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An Instructor’s Solutions Manual to Accompany th STEEL DESIGN, Edition WILLIAM T SEGUI www.elsolucionario.org ISBN-13: 978-1-111-57601-1 ISBN-10: 1-111-57601-7 © 2013, 2007 Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below For product information and technology assistance, contact us at Cengage Learning Academic Resource Center, 1-800-423-0563 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.com Cengage Learning 200 First Stamford Place, Suite 400 Stamford, CT 06902 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd For your course and learning solutions, visit www.cengage.com/engineering Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review 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be made available in the classroom and collected at the end of each class session, or Printed in the United States of America 15 14 13 12 11 posted electronically as described herein Any material posted electronically must be through a password-protected site, with all copy and download functionality disabled, and accessible solely by your students who have purchased the associated textbook for the Course You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you not accept these conditions, you must return the Supplement unused within 30 days of receipt All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool www.elsolucionario.org INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY STEEL DESIGN FIFTH EDITION William T Segui www.elsolucionario.org Contents Preface vi Chapter Introduction 1-1 Chapter Concepts in Structural Steel Design 2-1 Chapter Tension Members 3-1 Chapter Compression Members 4-1 Chapter Beams 5-1 Chapter Beam-Columns 6-1 Chapter Simple Connections 7-1 Chapter Eccentric Connections 8-1 Chapter Composite Construction 9-1 Chapter 10 Plate Girders 10-1 PREFACE This instructor's manual contains solutions to the problems in Chapters 1–10 of Steel Design, 5th Edition Solutions are given for all problems in the Answers to Selected Problems section of the textbook, as well as most of the others In general, intermediate results to be used in subsequent calculations were recorded to four significant figures, and final results were rounded to three significant figures Students following these guidelines should be able to reproduce the numerical results given However, the precision of the results could depend on the grouping of the computations and on whether intermediate values are retained in the calculator between steps In many cases, there will be more than one acceptable solution to a design problem; therefore, the solutions given for design problems should be used only as a guide in grading homework I would appreciate learning of any errors in the textbook or solutions manual that you may discover You can contact me at wsegui@memphis.edu A list of errors and corrections in the textbook will be maintained at http://www.ce.memphis.edu/segui/errata.html William T Segui August 15, 2011 vi www.elsolucionario.org CHAPTER - INTRODUCTION 1.5-1 (a) P 2067 1340 lb f P 1340 68 02 psi 19 A (b) Since E f , 68 02 f 35 10 −6 E 29, 000, 000 f 68 psi 35 10 −6 1.5-2 (a) L 9/ sin 45 ° 12 73 ft ΔL L 10 −4 12 73 12 136 in (b) ΔL 136 in f E 10 −4 29, 000 25 81 ksi P fA 25 811 31 33 kips P 33 kips 1.5-3 (a) (b) 0 5 A d 196 in 4 f P 5000 25, 470 psi 1963 A −3 ΔL 792 10 49 10 −4 L 25, 470 E f 10 psi 49 10 −4 F u P u 14, 700 74, 900 psi 1963 A E 30, 000 ksi F u 74 ksi [1-1] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1.5-4 