Question 12.12: Following data is given for a caliper disk brake with annula...

\text { Given } M_{t}=1500 N – m \quad R_{o}=150 mm .

R_{i}=100 mm \quad p_{a}=2 MPa \quad \mu=0.35 .

Number of pads = 2
Step I Actuating force
Since there are two pads, the torque capacity of one
pad is (1500/2) or 750 N-m.

From Eq. (12.29),

R_{f}=\frac{2}{3} \frac{\left(R_{o}^{3}-R_{i}^{3}\right)}{\left(R_{o}^{2}-R_{i}^{2}\right)}                    (12.29).

From Eq. (12.30),

M_{t}=\mu P R_{f}                   (12.30).

M_{t}=\mu P R_{f} .

Step II Angular dimension of pad
P = average pressure × area of pad

\left[2 MPa =2 N / mm ^{2}\right] .

From Eq. (12.31),

A=\frac{1}{2} \theta\left(R_{o}^{2}-R_{i}^{2}\right)         (12.31),

A=\frac{1}{2} \theta\left(R_{o}^{2}-R_{i}^{2}\right) .

\text { or } \quad 8458.42=\frac{1}{2} \theta\left(150^{2}-100^{2}\right) .

\theta=1.3533 \text { radians } .

\text { or } \quad \theta=1.3533\left(\frac{180}{\pi}\right)=77.54^{\circ} .

The angular dimension of the pad can be taken as 80°.