--- title: "Number of plates in a multiplate clutch" subject: "Machine Design II" chapter: 7 key: "semester_7/machine_design_ii:multiplate-clutch-power-numerical" source: https://www.bctengineer.com/q/semester_7/machine_design_ii/multiplate-clutch-power-numerical site: BCT Engineering --- # Number of plates in a multiplate clutch Machine Design II · Chapter 7: Clutches and brakes - Asked in: year not given, 8 marks - Video: [Numerical Problem based on Multi Plate clutch।। Outside & Inner Radius।। Mean Radius।। Axial thrust](https://www.youtube.com/watch?v=MQstEPxDdkc) — RAJLAXMI CLASSES Technical & Non Technical - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/multiplate-clutch-power-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-7 ## Question A multiplate clutch must transmit 30 kW at 1200 rpm. The friction surfaces have an outer diameter of 200 mm and an inner diameter of 120 mm. The coefficient of friction is 0.12 and the maximum permissible pressure is 0.3 MPa. Assuming uniform wear, find the axial force, the torque per friction surface, the number of friction surfaces and the number of discs on the driving and driven shafts. Find also the actual axial force and the maximum pressure for the number of surfaces chosen. ## Short answer Torque to transmit: 238.7 N·m. Using uniform wear, max axial force from pressure limit is 4524 N, torque per surface 43.4 N·m; 6 friction surfaces needed (4 and 3 discs). Actual axial force 4145 N, maximum pressure 0.275 MPa. ## Answer #### Data $r_o = 100$ mm, $r_i = 60$ mm, $\mu = 0.12$, $p_{max} = 0.3$ N/mm², $P = 30$ kW, $N = 1200$ rpm. Uniform wear. #### Torque to be transmitted $$ T = \frac{P}{\omega} = \frac{30\,000}{2\pi\times1200/60} = \frac{30000}{125.66} = 238.7\ \text{N m} $$ #### Axial force from the pressure limit For uniform wear, maximum pressure is at the inner radius and $p_{max}\,r_i = C$: $$ F = 2\pi C(r_o - r_i) = 2\pi p_{max} r_i (r_o - r_i) $$ $$ F = 2\pi\times0.3\times60\times(100-60) = 4524\ \text{N} $$ #### Torque per friction surface Mean radius $R_m = \dfrac{r_o+r_i}{2} = 80$ mm. $$ T_1 = \mu F R_m = 0.12\times4524\times0.080 = 43.4\ \text{N m} $$ #### Number of friction surfaces $$ n = \frac{T}{T_1} = \frac{238.7}{43.4} = 5.5 \;\Rightarrow\; n = 6 $$ #### Number of discs Number of friction surfaces $= n = n_1 + n_2 - 1$, where $n_1$ and $n_2$ are the numbers of discs on the driving and driven shafts. $$ n_1 + n_2 = 7 \;\Rightarrow\; n_1 = 4,\ n_2 = 3 $$ (or 3 and 4). #### Actual axial force and pressure with $n = 6$ $$ F = \frac{T}{n\mu R_m} = \frac{238.7}{6\times0.12\times0.080} = 4145\ \text{N} $$ $$ p_{max} = \frac{F}{2\pi r_i(r_o - r_i)} = \frac{4145}{2\pi\times60\times40} = 0.275\ \text{MPa} < 0.3\ \text{MPa} $$ **Answer:** 6 friction surfaces (4 discs on one shaft, 3 on the other); axial force 4145 N; $p_{max} = 0.275$ MPa. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With uniform wear, the axial force from the maximum pressure is A. F = pi p_max (r_o^2 - r_i^2) B. F = 2 pi p_max r_i (r_o - r_i) C. F = 2 pi p_max r_o (r_o - r_i) D. F = p_max r_i / (r_o - r_i) Answer: B. p_max r_i = C and F = 2 pi C (r_o - r_i), giving 4524 N. (Key point: Axial force from pressure limit) 2. T = 238.7 N m and T_1 = 43.4 N m per surface. Number of surfaces is A. 5 B. 7 C. 6 D. 5.5 Answer: C. n = 238.7/43.4 = 5.5, rounded up to 6 surfaces. (Key point: Number of friction surfaces) 3. Friction surfaces n = n_1 + n_2 - 1. For n = 6 the discs are A. 4 and 3 B. 3 and 3 C. 6 and 6 D. 5 and 2 Answer: A. n_1 + n_2 = 7, so 4 and 3 discs. (Key point: Discs on shafts) 4. Why does the actual axial force drop to 4145 N with 6 surfaces? A. Because 6 surfaces transmit more torque than needed, so less force is required: F = T/(n mu R_m) B. Because friction coefficient is higher C. Because p_max is raised D. Because T changes to 200 N m Answer: A. The torque is shared by 6 surfaces, F = 238.7/(6 x 0.12 x 0.08) = 4145 N, and p_max = 0.275 MPa < 0.3. 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