mill operations, such rolling, cooling, and so on. Any small imperfections will result in a eccentricity about one (or both) of the axes and produce lateral deflectio ,(buckling) due to the bending moment that is produced as the product ;load x eccentricity. 3. It is often difficult to apply a load through the center of area (i.e., appl
From these several considerations it is evident that if an ideal, isotropic
-------------1_____
x
=
A sin hy
+
B cos Xy
member. undary conditions of x = 0 at y = 0, we obtain B = 0 an
6-2 THE EULER COLUMN FORMULA One of the most popular column formulas ever proposed (and there are a ve large number) was derived in about 1759 by Leonhard Euler, a Swiss mathematician. The formula is readily derived (refer to Fig. 6-1) as follows. Writing the fundamental bending equation for moment, ,. , . .
x = A
sin hy
ection of the equation, so the solution must b sin ky = 0. This is only possible for values of k L as follows: kL
= n,
Zn, 3n, . . . , n7;
..,,..,...
we have for the Euler case that the moment is P x , so
dZx EI-= (iy2
-Px
Using Eqs. (e) and (c) to obtain P, we obtain
'
&
Ql, b I
This equation for the critical column buckling load P is generally called Euler .equation (and the load, the Euler load; the stress, the Euler stress). es of Eq. (6-2) by the column area A, noting that the ra ction r = , Fcr= P,,/A, and that n = 1 gves (or Fcr), we obtain
t
7 r 2 ~
. I 1
.
Fcr= (L/rI2 we obtain from Eq. (6-2), . I
t
4n2E Fc, = -- . ( ~ / r ) ~
I
a
which is equivalent to the column containing two sine waves in the length L. ~ G i is s called the second buckling mode, n = 1 is the first buckling mode (single sine wave), and from Eq. (6-2) it becomes evident that the minim m crit' . , b ~ c q load g (or stress) isiobtained in the first buckling mode. lnispection of:,Eq. (6-3) indicates that very large values of Fc;,'kfn b obtai'ried by using L / r + 0. Implicit in writing the differential, .equatibn bending [Eq. (a)], however, is stress being proportional to strain. Thus the uvv limit of validity is the proportional limit, which is often taken as Fcr-+ Fy
&,
Taking the derivative dx/& for the slope, we obtain
& Since dx/dy
5-3 COLUMNS WITH END CONDITIONS Rotation of the ends of columns in buildings is restrained by the beams that frame into them. The ends of truss members are similarly restrained. In both cases the design may be made on the basis of pinned ends. Note that the Euler equation derivation was for a perfectly straight, pin-ended column T h e dpnvgtion for the critical buckling load for columns with various end restraints can be done in a manner similar to that for the Euler case. This will be illustrated for the fixed-end column shown in Fig. 6-2a. The differential equation for bending now becomes J
2
=
cos hy - k sin ky P
sin k L
0 at y = 0, we have k M o 1 - cos k L P sin k L
The smallest root of this equatlon glves k L have
=
=
0
2 7 and with k
=
(P/EZ)'/', we
or the effective column length KL is L / 2 = 0.5 L and K = 0.5. We have now introduced one of the first attempts to adjust the column Iength for end conditions. The effective length K L or some equivalent is used in nearly aiI design formulas in all bullding specifications. When the top of the column :s f~xedagalnst translation and the base fixed against both translation and rotation as in Fig. 6-2b. we can rewrite Eq. ( f ) u
Using the standard method of solution as before and noting that we have simply added a constant, we obtain x = A sin ky 1
Mo + B cos ky - P
and in a similar manner obtain
where k = (P/EI)'/~, as before. With boundary conditions of x = 0 at bot y L'6 and L, we odtain the constants A and B : A,
MO 1 - cos kL . sin ky P sin k L
x=-(
a
+ cos I+ - 1
or K = 0.7. Several of the more common cases of column end f i t y are s h o ~ i l in Fig. 6-3.
b l K U L 1 UKAL
STEEL DESIGN
.AXIALLY LOADED
COLmNS .L\iD S<S 261
at accounts for a lessening effect of some of these variables as the (or K L / r ) decreases. For example, a small eccentricity is not a3 critical short column as for a long slender one, and so on. With these considria, let us investigate the AISC column equation for axially loaded columns.
.1 AISC Axially Loaded Column Design Stresses develop the AISC design stresses by taking the Euler equation for critical ss, and with n = 1 we have
F,,
T ~ E =
(KLI~)~
differentiating obtain Recotnineniled
d(F,,)
-
d(KL/r) Rotation fixed
conditions
Rotation free
.,,.
ce the maximum value of the Euler stress or any critical buckling stress is .to let us also define the critical stress in any re,mion where the Edzr tress is not valid (such as smail K L / r values):
5,
Translation free
Flgm 6-3 Theoretical and design values of K for columns with end conditions shown. ..
(a>
(KL/~))
Translation fixed Translation fixed
7
-277'E
.,...,...
Strictly, according to the derivation of these equations for critical buckling load, the tangent modulus of elasticity E, should replace the elastic modulus in tangent modulus concept was introduced in Sec. 3-7.
F,,
=
CY- "j7)
KL
(6)
uation can be differentiated to obtain d(Fcr)
-
-,+)
KL
P-'
d(KLlt.1 ake the slope d ( F , , ) / d ( K L / r ) = 0 at K L / r will arbitrarily define a parameter
(c> =
0.Also, at some point
KL
-=
6-4 ALLOWABLE STRESSES IN STEEL COLUMN The column stress obtained with the Euler equation of
r Cc lopes of Eqs. ( a ) and (c) will be equal (i.e., the two curves defined by these equations will have a common tangent). Also, experimental column test ta indicate that takingp = 2 is adequate. Now equating slopes at K L / r = C, d for p = 2, we obtain
- 271'~ - 2 r n ( ~ , )=~7 ( C,) m which obtain m as m=-
account for eccentricity, residual stresses, and the several other factors that complicate the theoi'y. It would also be appropriate to use a variable safety
T*E
(dl
c,* rom rearranging Eq. (b), inserting K L / r = C,,p
=
2, and using the Euler
,
,
.iAnwLIUn,u
.d
bltjEL DESIGN
AXIALLY LOADED C0LU;CNS
k i ~ ~ . for i c Fcr, we obtain
compression members may be designed for an allowable stress based on the following amplification factor using L / r ( K = 1) when L / r exceeds 120:
arid solving, we obtain for Cc,
We obtain the allowable deslgn stress uslng e~therEq. (6-5) or (6-7):
F; = F, x If
rLW STRUTS 2.63
value of m, p, and Cc are placed in Eq. (b), we obtain
the critical buckling stress becomes 0 . 5 q at K L / r = Cc under our acsumptions. In general, we have the buckling stress from Eq. (b): 711:!;
\k
Equations (6-5) to (6-7) are somewhat awkward for routine computations (even with the programmable desktop calculators), and ~t IS convenient to write a computer program to produce a table of Fa vs K L / r for various values of F, shown In Tables 11-5, 11-6, and VI-5. VI-6 The AISC manual has morz comp'ere tables using the several commonly used grades of I., , ~ncludlngthe amplrficairon factor In Eqs. (6-5) and (6-7) for secondarq members uhen L / r > 120.
6-4.2 AASHTO Axially Loaded Colunln Design Stresses
sv
-
the allowable stress is obtained for K L / r 5 Cc from Eq. ( j ) :
The AASHTO formulas for axlally loaded columns are derived similarly to the AISC values, but the SF tends to be somewhat more conservative, since the members of the bridge structure are, in general, in a more hostile environment than building members. The AASHTO formulas are as follows: For K L / r I Cc:
The AISC kas used the following variable safety factor since 1963: For K L / r for all values of K L / r I Cc. For K L / r > Cc, use a constant value of safety factor based on using K L / r = Cc in the equation above to give
When K L / r > Cc, the Euler equation with SF = 23/12 is used to obtain :he allowable column stress as
> Cc:
4'
F',=
n * ~
The SF = 2.12 for the AASHTO specifications and the values for the lens& factor K and Cc are computed the same as for the AISC spzc~fications.Tab!e 6-1 gives Cc for the several values of F, commonly used for columns. *
Table 6-1 Values of Cc according to the AlSC and AASHTO specifications for several values of F,; C, = ( 2 n 2 ~F,)'/~ /
With standard values of n 2 ~we , obtain:
.
..
fps:
Fa =
149 000
fps, ksi
ksi
(KL,'~)~ b
1
F',
=
1.03
x
lo6
Mpa (KL/~)~ Eqx~tlons(6-5) and (6-7) are used for main compression members. Secondary
f 6- 10)
SF(KL/~)'
SI, MPa
c,
264
STRUCTURAL STEEL DESIGN
643 AREA Axially Loaded Column Des;ign Stress
6-45 Column Design
~h~ AREA allowable column design stress formulas are somewhat similar to the
It is necessary to use an iterative process in the desiw of compression members using any of the AISC, U S H T O , or A3EA allowable stress equztioas. niusual design problem involves the following steps:
B
4
1. Detennlne column loads (unless the ~roblemonlv ~ n v n l v e
2 CAC~;R-
Allowable stress
Limitation
-
-, -
3. Make a tentative section selection (sometimes use can be made of bb!rs such
I>.
.
1U7 < % _ < = -
': fi
f
i
F , = O . ~ F , - (1k.2)
#,A,
.7
as Table 11-4 or Table VI-4 of SSDD). 4. Compute KL/r for the section selecied (now that r is known) and uit the appropriate stress equation to compute F, (or use tables such as Table 11-5 of
(6- 12)
ssnn
which
O~VPP
F
fnr r ~ ~ r n r oVT l
w
--+:--\
when Po = P,,,,,. In the AREA specifications, F, = ksi k = 0.75 for riveted, bolted, or welded compression
Two additional factors should be given consideration: .*
member end connections (and k , not K ) = 0.875 for pinned-end members
#.4
Net versus Gross Column Cross-sectional Area
q e net area (= gross area - loss for holes) is used in tension member design. In any connection design using mechanical fasteners (rivets or bolts), it is assumed that the fastener completely fills the hole. This assumption is very nearly met in riveted work, where the head fabrication enlarges the rivet shank, . and very nearly occurs in bolted work since the hole is only about 1.5 rnrn larger than the nominal bolt diameter. Under axial compression, although there are stress concentrations at the hole, it can be safely assumed that no loss qf net area occurs when a mechanical fastener fills the hole. When an intermediate open hole is in a compression member (as for utilities, erection, etc.), the designer must exercise judgment as to whether to use the net or the gross area. The AISC specifications give the allowable stresses as "On the gross 'section of axially loaded compression members . . ." Undoubtedly, .there will be adequate arching to transmit the load around $he hole if there is only a small an'iount of area lost at any section due to the holes. The most conservative procedure would be to use the gross section when the hole has a mechanical fastener that fills, or nearly fills, it and the net section for all other cases.
.
.
c
1. Commonly, only the,W8, W10, W12, and W14 sections and rectangulrrr tube and round pipe sections are used for columns, since the critical radiu of gyration is with respect to the Y axis. These sections have the best 5 vrtIues (and corresponding r , / r , ratios). In building design. practical consider-1+' IORS often necessitate use of a given nominal column size throughout the buildng. It is usual in building construction to run a single column t h ? o u ~ at I z a r two, and often three or more floors to avoid fleld column $ f i w . v ~ n c Iabor savings more than offset the Increased we~ghtof steel. 2. When KL/r > C, the AISC speclflcat~onreq6lres use of Eq. (6-7), the AASHTO specif~cation requlres Eq. (6-10). and ARE,-\ has a somewhat similar requirement for Eq. (6-13). In all these equations F, is independent of F,. Therefore, In column design one should use A-36 steel for all cases where K L / r exceeds C, or the AREA lirmtation, and even if F, > 36 ksi is being used for some of the other members. For example. if we are using F, = 60 ksi and a section is found where KL/r > 97.7 (refer to Table GI), we should try to specify A-36 steel instead of the more expenslve 60-ksi steel. The sligqtly larger safety factor in Eq. (6-7) allows for the transition between K L / r = C, = 126.1 (for A-36 steel) and the lower values of C, for the higher-strength steels.
Advantage should be taken of any ava~lable design aids in mdcilting a preliminary column selection. Tables such as Table 11-4 or VI-4 of SSDD c a often be used to make the final design. Tables 11-5 and 11-6 and the correcponding SI tables (Tables VI-5 and VI-6) can be used to advantage for any o h s r
266
STRUCTURAL STEEL DESIGN i
{md including W) shapes to quickly obtain Fa when KL/r is computed-partic-
< C,
ularly when KL/r KL/r.
60 lhlit! I I
+ 30 iblft'
rooi X 10 = I 8 k 1 0 (1
since both stress and safety factor now depend on
. .. 6-4.6 Design Examples The design of simple axially loaded columns and struts will be illustrated in the following examples.
Example 6-1 Design a column to be used in a one-story discount departstore building. Columns are spaced 20 ft on center both ways. The roof load is taken as 30 psf dead load and 60 psf snow. This gives a column load of 20(20)(0.090) = 36 kips. Use A-36 steel (C, = 126.1).
':.merit
SOLUTIONFrom inspection of Fig. E6- 1, take K = 1. By using Table 11-4 as a guide, the lightest W section (W8 X 18) for KL = 14 ft can carry 41 kips. This could be a iolution but is not very economical. We will first check this W8 section and then compare to using a tube column. Check the a W8 x 18 section:
KL - --=14(12) Y'
136.6
> Cc
[use Eq. (6-7)]
,?&
1.23 LI.Yb
Pa = 5.26(7.99) = 42.0
> 36 kips
-
Fa = -- 11.5 k s ~(vs. 11.54 in Table 11-5) 1.915 O.K. Pa = 1 lS(3.17) = 36.45 > 36 kips
Fa = 149 OoO = 7.99 ksi (136.6)~
O.K.
Use a W8 X 18 section (tentatively); the Table 11-4 value of 41 b p s is due to computer roundoff. Let us also investigate a pipe section (see Table 1-14). Now that we have some "experien~e" from just checking a W shape, let us investigate a 4-in-diameter pipe:
Use steel pipe 4 in x 10.79 lb/ft. Note that pipe and any other sections where r , / r , most economical shapes to use when KL, = KA,.
KL,
=
KL, = From Eq. (6'-5),
1 are geneidly
///
Example 6-2 Design a W section for the conditions shown in Fig. E6-2. Use F = 50 ksi steel and the AISC specifications.
1 I
4
16 f t 8 ft
11-4 to obtain an inltial estimate of column r:.:r. able 11-4 is dased on Fy = 36 ksi: use a ratio as 36 PCquiv= (750) 30 = 540 kips
Since
l ji I
AXIALLY LOADED COLUILCiS .LSD STRLTS
%3
I
This appears to be the lightest rolled section possible for this lozdisg situation. // /
Example 6-3 Check the pair of angles selected in Ertamplz 5-6 to be usid 23 vertical members in the main roof truss of Example 2-6. Use A-36 stet1 a d the AISC specifications. LC-1 (kN)
Member
LC-2 (kN)
L, m
The section selected for tension considsrarlons was 2L127 x 89 x 6.3: cl = 2.66 x
SOLUTION Take K
&
=
rn'
r,,,
= 37.5 mm
1. For member 39:
Figure E6-2
--
Since the table is based on KL/ry, we qote that rJr, 1.6 in the column sizes approaching P = 540 kips; therefore, look for L = 1.6 X 8 = 13 ft column length and 540 kips. Try W14 X 99:
For member 47: N.G. Revise the section for minimum r:
Compute HI
r, = 1.66ry = 1.66(3.71) -=-= KL
rx
16(12) 6.16
3 1.2
=
6.16 in
controls since it is largest
Note that a 12-mm gusset plate is used between angles at the connection, Possible angles are: 2L152 x 152 x 7.9: r,,, = 48.0 (required for tension) A = 4.71 > 0.5504 2 L152 x 102 x 6.3:
r,,,
A
C, = 107
(Table 6- 1)
=
41.5 mm
=
3 15 >0.5504
We will check the pair of angles 152 z: 102 x 6.3. slncs they are the !Ijl:est. Member 47: 8'2(1W0) = 197.6 -+ F, = 26.3 \(Pa 41.5 Member 39: -KL =
r
F,=--47'9 - 26.9 ksi
,
..
..
(= 27.0 in Table 11-6) 1.78 Po = 26.9(29.10) = 782.8 > 750 kips O.K.
P,, = 38.8(3.15) = 122.2 kN > 63.65 P4,=26.3(3.15)=82.8kN>37.6
(Table VI-5)
0.K O.K.
, XYIALLY LOADED
STRUCTURAL STEEL DESIGN
COLmfiS .<4ns n m s nl
By trial [and programming Eq. (6-9) on a pocket programmable ci!cuIator]:
Sectlon
A ~ , , ~in2 .
ry, I D
Fa. k s ~
W12X53 W12x58
15.60 17.00
2.48 2.5 1
9.35
P,,,,
=A
F*,Iups
N G , since ry too small
159.0> 99.3 O.K.
The r, value for the W12 x 58 IS 5.28 > 2.5, so the section is satisfactory. Also, this section is the same size as member 9. so the joint ullI be easy to fabricate. Use a W12 x 58 section. /// Example 6-5 Design the struts for the cable-supported roof of Examp!r 5-9. The strut load is 2.2 kips based on cables at 4 f t on centers and struts as shown in Fig. E6-5. The maximum strut length IS 2('33.3) = 46.6 ft.
Figure E65
SOLUTION The maximum K L / r for compression rnembsrs is limited to 2W. According to the AISC specifications, this would require a radius of a ~ a tion r of at least
+
Maximum column force = 56.75 42.56 = 99.3 kips (compression) Minimum column force = 56.75 - 16.11 = 40.6 kips (compression) P,, = P,,, - P,,, = 99.3 - ( + 40.6) = 58.7 kips (stress range)
1
The AASHTO specifications (Sec. 1-7.5) limits KL/r for compression nembers to KL/r = 120 for main members or members with both deadand live-load stresses. The minimum r is
Try a W12 section, since we have used a W12 x 53 in Example 5-7 for, amember 9, w p h frames into the joint on one end of this member. A gusset$ plate can covkr both members with a minimum of filler material if all the web members have the same nominal depth. We further note that Eq. (6-9) always determines the column stress for members using A-36 steel since K L / r I 120 and C, = 126.1 > 120.
Commonly, plpe struts are used with diameters rdnging from 4 to 6 ic. H e i ~ we have a rather large unbraced length. qo euarnindtlon of Table 1-14 (SSDD) indicates that we can use. Extra strong plpe: 8-in diameter: r = 3.88 > 2.80 in A
For KL/r
=
46.6(12)/2.88
F, P '
,
=
=
12.8 in'
194.2, (Table 11-5)
=
3.95 ksi
=
AF, = 12.8(3.95) = 50.6>>>2.3kips
Use 8-in-diameter extra-strong pipe. I t m a be advantageous to cse
2
4
272 STRUCTURAL STEEL DESIGN ,,
*
smaller section in the outer one-fourth of the span, where the value L = 2h is less than 46.6 ft.
/
6-5 DESIGN OF BUILT-UP COMPRESSION MEMBERS ,
A built-up section is a more practical design than using a rolled shape in many situations. This is particularly true when there is a very long unsupported column length involved such that to meet the L / r requirements would require one of the heavier rolled shapes. Another factor of primary importance is that the- radius oPgyration of built-up members can be controlled (see Table 6-2 for selected examples so that the value of rx can be made more nearly equal to r,, to produ
..
-up sections are very commonly'uskd for bridge trusses and..in co. r towers. Antennas are essentially built-up colu mns, although not onsidered ,as sudh. In any case, where compre:jsion (and tension) embers are used in large unsupported spans, a built-up m ember may need to e considered. Any of the sections considered in Chap. 5 may be used (refer to igs. 5-4, 5-7, and 5-8) in addition to any other section configuration, which ay, or could, be made appropriate for the design problem. It is somewhat more difficult to produce an optimum, or least-weish~, built-up section since there are several design parameters to satisfy, including:
. Types of members to use, including rolled angles; cha shapes, as well as plate segments. Arrangement of the basic members, including any size limitations for overz!I section dimensions. . . . . -.. . ._ 3. .The resulting computed values of I,, I.,, r,, and ry and K L / r,,,, ^whicil produce the allowable compression stress. 4. Producing an acceptable section area based on the allowcable stress from s:tp 3, which is not known until the area has already been established. i
One or more iterations are usually required to f~nallydevelop a satisfactory section. The number of iterations obviously wlll depend on 1. Engineering design versus material cost. 2. Number of sections to be fabricated; if 100 sections are to be fabricated, the material savings can be considerable, whereas in the fabrication of only focr or five sections, the cost of producing a refined design might cost more ti223 the material saved. 3. Reliability of load data a,nd intended use of the built-up members.
I
I
r..
rb7
= same as
r , = 0.4211
1
-
1
In general, a 5 to 15 percent "overdesign" of a built-up member will be satisfactory. Built-up sections may be built using rolled shapes as in Fig. 6-4, but more commonly are constructed using lacing, perforated cover plates, or batten plates, as shown in Fig. 5-7. A larger radius of gyration and more efficient use of steel is obtained by separating the load-carrying parts as much as practical (and as done with the flanges of W shapes). Where this is done, it is necessary to interconnect the parts so that these several parts act as a load-carrying unit. Several methods
t-7
-
---
----]l---I
I.oi2,L-s
U
/ r
y
7 7
i
Figure . 6-4 Built-up shapes using cornbinatlons of rolli.,! sh-psi. Sc.iti,,n
.
""
"A:-..
,--A
".-A
-1
""
,-,
---I..
-..A
-1
....
,I.,
-r~.,.L
!iri:lr:d
;;,:,n!ctr) 5-
"-2
? '
.L--.
.~:1:> :?j /.\
"-
.I of using lacing, single and double batten, and perforated plates (called, collectively, cover plates) are usually used. Where the steel is located inside a building, the cover plates may be solid and their use could reduce fabrication costs. In exterior environments, where corrosion is always a problem, it is necessary to have access to the interior of the section for maintenance and inspection; otherwise, the interior must be completely sealed. The "open" cover plates and lacing allow access to the interior of the section for cleaning and painting without the careful fabrication required to completely seal the interior. Presently, it appears that the economics of fabrication favors perforated cover plates to lacing, since automatic gas-cutting methods allow rapid cutting of the plate openings in a length of cover plate. The design of lacing and batten plates, in particular, requires attention to several details: Lack of proper attention to lacing design was believed to have caused the first' Quebec Bridge in Canada to fail in 1907. It is standard practice to allocate a portion of the axial load as the shear developed in the lacing or batten plate when the compression member "buckles," as shown in Fig. 6-5.
If we assume equal end moments, as shown in Fig. 6-5c, and use the differential equation for bending as used to develop the Euler column equation, and allow for boundary conditions, we obtain
The derivative at y
=
0 is d.~
kL 2
- = ke tan -
4
Now referring to Fig. 6-5h, we obtain for the $hear in the lacing,
where k
=
P / E I as in the Euler equation. AISC simply takes kL ke tan - = 0.02 2
The AASHTO and AREA specifications make the assumption that the end eccentricity e shown in Fig. 6-5c is equal and opposite on the two ends of the column (shown equal and with the same sign in Fig. 6-5c). With some a d d i c o d simplification of the preceding equation for V, we obtain
where
F,= steel y~eldstress, k s ~or lCf Pa L / r = value for entlre member wlrh rebpect to an axls perpendicular to the plane of lacing or perforated co\er plate as follo\~s: AASHTO
fps
(u)
(b)
(c)
Figure 6-5 Shear development for a laced (or battened) compression member. (a) Laced column. ( b ) Identification of laced column shear V. For lacing on both through faces, divide V equally on both lacing bars; for four-side lacing obtain 90" as above. (c) Basic concept of shear in lacing of built-up section.
SI
ARW
fps
SI
AASHTO and A ~ E Aspecifications also require that V be increased for any additional shear on the member, such as section weight or other tr, P-sverse loadings. Wind on bridge trusses would also contribute an increase in V ilndzr this interpretation. The value of V obtained for either the . U S H T O or XZSC computation may be either a tension or a compression value, and the l x i n g or batten plaie should be so designed. The spacing of lacing and batten plates must be such that the L / r of the main elements between fasteners is not greater than K L / r of the entire member; otherwise, local buckling might develop, particularly where angles are used, m d L / r , between fastener points may be critical. Ths XXSH-TO and =IIRE,I
End
AISC C':
L,, =
< (',
I
.:I-
[?j.
i..
\/F'J i i p s i
C',
=
2 if,50
5.;: Sl
.oillnrs~~t~~ri
;? L i 1 :
- -
.I I ,<>
ZL,L
:,"
2 ------ 110.:ai.-,
IML,,I(L:
iI
Figure 6-6 Design of perforated cover plates
where Iq,,=moment of inertia of the jth pdrt aith respect to the pciriicl r.ui I and through the centrold of the j t h part A,= cross-sectional area of the jth parr d,(,,= perpendicular d~stancefrom the centroid of the jth area to rile i axis 5. Compute the radlus of gyration wlth respect to both axes.
-
1. General outside dimensions and load to be carried. 2. Estimate the tentative compression area based on assuming an F, between 15 and 20 ksi or 100 to 140 MPa (based on F, = 36 ksi). This allows a modest reduction in the allowable stress from 0.6F, due to the K L / r of the built-up shape. 3, Dec~deon lacing, batten, or perforated cover plates (Fig. 5-7) or if sect~onis to be someth~nglike that of Fig. 6-4. 4. combute area, I,, I,, r,, and r,. The moment of inertla of the built-up section ,
#.
IS
I
t
I,
=
=o(J) + Z A , ~ ? , )
6. Compute KL/r, and KL/r, and obtaln the allo~~able compressive stress based on the largest KL/r.
,
7. Check P = AF, > P ,,, and iterate as necessarq. 8. Design lacing, perforated cover plates, batten plates. and/or tie (or SIZ:~) plates. This procedure will be illustrated by the following ex~mples.
4
AXIALLY LOADED COLLMXS .W STRL.?S
z9
Exampie 6-6 Design a laced section for the end post of the highway bridge truss of Example 6-4, which has seven panels at 25 ft each (see Fig. E6-6a). The unsupported length 4, = Lx is 24.04 ft. The computer output (with an impact factor of 0.17 included) is as follows: Member
LC-I I, kN
Dead load, kips
Compute the radius of gyration about the X and Y axes: About the X axis: locate a new X axls:
+
Assuming that F, mately
Total design load = 240.77 94.13 = 334.9 kips 14 ksi the area required in the section will be approxi-
--
I,,
=
Both
t/ and
t,
+ 211,d2 + i l p d 2
+
2(162) 2(8.82)(1.66)' = 324 + 48.6 + 134.4
=
Let us try two channels with a solid cover plate aqd lacing as shown in Fig. E6-6b. This configuration, with solid cover plate up, will provide some protection to the interior of the built-up section, and lacing will allow access for painting and cleaning. The spacing and configuration will be such that a reasonably easy framing of the W12 web sections can be made, as shown, using a pair of gusset plates. We note that filler plates will be required, since the W12's have depths greater than 12 in. C12 X 30 data:
2I,,
+ 6.375(4.59)'
= 507.0 in4 =
4.59 in
About the Y axis: Iyy = 2 1,
controls (after computing r, below)
+ 211,d2 + I,
> 0.23 in (Sec. 1-7.7)
334'9 22.5 =14.9 -
< 24.02 in2 furnished
O.K.
280
STRUCTURAL STEEL DESIGN
LSD STXbTS
.%XIALLY LOADED COLL-NS
Current practice is to use perforated cover plates welded to the rolled sections rather than lacing. We will design lacing for this example and use welding to attach it to the rolled sections. The only bolting will be the field conn~ecti-onsof the joints. The distance between flange holes (and the approximate center-tocenter weld distance) is shown in Fig. E6-6c.
L' = 12 - 2(1.75) = 8.5
<
15 in
Try t
-&
3l
in (0.313 in also rmnirnum AASHTO): L = r
0.288(0.3 13)
=
109
O.K.
Compute the lacing bar force. The bar force component psrpzndicu!~ to the member axis is computed using Eq. (6-14):
(also AISC)
100 62.8 + 10
6;ig)]
+ ---------
3.349(2.06)
=
=
6.9 kips
'?
We will increase this value 20 percent (author's decision) to allow for member weight, wind, and any other factors; thus Vdalgn is V, = 1.2(6.9) = 8.3 kips The axial force is
pd
=
8.3 ---- 9.6 kips
cos 30' 36[1 - 0.5(109/126.1)'] Fa = 2.12 btFa = 9.6 kips d/"
=
10.64 k s ~ q
Figure E6-6~
Use single k i n g bars at an angle of 60' to the member axis, as shown in the figure (refer also to Fig. 6-5a):
L=-'
5 - 9.81 in cos8 300
L' = 2(8.5)sin 30' = 8.5 in L'
[ -1"'
r = ,$if) t=
use 1 :-in plate
The final lacing bar dlmenslons wlll be taken as ( L
=
9.81 i- l.b9 in)
- 8.5 - 11.1 << 40 ---
0.763 Also, 11.1 is less than 0.67 X L / r ( = 41.9). Limit the L / r of the lacing to 130. The radius of gyration of a flat bar is ry(of channel)
Let us revise t to 7/ 16 in; Fa = 13.78 ksi: 9.6 = 1.58 in b = 0.44(13.78)
Design the end tie plates (AASHTO calls these stay plates). The AASHTO requirements are:
= 0.2881
9.8 1 = 0.262 in 130(0.288)
t > - -L= - - 981 - 0.245 in 40 40
Use a minimum of three fasteners (or equivalent weld) each side. Transllited into design (Fig. E6-6d):
t = - l2 = 50
0.24
use $-in, to match lacing bars
STRUCTURAL STEEL DESIGN
AXIALLY LOADED COLUMXS h 1 \3 STRUTS
Weld
\
Flgure E6-6d
Example 6-7 Design a built-up section using perforated cover plates for use as a..column in a water tower (see Fig. E6-7a). The unsupported column length is 24.7 m and the axial load for design is 1350 kN. Use F, = 250 MPa and the AISC specifications (noting that a water tower is not a "building" and may be located where a collapse is more of an expensive nuisance than a hazard to people, so it may not be necessary to use any specifications), since adherence to these specifications, although not necessary, will ensure a safe design. There is usually some wind bracing in water towers, but we will assume that the bracing is not sufficient to develop restraint against column buckling.
For this L / r value of 200, the allo~~nbie stress F, = 25.7 !vfPa. Tnerefoie we must use an r greater than this or the steel area WILL be excessive. For ar L / r = 100, the value of F, = 90 MPa and the area of steel will bc m'. approximately 1350/100 = 13.5 x Try 4 L152 x 152 x 14.3 (somewhat arbitrary choice). The data are:
,x = ). =
43.4 mrn
Place the angles in a symmetrical section with spacing as shown in Fig E6-7b. Compute I, = I, and r, = r, using section data:
- = 24'7(1000) = 1 10.26
r Pal,,,
and from Tab,? Y 1-5 obtzin F ,= 8 1 \fPr
. 224
=
81(4.184 x 4) = 1356
>
1350 kN required
O.K.
This cross section will be considered adequate and we will proceed to dzsis the perforated cover plates. The cover plate design is not so much "design'. as satisfying seIectec criteria and producing a hole spacing that will f i t the column Ien& Referring to Figs. E6-7c and 6-6, we obtain: Figure E67a
SOLUTION The minimum K L / r will be taken as 200 for main members. Therefore,
r =.75 mrn (arb~trarqselection of hole rzd~us) w = 2r = 150.mm a Use L, 2 1.25Lo + 0.45 m. Le,ngth of cover plate = 21.7 - 2(0.45) = 23. m. This choice 1s made so th5t bqstR' a hole length of L, = 3CYI rnm
AXIALLY LOADED COLUMNS AUD STRUTS
STRUCTURAL STEEL DESIGN
%
composite building construction, but more commonly the column is ferminated on a footing, pedestal, or pilaster. A pedestal is used to support the metal column above the ground to prevent corrosion when the footing is below ground level. A pilaster is an enlarge'd sect1)n of the basement wall used to transrmt the column load through the wall zone:to the footing. Sometimes, but not commonly, the column terminates6direc;tly o h the footing. A base plate 1s pzcessary when a steel column termmates on 'a 4. type of masonry to spread'tbe high intenslty of stress in the steel to a vaJu.s"Vlat can be safely carried by th? mqonry matenal. Masonry is, here deflned as concrete, concrete block, briGE<,,'a~:,tile; concrete IS most often used and will be the only material considered>&$ ' The base plate arjd the mating end of the column may be planed to affect load transfer by direct beanng. The base plate is seated on the foundation using a cement grout (thick sand-cement rmxture, often with an expansive a,vent to produce an intimate contact, since cement paste tends to shrink on dryins)Grout can take out up to about 1 in (25 mm) of footing-to-column mismatch as long as it is a "fill-in" discrepancy. Angles may be used to bolt or weld the base plate to the column. However, present practice tends to the use of a base plate that is shopwelde. to the column. Slightly oversize anchor bolt holes are drilled in the base p18 e to fit over anchor bolts placed in the foundation/footlng element during field construction. The overs~zedholes allow some misall~amentof thz pnchor bolts without redrilllng the base plate holes or taking out and resetting the anchors. The general method of fastening the column to the foundation IS illustrated in Fig. 6-7. The requ~redbase plate dimensions are based on the allouable unit contact pressure of the footing. The base plate thickness IS based on the base plitz contact pressure, producing bending on the cntical section uith dimensions as shown in Fig. 6-8c. When the column base resists a moment, the plate dimsnsions must be adjusted so that
d
\
Figure S 7 c
"r
Lb = 200 mm the length of 23.8 m gives 48 holes )I
=
48(0.30)
= 14.4
m
47 spaces = 47(0.20)= 9.4 m Total =23.8 m
Find the cover plate thickness (refer to Fig. 6-6):
Use 6 rnrn ( 1 / 4 in). Use welding and locate the plate as shown. Welding may be intermittent. If the welding is adequate to allow the cover plate and angles to act as a unit, AISC allows a contribution of the cover plate based on
( L - 2r)tp = [500 - 2(75)]0.006 = 2.1 x
where S is the sectlon modulus of the rectangular base plate kith respect to the moment axis. The plate th~cknessfor this case is also shown in Fig. 6-Sc. From Fig. 6-8a the area of the column base plate is
m2
of area to be used to increase the axial capacity of the section.
///
6-6 COLUMN BASE PLATES Steel columns are placed on some type of supporting member to interface the column and support. The supporting member may be a concrete column in
A number of combinations of side dimensions B and C can be obtained, bzt that combination should be used which produces r n n . Fabrication practice favors 3 and C in integers. The thickness of the base plate is obtalned by considering bznding on a critical section the distance m or n from the corresponding free edge (Fig. 6 - 3 ~ ) .
--
%.t
. AXUCTURALS n E L DESIGN
AXIALLY LOADED COLU;L[NS AND SIRUTS
Allowdble AISC stre.s
287
F-P ips: kipsiin2
S t . >L?:l
7.8
04 ii.25
Sandstone & l~rne.;tone Brick In cement InJtriu Full dred of conire[< ioiindarion
1.75
0.357;
L o s t h ~ nitill Jrru on iound~rion
dr
0.j5fc'
5 0.71;
A
lq.r n oi rn
,Lf =
d.v
With rnornenr (ci
S! Figure 6-8 General base plate dunensions and other desigo criteria. ( a ) Base plate dimensions. (b) Allowable stresses F,. (c) Base plate moment to compute base plate thickness.
For a uniform pressure and a strip tn or n
M
=
m q(m)Z
or
x
1 unit wide X
r thick, we have ?
n
!\I = q ( n ) 2-
Using the largest value of M (and noting that if m = n they are cqual), we have M fb -- Fb -- - = (for a strip one unit wide)
* t2
Solving for t , we obtain I'tgue 6-7 Column to foundation interfacing using base plates. (a) Use of angles to attach column I foundation. Method not widely used at present due to extra fabrication (cutting two angles and &"fully p l p n g column end and base plate). ( b ) Widely used shop-welded column-to-base plate pod f ~ attachihg r .columns to foundations [see also field photographs in (c), (d), and (e)]. (c) using shop-welded base plate. Gap for grouting to final grade can be easily seen. anchor bolts being used to attach column to footing. Diagonal member is a bracing element. (d) Column also using shopwelded base plate. Grouting space * The cable is being used to align frame. (e) Senes of interior columns fastened to Iirectly. Note again the use of shop-welded base plates.
= where
6M 'I2
(
1
3 X q X ( r n 2 0 r n 2 ) 'I2 =[
Fb
q = actual contact pressure Fb= 0.755 (AISC Sec. 1-5.1.4.3, based on bending on r e c t a r i p i ~ r
section)
-STRUCTURAL STEEL DESIGN
M A L L Y LOADED C O L L J S S AND STXLTS
Example 6-8 Design the base plate for a column as shown in Fig. E6-8. Use = 250 MPa, f,' = 20.7 MPa, and the AISC specifications.
F,
7s
Example 6-9 Redesign the column base plate of Example 6-8 to resist a bending moment of 265 kN . m in addition to the axial load (Fig. E6-9a).
Figure E6-8
SOLUTION The pedestal floor line dimensions will be the same as the bas plate. Thus F, = 0.35f,' = 7.245 MPa (Fig. 6-8b)
-
Let us make m n: From Table V-3 obtain d = 360 mrn; bf = 256 mm
(0.205
m = 0.106 m B = 205
C = 342
+ 2(106) = 417mm
+ 2(106) = 554mm
B X C = 0.417 X 0.554 The actual contact pressure q is q=--
231.0
4
.
After several side computations, let us assume that C
+ 2m)(0.342 + 2m) = 0.2305 shifting decimal m2 + 0.2735m = 0.04009
Check:
=
0.2310 m2 > 0.2305
O.K.
- 7.229 MPa
..
