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Bending/Shear Stress Elastic vs Plastic Deformation

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njm789

Automotive
Joined
Oct 9, 2019
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US
When calculating the max shear and bending stress in a beam using Sigma b = My/I and Tau = VQ/It, will the calculated stress be accurate past the yield stress for say a ductile steel or since the stress is distributed across more elements as it yields will the elements over yield experience less than the calculated stress. I am trying to compare hand calcs vs FEA. The part I am designing is able to exceed yield and still function safely. Thanks for the help!
 
Hi njm89

The answer is no, the formula are only valid for elastic stresses

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein
 
I am curious how any rigid part could still fulfill its job after surpassing the yield point. I know surpassing yield doesn't automatically mean immediate fracture, but it does mean the part can no longer safely withstand any load. Just curious.
 
A part that is loaded/unloaded once and then discarded.
 
We do that in seismic design.
 

The answer is YES.. the stresses at any point can not be higher than ULTIMATE TENSILE STRENGTH. The Euler- Bernoulli beam theory ( small deflection, plane sections remain plane, and linear distribution of stress ) specially linear distribution of stress is valid up to yield strength is reached.

The following picture depicts the stress strain distribution for elastic , partial plastic and full plastic states.

Plastic-Theory-0002_fn8fkm.png



Notice the similarity of typical stress -strain curve of a ductile material and stress distribution of a bending element.

Stress-Strain_Diagram_qq8otw.jpg
 
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