Find allowable axial compressive load

  • Thread starter jrizzle
  • Start date
  • #1
jrizzle
1
0

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
 

Answers and Replies

  • #2
minger
Science Advisor
1,496
2
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
Science Advisor
Homework Helper
Gold Member
7,179
513
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
mathmate
365
0
> 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
2,128
32
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.
 

Suggested for: Find allowable axial compressive load

  • Last Post
Replies
4
Views
566
  • Last Post
Replies
4
Views
968
  • Last Post
Replies
0
Views
331
Replies
6
Views
483
Replies
4
Views
164
Replies
3
Views
826
Replies
10
Views
317
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
496
Top