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Question

A single plate clutch has outer and inner radii 50 mm and 20 mm respectively. An axial clamping force required to engage the clutch is 2 kN and coefficient of friction between liner material is 0.4, the torque carrying capacity of the clutch using uniform wear theory is

The correct answer is

28 Nm

Understanding Clutch Torque Calculation Using Uniform Wear Theory

This solution details the calculation for the torque carrying capacity of a single-plate clutch, specifically applying the uniform wear theory. We will identify the given parameters, recall the appropriate formula, and perform the calculation step-by-step.

Given Parameters for the Clutch

  • Outer Radius ($R_o$): 50 mm
  • Inner Radius ($R_i$): 20 mm
  • Axial Clamping Force ($F$): 2 kN
  • Coefficient of Friction ($\mu$): 0.4

Concept of Uniform Wear Theory

In clutch design, the uniform wear theory is a common assumption. It posits that the wear rate across the friction surface is constant. Mathematically, this means the product of pressure ($p$) and radius ($r$) is constant ($p \cdot r = \text{constant}$). This condition influences the formula used to calculate the torque capacity.

Formula for Torque Calculation

For a single-plate clutch operating under the uniform wear theory, the torque ($T$) that the clutch can transmit is given by the formula:

$T = \frac{1}{2} \mu F (R_o + R_i)$

Where:

  • $T$ is the torque carrying capacity.
  • $\mu$ is the coefficient of friction.
  • $F$ is the axial clamping force.
  • $R_o$ is the outer radius of the friction surface.
  • $R_i$ is the inner radius of the friction surface.

Step-by-Step Calculation Process

First, we must ensure all measurements are in consistent SI units. Convert millimeters (mm) to meters (m) and kilonewtons (kN) to Newtons (N):

  • Outer Radius ($R_o$) = 50 mm = $\frac{50}{1000}$ m = 0.05 m
  • Inner Radius ($R_i$) = 20 mm = $\frac{20}{1000}$ m = 0.02 m
  • Axial Clamping Force ($F$) = 2 kN = $2 \times 1000$ N = 2000 N

The coefficient of friction ($\mu$) is given as 0.4.

Now, substitute these converted values into the torque formula:

$T = \frac{1}{2} \times \mu \times F \times (R_o + R_i)$
$T = \frac{1}{2} \times 0.4 \times 2000 \, \text{N} \times (0.05 \, \text{m} + 0.02 \, \text{m})$

Calculate the sum of the radii:

$R_o + R_i = 0.05 \, \text{m} + 0.02 \, \text{m} = 0.07 \, \text{m} $

Now, continue with the torque calculation:

$T = 0.5 \times 0.4 \times 2000 \, \text{N} \times 0.07 \, \text{m} $
$T = 0.2 \times 2000 \, \text{N} \times 0.07 \, \text{m} $
$T = 400 \, \text{N} \times 0.07 \, \text{m} $
$T = 28 \, \text{Nm} $

Final Result

The calculated torque carrying capacity for the single-plate clutch, based on the uniform wear theory, is 28 Nm. This value aligns with one of the options provided.

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Important Questions from Friction Clutches

  1. Which of the following statement is CORRECT about the dry clutch?

  2. In a multiple disc clutch, if there are 4 discs on the driving shaft and 3 discs on the driven shaft, then the number of pairs of contact surfaces will be equal to _______.
  3. In designing a plate clutch, assumption of uniform wear conditions is made because:

  4. Match List-I with List-II and select the correct answer using the codes given below the lists.

    List - IList - II
    J.Single-plate clutch 1.Scooters
    K.Multi-plate clutch 2.Rolling mills
    L.Centrifugal clutch 3.Trucks
    M.Jaw clutch 4.Mopeds
  5. The total normal force (F) acting on the area of contact of disc clutch (Assuming uniform distribution of interface pressure) is:

    (Where P is the intensity of pressure, r0 and ri outer and inner diameter of disc respectively)
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