Question

8. Find the necessary reinforcement for a square column (16 in x 16 in) to carry a factored load Pu = 265 kips and a factored moment M, of 320 ft-kip. Assume #4 ties, f'c = 4000 psi and fy = 60,000 psi. Include the spacing of the ties assuming that the acting shear is negligible (<<V/2).

Answer 1

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

side of square column(b)=16 in

factored load(Pu)=265 kips

factored moment(Mu)=320 kips

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

yield strength of steel(fy)=60000 psi

Puk265 kips 16 in 320 kips-ft -16 in

let the column is short axially loaded column

ultimate load carrying capacity of column(P^Iu)=0.4*f^Ic*Ac+0.75*fy*As

where, Ac=cross-section area of concrete

As=cross-section area of steel

Ac=Ag-As

Ag=gross-section area=16*16

Ag=256 in^2

Ac=256-As

values put in above equation-

ultimate load carrying capacity of column(P^Iu)=0.4*4000*(256-As)+0.75*60000*As

ultimate load carrying capacity of column(P^Iu)=409600-1600As+45000*As

ultimate load carrying capacity of column(P^Iu)=409600+43400*As

ultimate load carrying capacity of column(P^Iu)=factored load on column(Pu)

265000=409600+43400*As

As=-144600/43400=-3.3317 in^2

According above procedure the area of steel goes into negative but in column area of steel never goes negative into column, so we provide the minimum area of steel.

As(min)=0.8% of gross area

As(min)=(0.8/100)*256

As(min)=2.048 in^2

provide 0.875 in diameter bars

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

area of single bar=(3.14/4)*0.875^2

number of bars(n)=2.048/0.60

number of bars(n)=3.4133 $\approx$ 4

provide 0.875 in diameter 4 bars

provide 0.375 in diameter tie bars

spacing of tie bars(S)=minimum of {B or D, 16$\phi$, 12 in}

spacing of tie bars(S)=minimum of {16 in, 16*0.875=14 in, 12 in}

spacing of tie bars(S)=12 in

provide 0.375 in diameter tie bars of spacing 12 in center to center

provide 2 in clear cover all side

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