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Column buckling answer confirmation

by charger9198
Tags: buckling, column, confirmation
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charger9198
#1
Jan4-12, 07:08 AM
P: 60
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What is the minimum length of the column at which buckling is likely to occur?


The buckling equation is: F= ((∏^2)*E*I)/(K*L)^2

E is the modulus, I is the area moment of inertia, K is the buckling coefficient, and L is the length of the column.

the mass moment of inertia for a hollow tube is;

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(using meters)

I = (0.08^4-0.06^4)∏/64 = 1.37 x 10^-6 m^4


for the force, F, that will cause yielding.

The equation is:

σ= F/A

The cross sectional area of the column is:

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Need to convert MN/m^2 to GN/m^2 by multiplying by 10^-3.

Therefore, F = Aσ = 0.02 x 141 x 10^-3 = 3.08x10-4 GN

The value of K for fixed-fixed end conditions is 0.5

Therefore, rearange the equation and solve for L


L= (SQRT) ((∏^2)*E*I)/F*k^2 = 5.93 m

This means, if the column is longer than 5.93 meters, it will buckle due to slenderness. Otherwise, if it is shorter than 5.93 meters, it will not displace out of plane and fail in yielding.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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nvn
#2
Jan5-12, 09:23 AM
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charger9198: MN/m^2 is called MPa. And GN/m^2 is called GPa. Always use the correct, special name for a unit. E.g., 140 MPa, not 140 MN/m^2.

If you have written the problem statement in post 1 exactly as given, then it might indicate the instructor has a limited understanding of buckling, because the question is unclear. A column can buckle at almost any length, if the applied load is high enough (i.e., as we move toward, and into, the inelastic range). The question would be more clear if it were instead worded as follows. "What is the minimum length of the column at which elastic buckling is likely to occur?"

Currently, the best assumption is to assume the word "elastic" was intended to be inserted by the instructor. Also, we might assume the instructor wants you to pretend the column is theoretically perfectly straight.

Therefore, using these assumptions, your current answer, 5.93 m, is correct, or close to correct. (You rounded numbers in your intermediate calculations too much. If you had maintained at least four significant digits throughout all your calculations, which you should have done, then you would have obtained 5.94 m.) Also, you should use 140 MPa, not 141 MPa.
charger9198
#3
Jan5-12, 10:51 AM
P: 60
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

nvn
#4
Jan5-12, 11:02 AM
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Column buckling answer confirmation

charger9198: Once again, you rounded your numbers too much; therefore, your answer is slightly inaccurate. And your units are wrong. GPa is a unit of pressure or stress, not force.
charger9198
#5
Jan5-12, 12:14 PM
P: 60
Thanks nvn I have re wrote the full thing using the full answers and it looks a lot more clear now..

F=3.07875 x 10^-6
nvn
#6
Jan5-12, 09:19 PM
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charger9198: Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

Therefore, your final answer in post 5 should be rounded to three significant digits. You listed six significant digits in your final answer in post 5.

Also, your power (exponent) in post 5 is wrong. And no units are listed.
charger9198
#7
Jan6-12, 03:29 AM
P: 60
nvn: Thanks;

F = 3.078 x 10^-4 GN
nvn
#8
Jan6-12, 08:44 AM
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charger9198: It appears you might not know how to round numbers. When rounding numbers, 0 to less than 5 is truncated; 5 to 9 is rounded to the next higher digit.

E.g., 3.07875, rounded to four significant digits, is 3.079, not 3.078. Or, 3.07875, rounded to three significant digits, is 3.08.

Therefore, your final answer would be F = 3.08 x 10^-4 GN.

However, it is probably clearer to instead write 0.308e-3 GN, or 308e3 N, or 308 000 N, or 308 kN.

Here are some other, easier ways you can write the same number; 3.08e-4 GN, 0.308e-3 GN, 0.308 MN, 308 kN, 308e3 N, 0.308e6 N, 308 000 N.
pongo38
#9
Jan6-12, 03:31 PM
P: 696
Charger: you wrote "This means, if the column is longer than 5.93 meters, it will buckle due to slenderness. Otherwise, if it is shorter than 5.93 meters, it will not displace out of plane and fail in yielding."

Columns under load can displace out of plane without it being buckling. The test is whether on removal of the load, the column returns to its original position. One definition of buckling load is: If you displace the loaded column with a perturbation and it remains displaced, then it has buckled. NVN's advice is always good, but just don't believe that the results of the calculation are true to the number of significant figures you are using. In my experience you would be lucky if two sig figs were to correctly predict a column buckling load, there being very many variables that the theory ignores.
charger9198
#10
Jan6-12, 04:33 PM
P: 60
Thanks for all the help Pongo38 and NVN. chemistry is more of my field but my basic math principles are far from adequate. I'm trying to take all advice on board and apply to successfully gain better physics knowledge..thanks again
Nobby3601
#11
Jan18-13, 09:44 AM
P: 2
I am using the formula L= (SQRT) ((∏^2)*E*I)/F*k^2 and not coming out with anything like this answer. Would you mind breaking it down?

Thanks
nvn
#12
Jan19-13, 08:22 PM
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Nobby3601: You must use consistent units. Consistent units are, e.g., N, m, Pa.
Nobby3601
#13
Jan24-13, 01:56 PM
P: 2
Thanks for the help all. Got there in the end!
Carlo2986
#14
Feb8-13, 06:01 AM
P: 9
Your answer to 'At what load would you expect failure to occur?', Could'nt you have used the equation:

Fc = σcA

Fc being the axial load
σc being the critical stress which will cause buckling
A being the cross section area

It gives the same answer and is a lot shorter.

Or is it not as easy as that?
Rafeng404
#15
Mar28-13, 11:41 AM
P: 9
Hi all, I also have this question as part of my assignment and I was getting a different answer.. I was wondering how you know what shape the column is from the question? You can get rectangular columns too, can't you? :/
SteamKing
#16
Mar28-13, 12:00 PM
Emeritus
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HW Helper
Thanks
PF Gold
P: 6,334
The dimensions from the problem statement attachment list '80 mm dia.' for the outer dimension and '60 mm dia.' for the inner dimension. 'dia.' or diameter implies a circle rather than a square or rectangle.
Rafeng404
#17
Mar28-13, 12:06 PM
P: 9
So basically I need to be an English professor as well as an engineer to get this HNC... Haha. Thank you for your fast response, much appreciated.
Rafeng404
#18
Mar29-13, 08:25 AM
P: 9
Hello again, I have don't all ,y calculations and I have got the same as everyone else.. I am doing this distance learning and the books aren't very good. I'm struggling to see what values I should be comparing to determine what the mode of failure would be, ie if its buckling how have I proved it mathematically..

Not sure what all this has to to with electrical and electronic engineering!,, rant over haha.


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