Modeling torque as surface pressure

Question

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.

Answer 1

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:

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