Could someone please help with the problem below?

Edit 1:My guess on this question is: when I plot in the formulas, the stress rooting from the bending moment overcomes greatly over the other stresses; so if I neglect them, I will still come across a fairly accurate result, but still, please feel free to share your opinions.

Assume you have a 20 [mm] diameter 304 Stainless Steel shaft welded on a rigid surface horizontally (neglect weight). With the forces applied below, what could be the possible maximum length of this shaft?

  1. 20 [N.m] torque applied throughout the shaft
  2. 10 [kg] load is applied at the end of the shaft
  3. 50 [kg] loads of compression force is applied, again, from the end of the shaft
  4. Factor of safety is going to be 1.5

To make things easier I have calculated both $I$ and $J$ beforehand.



  • $\begingroup$ You're going to have to show at least an attempt at working this out yourself and showing where you got stuck, if you want an answer, otherwise your question is likely to get closed as a possible attempt to get others to do your homework for you ... $\endgroup$ – Gwyn Nov 17 '20 at 2:31
  • $\begingroup$ @Gwyn Yeah, it's actually from a homework question, but sincerely I'm stuck. I will post my attempt as soon as possible after I handle todays exam. $\endgroup$ – SirDancealot Nov 17 '20 at 5:24

This is a combined load case scenario. Since this is essentially a cantilever beam, you will find that the location that is most likely to fail is at the interface with rigid surface. So you need to consider what is happening at various points in the surface.

In the following analysis, I consider:

  • X: along the axis of the beam, (horizontal)
  • Y: vertical plane
  • Z: perpendicular to X and Y (on the horizontal plane).

enter image description here

I will also be naming points A, B, C, D for clarity later on.

More specifically:

  • Compression load:

This is the easiest. It will create a uniform Normal stress over the Cross-section. $$\sigma_{C} = \frac{F_{C}}{\pi d^2/4}$$

  • Bending: $F_1 =$10 [kg] load is applied at the end of the shaft

Bending also creates normal stresses:

$$\sigma_{bn}(y) = \frac{F_1*L}{I}d/2 $$

bending stress distribution

and shear stresses (they are minor though), and the maximum stress distribution isseen below.

enter image description here

The maximum shear stress can be observed at the middle of the section (at y=0), and its direction is downwards

$$\tau_{b,max } = \frac{4}{3} \frac{F_1}{\pi d^2/4} $$

  • Torsion

Torsion will produce only shear tresses. The shear stresses at a distance r from the center will be:

$$\tau_{t }(r) = \frac{M_t}{J }r $$

The stress distribution will have the following shape.

enter image description here

Please note that the left and the right side have different directions. This is in contrast with the shear stresses from the bending. So the stress in point A and C will in one case *sum up * and in the other case cancel out (depending on the direction of the torque).

Summing up the above.

As you've seen we have normal and shear stresses on each point of the beam cross-section. the Distribution will look slighlty more complicated than the following (because it also has the compression loading:

enter image description here

So what you need to is at least at points A,B,C and D.

  • sum all the normal forces
  • sum all the shear forces
  • find the equivalent stress according to the stress theory.

In all cases you will end up with a polynomial function of L.

Then equating this function of L to the allowable stress over the safety factor, you have an equation you can solve for L.


By using the concept shown by NMech, you shall be able to set up two equations as shown below, then solving for L.

enter image description here


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