Question

6. A plate clutch consists of 1 pair of contacting surfaces. The inner and outer diameter of the
friction disk is 100mm and 200mm respectively. The coefficient of friction is 0.2 and
permissible intensity of pressure is ISN/mm'. Assuming uniform wear theory, calculate the
operating force in the clutch
a) 15546N
b) 12344N
d 23562N
d) 24543N
​

Answer 1

Answer:

The operating force in the clutch using uniform wear theory is 15,707.96N

Step-by-step explanation:

Given a plate clutch that has a pair of contacting surfaces.

Inner  diameter of the friction disk, $d= 100mm$

Outer diameter of the friction disk, $D= 200mm$

Coefficient of friction, $\mu=0.2$ and

Permissible intensity of pressure, $P=1 N/mm^2$

In uniform wear theory, when the pressure is not uniform, the wear along the contact surface is assumed to be uniform.

i.e., $pr=constant$

where P is the intensity of pressure and r is the radius of the elemental ring.

And the operating force in the clutch is given by

$F=\dfrac{\pi Pd(D-d)}{2}$

Substituting the given values,

$F=\dfrac{3.14\times 1\times 100(200-100)}{2}$

$=\dfrac{\pi \times 100\times 100}{2}=15,707.96N$

Therefore, the operating force in the clutch according to uniform wear theory is 15.707.96N.

In the options given, no option has this value, only near by approximation  is option a. The options are more related to the power transmitting capacity of the clutch.

Answer 2

Answer:

The operating force in the clutch is option A

Step-by-step explanation:

• At least one clutch is a part of the powertrain and drivetrain systems in modern automobiles. Depending on the architecture and nature of the engine and powerplant, an AWD or 4WD vehicle may feature a number of clutches.
• To connect the internal combustion engine (ICE) to the gearbox, a clutch is a freestanding component. It can also be a sub-component or a key component in a torque converter, automatic transmission, transfer case, limited-slip differential (LSD), etc.

Given $\quad D=200 \mathrm{~mm} \quad d=100 \mathrm{~mm} \quad \mu=0.2 .$

$p_{a}=1 \mathrm{~N} / \mathrm{mm}^{2} \quad n=750 \mathrm{rpm} .$

Step I Operating force

From Eq. (11.6),

$\begin{aligned}&P=\frac{\pi p_{a} d}{2}(D-d) \\&P=\frac{\pi p_{a} d}{2}(D-d)=\frac{\pi(1)(100)}{2}(200-100) \\&=15707.96 \mathrm{~N}\end{aligned}$