= 50.9 mm
Use a column base plate 417
X
554
X
52 mm thick.
I
SOLUTION We will design the column-to-base plate weld in ExampIe 9-7. F, = 7.245 MPa 5 q
say 52 mm
=
750 rnm:
= 656 mm Figure E6-9b illustrates data so far including q = 7.245 and tained from using B and C in the preceding equatlon for q. t Along line x - x : q = 7.248 - 10 22x
ob-
STRUCTURAL STEEL DESIGN
AXIALLY LOADED COLLIMFS At?,
Usine three anchor rods for each side, the diameter is
The corresponding thickness is
Use three 32-mm-diameter anchor rods.
6-7 LATERAL BRACING OF COLUrvfNS
70.6 mm say 75 mm .*-om the other direction (line y-y) at point A : =
q = 7.248
- 10.22(0.219) = 5.01 MPa
A common and widely used empirical rule for lateral bracing for both compression flanges of beams and columns is to provide a bracing (z!so z compression member) element to carry a lateral brace force of P, = 0 . 0 2 P
(average for a unit width)
L-,e the largest thickness of 75 mm. The final column base plate dimensions nre 780 x 656 x 75 mm thick. For anchor bolts as shown in Fig. E6-9c, assume that the bolts will carry the full moment even though the axial force will reduce the effect of the moment considerably. This assumption provides some reserve capacity of the anchoring system to resist a considerable lateral force (colu.mn shifting laterally). I
1 :se
F,
= 345 for anchor bolts.
where P is the axial force in the compression member being braced; that is, P = A& for bending members, where A, = area of compression flange and f, = average (or maximum) bending compressive stress. For columns, use P = average axial force in column. This recommendation is given by the Structurzl Stability Research Council in Guide to Stabilip Design Criteria for hferal Structures, 3rd ed., edited by Johnston. A series of tests at Cornell University by Winter (see "Lateral Bracing of Columns and Beams," Transacfions, A S C E . Vol. 125. 1960) inclicates that very little lateral bracing is required to allow the compression element to develop Pie allowable design stress. This restraint could generally be developed by t5s weight of the floor system onto beams when full-length contact with the compression flange is made. Because of the variable nature of flooring (metal deckfine. concrete-to-beam. wood-to-beam. and so on). i t is suooested .,- that the 2 percent criterion be used. Winter also de'nved an analytical expression for the bracing requirement based on both. reitraint and the relative stiffness of the column and brace. If a SF of appro xi mat el^ 2.5 is used with this derived expression, the empirical rule of 2 percefft can be obtained " Z
P
Example 6-10 Determine the minimum-size spandrel (or girt) to brric;. tke W section with respect to the Y axis for the 730-kip a..tial load of Example 6-2. The distance between columns may be taken as IS ft. Fv for column = 50 ksi.
SOLUTIONThe axial force in the channel section used for the brace is
.-
P,
=
0.02 P
=
0.02(750)
=
15 kips
The maximum L / r (AISC) = 200 (compression member). The minimum radius of gyration, r, = 18(12)/200 = 1.08 in. For an L / r ratio = ZCO, Ci:: allowable axial stress F, = 3.73 ksi (Table 11-5. SSDD), applicable for all F,. F,
Areqd=----
-
l5
3.73
-j,02in'
rru
a i n u b i U K A I . blEEL DESIGN
AXIALLY
Simply search Tables 1-6 and 1-7 for this combination of A and r, and find MClO
X
28.5: A = 8.37 in2 ry = 1.17in
Using Example 6-1 as a guide, try a 4 in A = 3.17 in2 r = 1.51 in
X
LO.ADED COLUhNS .WD STRUTS 10.79 Ib/ft pipe column:
If bending or other requirements are also satisfied, this section can be used for the girj (or spandrel). !
1
'
1
///
*
1
6-8 COLUMN AND STRUT DESIGN USING LRFD The use of LRFDrequires the separation of dead and live loads on the column!? +t*. Once thls is done, obtain the ultimate column load in the following form: '4 .
+
P,,=$(FdD
i
+ FLL. . . )
Areqd
=
P" 54'12 = 3.79 in' > 3.17 OFc, - 0.65(21.94)
N.G.
Try a 5 in x 14.62 Ib/ft pipe:
Also, Here we need Table 3-1, since the value of @ ranges from 0.86 to 0.65 depending on q , which in turn depends on KL/r of the column as well as 5.We may also note that the value of F,, depends on the value of 11 as follows:
Fc, = 36[1 - 0.25(1.002)']
=
26.96 ksi
+ = 0.65 54'12 = 3.09 in2 < 4.30 0.65(26.96) Use a pipe column 5 in X 14.62 lb/ft. =
where 7 = value given in Table 3-1 and is repeated here:
Example 6-11 Given the columns of Example 6-1 spaced on 20 X 20 centers, dead load = 30 psf, live load = 60 psf (snow), column length = 14 ft, a d K = 1 (as in Example 6-1 and Fig. 6-3). Use A-36 steel and the LRF method. (A4 in x 10.79 lb/ft pipe column was selected in Example 6- 1 .) Redesign the column using LRFD as given in Sec. 3-7.
%
SOLUTION P,, = l . l ( l . 1 0
+ 1.5S)A
= 1.1 [ l.l(O.030)
+ 1.5(0.060)] (20 X 20) = 54.12 kips (vs. 36 kips P,)
O.K.
///
PROBLEMS 6 -1 Determine the allowable load that can be carried by a W14 steel and the AISC specifications if: (a) KL = 16 ft. (b) KL = 42 ft. (c) KL, = 68 ft and KL, = 4 4 ft.
X
21 1 column using
I;, = 50 hi
Make appropriate ,somments. Answer: (6) 602.7 kips. (c) 549.1 kips.
6 -2 Determine the allowable load for a W360 AISC specifications if: (a) KL = 5.1 m. ( b ) K L = 13.2 m. (c) KL, = 20.5 m and K L , = 13.6 m.
X
3 4 . 7 rolled ssc~lonusing
F, = 415 Sip2
md
Make appropriate comments. Answer: (c) 2669 kN.
6 -3 What is the lightest square tube section (see Table 1-15, SSDD) for a column loading of 121 kips and an unsupported length of 12.4 ft? Use the AISC specificarions and '2-36 steel. Answer: 6 X 6 x 0.375. 6 4 What is the allowable column load using the AISC specifications and iF, = 315 %(Pa for a rectangular tube section 300 X 200 X 9.52 mrn wall (Table V-16. SSDD) for an u n s u ~ p o n dl-.cgr;l of 4.8 m? 6 -5 What is the allowable load using the AREA spssificaiions lor s W l 4 X 145 cset L; a
% STRUCTURAL
STEEL DESIGN
ALLY LOADED C O L W S AND STRUTS
compression member in a bolted end connection for a bridge truss? The member is 15.5 ft long and cises A-36 steel. Answer: 771.6 kips. 6 -6 What is the allowable load using the AREA specifications for a W310 X 178.6 rolled section used as a compression member w t h a bolted end connection for a bndge truss? The member is 4.75 n? long and uses F, = 345 MPa steel. Answer: 3 163 kN. Q -7 What is the allowable load for the built-up section shown in Fig. P6-7 using the AREA speclfications and A-36 steel? Assume a bolted end connection and a length of 5.25 m.
i
6 -12 Design the lacing for the allowable load found in Prob. 6-10, allowing a 20 percent i n w e m
('310 X 44 6
V for wind, member weight, etc.
6 -U Design the perforated cover plates for the allowable load and wcuon w d in Prob. 6-1 1. 6 -14 Redo Example 6-1 if the contributory column arc3 u 30 X 20 instead ol 20 X 20 but dl the
5.380 X 74 4
other data are the same. Answer. 5 In at 14.6 Ib/ft.
Figure P67
-
6 -15 Design a laced column section (refer to Example 6-5) for a highway bndge truss end post The length is 7 . 2 ~ 10.18 m. The truss span 1s 50.4 m and the loads are dead load = - 1 I20 kY a d
6 -8 Whatlis the allowable load for the built-up member shown in Fig. P6-8, using the AASHTO Z speclfications, F, = 50 ksi, and an unsupported length of 18.7 ft?
"
4-L's 6" X 6" X
I - p l ~ t eI 2"
i
$"
J
x L"
.
a
"
.,i:;" ;: '>.
li
"
L
Figure P6-8
W
i -9 Referring to Fig. P6-7, place a second channel C310 X 44.64 on the bottom of the S shape to
-
zaake it symmetrical. What is the allowable column load using the AASHTO specifications if L 6.6 m and using A-36 steel? Answer: 1962 kN. 5 -10 What is the allowable column load for the built-up section shown in Fig. P6-10 if the member is of A-36 steel and the length is 14.5 ft? Use the AASHTO specifications.
w
1
>q
J
,
- 422 kN (wthout impact). Use F, = 250 hfPa and the AriSHTO spsdicatioos. live load Answer: Try two C380 X 50.4, 300 x 15 mm cover plate. 6 -16 Redesign the truss end post of Example 6-6 using a perforated cover plate for both sides of the channels. Note that AASHTO allows use of the net area of the perforated cover plate in computing the total section area and column capacity. Answer: 2C12 X 25, A,,, = 23.7, r,, = 4.58 m, ~ncludestwo 12 m X f plates with 3-in holes..
6 -17 Design a column base plate for the m a m u m capaclty of a W12 x 170 column with an unbraced length of 12.0 ft. Assume that K, = 5 = 1.0. Use F, = 50 k s ~ /,,' = 4 !GI, and Pis AISC speclficatlons. The column 1s Interfaced to a concrete psdesral 6 -18 Design the column base plate for a W14 X 120 secuon that carnes an m a 1 load of 5 0 ) kips and a base moment of 200 ft . lups. Use A-36 steel. /; = 3 ksi, and the base plate lntsriacrs 'Je column directly to the footmg, wh~chhas a total depth of 21 In. A w e r : 24: X 22; X 2;. 6 -19 Design a column base plate for the maximum capacity of a W3 10 X 117.6 rolled section wirh an unbraced length of 4.1 m. Assume that K, = K, = 1 .O. Use Fy = 3-45 ;LIP&f: = 28 blPa, and the AISC specifications. The column is interfaced to a concrete pedestal. 6 -20 Redo Example 6-1 1 for the lightest available W8 section. 6 -21 Redo Example 6-1 1 if the loads are as follows: dead load = 35 psf; live load = 75 psf. Answer: 5-in pipe at 14.6 Ib/ft. 6 -22 Redo Example 6-11 using the following data: dead load = 1.75 kPa; hve load = 3.75 LPa; column contributory area = 6.1 x 7.1 rn, column length = 4.98 m. and K, = K, = 1.0. Use eikrr a round pipe or a square structural tube for the column and Fy = 250 MPa steel.
6 -23 Design member 6 of Example 6-6 (refer to Fig. E6-60) if the dead load bar force = 283.9 Lips,
-
l6"-4
4 Figure P6-10
6 -11 What is the allowable column load for the built-up section shown in Fig. P6-1 I? The length is 5.3 m. Use the AISC specifications and A-36 steel. Neglect the contnbution of the perforated cover plate.
the maximum live load, including impact = 109.4 kips, and the minimum Live load = 0.0 kip. Compare the section to that obtained in Example 6-6. Take P, = +(pd DL + pL LL). wkere 4 1.3, 8, = 1.0, and BL = 1.67 (latest AASHTO). Also, P, = 0.85A Fc,, where F<, = Fa from Eq* (6-9) or Eq. (6-lo), without using the SF = 2.12. Answer: Use the same section as in Rob. 6-16.
-
t
BEAM-COLUhm
DESIGN
7-1 INTRODUCTION When a structural member 1s !oaded In a manner to produce more t#n oce stress mode, some adjustments must be made in the allowable stres$s- Vvnnere the stresses are produced as a combination of bending about the X,and Y axes as in Sec. 4-8, the final stresses used for deslgn are obtained by superposirioa, Figure W-1 Beam-columns and beams making up an industrial frame. Note alternate orientation of strong axis of beam-columns along sides of frame. A closeup of selected joints is shown in Fig. 9-15.
Since F,, may not be equal to F4, (part~cularlyin the case of FV shapes from flange geometry) the beam deslgn In Chap. 4 was obtalned by iteration. Accumulation of compressive (or tens~on)stresses at one edge of one of h e flanges was used In the following form of Eq ( a )
i
T /is equat~onwas obtained by d ~ v ~ d i nEq. g (a) by F,. A problem similar to this 1s often involved when the structural member is 1 , ded in a combination of bending and axlal load. These situtlons are always p oduced in rigid frame building construction- (i.e., the columns carry the building load axially as well as end moments from the girders that frame into them). In industnal buildings column brackets may be used to car:;, ::i~.e runway girders and, ultimately, the crane load. The resu!ting bracket eccsnti-xiry
298
STRUCTURAL STEEL DESIGN 4
produces a bending moment in addition to the axial loads in the column. In this case the column moment is not at the column ends. Similarly, wind pressuregon long vertical members may produce bending moments, since a large distahce between floors (or ground to roof) may disallow the concept of wind being carried, analogous to one-way slab action. In Examples 2-5 and 2-6 the framing of the side sheds to the columns of the main .bay produces large column moments which must be allowed for in their design (to be considered in a later section). Other design conditions produce bending in addition to axial forces. For example, the top chords of roof and bridge trusses are normally "pin-ended" compression members, but the weight of the member will produce bending as yell. Purlins placed between panel joints of a roof truss or rafter as a means of redu* both the purlin member size and roofing span will produce bending in \he chofd or rafter. In general, compression members are loaded with axial forces and moments. 'The moment(s) may be at the ends of the member, as in rigid framed buildings, or developed at an interior point from a bracket, local beam, cable attachment, or other loading. When the moment effect produces single curvature (see Fig. 7-1) a much more critical design condition is developed than when the moment(s) produce reverse curvature. Bending may also be produced in tension members such as the bottom :herds of bridge trusses where floor beams may frame into them. Bottom chords of building trusses may be used to attach hoisting devices; other temporary loads attached to the bottom chords of building trusses will produce bending in addition to the axial load present.
In many of these situations, particularly with truss members, the bendrng stresses are neglected. This may be a reasonable procedure where the bending stress results from the member weight, or even from purlins if they are relatively small (and light weight), and the resulting effects are possibly less than about 10 percent of the analyzed stresses. There are undoubtedly small bending stresses in the truss chord elements due to continuity across one or more panel points (a technique used to reduce fabrication costs). and additionally there is usually some overdesign, since it is a common practice to use a constant-sue top (and bottom) chord member (again to reduce fabrication costs). Exceptions to this may be obtained when it is safe to use only one bolt or rivet at the ends of the truss members, so that rotation-is less inhibited-it is. of course, not practical to fabricate "pinned" joints for the usual b.rfd~,e~o~~roof truss. The effects of actual 8 th'e &oi@length ~ivlth, respect to the joint bending can be minimized by kk'ipin~ member 1engths.into the joint .is"short as" practikal&l, coupled w i t h the fact that truss rnenibers usually have small E l / L rqJi&.,li!$:, Ion, member with smal! moment of inertia), the moment g;ad?@@,,isih&$ . .. .SF hnd much of the member length is essentially moment-free. ;."',; The moments that are prodixed-ig t;
:'
"'
or
Number of equations: 100 X 2
=
200
20d
=
40,000 words
CPU requirements:
n g ~ d :100 x 3
=
3 0
300'
=
90,020 words
Since each number (or word) requires approximately 4 bytes the requirements become: :b
Pinned:
40,000 x 4
=
160.000 = 160K bytes
90,000 x 4 = 360,000 = 360K bytes
Figure 7-1 Column loading curvature resulting. The single curvature of ( b ) is often the most critical. (a) Reversed curvature in building frame. ( b ) Single curvature interior loading. (c) Reversed cwature interior loading.
.
A capacity of 360K + program requirements would tax the capacity of some of the larger computers currently available had not efficient methods of matrix solution been developed so that with clever coding (reduce the bandwidth of the matrix to a minimum) and/or use of boundary conditions, the three-drser-offreedom truss might only require some 30 to 5OK. which is well urithin the capacity of all but the smaller desktop minicomputers.
300
STRUCTURAL STEEL DESIGN
BEAM-COLLhl?i
DESIGN
331
7-2 GENERAL CONSIDERATIONS OF AXLAL LOAD WITH
B~NDING
2
When tension axial load and bending occur simultaneously, the principal of superposition may be safely applied. This is because (see Fig. 7-2a) the tension load tends to reduce the bending effects, so that the value of A is reduced somewhat, and, of course, the actual stress is also somewhat less so that the maximum allowable stress conditions computed as -f,+ - < fb I F, Fb will be safe. This safety is partially obtained by neglecting the effects of P - A on fb, which could be computed (to be strictly correct) as Mc c fb = +-+ P,AI I When Pb is at midspan, we have (with the limitation A 2 0, see Fig. 7-2) pbL3 P,~L' A=+--(a> 48 EI 8 EI and the resulting bending stress is (b)
The actual bending stress can only be obtained by iteration of Eq. (a) until the A value used on the right side is sufficiently close to the A obtained on the left side. This iterative solution may be reasonably practical on a computer, but with hand computations, in only a few design situations (where the number of members is limited ) is this approach economically justified. Neglecting the P-A effect in Eq. (b) is an error, but on the conservative side. Inspection of Eq. (a) indicates that the tension stress reduces the deflection and also reduces the compression stress due to bending. With the allowable tension stress F, a constant value and recalling that Fb may depend on the unbraced length of the compression flange (and possibly reducing the allowable compressive stress), we see that use of -fb+ - - f<, I Fb -
FI
provides a satisfactory solution.
Pb
+
i=ll801n \ r = 21
Figure EY-1
L)O~rl #
Example 7-1 Glven the portlon of* b#hway bndge truss w i t h 102:s an2 members as in ~ l g E7-1, . what is t h i ' ~ & u m tension stress in the bwer *?" , chord? .SOLUTION (neglect the P-A effect) 0.0.10(25~)( 12) 8(5l 9)
WL?
jb=-=
8sr P
J = -AF
=
0 72 ksl
165 fb -- - +101782-k s 1
f, = 13.98 + 0.72 = 14.70 k s ~
(mau~rnum)
f, = 13.98 - 0.72 = 13.26 ksl
(m~n~mum)
///
When a compressive axla1 load acts together bb~th a bending morncrt. tks deflection is ampl~fiedand the compress~vestresb incre~seb S~ncrthe aJ!ow\.ab!e compression stresses take Into account poss~ble buckl~ng(l~teraldefIecuons), member design IS more senslt~veto t h ~ sloaciTnc rp&e than to one producing tension stresses. Referring to Flg. 7-26, we note that rhe ax131 load (absumed to be applrcd last as belng easler to vlsuallze) Increases the deflect~on.The order of Iorid application does not affect the outcome, honever. as long 3s jrelding is not produced. The Important concern IS that there IS an Increase In the detlection due to the P-A effect, wlth d correspond~ngIncrease In the bending stresses. T: value of the deflect~on1s (w~thPb at m~dspan)for t h ~ scase A
=
P ~ L P ~A L ' --- '+ 48 E l 8E l
and the resulting bendlng stresses w ~ t hP, =
(-)
(c)
are
(4 Figure 7-2 P-A effects on tension and compression members.
A critical evaluat~onof Eq. (c) indicates that an lterat~vesolution IS required as for the tenslon mode, but that the deflection "feeds" upon ltszll (dcGtctlcrr
2
- 1 STRUCTURAL STEEL DESIGN
B U M - C O L L W DESIGN
ch.ljes more deflection), and for members with an I too small or an L too large a I-l..,kling failure can occur. The P-A effect can also develop in tall buildings, as qualitatively shown in Fig. 7-3, when wind or earthquake forces or unsymmetrical loading produces lateral displacements of the upper floors with respect to the lower building eiencnts. A computer analysis can be made to analyze the P-A effect but requires iteration. The steps include: !. Vake a conventional computer analysis using the lateral forces. 2. Cjbtain ith-story lateral displacements Xi and for XI+, (next story above). 3. Compute an additional P-matrix moment as
Mi = ~ I + l ( X I +l Xi) arid a shear as indicated in Fig. 7-36 using
which is applied at the top and bottom of the story with due regard to signs.
,
Pi + = axial force in column between i and i
e. .
+ 1 floor levels
4. Compute new X, and compare to previous values used in step 3. Iterate until satisfactory convergence is obtained, such as, say, 0.02 ft or 0.006 m (app;oximately in).
4
r?ik P
A. Another stresses (from loads) by a factor such as 10 percent to a110 eloped from fwtor that tends to mitigate the P-h effect IS that it is;u'& stresses that wind or earthquake analysis where the designer can use a are increased by one-third. The P-A effect would only exce& this in rare cases. The analysis computer program in the Appendix h a gk ! eeily modified to automatically scan the deflection matns for the appropriate X values, recompute the P matrix using these values, then storlng them for comparison with the new values from the current cycle until convergence and exit.
7-3 EFFECTIVE LENGTHS OF COLUiLlNS I N BUILDfNG FRAMES The concept of effective column length L' = K L was introduced in Sec- 6-3 rind a value of K was obtained for several common cases. I t was observed that when the column ends were laterally restrained so that the P-A effect (as in Fig. 7-3) could not develop, the K factor was K <_ 1.0. With the "flagpole" of Fig. 6-3e or the pinned base of Fig. 6-3f, the K factor was 2.0 or more. We shall find that K in multistory buildings, which can translate. m ~ be y considerably more than 2.0, as illustrated in Fig. 7-40. For :he portion of the elastic frame shown in Fig. 7-46 and using Eq. (6-1): x = A sin hy
(6-1)
With k = ( P / E I ) ' / ~ and using the effective .+ength L' x = A
sin-
=
K L , we obtain
';Y
KL
Taking the origin of coordinates at a point of inflection as shown in Fig. 7 4 b , we have at the top of the column:
.' = ,I,
v, = y ,
?'I
s , = A sin-
KL
At the bottom of the column:
Noting that sin (a -
P ) = sin a
cos ,O - cos a sin ,B,
9
At the top and base of the column, the slope (a) and (6):
Flg-le 7-3 P-A effect for tall bulldings. ( a ) Structure with lateral loads. ( b ) ith story with deflections gt j exaggerated.
8,
(
=-
"TI
-
IS
d d t / l $ , and we obtain. from Eqs.
cos -cos - + sin -s!n KL K KL A' 11
304
STRUCTURAL STEEL DESIGN
Substituting, - we obtain 2
8, =
C o x , where Ga = 2 El, / LC 2 EIb / L,
Similarly,
Substituting for x , and x , [using Eqs. ( a ) and (b)], we obtain
8,
=
(z), A G sin-T I K a 7 KL
8,
=
- - -G
(:)iLc (
"Yl sm-cos-
7i
KL
-
K
1 . 77 cos-s~nKL K
At the column-beam intersection of rigid frames, the rotztion of the c o l u m equals the rotation of the beam, so equating the two values of 0,. we obtain nG,
TY~
..
-
K 6 KL Similarly equating Eqs. ( h ) and (d), we obtain Lu"
5
- ?I ( t a n 2 K 6 KL
-
tan
=
K Substituting Eq. ( i ) for tan v , / K L , we obtain
Figure 7-4 Elastic frame for derivation of G, and Gb terms to obtain effective column length KL. (a) Part of an elastic frame. KL defined as distance between points of inflection. ( b ) Column element isolated from (a) with terms used in derivation identified. (c) Conjugate beam and moment variation (assumed).
From the bending moment diagram for the assumed moment distribution along the beams linearly varying as shown in Fig. 7-4c, the slope of the beam at the juncture with the column (and using conjugate beam principles) gives '
where the summation ( c ) is taken because load and moment are coming from both directions. From the earlier derivation of the Euler equation and summing
1
"YI + tan -tan KL K 77
f, ' -A,/ 6(Ga + G,) tan s / ~ We may program Eq. (7-2) for increments of Ga and G, and Iterate untd a \ d u e of F = n / K 1s obtalned to sat~sfythe equality. The value of 7 / K thus obtalned is used to obtain K as
r
h,
A plot of Ga and Gb vs. K can be made as shown in Fig. 7-5a. is nomogiaph was first developed by Julian and Lawrence in unpublished lecture notes and as cited by several references, including the Structural Stability Research Couaci:. Guide to Stability Design Criteria for Meral Srnrctures, 3rd ed., edited by Johnston and published by John Wiley & Sons, Inc., New York. In a somewhat similar manner, since the boundary conditions are different. one may develop equations for Ga, Gb, and K for frames restrained ~ g a i m t lateral translation as G,
~b
($1'
+
+
2
Gb
-
tan s/ K
) + -_v2 tan K I T/
=
(7-3)
,
% STRUCTURAL STEEL DESIGN
(i0
j
1000-50 0 30.0:
I
I1
0°:
'5.8 8'0 -
1
K 70.0 1 00 5.0 -4 o--
3 0 --
1.0 6.0 -
so-
i
i
2.0 --
1
- CC
100.050 0 30.0 20.0 -
10.0 9.0 8.0 7.0 6.0 5.0
-
4.0
-
-
3.0 -
3.0:
1.5 --
2.0
-
1.0 -
1.0-
i
~
K
Go
Gh
0 -
5 0 . 0 3 ~ 10.0~ 5.0 4.0 3.02.0 -
10-
0.9 --
Gh
~ 0 . 0 ~ 1005.0 4.0 3.0
--
0.7 --
0.6
-
0.5 -
0.3 -
-
0.6 --
963. Since smaller K values had been used in structures that had an adequate ervice history, a new look was taken of the derivation for G,, Gb and the resulting K. Yura (see "The Effective Length of Columns in Unbraced Frames," A I S C Engineering Journal, April 1971) correctly pointed out that where the K L / r ratio was less than C,, inelastic buckling should be considered and E, should be used in Eqs. (7-4) and (7-5). The use of E, is equivalent to
-
0.2 -
-
-
0.1 -
0.1-
o-
The 'use of K factors obtained in the manner just described has been required by AISC since 1963 and by A A S H T O since 1974. The K factors tend to
-
0.3 -
-
a. The far end is hinged: multiply the E $ / L , ratio x 1.5. b. The far end is fixed: multiply the EI,/L, ratio x 7.0. 4. If beams are simply framed to columns, use Fig. 6-3 for K.
1.00.9 0.8 0.7 -
0.4 -
-
0.5 -
a beam or girder is used with adequate attachment to the coiurnn of
-
0.4 -
0.2 -
. When
2.0 0.8 --
I .O 0.9 0.8 0.7 0.6 0.5 -
suggested that one use G = 10 since there is some difficulty in prodlrcin~a true pinned connection. 2. When the column is (rigidly) fixed to an infinitely rigid base, the EIb/L, ratio -+ co and the corresponding G -. 0. For this situation it is suggested to use G = 1.0 since there is some difficulty in producins a tmIy rigid connec-
o-
- E, Gine~asilc- E Ge,,st,c ( a ) S ~ d e s w a ypermitted
( b ) No sidesway
Figure 7-5 Nomographs for the effective length of columns in continuous frames for lateral restraint conditions indicated.
This equation may also be programmed for values of Go and Gb and to find the corresponding value of F = a / K to satisfy the equality. A plot of this is shown in the +-iornograph in Fig. 7-5b. The use of both nornographs shown in Fig. 7-5 involves computing values of
2 EIC / LC 2 EIb / L b - 2 EIc/ LC G - 2EIb/Lb
Go = and
(at far end of column)
(75)
ince El = AE and A I 1, it follows that use of G,, ,,,,, ic gives a Kkchric < nce El is somewhat awkward to obtain and recalling in the derivation of &e ISC equations for Fa in the inelastic region,
-9;. ..
we used essentially a SF on FCr where
\
(7-4) and in the elastic region
(at near end of column)
(7-5)
From the derivation involving Go and G,, it is evident that if we call one of the values Go, the other end produces Gb (i.e., the values can be used interchangeably). When E = constant it may cancel from Eqs. (7-4) and (7-5); however, when inelastic buckling is developed, E, should be used for E in the E I c / L c ratio. Other considerations include: 1. When the column is pinned to the base, the E I b / L b ratio is zero, since the theoretical value of I --+ 0 that results is G + co. For this situation it is
Fa
=
SF(KL/~)' From these equations it follows (using i
=
inelastic. r
FO' G,= G,
=
elastic) that
(7-7) Fa, is computation neglects the variable SF. which ranges from 1.67 at K L / r = 0 23/12 (use 1.92 for hand computations) at K L / r = C'..For most columns in e range of K L / r = 40 to 60, the variation in SF is essentially negligible. The se of Eq. (7-7) requires values of Fa, for the same K L / r value as F,,, so i t is
3U8
BE.L\I-COLUbN
STRUCTURAL STEEL DESIGN
DESIGN
3@
necessary to compute a table of values such as Table 11-7 or VI-7 (SSDD) and using Foe= Fd to correspond to the AISC Specifications F,' =
1 2 - 7 ~ .' ~ 23(~~/r)~
Note thgtothis value is independent of 5. The values shown in the table of F,' > are for the purpose of using Eq. (7-7) and are not intended to be real stresses that might be used in a design. The use of Eq. (7-7) in an actual design requires iteration; that is, assume a column; compute G,, G,, and find K; compute KL/r and Fa,Fi;revise G,, G,; and find a new K for
6
1. Several cycles, or 2. To convergence, or 3. To an arbitrary limiting K, such as 1.2 or 1.5.
*
Of course, if the section chosen is not adequate, a new section must be selected and the computations repeated. Disque (see AISC Engineering Journal, No. 2, 1973) proposed that the iterations for K could be eliminated by using fa = P / A instead of F, to obtain
Figure E7-2 , ' ,l
Find K:use W14 x 99 through three floors:
=
3.94
Reduce for inelastic effects: In lieu of using f,, which might reduce Ge excessively, one may elect to use F, = 0 . 6 4 . Again carefully note that the reduced G is used only when KL/r < C,. It should be self-evident that values of K 5 1.0 are not to be adjusted. Example 7-2 Given a portion of a high-rise frame shown in Fig. E7-2 with sidesway permitted, assume adequate bracing perpendicular to bent so that K,, does not have to be considered (e.g., close contact with interior masonry walls) and F, = 36. Using AISC specifications, design columns CD. SOLUTION Assume that KL section W14 x 99:
fo
From Table 11-5 for compute Fd =
fa =
712E
23/12 x 5 j 2
570
= -z--= 17.86 ksi
29.1
17.9 ksi, obtain K L / r
=
49.4 ksi
=
5 5 . For K L / r = 55
(calculated also in Table 11-7)
The revised
= 14 ft and use Table 11-4 to obtain a tentative For G, we note that the column is "dly I, -+ KI:
attached to the base so that
This value is not reduced for ~nelastlcbehav~or.From Fig. 7 - j a , we obtain K = 1.38. With K = 1.38,
BE.L'.l-COLL3W
From Table 11-5 with this value of K L / r , we obtain
P
,
SOLUTIONMake an initial column-size estimate that KL/r = 40, for xxEc5 F, = 178.2 MPa. For Fa = 178.2 MPa, the tentative column area is
Fa = 19.37 ksi
=29.1(19.37)=563.6>520kips
DESIGN 311
O.K.
Si:lce this value is about 40 kips larger than needed, try a W14
x
90: Try W250
X
114.6:
520 KL = 34 (Table 11-5) - 19.62 ksi 26.50 r i c2m Table 11-7 obtain (note that this table is computer-generated and uses S F = 23/12): fa=--
Fd
=
129.18 ksi
F'
7
=
1 as before, and from Fig. 7-5a obtain K
Fa = 19.64 ksi Pa
=
1.25.
> 520 kips
r2(200 OoO) = 366.6 MPa 23/12~53~
(also Table V1-7). --- /
!
The adjusted G, = 0.673(164.6/366.6) = 0.302 Right column: G'b = 189.4/4.6 = 0 . use 1.0 (AISC rzcommendaticn) co Left column:
(Table 11-5)
26.50(19.64) = 520.5
Use a W 14 x 90 for column 1.
=
=
O.K.
189.4/4.6 0 Obtain from Fig. 7-5a:
Gb =
///
Example 7-3 Given the frame shown in Fig. E7-3, with sidesway possible, use the AISC specifications and Fy = 345 MPa to find the required column size. Use the same section for both.
=
K
co
=
1.2
use 10.0 (AISC recommendation)
for right column
K = 1.75 for left column Check the left column first, since K is much larger than the right column: 1.75(4.6 x 1000) = 70,6 r 114.05 Fa = 143.7 MPa (Table VI-6)
KL
-=
Pa
=
AFa
=
14.58 x 143.7 = 2095 kN
> 2OOO
0.K.
Now check the right column:
F, = 169.7 MPa Pa
=
14.58(169.7) = 2474.2
> 2400 ki.;
A solution: use W250 x 114.6 sections for the columns
O.K.
Example 7-4 Redo Example 7-3 with sidesway somehow restricted. SOLUTION One solution is to use K = 0.80 for the left column aqcording to Fig. 6-36; correspondingly, one would obtain K = 0.65 using Fig. 6-3a for the right column and in both cases, using the "recommended" design values. Alternatively, use Fig. 7-5b. Since the K must be less than 1.0, let us use the "experience" of the previous problem to check a somewhat lighter section. Try a W250 x 101.2:
SHTO, and AREA) interaction equation with bending about both ~ x e A,, s as well as axial load as fa
fbx
-fby
Fa
Fb
Fb
(7- 10)
1
-+----I--<
-
rior to 1963, the value of F,, -. F,, = Fb. Currently, we recall that F, depends Witeria, a d n several factors, including unbraced length and compact section .C '6 " general AISC allows ..> "$
Fbx = 0.66Fy
or
F,,
or
F,, = values from AISC Eq. (1.5-6a),( 1 3-6b), or ( 1.5-7)
=
0.60Fy
Fby = 0 . 7 5 5 for W shapes due to having solid rectangular flanges
F,' = 1032 MPa (interpolating Table VI-7)
Gb = 10
for right column
K = 0.72 K = 0.65 Check the right column:
KL
'4
for left column
Gb = 1.0 Using Fig. 7-56, we obtain
-=
Currently, Eq. (7-10) is used only in certain limited stress conditions. For t5z remaining stress cases, ther more complicated formulas based on reszarch, plastic design,..,and elastic stability concepts are uszd. These will be psr-tidiy developed in the next several paragraphs to indicare some of the limirations so that the practitioner will have some idea of how to follow through should tbs design vary from routine. Refer to Fig. 7-6 for a short ( L / r + 0) rectangular section of dimensions X d that is stressed with both an axial force and a moment sufficient to evelop a plastic hinge. The plastic moment in the presence of a compressive
for left cblumn for right column
0.65(4.6 X 1000) = 26.5 112.78
rx Pa = 12.9(190.3) = 2455
Fa
> 2400 kN
=
190.3 MPa O.K.
Check the left column:
-KL- - 0.75(4.6 X
1000) = 30.6 Fa = 186.8 MPa 112.78 rx Pa = 12.9 x 186.8 = 2410 > 2000 kN O.K. Use a W250 X 101.2.
///
7-4 DEVELOPING THE BEAM-COLUMN DESIGN FORMULAS Pnor to the sixth edition of the AISC Manual of Steel Construction in 1963, the des~gnof compression members subjected to bending was obtained as f, + fb
' -I,= b X
:I.,,
Fallow
Dividing this equation by Fa,,,, = Fa, one obtained the widely used (AISC,
Figure 7-6 Plastic hnge formation in a very short membsr subjected to both an 1G2! force .n\f moment.
314
STRUCTURAL STEEL DESIGN .i'
'* "
~uk$i?&in~ for yo (shown on Fig. 7-6), webobtain
Multiplying the P ratio by d 2 / d 2and noting that b2d2u.f= P.,', we obtain
However, from Sec. 3. Example 3-3, M,
=
u,bd2/4; thus we obtain
effect as Eq. (b) of Sec. 7-2. However, i t can be shown [see, for e:c~mp!e, Timoshenko and Gere, Theory of Elastlc Stablliry, 2nd ed. (New Yock: McGraw-Hill Book Company), Sec. 1-1 I ] that i t is sufficiently accurate (a 1 to 2 percent error) to amplify the moment for P-1:
This factor may be called an amplfl~cationjocioi. since its effect is !o aF$if-j or increase M, T h ~ svalue has been uced In the curvec shoun I n Flgs. 7-8 tb3 7-10. The P I P , ratio is the ratlo of the actual column load to the Euler column load and f,/F: is simply dlviding both loads by the column area. W ~ t hthis adjstment in bending moment, we may rewrlte Eq (e) to obtain
Although the development above has been made for a rectangular cross section, it is also valid for all (including W, S, and M) shapes. A plot of Eq. (7-1 1) is shown in Fig. 7-7. Also shown is the plot of a linear equation of the form
If one were to plot
for k L / r = 40 and for K L / r = 120, the straight lines shown also on Fig. 7-7 would be obtained. These curves will be somewhat in error, since the P-A effect $as been neglected. We could use an iterative approach to include the P-A
Figure 7-7 Plot of interaction equations as shown. P, and Ma = allowable values.
Shown in Flg. 7-8 is a plot of the loading situation u here tf, = .t12 = Jf and in Fig. 7-9 the case where M I = itt and itlz = 0. These t-o plots represent the extreme range of cases where a column is loaded w t h end moments, ma building frame. The curves shown in F ~ g s7-8 and 7-9 have been made using a modification of Eq. (e) proposed by Galambos and Ketter (Transacf~om,ASCE. Vol. 126, pp 1-25, 1961), whlch gives, for equal end moments.
F*
7-8 Influence of K L / r ar
BF.A\i-COLW
316 STRUCTURAL STEEL DESIGN
2 hlo
The coefficients are KL/r-dependent, and several values given by Galambos and Ketter are as follows: r
Figure 7-10 Plot of c o l u m Inr:raction for end moment on o a s end and with efiscr of using C,= (dashed lines) ior comp~fison.
