Discussion Groups

Simulation Mechanical and Multiphysics

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

Beam "Worst Stress" for Circular Cross-Sections

241 Views, 2 Replies
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.

 

Adam

Please use plain text.
Valued Mentor
AstroJohnPE
Posts: 494
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

16 years experience with Simulation Mechanical
Please use plain text.
New Member
StreamlineDesigns
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.
Please use plain text.