lecture23

# lecture23 - andTwowaySlabs August1,2003 CVEN444...

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Lecture 23 -  Slender Columns  Lecture 23 -  Slender Columns  and Two-way Slabs and Two-way Slabs August 1, 2003 CVEN 444

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Lecture Goals Lecture Goals Slender Column Design One-way and two-way 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 dead-load moments M 2 = 3500 k-in. on top and M 1 =2500 k-in. at the bottom. The service live load moments are 80% of the dead-load moments. The column carries a service axial dead-load P D = 200 k and a service axial live-load 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.

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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 k-in 1.6 0.8 2500 k-in 6200 k-in. 1.2 1.6 1.2 3500 k-in 1.6 0.8 3500 k-in 8680 k-in. 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 = + =

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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 k-in. 32.4 34 12 25.43 8680 k-in. 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 k-in. 8680 k-in. M P h M = + = + = =

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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 k-in E I EI β = = + + =

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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 k-in 12 in. 0.81 25 ft ft 7354.3 k EI P kl π π = = =
Design of Long Columns- Example Design of Long Columns- Example Compute the coefficient, C m , for the magnification δ coefficient 1 m 2 0.6 0.4 6200 k-in.

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