Figure 7-9 Plot of K L / r for 40 and 120 for a column with moment on one end.
and for unequal end moments a linear equation of the form
DESIGN 317
may note that the reciprocal of Eq. (7-12b) is used as the C, amplification or for laterally unbraced beams. Equation (7-12a) is used as C, in the AISC s when using t : i ~
! I I ~ / , I I=~ ( - )
Ill
\!!
= 1-1
%,
KL/r
C
D
F
G
0 20 40 80 120
0.42 0.70 0.99 1.81 3.16
0.77 0.46 0.17 - 0.72 -2.51
1.13 1.14 1.16 1.19 1.25
1.11 1.18 1.23 1.52 2.53
is in agreement with the definition for C, (i.e., single curvature bending is critical for buckling instability than reversed cufvature, and simiIarIy the The effect of using C, is to generally decrease the effect of the amplification factor and is illustrated in Fig. 7-10. The use of C, produces a n intersection of
The plots using Eqs. ( h ) and ( i ) are reasonably satisfactory for all of the cases with equal end moment, but it is rather conservative for those cases of unequal values other than at M , / M , = 1.0 and with a rapidity determined by the / M , ratio. To the right of the intersection of the two curves, Eq. (d) cor:;:i.!i. is requires use of two equations in design and using the most critical .of
researchers are as follows: Cm= 0.60 -
cm=
0.4Ml 2 0.4 M2
1.75 - 1.05M,/M2
+ o.~(M,/M,)~
We obtain for bending about the X axis the f o l l o ~ n g
' it
by
'llis is because it is easier to solve for the most critical value by using the two equstlons than to make a plot and locate the intersection and then use the goverli~ngequation. These equations are adjusted for design use by substitution sf PC,for Py and M,,,,, for M, and with section area and section modulus to ob , stresses. This gives
P
a,
+ C , ~ B ~ ~ a, ~ , (P(KL)? ~ )
ax= 0.149~i-2x lo6 ksi
where
PdliOW
-
(fps unlts)
P(KL)* = ksl #
'
By analogy, Eq. (7-13) becomes Fb = allowable bending stress
M
fb
,#@ E ~r
= -;; 3
For Eq. (7-14), we obtain in a similar manner:
CG iining terms and extending the case to biaxial bending, we obtain
Fa Fa Fa P0 . 6 5 + B, M,yFb.r + B,. M, - I Pal,,W
AISC Eq. 1.6- 1a
-fa+ Fa
'1
i
cmxfb~ f
a
F
'myfby
+
b
x
(l-fa/F&)Fby
< 1.0 -
(7- 13)
When j,/Fa (or P I P a )
(7-140)
*84
. Fby
.
< 0.15, we can obtain
and AISC Eq. (1.6-1b)
fa + - + - - < I fbx
fby
A careful analysis of these equations shoxvs that F, is based on P,,,which depends on the critical oallie of KL/r. Bending resistance and any ampiification/reduction of moment effects is with respect to the bending axis ~~iitfi nrnh)pmc+ h P n\\nwahlP 2uial stress F_will be ... -l. ,..., ;* S U v S C I 1 P L I I I ~> I I U W I I . 1 1 1 1 1 3 111 t/l .. --.-. .-. based on K L / r , , but the P(KL)' term will depend on the axis resisting bending Since the X axis is usually used, it is the P(KL,)~product that would usualIy be required. The modified form of these interaction equations is generally considered to can be tabulsted for a number be easier to use, since the right-hand side L---:-b:--
IIIUIIJ
Referring agaln to Fig. 7-7, we note no reduction in moment capacity until P/ P, > 0.18, so rounding 0.18 to 0.15 for convenience and to be conservative, we obtain: A
Limitation of f,/Fa 5 0.15: AISC Eq. (1.6-2)
Since Eqs. (7-13) to (7-15) are somewhat awkward to use except on a cqmputer, let us multiply through by AFa to [using Eq. (7-13) without fby for a particular illustration] obtain
but ,$A the actual column load P and AF, = maximum allowable column *(bad ${anbis not the same value shown on Figs. 7-8 and 7-9 and used to -devcin:; ,.the curves shown). Also noting that /, = M/S, define A / S = B, niult~ply/./F: by A/A, and take FiA = P, = 0.149 X IO~/(KL/~)'ksi (in fps).
u u r r ~ r t rr
~
r
r
of column sections and for several assumed column lengths based on kTL/ry. Similarly, the terms ax, a,, B,, and By can be computed and tabulated. These a x shown in Tables 11-4 ( 5 = 36 ksi) and VI-4 (Fy = 250 hipa) for the W shspcs commonly used as columns. The AISC manual also has these tabulations for F. = 50 ksi steel and includes use of S shapes and tube and pipe sections. -Y
7-5 DETERMINATION OF THE INTERACTION REDUCTION COEFFICIENT C, When a column in a structural frame is restrained against lateral translation with end moments, as illustrated in Figs. 7-8 to 7-10 and 7-1 la, the value of C,
320
STRUCTURAL STEEL DESIGN
BEh?f-COLL'4N DESIGN
331
en the column has transverse loadings as in Fig. 7-1 1 c, use C,,
=
1.0 + -7 Fe +la
(7-15)
here f, = actual column stress Fb= Euler stress as previously defined (including SF = 23/17) $= factor determined from Fig. 7-1 ld, which depends on end restraint and transverse loading ."?*
'7-6 A A S M O AND AREA REAM-COLUMN DESIGN F$&V~C'L~S ,,*
,,*$
AASHTO working stress design uses' essentially~thzsame equations as .A.,ISC, with some additional adjustment to the amplification factor. The C , factor is defined similar to AISC. '"Ib,
+
(l-fo/~;;)~bx
CmJby ( 1 --I;/F;)F~~
' "
+ -b - - + - fh ---I1 Fb, Fb
fo
0.472F; F"
_< 1
(7-17)
(7- 15)
n ' ~ F
2. I ~ ( K L / ~ ) ' Generally use C,,, = 0.85 for end conditions of Fig. 7-1 1 b and c; use C, = 1.0 hen the interior moment is greater than end values or with an interior moment The AREA equations are similar to the AISC equations. For,f;/F,
+
fbx
Figure 7-11 C, reduction factor for beam-column interaction equations. (a) No sidesway: C , = 0.6 - 0 . 4 M , / M 2 . (b) Sidesway: C , = 0.85. (c) Column with transverse loading: C, = 1 + +fa/F,'. ( d ) Several cases of transverse loading and factors shown.
1 -a
/F
.b
.
rb." ( 1 - .fa / 1";') Fbv
<
1
+
V'E
F;" = is computed using Eq. (7-12a):
1.431(~L/r)' lso, at points braced in the plane of bending:
fa with attention to signs (single curvature = - M,/M2). Note also that M, is defined as the smaller of the two end moment values. When sidesway (Flg. 7-1 16) is possible, AISC specification allows:
o.55Fy
fbx
jby
Fbx
Fb,y
+-+---I1
fo
fbx
fby
Fo
fby
Fg,
-+-+ 1 ---I
Cm = 0.85
(sidesway present)
> 0.15: (7- 19)
32.'; ... :UCTURAL STEEL DESIGN
BEAU-COLL3c.F
7-'? 9EAM-COLUMN DESIGN USING INTERACTION EQUATIONS
value of KL' to
i
,
The design of beam columns using the interaction equations is essentially a trial (i!.-i.;:tive) process. A section is tentatively selected and analyzed and if the i. se;.:ien is too small, a new section is selected and the process repeated until a sz:i:i'actory section (both strength and weight) is obtained. The steps may be ou: ,-.edas follows:
,.
Fb = 0.66Fy
gA 1.
nr
=
C,BM-
AP
=
BM-Fa Fb
Fa ,B Fb
[using a part of Eq. (7-13a)] [using a part of Eq. (7-14a)]
'::,pection of Table 11-4 (or VI-4) indicates that C, Bx-Fa /3
=
0.2 to 0.3
(6.5 to 9.0 in SI)
Bx- Fa Fbx
=
0.1 to 0.2
(3.3 to 6.5 in SI)
,B
=
estimation for
ax ax - P(KL,)~
...
, ' ,
:,
..
+ A P = P + (0.2 to 0.3)M = P + (6.5 to 9.O)M
@
: *6
(M in in . kips)
-
( M in kN m and SI)
ihk value of P, enter a Table such as 11-4 or YI-4 (or in AISC Manual o i Yables which have been prepared for design use using a computer) with the K4, value and obtain P,,,,. If Kx/K; I rx/ry, the KL/r,, values control; if P.',/K; > rx/ry, the KLx/rx ratio controls and one must use an adjusted r'
check for Ilghte~s ctlon
4
use a 1 a ~ ~ e r s e c t i 6to4sa:~sfqdesign
With care and some preliminary scrutlny ofv Table 11-4 (or Y I J ) , an adequate section can be obtained In one to three inals. Thls IS possible. since the tables show little change in r,/r, and B for the two or three sect~onson either side of the selected section.
&mate can be made similarly. An estimate for AP can be made in a s,.rnewhat similar manner for using the AASHTO and AREA equations if desired. 4. Compute the equivalent allowable column load as
':!
=
L,nb,acedI L, computed from Eq. (4-23), (4-26), or (4-27) if
+ P,, << P,,,,, P + Peq > P,,,,
~ h t values : for bending about the Y axis are considerably larger, but a AP
+w,(;&
Fb
0.6Fy
P
--,
P,, = P
=
One should attempt to achieve an equality as clossly as possible for one of the equations (it cannot be done simultaneously for both equations) and if:
fbx
;~ntl
Fb
if LUnbrace, I LC
i f L ,,r,, > L, The unbraced length IS the actual and not the KL value and IS taken with respect to the bending 2x1s. 7. Compute K L / r cntical, obtaln or compute F,, and compute Ju = P / A . Compare fa/ Fa I 0.15. If the ratio IS less. use Eq. (7-15) to see ~f the section is adequate. If fa/Fa > 0.15, ~t will be necessaryfto :heck barh Eqs. (7-130) and (7-140). In this case compute: P ( K L ) ~ use KL with respect to the bendlng a.uls. ~ h l c hmay be different from the cntical K L / r used to compute the allonable avlai stress Fa 8. Check both Eqs. (7-13a) and (7-14a) to sat~sfq
,
AP
" #*?
e tables, obtained as KL KL' = ---I rr/ rv
It is necessary to use r,/r,, slnce the ratlo IS fixed for a section but is cor known until a section is tentatively selected. 6. Record Ptable,A 1 ry, r,/ry, LC,L,, B,, and a,. L, and L, are needed, so a rspid determination of Fb can be made:
,-(erminethe axial. force and column moments. We note that this step is also 1 yitive,, since indeterminate frame values are not found until a tentative .,.!~tlon is used in the analysis. ''2.::>&ipute K to obtain KL. It may be necessary to determine both Kx and Y, !':bending on column end conditions and lateral bracing. 2 . .'?.,timate the moment contribution (and we will use bending about a single axis for illustration) as an equivalent axial load AP:
*
.*
'
'
Example 7-5 Given column and bending moments s h o \ n in Fig. E7-5 as part of building frame in which sidesway is possible, we will limit the column section to not over a W12. Use the AISC specifications and A-36 steel. Select a tentative section and use the basic equations (7-13), (7-i4), or (7-15) as applicable. SOLUTION TO keep the deslgn of a beam column using the ~nterzctian equations In perspective, we wlll assume that K, = 1.3 ~ n dX; = 1.0 2r,d
3.44
JIKUGIURAL STEEL DESIGN BE.4'4-COLL5B
DESIGN
By Eq. (7-13): fu ' d b r + ---5 1
Fu
PFb,
0.47+ O 85(7'88) = 0,796 << I 0 0 934(22)
By Eq. (7-14):
i;+f- L < 1 0.65 Figure E7-5
7.48 22
:
Try a smaller section-try A
>
=
1 1.80 1n2
LC = 8.5
<
L, .
12
L,,
=
19.6 > 12 ; . Fb = 0.6% = 22 ksi
=
16
&=z=-=
12'ft -+ Fb
F, = 15.98 k s ~
p
1.92(36.2)'
=
0.6F,
=
22 ksi
42.5( 12) jb=----- 9.83 ksi 5 1.9
.112~
=
'"
1.92(36.62)
=
1 1 I .3 ksl
9.32 I I 1.3
= 1 - ---- - 0.92
By Eq. (7-14): 0.58
a 2 ~
74.2
15.98
7.48 ksi
1.92(~~,/r,)'
KL
-+ - =
-f, -- - =9.32 0.58>0.15 F,
We must use both Eqs. (7-13) and (7-14):
Fel =
>
F'
lo 14.70
9.32 Lsl
S, = 51.9 in3
,
Sx = 64.7 in3 (Table 1-3)
P
l lo 1 1.80
= ----- =
Pa,,, = 236 kips
Fa = 16.06 ksi (Table 11-5)
Y'
ju
'7
5
x' = 2.64
x 40.
W 12
ry = 1.94 In
A = 14.70 in2 (use Table 11-4)
7.88 22
-+ - = 0.70 << 1 .O
that the K values do not change with section size.
Use C, = 0.85 with sidesway. After some study of Table 11-4 (with P 1 lo), let us try W12 X 50:
Fb,
= 113.8 ksi
+ 0.85(9.83) = 0.991 < 0.92(22)
O.K.
1.0
By Eq. (7-14):
ru
0.6%
932 + -9'83 -22 22 Use a W12 x 40 column.
+-<
F, -
1.0
- 0.87 < 1 0
1
O.K.
3 3
326. 5
bTURAL STEEL DESIGN
E;.awple 7-6 Given the column and bending moments shown for a building frame, with sidesway for K, restricted by use of bracing and shear walls, use the AISC specifications and F, = 250 MPa steel to select a tentative column section.
,
SOLUTION Refer to Fig. E7-6 and assume that 1
Kx=1.25
k;=1.0
Figure E7-6
Check Eq. (7-14a):
List. Cm = 0.95 (AISC actually allows Cm= 0.85 if desired). Tentatively:
4
+ pequv
pgwen
. 445.. + 7(41)
4
[estimate factor as 7 (between 6.5 and 9)]
E
= 872 kN Sc2.1 Table VI-4 and select W310
x 59.5:
P,,,,,
x
A = 7.61
l o v 3 m2
=
rx
-=
114.1 I 14.1 = 753 (445) -+ 8.94(61) --_ 150 130 Using Eqs. (7- 13) and (7- 14) would give: Eq. (7-13):
< 868.3 kY
860.5 kN 2.64
'Y
1.25 >I
O.K. Use as tentative section, W310 x 59.5.
Fa = 1 14.1 MPa
LC = 2.58 m Fb = 0.6%
4.87 m I~O'MP~
=
F ( x L ~ )=~445(1.25 fa
Actual
=
7.61 =
(Table VI-5) L,
=
> 3.45
X
3.45)2 = 8.27 X lo3 kN . mi
=
58.5 MPa
AFa
=
- --- 58'5 Fa
114.1
(in same form as ax)
-0.51>0.15
P + G,,,,BxMx-
\*@
Fa Fbx
445 +,0.95(8.94)(613-3
///
Example 7-7 The top chord member (No. 6) of the truss used in Examples 6-4 and 6-5 will be designed to include the member w e i a t and a temporary concentrated force of 2.2 kips that will be applied to the center of the chord during maintenance operations (see Fig. E7-7a). The bridge may be temporarily closed to traffic if the maintenance load is too large to be carried safely with traffic (live loading). Other data: Dead load = - 283.75 kips Live load = - 109.44 kips P = - 393.19 kips
7.61 (1 14.1) = 868.3 (vs. 860.6 interpolated) SOLUTION Since a pair of C15 x 40 channels was used in Example 6-4 for a somewhat smaller load, we will try a pair of Cl5 x 50 channels (Fig. E7-76):
Using Eq. (7-13a): vB '
O.K.
ax < 'allow ax - P(KLX) -
132'8 150 132.8 - 8.27
=
865.3
< 868.3 kN
A = 14.70 in2
O.K.
I,
=
11.00 in'
.
3%
.
A ~ C T U R A ISTEEL . DESIGN
e used to support the siding and for lateral bracing. Caution is nec&sary, owever, that girts should be continuous and, should building repair require emoval of a girt, that it be done in only one bay at a time so that the Isteraf. support is not lost. 3. There will be a column moment at the crane n l n w q l a e l due to the IongirudinoI thrust of the crane starting or stopping suddenly. This will also produce a moment at the base plate even if the analysis is made in such a manner as to ignore the moment at the crane level. This force will also produce a column shear that must be resisted by the anchor bolts at the base. The general design of a column of this type proceeds as follows:
. As in Examples 2-5 and 2-6, tentatively analyze the structure and revise until i&
Crane colurnns in an industrial building. ( a ) Column w t h bracket. ( b ) Stepped column. 4 ~ stepped p column.
';62
(c) 1 .
I
bo.,. l:;e X and Y axes. Now if we take the X axis of the main column member ' %Pic.r~~~d for bending in the plane of the bent and the Y axis for bending out of )iac,: (with respect to building length), we have the following considerations:
reasonable member sizes and deflections are obtained. Different member .sizes can be used for any member. This was not done initially in the two examples, as the work is still highly preliminarq.. 2. Based on the computer output. begin to redesisn the rnc.rnb.srs. Increesc or decrease sizes, depending on forces and deflections. . Reprogram and check output for forces and deflections. . Repeat as necessary.
-8.1 Modification of K for Stepped Columns
1. 5 x e d or pinned against rotation at the roof truss level. The roof truss of the ii~dustrialbents of Examples 2-5 and 2-6 provides rotation fixity (at least nearly so), but translation may take place. 2. Sidesway control. The side sheds in Examples 2-5 and 2-6 plus any bracing in the plane of the first and last bents will act to control sidesway. If it is still excessive, knee bracing may be required from the column to the roof truss. 3. There wiN likely be moments in the column at the roof truss. There will possibly be moments in the column at the crane girder level due to lateral thrust of the crane trolley against the rails making the track. There will be a column moment at the base due to assumed base fixity. These moments produce beam-column interaction for which Eqs. (7-13) and (7-14) must be used.
pin-ended column free to buckle and with a load Poand an interior Ioad Pi a in Fig. 7-13 is in a state of unstable equilibrium if the loads are sufficientlq. IrrrgeIf we use the differential equation
and allow for boundary conditions of different loads in the lower se-ment, no lateral displacement at the column ends, change in I for the lower segment and a common slope at the junction of the upper and lower sezgnents. we obtain
'
n*
Y Axis i
f
**
$1
d&iyi4!yss :. : I
" E I X ~or plnned at roof level. If we put some cross bracing in the plane of the chord and vertical cross bracing in one or more of the bays on 'op;osite s
where J = 1 + Po/ P, - a a = L,/(Lo+ L , ) This is a solution as given by Sandhu, "Effective Length of Columns nich Intermediate and Axial Load," A ISC Engirieeririg Journal, October 1977. Ths ki
32
STRUCTURAL STEEL DESIGN
BE.L.f-COLL?Ci
DESIGS
33
I
Example 7-8 Make a tentative design for the crane column for the industrid building in Example 2-6. Refer to Fig. E7-8a and assume the folio-kg:
I. Adequate bracing with respect to the Y axls of main column (don2 bays). 2. Sidesway in plane of bent with no top rotation and base fixed. 3, For initial tentative design, use only dead and live loads from computer output + crane loads. 4. Crane girder will be placed at a level with the bottom of the side sksd truss. 5. Use a stepped (built-up) section.
Figure 7-13 Figure for derivation of effective length coefficient for a stepped column.
i
value obtained as the solution of Eq. (7-22) is as used in Chap. 6:
Rearranging yields '1
I'
but and equating this load to the Euler load, we obtain
Solving for the equivalent length factor K, yields
This value of Ke is for a pin-ended column, and it is necessary to multiply this by the AISC value of K as obtained from Fig. 7-5 or 6-3, where the actual end conditions are taken into account. For general design office use, Eq. (7-22) should be programmed on a computer, so a plot of a' = Po/Ph versus K, can be made for selected ratios of a.
SOLUTION A single W690 x 264.9 was used to obtain prclimin~ryou:put shown in Example 2-6. This size section (area and moment of ~ n e r t ~ ais) necessary to reduce the lateral deflections at maln truss roof level acd at side shed-to-column intersection to tolerable values. The built-up section will require this or larger values to provide satisfactory lateral displacements. From the computer output (and for a tentative initial built-up section iteration) for LC = 1, the axial loads and moments (for left column) are: Member 26 (uppermost):
P
= 318.74 kN
moment
=
120.59 k3
c'
m
'4 STRUCTURAL STEEL DESIGN
Member 25 (intermediate): P = 387.1 1 kN Member 24 (lower section):
.
moment
=
338.99 kN . m
P=543.01kN moment=-313.6kN-m base moment = - 190.1 kN . m Let us somewhat arbitrarily try a section made up of one W360 x 314 main column and one W360 X 101.2 crane column, as shown in Fig. E7-86. .
,
A.
considered stable by adequate use of girts and siding). Use the largest force in the upper column, 387.1 1 kN.
Substituting values into Eq. (7-22) yields
Noting that the term to be evaluated as cotangent must be identifix! i$ radians, the problem can be quickly programmed on a pocket progammable calculator to obtain k L = 3.525. Figure E7-86
I'
W360
X
314 data:
PI
I, = 1107.2 x
m4
A = 40.0 X lo-' m2
W360
X
d
Zy = 428.7 X =
399 mm
bi
m4
PI =
+ 1262.11 = 1805.1 kN (actual) 12.426(1 + a') EI =- 15.82EI
543.01
L
= 401 mm
L~
Also,
101.2 data: 1, = 300.9 X l o v 6 m4
A = 12.9 X
=
m2
ly = 50.4 X
d = 357 mm
m4 bf = 255 mm
Compute r and ry. Use EMx, = 0 to locate the new Y- Y axis. (40
+ 1 2 . 9 ) ~= 910(12.90) X
=
910(12.9) = 22 1.9 mm 52.9
(use 222 rnm)
The actual K 1.2.
=
Ke
X
Kc,,
,,,,,,,, ., K
=
From Fig. 6-3c obtain k',,,
c,,,,,,,
-
0.79(1.2) = 0.95
From Table VI-5, obtain F, = 134.6 MPa From Table VI-7, obtain F,'= 9 732.8 MPa. Check the Interaction equation [Eq (7-13)]. A
Check allowable stress Fa based on K L / 5 (since the other direction is
736 STRUCTURAL
'
STEEL DESIGN
BE.+(-COLWW
DESIGN
337
Use ,C = 0.85. (We will not check for bending about the X axis at this time, since computations are very preliminary. After the next computer run, if this section is still satisfactory, we would check bending about both axes.) I
# U.
13'6(0.8 16) = 27.7 MPa ;# (note decimal shifted for 1 ) , &'= I 9.235 Assume that Fb = 0 . 6 5 . [We will have to compute this later using Eqs. 1.5-6 and 1.5-7 (AISC numbers) as appropriate.] =
Figure 7-14 Sldesway control. ( a ) Masonry shear wall ( b ) D~agonalbracing
By inspection, Eq. (7.14) will be satisfied for the lower column segment. The upper column segment should be checked for interaction to make sure that the W360 X 314 is adequate as a column for the full height of the bent. We can now reprogram this example with the new column sections and any revised sectlons for the truss members and see if the lateral deflections far the several load conditions is satisfactory and that the bending moments and axlal forces are compatible with the section being analyzed. This should be done prior to refining the final design, to keep the engineering calculatlons to a minimum and maxlmize use of the computer.
///
in
zcl
2. Diagonal bracing (essentially produce a vertically oriented truss) in one or more bays. It may not be necessary to use n g ~ djoints wlth shear walls or diagonzr bracing. However, some materlal economy may be obtalned via use of rigid joints as well as providing some addltlonal room for uncsrtalnties. 0 criter;2 Diagonal braclng requirements are usually small The controlhn, may be the L / r ratio rather than the cross-sectional area requirements. Gs!ambos ("Lateral Support for Tier Bullding Frames." A ISC Engtneering Jozirml. January 1964), using essentially the same method proposed by Winter for Iatsrzi bracing of columns and beams, develops an expression for the area of tF,c diagonal brace member:
7 9 CONTROL OF SIDESWAY
1
It is evident that the most efficient column design results when the frame is adequately braced against sidesway. With no sidesway: 1. C,,, can be less than 0.85 (but may be more in selected cases). 2. The effective length factor K is not greater than 1.0. A rigidly framed structure can translate laterally sufficiently to undergo "~idesway."Note also that the development of the equations for the G factors is based on a common slope at a joint that can only be obtained for a "rigid" joint. It is therefore necessary to provide specific resistance to sidesway to obtain the a ~ obtained (see Fig. 7-14) via: most efficient columns. ~ h i s ' m be
1. Shear walls (use rigid vertical walls of brick, tile, or concrete block to contain the lateral movement). If masonry walls are used, a close contact with the column should be provided so that the column cannot translate in the construction void between materials.
where
A,= area of bracing, in2 or m' C P = sum of all column loads at a story level In plane of bracing a = horizontal run of diagonal brace/length of column E= modulus of elasticity assuming column and brzce of s a ~ e material, ksi or MPa
Diagonal bracing is usually designed only for tension. It is assumed that 13bracing member is so small and flexible that it will buckle (with stresses we:! below under a very small compression load. For opposite-direction loading, the buckled member straightens with no damage due to small buckling stresses and prevents sidesway from occurring by carrying the necessary tension load.
6)
Example 7-9 Given the story of a tiered building shown in Flg. E7-9, design the diagonal bracing for the intenor bay. Assume that this wll be placed in alternate bents in the out-of-plane direction.
BE.k\f-COLb\fX
DESIGN
339
4%
Additionally, we must always satisfy the following
P,, 5 0'.pu When there is bending about only one axis we have [he following:
and
+
I n these equations = 0.86 (for beams) and o = value given in Table 3-! i ~ i columns a n d noting varies from 0.65 to 0.86 d e p s n d ~ n gon k Z / r . The C, terms are as in AISC. The values of P, are
+,
Pu=AFy(l-0.25r12)
P,
=
AFy 4
(7-30)
~i\'>
(but use appropriate bending axis)
T h e design forces a n d moments are as computed in Chaps. 6 and 4. T h e use of these LRFD equations will be tllustrated by the folIouing example.
Example 7-10 Given the beam column and loading shown in Fig. E7-IC. select the lightest W310 shape using F, = 350 bIPa and LRFD.
Figure E7-10
SOLUTIONUse F, Mu, P,,
=
1.5 (author's cholcc). l.l[1.1(120) + l.5(\80)]
=
1 . 1 [ 1.1(400)
=
+ Ii(6Cn'l:
=
42.2 L S i - rn
= Id7-!
'K"
U
BMU-COLUMX DESIGN
STRUCTURAL STEEL DESIGN
/
With bending only about the X axis, Eq. (7-28) applies, so that
PROBLEMS
-
4-1 Using the general solution for column buckling,
Inspection of this equation indicates that
x = A rin!iy
show that K
=
+ B cosky + Cy + D
x = Oaty = Oandy =
M=Oaty = O
Z, = 2.687 X
138.9
r,
L
-0aty -" Q-
=
L
7-2 Determine the effective length coeific~entsK lor the colurnns In the frame shown In Fig. Note that the far end of the girder of column A is pinned. Answer: AB: 1.75; FG: 0.78.
Using these values of A and Z , try a W310 x 157.7 section: rx =
=d-&
0.7 for the column of Fig. 6-,3b. Note the boundary conditions: ,
A = 20.13 X ry = 79.0 mm
X
and
controls
m3
from which qic =
Figure W-2
0.90 - 0.25(0.534) = 0.77
Pu = AF,[ 1 - 0.25(0.534)~]= 20.13(250)(0.9287) = 4674 kN 1
7-3 Determine the effective length coefficients K for the columns Ln the frame shown in Fig
-
M, = ZF, = 2.687(250) = 672 kN m Substituting values into Eq. (7-28), we obtain 1474 442 N.G. 0.86(4674) 1.18(0.86)(672) = 0.367 + 0.648 = 1.015 > 1.0 Check: + 3 u 2 p, 0.77(4674) = 3599 >> 1474 O.K. Since this column is just over, the next largest W310 section should work. Tabulating data so that a comparison can be made, we obtain
Annuer: BC = 0.65; EF = 0.63.
+
*
W310 x 178.6: A = 22.77 X m2 ry = 79.5 mm (and controls, as before) Zx = 3.055 X m3 By inspection of this data it is evident that the section is adequate. It is not necessary to check that Pud i +=PU,since this ratio is almost that obtained from the first term of Eq. (7.28). Use a W310 X 178.6 section.
///
-
Figure W-3
I
ill
t:
Figure P7-5
Figure W-8
I!I
344 STRUCTURAL STEEL DESIGN
I
,
7-9 Design the top chord of a railroad truss with a panel length of 27.583 ft for the flollowmg
conditions: Live load = 867.2 h p s (E-80 loading) Impact = 473.6 kips Dead load = 402 kips (estimated) .@
use a bullt-up section somewhat as shown in Fig. P7-9. Use the AREA specdications and A-36 steel. The chord member ends will be either riveted or bolted. A m e r : Two S24 X 100 and one S24 X 79.9 with a 30 x cover plate.
t
7-10 Check the section of the column above the crane runway girder of Example 7-8 and redesign as required. 7-11 Using Example 7-8 as a guide, make a tentative redesign of the main column of Example 2-5. Use A-36 steel and the AISC specifications. 7-12 Redo Example 7-2 with the column sue limited to W12. A m e r : W 12 X 96. 7-13 Design the diagonal bracing for the bent shown in Fig. E7-3 to inhibit sidesway. Use the lightest pair of angles with a 12-mm gusset plate. Annoer: Two L63 X 51 X 4.8. 7-14 Check the exterior columns of Example 2-3 using the computer output and resize the columns if necessary. Note that both exterior columns are to be the same sue. Use a single column (no sphc*for full building height. Use A-36 steel and the AISC specifications. 7-15 CHeck the interior columns of Example 2-3 using the computer output and resize the columns as necessary. Note that basement columns are not necessarily the same size as the upper column wbch is to be used for full building height. Use A-36 steel, the AISC specifications, and not aver W 10 columns. Answer: W8 X 48 bottom; W8 X 40 upper. 7-16 Use the computer output of Example 2-4 for the exterior columns as outlined in Prob. 7-14. Limit column size to W250. A m e r : W250 X 67.0. 7-17 Use the computer output of Example 2-4 for the interior columns as outlined in Prob. 7-15. Limit column size to W250. 7-18 Venfy w t h computahons that the W310 X 178.6 column section of Example 7-10 is adequate. $f@ 7-19 R & ~ OExample 7-10 d ML = 210 kN m and D = 425 kN. AU other data are the same. Answer: W310 X 178.6. 7-20 Redo Example 7-10 if My moments are present: M,, = 50 kN . m; MLy = 75 kN . m. Answer: W310 X 282.8.
.
BOLTED A N D RIVETED CONNECTIOhTS
8-1 INTRODUCTION A steel structure is produced as an assemblase of the structura1 members making up the framework. Connections are required where the various member ends must be attached to other members sufficiently to allow the load to continue an orderly flow to the foundation. Since the connection serve5 ru carry a d from or to adjoining members, i t must be adequately designeti. c : ; I onnection design involves producing a joint that is safe, economical of materiIs, and capable of being built (it must be practical). The more practical onnections are usually more economical. since fabrication costs greatly affec: economy of both connections (or joints) and the members thsmsc.!v:s, 2.i ustrated earlier, particularly concerning built-up tension and compression embers. Several structural connections are illustrated in Fig. 8-1. Connections (or structural joints) may he classified according to:
. Method of fastening, such as rivets (hardly ever), bolts, or welding. Connec-
FIgure Vm-1 High-strength bolted joints. ( a ) Splicing smaller-to-larger column using filler plates. (b) Splicing same-size column. ( c ) Diagonal bracing.
tions using bolts are further classified as bearing orjricrion-type cop. nsctions. 2. Connection rigidity, which may be simple, rigid (as produced by an indeterminate structure analysis), or of intermediate rigidity. The AISC, in See. 1.2 of the specifications, classifies joints based on connection rigidity as: Type 1: rigid connections that develop the full moment capacity of tilt connecting members and retain a constant relative ang!: between tI.5 connected parts under any joint rotation.
P (el
used as tension hanger.
Type 2: simple framing with no connected parts. Actually, a small amount of moment will be developed, but it is ignored in the design. Any joint eccentricity less than about 2; in (63 rnrn) is neglected. Type 3: semirigid connections with less than the full moment capacity of the connected members being transferred. Design of these connections requires assuming (with adequate documentation) an a;bitrary amount of moment capacity (e.g., 20, 30, 3. Type of forces transferred across the structural connection: a. Shear forces: common for floor beams and joists. b. Moment: either bending or torsion. c. Shear and moment: as in type 1 or 3 connections. . . , d. Tension or compression: as for column splices and for "pinned" truss members. e. Tension or compression with shear: as for diagonal bracing. 4. Connection geometry: a. Framing angles used to connect floor joists and stringers to beams and columns.
and a web angle for shear. Web angle is optional.
b. Welded connections using plates and angles. c. End plates on beams or rafters. d. Plates or angles used on one side of a floor joist or beam. e. Seat angles with or without stiffeners. Several of these connections are illustrated in Fig. 8-2. 5. Fabrication location: a. Shop connections: produced in the fabrication shop. b. Field connections: joint parts fabricated in the shop but assembled on 2-s ,
.
a. Friction connections. Connections designed as friction connections ha~:s
connections. Connections where the joint resistance is taken as ation of connector shear resistance and bearing of the c o m s c 1 against the connector. This mode of beh , r #evelo sufficient slip occurs to bring the connected matePlal id contact wi back projection of the connector pgar the w6king or 'ddrsipfocd. connector shear is a portion of the resistance in beanag$onnecfio analysis, the reduced shear .area available for threaded connectors w'cz the threads are in any slip plane requires 3, reduction in the design load. In actual practice, threads in the shear plane.result in a lower allow3S:s design shear stress for the fastener.
r'
The design of both friction and bearing corinections involves use of an wable shear stress. The value is much lower for friction connections, siccr nt'any joint slippage under working loads. The value is consideror bearing connections, since some small amount of relative moveeen the parts making up the joint can be t o l e ~ a t ~ dBoth . types of dition to being designed for "shear," are rout~nelychecked for on the net section and for bearing of the connected material against the fabrication practice tends to use of slotted over-sized holes connections. The slot allows easier field erection since more a l i m e n t e is available for temporary erection bolts.
2 RIVETS AND RIVETED CONNECTIONS
Figure 8-3 Several modes of joint resistance. (a) Bolt shear. (b) Plate shear or tear-out. (c) Bolt bearing. (d) Plate bearing. (e) Bolt tension failure. (A Tension on net section.
connections is not developed as the shear resistance of the connectors; rather, it is developed as the product of the clamping force produced by tightening the bolts (or driving the rivets) and the coefficient of friction between the clamped parts. It is expected that in load factor resistance design this will be directly used as the design parameter, producing an equation of the general form used in several other design codes (outside the United States) as
where
+ = performance factor (0.67 to 0.70) p= coefficient of friction x number of slip surfaces = total developed clamping force as the sum from all the bolts used in
2 A, F,
the connection
For many years rivets were the sole practical means of producing safe 2nd iceable metal connections. The process required piinching or drilling ho!:s roximately 1/16 in (1.5 mrn) oversize, assembling the parts using drift pins to ign the holes, and using one or more bolts to hold the parts togetbet mporarily. &vets were heated in a furnace (portabie for onsitz use) to a cherry ed color (approximately 980°C) and inserted into the aligned hole through tLe everal parts to be connected. One member of the riveting crew then applied a ng bar with a head die to the manufactured rivet head to hold the rivet 1I1 and to shape. Another crew member used a pneumatic driver with a h a 3 to forge the protruding rivet shank to produce the other head. The forg.ing ation simultaneously reworked the rivet metal and caused a shank enluget to very nearly fill the oversized hole. This reworking and shank enlargeent, together with the shrinking of the hot rivet, produced a substantial joint of the time. The rivet contraction during cooling is resisted by the joint rial and develops tension in the rivet so that a riveted joint is intermediate een a friction- and a bearing-type connection (a bearing type is commonly ed). This joint transmits the design load primarily by friction between the ed plates making up the joint. The riveted joint has had a long history of cess under fatigue stresses as in railroad bridges. Only recently has ,ARE.-: llowed use of high-strength bolts and welds in joints for railroad bridgts.
#
- 3 STRUCTURAL STEEL DESIGN
BOLTED &\TRIVETED COhC+TCnO?iS
355
41
t the procedure for the design of a riveted connection is exactly the sao;,e a for a bolted connection. Figure 8-4 illustrates several sizes of undriwn rivers 2nd structural applications using rivets.
8-3 HIGH-STRENGTH BOLTS There are two general classes of bolts used in structural appIications. These artre general-use A-307 (ASTM designation), sometimes called unfinished baits. These bolts have a somewhat rough shank and bearing surfaces, since not as at care is taken in their manufacture. The A-307 bolts are made of steel ~ i t I . ~ ultimate tensile strength F, on the order of 60 (grade A) to 100 ( g a d s B)'hi 15 to 690 MPa) and available from $ in (6 mm) to 4 in (102 mm) in diarnerer in lengths from 1 to 8 in, in increments of 1/4 in. and over 9 ir.. in ements of 1/2 in. A-307 bolts are available with several head a n d auf configurations, but the hexagonal and square head are most cornmody used. Several sizes of A-307 bolts are illustrated in Fig. 8-5. A-307 bolts are cheaper than A-325 and A-490 bolts and sbouid be use2 in tatic load structural applications whenever possible. Applications incIuds csc in mall structures, locations where the bolt installation is visibIe for i ~ s d ~ i serviceability checks, and in service loads which are relatively smaII. High-strength bolts are available in the ASTM classifications, sizes, sad ultimate tensile strength shown in the lower inset of Table 8-1. The general length, head, a n d nut configurations are the same IS for 11-307 olts except that larger diameters may not be available. The A-325 bolts can be btained with metallurgy for special purposes, such as high resistance to coxosion. A-325 bolts may also be obtained with a galvanizing coating, &LAG.?LS. When high-strength bolts were first introduced into structurd applic-': washers were required to spread the bolt load to a larger area of the sofie:. r;,ttir! of the fastened parts. This requirement was partially caused by the nut a n d head to dig (called galling) into the A-33 and A-7 (F,= 33 ksi) steel avrri!?ds time. Current high-strength bolting application; require that a hardraed asher be used under the turned element as follows: Method of tightening
A-325" A-490b
Figure 8-4 (a) Several sues of undnven structural nvets. ( 6 ) k v e t s in structural apphcations
Turn-of-nut
Specified torque
No Yes
Yes Yes
a Washers are required when uslng oversued bolt holes. Washers are requlred when flange slops is greater than 1 : 20 (S and C shapes both have a slope of inner flange face of approximately 1 : 6). Use washers on sloplug flange faces as above. Use two washers when the fastened material has F, < 40 ksi.
