Question

A circular column is fixed at the base and not supported at the top. If the column needs to be 15ft and hold 10kips, what is the required size of the column? Assume the material is aluminum.

Answer 1

The required size of column is length = 15 ft and diameter = 4.04 inches

Step-by-step explanation:

Given;

Length of the column, L = 15 ft

Applied load, P = 10 kips = 10 × 10³ Psi

End condition as fixed at the base and free at the top

thus,

Effective length of the column,
`\L_e` = 2L = 30 ft = 360 inches

now, for aluminium

Elastic modulus, E = 1.0 × 10⁷ Psi

Now, from the Euler's critical load, we have

`P =(\pi^2EI)/(L_e^2)`

where, I is the moment of inertia

on substituting the respective values, we get

`10*10^3 =(\pi^2*1.0*10^7* I)/(360^2)`

or

I = 13.13 in⁴

also for circular cross-section

I =
`(\pi)/(64)* d^4`

thus,

13.13 =
`(\pi)/(64)* d^4`

or

d = 4.04 inches

The required size of column is length = 15 ft and diameter = 4.04 inches