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