Following data is given for a caliper disk brake with annular pad, for the front wheel of the motorcycle:
torque capacity = 1500 N-m
outer radius of pad = 150 mm
inner radius of pad = 100 mm
coefficient of friction = 0.35
average pressure on pad = 2 MPa
number of pads = 2
Calculate the angular dimension of the pad.

Question

Accepted Answer

[latex] ext { Given } M_{t}=1500 N - m \quad R_{o}=150 mm [/latex].

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

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),

[latex] R_{f}=\frac{2}{3} \frac{\left(R_{o}^{3}-R_{i}^{3} ight)}{\left(R_{o}^{2}-R_{i}^{2} ight)} [/latex] (12.29).

[latex] R_{f}=\frac{2}{3} \frac{\left(R_{o}^{3}-R_{i}^{3} ight)}{\left(R_{o}^{2}-R_{i}^{2} ight)}=\frac{2}{3} \frac{\left(150^{3}-100^{3} ight)}{\left(150^{2}-100^{2} ight)}=126.67 mm [/latex]

From Eq. (12.30),

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

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

[latex] 750(10)^{3}=0.35 P(126.67) ext { or } P=16916.85 N [/latex]

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

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

[latex] 16916.85=2 A \quad ext { or } \quad A=8458.42 mm ^{2} [/latex]

From Eq. (12.31),

[latex] A=\frac{1}{2} heta\left(R_{o}^{2}-R_{i}^{2} ight) [/latex] (12.31),

[latex] A=\frac{1}{2} heta\left(R_{o}^{2}-R_{i}^{2} ight) [/latex].

[latex] ext { or } \quad 8458.42=\frac{1}{2} heta\left(150^{2}-100^{2} ight) [/latex].

[latex] heta=1.3533 ext { radians } [/latex].

[latex] ext { or } \quad heta=1.3533\left(\frac{180}{\pi} ight)=77.54^{\circ} [/latex].

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

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

This Question has been Answered.

Already have an account?

Login

Related Answered Questions