find allowable axial compressive load


by jrizzle
Tags: allowable, axial, compressive, load
jrizzle
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#1
May11-09, 11:51 PM
P: 1
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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minger
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#2
May12-09, 07:59 AM
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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?
PhanthomJay
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#3
May12-09, 08:05 AM
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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
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#4
May12-09, 01:59 PM
P: 366

find allowable axial compressive load


> Is this an Euler beam buckling problem?
Google "Euler buckling" and you will have a lot of relevant reading to get you started.
nvn
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#5
May12-09, 10:10 PM
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jrizzle wrote: "stress = P/A; P_all = P_u/f.s."

Excellent, jrizzle. That's correct. And your first equation, quoted above, can be written P_u = sigmacr*A. Therefore, can you find in your text book a formula for sigmacr or P_u? Notice the hints given by PhanthomJay and mathmate. Hint: P_u might sometimes be called Pcr.


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