Question

Design a nonslender column to support the following service loads and moments. The factored ultimate axial force P₁= 100 kips, PD= 100 kips, ML= 2500 in.-kips, and MD= 1000in.- kips,. The eccentricity e to geometric centroid = 6 in. Given: f'c = 4000 psi fy = 60,000 psi.

Please show all work and explain your steps, please.

Answer 1

Solution:- the values given in the question are as follows:

factored live load(P_L)=100 kips

factored dead load(P_D)=100 kips

bending moment due to live load(M_L)=2500 kips-in

bending moment due to dead load(M_D)

eccentricity(e)=6 in

characteristic strength of concrete(f^Ic)=4000 psi

yield strength of steel(fy)=60000 psi

let the area of concrete is Ac and area of stee(reinforcement) is As

let the column is short axially(with minimum eccentricity) loaded column

ultimate axial load on column(P)=factored live load+factored dead load

ultimate axial load on column(P)=100+100

ultimate axial load on column(P)=200 kips

ultimate axial load on column(P)=200*10^3 lb

Calculating load carrying capacity of column:-

Pu=0.4*f^Ic*Ac+0.75*fy*As , [Eq-1]

where, Pu=load carrying capacity of column

let provide the area of steel is 2% of gross area

area of steel(As)=0.02*Ag

where, Ag=gross area

area of concrete(Ac)=Ag-As

area of concrete(Ac)=Ag-0.02*Ag

area of concrete(Ac)=0.98*Ag

for design of column the factored axial load on column is equal to ultimate load carrying capacity of column.

Pu=P=200*10^3 lb

values put in equation-(1) and calculate the value of gross area(Ag)

200*10^3=0.4*4000*0.98Ag=0.75*60000*0.02Ag

2*10^5=2468*Ag

Ag=81.03727 in^2

let the column is square and side of column is B

required side of square column(Breq)=9.002 in

provide side of square column(B)=9.2 in

area of column provided(Ag)=9.2^2=84.64 in^2

area of steel(As)=0.02*84.64

area of steel provided(As)=1.6928 in^2

let provide diameter of bar is 0.63 in

number of bars(n)=As/area of one bar

area of 16 mm diameter one bar=(3.14/4*)0.63^2

area of 16 mm diameter one bar=0.31156 in^2

number of bars(n)=1.6928/0.31156=5.433

number of bars(n)=6

provide 0.63 in diameter of 6 number of main bars

let provide 2-legged 0.315 in dia tie bars

spacing of tie bars(S)=minimum of{B or D , 16 , 11.811 in}

spacing of tie bars(S)=minimum of{9.2 in , 16*0.63=10.08 in , 11.811 in}

spacing of tie bars(S)=9.2 in center to center.

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