Ebook Structural analysis (Eighth edition): Part 1

Part 1 of ebook Structural analysis (Eighth edition) provide readers with content about: types of structures and loads; analysis of statically determinate structures; analysis of statically determinate trusses; internal loadings developed in structural members; cables and arches;... Please refer to the part 1 of ebook for details!

L Table for Evaluating L0 m m؅ dx L L0 m m¿ dx mm¿L mm¿L m1m¿1 + m¿22L 2 mm¿L mm¿L mm¿L m1m¿1 + 2m¿22L mm¿L 12 m¿1m1 + m22L m¿1m1 + 2m22L 3m¿ 12m1 + m22 + m¿21m1 + 2m224L mm¿L mm¿1L + a2 m3m¿11L + b2 + m21L + a24 3a a2 mm¿ a3 + - bL 12 L L mm¿L mm¿L m12m¿1 + m¿22L mm¿L 3m¿13m1 + 5m224L 12 Beam Deflections and Slopes Loading v + c vmax = at x = L vmax = PL3 3EI MO L2 2EI at x = L u + g PL2 2EI at x = L umax = - umax = MO L EI at x = L Equation ϩ v = c ϩg P 1x3 - 3Lx22 6EI v = MO 2EI x2 Beam Deflections and Slopes (continued) vmax = - wL4 8EI at x = L vmax = - PL3 48EI at x = L>2 wL3 6EI at x = L umax = - PL2 16EI at x = or x = L u = at x = 5wL4 384EI P 14x3 - 3L2x2, 48EI … x … L>2 Pab1L + b2 v = - 6LEI Pab1L + a2 wL3 24EI Pbx 1L2 - b2 - x22 6LEI 0…x…a 6LEI umax = Ϯ w 1x4 - 4Lx3 + 6L2x22 24EI v = umax = Ϯ uL = - vmax = - v = - v = - wx 1x3 - 2Lx2 + L32 24EI L v = uL = uR = 3wL3 128EI 7wL3 384EI wx 116x3 - 24Lx2 + 9L32 384EI … x … L>2 v = - wL 18x3 - 24Lx2 + 17L2x - L32 384EI L>2 … x … L uL = vmax = - MO L2 923EI MO L 6EI v = - uR = MO L 3EI MO x 6EIL 1L2 - x22 STRUCTURAL ANALYSIS This page intentionally left blank STRUCTURAL ANALYSIS EIGHTH EDITION R C HIBBELER PRENTICE HALL Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Library of Congress Cataloging-in-Publication Data on File Vice President and Editorial Director, ECS: Marcia J Horton Acquisitions Editor: Tacy Quinn/Norrin Dais Managing Editor: Scott Disanno Production Editor: Rose Kernan Art Director: Kenny Beck Managing Editor, AV Management and Production: Patricia Burns Art Editor: Gregory Dulles Media Editor: David Alick Director, Image Resource Center: Melinda Reo Manager, Rights and Permissions: Zina Arabia Manager, Visual Research: Beth Brenzel Manager, Cover Visual Research and Permissions: Karen Sanatar Manufacturing Manager: Alexis Heydt-Long Manufacturing Buyer: Lisa McDowell Senior Marketing Manager: Tim Galligan Cover Designer: Kenny Beck About the Cover: Background Image: Orange Steel girders/zimmytws/Shutterstock; Inset image: Building under construction/Vladitto/Shutterstock © 2012 by R C Hibbeler Published by Pearson Prentice Hall Pearson Education, Inc Upper Saddle River, New Jersey 07458 All rights reserved No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher Pearson Prentice Hall™ is a trademark of Pearson Education, Inc The author and publisher of this book have used their best efforts in preparing this book These efforts include the development, research, and testing of the theories and programs to determine their effectiveness The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs Previous editions copyright © 2009, 2006, 2002, 1999, 1995, 1990, 1985 by R C Hibbeler Pearson Education Ltd., London Pearson Education Australia Pty Ltd., Sydney Pearson Education Singapore, Pte Ltd Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educación de Mexico, S.A de C.V Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte Ltd Pearson Education, Upper Saddle River, New Jersey Printed in the United States of America 10 ISBN-10: 0-13-257053-X ISBN-13: 978-0-13-257053-4 To The Student With the hope that this work will stimulate an interest in Structural Analysis and provide an acceptable guide to its understanding This page intentionally left blank PREFACE This book is intended to provide the student with a clear and thorough presentation of the theory and application of structural analysis as it applies to trusses, beams, and frames Emphasis is placed on developing the student’s ability to both model and analyze a structure and to provide realistic applications encountered in professional practice For many years now, engineers have been using matrix methods to analyze structures Although these methods are most efficient for a structural analysis, it is the author’s opinion that students taking a first course in this subject