Belt Drive

 

ENGD2005- Theory of Machines

Belt Friction Experiment Objective

This experiment has three objectives:

 1.  To show that a constant angle of lap θ has a constant tension ratio of    2.  To verify the relationship of 

=    



 3.  To determine the friction  between the CI pulley and the belt.





Theory

As given in the lab sheet; For a flat belt drive:

 =     

T1 = Tight side tension T2 = Slack side tension µ = Coefficient of friction θ = Angle of lap (radians) (radians)

If a dependence of T1 is plotted against T2 for a constant angle of lap θ then a straight line should pass

 through the origin because    is constant. T1 = k ..T T2   =    =   or The slope k  can  can be measured and it can yield a value for the coefficient of friction  .  =    =      ln   ln k =   =  = ln  

 If ln  is plotted against θ then the resulting curve should be a straight line that passes through the origin. 

 The slope for this line equals µ: ln 

 = µ 

Apparatus

       









Belt friction rig Dial test indicator Weight carrier 2 – 25  – 25 kg weights

Figure 1: Belt friction rig with weight carrier and weights 

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Figure 2: Dial test indicator

P15219444

 

ENGD2005- Theory of Machines

Procedure

As given in lab sheet; Cantilever calibration 1.  Turn angle of lap protractor to 0 2.  Hang belt and weight carrier on the cantilever 3.  Set the dial test indicator to 0 4.  Measure the deflection of the cantilever for masses from 0 – 0  – 24kg  24kg in steps of 3kg. 5.  Plot T2 against deflection for the cantilever 6.  Use the plot as a calibration curve to determine T2 on the slack side against deflection Constant angle of lap 1.  Set lap θ to 90⁰, attach the belt and set the dial indicator to 0  0  2.  Switch the pulley motor on 3.  Measure the deflection for masses m asses 0-20kg 4.  Using the calibration curve compute T2 and plot T1 against T2. 5.  Calculate µ using first method in theory section Variable angle of lap 1.  Attach belt and set the test indicator to 0 2.  Attach 20kg to the belt 3.  Switch pulley motor on 4.  Measure deflection for angle of lap θ from 240⁰ to 10⁰ 10⁰   5.  Determine T2 for each angle with aid of the calibration curve 6.  Plot

rad   T T against θ, rad 

7.  Record µ from the slope of the curve

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ENGD2005- Theory of Machines

Results

1.  Cantilever calibration m (kg) T₂ = mg (N)  (N) 

0

3

6

9

12

15

18

21

24

0

15

30.5

45

60

75

90

104

118

0

29.43

58.86

88.29

117.72

147.15

176.58

206.01

235.44

Table 1: Cantilever calibration results

Calculation T2 = m × g T2 for 3 kg = 3 kg × 9.81 (earth gravitational force) T2 for 3 kg = 29.43 N

Slack side (T2) against Deflection () 250 y = 1.9882x 1.9882x - 1.019 R² = 0.9998

200     ) 150    N     (   ₂    T 100

50 0 0

20

40

60

80

100

120

140



Figure 3: Calibration curve to show tension on the slack side against the cantilever deflection

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ENGD2005- Theory of Machines 2.  Constant angle of lap m (kg)

0

2

4

6

8

10

12

14

16

18

20

0

5

11

18

23

29

34

40

45

51

58

T₁ = mg (N)  (N) 

0

19.62

39.24

58.86

78.48

98.1

117.72

137.34

156.96

176.58

196.2

T₂ = k x δ  δ 

0

9.94

21.868

35.784

45.724

57.652

67.592

79.52

89.46

101.388

115.304

Table 2: Constant angle of lap results

Calculation T2 = k × δ  T2 = 1.988 × 5 T2 = 9.94 N

 Given the equation in the lab sheet 

 =

T1 = 19.62 N T2 = 9.94 N K = 1.988 θ = π/2 /2  

   =  

 