-F',.-'i;?,
i
BOLTED &\Q R
I
w C O h > % ~ O ~ 359 S ,J
r proper installation
;':'
re p= Slip coefficient (usually can use 0.35 for clean mill scale; most 0~~~~ surfaces are less than this value. and i t may be necessctry to deternine the value by test) m = number of slip surfaces N = number of fasteners T = proof load of each fastener (as in Table 5-2)
i.
.
p 4.
eration of Eq. (8-1) we can readily see that regardless of +e t_Vcof d (friction, or no slip, or bearing with some slip acce~&bIehthe eed P,,,, before either bolt shear or bolt (or m a t i t3.s may now see the rationale for P, = t(,K,
L
7
joint actually develops resistance as n cornbinstion of boir shear
e 8-1 What is the nominal safety factor against relative slip in a -type joint using 20-mm A-325 bolts? Take
,U =
0.35-
kN (Table 8-2). The allowable ess using AISC specifications (Table 8-1) is 120 hfPa with threads. ne. From Eq. (8-1): LunoN The bolt proof load is 141
Psli, = rn,uNT
iTU=
ultimate tensile strength (see Table 8-1).
=
k11(141)
since it is only necessary to consider one bolt and we will consider slip plane, as in a lap joint. The safety factor is always defined as
SF =
Q
~
one Y
Prrslsting Pallowable
~dequacy. A bolt tension of approximately 0.7Fu gives adequate reserve strengt the bolt be somewhat overstressed (say, 3/4 turn instead of 1/2 turn). The belt tension acts as a mawive spring in tension to hold the fastened parts in relati' ~ositil~n... This clamping effect also tends to hold the joint against nut loosening in fatigue load situations, SO that most of the time a locking nut is not If A-325 bolts have not been excessively overstressed (not more than 3/4 turn of nut) they may be reused one or more times. Tests on reuse indicate that A-490 bolts should not be reused in any situations.
Pallow
= F,Ab(nornina~)
Combining, we obtain the safety factor:
SF
=
0.35(141) 120(0.7854 X 0.0'0@)I @
=
1.31
The reader should note that this safety factor is against slip and is not safety factor of the joint, which is on the order of 1.67 (for a tension f// t and may depend on tension on net section).
FACTORS AFFECTING JOINT DESIGN has considered several factors involved in c @ n ~ e c : i ~ n pertains to fasteners. We *ill now consider ~evc;a1
S ! STRUCTURAL STEEL DESIGN
8-4.1 Joint Length
assuming reasonable bolt spacing on the order of 3 x diameter
One factor of considerable importance is joint size. Obviously, smaller joints economical of material. However, since an assumption is made that eac fastener in a joint carries a prorated share (equal for constant-size fasteners), problem arises for long joints. Referring to Fig. 8-8, we see that the distribution of strain is unequal from the frontmost bolt to the rear bolt. If the joint is too long (with "too" not being specifically defined here), it is evident that the first bolt will carry more than P / N of the load and the last bolt will carry nothing to almost nothing. With the base metal or plate designed to be adequate for tension in the net section, the plate does not pull apart but does stretch based.on P L / A E , so the forward bolts (or rivets) will either undergo compatible shear strains or will shear off if the strain and resulting forward displacement of the hole is too great. The loss of the forward bolt will transfer the load to the next bolt(s) in line, and the next bolt may shear, and so on-producing a progressive joint failure (a process called "unbuttoning"). Note, however, that with the large loads involved, this process is very nearly instantaneous. If the joint is short enough that all the bolts carry load, the first bolt strains with the plate. When strains corresponding to yield stress (shear) develop, the bolts continue to strain with no increase in load and the next bolt(s) in line will pick up the transfemed load. The ultimate joint load is reached when all the bolts have yielded. Strain compatibility analyses are seldom made, since the factors of safety used.together with the property of steel ductility are such that except for long joints, only the first bolts (if any) in a connection are yielded or are close to yield. We should note that the safety factor for the connection (particularly he fasteners) shocld be higher than for the members being connected. This is so +that a'member failure will always occur before a joint failure. A joint failure will generally be catastrophic, whereas a member fai,lure is likely to allow time for safety measures to be undertaken. Recalling that no joint (with holes and in tension) is more than 85 percent efficient, and based on the work of Bendigo, Hansen, and Rumpf ("Long Bolted Joints," Proceedings, ASCE, Vol. 89, ST6, December 1963), the efficiency of a
E
=
0.85 - C , ( L - C2)
fps: C, = 0.007
SI:
L 2 Cz
C, = 0.00275 C, = 406 mm
C, = 16 in
is equation indicates that connections with joint lengths up to 4% efficiency of 85 percent (i.e., no reduction in connection capacity for j greater lengths there is nearly a linear 10s~o f Joint ching a capacity of about 60 percent of the short joint whe~lthe 1sng:h is Qa order of 50 in (1250 mm). The cufient AISC specifications indirectly ailow for long joints by use of percent efficiency factor and adjusting the allowable fastener stresslues for fastener stress are considered valid (for bearing connections) up to a nt length of 50 in (1250 mm). Above this length the allowable shear stress is (0 reduced 20 percent. Recalling that the specifications provide minimum uirements, the structural designer has the option of using Eq. (8-2) for termediate joint lengths, between 16 and 50 in. -4.2 Edge Distance lifie of stress are located too close to the i.dgr. i t may be possible plate as shown in Fig. 8-3b and in the actual joi%t shown in Figbe avoided by using an edge distance obtained by equating s h c a es using F, = constant for both bolt and base metal, to obtain A,F, = dtF,
h the edge distance d is A
d =4 p
I
{
i
I
I
t
I
P
,Is
.'
4 y
Strain distribution (approx) in bottom plate ofjoint above
m 8-8 Load and strain distribution in bolted joints. .
. . ,,
, ., ,,, Oi*
,,. ....
..
distance is required by AISC in Sec. 1-16.4 and using an edge distance he bolt is in double shear. thing holes, it is necessary to have an adequate edge and end avoid warping damage to the material. The AISC specifications (Seee distances for this based on the nominal bolt diameter. For bolt 5 1: in (30 rnm) nominal dimension measured from center of
Rolled edge:
D, = 1.4
>(
diameter (rounded to n?ar?St
in Or 3 mm)
30-
.UC?URAL STEEL DESIGN BOLTED
I'
&\ID RIVETED COWEC~~OE~S 363'
and superposition of effects eventually produces a l~mitingfnction resistance (i.e., yN does not increase w~thoutbound w ~ t hlncreaslng N). A spacing that is too close can cause difficulty In ~nstallingthe fasteners, slnce the wench hezd requires a mlnimum working space Theae problems are rcsol~zdby usmg the mlnimum spaclng requirements ,,s,
=
2.67 x diameter (3 x diameter preferred)
AASHTO (Sec 1-7.22C): ,s,
=
3 x diameter
AISC (Sec 1-16.4): . .
-,
AREA (Sec. 1-9.3):
smin= 3
X
diameter
Maximum spacing of a single line of fasteners in the direction of sris:is should generally be limited to 12r, where t = thickness of thinnest part being clamped. This spacing can be used for AISC, AASHTO. and AREA. Fabrication practice coupled with wide usage of the through I-in-diaizster fasteners has led to certain standard gage distances. These values are shavn in the tables for rolled shapes (see Tables 1-3 to 1-7 and V-3 to V-7 of SSDD) and depend on width of angle' leg for angles. as in Tables 1-13 and V-13 and in the AISC manual. The user must also check the minimum edge distance for angles so that the fastener size does not result in a hole too close to the edge to satisfy the specifications. The rolled shapes, which have very wide flanges. may have a second gagz line. Values for the very yide flanges (e.g.. W14 larger than 142 lb/ft) are shown in the AISC manual, as well as in tables available from the steel producers, but are-not shown 'in the SSDD tables. The reader should note that the "standard" gage distances will generally result in a more economical fabrication cost, but these distances are not the only ones that can be used. The designer should, however. consult the fabricator if other than standard gages are contemplated so that an economical joint is produced.
:-
F b#:-. Several modes of joint failure. ( a ) Several joints. (6) Bolt shear failure. (c) Tension on net section iL.:-re. ( d ) Tear-out failure due to bolts being too close to end in the direction of the stress.
8-4.4 Minimum Joint Design
For bo!: diameters larger than I f in (or 30 mm) use: Sheared edge:
D, = 1.75 X diameter
Rolled edge:
De = 1.25 X diameter
?*A 8-42 Bolt Distribution and Gage Distances It is necessary to ensure a reasonably compact joint and one where the cted material is in reasonably good contact so that the developed friction is uniform between the parts. If the bolts are too close together, is obtained, since the maximum coefficient of friction is p 0.35,
--
AISC specifications require that all connections. except those in rrusses, c2rrying calculated stresses be designed for the design load but not less thrin 6 kips (or 77 kN). AISC specifications require that truss joints in either tension or compression be designed for the design load but not less than 50 percent of the effective strength of the member based on the type of desi,on stress. The AASHTO specifications require that connections be designed on the basis of the average of the design load and the effective strength of the member but not less than 75 percent of the effective strength of the member. This is because many of the structural members in U S H T O design are controlled by factors other than stress, such as L / r . At least two fasteners are :equired in any AASHTO-designed connection.
-
364
S~RUCRRAL STEEL DESIGN
BOLTED A\?) W ~ E CDO ~ X X C ~ O ~ S .c
tion.
I
-Pi
8-4.5 Shear Lag Long Joints are undesirable from the standpoi& of reduced efficiency (below 85 percent when L > 406 mm), but in cases where W, S, or C shapes are used with gusset plates on the flanges (see Fig. 8-10), ~t is necessary to produce a jo sufflclentl~long that the stress in the section at A-A can be transferred to a
used for strut
resulting progressive failure across the section. A measure of shear lag efflclenc~is based on the distance from the gravity axis of the member to the fastener (or gusset plate) plane. Munse and lChesson ("Riveted and Bolte Joints: Net Section Design," Proceedings, ASCE, STI, February 1963) give an equation for shear lag efficiency: L
=
SD
1 percent f o r each
where
& I P ( ~m)In eXCe,S
of
L = 450 Same as M S H T O
x= distance from gravity axis f 0 fastener plane (= d / 2 for w shapes and simply 2 given in tables for angles) L = joint length
For most well-designed sections, Eq. (8-3) should give a value of e , from 85 A
to 90 Percent. The AISC reductions for shear lag were presented in Set. 5-3.2.
I n ~ t ~ transfer dl of some flange
---
f
basic premise In connection design is that each fastener c a r 7 a prorated share the connection load. For a joint with constant-size fastener~in a s ~ m e t r i c z I
'fastener
'"%re 8-10 Shear 1% in a W shape connected to a pair of gusset plates.
?
Ptotai - number of fasteners
is assumpllon made for ellher fastener shear o i bcanns. BeannZ i s ccriided In some speclflcatlons as beans: of fastener on ha$ nlelal 37J G~ C a t +* * ,"cur
-
t
STRUCTURAL STEEL DESIGN
1
BOLTED .LXD RIbEXED COh3ZCTIOE;S
Table 8-3 Table to determine required bolt length.Fased on grip and thread length (use table to determine if threads will fall idshear plane; see example below) L,,, gnp + L; round Lrcqdto next larger f in or 6 rnm
-
Bolt sue, m I I ' i
Thread L
L, m
Bolt sue, rnm
Thread La
1
I1 16
12.5
25
If 1;
7 8
15 20
30 35
1
2 proximate and based on soft conversion. r,xample Gwen the connection shown m sketch and uslng a 2-~n-&meter bolt. Are tt : J ~ S In the shear plane (defmed by the plate junction)?
Grip=;+ Fiwi. the table, L
%: tqr
=
L,,,
+
& + ;=
1&1n
1 In:
* R ~ u n to d nearest larger
= I&
+ 1= 2 1 16
t
ble 8-4 Allowable bolt and rivet stresses for AASHTO specific~fions stress mm normnal wit area except l u r r \ - ~ v iwiu 211 (use glven b! on)B
/
r ~ i d w a b ~shear e stress. F-
Data from AASHTO specificat~ons,12th ed , Secs 1-7 22 and 1-7 41 Fb = allowable beanng stress on nvet or bol: from fastened material. F n c t ~ o nconnectlons are required for connectlons subject ro stress reversal. Bolts In beanng-type connections must be X-type (thresh excluded from s h a r nes). Beanng-type connections are to be used,@ condary members. Reduce h~gh-strengthbolt values 2 an 24 In and F, < 42 k s ~ N, threads m shear plane. X, threads exciudbd
Lreqd= 2f In
T h e a d length = 1; (from table):
*"\% . 2.25 - 1 375 = 0 875 In
Distance for two plates = + $ m = 0.8125 In Therefore, threads are out of the shear plane.
ineta1 on fastener. The shear area is obtained using the nominal fastener diameter. The bearing area is the projected fastener diameter x plate thickness: the fastener load Pi is
b.
P, = A , x F, (shear) Pb = D X t X Fb (bearing) No allowance is made for the hole size being in (1.5 mm) larger than the hnl shank in bearing. 4 $ n the fastener group is unsymmetrical or the load does not pass through oid of the fastener group, the fasteners are not equally - * stressed. This ii?,ibn is considered in detail in the next section. .We note that the allowable shear stress F, depends on the design assu ,,tie: (:.at the joint is either a friction or a bearing type, and whether the - 1 0
$?
1
ds are In the shear plane for h to determrne ~f the threads are In the S av be used to obtain the alIowabl -,ecificat~ons,respect~vely. Table 8-6 may be earing stress for base metal-to-bolt or his table are based on the ultlmate
- --
bection. Table 5-3 W!i be
!.XASHTO, and A E X $0 obta~nthe dIott-ahis XISC values s h o i t ~in , material, compcisd
No made for fasteners in either sin& . distinction - --- - -- - .. is . - . -
or doubIr shrzi. T i e
SHTO bearing stress based on metal to fastener is
F, = 1.225 nd will generally be limited by fastener-to-metal stresses. as shown in Tab12 8-4 der the column headed F,. The AREA bearing values are stipulated for the fastener type except th2t ring is not considered. in- the for connections using his!-strength WE. . . design - , ese values are shown In 1 able 8-6.
BOLTED hSD RIV-
Table 8-5 AREA fastener stresses (all connections are "friction"type) F, Fastener &vetsa Hand-driven Power-driven ~olts~ A-325 A490
MPa
ksi
-
-
11.0 13.5
248 248
20 27
36 36
,
ber 1965) give average values of ultimate shear in terms of ul rength F, (see Table 8-1) of
Fo
ksi
MPa
76 93 138 186
Use beanng stress on n y t s : single shear, 27 ksl or 185 MPa; double shear, 36 k s ~or i 2 5 0 MPa. Need not cons~derbeanng on bolts In frict~onconnections.
e
.
I
Table 8-6 Bearing values for rivets and bolts by several specifications (top part of table is metal-on-fastener,bottom part is fastener-to-metala Material ksi
AISC
FU
I
I
MPa
I
I
ksi
MPa
ksi
MPa
AASHTO
I
ksi
MPa
AREA
I
ksi
MPa
F=
0.62(825 MPa) 120
F
250 Not required
Top part: F, = 1.5F,,; AASHTO: Fb = 1.22Fy. Bottom part: depends on fastener; fastener bearing generally controls for bridge design.
=
4.26
int geometry reduces this value of F to something on the order of 3.3 rot mpact joints (in tests) and to around 2.0 for joints whose length is in excess of 70 mm (50 in). This value of F compares to the tension value on the cross ction for A-36 steel of
F = - - F" - 1.72 0.58 F,
A-307 bolts &vets A-501 and A-502 Power dnven slngle shear double shear A-325 and A-490 bolts
CO
=
(net section after slip)
---F" - 400 - 2.67 0.6Fy
150
(gross section before slip)
servations of joints in long service indicate that a safety factor of F 2 2.0 lor fasteners gives satisfactory service.
-4.10 Splices in Beams
a
8-4.9 Nominal Factor of Safety of Fasteners and Connections The nominal safety factor against slip for bolted joints is on the order of 1.25 to 1.30 (as computed in Example 8-1). One might ask what the nominal safety factor,d,the..jointand of the mechanical fasteners is against failure. The safety factor of the mechanical fasteners can be readily estimated based on the ultimate bolt tension and shear stress values divided by the allowable values given in Tables 8-1, 8-4, and 8-5. Shear strength tests on high-strength bolts (see Wallaert and Fisher, "Shear Strength of High Strength Bolts," Journal of Structural Division, ASCE, ST5,
ms are often spliced to produce continuous spans. Splices are usualIy p!aceiI lose to the location of zero shear in the span. Any use of mechanical fast*Pnzrs the tension flange will reduce the effective area somewhat. Based on otes:s, r~ss ISC and recent AASHTO specifications allow ths designer to use the , lange area for stress calculations as long as th? holes (in the tension flange) are s than 15 percent of the tension flange area. When the area of holes e x c r d s percent, the flange area is reduced by that portion of holes in excess oi 15 For example, if A, (of tension flange) = 20, A,,,,, = 4. the percent holes = 4 = 20 percent. The excess hole area = 20 - 15 = 5 percent and, the ension flange area is 20(1 - 0.05) = 19, not 16, as would be obtained by t-.l.ring t all the hole area (20 - 4 = 16). These computations should also be used for the flnngz spiics plates.
x 100/20
370
BOLTED
STRUCTURAL. STE
A-325 bolts a r i required for the tension sp = 250 MPa a be friction type, or bearing type.
. E8-2a if the metal is F,
+
-mm bolts, so the hole diameter = 22 3.0 = must be sheared twice for the beam web to splice plates, the load per bolt is
P,,,, = 2
I /
11 ,' For a frictidn-type connection:
X A,
x F,
I
P,,,,= 2(0/7854 x 0.022~)(120)= 91.2 kN/bolt
(double shear) he number of bolts required is N = 442.5/91.2 = 4.85. Use five bolts. the bolt pattern shown in Fig. E8-26, .so that the splice plate wihth will maximum but the maximum net section is obtained for the channel.
.OmRIVETED COh?iF,mOt;S 371
r a three bolt line and sheared edges, the minimum edge dis use 40 mm 1.7(22) = 37.4-mm Minimum bolt spacing = 3 0 = 3(22) = 66 mm 250 - 2(40) O.K. Use spacing = 2 For the forward two bolts: 85 Center bolts as 40 + - = 82.5 nim fronI edges 2 Spacing to edge is less than maximum allowed of 12t = 12 x 10 = 120 m a . Use the distance from the front bolt to the edge of splice plate at 40 (1.75D for sheared edges) and similarly from the back bolts to the edge of the W410. Set the gage distance so that-only two holes are deducted from the critical net section.
C310 X 44.6
s =\ / 2 1 z = 46.1 mm
use s
=
50 mm
(arbitrary choice)
Check the bearing: On splice plates:
=
F, = 5(0.01)(27)(1.50 X 400) 660 > 442.5 O.K.
=
5(0.013)(22)(1.50 x 400)
Ph = iV x ,+Ihx
On web of C3 10: Ph
O.K. 858 > 442.5 kY Use five bolts for the friction connection. For a bearing-ope corlrrecrlorr Grip = 2 x 5 x 13.0 = 23.0mm L,,,,=23.0+28=51 iise55rnrn =
Figure W2a
Figure E8-2b
Check the net section of the splice plates. With three bolts out, the are requirements are: Gross:
442.5 A, = = 1.475 X 2 x 0.6Fy
m2
Net:
Ap =
442'5 = 1.106 X OSF,
m2
2
X
A >-= "lo6 - 0.85
1.301
x
13 = 18.0 mrn
say 4 bolts
m2
> 1.106
+
Distance from bolt head to end of thread = 55 - 38 = I7 mm < 18 ma This computation shows that the threads are in the shear plane, so F, = 13 MPa (instead of 205) and P,,,, is
Try two plates 250 x 5 mm with three holes in the critical section. A,,, = [250 - 3(25.0)]0.01 = 1.75 X
L,,,,,, = 38 mm (Table 8-3) Width of one splice plate + beam,web = 5
O.K.
.=
2.6
(also use 4)
Use two columns of two bolts for bearing-type splice
STRUCTLTRAL STEEL DESIGN
Example 8-3 Design the connection for t (NO. 7) for the highway truss of Example E6-6 of Example 6-6 and to Fig. E8-3 (ref 8-46). The previous examples were used members 7 and 9. Let us use the same cro used for the end post (all in compression use the same sections as for 7 and 9 ( stress). In any case, let US design the conn (No. 7), which is in compression. Data f
II
BOLTED .kKD RIYETE# CObYECTIOhS
.-3
6-4, the allowable axla1 compressiv; stress was 9.35 ksi.
TODchord
=
AF, = 17.00(9.35) = 159.0 kips
159.0 -t 99.3 = 129.2 kips 2 The 75 percent member strength criteria give pa" =
or the top chord as was ng web members may controlled rather than e vertical web member
718 bolts as req'd. .
P,,,
, and 6-6. Refer to Fi
Use A-36 steel and the AASHTO specificatiJns. Use A-325 high-strength bolts. P,,, = - 99.3 kips (P,, = - 40.6 kips). The gusset plate t = 5/16 in (minimum t allowed by AASHTO for a plate).
1
373
/
Memno
I'
Po,,
=
159.0(0.75) = 119.3 kips
< 129.2
From Table 8-4, the allowable bolt shear stress F, = 13.5 ksi. Thz nuxSsr of bolts required in the connection to transfer 129.2 kips is 129.2 P N=---?= = 15.9 bolts AbF, 0.7854 x 0.875" 13.5 Use N = 16 bolts for symmetry and since 0.9 bolt is not possible. Tne use of 16 bolts requires four rows. Use a bolt spacing of 3D:
Use a minimum edge distance in l~neof stress,,= 1; in (1.750 for cut ecZ): ,. Total length of joint
=
4(2 625)
+ Z(1.5) =
13.5 in
W ~ t ha joint length of 13; in (nominal), shear lag IS not minimum required transverse edge distance = 1.25D rounded larger 1/8 in or 3 mm for flanges of beams and channels (but and other elements). Thls glves
4
I
Figure J33-3
SOLUTION The fastener design will be based on a friction-type connection. AASHTO does not allow a bearing-type connection in a main membe Stress range does not have to be considered for connection design. We will try four 7/8-in-diameter bolts at a section as shown in Fi E8-3, the same as assumed in the design of member 9 for tension Example 5-7. A deduction for net area does not have to be made f compression members unless a complete stress reversal occurs (this does not occur here). AASHTO (Sec. 1-7.16) requires that a connection be designed for th average of the design load and the full effective member strength but not less than 75 percent of the effective strength of the member. From Example
=
1 2 j ( 8I )
=
l $ in (1.125 in)
The distance furnished and based on the standard gags distance = 5.5 in (see Table 1-3) is computed as b, - 5.5 10.01 - 5.5 dlurn= ------- = = 2.26 ~n > I . 125 O.K. 2 2 Check the bolt beanng on the gusset plate. slnce I, = 0.313 < 0.640 of flange of a W12. O.K. x 0.3 13)(40) = 175.3 kips > 129.2 ade sufficiently wide that tension on a net section li be some f~llerplates needed between the gusset nce the W 12 sections are deeper than I2 in. /// '
.Jf\
Example 8-4'Design,.thg connection for the vertical members of the mSn roof truss of Example 2-6 that were designed in Exampls 5-6. F, = 250 MPa. Other data include: P = 70.18 kN (tension)
."a
STRUCTURAL STEEL DESIGN
'
,t*'
BOLTED .CYD WYETED. ~ O ~. &..T I O K S
This design was based on using a 12-mrn gusset plate and 25high-strength bdlts.
With the'threads in the shear plane, F, = 150 MPa P,,,,
SOLUTIONAISC requires that connections for truss members be designed for either the design load or 50 percent of the effective strength of the td member. L
P,, = 0.5(0.6<)~,
= 0.5(150)(2.66) =
199.5 kN
controls
=
2t0.7854 x 0.025')(150) l d
=
375
able 8-1): 147 kN
199.5 = 1.36 147 Use two bolts for shear, since a fraction is not possible. Check the beanng: N
=-
Fb = 1.5F, = 1 j(400) = 600 b1Pa ( Fu from Table 8-6)
The bolts through the angles and gusset plate will be in double shear, as illustrated in Fig. E8-4b. Assume that a bearing-type connection (slip can tolerated) is satisfactory (a designer's prerogative). Check the bolt len using Table 8-3 to see if the threads are in the shear plane.
0.E';On angles: Pb = 2(2 x 0.0063)(25)(600) = 375.0 > 175.9 kN On gusset plate: P, = 2(0.012 x 25)(600) = 360.0 > 175.9 !c3
Use two 25-mm A-325 bolts and a 12-mm gusset plate.
-5 RIVETS AND BOLTS S U B J E m D TO E C C E h i C LOXDlN
15-mrn bolts 1 2 7 X 89 X 6 . 3 I
:I -,.I
I
I
V\-FJ/v I
_ I -
(
I
Figure E846
Bolt grip
=
6.3 x 2 -t 12 = 24.6 mm
24.6 + 30 = 54.6 use 60 mm (Table 8-3) 45 mm One,aQgle + gusset = 6.3 12 = 18.3 mm Thread runout location = 60 - 45 = 15 mm 15 mm
<
L
=
L,,,,,,
=
+
nerally, when the eccentricity of the load on a bolt group is less than sboct in (60 mm), it is neglected. Joints such as the simple frame connection of Fig a, which is widely used, are in this category. The bracket connectiori of g. 8-12a is loaded with an eccentricity that is obviously too l a r ~ eto be glected. The framed beam connection may be large enough that the resuf5ng ccentricity,,isalsd too large to neglect. One may note that the standard frzrnstf connection angles in the AISC design manual neglect the eccentricity for vduls to about 3.7 in (one of: the standard framing angle connections with nearly value of maximum eccentricity is shown in Fig. 8-12b). A load to be resisted by a bolt group that is eccentric with respect to the id of the group pattern can be replaced with a force that has a'lineof = PP, through the pattern centroid and a moment with the magnitude Lti ere e is the eccentricity of the load. This is illustrated in Fig. 8-l3a a ~ 5.d considering that each bolt in a pattern that is centrally loaded carries its ed share of the total load, we have, for equal-sized bolts, P p.=-
"
!V
where Psi is the shear force on the ith bolt with a vector to resist the a p p l i d force P . An additional bolt force is develo~edbv the eccentric moment ,bf = ?r. ~ . Assuming a group of bolts acting as an elastic unit. we have a concept similx to that of beam resistance being developed and as related to the beam moment of inertia. Referring to Fig. 8-14, we have a bolt pattern \%-ithan applied moment .El which produces a resisting moment for rotational equilibrium that is equal to -
,=n
i- l
18.3 mm. ' .threads in the shear plane (see Fig. E8-4a)
f vbe assume that the value of R, is proportional to the distilnce f r o n the
1I
:
Z , 1.
3 73 STRUCTURAL STEEL DESIGN
BOLTED L\D RWEED
\
the action of P through the bolt group centroid produces an additi l'olt resistance R, which must be added to the R,,,,,, vector, it is better obtain the H and V components of R,,,,,,. Referring to Fig. components of Rl are Rh and R, can be obtained by proportion as: L::-IJ~
Rh - RL -
Rv = -
R,
COk>~~3 o77~8 ,s
he bolt pattern and plate adequate for the g~venload in a bearing-~?c-nnection assuming threads In the shear plane? SOLUTIONSince the bolt pattern 1s symrnetncal (as in most ~ractlclrpeob]ems), the centroid of the pattern is readllq located and marked 3s c-g-, 3s shown on sketch Compute z ( x 2 + y 2 )
Y - - MdY - -My R,=R-= I dl C.d:dl
S,rnllarly, from Fig. 8-14b, it is evident that
erld can be interpreted as the polar moment of inertia of a group of unit are :.ate that if we use the area A with the denominator of either Eq. (A or ( g ) , arLJR, are obtained as stresses. Multiplying the numerator by the area A of zth !~olt produces the force R, or Rh. With A in both the numerator denominator, it cancels, giving R as a force. For general design, the equat are
I
f 18.3 Example 8-5 Given the bracket connection shown in Fig. E8-5a and th the fasteners are 25-rnm A-325 bolts and the plate is = 250 MPa steel,
<
1l s.?
Figure E & 5 b
Com~ute
P = l I0 k N
shown on Fig. E8-56 to iesist applied P. compute e = 150 + 125/2 = 212.5 mm:
R,,,
=
0.5088~
Set up a table as follows and omit the signs of I and y (use veciors previously drawn on Fig. E8-5b to determine the direction of the ir afid c
vectors):
Placing these values on Fig. E8-5b, it is easy to see that bolts 1 and 5 are th most highly stressed (critical). Bolt 3 is loaded the least amount (13.47 kP., The resisting force on bolt 1 is computed as R =\/(31.8
+ 18.33)' + 38.2'
63.03 f, = 0.7854 X 0.025' Check the plate bearing:
'
=
1000
X
63'03 0.025(12)
=
=
63.03 kN
= 128.4 < 150 MPa
kips
O.K. x = 2.75 in I
=
210.1
< 1.5(400)
y, depends on the number of bolts Try 12 bolts Note t h ~ boIt t I is alw~ys the most stressed for a bolt pattern such as this.
O.K.
Check the possible tension rupture of the plate along the forward bolt line: Moment of inertia, I = 0'012(0'270)3 12
- 2(0.012 X
C(x2 + y2) = 12(2.7512+ 4(7.j2 + 4.j2 + 1.5') = 405.75 50 R, = 12 = 4.17 kips
J 8
0.025)(0.075)~
887.5
I
Section modulus S = - = C
16.308(2) = 0.12-.. . . ., 270
Moment at forward bolt line = Pe'
R, = --(7.5) = 16.40 kips 405.75 , , ,
= 110(0 15) = 1 6< LN
- 136.6 MPa < 0.6G f,=s=-0.1208
. .,
Q
Since 16.40/0.7854 = 20.88 > 17.5 ksi, the numbzr of bolts at a I-in diameter is too small without computing R. Try 16 bolts (bypassing 14, since 12 bolts were so hi@Iy stressed):
O.K.
Check plate buckling: - = - -150
- 12.5 < 2 5 0 / f i
887.5 405.75
R, = ---- (2.75) = 6.01 kips
4
C(x2 + y2) = 16(2.75)' O.K.
t 12 The joint is adequate for bolt shear, plate bearing on bolt, and bending.
R,
=
+ 4(10.j2 + 7.5' + 4.5' + 1.5')
= 877.0
50 = 3.12 kips 16
//J Example 8-6 Given a crane runway bracket that carries a load as shown in Fig. E8-6. Use A-36 steel and either 7/8- or 1-in-diameter A-325 bolts. Assume a friction-type connection. Find the number of bolts and the bracket plate thickness.
R, = 1.012(2.75) = 2.78 kips R =\/(2.78
+ 3.12)' + 10.62'
=
12.15 kips
Use sixteen 1-in diameter A-325 bolts.
< 0.7Sj-l( 17.5)
O.K.
'Wh6
'w
IS
Find the plate thickness for the bearing (assume that the column flang adequate):
~q 8-7 for the fastener eccenrnc:ty shov*n
tp(1)(1.5 X 58) = 12.15 kibs
tp=-=
12.15 87(1)
l + 5 0 625 In 2 ccentnclty (for 101nt performance). it is
e,,,
=
0.139 ~n
h=7x3+3=24in tph3 12
M=50(15)=750in.k1ps 2tp(lO.S2 + 7.5' C 4.52
$1
fi
Me
Fb=22ksl
+ 1 S 2 ) = 7745
3.625 -
~~~~~l~
S ~ L U T I ~Referring N to
Fig. E8-7, try twelve I-in-diameter bolts-
750(12) = 0.53 in t = -----774(22)
AIsC, Set. 1-9.1.2, for unstiffened elements under compres
-b< - - 95 - 15.8 t - v36
15.0 15.8
- 0.95 in
use tp = 1.O in
Summary: use sixteen 1-in-diameter A-325 bolts and a bracket plate In.
Figure W7
= 1.0
8-3.: AISC Reduced Eccentricity for Connections preceding method of joint analy~lswith eccentricity is widely used. Based on a series of tests which displayed that this method of joint analysis is ccrslderabl~~ ~ n s e ~ a t i AISC v e , uses (in tables for design of eccentrically lo . ? L J fastener groups in the design manual) a reduced eccentricity which is cc-.?~uted,for a single line of n fasteners: T h b
%&
'$c
eeff= e
I + 2n - --4
llne? of fasteners with n fastenen in any line: ' 8 ,
.
1 f i
eeff= e
l + n 2
--
to sre
8-7 Redo Example 8-6 taking into account the reduced ecceL;:::-
'6
t=--
Fig-
----- =
750(12) 774%
f b = F b = 2 2 + - = -1
1.
,333
ctlce of neglecting the eccentnclty for most f r ~ * d nable design procedure.
Check the plate bearing along the forward fastener line and neglect bolt $ales 1/16 In larger than bolt:
Ip=--
-
I
BOLTED f i RI\ETED. ~ CO~+XXOP.IS
STRUCTURAL STEEL DESIGN
M
=
jO(14.25)
C ( x 2 + y 2 ) = 405.75
=
712.5 In . kips
(Example 5-6)
R,
=
*" (2.75) = 4.83 kips 405.75
R,
=
1.756(7.5)
R =\/(4.17
=
13.17 kips
+ 4.83)'
j-,=-- 15'95 - 20.3 0.7854
i13.17'
> 17.5
=
15.95 kips
N.G.
% STRUCTURAL STEEL
BOL I E D L\I) R I L E r t D COh3ECTIOhS
DESIGN
x!?
Try fourteen I-in-diameter bolts:
8 e,,, = 17.75 - - = 13.75 in 2 M = 50(13.75) = 687.5 in . kips
C ( x 2 + y 2 ) = 14(2.75)'
H"
+ 4(92 + 62 + 3')
=
609.9
6873
--(2.75) = 3.10 kips 609.9 R, = 1.127(9) = 10.14 kips =
R =\/(3.57 + 3.10)~+ 10.14' = 12.14 kips 12.14 O.K. jv = l.O(O.7854) = 15.45 ksi < 17.5 Using the reduced eccentricity requires two fewer bolts in the connection. The plate thickness of 1.0 in is still necessary to satisfy b / t , and beardg IS not a problem.
i/.t
The AISC Design Manual gives tables based on one, two, and four vertical columns of fasteners which may be used to design eccentric connections. By assumlng the number of fasteners in a row, a computation for e,, is made, and with n and e,,, we obtain a coefficient that is multiplied by the allowable fastener load to give the total group eccentric load. One may readily derive an equation for the allowable eccentric load for a single vertical (or horizontal) line of fasteners. Equations are given in several textbooks which may be used in an attempt to reduce the computational effort in finding the number of fasteners for an eccentrically loaded connection. The author suggests that with the increased computational efficiency available to the designer with the pocket calculators, it is as easy to "punch it out'' as to try to use an equation developed by someone else. This is so particularly because no simple equation exists and most equations require some iteration anyway.
8-6 BEAM FRAMING CONNECTIONS Figures 8-2a and 8-15 illustrates the most common methods of framing steel structures for small buildings of five or fewer stories where the connections are AISC Type 2 (simple shear connections). The eccentricity of the beam shear is neglected with these connections, and Ao moments are assumed to be transferred across the connection. Frame stability is.provided by use of wind bracing pi masonry walls closely fitted to the columns as shear walls.
(dl
-
Q-15 Frarmng connecttons. (a) Moment-reslstlng connecbon using top and s a t m$a bckd o column flange. Web angles c a n y shear. View l w k m g down (b) M o m c n r - r e ~ t m gc o c n c c ~ ~~rz, : o
olumn web. Note that top and bottom plates are welded lnto column web a d flrrc&eto ~ ; i oiumn snlrclici>.(c) F r a m ~ n gfloor system In pgwer statlm Note >hear zufirnen 113 %eb of r ~1. der. Coping IS shown for small floor beams IJI nesr foreground (d) F r z m g for bndse Siriz,t,s.
F =A.
8 '
A number of the joints shown in Fig. 8-20 are ..~tandardizcd'~ as to b9uir ttern and with an angle length that depends on the beam size and as given in AISC design manual. The angle is selected based on bolt bearing and wirh ee dimensions that depend on producing adequate edge distance and without I I L L G L L G L G L L L G"etween bolts and wrench during installation if the boIt hokes are 3ligned both vertically and horizontally. Use of these mbles often prodcccr u :onnection that is overdesigned. However, the cost of ovcrdesip is gsncraiiy more than offset by reduced fabrication costs from using standard dimensions. The simple framing tonnection is used to connecr strin,oers to floor beams ~ n dfloor beams to -girders in bridge fabrication. In bridge design the conncctions should be standardized for the given bridge to reduce fabrication co,jE; neral standardization for bridges is not as easily done.
b
3PtL STRUCTURAL STEEL
DESIGN
BOLTED LL?)
