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1264_Mechanics Homework Mechanics of Materials Solution

# 1264_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: '10.86 A column with a 19.5an eﬁ'eaive length supports a centric load, with ratio of PROBLEM 10-“ dead to live load equal to 1.35. The dead load ﬁwtor' IS 79— — 1. 2, the live load factor «(L— = l .6 and the resistance factor ¢= 0.85. Use load and resistance factor design to determine the allowable centric dead and live loads If the column 13 made of the following rolled-steel shape: (0) W10 x 39, (1:) W 14 X 68. Use E= 29 x 10‘ psi and a, = 50 ksi. SOLUTION L4: : no.5 Pr : 23-9 8.. (on WIOxsq ll. 5 in“ if] = 1.93 in Lc/arJ = ll8.l8 Le/P/%__1_T_l__8fl18. go = \_5Q20 > LS 29000 0.37? ‘ . )o.377')(.5‘o) _ r. PU = A( )5", - w—H-S‘Gzo)‘ _ 206.6? k? . O -'-' Pu ‘ Papal-Kit. 90 PD 1. 73.7ki'lgs ‘ TE (mums PM + 1.6 PL =(o.ssi(20e,eﬂ Pf = 5.4 e laps —- '1' 4 ‘- (b) W was A = 20.0wz r3=2.46 :u Len} = 45.0. \E A; = 3151/2?» = Q’s-‘2 5’0 = n.2572 < 1.5 i' Ti E Tr 27000 EC‘ 5: = Lssos 1 A“ Lseor. __ _. . Pu = A (0.653) 6,, r (20.0)(o.658) (503 — we kups ‘E Y: P T Pu ~ 'G’Pp + L 90 Pp 133,“? kIP\$ ‘— E (L2 wuss P53 + L; a =(o.85‘)(5m) 4 ...
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