Structural Concepts and Systems

Draft DRAFT

Lecture Notes in: STRUCTURAL CONCEPTS AND SYSTEMS FOR ARCHITECTS

Victor E. Saouma Dept. of Civil Environmenta Environmentall and Architectural Architectural Engineering University of Colorado, Boulder, CO 80309-0428

Draft Contents 1 INTRO INTRODUC DUCTIO TION N 1.1 Scien Science ce and Tech echnology nology . . . . . . 1.2 Str Struct uctura urall Engi Enginee neerin ringg . . . . . . . 1.3 Struc Structures tures and their Surr Surroundi oundings ngs 1.4 Arc Architec hitecture ture & Engin Engineerin eeringg . . . . 1.5 Arc Architec hitectural tural Desig Design n Process . . . 1.6 Arc Archit hitect ectura urall Desi Design gn . . . . . . . . 1.7 Str Struct uctura urall An Analy alysis sis . . . . . . . . . 1.8 Str Struct uctura urall Des Design ign . . . . . . . . . . 1.9 Load Trans ransfer fer Mec Mechanis hanisms ms . . . . 1.10 Struc Structure ture Types Types . . . . . . . . . . 1.11 Struc Structural tural Engineering Engineering Courses . . 1.12 Refe Reference rencess . . . . . . . . . . . . . .

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2 LOAD ADS S 2.1 In Introd troduct uction ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Vert ertica icall Loa Loads ds . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2. 2. 2.11 De Dead ad Lo Load ad . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 2. 2.2 Li Liv ve Lo Load adss . . . . . . . . . . . . . . . . . . . . . . . . E 2-1 Liv Livee Load Load Reduc Reduction tion . . . . . . . . . . . . . . . . . . . 2.2. 2. 2.33 Sno now w. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.33 La 2. Late tera rall Lo Load adss . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3. 2. 3.11 Win ind d . . . . . . . . . . . . . . . . . . . . . . . . . . . E 22-22 Wi Wind nd Lo Load ad . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 2.3 .2 Ear Earthq thquak uakes es . . . . . . . . . . . . . . . . . . . . . . . . E 2-3 Ear Earthq thquak uakee Load on a Fram Framee . . . . . . . . . . . . . . E 2-4 Earth Earthquak quakee Load on a Tall Tall Building, Building, (Schueller (Schueller 1996) 1996) 2.44 Ot 2. Othe herr Lo Load adss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 2.4 .1 Hy Hydro drosta static tic and Ear Earth th . . . . . . . . . . . . . . . . . . E 2-5 2-5 Hy Hydro drosta static tic Loa Load d. . . . . . . . . . . . . . . . . . . . . 2.4.2 2. 4.2 Th Ther erma mall . . . . . . . . . . . . . . . . . . . . . . . . . . E 2-6 Ther Thermal mal Expansion/S Expansion/Stres tresss (Schuelle (Schuellerr 1996) . . . . . . 2.4.3 2. 4.3 Br Brid idge ge Lo Load adss . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 2. 4.4 Im Impa pact ct Lo Load ad . . . . . . . . . . . . . . . . . . . . . . . 2.5 Other Importa Important nt Consi Considerati derations ons . . . . . . . . . . . . . . . . 2.5.1 2.5 .1 Loa Load d Com Combin binatio ations ns . . . . . . . . . . . . . . . . . . . . 2.5.2 2.5 .2 Loa Load d Plac Placeme ement nt . . . . . . . . . . . . . . . . . . . . .

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Flexur Flex uree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Basic Kinem Kinematic atic Assum Assumption; ption; Curv Curvature ature . . . . . . . . . . 5.4.2 5.4 .2 Str Stress ess-St -Strai rain n Rel Relatio ations ns . . . . . . . . . . . . . . . . . . . . 5.4.3 Int Internal ernal Equili Equilibrium brium;; Sectio Section n Proper Properties ties . . . . . . . . . . 5.4.3 .3..1 ΣF x = 0; Neutral Axis . . . . . . . . . . . . . . . 5.4.3 .3..2 ΣM = 0; Moment of Inertia . . . . . . . . . . . 5.4.4 5.4 .4 Bea Beam m Form ormula ula . . . . . . . . . . . . . . . . . . . . . . . . E 5-10 Design Example . . . . . . . . . . . . . . . . . . . . . . . 5.4.5 5.4 .5 Ap Appro proxim ximate ate Ana Analys lysis is . . . . . . . . . . . . . . . . . . . . E 5-11 Approximate Analysis of a Statically Indeterminate beam

6 Case Study Study II: GEORGE GEORGE WASHIN WASHINGTON GTON BRIDGE BRIDGE 6.11 Th 6. Theo eory ry . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.22 Th 6. Thee Ca Case se St Stud udy y. . . . . . . . . . . . . . . . . . . . . . 6.2.1 6. 2.1 Ge Geom omet etry ry . . . . . . . . . . . . . . . . . . . . . 6.2. 6. 2.22 Lo Load adss . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 6.2 .3 Cab Cable le Forc orces es . . . . . . . . . . . . . . . . . . . 6.2.4 6. 2.4 Re Reac actio tions ns . . . . . . . . . . . . . . . . . . . . .

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7 A BRIEF HISTOR HISTORY Y OF STRUCTURAL STRUCTURAL ARCHITE ARCHITECTURE CTURE 7.1 Bef Before ore the Gre Greeks eks . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.22 Gr 7. Gree eeks ks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.33 Ro 7. Roma mans ns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Medie Mediev val Per Period iod (477-149 (477-1492) 2) . . . . . . . . . . . . . . . . . . . 7.5 The Re Renai naissa ssance nce . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 7.5 .1 Leo Leonar nardo do da Vinc Vincii 14521452-151 15199 . . . . . . . . . . . . . . . . 7.5.2 7.5 .2 Bru Brunel nelles lesch chii 137 1377-1 7-1446 446 . . . . . . . . . . . . . . . . . . . . 7.5.3 7.5 .3 Albe Alberti rti 1404 1404-14 -1472 72 . . . . . . . . . . . . . . . . . . . . . . 7.5.4 7.5 .4 Pa Pallad lladio io 150 1508-1 8-1580 580 . . . . . . . . . . . . . . . . . . . . . . 7.5. 7. 5.55 St Stev evin in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.6 7.5 .6 Gal Galile ileoo 1564 1564-16 -1642 42 . . . . . . . . . . . . . . . . . . . . . . . 7.6 Pre Modern Modern Period, Period, Seve Seventee nteenth nth Cen Century tury . . . . . . . . . . . . . 7.6.1 7.6 .1 Hook Hooke, e, 1635 1635-17 -1703 03 . . . . . . . . . . . . . . . . . . . . . . . 7.6.2 7.6 .2 New Newton ton,, 164 1642-1 2-1727 727 . . . . . . . . . . . . . . . . . . . . . . 7.6.3 Bern Bernoulli oulli Family 1654-1 1654-1782 782 . . . . . . . . . . . . . . . . . 7.6.4 7.6 .4 Eul Euler er 170 1707-1 7-1783 783 . . . . . . . . . . . . . . . . . . . . . . . 7.7 The pre-Mode pre-Modern rn Period; Period; Coulom Coulomb b and Nav Navier ier . . . . . . . . . . . 7.8 The Modern Per Period iod (1857-P (1857-Presen resent) t) . . . . . . . . . . . . . . . . . 7.8.1 Stru Structure ctures/Mec s/Mechanic hanicss . . . . . . . . . . . . . . . . . . . . 7.8.2 7.8 .2 Eiff Eiffel el Tow ower er . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.3 7.8 .3 Sul Sulliv livan an 1856 1856-19 -1924 24 . . . . . . . . . . . . . . . . . . . . . . 7.8.4 7.8 .4 Roeb Roebling ling,, 180 1806-1 6-1869 869 . . . . . . . . . . . . . . . . . . . . . 7.8.5 7. 8.5 Ma Maill illar artt . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.6 7.8 .6 Ner Nervi, vi, 189 1891-1 1-1979 979 . . . . . . . . . . . . . . . . . . . . . . . 7.8. 7. 8.77 Kha han n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.8 et al. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Victor Saouma

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Structural Concepts and Systems for Architects

Draft CONTENTS

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12 PRESTRESSED CONCRETE 12.1 Intr Introduction oduction . . . . . . . . . . . . . . 12.1.1 Mater Materials ials . . . . . . . . . . . 12.1.2 Pres Prestress tressing ing Forces Forces . . . . . . 12.1.3 Assu Assumption mptionss . . . . . . . . . 12.1.4 Tendon Configur Configuration ation . . . . 12.1.5 Equiv Equivalen alentt Load . . . . . . . 12.1.6 Load Defor Deformation mation . . . . . . 12.2 Flexu Flexural ral Stresses Stresses . . . . . . . . . . . E 12-1 Prestressed Concrete I Beam 12.3 Case Study: Study: Walnu alnutt Lane Bridge . . 12.3.1 Cross Cross-Sect -Section ion Properties Properties . . . 12.3.2 Pres Prestress tressing ing . . . . . . . . . . 12.3.3 12. 3.3 Loa Loads ds . . . . . . . . . . . . . 12.3.4 Flex Flexural ural Stress Stresses es . . . . . . .

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13 ARCHES and CURVED STRUCTURES 13.1 Arc Arches hes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1.1 Static Statically ally Determi Determinate nate . . . . . . . . . . . . . . . . . . . . . . . . . E 13-1 Three Hinged Arch, Point Loads. (Gerstle 1974) . . . . . . . . . . E 13-2 Semi-Circular Arch, (Gerstle 1974) . . . . . . . . . . . . . . . . . . 13.1.2 Static Statically ally Indeterm Indeterminate inate . . . . . . . . . . . . . . . . . . . . . . . . E 13-3 Statically Indeterminate Arch, (Kinney 1957) . . . . . . . . . . . . 13.2 Curv Curved ed Space Structures Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-4 Semi-Circular Box Girder, (Gerstle 1974) . . . . . . . . . . . . . . 13.2.1 13. 2.1 The Theory ory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.1. 13. 2.1.11 Geo Geomet metry ry . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.1.22 Equili 13.2.1. Equilibrium brium . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)

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14 BUILDING STRUCTURES 229 14.1 Intr Introduction oduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 14.1.1 Beam Column Column Connectio Connections ns . . . . . . . . . . . . . . . . . . . . . . . . . . 229 14.1.2 Beha Behavior vior of Simple Frames Frames . . . . . . . . . . . . . . . . . . . . . . . . . . 229 14.1.3 Ecce Eccentr ntricity icity of Applied Applied Loads . . . . . . . . . . . . . . . . . . . . . . . . . 230 14.2 Buildin Buildings gs Structures Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 14.2.1 Wall Subsystem Subsystemss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 14.2.1.11 Examp 14.2.1. Example: le: Concr Concrete ete Shear Shear Wall Wall . . . . . . . . . . . . . . . . . . . 233 14.2.1.22 Examp 14.2.1. Example: le: Trusse russed d Shear Wall Wall . . . . . . . . . . . . . . . . . . . 235 14.2.2 Shaf Shaftt Systems Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 14.2.2.11 Examp 14.2.2. Example: le: Tube Subsy Subsystem stem . . . . . . . . . . . . . . . . . . . . . 23 236 14.2.3 Rigid Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 14.3 Appr Approxim oximate ate Analysis of Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . 238 14.3.1 Vertic ertical al Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 14.3.2 Horiz Horizonta ontall Loads Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 14.3.2. 14. 3.2.11 Po Porta rtall Method Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 E 14-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads243 Loads 243 14.4 Later Lateral al Deflections Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Victor Saouma


Draft List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

Types of of Forces Forces in Structura Structurall Elements Elements (1D) (1D) . Basic Bas ic Aspec Aspects ts of Cab Cable le Syst Systems ems . . . . . . . . Basic Bas ic Aspec Aspects ts of of Arc Arches hes . . . . . . . . . . . . Types Ty pes of Trus russes ses . . . . . . . . . . . . . . . . Variati ariations ons in Post Post and and Beams Configu Configuration rationss Differe Diff erent nt Bea Beam m Types Types . . . . . . . . . . . . . Basic Bas ic Form Formss of Fram Frames es . . . . . . . . . . . . . Examples Examp les of Air Suppor Supported ted Struc Structures tures . . . . Basic Bas ic Form Formss of She Shells lls . . . . . . . . . . . . . .

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2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18

Approximation Approxim ation of of a Series Series of Closel Closely y Spaced Spaced Loads Loads . . . . . . . . . . . . . . . . . 30 Snow Sno w Map of of the Uni United ted Stat States, es, ubc ubc . . . . . . . . . . . . . . . . . . . . . . . . . 33 Loads Loa ds on on Project Projected ed Dime Dimensi nsions ons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Vertic ertical al and Normal Normal Loads Acting Acting on Inclined Inclined Surfac Surfaces es . . . . . . . . . . . . . . 34 Wind Win d Map of the the Unite United d States, States, (UBC (UBC 1995) 1995) . . . . . . . . . . . . . . . . . . . . 35 Effect of of Wind Load Load on Structu Structures(S res(Sch chueller ueller 1996) . . . . . . . . . . . . . . . . . 36 Approxim Appr oximate ate Design Wind Pressure Pressure p for Ordinary Ordinary Wind Force Force Resisting Resisting Building Building Structures Structures 39 Vibrat Vib ration ionss of a Build Building ing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Seismi Sei smicc Zones Zones of the Unite United d States, States, (UBC (UBC 1995) 1995) . . . . . . . . . . . . . . . . . . 41 Earth and Hydrostatic Hydrostatic Loads on Structures Structures . . . . . . . . . . . . . . . . . . . . . 47 Truc ruck k Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Load Placement Placement to Maxim Maximize ize Moments Moments . . . . . . . . . . . . . . . . . . . . . . . . 51 Load Life of a Struc Structure, ture, (Lin and Stotesbury Stotesbury 1981) . . . . . . . . . . . . . . . . 51 Concept Conce pt of Tribut ributary ary Areas for Struc Structural tural Member Loading Loading . . . . . . . . . . . . 52 One or Two Way Way actions actions in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 Load Transf Transfer er in R/C Build Buildings ings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Two Way Way Actions Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Example Examp le of Load Transfer Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.1 3.2 3.3 3.4 3.5 3.6 3.77 3. 3.8

Stress Strain Stress Strain Curves Curves of Concrete Concrete and Steel Steel . . . . . . . . . . . Standa Sta ndard rd Rol Rolled led Sec Sectio tions ns . . . . . . . . . . . . . . . . . . . . Residual Resid ual Stres Stresses ses in Rolled Rolled Sectio Sections ns . . . . . . . . . . . . . . Residual Resid ual Stress Stresses es in Welded Secti Sections ons . . . . . . . . . . . . . . Influence Influe nce of Residual Residual Stress Stress on Average Average Stress-St Stress-Strain rain Curve Curve of Concre Con crete te mic microcr rocrac ackin kingg . . . . . . . . . . . . . . . . . . . . . W and and C sec secti tion onss . . . . . . . . . . . . . . . . . . . . . . . . . prefab pre fabric ricate ated d Steel Steel Joi Joists sts . . . . . . . . . . . . . . . . . . . .

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7.12 7.13 7.14 7.15 7.16 7.17

Experimental Experiment al Set Up Used by Hooke Hooke . . . . . . . . . . . . Isaac Newton Newton . . . . . . . . . . . . . . . . . . . . . . . . . Philosophiae Naturalis Principia Mathematica , Cover Page Leonhard Leonh ard Euler Euler . . . . . . . . . . . . . . . . . . . . . . . . Coulomb Coulom b. . . . . . . . . . . . . . . . . . . . . . . . . . . . Nervi’s Nerv i’s Palazetto Palazetto Dello Sport . . . . . . . . . . . . . . . .

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

Magazzini Generali; Magazzini Generali; Overall Overall Dimensio Dimensions, ns, (Billington (Billington and and Mark 1983) 1983) . . . . . . . 1 5 8 Magazzini Magazz ini Generali; Generali; Support Support System System,, (Billington (Billington and Mark 1983) 1983) . . . . . . . . . 158 Magazzini Magazz ini Generali; Generali; Loads (Billin (Billington gton and and Mark Mark 1983) . . . . . . . . . . . . . . . 15 159 Magazzini Magazz ini Generali; Generali; Beam Beam Reactions Reactions,, (Billington (Billington and Mark Mark 1983) 1983) . . . . . . . . . 159 Magazzini Magazz ini Generali; Generali; Shear Shear and Moment Moment Diagram Diagramss (Billington (Billington and Mark Mark 1983) . . 160 Magazzini Magazz ini Generali; Generali; Interna Internall Moment, Moment, (Billington (Billington and Mark Mark 1983) . . . . . . . . 1 6 0 Magazzini Magazz ini Generali; Generali; Similarities Similarities Betwee Between n The Frame Frame Shape and its Moment Moment Diagram, (Billington (Billington and Magazzini Magazz ini Generali; Generali; Equilibrium Equilibrium of Forces Forces at the Beam Beam Support, (Billingto (Billington n and Mark 1983)161 1983)161 Magazzini Magazz ini Generali; Generali; Effect Effect of Lateral Lateral Supports, Supports, (Billington (Billington and Mark Mark 1983) . . . 162

9.1 9.2 9.3 9.4 9.5

Load Life Load Life of a Struc Structur turee . . . . . . . . . . . . . . . . . . . . . . . . . . Normalized Norma lized Gauss Gauss Distribution, Distribution, and Cumulativ Cumulativee Distribution Distribution Functi Function on Frequ requency ency Distr Distributio ibutions ns of Load Q and Resistance R . . . . . . . . . . Definit Defi nition ion of of Reliab Reliabilit ility y Index Index . . . . . . . . . . . . . . . . . . . . . . . Probab Pro babilit ility y of Failu Failure re in terms terms of β of β . . . . . . . . . . . . . . . . . . . .

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. . . . .

. 164 . 16 166 . 167 . 167 . 168

10.1 10.2 10.3 10.4 10.5 10.6 10.7

Lateral Bracing for Steel Lateral Steel Beams . . . . . . . . . . . . . . . . . . . Failure of Steel beam; Plastic Hinges Hinges . . . . . . . . . . . . . . . . Failure of Steel beam; Local Buckling Buckling . . . . . . . . . . . . . . . Failure of Steel beam; Later Lateral al Torsion orsional al Buc Buckling kling . . . . . . . . . Stresss distribution Stres distribution at differ different ent stages stages of loading . . . . . . . . . . Stress-str Stres s-strain ain diagram for most structural structural steels . . . . . . . . . . Nominall Momen Nomina Moments ts for Compac Compactt and Partially Partially Compact Sections .

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. 173 . 175 . 175 . 17 176 . 176 . 177 . 179

11.1 11.2 11.3 11.4 11.5

Failure Modes for R/C R/C Beams . . . . . . . Internal Inte rnal Equilibriu Equilibrium m in a R/C Beam . . . Cracked Crac ked Section, Section, Limit State . . . . . . . Whitney Whitne y Stress Block Block . . . . . . . . . . . Reinforcem Reinf orcement ent in Continuous Continuous R/C Beams

. . . . .

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. . . . .

. 184 . 185 . 187 . 187 . 194

12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9

Pretensioned Pretension ed Pres Prestress tressed ed Concrete Beam, (Nilso (Nilson n 1978) . . . . . . . . . . . . . . 198 Posttensi Post tensioned oned Prestressed Prestressed Concrete Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . 198 7 Wire Pres Prestress tressing ing Tendon Tendon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Alternativ Alter nativee Sch Schemes emes for Pres Prestress tressing ing a Rect Rectangula angularr Concr Concrete ete Beam, (Nilso (Nilson n 1978) 1978)201 201 Determinati Deter mination on of Equivalen Equivalentt Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Load-Deflec LoadDeflection tion Curve and Corre Correspondin spondingg Int Internal ernal Flexural Flexural Stres Stresses ses for a Typ Typical ical Prestressed Prestressed Concrete Flexural Flexu ral Stress Distribution Distribution for a Beam with Variabl ariablee Ecce Eccentr ntricity icity;; Maxim Maximum um Momen Momentt Secti Section on and Sup Walnu alnutt Lane Bridge, Plan View . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 Walnu alnutt Lane Bridge, Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . 208

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. . 145 . . 146 . . 146 . . 148 . . 149 . . 153

13.1 Momen Momentt Resisting Force Forcess in an Arch or Suspension System System as Compared Compared to a Beam, (Lin and Stotesbur Stotesbur 13.2 Statics of a Three Three-Hing -Hinged ed Arch, (Lin and Stotesbury Stotesbury 1981) . . . . . . . . . . . . 212 Victor Saouma


Draft List of Tables 1.1 Structural Structural Eng Engineer ineering ing Cov Coverage erage for for Architect Architectss and Engine Engineers ers . . . . . . . . . . . 26 1.22 tab 1. tab:s :sec ecae ae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13

Unit Weig Unit Weight ht of Mate Materia rials ls . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Weigh eights ts of Buildi Building ng Mater Materials ials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 Avera Av erage ge Gross Gross Dead Dead Load Load in Buildin Buildings gs . . . . . . . . . . . . . . . . . . . . . . . . 31 Minimum Minim um Uniformly Uniformly Distrib Distributed uted Live Live Loads, Loads, (UBC (UBC 1995) . . . . . . . . . . . . . 32 Wind Velocit Velocity y Variatio Variation n above above Ground Ground . . . . . . . . . . . . . . . . . . . . . . . . 36 C e Coefficients for Wind Load, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . 37 Wind Win d Press Pressure ure Coeffi Coefficie cient ntss C q , (UBC 1995) . . . . . . . . . . . . . . . . . . . . . 37 Importance Importan ce Factor Factorss for Wind and and Earthquak Earthquakee Load, (UBC (UBC 1995) . . . . . . . . . 38 Approxim Appr oximate ate Desig Design n Wind Wind Press Pressure ure p for Ordinary Wind Force Resisting Building Structures 38 Z Factors for Different Seismic Zones, ubc . . . . . . . . . . . . . . . . . . . . . . 42 S Site Coefficients for Earthquake Loading, (UBC 1995) . . . . . . . . . . . . . . 42 Partial Par tial List of R of RW for Various Structure Systems, (UBC 1995) . . . . . . . . . . 43 Coefficients Coefficien ts of Ther Thermal mal Expansion Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.1 3.2 3.3 3.4

Properties Propert ies of of Major Struc Structur tural al Steels Steels Propert Pro perties ies of of Reinfo Reinforci rcing ng Bars Bars . . . . Joistt Serie Jois Seriess Chara Characte cteris ristics tics . . . . . Joistt Pro Jois Propert perties ies . . . . . . . . . . . .

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59 61 73 75

5.1 Equ Equatio ations ns of Equi Equilib libriu rium m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 5.2 Static Deter Determinacy minacy and Stabilit Stability y of Trusse Trussess . . . . . . . . . . . . . . . . . . . . . 94 5.3 Sec Section tion Pro Propert perties ies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 9.1 Allowabl Allowablee Stresses Stresses for Steel Steel and and Concret Concretee . . . . . . . . . . . . . . . . . . . . . . 165 9.22 Se 9. Sele lect cted ed β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . 169 9.3 Stren Strength gth Reduction Reduction Factors actors,, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 14.1 Columns Combined Approximate Approximate Vertical Vertical and Horizontal Loads . . . . . . . . . 254 14.2 Girders Combined Approximate Approximate Vertical Vertical and Horizontal Loads . . . . . . . . . . 255

Draft Chapter 1

INTRODUCTION 1.1 1.1

Scie Scienc nce e and and Techn echnolo ology gy

“There “There is a fundam fundamen ental tal differe difference nce betwe between en scienc sciencee and and techno technolog logy y. Engine Engineeri ering ng or technology is the making of things that did not previously exist, whereas science is the discovering of things that have long existed. existed. Technologi echnological cal results are forms that exist only because people want to make them, whereas scientific results are informations of what exists independently dently of human human intentions. intentions. Technology echnology deals with the artificial, artificial, science science with the natural.” (Billington (Billington 1985) 1

1.2 1.2 2

Stru Struct ctur ural al Engi Engine neer erin ing g

Structural engineers are responsible for the detailed analysis and design of:

Architectural structures: Buildings, houses, factories. They must work in close cooperation with an architect who will ultimately be responsible for the design. Civil Infrastructures: Bridges, Bridges, dams, pipelines, pipelines, offshore structures. structures. They work with transportati portation, on, hydra hydrauli ulic, c, nucle nuclear ar and other other engine engineers ers.. For those those struct structure uress they they play play the leading role. Aerospace, Mechanical, Naval structures: aeroplanes, spacecrafts, cars, ships, submarines to ensure the structural safety of those important structures.

1.3 1.3 3

Stru Struct ctur ures es and and their their Surr Surrou ound ndin ings gs

Structural design is affected by various environmental constraints: 1. Major movements movements:: For example, elevator elevator shafts are usually shear walls good at resisting lateral load (wind, earthquake). 2. Sound and structure structure interact: interact:

• A dome roof will concentrate the sound • A dish roof will diffuse the sound

Draft

1.6 Architectural Design

1.6 1.6 12

17

Arc Architec hitectu tura rall Desi Design gn

Architectural design must respect various constraints:

Functionality: Influence of the adopted structure on the purposes for which the structure was erected. Aesthetics: The architect architect often imposes his aesthetic concerns concerns on the engineer. engineer. This in turn can place severe limitations on the structural system. Economy: It should be kept in mind that the two largest components of a structure are labors and materials. Design cost is comparatively negligible. 13

Buildings may have different functions:

Residential: housing, which includes low-rise (up tp 2-3 floors), mid-rise (up to 6-8 floors) and high rise buildings. Commercial: Offices, retail stores, shopping centers, hotels, restaurants. Industrial: warehouses, manufacturing. Institutional: Schools, hospitals, prisons, chruch, government buildings. Special: Towers, stadium, parking, airport, etc.

1.7 1.7

Stru Struct ctur ural al Anal Analys ysis is

Given an existing structure subjected to a certain load determine internal forces (axial, shear, flexural, torsional; or stresses), deflections, and verify that no unstable failure can occur.

