Question 12.7: A pivoted double-block brake, similar to that in Fig. 12.12,...

\text { Given } \quad D=500 mm \quad w=100 mm \quad \mu=0.2 .

p_{\max .}=0.5 N / mm ^{2} \quad 2 \theta=100^{\circ} .

Step I Distance of pivot from axis of drum
From Eq. (12.12),

h=\frac{4 R \sin \theta}{2 \theta+\sin 2 \theta}               (12.12).

h=\frac{4 R \sin \theta}{2 \theta+\sin 2 \theta}=\frac{4(250) \sin \left(50^{\circ}\right)}{\left(\frac{100 \pi}{180}\right)+\sin \left(100^{\circ}\right)} .

= 280.59 mm              (i)
Step II Torque capacity of each shoe
From Eq. (12.13),

M_{t}=2 \mu R^{2} w p_{\max } \sin \theta             (12.13).

M_{t}=2 \mu R^{2} w p_{\max .} \sin \theta .

=2(0.2)(250)^{2}(100)(0.5) \sin \left(50^{\circ}\right) .

= 957 555 N-mm                (ii)
Step III Reactions at pivot
From Eqs (12.14) and (12.15),

R_{X}=\frac{1}{2} R w p_{\max }(2 \theta+\sin 2 \theta)           (12.14).

R_{Y}=\frac{1}{2} \mu R w p_{\max }(2 \theta+\sin 2 \theta)         (12.15).

R_{X}=\frac{1}{2} R w p_{\max }(2 \theta+\sin 2 \theta) .

=\frac{1}{2}(250)(100)(0.5)\left[\left(\frac{100 \pi}{180}\right)+\sin \left(100^{\circ}\right)\right] .

= 17063 N.

R_{Y}=\frac{1}{2} \mu R w p_{\max }(2 \theta+\sin 2 \theta) .

=\mu R_{x}=0.2(17063)=3412.6 N             (iii).