Question

A hollow circular column is made of AISI C1015, as rolled and its Outer and inner diameter is 4.5 in and 4.021 in.

I = 7.2 in^4, k= 1.5 in, L= 15 ft, effective length= 0.65*L, N=2.5.

Single column area= 3.174 in^2

total load,F= 200 kips

Determine minimum number of column and equivalent stress.

Answer 1

Outer diameter=d_o=4.5 in, inner diameter=d_i=4.021 in, I= 7.2 in⁴, k= 1.5 in= radius of gyration, L=15 ft, effective Length=0.65 L, N=2.5, single column area= 3.174 in²,Total load= F=200 kips

section modulus=Z= $\frac{I}{y}= \frac{7.2}{2.25}=3.2 in^3$

Critical load for one coloumn

P=$\frac{\pi^2EI}{L^2}=\frac{3.14^2\times 27557\times 7.2}{180^2}=60.377 kips$

(i) Number of column=$\frac{200}{60.377}=3.31$ , number of column will be 4

(ii) Now value of equivalnt stress= $\frac{60.377}{3.174}=19.022 ksi$