Given,
A36 steel
W14*211 Column
d = 15.72 in
bf = 15.8 in
Pu = Pr = 1800 kips
B = 24 in
N = 30 in
l = max {m, n, }
m = (N-0.95 d)/2 = (30 - 0.95 * 15.72)/2 = 7.533 in
n = (B-0.8 bf)/2 = (24 - 0.8 * 15.8)/2 = 5.68 in
=
= 3.94 in
l = max {7.533, 5.68, 3.94} = 7.533 in
= 3.1164 in
Determine the required thickness (inches) for a base plate of A36 steel support a W14x211 column...
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From the text "Applied Structural Steel Design"
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i need clear and right solution pleaee.
pleaes
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