+
Zxample 8-8 Design a beam framing connection for a W 14 to frame into a W18 X 50 girder. The W14 x 3 0 carries (live + dead) = 2.5 kips j f t and the span is L = 18.0 ft. specifications, A-36 steel, and A-325 high-strength bolts.
;
1/16) as in Fig. ES-8-this .@'?s
designed in Examples 4-1 rength bolts. F, = 138 hfPa ction-type connections.
SOLUTION From Example 4- 15, obtain V,,,,
78.3 kN
=
v , , ~=,140.5 kX Figure W 8
V,,, = 178.0 x 0.795 = 1 1 1.7 k N
0 should be coped as shown in Fig. E8 ch side) as shown, but we may note t angle on only one side of the web. sides it is evident that the bolts in t ii-ebs of both the floor 'girder and the floor beani will be in double sheay.
.
v = -w=L
(2.5 + 0.0')(18) = 22-77 lups 2 2 LTsc. 3/4-in A-325 bolts in a friction connection. ,.. ",_.... P b , , , = 15.46 kips double shear (Table 11-7, SSDD) I!! a W14 x 30:
~h~ floor beam is a W760
X
Total shear = 280.5 k s 160.7 and frames into a plate girder (Example I,.
=
16.0 mm
t,.,
=
13.8 mm
~ e ust use 22-mm-diameter bolts: p,,,, = 0.7854(0.022~)(135) jId) = 52.3 k s
.'I....
3.
22.77 use two bolts Nshear = -- 1.47 15.46 22.77 = 1.29 Nbrg - 1.5 X 58 X 0.75 x 0.27 In the web of a W18 x 50: 22.77(2) Nbrg
= 1.5 X 58 X 0.75 X 0.355
use two bolts
= 1.96
requires two bolts
=
2t(0.75)(1.5 X 58)
=
=
22.77
2.7
use three bolts (double shear)
or bearing with bolts in double shear, Fb 280.5 = 3.7 N b r g - 0.022 x 250 x 13.8
F~~the web of a plate value); Table 8-5)):
22.77(2) use four bolts for symmetry 15.46 = 2.95 Use a web angle with a length of 6 in and t to be determined: For bearing: Nshear
For the web of a W760:
Nbrg =
Nshear =
150 hfpa (Table S-5): use bolts as required
[bolts in single shear and Fb
280.7 0.022 x 185 x 16 = 4.3 1 280'7 -- 5.36 52.3
Place bolts as shown in Fig. E8-9. ~ 7 6 x0 160.7
=
use six bolts
, *L
-
2t(6 - 2(0.75)(0.4~,)= 22.77
ii
tz0.176
.r (1 '
uset=$in
an angle with a length of 6 in. Use two L4 x 4 x 1/4 x 6 in long with ,
I
,I:-
- 1 1 ; ~
1
-
; 1 , ; +(-1;- 1
!
,
185 hfPa (rivet
use six bolts
T , -'77
t = 0.174 in For shear:
=
#
!!-
.. ,
![;a
STRUCTURAL STEEL DESIGN
4
Design the framing angle:
'
'
'
L=5
30
+2X
1.750 = 18.50 = 18.5(22) = 407rnrn
use410 The angle thickness will be controlled by the bearing of the three in the outstanding leg (o.s.1.) fastening the beam to the girder; therefore, t 2 16 mm. X
2t(0.022)(3)(185) = 280.5 i'
,!fx;& Checking the tables, try: L127
X
t 2 ll.5^'mm 127 X 11.1 (Table V-9):
g = 75 mm (Table V- 13) Effective edge distance de = 127 - 75 = 52 mm ."
I
d,,,,,,, 2 1.5D = 33 mm
4
A L102 X 102 X 11.1 will not give sufficient clearance between the -- - W7hO -bolts and the bolts through the girder web in the o.s.1. The small amount of overdesign is rather negligible in any case. The design is summarized in Fig. E8-9. Note that AREA does not require a "bearing" check using A-325 bolts. This check established the approximate angle t and is recommended whether required or not.
8-7 FASTENERS SUBJECTED TO TENSION ~.~ .,,, ... .,," ,
1
Figure 8-16 illustrates the usual conditions for bolts in tension. When the bolt ;c tightened to develop the proof load, the shank elongates. Simultaneously tLclamped plates are compressed. When we apply a load to the connection, we have the free body of Fig. 8-16c, which gives
Obviously, as P becomes larger and larger, two events occur simultaneouslv: 1. The bolt tension T in9reases slightly, producing a slight shank. elongation. 2. Shank elongation reduces the plate clamping pressurk u, since the plate compression e, = 2T$,/ApE is small and caused by r, being relatively small q npnrlv nf and A, relatively large. On the other hand. e,-,, = A - -T-r ,/ / A E i-constant magnitude as long as the plates remain in contact. ""ll
------
In equation form and now considering a single bolt as in Fig. 8-16d. we *
-$4
j-
,$
- Ae,
= Aebolt
he two clhmped plates expand, and taking P' = prorated part of total P
3%
STEEL DESIGN
where terms
BOLTED hVD RIVETED
~reviouslydefined in Eqs. (8-9) and (8-10) and ~ for A-36 steel,
A,,, = 0.7854(0.875)~= 0.601 ln' F, = 22 ksl (bearing-type connection and threads In shear plane)
r = (36 - 22)/36 = 0.3889
62.4 0.601 X 22
'a = thickness of beam flange supporting load
= X
Check the bearing: pb = 3(2 x 0.25
8110 Design a hanger using a WT section for a load to be
by a palr Of The load Is E8- I Oa.
*
i 8-17 ~ are .
force carried by flange on one side of beam web =stress ratio, defined as (5 - F,)/F,;
COh%xmNS 3%
Use A-325 bolts and A-36 steel; the load is 62.4 kips. from the bottom of a W33 x 221 beam as shown in Fig.
W33 X 201
2
2.36
use three bolts
x 1.50 X 58) = 131 < 62.4 kips
~~t~ if thickness of the WT web 2 2 x
land,
o.K-
bearing and shear
a
Step 3. Design WT. ~h~~ step simply lnvoIves studying tables after making a computation approximate depth based on edge distances and bolt spacing in d, 2 1.5 + 2 x 3 + 1.5 + k
= 9 + kln a WT12
X
47:
tf
-
-
2L's
%dge
b, = 9.065
d = 12.15 in =
0.875
- 9.qo-c - :5. = - -
t, = 0.515 5 - *I
-& \
' 7
;
k = 1-53 In
&?\K
\ '??l+.
S>L3)
7-
I
6 2 4k
Required depth 3 = 9 + 1 53 = 1053 > 12.15 ln Check the bending moment at the toe of the fillet In web.
Figure =lob
Firwe m-loo >
*>
.
SoLuT1oN The hanger design will requlre design of a WT to attach the for the load and to select enough bolts of the proper sire to carry the load in tension. Step 1. Design the angles. Assume that L / r is not critical and choose 7/8-in bolts:
A, =
2.15in2
2.15 0.85
- 2.53 in2
A&---
= (3'38
+ 1/8)]
r
~t~
+ 41.92 in - kips ~(0.875)~
= 0.1276L
Fb
0.50 in2
- 0.50)0.90 = 2.60 > 2.15 required
M = )I.*(?)
'=6= 6 s =M
at least 1/4 in thick, SO that bearing is not a problem. T~ two 1/4 (long legs back to back): = 2[1/4 X (7/8
55 1 b = -L -= 2.6875 in 2 16 T = -62.4 - - 3 1.2 kips 2
Fb = 0.75Fy = 27 ksl
62.4 = 2.84 in2 22
A , = - 62.4 =29
0.K-
O.K,
41.92 =12.161n useL=12.5in 0.1276(27) ~~~i~~~~the number of bolts to can7 the hanger force. Use 7/8-in-diameter bolts for the hanger to connect the WT to the beam: 62.4 = 2.35 N= 0.601(4) L =
' 4 STRUCTURAL STEEL DESIOFI
side of the flange is P = 36(1.15)'(2 x 0.3889)+(1 +
pf
=
P 1.70
61.1 1.7
s) =
62.4 -= -= 35.9 kips > 7
-
61.1 kips
O.K.
///
The flange does not require stiffening for this load.
4
Figure WlOc
Use four bolts for symmetry. Also, data for a W33 b,
=
X
201:
15.745 in
tf = 1.15 in g = 5.50 in Step 4. Check the "prying action" using Eq. (8-9):
F = - -62.4 - 15.6 kips 4 a = dedge= 1.78 > 2(0.875) use a = 2(0.875)
--.' ( 70(1.75)(0.875)'
Example 8-11 A stairway hanger rod is attached to the lower flange of a W410 x 59.5 floor beam. The load to be camed by the hanger rod is estimated to be 38.75 kN. F, = 250 MPa for both rod and beam. Design a hanger rod and check if the beam can carry thls load without a web-toflange stiffener.
SOLUTION Design the rod:
1.75 in
=
+ 21(6:25)(0.875)'
By trial, obtain D = 22 mm (at 10 threads/25 mm), or
x l o T 3= 0.299 X 10-3 mi > 0.2583
=: I 1.1 kips
0.K.
The posslble moment at the bolt is 1 75 1l.l(t)=9.71<41.92in.kips \
I
\
O.K. --
.
--L--/
Since this value is only 300 lb/bolt over the allowable, take it as 0.K WT12 x 47 with four 7/8-in-diameter A-325 bolts to the flange . Use a of the W33 X 201 beam.
L = 12.5 in 2L4 X 3 X 1/4 with long legs back to back Step 5. Check the beam flange for adequacy without using stiffeners. tb = 1.15 in (tnanee - of W33 X 201)
t... = 0.715 in - --w
Check the beam for the hanger force. Neglect the torsion produced bythis hanger rod on only one side of the beam flange, since it is located very near the end of the beam. Take the hanger rod hole at the standard gage distance of 88.9 mm for a W410 x 59.5 section. Also, r, = 12.8 rnrn = t, tw = 7.7 mm bf = 178 mm b=
89 - 7.7 - 1.5 2
=
39.9 mrn
By Eq. (8-13) and directly incorporating L F = 1.7, we obtain
.
= -.5 .F in
From which
P
250(12.8)'[0.389(1 =
+ 44.5/39.9)] + ( lo-') 1.7
=
19.8 kN
< 38.75
N.G.
Since 19.8 < 38.75 kN, the flange of the W410 x 59.5 is too thin and we must either use a section with a thicker flange or reinforce the flange of this section. It will be about as economical to use a section with a thicker flange.
///
3%
FJOL.TED AND RIVETED COhXECTI
STRUCI'UIUL STEEL DESIGN
8-8 CONNECTIONS SUBJECTEDTO TENSION
COMBINED
SH
Figure 8-18 illustrates several cases of connections subjected to sion is adequately conservative if designed as either a friction or a beari connection. It is evident that the compressive force on the connectiqn increa the slip resistance, whereas a tension force tends to decrease it.
of the nominal fastener area are
f, + f, r fa,,,, A better fit is obtained using a quadrant of a stress ellipse, since F,,,,, has separate values for tension and shear F,, whi& give
(8-ljj
ere
T,= bolt proof load as obtained from Table 8-2 J A , = nominal bolt tension force (f,= T / A d
Tfie ~ S H T O equation for the combined stress cases are the same as AISC some additional conservatism and simplification, giving, for friction'
(c)
(d)
,,
F; 5 13.5 - 0.221
ksi
(8- 17)
MPa
(8-17a)
Fm
Connections where fasteners are subjected to c0mbio.d & a r and t e b . Moment-resistin~ c o ~ e c t i o nusing WT and column stjffeners. (b) Type 2 ( k p l e ) or,tm 3 (scm%d) wnnection If angle is used with clip angles, assvme &at it cames of shear. (c) connection using a Pair of angles back to back or a WT ( d ) Co-0~ type for wind bracing. One may use a cable instead of the angle showd.
F:
93 - 0.221
3.
STRUCTURAL STEEL DESIGN
'?'-..,le8-7 Coefficients for allowable fastener stresses for combined shear 1
2.
'
tension cohnections AISC Bearing
c,
,-
a-astener
.; 502 grade 1 grade 2
.
c2
f ~ q
SI
30 38
207 260
26 55
180 380
c 3
fps
SI
1.3 1.3
23 29
158 200
1.8 1.8
20 44
138 303
Friction
AASHTO
cqa
c5
fps
SI
-
0.75 0.75
-
i(S
A.307 .". .325Nb
17.5
121
-
P
Set F,
=
FigurrF.8-12
303 MPa and find P based on shear limitations. sin 40 Iv= lO(O.4908) 303 P
" Standard holes 1/16 in (or 1/5 mm) larger than nominal bolt diameter; see the AISC ~ ~ (Sec. u l1-6.3) for slotted or oversize holes.
ILL
" N,'threads In shear plane.
=
For bolt diameters larger than If in (38 mm); F , = 90 ksi (620 MPa). Only static tensile loads allowed.
For bearing-type connections AASHTO uses:
Check P
=
P=1941>>420kN 420 kN. =
wi.~:..fv and f , are the actual computed stresses and C, is a coefficient from Table 8-7. The AREA currently has no provisions for fasteners in combined shear and ter:sii:n.
Example 8-12 Given the tension-shear connection of Fig. E8-12,~whatis th allowable load P for the W T to column connection using the AISC specif :ations, F, = 250 MPa steel, and 25-mm-diameter A-325 bolts? SOLUTIONAssume a bearing-type joint with threads included in shear plan F/ = C , - C2f,I C, From Table 8-7 (A-325N). obtain
0.13094P
380 - 1.4(0.13094P) 420 kN
Find P based on maximum tension stress. P cos 40" F, = 303 = 1O(0.4908)
- X, threads excluded from any shear plane. *
=
=
85 MPa
F,' = 380 - 1.4(85) = 260
<
useP=420kN
< 303 b1Pa
0.K. * ,. ,:*
4 fl
0.K.
150 hlPa of Table 8-1
//I
*
.
..
) CONNECTIONS
o c are commonly used building connections, with e of Figs. 8-160 and 8-180 and b used for rigid (or AISC tlpc 1) connections g mechanical fasteners. Figure 8-18b (see also Fig. 8 - l j a and b) is cononly used for both simple (type 2) and semirigid (type 3) connections. The esignations "simple" and "semirigid" for this connection are determined to me degree by the thickness of the top clip angle. For simple connections this imited to 1/4 in (or 6.3 mm), so that the an& can t a moment is not developed. The L V T for the ay be designed similar to Example 8-10. The clip angle for the connection of Fig. 8-186 requires a design for bending that is somewhat similar to the W T of Fig. 8-180. Critical sections and assumptions for the clip and seat angle design are shown in Fig. 8-20.
Find the tension force to be resisted by the clip a n d e (a = 0.6Fy = 150 MPa):
M
=
0.5(0.6FV)X 1.46 = 109.5 kN . m
T = -M = - - 109.5 - 239.6 kN d 0.457 number of bolts in tension in the clip angle is based on F,= 3 A, = 0.4908 x lo-' m2. =
1.6
use two bolts
We note that it is good to not have to use more than two bolts, since
I
a
Distance xo = L / 2 w ~ t hsolne XO
des~gners = N / 2 w ~ t hsome
= 3.25
designers
F i N 8-24) Critical sections and dimensions for clip (top) and scat angle d e s i a for use in beams to columns.
use four for symmetry
frarmng
Example 8-13 Design a semirigid (type 3) connection to resist one-half the capacity of a W460 X 74.4 section, as shown in Fig. E8-130. Use 25-mm-diameter A-325 bolts, 5 = 250 MPa steel, and the AISC specifications.
Figure B 1 3 c
205 mm use 200 mm long leg outstanding. The long Icg will have to be long enough to place two bolts at 3 D + edge distance + vaIuc of g, from Table 1-13 of SSDD. L > 28 + 75 + g, = 28 + 75 + 65 = 168 mm Try L178 X 102 X 22.2 (refer to Fig. E8- 13c): a = 50 - 22.2 = 27.8 mm .P
-
SOLUTIONW460
X
74.4 data: =
Assume that web angles will be used to c a q shear so that the seat angle carries only compression due to moment and the clip angle carries only tension.
Fb
=
0.755
Fb
=
0.75(250) = 187.5 MPa
0.01643 x
lo-'
m'
due to type of connection, location, and due to rectangular shape of cross section
402 STRUCTURAL STEEL DESIGN
Since this 'is thickest angle in this group, go to the top of the next since thickness controls. We would not make L > 205 mm, b extend.outside the column flange and also because the ben& b i ~ o m equestionable with a very long width and only two fastener TV L203 X 102 x 25.4 mm. a = 50 - 25.4 = 24.6 mm 0.0246 M = (239.6)= 2.947 kN . m 2 S = 200(0.0254)2 = 0.02 15 10- 3 m3 6 2.947 f = 137 < 187 MPa 0.K. (and no angle th -0.0215 A routine check for tension shows the section to be adequate. ~~t us see if it necessary to use a web angle for shear, since this angl thick. Refer to Fig. E8-13d for critical dimensions and other da analysis. R = 275 k~
ole cames one-half of an=
This is not :atisfactory even if we assumed the shear; therefore, web shear angles are required.
///"
~h~ moment connection shown in Fig. 8- 1 8 may ~ be treated in one of two
1. Assuming initial tension in the bolts (which is alivays developed with strength bolt connections). 2. Assuming no initial tension. .."
hi@~h~~~ two assumptions are illustrated in Figs. 8-21 and 8-22. strength bolts the assumption of initial bolt tension allows the connection to bending as an elastic unit and the stresses can be calculated using the moment equation, fb = M ~ / Iwhich , is valid up until the cOnnection plates separate int is never designed for a moment large eno"@ desi@ d shown in Fig. 8-21 is adequately conservative ry to design or a n a l ~ z ea j o i n t
.
1.4 3.
,,"
.'....,, .k .I/
j; = -
S
-- 0 ~'I;I:\
lL
---
,.'p,lrl
LIL! j;~)t [t
Figure E8-13d
With 12-mm standard beam end clearance, R = ity of
v acts
203 - 12 = 95.5 mm 2 Since this value is so large and because we are using the long leg out ing to provide adequate length for two rows of bolts, let us distance required for the beam and use that value to &termin in the seat angle. Assume that the reaction is concen from the end of the beam for computiilg the moment. e =
(N+k)t,Fb=R
k = 27.8 mm (N N = 135.2 mm e = - =135.2 67.6 mm 2
Re' = 275(0.0415)
ection with initial bolt tension assumed.
-
[Eq.(4-5)] tw = 9.0
+ 27.8)(0.009)(187.5)
= 275 (.OITI~!II:
1:
.I[, or bolt> t: = --
f
col:~p!l[e I L
('he.h I n r < r l ~ r l o ne ~ l u ~ [ i o n
e ' = e + 12 - kangle=67.6 =
--
=
+ 12 - 38.1 =41.5mm
11.412 >> 2.947 kN . m as used for top angle
STRUCTURAL STEEL DESIGN
BOLTED .&\TI RIVETED COhXECnONS
Example 8-14 The bracket connection shown in Fig. E8-14a uses a piece o WT and two pieces of angle to make a stiffened beam seat. The fasteners are A-325 high-strength bolts and A-36 steel. Is the connection adequate for the shear and moment to be resisted? Use the AISC specifications.
too two bolts:
Fi
8 2 5k +I
&
=
( "f) i
17.5 1 - -
=
17.5 1 -
= 11.08
-
);I
<
13.7 ksi
N.G.
It will be necessary to redesign the connection using either larger bolts or more bolts. ///
1.?0"+
* . .
1.5"
-
-10 LOAD RESIS'I'ANCF: FAC'I'OK DESI(;N (1,KFD) FOR ONNECTIONS
V)
%' Q'
1.5"
Figure ES-14a
SOLUTION Assume a friction-type joint. For 7/8-in-diameter bolts, T, = 39 kips/bolt (Table 8-2).
-
--
A
'
=
2-32 ksi
15(11.20) 82.5
= 13.72 ksi
= lO(0.601)
11.20(15)~ = 420.0 in3 6
S = -bh2 = 6
M = 82.5(1.47
+ 3.95) = 447.15 in . kips
f , = f =-=--
S
447'15 420
- 1.06 ksi
The resulting stress diagram is shown in Fig. E8-146. The hatched part pf the M / S diagram is tension due to moment, which must be carried by the
a current form of the LRFD equation for connection design using A-325 and -490 high-strength bolts (and with size limited to diameter 2 1.5 in) is R, = 1.1(1.1D + 1.4L) is value is compared to the fastener resistance or plate-to-bolt bearing as Bearing-type connection: R, = 0.625A,FU Friction-type connection: R, = 0.7rny A,Fu Combined shear and tension: 2
o
=
0.70
+ = 1.00
(pdhear))+ (0.6~u[iens,uni)' +(0.6"b~")' Plate bearing on bolt: Rn=3tdF,,,, ,,,, +=0.&
here Ab= A,= Fu= m= p=
6
=
0.75
nominal area of bolt bolt tension area (in thread zone) [see Eq. (5-4)] ultimate tensile strength (nominal) of bolt number of slip surfaces ( = 1 for single shear lap joints) coefficient of friction (commonly 0.35 for clean mill scale)
Example 8-15 Given the connection shown in Fig. E8-15, determine if the plate thickness shown is adequate and find the number of 20-mm A-325 bolts using a friction-type connection, and a bearing-type connection.
Figure B 1 S
406
STRUCTURAL STEEL DESIGN
BOLTED AND RIVETED COhTcZCTIOSS I1
SOLUTION
e connection is "safe" with either two or four bolts. Choose th ased on the final decision as to connection type, "friction" or '
+ 1.4L) = 1.1[1.1(45) + 1.4(75)] = 170kN
R, = I.l(l.10
I
Checking the plate dimensions.
A,+F,
= 170 (see Sec. 5- 10) shown in Fig. P8-1 wing 3/4-in bolts. Set the gage so that the critical net section is a minimum. Also determine the ess of cover plates to the nearest multiple of 1/16 in. Use the AISC spxifications an eel. Show a neat final sketch with all critical items required. Use a bearing connection. Answer: P = 159 kips.
does not The A, furnished is Find the number of bolts required in a friction-type connection (n tolerated, but note that after plates slip, !the bolts must still shear for connection failure). Assume that y = 0.35 (clean mill scale on fay' surface). A, = 0.7854(0 -
9743)' (fps) n
P
Figure P 8-1
m i n e the number of 22-mm-diameter A 4 9 0 bolts needed to develop the full effscti3-eT of the tension connection shown using a pair of angles and the usset plare shown in Fig a bolt spacing of at least 3D. Make a near sketch of the finj$ainr design rhoviag the ired, length, and any other critical design information. Usi:the AISC specifications, = 345 MPa steel, and a friction connection.
[Eq. (5-4)]
Assume 10 threads/25.4 mm for the 20-mm-diameter bolt. +R, = 0.7mp4,Fu = 170 kN m = 1 for single shear 0.70 x 1 x 0.35 X 0.24122 x 725 x N = 170 kN N = 3.97 use four bolts Flnd the number of bolts required in a bearing-type connection (slip tolerated and joint failure by bolt shear):
use two bolts Check the plate bearing using two bolts: R, = 3tdF,
qRn I
I
I
4
= 0.64
F, = 400 MPa
= 0.64(3)(0-012 x 20)(400)(2) = 368.6 kN
\ L as rzq'd.
+R, = +(0.625AbF,) N = 170 kN
>> 170 kN
.
O.K.
-
Figure P8-2
Redo Rob. 8-2 using F, = 250 MPa and the AASHTO spzcificarions for the full pacity of the angles. Use A-325 bolts and a friction connection. Answer: 18 bolts, L = 610 rnm just under L for no reduction. DO rob. 8-2 for the fps equivalent of the pair of angles and a 3/5-in _wset plate for an axid oad of 240 kips, A-36 steel, and A-325 bolts in a bearing connection. Answer: Eight 7/8-in-diameter bolts.
5 Given the beam splice shown in Fig. P8-5, use the AASHTO specifications to (a) W i m b o l ~ d w v e r plates for the full moment capacity of the beam. @) D e s i p b l o : and web plzta for the
&
m
3Ol.TED .\.\TIRIVETED COh3ECTIOh.S Jpjg
iull web shear capacity of the beam. Use 7/8-in-diameter A-325 bolts for all splice parts, and A-3 steel. Parr~alanswer. M = 1950 In . laps, T = 107 laps. I
are adequate for tearing but that the bearing should be checked as appropriate. .UI boIt hes angles are on standard gage distances, as in Table 1- 13 of SSDD.
ikfj+~f*\ Figure P8-5
8-6 Design the eccentrically loaded bracket connection for the load shown in Fig. P8 25-mm-diameter A-325 bolts in a friction connection. Determine the plate thickness for both and tear along the forward row of bolts.
P = 225
kN 9 Redo Example 8-14 for twelve 7/8-in-diameter ,A-325 bolts.
1 Redesign the connection of Example 8-8 if the floor beam on the left docs not frame into 8 x 50 section.
the
Answer: Yes.
8-7 Determine the number and placlng of 22-mm-d~ameterA-325 bolts for a cable connection to a W360 x 314 column as shown In Fig. P8-7. Assume any needed T d ~ m e n s ~ oton produce L that is '~va~lable. Check the tenslon in the stem of the T so that a large enough sectlon is used. Assume that the hole for the cable attachment ulll be reinforced so that capacity 1s not limted at that point. Use and F, = 250 MPa for the T section. the AISC specif~cat~ons Answer. 10 bolts.
Figure P8-7
Answer: P,
=
217 kips, N
=
six f -in bolts.
7 Do Prob. 8-2 using LRFD. 18 Do Prob. 8-5 using LRFD. Assume that the 225-k?J load is mads up of D = 125 kN. Check only for number of bolts. Answer: 16 bolts with I percent overstress.
=
ICO kN acd
approach sawed cuts in smoothness. As stated earlier, most welding uses an electric current. The current is used o heat a n electrode to a liquid state, which is then deposited as a filler along b e nterface of the two or more pieces of metal being joined. The process simuItaeously melts a portion of the base metal (metal being joined) at the interface so hat the electrode intermixes with the base metal and develops continuity of 411
412
STRUCTURAL STEEL DESIGN
material at the joint when cooling takes place. If the qua electrode 'is small relative to the thickness of the joined parts, the proce to be unreliable (i.e., insufficient melting of the base metal occurs so weld may pop off:or not fully join). This event can be preheating the base metal or limiting the minimum size of the weld. welding operation takes place in a very cold environment, it may be ne preheat the parts, particularly where the parts are thick so temperature differential does not develop in a short dista resulting thermal stresses are so high that the weld zone fails. Electric welding involves passing either dc or ac current through an :rode. By holding the electrode 'a very short distance fro is connected to one side of the circuit, an arc forms as the "shorted." With this "shorting" of thencircuit, a very large current flow t place which melts the electrode tip (at the arc) and the base metal in the vici of the arc. The electron flow making the circuit "carries" metal to the base metal to build up the joint. Careful con and current are necessary to produce a .quality weld wi define an adequate melt zone and while keeping electrode s mum. The electrode may be either the anode (+) side of the circ ( - ) side. Most commonly, the electrode is the anode and t is conducted using "reversed polarity." When the weld electrode is the cath (-), the circuit uses straight polarity. Most welding is done using dc current ac is used as the power supply, it is first transformed to dc. Of the numerous welding processes available, the foll to be used for structural applications: 1. Shielded metal arc welding (SMA W). This is the most common welding method using stick electrodes. The electrodes are available in leng!hs of 9 to 18 in and are coated with a material that produces an inert gas and slag when the welding current melts the metal. This gas surrounds the weld zone to prevent ,jxidation (see Fig. 9-la), which is a critical factor if more than one weld pass is Iiecessary to build the weld to the required size. The slag, being lighter than the metal, floats to the top of the weld and can be brushed away. On subsequent passes it is necessary to brush the earlier passes to remove any slag, dirt, or other foreign material whose presence might cause a flaw in t welding is the most commonly used field method using The maximum size of weld produced in one pass is about 5 / 16 in or 8 2. Gas-shielded metal arc welding (GMA W). This method of.weldi often used in shop welding, where uncoated electrodes ar welding unit. The unit controls electrode spacing and we inert gas source to shield the weld against the surrounding atmosphere. 3. Submerged arc welding (SAW). This method of welding is also used in fabrication shops. The joint is aligned and covered with a blanket of granular fusible material containing alloying and fluxing agents as well as inert gas, producers. The electrode is inserted into the granular material, the arc produced, and the melting of electrode and base metal takes place. The heat fuses the
Electrode covering Weld f ~ l l e rrneral Shielding atmosphere
-
Molten weld pool
Ih
)
s:Fcctj develop the gas shield and to obtain any n e slag is later brushed away to expose the g.This welding process is vew similar to the submerged an electrocondu~ti~e slag which is held in position between etal to be joined by water-cooled retaining plates (see Figrial is melted and current passed throu& it to maintain the and filler metal. The filler is obtained from the ed into the slag. The process is generally done in the filler melts the retaining plates are slowly rnpleted, partly cooled weld. which has a thin 'la=u cover in^ must be brushed away. ~ l welding is~ used to shop-weld ~ thick ~ plates ~together* I~t ha\ been~ er o p u l a r in bridge work to weld girder plates and floor platesthe order of 20 to 450 mm can be welded by this process in one pass-
9-3.1 Allowable Weld Stresses
I
II
The butt joint is the only joint likely to be ill direct tension. The a110 tension or compression stresses for weld metal is given in Table 9allowable stresses in tension or com~ressionfor the weld metnl mnv a into this by further limiting the F, of the base metal to 42 ksi for E60 elec and'to 55 ksi for E70 electrodes in structural grade steel. The allowable shear stress for fillet welds is limited to Fo = Oa3Fu(electrode) in the AISC specifications, but it is always necessary to check that th sufficient base metal to resist the same shear stresses. Tn penern1 t h e ~ P I A
checked against the specif;cations being used. The max~mumshear stress in the weld metal is the limiting v :onnectlons where the weld is subjected to combined shear and
'1-3.2Fillet Welds
I
The fillet weld shown in Fig. 9-4 is approximately triangular in crdss section. Care must be taken that the throat dimension shown in Fig. 9-4c & built ou adequately. In most cases the legs of the weld D are made eoual hiid t h i ~i s nnt *
..-2
,A
m
cj
E E I
computed as
T = D X cos 45" =
I D ~"0.70711
where T= throat dimension D = nominal leg dimension The leg dimension for fillet welds D should be taken to the nearest 1/16 in or 1
Figure 9-4 Critical shear area for fillet welds. (a) Fillet weld for tee joint. (b) Fillet weld for lap joint. !'c) Throat dimension for minimum shear area.
% 8
:32 u3 .L
L
0 .2
0
$*,a%g %
+
m m m
-.xu
alnuLr UKAL. >[EEL DESIGN
9-33 Plug and Slot Welds
is because the quantity of molten weld metal is much smaller and the resultin shrinkage is much less.
>
< ,'
t ~ ;~ r 7
In1
9-3.4 Minimum Weld Size 9-5 AISC
for plug and slot welds.
(0)
Slot
(b)
0
welds.
9-3.5 Maximum Weld Size The maximum size of fillet welds along the edges of connected parts is: 0.707 1 1D(0, F,) 5
here
Part
( 5 1/4 f>1/4
SI, mm'
AISC and AASHTO
t i 6 t>6
Use D = thickness of part r Use D = t - 1/16 in or r - 1 ma
Unless the throat dimension is specifically built out to use the full value of r. ~
.,
,,,
.,.. ,.
..
It is possible in the fabrication shop when using the submerged arc welding process to specifically produce a fillet throat thickness by rounding the wela out so that for b
D 5 3/8 in D > 3/8 in
x 0.7071 1 (in) use throat = D x 0.7071 1 + 0.1 1 (in) use throat = D
If this is contemplated, care must be taken to differentiate between the shop
[,,,I(
Pz F ~ )
p, =0.3, 0.27, etc. from the appropriate code for p,= 0.33, 0.4, etc. from appropriate code for the
""' rod
---a
-.-"<<,A,
1v 5 8" or betweet,
A I S ( ' S c c . I . I 7.4 Flat bar ro gusset plate
Improieil
Weld both
loin[ p l l l l
llc)l
'i17rl11~l
I O rollins .I\!,
Ii.1
at a welded joint. Also shown 1s one solution lo of a lamella tear occufing. ( a ) Lamella tear in base metal. (6) Joint %eomstV
the reduce
A[SC Set. 1. I 7 . 6 Lap joints
a large weld (or welds from both sides) is imposed on a thick piece b2se a lamella tear can occur. The tear can occur because the shrinkage strains om [he welding operation will be large and restrained. T h e restraint may be he far side or from the member rhickness Or a tch of electrode and base metal in a f u l l - ~ e n e ~ ~ ~ butt weld tends to increase the possibility of tearing (i.e.. from using an with A-36 base metal). A thin, stiffened ~f
Figure 96
Stnrcturai
AISC welded connection specifications. sine Code, they should be generally followed for
Other AISC specifications conc building consfru~tionare shown in ~i~ 9-6.
9-4 L M E L L A TEARING 'eaMg tsee Fig. 9-7) is a phenomenon that may occur in certain welded Joi'ts It is not a common condition because it involves several
must be large relative strains in the base metal (lamella not does in the weld metal). These straihs occur where large localked stre:& occur. . 2:: Lohc:; is generally perpendicular to the mill rolling direction that produced the '"lnber being welded. Beams welded (0 flanges produce this type loading in the column flanges p u t not in the beam flange). 3. :nust be strain restraint in the base metal,
to the column flange produce restraint. use of fillet welds, a joint design that allows strain relief or loading g direction, and the order of welding to minimize methods used to avoid lamella tearing- The AISC ly Restrained Welded Connections" (AISC Enineering Journal, vol. 3, 1973) discusses lamella tearing in some detail and gives a number joint alternatives that may be used to reduce lamella
5 ORIENTATION OF WELDS aboratory tests on small to medium-size joints show that butt welds limit joint capacity where the electrode has been "matched" to the base ~h~ orientation of the applied stresses does not have a si~dificanteffect butt joint strength. ~h~ orientation of stresses for fillet welds is a significant factor ultimate joint strength. Tests (see Butler and Kulak. "Strength of Filiet
,422 STRUCWRAL STEEL.DESIGN
wioU, C O ~ E C X ' I O N S
joint shown in Fig- E9-l a wing Pa, and the AlSC specifications-
(?
. ..
jr'
d
Figure E9- 1a
Joint efficiency = 100 percent
9-6 - WELDED CONNECTIONS
p
=
0.015(150)(0.6 F")= 337.5 kN
Use D = 15 - 2 . 0 13 ~ mm. F, = 0.3 x F, = 0.3(415)
=
124.5 MPa
se a 150-mm weld on each side of joint as s h o r n in FigSet. 1-17*7satisfies AISC set. 1-17.4). Use 26-mm end returns Per allowable shear strength of the weld. It be noted that a butt weld used to resist a moment develops stresse Figure E9- 1b
eld and 75 mm along
of f b is compared to the allowable weld shear stress.
Figure 9-8 Welded Imment connections. ( a ) Butt-welded moment come connection.
side'?
a 3/gin ample 9-2 ~~~i~~ the welds for connecting an L4 3i 20. Use the AISC specifications7 'I8 r static loading and dbnamic loading-
A,,@, = 1.8 1 in'
p = 1.81(22) = 39.8 kips
'-
- ----
"CI'U-L.
Ufi,.,,UN
.,,UO,.
Substituting L,
,.
"A,.-.,.
..
0.707 1 1 0(0.3 Fu)5 to( 0 . 4 ~ )
D 5
0.25(0.4 x 36) - 3.6 -0.70711X21 14.84
Use D = 3/16 in.
39.8 (0.707 1 1 X 0.1875)(21) Use the weld shown in Fig. ~ 9 - 2 b . =
Figure E9-26
I
Figure E9-2c
For loading it is necessary to balance the weld neutral axis of the angle (AISC, s Referring to Fig. E 9 - 2 ~and placing the weld across reduce the joint length, we have L1
+ L2 + 4 = 14.5 Ll + L2 = 10.5 L l = 10.5 - L,
Take the sum of moments about t eliminated; also, Pw = 0.707 1 1 X 0.1875
x
21 = 2.78 kips/in
L2(2.78)(4 - 1.16) + 4(2.78 Canceling Pw, we obtain
2.84L2 - 1.16Ll = 3.36 Substituting L l = 10.5 - L2, we obtain 2.84L2 - 1.16(10.5 - L2) = 3-36 L2 = 3.89 in
heck: 6.62
=
10.5 - L , into Eq. ( a ) yields
+ 3.89 + 4 = 14.5 in.
.1 Rigid Beam-Column Connections
ewlin, "Column Web Strength in Beam-to-Column Connections," 0 ves tructural Division, ASCE, ST9, September 1973) 3 4100t;fi PC, =
J ,
"c
we equate PC, to the beam flange compressive force (Pbf = A,&) and Iook he dimension d,, we obtain the current AISC equation [Eq. (I.15-2)l: dc
<
4
100r2\/1;
(fps)
Pbf 7
dc
10.73t:\jF,
<
(SO
Pb/
(9- 1m)
ere Pb,= beam flange force (compressive) x F F= 5/3 for dead and live and 4/3 for dead + live + wind, kips or kN t,= column web thickn~shan 0, d, = required colum~'.w,&'#h'i~$?@~s as d - 2k, in or mm yield stresk of kglur~fiSee 9: sl. or MPa
c=
.
