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Find allowable axial compressive load 
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#1
May1109, 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 ST aluminumalloy angle 43in long. It acts as a pinended column. Assume: Factor of safety = 2.5 Least radius of gyration, r = .43 Area = 1.19 in^2 2. Relevant equations 17 ST 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 


#2
May1209, 07:59 AM

Sci Advisor
P: 1,498

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?



#3
May1209, 08:05 AM

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PF Gold
P: 6,050

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



#4
May1209, 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. 


#5
May1209, 10:10 PM

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P: 2,124

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 = sigma_{cr}*A. Therefore, can you find in your text book a formula for sigma_{cr} or P_u? Notice the hints given by PhanthomJay and mathmate. Hint: P_u might sometimes be called P_{cr}. 


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