# BELT DRIVES: MINI PROJECTS, engineering homework help

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*timer*Asked: Jul 4th, 2017

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### Question Description

**Subject:** ASSIGNMENT -- MINI PROJECTS

The assignments below were due this Wednesday.

In addition, please complete the following mini projects:

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.

## Tutor Answer

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

1.11492

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 Dd

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

Dd

2C

1800 250

sin 1

2 2050

sin 1

22.21o or 0.3876 rad

2

2 0.3876

2.3663 rad or 135.58o

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

Sut Ad b

1753.3 50.1822

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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