Spreadsheet results: (a) (b) Load Stress (lb) (psi) microstrain 2,000 10,186 2,500 3,000 3,500 4,000 4,500 5,000 12,732 15,279 17,825 20,372 22,918 25,465 47 220 500 950 1,111 1,200 1,702 30,000 Stress (psi) 25,000 20,000 15,000 10,000 5,000 0.000000 0.000500 0.001000 0.001500 0.002000 Strain (c) Slope 9,210,000 psi modulus of elasticity [1-2] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org 10.7-5 From Problem 10.8-4, use a 16 45 web and 12 12 flanges Reaction V L 168 kips Design the bearing stiffeners at the supports and use the same design for the interior stiffeners 12 − 3/16 906 in Maximum stiffener width b f − t w 2 Try b in For b t st ≤ 56 Try two plates, E , t≥ F yst b 56 E F yst 56 29000 50 371 in in in., with 1-in cutouts Bearing strength: A pb 5 − 13/8 in R n 8F y A pb 8503 270 kips R n 75270 203 kips V L 168 kips (OK) Compressive strength: The maximum permissible length of web is 12t w 123/16 25 in Compute the radius of gyration about an axis along the middle of the web: I 2 253/16 3/85 53/82 3/32 12 12 33 04 in A 253/16 253/8 172 in r I A 33 04 814 in 172 Compute the compressive strength: KL Kh 7545 11 99 25 r r 814 [10-38] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part F cr F y 50 ksi c P n 90A g F cr 904 17250 188 kips 168 kips Use PL (OK) 5, with 1-in cutouts 10.7-6 (a) w u 2w D 6w L 21 0 62 4 kips/ft P u 6P L 6500 800 kips 480 800 Left reaction V L w u L P u 576 kips 2 I x t w h 2A f h t f 12 2 0 578 23 22 78 12 2 363 10 in The shear flow is Q Af h tf 2 At the support, 322 78 2 2673 in VuQ 5762673 516 kips/in Ix 236, 300 Minimum weld size 16 Minimum length 16 in (AISC Table J2.4) 75 in 1.5 in., use 1.5 in Use E70 electrodes, R n 392D kips/in., where D weld size in sixteenths Try 16 -in 12 -in intermittent fillet welds For two welds, Weld strength 3923 352 kips/in The base metal shear yield strength (web controls) is 6F y t 650 15 kips/in [10-39] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org Shear rupture strength is 45F u t 4565 14 63 kips/in Weld strength controls For a 1.5-in length, R n 58 352 12 53 kips Required spacing: R n VuQ s Ix 12 53 516 s s 92 in This spacing is too small for intermittant welds Use a continuous weld Maximum clear spacing: From AISC E6, d ≤ 75 E t 75 29, 000 3 54 in (or 12 in.; 12 in.; controls.) Fy f 50 Maximum s 12 1.5 in 13.5 in For s 13.5 in., R n VuQ s Ix 12 53 V u 2673 13 236, 300 V u 81 95 kips Shear at mid-span, left of load, 576 − 440 400 kips, so maximum spacing will never be used Spacing required at mid-span R n I x 12 53236300 77 in VuQ 4002673 This spacing is too small for intermittant welds Use continuous 16 -in E70 fillet welds (b) w a w D w L kips/ft P a 500 kips 380 500 Left reaction V L w a L P a 370 kips 2 I x t w h 2A f h t f 12 0 578 23 22 78 12 363 10 in [10-40] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Shear flow: Q Af h tf 2 Minimum weld size 16 Minimum length 16 16 2673 in VaQ 3702673 185 kips/in Ix 236300 At the support, Try 322 78 2 in (AISC Table J2.4) 75 in 1.5 in., use 1.5 in -in 12 -in intermittent fillet welds For two welds and E70 electrodes, weld strength 92793 567 kips/in Base metal shear yield strength (web plate controls) is 4F y t 450 10 kips/in Shear rupture strength is 3F u t 365 75 kips/in Weld strength controls For a 1.5-in length, R n 55 567 351 kips Required spacing: R n / V Q a s Ix 351 185 s s 00 in This spacing is too small for intermittant welds Use a continuous weld Maximum clear spacing: From AISC E6, d ≤ 75 E t 75 29, 000 3 54 in (or 12 in.; 