Question

A single-plate clutch has a friction disc with inner and outer radii of 20 mm and 40 mm respectively. The friction lining in the disc is made in such a way that the coefficient of friction $\mu$ varies radially as $\mu$ = 0.01 r, where r is in mm. The clutch needs to transmit a friction torque of 18.85 kN mm. As per uniform pressure theory the pressure (in MPa) on the disc is

• A. 0.5

Answer 1

The correct option is A 0.5
$R_i=20mm$

$R_0=40mm$

$\mu =0.01r$

$T_f=18.81kN/mm$

Let p= pressure on the disc

$d_A=2\pi r\left(dr\right)$


Normal load on elemental ring $\left(dW\right)=p2\pi rdr$

Frictional force on elemental ring $\left(d{F}_f\right)=\mu dW$

Frictional torque transmitted by elemental ring,

$d{T}_f=\left(d{F}_f\right)r$

Hence, $d{T}_f=\mu p2\pi r^{2}dr$

Total frictional torque transmitted by clutch plate

$\left(T_f\right)={\int }_{R_i}^{R_0}d{T}_f$

$T_f={\int }_{R_i}^{R_0}\mu p2\pi r^{2}dr$

As per U.P.T

$T_f=2\pi p{\int }_{R_i}^{R_0}\mu r^{2}dr$

$T_f=2\pi p{\int }_{R_i}^{R_0}\left(0.01r\right)r^{2}dr$ $(T_f{)}_{U.P.T}=(2\pi r\left)\right(0.01)\left[\frac{R_0^4-R_i^4}{4}\right]$

$18.85\times {10}^3=2\pi p\left(0.01\right)\left[\frac{{40}^4-{20}^4}{4}\right]$

$p=0.5Mpa$