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In checking for self-locking, use a value for f that is 25 to 50 percent greater than the maximum value cited by the lining manufacturer. Example 8. A band brake like that in Fig. 8.19 is needed to exert a braking torque of 3100 lb in (350 N m) on a drum with 10-in (0.254-m) diameter. The actuating force Q (exerted by the operator s foot) should not have to exceed 25 lb (111 N). Limit the maximum contact pressure to 60 pounds per square inch (psi) [0.414 megapascal (MPa)]. The value for f is 0.31 0.03. (1) Make sure the brake will not be self-locking for an f value 30 percent above the maximum value. (2) Calculate the bandwidth b to limit the contact pressure. (3) Find the length for the operating lever. (4) For the same actuating force, what is the braking torque if the drum s rotation is reversed Solution. A scale layout indicates that a = 1 in (0.0254 m) and = 200 are feasible values when the lever s pivot point A is placed directly below the center of the drum. Then c = 5 in (0.127 m), corresponding to the drum s radius. 1. First make sure that the brake will not be self-locking. Use f = 1.3(0.31 + 0.03) = 0.442 and = 200 /57.296 = 3.491 rad. Then, from Eq. (8.47), a< 5 0.442(3.491) or a < 1.069 in

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So the dimension a = 1 in will do nicely. 2. Now select a bandwidth b so that pmax does not exceed 60 psi (0.414 MPa). By Eqs. (8.42) and (8.43), P1 = e 0.28(3.491) = 2.658 P2 3100 = (P1 P2) 10 2

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Solving these for P1 and P2 gives P1 = 994 lb (4420 N) and P2 = 374 lb (1664 N). So, by Eq. (8.44), the bandwidth is b= 2 P1 2(994) = = 3.31 in (say 3.5 in) p max D 60(10)

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3. Next the moment arm length Eq. (8.45): =

Summary

4. Finally, the torque capacity of the brake for the same actuating force, but with the drum rotating counterclockwise, can be evaluated. Since the drum turns counterclockwise, the braking torque must be clockwise. Thus, forces P1 and P2 are interchanged. The larger force P1 is applied at point C, and the force P2 is applied at point B. Equation (8.45), suitably rewritten, is P 1c P 2 a Q = 0 But Eqs. (8.42) and (8.43) do not need to be changed: P1(5) P2(1) 25(35) = 0 or 5P1 P2 = 875 P1 = e f = 2.658 P2 Then P1 = 189 lb and P2 = 70 lb. And the net braking torque is T = (P1 P2) D 10 = (189 70) = 595 lb in 2 2

This is considerably less than the 3100-lb in capacity for clockwise rotation of the drum. 8.6.2 Cone Brakes and Clutches Two mating cones kept in contact by an axial force can be used as a clutch, as in Fig. 8.20, or as a brake, as in Fig. 8.21. A small cone angle produces a wedging action, and a large torque capacity is achieved for a small actuating force. But if the cone angle is too small, it becomes difficult to disengage the cones. A cone angle of 10 to 15 is a reasonable compromise. Basic Relationships. A cone is shown schematically in Fig. 8.22 with three elementary quantities indicated: area dA, normal contact force dP, and actuating force dF. From these it follows that the elementary torque dT is dT = fr dP = frp 2 r dr 2 f pr 2 dr = sin sin (8.48)

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Similarly, the elementary actuating force dF is dF = dP sin = 2 pr dr (8.49)

The actuating force F and the torque capacity T are found by integrating in Eqs. (8.49) and (8.48) from the inside radius d/2 to the outside radius D/2:

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