Question

Design a square tied column to support an axial dead load of PD-260 k, P-450 k Include the design of ties or spirals and a sketch of the cross sections selected, including bar arrangements fc = 4000 psi, and fy = 60,000 psi. Initially assume Ps= 4%.

Answer 1

Pol - 260 kips (Dead lead) P = 450 kips (live load) f = 4000 pi workei by = 60000 pri = 60ksi y = 4% = 0.04 Max sowice load, Pu 12 Pol + 16 Pu Pu= 1032 kips Now, Pg - Ast e. Ast - () (Ag) - (i) * Pu - O Promax) = 0.80 (0.85 he (Ag-Ast) + fy Ast) O Panor - Maximum design axial load. strength reduction factor . 1032 - 08*0.65 (085** (Ag - 0.04 Ay) + (60x0.06x Ag) Ag = 350.4 in 351in2 consider the Column size of 19 in x 19in with grass Cross sectional area of Column of 36 1in2 Load carried by Concrete, pe 0.8 0.65 (0.85+ 4x (361-10-04 x 3417) Pc = 612.7 kips. Load carried by steel - Ps = Pu-le 1032-612.7 b-414-3 kips

1 I . P. - 0.80 by Art 419.3 = 0.2 x 0.65 x 60 x Ast (Ast= 13.44 in Ast = 14 in Provide a number of bars and #11 bar number ( 14116) Ich core, the maximum number Ef a bars is allowable. Thus, the number of longitudinal kans is sokefactory. check for ties- size of No.& is acceptable for longitudinal bars upts to the No llbar spacing 48 Pri 48x0.5 - 24in 168 rain = 16+1:41 - 22.5616 least lateral dimension = 19 in This the size of bars is #4 19 in. clear distance - 19-2 (1.5)- 2(0.5)-3 (1.4) Minimum 2 = 5.385in bin Thus, the extra tie is not reorsry.

+#4 @ 19" 1.5 in clear cover TH9-#ll bars 19 inch