should also be well versed in some of the more important classicial methods Practice in applying these methods will develop a deeper understanding of the basic engineering sciences of statics and mechanics of materials Also, problem-solving skills are further developed when the various techniques are thought out and applied in a clear and orderly way By solving problems in this way one can better grasp the way loads are transmitted through a structure and obtain a more complete understanding of the way the structure deforms under load Finally, the classicial methods provide a means of checking computer results rather than simply relying on the generated output New to This Edition • Fundamental Problems These problem sets are selectively located just after the example problems They offer students simple applications of the concepts and, therefore, provide them with the chance to develop their problem-solving skills before attempting to solve any of the standard problems that follow You may consider these problems as extended examples since they all have solutions and answers that are given in the back of the book Additionally, the fundamental problems offer students an excellent means of studying for exams, and they can be used at a later time to prepare for the exam necessary to obtain a professional engineering license • Content Revision Each section of the text was carefully reviewed to enhance clarity This has included incorporating the new ASCE/ SEI 07-10 standards on loading in Chapter 1, an improved explanation of how to draw shear and moment diagrams and the deflection curve of a structure, consolidating the material on structures having a variable moment of inertia, providing further discussion for analyzing structures having internal hinges using matrix analysis, and adding a new Appendix B that discusses some of the common features used for running current structural analysis computer software 248 CHAPTER INFLUENCE LINES S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S FOR EXAMPLE 6.19 Determine the maximum positive moment created at point B in the beam shown in Fig 6–31a due to the wheel loads of the crane kN kN kN 3m 2m A B 2m MB kN kN 3m 3m 1.20 kN 0.4 2m x A 2m Ϫ0.8 influence line for MB C B 3m 2m (a) (b) Fig 6–31 SOLUTION The influence line for the moment at B is shown in Fig 6–31b 2-m Movement of 3-kN Load If the 3-kN load is assumed to act at B and then moves m to the right, Fig 6–31b, the change in moment is ¢MB = - 3a 1.20 1.20 b122 + 8a b122 = 7.20 kN # m 3 Why is the 4-kN load not included in the calculations? 3-m Movement of 8-kN Load If the 8-kN load is assumed to act at B and then moves m to the right, the change in moment is 1.20 1.20 1.20 b132 - 8a b132 + 4a b122 3 = - 8.40 kN # m ¢MB = - 3a Notice here that the 4-kN load was initially m off the beam, and therefore moves only m on the beam Since there is a sign change in ¢MB, the correct position of the loads for maximum positive moment at B occurs when the 8-kN force is at B, Fig 6–31b Therefore, 1MB2max = 811.202 + 310.42 = 10.8 kN # m Ans 6.6 249 MAXIMUM INFLUENCE AT A POINT DUE TO A SERIES OF CONCENTRATED LOADS EXAMPLE 6.20 Determine the maximum compressive force developed in member BG of the side truss in Fig 6–32a due to the right side wheel loads of the car and trailer Assume the loads are applied directly to the truss and move only to the right H kN kN G F FBG 1.5 kN 4m 0.3125 3m 2m A B 3m C 3m D 3m 3m 0.3125 b11 m2 3m = - 0.729 kN 4-kN Load at Point C By inspection this would seem a more reasonable case than the previous one -0.625 b14 m2 + kN10.31252 6m = - 2.50 kN 2-kN Load at Point C In this case all loads will create a compressive force in BC FBG = kN1- 0.6252 + kN a = - 2.66 kN x (b) In this case FBG = kN1 - 0.6252 + 1.5 kNa 12 influence line for FBG SOLUTION The influence line for the force in member BG is shown in Fig 6–32b Here a trial-and-error approach for the solution will be used Since we want the greatest negative (compressive) force in BG, we begin as follows: FBG = 1.5 kN1-0.6252 + 4102 + kN a Ϫ0.625 Fig 6–32 1.5-kN Load at Point C E (a) -0.625 -0.625 b13 m2 + 1.5 kNa b11 m2 6m 6m Ans Since this final case results in the largest answer, the critical loading occurs when