14

15

Thus the basic structural requirements are:

Strength: stresses should not exceed critical values: σ < σ f Stiffness: deflections should be controlled: ∆ < ∆max Stability: buckling or cracking should also be prevented

1.8 1.8 16

Stru Struct ctur ural al Desi Design gn

Given a set of forces, dimension the structural element.

Steel/wood Steel/wood Structures Structures Select appropriate section. Reinforced Reinforced Concrete: Concrete: Determine dimensions of the element and internal reinforcement (number and sizes of reinforcing bars). For new structures, structures, iterative process between analysis and design. A preliminary design is made using rules of thumbs (best known to Engineers Engineers with design experience) experience) and analyzed. analyzed. Following design, we check for 17

Victor Saouma


Draft

1.10 Structure Typ es

19

Tension & Compression Structures: only only,, no shear, shear, flexu flexure re,, or torsio torsion. n. Thos Thosee are are the the most efficient types of structures. Cable (tension only): The high strength of steel cables, combined with the efficiency of simple tension, makes cables ideal structural elements to span large distances such as bridges, and dish roofs, Fig. 1.2 1.2.. A cable structure develops its load carrying

Figure 1.2: Basic Aspects of Cable Systems capacity by adjusting its shape so as to provide maximum resistance ( form follows function ). ). Care should be exercised in minimizing large deflections and vibrations. Arches (mostly (mostly compressio compression) n) is a “reversed “reversed cable structure”. structure”. In an arch, arch, flexure/sh flexure/shear ear is minimized and most of the load is transfere transfered d through axial forces only. only. Arches Arches are Victor Saouma


Draft


21

Figure 1.4: Types of Trusses

Victor Saouma


Draft


23

VIERENDEEL TRUSS

TREE-SUPPORTED TRUSS

OVERLAPPING SINGLE-STRUT CABLE-SUPPORTED BEAM

BRACED BEAM

CABLE-STAYED BEAM

SUSPENDED CABLE SUPPORTED BEAM

BOWSTRING TRUSS

CABLE-SUPPORTED STRUTED ARCH OR CABLE BEAM/TRUSS

CABLE-SUPPORTED MULTI-STRUT BEAM OR TRUSS

GABLED TRUSS

CABLE-SUPPORTED ARCHED FRAME

CABLE-SUPPORTED PORTAL FRAME

Figure 1.6: Different Beam Types

Victor Saouma


Draft

1.11 Structural Engineering Courses

25

Folded olded plates plates are used mostly mostly as long long span span roofs. roofs. Ho Howe weve ver, r, they they can also be used used as vertical walls to support both vertical and horizontal loads. Membranes: 3D structures composed of a flexible 2D surface resisting tension only. They are usually cable-supported and are used for tents and long span roofs Fig. 1.8 1.8..

Figure 1.8: Examples of Air Supported Structures Shells: 3D structures composed of a curved 2D surface, they are usually shaped to transmit compressive axial stresses only, Fig. 1.9 1.9.. Shells are classified in terms of their curvature.

1.11

Struct Structura urall Engine Engineeri ering ng Courses Courses

Structural engineering education can be approached from either one of two points of views, depending on the audience, ??. ??.

22

Architects: Start from overall design, and move toward detailed analysis. Emphasis on good understanding of overall structural behavior. Develop a good understanding of load trans-

Victor Saouma


Draft

1.12 References

27

fer mechanism for most types of structures, cables, arches, beams, frames, shells, plates. Approximate analysis for most of them. Engineers: Emphasis is on the individual structural elements and not always on the total system. Focus on beams, frames (mostly 2D) and trusses. Very seldom are arches covered. Plates and shells are not even mentioned.

1.12 1.12

Refe Refere renc nces es

Following are some useful references for structural engineering, those marked by were consulted, and “borrowed from” in preparing the Lecture Notes or are particularly recommended.

†

23

Structures for Architect 1. Ambrose, Ambrose, J., J., Building Structures, Structures, second Ed. Wiley, 1993. 2. Billington, Billington, D.P. D.P. Rober Maillart’s Bridges; The Art of Engineering , Princeton University Pres, 1979. 3. Billington, D.P., The Tower and the Bridge; The new art of structural engineering , Princeton University Pres,, 1983.

†

4. Billington, D.P., Structures and the Urban Environment , Lectures Notes CE 262, Department of Civil Engineering, Princeton University, 1978

†

5. French, rench, S., Determinat Determinatee Structur Structures; es; Statics, Statics, Strength, Strength, Analysis Analysis,, Design Design , Delmar Delmar,, 1996. 6. Gordon Gordon,, J.E., J.E., Structures, or Why Things Do’nt Fall Down , Da Capo paperback, New York, 1978. 7. Gordon, Gordon, J.E., J.E., The Science of Structures and Materials, Materials, Scientific American Library, 1988. 8. Hawkes Hawkes,, N., Structures, the way things are built , MacMillan, 1990. 9. Levy, Levy, M. and Salvadori, Salvadori, M., Why Buildings Fall Down , W.W.Norton, 1992. 10. Lin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engineers, Engineers, John Wiley, 1981.

†

11. Mainstone, R., Developments in Structural Form , Allen Lane Publishers, 1975.

†

12. Petroski Petroski,, H., To Enginer is Human , Vintage Books, 1992. 13. Salvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, Buildings, Prentice Hall, Third Edition, 1986.

†

14. Salvadori Salvadori,, M. and Levy, Levy, M., Structural Design in Architecture, Architecture, Prentice hall, Second Edition, Edition, 1981. 15. Salvadori Salvadori,, M., Why Buildings Stand Up; The Strength of Architecture, Architecture, Norton Paperack, 1990. 16. Sandaker, B.N. and Eggen, A.P., The Structur Structural al Basis of Archite Architectur cturee, Whitney Library of Design, 1992.

†

17. Schueller, W., The design of Building Structures, Structures, Prentice Hall, 1996.

†

Structures for Engineers Victor Saouma


Draft Chapter 2

LOADS 2.1 2.1

Intr Introdu oduct ctio ion n

The main purpose of a structure is to transfer load from one point to another: bridge deck to pier; slab to beam; beam to girder; girder to column; column to foundation; foundation to soil.

1

There There can also be second secondary ary loads loads such such as therma thermall (in restra restraine ined d struct structure ures), s), differe different ntial ial settlement of foundations, P-Delta effects (additional moment caused by the product of the vertical force and the lateral displacement caused by lateral load in a high rise building).

2

3

Loads are generally subdivided into two categories

Vertical Loads or gravity load 1. dead load (DL) 2. live load (LL) also included are snow loads. Lateral Loads which act horizontally on the structure 1. Wind load (WL) 2. Earthquake load (EL) this also includes hydrostatic and earth loads. This distinction is helpful not only to compute a structure’s load, but also to assign different factor of safety to each one.

4

5

For a detailed coverage of loads, refer to the Universal Building Code (UBC), (UBC 1995).

2.2 2.2

Vertic ertical al Loads Loads

For closely spaced identical loads (such as joist loads), it is customary to treat them as a uniformly distributed load rather than as discrete loads, Fig. 2.1

6

Draft

2.2 Vertical Loads

31

lb/ lb/ft2

Material

Ceilings Channel suspended system Acoustical fiber tile Floors Steel deck Concrete-plain 1 in. Linoleum 1/4 in. Hardwood Roofs Copper or tin 5 ply felt and gravel Shingles asphalt Clay tiles Sheathing woo d Insulation 1 in. poured in place Partitions Clay tile 3 in. Clay tile 10 in. Gypsum Block 5 in. Wo od studs 2x4 (12-16 in. o.c.) Plaster 1 in. cement Plaster 1 in. gypsum Walls Bricks 4 in. Bricks 12 in. Hollow concrete block (heavy aggregate) 4 in. 8 in. 12 in. Hollow concrete block (light aggregate) 4 in. 8 in. 12 in.

1 1 2-10 12 1 4 1-5 6 3 9-14 3 2 17 40 14 2 10 5 40 120 30 55 80 21 38 55

Table 2.2: Weights of Building Materials

Material Timber Steel Reinfo Reinforce rced d concre concrete te

lb/ lb/ft2 40-50 50-80 100-150 100150

Table 2.3: Average Gross Dead Load in Buildings

Victor Saouma


Draft

2.3 Lateral Loads

33

Floor Cumulative R (%) Cumulative LL Cumulative R× LL

Roof 8.48 20 18.3

10 16.96 80 66.4

9 25.44 80 59.6

8 33.92 80 52.9

7 42.4 80 46.08

6 51.32 80 38.9

5 59.8 80 32.2

4 60 80 32

3 60 80 32

2 60 80 32

Total 740 410

The resulting design live load for the bottom column has been reduced from 740 Kips to 410 Kips . 5. The total dead load is DL = (10)(60) = 600 Kips, thus the total reduction in load is 740−410 100= 25% . 740+600

×

2.2. 2.2.3 3

Sno Snow

geographic location and elevation elevation.. They Roof snow load vary greatly depending on geographic range from 20 to 45 psf, Fig. 2.2 2.2..

19



Figure 2.2: Snow Map of the United States, ubc 20

Snow loads are always given on the projected length or area on a slope, Fig. 2.3 2.3..

The steeper the roof, the lower lower the snow snow retention. retention. For snow loads greater than 20 psf and ◦ roof pitches α more than 20 the snow load p may be reduced by 21

R = (α 22

−



p 20) 40

− 0.5



(psf )

(2.2)

Other examples of loads acting on inclined surfaces are shown in Fig. 2.4 2.4..

2.3 2.3 2.3. 2.3.1 1

Late Latera rall Lo Load adss Wind ind

Wind load depend on: velocity of the wind, shape of the building, height, geographical location, texture of the building surface and stiffness of the structure. structure.

23

Victor Saouma


Draft

2.3 Lateral Loads

24

35

Wind loads are particularly significant on tall buildings1 .

When a steady streamline airflow of velocity V is completely stopped by a rigid body, the stagnation pressure (or velocity pressure) qs was derived by Bernouilli (1700-1782) 25

1 qs = ρV 2 2

(2.3)

where the air mass density ρ is the air weight divided by the acceleration of gravity g = 32 32..2 2 o o 3 ft/sec . At sea level and a temperature of 15 C (59 F), the air weighs 0.0765 lb/ft this would yield a pressure of 2 1 (0. (0.0765)lb/ft3 (5280)ft/mile (2.4) qs = V 2 (32. (3600)sec/hr (32.2)ft/sec2 or





qs = 0. 0 .00256 00256V V 2

(2.5)

where V is the maximum wind velocity (in miles per hour) and qs is in psf. V can be obtained from wind maps (in the United States 70 V 110), Fig. 2.5 2.5..

≤ ≤



Figure 2.5: Wind Map of the United States, (UBC 1995) During storms, wind velocities may reach values up to or greater than 150 miles per hour, which corresponds to a dynamic pressure qs of about 60 psf (as high as the average vertical occupancy load in buildings).

26

27

Wind pressure increases with height, height, Table 2.5 2.5..

28

Wind load will cause suction on the leeward sides, Fig. 2.7

29

This magnitude must be modified to account for the shape and surroundings of the building. 1

The primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and possibly earthquakes).

Victor Saouma


Draft

2.3 Lateral Loads

37

Thus, the design pressure p (psf) is given by (2.6)

p = C e C q Iq s The pressure is assumed to be normal to all walls and roofs and

C e Velocity Pressure Coefficient accounts for height, exposure and gust factor. It accounts for the fact that wind velocity increases with height and that dynamic character of the airflow (i.e the wind pressure is not steady), Table 2.6 2.6.. C e 1.39 1.39-2 -2.3 .344 1.06 1.06-2 -2.1 .199

Exposure D C

0.62 0.62-1 -1.8 .800

B

Open pen, flat terr terrai ain n facin acingg lar large bodi bodies es of water ater Flat Flat open open terr terrai ain, n, exte extend ndin ingg oneone-ha half lf mile mile or open open from from the the site in any full quadrant Terra errain in with with buil buildi ding ngs, s, fore forest st,, or surf surfac acee irre irregu gula lari riti ties es 20 ft or more in height

Table 2.6: C e Coefficients for Wind Load, (UBC 1995) C q Pressure Pressure Coefficient Coefficient is a shape factor which is given in Table 2.7 for gabled frames. Windw Windward ard Side Side Gabled Frames (V:H) Roof Slope <9:12 0.7 9:12 to 12:12 0.4 >12:12 0.7 Walls 0.8 Buildings (height < 200 ft) Vertic ertical al Projec Projectio tions ns heigh heightt < 40 ft 1.3 height > 40 ft 1.4 Horizontal Projections 0.7

−

−

Leewa Leeward rd Side Side

−0.7 −0.7 −0.7 −0.5 −1.3 −1.4 −0.7

Table 2.7: Wind Pressure Coefficients C q , (UBC 1995) I Importance Factor as given by Table 2.8 2.8.. where I Essential Facilities: Hospitals; Hospitals; Fire and police stations; Tanks; Emergency Emergency vehicle vehicle shelters, standby power-generating equipment; Structures and equipment in government. communication centers. II Hazardous Facilities: Structures housing, supporting or containing sufficient quantities of toxic or explosive substances to be dangerous to the safety of the general public public if released. released. III Special occupancy structure: Covered Covered structures whose primary occupancy is public assembly, capacity > 300 persons. persons. Victor Saouma


Draft

2.3 Lateral Loads

39

400 Exposure B, 70 mph Exposure B, 80 mph Exposure C, 70 mph Exposure C, 80 mph

350

300 ) t f ( e 250 d a r G e v 200 o b A t h g 150 i e H

100

50

0

0

5

10

15 20 25 30 35 40 Approximate Design Wind Pressure (psf)

45

50

Figure 2.7: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures

Exampl Example e 2-2: 2-2: Wind Wind Load Load Determine the wind forces on the building shown on below which is built in St Louis and is surrounded by trees. Solution: 1. From Fig. 2.5 the maximu maximum m wind wind velocit velocity y is St. Louis Louis is 70 mph, since since the buildin buildingg is protected we can take C e = 0. 0 .7, I = 1.. The base wind pressure is qs = 0.00256 (70)2 = 12 12..54 psf.

×

Victor Saouma


Draft

2.3 Lateral Loads

41

Figure 2.8: Vibrations of a Building 36

The horizontal force at each level is calculated as a portion of the base shear force V V =

ZI C W RW

(2.8)

where: Factor: to be determined from Fig. 2.9 and Table 2.10 Z : Zone Factor: 2.10..

Figure 2.9: Seismic Zones of the United States, (UBC 1995) I : Importance Factor: Factor: which was given by Table 2.8 2.8.. Victor Saouma


Draft

2.3 Lateral Loads

Structural System

43

RW

Bearing wall system Light-framed walls with shear panels Plywo od walls for structures three stories or less 8 All other light-framed walls 6 Shear walls Concrete 8 Masonry 8 Building frame system using trussing or shear walls) Steel eccentrically braced ductiel frame 10 Light-framed walls with shear panels Plywo od walls for structures three stories or less 9 All other light-framed walls 7 Shear walls Concrete 8 Masonry 8 Concentrically braced frames Steel 8 Concrete (only for zones I and 2) 8 Heavy timber 8 Moment-resisting frame system Special moment-resisting frames (SMRF) Steel 12 Concrete 12 Concrete intermediate moment-resisting frames (IMRF)only for zones 1 and 2 8 Ordinary moment-resisting frames (OMRF) Steel 6 Concrete (only for zone 1) 5 Dual systems (selected cases are for ductile rigid frames only) Shear walls Concrete with SMRF 12 Masonry with SMRF 8 Steel eccentrically braced ductile frame 6-12 Concentrically braced frame 12 Steel with steel SMRF 10 Steel with steel OMRF 6 Concrete with concrete SMRF (only for zones 1 and 2) 9

H (ft)

65 65 240 160 24 240 65 65 240 160 160 65

N.L. N.L. 160 -

N.L. 160 160-N.L. N. L. N.L. 160 -

Table 2.12: Partial List of R of RW for Various Structure Systems, (UBC 1995)

Victor Saouma


Draft

2.3 Lateral Loads

45

5. The total vertic vertical al load is W = 2 ((200 ((200 + 0.5(400)) (20) = 16000 16000 lbs 6. The total seismic seismic base shear shear is ZI C (0. (0.3)(1. 3)(1.25)(2. 25)(2.75) V = = = 0.086 086W W RW 12 = (0. (0.086)(16000) = 1375 lbs

(2.17)

(2.18-a) (2.18-b)

7. Since Since T < 0.7 sec. there is no whiplash. 8. The load on each floor is thus given given by F 2 = F 1 =

(1375)(24) = 916.7 lbs 12 + 24 (1375)(12) = 458.3 lbs 12 + 24

(2.19-a) (2.19-b)

Example Example 2-4: Earthquak Earthquake e Load on a Tall Building, Building, (Schueller (Schueller 1996) Determine the approximate critical lateral loading for a 25 storey, ductile, rigid space frame concrete concrete structure structure in the short direction. direction. The rigid frames are spaced spaced 25 ft apart in the cross section section and 20 ft in the longitudinal longitudinal direction. direction. The plan dimension dimension of the building building is 175x100 ft, and the structure structure is 25(12)=300 25(12)=300 ft high. This office building building is located in an urban environ environmen mentt with a wind velocity of 70 mph and in seismic zone 4. For this investigation, an average building total dead load of 192 psf is used. Soil conditions are unknown. 470 k

2638 k ’ 0 0 3 = ) 2 1 ( 5 2

1523 k

’ 0 5 1 = 2 / 0 0 3

’ 0 0 2 = 3 / ) 0 0 3 ( 2

84000 k

’ 5 7 1 = ) 5 2 ( 7

3108 k

5(20)=100’

Victor Saouma


Draft

2.4 Other Loads

2.4 2.4

2.4.1 2.4.1

47

Othe Other r Lo Loa ads Hydro Hydrosta static tic and and Eart Earth h

Structures below ground must resist lateral earth pressure. pressure.

39

(2.28)

q = Kγ h 1−sinΦ where γ is the soil density, K = 1+sin pressure coefficient, coefficient, h is the height. 1+sin Φ is the pressure

For sand and gravel γ = 120 lb/ lb/ ft3 , and Φ

40

≈ 30◦.

If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.10 2.10.. 41

Figure 2.10: Earth and Hydrostatic Loads on Structures (2.29)

q = γ W W h where γ W 62..4 lbs/ft3 . W = 62

Exampl Example e 2-5: Hydros Hydrostat tatic ic Load Load The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade, what thickness concrete slab is required to exactly balance the hydrostatic uplift? Solution: The hydrostatic pressure must be countered by the pressure caused by the weight of concrete. Since p = γh we equate the two pressures and solve for h the height of the concrete slab (62. (62.4) lbs/ft3



15.0 inch

3

(62.4) lbs/ft × (12 − 9) ft = (150) lbs/ft3 × h ⇒ h = (62. (3) ft(12) in/ft = 14 14..976 in  (150) lbs/ft



water

Victor Saouma

3

    concrete


Draft

2.5 Other Imp ortant Considerations

2.4. 2.4.3 3

49

Brid Bridge ge Lo Load adss

For highway bridges, design loads are given by the AASHTO (Association of American State Highwa Highway y Transportat Transportation ion Officials). The HS-20 truck is used for the design design of bridges on main highways, Fig. 6.3 6.3.. Either the design truck with specified axle loads and spacing must be used or the equivalent uniform load and concentrated load. This loading must be placed such that maximum stresses are produced.

Figure 2.11: Truck Load

2.4. 2.4.4 4

2.5 2.5.1 2.5.1 45

Impa Impact ct Lo Load ad

Other Other Importan Importantt Consid Considera eratio tions ns Load Load Com Combina binatio tions ns

Live loads specified by codes represent the maximum possible loads.

The likelihood likelihood of all these loads occurring occurring simultaneous simultaneously ly is remote. remote. Hence, Hence, building codes allow certain reduction when certain loads are combined together.

46

47

Furthermore, structures should be designed to resist a combination of loads. loads.

Denoting D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil: 48

For the load and resistance factor design (LRFD) method of concrete structures, the American Concrete Institute (ACI) Building design code (318) (318 n.d.) requires that the following load combinations be considered:

49

1. 1.4D+1.7L 1.4D+1.7L

Victor Saouma


Draft


51

Figure 2.12: Load Placement to Maximize Moments

Figure 2.13: Load Life of a Structure, (Lin and Stotesbury 1981)

Victor Saouma


Draft


53

1. The section section is part of a beam or girder. girder. 2. The beam or girder girder is really part part of a three dimensional dimensional structure structure in which which load is transtransmitted from any point in the structure to the foundation through any one of various structural forms. Load transfer in a structure is accomplished through a “hierarchy” of simple flexural elements which are then connected to the columns, Fig. 2.16 or by two way slabs as illustrated in Fig. 2.17 2.17.. 57

58

An example of load transfer mechanism is shown in Fig. 2.18 2.18..

Victor Saouma


Draft


55

Figure 2.17: Two Way Actions

Victor Saouma


Draft Chapter 3

STRUCTURAL MATERIALS Proper understanding of structural materials is essential to both structural analysis and to structural design. 1

2

Characteristics of the most commonly used structural materials will be highlighted.

3.1 3.1.1 3.1.1

S t e el Struc Structu tural ral Steel Steel

Steel is an alloy of iron and carbon. carbon. Its properti properties es can be greatly greatly vari varied ed by alteri altering ng the carbon content (always less than 0.5%) or by adding other elements such as silicon, nickle, manganese and copper. 3

Practically all grades of steel have a Young Modulus equal to 29,000 ksi, ksi, density of 490 − 5 lb/cu ft, and a coefficient of thermal expansion equal to 0. 0 .65 10 /deg F.

4

×

The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly commonly used structural structural steel are A36 (σyld = 36 ksi) and A572 (σ (σyld = 50 ksi), Fig. 3.1 5

Struct Structura urall steel steel can be rolled rolled into a wide wide variet ariety y of shapes and sizes. sizes. Usually Usually the most desirable members are those which have a large section moduli (S ( S ) in proportion to their area (A), Fig. 3.2 3.2.. 6

7

Steel can be bolted, riveted or welded.

Sections are designated by the shape of their cross section, their depth and their weight. For example W 27 114 is a W section, 27 in. deep weighing 114 lb/ft. 8

×

9

Common sections are:

S sections were the first ones rolled in America and have a slope on their inside flange surfaces of 1 to 6. W or wide flange sections have a much smaller inner slope which facilitates connections and rivetting. W sections constitute about 50% of the tonnage of rolled structural steel. C are channel sections MC Miscellaneous channel which can not be classified as a C shape by dimensions.

Draft 3.1 Steel

59

HP is a bearing pile section. M is a miscellaneous section. L are angle sections which may have equal or unequal sides. WT is a T section cut from a W section in two. 10

The section modulus S x of a W section can be roughly approximated by the following formula S x

≈ wd/1 wd/10

or I x

≈ S x d2 ≈ wd2/20

(3.1)

and the plastic modulus can be approximated by Z x

11

(3.2)

≈ wd/9 wd/9

Properties of structural steel are tabulated in Table 3.1 3.1.. ASTM Desig. A36

Shap es Available

A500

Cold formed welded and seamless sections;

A501

Hot fo formed welded an and seamless sections; Plates and bars 12 in and less thick;

A529

Shap es and bars

A606 A606

Hot Hot and and cold cold roll rolled ed shee sheets ts;;

A611

Cold rolled sheet in in cu cut lengths Stru Struct ctu ural ral shap shapes es,, plat plates es and bars

A 709 709

σy (kksi)

Use Riveted, bolted, welded; Buildings and bridges Gene Genera rall stru struct ctur ural al purpos pose Riveted, welded or bolted; Bolted and welded Buildi Building ng frames frames and trus trusse ses; s; Bolt Bolted ed and and welded Atmo Atmosp sphe heri ricc corr corros osio ion n resistant Cold Cold form formed ed secti section onss Bridges

36 up thr throu ough gh 8 in. in. above 8.)

σu (kksi) (32 (32

Grade A: 33; Grade B: 42; Grade C: 46 36 42

45-50 Grad Gradee C 33; Grad Gradee D 40; Grade E 80 Grade 36: 36 (to 4 in.); Grade Grade 50: 50; Grade Grade 100: 100 (to 2.5in.) and 90 (over 2.5 to 4 in.)

Table 3.1: Properties of Major Structural Steels Rolled sections, Fig. 3.3 and welded ones, Fig3.4 Fig3.4 have residual stresses. stresses. Those origina originate te during the rolling or fabrication of a member. The member is hot just after rolling or welding, it cools unevenly unevenly because of varying varying exposure. exposure. The area that cool first become stiffer, stiffer, resist resist contracti contraction, on, and develop develop compressive compressive stresses. stresses. The remaining remaining regions regions continue continue to cool and contract in the plastic condition and develop tensile stresses. 12

Due to those residual stresses, the stress-strain curve of a rolled section exhibits a non-linear segment prior to the theoretical yielding, Fig. 3.5 3.5.. This would have have important important implications implications on the flexural and axial strength of beams and columns.

13

Victor Saouma


Draft

3.2 Aluminum

3.1.2 3.1.2

61

Rein Reinfor forci cing ng Stee Steell

Steel is also used as reinforcing bars in concrete, Table 3.2 3.2.. Those bars have a deformation on their surface to increase the bond with concrete, and usually have a yield stress of 60 ksi 1 .

14

Barr Desi Ba Design gnat atio ion n No. 2 No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 No. 10 No. 11 No. 14 No. 18

Diam Diamet eter er (in.) 2/8=0.250 3/8=0.375 4/8=0.500 5/8=0.625 6/8=0.750 7/8=0.875 8/8=1.000 9/8=1.128 10/8=1.270 11/8=1.410 14/8=1.693 18/8=2.257

Area Area ( in2 ) 0.05 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1 .27 1. 1 .56 1. 2 .25 2. 4 .00 4.

Perim erimet eter er in 0.79 1.18 1.57 1.96 2.36 2.75 3.14 3.54 3.99 4.43 5.32 7.09

Weigh eightt lb/ft 0.167 0.376 0.668 1.043 1.5202 2.044 2.670 3.400 4.303 5.313 7.650 13.60

Table 3.2: Properties of Reinforcing Bars Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected from fire), and becomes brittle at 30 deg. F

15

−

Steel is also used as wire strands and ropes for suspended roofs, cable-stayed bridges, fabric roofs and other structural structural applications. applications. A strand is a helical arrangement of wires around a central wire. A rope consists of multiple strands helically wound around a central plastic core, and a modulus of elasticity of 20,000 ksi, and an ultimate strength of 220 ksi. 16

17

Prestressing Steel cables have an ultimate strength up to 270 ksi.

