Chapter 18 Clutches and Brakes

Equations and Calculation 1. Assume uniform distribution of interface pressure 2. Assume uniform rate of wear at interface 3. Sample Problem: Multiple Disk Wet Clutch 4. Short Shoe Drum Brakes 5. Sample Problem: Two shoe External Drum Brake

Assume uniform distribution of interface pressure: 1. Normal Force acting on a differential ring element of radius r: 𝑑𝐹= 2𝜋𝑟𝑑𝑟 𝑝 𝐹=𝑝 𝑟 𝑖 𝑟 𝑜 2𝜋𝑟𝑑𝑟=𝜋𝑝( 𝑟 0 2 − 𝑟 𝑖 2 ) 2. Friction Torque: 𝑑𝑇= 2𝜋𝑟𝑑𝑟 𝑝𝑓𝑟 𝑇= 𝑟 𝑖 𝑟 𝑜 2𝜋𝑝𝑓 𝑟 2 𝑑𝑟 = 2 3 𝜋𝑝𝑓( 𝑟 𝑜 3 − 𝑟 𝑖 3 ) 3. N friction interfaces: 𝑇= 2𝐹𝑓( 𝑟 0 3 − 𝑟 𝑖 3 ) 3( 𝑟 𝑜 2 − 𝑟 𝑖 2 ) 𝑁 𝑇= 2 3 𝜋𝑝𝑓 𝑟 𝑜 3 − 𝑟 𝑖 3 N

Assume uniform rate of wear at interface New Clutch with uniform distribution of interface pressure Question: Where is the greatest initial wear ? Answer: Outer radius Why? Wear Rate ∝ Rate of friction work 𝐹 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑉 𝑟𝑢𝑏𝑏𝑖𝑛𝑔 𝑉 𝑟𝑢𝑏𝑏𝑖𝑛𝑔 ∝ 𝑟 𝑐𝑙𝑢𝑡𝑐ℎ 𝑓𝑎𝑐𝑒 After initial wear, the friction lining tends to wear at a uniform rate from a uniform rate of friction work 𝑝𝑟=𝐶= 𝑝 𝑚𝑎𝑥 𝑟 𝑖 Normal Force: 𝐹= 𝑟 𝑖 𝑟 𝑜 2𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 𝑑𝑟=2𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 ( 𝑟 𝑜 − 𝑟 𝑖 ) Friction Torque: T= 𝑟 𝑖 𝑟 𝑜 2𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 𝑓𝑟𝑑𝑟𝑁=𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 𝑓 𝑟 𝑜 2 − 𝑟 𝑖 2 𝑁 𝑇=𝐹𝑓( 𝑟 𝑜 + 𝑟 𝑖 2 )

Calculated clutch capacity(uniform wear rate) < Calculated clutch capacity (uniform pressure) ∵𝑆𝑚𝑎𝑙𝑙𝑒𝑟 𝑡𝑜𝑟𝑞𝑢𝑒 𝑎𝑟𝑚 (ℎ𝑖𝑔ℎ𝑒𝑟 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑤𝑒𝑎𝑟 𝑡𝑜𝑤𝑎𝑟𝑑 𝑡ℎ𝑒 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑠ℎ𝑖𝑓𝑡𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑡𝑜𝑤𝑎𝑟𝑑 𝑡ℎ𝑒 𝑖𝑛𝑠𝑖𝑑𝑒 A parameter in the design of clutches – the ratio of inside to outside radius 𝑇=𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 𝑓 𝑟 𝑜 2 − 𝑟 𝑖 2 𝑁 18.6 𝑑𝑇 𝑑 𝑟 𝑖 = 𝜋𝑝 𝑚𝑎𝑥 𝑓𝑁 𝑟 0 2 −3 𝑟 𝑖 2 =0 (𝑟 0 2 −3 𝑟 𝑖 2 )=0 𝑟 𝑖 =0.58 𝑟 0 Usually choose 𝑟 𝑖 =0.45 𝑡𝑜 0.8 𝑟 𝑜

