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New Member
Posts: 2
Registered: ‎10-28-2010

Beam "Worst Stress" for Circular Cross-Sections

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06-11-2013 10:28 AM

Hi all,


I do a lot of beam-based analysis of structures with circular tube cross-sections. As I've been thinking about the definition of "worst stress" for beams, it seems that this may not be totally accurate for a circular or hollow circular cross-sections.


The help documentation indicates that mathematical worst-stress could be at a non-existent location for asymmetrical beams. For a round beam, despite it being symmetrical, it seems that the worst stress reported may not be exactly correct either. It would seem that in-between the '2' and '3' axes on a circular cross-section (say at a 45° angle) that the stress would be somewhere between the values of |P/A| + |M2/S2| and |P/A| + |M3/S3|.


Does this make sense, or am I thinking about this incorrectly?


I'd appreciate any feedback.



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*Expert Elite*
Posts: 575
Registered: ‎08-30-2012

Re: Beam "Worst Stress" for Circular Cross-Sections

06-12-2013 07:06 AM in reply to: StreamlineDesigns

I agree with your thinking Adam. The highest stress on the surface of a circle or tube is really lower than |M2/S2| + |M3/S3|.


So, if you want to have some fun today :smileywink:, you can check my math. The stress at a point depends on the distance from the neutral axis, it seems reasonable (but perhaps not correct!) that the

          worse stress = stress2*cosine + stress3*sine.

where the angle is measured from one of the axes. The maximum result occurs at the angle where the derivative equals zero, or

         change in worse stress/change in angle = -stress2*sine + stress3*cosine = 0

So the angle depends on the values for the maximum stress about axes 2 and 3. If we look at the ratio stress3/stress2 = X, the equation simplifies to

       -stress2*sine + (X*stress2)*cosine = 0

        stress2*(X*cosine - sine) = 0

Of course, stress2 could be 0, but the more interesting result is when

        X*cosine - sine = 0

        X*cosine = sine

        X = sine/cosine = tangent

So by calculating a few ratios of X, the angle where the maximum is located can be calculated, which can then be used to calculate the maximum worse stress of stress2*(cosine+X*sine). Here are a few examples:


X =

stress3/stress2    Angle    worse stress/stress2   software's (|M2/S2| + |M3/S3|)/stress2

0                                   0                        1                                                         1

0.578                         30                       1.16                                                   1.56

1                                 45                       1.41                                                    2

1.5                             56.3                     1.80                                                  2.5

2                                63.4                     2.24                                                   3

3                                71.6                    3.16                                                    4


If you do not want to use the conservative values from the software, you might be able to write a custom formula to do the above calculation. That would be neat to see if it works!

John Holtz, PE
Mechanical Engineer
Pittsburgh, PA

Simulation Mechanical user since Dec 1997
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New Member
Posts: 2
Registered: ‎10-28-2010

Re: Beam "Worst Stress" for Circular Cross-Sections

06-12-2013 07:32 AM in reply to: AstroJohnPE
Thanks for the thoughtful response, John. I'll have to spend some time thinking about this.
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