3.2 3.2

Alum Alumin inum um

Aluminum is used whenever light weight combined with strength is an important factor. Those properties, along with its resistance to corrosion have made it the material of choice for airplane structures, light roof framing.

18

19

Aluminum members can be connected by riveting, bolting and to a lesser extent by welding.

Aluminum has a modulus of elasticity equal to 10,000 ksi (about three times lower than steel), a coefficient of thermal expansion of 2. 2.4 10−5 and a density of 173 lbs/ft3 .

20

×

The ultimate strength of pure aluminum is low (13,000 psi) but with the addition of alloys it can go up. 21

1

Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.

Victor Saouma


Draft 3.4 Masonry

63

30

Density of normal weight concrete is 145 lbs/ft3 and 100 lbs/ft3 for lightweight concrete.

31

Coefficient of thermal expansion is 0. 0.65

× 10−5 /deg F for normal weight concrete.

When concrete is poured (or rather placed), the free water not needed for the hydration process evaporates over a period of time and the concrete will shrink. shrink. This shrinkage is about 0.05% after one year (strain). Thus if the concrete is restrained, then cracking will occur3 .

32

Concrete Concrete will also deform with time due to the applied applied load, this is called creep. creep. This should be taken into consideration when computing the deflections (which can be up to three times the instantaneous elastic deflection).

33

3.4 3.4

Masonr onry

Masonry consists of either natural materials, such as stones, or of manufactured products such as bricks and concrete blocks 4 , stacked and bonded together with mortar.

34

As for concrete, all modern structural masonry blocks are essentially compression members with low tensile resistance. resistance. 35

The mortar used is a mixtur mixturee of sand, sand, masonr masonry y cemen cement, t, and either either Portl Portland and cement cement or hydrated lime. 36

3.5 3.5

Timber ber

Timber is one of the earliest construction materials, and one of the few natural materials with good tensile properties.

37

38

The properties of timber vary greatly, and the strength is time dependent.

Timber is a good shock absorber (many wood structures in Japan have resisted repeated earthquakes). 39

The most commonly used species of timber in construction are Douglas fir, southern pine, hemlock and larch. 40

Members can be laminated together under good quality control, and flexural strengths as high as 2,500 psi can be achieved.

41

3.6 3.6 42

Stee Steell Sect Section ion Proper Properti ties es

Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.7 3.7.. ==============

3

For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, can reduce the shrinkage by 75%. 4 Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos...

Victor Saouma


Draft

3.6 Steel Section Properties Designation

A in2 W 27x539 158.0 W 27x494 145.0 W 27x448 131.0 W 27x407 119.0 W 27x368 108.0 W 27x336 98.7 W 27x307 90.2 W 27x281 82.6 W 27x258 75.7 W 27x235 69.1 W 27x217 63.8 W 27x194 57.0 W 27x178 52.3 W 27x161 47.4 W 27x146 42.9 W 27x129 37.8 W 27x114 33.5 W 27x102 30.0 W 27x 94 27.7 W 27x 84 24.8 W 24x492 144.0 W 24x450 132.0 W 24x408 119.0 W 24x370 108.0 W 24x335 98.4 W 24x306 89.8 W 24x279 82.0 W 24x250 73.5 W 24x229 67.2 W 24x207 60.7 W 24x192 56.3 W 24x176 51.7 W 24x162 47.7 W 24x146 43.0 W 24x131 38.5 W 24x117 34.4 W 24x104 30.6 W 24x103 30.3 W 24x 94 27.7 W 24x 84 24.7 W 24x 76 22.4 W 24x 68 20.1 W 24x 62 18.2 W 24x 55 16.2 W 21x402 118.0 W 21x364 107.0 W 21x333 97.9 W 21x300 88.2 W 21x275 80.8 W 21x248 72.8 W 21x223 65.4 W 21x201 59.2 W 21x182 53.6 W 21x166 48.8 W 21x147 43.2 W 21x132 38.8 W 21x122 35.9 W 21x111 32.7 W 21x101 Victor Saouma 29.8 W 21x 93 27.3 W 21x 83 24.3 W 21x 73 21.5

d in 32.52 31.97 31.42 30.87 30.39 30.00 29.61 29.29 28.98 28.66 28.43 28.11 27.81 27.59 27.38 27.63 27.29 27.09 26.92 26.71 29.65 29.09 28.54 27.99 27.52 27.13 26.73 26.34 26.02 25.71 25.47 25.24 25.00 24.74 24.48 24.26 24.06 24.53 24.31 24.10 23.92 23.73 23.74 23.57 26.02 25.47 25.00 24.53 24.13 23.74 23.35 23.03 22.72 22.48 22.06 21.83 21.68 21.51 21.36 21.62 21.43 21.24

bf

2tf

2.2 2.3 2.5 2.7 3.0 3.2 3.5 3.7 4.0 4.4 4.7 5.2 5.9 6.5 7.2 4.5 5.4 6.0 6.7 7.8 2.0 2.1 2.3 2.5 2.7 2.9 3.2 3.5 3.8 4.1 4.4 4.8 5.3 5.9 6.7 7.5 8.5 4.6 5.2 5.9 6.6 7.7 6.0 6.9 2.1 2.3 2.5 2.7 2.9 3.2 3.5 3.9 4.2 4.6 5.4 6.0 6.5 7.1 7.7 4.5 5.0 5.6

65 hc tw

I x S x I y S y 4 3 4 in in in in3 12.3 25500 1570 2110 277 13.4 22900 1440 1890 250 14.7 20400 1300 1670 224 15.9 18100 1170 1480 200 17.6 16100 1060 1310 179 19.2 14500 970 1170 161 20.9 13100 884 1050 146 22.9 11900 811 953 133 24.7 10800 742 859 120 26.6 9660 674 768 108 29.2 8870 624 704 100 32.3 7820 556 618 88 33.4 6990 502 555 79 36.7 6280 455 497 71 40.0 5630 411 443 64 39.7 4760 345 184 37 42.5 4090 299 159 32 47.0 3620 267 139 28 49.4 3270 243 124 25 52.7 2850 213 106 21 10.9 19100 1290 1670 237 11.9 17100 1170 1490 214 13.1 15100 1060 1320 191 14.2 13400 957 1160 170 15.6 11900 864 1030 152 17.1 10700 789 919 137 18.6 9600 718 823 124 20.7 8490 644 724 110 22.5 7650 588 651 99 24.8 6820 531 578 89 26.6 6260 491 530 82 28.7 5680 450 479 74 30.6 5170 414 443 68 33.2 4580 371 391 60 35.6 4020 329 340 53 39.2 3540 291 297 46 43.1 3100 258 259 41 39.2 3000 245 119 26 41.9 2700 222 109 24 45.9 2370 196 94 21 49.0 2100 176 82 18 52.0 1830 154 70 16 50.1 1550 131 34 10 54.6 1350 114 29 8 10.8 12200 937 1270 189 11.8 10800 846 1120 168 12.8 9610 769 994 151 14.2 8480 692 873 134 15.4 7620 632 785 122 17.1 6760 569 694 109 18.8 5950 510 609 96 20.6 5310 461 542 86 22.6 4730 417 483 77 24.9 4280 380 435 70 26.1 3630 329 376 60 28.9 3220 295 333 54 31.3 2960 273 305 49 34.1 2670 249 274 44 37.5 2420 227 248 0 Structural Concepts a4n d 32.3 2070 192 93 22 36.4 1830 171 81 20 41.2 1600 151 71 17

Z x Z y 3 in in3 1880.0 437.0 1710.0 394.0 1530.0 351.0 1380.0 313.0 1240.0 279.0 1130.0 252.0 1020.0 227.0 933.0 206.0 850.0 187.0 769.0 168.0 708.0 154.0 628.0 136.0 567.0 122.0 512.0 109.0 461.0 97.5 395.0 57.6 343.0 49.3 305.0 43.4 278.0 38.8 244.0 33.2 1550.0 375.0 1410.0 337.0 1250.0 300.0 1120.0 267.0 1020.0 238.0 922.0 214.0 835.0 193.0 744.0 171.0 676.0 154.0 606.0 137.0 559.0 126.0 511.0 115.0 468.0 105.0 418.0 93.2 370.0 81.5 327.0 71.4 289.0 62.4 280.0 41.5 254.0 37.5 224.0 32.6 200.0 28.6 177.0 24.5 153.0 15.7 134.0 13.3 1130.0 296.0 1010.0 263.0 915.0 237.0 816.0 210.0 741.0 189.0 663.0 169.0 589.0 149.0 530.0 133.0 476.0 119.0 432.0 108.0 373.0 92.6 333.0 82.3 307.0 75.6 279.0 68.2 253.0 Systems61f.o7 r 221.0 34.7 196.0 30.5 172.0 26.6

Architects

Draft

3.6 Steel Section Properties Designation W 14x 74 W 14x 68 W 14x 61 W 14x 53 W 14x 48 W 14x 43 W 14x 38 W 14x 34 W 14x 30 W 14x 26 W 14x 22 W 12x336 W 12x305 W 12x279 W 12x252 W 12x230 W 12x210 W 12x190 W 12x170 W 12x152 W 12x136 W 12x120 W 12x106 W 12x 96 W 12x 87 W 12x 79 W 12x 72 W 12x 65 W 12x 58 W 12x 53 W 12x 50 W 12x 45 W 12x 40 W 12x 35 W 12x 30 W 12x 26 W 12x 22 W 12x 19 W 12x 16 W 12x 14 W 10x112 W 10x100 W 10x 88 W 10x 77 W 10x 68 W 10x 60 W 10x 54 W 10x 49 W 10x 45 W 10x 39 W 10x 33 W 10x 30 W 10x 26 W 10x 22 W 10x 19 W 10x 17 W 10x 15 W 10x 12

Victor Saouma

A in2 21.8 20.0 17.9 15.6 14.1 12.6 11.2 10.0 8.9 7.7 6.5 98.8 89.6 81.9 74.1 67.7 61.8 55.8 50.0 44.7 39.9 35.3 31.2 28.2 25.6 23.2 21.1 19.1 17.0 15.6 14.7 13.2 11.8 10.3 8.8 7.7 6.5 5.6 4.7 4.2 32.9 29.4 25.9 22.6 20.0 17.6 15.8 14.4 13.3 11.5 9.7 8.8 7.6 6.5 5.6 5.0 4.4 3.5

d in 14.17 14.04 13.89 13.92 13.79 13.66 14.10 13.98 13.84 13.91 13.74 16.82 16.32 15.85 15.41 15.05 14.71 14.38 14.03 13.71 13.41 13.12 12.89 12.71 12.53 12.38 12.25 12.12 12.19 12.06 12.19 12.06 11.94 12.50 12.34 12.22 12.31 12.16 11.99 11.91 11.36 11.10 10.84 10.60 10.40 10.22 10.09 9.98 10.10 9.92 9.73 10.47 10.33 10.17 10.24 10.11 9.99 9.87

bf

2tf

6.4 7.0 7.7 6.1 6.7 7.5 6.6 7.4 8.7 6.0 7.5 2.3 2.4 2.7 2.9 3.1 3.4 3.7 4.0 4.5 5.0 5.6 6.2 6.8 7.5 8.2 9.0 9.9 7.8 8.7 6.3 7.0 7.8 6.3 7.4 8.5 4.7 5.7 7.5 8.8 4.2 4.6 5.2 5.9 6.6 7.4 8.2 8.9 6.5 7.5 9.1 5.7 6.6 8.0 5.1 6.1 7.4 9.4

67 hc tw

25.3 27.5 30.4 30.8 33.5 37.4 39.6 43.1 45.4 48.1 53.3 5.5 6.0 6.3 7.0 7.6 8.2 9.2 10.1 11.2 12.3 13.7 15.9 17.7 18.9 20.7 22.6 24.9 27.0 28.1 26.2 29.0 32.9 36.2 41.8 47.2 41.8 46.2 49.4 54.3 10.4 11.6 13.0 14.8 16.7 18.7 21.2 23.1 22.5 25.0 27.1 29.5 34.0 36.9 35.4 36.9 38 3 8.5 46 4 6.6

I x in4 796 723 640 541 485 428 385 340 291 245 199 4060 3550 3110 2720 2420 2140 1890 1650 1430 1240 1070 933 833 740 662 597 533 475 425 394 350 310 285 238 204 156 130 103 89 716 623 534 455 394 341 303 272 248 209 170 170 144 118 96 82 69 54

S x in3 112 103 92 78 70 63 55 49 42 35 29 483 435 393 353 321 292 263 235 209 186 163 145 131 118 107 97 88 78 71 65 58 52 46 39 33 25 21 17 15 126 112 98 86 76 67 60 55 49 42 35 32 28 23 19 16 14 11

I y in4 134 121 107 58 51 45 27 23 20 9 7 1190 1050 937 828 742 664 589 517 454 398 345 301 270 241 216 195 174 107 96 56 50 44 24 20 17 5 4 3 2 236 207 179 154 134 116 103 93 53 45 37 17 14 11 4 4 3 2

S y in3 27 24 22 14 13 11 8 7 6 4 3 177 159 143 127 115 104 93 82 73 64 56 49 44 40 36 32 29 21 19 14 12 11 7 6 5 2 2 1 1 45 40 35 30 26 23 21 19 13 11 9 6 5 4 2 2 1 1

Z x in3 126.0 115.0 102.0 87.1 78.4 69.6 61.5 54.6 47.3 40.2 33.2 603.0 537.0 481.0 428.0 386.0 348.0 311.0 275.0 243.0 214.0 186.0 164.0 147.0 132.0 119.0 108.0 96.8 86.4 77.9 72.4 64.7 57.5 51.2 43.1 37.2 29.3 24.7 20.1 17.4 147.0 130.0 113.0 97.6 85.3 74.6 66.6 60.4 54.9 46.8 38.8 36.6 31.3 26.0 21.6 18.7 16.0 12.6

Z y in3 40.6 36.9 32.8 22.0 19.6 17.3 12.1 10.6 9.0 5.5 4.4 274.0 244.0 220.0 196.0 177.0 159.0 143.0 126.0 111.0 98.0 85.4 75.1 67.5 60.4 54.3 49.2 44.1 32.5 29.1 21.4 19.0 16.8 11.5 9.6 8.2 3.7 3.0 2.3 1.9 69.2 61.0 53.1 45.9 40.1 35.0 31.3 28.3 20.3 17.2 14.0 8.8 7.5 6.1 3.3 2.8 2.3 1.7


Draft

3.6 Steel Section Properties Designation C 15.x 50 C 15.x 40 C 15.x 34 C 12.x 30 C 12.x 25 C 12.x 21 C 10.x 30 C 10.x 25 C 10.x 20 C 10.x 15 C 9.x 20 C 9.x 15 C 9.x 13 C 8.x 19 C 8.x 14 C 8.x 12 C 7.x 15 C 7.x 12 C 7.x 10 C 6.x 13 C 6.x 11 C 6.x 8 C 5.x 9 C 5.x 7 C 4.x 7 C 4.x 5 C 3.x 6 C 3.x 5 C 3.x 4

Victor Saouma

A in2 14.7 11.8 10.0 8.8 7.3 6.1 8.8 7.3 5.9 4.5 5.9 4.4 3.9 5.5 4.0 3.4 4.3 3.6 2.9 3.8 3.1 2.4 2.6 2.0 2.1 1.6 1.8 1.5 1.2

d in 15. 15. 15. 12. 12. 12. 10. 10. 10. 10. 9. 9. 9. 8. 8. 8. 7. 7. 7. 6. 6. 6. 5. 5. 4. 4. 3. 3. 3.

bf

2tf

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

69 hc tw

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

I x in4 404.0 349.0 315.0 162.0 144.0 129.0 103.0 91.2 78.9 67.4 60.9 51.0 47.9 44.0 36.1 32.6 27.2 24.2 21.3 17.4 15.2 13.1 8.9 7.5 4.6 3.8 2.1 1.9 1.7

S x in3 53.8 46.5 42.0 27.0 24.1 21.5 20.7 18.2 15.8 13.5 13.5 11.3 10.6 11.0 9.0 8.1 7.8 6.9 6.1 5.8 5.1 4.4 3.6 3.0 2.3 1.9 1.4 1.2 1.1

I y in4 11. 9.23 8.13 5.14 4.47 3.88 3.94 3.36 2.81 2.28 2.42 1.93 1.76 1.98 1.53 1.32 1.38 1.17 0.97 1.05 0.87 0.69 0.63 0.48 0.43 0.32 0.31 0.25 0.20

S y in3 3.78 3.37 3.11 2.06 1.88 1.73 1.65 1.48 1.32 1.16 1.17 1.01 0.96 1.01 0.85 0.78 0.78 0.70 0.63 0.64 0.56 0.49 0.45 0.38 0.34 0.28 0.27 0.23 0.20

Z x in3 8.20 57.20 50.40 33.60 29.20 25.40 26.60 23. 19.30 15.80 16.80 13.50 12.50 13.80 10.90 9.55 9.68 8.40 7.12 7.26 6.15 5.13 4.36 3.51 2.81 2.26 1.72 1.50 1.30

Z y in3 8.17 6.87 6.23 4.33 3.84 3.49 3.78 3.19 2.71 2.35 2.47 2.05 1.95 2.17 1.73 1.58 1.64 1.43 1.26 1.36 1.15 0.99 0.92 0.76 0.70 0.57 0.54 0.47 0.40


Draft

3.6 Steel Section Properties Designation L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L

5.0x5.0x0.875 5.0x5.0x0.750 5.0x5.0x0.625 5.0x5.0x0.500 5.0x5.0x0.438 5.0x5.0x0.375 5.0x5.0x0.313 5.0x3.5x0.750 5.0x3.5x0.625 5.0x3.5x0.500 5.0x3.5x0.438 5.0x3.5x0.375 5.0x3.5x0.313 5.0x3.5x0.250 5.0x3.0x0.625 5.0x3.0x0.500 5.0x3.0x0.438 5.0x3.0x0.375 5.0x3.0x0.313 5.0x3.0x0.250 4.0x4.0x0.750 4.0x4.0x0.625 4.0x4.0x0.500 4.0x4.0x0.438 4.0x4.0x0.375 4.0x4.0x0.313 4.0x4.0x0.250 4.0x3.5x0.500 4.0x3.5x0.438 4.0x3.5x0.375 4.0x3.5x0.313 4.0x3.5x0.250 4.0x3.0x0.500 4.0x3.0x0.438

Victor Saouma

A in2 7.98 6.94 5.86 4.75 4.18 3.61 3.03 5.81 4.92 4.00 3.53 3.05 2.56 2.06 4.61 3.75 3.31 2.86 2.40 1.94 5.44 4.61 3.75 3.31 2.86 2.40 1.94 3.50 3.09 2.67 2.25 1.81 3.25 2.87

wg t k/f t 27.20 23.60 20.00 16.20 14.30 12.30 10.30 19.80 16.80 13.60 12.00 10.40 8.70 7.00 15.70 12.80 11.30 9.80 8.20 6.60 18.50 15.70 12.80 11.30 9.80 8.20 6.60 11.90 10.60 9.10 7.70 6.20 11.10 9.80

71 I x in4 17.8 15.7 13.6 11.3 10.0 8.7 7.4 13.9 12.0 10.0 8.9 7.8 6.6 5.4 11.4 9.4 8.4 7.4 6.3 5.1 7.7 6.7 5.6 5.0 4.4 3.7 3.0 5.3 4.8 4.2 3.6 2.9 5.1 4.5

S x in3 5.2 4.5 3.9 3.2 2.8 2.4 2.0 4.3 3.7 3.0 2.6 2.3 1.9 1.6 3.5 2.9 2.6 2.2 1.9 1.5 2.8 2.4 2.0 1.8 1.5 1.3 1.0 1.9 1.7 1.5 1.3 1.0 1.9 1.7

I y in4 17.80 15.70 13.60 11.30 10.00 8.74 7.42 5.55 4.83 4.05 3.63 3.18 2.72 2.23 3.06 2.58 2.32 2.04 1.75 1.44 7.67 6.66 5.56 4.97 4.36 3.71 3.04 3.79 3.40 2.95 2.55 2.09 2.42 2.18

S y in3 5.17 4.53 3.86 3.16 2.79 2.42 2.04 2.22 1.90 1.56 1.39 1.21 1.02 0.83 1.39 1.15 1.02 0.89 0.75 0.61 2.81 2.40 1.97 1.75 1.52 1.29 1.05 1.52 1.35 1.16 0.99 0.81 1.12 0.99

Z x in3 9.33 8.16 6.95 5.68 5.03 4.36 3.68 7.65 6.55 5.38 4.77 4.14 3.49 2.83 6.27 5.16 4.57 3.97 3.36 2.72 5.07 4.33 3.56 3.16 2.74 2.32 1.88 3.50 3.11 2.71 2.29 1.86 3.41 3.03

Z y in3 9.33 8.16 6.95 5.68 5.03 4.36 3.68 4.10 3.47 2.83 2.49 2.16 1.82 1.47 2.61 2.11 1.86 1.60 1.35 1.09 5.07 4.33 3.56 3.16 2.74 2.32 1.88 2.73 2.42 2.11 1.78 1.44 2.03 1.79


Draft 3.7 Joists

3.7

73

Joists

Steel joists, Fig. 3.8 look like shallow trusses (warren type) and are designed as simply supported uniformly loaded beams assuming that they are laterally supported on the top (to prevent prevent lateral torsional torsional buckling). buckling). The lateral support is often profided profided by the concrete slab it suppors. 43

The standard open-web joist designation consists of the depth, the series designation and the chord type. Three series are available for floor/roof construction, Table 3.3

44

Seri Series es K LH DLH

Dept Depth h (in) (in) 8-30 18-48 52-72

Span Span (ft) (ft) 8-60 25-96 89-120

Table 3.3: Joist Series Characteri Characteristics stics ] . T F 3 3 . 0 – n a p S = h t g n e L n g i s e D [

" 4

" 4

" 4

n a p S

Figure 3.8: prefabricated Steel Joists Typical joist spacing ranges from 2 to 4 ft, and provides an efficient use of the corrugated steel deck which itself supports the concrete slab.

45

For preliminary estimates of the joist depth, a depth to span ratio of 24 can be assumed, therefore 46

d

≈ L/2 L/2

(3.5)

where d is in inches, and L in ft. Table 3.4 list the load carrying capacity of open web, K-series steel joists based on a amximum allowable allowable stress of 30 ksi. For each span, the first line indicates indicates the total safe uniformly uniformly 47

Victor Saouma


Draft 3.7 Joists

75

Join Jointt 8K1 8K1 10K1 10K1 12K 12K1 1 12K3 12K3 12K5 12K5 Desig. Depth 8 10 12 12 12 (in.) 5.1 5 5 5.7 7.1 ≈ W (lbs/ft) Span (ft.) 8 550 550 9 550 550 10 550 55 550 480 550 11 532 55 550 377 542 12 444 550 550 55 5 50 55 5 50 288 455 550 550 550 13 377 479 550 55 5 50 55 5 50 225 363 510 510 510 14 324 412 500 550 550 179 289 425 463 463 15 281 358 434 543 550 145 234 344 428 434 16 246 313 380 476 550 119 192 282 351 396 17 277 336 42 4 20 55 5 50 159 234 291 366 18 246 299 37 3 74 50 5 07 134 197 245 317 19 221 268 33 3 35 45 4 54 113 167 207 269 20 199 241 30 3 02 40 4 09 97 142 177 230 21 218 27 273 37 370 123 153 198 22 199 24 249 33 337 106 132 172 23 181 22 227 30 308 93 116 150 24 166 20 208 28 282 81 101 132 25 26 27 28 29 30 31 32

Victor Saouma

14K 14K1 1 14K3 14K3 14K4 14K4 14K6 14K6 16K2 16K2 16K3 16K3 16K4 16K4 16K5 16K5 16K6 16K6 16K7 16K7 16K9 16K9 14

14

14

14

16

16

16

16

16

16

16

5.2

6

6.7

7.7

5.5

6.3

7

7.5

8.1

8.6

10.0

550 550 511 475 448 390 395 324 352 272 315 230 284 197 257 170 234 147 214 128 196 113 180 100 166 88 154 79 143 70

550 550 550 507 550 467 49 4 95 404 44 4 41 339 39 3 95 287 35 3 56 246 32 322 212 29 293 184 26 268 160 24 245 141 226 124 209 110 193 98 180 88

550 550 550 507 550 467 55 5 50 443 53 5 30 397 47 4 75 336 42 4 28 287 38 388 248 35 353 215 32 322 188 29 295 165 272 145 251 129 233 115 216 103

550 550 550 507 550 467 55 5 50 443 55 5 50 408 55 5 50 383 52 5 25 347 47 475 299 43 432 259 39 395 226 36 362 199 334 175 308 56 285 139 265 124

550 550 512 488 456 409 408 347 368 297 333 255 303 222 277 194 254 170 234 150 216 133 200 119 186 106 173 95 161 86 151 78 142 71

550 550 55 5 50 526 50 5 08 456 45 4 55 386 41 4 10 330 37 371 285 33 337 247 30 308 216 28 283 189 260 167 240 148 223 132 207 118 193 106 180 96 168 87 158 79

550 550 55 5 50 526 55 5 50 490 54 5 47 452 49 4 93 386 44 447 333 40 406 289 37 371 252 34 340 221 313 195 289 173 268 155 249 138 232 124 216 112 203 101 190 92

550 550 55 5 50 526 55 5 50 490 55 5 50 455 55 5 50 426 50 503 373 45 458 323 41 418 282 38 384 248 353 219 326 194 302 173 281 155 261 139 244 126 228 114 214 103

550 550 55 5 50 526 55 5 50 490 55 5 50 455 55 5 50 426 54 548 405 49 498 351 45 455 307 41 418 269 384 238 355 211 329 188 306 168 285 151 266 137 249 124 233 112

550 550 55 5 50 526 55 5 50 490 55 5 50 455 55 5 50 426 55 550 406 55 550 385 50 507 339 46 465 298 428 263 395 233 366 208 340 186 317 167 296 151 277 137 259 12 1 24

550 550 55 5 50 526 55 5 50 490 55 5 50 455 55 5 50 426 55 550 406 55 550 385 55 550 363 55 550 346 514 311 474 276 439 246 408 220 380 198 355 178 332 161 311 147

Structural Concepts and Systems for Architects Table 3.4: Joist Properties

Draft Chapter 4

Case Study I: EIFFEL TOWER Adapted Adapted from (Billin (Billingto gton n and Mark Mark 1983) 1983)

4.1 4.1

Mater Materia ials ls,, & Geome Geometr try y

The tower was built out of wrought iron, less expensive than steel,and Eiffel had more expereince with this material, Fig. 4.1

1

           

Figure 4.1: Eiffel Tower (Billington and Mark 1983)

Draft 4.2 Loads

79

Lo cation Support First platform second platform Intermediate platform Top platform Top

Height 0 186 380 644 906 984

Width/2 164 108 62 20 1 0

Width Estimated Actual 328 216 240 12 3 110 40 2 0

dy dx

.333 .270 .205 .115 .0264 0.000

β 18.4o 15.1o 11.6o 6.6o 1.5o 0o

The tower is supported by four inclined supports, each with a cross section of 800 in 2 . An idealization of the tower is shown in Fig. 4.2 4.2..

