Download presentation

1
**Chapter 5 Efficiency and Self-lock of Machineries**

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

2
**§5－1 Friction in Kinematic Pairs**

一、 Friction in the sliding pairs FN21 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 FN21, thus the friction Ff21 is v12 2 1 Ff21 F G Ff21＝f FN21 If the load G is constant , the friction Ff21 is related with the geometry of elements of pairs. G 2 1 θ 1 FN21 G 2 1 2 Planar Vee-slot Cylinder G FN21 2 FN21= G Ff21=f G/sinθ FN21= kG Ff21=f G Ff21= kf G FN21=G/sinθ k =1~π/2

3
**fv —equivalent coefficient of friction**

The friction of sliding can be calculated by the following general formula fv —equivalent coefficient of friction Ff21= f FN21= fv G φ FR21 Planar friction： fv = f v12 G F Ff21 2 1 FN21 Vee-slot friction： fv = f / sin θ Cylinder friction： fv = f Determination of total reaction force in sliding pairs： Total reaction force FR21—vector sum of FN21 and Ff21 Frictional angleφ—is the angle between FR21 and FN21. φ= arctan f The angle between FR21 and v12 is constant （90＋）.

4
**For example （一）Slope friction**

The slider 1 moves at a constant velocity on slope 2, let us consider the drive force F. n φ FN FR21 Ff v12 G F 1 ．Slider moves up 1 2 α F According to equilibrium conditions of force: F + FR21+G = 0 FR21 α+φ G From figure, F = Gtan(α+φ) 2 ． Slider moves down 1 2 α n v12 G F ’ φ F’R21 F’ According to equilibrium conditions of force: F’ + F’R21+G = 0 F’R21 G α-φ 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
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 G/2 1 2 d2 d3 d1 F 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 2 πd2 l α

7
**fv = f / sin(90°-β) = f / cosβ**

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 fv and equivalent frictional angle φv to solve this problem. fv = f / sin(90°-β) = f / cosβ φv = arctan fv Torque required in tightening a nut Torque required in unscrew off a nut

8
**二、Friction in the Revolute Pairs**

（一）Friction in shaft neck In Fig. , radial load G and drive moment of couple Md act on shaft neck 1, which turns at a constant angular speed 12 . ω12 Md G r 1 2 Mf FR21——is vector sum of FN21and Ff21 According to equilibrium conditions of force: FR21 FN21 Ff21 ρ FR21=- G Md =－Mf = FR21 ρ =Gρ

9
**Mf= F21 r = fv G r fv = kf = (1~π/2 )f**

ω12 Md G r 1 2 Mf ρ= fv r Mf= fv G r = Gρ To a specific shaft neck-bearing pair, because fv and r are constant, is also constant. is the radius of frictional circle. FR21 FN21 Ff21 ρ r 1 2 Mf ω12 Md FR21 FN21 Ff21 G Now we know that if shaft neck turns in bearing, the total reaction force FR21 will be always tangent to the frictional circle.

10
**（二）Friction at the shaft end**

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

11
**§5－2 Efficiency of Machineries**

When a machine is running: Wd= Wr+Wf Efficiency of Machineries is the ratio of output work to input work, usually denoted by η. F vF F0 vF η＝Wr / Wd ＝1－Wf /Wd If Nd ,Nr and Nf represent input power, output power and lost power respectively, we have η＝Nr / Nd ＝1－Nf /Nd Ideal machineryη0 Actual machineryη G vG η＝ F0 / F ＝M0 / M η＝Nr / Nd ＝G vG /F vF η0＝Nr / Nd ＝G vG /F0vF =1

12
**Example 1. Slope F = Gtan(α+φ) F0 = Gtanα η= F0 / F = tanα / tan(α+φ)**

η ’ = G / G0 = F0 / F ’ = tan(α-φ) / tanα 2. Helical pairs screw unscrew 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**

1. Serial structure ＝123…k 2.Parallel structure 3. Parallel-serial structure Its total efficiency can be calculated by combining above two equations.

Similar presentations

OK

Two-Dimensional Rotational Dynamics W09D2. Young and Freedman: 1

Two-Dimensional Rotational Dynamics W09D2. Young and Freedman: 1

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on history of atoms Ppt on you can win pdf Ppt on computer animation English 8 unit 12 read ppt online Ppt on switching network environment Download ppt on sets for class 11 Ppt on job rotation program Ppt on atrial septal defect Ppt on project tiger for class 10 Ppt on information communication technology