# Modeling torque as surface pressure

Consider a rod with a length of 150 [mm] and a diameter of 44 [mm]. At the left boundary the rod is fixed to a wall. At the right end a torque of 1000 [N m] is applied. The remaining boundaries are free to move. As a material a S275 steel is used. Thus Young's modulus is given as $$Y=210000$$ [N/$$mm^2$$] and Poisson's ratio is $$\nu=0.3$$.

A reference solution for the maximal displacement suggests it should be around 0.1103 [mm] and the maximal von Mises stress should be about 103.6 [N/mm^2]. In this Solidworks tutorial the same thing is done, but with different values than the reference I used (which is also based on a Solidworks example).

The problem I have is that the finite element simulation tool I use does not have a direct way to specify a torque as a boundary condition. One can, however, specify a boundary surface pressure. I'd like to convert the given torque into a surface pressure. Here is how I did the conversion. I created a vector field that gives the direction of the pressure load in the $$x$$, $$y$$ and $$z$$ direction and converted the torque value into a pressure. As a vector field I used:

{0, Sin[ArcTan[y, z]], -Cos[ArcTan[y, z]]}


A plot of the vector field looks like this:

VectorPlot[{Sin[ArcTan[y, z]], -Cos[ArcTan[y, z]]}, {y, -1, 1}, {z, -1, 1}]


Then I converted the $$1000$$ [N m] into $$1000*10^3$$ [N mm]. To get a pressure I think I'd need to use something of this sort:

$$M_t / q * 1 / (\pi r^2)$$

where $$M_t$$ is the given torque value. I would like to convert the torque into a force by dividing the force by a length. The $$q$$ is the part I am uncertain about what to use. I tried $$q=\sqrt{y^2 + z^2}$$ such that it gives the radius to the axis but that does not give correct values. Then I divide by the area of the surface to get a pressure.

I think I am making a mistake in the conversion. If someone could point me in the right direction...

Update:

Here is a different approach: Based on this we start from

$$M_t = \frac{J_T}{r} \tau$$

where $$J_T$$ is the torsion constant, $$r$$ the radius and $$\tau$$ the maximum shear stress. We solve for $$\tau$$ to get

$$\tau = \frac{M_t r}{J_T}$$

The torsion constant can be computed with

$$J_{zz} = J_{xx} + J_{yy} = \frac{\pi r^4}{2}$$

With this approach I get a maximum displacement of $$0.111035$$ [mm] and a maximum von Mises stress of $$104.277$$ [N/mm^2], both close enough for government work.

A plot of the von Mises stress:

In principal, I think, this approach could be used for other surfaces, then one would need to compute the area moments of inertia for those surfaces.

• If I understand correctly, you should divide the torque, T, by radius to get the tangential force, Ft, then divide it by 2*pi*r (circumference of the shaft) to get shear force per unit length, fv (force/length). – r13 May 25 at 16:03
• @r13, thank you for your suggestion. I was not able to get it to work though. – user21 May 26 at 5:40

I'll hazard a guess. From what I undestand what you are after is given a specific torque - I'll call it $$M_t$$ (instead of $$F_t$$)- you want to calculate what is the uniform pressure that you need to apply on the face of a cylinder to obtain as a result $$M_t$$.

Assume a thin annular stripe at distance r with with dr.

The area of that stripe will be: $$dA = 2\pi r dr$$

Assuming a constant pressure q [$$N/m^2$$]is applied, the total force on that stripe is

$$dF = q dA = 2\pi qr dr$$

The torque of that thin stripe $$dM_t$$ will be

$$dM_t = r\cdot dF = r q dA = 2\pi q r^2 dr$$

By integrating from 0 to $$R$$ (: external radius).

$$M_t = \int_0^{R} 2\pi q r^2 dr$$ $$M_t = 2\pi q \int_0^{R} r^2 dr$$ $$M_t = 2\pi q \left[ \frac{r^3}{3} \right]_0^{R}$$ $$M_t = \frac{2}{3}\pi q R^3$$

Then the magnitude of the constant pressure $$q$$ should be:

$$q= \frac{3}{2}\frac{M_t}{\pi R^3}$$

• Thank you for your suggestion. Unfortunately, it does not give correct values but I am not able to say why. Your idea looks correct to me. See my update on how I did it now. – user21 May 26 at 6:08
• @user21 If you do it that way then the pressure (or more precisely stress) should be increasing with the radius. So at the center you have no stress, and at the outer rim you have the highest stress. – NMech May 26 at 6:10
• OK, I had another go at this and replaced the sqrt with the radius proper. Then your approach works. Great. I was so captured in integrating over the region - this lead me to the wrong conclusion that I needed to integrate over your suggested approach. That was not needed. – user21 May 31 at 6:31
• The discrepancy is because you are applying forces (pressure) on the surface, that creates an extra shear stress component. So if you did not use a uniform pressure you would have had a different von Mises distribution, (same on the displacement). – NMech May 31 at 7:01
• Regarding your latter question (i.e.**how to make my post to make it more attractive for upvotes**) there is no real answer. I'd hazard a guess that people usually upvote if they feel they can relate to a question (for example your question is on a specific solver which you don't name), so its a niche question. In any case, I wouldn't advice going down the rabbit hole of chasing after reputation points, because you might feel disappointed. Just try to write clean questions and answers, is my advice (if you'll have it). – NMech May 31 at 7:04