4

ACTUAL CONTINUOUS CONNECTION

IDEALIZED CONTINUOUS CONNECTION

ACTUAL POINTS OF CONNECTION

Figure 4.2: Eiffel Tower Idealization, (Billington and Mark 1983)

4.2 5 6

Loads

The total weight of the tower is 18, 18, 800 k. The dead load is not uniformly distributed, and is approximated as follows, Fig. 4.3 4.3::         

Figure 4.3: Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983)

Victor Saouma


Draft

4.3 Reactions

81

LOADS

TOTAL LOADS

P

Q

P=2560k

H0

Q=22,280k

L/2

V0

REACTIONS M0

Figure 4.5: Eiffel Tower, Wind Loads, (Billington and Mark 1983)

+

=

WINDWARD SIDE

VERTICAL FORCES

WIND FORCES

LEEWARD SIDE

TOTAL

Figure 4.6: Eiffel Tower, Reactions; (Billington and Mark 1983)

Victor Saouma


Draft

4.4 Internal Forces

83

β=18.40 β=18.4 INCLINED INTERNAL FORCE: N

CONSEQUENT HORIZONTAL COMPONENT: H

N

KNOWN VERTICAL COMPONENT: V

V

H FORCE POLYGON

Figure 4.7: Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983) Gravity load are first considered, remember those are caused by the dead load and the live load, Fig. 4.7 4.7:: 12

V V N = N cos β 11 11,, 140 k N = = 11,730 kip cos18. cos18.4o H tan β = H = V tan β V H = 11 11,, 140 k(tan (tan 18. 18.4o ) = 3,700 kip cos β =

⇒ ⇒

(4.12-a) (4.12-b) (4.12-c) (4.12-d)

The horizontal forces which must be resisted by the foundations, Fig. 4.8 4.8..

H

H 3700 k

3700 k

Figure 4.8: Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983) 13

Because the vertical load decreases with height, the axial force will also decrease with height.

At the second platform, the total vertical load is Q = 1, 100 + 2, 2, 200 = 3, 3, 300 k and at that o height the angle is 11. 11.6 thus the axial force (per pair of columns) will be 14

N vert = vert

3,300 2

k

= 1, 685 k (4.13-a) cos11. cos11.6o 3, 300 k H vert = (tan (tan 11. 11.6o ) = 339 k (4.13-b) vert 2 Note that this is about seven times smaller than the axial force at the base, which for a given axial strength, would lead the designer to reduce (or taper) the cross-section. Victor Saouma


Draft Chapter 5

REVIEW of STATICS To ev ever ery y acti action on ther theree is an equ qual al and opposite reaction . Newton’s third law of motion

5.1 5.1

Reac Reacti tion onss

In the analysis of structures (hand calculations), it is often easier (but not always necessary) to start by determining the reactions.

1

Once Once the reacti reactions ons are determ determine ined, d, inter internal nal forces forces are determ determine ined d next; next; finally finally,, inter internal nal 1 stresses stresses and/or and/or deformation deformationss (deflections (deflections and rotations) rotations) are determined determined last . 2

3

Reactions are necessary to determine foundation load. load.

Depending on the type of structures, there can be different types of support conditions, Fig. 5.1.. 5.1

4

Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may provide restraint in one or two directions. A roller will allow rotation. Hinge: allows rotation but no displacements. Fixed Support: will prevent rotation and displacements in all directions. 5.1.1 5.1.1

Equi Equilib libriu rium m

5

Reactions are determined from the appropriate equations of static equilibrium.

6

Summation of forces and moments, in a static system must be equal to zero 2 . 1

This is the sequence of operations in the flexibility method which which lends itself to hand calculation. calculation. In the stiffness method, we determine displacemen displacements ts firsts, then internal internal forces and reactions. This method is most suitable to computer implementation. 2 In a dynamic system ΣF ΣF = ma where m is the mass and a is the acceleration acceleration..

Draft

5.1 Reactions

87

Structure Typ e Beam, no axial forces 2D Truss russ,, Frame rame,, Beam Beam Grid 3D Truss, Frame Beams, Beam s, no axia axiall Forc orce 2 D Truss russ,, F Fra rame me,, B Bea eam m

Equations ΣF x

ΣF y ΣF y

ΣM z ΣM z ΣF z ΣF z

ΣF x ΣF y Alternate Set ΣM zA ΣM zB ΣF x ΣM zA ΣM zB ΣM zA ΣM zB ΣM zC

ΣM x ΣM x

ΣM y ΣM y

ΣM z

Table 5.1: Equations of Equilibrium 2. Assume Assume a direction direction for the unknown unknown quantities quantities 3. The right right hand side of the equation should should be zero If your reaction is negative, then it will be in a direction opposite from the one assumed. Summation of external forces is equal and opposite to the inte intern rnal al ones. ones. Thus Thus the net force/moment is equal to zero.

16

17

The external forces give rise to the (non-zero) shear and moment diagram.

5.1.2 5.1.2

Equa Equatio tions ns of of Cond Conditi itions ons

If a structure has an internal internal hinge (which may connect two or more substructures), then this will provide an additional equation (ΣM (ΣM = 0 at the hinge) which can be exploited to determine the reactions.

18

Those equations are often exploited in trusses (where each connection is a hinge) to determine reactions. 19

In an inclined roller support with S x and S y horizontal and vertical projection, then the reaction R would have, Fig. 5.2 5.2..

20

Rx S y = Ry S x

(5.3)

Figure 5.2: Inclined Roller Support

Victor Saouma


Draft

5.1 Reactions

89

Figure Figure 5.4: Geometric Geometric Instability Instability Caused Caused by Concurrent Concurrent Reactions Reactions 5.1. 5.1.5 5

Exam Exampl ples es

Exampl Examples es of reaction reaction calcul calculatio ation n will be shown shown next. Each Each exampl examplee has been carefu carefully lly selected selected as it brings brings a different different “twist” “twist” from the preceding preceding one. Some of those same problems will be revisited revisited later for the determinati determination on of the internal internal forces and/or deflections. deflections. Many Many of those problems are taken from Prof. Gerstle textbok Basic Structural Analysis. Analysis. 29

Exampl Example e 5-1: Simply Simply Support Supported ed Beam Determine the reactions of the simply supported beam shown below.

Solution: The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate. (+ - ) ΣF x = 0; (+ 6) ΣF y = 0;  c (+  ¡) ΣM = 0; z

⇒ ⇒ ⇒

Rax 36 k = 0 Ray + Rdy 60 k (4) k/ft(12) ft = 0 12 12R Ray 6Rdy (60)(6) = 0

−

−

−

−

−

or through matrix inversion (on your calculator)

 

1 0 0 1 0 12

0 1 6

−

   

Rax Ray Rdy

 

=

 

36 108 360

   ⇒ 

Rax Ray Rdy

Alternatively we could have used another set of equations:  a (+  0;; (60)(6) (60)(6) + (48)(12) (48)(12) (Rdy )(18) = 0 ¡) ΣM z = 0  d (+  (60)(12) (48)(6) = 0 ¡) ΣM z = 0; (Ray )(18)

−

Victor Saouma

−

−

 

=

⇒ ⇒

 

36 k 56 k 52 k

 

Rdy = 52 k 6 Ray = 56 k 6


Draft

5.1 Reactions

91

2. Isolat Isolating ing bd:  (+  ¡) ΣM d = 0;

 (+  ¡) ΣM c = 0;

−(17. (17.7)(18) − (40)(15) − (4)(8)(8) − (30)(2) + Rcy (12) = 0 ⇒ Rcy = 1,12236 = 10 1033 k 6 −(17. (17.7)(6) − (40)(3) + (4)(8)(4) + (30)(10) − Rdy (12) = 0 201.3 ⇒ Rdy = 201. 16.7 .7 k 6 12 = 16

3. Check Check ΣF y = 0; 6; 22 22..2

√ − 40 − 40 + 103 − 32 − 30 + 16. 16.7 = 0

Exampl Example e 5-3: Three Three Hinged Hinged Gable Fram Frame e The three-hinged gable frames spaced at 30 ft. on center. Determine the reactions components nents on the frame due to: 1) Roof dead load, load, of 20 psf of roof area; 2) Snow load, of 30 psf of horizontal projection; 3) Wind load of 15 psf of vertical vertical projection. Determine Determine the critical design values for the vertical and horizontal reactions.

Solution: 1. Due to symmetry, symmetry, we will consider consider only the dead load on one side of the frame. Victor Saouma


Draft 5.2 Trusses

5.2 5.2

5.2.1 5.2.1

93

Truss russes es Assu Assump mptio tions ns

Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimensional components components which which transfer transfer only axial forces along their axis. 30

31

Trusses are extensively used for bridges, long span roofs, electric tower, space structures.

32

For trusses, it is assumed that 1. Bars Bars are pin-connected 2. Joints Joints are frictionless frictionless hinges4 . 3. Loads are are applied applied at the joints only. only.

A truss would typically be composed of triangular elements with the bars on the upper chord under compression and those along the lower chord under tension. Depending on the orientation of the diagonals, diagonals, they can be under either tension or compression. 33

In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.5 34

Figure 5.5: Bridge Truss

5.2.2 5.2.2

Basic Basic Relat Relation ionss

Sign Convention: Tension ension positive, positive, com compre pressi ssion on negativ negative. e. On a truss truss the axial forces forces are indicated as forces acting on the joints. Stress-Force: σ =

P A

Stress-Strain: σ = Eε Force-Displacement: ε =

∆L L

4

In practice the bars are riveted, bolted, or welded directly to each other or to gusset plates, thus the bars are not free to rotate and so-called secondary bending moments are developed at the bars. Another source of secondary moments is the dead weight of the element.

Victor Saouma


Draft 5.2 Trusses

95

Figure 5.6: A Statically Indeterminate Truss 3. Sketch Sketch a free body diagram showing showing all joint joint loads (including (including reactions) reactions) 4. For each each joint joint,, and starting starting with the loa loaded ded ones, apply apply the appropri appropriate ate equations equations of equilibrium (ΣF (ΣF x and ΣF ΣF y in 2D; ΣF ΣF x , ΣF y and ΣF ΣF z in 3D).  = F  x + F  y ) must 5. Because truss elements elements can only carry axial forces, the resultant resultant force (F be along the member, Fig. 5.7 5.7.. F F x F y = = l lx ly

(5.4)

Always keep track of the x and y components of a member force (F (F x , F y ), as those might be needed later on when considering the force equilibrium at another joint to which the member is connected. 44

Figure 5.7: X and Y Components of Truss Forces 45

This method should be used when all member forces should be determined.

convention. A member is assumed to be under tension In truss analysis, there is no sign convention. (or compression) compression).. If after analysis, analysis, the force is found found to be negative, negative, then this would would imply that the wrong assumption was made, and that the member should have been under compression (or tension). 46

Victor Saouma


Draft 5.2 Trusses

97

Node A: Clearly AH is under compression, and AB under tension.

(+ 6) ΣF y = 0;

⇒

F AH 58 = 0 AH y l F AH AH = ly (F AH AH y ) ly = 32 l = 322 + 242 = 40 40 F AH 72.5 Compression AH = 32 (58) = 72. F AH AH x + F AB AB = 0 lx 24 F AB 43.5 Tensi ension on AB = ly (F AH AH y ) = 32 (58) = 43.

−

√

(+ - ) ΣF x = 0;

⇒ ⇒ −

Node B:

(+ - ) ΣF x = 0; (+ 6) ΣF y = 0;

⇒ ⇒

F BC 43..5 Tensi ension on BC = 43 F BH Tension BH = 2 0

Node H:

(+ - ) ΣF x = 0; (+ 6) ΣF y = 0;

⇒ ⇒

F AH F HC F HG AH x HC x HG x = 0 24 √ 24 43 43..5 √242 +322 (F HC (F HG (I) HC ) HG ) = 0 242 +102 F AH 12 F HG 20 = 0 AH y + F HC HC y HG y 32 58 + √242 +322 (F HC 12 √2410 (F HG 20 = 0 (II (II) HC ) HG ) 2 +102

− −

−

−

− − − − −

−

Solving for I and II we obtain F HC 7.5 Tensi ension on HC = F HG Compression HG = 52

−

Victor Saouma


Draft

5.3 Shear & Moment Diagrams

99

+ Axial Force

+ve Load

+

+ve Shear

+ve Moment

Figure 5.9: Shear and Moment Sign Conventions for Design Load Positive along the beam’s local y axis (assuming a right hand side convention), that is positive upward. Axial: tension positive. Flexure A positive moment is one which causes tension in the lower fibers, and compression in the upper ones. For frame members, members, a positive moment moment is one which causes causes tension along the inner side. Shear A positive shear force is one which is “up” on a negative face, or “down” on a positiv positivee one. one. Altern Alternativ atively ely,, a pair pair of positiv positivee shear shear forces forces will cause cause clockwi clockwise se rotation. Torsion Counterclockwise positive

Draft

3D: Use double double arrow arrow vecto vectors rs (and (and NO NOT T curve curved d arrow arrows). s). Forces orces and mom momen ents ts (inclu (includin dingg torsions) are defined with respect to a right hand side coordinate system, Fig. 5.10 5.10..

Figure 5.10: Sign Conventions for 3D Frame Elements

5.3.1. 5.3.1.2 2

Load, Load, Shear, Shear, Momen Momentt Relati Relations ons

Let us (re)derive (re)derive the basic relations relations between between load, shear and moment. moment. Considering Considering an in6 finitesimal length dx of a beam subjected to a positive load w(x), Fig. 5.11 5.11.. The infinitesimal section must also be in equilibrium. 50

There are no axial forces, thus we only have two equations of equilibrium to satisfy ΣF Σ F y = 0 and ΣM ΣM z = 0. 51

Since dx is infinitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx. dx. 52

6

In this derivation, as in all other ones we should assume all quantities to be positive.

Victor Saouma


Draft


101

The change in moment between 1 and 2, ∆M 21 21 , is equal to the area under the shear curve between x1 and x2 . 57

Note that we still need to have V 1 and M 1 in order to obtain V 2 and M 2 respectively.

Fig. 5.12 and 5.13 further illustrates the variation in internal shear and moment under uniform and concentrated forces/moment.

58

Figure 5.12: Shear and Moment Forces at Different Sections of a Loaded Beam Positi Positive ve Consta Constant nt

Negati Negative ve Consta Constant nt

Positive Positive Increas Increasing ing Positive Positive Decrea Decreasing sing Negative Negative Increasing Increasing Negative Negative Decreasing

Positive Constant

Negative Constant

Positive Positive Increas Increasing ing Positive Positive Decrea Decreasing sing Negative Increasing Negative Decreasing

Load

Shear

Shear

Moment

Figure 5.13: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment

Victor Saouma


Draft


103

Reactions are determined determined from the equilibrium equilibrium equations (+ )ΣF x = 0;  (+  ¡) ΣM A = 0; (+ 6) ΣF y = 0;

⇒ −RA + 6 = 0 ⇒ RA = 6 k ⇒ (11)(4) + (8)(10) + (4)(2)(14 + 2) − RF (18) = 0 ⇒ RF ⇒ RA − 11 − 8 − (4)(2) + 14 = 0 ⇒ RA = 13 k x

x

y

y

y

= 14 14 k

y

Shear are determined next. 1. At A the shear is equal to the reaction and is positive. 2. At B the shear drops (negative load) by 11 k to 2 k. 3. At C it drops again by 8 k to

− 6 k.

4. It stays stays consta constant nt up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 k per linear foot up to 14 k. 5. As a check check,,

−

−14 k is also the reaction previously determined at F . F .

Moment is determined last: 1. The momen momentt at A is zero (hinge support). 2. The change change in moment moment between between A and B is equal to the area under the corresponding shear diagram, or ∆M ∆M B −A = (13)(4) = 52. Victor Saouma


Draft


Victor Saouma

105


Draft


107

B C 3. If we need need to determine the maximum maximum moment along B C , we know that dM dx =0 64. 64.06 at the point where V B−C = 0, that is V B −C (x) = 64. 64.06 3x = 0 x= 3 = 25 25..0 ft. In other words, maximum moment occurs where the shear is zero.

−

2

−

−

⇒

(25.0) Thus M Bmax 64.06(25. 06(25.0) 3 (25. = 432 + 1, 1, 601 601..5 937 937..5 = 232 k.ft −C = 432 + 64. 2 4. Finally Finally along along C D, the moment varies quadratically (since we had a linear shear), the moment first increases (positive shear), and then decreases (negative shear). The moment along C D is given by

− − −

−



−

x M C 139.8 + C −D = M C C + 0 V C C −D (x)dx = 139. x2 = 139.8 + 13. 13.22 22x x 32

−

−



x (13.22 0 (13.

− 3x)dx

which is a parabola. Substituting for x = 15, we obtain at node C M C 139.8 + 13. 13.22(15) C = 139. 139..8 + 198. 139 198.3 337 337..5 = 0

−

√

− 3 152

2

=

Example Example 5-7: Frame Shear and Moment Moment Diagram; Diagram; Hydrostatic Hydrostatic Load The frame shown below is the structural support of a flume. Assuming that the frames are spaced 2 ft apart along the length of the flume, 1. Determine Determine all internal member member end actions actions 2. Draw Draw the shear and moment diagrams diagrams 3. Locate and compute maximum maximum internal internal bending moments 4. If this is a reinforced reinforced concrete frame, frame, show the location of the reinforcem reinforcement ent..

Solution: The hydrostatic pressure causes lateral forces on the vertical members which can be treated as cantilevers fixed at the lower end. Victor Saouma


Draft


109

1. Base at B the shear force was determined earlier and was equal to 2. 2 .246 k. Ba Base sed d on the orientation of the x y axis, this is a negative shear.

−

2. The vertical vertical shear shear at B is zero (neglecting the weight of A of A 3. The shear shear to the left left of C of C is V = 0 + ( .749)(3) = 4. The shear shear to the right right of C is V =

−

− B)

−2.246 k.

−2.246 + 5.5.99 = 3.3.744 k

Moment diagrams 1. At the the base: base: B M = 4.493 k.ft as determined above.

−4.493 + (−.749)(3)( 32 ) = −7.864 k.ft 3. The maximum maximum moment moment is equal to M max (.749)(5)( 52 ) = 1.50 k.ft max = −7.864 + (.

2. At the suppor supportt C , M c =

Design: Reinforcement should be placed along the fibers which are under tension, that is on the side of the negative moment7 . The figure below schematically illustrates the location of the flexural flexural8 reinforcement.

Example Example 5-8: Shear Moment Moment Diagrams Diagrams for Frame Frame 7

That is why in most European countries, the sign convention for design moments is the opposite of the one commonly used in the U.S.A.; Reinforcement should be placed where the moment is “postive”. 8 Shear reinforcement is made of a series of vertical stirrups.

Victor Saouma


Draft


111

Example Example 5-9: Shear Moment Moment Diagrams Diagrams for Inclined Frame Frame

26k

26k

10’

13’

10’

13’ 13’

20k C

13 5

5

3

12

15’

4

D

B 2k/ft

20’ Ha

A

E

36’

20’

Ve

Va 2 / ) 0 2 ( ) 0 2 ( ) 2 ( ) 0 2 ( ) 0 6 (

19.2k

’ k 0 0 8

20k

)

20k 0 2 ( ) 2 ( 0 6

2k/ft

60k

48.8k

2 / ) 0 2 ( ) 0 2 0 6 ( + ) 0 2 ( ) 0 2 (

Fx

800’k

F

0k’

60k 2

26k

10k

26k

10k

k 4

778k’ 0k 2 0 1 0 1 0 2

26.6k 11.1k 7.69k

28.8k

20k

18.46k 7.38k

17.72k

3

k

4 2

800k’

. + 2 5

k 1 . 3 2

48.8k

778k’

1,130-(.58)(13) 2 k ’ 800+(25.4)(13) 1 1 2

k 0 ’ k 1 1 3

8 B-C ) 3 1 ( ) 6 . 6 2 ( 2 2 1 , 1

’ 7 7 7 k

9 B-C

13

k 0 2 +

- 2 23 .1 k 6 k 5 8 k - 2 6. 6 - 0. 5 - 3 39 .1 k

(20)(15)/13=7.7

0 2 k 6 1

(20)(12)/(13)=18.46 (19.2)(5)/(13)=7.38

0k’ 39.1k

29.3k

48.9k

- 2 3 .1 k - 3 - 2 9 3 .1 .1 k - 1 1 6

’ k 7 7 7

) 5 . 2 1 ( ) 1 . 3 2 ( + 8 8 4

10

11 C-D

(19.2)(12)/(13)=17.72 (26)(12)/(13)=24 (26.6)(13)/(12)=28.8 (26.6)(5)/(12)=11.1 (28.8)(4)/(5)=23.1 (28.8)(3)/(5)=17.28 (20)(4)/(5)=16

(39.1)(12.5)

488’k

(20)(3)/(5)=12 (39.1)(5)/(4)=48.9 12 C-D

14

. 4 k + 2 5

6

1 7 . 2 8 k 1 2 k k

0k CD

6 k . 6 8 k - 2 6 - 0. 5 - 2 6 - 0. 6 - 2 6

5

28.8k

7

2 2 5. 4

k ’ k 8 0 0

4

BC

19.2k 7 + 7. 7 1 7. 7 4 k

F/Fy=z/x F/Fx=z/y Fx/Fy=y/x

ED 1

y

y F

x

AB 19.2k

z

k ’ 1 2 2 1 k 0 ’ k 1 1 3

(39.1)(3)/(4)=29.3 777k’ 4 8 8 ’ k

800’k

k 0 6 +

Victor Saouma


Draft 5.4 Flexure

1 13

where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to the distance from the neutral axis. ρ (Greek letter rho) rho) is the radius of curvature. curvature. In some textbook, the curvature κ (Greek letter kappa ) is also used where 1 κ= (5.14) ρ 64

thus, εx = 5.4.2 5.4.2

−κy

(5.15)

StressStress-Stra Strain in Relatio Relations ns

So far we considered the kinematic of the beam, yet later on we will need to consider equilibrium in terms of the stresses. Hence we need to relate strain to stress.

65

66

For linear elastic material Hooke’s law states σx = Eε x

(5.16)

where E is Young’s Modulus. Modulus. 67

Combining Eq. with equation 5.15 we obtain σx =

5.4.3 5.4.3

−Eκy

(5.17)

Intern Internal al Equilib Equilibrium rium;; Section Section Properti Properties es

Just as external external forces acting on a structure structure must be in equilibrium, equilibrium, the internal internal forces forces must also satisfy satisfy the equilibrium equilibrium equations. equations. 68

The internal forces are determined by slicing the beam. The intern internal al forces forces on the “cut” section must be in equilibrium with the external forces.

69

5.4.3.1 70

ΣF x = 0; Neutral Axis

The first equation we consider is the summation of axial forces.

Since there are no external axial forces (unlike a column or a beam-column), the internal axial forces must be in equilibrium.

71

ΣF x = 0

 ⇒

A

σx dA = 0

(5.18)

where σx was given by Eq. 5.17 5.17,, substituting we obtain



A

Victor Saouma

σx dA =

 −

A

EκydA = 0

(5.19-a)


Draft 5.4 Flexure

1 15

Y

Y

A x y I x I y

x

h

X y

= bh = 2b = h2 3 = bh 12 3 = hb 12

h h’

X y b’

b

= bh b h = 2b = h2 3 3 = bh −12b h 3 3 = hb −12h b

A x y I x I y

x

−









b

c

Y

Y

a

h

A = y =

X y

I x =

h(a+b) 2 h(2a (2a+b) 3(a 3(a+b) h3 (a2 +4ab +4ab+ + b2 36(a 36(a+b)

x

h y

X

b

b

A x y I x I y

= = = = =

bh 2 b+c 3 h 3 bh3 36 bh 2 36 (b

− bc + c2)

Y

Y

r X

A = πr 2 = 4 I x = I y = πr4 =

πd 2 4 πd 4 64

t

r

X

A = 2πrt = πdt 3 I x = I y = πr 3 t = πd8 t

Y

b X b a

A = πab 3 I x = πab 3 3 I y = πba 4

a

Table 5.3: Section Properties

Victor Saouma


Draft 5.4 Flexure

1 17

5. We now set those two values values equal to their respective respective maximum maximum ∆max σmax

5.4.5 5.4.5

(20) ft(12) in/ft 65 65..65 L = = = 0.67 in = 360 360 r3 764 764 = (18) ksi = 2 r= = 6.51 6.51 in 18 r

⇒



3

65 65..65 = 4.61 (5.31-a) in 0.67 (5.31-b)

Appro Approxim ximate ate Analysi Analysiss

M From Fig. 5.14 5.14,, and Eq. 5.25 ( EI = κ= proportional to the curvature κ.

