# Slenderness ratio, mode of failure

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1. Oct 26, 2015

### Jayohtwo

1. The problem statement, all variables and given/known data
Hi guys (first post here let me know if I can fix anything that I've done wrong),

I'm struggling a little with the second part of this question and seem to be going round in circles

A column has the dimensions shown in the diagram below.
(a) What is the minimum length of the column at which buckling is
likely to occur?
(b) If the column is the length determined in (a),
(i) What will be the mode of failure?
(ii) At what load would you expect failure to occur?
(c) If the column is half the length determined in (a) :
(i) What will be the mode of failure?
(ii) At what load would you expect failure to occur?
2. Relevant equations
A = 8.796x10-3 m2
I = 2.199x10-5 m4
K = 0.05 (least value of radius of gyration)
Slenderness ratio (using length not effective length unsure if this is correct) = 237.4

3. The attempt at a solution
Part a: I have determined the effective length to be 5.94m as the column is fixed at both ends and Le=L/2

Part b: Here's where my google skills fall short and I'm not sure how to determine what failure mode. I've also calculated the effective slenderness ratio using ESR = √(π2E)/(σ) = 118.74.
As this is less then the slenderness ratio does this mean that the failure mode is crushing?

Part c: Will be straightforward enough to solve when I can figure ou how to do part b

2. Oct 26, 2015

### PhanthomJay

If the effective length is as calculated, what is the minimum length of the column for buckling to occur?

3. Oct 27, 2015

### Nidum

Has this exact same problem has been discussed previously on PF ?

4. Oct 27, 2015

### Jayohtwo

Yes it has but the thread had died about 2 years ago and only the first part of the question was discussed. I'll put some more work into this question later today and see if I can come to a conclusion.

5. Oct 27, 2015

### Jayohtwo

I've gone back over the question and noticed I've used diameter instead of radius for area so A= 2.199x10-3 m2
this makes Le = 2.96 m and L = 5.93 m and this seems to be the minimum length that buckling will occur.

Part b: Failure mode is buckling (not 100% on this)
ii: Failure Load Fc = σcA=307876.08 N

Part c: if min length for buckling was 5.94 failure mode will be crushing (again not 100% certain)
ii: Failure load for 1/2 length Fc = π2EI/Le2 = 1234601.72 N

6. Oct 27, 2015

### PhanthomJay

I am not sure how you arrived at your numbers for I and A etc but aside from the math, your approach for finding the minimum length is correct, above which failure is likely by buckling and below which failure would be by crushing or yielding. But you should check again about the failure load if the column was halved...your answer assumes buckling is the failure load, so recheck that......

7. Oct 28, 2015

### Jayohtwo

I got my answer for I and A using :

I = (D4 - d4)π/64
= (0.084 - 0.064)π/64
= 1.37x10-6 m4

A = π(R2 - r2)
= π(0.042 - 0.032)
= 2.199 x10 -3 m2

I'll recheck my notes later today and read up on the failure modes.

8. Dec 1, 2015

### Bulby

Hi I am on the same question and its not very clear I don't think. I am finding the units are tricky. How do you work out if it will buckle or crush ?