Question 12.13: The following data is given for a caliper disk brake, with c...

\text { Given } M_{t}=1500 N – m \quad p_{a}=2 MPa \quad \mu=0.35 .

(R/e) = 0.2 Number of brakes = 3
Number of pads on each brake = 2
Step I Radius of pad
Since there are three caliper brakes, each with two pads, the torque capacity of one pad is (1500/6) or 250 N-m.

\left(\frac{R}{e}\right)=0.2 \quad \text { from Table 12.1A, } \delta=0.9693 .

From Eq. (12.33),

R_{f}=\delta e             (12.33).

R_{f}=\delta e=0.9693 e=0.9693\left(\frac{R}{0.2}\right) .

= (4.8465)R mm.

P=\text { average pressure } \times \text { area of pad }=2\left(\pi R^{2}\right) .

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

From Eq. (12.30),

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

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

250\left(10^{3}\right)=0.35\left(2 \pi R^{2}\right)(4.8465 R) .

R = 28.63 mm
The radius of the pad can be taken as 30 mm.