An Instructor's Solutions Manual to Accompany STEEL DESIGN, 5 th Edition PDF

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An Instructor’s Solutions Manual to Accompany th STEEL DESIGN, 5 Edition WILLIAM T. SEGUI       © 2013, 2007 Cengage Learning ISBN-13: 978-1-111-57601-1 ISBN-10: 1-111-57601-7 ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or Cengage...

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An Instructor’s Solutions Manual to Accompany th

STEEL DESIGN, 5 Edition WILLIAM T. SEGUI  

 

 

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.

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INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY

STEEL DESIGN FIFTH EDITION

William T. Segui

Contents Preface

vi

Chapter 1

Introduction

1-1

Chapter 2

Concepts in Structural Steel Design

2-1

Chapter 3

Tension Members

3-1

Chapter 4

Compression Members

4-1

Chapter 5

Beams

5-1

Chapter 6

Beam-Columns

6-1

Chapter 7

Simple Connections

7-1

Chapter 8

Eccentric Connections

8-1

Chapter 9

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 [email protected]. 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 

CHAPTER 1 - INTRODUCTION

1.5-1 (a)

P  2067  1340 lb

f  P  1340  68. 02 psi 19. 7 A (b) Since E  f , 68. 02  f   2. 35  10 −6 E 29, 000, 000

f  68. 0 psi

  2. 35  10 −6

1.5-2 (a)

L  9/ sin 45 °  12. 73 ft ΔL  L  8. 9  10 −4  12. 73  12  0. 136 in.

(b)

ΔL  0. 136 in.

f  E  8. 9  10 −4  29, 000  25. 81 ksi P  fA  25. 811. 31  33. 8 kips

P  33. 8 kips

1.5-3 (a)

(b)

2 0. 5 2 A  d   0. 196 3 in. 2 4 4 f  P  5000  25, 470 psi 0. 1963 A −3   ΔL  6. 792  10  8. 49  10 −4 L 8 25, 470 E  f   3. 0  10 7 psi 8. 49  10 −4 F u  P u  14, 700  74, 900 psi 0. 1963 A

E  30, 000 ksi F u  74. 9 ksi

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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 0.000000

0.000500

0.001000

0.001500

0.002000

Strain

(c)

Slope  9,210,000 psi  modulus of elasticity

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1.5-5 (Note: These results are very approximate and depend on how the curves are drawn.) (a) 80

Stress (ksi)

70 60 50 40 30 20 10 0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.008

0.01

0.012

Strain

60

Stress (ksi)

50 40 30 20 10 0 0

0.002

0.004

0.006 Strain

(b)

F prop ≈ 47 ksi

(c)

E ≈ 40/0. 004  10, 000 ksi

(d)

F y ≈ 52 ksi

(e)

F u ≈ 70 ksi

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(f)

2 0. 5 2 A  d   0. 196 3 in. 2 4 4 10 f P   50. 94 ksi 0. 1963 A

 r ≈ 0. 0015,

 r   r L  0. 00158  0. 012 in.

 r  0. 012 in.

1.5-6 Spreadsheet results: (a) Load

Elongation

Stress

(kips)

(in.)

(ksi)

0 1.0 2.0 2.5 3.5 5.0 6.0 7.0 8.0 9.0 10.0 11.5 12.0

0 0.0010 0.0014 0.0020 0.0024 0.0036 0.0044 0.0050 0.0060 0.0070 0.0080 0.0120 0.0180

0 5.094 10.19 12.74 17.83 25.47 30.57 35.66 40.75 45.85 50.94 58.58 61.13

0.002

0.004

Strain 0 0.0005 0.0007 0.0010 0.0012 0.0018 0.0022 0.0025 0.0030 0.0035 0.0040 0.0060 0.0090

(b) 70

Stress (ksi)

60 50 40 30 20 10 0 0

0.006

0.008

0.01

Strain

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E≈

(c)

38  13, 600 ksi 0. 0028

E ≈ 13, 600 ksi

(d)

F pl ≈ 38 ksi

(e)

F y ≈ 60 ksi

1.5-7 Spreadsheet results: (a)

Load Elongation x 103 (kip s) ( in.) 0 0 0.5 0.16 1 0.352 1.5 0.706 2 1.012 2.5 1.434 3 1.712 3.5 1.986 4 2.286 4.5 2.612 5 2.938 5.5 3.274 6 3.632 6.5 3.976 7 4.386 7.5 4.64 8 4.988 8.5 5.432 9 5.862 9.5 6.362 10 7.304 10.5 8.072 11 9.044 11.5 11.31 12 14.12 12.5 20.044 13 29.106

