Second moment opf area/buckling

I'm an optical/electronic engineer/physicist trying to do a structural calculation for experimental equipment design.

Which is probably not a good idea, tho' no one gets hurt if I get it wrong!

I need to calculate the buckling instability limit of thin columns.

I'm using Euler's formula as a start & comparing to COMSOL/FEMLAB calculations. I want to add some complications later when I understand this case!

The formula involves the second moment of area of the shape, initially a circle.

*Several* references on the web give it as (pi*D^4)/64 whilst one gives /32 When I do it from the definition of second moment of area I get /32

And /32 gives a result which agrees well with FEMLAB/COMSOL, whereas /64 is a factor of two wrong - but that might be luck, I dont think I should

*expect* very good agreement should I?

So.... Is it /32 or /64? And if /64 what schoolboy mistake am I making? Integral 0-2Pi, 0-R of (r^2) r dr dtheta seems easy enough! Too much Christmas CH3CH2OH perhaps lingering.........

And how good a guide is Eulers formula, how good an agreement might I expect with COMSOL/FEMLAB? (1cm diameter 20 cm long, bottom constrained x,y,z & top constrained x,y loaded z. Default 316 properties. Side load in x of 10^4N/m^2 to break the symmetry. Nonlinear solver, large deformations enabled. Fails to converge above about 6.5e8N/m^2 in z, which I take as the buckling limit. Just below that slowly converges to deflections of several mm near the centre.)

Harvey

Reply to
Harvey Rutt
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The URL below is a very concise summary of Euler style buckling, which occurs at lower values than the yield value for the column.

Encased ends to which you refer, lead to more buckling resistance than the pinned ends mentioned in this URL.

For a circular column: I = pi * r^4/4 or I = pi * D^4/64

So for the Euler formula where D = 0.01m., L = 0.2 m., E = 210GPa, pi = 3.142

The critical force for buckling, Fcrit = pi^2*E*I / L^2 = pi^2* E* pi * D^4 / (64 * L^2) = pi^3 * E * D^4 / (64 * L^2) = pi^3 * 2.10E11 * (0.1E-1)^4 / ( 64 * 0.2^2) = 31 * 2.1E11 * 1.56E-10 = 25430N

Euler supposes that the column fails by buckling at LESS than the compressive strength of the material.

You take the critical load at the buckling limit to be 51kN A factor of two greater, as you note. I suspect the factor is due to the more robust end fixings or even more likely, due to the relative shortness of the column. A short column fails by a combination of compression and eccentric loading, which allows part of the surface to yield.

Before there were finite element methods, there was Marks Handbook. So, for comparison, I show Rankine's formula for short columns from Marks Handbook

Fcrit = Sc * A / (1 + K (l^2 / rg^2))

Where Sc is compressive strength = (say) 860MPa = 8.6E8 Pa A is cross section area = 7.85E-5 m^2 K is an empirical coefficient: for steel, both ends fixed = 1/25000 l length = 0.2 m rg radius of gyration = sqroot(I/A) or =sqrt( pi*r^4/ (4 * pi * r^2)) = r/2 = 2.5E-3 m r = 0.005m = 5E-3 m

So, by Rankine, Fcrit = 8.6E8 * 7.85E-5 / ( 1 + 4E-5 * ( 0.2^2 / (2.5E-3)^2)) = 53750 N

This value is reasonably close to the value you obtained by FE means. The difference is probably associated with part of the material going to yield. I suppose the yield value you used was in the region of 810 MPa ??

Brian Whatcott Altus OK

Reply to
Brian Whatcott

The polar moment of inertia (torsion), J = Ix + Iy = 2 * I = pi * D^4/32

That's probably where the /32 comes from.

If yielding occurs use the tangent modulus, Et.

Reply to
Jeff Finlayson

If anyone is interested, it was a 'school boy error' in getting the second moment. I forgot the cos(theta) squared. But I dont seem to be the only person to make that mistake :-)

Harvey

Reply to
Harvey Rutt

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