I'm doing a static analysis on a uniformly loaded cantilever beam in ANSYS APDL. I'm using BEAM188 with all the default options, so it uses linear form. for the element behaviour, which I think means the bending moments and tensions are linear within the element.

I want to plot the bending moments along the beam with an area contour, for that I'm using:

  • PLLS,MI,MJ,1,0

This are the axial stresses due to bending, it looks very similar to the bending moment, it shows discontinuities between elements and for each element it shows the same value on both nodes.

enter image description here

Why does MI=MJ? Shouldn't there be a linear variation of the moment in the elements? Does it make sense for it to be discontinuous between elements? Or is this just a bad plot and I should plot the values on the nodes I get with PRESOL,M as opposed the plot each I and J node values for each element.

The results overall seem good, but I'm not sure if this is a correct representation of the behaviour of the elements.

I'm sorry for such a basic question. I hope I wasn't too confusing.

  • $\begingroup$ Hi Carol, welcome to engineering.SE. It is difficult to interpret your question without being able to see the plots. Could you please add them to your question? $\endgroup$ Jun 24, 2015 at 19:03
  • $\begingroup$ !the best i could do, I don't know how to upload images here. Ups, those are actual the stress due to bending, but it has the same problem, naturally. $\endgroup$
    – Carol
    Jun 24, 2015 at 19:37
  • $\begingroup$ You can add images directly to your post. I've added this one for you, but you may want to add a description. $\endgroup$ Jun 24, 2015 at 19:58
  • $\begingroup$ @Carol - I get it now. Essentially, in between each element, it adds the $\frac{wL^2}{12}$, which replaces your nice continuous loaded beam to a beam which continues to add a bunch of discontinuous moments. Like I said, it only adds new loads at each node - so in between, the load will have constant moment. $\endgroup$
    – Mark
    Jun 24, 2015 at 20:30
  • $\begingroup$ @Mark - Yeah, that must be it. I guess it doesn't treat the SFBEAM load the way I thought. Thank you so much for your help/time! Is there anything I should do to this question now? $\endgroup$
    – Carol
    Jun 24, 2015 at 20:46

2 Answers 2


Short Answer: The finite element method means uniform loads are not treated as uniform - and varying moments are not treated as varying.

Long Answer:

This is an interesting highlight of the differences between the finite element method and the analytical method. A uniformly distributed beam can't be treated uniformly on a finite element. Instead, it breaks up the load into two separate loads on each end, as it only analyzes each node.

Breaking up the beam

Between each of the nodes, it treats the load as constant. Only at each node, does it perform a new "sum of the forces and moments = 0", and spits out the brand new moments - hence the disjointed graph

Thus, it will treat shear loads as constant, moment loads as linear, rotations as quadratic, and displacements as cubic:

Beam comparison between single element and theory

Meanwhile, analytical theory will treat shear loads as linear, moments as quadratic, rotations as cubic, and displacements as quartic. This is why you need so many elements. A disconnected cubic will, with enough elements, resemble a quartic - just like a flight of stairs, with enough stairs, resembles an inclined plane.

  • $\begingroup$ Thank you for answering! I guess my question is more.. If on both ends of the finite element you have the same moment, isn't it being treated as constant instead of linear? $\endgroup$
    – Carol
    Jun 24, 2015 at 18:51
  • $\begingroup$ Well, the exact change from uniform load is to put $\frac{wL^2}{12}$ clockwise moment on the right and counterclockwise moment on the left. Are you sure the sign didn't flip? $\endgroup$
    – Mark
    Jun 24, 2015 at 19:04
  • $\begingroup$ Yeah, if i understood you correctly.. If I use one finite element alone on the cantilever beam, shouldn't I get some value on the wall(the "I" node) and a zero bending moment on the free end(the "J" node)? They're the same. $\endgroup$
    – Carol
    Jun 24, 2015 at 19:11
  • $\begingroup$ Another fun one with using FEA - the moments at the end rarely have 0 bending moment. I've included an image from my book that would be useful in the answer. I don't think that they should be the same though, based upon this image. $\endgroup$
    – Mark
    Jun 24, 2015 at 19:14
  • $\begingroup$ Just for the sake of clarity, you could have finite elements that treat distributed loads as opposed to treating it as point loads on nodes, but I guess not this one. :) $\endgroup$
    – Carol
    Jun 24, 2015 at 21:09

You might have got the answer if not, come to your direct question for plotting actual bending moment. Just choose the element option key for "quadratic" instead of linear and re run the problem and plot the BMD.


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