The following data is given for a caliper disk brake, with circular pad, for the lightweight two-wheeler, torque capacity = 1500 N-m number of caliper brakes on the wheel = 3 number of pads on each caliper brake = 2 coefficient of friction = 0.35 average pressure on pad = 2 MPa The ratio of pad

Question

The following data is given for a caliper disk brake, with circular pad, for the lightweight two-wheeler, The following data is given for a caliper disk brake, with circular pad, for the lightweight two-wheeler,

torque capacity = 1500 N-m

number of caliper brakes on the wheel = 3

number of pads on each caliper brake = 2

coefficient of friction = 0.35

average pressure on pad = 2 MPa

The ratio of pad radius to the distance of the pad center from axis of disk is 0.2. Calculate the radius of the pad.

Accepted Answer

[latex] ext { Given } M_{t}=1500 N - m \quad p_{a}=2 MPa \quad \mu=0.35 [/latex].

(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.

[latex] \left(\frac{R}{e} ight)=0.2 \quad ext { from Table 12.1A, } \delta=0.9693 [/latex].

From Eq. (12.33),

[latex] R_{f}=\delta e [/latex] (12.33).

[latex] R_{f}=\delta e=0.9693 e=0.9693\left(\frac{R}{0.2} ight) [/latex].

= (4.8465)R mm.

[latex] P= ext { average pressure } imes ext { area of pad }=2\left(\pi R^{2} ight) [/latex].

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

From Eq. (12.30),

[latex] M_{t}=\mu P R_{f} [/latex] (12.30).

[latex] M_{t}=\mu P R_{f} [/latex] .

[latex] 250\left(10^{3} ight)=0.35\left(2 \pi R^{2} ight)(4.8465 R) [/latex].

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

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

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