Engineering Mechanics: Statics

Transcription

1 Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines Wedges Used to produce small position adjustments of a body or to apply large forces When sliding is impending, the resultant force on each surface is inclined from the surface normal by an amount equal to the friction angle. The component of the resultant along the surface is the friction force - direction to oppose the motion of the wedge. Force P required to lift up a large mass m can be obtained from force equilibrium R 1 and R 2 make an angle f to the surface normal 1

2 Wedges If P is removed and the wedge is sliding out, slippage must occur at both upper and lower surfaces simultaneously Fig (a) upper surface slips R 2 is of angle f to the inclined surface Fig (c) lower surface slips R 1 is of angle f to the surface Wedges Therefore, the wedge will be self-locking (not sliding out when P is removed) if the wedge angle is in between the two cases or a < 2f -- self-locking condition If the wedge is self-locking and is to be withdrawn, a pull P on the wedge will be required. 2

3 Sample Problem 6/6 m s for both pairs of wedge surfaces = 0.30 m s between the block and the horizontal surface = 0.60 Determine the least P to move the block Screws Used to fastening and for transmitting power or motion For transmitting power, square thread is more efficient than the V-thread o Consider a square-threaded jack under an axial load W and a moment M, screw lead = L (advancement per revolution) mean radius = r helix angle -- To raise load W = SRcos(a + f) M = rsin(a + f) SR M = Wr tan(a + f) 3

4 Screws Conditions for unwinding If M is removed, the friction force changes direction. The screw will be self-locking if a < f. An equivalent force P = M/r must be applied to pull the thread down. For a > f, the moment is required to prevent unwinding. If a < f, M = Wr tan(f - a) If a > f, M = Wr tan(a - f) Problem 6/56 The bar clamp is being used to clamp two boards together while the glue between them cures. What torque M must be applied to the handle of the screw in order to produce a 400-N compression between the boards? The single-thread screw has square threads with a mean diameter of 10 mm an d a lead (advancement per revolution) of 1.5 mm. The effective coefficient of friction is 0.2. Neglect any friction in the pivot contact at C. What torque M is required to loosen the clamp? 4

5 Journal Bearings give lateral support to a shaft (not axial) As the shaft begins to turn in the direction shown, it will roll up the inner surface of the bearing until it slips Reaction R (caused by radial load L and torque M) made an angle f to normal SF y = 0; SM = 0; R = L M = Lr f = Lr sin f radius of the friction circle For small m, (m = tan f ~ sin f) M = m Lr M = Applied moment to overcome friction Journal Bearings Unwinding the cable from this spool requires overcoming friction from the supporting shaft FBD of the shaft 5

6 Example A torque M of 1510 N.m is applied to the 50-mm-diameter shaft of the hoisting drum to raise the 500-kg load at constant speed. The drum and shaft together have a mass of 100 kg. Calculate the coefficient of friction m for the bearing. Sample Problem The diameter of the bearing for the upper pulley is 20 mm and that for the lower puller is 12 mm. For m = 0.25 for both bearings, calculate T, T 1 and T 2 if the block is being raised slowly. 6

7 Thrust Bearings; Disk Friction Friction between circular surfaces under distributed normal pressure Pivot bearings, clutch plates, disk brakes 1. New surface p = uniform = P/ R 2 M = mprda = = 2/3 mpr -- equivalent to moment due to friction force mp acting at 2/3 R from the shaft center For ring friction disks (ex. collar bearing) 2 R - R M mp 3 R R 3 3 o i 2 2 o - i Thrust Bearings; Disk Friction 2. After wearing-in period, further wear is constant over the surface Wear depends on - circumferential distance (proportional to r) - pressure p \ rp = K P = pda = M = mprda = For rings, M = ½ mp(r o + R i ) = 2 KR = ½ mpr = ¾ times of the new plate 7

8 Problem 6/73 Circular disk A is placed on top of disk B and is subjected to a compressive force of 400 N. The diameters of A and B are 225 mm and 300 mm, respectively, and the pressure under each disk is constant over its surface. If the coefficient of friction between A and B is 0.4, determine the couple M which will cause A to slip on B. Also, what is the minimum coefficient of friction m between B and the supporting surface C which will prevent B from rotating? 400 N Flexible Belts: cable, rope If friction is neglected, T 1 = T 2 If friction is consider, M (CW) -- T 2 > T 1 Equilibrium SF t = 0 Frictional moment to resist rotation mdn = dt 1 SF n = 0 dn = Tdq 2 8

9 Flexible Belts: cable, rope Substitute 2 into 1 T 2 /T 1 = e mb b = total angle of belt contact (-- in radians!) If a rope were wrapped around a drum n times: b = 2 n radians Can also be used for a noncircular contact where the total angle of contact is b Sample Problem 6/9 Let m between the cable and the fixed drum be 0.30 (a) For a = 0, determine the maximum and minimum values which P may have in order not to raise or lower the load (b) for P = 500 N, determine the minimum value of a before the load begins to slip 9

10 Sample Problem Calculate the horizontal force P required to raise the 100-kg load. The coefficient of friction between the rope and the fixed bars is