I did some calculations to find out the moment of inertia and physical qualities of some metal.

I am using typical medium carbon steel.

I'll use a density of 7.8 g/m^3 until I get better numbers. There's a solid circular shaft that's 1000mm long The diameter of this shaft is 40mm

This gives me: Polar moment of area 251327 mm^4

The next part is the motor.

This will be a shaft connected to a 15HP 1200RPM motor but I'll be adding a gearbox to bring the RPM down to about 85RPM this will give me about 1256 newtons/meter of force.

I am a bit lost at this junction.

Do I need to convert the newtons to MPa or the Polar moment to some other units?

How do I compare the 1256 newtons to the amount of force this cylinder can withstand 251327 mm^4

would this pipe bend or break under the max torque load?

How can I relate 1256N.m to 251327mm^4 to see if the pipe would break?


2 Answers 2


Andrew has provided some good information with regard the formula for shear stress.

What I disagree with is the standard yield strength ($\sigma_y$) values for a medium carbon steel. I think Andrew has forwarded UTS values. For this I think you should be working in the $275MPa$ for low strength, to $450MPa$ for high strength range, with most structural steel being $355MPa$.

To add apply this to your query, you have:

  • $T = 1256 N m$ (torsional moment)
  • $r = 0.02 m$ (radius)
  • $J = 25.133 cm^4$ (polar moment of area)
  • $\tau_{allow} = \frac{\sigma_y}{2} $

So, if the calculation results in $\tau_{max} > \tau_{allow}$ then you know you are in trouble regarding any deformations / damage to the shaft.

  • $\begingroup$ I agree with you on the standard yield strength; this is the actual steal that I am working with: matweb.com/search/… it has a yield strength of 490MPa. I thought it would be wise subtract 25MPa which by rounding up to 500MPa and taking 5%; maybe using 275 would definitely keep me in the green. $\endgroup$ Aug 24, 2017 at 14:34

If you don't want plastic deformation to occur in the shaft you need to make sure that you don't exceed the yield strength $\sigma_y$ of the material throughout the cross-section of the material. To calculate the maximum elastic shear stress in the (solid circular) cross-section you can use the following formula: $$ \tau_{max}=\frac{Tr}{J} $$ $ T\quad…\quad $applied Torsional Moment
$r \quad…\quad$Radius
$J=\frac{\pi}{2}r^4 \quad…\quad $Polar Moment of Area

Now, as discussed in this post, for ductile materials subjected to pure shear stress $t_{y}\approx\frac{\sigma_y}{2}$. For medium carbon steel $\sigma_y$ is in the range of $500-700$ $MPa$


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