Sample Problem: A multiple disk wet clutch is to be designed for transmitting a torque of 85 Nm. Space restrictions limit the outside disk diameter to 100 mm. Design values for the molded friction material and steel disks to be used are f=0.06 (wet) and 𝑝 𝑚𝑎𝑥 = 1400𝑘𝑃𝑎. Determine appropriate values for the disk inside diameter, the total number of disks and the clamping force. Outside disk diameter = 100 mm  Radius = 50mm Select 𝑟 𝑖 =29𝑚𝑚 (2) Use Equation 18.6: 𝑁= 𝑇 𝜋 𝑝 𝑚𝑎𝑥 𝑟 𝑖 𝑓( 𝑟 0 2 − 𝑟 𝑖 2 ) =6.69 (𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝐼𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒𝑠) N must be even integer  N=8 Total number of disks: 9 disks =4+5

Clamping Force: 𝑇=𝐹𝑓 𝑟 0 + 𝑟 𝑖 2 𝑁 85=𝐹 0.06 0.05+0.029 2 8 𝐹=4483𝑁

Short-shoe Drum Brakes 1. External shoes that contract to bear against the outer drum surface 2. Internal shoes that expand to contact the inner drum surface Shoe only contacts a small segment of the drum periphery; Force F at the end of the lever applies the brake; Normal force N and friction force fN are distributed continuously over the contacting surface, but assume these force are concentrated at center of contact Figure 18.6 External drum brake with"Short-shoe"

Taking moments about pivot A for the shoe and lever assembly: Fc + fNa - Nb=0 Summation of moments about O for drum: T=fNr Inertial or load torque = friction torque If drum rotate clockwise(counter clockwise), friction force assist (prevent) applied force F to brake, self-energizing(self-deenergizing) T=fFcr/(b+fa) T=fFcr/(b-fa) self-energizing Reverse rotation of drum: T=fFcr/(b+fa) self-deenergizing

T=fFcr/(b-fa) self-energizing 𝑏≤𝑓𝑎 self-locking Eg: If f=0.3, b<=0.3a. Self-locking requires that shoe be brought in contact with the drum(with F=0) for the drum to be locked against rotation in one direction.

Two Shoe External Drum Brake The two shoe external drum brake shown in Figure 18.7 has shoes 80 mm wide that contact 90 degree of drum surface. For a coefficient of firction of 0.20 and an allowable contact pressure of 400 kN per square meter of projected area, estimate (a) the maximum lever force F that can be used, (b) the resulting braking torque and (c) the radial load imposed on the shaft bearings. Use the derived short shoe equations.

- No Vertical Force for component 4 Moments about point 2,5 𝐹∗400= 𝐻 45 ∗100 - Horizontal Force 𝐻 45 =4𝐹 - No Vertical Force for component 4 𝐻 45 = 𝐻 43 =4𝐹 Summation of moments about 𝑂 13 : 4𝐹∗700+0.2 𝐻 63 ∗170− 𝐻 63 ∗300=0 𝐻 63 =10.53𝐹 𝐻 25 =4𝐹, 𝑉 25 =𝐹 Summation of moments about O 12 : 4𝐹 600 −𝐹40− 𝐻 62 300 −0.2 𝐻 62 170 =0 𝐻 62 =7.07𝐹 R=250 L= 250-80=170 Assume the drum angular acceleration is zero: Load Torque T: 𝑇= 2.11𝐹+1.41𝐹 ∗ 𝑅 𝑑𝑟𝑢𝑚 T = 880F Force applied at fixed pivot O: 𝐻 16 = 𝐻 36 − 𝐻 26 =3.46𝐹 𝑉 16 = 𝑉 36 − 𝑉 26 =0.70𝐹

𝐴=80 2 250𝑠𝑖𝑛 45 𝑜 =28284 𝑚𝑚 2 P=10.53F/28284=0.0003723F N/mm^2 𝑝 𝑚𝑎𝑥 =0.40𝑁/ 𝑚𝑚 2 0.0003723F=0.40  F=1074N Brake Torque: T=880F=880(1074) = 945 N m Resultant radial load transmitted to the bearings is: 0.7 2 + 3.46 2 𝐹=3.53𝐹=3791𝑁