Shear stress calculations and torsion - need clarification on something

[BM44]

Greetings,

I am studying shear stresses and torsion. Assuming the specimen is a cylindrical solid shaft, I've had multiple sources (professors, etc) state that the outer-most circumference experiences the most shear stress; which they further state that one would expect a cylindrical shaft to have a failure in shear from the outside dimension first and propagate inward.

I am not understanding this when I apply the math for calculating shear stresses as it relates to specimen diameter, applied torque, etc.

For instance, my formula is:

stress = (torque/J)*r where J is the 2nd moment of inertia (solid cylinders) that equates to (pi*d^4)/32, where d is diameter, r is radius of specimen

I plot this in excel to try and understand the relationship between radius and shear stress. When I plot a constant torque and a varying radius, the shear stress decreases as radius increases. That makes sense. Stress would be lower as the shaft gets larger (its stiffer).

So, I don't quite understand what my sources mean when they say a cylindrical component experiences the most shear stress at a larger diameter. Could anyone explain this concept to me and perhaps what I am missing?

Thanks,
Regards

 

[CWB1]

Per your thought exercise and first equation:

torque = constant
J = constant for a given shaft
r = point at which stress is measured within the shaft
stress = torque/J*r
stress = c1/c2*r ---> stress = c*r ---> stress increases with diameter, not conversely.
c*0 = 0 stress at center axis. c*.000000000000001 = tiny stress near center c*1M = big stress far from center

 

[desertfox]

BM44

For any given shaft size the shear stress is at a maximum at the outer surface.
So if you have 10Nm of torque applied to a 100mm diameter shaft the maximum torque will occur on the outer diameter. Now apply the same torque to a 200mm diameter shaft, the stress will still be greater on the outer diameter but it will be a lot less than the stress on the 100mm diameter.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[3DDave]

Very simply - the greater the deformation the greater the shear and the greater the shear stress. Along the center of a rod there is no deformation from torsion; the atoms stay where they are, for small amounts of twist. Deformation increases with the distance outward for a given angle of twist.

The total deformation has to be enough to absorb the work done in twisting the shaft. The smaller the diameter is the smaller the amount of material, so larger stresses are required. And it gets worse because the smaller the diameter the greater the angle the torque is applied though so the greater the amount of work that is done on the shaft.

 

[dhengr]

BM44:
Dig out a couple of good Strength of Materials, Theory of Elasticity, or Machine Design textbooks, and do a little reading on shear stress due to torsion and on shafts of various shapes. Learning it this way, it will stick with you much longer, and the texts will provide much more detail than we have the time for here on E-Tips.

 

[BM44]

Hi all,

thanks for the replies. The theory that the atoms on the outside of the shaft experience the most deformation is easily understood. I am more so trying to understand this mathematically with my calculations.

CWB1 walked through the math. The problem is, I don't understand why the 2nd moment of inertia is constant. When this J value is a function of radius (or diameter). For solid shaft, J = (pi*d^4)/32

Where (to my understanding) d is the diameter of the shaft which we are interested in measuring the shear stress at, not, however, the entire diameter of the shaft. But as CWB1 points out, J is in fact the diameter of the shaft. That must be where my math is wrong. Any comments?

Thanks again

 

[3DDave]

J is the integral of the contributions of individual layers. If you want to find the contribution of each layer, take the first derivative with respect to radius.

 

[CWB1]

You wanted proof the stress is highest on the outer surface of a shaft vs the inside. My method was to simply evaluate the stress at various distances (r) from the center axis of any single given shaft. Via my method the stiffness of the shaft (J/area moment/2nd moment of inertia) remains constant bc I'm not changing anything about the shaft, J remains constant because diameter remains constant bc we are only looking at a single shaft, not multiple shafts.

Via your method you are comparing the stress at the outer surface of multiple shafts. You found that as diameter increases so does stiffness (J), and stiffer shafts will have lower stress on their outer surface - all very true. Unfortunately this does not tell you anything about the stress level below the outer surface of the shaft, varying shaft diameter and consequently stiffness skews the math significantly.