80

81

⇒r=



1 ρ ),

we recall that that the moment is directly

Thus, 1. A positive positive and negative negative moment moment would would correspond correspond to positive positive and negative curvature curvature respectively (adopting the sign convention shown in Fig. 5.14 5.14). ). 2. A zero moment correspn correspnds ds to an inflection inflection point in the deflected shape.

82

Hence, Hence, for

Statically determinate structure, we can determine the deflected shape from the moment diagram, Fig. 5.15 5.15.. Statically indeterminate structure, we can: 1. Plot the deflected deflected shape. shape. 2. Identify Identify inflection points, approximate approximate their location. 3. Locate Locate those those inflect inflection ion points points on the struct structure ure,, which which will will then then become become static statically ally determinate. 4. Perform Perform an approximate approximate analysis. analysis.

Example Example 5-11: Approximate Approximate Analysi Analysiss of a Statically Indetermi Indeterminate nate beam Perform an approximate analysis of the following beam, and compare your results with the exact solution.

20k 16’ 12’ 28’ Victor Saouma

28’


Draft 5.4 Flexure

1 19

Figure 5.16: Approximate Analysis of Beams

Victor Saouma


Draft 5.4 Flexure

1 21

6. Check Check  (+  (20)(16 16)) ¡) ΣM A = 0; (20)(

− (RC )(28) + (R (RD )(28 + 28) = 320 − (17. (17.67)(28) + (3. (3.12)(56) = √ 320 − 494 494..76 + 174. 174.72 = 0

(5.34-a)

7. The moments moments are determined determined next M max (5.45)(16) = 87.2 max = RA a = (5.

(5.35-a)

M 1 = RD L = (3. (3.12)(28) = 87.36

(5.35-b)

8. We now now com compar paree with with the exact exact solutio solution n from from Sectio Section n ??, ??, solution 21 where:L where:L = 28, a = 16, b = 12, and P = 20 R1 = RA = = R2 = RB = = R3 = RD = =

Pb 4L2 a(L + a) 3 4L (20)(12) 4(28)2 (16)(28 + 16) = 6.64 4(28)3 Pa 2L2 + b(L + a) 2L3 (20)(16) 2(28)2 + 12(28 + 16) = 15.28 2(28)3 P ab (L + a) 4L3 (20)(16)(12) (28 + 16) = 1.92 4(28)3

 



−

 



−



− −



−

M max (6.64)(16) = 106.2 max = R1 a = (6. M 1 = R3 L = (1. (1.92)(28) = 53.8

(5.36-a) (5.36-b) (5.36-c) (5.36-d) (5.36-e) (5.36-f) (5.36-g)

9. If we tabulate the results results we have Value alue RA RC RD M 1 M max max

Appr Approoxima ximate te 5.45 17.67 3.12 87.36 87.2

Exac Exactt 6.64 15.28 1.92 53.8 106.2

% Erro Errorr 18 -16 63 62 18

10. Wherea Whereass the correl correlati ation on betwe between en the appro approxim ximate ate and exact exact result resultss is quite quite poor, one should not underestimate the simplicity of this method keeping in mind (an exact analysis of this structure structure would would have have been computationally computationally much more involv involved). ed). Furthermore urthermore,, often one only needs a rough order of magnitude of the moments.

Victor Saouma


Draft Chapter 6

Case Study II: GEORGE WASHINGTON BRIDGE 6.1 6.1

Theor eory

Whereas the forces in a cable can be determined from statics alone, its configuration must be derived derived from its deformation. deformation. Let us consider a cable with distributed distributed load p(x) per unit horizontal horizontal projection of the cable cable length length (thus (thus neglecti neglecting ng the weigh weightt of the cable) cable).. An infinitesimal portion of that cable can be assumed to be a straight line, Fig. 6.1 and in the absence of any horizontal load we have H =constant. Summation of the vertical forces yields 1

(+ ?) ΣF y = 0

⇒ −V + wdx + (V (V + dV ) dV )

= 0

(6.1-a)

dV + wdx = 0

(6.1-b)

where V is the vertical component of the cable tension at x (Note that if the cable was subjected to its own weight then we would have wds instead of wdx). wdx). Because the cable must be tangent to T , T , we have V tan θ = (6.2) H Substituting into Eq. 6.1-b yields d(H tan H tan θ) + wdx = 0 2

⇒ − dxd (H tan H tan θ) = w

(6.3)

But H is constant (no horizontal load is applied), thus, this last equation can be rewritten as

− H dxd (tan θ) = w Written in terms of the vertical displacement v , tan θ = 6.4 yields yields the governing equation for cables

3

− H v = w

(6.4) dv dx

which when substituted in Eq. (6.5)

For a cable subjected to a uniform load w, we can determine its shape by double integration of Eq. 6.5 4

− H v

= wx + C 1

(6.6-a)

Draft 6.1 Theory

1 25

wx2 + C 1 x + C 2 (6.6-b) 2 and the constants of integrations C 1 and C 2 can be obtained from the boundary conditions: v = 0 at x = 0 and at x = L C 2 = 0 and C 1 = wL 2 . Thus w (6.7) v= x(L x) 2H This equation gives the shape v (x) in terms of the horizontal force H ,

−H v

=

⇒

5

− −

Since the maximum sag h occurs at midspan (x (x =

L 2)

we can solve for the horizontal force

wL2 8h

H =

(6.8)

we note the analogy with the maximum moment in a simply supported uniformly loaded beam 2 M = H h = wL 8 . Furthermore, this relation clearly shows that the horizontal force is inversely proportional to the sag h, as h H . Finally, we can rewrite this equation as

 

r

def

=

h L

(6.9-a)

=

8r

(6.9-b)

wL H 6

Eliminating H from Eq. 6.7 and 6.8 we obtain v = 4h



−

x2 x + L2 L



(6.10)

Thus the cable assumes a parabolic shape (as the moment diagram of the applied load). Whereas the horizontal force H is constant throughout the cable, the tension T is not. The maximum tension occurs at the support where the vertical component is equal to V = wL 2 and the horizontal one to H , thus 7

T max max =



V 2 + H 2 =

   wL 2

  

2

+ H 2 = H 1 +

wL/2 wL/2 H

2

(6.11)

Combining this with Eq. 6.8 we obtain1 .



T max 16r 2 max = H 1 + 16r

≈ H (1 (1 + 8r 8r 2 )

(6.12)

Had we assumed a uniform load w per length of cable (rather than horizontal projection), the equation would have been one of a catenary 2 .

8

  − 

H w L v= cosh w H 2

x

(6.13)

+h

The cable between transmission towers is a good example of a catenary. 2

Recalling that (a (a + b)n = an + nan 1 b + n(n2! 1) an 2 b2 + · or (1+b (1+ b)n = 1 + nb+ nb + n(n 2!1)b + n(n √ Thus for b2 << 1, 1 + b = (1 + b) ≈ 1 + 2b 2 Derivation of this equation is beyond the scope of this course. 1

−

1 2

Victor Saouma

−

−

−

1)(n−2)b3 3!

−

+ · · ·;


Draft

6.2 The Case Study

12 7

tower supports and are firmly anchored in both banks by huge blocks of concrete, the anchors. Because the cables are much longer than they are thick (large LI ), they can be idealized a perfectly flexible members with no shear/bending resistance but with high axial strength.

14

The towers are 578 ft tall and rest on concrete caissons in the river. Because of our assumption regarding the roller support for the cables, the towers will be subjected only to axial forces.

15

6.2. 6.2.2 2

Load Lo adss

The dead load is composed of the weight of the deck and the cables and is estimated at 390 and 400 psf respectively for the central and side spans respectively. Assuming an average width of 100 ft, this would be equivalent to

16

DL = (390) psf (100) (100) ft

k

(1, (1, 000) lbs

= 39 k/ft

(6.14)

for the main span and 40 k/ft for the side ones. For highway bridges, design loads are given by the AASHTO (Association of American State Highwa Highway y Transportat Transportation ion Officials). The HS-20 truck is often used for the design of bridges bridges on main highways, Fig. 6.3 6.3.. Either the design truck with specified axle loads and spacing must be used or the equivalent uniform load and concentrated load. This loading must be placed such that maximum maximum stresses stresses are produced. produced. 17

Figure 6.3: Truck Load 18

With two decks, we estimate that there is a total of 12 lanes or LL = (12)Lanes(. (12)Lanes(.64) k/ ft/Lane = 7. 7.68 k/ft

≈ 8 k/ft

(6.15)

We do not consider earthquake, or wind loads in this analysis. 19

Final DL and LL are, Fig. 6.4 6.4:: T L = 39 + 8 = 47 k/ft

Victor Saouma


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6.2 The Case Study

6.2.3 6.2.3

12 9

Cable Cable Forces orces

The thrust H (which is the horizontal component of the cable force) is determined from Eq. 6.8 2 wLcs H = 8h

20

(47)

2 (3,500)2 ft k/ft(3,

= (8)(327) ft = 220, 000 k From Eq. 6.12 the maximum tension is

r = Lh = 3327 ,500 = 0.0934 cs T max 16r 2 max = H 1 + 16r = (2, (2, 200) k 1 + (16)(0. (16)(0.0934)2 = (2, (2, 200) k(1. (1.067 0675) 5) = 235 235,000 ,000 k

√



6.2.4 6.2.4 21

Reac Reactio tions ns

Cable reactions are shown in Fig. 6.5 6.5.. POINTS WITH REACTIONS TO CABLES

Figure 6.5: Location of Cable Reactions The vertical force in the columns due to the central span (cs) is simply the support reaction, 6.6

22

w TOT= 39 + 8 = 47 k/ft B

A

REACTIONS AT TOP OF TOWER

POINT OF NO MOMENT

L = 3,500 FT

Figure 6.6: Vertical Reactions in Columns Due to Central Span Load 1 1 V cs (3, 500) ft = 82 82,, 250 k cs = wLcs = (47) k/ft(3, 2 2 Victor Saouma

(6.16)


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6.2 The Case Study

27

13 1

The cable stresses are determined last, Fig. 6.8 6.8:: Awire = Atotal = Central Span σ = ss Side Span Tower σtower = ss Side Span Anchor σtower =

πD 2 (3. (3.14)(0. 14)(0.196)2 = = 0.03017 in2 (6.22-a) 4 4 (4)cab (4)cables les(26 (26,, 474)wires/cable(0. 474)wires/cable(0.03017) in2 /wire = 3, 3, 200 (6.22-b) in2 H (220, (220, 000) k = = 68 68..75 ksi (6.22-c) A (3, (3, 200) in2 ss T tower (262, (262, 500) in2 = = 82 ksi (6.22-d) A (3, (3, 200) in2 ss T anchor (247, (247, 000) in2 = = 77 77..2 ksi (6.22-e) A (3, (3, 200) in2

73.4 ksi

81.9 ksi

68.75 ksi

77.2 ksi

Figure 6.8: Cable Stresses If the cables were to be anchored to a concrete block, the volume of the block should be at (112,000) k(1, (1,000) lbs/ k least equal to V = (112, = 747, 747, 000 ft3 or a cube of approximately 91 ft 3 28

150

lbs/ft

The deck, for all practical purposes can be treated as a continuous beam supported by elastic springs with stiffness K = AL/E (where L is the length of the supporting cable). This is often idealized as a beam on elastic foundations, and the resulting shear and moment diagrams for this idealization are shown in Fig. 6.9 6.9.. 29

Victor Saouma


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A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE If I have been able to see a little farther than some others, it was because I stood on the shoulders of giants . Sir Isaac Newton

More than any other engineering discipline, Architecture/Mechanics/Structures is the proud outcome of a of a long and distinguished history. Our profession, second oldest, would be better appreciated if we were to develop a sense of our evolution.

1

7.1 7.1

Befo Before re the the Gree Greeks ks

Throughout Throughout antiquity antiquity,, structural structural engineering engineering existing as an art rather than a science. science. No record exists of any rational consideration, either as to the strength of structural members or as to the behavior behavior of structural structural materials. materials. The builders builders were guided by rules of thumbs thumbs and experience, which were passed from generation to generation, guarded by secrets of the guild, and seldom supplemented by new knowledge. Despite this, structures erected before Galileo are by modern standards quite phenomenal (pyramids, Via Appia, aqueducs, Colisseums, Gothic cathedrals to name a few). 2

Imhotep, one of only two comThe first structural engineer in history seems to have been Imhotep, moners to be deified. He was the builder of the step pyramid of Sakkara about 3,000 B.C., and yielded great influence over ancient Egypt.

3

Hamurrabi’s code in Babylonia (1750 BC) included among its 282 laws penalties for those “architects” whose houses collapsed, Fig. 7.1 7.1..

4

7.2 7.2

Greeks

The greek philosopher Pythagoras (born around 582 B.C.) founded his famous school, which was primarily a secret religious society, at Crotona in southern Italy. At his school he allowed

5

Draft 7.3 Romans

1 35    

Figure 7.2: Archimed conqueror of Syracuse.

7.3 7.3

Romans

Science Science made much less progress progress under the Romans than under the Greeks. The Romans apparently were more practical, and were not as interested in abstract thinking though they were excellent fighters and builders.

10

As the roman empire expanded, the Romans built great roads (some of them still in use) such as the Via Appia, Cassia, Aurelia; Also they built great bridges (such as the third of a mile bridge over the Rhine built by Caesars), and stadium (Colliseum).

11

12

One of the most notable Roman construction was the Pantheon, Pantheon, Fig. 7.3 7.3.. It is the best-

Figure 7.3: Pantheon preserved major edifice of ancient Rome and one of the most significant buildings in architectural histor history y. In shape it is an imm immens ensee cylind cylinder er concealin concealingg eight eight piers, piers, topped topped with a dome dome and fronted by a rectangular colonnaded porch. The great vaulted dome is 43 m (142 ft) in diameter, and the entire structure is lighted through one aperture, called an oculus, oculus, in the center of the dome. The Pantheon was erected by the Roman emperor Hadrian between AD 118 and 128. Victor Saouma


Draft

7.4 The Medieval Perio d (477-1492)

137

Figure 7.5: Hagia Sophia dieval masons’ efforts to solve the problems associated with supporting heavy masonry ceiling vaults over over wide spans. spans. The problem was that the heavy heavy stonework stonework of the traditional traditional arched barrel vault and the groin vault exerted a tremendous downward and outward pressure that tended to push the walls upon which the vault rested outward, thus collapsing them. A building’s vertical supporting walls thus had to be made extremely thick and heavy in order to contain the barrel vault’s outward thrust. Medieval masons solved this difficult problem about 1120 with a number of brilliant innovations. ations. First and foremost they developed developed a ribbed vault, vault, in which which arching and intersecting intersecting stone ribs support support a vaulted ceiling ceiling surface that is composed of mere thin stone panels. This greatly reduced the weight (and thus the outward thrust) of the ceiling vault, and since the vault’s weight was now carried at discrete points (the ribs) rather than along a continuous wall edge, separate widely spaced vertical piers to support the ribs could replace the continuous thick thick walls. walls. The round arches arches of the barrel barrel vault ault were were replac replaced ed by pointed pointed (Gothi (Gothic) c) arches arches which distributed thrust in more directions downward from the topmost point of the arch. Since the combination combination of ribs and piers relieved relieved the interv intervening ening vertical vertical wall spaces of their supportive function, these walls could be built thinner and could even be opened up with large windo windows ws or other other glazing. glazing. A crucia cruciall point point was was that that the outward outward thrust thrust of the ribbed ceiling ceiling vaults was carried across the outside walls of the nave, first to an attached outer buttress and then then to a freest freestand anding ing pier by means means of a half half arch arch known known as a flying flying buttress buttress.. The flying flying buttress leaned against the upper exterior of the nave (thus counteracting the vault’s outward thrust), crossed over the low side aisles of the nave, and terminated in the freestanding buttress pier, which ultimately absorbed the ceiling vault’s thrust. These elements enabled Gothic masons to build much larger and taller buildings than their Romanesqu Romanesquee predecess predecessors ors and to give their structures structures more complicated complicated ground plans. The skillful use of flying buttresses made it possible to build extremely tall, thin-walled buildings whose interior interior structural structural system of columnar columnar piers and ribs reinforced reinforced an impression impression of soaring soaring verticality. 19

Vilet-Le-Duc classical book, (le Duc 1977) provided an in depth study of Gothic architecture.

Victor Saouma


Draft

7.5 The Renaissance

13 9

Unfortunatly, these important findings, were buried in his notes, and engineers in the fifteenth and sixteenth centuries continued, as in the Roman era, to fix dimensions of structural elements by relying on experience and judgment.

31

7.5.2 7.5.2

Brunelle Brunellesc schi hi 1377-144 1377-1446 6

Brunelleschi was a Florentine architect and one of the initiators of the Italian Renaissance. His revival of classical forms and his championing of an architecture based on mathematics, proportion, and perspective make him a key artistic figure in the transition from the Middle Ages to the modern era.

32

He was born in Florence in 1377 and received his early training as an artisan in silver and gold. In 1401 he entered, and lost, the famous design competition for the bronze doors of the Florence Baptistery Baptistery.. He then turned turned to architect architecture ure and in 1418 receive received d the commission commission to execute the dome of the unfinished Gothic Cathedral of Florence, also called the Duomo. Duomo. The dome, Fig. 7.6 a great innovation both artistically and technically, consists of two octagonal vaults, one 33

Figure 7.6: Florence’s Cathedral Dome inside inside the other. other. Its shape was dictated dictated by its struct structura urall needs needs one of the first first exampl examples es of architectural functionalism. functionalism. Brunelleschi made a design feature of the necessary eight ribs of the vault, carrying carrying them over to the exterior of the dome, where where they provide provide the framework framework for the dome’s decorative elements, which also include architectural reliefs, circular windows, and a beautifully proportioned proportioned cupola. cupola. This was the first time that a dome created the same strong effect on the exterior as it did on the interior. Completely different from the emotional, elaborate Gothic mode that still prevailed in his time, Brunellesc Brunelleschi’s hi’s style style emphasized emphasized mathematical mathematical rigor in its use of straight lines, flat 34

Victor Saouma


Draft

7.5 The Renaissance

14 1

Figure 7.7: Palladio’s Villa Rotunda

Victor Saouma


Draft

7.5 The Renaissance

14 3



Figure 7.9: Galileo was born. His contract was not renewed in 1592, probably because he contradicted Aristotelian professors professors.. The same year, he was appointed appointed to the chair of mathematics at the Universit University y of Padua, where he remained until 1610. In Padua he achieved great fame, and lecture halls capable of containing 2,000 students from all over over Europe were used. In 1592 he wrote Della Scienza Meccanica in Meccanica in which various problems displacement. He subsequen of statics were treated using the principle of virtual displacement. subsequently tly became interested in astronomy and built one of the first telescope through which he saw Jupiter and became an ardent proponent of the Copernican theory (which stated that the planets circle the sun as opposed to the Aristotelian and Ptolemaic assumptions that it was the sun which was circling Earth). This theory being condemned by the church, he received a semiofficial warning to avoid theology theology and limit himself to physical physical reasoning. reasoning. When he published his books dealing dealing with the two ways of regarding the universe (which clearly favored the Copernican theory) he was called to Rome by the Inquisition, condemned and had to read his recantation (At the end of his process he murmured the famous e pur se muove). muove). 50

When he was almost seventy years old, his life shattered by the Inquisition, he retired to his villa near Florence and wrote his final book, Discourses Concerning Two New Sciences, Sciences, (Galilei (Galilei 1974), Fig. 7.10 7.10.. His first science was the study of the forces that hold objects together 51

Figure Figure 7.10: Discourses Concerning Two New Sciences, Sciences, Cover Page Victor Saouma


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7.6 Pre Mo dern Perio d, Seventeenth Century

145

appointed Gresham Professor of Geometry at Oxford in 1665. After the Great Fire of London in 1666, he was appointed surveyor of London, and he designed many buildings. Hooke anticipated some of the most important discoveries and inventions of his time but failed to carry many of them through through to completion. completion. He formulated formulated the theory of planetary motion as a problem in mechanics, and grasped, but did not develop mathematically, the fundamental theory on which Newton formulated the law of gravitation. 57

His most important contribution was published in 1678 in the paper De Potentia Restitutiva . It contained results of his experiments with elastic bodies, and was the first paper in which the elastic properties of material was discused, Fig. 7.12 7.12..

58



Figure 7.12: Experimental Set Up Used by Hooke “Take a wire string of 20, or 30, or 40 ft long, and fasten the upper part thereof to a nail, and to the other end fasten a Scale to receive the weights: Then with a pair of compasses take the distance of the bottom of the scale from the ground or floor underneath, and set down the said distance,then put inweights into the said scale and measur measuree the several several stretc stretchin hings gs of the said string string,, and set them them down. down. Then Then compare the several strtchings of the said string, and you will find that they will always bear the same proportions one to the other that the weights do that made them”. them”. This became Hooke’s Law σ = Eε. Eε . Because he was concerned about patent rights to his invention, he did not publish his law when first discovered it in 1660. Instead he published it in the form of an anagram “ceiinossst“ ceiinosssttuu ” in 1676 and the solution was given in 1678. Ut tensio sic vis (at the time the two symbos u and v were employed interchangeably to denote either the vowel u or the consonant v ), ), i.e. extension varies directly with force. force. 59

7.6.2 7.6.2 60

Newt Newton, on, 164 1642-1 2-1727 727

Born on christmas day in the year of Galileo’s death, Newton, Fig. 7.13 was Professor of

Victor Saouma


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7.6 Pre Mo dern Perio d, Seventeenth Century

147

The Principia’s appearance appearance also involv involved ed Newton in an unpleasant unpleasant episode with the English philosopher and physicist Robert Hooke. In 1687 Hooke claimed that Newton had stolen from him a central idea of the book: that bodies attract each other with a force that varies inversely as the square square of their their distan distance. ce. Ho Howe weve ver, r, most most histor historian ianss do not accept accept Hooke’s Hooke’s charge charge of plagiarism. 63

Newton Newton also also engage engaged d in a violen violentt disput disputee with Leibniz Leibniz over over priori priority ty in the inve invent ntion ion of calculus. calculus. Newton Newton used his position as president president of the Royal Royal Society to have a committee of that body investigate the question, and he secretly wrote the committee’s report, which charged Leibniz Leibniz with deliberate deliberate plagiarism. plagiarism. Newton Newton also compiled compiled the book of evidence evidence that the society society published. The effects of the quarrel lingered nearly until his death in 1727. 64

In addition to science, Newton also showed an interest in alchemy, mysticism, and theology. Many pages of his notes and writings particularly from the later years of his career are devoted to these topics. Howev However, er, historians historians have found little connection connection between between these interests interests and Newton’s scientific work. 65

7.6.3 7.6.3

Bernoull Bernoullii Fami Family ly 1654-178 1654-1782 2

The Bernouilli family originally lived in Antwerp, but because of religious persecution, they left Holland and settled in Basel. Near the end of the sevente seventeenth enth century century this family produced outstanding outstanding mathematician mathematicianss for more than a hundred hundred years. years. Jacob and John were were brothers. John was the father of Daniel, and Euler his pupil. 66

Whereas Galileo (and Mariotte) investigated the strength of beams (Strength), Jacob Bernoulli (1654-1705) made calculation of their deflection (Stiffness) and did not contribute to our knowledge of physical properties. Jacob Bernouilli is also credited in being the first to to have assumed that a bf plane section of a beam remains plane during bending, but assumed rotation to be with respect to the lower fiber (as Galileo did) and this resulted in an erroneous erroneous solution (where is the exact location of the axis of rotation?). He also showed showed that the curvature curvature at any point along a beam is proportional to the curvature of the deflection curve. 67

Bernoulli made the first analytical contribution to the problem of elastic flexure of a beam. In 1691 he published a logogriph Qrzumubapt dxqopddbbp ... whose secret was revealed in 1694. A letter is replaced by the next in the Latin alphabet, the second by the letter three away, and the third by the letter six away, so that aaaaa would aaaaa would be encoded as bdgbd . The logogriph reads Portio axis applicatem... and the decoded is that the radius of curvature at any point of an initially straight beam in inversely proportional to the value of the bending moment at that point. 68

Daniel Bernoulli (1700-1782) first postulated that a force can be decomposed into its equivalent (“Potentiis (“Potentiis quibuscunque possunt substitui earundem aequivalentes”. aequivalentes”. Another hypothesis defined the sum of two “conspiring” forces applied to the same point. According to Bernoulli, this “necessary truth” follows from the metaphysical principle that the whole equalts the sum of its parts, (Penvenuto 1991). 69

7.6.4 7.6.4

Eule Euler r 170 1707-1 7-1783 783

Leonhard Euler was born in Basel and early on caught the attention of John Bernoulli whose teaching teaching was attracting attracting young mathematicians mathematicians from all over over Europe, Fig. 7.15 7.15.. He 70

Victor Saouma


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7.7 The pre-Mo dern Perio d; Coulomb and Navier

7.7 7.7 75

1 49

The The pre-Mod pre-Moder ern n Period; Period; Coul Coulom omb b and Navi Navier er

Coulomb (1736-1806) was a French military engineer, Fig. 7.16 7.16,, as was the first to publish

Figure 7.16: Coulomb the correct analysis of the fiber stresses in flexed beam with rectangular cross section (Sur ( Sur une Applica Application tion des R´ egles egles de maximis maximis et minimis minimis a` quelques probl` problèmes emes de statique relatifs relatifs a` l’architecture in 1773). He used Hooke’s Hooke’s law, placed placed the neutral neutral axis in its exact exact positio position, n, developed the equilibrium of forces on the cross section with external forces, and then correcly determined the stresses. He also worked on friction (“Coulomb friction”) and on earth pressure. Coulom Coulomb b did also resear research ch on mag magnet netism ism,, frictio friction, n, and electricit electricity y. In 177 17777 he inve invent nted ed the torsion torsion balanc balancee for measurin measuringg the force of mag magnet netic ic and electrica electricall attract attraction ion.. With With this this invention, Coulomb was able to formulate the principle, now known as Coulomb’s law, governing the interaction interaction between between electric charges. charges. In 1779 Coulomb Coulomb published published the treatise treatise Theorie des machines simples (Theory of Simple Machines), an analysis of friction in machinery. After the war Coulomb came out of retirement and assisted the new government in devising a metric system system of weights weights and measures. measures. The unit of quantity quantity used to measure measure electrical electrical charges, charges, the coulomb, was named for him. 76

Navier (1785-1836) Navier was educated at the Ecole Polytechnique and became a professor there in 1831. Whereas Whereas the famous memoir of Coulomb Coulomb (1773) contained contained the correct solution solution to numerous important problems in mechanics of materials, it took engineers more than forty years to understand them correctly and to use them in practical application 77

In 1826 he published his Le¸ cons cons (lecture notes) which is considered the first great textbook in mechanics for engineering. In it he developed the first general theory of elastic solids as well as the first systematic treatment of the theory of structures.