the 2-kN load is at C 250 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S 6.7 Absolute Maximum Shear and Moment In Sec 6–6 we developed the methods for computing the maximum shear and moment at a specified point in a beam due to a series of concentrated moving loads A more general problem involves the determination of both the location of the point in the beam and the position of the loading on the beam so that one can obtain the absolute maximum shear and moment caused by the loads If the beam is cantilevered or simply supported, this problem can be readily solved Shear For a cantilevered beam the absolute maximum shear will Vabs occur at a point located just next to the fixed support The maximum shear is found by the method of sections, with the loads positioned anywhere on the span, Fig 6–33 For simply supported beams the absolute maximum shear will occur just next to one of the supports For example, if the loads are equivalent, they are positioned so that the first one in sequence is placed close to the support, as in Fig 6–34 max Fig 6–33 Moment The absolute maximum moment for a cantilevered beam occurs at the same point where absolute maximum shear occurs, although in this case the concentrated loads should be positioned at the far end of the beam, as in Fig 6–35 For a simply supported beam the critical position of the loads and the associated absolute maximum moment cannot, in general, be determined by inspection We can, however, determine the position analytically For purposes of discussion, consider a beam subjected to the forces F1, F2, F3 shown in Fig 6–36a Since the moment diagram for a series of concentrated forces consists of straight line segments having peaks at each force, the absolute maximum moment will occur under one of the forces Assume this maximum moment occurs under F2 The position of the loads F1, F2, F3 on the beam will be specified by the distance x, measured from F2 to the beam’s centerline as shown To determine a specific value of x, we first obtain the resultant force of the system, FR, and its distance Vabs max Fig 6–34 Mabs max Fig 6–35 FR F1 F2 _ F3 (x¿Ϫx) x F1 d1 A _ x¿ d1 L — M2 B L Ϫ x) (— d2 L — Ay V2 Ay By (a) Fig 6–36 (b) 6.7 ABSOLUTE MAXIMUM SHEAR AND MOMENT 251 x¿ measured from F2 Once this is done, moments are summed about B, which yields the beam’s left reaction, Ay, that is, ©MB = 0; Ay = L 1FR2c - 1x¿ - x2 d L If the beam is sectioned just to the left of F2, the resulting free-body diagram is shown in Fig 6–36b The moment M2 under F2 is therefore ©M = 0; M2 = Ay a L - x b - F1 d1 = L L 1F 2c - 1x¿ - x2 d a - xb - F1 d1 L R 2 = FR L FRx¿ FRx2 FR xx¿ + - F1 d1 L L The absolute maximum moment in this girder bridge is the result of the moving concentrated loads caused by the wheels of these train cars The cars must be in the critical position, and the location of the point in the girder where the absolute maximum moment occurs must be identified For maximum M2 we require -2FRx FRx¿ dM2 = + = dx L L or x = x¿ Hence, we may conclude that the absolute maximum moment in a simply supported beam occurs under one of the concentrated forces, such that this force is positioned on the beam so that it and the resultant force of the system are equidistant from the beam’s centerline Since there are a series of loads on the span (for example, F1, F2, F3 in Fig 6–36a), this principle will have to be applied to each load in the series and the corresponding maximum moment computed By comparison, the largest moment is the absolute maximum As a general rule, though, the absolute maximum moment often occurs under the largest force lying nearest the resultant force of the system Envelope of Maximum Influence-Line Values Rules or formulations for determining the absolute maximum shear or moment are difficult to establish for beams supported in ways other than the cantilever or simple support discussed here An elementary way to proceed to solve this problem, however, requires constructing influence lines for the shear or moment at selected points along the entire length of the beam and then computing the maximum shear or moment in the beam for each point using the methods of Sec 6–6 These values when plotted yield an “envelope of maximums,” from which