COLUMN EFFECTIVE LENGTH 2

[SAIL3]

I have a column with a distributed uniform axial load
for the full length.
The column is pinned at both ends.
My question is how would one calculate the effective length
using the total axial load.
Or using the actual length, what would be an equivalent
axial load to use when designing the column.
Tanks for your help

 

[apsix]

The question doesn't make sense, is this a homework question?

Effective length is equal to actual length for a pin-pin ended column.

 

[SAIL3]

Not a student..at 65years old, I'm well done with college.
The column is already designed.
Designing the col. for the full total axial load as if
it were applied at the top and using the full length as
the effective length would seem to me overly conservative.
Suppose the total axial load was applied at mid-height of
the column, what effective length to use?
Thanks again

 

[hokie66]

Maybe if you told us how an axial load is applied in a distributed manner, the question would make more sense.

 

[csd72]

I understand where you are coming from as I asked a similar question as a fairly young engineer.

The answer that I got from those wiser than me was that it was best to treat it as though the full length was the length for buckling under the full axial load.

But as others have said, it would be useful to know the situation.

 

[Lion06]

Here is a paper on calculating effective length of a column with an intermediate axial load. I thought I had one that applied exactly to your situation, but apparently not.

 

[IDS]

The link below is about buckling of columns under self weight, but buckling under uniformly distributed load is exactly the same problem. The link includes a formula for critical load, and a spreadsheet doing the calculation with a numerical analysis, which gives the same answer.

http://newtonexcelbach.wordpress.com/2010/06/27/more-on-buckling-columns/

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[SAIL3]

Thanks for all the replies and help.
Intuitively, without checking any backup theory I would
tend to use the following:
For a concentrated load at mid-height of column,I would
assume an effective length of approx. 0.75xTotal length.
For a uniformly distributed axial load, I would assume
an effective length of approx. 0.6xTotal length.
I will check the references suggested, and do appreciate
them.

 

[csd72]

Sail3, I dont think the udl makes that much difference, I am sure that it came out as something like 80 something percent but that is about a decade ago so...

 

[msquared48]

Kinda depends on whether or not the intermediate loading points down the length of the column provide lateral stability too as to that the unsupported length actually is.

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[BAretired]

The problem is solved approximately in "Theory of Elastic Stability" by Timoshenko and Gere - 2nd edition - p. 107

BA

 

[IDS]

The critical length of a free ended column subject to an end load F is given by the Euler buckling equation as:

L(e) = pi/2*(EI/F)^.5

The critical length of a column under self weight, of total weight W, is given by Timoshenko's equation as:

L(d) = 3B/2(EI/W)^.5

where B is the first zero of the Bessel function of the first kind of order -1/3, which is equal to 1.86635085876156

http://en.wikipedia.org/wiki/Buckling

The effective length factor for the column under self weight (or uniformly distributed load) is therefore:

L(e)/L(d) = pi / (3B) = 0.561

This assumes that the applied load is equivalent to self weight, i.e. it moves with the column and does not provide any restraint.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[SAIL3]

Thanks for all the responses, they have been a great help.