Maximum torsion shear stress in cylinder

Question

I am trying to figure out am I doing this calculation correctly.

To get the maximum shear stress for a solid cylindrical pipe I need two formulas:

Moment of inertia = pi/2*r^4

Not sure what this formula is called but its:

max = T*r/ Moment of inertia

I tried to calculate these numbers for a solid shaft that is 20mm in diameter with a torque of 1255nm.

1255*0.02/((pi/2)*0.02^4)

Pascals = 99869726.7902 MegaPascals = 99.87

The material that I have has a yield strength of 490 MPa.

Is the calculation saying that even at 20mm radius the stress will be below the 490MPa that would cause this pipe fail elastically?

Answer 1

I have not repeated your maths, but the formulae are shown in the image. If you have the numbers correct the you have a shaft that will support nearly 5 times the load you are applying...

Source : http://www.engineersedge.com/material_science/shear_stress_in_shafts_13117.htm where you can enter your values and have them confirmed.

Answer 2

Your calculation is almost correct. The only problem is that you confused diameter with radius in the calculation.

If the diameter is 20[mm] (i.e. radius =10mm) then the torsional stress is 798.95 MPa.

If the diameter is 40[mm] (i.e. radius =20mm) then the torsional stress is 99.87 MPa.

Please notice that although you have unwittingly only doubled the diameter, the stresses decreased 8 fold ($2^3$), due to the non linearity of the problem. So, you can easily get it wrong.