bending stress equation 1

[BrianE22]

I see for 3 point bending of plastics the bending stress is given as:

3PL/2bd^2.

For metals, the bending stress (from Mc/I) would be:

3PL/bd^2.

So they differ by a factor of 2. Is this correct? Why the difference in calculated stress for plastics as compared to metals?

 

[btrueblood]

The plastic section is assumed to have constant stress across each half section, vs. a triangular distribution for elastic stresses in metals.

 

[rickfischer51]

Brian, I think maybe the use of the term plastic to mean yielding and plastic to mean polymer may be causing some confusion. The maximum bending stress due to load P on a beam L long, b wide and d deep for a center loaded simple beam is Mc/I = (PL/4)* (d/2) / (1/12(b*d^3)) = 3PL/2bd^2. This is true for all materials. This gives the maximum bending stress at the top and bottom surfaces of the beam midway between the supports. When the stress so calculated is equal to the yield strength, the load P represents the onset of yielding. The stress distrubution is triangular, as btrueblood stated. If the material is assumed to be elastic, perfectly plastic, increasing P does not increase the stress, but it does change the stress distribution. At a load equal to 1.5P, the stress distribution is rectangular. This results in the so-called "plastic moment," where yielding occurs across the entire section. This is all based on assumed stress distributions, geometry, and static equilibrium, and is independent of material.



Rick Fischer
Principal Engineer
Argonne National Laboratory

 

[BrianE22]

Rick -

You're correct. I wasn't very careful reading the info I was finding on the web. Thanks to the both of you!