78

It should be noted that no clear division existed between the theory of elasticity and the theory of structures until about the middle of the nineteenth century (Coulomb and Navier would today be considered professional structural engineers).

79

Three other structural engineers who pioneered the development of the theory of elasticity from that point on were were Lam Lam´é, e, Clapeyron Clapeyron and de Saint-Venan Saint-Venant. t. Lam’e published published the first book on elasticity in 1852, and credited Clapeyron for the theorem of equality between external and inter internal nal work work.. de Saint Saint-V -Vena enant nt was was perhaps perhaps the greate greatest st elastic elastician ianss who accord according ing to Southwell “... combined with high mathematical ability an essentially practical outlook which gavee direction to all his work”. gav work”. In 1855-6 he published published his classical work on torsion, torsion, flexure, flexure, 80

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7.8 The Mo dern Perio d (1857-Present)

151

His famous axiom, Form follows function became the touchstone for many in his profession. Sullivan, Sullivan, however, however, did not apply it literally literally.. He meant that an architec architectt should should consider the purpose of the building as a starting point, not as a rigidly limiting stricture. 88

He also had tremendous respect for the natural world which played an enormous role in forging his theories about architecture (he spent all of his first summers on his grandparents’ farm in Massachusetts where he developed this love and respect for nature) expressed in his Autobiography of an Idea), Idea), 1924).

89

7.8.4 7.8.4

Roebli Roebling, ng, 180 18066-186 1869 9

engineer, who was one of the pioneers John Augustus Roebling was an American civil engineer, in the construc construction tion of suspens suspension ion bridges. bridges. He was was born in German Germany y, educat educated ed at the Royal Royal Polytechnic School of Berlin and immigrated to the States in 1831. 90

In his first first job he was was emplo employe yed d by the Penns Pennsylv ylvani aniaa Railro Railroad ad Corp. Corp. to surve survey y its route across across the Allegheny Allegheny Mountains Mountains between between Harrisbur Harrisburgg and Pittsburgh. Pittsburgh. He then demonstrated demonstrated the practicability of steel cables in bridge construction and in 1841 established at Saxonburg the first factory to manufacture steel-wire rope in the U.S. 91

Roebling utilized steel cables in the construction of numerous suspension bridges and is genera generally lly consid considere ered d one of the pionee pioneers rs in the field field of suspens suspension ion-br -bridge idge construc construction tion.. He built railroad suspension bridges over the Ohio and Niagara rivers and completed plans for the Brookly Brooklyn n Bridge Bridge shortly shortly before his death. death. Roeblin Roeblingg was was the author author of Long Long and Short Short Span Span Railway Bridges (1869). 92

7.8. 7.8.5 5

Mail Mailla lart rt

From From (Billin (Billingto gton n 1973) 1973)

Robert Maillart was born on February 6, 1872, in Bern, Switzerland, where his father, a Belgian citizen, was a banker. He studied civil engineering at the Federal Institute of Technology in Zurich and graduated in 1894. Ironically, one of his lowest grades was in bridge design, even though he is regarded today as one of the half dozen greatest bridge designers of the twentieth century. 93

For eight years following his graduation, he worked with different civil engineering organizations. In 1902, he founded his own firm for design and construction; thereafter, his business grew grew rapidl rapidly y and expand expanded ed as far as Russia Russia and Spain. Spain. In the summer summer of 1914 1914,, he took his wife and three children children to Russia. Russia. Since the World War prevent prevented ed their return to Switzerland Switzerland,, Maillart stayed and worked in Russia until 1919, when his business was liquidated by the Revolution. olution. Forced to flee, he returned returned to Switzerland Switzerland penniless and lonely, lonely, his wife having having died in Russia. 94

Because of these misfortunes Maillart felt unable to take up the construction business again and hencefo henceforth rth concen concentra trated ted on design design alone. alone. He opened an offic officee in Geneva Geneva in 1919 and branches in Bern and Zurich in 1924. 95

During the twenties he began to develop and modify his ideas of bridge design; and from 1930, when the Salginatobel and Landquart Bridges were completed, until his death in 1940,

96

Victor Saouma


Draft


153

Figure 7.17: Nervi’s Palazetto Dello Sport needed needed for high high towe towers, rs, elimina eliminated ted the need need for inter internal nal wind wind bracin bracingg (since (since the perimet perimeter er columns carried the wind loadings), and permitted freer organization of the interior space. His later projects included the strikingly different Haj Terminal of the King Abdul Aziz International Airport, Jiddah, Saudi Arabia (1976-81), and King Abdul Aziz University, also in Jiddah (1977-78). 7.8.8

et al.

To name just a few of the most influential influential Architects/En Architects/Engineer gineers: s: Menn, Menn, Isler, Isler, Candella, Candella, Torroja, Johnson, Pei, Calatrava, ...

102

Victor Saouma


Draft


155

than my bare hands and no further further addition to my academic academic background. background. After After several years years of general practice in Mexico, as draftsman, designer and contractor, I recalled my old fancy with shells and began to collect again papers on the subject. Whatever I learned from then on was to be the hard way, working alone, with no direct help from any university or engineering office. But I am indebted to many people who did help me through their writings and Maillart was one of the foremost. I discovered discovered him in Giedion’s Giedion’s Space, Time and Architec Architecture; ture; and then I got Max Bill’s book with its invaluab invaluable le collection of Maillart’s Maillart’s essays. essays. I devoured devoured his articles articles about ”Reinforced ”Reinforced Concrete Design and Calculation” (he was very careful to differentiate the meaning of such words and to avoid the more than semantic confusion prevalent nowadays in English-speaking countries)., ”The Engineer and the Authorities” which expresses his position in front of the establishment and ”Mass and Quality in Reinforced Concrete Structures.” Very short papers, indeed, but well provided with opinions, something I could rarely find in other engineering articles. ticles. I learned later that to express personal personal opinions opinions is considered considered bad taste among technical technical writers. writers. Any discussion discussion should be restricted restricted to insignifican insignificantt details, details, but never touch touch fundamental dogmas, in a fashion curiously similar to what could be expected of the councils of the Church or the meetings of any Politbureau. But my attitude with respect to calculations of reinforced concrete structures was becoming unorthodox, being tired perhaps of performing long and tedious routines whose results were not always always meaningful. meaningful. Therefore, Therefore, I found Maillart’s Maillart’s thoughts delightfully delightfully sympathetic sympathetic and encouragin encouraging. g. If a rebel was able to produce such beautiful beautiful and sound structure structuress there could not be anything wrong with becoming also a rebel, which was besides, my only way to break the mystery surrounding shell analysis. Thus, Thus, I starte started d to follo follow w the bibliog bibliograp raphic hic tread tread and met, throug through h their their writin writings, gs, with Freudenthal, Johansen, Van der Broek, Kist, Saliger, Kacinczy and so many others who showed me there was more than a single and infallible infallible manner to approach approach structural structural analysis. The discovery of rupture methods, with their emphasis on simple statics and their bearing on the actual properties of construction materials and their behavior in the plastic range, allowed me to trust in simplified procedures to understand and analyze the distribution of stresses in shell struct structure ures. s. It also helped me to get out of my naive naive belief belief in the indispu indisputab table le truth of the printed word and to start reading with a new critical outlook. No longer did I need to believe whatever was in print, no matter how high-sounding the name of the author. I could make my own judgements about what methods of stress analysis were better suited for my practice. Since I was working practically alone, I could not afford nor had time for complex calculations and did welcome Maillart’s advice that simpler calculations are more reliable than comple com plex x ones, ones, especia especially lly for somebody somebody who builds builds his own own struct structure ures. s. This This was was exactl exactly y my case and, since most structures I was building were of modest scale, I could control what was happening, check the results and confirm the accuracy of my judgement or correct my mistakes. In a way, I was working with full scale models. I understand that this was also true of Maillart who in many cases was the actual builder of his designs. Following the general trend to mess up issues, there has been a lot of speculation about the engineer as an artist and in some instances, like in the case of Nervi, about the engineer as an architect (as if the title of architect could confer, per se, artistic ability to its holder); but few people realize that the only way to be an artist in this difficult specialty of building is to be your your own contractor. contractor. in countries countries like this, where the building building industry has been thoroughly thoroughly and irreversibly fragmented and the responsibility diluted among so many trades, it may be shocking to think of a contractor as an artist; but it is indeed the only way to have in your hands the whole set of tools or instruments to perform the forgotten art of building, to produce Victor Saouma


Draft Chapter 8

Case Study III: MAGAZINI GENERALI Adapted Adapted from (Billin (Billingto gton n and Mark Mark 1983) 1983)

8.1 8.1

Geom Geomet etry ry

This sotrage house, built by Maillart in Chiasso in 1924, provides a good example of the mariage between aesthetic and engineering.

1

The most strking feature of the Magazini Generali is not the structure itself, but rather the shape of its internal supporting frames, Fig. 8.1 8.1..

2

The frame can be idealized as a simply supported beam hung from two cantilever column supports. Whereas the beam itself is a simple structural idealization, the overhang is designed in such a way as to minimize the net moment to be transmitted to the supports (foundations), Fig. 8.2 8.2.. 3

8.2

Loads

The load applied on the frame is from the weights weights of the roof slab, and the frame itself. Given Given the space between adjacent frames is 14.7 ft, and that the roof load is 98 psf , and that the total frame weight is 13.6 kips, the total uniform load becomes, Fig. 8.3 8.3::

4

qroof = (98) psf (14. (14.7) ft = 1.4 k/ft (13. (13.6) k qframe = = 0. 0 .2 k/ft (63. (63.6) ft qtotal = 1.4 + 0. 0 .2 = 1.6 1.6 k/ft

(8.1-a) (8.1-b) (8.1-c)

Draft

8.3 Reactions

1 59 q ROOF = 1.4 k/ft + q FRAME = 0.2 k/ft q ROOF = 1.4 k/ft + q FRAME = 0.2 k/ft q TOTAL= 1.6 k/ft

Figure 8.3: Magazzini Generali; Loads (Billington and Mark 1983)

8.3 8.3 5

Reac Reacti tion onss

Reactions for the beam are determined first taking advantage of symmetry, Fig. 8.4 8.4:: W = (1. (1.6) k/ft(63. (63.6) ft = 102 k W 102 R = = = 51 k 2 2

(8.2-a) (8.2-b)

We note that these reactions are provided by the internal shear forces. q TOTAL = 1.6 k/ft

63.6 ft

51 k

51 k

Figure 8.4: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)

8.4 8.4

Forc orces

The internal forces are pimarily the shear and moments. Those can be easily determined for a simply simply supporte supported d unifor uniformly mly loaded loaded beam. beam. The shear shear varies aries linearly linearly from 51 kip to -51 kip with zero at the center, and the moment diagram is parabolic with the maximum moment at the center, Fig. 8.5 8.5,, equal to: 6

M max max

qL2 (1. (1.6) k/ft(63. (63.6) ft2 = = = 808 k.ft 8 8

(8.3)

The externally induced moment at midspan must be resisted by an equal and opposite internal moment. moment. This can be achieve achieved d through through a combination combination of compressive compressive force on the upper fibers, and tensile ones on the lower. lower. Thus Thus the net axial force is zero, zero, howeve howeverr there is a net internal internal couple, couple, Fig. 8.6 8.6.. 7

Victor Saouma


Draft 8.4 Forces

1 61 ext ⇒ C = M dext (808) k.ft = ± 88 (9. (9.2) ft

M ext ext = Cd T = C =

(8.4-a) (8.4-b)

k

Because the frame shape (and thus d(x)) is approximately parabolic, and the moment is also parabolic, then the axial forces are constants along the entire frame, Fig. 8.7 8.7..

8

d

M MOMENT DIAGRAM

FRAME

CABLE :

FRAME :

CURVE OF DIAGRAM

SHAPE OF DIAGRAM

Figure Figure 8.7: Magazzini Magazzini Generali; Similarities Similarities Between Between The Frame Shape and its Moment Moment Diagram, (Billington and Mark 1983) The axial force at the end of the beam is not balanced, and the 88 kip compression must be transmitted to the lower chord, Fig. 8.8 8.8.. Fig. 8.9 This is analogous to the forces transmiited

9

88 k

Tension 88 k

88 k

88 k

Compression Horizontal Component Tied Arch

Cable Force Axial Force Vertical Reaction

Figure Figure 8.8: Magazzini Magazzini Generali; Equilibrium Equilibrium of Forces at the Beam Support, (Billington (Billington and Mark 1983) to the support by a tied arch. It should be mentioned that when a rigorous computer analysis was performed, it was determined that the supports are contributing a compression force of about 8 kips which needs to be superimposed over the central values, Fig. 8.9 8.9..

10

Victor Saouma


Draft Chapter 9

DESIGN PHILOSOPHIES of ACI and AISC CODES 9.1 9.1

Safe Safetty Pro Provisi vision onss

Structures and structural members must always be designed to carry some reserve load above what is expected under normal use. This is to account for

1

Variability in Resistance: The actual strengths (resistance) of structural elements will differ from those assumed by the designer due to: 1. Variability ariability in the strength of the material (greater variabilit variability y in concrete concrete strength than in steel strength). 2. Difference Differencess betw b etween een the actual actual dimensions dimensions and those specified (mostly in placemen placementt of steel rebars in R/C). 3. Effect of simplifying simplifying assumptions assumptions made in the derivation derivation of certain certain formulas. formulas. Variability in Loadings: All loadings loadings are varia variable ble.. There There is a greate greaterr variatio ariation n in the live loads than in the dead loads. Some types of loadings are very difficult to quantify (wind, earthquakes). Consequences of Failure: The consequence of a structural component failure must be carefully assessed. The collapse of a beam is likely to cause a localized failure. Alternatively the failure of a column is likely to trigger the failure of the whole structure. Alternatively, the failure of certain components can be preceded by warnings (such as excessive deformation), whereas other are sudden and catastrophic. Finally, if no redistribution of load is possible (as would be the case in a statically determinate structure), a higher safety factor must be adopted. The purpose of safety provisions is to limit the probabilit probability y of failure failure and yet permit economical structures. 2

3

The following items must be considered in determining safety provisions: 1. Seriousnes Seriousnesss of a failure, either to humans humans or goods. go ods.

Draft

9.3 Ultimate Strength Metho d

165

σ < σall =

σyld F.S.

(9.1)

where F.S. is the factor of safety. 10

Major limitations of this method 1. An elastic analysis analysis can not easily account account for creep creep and shrink shrinkage of concrete. concrete. 2. For concrete structures structures,, stresses stresses are not linearly proportional proportional to strain beyond beyond 0. 0.45 45f f c . 3. Safety Safety factors are not rigorously rigorously determined determined from a probabilist probabilistic ic approach, approach, but are the result result of experience experience and judgment judgment..

11

Allowable strengths are given in Table 9.1 9.1.. Steel, AISC/ASD Tension, Gross Area F t = 0.6F y ∗ Tension, Effective Net Area F t = 0. 0 .5F u Bending F b = 0.66 66F F y Shear F v = 0.40 40F F y Concrete, ACI/WSD Tension 0 Compression 0.45 45f f c

∗ Effective net area will be defined in section ??. ??. Table 9.1: Allowable Stresses for Steel and Concrete

9.3 9.3 9.3.1 9.3.1

Ulti Ultima mate te Stren Strength gth Meth Method od The The Norm Normal al Dist Distrib ributi ution on

The normal distribution has been found to be an excellent approximation to a large class of distributions, and has some very desirable mathematical properties:

12

1. f ( f (x) is symmetric with respect to the mean µ. 2. f ( f (x) is a “bell curve” with inflection points at x = µ

± σ.

3. f ( f (x) is a valid probability distribution function as:



∞

f ( f (x) = 1

(9.2)

−∞

4. The probability that xmin < x < xmax is given by: P ( P (xmin < x < xmax ) = Victor Saouma



xmax

xmin

f ( f (x)dx

(9.3)


Draft


167

Figure 9.3: Frequency Distributions of Load Q and Resistance R Failure would occur for negative values of X The probability of failure P f f is equal to the ratio of the shaded area to the total area under the curve in Fig. 9.4 9.4.. 19

Figure 9.4: Definition of Reliability Index

 

R If X is assumed to follow a Normal Distribution than it has a mean value X = ln Q and a standard deviation σ .

20

21

We define the safety index (or reliability index) index) as β =

m

X σ

For standard distributions and for β = 3.5, it can be shown that the probability of failure is 1 P f 10−4 . That is 1 in every 10,000 structural members designed with β = 3.5 f = 9,091 or 1.1 will fail because of either excessive load or understrength sometime in its lifetime.

22

×

Reliability indices are a relative measure of the current condition and provide a qualitative estimate estimate of the structural structural performance. performance. 23

Structures with relatively high reliable indices will be expected to perform well. If the value is too low, then the structure may be classified as a hazard.

24

25

Target values for β are shown in Table 9.2 9.2,, and in Fig. 9.5

Victor Saouma


Draft


169

Type of Load/Member AISC DL + LL; Members DL + LL; Connections DL + LL + WL WL;; Me Mem mbers bers DL + LL +EL; +EL; Me Mem mbers bers ACI Ductile Failure Sudden Failures

β 3 .0 4.5 3.5 3.5 1.75 1.75 3-3.5 3.5-4

Table 9.2: Selected β values for Steel and Concrete Structures Φ is a strength reduction factor, factor, less than 1, and must account for the type of structural element, Table 9.3 9.3.. Type of Member

Φ ACI

Axial Tension Flexure Axial Compression Compression,, spiral spiral reinforcem reinforcement ent Axial Compression, other Shear and Torsion Bearing on concrete AISC Tension, yielding Tension, fracture Compression Beams Fasteners, Tension Fasteners, Shear

0.9 0.9 0.75 0.70 0.85 0.70 0.9 0.75 0.85 0.9 0.75 0.65

Table 9.3: Strength Reduction Factors, Φ Rn is the nominal resistance (or strength). ΦRn is the design strength. strength. αi is the load factor corresponding to Qi and is greater than 1. Σαi Qi is the required required strength strength based on the factored load: load: i is the type of load 32

The various factored load combinations which must be considered are

AISC

Victor Saouma


Draft 9.4 Example

9.4 9.4

1 71

Example

Exampl Example e 9-1: 9-1: LRFD LRFD vs ASD To illustrate the differences between the two design approaches, let us consider the design of an axial member, subjected to a dead load of 100 k and live load of 80 k. Use A36 steel. ASD: We consider the total load P = 100 + 80 = 180 k. From Table Table 9.1 9.1,, the allowable stress is 0.6σyld = 0.6 36 = 21. 21.6 ksi. Thus the required cross sectional area is

∗

A=

180 = 8.33 in2 21 21..6

USD we consider the largest of the two load combinations Σαi Qi : 1.4D = 1.4(100) = 140 k 1.2D + 1. 1.6L = 1.2(100) + 1. 1.6(80 6(80)) = 24 2488 k  From Table 9.3 Φ = 0.9, and ΦR ΦRn = (0. (0.9)Aσ 9)Aσyld. He Henc nce, e, applyi applying ng Eq. 9.9 the cross sectional area should be A=

Σαi Qi 248 = = 7.65 in2 Φσyld (0. (0.9)(36)

Note that whereas in this particular case the USD design required a smaller area, this may not be the case for different ratios of dead to live loads.

Victor Saouma


Draft Chapter 10

BRACED ROLLED STEEL BEAMS This chapter deals with the behavior and design of laterally supported steel beams according to the LRFD provisions.

1

A laterally stable beam is one which is braced laterally in the direction perpendicular to the plane of the web. Thus Thus overall overall buckling buckling of the compression compression flange as a column column cannot occur prior to its full participation to develop the moment strength of the section. 2

3

If a beam is not laterally supported, Fig. 10.1 10.1,, we will have a failure mode governed by lateral

A) COMPOSITE BEAM

B) OTHER FRAMING

C) CROSS BRACING

Figure 10.1: Lateral Bracing for Steel Beams torsional buckling. By the end of this lecture you should be able to select the most efficient section (light weight with adequate strength) for a given bending moment and also be able to determine the flexural strength of a given beam.

4

Draft

10.2 Failure Mo des and Classification of Steel Beams

1 75

w

σy

-

M=(wL2 )/8

+

σy

wu -

σy

+

Mp

σy 2

M p=(wL )/8

Figure 10.2: Failure of Steel beam; Plastic Hinges

tw

bf

h f hc

COMPACT

FLANGE BUCKLING

WEB BUCKLING

Figure 10.3: Failure of Steel beam; Local Buckling

Victor Saouma


Draft

10.3 Compact Sections

17 7

Figure 10.6: Stress-strain diagram for most structural steels When the yield stress is reached at the extreme fiber, the nominal moment strength M n , is referred to as the yield moment M y and is computed as 13

(10.3)

M n = M y = S x F y (assuming that bending is occurring with respect to the x

− x axis).

When across the entire section, the strain is equal or larger than the yield strain (ε ( ε εy = F y /E s ) then the section is fully plastified, and the nominal moment strength M n is therefore referred to as the plastic plastic moment moment M p and is determined from

≥

14

M p = F y



A

ydA = F y Z

(10.4)

where def

Z =



ydA

(10.5)

is the Plastic Section Modulus. Modulus. The plastic section modulus Z should not be confused with the elastic section modulus S defined, Eq. 5.23 as 15

S = def

I =

16

I d/2 d/2

(10.6-a)



(10.6-b)

A

y2 dA

The section modulus S x of a W section can be roughly approximated by the following formula S x

≈ wd/1 wd/10

or I x

≈ S x d2 ≈ wd2/20

(10.7)

and the plastic modulus can be approximated by Z x

Victor Saouma

≈ wd/9 wd/9

(10.8)


Draft

10.5 Slender Section

17 9

Draft

Figure 10.7: Nominal Moments for Compact and Partially Compact Sections where: M r Residual Moment equal to (F (F y F r )S λ bf /2tf for I-shaped member flanges and hc /tw for beam webs.

−

All other quantities quantities are as defined earlier. earlier. Note that we use the λ associated with the one being violated (or the lower of the two if both are).

22

10.5 10 .5

Slen Slende der r Sect Sectio ion n

If the width to thickness ratio exceeds λr values of flange and web, the element is referred to as slender compression compression element. Since the slender sections involve involve a different different treatment, treatment, it will not be dealt here. 23

10.6 10 .6

Exam Exampl ples es

Example 10-1: Z for Rectangular Section Determine the plastic section modulus for a rectangular section, width b and depth d.

Victor Saouma


Draft

10.6 Examples

18 1

2. Comput Computee the factor factored ed load moment moment M u . For a simply simply support supported ed beam carrying carrying uniformly distributed load, M u = wu L2 /8 = (1. (1.52)(20)2 /8 = 76 k.ft Assuming compact section, since a vast majority of rolled sections satisfy λ both the flange and the web. The design strength φb M n is

≤ λ p for

φb M n = φb M p = φb Z x F y The design requirement is φb M n = M u or, combing those two equations we have: φb Z x F y = M u 3. Required Required Z x is

M u 76(12) = = 28 28..1 in3 φb F y 0.90(36)

Z x =

From the notes on Structural Materials, we select a W12X22 section which has a Z x = 29 29..3 in3 (22)(12) Note that Z x is approximated by wd = 29 29..3. 9 = 9 4. Check Check compact section section limits λ p for the flanges from the table λ=

bf 2tf

= 4.7 = 65 = √6536 = 10 10..8 > λ

and for the web:

λ=

√

√F

λ p

y

hc tw

λ p

= 41 41..8 640 = 640 = √ = 107 36

√F

y

√

5. Check the Strength Strength by correcting correcting the factored moment moment M u to include the self weight. Self weight of the beam W12X22 is 22 lb./ft. or 0.022 kip/ft wD = 0.2 + 0. 0.022 = 0. 0.222 k/ft wu = 1.2(0. 2(0.222) + 1. 1.6(0. 6(0.8) = 1. 1.55 k/ft 2 M u = (1. (1.55)(20) /8 = 77. 77.3 k.ft M n = M p = Z x F y = φb M n =

(29. (29.3) in3 (36) (12) in/ft

ksi

√ 0.90(87. 90(87.9) = 79. 79.1 k.ft > M u

= 87 87..9 k.ft

Therefore Therefore use W12X2 W12X222 section. section. 6. We finally check for the maximum maximum distance between between supports. ry = L p = =

Victor Saouma

  

I y 5 = = 0.88 in A 6.5 300 ry F y 300 0.88 = 43 ft 36

√

(10.16-a) (10.16-b) (10.16-c)


Draft Chapter 11

REINFORCED CONCRETE BEAMS 11.1 11.1

Intr Introduc oducti tion on

Recalling that concrete has a tensile strength (f (f t ) about one tenth its compressive strength (f c ), concrete by itself is a very poor material for flexural members. 1

To provid providee tensile tensile resista resistance nce to concre concrete te beams, beams, a reinfo reinforce rcemen mentt must must be added. added. Steel Steel is almost universally used as reinforcement (longitudinal or as fibers), but in poorer countries other indigenous materials have been used (such as bamboos).

2

The following lectures will focus exclusively on the flexural design and analysis of reinforced concrete concrete rectangular rectangular sections. sections. Other concerns, concerns, such such as shear, shear, torsion, torsion, cracking, cracking, and deflections deflections are left for subsequent ones. 3

Design Design of reinfo reinforce rced d concre concrete te struct structure uress is go gove verne rned d in most most cases cases by the Building Building Code Code Requirements for Reinforced Concrete, Concrete, of the American Concrete Institute (ACI-318). Some of the most relevant provisions of this code are enclosed in this set of notes.