S tress ( ksi) 0 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30 32.5 35 37.5 40 42.5 45 47.5 50 52.5 55 57.5 60 62.5 65

St rain x 10 3 (in./in .) 0 0.080 0.176 0.353 0.506 0.717 0.856 0.993 1.143 1.306 1.469 1.637 1.816 1.988 2.193 2.320 2.494 2.716 2.931 3.181 3.652 4.036 4.522 5.655 7.060 10.02 14.55

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(b) 70

S tress (ksi)

60 50 40 30 20 10 0 0

2

4

6

8

10

12

14

16

M icro stra in

(c)

Using the dashed line, E ≈

56  15, 600 ksi 5. 6 − 2  10 −3 E ≈ 16, 000 ksi

(d)

F pl ≈ 42 ksi

(e)

F y ≈ 58 ksi

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CHAPTER 2 - CONCEPTS IN STRUCTURAL STEEL DESIGN

2-1 D  30. 8 kips, L  1. 7 kips, L r  18. 7 kips, S  19. 7 kips Combination 1:

1. 4D  1. 430. 8  43. 12 kips

Combination 2:

1. 2D  1. 6L  0. 5S  1. 230. 8  1. 61. 7  0. 519. 7  49. 53 kips

Combination 3:

1. 2D  1. 6S  0. 5L  1. 230. 8  1. 619. 7  0. 51. 7  69. 33 kips P u  69. 3 kips

(a) Combination 3 controls. (b) Since P u ≤  c P n , (c) P n 

 c P n  69. 3 kips

cPn 69. 33  c  0. 90  77. 03 kips

P n  77. 0 kips

(d) Combination 3 controls. P a  D  L r or S or R  D  S  30. 8  19. 7  50. 5 kips (e) P a ≤ P n , P n  P a  1. 6750. 5  84. 34 kips 

P a  50. 5 kips P n  84. 3 kips

2-2 D  26 kips, L  15 kips, L r  5 kips, S  8 kips, R  5 kips, W  8 kips Combination 1:

1. 4D  1. 426  36. 4 kips

Combination 2:

1. 2D  1. 6L  0. 5S  1. 226  1. 615  0. 58  59. 2 kips

Combination 3:

1. 2D  1. 6S  0. 5L  1. 226  1. 68  0. 515  51. 5 kips

Combination4:

1. 2D  1. 0W  0. 5L  0. 5S  1. 226  1. 08  0. 515  0. 58  50. 7 kips P u  59. 2 kips

(a) Combination 2 controls. (b) Since P u ≤  c P n ,

 c P n  59. 2 kips [2-1]

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(c) P n 

cPn 59. 2  c  0. 90  65. 78

P n  65. 8 kips

(d) Combination 6a controls. P a  D  0. 750. 6W  0. 75L  0. 75S P a  46. 9 kips

 26  0. 750. 68  0. 7515  0. 758  46. 85 kips (e) P a ≤ P n , P n  P a  1. 6746. 85  78. 24 kips 

P n  78. 2 kips

2-3 D  0. 2 kips/ft, L r  0. 13 kips/ft, S  0. 14 kips/ft Combination 1:

1. 4D  1. 40. 2  0. 28 kips/ft

Combination 2:

1. 2D  1. 6L  0. 5S  1. 20. 2  1. 60  0. 50. 14  0. 31 kips/ft

Combination 3:

1. 2D  1. 6S  1. 20. 2  1. 60. 14  0. 464 kips/ft P u  0. 464 kips/ft

(a) Combination 3 controls.

(b) Combination 3 controls: P a  D  S  0. 2  0. 14  0. 34 kips/ft P a  0. 34 kips/ft

2-4 (a) LRFD Roof: D  30 psf, L r  20 psf, S  21 psf, R  4 62. 4  20. 8 psf 12 Combination 1:

1. 4D  1. 430  42. 0 psf

Combination 2:

1. 2D  1. 6L  0. 5S  1. 230  1. 60  0. 521  46. 5 psf

Combination 3:

1. 2D  1. 6S  1. 230  1. 621  69. 6 psf P u  69. 6 psf

Combination 3 controls.

