Question 12.8: A steel shaft supported in bearings A and B at its ends carr...

Clearly from Figure 12.18 given in the problem, portion AC of the shaft is subjected to the torque:

T = (1250 − 250)(300)Nmm = 300 ×10³ Nmm

Shear stress developed on any element placed on the shaft periphery is

\tau=\frac{16 T}{\pi d^3}=\frac{(16)(300)\left(10^3\right)}{\pi d^3}  N mm =\frac{4800\left(10^3\right)}{\pi d^3}  N mm

Again, the shaft AB is subjected to downward bending load at point C of magnitude = 1500 N.
The reaction forces at the bearings are

R_{ A }=\frac{1500(1.5)}{2.5} N =900  N \text { and } R_{ B }=(1500-900) N =600  N

Bending moment at C is

M = (900)(1.0)N m = 900 N m = 900 × 10³ N mm

Normal stress on the shaft element at AC is

\sigma_{x x}=\frac{32 M}{\pi d^3}=\frac{(32)(900)\left(10^3\right)}{\pi d^3}

Principal normal stress on the element is

\left(\sigma_n\right)_{\max }=\frac{\sigma_{x x}}{2}+\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau^2}

=\frac{(16)(900)\left(10^3\right)}{\pi d^3}+\sqrt{\left[\frac{(16)(900)\left(10^3\right)}{\pi d^3}\right]^2+\left[\frac{4800\left(10^3\right)}{\pi d^3}\right]^2}

=\frac{(16)(1848.68)\left(10^3\right)}{\pi d^3} \frac{ N }{ mm ^2}

Therefore,

\frac{(16)(1848.68)\left(10^3\right)}{\pi d^3}=85

or                  d = 48.03mm

and maximum shear stress on the element is

\tau_{\max }=\sqrt{\left[\frac{(16)(900)\left(10^3\right)}{\pi d^3}\right]^2+\left[\frac{4800\left(10^3\right)}{\pi d^3}\right]^2}=\frac{(16)(948.68)\left(10^3\right)}{\pi d^3}

Therefore,

\frac{(16)(948.68)\left(10^3\right)}{\pi d^3}=42.5

or                d = 48.44mm

We select the shaft diameter as d = 48.44 mm.