# Stress/buckling question

1. Nov 13, 2015

### Niall11

1. The problem statement, all variables and given/known
I have been asked to calculate the minimum length of this column (attatched) at which buckling is likely to occur

2. Relevant equations
E.S.R = sqrt(π^2*E / σ)

2nd moment of Area I = AK^2

E.S.R = L/K

I= π/32*(D^4-d^4)

3. The attempt at a solution

so E.S.R = sqrt(π^2*200*10^9 / 140*10^6) = 118.74

Since I = AK^2 and E.S.R = L/K
L= (E.S.R)*K = (E.S.R)*sqrt(I/A)

so I = π/32 * (0.08^4-0.06^4) = 2.75*10-6

Now L = 188.7 * sqrt(2.75*10-6 / Area)

ive got that far but don't know a suitable equation for area, I'm not looking for answers just for someone who can tell me if I'm on the right track, any help would be appreciated!!

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2. Nov 13, 2015

### SteamKing

Staff Emeritus
They're talking about the area of the cross section of the column. You are given the outer diameter and the inner diameter of the column.

What else do you need? Do you know how to calculate the area of a circle?

3. Nov 14, 2015

### Niall11

thanks for the fast reply, I was thinking I needed the area of the whole column for some reason.

yes area of circle = π*r^2 so area of outer circle - area of inner circle = cross sectional area of column;

π (0.04)^2 - π(0.03)^2 = 2.199*10^-3 m^2

meaning L = 118.7 * sqrt(2.75*10^-6 / 2.199*10^-3) = 4.19m

effective length is 1/2 of the column length for a column which both ends fixed therefore length = 4.19m * 2 = 8.38m

4. Nov 14, 2015

### SteamKing

Staff Emeritus
This formula for the second moment of area of the column is incorrect. You used the polar moment, which is Ix + Iy

You are supposed to use the least value of the second moment of area, which for a circle is I = (π/64)D4

Since this is a hollow cylinder, I = (π/64)*(Do4 - Di4)

You'll have to re-do your calculations of the length of the column accordingly.

5. Nov 14, 2015

### Niall11

oh right I have on my notes that least 2nd moment of area is π/32(D^4-d^4)!

so least 2nd moment of area = π/64 (D^4-d^4) = π/64 (0.08^4-0.06^4) = 1.374 * 10^-6

now effective L comes out at 2.94m and length 5.94m!
I'm hoping that is correct if i've used all the right formulae

6. Nov 14, 2015

### SteamKing

Staff Emeritus
Yes, your length calculation looks correct now.

FWIW, this same problem has cropped up on PF several times before.

7. Nov 14, 2015

### Niall11

Thanks a lot that was a big help, oh right I didn't realise ill look next time!

8. Nov 16, 2015

### Marv K

If the effective length is half of the actual length, are those figures not incorrect? I only ask cause I'm doing the same course and wanted to check. Is the effective length not 2.968m?

9. Nov 16, 2015

### SteamKing

Staff Emeritus
The effective length is approx. 2.94 m, but the OP made a slight error in doubling this figure to find the actual length, which should be approx. 5.88 m.

10. Nov 16, 2015

### Marv K

Ok, sorry to be annoying, but worried Ive done it wrong now.

If E.S.R. = 118.74

I = 1.374*10^-6

A = 2.199*10^-3

Then the equation is Le = 118.74*sqrt(1.374*10^-6)/(2.199*10^-3)

Which equals 2.968m?

Also, sorry if I've written that equation terribly.

11. Nov 16, 2015

### SteamKing

Staff Emeritus
I'm not sure what you point is here. Le = 2.968 m or Le = 2.94 m is the same number for all intents, given that this is a buckling problem.

Depending on how you round the intermediate computations, there will be some variation in the fractional portion of the final result.