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Euler Beam - What is total displacement?

Posted Jan 25, 2014, 3:52 p.m. EST Structural Mechanics Version 4.4 2 Replies

John
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Hi, I just ran a simple cantilevered beam supported at one end and output the displacements "u", "v" and compared them to standard analytic formula for total displacement.

z u (m) v (m) u+v (m) -z^2*beam.rho*beam.area*g_const/24/beam.E/beam.Izz*(z^2+6*(5[m])^2-4*5[m]*z) (m)
0 0 0 0 0
0.5 1.03E-04 -3.66E-04 -2.64E-04 -2.64E-04
1 3.83E-04 -0.001368097 -9.85E-04 -9.85E-04
1.5 8.04E-04 -0.002872669 -0.002068407 -0.002068533
2 0.001333305 -0.004762371 -0.003429067 -0.003429224
2.5 0.001941658 -0.006935375 -0.004993717 -0.004993902
3 0.0026052 -0.009305517 -0.006700317 -0.006700523
3.5 0.003304193 -0.011802297 -0.008498104 -0.008498328
4 0.004023288 -0.014370886 -0.010347598 -0.010347834
4.5 0.00475152 -0.016972117 -0.012220597 -0.012220841
5 0.00548231 -0.019582491 -0.014100182 -0.014100428

So as you can see, absolutely gorgeous agreement between u+v and the standard analytic formula.

But why is u+v the total displacement? I was thinking it should be sqrt(u^2+v^2).

But that just gets really to my next problem. Why is u not zero anyway? I've posted the MPH.

Thanks, John

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2 Replies Last Post Jan 27, 2014, 11:18 a.m. EST
Henrik Sönnerlind COMSOL Employee

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Posted: 1 decade ago Jan 27, 2014, 7:30 a.m. EST
Hi,

You have different stiffness (Iyy and Izz) in the two principal directions. The principal axes are directed at 45 degrees from your loading direction, so the bending will be 'skewed'. To get the deflection in the same direction as the load, you could either align the principal axes of the section with the loading direction (using the Section Orientation subnode to Cross Section Data), or use an 'isotropic' cross section.

And yes, you are right about total displacement. The built in variable beam.disp is defined as sqrt(u^2+v^2+w^2).

Regards,
Henrik
Hi, You have different stiffness (Iyy and Izz) in the two principal directions. The principal axes are directed at 45 degrees from your loading direction, so the bending will be 'skewed'. To get the deflection in the same direction as the load, you could either align the principal axes of the section with the loading direction (using the Section Orientation subnode to Cross Section Data), or use an 'isotropic' cross section. And yes, you are right about total displacement. The built in variable beam.disp is defined as sqrt(u^2+v^2+w^2). Regards, Henrik
John
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Posted: 1 decade ago Jan 27, 2014, 11:18 a.m. EST
Duh :-)

Thanks Henrik.

Attached shows the anticipated agreement.
Duh :-) Thanks Henrik. Attached shows the anticipated agreement.

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