5
$\begingroup$

I'm designing a circular shaft that's going to carry a spread radial load of about 200kg and also transmit a 160Nm torque. I'm not very familiar with materials but what I'm guessing I'll be able to get at a reasonable price locally is some lower grade steel.

Replacing the spread load with a single force in the middle of the shaft I guess will look something like this

enter image description here

Both points A and B are symmetrical so I'm taking in consideration only half of the shaft and calculating for that.

$$T = F\ell = 200 \cdot 0.5 = 100\text{ kNm}$$

Where $T$ is the torque at point $A$ created by the force $F$.

The tensile stress on the shaft will then equal

$$t = \dfrac{T}{W}$$

Where $t$ is tensile stress and $W$ is section modulus.

For a solid circular face the section modulus equals

$$W = \dfrac{\pi d^3}{32}$$

Now I have to assume either $t$ or $d$ in order to solve. I looked it up in this table and it looks like the tensile yield strength of steel is 350 kPa so I'm taking a conservative estimate of 300 kPa for $t$.

$$\begin{align} d^3 &= \dfrac{32(T / t)}{\pi} \\ &= \dfrac{32 \cdot (100e3 / 300e6}{\pi} \\ &= 0.003397028 \\ \therefore d &= 0.150\text{ m} \end{align}$$

Obviously there is something wrong either in my logic or calculations because I'm getting unrealistic results.

I went ahead and calculated the torsion stress as well with this formula. I found that shear stress for steel is about 200-300 kPa so I'm again taking a conservative 200.

$$\begin{align} d &= 1.72 \cdot \left(\dfrac{M}{\tau_{max}}\right)^{1/3} \\ &= 1.72 \cdot \left(\dfrac{160}{200e6}\right)^{1/3} \\ &= 0.0159 ~ 16mm \end{align}$$

which seems about right.

I was thinking of getting a 20 mm diameter profile but I would rather play it safe and ask for some help instead of just assume it's going to hold.

$\endgroup$
  • 1
    $\begingroup$ A very important thing on 'Calculating shaft diameter' is fatigue failure. You should design your rotating shaft taking that into consideration, and not considering only a static load. This calculation is very tedious to do by hand, and the use of a computer should help in the iterative calculations. $\endgroup$ – Thales Oct 8 '19 at 19:15
  • $\begingroup$ Two books you could look up to these calculations are Norton and Shigley $\endgroup$ – Thales Oct 8 '19 at 19:16
2
$\begingroup$

According to a couple of textbooks on machine design and assuming that your FBD is correct, the maximum bending moment occurs at the centre of the shaft.

For a point load, it is equal to:

$WL/4 = M$

Where W is the point load and L is the length of your beam.

$(200*10^3)*1/4 = 50*10^3 Nm = 50*10^6 Nmm = M$

According to the maximum shear stress theory the combined torsional and bending stress is given by:

$T_e = \sqrt{M^2 + T^2}$

But we also know that

$ T_e = (\pi\div 16)*\tau*d^3$

Where tau is the allowable shear stress for each material.

Substituting the numbers into the first equation and then equate both of the $T_e$ values. You can then solve for d. This is one of the answers.

But also taking into consideration the bending moment using maximum normal stress theory,

$ M_e = 0.5(M + \sqrt{M^2 +T^2}) $

When designing the shaft for the bending moment, we can apply the formula shown:

$M_e = (\pi \div 32)*\sigma_b*d^3 $

Where $\sigma_b$ = Ultimate/Allowable Stress

Substituting the numbers into the first equation $M_e$ equation and then equate both of the $M_e$ values.

You can then solve for d. This is one of the answers. Find the largest diameter out of the first and second equation and use it. You should also apply some safety factor in the allowable stresses and maybe the diameter. If the load which you stated is uniformly distributed the maximum bending moment will also change which is contrary to your point load diagram. I am also unsure about the action of fatigue.

| improve this answer | |
$\endgroup$
1
$\begingroup$

@php_nub_qq, Even if its a bit late, my main comment is that you probably have messed up with the engineering units.

The tensile stress you are referring to, is probably 3 order of magnitude lower than the actual UTS even of a common grade steel (300MPa instead of 300 kPa). That allow would mess up and produce significant differences in the loads.

Also I concur with @Amit in that your method you describe is valid only for the bending moment imposed from the 200kN load. In your calculation you do not account for the shear stress due to the torque. Due to my late answer I will only point you to the following link https://www.engineeringtoolbox.com/torsion-shafts-d_947.html, so that you can calculate $\tau_{xz}$

After that, you could use the maximum stress theory or Von Mises (it will be much more straightforward in your case and also differences in most cases are negligible).

$$\sigma_{VM} = \sqrt{\sigma_{xx}^2 + 3 \cdot \tau_{xz}^2}$$

For justification about that, you best do a bit of reading at Norton or Shigley's books, and look at design of shafts under combined loading.

EDIT 1: I performed a rough analysis (linear) and the estimated normal stresses (due to bending) compared to the shear stresses (induced by torsion) are approximately 50 times larger in the scenario you are describing.

E.g. assuming a maximum tensile stress of 200 MPa, the required diameter is approximately 4.7[cm]. For that diameter the torsionally induced shear stress is about 8[MPa].

Bottom line, from an engineering point of view you are ok just considering the bending stresses only.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. I also noticed not only have I messed up the stress units but also 200kg is 2kN and not 200. $\endgroup$ – php_nub_qq Aug 5 at 10:50
  • $\begingroup$ Another thing I am not certain about upon is you mention spread radial load of 200 but your diagram implies concentrated load. The Bending moment differs in that case. The concentrated load is more severe to the structure. $\endgroup$ – NMech Aug 5 at 11:51
  • 1
    $\begingroup$ That was exactly the logic behind the representation, if it can withstand a concentrated load then my spread load will be like a walk in the park. Keeping in mind that is not a static load and also taking safety factor into account it's probably not the most optimal assumption but is fair. $\endgroup$ – php_nub_qq Aug 5 at 21:42
  • $\begingroup$ If its not a static load then you'd have to consider fatigue, and also stress concentration factors. Additionally, if the rpm is high you might need to look at Jeffcott motor (no need for anything fancier like Rayleigh-Ritz method approximations since your problem is the definition of the problem). $\endgroup$ – NMech Aug 6 at 7:54
  • $\begingroup$ I couldn't thank you enough for your time and references, they are of great value! Regarding the RPM, the maximum is 550 but I couldn't tell if that would qualify as high, for an engine certainly not but for a merry-go-round on the other hand.. $\endgroup$ – php_nub_qq Aug 6 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.