Shear stress soo small?! WHY?! HELP!?

Question

Hey,

What's the engineering reasoning behind why sometimes shear stress is neglected if it is significantly smaller than,say, bending stress. So I guess what I am asking is that why does shear stress come out soo small?

like for example, a bike pedal with crank has 300lbs vertical only force applied to it, and made a cut and analyzed it. The bending stress due to bending moment was 50kpsi while shear stress was 1.7 kpsi.

We also made that assumption when I took streghts of materials for many of the cases. I still don't know why it is safe to make that assumption.

Thanks

Answer 1

It's not safe to always make the assumption that shear stresses are small. They probably did it in class to simply the problem and get to the point that they want to teach instead of spending time transforming stresses.

When you do a calculation of stress state on an object, the stress state could have large shear stresses or or could have very little to none. It all depends on how the coordinate axes or your calculation are oriented compared to the principal directions of the stress state. When the coordinate axes are such that they are oriented along the principal directions, shear stresses will disappear and all you get are the three principal stresses along the three principal directions. When you rotate your frame of reference off the principal axes, the stresses described in the rotated frame will include shear stresses.

The reason that analysis of material and evaluating whether the stress state will exceed the strength of the material are done assuming no to low shear is because the simpler engineering failure criteria for materials are all described in principal stresses. To make an evaluation whether a material has exceeded its strength requires you to do a step of stress transformation to transform the stress state into the principal axes. Otherwise, you can't use the failure criteria. When the teacher wants to simply illustrate the failure criteria, he's not going to want to spend 20 minutes transforming stress states. So he's going to pick an object and a loading such that the intuitive coordinate system ends up to be the principal directions. That way, the stresses he gets are already in the principal direction and he doesn't have to transform. Since the streses he gets are already in the principal directions, it going to end up with very low or no shear stress. That's what the principal directions do for you: it finds you a coordinate reference system where the shear stresses disappear.

As far as the example you have, the axes you used to calculate stresses probably are close to the principal directions as well. That's why shear stresses are low. Take that same example, except calculate the stresses along directions that are rotated 45 degrees from the axes you have now. Then recalculate the stress state. You'll find that shear stresses will appear when stresses are stated in the new coordinate system.

If you haven't at least studied two dimensional Mohr's circle and stress transformation, this is going to be a confusion topic. Look up Mohr's circle to help you understand why and how you can rotate axes to and from the principal directions and make shear stresses appear and disappear according to your axes.