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, = 2L = 30 ft = 360 inches
now, for aluminium
Elastic modulus, E = 1.0 × 10⁷ Psi
Now, from the Euler's critical load, we have
where, I is the moment of inertia
on substituting the respective values, we get
or
I = 13.13 in⁴
also for circular cross-section
I =
thus,
13.13 =
or
d = 4.04 inches
The required size of column is length = 15 ft and diameter = 4.04 inches