both the absolute maximum value of shear or moment and its location can be found Obviously, a computer solution for this problem is desirable for complicated situations, since the work can be rather tedious if carried out by hand calculations 252 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S EXAMPLE 6.21 Determine the absolute maximum moment in the simply supported bridge deck shown in Fig 6–37a FR ϭ 4.5 k 2k 1.5 k _ x ϭ 6.67 ft B A 1k C 10 ft ft 30 ft (a) SOLUTION The magnitude and position of the resultant force of the system are determined first, Fig 6–37a We have FR ϭ 4.5 k 2k + T FR = ©F; 1.5 k k e +MRC = ©MC; FR = + 1.5 + = 4.5 k 4.5x = 1.51102 + 11152 x = 6.67 ft A B Ay 6.67 ft 6.67 ft By ft 1.67 ft 15 ft 15 ft Let us first assume the absolute maximum moment occurs under the 1.5-k load The load and the resultant force are positioned equidistant from the beam’s centerline, Fig 6–37b Calculating Ay first, Fig 6–37b, we have d + ©MB = 0; - Ay1302 + 4.5116.672 = Ay = 2.50 k (b) Now using the left section of the beam, Fig 6–37c, yields d + ©MS = 0; - 2.50116.672 + 21102 + MS = MS = 21.7 k # ft 2k 1.5 k MS 6.67 ft 10 ft Ay ϭ 2.5 k (c) Fig 6–37 VS 6.7 ABSOLUTE MAXIMUM SHEAR AND MOMENT 253 There is a possibility that the absolute maximum moment may occur under the 2-k load, since k 1.5 k and FR is between both k and 1.5 k To investigate this case, the 2-k load and FR are positioned equidistant from the beam’s centerline, Fig 6–37d Show that Ay = 1.75 k as indicated in Fig 6–37e and that MS = 20.4 k # ft By comparison, the absolute maximum moment is MS = 21.7 k # ft Ans which occurs under the 1.5-k load, when the loads are positioned on the beam as shown in Fig 6–37b FR ϭ 4.5 k 2k 1.5 k 1k 11.67 ft Ay 3.33 ft By 15 ft (d) 2k MS 11.67 ft Ay ϭ 1.75 k (e) VS 254 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S EXAMPLE 6.22 The truck has a mass of Mg and a center of gravity at G as shown in Fig 6–38a Determine the absolute maximum moment developed in the simply supported bridge deck due to the truck’s weight The bridge has a length of 10 m SOLUTION As noted in Fig 6–38a, the weight of the truck, 211032 kg19.81 m/s22 = 19.62 kN, and the wheel reactions have been calculated by statics Since the largest reaction occurs at the front wheel, we will select this wheel along with the resultant force and position them equidistant from the centerline of the bridge, Fig 6–38b Using the resultant force rather than the wheel loads, the vertical reaction at B is then 19.62 kN d + ©MA = 0; By1102 - 19.6214.52 = By = 8.829 kN G 2m 6.54 kN The maximum moment occurs under the front wheel loading Using the right section of the bridge deck, Fig 6–38c, we have 1m 13.08 kN 8.82914.52 - Ms = d + ©Ms = 0; Ms = 39.7 kN # m (a) Ans 19.62 kN 13.08 kN 6.54 kN 0.5 m 0.5 m A B 5m 5m Ay By (b) 13.08 kN Ms Vs (c) Fig 6–38 4.5 m 8.83 kN 6.7 255 ABSOLUTE MAXIMUM SHEAR AND MOMENT PROBLEMS 6–59 Determine the maximum moment at point C on the single girder caused by the moving dolly that has a mass of Mg and a mass center at G Assume A is a roller 6–62 Determine the maximum positive moment at the splice C on the side girder caused by the moving load which travels along the center of the bridge kN G A B C kN 0.5 m 1.5 m A C B 4m 5m 5m 8m 5m 8m 8m Prob 6–62 Prob 6–59 *6–60 Determine the maximum moment in the suspended rail at point B if the rail supports the load of 2.5 k on the trolley 6–63 Determine the maximum moment at C caused by the moving load 2400 lb ft ft ft ft A ft B ft C A B C 15 ft 15 ft ft ft Prob 6–63 2.5 k Prob 6–60 6–61 Determine the maximum positive shear at point B if the rail supports the load of 2.5 k on the trolley ft ft A ft B ft C *6–64 Draw the influence line for the force in member IH of the bridge truss Determine the maximum force (tension or compression) that can be developed in this member due to a 72-k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck, so that half its load is transferred to each of the two side trusses Also assume the members are pin connected at the gusset plates 32 k 32 k 8k J A I L K B H G 10 ft F 10 ft M C D E 25 ft 15 ft ft ft 20 ft 20 ft 20 ft 20 ft 20 ft 2.5 k Prob 6–61 Prob 6–64 256 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S 6–65 Determine