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 Calculation:
Given:
(1) Mx = 115 k-ft
(2) My = 0
(3) P = 53.5 k
(4) V = 15.1 k
(5) Span length, L = 17'-0"
(6) Unbraced lengths, *lx = 17'-0" and *ly = 17'-0"
(7) ASTM A36 steel, Fy = 36 ksi, and c = 1.1
Fdy = cFy = 1.1(36) = 39.6 ksi
Solution:
(1) S = Mx/Fdy = 115(12)/39.6 = 34.8 in3
Try W12 x 35 (S = 45.6 in3)
d/tw = 41.7
bf/2tf = 6.31 < 8.5
O.K.
(2) Py = AFy = 10.3(36) = 371 k
P/Py
= 53.5/371 = 0.14 < 0.27
d/tw
= (412/F1/2y)[1 - 1.4(P/Py)]
= (412/361/2)[1 - 1.4(0.14)]
= 55.2 > 41.7 O.K.
(3) Vp
= FdvAw
Fdv = 0.55Fdy = 0.55(39.6) = 21.8 ksi
Aw = tw(d - 2tf) = 0.305(12.50 - 2(0.520)]
= 3.50 in2
Vp = 21.8(3.50) = 76.3 k > 15.1 k
O.K.
(4) rx
= 5.25 in.
ry = 1.54 in.
Z = 51.2 in3
(5) Mp
= FdyZ
= 39.6(51.2 x 1/12) = 169.0 ft-k
K = 0.75 (Section 1.8, Commentary on AISC Specification)
2.08-183
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