Answer to Question #162295 in Mechanical Engineering for Zawad

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 d₀ = 4.5 in

Inner diamer d_i = 4.021 in

I = 7.2 in⁴

k = 1.5 in

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

Number of column = 60.377 ​= 3.31

number of column will be 4

Value of equivalnt stress = 

$\frac{60.377}{3.174}=19.022 ksi$