Question

Design a circular column, using approximate methods, for a factored load of 80 Kips and a
factored moment of 40 Kip-ft. about x and y axes each. The diameter of column is 18".
Material strengths are fc' = 4Ksi and fy = 60Ksi. Use appropriate column interaction diagram.

Answer 1

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

factored load(Pu)=80 kips

factore moment about x and y=40 kip-ft

diameter of column(D)=18^II , or 18 in

characteristic strength of concrete(fc)=4 ksi

yield strength of steel(fy)=60 ksi

let the column is short column and we use the spiral reinforcement

load carrying capacity of spiral column(P)=1.05[0.4*fc*Ac+0.75*fy*As] , [Eq-1]

helical reinforcement increase the 5% load carrying capacity of column

where, Ac=cross-section area of concrete

As=cross-section area of steel reinforcement

let provide the 4% of gross area column, steel reinforcement.

gross area(Ag)=(Pi/4)*D^2=(3.14/4)*18^2

gross area(Ag)=254.34 in^2

area of concrete(Ac)=Ag-As

area of concrete(Ac)=254.34-As

Column design to carry factored load 80 kips, and moment at 40 kip-ft in X and Y-direction.

values put in above equation-(1) and calculate the value of area of reinforcement

80=1.05[0.4*4*(254.34-As)+0.75*60*As]

76.1905=406.944-1.6As+45As

76.1905=406.944+43.4*As

As=-(76.1905/406.944)

according above equation the area of reinforcement is goes in negative, but area of reinforcement never goes in negative in the column. so we provide area of steel is 4% of total groass area.

area of steel(As)=(4/100)*Ag

area of steel(As)=0.04*254.34

area of steel bars(As)=10.1736 in^2

let provide 1.128 in diameter reinforcement bar

number of bar(n)=As/area of single bar

number of bar(n)=10.1736/0.8859=11.484 12

provide number of bar(n)=12

let provide 0.375 in diameter helical reinfocement

let provide the clear cover is 1 in

core diameter(dc)=D-2*cover

core diameter(dc)=18-2*1

core diameter(dc)=16 in

diameter of helical reinforcement(dh)=dc-dia of helical reinforcement

diameter of helical reinforcement(dh)=16-.375

diameter of helical reinforcement(dh)=15.625 in

Calculating area of helical reinforcement(Vh) and spacing of helical reionforcement(S)

, [Eq-2]

0.36[(254.34/244.1664)-1]*(4/60)=Vh/Vc

where, Vc=(3.14/4)*16^2*1=200.96

put value of Vc in above equation-(2)

0.001=Vh/200.96

volume of helical reinforcement(Vh)=0.20096 in^3/ft

Vh=(1/S)*(pi*dh)*(Pi/4)*h^2

0.20096=(1/S)*(3.14*15.625)*(3.14/4)*0.375^2

S=5.4158/0.201

S=26.7 in

spacing S is very large so we provide spacing of helical reinforcement is 11 in

spacing of helical reinforcement(S)=11 in

[Ans]