Bending/Shear Stress Elastic vs Plastic Deformation

[njm789]

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!

 

[desertfox]

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

 

[Jboggs]

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.

 

[njm789]

A part that is loaded/unloaded once and then discarded.

 

[Jboggs]

Okee dokee!

 

[BUGGAR]

We do that in seismic design.

 

[HTURKAK]

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.

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