Question

Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its length is 64 in and its ends are fixed. Use a design factor of 3.0. Use aluminum 6061-T6.

Answer 1

Answer:

The dimensions are 0.95 in

Explanation:

Given:

l = length = 64 in

lc/2 = 64/2 = 32 in

FOS = factor of safety = 3

E = 10.6x10⁶ psi

σ = 40000 psi

The square section is

I = a⁴/12

$P=\frac{\pi^{2}*E*I }{lc^{2} }\\6500=\frac{\pi ^{2}*10.6x10^{6}*a^{4} }{12*32^{2} } \\a=0.95in$

σallowable = σ/3 = 40000/3 = 13333.3 psi

Area = a² = 0.95² = 0.902 in²

σactual = 6500/0.902 = 7206.2 psi

like σactual < σallowable, the dimensions are 0.95 in

Answer 2

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches