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Beam under Biaxial transverse load - Von Mises Stress calculation

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Astrostructural

Structural
Mar 26, 2020
2
Hi everyone,
It might seem like too simple a Question for many but I would like to calculate the Von Mises stress for a beam in biaxial bending. The bending stresses run along the beam axis so they can be added up at the Cross section Corners, but I would like to find the maximum Von mises stress including the components of shear stress as well. There is also an axial load on the beam, that generates its own normal stress. I have attached a diagram for explanantion.

At which Point on the Body should the Von mises stress be calculated to obtain a maximum value?
My bending stresses are 303 MPa for Sigma bxk, and 408 MPa for Sigma bxs(Qz,) and maximum txz (Qz) is 16 MPa, and maximum txy is 132 MPa.

Thank you all in advance

Regards
Asutosh
beam_vzll47.png
 
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Concrete beam ?

doubly symmetric beam ? (so the orthogonal axes are also the principal axes)

what's the stress state in the corners of the beam ? sum of the bending and axial normal stresses with no shear ?

another day in paradise, or is paradise one day closer ?
 
You need to determine through trial and error where the maximum occurs. There's no magic answer here. Start with the obvious points where the respective maximums occur and go from there.

If this were practical steel design, then von mises is not relevant as some degree of plasticity is acceptable and encouraged by codes when undertaking design checks.
 
I have made the stress distribution equations, guess I will have to go by trial and error where the maximum lies.
Just wanted to know what the engineering community does for such cases.
Thanks :)
 
The vM stress formula is basically a sum of the squareds thing like pythagoras so usually the peak stress in one dimension will govern. Generally speaking we only use von Mises for FEA of shell and solid structures, though occasionally for bending in a pin. Also the new version of Strand7 does vM calculations in beam elements.
 
what sort of beam do you have ? metal or concrete ?

if metal, why are you bothering with von Mises ? sum the normal stresses from bending and axial, if you really want, look at principal stress at a section with less than maximum bending stress (probably not critical, or significantly different to the peak bending stress. for max shear, on CL/centroid, make a principal stress from axial normal and bending shear stress. Von Mises combines different (orthogonal) principal stresses together, something you get in a prismatic section … you have a prismatic metal beam ??

another day in paradise, or is paradise one day closer ?
 
Calculate the von Misses stress at all critical locations of the examined cross section, ie at 4 corners and the center as well and take the absmax.
 
rb1957 said:
Von Mises combines different (orthogonal) principal stresses together

Agreed that von Mises is usually not useful in beam design in a practical sense. von Mises to me is most useful because it combines shear and axial stress and gives you a stress which is comparable to the axial yield strength. We normally only use principal stresses for brittle materials like glass.
 
von Mises may not be common for typical wide flange beam design (shear stresses on webs usually considered independent of the flange, and biaxial tension is rare in structural steel construction), but vm is the right solution for thick steel plates under biaxial or triaxial stress.
I would say the problem proposed is a bit funny; more theoretical than practical.

 
we're assuming a metal beam (unlike his sketch), and he hasn't answered by question (twice) concrete or metal ? important as the approach is different.

for me, vm for this example collapses to max principal which collapses to max normal stress (bending + axial). why make things more complicated than they need to be ?

this would probably not be true for a concrete beam.

another day in paradise, or is paradise one day closer ?
 
rb1957: do you ever use principal stresses for ductile metals? I mean other than the case of a uniaxial stress. Maybe some sort of post yield failure criteria?
 
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