4

We will focus on determining the amount of flexural (that is longitudinal) reinforcement required at a given given section section. For that section section,, the moment moment which which should should be consid considere ered d for design is the one obtained from the moment envelope at that particular point. 5

11.1 11 .1.1 .1 6

Nota Notati tion on

In R/C design, it is customary to use the following notation

Draft

11.1 Intro duction

11.1.3 11.1.3 11

18 5

Analy Analysi siss vs vs Desig Design n

In R/C we always consider one of the following problems:

Analysis: Given Given a certain certain design, design, determ determine ine what is the maximum maximum mom momen entt which which can be applied. Design: Given an external moment to be resisted, determine cross sectional dimensions (b ( b and h) as well as reinforcement (A (As ). Note that in many cases the external dimensions of the beam (b (b and h) are fixed by the architect. 12

We often consider the maximum moment along a member, and design accordingly.

11.1.4 11.1.4

Basic Basic Relat Relations ions and Assump Assumption tionss

In developing a design/analysis method for reinforced concrete, the following basic relations will be used, Fig. ??: ??: 13

Compatibility

Equilibrium

C d

εy

T

T=C M_ext=Cd Figure 11.2: Internal Equilibrium in a R/C Beam 1. Equilib Equilibriu rium: m: of forces forces and moment moment at the cross cross section. section. 1) ΣF ΣF x = 0 or Tension in the reinforcement = Compression in concrete; and 2) ΣM Σ M = 0 or external moment (that is the one obtained from the moment envelope) equal and opposite to the internal one (tension in steel and compression of the concrete). 2. Material Material Stress Strain: Strain: We recall that all normal normal strength concrete concrete have a failure strain  u = .003 in compression irrespective of f c . 14

Basic assumptions used:

Compatibility of Displacements: Perfect Perfect bond between steel and concrete concrete (no slip). Note that those two materials do also have very close coefficients of thermal expansion under normal temperature. Plane section remain plane Victor Saouma

⇒ strain is proportional to distance from neutral axis. Structural Concepts and Systems for Architects

Draft

11.2 Cracked Section, Ultimate Strength Design Metho d

1 87

Figure 11.3: Cracked Section, Limit State

Figure 11.4: Whitney Stress Block

Victor Saouma


Draft

11.2 Cracked Section, Ultimate Strength Design Metho d 11.2 11 .2.3 .3

1 89

Anal Analys ysis is

Given As , b, d, f c , and f y determine the design moment: 1. ρact =

As bd 

87 2. ρb = (.85)β 85)β 1 f f yc 87+f 87+f y

3. If ρ If ρact < ρb (that is failure is triggered by yielding of the steel, f s = f y ) As f y .85f 85f c b

a = From Equilibrium M D = ΦAs f y d a2

−

Combining this last equation with ρ =

−  

M D = Φ As f y d

As f y 0.59  f c b

M n

As bd

yields 2

M D = Φρf Φ ρf y bd 4.



− 1

f y .59 59ρ ρ  f c



  (11.9)

† If ρact

> ρb is not allowed by the code as this would be an over-reinforced section which would fail with no prior warning. However, if such a section exists, and we need to determine its moment carrying capacity, then we have two unknowns: (a) Steel strain εs (which was equal to εy in the previous case) (b) Location of the neutral neutral axis c. We have two equations to solve this problem

Equilibrium: of forces c=

As f s .85 85f f c bβ 1

(11.10)

Strain compatibility: since we know that at failure the maximum compressive strain εc is equal to 0.003. Thus from similar triangles we have c .003 = d .003 + εs

(11.11)

Those two equations can be solved by either one of two methods: (a) Substitute Substitute into into one single equation equation (b) By iteration iteration Once c and f s = Eε s are determined determined then

− 

M D = ΦAs f s d

Victor Saouma

β 1 c 2

(11.12)


Draft

11.2 Cracked Section, Ultimate Strength Design Metho d

1 91

5. Check Check equilibrium equilibrium of forces in the x direction direction (ΣF (ΣF x = 0) a=

As f s .85 85f f c b

(11.16)

6. Check Check assumption assumption of f of f s from the strain diagram εs d where c =

a β 1 .

−c

=

.003 c

⇒ f s = E s d −c c .003 < f y

(11.17)

7. Iterate Iterate until until conver convergence gence is reached. reached.

Example Example 11-1: Ultimate Ultimate Strength Strength Capacit Capacity y Determine the ultimate Strength of a beam with the following properties: b = 10 in, d = 23 in, As = 2.35 in2 , f c = 4, 000 psi and f y = 60 6 0 ksi. Solution: ρact = ρb =

As bd

2.35 (10)(23) = .0102  87 4 87 .85 85β β 1 f f yc 87+f ( .85)(. 85)(.85) 60 87+f y = (. 87+60 = As f y (2. (2.35)(60) .85f 85f c b = (.85)(4)(10) = 4.147 in (2. (2.35)(60)(23 4.147 2 ) = 2, 950 k.in

=

.02885 > ρact

√

a = M n = M D = ΦM n = (. ( .9)(2, 9)(2, 95 950) 0) = 2, 660 k.in

−

Note that from the strain diagram c=

a 4.414 = = 4.87 in 0.85 0.85

Alternative solution

−

M n = ρact f y bd2 1

−

= As f y d 1

.59 59ρ ρact f f y 

.59 59ρ ρact f f y c

−

= (2. (2.35)(60)(23) 1



c

 

(.59) 60 4 (.01021) = 2, 950 k.in = 245 k.ft

M D = ΦM n = (.9)(2, 9)(2, 95 950) 0) = 2, 660 k.in

Exampl Example e 11-2: 11-2: Beam Beam Design Design I

Victor Saouma


Draft

11.3 Continuous Beams

1 93

6. Check Check equilibrium equilibrium of forces: As f y (2. (2.42) in2 (40) ksi a= = = 3.3 in 85f (.85)(3) ksi(11. (11.5) in .85 f c b

√

7. we have have converged converged on a. 8. Actual Actual ρ is ρact =

2.42 (11. (11.5)(20)

= .011

9. ρb is equal to ρb = .85 85β β 1

f c 87 3 87 = (.85)(. 85)(.85) = .037 f y 87 + f y 40 87 + 40

10. ρmax = .75 75ρ ρ = (0. (0.75)(0. 75)(0.037) = .0278 > 0.011

11.3 11.3

√ thus f s = f y and we use

As = 2.42 in2

Con Contin tinuous uous Beams Beams

Whereas coverage of continuous reinforced concrete beams is beyond the scope of this course, Fig. ?? illustrates a typical reinforcement in such a beam. 28

11.4 1.4

ACI Code ode

Attached is an unauthorized copy unauthorized copy of some of the most relevant ACI-318-89 design code provisions. 8.1.1 - In design of reinforced concrete structures, members shall be proportioned for adequate strength in accordance with provisions of this code, using load factors and strength reduction factors Φ specified in Chapter 9. 8.3.1 - All members of frames or continuous construction shall be designed for the maximum effects of factored loads as determined by the theory of elastic analysis, except as modified according to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used. 8.5.1 - Modulus of elasticity E c for concrete may be taken as W c1.5 33 f c ( psi) for values of W c betwe between en 90 and 155 lb per cu ft. For normal normal weigh weightt concre concrete, te, E c may be taken as  57 57,, 000 f c . 8.5.2 - Modulus of elasticity E s for non-prestressed reinforcement may be taken as 29,000 psi. 9.1.1 - Structures and structural members shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code. 9.2 - Required Strength 9.2.1 - Required strength U to resist dead load D and live load L shall be at least equal to





U = 1.4D + 1. 1.7L 9.2.2 - If resistance to structural effects of a specified wind load W are included in design, the following combinations of D, L, and W shall be investigated to determine the greatest required strength U U = 0.75(1. 75(1.4D + 1. 1.7L + 1. 1.7W ) W ) Victor Saouma


Draft

11.4 ACI Co de

1 95

where load combinations shall include both full value and zero value of L to determine the more severe condition, and U = 0.9D + 1. 1.3W but for any combination of D, L, and W, required strength U shall not be less than Eq. (9-1). 9.3.1 - Design strength provided by a member, its connections to other members, and its cross sections, in terms of flexure, axial load, shear, and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code, multiplied by a strength reduction factor Φ. 9.3.2 - Strength reduction factor Φ shall be as follows: 9.3.2.1 - Flexure, without axial load 0.90 9.4 - Design strength for reinforcement Designs shall not be based on a yield strength of reinforcement f y in excess of 80,000 psi, except for prestressing tendons. 10.2.2 - Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except, for deep flexural members with overall depth to clear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a non-linear distribution of strain shall be considered. See Section 10.7. 10.2.3 - Maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003. 10.2.4 - Stress in reinforcement below specified yield strength f y for grade of reinforcement used shall be taken as E s times steel strain. For strains greater than that corresponding to f y , stress in reinforcement shall be considered independent of strain and equal to f y . 10.2.5 - Tensile strength of concrete shall be neglected in flexural calculations of reinforced concrete, except when meeting requirements of Section 18.4. 10.2.6 - Relationship between concrete compressive stress distribution and concrete strain may be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction prediction of strength strength in substanti substantial al agreemen agreementt with results of comprehens comprehensive ive tests. 10.2.7 - Requirements of Section 10.2.5 may be considered satisfied by an equivalent rectangular angular concrete concrete stress stress distribution distribution defined by the following: following:  10.2.7.1 - Concrete stress of 0. 0.85 85f f c shall be assumed uniformly distributed over an equivalent lent compressio compression n zone bounded by edges of the cross section and a straight straight line located parallel to the neutral axis at a distance (a (a = β 1 c) from the fiber of maximum compressive strain. 10.2.7.2 - Distance c from fiber of maximum strain to the neutral axis shall be measured in a direction perpendicular to that axis. 10.2.7.3 - Factor β 1 shall be taken as 0.85 for concrete strengths f c up to and including 4,000 psi. For strengths above 4,000 psi, β 1 shall be reduced continuously at a rate of 0.05 for each 1000 psi of strength in excess of 4,000 psi, but β 1 shall not be taken less than 0.65. 10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcement reaches the strain corresponding to its specified yield strength f y just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3 - For flexural members, and for members subject to combined flexure and compressive axial load when the design axial load strength (ΦP (ΦP n ) is less than the smaller of (0. (0.10 10f f c Ag ) or (ΦP (ΦP b ), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio ρb that would produce balanced strain conditions for the section under flexure without axial load. For members with compression reinforcement, the portion of ρb equalized by compression reinforcement need not be reduced by the 0.75 factor. 10.3.4 - Compression reinforcement in conjunction with additional tension reinforcement may be used to increase the strength of flexural members. Victor Saouma


Draft Chapter 12

PRESTRESSED CONCRETE 12.1 12.1


Beams with Beams with longer longer spans spans are archit architect ectura urally lly more more appealin appealingg than than those those with with short short ones. ones. However, for a reinforced concrete beam to span long distances, it would have to have to be relatively deep (and at some point the self weight may become too large relative to the live load), or higher grade steel and concrete must be used.

1

However, if we were to use a steel with f y much higher than 60 ksi in reinforced concrete (R/C), then to take full advantage of this higher yield stress while maintaining full bond between concrete concrete and steel, will result result in unacceptabl unacceptably y wide crack widths. Large crack crack widths will in turn result in corrosion of the rebars and poor protection against fire.

≈

2

One way to control the concrete cracking and reduce the tensile stresses in a beam is to prestress the beam by applying an initial state of stress which is opposite to the one which will be induced by the load. 3

For a simply supported beam, we would then seek to apply an initial tensile stress at the top and compressive compressive stress at the bottom. In prestressed prestressed concrete concrete (P/C) this can be achieve achieved d through prestressing of a tendon placed below the elastic neutral axis. 4

Main advantages of P/C: Economy, deflection & crack control, durability, fatigue strength, longer spans. 5

6

There two type of Prestressed Concrete beams:

Pretensioning: Steel is first stressed, stressed, concrete is then poured around the stressed stressed bars. When enough concrete strength has been reached the steel restraints are released, Fig. 12.1 12.1.. Postensioning: Concrete is first poured, then when enough strength has been reached a steel cable is passed thru a hollow core inside and stressed, Fig. 12.2 12.2.. 12.1.1 12.1.1 7

Mater Materia ials ls

P/C beams usually have higher compressive strength than R/C. Prestressed beams can have

f c as high as 8,000 psi. 8

The importance of high yield stress for the steel is illustrated by the following simple example.

Draft

12.1 Intro duction

19 9

If we consider the following: 1. An unstressed unstressed steel steel cable of length ls 2. A concrete concrete beam of length length lc 3. Prestr Prestress ess the beam with the cable, cable, result resulting ing in a stress stressed ed length of concre concrete te and steel   equal to ls = lc . 4. Due to shrinkage shrinkage and creep, there will be a change change in length ∆lc = (εsh + εcr )lc

(12.1)

we want to make sure that this amout of deformation is substantially smaller than the stretch of the steel (for prestressing to be effective). 5. Assuming Assuming ordinary ordinary steel: f s = 30 3 0 ksi, E s = 29 29,, 000 ksi, εs = 6. The total steel steel elongation elongation is εs ls = 1.03

30 29, 29,000

= 1. 1 .03

× 10−3 in/ in

× 10−3ls

7. The creep and shrink shrinkage strains are about εcr + εsh

 .9 × 10−3

8. The residual residual stres which is left in the steel after after creep and shrinkage shrinkage took place is thus thus (1. (1.03 Thus the total loss is

30−4 30

− .90) × 10−3(29 × 103) = 4 ksi

(12.2)

= 87% which is unacceptably too high.

9. Alternativ Alternatively ely if initial stress stress was 150 ksi after losses we would be left with 124 ksi or a 17% loss. 10. Note Note that the actual actual loss is (. (.90 9

× 10−3)(29 × 103) = 26 ksiin each case

Having shown that losses would be too high for low strength steel, we will use

Strands usually composed of 7 wires. Grade 250 or 270 ksi, Fig. 12.3 12.3..

Figure 12.3: 7 Wire Prestressing Tendon Tendon have diameters ranging from 1/2 to 1 3/8 of an inch. Grade 145 or 160 ksi. Wires come in bundles of 8 to 52. Note that yield stress is not well defined for steel used in prestressed concrete, usually we take 1% strain as effective yield. Steel relaxation is the reduction in stress at constant strain (as opposed to creep which is reduction reduction of strain at constant constant stress) occrs. Relaxation Relaxation occurs indefinitely indefinitely and produces 10

Victor Saouma


Draft

12.1 Intro duction

20 1

W

h f’y

Q

f c

P

P

2f c

f c

+

h/2 f c

= fc =f t

2Q

2f c

0

2f c

+ P

P

=

2h/3 2f c

2f =2f t c

0

0

2f c

2f c

+

2Q P

P

=

2f c f c

2f t =2f c

+

h/2 h/3

0

Midspan 0

=

Ends

f c

f c

f c

0

Q

f c

+ 2f c

P

P

h/2

= ft =f c

f c f c

Midspan

f c

+

h/3

0

=

Ends

fc

f c

f c

Figure 12.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978) Member

Equivalent Equivalent load on concrete concrete from tendon Moment from from prestressing prestressing

(a) P

P sin θ

P

θ

P sin θ

P cos θ

P cos θ 2 P sin θ

(b) P

P sin θ

P

θ

P sin θ

P cos θ

P cos θ

(c) Pe P

e

Pe P

P P

(d) P

P

θ M

P sin θ

P sin θ

M

e P cos θ

P cos θ

(e) P

P P

P sin θ

P sin θ None

P cos θ

P cos θ 2 P sin θ

(f) P P

None

(g) P

P

Figure Figure 12.5: Determinati Determination on of Equivalen Equivalentt Loads Victor Saouma


Draft

12.2 Flexural Stresses

2 03

4. P e and M 0 + M DL DL + M LL LL

− AP − AP

f 1 =

e c

f 2 =

e c

 − −   1

1+

ec1 r2 ec2 r2

+

M 0 +M DL DL +M LL LL S 1 M 0 +M DL DL +M LL LL S 2

(12.7)

The internal stress distribution at each one of those four stages is illustrated by Fig. 12.7 12.7.. Pi

Pi e c 1

Pi

Ac

Ic

Ac

e c1 ) r2

(1-

c1 e

c2

Stage 1

Pi Ac

Ac

Pe Ac

(1-

Pe

Stage 4

Ac

Pi e c 2

Pi

Ac

Ic

Ac

e c1 ) r2

(1-

Pi

Stage 2

Pi

(1+

e c2 ) r2

-

+

e c1 Mo )r2 S1

(1+

e c2 Mo )+ r2 S2

Pi

S1

Ac

Pi

S2

Ac

Md + Ml

Pe

S1

Ac

Md + Ml

Pe

S2

Ac

e c2 ) r2

e c1 Mo )r2 S1

(1-

Mo

-

+

Mo

(1+

(1+

(1-

(1+

e c2 Mo )+ r2 S2

e c1 Mt )r2 S1

e c2 Mt )+ r2 S2

Figure Figure 12.7: Flexural Flexural Stress Distribution Distribution for a Beam with Variable Variable Eccentricit Eccentricity; y; Maximum Maximum Moment Section and Support Section, (Nilson 1978) Those (service) flexural stresses must be below those specified by the ACI code (where the subscripts c, t, i and s refer refer to compression compression,, tension, tension, initial and service service respective respectively): ly):  f ci permitted concrete compression stress at initial stage .60 60f f ci ci f ti permitted concrete tensile stress at initial stage < 3 f ci ti f cs permitted concrete compressive stress at service stage .45 45f f c cs f ts perm permit itte ted d conc concre rete te tens tensil ilee stre stress ss at init initia iall stag stagee 6 f c or 12 f c ts  Note that f ts ts can reach 12 f c only if appropriate deflection analysis is done, because section would be cracked. 17

  



18

Based on the above, we identify two types of prestressing:

Victor Saouma


Draft

12.2 Flexural Stresses

2 05 (.183)(40)2 = 36 36..6 k.ft 8

M 0 =

(12.9-b)

The flexural stresses will thus be equal to: f 1w,20 =

(36.6)(12, 6)(12, 000)  S M 1,02 =  (36. = 439 psi 1, 000

 − −

P i ec1 M 0 1 2 Ac r S 1 83 439 = 522 psi

f 1 =

(12.11-a)

− − −

=

f ti ti = 3

(12.11-b)

− √ f  = +190



(12.11-c)

c





0 − AP ic 1 + ecr22 + M S 2 −1, 837 + 439 = −1, 398 psi √ .6f c = −2, 400

f 2 = = f ci ci =

3. P e and M 0 . If we have 15% losses, then the effective force P e is equal to (1 144 k f 1 = = = f 2 = = = note that 71 and respectively.

−

−

− − − AP ec − −

 − − − −  − 

P e ec1 M 0 1 Ac r2 S 1 144,, 000 144 (5. (5.19)(12) 1 176 68 68..2 71 439 = 510 psi ec2 r2

(12.11-d) (12.11-e) (12.11-f)

− 0.15)169 = (12.12-a)

−

439

M 0 S 2 144,, 000 144 (5. (5.19)(12) 1+ + 439 176 68 68..2 1, 561 + 439 = 1, 122 psi 1+

(12.10)

+



−

(12.12-b) (12.12-c) (12.12-d) (12.12-e) (12.12-f)

−1, 561 are respectively equal to (0. (0.85)(−83) and (0. (0.85)(−1, 837)

4. P e and M 0 + M DL DL + M LL LL (0. (0.55)(40)2 = 110 k.ft 8

(12.13)

(110)(12, 000)  (110)(12, = 1, 320 psi 1, 000

(12.14)

M DL DL + M LL LL = and corresponding stresses f 1,2 = Thus, f 1 = Victor Saouma

−

 − −

P e 1 Ac

ec1 r2

M 0 + M DL DL + M LL LL S 1

(12.15-a)


Draft

12.3 Case Study: Walnut Lane Bridge

207

80 ft CENTER LINE

ELEVATION OF BEAM HALF

9.25’

44 ’

ROAD

9.25’

SIDEWALK

BEAM CROSS SECTIONS

TRANSVERSE DIAPHRAGMS

CROSS - SECTION OF BRIDGE

52" 10" 7"

3"

TRANSVERSE DIAPHRAGM 10"

7"

3’-3"

6’-7" SLOTS FOR CABLES

6 1/2" 3 1/2" 7" 30"

CROSS - SECTION OF BEAM

Figure 12.8: Walnut Lane Bridge, Plan View

Victor Saouma


Draft

12.3 Case Study: Walnut Lane Bridge 12.3 12 .3.3 .3

26

209

Load Lo adss

The self weight of the beam is q0 = 1. 1 .72 k/ft.

The concrete (density=.15 k/ ft3 ) road has a thickness thickness of 0.45 feet. Thus Thus for a 44 foot width, the total load over one single beam is

27

qr,tot = 28

1 (44) ft(0. (0.45) ft(0. (0.15) k/ ft3 = 0.23 k/ft 13

(12.20)

Similarly for the sidewalks which are 9.25 feet wide and 0.6 feet thick: qs,tot =

1 (2)(9. (2)(9.25) ft(0. (0.60) ft(0. (0.15) k/ ft3 = 0.13 k/ft 13

(12.21)

We note that the weight can be evenly spread over the 13 beams beacause of the lateral diaphragms. 29

The total dead load is qDL = 0.23 + 0. 0.13 = 0. 0.36 k/ft

(12.22)

The live load is created by the traffic, and is estimated to be 94 psf, thus over a width of 62.5 feet this gives a uniform live load of

30

wLL = 31

1 (0. (0.094) k/ft /f t2 (62. (62.5) ft = 0.45 k/ft 13

(12.23)

Finally, the combined dead and live load per beam is wDL+ 0.45 = 0. 0.81 k/ft DL+LL = 0.36 + 0.

12.3.4 12.3.4

(12.24)

Flexu Flexural ral Stres Stresse sess

1. Prestress Prestressing ing force, P i only f 1 =

−

=

−

f 2 =

−

=

−

−  ×  −   × 

P i ec1 1 Ac r2 (2 106 ) (31. (31.8)(39. 8)(39.5) 1 1, 354 943. 943. P i ec2 1+ 2 Ac r 6 (2 10 ) (31. (31.8)(39. 8)(39.5) 1+ 1, 354 943.. 943

(12.25-a)



= 490 490.. psi

(12.25-b) (12.25-c)



=

−3, 445 445.. psi

(12.25-d)

2. P i and the self weight of the beam M 0 (which has to be acconted for the moment the beam cambers due to prestressing) (1. (1.72)(160)2 M 0 = = 5, 504 k.ft 8

(12.26)

The flexural stresses will thus be equal to: f 1w,20 = Victor Saouma

(5, 50 50..4)(12, 4)(12, 000)  S M 1,02 =  (5, = 2, 043 psi 943.. 943

(12.27)


Draft Chapter 13

ARCHES and CURVED STRUCTURES 1

This chapter will concentrate on the analysis of arches.

The concepts used are identical to the ones previously seen, however the major (and only) difference is that equations will be written in polar coordinates.

2

Like Like cables, arches arches can be used to reduce reduce the bending moment in long span structures structures.. Essentially, an arch can be considered as an inverted cable, and is transmits the load primarily through axial compression, but can also resist flexure through its flexural rigidity.

3

4

A parabolic arch uniformly loaded will be loaded in compression only.

A semi-c semi-circ ircula ularr arch arch unifirm unifirmly ly loaded loaded will have have some some flexura flexurall stress stresses es in additi addition on to the compressive ones. 5

13.1 13 .1

Arc Arches hes

In order to optimize dead-load efficiency, long span structures should have their shapes approximate the coresponding moment diagram, hence an arch, suspended cable, or tendon configuration in a prestressed concrete beam all are nearly parabolic, Fig. 13.1 13.1..

6

Long span structures can be built using flat construction such as girders or trusses. However, for spans in excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspended cable or thin shells.

7

Since Since the dawn dawn of histor history y, mankin mankind d has tried tried to span span distan distances ces using using arch arch constr construct uction ion.. Essentially this was because an arch required materials to resist compression only (such as stone, masonary, bricks), and labour was not an issue.

8

The basic issues of static in arch design are illustrated in Fig. 13.2 where the vertical load is per unit horizontal projection (such as an external load but not a self-weight). Due to symmetry, the vertical reaction is simply V = wL 2 , and there is no shear across the midspan of the arch (nor a moment). Taking moment about the crown,

9

M = H h

−

wL 2

− L 2

L 4

=0

(13.1)

Draft 13.1 Arches

2 13

Solving Solving for H H =

wL2 8h

(13.2)

We recall that a similar equation was derived for arches., and H is analogous to the C T forces in a beam, and h is the overall height of the arch, Since h is much larger than d, H will be much smaller than C T in a beam.

−

−

Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoretically result in no moment on the arch section.

10

Three-hinged arches are statically determinate structures which shape can acomodate support settlements and thermal expansion without secondary internal stresses. They are also easy to analyse through statics.

11

An arch carries the vertical load across the span through a combination of axial forces and flexural flexural ones. A well dimensione dimensioned d arch arch will have have a small to negligible negligible moment, moment, and relatively relatively high normal compressive stresses. 12

An arch is far more efficient than a beam, and possibly more economical and aesthetic than a truss in carrying loads over long spans.

13

If the arch has only two hinges, Fig. 13.3 13.3,, or if it has no hinges, then bending moments may exist either at the crown or at the supports or at both places.

