CAE303_HW_6_Solution_F07

CAE303_HW_6_Solution_F07 - CAE303 FALL 2007 STRUCTURAL...

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CAE 303 HW6 Solution_ F07 Page 1 of 17 Solution Set # 6 Problem (1) Approach 1: without column tables in AISC 13 th Edition Manual 1. Check W12×72 and modify if necessary Required design load P r = 1.2D + 1.6L = (1.2×0.60 + 1.6×0.40) ×550 = 748 kips (LRFD) P r = D + L = (0.60 + 0.40) ×550 = 550 kips (ASD) Determine nominal compression strength P n W12×72, A g = 21.1 in 2 , r x = 5.31 in, r y = 3.04 in., F y = 50 ksi K x = 1.0, K y = 1.0; L x = 18 ft, L y = 18ft 1.0 18 12 ( ) 40.7 5.31 KL x r ×× == 1.0 18 12 ( ) 71.1 3.04 KL y r (Controls) 29000 ( ) 71.1 4.71 4.71 113 50 KL E rF y =< = = 2 2 () E F e KL r π = = 2 3.14 29000 2 71.1 × = 56.6 ksi (0.658 ) y e F F F F cr y = = 50 56.6 (0.658 ) 50 × = 34.5 ksi P n = F cr A g = 34.5×21.1 = 728 kips Check strength requirement We need to have: P c P r LRFD: P c = φ c P n = 0.90×728 = 655 kips < P r =748 kips. ASD: P c = P n / c Ω = 728/1.67 = 436 kips. < P r =550 kips For both LRFD and ASD, W12×72 is inadequate, we need to modify it. Try a heavier W12 shape, W12 × 87 Simple Connection Structural Analysis Model f g Deformed shape when column fgh buckles about both x- and y-axes 18’ 12’ h P 2 P 1 CAE303 – FALL 2007 STRUCTURAL DESIGN I: STEEL STRUCTURES Instructor: J. Shen, Ph.D., P.E., S.E. E-mail: Shen@iit.edu
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CAE 303 HW6 Solution_ F07 Page 2 of 17 2. Check W12 × 87 and modify if necessary Determine nominal compression strength P n W12×87, A g = 25.6 in 2 , r x = 5.38 in, r y = 3.07 in., F y = 50 ksi K x = 1.0, K y = 1.0; L x = 18 ft, L y = 18ft 1.0 18 12 () 4 0 . 1 5.38 KL x r ×× == 1.0 18 12 ( ) 70.4 3.07 KL y r (Controls) ( ) 70.4 4.71 113 KL E rF y =< = 2 2 E F e KL r π = = 2 3.14 29000 2 70.4 × = 57.7 ksi (0.658 ) y e F F F F cr y = = 50 57.7 (0.658 ) 50 × = 34.8 ksi P n = F cr A g = 34.8×25.6 = 891 kips Check strength requirement We need to have P c > P r LRFD: P c = φ c P n = 0.90×891 = 802 kips > P r =748 kips. ASD: P c = P n / c Ω = 891/1.67 = 534 kips < P r =550 kips W12×87 is adequate for LRFD, but, inadequate for ASD. We need to choose a heavier W12 shape for ASD, say, W12×96 3. Check W12 × 96 and modify if necessary Determine nominal compression strength P n W12×96, A g = 28.2 in 2 , r x = 5.44 in, r y = 3.09 in., F y = 50 ksi K x = 1.0, K y = 1.0; L x = 18 ft, L y = 18ft 1.0 18 12 ( ) 39.7 5.44 KL x r 1.0 18 12 ( ) 69.9 3.09 KL y r (Controls) ( ) 69.9 4.71 113 KL E y = 2 2 E F e KL r = = 2 3.14 29000 2 69.9 × = 58.5 ksi
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CAE 303 HW6 Solution_ F07 Page 3 of 17 (0.658 ) y e F F F F cr y = = 50 58.5 (0.658 ) 50 × = 35.0 ksi P n = F cr A g = 35.0×28.2 = 987 kips Check strength requirement ASD: P c = P n / c Ω = 987/1.67 = 591 kips > P r =550 kips W12×96 is adequate for ASD. 4. Summary: LRFD W12×87; ASD W12×96
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CAE 303 HW6 Solution_ F07 Page 4 of 17 Approach 2: with use of AISC 13 th Edition Manual Table (Only for F y = 50 ksi steel) 1. Check W12×72 and modify if necessary Required design load The same as Approach 1 Determine nominal compression strength P n (KL) y = 1.0×18 = 18 ft. (controls) () KL x r x r y = 1.0 18 1.75 × = 10.3 With Effective (KL) = the larger of (KL) y and KL x rr x y = 18 ft, enter Table 4-1 of AISC 13 th Edition Manual (F y = 50 ksi) for W 12 x 72: LRFD: P c = φ c P n = 657 kips ASD: P c = P n / c Ω = 437 kips. Check strength requirement LRFD: P c = c P n = 657 kips < P r =748 kips. ASD: P c = P n / c Ω = 437 kips. < P r =550 kips For both LRFD and ASD, W12×72 is inadequate, we need to modify it. Try W12 × 87 2 Check W12 × 87 and modify if necessary Determine nominal compression strength P n (KL) y = 1.0×18 = 18 ft. (controls) KL x r x r y = 1.0 18 1.75 × = 10.3 With Effective (KL) = the larger of (KL) y and KL x x y = 18 ft, enter Table 4-1 of AISC 13 th Edition Manual (F y = 50 ksi) for W 12 x 87: LRFD: P c = c P n = 801 kips ASD: P
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This homework help was uploaded on 04/07/2008 for the course CAE 303 taught by Professor Shen during the Fall '08 term at Illinois Tech.

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CAE303_HW_6_Solution_F07 - CAE303 FALL 2007 STRUCTURAL...

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