Solution Manual for Steel Design 6th Edition Segui

CHAPTER 1 - INTRODUCTION


1.5-1
(a) A   (0.550)  0.2376 in.2
Pu 28,500
Fu    120,000 psi = 120 ksi
A 0.2376
Fu  120 ksi

2.300  2.030
(b) e  100  13.3%
2.030
e  13.3%
(c)
A f   (0.430)  01452
. in.2

A f  Ao 01452
.  0.2376
Change =  100   100  38.9%
Ao 0.2376
Reduction = 38.9%


1.5-2
A   (0.5)  01963
. in.2
P 135
.
f    68.77 ksi
A 01963
.
L 4.66  10 3
=   2.33  10 3
L 2
f 68.77
E   29,500 ksi
 2.33  10 3
E = 29,500 ksi


1.5-3
A   (0.510)  0.2043 in.2
P 250
For P  250 lb, f    1224 psi
A 0.2043




[1-1]
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or posted to a publicly accessible website, in whole or in part.

, Spreadsheet results:

(a) Load Stress Strain x 106
(lb) (psi) (in./in.)
0 0 0
250 1224 37.1
500 2447 70.3
1000 4895 129.1
1500 7342 230.1
2000 9790 259.4
2500 12237 372.4
3000 14684 457.7
3500 17132 586.5


(b)


20000


15000
Stress (psi)




10000


5000


0
0 100 200 300 400 500 600
Strain x 10^6 (in./in.)


(c) E = slope = 30,100 ksi


1.5-4
A   (0.5)  01963
. in.2

f P/ A P L
E     1392(
. )  28,400 ksi
 L / L L A
E = 28,400 ksi




[1-2]
© 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.