* ,Q
tTpTi:r
ner is required oppofite the"co+ression flange if the actual column reviously defined is greater than that given by the right side of Eq. (9-1). Stiffeners (for the column web) are required opposite the beam ge (see Figs. 9-lob and 9- 15a) as follows:
be concerned with determination of whether these column web and fl stiffeners are required.
as
this inequality is satisfied as shown (right side equal or larger), no c o l u m is required. A more convenient method to determine the stiffener ents is to equate p,, + pm = b'f ith P,, = A,,F ,, we obtain 'A,, =
The earlier AISC specifications rounded 183 to 180. Th substitution of values
;+
,"here the additional term (~,/36)"' is used to adjust for other grades of steel. A forther adjustment to the factor 33 400 to incorporate test results (see Chen
Pb, - ~,,t,(tb/ + 5kc)
(9-2)
F,,,
AISC equation for column stiffeners opposite the beam that only a positive stiffener area is valid. The column flange must be of sufficient thickness to resist the beam flange t excessive deformation. A yield line analysis his given the
, I
#
"
i-'
~
~ U C I T J R N . STEEI; DESIGN
,
wELUE;
.
'.a""r,'.',glq: ' . .
.W
j ~. ;. , .: ..,'*" .
CONNECI~&$~ ( :.<
. . .
. '
I' the w1umn flange thickness tcf is less than that given on the right sides of eqyations above, flange stiffeners are required. neMSC sp&fications req column web or flange stiffeners meet the following criteria: I
"
, . S f . - Eq. (9-1) if this equation is applicable. 2. Width of both siiffeners + t, 2 0.67%. >'/+A, . -' Stiffener thickness 1, 2 $3/2 (also, the b/t ratio must be satisfied). pzr beams on one side of the column, the stiffeners may extend onlyone-h ::i the column depth. ." &'hew e l d j o c n g the stiffeners to the column web must be sked to carry i-~balancedmoments on each side of the celumn. 6 siiffeners for tension requirements must be welded to the column fla
*a,,)!
"
dfficient to carry A,tF,, (i.e., use full penetration butt welds). :ompression stzfeners must be welded or accurately fitted to the opposite the beam flange delivering the compression load.
,
compute the plate length. Note that the plate will have to be long enough to allow placing 5 in of weld at D = 0.5 in. I t will also have the some length between the end of the beam welds and the butt column in order to develop adequate strain. BY proportion:
% m ~ l e9-3 Design the moment connection shown in Fig. ~ 9 - j usi u
the AISC specifications, and A-36 steel.
e a full-penetration butt weld using the -3c), use a plate 12 in deep on one side. .,
,
, , ..
.,
1 :
F
. ,.. ...
b E9-3a
b o ~ u m oDesigning ~ for full moment capacity: M = FbS, = 24 I
X
64.7 = 1552.8 in . laps
T = C=-=
'd
-=
16.01
" ..
',Se a fop plate that is 8 in wide at the column weld and tapered to 3 in, zhown in Fig. E9-36.
'P
Figure E9-3c
'
97
= --- = 0.73 22 X 6
26.4
use 3/ 16-in plate = 0.1 j 12x0.4FY 26.4 use3/16in Weld D = 0.70711 X 12 X 21 =0.14gin check if the column needs reinforcing (AISC Set. '-I5.'): t = p
use 3/4in plate
9 -,TakeD for fillet welds at 1/2 in. P,
= 0.7071 1(0.5)(21) = 7.42 kips/in
& = - =97
7.42
13 in and distribute as shown
97( 5/3) - 36(0.295)[0.505 + 5 ( 1.2j)] - 161.7 - 71-74 = 2.49 b 36
2
Therefore, a pair of stiffener plates are required opposite the tension flangeu s e plates 3 in wide (2 x 3 > 0.67 x 8 o.K.).
'
t430 STRUCTURAL4 STEEL DESIGN
Neglect ttus zone
Check the compression flange:
1 <,r. line = L ; 3 t ~ b o u bast) t / o r line = [.' I l iccntrord)
9-11Eccentrically loaded beam framing an!&s
also Fig. 9-IZd).
Figure E93d
to carv97 kips and across the ends to the flange adjacent The stiffener may be only one-half of the column depth, since the eccentricit4-
Use a f -in
1
e,
(16.91 X 2)(0.70711 x 21) on both top and bottom of each stiffener plate.
flange for der web), the 3-in leg should be welded to the beam and bolts in the 44x3 Q-sJthat there will be adequate erecti~nclearance and edge distanceangle is welded to the beam web and field-bolted to the colu-
% = (R:oment + R : ~ ~ ~ ~ ) ~ ' ~ where :.:\
a:
,
4
.
8
...
.i
= lorceor stress/unit df length (or area). The value of 4 is lifited
-< Rallowable
to
in ~~~~~l~ 9-4 Design the web ande connections for a W18 x 55 as framed beam-to-column flange connection cqi ~ i ~ g~- 4 .afor a a 55-kip end reaction. u s e two ~3 x 3 x t in x 12 in long. Use the specifications, E70 electrodes, and A-36 steel.
o ~ u n o NFor the beam web weld, locate neutral aGs as (12
+2
2.5 x 2.5)s = 2 C 2 . 5 ) ~
-
.X = 0.367 in
4-
- L K U C T U WSTEEL DESIGN
4
Figure E9-40
'~nsideronly one side, so
R = - - = -55=
27.5 kips 2 2 /' /~ / 3 ) : : 'le polar moment of inertia is (I of line about base = ' ~
3.11 ' a = - - 14.4
7 he weld shear resistance
.2 Welded Beam Seat Angles 27.5
, r , - Y = 1.617 kips/in ioint 1 of the weld is critical by inspection (or drawing rays from c.g.) M = Re = 27.5(3.0 - 0.367) = 72.4 in . kips
4. = ((0.465 +
5&&,
:ompu t.ing the odulus S = b
-
ere b = seat angle width. The value of r is obtained by trial. We may initialIy timate the eccentricity (assuming the k distance for any angle is t + 0.375 in t 9 mm) to obtain
+
N
e.1n111al .. =2 b
R,=-=-= R L 27'5 12
2.29 kips/in
N
+ 0.5 - ( t + 0.375)
einlt~al . . . = - 2+ 1 2 - ( t + 9 )
in rnrn
(and neglecting end returns of 2 0 ) - .
3.38 D = 0.70711 x 21 = 0.227 in
la1 bending str the required
use 3/ 16-in weld
2(3/16 X 0.7071 1)(21) 5 0.39(0.4 X 36) For the column flange weld (assume 1/4 5 r, 5 3/4: why??) 54Pe, Rh = (from Fig. 9-1 1, refer also to Fig E9-46) 25 L~
I
f
,
+
2.46 0.7071 1 X 0.3 X 70 = 0.165 in Check the web shear capacity:
O.K.
e beam seat angle shown in Fig. 9-12 must be designed for bending stability must be of sufficient thickness and leg length that an adequate fillet weld be placed along the vertical legs to carry the shear and.moment due to the ccentricity of the reaction. The angle is checked for bsnding at the fillet runout k distance from tables) as in Fig. 9-12a. The allowable bending stress is taken
1.617)~ 1.308~)"~ = (6.b45)'I2 = 2.46 kips/in
D=
=D
C
+*
,
0.257 in
Use L3 x 3 x 5/16 x 12 in long.
& = -R-
,
-
I 23 + 2(2.5)(6)2 + 12(0.367)' + 2(0.367~+ 2.133~) LP = Zx + 1 , =12 3 = 144 + 180 + 1.62 + 6.49 = 332.1 in4/width of weld
use 1/4 in weld
actual required angle thickness t is very sensitive to the k value, so the ative angle should always be accurately checked.
Example 9-5 Design a beam seat angle and weld for the conditions shown in Fig. E9-5a. Use F, = 250 MPa and E70 electrodes. Data for a W4lO X
STRUCTURAL STEEL
WELD W COWECTI
Use N = 120 mm: -N= - -120 - 6 0 m m 2 2 of the WT at least as large as f+ the
Use the Also,
- < - = 335 21.2' t - m an inspection of Table V- 18 (S b, = 395.4 mm
r/
Figure E 9 d b
Shear Stress due to load:
= 30.2 mm
M
Find the maximum weld size D for s on each side. loot, = = 9.19 205.8 Check the bending stress in the WT
D
fb=--
6M twd2
-.
.,,.
1
,..
-
6(15.16) = 132.9 MPa 18.9(0.19022)
870
0.842
> 8 as minimum for r/
-= 0.3 x 485 = 145.5 < 150
...~.,. y=--= 50 040
15.16
L=~=o.lsoo,=~,
57.5 mm
0.~.
Use D = 9 mm.
/li
WELDED COLUMN BASE PLATES
= 2.4733 D, x 10-6 m4
ob site. Several situations using CO~UIIIII end plates are sho"n in Fig* '-I4. base plate may be either butt- or fillet-welded to the column- The decision is
quires additional fabrication. The general base plate d e s i 9 for dimensions width x length x thickness) is as outlined in Set. 6-6.
-
r
W
STRUCTURAL STEEL DESIGN
Beam: W610 X 241.1 d = 635 m m , b, = 329 m m l / = 3 1 mm,r,= 17.9mm
Since d,, furnished = 538.2 > 332.2 required, a web stiffener is rtq&& ~hi'stiffener(a pair with one on -each side of the web op&site the column compression flange) only has to be one-half of h a m depth, since the load is only on one side. 33.3
tc/
Is,,"
> T = -7(0.67
Column: W360 X 261.9
SOLUTION Step I. Design vertical stiffeners 1 and 2. For stiffener I, opposite the tension flange of the column: 610 : f b ' - - M --- 132.6 MPa S, 4.60 AfC = 6,$ = 0.398 X 33.3 = 13.2534 x 10-
-
(
G c tcw tbf
17 m m
=
329) 2
use tJt = 20 m m
+ 17.9 = 1 19.2 rnrn
use b,,
=
120 mm
Step 2. Check i f a d~agonalst~ffener1s required. Use plastic design [Eq. (9-7)]: M, = 1.7 x IM = 1.7 x 610 = 1037 kN . m
Since the furnished t,, diagonal stiffener. 8
Pw = ; f b ~ = / j(132.6 X 13.2534) = 2929 kN 'bf
X
bst,ff =
d Z 3 8 7 mm, b f = 3 9 8 mm
As, 2
I
WELDED C
=
17.9 < 31.6 mm required, it is necessary to use a
635 -- tan-' = 58.64' 387
cos 8
=
0.52042
= k,
F,,
Use the beam for the "column" dimension and 2.5 instead of 5k, sin column tension flange is at the end and not centered on the 5k zone 2929 - 250 X 0.0179(33.3 250 = 10.146 X m2
As, =
- 2.5 x 48.4)
=
6.2615 x l o p 3m2
Use two plates 20 mm x 160 mm wide:
Use end plate:
A,
=
6.4 x low3m 9 . K .
b, = 398 mm (width same as column) Step 3. Design the welds. The weld for plate 1:
t, = 28mm A , = 0.028 X 398 = 1 1.14 x lo-' m2
0.K. For stiffener 2, opposite the compression flange of the column: db(,+,.s, = d - 2k = 635 - 2(48.4) = 538.2 mm A stiffener is required if bb > dbw:
, , ,..
.,.,..,...
dbw
> >
F,
=
0.3 Fu x 1.7 = 247 MPa (plastic design)
Using inside fillet welds (see Fig. E9-9b), we obtain D,, = 8 mm, since = 25 > 20 mrn. Check the effective D of the beam web for shear:
t,
10,73t & f i pbf
i
PbJ = 2929 kN
.
.
10.73(17.9)~a = 332.2 mm 2929
So only 7.4 mm of the weld on the beam web is effective, since the shszr of the base metal controls. Use D along the inside flange as required and an
WELDED C O h W C
Ssuming that 600 mm is effective, check P: pWe, = 2(600)(0.0074)(0.70711)(247) = 1550.9 kN >> 600
Use flange D = 14 mm. esign using LRFD is similar to load resistance factor- D ctions. ~t is necessary to compute the.factored load R, = l.l(l.10 + 1.4L)
R, = 00.6 FEmA,
+= 0.80 [currently (197811 I\'(J/~.
All plates but I ore 20 mm thick. ~ l fillet l are D = 8 lnln except b e a m - t o - ~ o e]n~d ~plate, ~
Figure E9-96
The web and
=
for plate 2: with maximum weld D = 7.40 nM in required for 20-mm stiffener plates and four lengths
= 250(0.0179)[33.3 Pweld
+ 2(20 + 48.4)]
= 761.2 kN
"- P b f - PWeb = 2929 - 761.2 = 2167.8 kN
4L x 0.0074 x 0.707 11 X 247 = 2 167.8 kN
Weld for diagonal stiffener:
,
allowable for the base metal. r: p = 8 1 kips, L, = 12 in, using D = 9 / 16.
,
.
i
. WELDED C O W E
9 ,
,,
W410
..
84.8 beams frame into a W530 x 123.5 girder as in Fig. P9-19 in trcdes, Fy = 250 and
-
9
'sign the welds and gusset plate for a pair of L152 x 102 x 19 for ctive angle capacity in static tension. Use F, = 250 MPa, Ej'0 electr pi.,. . : , : a t i ~ t ~ Keep . the joint length to a minimum, 9. - 30Prob. 9-9 for a dynamic load. . , 9. Design a welded framed simple bea+ connection to ca 22 ).: 50 beam to a W12 x 53 column. Use ai.2 Li X 3 X t angles with a length of 12 in. dnnuer:tP5/16in;Dwcb=3/16;Dcol=1/4. , 9-.: : Design the framing angles and weld for 1:
--
hear due to uniform load is steel, A-325 bolts, and E70 A
9
,
9":
'
YO specifications. .~.moer:1, = 5/16; D = 1/4 in. Design the welded end connec ,
9 - f . ~Design a framed beam c o ~ e c t i o nto carry a beam shear of 152 k~ from a U s e a pair of angles 76 X 76 x t with L and E70 electrodes. 9-15 Design the weld and plate for Answer: D = 3/16 in; r = 1/2 in. 20
Figure P9-15
(.
simple :Pan carrying a uniform load of 1 ;. design sketch. Use E70 electr fram. ::-;a W12 x 53 column. 9-17 J-ai@ a stiffened b e ~ mseat using a WT fo
. neat
':
.: .
,
,
Built-up rolled sections may be used where the overall depth is limited in
section is to use two lighter rolled shapes in parallel. Even where the to& weight
r the built-up section.
g 10-10 is by far the most common use of girders in building coqtmction. Plate girders are commonly fabricated by welding two flmge plates to a we1 e as illustrated in line details in Fig. 102. The most common materids i r l
. PLATE GIRDERS
4g ,,
."+ i
,,
..,
( C)
epth D and clear distance hatio h,rb is a significant parameter in plate girder design. (a) Welded plate girder- ( b ) in drsim).,( Girder elevation (side view) illusirating other plate
the reduction7 or and larger clear web
rn.ical for spans UP 15 f t ( 5 m) Or morePan (D/Q ratio, Ge of to Larger
A
A-
n exceeds 50 to 00 ft for spans up to 300 00 m) or more. Roued beams are generally more economical for bridge less than 50 to 60 ft and are used in a deck stringer confisrationough girders are generally more economical than trusses, the latter are sm in many situations for esthetics, particularly where additiond lanes ired and the existing facility is a truss-
I
460
I
STRUCTURAL STEEL DESIGN
However, the current AREA specifications do not rather, the computed deflection of the girdei under including impact, is limited to A/span 5 1/640. ;Girders are fabricated' in segments limited by the ere4tion/transport equipment and rolling mill capaci plate girders are almost exclusively used in American the reduced fabrication costs and the fact that the we1 the web can be almost fully automated. A-36 steel is most commonly used for plate gird for webs and stiffeners. .Continuous gird sections in the positive moment region and d moments regions over reactions. An increa esthetically pleasing; however, fabrication costs are increased, since web must be cut and additional stiffeners are usually required,for the $1 rati9:prod~ced.~ An alterriative to an incre strength steels in the zones of increased moment. for the flange either throughout the span or in produces a hybrid plate girder. An inspection of Figs. 10-1 to 10-3 sel-vice readily indicates that there is n problem. Within the constraints of D / L possibilities exist for cross-sectional area. If we define optimization o f a pl girder design as that producing the least overal ilsing a computer program, to "optimize" the gir necessarily (and probably will not) mean least ov considerations. What one should attempt to obt and fabrication cost to produce as economical ( effort, which is also a cost factor) a solution as possible. 20-2 LOADS plate girder loads are obtained similarly to those for be building design the loads are often moving, as for crane industrial buildings or warehouses. AISC specif factor for these types of loadings. Plate girder design for highway bridges involves a girdqrs are to be used per lane, using the AAS amount o f truck or lane loading to the s influence-line type analysis using either stan lpading to find 'the maximum moment and specifications allow use of only one truck per lane for simple spans, in general, although as shown in Fig. 10-5, more than one truck may be on: the bridge at the critical location at a given instant. When more than one truck is on a lane, the \
,
s [ ~ i r ~ Jt . lr (t Jl c o ri I J I I C
I o ~ I J\$kith I~~
I O J ~or
L~~~ = .![her
j r ~ n d ~ truck r d
because of truck length factors and positioning of wheels. When the 'pan is a line long, the lane loading with single concentrated load tvine to find lhc enough and is more convenient to use than of trucks number of sucks and their positioning for maximum stress
computer program. me ln all girder design the standard Iane/tmck loads are incrcased ,impact factor, which is a function of span length. Figure 10-5 illustrates the placing of the standard truck loading On two-lane bridge to obtain the contribution of truck loads to either of ths lin2' girders. The designer must make a similar type of 3 n n l ~ for ~i~ using more than two girders. Example 10-1 what are the design moment and shear values for the *d'r! of a highway bridge as shown in Fig. 10-5? HS 20 loading, s = 19 it, w = 28 ft, span = 110 f t - D a d load due " deck, sidewalk, and so on = 3.1 kips/ft. S o L u n o ~nelane factor L, is obtained by 9"1 about the left
X+ Y Lf = S
where X = S / 2 - 5 = 19/2 - 5 = 4.5 f t y= x + ~ / =2 4.5 + 2812 = 18.5 f t
aderFig
,
rWrL STRUCTURAL STEEL DESIGN
PLATE G
Substituting values, we obtain
ue to the moving train either at the centerline or very close to it. made based on a combination of use of d M / d x = 0 for short sp& I where only three or four wheels are simultaneously on the span and an influ ine with apex at the centerline for a larger number of wheels on the span. given in Table 1-2 are the end reactions (shear) and the maximum reactio (shear) developed in a floor beam into which two stringers of lengths s h o \ v ~i ~ the first column frame. [For example, for a floor beam betwee stringers, the maximum reaction is obtained when one 10-kip wheel load is on the floor beam and the other two wheels are 5 it away, giving R = 10 i 2(2/7)10 = 15.7 kips; similarly, for I! = 10 ft, thc reaction R = 10 + 2(5/10)IO = 20.0 kips; etc.] The values in the table are shown for Cooper's E-80 loading; any other loading X is X/80 x table value.
The impact factor is computed (see Sec. 1-10) as 50
I =
L The moments are
+ 125
-
50 110
+ 125 = 0.21 < 0.30
O.K.
Use the equations given in Sec. 1-9 for L = 110 < 145.6 ft.
10-3 FLANGES AND WEBS OF GIRDERS - - - PROPORTIONING - BUILT-UP SECTIONS
Substituting values yields ,
-
. -..
40 110 The shear values are
= ~ ( 5 9 1 6 . 8- 1232.0) = 1703.6 ft
The flange cover plate of a built-up section usmg a rolled section as the bas may be proportioned as follows (refer to Fig. 10-6). Assurmng that the maxim - bending stress fb is the allowable value of Fb,the average flange value is abo 0.95Fb. At the junction of the cover plate and the rolled section flange the stres s about 0.90Fb. Therefore, let M, = = O.90Fb.Sr
kips
-
Vd = W L = 3.1(110) = 170.5 kips 2 2 'i'he live-load shear (one 32 k p wheel on reaction) at reaction is
M , = afAC = (d + t;)(t;bj x 0.95Fb) The required section capacity is lvftou, = lbf, 4- lbf2
We note that Md is at the centerline of the span, whereas ML is 2.33 ft to the left of the centerline (truck moving left to right). The difference is so small that we will simply add the two values as if they both occur at the centerline to obtain a slightly conservative design M :
M,,,,,
Substitution of Eqs. (a) and ( b ) into (c) and taking (d + C) = (d + q) definin'g the flange area A , = bit;, and taking the ratio 0.90/0.95 as approximately 1, we
+ impact + Md = 1703.6(1.21)(1 + 0.21) + 4689 = 7183 ft . kips = ML(Lf)
The design shear is
+ Vd = 65.9(1.21)(1.21) + 170.5 = 267 kips
Vdeslgn = VL(Lf)+ impact
///
Railroad girders are more difficult to design, since there are 18 wheels in the Cooper E-series standard loading. Design aids are available, such as those shown in Table 1-2, which gives the approximate (to exact) maximum moments
(c)
Figure 1 0 4 Bending stress distribution on a built-up beam cross sccrlon
PLATE GIRDERS
-
Mtota~ - sx ( d + C)0.95Fb d + C .. 17 5" )u!d be evident that Eq. (10-1) does not give a uniqr.. r b ovluLlun a n a the final f!$P.:;- cover plates must be investigated using f, = M ,/ , q. A reasonabl m e$\iliit' of C = 1 in or 25 -. Af
ZX
--'--a:-
,
' , + ~ m p l e10-2 Design the cover plates for a W92O x 34" ,Ii ,-250 MPa for a moment of 3630 kN . m L~ U I irle conditions shown in : EIO-20. Use the AISC specifications.
465
We could use flange cover plates 25 X 548 mm; however, let us arbitrarily use a plate width 52 mm larger than the bf of the beam, which allows a 26-mm overhang on each side for welding the plate to the flange. This gives b; = 418 + 52 = 470 mm
t;=-=13'82 0.47
29.4 rnm
use 30 mm
^
&'-
Now check if cover plates 30 x 470 mm are adequate: I, = I,
r:;t
5..
Data for a W920 x 342.3:
c'JTION
wt
=
3.36 kN/in
d = 912 rnm
12 499.4 x The section modulus is =
M L - . . .wL2 -= 8 Md "
72.6(20)' 8
.,
f b = 2 5 .3855 719=
Mdes~gn
=3800 kN . m
+
3800 0.025)(0.95
x 0.6F,)
- "'I. =
rrFh,
7
20
- 1
IS
ill
---I
<
150
O.K.
13.72 0.912 + 0.025
"
by- 418
t;
1
,'
726 KC 1
'
-
w- 1)J
149.9 MPa
?'he cover plates will be welded as shown in Fig. E10-2b. This requires checking by/$ for acompliance with AISC Sec. 1-9.2.2 (stiffened edge element) :
72.6 kN11n
i\
- 1
ruqina Fn i ~ 1n1 1 :. ------a -y, (1
(0.912
.m
= 17OkN.b
.... ...
A, =
i-
= 3630 kN
3.36(20)~ 8
Te~t~-:l"ely,the area of each cover plate , ,..
m4
The additional moment due to the flange cover plates is proportional to the flange area:
,'noments are:
is?
+2~d'
30
-
13.93 << 42.4
It is not necessary to check b/29 of beam since plates 30 x 470 mm.
Figure E10-26
O.K.
9 = 32.0
mm. Use cover
///
466
STRUCTURAL STEEL DESIGN
10-4 PARTIAL-LENGTH COVER PLATES
acing of bolts or welds can be found by equatlng the weld or bolt capacity and the spacing s as resistance capaclty S =
us
The weld capacity at the theoretical cover plate cutoff 1s obtalned by e the moment of the total weld capacity to the moment at the theoretic lance.
I 1
v$ =
vQ I
(kips/in or kN/m)
where us = force/length of developed shear Q = statical moment of cover plate with respect to neutral axis I = moment of inertia of section, including cover plates AlSC Sei 1 1 0 4
AASIlTO Set 1 7 12
lLfQ =-
(hips or kY) I where F,= total force to be carried by the weld In the cutoff length a M = bending moment at the theoretical cutoff point Q, I= terms previously defined
F,
I* Example 10-3 What 1s the theoretical cutoff dlstance for the cover-plat beam of Example 10-2? Also, design the weldlng to fasten the ~ l a t e sto the beam using E60 electrodes
S~LTJTION Obtain the cutoff points (Fig E10-3a) The capacity of the beam without cover plates is M, = SxF, = 13.72(150) = 2058 kN . m
-b
20
-----------+I
AASHTO rnlnlmurn Lover plate length L,,,, fps 3 d b + 3 ft SI 3 d b + 0 9 1 ~n
Figure 10-7 Code requirements for theoretical cutoff d~stancesfor cover plates. Requireme same for a v e r plates narrower (shown above) or wder than next underlying plate or beam
Figure E10-3a
a*
Since the moment diagram is a ~ a f a b o l awith slope equation for moment IS ,+fx
=
M, - k,~'
=
0 at midspan,
+
" . ddS STRUCTURAL. S'I'EBL DESlG ii
f
,' x = ;i :
:$ 9
'8
?.
4% x 2 M, - L With M, = M, = 2058 kN . m at the distance x from midspan where beam capacity is adequate without cover the theoretical cut is M,
g : .
L/2.wehave =
,, :,,, v :,
(3855 - 2058) [ The theoretical length of the cover plate x =
,:
,
,
= 6.828 m ]'I2 = 2x = 2(6.865) = 13.73 m. 470 rnm; I, = 12 499.4 x
3855 L2
l ' o t ~ lengrh i CP= I i . l i
Design the welds (cover plate 30 x at the cutoff point x' = 10 - 6.83 = 3.17 m. Shear V
Q
=
771 - 77.1(3.17)
=
= AJ = 0.030(0.470)
nl-
1,=:0~,--
r:=;O;iirn
530 k ~
(Y
- + 15
)
/I/ = 6.41
x
m3
10-5 GENERAL PROPOR1[1IONS OF PLATE GIRDERS
us = 530(6'41) = 281.6 kN/m
12499.4
(Noting that I is and Q = low3,we obtain 281.6 k N / h and not the direct calculator reading of 0.28 16.) The allowable stress using E60 electrodes:
Fw = 0.3(415) = 124.5 MPa We will arbitrarily use D = 8 mm for the weld (tf > 20 mm per AISC Sec. 1-17.2) so that the theoretical spacing at the cutoff point to the interior of the span is
follows. The moment camed by the flange is The moment camed by the web is t,h2
-Pwm-- 2(0.008 X 0.7071 1)(124.5) = 1.4085 Tentatively, use a 40-mm length of weld (40 mm = 5D carry
Pi
iVf, =
> 2D),
which will
flSrw= f;7
The total moment is M
=
IMJ
+
.\/,
= 40(1.4085) = 56.34 kN
This corresponds to a distance along the beam of 56.34 X lo3 = 200 mm O.K. 281.6 Use 40-nun lengths of intermittent weld alternating on each side for each 2d0-mm increment of length. The weld in the distance a (a = 1.5b, since D < 0.75t;) is continuous. nd the effective moment Me is S =
'
Me = [0.705(2)
+ 0.418](1.4085)[912 + 2(30)]
= 2503 kN m > 2058 O.K. If Me had been less than the moment at the theoretical cutoff point, it would have been necessary to either increase D or the distance a (or both). de final sketch is as shown in Fig. E10-3b.
Figure 10-8 ~
i section ~ to dobtain~ an approximate ~ expression for the
flan~ area-
hd, ..
472 STRUC.IUK~.L y
i b c ~ L)L!SII,~J
4
Figure 10-10 illustrates the assumed girder web plate loading on a strip dx(h). From Chap. 3 the critical buckling stress was found to be $! .,-,," ~,T~E Fcr = 12(1 - p2)(h/t)2 ..l 3 I
,,&
e AISC rounds this to obtain the limiting h / t ratio as 14 000 h fps - I [~,(~~+16.j)]'/'
\' '
SI:
where the ratiogh/t has been substituted for b / t as given in Eq. (3-5) The force exerted on the web of the web segment shown in Fig. 10-lq$ A f f sin @
r A,fb
+
since (in radians) is a very small angle. The stress f/ must be of suffi magnitude to overcome any residual stress Fr in the web; thus the web strai units of F L - 2 ) at yield is f =er+ey=-
&
I
t and the angle @:is
Fr + E
97 100
h
-5
[F,(F,
The AISC specification allows a somewhat larger h / t ratio if transvers feners are used at a spacing ratio of a / h I 1.5 of
F, .
. I
I
v
+=
d~ &-
=
The maximuq h / t ratios for several grades of steel are as follows:
2 ( F r + F , ) dx
-
fi/2
+ 114)11/*
E
h
At web yield the applied force is .. 2
, ..
.,. .. , .. . .
lor small angles and
A d tan
+
=
AJffl
but
+ in radians, giving Afffl
=
MPa
dx 2F,(F, + F')Eh
h t
<
1.5
259 248
kcT2E
( L ) 2 ( t )dx 12(1 - p2) h
Equating Eqs. ( a ) and (b) and solving for h / t , we obtain -=[
a/h
333 284
This value'should not be larger than the critical web buckling force FCrtdx=
[by Eq. (10-4)]
kCnZE A, 24(1 - p2)
1
F,(< + F,)
I
When A-7 steel was used, it was assumed that the residual stress Fr could be ~dequatelyapproximated as Fr = f , / 2 = 33/2 = 16.5 ksi. This value !is cur.-ently being',used for all values o f . steel F,. It I is! also',assumed 1 .4,/A, = ,I Poisson's ratio for steel p = 0.3, E = 29 000 ksi, and for '2"Yjtain h 13 784 t [ F ~ ( F+, 1 6 . 5 ) ] ' / ~
Experimental studies on full-size girders have shown that the web the compression zone deflects laterally by small amounts at very early bending, with the resulting transfer of stresses from the compressio? web l r to the compression flange. This results in an increase in the flange strebs.'over that amount indicated by theory as shown in Fig. 10-10c. This increme in compression flange stress requires a reduction of the allowable compressive stress so that the stress actually developed does not cause a flange failure. The experimental studies indicated that this flange stress reduction could be expressed in terms of A,, A,, h / t , and Fy. A possible equation in terms of ultimate moment is
1
Since the ratio of M u / M v = Fu/Fv = F,'/F,, we may rewrite this equation ih
474 STRUCTURAL STEEL DESIGN
P U T E GIRDERS
'47
terms of stresses as long as the compression flange is sufficiently stable with respect to L / r , and d / A P This is done by limiting the maximum allowable stress io F, as defined by AISC Eqs. (1.5-6) and/or (1.5-7). The value of 8, depends on the flange restraint effects on the web, and if we assume partial restraint,' ,& may be taken as 5 . 7 v m (see Basler and Thurlimann, "Strength of Plate Girders in Bending," Journal of Structural Division, ASCE, August 19611, where F,, is the critical web buckling stress. If we replace F,. = 1.65Fb,we obtain (with slight rounding):
T l ii I
i bl
Inspection of Eq. (10-5) indicates that if
girder to Pratt truss. ( a ) R a t [ truss ( - ) ate girder with tension fields.
or the SI equivalent, no flange stress reduction is necessary (i.e., F; = Fb).
t the yield shear stress
F, according to
vu =
Vb +
(b)
the van ~Misescriteria is
v"j
(4
e obtain A , and substitution into ( b ) , Vb =
fi (V,)Fr,
5
. g ~ , , , the value of FA is, from Eq. (3-5)-
( e>
here
(3-5) ~ q (10-6), . the critical stress is taken as the mean of q. (10-6) and the value of 0.8Fy,, to obtain (10-7)
v,
The beam shear contribution is
cornpression
FYs = 5
10-6.3 Shear and Stiffener Requirements Where the l i l t ratio is sufficiently small, web buckling will not occur under shear before shear yielding occurs. Actually, a beam web as part of a flexural member subjected to a bending moment is canying the shear in a "tension field" mode. Some researches have likened this to,a Pratt truss, where the stiffeners (if uszd) are the compresbion members and the web segment between a Pair of stiffeners is the tension element, as illustrated in Fig. 10-11a. Actual girder tests carried to yield so that mill scale forms stress lines displays a stress pattern similar to that idealized in Fig. 10-11b. Girder webs are designed by AISC specifications assuming that the shear is, carried by beam action shear-as for rolled shapes-until shear web buckling stresses are reached; then the additional shear is carried by tension field action. The ultimate,shear resistance in a web panel (space between two stiffeners) is the sum of the two shear (beam, Vb, and tension field, V,) contributions, or
=
d where as previously defined (see Table 3-21;
! V; = Fc,(htw) = FcrAw ultimate shear force capacity of a girder for web plastification (plastic capacity) is
., y,
v-
= -ysAw
k = 4.00
+(10-8)
P L i T E GLRl
-t:;d
#
4. In girders designed on the basis of tensional;! action, the spacing of the interior stiffener from the end, bea,ri&;ifi'ffener, or any stiffgners adj large holes, will be based on limiting h e shear stress to AISC Eq. (1. appropriate to obtain: For C, 2 0.8:
The AISC uses
''
10-6.4 Combined Bending and Shear (Interaction Check) When a girder is simultaneously subjected to a large value of shear and bzn simultaneously, the allowable shear stresses may have to be reduced. interaction solution was developed by Basler (see "Strength of Plate Girde Under Combined Bending and Shear," Journal of Stndcrziral Dioision, ASCE October 1961) based on girder tests at Lehigh University. AISC has slight1 rounded the values to obtain 0.375f,
5, 2
0.6Fy
eneral, no interaction check is required if (a) f, < 0.6F" and f, 2 0 . 6 5 (b) f, = F, butf, 5 0.75Fb. en either one of these conditions is not met, the interaction using Eq. (10ust be checked for possibly reducing the allowable flange bending stress F, e of this equation.
-6.5 Intermediate Stiffeners ffener design is. based on obtaining the vertical stiffener = 0 from Fig. 10-12c to obtain P, = f,t,a sin + sin
+
Substitution for f,, $, and A,, = ~',/F,II,, gives
) .
19'A i*";,:
<'.. i
are always required. 2. Intermediate transverse Stiffeners (AISC does not have a specification for longitudinal stiff€mx'S) are not required if bottf~of the following are met:
Intermediate stiffeners are required for any other shear stress condition. When stiffeners are required, the spacing is limited to :J
but a / h I 3.0.
[ I + (o/h)']'/'
YDht,
(10.14)
P
which is AISC Eq. 1.10-3. Where Y = F,,/F,,, to account for ~ossibleusciof a different yield-grade steel for the stiffener'than ior the web, D = factor to account for reduced efficiency for a stiffener plate on one side of the web or
D = 1.0 for stiffener plates on both sides of web plate D = 1.8 for angle used as stiffener on one side of web D = 2.4 for stiffener plate on one side of web plate The required area of stiffener is often very small. To ensure a stiffener ufficiently large and stiff to maintain the girder web shape, the moment of
.JCTURAL STEEL I)i,%k
*&"
P U T E GI1U)EY.S
$4
:. .'*
inei:.~cr ust be at least4 ,$
AlSC a n d A R E A
.A.iStlTO
1
h
6
4, 2 (%)
(10-15)
= in4 or m4. Since the stiffener is a compression
,.:.the minimum
io should be maintained. he stiffener must be fastened to the web to carry some vertical force. ! i'ifferentiation of the equation for P, with respect to a / h gives
{
,
*":I
2
P, = r 0.015~,h~< "LJsiui ?, = F,/ E, a safety factor of 1.67, and assuming that the stiffener force is debe1;qxd in the distance h/3 from the compression flange, we obtain .B;
I
h
6
a
ha&u tips
N
1.67
D o r fillet
h some additi* obtained as
.I
hfr
C o m p u t e L'lr L 1 = O75h r = [11'4~"~
Cornpure L r
=
h r
~ ~ ( f i Check :
as column based o n cross-hatched area shown above Check bearing: use on!y area of plates as .Ah = birr. Check b,/r, ratlos.
Figure 10-13 Bearing stiffener design requirements for speclficatlons ~nlcated.
Use t;
1096.f
=
+
in or m. The allowable bearing stress F,,, = 0.90FY (AISC Sec. 1-5.1.j.1) based on the lesser F, if the flange and stiffener are of different y~eldgrades of steel.
w i n g Stiffeners b
Beanr,, tdfeners are always required in pairs over the reactions. Bearing stiffece-s mgy be required beneath concentrated loads carried by plate girders. These s:,:feriers must extend the full flange to flange distance and have a close bearir~,:against the flange delivering the load. The stiffener width must be such as to extend approximately to the outer edges of the flange or angles. Bear'ng stiffeners are designed as columns with an area that includes the stiff n ~ r sand a central weblarea of 12twfor end and 25tw for interior bearing ? stiffeners (see Fig. 10-13). This area is used for computing the radius of gyration andrfo. checking the coludn stresses. The effective length of the stiffener may be taken as 0.75h because of being securely connected to the web. The effective bearing area A: is taken as the area outside the flange angle 'ffiet oi ihe flange-to-web &~ds. design requires co ing L r / r to find the allowable column stress Fa
e
.
$$@$$:eking
<
PY $4
fif
P = AF, I applied load or reaction Al5a.check using the effective bearing area A, = bits (Fig. 10-13) to obtain 6
P
= AbFbrg I applied load or reaction
:i r
10-6.7 Web Crippling Webs of plate girders are required to be proportioned so that web crippling (same phenomena as for rolled beams) does not occur. This 1s accounted for at reactions by bearing stiffeners. Where the compression flange supports a uniform load or concentrate ds for which bearing stiffeners may not be requirgd2 compressive delivered by the flange to the girder web must be iuff.iciehtly low that c (Fig. 10-96 illustrates buckling) does not occur. This is an cage-loaded plate 'stabihty problem, and agaln Eq. (3-5) IS used (see Basler, '.'New Provisions for Plate Glrder Design,'' Proceedings 1961 AISC National Conference, AISC,, New York), SF = 2 65. E = 19 000. and some rounding, to obtain the allowable web compressive stress as 10 OOOk, fps: F = (hllw12
and for a flange delivering compressive load restrained against rotation.
