Solutions Manual for Steel Design 6th Edition Segui

SOLUTIONS MANUAL FOR STEEL
DESIGN 6TH EDITION SEGUI

,Steel Design 6th Edition Segui Solutions Manual




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 rupture of the net section,

A e  3/8 7 − 1  3  2. 180 in. 2
16
P n  F u A e  582. 108  122. 3 kips
a) The design strength based on yielding is
 t P n  0. 9094. 5  85. 05 kips
The design strength based on rupture is
 t P n  0. 75122. 3  91. 73 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 rupture is
P n  122. 3  61. 15 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
For rupture,
F t  0. 5F u  0. 558  29. 0 ksi
and the allowable load is F t A e  29. 02. 180  63. 22 kips
The allowable service load is the smaller value  56. 7 kips




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,3.2-2
For yielding of the gross section,
A g  63/8  2. 25 in. 2
P n  F y A g  502. 25  112. 5 kips
For rupture of the net section,
A e  A g  2. 25 in. 2
P n  F u A e  652. 25  146. 3 kips
a) The design strength based on yielding is
 t P n  0. 90112. 5  101 kips
The design strength based on rupture is
 t P n  0. 75146. 3  110 kips
The design strength for LRFD is the smaller value:  t P n  101 kips
b) The allowable strength based on yielding is
P n  112. 5  67. 4 kips
t 1. 67
The allowable strength based on rupture is
P n  146. 3  73. 2 kips
t 2. 00
The allowable service load is the smaller value: P n / t  67. 4 kips
Alternate solution using allowable stress: For yielding,
Ft 0. 6F y 0. 650  30. 0 ksi
and the allowable load is
FtAg 30. 02. 25  67. 5 kips
For rupture,
Ft 0. 5F u 0. 565  32. 5 ksi
and the allowable load is
F tAe 32. 52. 25  73. 1 kips
The allowable service load is the smaller value 67. 5
kips




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, 3.2-3
For yielding of the gross section,
P n  F y A g  503. 37  168. 5 kips
For rupture of the net section,

A n  A g − A holes  3. 37 − 0. 220 7  1  2 holes  2.930 in. 2
8 8
A e  0. 85A n  0. 852. 930  2. 491 in. 2
P n  F e A e  652. 491  161. 9 kips
a) The design strength based on yielding is
 t P n  0. 90168. 5  152 kips
The design strength based on rupture is
 t P n  0. 75161. 9  121. 4 kips
The design strength is the smaller value:  t P n  121. 4 kips
Let P u   t P n
1. 2D  1. 63D  121. 4, Solution is: D  20. 23
P  D  L  20. 23  320. 23  80. 9 kips P  80. 9 kips
b) The allowable strength based on yielding is
P n  168. 5  100. 9 kips
t 1. 67
The allowable strength based on rupture is
P n  161. 9  80. 95 kips
t 2. 00
The allowable load is the smaller value  80. 95 kips P  81. 0 kips
Alternate computation of allowable load using allowable stress: For yielding,
F t  0. 6F y  0. 650  30. 0 ksi
and the allowable load is
F t A g  30. 03. 37  101. 1 kips
For rupture,
F t  0. 5F u  0. 565  32. 5 ksi
and the allowable load is
F t A e  32. 52. 491  80. 96 kips


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© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated,
or posted to a publicly accessible website, in whole or in part.