# Find allowable axial compressive load

jrizzle

## Homework Statement

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

## Homework Equations

17 S-T gives modulus of elasticity of:
E = 10,600 ksi

some formulas (im sure I am missing some)
f.s. = P_u/P_all
Stress = P/A

## The Attempt at a Solution

P_all = P_u/f.s = P_u/2.5

im really confused. i don't 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

I'm assuming one leg is 3", the other is 2" and the bar is 1/4" thick. The whole thing is then 43" long. Is this an Euler beam buckling problem?

Homework Helper
Gold Member
The most important formula you are missing is the formula for ultimate buckling stress. Are you familiar with it? Also, regarding the area of the angle, do you know what a 3 x2 x 1/4" angle looks like? See the site below for a cross section. The thickness is 1/4", and the legs are 3" and 2" , respectively. The cross sectional area works out to 1.19 in^2. The length is 43 inches (into the plane of the page). http://www.engineersedge.com/angle_unequal.htm

mathmate
> Is this an Euler beam buckling problem?
Google "Euler buckling" and you will have a lot of relevant reading to get you started.