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Unformatted text preview: Lecture 23  Slender Columns Lecture 23  Slender Columns and Twoway Slabs and Twoway Slabs August 1, 2003 CVEN 444 Lecture Goals Lecture Goals Slender Column Design Oneway and twoway slab Slab thickness, h Design of Long Columns Example Design of Long Columns Example A rectangular braced column of a multistory frame building has floor height l u =25 ft. It is subjected to service deadload moments M 2 = 3500 kin. on top and M 1 =2500 kin. at the bottom. The service live load moments are 80% of the deadload moments. The column carries a service axial deadload P D = 200 k and a service axial liveload P L = 350 k. Design the cross section size and reinforcement for this column. Given Ψ A = 1.3 and Ψ B = 0.9. Use a d’=2.5 in. cover with an sustain load = 50 % and f c = 7 ksi and f y = 60 ksi. Design of Long Columns Example Design of Long Columns Example Compute the factored loads and moments are 80% of the dead loads ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 u D L 1u D L 2u D L 1.2 1.6 1.2 200 k 1.6 350 k 800 k 1.2 1.6 1.2 2500 kin 1.6 0.8 2500 kin 6200 kin. 1.2 1.6 1.2 3500 kin 1.6 0.8 3500 kin 8680 kin. P P P M M M M M M = + = + = = + = + = = + = + = Design of Long Columns Example Design of Long Columns Example Compute the k value for the braced compression members Therefore, use k = 0.81 ( 29 ( 29 ( 29 A B min 0.7 0.05 0.7 0.05 1.3 0.9 0.81 1.0 0.85 0.05 0.85 0.05 0.9 0.895 1.0 k k = + Ψ + Ψ = + + = ≤ = + Ψ = + = ≤ Design of Long Columns Example Design of Long Columns Example Check to see if slenderness is going to matter. An initial estimate of the size of the column will be an inch for every foot of height. So h = 25 in. ( 29 ( 29 ( 29 n 0.81 25 ft 12 in./ft 32.4 r 0.3 25 in. 6200 kin. 32.4 34 12 25.43 8680 kin. kl = = ≥= We need to be concerned with slender columns Design of Long Columns Example Design of Long Columns Example So slenderness must be considered. Since frame has no side sway, M 2 = M 2ns , δ s = 0 Calculate the minimum M 2 for the ratio computations. ( 29 ( 29 ( 29 2,min u 2 0.6 0.03 800 k 0.6 0.03 25 in. 1080 kin. 8680 kin. M P h M = + = + = ⇒ = Design of Long Columns Example Design of Long Columns Example Compute components of concrete The moment of inertia of the column is ( 29 1.5 1.5 c c 6 3 33 33 150 7000 5.07x10 psi 5.07x10 ksi E w f = = = → ( 29 ( 29 3 3 g 4 25 in. 25 in. 12 12 32552 in bh I = = = Design of Long Columns Example Design of Long Columns Example Compute the stiffness, EI ( 29 ( 29 3 4 c g d 7 2 0.4 5.07x10 ksi 32552 in 0.4 1 1 0.5 4.4x10 kin E I EI β = = + + = Design of Long Columns Example Design of Long Columns Example The critical load (buckling), P cr , is ( 29 ( 29 ( 29 2 7 2 2 cr 2 2 u 4.4x10 kin 12 in....
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This note was uploaded on 10/03/2011 for the course CVEN 444 taught by Professor Staff during the Summer '08 term at Texas A&M.
 Summer '08
 Staff

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