Question 12.9: An internal-expanding brake with four identical shoes is sho...

\text { Given } w=50 mm \quad \mu=0.3 \quad p_{\max }=1 N / mm ^{2} .

R = 250 mm h = 200 mm
Step I Actuating force
It is assumed that the maximum normal pressure will occur between the drum and the lining on right-hand shoe of the upper half. For the righthand shoe,

\theta_{1}=15^{\circ} \quad \theta_{2}=75^{\circ} \quad \phi_{\max }=75^{\circ} .

\sin \phi_{\max .}=0.9659 \quad h=200 mm \quad c=200 mm .

where h is the distance of the pivot axis from the axis of the brake drum.
From Eq. (12.19),

M_{f}=\frac{\mu p_{\max } R w\left[4 R\left(\cos \theta_{1}-\cos \theta_{2}\right)-h\left(\cos 2 \theta_{1}-\cos 2 \theta_{2}\right)\right]}{4 \sin \phi_{\max .}}           (12.19).

M_{f}=\frac{\mu p_{\max } R w\left[4 R\left(\cos \theta_{1}-\cos \theta_{2}\right)-h\left(\cos 2 \theta_{1}-\cos 2 \theta_{2}\right)\right]}{4 \sin \phi_{\max }} .

=\frac{0.3(1)(250)(50)\left[4(250)\left(\cos 15^{\circ}-\cos 75^{\circ}\right)-200\left(\cos 30^{\circ}-\cos 150^{\circ}\right)\right]}{ 4 (0.9659)} .

= 350 091.19 N-mm
From Eq. (12.20),

M_{n}=\frac{p_{\max } R w h\left[2\left(\theta_{2}-\theta_{1}\right)-\left(\sin 2 \theta_{2}-\sin 2 \theta_{1}\right)\right]}{4 \sin \phi_{\max }}            (12.20).

=\frac{1(250)(50)(200)\left[2\left(\frac{60 \pi}{180}\right)-\left(\sin 150^{\circ}-\sin 30^{\circ}\right)\right]}{4(0.9659)} .

= 1 355 209.59 N-mm.

From Eq. (12.22),

P=\frac{M_{n}-M_{f}}{C}             (12.22).

=\frac{1355209.59-350091.19}{200}=5025.59 N          (i).

Step II Torque-absorbing capacity
From Eq. (12.21), the torque \left(M_{t}\right)_{R} for the right hand shoe in the upper half is given by

M_{t}=\frac{\mu R^{2} p_{\max .} w\left(\cos \theta_{1}-\cos \theta_{2}\right)}{\sin \phi_{\max }}                   (12.21).

\left(M_{t}\right)_{R}=\frac{\mu R^{2} p_{\max } w\left(\cos \theta_{1}-\cos \theta_{2}\right)}{\sin \phi_{\max }} .

=\frac{0.3(250)^{2}(1)(50)\left(\cos 15^{\circ}-\cos 75^{\circ}\right)}{0.9659} .

= 686 315.98 N-mm.

The maximum intensity of pressure for the left-hand shoe in the upper half is unknown. For identical shoes, it can be seen from the expressions of M_{n} \text { and } M_{f} that both are proportional to \left(p_{\max }\right) .
For the left-hand shoe, the maximum intensity of pressure is taken as \left(p_{\max }^{\prime}\right) . Therefore, for the left-hand shoe,

M_{f}^{\prime}=\frac{350091.19 p_{\max .}^{\prime}}{p_{\max }}=\frac{(350091.19) p_{\max .}^{\prime}}{(1)} .

=350091.19 p_{\max }^{\prime} .

M_{n}^{\prime}=\frac{1355209.59 p_{\max }^{\prime}}{p_{\max }}=\frac{(1355209.59) p_{\max }^{\prime}}{(1)} .

=1355209.59 p_{\max }^{\prime} .

Using Eq. (12.23) for the left-hand shoe in the upper half,

P=\frac{M_{n}+M_{f}}{C}                     (12.23).

P=\frac{M_{n}^{\prime}+M_{f}^{\prime}}{C} .

\text { or } \quad 5025.59=\frac{(350091.19+1355209.59) p_{\max }^{\prime}}{200} .

\therefore \quad p_{\max .}^{\prime}=0.59 N / mm ^{2} .

\left(M_{t}\right)_{L}=686315.98\left(\frac{p_{\max }^{\prime}}{p_{\max }}\right)=686315.98\left(\frac{0.59}{1}\right) .

=404926.43 \quad N – mm .

The total torque-absorbing capacity of the pair of shoes in the upper half of the brake drum is given by,

\left(M_{t}\right)_{u h}=\left(M_{t}\right)_{R}+\left(M_{t}\right)_{L}=686315.98+404926.43 .

= 1 091 242.41 N-mm
Since all four shoes are identical, the two pairs are also identical. Therefore,

M_{t}=\left(M_{t}\right)_{u h}+\left(M_{t}\right)_{f f}=2\left(M_{t}\right)_{u h}=2(1091242.41) .

= 2 182 484.82 N-mm
or,          M_{t}=2182.48 N – m          (ii).