# EX11.3 - CEE 415 Example 11.3 Design of column section(tied...

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CEE 415 1 Example 11.3 – Design of column section (tied) for combined axial load and bending. Use an R-type arrangement of longitudinal steel and start with ρ g = ρ t = 0.025. Material properties: f c ' = 5000 psi, f y = 60 ksi Given loads: P D = 300 k M D = 60 k-ft. P L = 200 k M L = 120 k-ft. P W = ± 60 k M W = ± 100 k-ft. Load Case 1 : Gravity P u = 1.2 (300 k) + 1.6 (200 k) = 680 k M u = 1.2 (60 k-ft) + 1.6 (120 k-ft) = 264 k-ft Design eccentricity, e = M u /P u = 4.66 in. Load Case 2 : Gravity plus Wind P u = 1.2 (300 k) + 0.5 (200 k) + 1.6 (60 k) = 556 k M u = 1.2 (60 k-ft) + 0.5 (120 k-ft) + 1.6 (100 k-ft) = 292 k-ft Design eccentricity, e = M u /P u = 6.30 in. Load Case 3 : Low Gravity combined with Wind P u = 0.9 (300 k) – 1.6 (60 k) = 174 k M u = 0.9 (60 k-ft) + 1.6 (100 k-ft) = 214 k-ft Design eccentricity, e = M u /P u = 14.8 in. Initial design : I will use Eq. (11-21a) from the textbook to get an initial size for the column. For this initial estimate, use P u from the gravity load case. () 2 680 kips (trial) 0.40 5 ksi 0.025 60 ksi 0.40 262 in. u g ct y P A ff ρ == ′+ = Select b 0.75 h, so h 262 18.7 in. 0.75 = ; thus, round up to use h = 20 in. and b = 15 in., giving us A g = 300 in.

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## This note was uploaded on 03/29/2011 for the course CEE 415 taught by Professor Wight during the Spring '11 term at University of Michigan.

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EX11.3 - CEE 415 Example 11.3 Design of column section(tied...

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