 = 1.988  ln(1.988)   × () = ln(1.988)  =   ×0.687   = 0.437  Graph of T1 against T2 250 200

y = 1.719x + 0.5495

    )    N     (    g 150    m   = 100   ₁    T

50 0 0

20

40

60

80

100

120

140

T₂ = k x δ (N) Figure 4: T1 against T2 graph

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ENGD2005- Theory of Machines 3.  Variable angle of lap Θ, deg δ 

240

220

200

180

160

140

120

100

80

60

40

20

10

25

26

29

33

37

41

47

53

60

67

76

86

92

T2, N

49.7

51.69

57.65

65.60

73.56

81.51

93.44

105.36

119.28

133.2

151.09

170.97

182.9

4.19

3.84

3.49

3.14

2.79

2.44

2.09

1.75

1.4

1.05

0.7

0.35

0.17

3.95

3.80

3.40

2.99

2.67

2.40

2.10

1.86

1.64

1.47

1.30

1.15

1.07

1.37

1.34

1.22

1.10

0.98

0.88

0.74

Θ, rad  

     

0.62

0.49

0.39

0.26

0.14

Table 3: Variable angle of lap results

 =  . .   = 0.327 

T1 = m × g T1 = 20 kg × 9.81 T1 = 196.2 N

T2 = K × δ  T2=1.988 x 25 T2=49.7

Graph of T1/T2 against angle of lap θ 4.5 4 3.5

y = 0.7395x + 0.7337

3        2        T2.5         /        1 2        T1.5 1 0.5 0 0

1

2

3

4

5

Angle of Lap θ (radians) Figure 5: Graph of T1/T2 against angle of lap

Graph of LnT1/T2 against angle of Lap θ(radians) 1.6 1.4

y = 0.3364x + 0.0294

1.2        2        T         /        1        T      n        L

1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

Angle of Lap θ (radians) Figure 6: Graph of ln(T1/T2) against angle of lap(rad)

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0.068

 

ENGD2005- Theory of Machines

Conclusion 1.  Is the tension ratio

 constant for a given angle of lap? 

The tension ratio for a given angle of lap is constant. This can be seen from the results from table 2. This can be further proved as when experiment 2 was conducted, the value of  stayed fairly

 constant through all masses.  = 



  shows that   is a constant. From From graph 2 we can see the

relationship between T1 and T2 gives us a gradient of 1.719 and the line is fairly linear. From experiment 3 when the angle is changed, the ratio of T1/T2 begins to decrease as the angle of lap decreases.



2.  What is the value of  between the belt and cast iron pulley and by considering the two values obtained, what is the experimental error? From the second experiment the value for  is 0.437 From the third experiment the value for  is 0.327







The  value between the belt and the pulley should be the same. However, as you can see from the above two values this is not the case. The difference in values is 0.11. This has a percentage error of 25%. This error is quite significant, but there are a few errors that could have caused this. Apparatus error could have occurred as the weights used may have not been the weight stated by the manufacturer. Another error could be hysteresis error which is the error that is a result of the histories that the piece of apparatus has gone through and apparatus error which is due to the inaccuracy of the apparatus by default. Random error may have also played a part in the experimental values not meeting the theoretical value as when the weight was placed in the middle of the beam, it was placed using the naked eye, which could have been misread. There are many ways in which this experiment could have been more accurate and minimise the risk or errors. However, there is not one specific method that could eliminate elim inate all errors and give the perfect reading. Below you will see a brief description on how the errors mentioned earlier could have been minimised. Apparatus error could be minimised by either getting more accurate measurement devices, or having the equipment maintained on a regular r egular basis. Hysteresis error could have been prevented by using a new belt Random error could have been prevented if an electronic device recorded the results.

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ENGD2005- Theory of Machines

 3.  Have you verified the relationship 

 =

?

The experiment conducted shows a strong correlation between T1/T2 and  ; The value for the two are very close to each other. From table 3 we can use 4.19 radians as an example, the value for T1/T2 is 3.95.



eµθ= e0.327 x 4.19 = 3.936. The equation above shows that the value from the equation is 3.94. The difference between the two values is 0.25, which gives a percentage error of 10.2%. This error is not as significant as the difference of , but could still have been avoided as mentioned earlier.



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