Find allowable axial compressive load

  • Thread starter jrizzle
  • Start date
  • #1
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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 im 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 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
 

Answers and Replies

  • #2
minger
Science Advisor
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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?
 
  • #3
PhanthomJay
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Homework Helper
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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
 
  • #4
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> Is this an Euler beam buckling problem?
Google "Euler buckling" and you will have a lot of relevant reading to get you started.
 
  • #5
nvn
Science Advisor
Homework Helper
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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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