14

w

h’

APPARENT LINE OF PRESSURE WITH ARCH BENDING EXCEPT AT THE BASE

M

h

h

2

H’
H’=wl /8h’< 2

w

h’ M base

H’ wl /8h

PRESSURE WITH ARCH BENDING INCLUDING BASE

V

V

H’
M crown

M base

h

L V

V

Figure 13.3: Two Hinged Arch, (Lin and Stotesbury 1981) Since H varies inversely to the rise h, it is obvious that one should use as high a rise as possible. possible. For a combinatio combination n of aesthetic aesthetic and practical practical considerations, considerations, a span/rise span/rise ratio ranging ranging from 5 to 8 or perhaps as much as 12, is frequently used. However, as the ratio goes higher, we may have buckling problems, and the section would then have a higher section depth, and the arch advantage diminishes. 15

In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in practi practice ce an arch arch is not subjected subjected to unifor uniform m horizon horizontal tal load. First, First, the depth (and thus the weight) of an arch is not usually constant, then due to the inclination of the arch the actual self weight weight is not constant. constant. Finally Finally, live loads may act on portion of the arch, thus thus the line of action will not necessarily follow the arch centroid. This last effect can be neglected if the live load is small in comparison with the dead load. 16

Victor Saouma


Draft 13.1 Arches

2 15

Solving those four equations simultaneously we have:

  

140 26.25 0 1 1 0 80 60

0 0 1 0

0 1 0 0

     

RAy RAx RCy RCx

  

=

  

2, 900 80 50 3, 000

  ⇒  

RAy RAx RCy RCx

  

=

  

15 15..1 k 29 29..8 k 34 34..9 k 50 50..2 k

  

(13.4)

We can check our results by considering the summation with respect to b from the right:  B (+  ¡) ΣM z = 0; (20)(20)

−

√ − (50. (50.2)(33. 2)(33.75) + (34. (34.9)(60) = 0

(13.5)

Example Example 13-2: Semi-Circula Semi-Circularr Arch, Arch, (Gerstle (Gerstle 1974) Determine the reactions of the three hinged statically determined semi-circular arch under its own dead weight w (per unit arc length s, where ds = rdθ). rdθ ). 13.6 dP=wRd θ

θ

B

B

r

R

R

θ

A

θ

A

C

R cos θ

Figure 13.6: Semi-Circular three hinged arch Solution: I Reactions The reactions can be determined by integrating the load over the entire structure 1. Vertical ertical Reaction Reaction is determined determined first: (+ ) ΣM A = 0; (C y )(2R )(2R) +

−

 ¡



θ =π

= = =

(13. (13.66-a) a)

dP

moment arm wR =π (1 + cos θ)dθ = [θ sin θ] θθ=0 2 θ =0



θ =π

⇒ C y

wRdθ R(1 + cos θ) = 0

     

θ =0

wR 2 wR [(π [(π 2 π 2 wR

−

|

− sin π) − (0 − sin sin 0)] 0)] (13.6-b)

2. Horizonta Horizontall Reactions Reactions are determined next (+ ) ΣM B = 0; (C x )(R )(R) + (C ( C y )(R )(R)  ¡

Victor Saouma

−

θ = π2

 −

wRdθ

R cos θ

dP

moment arm

       

θ =0

=0

(13.7-a)


Draft 13.1 Arches

2 17



θ

α=0

⇒

wRdα R(cos α

− cos θ) + M = 0 π (θ − π2 )cos θ 2 (1 − sin θ ) + (θ

(13.12)

·

M = wR 2

III Deflection are determined determined last





(13.13)

1. The real curvature curvature φ is obtained by dividing the moment by EI



M wR2 π φ= = (1 EI EI 2

− sin θ) + (θ (θ −

π )cos θ 2



(13.14)

1. The virtual force δP will be a unit vertical point in the direction of the desired deflection, causing a virtual internal moment δM =

R [1 2

− cos θ − sin θ]

0

≤ θ ≤ π2

(13.15)

p 2. Hence, Hence, application application of the virtual work equation equation yields: π

1



∆ = 2

 · δP

13.1.2 13.1.2

2

θ =0



wR2 π (1 EI 2

  −

=

wR 4 7π 2 16 16EI EI

=

.0337 wR EI

− sin θ) + (θ (θ −



M =φ EI

18 18π π

4



− 12

π )cos θ 2

· ·  

R [1 2

− cos θ − sin θ] Rdθ

   dx

δM

(13.16-a)

Statical Statically ly Indeter Indetermin minate ate

Example Example 13-3: Statically Statically Indetermina Indeterminate te Arch, (Kinney (Kinney 1957) Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch, Fig. 13.8 as caused by a concentrated vertical load of 10 k at the center line of the span. Consider shearing, axial, and flexural strains. Assume that the rib is a W24x130 with a total area of 38.21 in2 , that it has a web area of 13.70 in2 , a moment of inertia equal to 4,000 in4 , E of 30,000 k/in2 , and a shearing modulus G of 13,000 k/in2 . Solution: 1. Consider Consider that end end C is placed on rollers, as shown in Fig. ?? A unit fictitious horizontal force is applied at C . The axial and sheari shearing ng component componentss of this fictitio fictitious us force and of the vertical reaction at C , acting on any section θ in the right half of the rib, are shown at the right end of the rib in Fig. 13-7. Victor Saouma


Draft 13.1 Arches

2 19

2. The expression expression for the horizontal horizontal displacement displacement of C of C is



B

1 ∆Ch = 2

 δP

C

M δM ds + 2 EI



B

C

V δV ds + 2 Aw G



B

C

δN

N ds AE

(13.17)

3. From Fig. 13.9 13.9,, for the rib from C to B , M = δM = V = δV = N = δN =

P (100 R cos θ) 2 1(R 1(R sin θ 125. 125.36) P sin θ 2 cos θ P cos θ 2 sin θ

(13.18-a)

−

(13.18-b)

−

(13.18-c) (13.18-d) (13.18-e) (13.18-f)

−

ds = Rdθ

(13.18-g)

4. If the above values values are substituted substituted in Eq. 13.17 and integrated between the limits of 0.898 and π/2, π/2, the result will be ∆Ch = 22 22..55 + 0. 0.023

− 0.003 = 22. 22.57

(13.19)

5. The load P is now assumed to be removed from the rib, and a real horizontal force of 1 k is assumed to act toward the right at C in conjunction with the fictitious horizontal force of 1 k acting to the right at the same point. The horizontal displacement of C will be given by





B

δChCh

B M V = 2 δM ds + 2 δV ds + 2 EI Aw G C C = 2.309 + 0. 0.002 + 0. 0.002 = 2. 2.313 in



B

C

δN

N ds AE

(13.20-a) (13.20-b)

6. The value value of the horizontal horizontal reaction component component will be H C C =

∆Ch 22 22..57 = = 9.75 k δChCh 2.313

(13.21)

7. If only flexural strains strains are considered, considered, the result would would be H C C =

22 22..55 = 9.76 k 2.309

(13.22)

Comments 1. For the given rib and the single concentr concentrated ated load at the center center of the span it is obvious that the effects of shearing and axial strains are insignificant and can be disregarded.

Victor Saouma


Draft

13.2 Curved Space Structures

2 21

with respect to the x and y axis are BP and AB respectively. respectively. Applying three equations of equilibrium we obtain F zA

 −  −

M yA

θ=0

θ =π

wRdθ = 0

⇒

F zA = wRπ

(13.23-a)

(wRdθ)( wRdθ)(R R sin θ) = 0

⇒

M xA = 2wR 2

(13.23-b)

M yA =

(13.23-c)

θ =π

M xA

 −

θ =0

θ =π θ =0

(wRdθ) wRdθ)R(1

− cos θ) = 0 ⇒

−wR2π

II Internal Forces are determined next 1. Shear Force: Force:

 −

θ

(+ 6) ΣF z = 0

⇒ V

0

wRdα = 0

⇒

V = wrθ

(13.24)

2. Bending Moment: Moment:

 −

θ

ΣM R = 0

⇒ M

0

(wRdα)( wRdα)(R R sin α) = 0

M = wR 2 (1

⇒

− cos θ)

(13.25)

−wR2(θ − sin θ)

(13.26)

3. Torsion:



θ

ΣM T T = 0

⇒+

0

(wRdα) wRdα)R(1

− cos α) = 0 ⇒

T =

III Deflection are determined last we assume a rectangular cross-section of width b and height d = 2b and a Poisson’s ratio ν = 0.3. 1. Noting Noting that the member member will be subjecte subjected d to both flexural flexural and torsiona torsionall deform deformaations, we seek to determine the two stiffnesses. 3

3

b(2b (2b) 2. The flexural flexural stiffness stiffness EI is given by EI = E bd 12 = E 12 =

2Eb 4 3

= .667 667Eb Eb 4 .

3. The torsional torsional stiffness of solid rectangular rectangular sections sections J = kb 3 d where b is the shorter side of the section, d the longer, and k a factor equal to .229 for db = 2. He Henc ncee E E 4 4 G = 2(1+ν 385E E , and GJ = (. ( .385 385E E )(. )(.229 229bb ) = .176 176Eb Eb . 2(1+ν ) = 2(1+. 2(1+.3) = .385 4. Considering Considering both flexural flexural and torsional torsional deformations, deformations, and replacing replacing dx by rdθ: rdθ :

              π

δP ∆ δP ∆ = ∗

δW

0

δM

M Rdθ + EI z

π 0

δT

T Rdθ GJ

(13.27)

Torsion

Flexure

∗

δU

where the real moments were given above.

5. Assuming Assuming a unit virtual downw downward ard force δP = 1, we have δM = R sin θ δT =

Victor Saouma

−R(1 − cos θ)

(13.28-a) (13.28-b)


Draft


2 23

Figure 13.11: Geometry of Curved Structure in Space The weak bending axis is normal to both N and S , and thus its unit vector is determined from w = n×s (13.33) 24

13.2.1 13.2.1.2 .2

Equili Equilibri brium um

For the equilibrium equations, we consider the free body diagram of Fig. 13.12 an applied load P is acting at point A. The result resultan antt force force vector vector F and resultant moment vector M acting on the cut section B are determined from equilibrium 25

ΣF = 0;

P + F = 0;

ΣMB = 0; 0 ; L×P + M = 0;

F=

−P M = −L

(13.34-a) P

×

(13.34-b)

where L is the lever arm vector from B to A. The axial and shear forces N, V s and V w are all three components of the force vector F along the N, S , and W axes and can be found by dot product with the appropriate unit vectors: 26

Victor Saouma

N = F·n

(13.35-a)

V s = F·s

(13.35-b)

V w = F·w

(13.35-c)


Draft


2 25

Figure 13.13: Helicoidal Cantilevered Girder

Victor Saouma


Draft


2 27

6. Finally Finally,, the component componentss of the force force F = P k P k and the moment M are obtained by appropriate dot products with the unit vectors

−

N = F·n =

− K 1 P πRH

V s = F·s = 0

(13.47-b)

− K 1 P M n = − PK R (1 − cos θ)

V w = F·w = T =

·

M s = M·s = P R sin θ M w = M·w =

Victor Saouma

(13.47-a)

P H πK (1

− cos θ)

(13.47-c) (13.47-d) (13.47-e) (13.47-f)


Draft Chapter 14

BUILDING STRUCTURES 14.1 14.1


14.1.1 14.1.1

Beam Column Column Connec Connection tionss

1

The connection between the beam and the column can be, Fig. 14.1 14.1::

θb

θb

θb

θc

θb

=

θc

θc

θb

=

θc

Rigid

Flexible

θc

θ - θc ) M=K( s s b θ b = θc Semi-Flexible

Figure 14.1: Flexible, Rigid, and Semi-Flexible Joints Flexible that is a hinge which which can transfer transfer forces only. only. In this case we really have have cantiliver cantiliver action only. only. In a flexible flexible connection connection the column and beam end moments moments are both equal to zero, M col col = M beam beam = 0. The end rotation are not equal, θcol = θbeam .



Rigid: The connection is such that θbeam = θcol and moment can be transmitted through the connection. In a rigid connection, the end moments and rotations are equal (unless there is an externally applied moment at the node), M col col = M beam beam = 0, θcol = θbeam .



Semi-Rigid: The end moments are equal and not equal to zero, but the rotation are different. θbeam = θcol , M col urthermore,, the difference difference in rotation rotation is resisted resisted by col = M beam beam = 0. Furthermore the spring spring M spring θbeam). spring = K spring spring (θcol



14.1.2 14.1.2



−

Behavior Behavior of Simp Simple le Fram Frames es

For vertical load across the beam rigid connection will reduce the maximum moment in the beam (at the expense of a negative moment at the ends which will in turn be transferred to

2

Draft

14.1 Intro duction

23 1

2

W=wL, M=wL/8, M’=Ph

Frame Type

Shear

Deformation

L

w

Moment

w/2

Axial

M -w/2 w/2

w/2

a h

b

P p

POST AND BEAM STRUCTURE M’

w/2

M -w/2 w/2

w/2

c -M’/L

d

M’ L / ’ M

L / ’ M -

p

SIMPLE BENT FRAME

w/2

M -w/2 w/2

w/2

e -M’/L

f

M’ L / ’ M

L / ’ M -

p THREE-HINGE PORTAL

w/2

-w/2

M

M M/h

g

-M/h

w/2

M/h M’/2

-M/L

h

M’/2 p/2

p/2

p/2

w/2

-w/2

0.4M

0.4M

w/2

M’/2

-M/L

j

0.4M/h

0.64M

h / M 6 3 . 0

h / M 6 3 . 0 -

p/2 p/2

w/2

l

w/2

M’/2

p/2

-w/2

h / M 8 6 . 0 -

0.45M

h / M 8 6 . 0

-0.5M’/L

L / ’ M

L / ’ M -

TWO-HINGE FRAME

k

L / M

L / M -

THREE-HINGE PORTAL

i

w/2

0.45M 0.68M/h

0.55M

w/2

M’/4

w/2

M’/4 p/2

p/2

p/2

RIGID FRAME M’/4

M’/4

L 2 / ’ M -

L 2 / ’ M

Figure Figure 14.3: Deformation, Deformation, Shear, Moment, Moment, and Axial Diagrams for Various Various Types of Portal Portal Frames Subjected to Vertical and Horizontal Loads

Victor Saouma


Draft

14.2 Buildings Structures

14.2 14.2 11

233

Build Buildin ings gs Struc Structu ture ress

There are three primary types of building systems:

Wall Subsytem: in which very rigid walls made up of solid masonry, paneled or braced timber, or steel trusses constitute a rigid subsystem. This is only adequate for small rise buildings. Vertical Shafts: made up of four solid or trussed walls forming a tubular space structure. The tubular tubular structure structure may be b e interior interior (housing elevators elevators,, staircases staircases)) and/or exterior. exterior. Most efficient for very high rise buildings. Rigid Frame: which consists of linear vertical components (columns) rigidly connected to stiff horizontal ones (beams and girders). This is not a very efficient structural form to resist lateral (wind/earthquake) loads. 14.2.1 14.2.1

Wall Subsyst Subsystems ems

Whereas exterior wall provide enclosure and interior ones separation, both of them can also have a structural role in trnsfering vertical and horizontal loads.

12

13

Walls are constructed out of masonry, timber concrete or steel.

If the wall is braced by floors, then it can provide an excellent resitance to horizontal load in the plane of the wall (but not orthogonal to it).

14

When shear-walls subsytems are used, it is best if the center of orthogonal shear resistance is close to the centroi centroid d of latera laterall loads loads as applied applied.. If this is not the case, case, then then there will be torsional design problems. 15

14.2.1 14.2.1.1 .1

Exampl Example: e: Concret Concrete e Shear Shear Wall Wall

From From (Lin (Lin and Stotesb Stotesbury ury 1981) 1981)

We consider a reinforced concrete wall 20 ft wide, 1 ft thick, and 120 ft high with a vertical load of 400 k acting acting on it at the base. As a result result of wind, we assume assume a unifor uniform m horizo horizont ntal al force of 0.8 kip/ft of vertical vertical height height acting on the wall. It is required required to compute the flexural stresses and the shearing stresses in the wall to resist the wind load, Fig. 14.5 14.5.. 16

1. Maximum Maximum shear force and bending moment moment at the base V max (0.8) k.ft(120) ft = 96 k max = wL = (0. wL2 (0. (0.8) k.ft(120)2 ft2 M max = = = 5, 760 k.ft max 2 2

(14.8-a) (14.8-b)

2. The resulting resulting eccentricit eccentricity y is eActual =

M (5, (5, 760) k.ft = = 14 14..4 ft P (400) k

(14.9)

3. The critical critical eccentricit eccentricity y is L (20) ft = = 3.3 ft < eActual N.G. 6 6 thus there will be tension at the base. ecr =

Victor Saouma

(14.10)


Draft


235

9. The compress compressiv ivee stress stress of 740 psi can easily easily be sustain sustained ed by concret concrete, e, as to the tensile tensile stress of 460 psi, it would have to be resisted by some steel reinforcement. 10. Given Given that those stresses stresses are service stresses and not factored ones, we adopt the WSD approach, and use an allowable stress of 20 ksi, which in turn will be increased by 4/ 4/3 for seismic and wind load, 4 σall = (20) = 26. 26.7 ksi (14.16) 3 11. The stress stress distribution distribution is linear, compression compression at one end, and tension at the other. The length of the tension area is given by (similar triangles) x 20 = 460 460 + 740

⇒ x = 460460 (20) = 7. 7.7 ft + 740

(14.17)

12. The total tensile force force inside this triangular triangular stress block is 1 T = (460) ksi(7. (7.7 2

× 12) in (12) in = 250 k

(14.18)

     width

13. The total amount of steel reinforcemen reinforcementt needed needed is As =

(250) k = 9.4 in2 (26. (26.7) ksi

(14.19)

This amount of reinforcement should be provided at both ends of the wall since the wind or eartquake can act in any direction. In addition, the foundations should be designed to resist tensile uplift forces (possibly using piles). 14.2.1 14.2.1.2 .2

Exampl Example: e: Trussed russed Shear Shear Wall Wall

From From (Lin (Lin and Stotesb Stotesbury ury 1981) 1981)

We consider the same problem previously analysed, but use a trussed shear wall instead of a concrete one, Fig. 14.6 14.6.. 17

1. Using the maximum maximum moment moment of 5, 5, 760 kip-ft (Eq. 14.8-b 14.8-b), ), we can compute the compression and tension in the columns for a lever arm of 20 ft. F =

(5, 760) k.ft ± (5, = ±288 k (20) ft

(14.20)

2. If we now add the effect effect of the 400 kip vertical vertical load, the forces forces would be C = T =

k − (400) − 288 = −488 k 2 k − (400) + 288 288 = 88 k 2

(14.21-a) (14.21-b)

3. The force in the diagonal which which must resist a base shear of 96 kip is (similar triangles) triangles) F = 96



(20)2 + (24)2 20

⇒ F =



(20)2 + (24)2 (96) (96) = 154 154 k 20

(14.22)

4. The design could be modified to have have no tensile tensile forces in the columns columns by increasing increasing the width of the base (currently at 20 ft). Victor Saouma


Draft


237 ~ 20 ’

20 ’

20 ’

~ 20 ’

w = 0.8 k/ft

120 ’ H = 96 k N.A.

60 ’

Figure 14.7: Design Example of a Tubular Structure, (Lin and Stotesbury 1981) 4. The maximum maximum flexural stresses: stresses: σf l =

(5, 760) k.ft(20/ (20/2) ft ± MI C = ± (5, = ±12 12..5 ksf = ±87 psi (4, (4, 600) ft4

(14.25)

5. The average average shear shear stress is τ =

V (96) k = = 2.4 ksf = 17 psi A 2(20)(1) ft2

(14.26)

6. The vertical vertical load of 1,600 k produces an axial stress stress of σax =

P (1, (1, 600) k = = A (4(20)(1) ft2

−

−20 ksf = −140 psi

(14.27)

7. The total stresses stresses are thus thus σ = σax + σf l σ1 = σ2 =

−140 + 87 = −53 psi −140 − 87 = −227 psi

(14.28-a) (14.28-b) (14.28-c)

thus we do not have any tensile stresses, and those stresses are much better than those obtained from a single shear wall. 14.2.3 14.2.3

Rigid Rigid Frame ramess

Rigid frames can carry both vertical and horizontal loads, however their analysis is more complex than for tubes. 21

Victor Saouma


Draft

14.3 Approximate Analysis of Buildings

14.3.1 14.3.1

239

Vertic ertical al Loads Loads

The girders at each floor are assumed to be continuous beams, and columns are assumed to resist the resulting unbalanced moments from the girders.

30

31

Basic assumptions 1. Girders Girders at each floor act as continous continous beams supporting supporting a uniform uniform load. 2. Inflection Inflection points points are assumed assumed to be at (a) One tenth tenth the span from both ends of each girder. girder. (b) Mid-heigh Mid-heightt of the columns 3. Axial forces forces and deformation deformation in the girder girder are negligibly small. 4. Unbal Unbalanc anced ed end moments moments from from the girders girders at each each joint joint is distri distribut buted ed to the columns columns above and below the floor.

Based on the first assumption, all beams are statically determinate and have a span, L s equal to 0.8 the original length of the girder, L. (Note that for a rigidly connected member, the inflection inflection point is at 0.211 L, and at the support for a simply supported beam; hence, depending depending on the nature of the connection one could consider those values as upper and lower bounds for the approximate location of the hinge). 32

33

End forces are given by

Maximum positive moment at the center of each beam is, Fig. 14.9 w

rgt

M

lft

M

rgt

V lft

V

0.1L

0.8L

0.1L

L

Figure 14.9: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments 1 1 M + = wL2s = w (0. (0.8)2 L2 = 0.08 08wL wL2 8 8

(14.29)

Maximum negative moment at each end of the girder is given by, Fig. 14.9 M left = M rgt =

Victor Saouma

w − w2 (0. (0.1L)2 − (0. (0.8L)(0. )(0.1L) = −0.045 045wL wL2 2

(14.30)


Draft


241

Column Shear Points of inflection are at mid-height, with possible exception when the columns on the first floor are hinged at the base, Fig. 14.11 V =

M top

(14.34)

h 2

Girder axial forces are assumed to be negligible eventhough the unbalanced column shears above and below a floor will be resisted by girders at the floor. 14.3.2 14.3.2 34

Horiz Horizon onta tall Loads Loads

We must differentiate between low and high rise buildings.

Low rise buidlings, where the height is at least samller than the hrizontal dimension, the deflected shape is characterized by shear deformations. High rise buildings, where the height is several times greater than its least horizontal dimension, the deflected shape is dominated by overall flexural deformation. 14.3.2 14.3.2.1 .1

Porta Po rtall Method Method

Low rise buildings under lateral lateral loads, have have predominan predominantly tly shear deformations. deformations. Thus, Thus, the approximate analysis of this type of structure is based on

35

1. Distributio Distribution n of horizontal horizontal shear forces. 2. Location of inflection inflection points. points. 36

The portal method is based on the following assumptions 1. Inflection Inflection points points are located at (a) Mid-heigh Mid-heightt of all columns above above the second floor. (b) Mid-heigh Mid-heightt of floor columns if rigid support, or at the base if hinged. (c) At the center center of each girder. 2. Total horizon horizontal tal shear at the mid-heig mid-height ht of all columns columns at any any floor level level will will be disdistributed among these columns so that each of the two exterior columns carry half as much horizontal shear as each interior columns of the frame.

37

Forces are obtained from

Column Shear is obtained by passing a horizontal section through the mid-height of the columns at each floor and summing the lateral forces above it, then Fig. 14.12 V

Victor Saouma

ext

=



F lateral

2No. of bays

V int = 2V ext

(14.35)


Draft


243 above

P

rgt i-1

lft

V

V

i

Figure 14.14: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force below

P Column Axial Forces are obtained by summing girder shears and the axial force from the column above, Fig. ??

P = P above + P rgt + P lft

(14.39)

Example Example 14-1: Approximate Approximate Analysis Analysis of a Frame subjected to Vertical ertical and Horizonta Horizontall Loads

Draw the shear, and moment diagram for the following frame. Solution:

0.25

K/ft

K

15

5 30

12

6

13 K/ft 0.50

7

14

8

14’

K

10

9

3

2

1

20’

30’

11

4

16’

24’

Figure 14.15: Example; Approximate Analysis of a Building

Vertical Loads

Victor Saouma


Draft


0.25

12

5

6

245

K/ft

13 K/ft 0.50

9

10

20’

30’

+11.5’

k

-10.1’

-9.0’

+23.0’

k

k

-9.0’

k

-4.5’

k

-20.2’

k

+3.6’

k

k

-5.6’

k

k

-3.6’ -5.6’

+5.6’

k

+6.5’

k

k

-4.5’

-13.0’

-13.0’

k

+5.6’ k

k

k

-20.2’

k

k

+32.0’ k

+4.5’

k

-6.5’

k

k

-10.1’ -6.5’

k

+16.0’

16’

k

k

-4.5’

k

k

4

k

k

14’

24’

+18.0’

k

+8.0’

+4.5’

8

11

3

2

1

-4.5’

14

7

-6.5’ k

+3.6’

+6.5’ k

-3.6’

k

-6.5’

Figure 14.16: Approximate Analysis of a Building; Moments Due to Vertical Loads

Victor Saouma


Draft


247

K

+3.75

K

+2.5

K

+3.0

K

-2.5

K

-3.75

K

+7.5

K

+5.0

K

-3.0

K

+6.0

K

-5.0

K

-6.0

K

-7.5

K

K

-0.64

K

-0.80

K

K

-0.56

K

+0.51

-0.70

K

+0.45

+0.93

K

+0.81

Figure 14.17: Approximate Analysis of a Building; Shears Due to Vertical Loads Horizontal Loads, Portal Method

1. Column Shears Shears

V 5 V 6 V 7 V 8 V 1 V 2 V 3 V 4

= = = = = = = =

15 (2)(3)

2(V 2(V 5) = (2)(2. (2)(2.5) 2(V 2(V 5) = (2)(2. (2)(2.5) V 5 15+30 (2)(3)

2(V 2(V 1) = (2)(7. (2)(7.5) 2(V 2(V 1) = (2)(2. (2)(2.5) V 1

= = = = = = = =

2.5 k 5k 5k 2.5 k 7.5 k 15 k 15 k 7.5 k

2. Top Column Moments Moments M 5top M 5bot M 6top M 6bot

= = = =

M 7top = M 7bot = M 8top M 8bot

Victor Saouma

= =

(2.5)(14) V 1 H 5 = (2. 2 2 M 5top V 6 H 6 = (5)(14) 2 2 M 6top V 7up H 7 = (5)(14) 2 2 M 7top V 8up H 8 (2.5)(14) = (2. 2 2 M 8top

− − − −

= = = = = = = =

− − − −

17 17..5 17 17..5 35 35..0 35 35..0

k.ft k.ft k.ft k.ft

35 35..0 k.ft 35 35..0 k.ft 17 17..5 k.ft 17 17..5 k.ft


Structural Concepts and Systems

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