CAE 307 HW4 Solution

CAE 307 HW4 Solution - CAE 307 Homework 4 Solution Spring...

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Spring 2005 E c = 57000 4000 = 4 . 031 × 10 6 psi I g = 1 12 (6)(12) 3 = 864 in 4 n = E s E c = 29000 4031 = 7 . 19 use 7 Calculate I cr : Q A = Q B 6 x 2 2 = nA s (10 . 6875 - x ) 3 x 2 = (7)(0 . 62)(10 . 6875 - x ) x = 3 . 275 in I cr = 1 3 (6)(3 . 275) 3 + 7(0 . 62)(10 . 6875 - 3 . 275) 2 = 308 . 7 in 4 Beam Dead Load (self weight): w DL = 0 . 150(6 / 12)(12 / 12) = 0 . 075 kips/ft Initial Dead Load deFection: DL ) i = 5 w DL L 4 384 E c I g = 5(0 . 075)(9) 4 (12) 3 384(4031)(864) = 0 . 0032 in LOAD Ma Ie LL + DL ) i LL ) i (kips) (kips-ft) (in 4 ) (in) (in) 0 0.759 864 0.00318 0 0.378 1.89 864 0.00803 0.00485 0.756 3.03 446.0 0.0250 0.0218 1.134 4.16 361.6 0.0424 0.0392 1.512 5.30 334.4 0.0584 0.0552 1.89 6.43 323.0 0.0734 0.0702 2.835 9.26 313.5 0.109 0.106 3.78 12.1 310.9 0.144 0.141 4.725 14.9 309.8 0.178 0.175 5.67 17.8 309.4 0.212 0.209 Recalculate immediate dead load deFection due to decrease in e±ective sti±ness DL ) i = 5 w DL L
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CAE 307 HW4 Solution - CAE 307 Homework 4 Solution Spring...

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