Chapter 2 Specifications, Loads, and Methods of Design Problem 2-1 1.4 D 1.4(40) = 1. =
___________________________________________________________ 56 psf
2. 1.2 D + 1.6 L + (0.5 Lr or 0.3S or 0.5 R) ________________________________________________ 58 psf a. 1.2(40) + 1.6(0) + 0.5(20) = ________________________________________________ 57 psf b. 1.2(40) + 1.6(0) + 0.3(30) = ________________________________________________ 55 psf c. 1.2(40) + 1.6(0) + 0.5(14) = 3. 1.2D + (1.6Lr or 1.0S or 1.6R) + (L * or 0.5W ) a. 1.2(40) +1.6(20) + 0 =____________________________________________________ 80 psf
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b. 1.2(40) +1.0(30) + 0 =____________________________________________________ 78 psf c. 1.2(40) +1.6(14) + 0 =____________________________________________________ 70.4 psf 4. 1.2D +1.0(W + WT ) + L * + (0.5Lr or 0.3S or 0.5R) a. 1.2(40) +1.0(0) + 0 + 0.5(20) =_____________________________________________ 58 psf ______________________________________________ 57 psf b. 1.2(40) +1.0(0) + 0 + 0.3(30) = c. 1.2(40) +1.0(0) + 0 + 0.5(14) =______________________________________________ 55 psf 5. 0.9D +1.0 (W or WT ) 0.9(40) +1.0(0) = __________________________________________________________ 36 psf
(No Uplift) 6. 1.2D + E + L * + 0.15S 1.2(40) + 0 + 0 + 0.15(30) =___________________________________________________ 52.5 psf
7. 0.9D +1.0 E 0.9(40) +1.0(0) = __________________________________________________________ 36 psf (No Uplift) Governing factored load = 80 psf (Case 3a)
Problem 2-2 = = 1. 1.4D 1.4(15,000)
________________________________________________________ 21,000 lb
2. 1.2 D + 1.6 L + (0.5 Lr or 0.3S or 0.5 R) 1.2(15,000) + 1.6(0) + 0.3(18,000) = _____________________________________________ 23,400 lb
3. 1.2D + (1.6Lr or 1.0S or 1.6R) + (L * or 0.5W ) 1.2(15,000) + 1.0(18,000) + 0.5(36,000) = ________________________________________ 54,000 lb
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