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Designing of belts, Clutch and Gear Assignment


Question No 1
A pulley system uses a flat belt of cross section area 1000 mm square and density 1150 kg/m^3.  The angle of the lap on the smaller wheel is 130, the coefficient of friction is 0.32 and the maximum force allowed in the belt is 550N and velocity is 9 m/sec. Calculate 

  • The maximum power when centrifugal force is not included
  • The maximum power when centrifugal force is included 
  • The initial belt tensions 


Solution

  • a. Maximum Power without centrifugal force

Power= P=(T_1- T_2 )v

T_1/e^uθ = T_2

550/e^(0.32×3.14) = T_2

T_2=201.36 N 

P=(550- 201.36)9

P= 3137.71 WAns


  • b. Maximum Power with centrifugal force

Power= P=(T- T_c )  v ×C

C=1-  1/e^uθ 

C=1-  1/e^(0.32*3.14) 

C=0.63

T_c=mv^2

m=area ×length ×density

m=0.001×1×1150=1.15 kg

T_c= 1.15×9^2=93.15 N

P=(550- 93.15)  9 ×0.63=2590.33 WAns

  • c. initial belt tensions

Without Centrifugal force

T_o=((T_1+ T_2 ))/2  

T_o=(550+201.36)/2=375.68 N 

With centrifugal Force

T_o=((T_1+ T_2+2T_c ))/2

T_o=(550+201.36+2(93.15))/2  =468.83 N Ans 


Question 2
The following data is for a conical clutch
Inside diameter 35 mm
Outside diameter 150 mm
Coefficient of friction 0.35
Included angle 120
Speed 2500 rev/min

Calculate the axial forced needed to allow the transmission of 1000 watts without slipping using

  • The uniform pressure theory
  • The uniform wear theory
  • If the clutch is replaced by a multi-plate clutch with 3 contact surface for the same axial force what power can be transmitted by 


  1. The uniform wear theory
  2. Uniform pressure theory

Solution
Power= P= τω

P/ω= τ

τ=1000/((2 ×3.14 ×2500)/60)=3.82 Nm Ans

Uniform pressure theory

T_r=2 π u p_n  cosec α ((〖ro〗^3- 〖ri〗^3)/3)

3.82=2 ×3.14×0.35 × p_n× cosec 60 ×((〖0.075〗^3- 〖0.0175〗^3)/3)^

p_n=3.82/0.000152=25035 Pa

Axial Force = W = 2 π p_n  r dr

   W= 2 π ×25035×((0.075+ 0.0175)/2)×(0.075-0.0175)

W = 418.10 N Ans
Uniform Wear theory 

T_r=2 π u C cosec α ((〖ro〗^2- 〖ri〗^2)/2)

3.82=2 ×3.14×0.35 × C× cosec 60 ×((〖0.075〗^2- 〖0.0175〗^2)/2)^

C = 3.82/0.002922 = 1307.03

Axial Force = W =2 π C dr

W = 2 π ×1307.03× (0.075-0.0175)

W = 471.96 N Ans
Multi disc clutch
Uniform pressure 

Torque= T=uW/3  ×(Do^3-Di^3)/(Do-Di)  ×n 

T=  (0.35×418)/3  ×(〖0.15〗^3-〖0.035〗^3)/(0.150-0.035)  ×3 

T=48 ×0.029×3=4.17 Nm

Power=P=((2×3.14×N))/60×T

P=1091.15 WAns
Uniform wear

Torque= T=uW/3  ×(Do+Di) ×n 

T=(0.35×471.96 )/4  ×(0.15+0.035)×3

T=22.91 N

 Power=P=((2×3.14×N))/60×T

P=5994.78 WAns 

Question 3
3) The following gear train has an efficiency of 75% and the drum rotates at 75 rpm. If the torque input from the motor is 8Nm.
Calculate 

  • The number of teeth on the gear X
  • The speed at which the load is moving up as the rope on the drum
  • The power input by the motor
  • The power output by the gear train
  • The load on the drum

Solution

a. Number of Teeth
Gear ratio=  50/x  ×  100/30  ×  80/25=75/2000

50/x  ×3.34 ×3.2=0.0375

50/x=0.0375/10.688

50/x=0.003508

x=50/0.00350=14222.22=14223 Teeths Ans

b. Speed

v=r × ω

v=0.2/2  ×  (2 ×3.14 ×75)/60

v= 0.785 m/sec Ans

c. Power of motor
P= τω

P=8 ×(2 ×3.14 ×2000)/60

P = 1674.66 W Ans

d. Power output
As efficiency of the train is 75% so

P_out=0.75 × P_motor  

P_out=0.75 ×1674.66 

P_out=  1256 W Ans

e. load torque
A power at the drum is the power of gear train so

P_out= τω

τ=P_out/ω

τ=1256/((2 ×3.14 ×75)/60)

τ=160 Nm Ans 


Question 4

4) The diagram below shows two masses on two rotors in planes B and C. Determine the masses to be added to the rotor in plane A and D at a radius of 50 mm which will produce static and dynamic balance.

Solution


M mass kg
R distance mm
M*r
X distance from A
M*r*x
A
Ma
50
50Ma
0
0
B
5
20
100
80
8000
C
3
50
150
150
22500
D
Md
50
50Md
250
12500Md

From the polygon for M*r*x we have

12500Md = 25000
Md = 2 Kg Ans

Mr of point D = 50*2 = 100 at an angle of 58 degree Ans




From the polygon for M*r we have

50 Ma = 105


Ma = 2.1 Kg at an angle of 35 degree Ans







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