Answer:
The required size of column is length = 15 ft and diameter = 4.04 inches
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, [tex]\L_e[/tex] = 2L = 30 ft = 360 inches
now, for aluminium
Elastic modulus, E = 1.0 × 10⁷ Psi
Now, from the Euler's critical load, we have
[tex]P =\frac{\pi^2EI}{L_e^2}[/tex]
where, I is the moment of inertia
on substituting the respective values, we get
[tex]10\times10^3 =\frac{\pi^2\times1.0\times10^7\times I}{360^2}[/tex]
or
I = 13.13 in⁴
also for circular cross-section
I = [tex]\frac{\pi}{64}\times d^4[/tex]
thus,
13.13 = [tex]\frac{\pi}{64}\times d^4[/tex]
or
d = 4.04 inches
The required size of column is length = 15 ft and diameter = 4.04 inches
