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Question: A multi-disk clutch consists of two steel disks with one bronze disk. The inner and outer diameters of the contacting surfaces are 200 and 250 mm respectively. The coefficient of friction is 0.1 and the maximum pressure between the contacting surfaces is limited to 0.4 MPa. Assuming uniform wear theory, design the required force to engage the clutch and the power transmitting capacity at 720 rpm.
To design the required force to engage the clutch and the power transmitting capacity at 720 rpm, we need to follow these steps: Step 1: Calculate the mean radius (R) of the contacting surfaces: R = (Outer Diameter + Inner Diameter) / 2 / 2 R = (250 + 200) / 2 / 2 R = (450) / 4 R = 112.5 mm Step 2: Calculate the contact area (A) between the two steel disks and the bronze disk: A = π * (Outer Diameter² - Inner Diameter²) / 2 A = π * (250² - 200²) / 2 A = π * (62500 - 40000) / 2 A = π * 22500 / 2 A = π * 11250 A ≈ 35343.3 mm² Step 3: Calculate the maximum force (F_max) acting at the contacting surfaces using the maximum pressure value: F_max = Maximum Pressure * Contact Area F_max = 0.4 MPa * 35343.3 mm² F_max = 400 N/mm² * 35343.3 mm² F_max = 14137320 N Note that since there are two contact surfaces in a multi-disk clutch, the force acting on both surfaces is: Total F_max = 2 * F_max Total F_max = 2 * 14137320 N Total F_max = 28274640 N Step 4: Calculate the required force (F_required) to engage the clutch using the coefficient of friction (μ): F_required = F_max * μ F_required = 28274640 N * 0.1 F_required = 2827464 N Step 5: Calculate the torque (T) that can be transmitted by the clutch: T = F_required * Mean Radius T = 2827464 N * 0.1125 m T = 318088.8 Nm Step 6: Calculate the power transmitting capacity (P) at 720 rpm: P = 2 * π * Torque * RPM / 60 P = 2 * π * 318088.8 Nm * 720 / 60 P = 6.2832 * 318088.8 Nm * 12 P ≈ 23886239.3 W or 23.89 MW So, the required force to engage the clutch is approximately 2827464 N, and the power transmitting capacity at 720 rpm is around 23.89 MW.
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