In the AISC specifications, Eq. (10-17) is directly combined with the appropriate kc term for displaying the design equations. Ir
Example 10-4 Design a welded plate girder to support two columns span ning an auditorium space in a high-rise building. Floor loads deliver cequivalent uniform load to the top flange of 2.8 kins/ft- - (not i.. n.. r. l.i i d i n a t&he ,---" girder weight): General span and loading is a's shown in Fig. E10-4a. We f l n n a ~st, the ends and at the will assume lateral bracing" of the comnressinn - --r-------I- - concentrated loads. Other design data: E60 electrodes, . ---- ,n,..~.,,,r,,i,.f i r tai n n c ATSC s, A-36 steel, and girder depth limited to 84 in.
Figure ElO4b
-
90k
FO;
h/r
For
t, = 0.5 in, h/t = 78/0.5 = 156. Step 2. Make a preliminary flange design.
=
162, obtain t
78/162 h Maximum - = =
=
0.48 in. Try tentative t,,, 14 000 = 322
=
0.5 in.
105'
--
21 ksi and an average distance to the center of the flange (Assume that fb area of 79 in.) Try a flange plate 7/8 X 18 in. b = 18 9j = 10.3 < ---(AISC Sec. 1-9.1.7) 2rj 2(0.875)
fi
2R
Step 3 . Compute the actual moment of inertia and section modulus of the trial section and revise the dimensions as required.
= 343.4 kips
bh3 I,, = 12 + 2,4d2
Now draw shear and moment diagrams as shown in Fig. E10-4b. Step 1. Make a preliminary web plate design. Assume that web depth = 78 in. The limiting h / t w for no reduction in flange bending stress is h 760 --=-=
$8
760
m
-,It
162 (approximately)
(neglect I, of flange plates about cenlioid axis
31 8L'C 1 U~U'L b 4 LLi. UIjblbiu
Gxnpute the weight of the girder:
.,
Check the allowable stress F,. te L / r , with r, computed using I, of the compressi f gyration is obtained using A, + A,/6:
The approximate additional 'bending moment (conservatively compute ilnce the moment due to other loads is slightly off center) is
I
7-
0.24(53)~ = 84.3 ft - kips 8 The total bending moment = 2722 84.3 = 2806.3 ft . kips. - M 2806.3f12'1 JbC-=: 19.53 k s ~ S 1724.5
M=
+
I
\
-/
I
Since this is considerably less than 21 or 22 ksi, let us tentatively revise the web thickness 2, to $ in.
, I
-
-w
-.
Check the web shear SO that the plate is not too thin (neglect beam weight at j'lls pmnt): 1) 1 O(U.3
-. .-
13)
v.n.
step 4 R e c o ~ n ~ uIt eand&, I
I
= 14 829.8
+ 48
We must investigate both th; end panel gnd the Intenor (between two columns) panel smce C, is different for each locat~on.For the end panel: L = 18 f t (largest value), and m , = 0 and ibf, = 2664 f t . klps
992 = 6 822 in4
For the interior panel: C, ft . kips):
=
1.0 (since Interior moment of 2722
> 2664
s, = 63-822
1600 ln3 39.875 0.490 New girder weight = (3 1.5 29.3) = 0.207 kip/ft 144 'be approximate additional moment due to girder weight is
-
M=
+
For this L / r , ratio, F, = 0 . 6 5 = 22 ksi for both the end and interior panels. Now check AISC Eq. 1.10-5, since h / t w = 208 > 7 6 0 / a :
0.207(53)~
MtOw = 2722
= 72.7
+ 72.7 = 2795 ft . kips
2795(12) "4, = 1600 = 20.96 ksi Continue the design. Use two flange plates x 18 in and a web plate 3 y X 78.
==
2 1.05
> 10.96 ksi
O.K.
- At this point the bending stress and slider proportions are adequaie unless a later interaction check requlres a revlsion of girder section. Step 6. Compute the stiffener requirements. -AISC specifications require bearing stiffeners under reactions and the two concentrated column loads. In the end panel the actual shear stress at the reaction is -
h = 208 tw
D
= 79.75 m
SX=1600in3
A, = 29.3 in2 Af=15.75in2
V
f =-=-
A,
173.25 + 0.207(53/2) - = 6.11 ksi 0.375(78)
According to AISC Sec. 1-10.5.3, intermediate stiffeners (other than the end
486 STRUCTURAL STEEL DESIGN
beanng stiffener) are not required if -,E;.,,C.," h - I L6U and j, f, FF,= t 2.89 We will provide bearing stiffeners under the column loads, which leaves clear distance in the 18-ft end panel of 18 x 12 = 216 in.
been obtained from Table 11-8 of SSDD but requires double i Step 7. Check the interaction at concentrated loads [Eq. (10-13) At dx to the right of the right column, the shear V is
V = 178.4 - 18(2.80 + 0.207) = 124.6 kips (including girder wei For V = 124.6 kips, the web shear stress is 124.6 j" = = 4.25 ksi 29.3
-
We have just found F, = 7.67 ksi in step 6.
(
I]
4'25 36 0.825 - 0.375 7.67
Assuming that C, will compute
C" =
45 OOOk - 45 q ( h / tJ2 36(20gL) is limited
t
36(0.169) = 2.1 1 ksi << 6.1 1 N.G. stiffeners required 2.89 Try one stiffener at half the distance (noting that if this works for the 18-ft panel it will also work on other end in 17.5-ft ane el):
F" =
a h
108 78
--- =
1.38 < 3
22.2
> 22 ksi
O.K.
Since the allowable stress in interaction is 22 ksi (maximum) and the a c d and allowable values based on flange stability are less, the lesser values control. _Tb = 20.96 k s ~ F, = 21.05 ksi
< 0.8, we will try Eq. (10-1 1):
The allowable shear stress if no intermediate stiffeners are used
=
Step 8. Check the stiffener requirements of the 17.5-ft interior span between columns. The maximum shear is obtained coming from the left:
(Y1
V = 170.15 + 0.207 - - 17.5(2.8 + 0.207) - 90
=
33.01 kips
j,=--33'01 - 1.13 ksi 29.3 Try no stiffeners:
k
O.K. (less tha maximum a / h allowed) = 1.56
> 1.38
O.K.
F, = 36(0'170) 2.89 = 2.12 > 1.13 ks, With intermediate stiffeners and C, Eq. (10-9):
< 1, we
can use AISC Eq. (1.10-2) or
Since 7.67 ksi is greater than the actual shear stress at the bearing stiffener, 6.1 1 ksi, and the shear is less at the interior points, it is not necessary to check the shear stress further for stiffener analysis. Note that Fv could have
O.K.
no st~ffenzrs required
Step 9. Check the web cnppllng under the compression tlange due ti uniform load. Assume that the flange is restrained against rotation (since it carria uniform load). The load carried in compression to the web is 2.8 kipsift -t weight of top flange. We will neglect flange weight, so that the compressiv stress is
2.8
jc = 0.375(12)
= 0.622 ksi
I
STRUcrCnW, SIEliL DfiSICb*
The allowable compressive stress (checking at the location where a / h is critical is
1
F = IO"*[5.5 ( h j tWl2
4
1.4
> 0.622 h i
O.K. 208' 2.69' Note that if F
+
I
=
0.215 and a / h = 1.38 from step 6.
[Eqk (10- 17) combined with kc]
+
- &(5.5
'
+
for web and stiffener); C,
In terms of A,, x 100/A, this value could have been obtained from TabIe 11-8 by using double interpolation. Try two bars 3/8 x 6 in: A
=
2 ( : x 6)
=
4.5 1n2 > 3.02
O.K.
-16 O.K. 0.375 The minimum moment of inertia of the stiffeners is -=--
t
fi
oS(16.35)~ = 182.9 in4 12 The "effective" column area for the radius of gyration (see Fig. 10-13): I =
A = 16 x 0.5
r
+ 12(0.375)(0.375) = 9.69 in2
=dF 182.9
9.69
76.5 in
Step 11. Design welding to fasten the stiffeners to the web.
jU= h[(&y]l/'
- 18.45 > 20.98 ksi
O.K.
Check the bearing stress; assume a 5/16-in weld, so that the effective bearing area = (8 - 0.3125)(0.5)(2) = 7.7 in2:
F,,, = 0.9G = 32.4 ksi
P, = 32.4(7.7)
=
kips/in
[ Eq. (10-16)]
= 4.34 in
L = 0.75h = 0.75(78) = 58.5 in L 58.5 ---- 13.5 r 4.34 From Table 11-5, the allowable column stress is Fa = 20.98 ksi. The column stress is fa=-----178'74
Use a stiffener plate length h - 4 t , = 78 - 4(0.375)
= 249.5 >> 178.74 kips
O.K. For intermediate stiffeners, with only one intermediate stiffener use two plates.
For a plate on both sides of the beam web, D
=
.
1.0 and Y
=
1 (A-36 steel
For a pair of stiffeners, we have 2.7 kips/in; for each side this becorn&; 2.7/2 = 1.35 kips/in. For 1/2-in bearing stiffeners, use a 3/ 16-in weld and E60 electrodes. F,,, = 0.1875(0.70711)(0.3 X 60) = 2.39 kips/in Use a 3/16-in fillet weld continuous for both bearing and intermediate stiffeners. Rationale: Few stiffeners and the weld can be made in one pass. I t 2 too difficult to measure and set up alternating weld distances and gaps. Step 12. Design welding to fasten the flange plate to the web. Check the end for maximum shear:
Use a 5/16-in (t, > 3/4 in) continuous weld on both sides at F, = 392 kips/in. Note that the weld 1s considerably overdesigned, but for an im portant girder the use of intermittent welds is not worth the savingsularly if the weld can be made in one pass, as here. Figure E10-4c illustrates the design summary for the girder.
I
490
STRUCTURAL STEEL DESIGN 9ok
105~ 17 5'
AREA: When cross-ties rest directly on the girder flange:
18' 16
? - 8 X j t
2 - 6 x i 8 ' t
Web
X 78"
1 2-8X+"e
1-2
-
6X
:"t
X 18"
The limiting b/2$ = 10 for A-36 steel When cross-ties do not rest directly on the girder flange:
The limiting b/29 = 12 for A-36 steel.
10-7 BL'km GIRDER DESIGN THEORY-AASHTO
AND AREA
Z1
8
10-7.2 Web Plate Design
d
Plate girder design using AASHTO and AREA specifications is very similar but more conservative than AISC because of the more hostile environment to which the girder will be subjected. In general, however, the same general considerations apply: 1. The girder is proportioned by the moment-of-inertia method. 2. No uniq'ue solution is possible. 3. Shear and stiffener requirements are more rigid. Bearing stiffeners are always required, but AASHTO allows use of longitudinal stiffeners.?
10-7.1
'Firder Flanges (AASHTO and AREA)
,,
In general, the basic allowable bending stress is
Fb 1 0.555 However, if the compression flange is laterally unsupported in a length L, the stresses must be reduced as in Part I11 of SSDD and in the appropriate specification. The b/2$ ratio for the flanges is also limited: AASHTO:
AASHTO also uses Eq. (3-5):
and solving for h/tw, substituting k c obtain
=
23.9, E = 29 000, and F ,
=
1.19Fb, we
'Xlis gives h/tw = 163 for A-36 steel. The AREA limiting h/t, ratio is obtained from Eq. (3-5) by substitution of kc = 23.9, E = 29 000, and Fcr = 0 . 6 5 to obtain (with slight rounding) h 1030 2700 SI: - - fps: - = lw
fi
lw
6
For A-36 steel, the limiting h/t, = 171. When the value of fb compression flange, AREA allows an increase in the h/t, ratio:
< Fb in
the
with the ratio (F,/ fb)'I2 I 2.0. AASHTO and AREA limits the web thickness of fabricated plate girders to: WSHTO:
Ths limiting 6/29
t
Alsg
= 12 for A-36 steel when fb =
ARE AREA, but for continuous girders.
Fb = 0.55Fy.
tw 2 5/ 16 in (8.0 mm)
AREA:
t, 2 0.335 in (8.5 rnm)
AASHTO allows the limiting h / t , to be increased with a longitud stiffener at 1/5 of the clear web depth from the compression flange. This value is based on Eq. (3-5) using kc = 129 (since a more substantial edge fixity is ?
C
*
."a
492 S T R U C ~ U I L A I:,. 1 ,XL u~.zl,.;:-r 8
..
PWTE GI&LRS
4 ~ 3
.
obtained) and F,,
=
'
1.64, to obtain
fps:
v3
'W
6
i 3 e limiting h / t w for A-36 steel; is 327.
fi
:he AASHTO requirement for stiffeners is also computed using E ~ (3-5) . and the k factors given in Eq. (10-8) and with varying safety factors. If the such that a / h -+ (no stiffeners), kc = 5.34 from ~ q (10-8), . and taking F,,-2.5 f,, we obtain h
234
h 618 -5tw lw The limiting h / t w = 68 for A-36 steel when f, = F, = 0 . 3 3 ~ ~ . AREA requirements are slightly more conservative: h 360 h 950' fps: - 2 SI: -
<-
vx
w
S1:
ips:
332t
a = - (in)
fi
875r
SI: a=-----
fl
(mm)
In these equations t has units of in or mm. The stiffener spacing is limited to not more than 72 in or 1800 mm when using Eq. (10-22).
fi
6
fi
where t has units of in or mm. The previous two editions of the AfSC specifications included this requirement for spacing the first interior stiffener from the bearing stiffener. Currently, the XASHTO specifications for intermediate stiffeners is somewhat more indirect and is given in Sec. 10-7.4 foI10wing under "Intermediate Stiffeners." The AREA specifications reduce the 348 factor slightly, to obtain
10-73 Shear and Stiffener Requirements
fps:
.I_
From an approximate aqeraging of fi we obtain 348, which if used to back-conpute the SF gives 1.37 at a / h = 0.5 to 2.02 at a / h . 1.0. In earlier AASHTO specifications the stiffener spacing could be computed as 348 t 914t fps: a = - (in) SI: a = - (mm)
h <3840 1455 ' S I : -5 -
"
>
.
5
I
10-7.4 Stiffener Design .
Longitudinal stiffenem The AASHTO value for longitudinal s$feners in terms of moment bf inertia (see Erickson and Eenam, "Application and Development of AASHTO Specifications to Bridge Design," Journal of Structural Dit'inbn ASCE July 1957) is (10-23
IJ = 11t1.[24(%)2- 0.131
I
with a plate thickness limited to b 71 fps: -5---
from which =
a
-
kT ~ E 12(1-p2)SF
I
--
- 0-
t
tw
'JsiM,S.F..= 1.5, several a / h ratios from 0.5 to 1.0, and computing kc fr (10-8), we typically obtain a/h
k
P
0.5 0.6 0.7 0.8 0.9 1 .o
25.36 18.83 14.90 12.30 10.58 9.34
332.8 344.4 357.2 370.9 387.2 404.0
a
SI:
b -
188
5 -----
(1 0-24
where f, = calculated bending stress in the compression flange. The stiffener may be placed on one side of the web at the 1/5 point from the compressio~ flange. The AREA allows longitudinal stiffeners for continuous girders in d l negative-moment regions and the specifications are exactly the same as fo AASHTO for I, and b/t. Bearing stiffeners are always required for both AASHTO Bearing AREA girders. They are designed as columns using an area as shown in Fig 10-13. The allowable column stress depends on the L / r ratio computed a shown on Fig. 10-13. The bearing stiffeners must be checked for bearing as we! as acting as columns. The bearing area is as shown on Fig. 10-13 allowable bearing stress is USHT~:
Fbrg = 0.80Fy
AREA:
Fbrg= 0.835
494 STRUCTURAL STEEL DESIGN
AREA requirements for the b,/t, ratio of bearing for any other compression member.
feners if
h -
and
f,5Fo
where I
but!,F, I 5 / 3 . This criterion gives the limiting h / t w = 68 for F, = ~ , / 3for A-36 steel. Thus any h / t , ratio that is less than 68 does not require transverse stiffeners if fo < 5 / 3 . Any h / t w < 150 does not require stiffeners if f , < F, as givdn in Eq. (10-26). When intermediate stiffeners are required or when j, > F~of E ~ (10-2 . the maximum spacing is limited to a 2 1.5h and the allowable shearing limited to
lnediate stiffeners(including other than reaction bearing stiffeners) of
ti
& [ I+ ( h / a ) 2 ] where C= -I I ~,(h/t,)~ ' B = 222 000 in fps units = 1 520 000 in SI units
It is often convenient to plot the limits of these stresses as in Fig. 10-14. When ~ m s v e r s estiffeners are required, the first stiffener is placed sue a / h 5 0.5 and the actual shearing stress, f, 5 Fi, is obtained from the followi equation:
where B' = 70 000 in fps units = 483 000 in SI units The U S H T O specifications require a minimum moment of inertia of inter-
ere I
p=0.3 for steel a = actual stiffener spacing tw= thickness of girder web plate J = 25(h/a)l - 20 but J 2. 5.0
AREA requires intermediate stiffeners if the h / t , ratio is Wate values defined by ~ q (10-21). . ~f stiffeners are required the spacing is
but a I 72 in (or 1.829 m), where t,= girder web thickness, in or mm f v = actual web shear stress = V/dr,, ksi or blPa
SSTRUCTUKRLb 1LLL. UEa~crr.r
The stiffener dimensions must be at least as follows: AASHTO:
t
AREA:
+ D/30 and width 2 bf/4 2 51 mm + D/30 and width 2 bf/4
width 2 2 in
thickness 2 width/ 16 same as for AASHTO
10-7.5 Interaction .\ASHTO specifications include a bending stress reduction if the shear f, I 0.6Fv according to
id-7.6 Lateral Bracing and Diaphragms Figures 10-15 and 10-16 illustrate bracing and diaphragms in current practice. .AASHTO specifications require either cross frames or diaphragms at the ends ;-nd at intervals along the span not to exceed 25' ft (7.62 m). Diaphragms (cross beams) must be at least one-third and preferably one-half or more of the gird depth. When spans exceed 125 ft (38.1 m) for concrete deck bridges, an additional ,:ystem of lateral bracing will be used parallel to and as close as practical to the bottom flange and in at least one-third of the bays (when multiple girders are ysed). The smallest angle permitted by AASHTO for this is the 3 X 2; (76 X 64 ,mm) fastened with at least two bolts or the kquivalent weld in each end konnection of the angles.
Fi;rve 10-15 Bridge bracing using diaphragms. (a) Diaphragms used with five plate girders. (b) E 6i:.phragm. Must be sufficiently set back from adjacent diaphragm or abutment so that it can i*.?ected and periodically painted.
Figure 10-16 Longitudinal and cross bracing. Note use of verrical and longitudinal stiffeners. Bridge in background of (c) is older and uses riveted construction where bridge in foreground is alI-welded The three girders carry a four-lane roadway plus shoulders and walkway for both bridges. ( a ) Cross bracing. (b) Fabrication details. Note that the end fastener plates ire welded to both web and fiansOn continuous bridges these plates provide flange brace points. (c) Use of both cross bracing and longitudinal truss bracing. Note longitudinal stiffener along compression flange, where h / is ~ excessive.
AREA requires the following: 1. For beams or girders less than 42 in (1066 mrn), use of I-shaped diaphragms (cross beams) as deep as practical and asin:, double-angle beam framing connections. 2. For girders deeper than 42 in and spaced more than 48 in (1220 mm) on centers, a system of cross-frame bracing in which the angle of the cross-frame diagonals with the vertical is not greater than 60". 3. Spacing of diaphragms or cross frames is 18 ft (5.5 m) for open-deck construction 12 ft (3.65 m) for ballasted-deck bridges 4. Where ballasted decking is carrled on Gross beams without stringers (as in Fig. 10-5), at least one line of longitudinal d ~ a p h r a p s1s to be used, as shown in Fig. 10-3d. 5. Knee bracing is required to support the compression flange for through deck girders. This bracing (see Fig. 10-17) is usually placed at selected intermediate stiffeners for ease of attachment and ranges in slope from about 3 V : l H to somewhat less. The maximum spacing is limited to 12 f t (3.66 m). It may be
Cross b a r n s a t 2 . 3 f r 760 min
Figure E10-5a
The impact factor is computed (see Fig. E10-5a) based on L
:IR+
16
71
Figure 10-17 Top flange bracing using web plate and bracket for the through-girder ballasted-deck rallroad bridge shown m Fig. 10-4.
designed as a column for an axial force based on the horkontal corn 2; percent of the compression flange force =
H/ V
The approximation of the axial force using the sin defined by H and than the actual length is sufficiently accurate for design. ----
use 28 percent
(Note that tlphimpact factor is not the same as used in Example 4-15 design momecf
..
M, = 9984(1.28)
=
12 779.5 kN . m
The design shear
V,
=
1680(1.28)
=
2 150.4 kN
We will neglect the live load on the pedestrian/semce walkway. With rhis assumption the dead-load estlmate is
0.025&Af Pkb
+ 27.5 - 9 = 27.6
v rather
Example 10-5 Design a welded railroad bridge girder for a 27.5-m-span single-track ballasted-through deck bridge. Use a Cooper E-110 loading, A-36 steel, E70 electrodes, and the AREA specifications. The general bridge configuration is as shown in Fig. E10-5a. S ~ L ~ TFrom I ~ NTable 1-2, the design live load for one-half the track (one rail load to each girder) and adjusted for E-110 loading is:
Girder weight (including stiffeners and weld)
= 10.0 kll
Ballast, including ties at 350 rnrn depth and a reduced effective width: 120(9.807/62.4)(0.3j0)(4.42/2)
= 14.6
Steel deck and cross beams and factor of 1.10 From Example 4-15, the beam is a W760 X 160.7/ 1.58 Deck: 0.015(77.0)(6.49>(1.10)/2.0
x 4.45 = 1680 k~
=4.1 kN
kN
= 6.2
Track at 200 lb/ft: 0.200 x 14.59 (AREA spzc~f~catlon)
=2.9 kN
=
kN 42.2 kN/n
= 4.4
0.300 x I4 59 Total
= 377.4 kips
kN
Cross beams: 37(1.58)(5.79)/(27.5 x 2 )
Walkway: estimate 300 Ib/ft
V = 274.5(%)
> 25 m as
=
Note that some approximation 1s used for the steel deck to allow for forming the trough to hold the ballast to g v e an effective length of 6.49 m, Note also that an estimat~ngfactor of 10 percent is applled for
<
S T R U C T n k L Ki'W:L3
1 )I: , I C ~ ? ;
P'LATE G~RDERS
mcertainties. The dead load moment is M = - =wL2
42'2(27'5)2 = 3990 kN 8 8 Tne dead load shear (at reaction) is
v = -WL =
The section modulus is
.
The corresponding maximum bending stress is M 16770 fb = - = = 135.4 MPa < 137.5 Sx 123.81
40.2Q7.5) = 553 kN 2
2
-
The total design moment is
+ Mdead= 12 780 + 3990 = 16 770 kN . m VdeSi,,= V L + Vdead= 2150.4 + 580 = 2730.4 kN
Step 3. Check the girder weight.
MdeSi,, = M L
Step 1. Find the girder proportions. Take h 2134 - 2(50) 2034 mm. Also the web must be at least
--
--
t
v
"
A,,,,, = 2(0.070)(0.815) + 0.016(1.994) = 0.1460 m2 '
hF,
Ihe,maximum ,.,. , , . ..,... shear is at the reaction, but with a moving load contributing .he major effects (with impact) will not change much for some, distance along the girder, so we will take t,,, = 16 mm. ~
- =-= tw
Try one web plate 16 x 1994 mm and two flange plates 70
X
815 rnrn.
2730.4 2.034(0.35 x 250) = 15.3 mm
N-L1:
1:
(O.K. so far)
127 << 170
2034 16
O.K.
Step 4. Compute the allowable bending stress F,. The compression flange will be laterally braced using knee brackets in Fig. 10-17 at every fourth cross-beam supporting the deck. These intern are approximately 4(0.76) = 3.04 m < 3.66 (maximum by A m ) . radius of the gyration of the compression flange is
'fie tentative web area A, = 0.016(2034) = 32.54 X lop3 m2 Wsing Eq. (10-2), we find the tentative flange area as , r'
I
,
I
'
16 580 32.54 --- 2034 X 0 . 5 c 6
A -
Flange area
- 65.21 - 5.42 = 59.79 x
m2
=
0.070 x 0.815 = 0.05705 m2
A w - 1.994 x 0.016 -
6
6
=
0.0052 m~
total
= 0.06237
m2
i i y two flange plates 70 X 815 mrn ( t available only in 10-mm increments): h = 2134 - 2(70) = 1994 rnm ---815 - 5.82 << 190 O.K. 25 2 x 7 0
-b -
fi
- = - =1994
16
tw
125 << 170
O.K.
Step 2. Compute the moment of inertia of girder and determine the a-tual bending stress fb so that we can checkf, I Fb.
The allowable bending stress is
+
bftf3 Ix = Iweb+ 2 ~ # ~ 12
(
720)' +
- 106(0'016)(1'994)3+ 2(0.070 x 0.8 15) 997 + -
12 = 10 571 + 121 510
+ 24 = 132 114 x
lOW6m4
815(0,070)~10~ 12
Step 5. Design the bearing stiffeners. We will wrap the top flange around the end of the girder web to provide additional tension field resistance for the erid post as shown in both Fig,
P
The,clear distance a is limited to 0.8721, a = -------- 5 1.828 m (AREA 72-in limitation on spacing
fI
At the bearink stiffener '
v
jv=-= ht
'
2730.4 = 85.6 MPa 1994(0.0 16)
< 0.355
O.K.
This value of j, results in an allowable spacing of a =
0.872(16)
m
=
1.5 1 m
<
1.828
O.K.
We can now do one of two things: 1. Space all stiffeners (but so that they come out as an integer) at mately this spacing. 2. Use this spacing for each end part of the bridge and a Iarg the interior part, since the live-load shear (without impac (Table 1-2). Similarly, the dead-load shear is only 1/2 of Let us investigate the spacing for the approximate center h span: Live load
=
764
x 1.28 = 1234 kN
Dead load = 580/2 Total
=
290 kN
= 1524 kN
and 1524 jv = 1994(O.Q16) = 47.8 MPa
We will use the spacing as shown in Fig. ElO-5c. 1
Figure E10-5c 9
a SmuC?UiUU S'TYEL DESiON Use a pair of stiffener plates for each intermediate stiffener. D 2134 6, 2 50 + - = 50 + -- = 121 mm use b, = 125 mm 30 30
Step 7. Design the flange-to-web weld. The AREA specifications allow either full-penetration groove welds o fillet welds; either must be continuous. We will arbitrarily use full-penetra tion groove welds. For this type of weld we will only have to check the she produced and compare it to the maximum value allowed for E70 electrod
Q = A,-- = 0.070(0.8 15) 997
(
= 58.88
x
+-
;"I
m3
j-
i
u = - =V Q
2730.4(58.88 X I 0.132114 , The shear resistance is limited to
I Field bolt
= 1216.6 kN/m
1
1
~ I O O IO V J ~ ~r I
ou
A I O U . I I 1.30
s h o p weid Framing angle, designed ln Example 8-9
F, = 0.35F,ASh,, = 0.35(250)(0.016>(1rn) ---/ x lo3 -" - / \ -
-~-twnlu/rn~1~10.0
"
Figure E10-5d
0.K.
Step 8. Check thc:deflection. . that . the . live-load moment is due to an equivalent uniform nssumlng loa~ d we , obtain WL =
8M
L~
and the total equivalent uniform load we is we = wD
I
+ wL = 42.2 + 8(12 780) = 177.3 kN/m 27.5'
@
.The.dsflection is approximately
A = - - =5wL4 384EI
force
5(0' 1773)(27'5)4 = 0.04996 m (say 50 -) 384(200 000)(0. 132114)
27.5(1000) = 43 mm = 50 mm (take as O.K.) Amax = 640 Step 9. Design the upper flange knee bracing. We shall arbitrarily place a brace on every fourth floor beam f spacing of q0.76) = 3.04 m. Where this coincides with or is "close eno to a stiffener, we will increase the stiffener thickness from 8.5 mm to 1 For any bracing plate alone, we will use a 12-mm plate. Shop-we stiffener or brace plate to girder web and compression flange as shown i Fig. E10-5d. Field-weld the bottom of the stiffener brace plate to the c beam. Shop-weld the angle for the bracket and punch for three 22high-strength bolts at each end. Take the horizontal component of stiffener
The number of 22-mm A-325 bolts in double shear required horizontal component of stiffener force to attach stiffener to floor 193 = 1.83 N = (0.7854 x 0.022*)(138 x 1d)(2) Use three fasteners (AREA minimum number of fasteners in a co For the stiffener try a WT 385 x 73.7 A = 9.39 X m2 t, = 17.02 m m d = 376.6 mm
Pa
=
t,
=
13.21 m m
AFa = 9.39 x 133 = 1249 kN >> 579
O.K.
-
Annvec 1638 kN m.
given in Sec. 1-9. I
Wind on girder: .1.5(0.030 ksf)(47.88)(2.134 rn dept
,
Wind on train: 0.300 klf(14.59) Design the channel-to-beam weld for the girder of Prob. 10-3. Design the channel-to-beam weld for the girder of Frob. 10-4.
i
An additional concentrated force of one-fourth of the heav
applied at each critical point;. this gives a force of
& d!,S
;
,
1 10(4.45)(0.25) = 122.4 kN
I
End panel:
)
Interior panels:
122.4 +
90 = 143.01 k~ 2
122.4 + 9.0(4.58) = 163.6 kN
ave a separate set of design data of length depth and load. Be sure to w th
P
-
175 to 250 kips, L = 40 to 60 ft, and D = 60 to 96 in, as assigned. Use A-36 stre electrodes, and the AISC specifications. Assume lateral support at the ends and load points. If no specific data are assigned, take P = 2 15 kips, L = 54 ft, and D = 80 in.)
-L-
welded plate,girder for a crane runway for the conditions (-urn loads o n in Fig. P10-9. Use the AISC specifications, A-36 steel, and E7O electrodes. Limit
STRUc'nRAL STEIiL DESIGN
gipder depth to 84 in. Lateral support is only at girder ends. Use W / L aOft, and S = 15 ft. 19-10 Desi~na welded plate girder for a crane runway for the condit shown in Fig. P10-9. Use the AISC specifications, A-36 steel, a @der depth to 2.15 m. Use W/4 = 110 kN, P/2 = 950 k~
*dtr) bC (,vt,
D ~ S ~a W welded plate girder for a single-track open-deck railr ,rrack contributes 0.25 kip/ft loading, including rail, ties, and so on. loading and span from 40 to 110 ft as assigned. Use A-36 stee s~--.zications.Limit the overall girder depth to L/10 2 D 2 ~ / 1 5 , :t. Take the girder spacing S 2 L/15.Design sway bracing and/or diaphragms in stby flanges, stiffeners, and welds. (Note: If specific problem data are not assigne E-1 i0 loading, L 90 ft, D = 96 in, and s = 6.5 ft.) &-12 Denim a welded plate girder for a singletrack open-deck railroad bridge. A tr7;k contributes 3.65 kN/m of loading, including rail, ties, and so on. ~~~i~~for C-l live loading and span from 14 to 34 m, as assigned: Use the a-d depth D but limit D to between 1.22and 2.45 m. If specific problem CooPr E-80 loading, L 23 m, D = 2.3 m, and S = 1.6 m. '
SELECTED COMPUTER PROG
-
-
. ,.,.
.
A-1 FRAME ANALYSIS PROGRAlM The frame may be This computer program will analyze any plane pinned, or a combination. It must have a constant modulus of elasticity, E.
ated loads are on the member. The program computes fixed-end r a uniform beam loading (and orientation from horizontal to vertic so computes fixed-end moments for up to two concentrated loads on
The following steps are required to run the program, with tenninol matching the computer program listing. 1, Code the structure according to rigid frame, truss, or combination- T account any hinges, as in chap. 2. 1Determine NP. The computer program computes NPPl = NP
,, \$.510 STRUCTURAL STEEL DESIGN
i
S E U C I ED
I
5. Determine NLC that uses wind the first time as.NLW. 6. Determine NLC that uses D + L, which is to be later comb NLC that uses wind. 7. Determine the NLC that has wind on any beams or columns. Call this values, which should be only wind. 8. Find NBAND using the method outlined in the text. 9. Make sure that ISTIFF is dimensioned greater or at
1 = NP value of the entry This inputs the NLC nodal forces at 1 , ~ L) ~ = 1,NLC for NLC load conditions using 8FlO. Note that this requires at least two cards NNZP. 5. Add your system control cards. Note that you may (but not recommended) read nodal moments hove. If you do read nodal moments, the output moments must be Correc and to get the final design moments as Ffinal = F ~ ~ t p + ~ tMrrad
careful attention to signs. The output signs are interpreted using Fig e program computes the fixed end moments, the output is automati corrected for the FEMs to obtain the design values. Member forces do not to be corrected for any node forces read in other than moments-
'
1. Punch cards for lines 4, 8, and 10 of computer listing. 2. Punch member data cards according to the FORMA MAT(7 14, 4G 10.4, 12) The G format allows use of either F10.4 or xxxE-3 or xxx NPE(7) uses the I2 format and goes in column 70.
f input units are as,identified here, the output will be
X = displacements = in, mm, or rad F = member forces Truss members: kips or kN Beam members:
Use NPE(7) = 1 for all members without transverse loads. For any members with transverse loads: NPE(7) = 0 for normal gravity beam loadings. NPE(7) = - 1 for wind loads on beam or transverse loa columns.
kips or kN for axial for f t . kips or kN
m for mom
entification of Program Variables
of gravity X 1, X2 = horizontal distance from origin end of member of P1 a or m Vl, V2 = column shears from alternate direction for a space frame, V1 = near end shear; use kips or kN and sign same as P1, P2 3. Put a blank card at the end of the member data.
any alphanumeric data using not more than SO c Recommend using your name, problem number, o r , UNIT3, U N I T 4 units identification for fps an UNIT4 UNIT2
FT or M (start in column 1 of card) I N or MM (start in column 5 ) K-IN or K-MM (start in column 11-unit
used for
K-FT or K N . M (start in column 21) number of P's coded = 2 x number of nodes - number reactions
JllAL STEEL DESIGN
~LW; NNZP
*
SELECTED C O W
number of members (NM = N P for determinate trusses number of load conditions 1 to list band matrix (part of total AS finding X's) first load condition with wind number of nonzero P-matrix entries to 0 for fps problems = 1 for SI problems 1 for all truss members; = 0 if any present band matrix and must be dimensioned NP x NBAND at least band width used. Is difference betwe NP on ends of any member? Valu needed but not larger than NP. If valu numbers too small, tends to give a sometimes-other times gives overflow messages. NLC with D + L which is saved on disk file to add to win NLC NLC which uses D L from disk file with current force 1 to write X (deflection) matrix 0 to obtain output with few members 1 to obtain output when large number of NLC a modulus of elasticity, ksi or MPa member number assigned to element during cod NP values for the element; truss has 1 through through 6 horizontal distance to far end of member including si or m vertical distance to far end of member m cross-sectional area of member, in2 or m2 moment of inertia of member, in4 or m4 (not need truss members) control switch for members with trans 0 for members with transverse loading 1 for members without transverse loading MENT cards in computer program) BE SURE TO PUT BLANK CARD AT BEK DATA NP value of P matrix entry where a con moment) is located value of corresponding P-matrix entry. The are in a DO loop, so use as many pai
+
IWINRE IOLPL
UNIT^ UNIT,,
..
..
COUNTEL! TO ADO WINO LOAD OF&r'$E>ll C O U N T E R F O R D E A D + LTVE.C?~O~-LI(:
-- UNITZ . IN-K OR UN-PM .AU U I T 3 $ 1 4 011 UN11M -- * * * NOTE C ~ E F U L L YU N I T S OF U M I T ~ I IFOR I N P U J
"
,
-Y9
K OR K N
I N OR
9
* ,
lrPMIN(I15I DATA F F U 1 1 2 ~ r 1 0 0 0 . r 1 . ~ 1 0 0 0 . ~ 1 ~ ~ 1 0 0 0 0 0 0 0 1 1 2 ~ t ~ ~ ~
READ CARD 2 R E A D ( ~ , ~ ~ O O ) L Z Z , U N I T I ~ U H I T ~ ~ U ~ ~ ~ Z ~ U ~ ~ T ~ 000 F O R M A T ( Z O A ~ , I , A Z S Z X S A Z ~ ~ X ~ ~ ~ A ~ S ~ X ~ ) IF(EOF(lll750~3 3 WRITE ( 3 r Z 0 0 0 ) ZZZ 2000 F O R M A T ( ~ ~ ' J ~ ~ S ~ X ~ Z O A ~ ~ ~ ~ REAO CARO
R E A O ( ~ . ~ ~ ~ ~ ) N P J N ~ ~ ~ N L C P L I S T R , N L Y ~ N N Z P J I ~ E ~ # ~IDLPL,IYINBE~IWRITX~IW~ITP 1 0 0 1 FORMAT(lbI5l L++*+* REAO MODULUS OF E L A S T I C I T Y - K S I READ(lr1002lE
1 0 0 2 FORMAT(BF1O.C)
- *
-
D E F I N E U N I T CONVER$ION FACTORS IUNIT 1 IF(IMET.GT.O)IUNIT 2 0 0 4 0 1 . 1 ~ 6 FU(I) FFU( I U M I T s I I 4 0 CONTINUE
-
NPP1
NP
OR B P I
***********
I
1
1
I I
5 STRUCTURAL STEEL DESIGN FASAT(Ze3) = -EASATII.Z) FASAT12s4) = -EASAT(2,2) EASAT(391) = -EASATf1.1) E d S A T f 3 r 2 ) = -EASAT(1,2) EASAT(3131 * EASATf1.1) EASAT(3r4) = EASATf1.2) EASAT(4rl) -EASAT(I.Z) FASATf4.21 = -EASATf?,Z) EASAT(4.31 EASAT(1.2) E A S A T ( 4 * 4 1 = EASAT(2.7) 0 0 1 7 9 111.4 IF(NPE(I).GE.NPP~IGo TO 1 7 9 N t l = (NPE(I )-~)*NRANO 00 178 J = l r 4 I F ~ N P E ( J I . G E . N P P I ) G o TP 1 7 9 ~F(NPE(J).LT.NPE(I))G~ rn 178 NSZ NPE(J) NPE(1) t 1 STTFF(NSl NS2) S T I F F I N S I t N S Z ) + EASAT(I,J) 7 8 CONTINIJF 1.9 CONTINLJF G n TO 5 5 0 0 1 0 5 EAOL = E * A / ~ L E I O L = (E*xIIxL)*Fu(?) S I N O L = XSINIXL COSOL XCOSlXL CA(1.l) 0. EAlloZ) 1. EA(lv3) = 01 FAf2rl) -XCOS FA(2rZI SINOL Eb(2r31 SINOL EA(3.1) -XSIN EA(3.21 -COSOL E A ( 3 r 3 ) = -COSOL E A ( 4 . 1 1 = 0. EAf4.2) 0. EAf413) 1. EA(5.1 = XCOS FA(5.2) = -SIYOL EA(5.31 -SINOL EA(6.1) = XSIN E b ( b t 2 l = COSOL E A t 6 r 3 ) = COZOL ES(lr1) EAOL E S ( 1 ~ 2 1 = 0. ES11.3) = 0. ET(2.1) 0. ES12.2) 4.*EIOL ES(2.3) = Z.*EIOL ESf3rl) 0. ES(3.2) = ES(2.31 ES(3.3) = ES(2.21 DO 2 0 2 I = 1.3 0 0 202 J 1.6 E S A T ~ I P J )= 0. 0 0 1 8 6 K 1 1.3 ESAT(I*J) * ESATfIpJl + E S f I ~ Y ) * E A f J s ~ ) 1 - - CONTINUF to.? CONTINUE
SELECTED CObIP 2 0 4 CONTINUE DO 2 0 6 1 = l r 6 IFfUPE(I).GE.NPPl)GO
-
-
-
'
.