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Floor: D  62 psf, L  80 psf Combination 1:

1. 4D  1. 462  86. 8 psf

Combination 2:

1. 2D  1. 6L  1. 262  1. 680  202 psf P u  202 psf

Combination 2 controls. (b) ASD Roof: Combination 3 controls: D  S  30  21  51. 0 psf

P a  51. 0 psf

Floor: Combination 2 controls: D  L  62  80  142. 0 psf

P a  142 psf

2-5 D  13. 3 kips, L  6. 9 kips, L r  1. 3 kips, S  1. 3 kips, W  150. 6 kips, E  161. 1 kips (a) LRFD Combination 1:

1. 4D  1. 413. 3  18. 62 kips

Combination 2:

1. 2D  1. 6L  0. 5L r  1. 213. 3  1. 66. 9  0. 51. 3  27. 65 kips

Combination 3:

1. 2D  1. 6S  0. 5W  1. 213. 3  1. 61. 3  0. 5150. 6  93. 34 kips

Combination 4:

1. 2D  1. 0W  0. 5L  0. 5L r  1. 213. 3  1. 0150. 6  0. 56. 9  0. 51. 3  170. 7 kips

Combination 5:

1. 2D  1. 0E  0. 5L  0. 2S  1. 213. 3  1. 0161. 1  0. 56. 9  0. 21. 3  180. 8 kips P u  181 kips

Combination 5 controls.

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(b) ASD Combination 5 controls: D  0. 6W or 0. 7E  13. 3  0. 7161. 1  126. 1 kips P a  126 kips

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CHAPTER 3 - TENSION MEMBERS

3.2-1 For yielding of the gross section, A g  73/8  2. 625 in. 2 ,

P n  F y A g  362. 625  94. 5 kips

For fracture of the net section, A e  3/8 7 − 1  1 8

 2. 203 in. 2

P n  F u A e  582. 203  127. 8 kips a) The design strength based on yielding is  t P n  0. 9094. 5  85. 05 kips The design strength based on fracture is  t P n  0. 75127. 8  95. 85 kips The design strength for LRFD is the smaller value:

 t P n  85. 1 kips

b) The allowable strength based on yielding is P n  94. 5  56. 59 kips t 1. 67 The allowable strength based on fracture is P n  127. 8  63. 9 kips t 2. 00 The allowable service load is the smaller value:

P n / t  56. 6 kips

Alternate solution using allowable stress: For yielding, F t  0. 6F y  0. 636  21. 6 ksi and the allowable load is F t A g  21. 62. 625  56. 7 kips

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For fracture, F t  0. 5F u  0. 558  29. 0 ksi and the allowable load is F t A e  29. 02. 203  63. 89  63. 89 kips The allowable service load is the smaller value  56. 7 kips

3.2-2 For A242 steel and t  ½ in., F y  50 ksi and F u  70 ksi. For yielding of the gross section, A g  81/2  4 in. 2 P n  F y A g  504  200 kips For fracture of the net section, A n  A g − A holes  4 − 1/2 1  1 8

 2 holes  2.875 in. 2

A e  A n  2. 875 in. 2 P n  F u A e  702. 875  201. 3 kips a) The design strength based on yielding is  t P n  0. 90200  180 kips The design strength based on fracture is  t P n  0. 75201. 3  151 kips The design strength for LRFD is the smaller value:

 t P n  151 kips

b) The allowable strength based on yielding is P n  200  120 kips t 1. 67 The allowable strength based on fracture is

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P n  201. 3  101 kips t 2. 00 P n / t  101 kips

The allowable service load is the smaller value: Alternate solution using allowable stress: For yielding, F t  0. 6F y  0. 650  30 ksi and the allowable load is F t A g  304  120 kips For fracture, F t  0. 5F u  0. 570  35 ksi and the allowable load is F t A e  352. 875  101 kips The allowable service load is the smaller value  101 kips

3.2-3 Gross section: Net section:

P n  F y A g  508. 81  440. 5 kips Hole diameter  1 

1 8

 1 18 in.

A n  A g − 2t f d h  8. 81 − 20. 5011. 125  7. 683 in. 2 A e  0. 9A n  0. 97. 683  6. 915 in. 2 P n  F u A e  656. 915  449. 5 kips a. Gross: Net:

 t P n  0. 90440. 5  396 kips  t P n  0. 75449. 5  337 kips  t P n  337 kips

Net section controls: b. Gross:

P n  440. 5  264 kips t 1. 67 [3-3]

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Net:

P n  449. 5  225 kips t 2. 00 P n / t  225 kips

Net section controls:

3.2-4 For yielding of the gross section, A g  63/8  2. 25 in. 2 P n  F y A g  362. 25  81. 0 kips For fracture of the net se...