12 in.; controls.) Fy f 50 Maximum s 12 1.5 in 13.5 in For s 13.5 in., R n / V Q a s Ix 351 V a 2673 13 236300 V a 54 69 kips Shear at mid-span, left of load, 370 − 340 250 kips, so maximum spacing will [10-41] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org never be used Spacing required at mid-span R n /I x 351236300 95 in VaQ 2502673 This spacing is too small for intermittant welds Use continuous 16 -in E70 fillet welds 10.7-7 (a) Assume a girder weight of 400 lb/ft w u 2w D 6w L 20 400 64 88 kips/ft P u 6P L 6175 280 kips 8870 Vu VL wuL Pu 280 520 kips 2 M u 520 835 − 8835 /2 − 28070/6 10, 750 ft-kips Try t f in For a slender web, ∴ tw ≤ h 712 − 21 5 81 in h ≥ 70 E 70 29, 000 137 tw Fy 50 h 81 589 in 137 137 For a ≤ 5, h ≤ 12 E 12 29000 289 tw Fy 50 h 429000 For a 5, h ≤ 4E 232 tw Fy 50 h tw ≥ Try a h 81 349 in 232 232 in 81 in web h 81 162, tw A w 581 40 in Estimate required flange size [10-42] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Af M u − A w 1075012 − 40 28 64 in 6 9hF y 98150 b f ≥ 28 64 19 09 in Try a 12 -in 24-in flange, A f 524 36 in Girder weight 40 236 490 144 (OK) I x t w h 2A f h t f 12 382 kips/ft 0.400 kips/ft estimate 0 581 236 81 12 447 10 in Ix 144700 S x Icx 3445 in h/2 t f 81/2 5 Compression flange: Check FLB: b f 24 0, 2t f 21 5 p 38 E 38 29, 000 152 Fy 50 F cr F y 50 ksi Since p , Check lateral-torsional buckling h 81 13 in., 6 I 13 50 5 1 524 1728in 12 12 A 13 50 5 241 5 42 75 in , rt I A 1728 358 in 42 75 L b 23 33 ft L p 1r t L r r t E 16 358 Fy E 6 358 7F y 29000 168 in 14.03 ft 50 29000 575 in 47.92 ft 750 Since L p L b L r , [10-43] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org F cr C b F y − 3F y Lb − Lp Lr − Lp ≤ Fy 50 − 350 23 33 − 14 03 47 92 − 14 03 45 88 ksi (controls) (C b is a slightly conservative estimate.) Compute the plate girder strength reduction factor R pg − aw 1200 300a w h c − E tw Fy ≤ 811/2 aw hctw 125 10 b fc t fc 241 5 R pg − 125 162 − 29000 50 1200 3001 125 981 M n R pg F cr S xc 981945 883445 552 10 in.-kips b M n 901 552 10 /12 11, 640 ft-kips 10,750 ft-kips Try a (OK) 81 web and 12 24 flanges Shear: At left end (end panel), Required vVn 520 12 86 ksi, 40 Aw h 162 tw From Table 3-17a in the Manual, a 0.72 by interpolation h a 72h 7281 58 32 Use a 58 in At 58 in from left end, V u 520 − 88 58 12 487 kips vVn 487 12 04 ksi 40 Aw From Table 3-17b, a 2.6 h a 681 210 Use a 210 210 in At 70/3 23 33 ft from the left end (to the right of the concentrated load), [10-44] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part V u 520 − 8823 33 − 280 80 29 kips vVn 80 29 982 ksi 40 Aw Required V For a 23 33 12 46 3, v n ksi, ∴ stiffeners not needed in middle 81 Aw h 1/3 Use intermediate stiffeners spaced from each end as follows: at 58 in and at 111 in Use a 81 web and 12 24 flanges (b) Intermediate stiffener size: b t st E Fy ≤ 56 Available width: b f − t w 24 − 11 75 in Try b in 2 t≥ b 56 29000 50 56 E Fy I st ≥ I st1 I st2 − I st1 I st1 j a 444 V r − V c1 V c2 − V c1 t 3w j h − − 876 5, ∴ use j 876 58/81 a/h I st1 I st2 h 1.3 st 40 58 81 F yw E 1/2 2 876 20 85 in 1.5 , st max F yw /F yst 1 [10-45] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org I st2 81 1 1.3 40 50 29000 1.5 77 04 in For h/t w 162 and a/h 111/81 37, vVn ksi for the no tension field case and Table 3-17a Aw V c1 v V n 7A w 781 5 271 kips vVn 17 ksi for the tension field case and Table 3-17b Aw V c2 v V n 17A w 1781 5 688 kips V r 520 kips I st ≥ I st1 I st2 − I st1 V r − V c1 V c2 − V c1 20 85 77 04 − 20 85 520 − 271 688 − 271 