the maximum positive moment at point C on the single girder caused by the moving load kN kN kN *6–68 Draw the influence line for the force in member IC of the bridge truss Determine the maximum force (tension or compression) that can be developed in the member due to a 5-k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck, so that half the load shown is transferred to each of the two side trusses Also assume the members are pin connected at the gusset plates ft 3k 2k 1.5 m 2m A 5m C B 5m J I H G F Prob 6–65 15 ft E B A 20 ft 6–66 The cart has a weight of 2500 lb and a center of gravity at G Determine the maximum positive moment created in the side girder at C as it crosses the bridge Assume the car can travel in either direction along the center of the deck, so that half its load is transferred to each of the two side girders C 20 ft D 20 ft 20 ft Probs 6–67/6–68 6–69 The truck has a mass of Mg and mass center at G1, and the trailer has a mass of Mg and mass center at G2 Determine the absolute maximum live moment developed in the bridge 1.5 ft ft G2 G1 G A ft 1.5 m 1.5 m B C 0.75 m ft A 8m B Prob 6–69 Prob 6–66 6–70 Determine the absolute maximum live moment in the bridge in Problem 6–69 if the trailer is removed 6–67 Draw the influence line for the force in member BC of the bridge truss Determine the maximum force (tension or compression) that can be developed in the member due to a 5-k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck, so that half the load shown is transferred to each of the two side trusses Also assume the members are pin connected at the gusset plates G2 G1 1.5 m 1.5 m 0.75 m A 8m Prob 6–70 B 6.7 6–71 Determine the absolute maximum live shear and absolute maximum moment in the jib beam AB due to the 10-kN loading The end constraints require 0.1 m … x … 3.9 m 257 ABSOLUTE MAXIMUM SHEAR AND MOMENT 6–73 Determine the absolute maximum moment in the girder bridge due to the truck loading shown The load is applied directly to the girder 15 k 10 k 7k 3k 20 ft ft ft 4m A A B 80 ft B x Prob 6–73 10 kN 6–74 Determine the absolute maximum shear in the beam due to the loading shown 20 kN 25 kN 40 kN Prob 6–71 4m A 1.5 m B 12 m Prob 6–74 *6–72 Determine the maximum moment at C caused by the moving loads 6–75 Determine the absolute maximum moment in the beam due to the loading shown 20 kN 25 kN 40 kN 6k 4k 2k 2k A A 1.5 m B 12 m B C 20 ft 4m 30 ft Prob 6–72 ft ft ft Prob 6–75 258 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S *6–76 Determine the absolute maximum shear in the bridge girder due to the loading shown 6–79 Determine the absolute maximum shear in the beam due to the loading shown 6k 10 k 6k k4 k 3k ft A ft ft ft B 30 ft 30 ft Prob 6–79 Prob 6–76 *6–80 Determine the absolute maximum moment in the bridge due to the loading shown 6–77 Determine the absolute maximum moment in the bridge girder due to the loading shown 6k 3k 10 k 6k k4 k ft ft ft 30 ft Prob 6–80 ft A B 30 ft Prob 6–77 6–81 The trolley rolls at C and D along the bottom and top flange of beam AB Determine the absolute maximum moment developed in the beam if the load supported by the trolley is k Assume the support at A is a pin and at B a roller 20 ft 6–78 Determine the absolute maximum moment in the girder due to the loading shown D A B C ft 10 k 8k 0.5 ft k4 k ft ft ft 25 ft Prob 6–78 Prob 6–81 259 PROJECT PROBLEMS PROJECT PROBLEMS 6–1P The chain hoist on the wall crane can be placed anywhere along the boom (0.1 m x 3.4 m) and has a rated capacity of 28 kN Use an impact factor of 0.3 and determine the absolute maximum bending moment in the boom and the maximum force developed in the tie rod BC The boom is pinned to the wall column at its left end A Neglect the size of the trolley at D 6–2P A simply supported pedestrian bridge is to be constructed in a city park and two designs have been proposed as shown in case a and case b The truss members are to be made from timber The deck consists of 1.5-m-long planks that have a mass of 20 kg>m2 A local code states the live load on the deck is required to be kPa with an impact factor of 0.2 Consider the deck to be simply supported on stringers Floor beams then transmit the load to the bottom joints of the truss (See Fig 6–23.) In each case