________________________________________________________________________
PROBLEM (*7.40) Based upon a factor of safety of ns = 1.5, determine the minimum
permissible width b of the beam shown in Fig. P7.38
Given: P = 90 kips, all
σ
= 18 ksi, r = 5/16 in.
*SOLUTION
At a section through B
.12(90) 1080
Blb in== ⋅, 33
1( )(1.5) 281.25 10 .
12 4
I
bb
−
==×
M
in
18 1.5 12
all ksi
σ
==
0.9375 2.25
0.625 1.5
1.5 1.5
rD
dd
== ==
, 1.15K
=
(Fig. 7.13)
3
3
1080(0.75) 2.881 10
281.25 10
nom bb
σ
−
×
==
×
Thus,
33
12 10 1.15(2.881 10 )
all b
σ
=× = ×
yields
0.276 .bi=n
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