Try two plates 54 in 6 I st ≈ 3/86 6 61 in 54 in 12 (OK) Length: From Figure 10.9 in the textbook, c ≥ 4t w 40 5 in., and c ≤ 6t w 60 5 in Assume a flange-to-web weld size of w 16 in (minimum size) and c in Length h − w − c 81 − − 77 81 in., say 78 in 16 c 81 − 78 − 813 in 16 OK) Use two PL 6´-06´´ for intermediate stiffeners Design the bearing stiffeners at the supports for a load of V L 520 kips Maximum stiffener width b f − t w 24 − 11 75 in 2 [10-46] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part b Try b in and t ≥ E F yst 56 Try two plates, 8 56 29000 50 593 in in in., with 1-in cutouts A pb 8 − 15/8 75 in Bearing strength: R n 8F y A pb 8508 75 787 kips R n 75787 5 590 kips V L 520 kips (OK) Compressive strength: The maximum permissible length of web is 12t w 120 5 in Compute the radius of gyration about an axis along the middle of the web: I 60 5 5/88 85/84 1/4 12 12 234 in A 60 5 285/8 13 in r 234 243 in 13 I A Compute the compressive strength: KL 7581 14 32 25 ∴ F cr F y 50 ksi r 243 c P n 90F cr A g 905013 585 kips 519.1 kips Use PL (OK) with 1-in cutouts for bearing stiffeners at the supports Because there is a large difference between the reactions and the interior concentrated loads, use a different size for the interior bearing stiffeners Design the interior bearing stiffeners for a load of P u 280 kips Try b in and t ≥ Try two plates, b 56 E F yst 56 29000 50 593 in in in., with 1-in cutouts [10-47] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org A pb 6 − 15/8 25 in Bearing strength: R n 8F y A pb 8506 25 562 kips R n 75562 5 422 kips P u 280 kips (OK) Compressive strength: The maximum permissible length of web is 25t w 250 5 12 in Compute the radius of gyration about an axis along the middle of the web: I 12 50 5 5/86 65/83 1/4 12 12 101 in A 12 50 5 265/8 13 75 in r I A 101 721 in 13 75 Compute the compressive strength: KL 7581 22 33 25 ∴ F cr F y 50 ksi r 721 c P n 90F cr A g 905013 75 618 kips 519.1 kips Use PL (OK) with 1-in cutouts for the interior bearing stiffeners (c) Design the flange-to-web welds Q Af h tf 2 The shear flow is At the support, 1 24 81 2 1485 in VuQ 520 81485 345 kips/in Ix 144700 Minimum weld size 16 Minimum length 16 in (AISC Table J2.4) 75 in 1.5 in., use 1.5 in Use E70 electrodes, R n 392D kips/in., where D weld size in sixteenths Try 16 -in 12 -in intermittent fillet welds For two welds, Weld strength 3923 352 kips/in [10-48] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Base metal shear yield strength (web plate controls) is 4F y t 450 10 kips/in Shear rupture strength is 3F u t 365 75 kips/in Weld strength controls For a 1.5-in length, R n 58 352 12 53 kips Required spacing: R n VuQ s Ix 12 53 345 s s 34 in Since this is less than twice the length of the weld, use a continuous weld For s 21 5 in., V u R n I x 12 53144700 407 kips sQ 31485 This occurs when 520 − 88x 407, Solution is: x 16 54 ft Maximum clear spacing: From AISC E6, d ≤ 75 E t 75 29, 000 1 5 27 in (or 12 in.; 12 in.; controls.) Fy f 50 Maximum s 12 1.5 in 13.5 in For s 13.5 in., R n VuQ s Ix 12 53 V u 1485 13 144700 V u 90 kips Shear at first interior load, left of load, 520 − 8823 33 360 kips, so maximum spacing will not be used in the first third of the span Spacing required at left side of first interior load is R n I x 12 53144700 389 in VuQ 360 31485 Check middle third of span Shear on right side of load 360 − 280 80 kips s R n I x 12 53144700 15 in 13.5 in maximum ∴ use s 13 12 in VuQ 80 31485 [10-49] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org