Column buckling answer confirmation

In summary, the buckling equation for a column is F= ((∏^2)*E*I)/(K*L)^2, where E is the modulus, I is the area moment of inertia, K is the buckling coefficient, and L is the length of the column. For a hollow tube with a mass moment of inertia of I = (0.08^4-0.06^4)∏/64 = 1.37 x 10^-6 m^4, the minimum length for buckling to occur is 5.93 meters. The value of K for fixed-fixed end conditions is 0.5. The question about the mode of failure and the load at which failure would occur can be solved
  • #36
charger9198 said:
Nvn: thanks I've altered to suit and I used all the numbers without rounding to come up with 5.94.

The question given was exactly as wrote.

To go with the question I got asked 'what will be the mode of failure' which I got as buckling

The question after that asked 'at what load would you expect the failure to occur'

I used the equation (Pi^2)*E*I/(K*L)^2
=
(Pi^2)*200*1.374x10^-6/(0.5*5.94)^2
Giving me 3.07 x 10^-4 GPa
 
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  • #37
Gavin Cooper said:
can someone explain what units you are supposed to use?
Is putting 200 into the equation correct as it is 200GNm-2.
I am confused

Remember, G = * 10^9. You want to express all quantities with K, M, or G, or m, or c in terms of their numeric values multiplied by the value of these unit prefixes.
 
  • #38
Steam king Thanks for the reply, when using Eulers equation does it matter what units you use, should E be GNm-2 and I be Meters, or E be Nm-2 an I be mm, just not quite got my head round this bit.
 
  • #39
Gavin Cooper said:
Steam king Thanks for the reply, when using Eulers equation does it matter what units you use, should E be GNm-2 and I be Meters, or E be Nm-2 an I be mm, just not quite got my head round this bit.

Yes, it does matter which units you use, not only for Euler's formula, but for any formula. No, you can't throw a bunch of different units into a formula and expect things to sort themselves out properly.

It sounds like you have some basic misconceptions about applying units, especially those in SI.

Take, for example, a very simple formula, like the area of a rectangle, A = b * h. If I have measured the breadth in kilometers and the height in centimeters, I can't simply multiply these two different measurements together and expect the area to come out in square meters. If I want to calculate an area measured in square meters using the formula A = b * h, then the measurements of b and h must be converted to meters before I plug them into the formula. Failing this, then I have to calculate a conversion factor to determine how many kilometer-centimeters are in 1 square meter.
 
  • #40
Thanks Steam Kining, think I've got it, just didn't read things correctly.
My last problem is that I've worked ot the SR to a colulumn as being equal to the ESR. My notes do not tell me how the column would fail in a case like this. Is there any other equation I should be looking at to find this answer.
 
  • #41
Gavin Cooper said:
Thanks Steam Kining, think I've got it, just didn't read things correctly.
My last problem is that I've worked ot the SR to a colulumn as being equal to the ESR. My notes do not tell me how the column would fail in a case like this. Is there any other equation I should be looking at to find this answer.
Disregard this comment. I know what I done wrong. Doh!
 
  • #42
Good afternoon, i am working on the same assignment and I'm having a couple of issues. Any help would be greatly appreciated.
To calculate to load i would expect failure at the first length(5.92m), my numbers are as follows:

F= Pi^2*200*(1.37*10^-6)/2.96^2
F=3.087*10^-4
Firstly, could somebody confirm whether this is in GN? The reason i ask is because the value i put in for E is 200GNm^-2.

With regard to the next question, i saw that mamike wrote 'Basically halving the length of the column increases the load failure by approx 4 times'

My calculation suggests otherwise:
F=Pi^2*200*(1.37*10^-6)/1.48^2
F= 1.234*10^-3

Lastly, with regard to finding the mode of failure.
I have calculated the ESR, and achieved 118.74 like everybody else. Now when i calculate the S.R. i get the same again (using SR=Le/k).
Gavin cooper was having the same issue, but found the solution and didn't share it.
There is nothing in my material to suggest how i can calculate this. Frustrating.
Thanks in advance.
 
  • #43
js3 said:
To calculate to load i would expect failure at the first length(5.92m), my numbers are as follows:

F= Pi^2*200*(1.37*10^-6)/2.96^2
F=3.087*10^-4

With regard to the next question, i saw that mamike wrote 'Basically halving the length of the column increases the load failure by approx 4 times'

My calculation suggests otherwise:
F=Pi^2*200*(1.37*10^-6)/1.48^2
F= 1.234*10^-3

If F = 3.087×10-4, then what do you calculate for 4*F ? Compare with 1.234×10-3. :wink:
 
  • #44
SteamKing said:
If F = 3.087×10-4, then what do you calculate for 4*F ? Compare with 1.234×10-3. :wink:

*hides red face* Haha thanks for that, i'll never know how i didn't see that...?:)
As for units, am i correct in thinking we are talking gigaNewtons?
Thanks again.
 
  • #45
js3 said:
*hides red face* Haha thanks for that, i'll never know how i didn't see that...?:)
As for units, am i correct in thinking we are talking gigaNewtons?
Thanks again.
The units of F appear to be gigaNewtons.
 
  • #46
Much appreciated. Thanks for your help steam king.
I just need to try and get my head around how to decide whether the column will fail by buckling or yielding.
I don't understand the way it is worded in my notes. It does not seem to give a calculation for which you can determine the type of failure.
 
  • #47
Hi
Js3, did you get the answer to the last part of your post above. I am also struggling here.
If I have this right, we need to find the SR of our tube. Then if the ESR is less than that, then the failure is crushing, if its more then the failure is buckling. Is that right?

I get the ESR to 118.74, but get the SR to 449.02. So I guess I have done something wrong here? As we have just calculated the minimum length at which buckling can occur, then I would expect the ESR and SR to be close (if not the same). Using my theory above it would suggest the failure is crushing. It does not add up to me

for SR, I used L/k.
k=(1/2)√((R2-r2)=0.5 x √(0.042-0.032) = 0.013229

so SR=5.94/0.013229=449.02

Kr
C
 
  • #48
cjm181 said:
Hi
Js3, did you get the answer to the last part of your post above. I am also struggling here.
If I have this right, we need to find the SR of our tube. Then if the ESR is less than that, then the failure is crushing, if its more then the failure is buckling. Is that right?

I get the ESR to 118.74, but get the SR to 449.02. So I guess I have done something wrong here? As we have just calculated the minimum length at which buckling can occur, then I would expect the ESR and SR to be close (if not the same). Using my theory above it would suggest the failure is crushing. It does not add up to me

for SR, I used L/k.
k=(1/2)√((R2-r2)=0.5 x √(0.042-0.032) = 0.013229

so SR=5.94/0.013229=449.02

Kr
C
cjm181:
The policy at PF is not to encourage posting to old threads, also called necroposting. The member you are trying to engage has moved on and not posted anything in 6 months or so.

If you have a specific question about a problem, create your own thread. You can always refer to other threads at PF for clarity.
 

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