# Thread: Can anyone help with this problem on Force??

1. ## Re: Can anyone help with this problem on Force??

I have the modulus of elasticity = 140MN^-2 and I came to 140x10^6 is this the problem? sorry not modulus of elasticity I meant yield stress. Anyway very confused now.....

2. ## Re: Can anyone help with this problem on Force??

Originally Posted by Jock
A circulur column has a diameter of 80mm outside edge and 60mm inside edge.
Both ends are fixed with equal force being applied top and bottom.
Young's Modulus E = 200GN m-2
Yield Stress deltay = 140 MN m-2

What is the minimum length of column at which buckling is likely to happen?
If the column was this length what load would you expect failure to occur?
If the column is half this length what load would you expect failure to occur?

Really struggling with this one if anyone can help id really appreciate it along with the working of how they came to their final answers......
Fcr = 4(Pi^2)EI/(4l^2) Roark, Formulas for Stress and Strain

MIn length for buckling: substitute Fcr = AxYS and solve for l.
If length is less than this, you will reach compressive yield strength before buckling.

At min length, Fcr=AxYS.
At half min length, Fcr=AxYS

3. ## Re: Can anyone help with this problem on Force??

Still not really following it plus I don't know the final answer so therefore don't know what I am actually looking for as a final result to see if I am right or not....

4. ## Re: Can anyone help with this problem on Force??

Originally Posted by Jock
Still not really following it plus I don't know the final answer so therefore don't know what I am actually looking for as a final result to see if I am right or not....
Suppose the column is 2mm high. You will never buckle it no matter how much force you apply, just like you could never buckle a penny no matter how much force you apply. How much force could you apply before “mushing” it: F=AxYS.

OK, now suppose you tried to apply the same force, F=AxYS, to the same column that was 100meters high? You would buckle way before you ever got there.

So at what point do you switch from crushing to buckling? Substitute F=AxYS in the buckling formula to get the length at which this happens.

If you understand that, getting the right answer is just a matter of some algebra and using the right units.

If I were a teacher and a student got the right answer with no explanation, I would give him 0/10. If he had the right method showing how he got it but the wrong answer, I would give him 10/10- no, 11/10 for not wasting his time on the algebra.

OK, yes, not knowing if you got the right answer can be frustrating, but only if you are not sure you are going about it right.

5. ## Re: Can anyone help with this problem on Force??

I want to do it right and show all my working but the explanations aren't making much sense and have never came across F=AxYs so don't know what it all stands for and all the components of it are. Just really wish I could see it in practice for the equation then follow it.........

6. ## Re: Can anyone help with this problem on Force??

Originally Posted by Jock
I have the modulus of elasticity = 140MN^-2 and I came to 140x10^6 is this the problem? sorry not modulus of elasticity I meant yield stress. Anyway very confused now.....
I gave you the formula to use back in post #13 of this thread. You tried to use that formula but made the mistake of plugging in yield stress instead of area moment of inertia for the 'I' term. Remember you have already calculated $\displaystyle I = 1.37 \times 10^{-6} \ m^4$ in post #6. Use that value in the formula for the 'I' term and you should get the right answer.

7. ## Re: Can anyone help with this problem on Force??

Just done it and calculator threw back 307610.388 which I hope is 3.17x10^6 so hope this is right!!!....

8. ## Re: Can anyone help with this problem on Force??

Simply having the formula doesn't do it. You have to understand that you have to substitue F=AxYS in the formula for Fcr and solve for l. Why is that a problem? The actual solving is just algebra, if you use consistent units, say N and m.

Just saw your post. What did you do and why? Saying you got some number is meaningless

9. ## Re: Can anyone help with this problem on Force??

Post 13 used the formula there and amended one of the figures in the equation to get that answer. Is the answer correct??? pi^2x200x10^9x1.37x10^-6/(0.5x5.93)^2 was what I did....

10. ## Re: Can anyone help with this problem on Force??

Originally Posted by Jock
Just done it and calculator threw back 307610.388 which I hope is 3.17x10^6 so hope this is right!!!....
Yes - that's the critical load in newtons.

11. ## Re: Can anyone help with this problem on Force??

so if I half the 5.93 the next answer will work by plugging in the correct info? Also how do I know what mode of failure these findings produce?....

12. ## Re: Can anyone help with this problem on Force??

so if I have done the last one correct my answer of 6.15x10^6 will be correct also?? if this is all right I am only left with mode of failure for them both if anyone can help?....

13. ## Re: Can anyone help with this problem on Force??

Would I be right in saying the E.S.R for the 5.93m length is 118.74 which equals buckling and E.S.R for the half length is 59.38 which equals yielding?

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