Question 12.8: An automotive type internal-expanding double-shoe brake is s...

\text { Given } w=40 mm \quad \mu=0.32 \quad p_{\max }=1 N / mm ^{2} .

R = 125 mm
Step I Actuating force
It is assumed that the maximum normal pressure will occur between the lining on the right-hand shoe and the brake drum. For the right-hand shoe,

\theta_{1}=0 \quad \theta_{2}=120^{\circ} \quad \phi_{\max .}=90^{\circ} \quad \sin \phi_{\max .}=1 .

The distance h of the pivot from the axis of the brake drum is given by

h=\sqrt{86.6^{2}+50^{2}}=100 mm .

From Eq. (12.19),

M_{f}=\frac{\mu p_{\max } \cdot 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 } \cdot 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{0.32(1)(125)(40)\left[4(125)\left(1-\cos 120^{\circ}\right)-100\left(1-\cos 240^{\circ}\right)\right]}{4(1)} .

= 240 000 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).

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 .}} .

=\frac{1(125)(40)(100)\left[2\left(\frac{120 \pi}{180}\right)-\sin \left(240^{\circ}\right)\right]}{4(1)} .

= 631 851.95 N-mm
From Eq. (12.22),

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

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

=\frac{631851.95-240000}{100.9+86.6}=2089.88 N         (i).

Step II Torque-absorbing capacity
From Eq. (12.21), the torque \left(M_{t}\right)_{R} for the righthand
shoe 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 } \cdot w\left(\cos \theta_{1}-\cos \theta_{2}\right)}{\sin \phi_{\max }} .

=\frac{0.32(125)^{2}(1)(40)\left(1-\cos 120^{\circ}\right)}{1} .

= 300 000 N-mm.

The maximum intensity of pressure for the left-hand shoe 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 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{240000 p_{\max }^{\prime}}{p_{\max }}=\frac{(240000) p_{\max }^{\prime}}{(1)} .

=240000 p_{\max }^{\prime} .

Similarly,

M_{n}^{\prime}=\frac{631851.95 p_{\max }^{\prime}}{p_{\max }}=\frac{(631851.95) p_{\max .}^{\prime}}{(1)} .

=631851.95 p_{\max }^{\prime} .

For the left-hand shoe,

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

\text { or } \quad 2089.88=\frac{(240000+631851.95) p_{\max }^{\prime}}{(100.9+86.6)} .

∴          p_{\max .}=0.45 N / mm ^{2} .

Since the shoes are identical,

\left(M_{t}\right)_{L}=300000\left(\frac{p_{\max }^{\prime}}{p_{\max }}\right)=300000\left(\frac{0.45}{1}\right) .

= 135000 N-mm.

The total torque-absorbing capacity of the brake is given by,

M_{t}=300000+135000=435000 N – mm .

or          435 N-m             (ii).