(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the allowable axial compressive load for a 3in by 2in by .25in 17 S-T aluminum-alloy angle 43in long. It acts as a pin-ended column.

Assume: Factor of safety = 2.5

Least radius of gyration, r = .43

Area = 1.19 in^2

2. Relevant equations

17 S-T gives modulus of elasticity of:

E = 10,600 ksi

some formulas (im sure im missing some)

f.s. = P_u/P_all

Stress = P/A

3. The attempt at a solution

P_all = P_u/f.s = P_u/2.5

im really confused. i dont understand how there could be 3inX2inX.25in and then 43in long for one. then there is a weird area of 1.19 in^2

can someone help me with this? they didnt give a diagram either

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# Find allowable axial compressive load

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