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| Jeff Finlayson |
 Posted: Thu Dec 29, 2005 3:45 pm |
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Guest
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Brian Whatcott wrote:
[quote:ca93e7008b]Harvey Rutt wrote:
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
Is it /32 or /64? And if /64 what schoolboy mistake am I making?
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
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.
http://www.twi.co.uk/j32k/protected/band_3/ksgs001.html
For a circular column:
I = pi * r^4/4 or I = pi * D^4/64
[/quote:ca93e7008b]
The polar moment of inertia (torsion), J = Ix + Iy = 2 * I
= pi * D^4/32
That's probably where the /32 comes from.
[quote:ca93e7008b]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 ??
[/quote:ca93e7008b]
If yielding occurs use the tangent modulus, Et. |
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| Harvey Rutt |
| Posted: Fri Dec 30, 2005 5:56 am |
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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 |
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