L = 10 in, $T_{max}$ = 153 in-lb, $T_{min}$ = 13 in-lb, Carbon Steel 1020 CR, $f_y$ = 57 ksi, $f_{ut}$ = 68 ksi

I am not sure how to solve if I do not have the weight of the shaft. I really appreciate any help.

  • 1
    $\begingroup$ Mass of shaft will be based on the volume and density. Density probably listed in table suggested. $\endgroup$
    – Solar Mike
    Commented Nov 4, 2021 at 4:51
  • $\begingroup$ The 1020 cold rolled will be less than 57,000 yield in a shaft. It will be that high in flat rolled sheet. I would guess about 40,000 psi. $\endgroup$ Commented Nov 4, 2021 at 14:54
  • $\begingroup$ Can you provide the equation from lecture 21? I think the design will be affected by two factors - torsional stress of the shaft AND the allowable deflection of the shaft due to the pull-down force and weight of the assembly. However, as the thickness of the sprocket and the force are not defined, I don't think there is a simple solution. A better bet is to set the limit of the deflection (based on strength of the shaft), then back calc the permissible weights and force, and proceed from there to check the resulting stress. $\endgroup$
    – r13
    Commented Nov 4, 2021 at 17:09
  • $\begingroup$ Based on your latest edit, you can ignore the self-weight of the shaft, which makes the finding of shaft diameter straightforward - controlled by the stress (TL/J) caused by the net torque, $\endgroup$
    – r13
    Commented Nov 5, 2021 at 1:12

1 Answer 1


You can either go

  1. full analytical by setting a symbol d for the diameter.

Then you'd calculate weight as $\rho \cdot \frac{\pi\cdot d^2L}{4} \cdot g$ and the forces, then calculate transverse forces, bending moment and finally the stresses and the safety factor(again d will be in the formulas). At the end you should arrive at an equation (or several representing the different points of check on the shaft) of the safety factor that is a function of the diameter. You can then solve for d.

This would be the most complicated in term of length of equations, and the most convoluted because of the nonlinear terms (square roots etc).

  1. Do an iterative approach:

Select a diameter e.g. 10[mm], then calculate the weight $\rho \cdot \frac{\pi\cdot (10\; \text{mm})^2L}{4} \cdot g$, then calculate transverse forces, bending moment and finally the stresses and the safety factor. If the safety factor is not betwen 3 or 4, then guess another diameter and try again (e.g. if the safety factor is 2, then select larger diameter e.g. 16 mm).

if you automated this in an excel sheet, its very easy to do the calculation and arrive at a solution.

  1. Do the iterative in a smart way

What you can do is do the first iteration, by assuming the weight of the shaft is negligible. The equations are much simpler that way. Solve for a high safety factor (say 4), and then you arrive at a diameter. Use that diameter as a starting point. Chances are that unless there is no external load on the shaft then you'll be within the 3 to 4 margin for the safety factor.

  • $\begingroup$ Offcourse if you write your calculations to a computer arithmetic system (CAS) then iteration is basically a free action, should something change making a new calc is easy too. $\endgroup$
    – joojaa
    Commented Nov 4, 2021 at 16:50

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