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Engineering Mechanics: Statics

Download Presentation Engineering Mechanics: Statics

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1. Engineering Mechanics: Statics Chapter 6B: Applications of Friction in Machines

2. 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 R1 and R2 make an angle f to the surface normal

3. 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 – R2 is of angle f to the inclined surface Fig (c) – lower surface slips – R1 is of angle f to the surface

4. 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 2f-- self-locking condition If the wedge is self-locking and is to be withdrawn, a pull P on the wedge will be required.

5. Sample Problem 6/6 ms for both pairs of wedge surfaces = 0.30 ms between the block and the horizontal surface = 0.60 Determine the least P to move the block

6. Screws Used to fastening and for transmitting power or motion For transmitting power, square thread is more efficient than the V-thread • 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)

7. 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)

8. Sample Problem 6/7 Screw has a square thread with mean diameter 25 mm and a lead of 5 mm. ms in the threads = 0.2 Apply a 300-N pull at A. This produces a clamping force of 5 kN between the jaws of the vise (a) Find frictional moment MB (b) Find force Q applied normal to the handle at A required to loosen the vise 150 mm 200 mm 250 mm

9. 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 SFy = 0; R = L SM = 0; M = Lrf = Lr sin f radius of the ‘friction circle’ For small m, (m= tan f ~ sinf) M = m Lr M = Applied moment to overcome friction

10. Journal Bearings • Unwinding the cable from this spool requires overcoming friction from the supporting shaft  • FBD of the shaft

11. 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.

12. 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, T1 and T2 if the block is being raised slowly.

13. Thrust Bearings; Disk Friction Friction between circular surfaces under distributed normal pressure Pivot bearings, clutch plates, disk brakes 1. New surface p = uniform = P/R2 M = prdA = = 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)

14. 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 = = 2KR M = prdA = = ½ mPR = ¾ times of the new plate For rings, M = ½ mP(Ro + Ri)

15. Flexible Belts: cable, rope If friction is neglected, T1 = T2 If friction is consider, M (CW) -- T2 > T1 Equilibrium SFt = 0 mdN = dT SFn = 0 dN = Tdq 1 2 Frictional moment to resist rotation

16. Flexible Belts: cable, rope  Substitute into T2/T1 = emb 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 1 2

17. 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

18. 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 0.40