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§5 - 1 Friction in Kinematic Pairs §5 - 1 Friction in Kinematic Pairs §5 - 2 Efficiency of Machineries §5 - 2 Efficiency of Machineries Chapter 5 Efficiency.

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Presentation on theme: "§5 - 1 Friction in Kinematic Pairs §5 - 1 Friction in Kinematic Pairs §5 - 2 Efficiency of Machineries §5 - 2 Efficiency of Machineries Chapter 5 Efficiency."— Presentation transcript:

1 §5 - 1 Friction in Kinematic Pairs §5 - 1 Friction in Kinematic Pairs §5 - 2 Efficiency of Machineries §5 - 2 Efficiency of Machineries Chapter 5 Efficiency and Self-lock of Machineries

2 The slider 1 acted on by drive force F moves forward at a constant velocity, considering vertical load G acting on the slider, normal reaction force F N21, thus the friction F f21 is v 12 一、 Friction in the sliding pairs F f21 = f F N21 G F F f21 2 1 If the load G is constant, the friction F f21 is related with the geometry of elements of pairs. 1 F N21 G 2 Planar Vee-slot F N21 = G F f21 =f G F N21 =G/sin θ F f21 =f G/sin θ G 2 1 θθ G 1 2 F N21 = kG F f21 = kf G k =1~ π /2 F N21 2 F N21 2 § 5 - 1 Friction in Kinematic Pairs Cylinder

3 The friction of sliding can be calculated by the following general formula f v —equivalent coefficient of friction Planar friction : f v = f Total reaction force F R21 —vector sum of F N21 and F f21 F f21 = f F N21 = f v G Vee-slot friction : f v = f / sin θ Cylinder friction : f v = f Determination of total reaction force in sliding pairs : Frictional angleφ—is the angle between F R21 and F N21. φ= arctan f The angle between F R21 and v 12 is constant ( 90  +  ). v 12 G F F f21 2 1 F N21 φ F R21

4 1 2 α For example (一) Slope friction The slider 1 moves at a constant velocity on slope 2, let us consider the drive force F. 1 . Slider moves up According to equilibrium conditions of force: F + F R21 +G = 0 F R21 G F n n v 12 G F φ FNFN F R21 FfFf  From figure, F = Gtan(α+φ) 2 . Slider moves down F’F’ F’ R21 G 1 2 α n n v 12 G F ’ φ F ’ R21  According to equilibrium conditions of force: F ’ + F ’ R21 +G = 0 From figure, F ’ = Gtan(α-φ)

5 (二) Friction in the Helical Pairs 1.Friction in the rectangular-thread helical pairs A rectangular-thread helical pair consists of bolt 1 and nut 2. We can suppose that forces produced between bolt 1 and nut 2 all act on the cylinder surface. If we expand this cylinder surface the thread will form an inclined plane with a slope angle α. Now friction in helical pairs is simplified into friction of slider on the slope.

6 1 2 d2d2 d3d3 d1d1 G/2 πd2πd2 l α F F If nut 2 moves up at a constant speed under the acting of the torque M and axial load G, this is similar to the slider moving along the slope at a constant speed. So we obtain 2 G G 1 If nut 2 moves down at a constant speed under the acting of the torque M and axial load G, this is similar to the slider moving along the slope at a constant speed. So we obtain

7 2. Friction in the triangular-thread helical pairs Friction in the triangular-thread helical pairs is very similar to friction in the rectangular-thread helical pairs. We can introduce equivalent coefficient of friction f v and equivalent frictional angle φ v to solve this problem. f v = f / sin(90°- β) = f / cos β φ v = arctan f v Torque required in tightening a nut Torque required in unscrew off a nut

8 r 1 2 二、 Friction in the Revolute Pairs (一) Friction in shaft neck In Fig., radial load G and drive moment of couple M d act on shaft neck 1, which turns at a constant angular speed  12. F R 21 ——is vector sum of F N21 and F f 21 According to equilibrium conditions of force: MfMf F R21 F N21 F f21 ω 12 MdMd G F R21 = - G M d = - M f = F R21 ρ =G ρ ρ

9 r 1 2 MfMf F R21 F N21 F f21 ω 12 MdMd G ρ M f = F 21 r = f v G r f v = kf = ( 1~ π /2 )f M f = f v G r = G ρ ρ = f v r To a specific shaft neck-bearing pair, because f v and r are constant,  is also constant.  is the radius of frictional circle. Now we know that if shaft neck turns in bearing, the total reaction force F R21 will be always tangent to the frictional circle. r 1 2 MfMf ω 12 MdMd F R21 F N21 F f21 G

10 (二) Friction at the shaft end Shaft end is the part of shaft that sustains axial load. When shaft end 1 turns on thrust bearing 2, frictional torque M f is generated between them. 1.For new shaft end To new shaft end, we may suppose the pressure on the contact surface p is constant, so we have To old shaft end, p  = constant. We have 2r 2R 2 1 G ω M MfMf 2. For old shaft end

11 Efficiency of Machineries 一、 Efficiency of Machineries When a machine is running: W d = W r +W f Efficiency of Machineries is the ratio of output work to input work, usually denoted by Efficiency of Machineries is the ratio of output work to input work, usually denoted by η. If N d,N r and N f represent input power, output power and lost power respectively, we have η = N r / N d = 1 - N f /N d η = W r / W d = 1 - W f /W d F vFvF Actual machinery η G vGvG η = N r / N d = G v G /F v F η 0 = N r / N d = G v G /F 0 v F =1 Ideal machinery η 0 η = F 0 / F = M 0 / M F0F0 vFvF § 5 - 2 Efficiency of Machineries

12 Example 1. Slope F = Gtan(α+φ)F 0 = Gtanα η= F 0 / F = tanα / tan(α+φ) F ’ = Gtan(α-φ) 2. Helical pairs screw unscrew F 0 = Gtanα η ’ = G / G 0 = F 0 / F ’ = tan(α-φ) / tanα Efficiency of machineries may also be calculated by experiment method. The efficiencies of some general mechanisms and pairs are listed in Table 5 - 1.

13 二、 Efficiency of a group of mechanisms 3. Parallel-serial structure Its total efficiency can be calculated by combining above two equations. 2.Parallel structure =123…k=123…k 1. Serial structure


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