Question 9.3: A line shaft supporting two pulleys A and B is shown in Fig....

\text { Given } S_{u t}=650 N / mm ^{2} \quad S_{y t}=380 N / mm ^{2} .

k_{b}=1.5 \quad k_{t}=1.0  .

\text { For belt drive, } P_{1} / P_{2}=3 .

Maximum belt tension = 2.7 kN
Step I Permissible shear stress

0.30 S_{y t}=0.30(380)=114 N / mm ^{2} .

0.18 S_{u t}=0.18(650)=117 N / mm ^{2} .

The lower of the two values is 114 N/mm² and there are keyways on the shaft.

Step II Torsional moment
The maximum belt tension is limited to 2.7 kN. At this stage, it is not known whether P_{1} is maximum or P_{3} is maximum. The torque transmitted by the pulley A is equal to torque received by the pulley B. Therefore,

\left(P_{1}-P_{2}\right)\left(\frac{250}{2}\right)=\left(P_{3}-P_{4}\right)\left(\frac{450}{2}\right) .

\left(P_{1}-P_{2}\right)=1.8\left(P_{3}-P_{4}\right)             (a).

\text { Also, } \quad P_{2}=\frac{1}{3} P_{1} \quad \text { and } \quad P_{4}=\frac{1}{3} P_{3} .

Substituting the above values in Eq. (a),

\left(\frac{2}{3} P_{1}\right)=1.8\left(\frac{2}{3} P_{3}\right) \quad \text { or } \quad P_{1}=1.8 P_{3} .

Therefore, the tension P_{1} in the belt on the pulley A is maximum.

P_{1}=2700 N \text { and } P_{2}=2700 / 3=900 N .

P_{3}=\frac{P_{1}}{1.8}=\frac{2700}{1.8}=1500 N \quad P_{4}=1500 / 3=500 N .

A simple way to decide the maximum tension in the belt is the smaller diameter pulley. Smaller the diameter of the pulley, higher will be the belt tension for a given torque.
The torque transmitted by the shaft is given by,

M_{t}=(2700-900)\left(\frac{250}{2}\right)=225000 N – mm .

Step III Bending moment
The forces and bending moments in vertical and horizontal planes are shown in Fig. 9.5(b). The maximum bending moment is at A. The resultant bending moment at A is given by,

M_{b}=\sqrt{(810000)^{2}+(250000)^{2}}=847703 N – mm .

Step IV Shaft diameter
From Eq. (9.15),

\tau_{\max .}=\frac{16}{\pi d^{3}} \sqrt{\left(k_{b} M_{b}\right)^{2}+\left(k_{t} M_{t}\right)^{2}}                  (9.15).

d^{3}=\frac{16}{\pi \tau_{\max }} \sqrt{\left(k_{b} M_{b}\right)^{2}+\left(k_{t} M_{t}\right)^{2}} .

=\frac{16}{\pi(85.5)} \sqrt{(1.5 \times 847703)^{2}+(1.0 \times 225000)^{2}} .

or d = 42.53 mm.