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Trying to determine buckling?
L = 8 in A-36 steel rod
rod dia = 0.249 in
I = 0.0001887 in^4
E = 30,000 lbs/in^2 (psi)
Using the buckling formula P=[3.14^2 x E x I]/ L^2
I get [(3.14)^2(30000)(0.0001887)]8^2
which is 0.87299 lbs.
This can't be a buckling load. It should be something around 1200 lbs. according to the instructor.
What am I doing wrong?
Thanks Karl. I knew I did it right and I should stop listening to the instructor and trusting this damn book. Both said to use 30,000 for E.
Well the instrctor may have tripped over his tounge, but the book clearly states:
Modulus of Elasticity for A-36 steel is 29.0 (10^3) ksi
Khan, we're only working with steel in the 30 range.
3 Answers
- 1 decade agoFavorite Answer
The obvious error is in the modulus of elasticity for steel, it's 30x10^6 psi, i.e., 30 million psi.
The equation you are using is only correct for simply-supported end boundary conditions, pinned, and in addition one end with a roller.
If both ends are clamped the buckling capacity is much higher.
In your equation there should be a factor K in the denominator, which is determined by the end (boundary) conditions. Your eq. is only correct for K=1, which are the pinned conditions I am referring to. Pinned means that the column can rotated at both ends, and in addition one end is free to move in the vertical direction..
So, using the correct E, your answer would be 873 lbs.
Source(s): I am a structural engineer. - 1 decade ago
Use the Correct E value,it should be 4.35 x10^7psi ( 3x10^5 N/mm2)
If u use the above E value u will get the force as 1200lbs.
