Find allowable axial compressive load

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    Axial Load
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Discussion Overview

The discussion revolves around determining the allowable axial compressive load for a specific aluminum-alloy angle acting as a pin-ended column. Participants explore the relevant formulas and concepts related to buckling and stress, while addressing confusion regarding the geometry and properties of the angle section.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the dimensions of the angle and the provided area, questioning how they relate to the problem.
  • Another participant interprets the dimensions, suggesting that one leg is 3 inches, the other is 2 inches, and the thickness is 1/4 inch, while confirming the length is 43 inches.
  • A participant emphasizes the importance of the ultimate buckling stress formula, asking if the original poster is familiar with it and providing a link to a resource for visualizing the cross-section of the angle.
  • There is a suggestion to research "Euler buckling" for additional context and understanding of the problem.
  • A participant confirms the correctness of the stress equations provided by the original poster and encourages finding a formula for critical load or critical stress, hinting that P_u may also be referred to as Pcr.

Areas of Agreement / Disagreement

Participants generally agree on the need for clarification regarding the geometry and the relevant formulas for buckling stress. However, there is no consensus on how to proceed with the calculations or the interpretation of the problem's parameters.

Contextual Notes

There are unresolved aspects regarding the assumptions made about the angle's geometry, the application of the buckling formulas, and the specific definitions of terms like P_u and Pcr. The discussion lacks a definitive resolution on these points.

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