find the member subjected to the largest tension and largest compression load and suggest why you would choose one design over the other Neglect the weights of the truss members H G F 1.25 m A E B 1.25 m C 1.25 m D 1.25 m E 1.25 m case a H F 1.25 m A C 0.75 m G E B A B C D 0.1 m 1.25 m D x 28 kN 3m Prob 6–1P 0.5 m 1.25 m 1.25 m case b Prob 6–2P 1.25 m 260 CHAPTER INFLUENCE LINES FOR S TAT I C A L LY D E T E R M I N AT E S T R U C T U R E S CHAPTER REVIEW An influence line indicates the value of a reaction, shear, or moment at a specific point on a member, as a unit load moves over the member Once the influence line for a reaction, shear, or moment (function) is constructed, then it will be possible to locate the live load on the member to produce the maximum positive or negative value of the function A concentrated live force is applied at the positive (negative) peaks of the influence line The value of the function is then equal to the product of the influence line ordinate and the magnitude of the force Ay F x ϭ L –– B A –– L Ay ϭ L L x (––12 )F Ay A uniform distributed load extends over a positive (negative) region of the influence line The value of the function is then equal to the product of the area under the influence line for the region and the magnitude of the uniform load Ay w0 A B Ay ϭ –– (1)(L)(w0) L L x Ay The general shape of the influence line can be established using the Müller-Breslau principle, which states that the influence line for a reaction, shear, or moment is to the same scale as the deflected shape of the member when it is acted upon by the reaction, shear, or moment deflected shape A Ay (a) (b) 261 CHAPTER REVIEW Influence lines for floor girders and trusses can be established by placing the unit load at each panel point or joint, and calculating the value of the required reaction, shear, or moment When a series of concentrated loads pass over the member, then the various positions of the load on the member have to be considered to determine the largest shear or moment in the member In general, place the loadings so that each contributes its maximum influence, as determined by multiplying each load by the ordinate of the influence line This process of finding the actual position can be done using a trial-and-error procedure, or by finding the change in either the shear or moment when the loads are moved from one position to another Each moment is investigated until a negative value of shear or moment occurs Once this happens the previous position will define the critical loading Vabs max Absolute maximum shear in a cantilever or simply supported beam will occur at a support, when one of the loads is placed next to the support Vabs max Absolute maximum moment in a cantilevered beam occurs when the series of concentrated loads are placed at the farthest point away from the fixed support Mabs max FR To determine the absolute maximum moment in a simply supported beam, the resultant of the force system is first determined Then it, along with one of the concentrated forces in the system is positioned so that these two forces are equidistant from the centerline of the beam The maximum moment then occurs under the selected force Each force in the system is selected in this manner, and by comparison the largest for all these cases is the absolute maximum moment F1 F2 _ x¿ — _ x¿ — F3 _ x¿ L — Ay L — By The portal to this bridge must resist loteral loads due to wind and traffic An approximate analysis can be made of the forces produced for a preliminary design of the members, before a more exact structural analysis is done ... Douglas Fir Wood, Southern Pine Wood, spruce 17 0 10 8 14 4 11 1 15 0 63 11 0 10 0 12 0 10 5 13 5 36 492 34 37 29 kN͞m3 26.7 17 .0 22.6 17 .4 23.6 9.9 17 .3 15 .7 18 .9 16 .5 21. 2 5.7 77.3 5.3 5.8 4.5 *Reproduced with... Structures 429 10 .10 Influence Lines for Statically Indeterminate Beams 435 10 .11 Qualitative Influence Lines for Frames 439 Problems 446 Chapter Review 448 11 11 .2 11 .3 11 .4 11 .5 12 .3 12 .4 12 .5 Displacement... mm¿L mm¿L m1m? ?1 + m¿22L 2 mm¿L mm¿L mm¿L m1m? ?1 + 2m¿22L mm¿L 12 m¿1m1 + m22L m¿1m1 + 2m22L 3m¿ 12 m1 + m22 + m¿21m1 + 2m224L mm¿L mm¿1L + a2 m3m? ?11 L + b2 + m21L + a24 3a a2 mm¿ a3 + - bL 12 L L mm¿L