Question

Problem 2 Assume normal weight concrete with fe' = 3,500 psi and steel bars of Grade 60 (fy = 60,000 psi). For design loads assume dead load Po = 300 kips (including member self-weight) and live load P. = 400 kips. Design a short round spiral column for axial load only, assuming initially po = 4%. Form sizes are available in 2 in increments.

Answer 1

ANSWER ::

Reinforced concrete columns

The design of reinforcement can be either spiral or tied. Spiral columns are cylindrical with a continuous helical bar wrapped around the column. This spiral provides support in the transverse direction. Tied columns have closed lateral ties spaced approximately uniformly across the column.

Given that fc' = 3,500 psi fy = 60,000 psi "Dead load (PD) = 300 kips Live Load (PL) = 400 kips assume P9244 loads (Puli- maximum service Pu= 1.2 PD + 1.6PL = 12 (300) +16 (400) = 360 +640 = 1000 kips using the formula By find gross cross sectional area (Ag). . Hxx Pu= 0 Pn (mah) maximum design axial load .. Pu = 0.858 [0.85 fc. (Ag - Ast) + fy nst] : Opnmax. 3 max design axial load . en mex? fob circular column (spital) Cg = o.ou , q=0.75

1000 =0.85 x 0.75 8543.5 x(Ag - Doung) + 60 x 0.04 AS 1000 = 0.6315 [ 2.915 (Ag - 004 ng ) + 9.493] 1000 = 0.6375 [ 2.915A970-19 Ag +2.4Ag] 1000 = 0.6315 [5.956 Ag] 1000 = 3.3507 AS Ag = 3.3507 Ag -298.44 "in? diameter of for gross sectional area Ag = 17 d² T2 9.98.4 d? 4x298-4 Tr d? = 379,78 d (379.78) da 19.48 I d = zo in area of tension steel (Ast). section I find total cross Ast = fg tag 30.04 X 298-4 in? Ast Rap = 14.936 in)