Summary for flange-to-web welds: Use 16 in continuous fillet welds for the first 17 feet Use 163 in 12 in intermittent E70 fillet welds at in c.c from 17 ft until the first interior bearing stiffener Use 163 in 12 in intermittent E70 fillet welds at 13 12 in c.c between interior bearing stiffeners Welds for intermediate stiffeners ( 12 ): Minimum weld size 16 Minimum length 16 in (AISC Table J2.4) 75 in 1.5 in., use 1.5 in Use E70 electrodes, R n 392D kips/in., where D weld size in sixteenths Try 16 in 12 in intermittent fillet welds For four welds, the weld strength is 3923 16 kips/in The base metal shear yield strength is 6F y t 650 30 kips/in Shear rupture strength is 45F u t 4565 29 25 kips/in Weld strength controls For a 1.5-in length, R n 516 7 25 05 kips From Equation 10.4, the shear to be transferred is f 045h F 3y 04581 E 25 05 568 kips/in s 50 568 kips/in 29, 000 s 31 in A center-to-center spacing of in is equal to twice the length of the weld segment, so either a continuous weld or an intermittent weld can be used Use intermittent welds Maximum clear spacing: From AISC E6, [10-50] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part d ≤ 75 E t 75 29, 000 1 5 27 in (or 12 in.; 12 in.; controls.) Fy f 50 Maximum s 12 1.5 in 13.5 in Use 16 in 12 in E70 fillet welds spaced at in c-c for the intermediate stiffeners Welds for bearing stiffeners at the supports (5/8 8): Minimum weld size 16 Minimum length 16 in (AISC Table J2.4, based on web thickness of 1/2 in.) 75 in 1.5 in., use 1.5 in Use E70 electrodes, R n 392D kips/in., where D weld size in sixteenths Try 16 in 12 in intermittent fillet welds For four welds, the weld strength is 3923 16 kips/in The base metal shear yield strength (web controls) is 6F y t 650 30 kips/in Shear rupture strength is 45F u t 4565 29 25 kips/in Weld strength controls For a 1.5-in length, R n 516 7 25 05 kips The shear to be transferred is Reaction length available for weld 25 05 592 kips/in s Use 16 Minimum weld size s 80 in in 12 in E70 fillet welds spaced at 12 in c-c for bearing stiffener at support Welds for interior bearing stiffeners ( Try 520 592 kips/in 81 − 21 0 16 6): in and minimum length 1.5 in in 12 in intermittent fillet welds For four welds, the weld strength is [10-51] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part www.elsolucionario.org 3923 16 kips/in The base metal shear yield strength (web controls) is 6F y t 650 30 kips/in Shear rupture strength is 45F u t 4565 29 25 kips/in Weld strength controls For a 1.5-in length, R n 516 7 25 05 kips The shear to be transferred is P u length available for weld 25 05 544 kips/in s Use 16 280 544 kips/in 81 − 21 0 s 07 in in 12 in E70 fillet welds spaced at in c-c for interior bearing stiffeners 10.7-8 See solution to problem 10.7-7 for an example of the procedure 10.7-9 See solution to problem 10.7-7 for an example of the procedure [10-52] © 2013 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... SOLUTIONS MANUAL TO ACCOMPANY STEEL DESIGN FIFTH EDITION William T Segui www.elsolucionario.org Contents Preface vi Chapter Introduction 1-1 Chapter Concepts in Structural Steel Design 2-1 Chapter Tension... This instructor's manual contains solutions to the problems in Chapters 1–10 of Steel Design, 5th Edition Solutions are given for all problems in the Answers to Selected Problems section of the... assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool www.elsolucionario.org INSTRUCTOR'S SOLUTIONS
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