 BELT DRIVES: MINI PROJECTS, engineering homework help Anonymous
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Question Description

Subject: ASSIGNMENT -- MINI PROJECTS

The assignments below were due this Wednesday.

1. BELT DRIVES: A 2-hp electric motor running at 1720 rpm is to drive a blower at a speed of 240 rpm. Select a V-belt drive for this application and specify standard V belts, sheave sizes, and the resulting center distance. The motor limits the center distance to at least 22 in.

2. CHAIN DRIVES: A roller chain is to transmit 90-hp from a 17-tooth sprocket to a 34-tooth sprocket at a speed of 300 rpm. The load characteristics are moderate shock with abnormal service conditions (poor lubrication, cold temperature, and dirty surroundings). The equipment is to run 18 h/day. Specify the length and size of chain for a center distance of about 25 pitches.

3. SPRING: Design a helical compression spring for a static load of 400 N at a deflection of 45 mm with a safety factor of 2.5. Use C=8. Specify all necessary parameters to manufacture the spring. State and justify all assumptions.

4. GEAR: Design a spur gear set transmitting 5-hp with a pinion speed of 1600 rpm and a speed ration of 3.5 for use in a propeller gear box. The center distance should be no more than 18 in. Provide justification for your decisions.

5. BEARING: A bearing is to carry a radial load of 2875 lb and a thrust load of 1350 lb at a shaft speed of 600 rpm. Assume a design life of 15, 000 hours and a bore diameter of 2.5 in.

The reports should be done professionally and typed with well-illustrated drawing using Creo or ACAD.

Ace_Tutor
School: Duke University   attached with the corrections

Question 1:
The ACAD drawing for belt drives is shown below:

From the problem it is given that

P  2 hp  2  746  1492 W
n  1720 rpm
N  240 rpm
C  22 in  558.8 mm
Assuming the light duty application working is for 10-16 hours. The design power is then
calculated by

Pd  Fa  P
 1.11492
 1641.2 W
 1.641 kW
Based on power input possible Belt type is B-section. Assumed that the bell speed is 21
m/s, then the pulley diameter is calculated by

 dn

V

60 103
60 103  V
d 
n
60 103  21
d 
3.14 1720
 d  233.18 mm  250 mm

Moreover,
n D

N d
nd
N
1720  250
D
240
 D  1791.66 mm  1800 mm
D

From the data table, we can select the standard diameters as d  250 mm and

D  1800 mm . Hence, the correct belt speed is calculated by
V

 dn

60  103
3.14  250 1720
V 
60  103
 V  22.51 m / s

The center to center distance is approximately equal to

C  Dd
 C  1800  250
 C  2050 mm
The pitch length is calculated by

L  2C 

2

D  d 

 L  2  2050 

2
 L  7613.12 mm

D  d 

2

4C

1800  250 

1800  250 

Hence, the exact pitch length is equal to L  7613 mm .
The exact center distance is then given by

4  2050

2

2
 

2

C  0.25   L   D  d     L   D  d    2  D  d  
2
2

 

2
 

2

 0.25  7613  1800  250    7613  1800  250    2 1800  250  
2
2

 


 1098  10982  300312.5
 2050 mm

Moreover, from data table, the power rating of B-section V-belt for faster pulley of 250
mm diameter is taken by the following linear interpolation
P  12.32
1720  1700

12.68  12.32 1800  1700
 P  12.392 kW

The arc of contact correction factor  Fd  is calculated by

 Dd 

 2C 
 1800  250 
 sin 1 

 2  2050 

  sin 1 

    2
   2  0.3876
With the assumption of linear interpolation, the correction factor  Fd  is calculated from
the data table as

Fd  0.87 135.58  133

0.88  0.87
136  133
 Fd  0.8786
The belt length correction factor is then calculated by, assumed the linear interpolation
holds

Fc  1.12 7613  7100

1.14  1.12 8000  7000
 Fc  1.1314
The modified power rating is calculated by

P  P  Fd  Fc
 12.392  0.8786 1.1314
 12.3182 kW
Now, since the design power is less than the modified power rating just calculated above,
it follows that the number of belt required is z  1 , as the selected B-section of V-belt is
summarized as follow:
Pitch length = 7613 mm
Diameter of small pulley = 250 mm
Diameter of large pulley = 1800 mm
Center to center distance = 2050 mm
Thus, the belt selected is B7613.

Question 2:
The ACAD drawing for chain drives is shown below:

From the problem it is given the followings: horse power = 90 hp, tooth sprocket = 17
tooth sprocket to 34 tooth sprocket, speed = 300 rpm, pitches = 25.
Therefore, the chain length in pitches is calculated by

L

N n
k
 2C 
2
C

 N  n
N n
 L
 2C 
2
39.5C
 34  17  
L

 2  25
2
 L  25.5  50  0.29
 L  75.79

2

 34  17 

2

39.5  25

Hence, L = 75.79 pitches
where

N is the number of teeth on large sprocket

n is the number of teeth on small sprocket
C is chain pitches

 N  n
k

2

39.5

is the number of teeth (large - small)

Since a fractional link cannot be used, 76 links of chain are required for the drive. Hence,
the required chain size is calculated by
24  25
12
 Size  175
Size 

Hence, the required chain size is 175 pitches.
Question 3:

The ACAD drawing for helical compression spring is shown below:

Consider diameter of the wire d  5 mm. Assume the material is ASTM A227. Ends are
parallel, and square ends. From coefficients and exponents table, we obtain that the
exponent is b  0.1822 and the coefficient values, A  1753.3 MPa .
Substituting 1753.3 MPa for A, and 0.1822 for b in Sut  Ad b , the ultimate tensile
strength is calculated using the following formula

 1753.3  50.1822
 1753.3  0.7458
 1307.69 N / mm 2
The yielding shear strength is calculated as

S ys  0.6  Sut
 0.6 1307.69
 784.61 N / mm2
Calculate the working stress using the following equation

 working 

S ys

F .O.S
784.61

2.5
 313.84 N / mm 2

The spring rate is calculated using the following equation

Force
Deflection
400

45
 8.89 N / mm

k

The number of active coils is calculated using the following equation
d4 G
8  D3  k
54  79289.71

8  403  8.889
 10.887
 N a  11
Na 

As the ends are square, the total number of coil is calculated by
Nt  N a  2
 N t  11  2
 N t  13

The shut height  Lx  is calculated as

Ls  d  N t
 5 13
 65 mm
The initial deflection is calculated by using the following equation
Initial force
k
0

36
0

yinitial 

The allowance is thus calculated by

 Lclash   0.15  yworking
 0.15  45
 6.75 mm
The free length  L f  is calculated by
L f  Ls  Lclash  yworking  yinitial
 65  6.75  45  0
 116.75 ...

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