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.
Engineering Stack Exchange is a question and answer site for professionals and students of engineering. It only takes a minute to sign up.
Sign up to join this communityL = 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.
You can either go
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).
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.
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.