-.
.
----
--
-
-
NS2 = N P E f J l NPE(II + 1 S T I F F I N S 1 4 NSZ) * STIFF(NSI+NSZ) 2 0 5 CONTINUE
c c
-
t
EASAT(1,J)
.
5 5 0 0 GO TO 1 0 6 THE B A N D M A T R I X IS N O W F O R M E D F O R R F O U C T I O N WRITE BAN0 R A T Q I X I F L I S T 4 > 0
C
c
72R IF(I1.GT.O)GO C H2 C
.
IN
ISTIFF
IN
COPE
TO 6 0 1
NSANO*NP
IF(LIST8.LE.O)GO TO A 9 R 9 WRITE(3r2009) 2 0 0 9 FORMAT( 1 1 0 1 ' T H E RAND " A T Q I X W I T H 1 0 0 0 F A C T O R E D ' n I l ni = 1 R2 N8ANO MCOU NP*NBAHO On 3 0 5 I = 1.NP W R I T E ( 3 r Z O l O ) I r ( S T I F F 1 II?.;~.MI,MZ 1 2 0 1 0 F O R M A T ( ~ X I I ~ , ~ X I - ~ P ( ~ F 11 ~ " ~5 .X ~~ )- 3~ P ( 9 F l Z . 2 ) ~ / , 5 X 1 - 3 P ( 9 F 1 2 . 2 ) 1 1FIMZ.GE.NCOUIGO TO 8 8 4 9 11 = M2 t 1 M2 = M2 t NBANO r 3 0 5 CONTINUE 8 8 8 9 WRITE(3.20111MZ~NRANO 2 0 1 1 F O R M A T ( I / ~ T ~ I ' M O S T I F F ( 1 I ENTQIES ' ~ I ~ . ~ O X P ' S A NY TOD T H . ' r I 4 t l / l C C C NOTE--DO NOT READ P - U A T P I X E N T Q I E S COW UNFOQM LOADS OM 3 E I N S C DO NOT READ P - M A T Q I X E N T R I E S FOR FEM--INOUT SO COUPUTEQ C COHPUTES F E * 5 0 F I N A L MOMEMTS A Q E COQDECTEO FOP FEM * * * * * a *
--
.
c
TO 4 2 6 IF(NNZP.EO.OIGO 0 0 9_91 NN = 1,NNZP
c
READ( l r 1 0 1 0 l t ! l C 1 0 1 0 FORMATf1615) C
READ CARD******
READ(1,1011)(PRfLI~L*l~NLC) C
.
IF(II.GT.09GO TO 6 0 5 203 0 0 204 I = l,b 0 0 204 J = 1.6 EASAT(1.J) 0. DO 1 8 7 K 9 1 . 3 E A S A T f I # J ) * E A S A T f I p J ) + EAfI,K)*ESATfK.JI 1 8 7 CONTINUE
TO 2 0 6
c
1011 F O R M A T ( 8 F 1 0 . 4 1 OD 9 9 0 L = 1,NLC P(n1.L) = Pfn1.L) + P R ( L ) 9 9 0 CONTINUE 9 9 1 CONTINUE 4 2 6 CONTINUE DO 9 9 2 NS1 9 l r N P DO 9 9 2 N S 2 9 l r N L C 9 9 2 T F f N S l r N S Z l = 0.
,.
************~*****;~******T~***lOtOOO~OOO.**
L
4 0 6 URITE(3,2012)UNITl~UNIT2 2 0 1 2 F O R M A T ( / * Z X s ' T H E P-MATPIX, ' r A 2 r 1 AN0 ' , A 4 r / ) NS1 = 1 NS2 = 1 0 4 2 7 IF(NSZ.GT.NLCINS2 = NLC DO 4 0 8 I = ltNP 408 V P I T E ( ~ ~ ~ O ~ + ~ I ~ ( P ( I ~ J I ~ J J N S ~ ~ N S Z ~
I
5 1%,STRUCTURAL STEEL
DESIGN
-
c
i 0 1 4 F O R M A T l T 5 r 'MP ',13~1X,lOFll.2) IF(NS2.EQ.NLC)GO TO 4 2 8 N S l = NSZ + 1 I NS2 9 NS2 + 1 0 URITE13r2016)NSl 2 0 2 6 FORHbT(11,5Xr1THE P-MATRIX CONTINUED BEGINNING U I T H NLC GO TO 4 2 7 6 2 8 CONTINUE
C'
.
SUBROUTINE TO REOUCE THE RAND MATRIX CO N 1 1' 0080 N-1pNP I - N DO 7 0 L = ZtNBANO NL (N-l)*NBAND + L I = I+1 IFISTIFF(NLI.EQ.O.IG0 TO 7 0 8 STIFF(NL)lSTIFF(Nl) J - 0 00 68 K L,NBANO J = J + 1 I J = 11-l)*NBAND 4 J NK (N-lI*NBAND + K
-
-
C
67 70
66 80
C
C C
-
-
-
= P(NtM)
IFfSTIFF(NK).NE.O.)P(NrM)
ENO OF MATRIX
REDUCTION
c C C
F(Z)-F(Zl*3~/4. F(3l*F(31*3.14. IF(JJ.NE.IDLPLIG0 Tn 973 C PUT 0 t L I N T F - M A T Q I X 00 970 I 193 9 7 0 TF(YEMNO.11 * F(I)
--
-
SOLUTION
IS
IN
P(I.JI
1.58 CONTINUE I F ( 1TRUSS.EQ.O)GO
TO LATER A D O TO Y I V O L O I D COHOIT.
TO 1 8 9
'
4 3 1 CONTINUE
C
I F AOOEO TO WINO
STIFF(NK)*P(L*KpM)
I C 1 7 F O R M A T ( I I r 5 X r ' T H E X-HATRIXp ',A29 OR RADIANS',/) NS1 = 1 NS2 10 4 2 9 IF(NSZ.GT.NLCbNS2 = NLC DO 5 0 3 1 = lrNP 503 U R I T E ( ~ ~ ~ O ~ ~ ) I ~ ( P ( I ~ J ) P J * ~ ( S ~ , N S Z ) L O 1 8 FORMAT(6X,'NX 9 *rI3~1XrlUFll.5) IF(NSZ.EQ.NLC)GO TO 4 3 1 N S 1 = NSZ + 1 NS2 = NSZ 4 1 0 ~ ~ 1 T E 1 3 r 2 0 1 9 ~ ~ S l 2 0 1 9 F O R M A T ( l l r 5 X r ' T H E X-MATRIX CON1 D BEGINNING WITH NLC
.
REDUCE MOMENTS FOP WINO ALSO NECESSARY TO REDUCE 0 t L MOMENTS B Y 0 . 7 5
-
9 6 CONTINUE GO TO 8 5 \fi CONTINUE C
NOTE I F YOU PEA0 D-HATQIY ENTRIES FOR FE* THEN OUTPUT *UST 5 E AOJUSTEO BY HAHO TO ACCOUNT FOP F I X E D END MOMENTS DM 9 E A M ENDS
6 0 7 CONTINUE Fl2l-(F(ZI-FEMlI/FU(l)
INCLUDE LOAD MATRIX I N REDUCTION 00 67 H 1,NLC P(IsM) = P(IPM) B*P(NrM) CONTINUE M = 1,NLC 0 0 66 P(H,M) = P(NrM)lSTIFF(Nl) CONTINUE N 1 = N 1 4 NBAND COMPLETE SOLUTION BY BACK S U B S T I T U T I O N IF(N.LE.O)GO TO 9 0 L = N - 1 0 0 06 K = 2, NBANO NK * ( N - l ) * N B A N D 4 K 0 0 86 H 1pNLC
t
.
-
-
602 IF(ITRUSS.EO.O.OR.IWPITP.E~~~)WPITE(~~~O~OIJJPUNIT~~UNIT~ 2 0 2 0 F O R M A T ( I P 4 x 9 'LOADING CONOITION NO 9 ' r I 3 r l r 5Xr *MEMBERit 4 Y p ' A X I P L 1 37X. 'DESIGN END MOMENTS C0RRECTEO'r/, 7FOWCFs ' r A29 8Yr'FOR FFM AN0 WINO (NEAR FNO F I P S T l r ' , A L P I ~ , RFWINO 5 GO T O 5 5 5 6 0 5 P ( N P P 1 r JJ).O. FEU1 = F M I I J J I
INCREMENT COUNTERS I 1 AND J J - - J J
COUNTS NL
IF(NLC.EO.1)GO
TO 1 9 5
IF(F(1I.GT.PMAX(ME*NOIIDMA~(MFRNO)
IF(F(1I.LT.PMIN(MEHNOIIPMIN~~EMNOI
c
-
* F(11
TF(HEMNO.JJ) F(1) I F ~ I W R I T P . L E . O ) G O TO 4 4 1 WRITE 9AR FORCES FOR NLC
F(11
9
1 AND r V Q I T P
IF(MEHNO.EO.~IURITE~~PZOZ~~JJ FORCES [ K I P $ O Q K N I FOR NLC
2 0 2 5 FORHAT(5X,'BAR
.
'rI3*11
.
1
5U)
. . STRUCTURAL STEEL DESIGN
SELECfED COMPUTEX
A-2 LOAD MATRIX GENERATOR FOR AASHTO TRUC LOADING ON A TRUSS BRIDGE
JJ = JJ 4 1 GO T O 1 8 9 195 JJ 1
-
C
4
1 9 1 GO T n 5 5 5 1 9 2 I F ( I T R U S S . E ~ . O I G O TO 1 9 3 IF(NLC.EO.l)GO TO 1 9 3 REWIND I C I T R U S S = 1 AND N L C I S L A R G E U S E I W R I T 0 FP O R M A X AND IN B A R F O R C E S ONLY---USE I W R I T P = -1 F O R C O ~ P L E T E L I S T I N G B A R F O R C E S WHEN N L C IS L A R G E R THAN 1
.
C VRIT~(3,2033) 2 0 3 3 F 0 R M A T ( 1 1 ' 8 X * ' T ~ ~n A Y I n U n L I V E L O A D B A R F O R C E S & N O D E A D L O A D v A L U E 1S"lr5Y'qnEf4 N O ' P ~ X Y ' M A X L L 1 , 9 X y * M f N L ~ q , f j x ~ rL O ~ A~D ~ORo L ~ S TN L C
z',/T
0 0 447
I
. 7
lrNn
~ R I T ~ ( 3 r 2 0 3 5 ) I ~ P n ~ ~ ( ~ ) . ~ n ~ ~ ( ~ ) , ~ ~ ( ~ y ~ ~ ~ )
c
FORHAT(6X*T ~ ~ X ~ F ~ O . ~ Y ~ X , F ~ O . Z ~ ~ X , F ~ O . ~ ] 4 4 2 CONTINUE W R I T E B A R F O P C F n A T a I v IF I W R I T P -I IF( I W R I T P . N E . - ~ ) G O T O 1 9 3
WRITE ( 3 1 z 0 3 z ) ~ ~ 1 2 0 3 2 F O R n A T ( I I * 5 X ~ ' T B~ A~R F O R C E M A T R I X 4 4 3 DO 4 4 4 1 . 1 , y n
.
S T A R T I N G WITH
NLC
.
qy13y11,
OF
This program steps a movlng load consisting in either two or thre loads along a truss. At each step the panel loads are computed as with concentrated loads. These panel loads are the P-matrix entries f weightless truss to use in the frame analys~sprogram. The loads move from left to right. If the larger wheel loads ar truck is "backed" across the truss. The llmtatlons are: 1. The user must use at least two and not more than three loads2. The "step" must be an integer multiple of the panel length. Since this is computed, the step should be input suffic~entlyaccurately and smaller so that in computer truncation the correct integer is obtained. 3. All panels must be of the same length.
An impact factor is automatically computed and the panel 1 impact factor in the output. MET is used to compute the impac
444 ~ R I T E ( ~ ~ ~ ~ ~ ~ ) I I ( T F ( I ~ J I ~ J ~ N ~ ~ ~ N ~ ~ ) 2028 F O R ~ A T ( ~ X ~ I ~ , ~ X Y ~ D F ~ ~ . ~ ) IF(NSZ.EO.NLC) GO TO 1 9 3 NS1-NSZ41 NSE=NSZ+~O IF(NS2.GT.NLC)NSZ.NLC URITE(3r2032)NSl GO T O 4 4 3 1 9 3 GO T O 2 C 7 5 0 STOP END
Output is automatically punched onto cards m the correct format for input as the P matrix in the frame analysis program. The user may wish to an optional punch control to inspect the output prior to having it punche
Variable Identification A sample set of data cards are llsted ulth thls program for both an fps and an bridge truss using three wheel loads, seven truss panels (25 ft or 7 5 m), heels 5 ft or 1.5 m. The coding is such that the panel NP are 26,2, 6, . . . and NPPI = 26 here and is not punched number of wheel loads number of truss panels step length, ft or m increment of wheel movement, f t or m STEP 0 for fps; = 1 for SI bndges. This is used to proper1 MET impact factor. wheel loads. Note the order of using loads on the sample cards. U s h g 8. 32. 32. runs the truck forward on the bndge. cumulative wheel spaclng as 14. 28. for using 14 ft between each set of X(1) wheels. This spacing or the SI equivalent wdl usually be the most critical spacing. NP numbers in order from left to nght where the panel loads are Ml(1) placed. M1 has the same meanlng here as in the frame analysis program. Note that NPPl is not used. For a through deck truss, this section of the program requlres a sllght modiflcatlon to get loads onto the truss where NPPl may be, since the current method of use omits the flrst and last Ml(1) values (they are input for ~dentdication).
922 r
S T R U C T U R A L STEEL DESIGN
NP IS THE NUMBER OF LOADS THAT A R E WANTED ON THE TRUSS SL IS THE LENGTH OF EACH PANEL I N THE TRUSS NS I S T H E NUMBER OF PANELS I N THE TRUS STEP I S THE MOVEMENT THAT I S WANTED FOR THE LOADS BETWEEN EACH LOAD CONDITION P t I ) I S THE WEIGHT OF EACH LOAD I N THE S E R I E S X t I I I S THE O f S1 ANCE THE LOAO I S B E H I N D THE F I R S T L O A D MET=O.O FBS ~ 1 . 0 M E T R I C D I M E N S I O N T~MPt20~70)rREACTtZO170),Pt5lvXt5Is M l ( 2 0 1
C (
r-
'
1 FORMATt215r2Fl0.4,15)
i,
WRITEt3v5)NPrSL,NSvSTEP
5 F O R M A T t ' l N U M B E R OF LOADS ON T R U S S 1 ~ 2 X , 1 3 ~ / / ~ ' SPAN LENGTH O F EACH lPANEL'vFB.2 v//v' NUMBER O F PANELS I N TRllSS'v2X113. I/.' LENGTH OF
I
2MOVEMENT OF THE LOA0S1,F8.2 DO 2 0 I = l . Z O DO 2 0 J = 1 . 7 0 TEMP( I,J)=O.O 70 REACTtI.Jl=O.O DO 1 0 0 1 - 1 1 5 Xt Il=O.O L') P t Il=O.O NLS=SL/STEP MX=NP-1 NSPl=NS+1 NSMl=NS-1 FORMATt1OFf3.21
WRITE(~~~)~IIP~I)~I=~~NP) b FORMATt///.*
P'.Ilr'
=lrF8.3)
WRITE~~.~)~JIX(J)~J=I~NXI
I
I
'
,
REA0(1.81(MltI)rI*lrNSPl
r//)
R E A O ~ ~ ~ ~ ) ~ P ~ I ) ~ I = ~ ~ N P ) ~ ~ X ~ J ) I J = ~ ~ N X ) '
2 0 1 CONTINUE I F (XIMPAT.GT.0.30) XIMPAT-0.30 00 1 0 3 I-1.MZ W 1 0 3 J-1,NSPl 103 REACT~J~II=REACTIJ,II*~~+XIMPAT) WRITEt3r4) 4 FORMAT( I / / / / *LOAOm,/* ' CONOITIOW9.TZO.' THE R E A C T I O N S ' ) 0 0 1 0 6 J=11M2 a 1 0 6 W R I T E t 3 . 3 1 Jt ~ R E A C T ~ I I J ) ~ I - ~ , N S P ~ ) " *. t 3 FORMATtT5,13qlOt5XvF6.1)) PUNCH M l FROM L E F T TO R I G H T ACROSS fk T R U S S " C PUNCH FORMAT I S COMPATIBLE W I 7 H A'NALYSIS PROGRAM FOR DIRE$$ C
7 FORMATl///r9 X 1 1 I 1 v ' =',F8.31 SK=O.O SK2=0.0 DO 1 1 0 M = l v N P DO 1 0 9 I = l r N S IPl=I+l DO 1 0 8 J = l , N L S RR=P(M)*tJ*STEP-SK-SK2)/SL ALR=PtMI-RR
I F .LJ*SJEP+STEP .GT. SLI SK-SL-J*STEP JZ= tSL/STEPI*tI-11 J3=J+J2 TEMPERARY STORAGE OF R E A C T I O N VALUES TEMPtIvJ3)=A1R TEMPfIPlrJ3)-RR 1 0 8 CONTINUE 1 0 9 CONTINUE L PLACE CONSECUTIVE LOADS I N T H E I R PROPER P O S I T I O N IX=XtM)/STEP IST=IX*STEP SK2=Xt M I - I S T INT=O I F tM.EQ.1) GOT0 1 0 5 DO 1 0 7 I=lrNSP1 DO 1 0 7 J = l r J 3 n n i = n-1 INT=XfMMl)/STEP M2=J+INT C ADO TEMPERARY STORAGE TO PERMANENT STORAGE 107 REACTII~MZ)=REACT~IIM~)-TEMP~I~J~ GOT0 1 1 0 105 00 104 I=ltNSPl DO 1 0 4 J = l r J 3 1 0 4 REACTtI.J)=-TEMP(1.J) 1 1 0 CONTINUE IFtMET.EQ.1IGOTO 200 XIMPAT=5O./tNS*SL*125.1 GOT0 2 0 1 700 X I M P A T = ~ ~ . ~ ~ / ~ N S * S L + ~ ~ . ~ ~ I
)
,
FORMAT( 2 0 1 4 I 00 1 1 2 I=Z.NS WRITE(2v9)MltII FORMATt15)
8 9
112 W R I T E ~ ~ ~ ~ O ) I R E A C T I I ~ J I ~ J J ~ ~ M ~ I 10 FORMATtBF10.41 4000
STOP EN0
****
FOLLOWING 3 CARDS ARE SET FOR FPS OUTPUT 5. 0 7 25. 14. 28. 32. 32. 8. 26 2 6 1 0 1 4 1 8 22 26 F O L C M I N G 3 CARDS ARE SET FOR S I OUTPUT C 1.5 1 3 7 7.5 36.0 142.0 142.0 4.25 8.5 26 2 6 1 0 1 4 1 8 22 26
C
3
****
I*
INPUT
5k-4smUclVbU.. STEEL DESIGN "ft
SELECTED COLLPUTER. PROG
,$i
A-3 LOAD MATRIX GENERATOR FOR LOADING ON A TRUSS BRIDGE n l s program steps Cooper's E-80 train lo (DX) the panel loads are computed as a loads are the P-matrix entries for a weight1 Program. Note that the program does not bl;idge,pnd only the uniform load is on. The only limitation to this program is The Cooper's E-80 loads and wheel sp rated into the program as DATA FFP/
1 FaR M E T R I ISUIT NO OF SPANS OR GIRDER SEGMENTS: OX = INCR. OF LOAD MOVEMENT. F T OR M; SPAN LEN FAC = L O I D R A T I O I F F P = € 8 0 LOAD TRUSS SPAN OR GIRDER SEGMENT: F A C = 0.75 FOR E ~ W - 0 . 9 FOR E72--1.00 FOR E B O - - L . ~ ~F~~ ~ ~ 1 1 0 NOSPAN
c c c
=
= 0 FOR FPS:
WIOH
IWRIT
CENTER T O CENTER SPACING OF T R U S S E S SELECTED COMPUTATIONS FOR DEBUGGING
= 1 TO WRITE
'
n
Variable Identification (see sample data cards listed at end of program listing) TITLE up to 80 columns of alphanumeric data for probl NOSPAN number of panels ISWIT 0 for fps; = 1 for SI prob spacing to SI and to IPUNCH 0 for not punching O U ~ P U = 1 to punch output in f frame analysis program for no impact factor; to write selected inter am; = 0 when pro crement of wheel movement left to right, f t or m ISPAN panel length, ft or m WIDH width between two trusses making up bridge an computing the impact factor, ft or m FAC factor to convert E-80 to E-110, E-60, etc.; use F E-80 loading NPs(J) NP numbers (same a established (note that these entries do not inc cation may be required for a through deck b
w.
@ i
i$ %
"r,Tt,
~ E ~ 0 1 1 ~ 1 0 0 2 1SPAN,UIOHHFAC 0 X ~
'
1 0 0 2 FORMAT(BF10.41 N P S ( J ) = NP N U M B E R S F O R NON-ZERO
*-MATRIX
ENTRIES
-
NSPN = NOSPAN 1 REAO(lr100611NPSIJI .J=lqNSpNI 0 6 FORHATl1615) I F 1 1SWIT.GT.OIDIS = 25. 8 = NOSPAN TOTSPN = B*SPAN SET WHEEL LOAOS AND WHEEL SPACING SUHXL = 0. 0 0 5 I-1.18 X I 1 I = FFXI I 1 IF(ISWIT.GT.OIXII I = x111*0.3048 P I 1 I = FFPI II*FAC F I( ISWIT.GT.OIPII I = PlI)*4.44822 SUMXL = SUHXL + XI T I 5 CONTINUE
7 PAC = 1. + PAC ~ ~ ~ ~ ~ ~ ~ , ~ ~ ~ ~ I T I T L E . N P , H O ~ P A N , O X , S P A N ~ ~ L O A O ~ S U ~ X L ~ P A C ~ F A
2000
F O R ~ A T ( / / / ~ ~ X , Z O A ~ , I , ~ X ~I ' N* PV I ~ V ~ X * ' N O S P = ~ ' N* 1 5 * 3 X * 1 2 x 1 I D X = o r ~ 7 . ~ l / r 5 ~ , ' S P A N LENGTH = ' . F 8 . 3 r 3 X v ' U N I F LOAO/LENGTH 2~,~6.l, /, ~ X , * S U H OF X ( L 1 OF CONC. LOADS a ' t F ~ . Z I ~ X V ' I M P & C T ~ A C T D R= * , ~ 6 . 3 , / , 5 ~ , IFACTOR FOR E-LOADS--1.0 FOR E - 8 0 = ' * 465.3, ~ x , S O I S T C-TO-C OF TRUSSES 'rF7-31//)
5Ad S T R U C T U R A L STEEL DESIGN URITE13r2005~1P(Il,I=1,NP) = 'rl2F9.l~lrl2X.
,145 F O R M A T I S X v ' P l I i
*
WRI T E ( 3 r 2 0 0 6 ) 2906 FOPMAT(lOX.'***
r
4 I.,
E
N
6F9.1.//)
LOAOS M I 0 IMPACT--USE
*
,
FOR FLOOR BEAM D E S I G N AN0 CH
F I N O NO OF LOAOS IN. CURRENT SPAN
.TOTSPNIXILI IFlL.EQ.NP+1.WO.AOXH.GT-TOTSPNIXIL)
IFIL.GT.NP+l.OR.XlLI-GTTSPANIXIL)
'
j4$!JZSTOP
= TOTIP. + SUMXL + o x 0 0 8 3 K=1,20 DO 7 3 L = 1 . 2 1 0 7 ' PR1K.L) = 0, ~ ' iCONTINUE
i
DO 9 5 L * NCIZOO IFIL .EQ.NP+I.ANO.AOXH.LE
-
= Z*SPAN SUMXL = O I F F + Z*SPAN SPAN
-
SWXL
IFlXfL)..LT.O.tGO TO 1 4 8 IF(XIL).NE.SPANIGO TO 2 2 0 1 = SPAN DZ = 0. 2 2 sun1 = sun1 + XILI IFlSUMl.GE.01-0.0001)GO TO 9 6 IFIL.EO.NP+lIZZX = SPAN (OZ + 0 . 0 0 5 ) IFIL-€O.NP+1~ANO.SUMl.GE.ZZXlGO TO 9 6 9 5 CONTINUE 96 C O N T I N U E SUM3 = SUM3 + SUM1 TOTSUM = SUM3 + SUM2 N C = L * l 02 = 01 RR = 0. R L = 0. I F 1 IWRIT.LE.OIG0 TO 5 5 0 u R I T E I ~ ~ ~ ~ ~ ~ ~ T O T S P N ~I MNr C 0~2 A 1 O O lXt HS~U M l t O Z t NNv L.SPAN.SUH 1 7 SUM21 O I F F t T O T S U M v J J v X 1 1 9 1 ~ X 1 2 0 ) ~ X 1 2 1 1 t X f L ~ 3 0 0 5 FORMATl3X.'TOTSPN = 1 , F 8 . 1 , 3 X ~ e N C = ' 1 1 5 r 3 X , 1 A 0 X X H = *,F7.3, 2 3 X V 1 1 M = l.15, 5 X v e O Z = 1.F7.312X.'01 = '~F7.3,' SUM1 *@,F7.3. 3/r5X* 'OZ a ' ~ F 7 . 3 r 3 X ~ ' N N - ' ~ 1 5 . 2 X , v L ~ 1 , 1 5 t 2 X T * S P A N = ~ , F 7 . 3 , 4 3 x 1 'SUM3 = ',F7.3,3X,'SUMZ = .F7.3r3X1'01FF~',F7.3,3Xv 'TOT 5'rF6.lr2Xv'JJ~',IZ1. /r5X1 'XI9 ~ ' ~ F 7 . 3 ~ 3 X ~ e X 2 0 = 1 ~ F 7 . 3 3 3 X ~ * X 2 1 = * 7 6F7.3v3XvQXL='.F7.31II CONTINUE ACCUMULATE EFFECTS OF LOAOS I N U I Y SPAN 0 0 9 9 L L = NN1L IFILL.LE.NP)GO TO 9 7 P I L L ) = WLOAO*XlLLl IFILL.EQ.NP+l.OR.NN.EO.L)OZ = 02 X1LL)IZ. 9 7 RR = O Z * P f L L ) / S P A N RL = P I L L 1 RR PTIKC) = PTIKCI RR PTIKC-21 = PTIKC-21 - RL 02 = 02 XILLI SUMP = SUMP + P I L L ) 9 9 CONTINUE SUMPT = SUMPT + P T I K C ) 1FlJJ.EO.IM)SUMPT = SUUPT + P T I K C - 2 ) Z = 1M JJ OZ = Z*SPAN + TOTSUM AOXH NN = L L SUM1 = 0. 1 0 2 CONTINUE SO CAN WRITE ONLY V E R T I C A L VALUES AT PANEL P O I N T S I H O R I Z = 0.) U S E P R I N T E D OUTPUT TO OESIGN TRANSVERSE FLOOR BEAMS BASED ON LARGEST PANEL LOAD * FLOOR BEAM REACTIONS AN0 DEPENDING ON NUCBER OF TRACKS ON BRIOGE OOES M T I N C L U D E ANY IMPACT FACTOR *a*+****+ WRITEl3rZOO8)I~lPTIJl~J,Z~NY~2~ 2 0 0 8 FORMAT(//.2Xv15v12F10.2v/, 5X,12F10.2,/1 WRITEI~~~OO~ISUMP~SU*PT 2 0 0 9 F O R M A T f 5 0 X , ' S U M TRUSS LOAOS = ' , F 1 0 . 2 . 3 X , 'SUM NODE LOAOS = * , 1F10.2,/) 0 0 1 3 8 J = 1,NY PRIJIII = PTIJI 1 3 8 CONTINUE 1 4 0 CONTINUE 141 U R I T E I 3 ~ Z O l l ) I 2 0 1 1 FORMATI/I~SXI 'ONLY UNIFORM LOAO ON BRIDGE. I = ' , I 3 , / / ) IFlIPUNCH.LE.OIG0 TO 1 4 9
-
DO 1 4 0 A
=
I = 1.210
M A I N DO LOOP FOR INCREMENTING LOADS
l
D I F F = 0. ADX = A*OX AOXH AOX C
******
1
TO STOP COMP T A T I O N S WHEN ONLY UNIFORM LOAD ON B R I D G E IFIA0X.GT.ZSTOP GO TO 1 4 1 I A O X = TOTSPN SUM = 0. IFI AOX.GE.TOTSP
Y
F I N O NO OF SPANS TO USE , 5 SPUSO = A O X l S P A N
C
c
B
ZERO DO 8 6
b
CONTINUE
I
MATRICES ARBITRARY K = 19.200
AMOUNTS
X I K I = 0. 1
L
i i
I b
4..
0 0 8 8 K = 1,50 T I K I = 0. CONTINUE I M = SPUSO RM = I M IFI8M.LT.SPUSOIIM = IM + 1 BM = I M 1FIAOXH.GT.TOTSPN)OIFF = AOXH ;UM2 = 0. ? F I A B S ( O I F F ~ . L E . O . O O ~ ) G ~ TO 1 5
DO 1 2 MM = l v N P >OM2 = SUMZ + x t n n ) :~lSUM2.GE.OIFFlGO TO 1 5 1 2 CONTINUE t K NC = tin + 1 fin = I M 6 7 = BM*SPAN AOXH + SUM2 'WR = NC LUM1 = 0. 5UM3 = 0. cUHP = 0. \UMPT = 0. I~~~~.EO.NP.ANO.OZ.LE.O.)OZ
-
-
i!
-
TOTSPN
-
-
3 1 = SPAN OZ = 2*lIM-JJ) 7 = JJ
KC
+ 4
-
**
= 0.
FOR NUMBER OF SPANS U S E 0 FOR CURRENT LOAD P O S I T I O N S J J LOOP = 1.TM
DO 1 0 2
-
-
-
C C
.
In1 = I- 1
528 STRUCTURAL STEEL DESIGN C
142 143 25
2014 145 149 148 2016 150
**
C ...,,,..
4
4
**
J E 9 1.05 4
***I,
FOLLOWIHG 4 CARDS T Y P I C A L SET O F F P S DATA FOR 9 PANEL TRUSS AREA B R I O G E FOR T E X T U S I N G COOPER E - 0 0 L O A D I N G AN0 I 0' 1 0 27.60 17.00 1.0 8 12 16 20 24 28 32 F O L L O W I N G 4 CARDS T Y P I C A L S E T O F S 1 0 A T A . F O R 9 P A N BOULES AREA B R I D G E FOR T E X T U S I N G COOPER'Ei80 LOAOIN 1 1 1 0 8 -40 5.20 8 12 i l l 20 2 4 lo& 32
J E BOYLES 3 . ,... 9 8 ..... o...
3.45 C
A P P L Y I M P A C T FACTOR TO LOADS I F I M P > o IF(IMP.LE.O)GO TO 2 5 DO 1 4 3 J = 1 v I M l DO 1 4 2 K K = Z v N Y v 2 PR(KK.J) = PR(KK,J!*PAC b, CONTINUE KK = 2 DO 1 4 5 J = l r N S P N KK = K K + 2 JJ = N P S ( J ) WRITE12vlOOb)JJ WRITE(2v2014)(PR(KKvM)vM=l,W FORMAT(BF10.4) CONTINUE GO TO 5 0 0 0 WRITE(3r2016) FORMAT(II/v5X***** PROBLEM T E R M I N A T E D - - X ( L ) I S NEGATIVE STOP EN0
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AASHTO column formulas, 263 with bending, 32 1 ' AASHTO minimum thickness of metal, 491 AISC column f o h u l a , 262 with bending, 3 18 Angle tension members, 777 Approximate analysis of frames, 54 cantilever method, 55 portal method, 54 AREA column formulas, 264 . with bending, 32 1 A d E A minimum thickness of
plates, column, 285 bending stresses in, 287 design criteria for, 287
BASLER, K.,
474, 479, 48 1
Beam analysis, 57 biaxial bending, 164
Beam analysis: differential equation, 58 for unsymmetrical bending. 167 Beam columns, 297 design formulas: XXSHTOand AREA, 311 AISC, 313, 318-319 design methods, 322 effective length of, 303, 306 G factors for, 306 inelastic eRect/reduction, 307 K factor chart, 306 Beam framing connections, 385 Beam stresses, 146 allowable: Ar\SHTO and AREA, 15 1 AISC, 149, 174 laterally unsupported, 174 bending, 146 biaxial bending, 164 elastic design, 148 shear, 146, 159 Beams: compact section criteria, 149 deflections, 146 laterallv unsu~norted.174
tearing of welds, 420 .' F tamella ,
Lateral bracing: . for beams, 150, 155, 174 for columns, 291 for girders, 496 knee bracing, 497 Live loads, 18 reduction for, 19 Load conditions, 70 Load resistance factor design (LRFD), 132 table of,@factors, 132 Loads, 18 bridge: AASHTO, 26, 27 AREA, 28 equations for shear and moment, 26 impact, 32 tabulated shear and moment, AREA, 29 dead, 18 earthquake, 33 hi: eccentric, on fasteners, 375 impact, 32 live, 18, 19 ponding, 25 snow, 24 map for, 25 , wind,21,31 Low-temperature effects, 13 Llr ratio: compression members, 258, 260, 277 tension members, 231, 237, 238, 242 LRFD, 132 beam column desig*, 338 beam design, 208 column design, 292 connections, 405 tension member design, 248 Lug angles, 3,65.. Mass density, steel, 7
tdius of gyration., 175, 237, 255,
MUNSE, W . ,
364
embers, tables, 272 ompression flange of beams, 179 Euler formula used in, 260 lacing for, 276 alh, 473, 478, 494 blt, 150, 465, 47 I , 490, 493
Net area: column, 264 in tension, 219, 225, 232 effective hole diameter for;233 at thre'dd root, 219 use of s2/4g, 233 N E W L I N , D. E., 427 NEWMARK, N. M., 185
hlt, 456, 48 1 , 49 1 Llr, 263, 27 1 , 275, 307 Reaction distance for beams, 157 bracing for, 158 Reduced eccentricity of fasteners.
Plastic section modul
stresses: bendin
AISC specifications fo
, computation: for beams, 146,
joint length as factor, 360 Shear stress, allowable: in beams,
S1 conversion factors. 39 Sidesway, control of, 337 Slot weld, 414, 419 AISC specifications for, 4 Spacing of bolts, 363 Stepped columns, 329 K factor for, 33 1 Stiffeners, bearing, 158, 37 493 column flange, 427 web. 4'76
diagonal, for corner connections, St6 glrder, requlred for, 456 ~ntermediate,479, 493, long~tudinal,AASHTO, moment of Inertia of, 480, spacing for, 493,495 seat angks for, 436 St~ffnebsmethod of analysis, 54 Stress, allowable: base plate des 287 bending, 149, 174, 469, 471 4 compression: AASHTO, 2 AISC, 261 AREA, 26-1. fasteners, tables, 353, 3 366-368 interaction, 3 12 shear, 146, 159 in girder webs, 478, 492 tension, 219 table of values, 221 weld, 416 Stress range: defined, 40 table of values, 41 Stress-strain curves, 1 1 , 113 Structural codes, list of, 17 Structural shape data, table, Structural shapes, 7-10
536
INDEX
*
Tapered flange beam, 182 Temperature coefficient, 5 Temperature e r s u s strength, 13 Tension stress&, 219 allowable values, 219 effective net area for, 225 Threaded area in tension, 2 19 table of design data for, 230 THURLIMANN, B., 474 rIMOSHENK0 and GOODIER, 133 Torsio&pnstant, J, 171, 174 rruss analysis, 59
$.% $& ', @j "able
($.,&"!;I &,
d ,
.
*'
3
Ult~rnatestrength. 7 of values, 6 Unhraced length, I75 LL'.1-51, 323 L", 151 Unlt welght of steel, 7
. VALLAERT, J . , 368
Warping constant C, , 174