# Maximum torsion shear stress in cylinder

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?

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.

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.