Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

Holooly Tables

All the data tables that you may search for.

Holooly Help Desk

Need Help? We got you covered.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Holooly Help Desk

Need Help? We got you covered.

Q. 10.8

A flywheel of mass 18 kg and radius of gyration 250 mm is mounted at the midpoint of a solid steel shaft of diameter 50 mm and length, L. The shaft rotates in the bearings A and B at its ends with an angular speed of 120 rpm. If both the bearings suddenly freeze so that the ends of the shaft become locked, the shaft will have to absorb the kinetic energy of the flywheel. Calculate the minimum length L of the shaft for which this can be done without exceeding a maximum shear stress $\tau_{\max }=80$ MPa. Assume modulus of rigidity, G = 82 GPa.

Verified Solution

Angular speed of flywheel, ω = 120 rpm = (2π )(120)/60 rad/s = 4π rad/s.
If the flywheel is suddenly stopped, the change of kinetic energy of flywheel $=(1 / 2) \bar{I} \omega^2 ; \text { where } \bar{I}$ is the mass moment of inertia of flywheel about axis of rotation:

$\bar{I}=m \bar{K}^2$

where $\bar{K}$ is the centroidal radius of gyration of the flywheel. Therefore, change in rotational kinetic energy of the flywheel is

$\frac{1}{2} m \bar{K}^2 \omega^2=\frac{1}{2}(18)(0.25)^2(4 \pi)^2 J =88.83 J$

If the flywheel is symmetrically placed at the shaft middle point, assuming no energy loss, we can say that

$2\left\lgroup\begin{array}{l} \text { Torsional strain energy } \\ \text { of one-half of shaft } \end{array} \right\rgroup =\text { Change in rotational kinetic energy of flywheel }$

or            $2 \frac{T^2 L}{4 G J}=88.83 \Rightarrow \frac{T^2 L}{G J}=177.66$                (1)

But τ = 16T/π d³ for solid shaft, therefore

$T_{\max }=\frac{\pi d^3}{16} \tau_{\max }=\frac{\pi}{16}(0.05)^3 80\left(10^6\right) N m$

or              $T_{\max }=1963.5 Nm$              (2)

From Eqs. (1) and (2), we get

$L_{\min }=\frac{(82)\left(10^9\right) \frac{\pi}{32}(0.05)^4(177.66)}{(1963.5)^2}=2.32 m$

The required minimum shaft length is 2.32 m.