Static engineering v -- stress/buckling

AI Thread Summary
The discussion revolves around calculating the minimum length of a column at which buckling is likely to occur, using given material properties and formulas. Participants share their attempts at solving the problem, with one user arriving at an effective length of 5.94 m based on their calculations. There is confusion regarding the correct formulas and the shape of the column, leading to differing answers among users. Additionally, users emphasize the importance of using proper unit nomenclature and formatting in their calculations. The thread concludes with a note that it is no longer active, suggesting users create new threads for ongoing questions.
oxon88
Messages
176
Reaction score
1

Homework Statement



A column has the dimensions shown below...

untitled.GIF



Column material properties:

young's Modulus E = 200 GN m^-2

Yield stress = 140 MN m^-2




Homework Equations




a) what is the minimum length of the column at which buckling is likely to occur?



The Attempt at a Solution




i got an answer of: 2*((118.74 * Sqrt((2.33x10^3)/(2800x10^-3))) = 6850.56 M
 
Physics news on Phys.org
oxon88: Your current answer is wrong. Try again. First, list relevant equations. And then show how you computed each numerical value used in your final equation.

By the way, MN*m^-2 is called MPa. And GN*m^-2 is called GPa. Always use the correct, special name for a unit. E.g., 140 MPa, not 140 MN*m^-2.
 

Homework Statement



a) what is the minimum length of the column at which buckling is likely to occur?


The Attempt at a Solution




least second moment of area. I = (B^4 - b^4)/12 = (80^4 - 60^4)/12 = 2.33x10^6

area. A = B^2 - b^2 = 80^2 - 60^2 = 2800

σ = 140x10^6

E = 200x10^9

effective slenderness ratio. E.S.R = sqrt(pi^2.E/σ) = sqrt(pi^2.(200x10^9)/(140x10^6)) = 118.74

effective length. L = (E.S.R). Sqrt(I/A) = 118.74(sqrt(2.33x10^3)/(2800)) = 3427.7


effective length = 1/2 column length. = 3427 x 2 = 6854 mm


--------------------------------------------------------------------------------------


I also tried another method and got the same answer...


Least value of radius of gyration. K = (1/2).sqrt((B^2+b^2)/3) = 1/2 sqrt((80^2+60^2)/3) = 28.868

effective length. L = (E.S.R x k) = 28.868 x 118.74 = 3427.73

effective length = 1/2 column length. = 3427 x 2 = 6854 mm
 
ok. so i worked it out.

the answer is 5.94m
 
Last edited:
ok. So if the column was to be 5.94 m. What would be the mode of failure? and at what load would you expect the failure to occur?

I guess it would be buckling?
 
any ideas?
 
useless forum...
 
We are not allowed to answer your homework questions for you.
 
  • #10
Hi all, I have the same question and was getting the same answers originally. I assume that your new answer is based on changing the formula to a hollow tube. I was wondering without being told in the question how we are supposed to determine what shape it is... Or am I being a bit thick...
 
  • #11
Wrong formula

The least second moment of area I about the xx axis is not what you stated so your value is incorrect. I get L = 5.236389088 m
 
  • #12
I am just wondering if this thread is still active as I am studying this question and facing similar problems to those on here
 
  • #13
David J said:
I am just wondering if this thread is still active as I am studying this question and facing similar problems to those on here
The last post was two years ago, so, no, it's not still active. :frown:

However, you can and should create your own, fresh thread. It's not like you can create only so many threads at PF before you run out. :wink:
 
Back
Top