Project Sample

 

330:148 (g) Machine Design Design Project

Design statement: Design a power transmission for an industrial saw that will be used to cut tubing for vehicle exhaust pipes to length prior to the forming process. The saw will

receive 25 hp from the shaft of an electric motor rotating at 1750 rpm. The drive shaft for the saw should rotate at approximately 500 rpm. Functions of the project:

1. To receive the the power from from an electric electric motor through a rotating rotating shaft 2. To transmit the power through machine elements that reduces the rotational speed to a desirable value 3. To deliver the power at the lower speed to an output shaft which ultimately drives the saw Design requirements

1. The reducer must transmit 25 hp 2. Input shaft speed is 1750 rpm. Use a motor motor frame size of 284T with a shaft size of 1.875 in and keyway of 0.5 × 0.5 in 3. The output speed should be around 500 rpm 4. High mechanical efficiency of say 95% 5. Minimum torque delivered delivered should be 2950 lb-in 6. Cutting operation is is smooth, but moderate shock when saw saw teeth are engaged 7. Flexible coupling are used 8. Compact size may be important since it will be mounted close to the the base of the saw. 9. Saw will operate for for 16 hours per day, 5 days a week and a life of 5 years. That means a total of 20,000 hours of operation. 10. Moderate cost for the reducer reduc er to be successful in the market 11. Production quantity may be of the order of 5000 units per year.

Selection criteria used

1. 2. 3. 4. 5. 6.

 

Safety Cost Small size High reliability Low maintenance Smooth operation

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Design alternatives examined for the various components

Belt drive Chain drive Single stage gear reducer with couplings Criteria Safety Cost Small size High reliability Low maintenance Smooth operation

Belt 6 9 5 7 6 8

Chain 6 8 6 6 5 6

Gear reducer 9 7 9 10 9 9

Based on the above, Single stage gear reducer with couplings is chosen. Detailed design of the individual components in the system such as housing, gears, shafts, bearings, seals, couplings, etc.

Gears

Start from Table 7.5 General purpose industrial drive; Pinion – 350 BHN; Gear – 300 BHN Pitch line speed = 1000 fpm; X-factor = 350; m = 3.5 Let the face width, W = 0.5 in 31,500 25 (3.5 + 1) 3 2 × ×  = 33.47 Centre distance, C  × (0.5) = 350 1750 3.5 Centre distance, C = 8.18 in Centre distance, C = 8.18 = (D p + Dg)/2 = 4.5 Dp / 2 D p = 3.636 in; Dg = 3.5 D p = 12.728 in Assuming a diametral pitch of 5, No of teeth in the pinion, Np = 5 × 3.636 = 18  No of teeth in the gear, Ng = 3.5 × 18 = 63 The actual diameter of Gear, Dg = 63 / 5 = 12.6 in The actual diameter of pinion, D p = 18 / 5 = 3.6 in The actual centre distance, C = (12.6 + 3.6 ) / 2 = 8.1 in 2 Dg Check for wear strength, F W = D p W K D p +  D g  = 3.6 × 0.5 × 196 ×

2 × 12.6 3.6

+ 12.6

 = 549 lb

Tangential force coming on the gear can be calculated from the HP to be transmitted

 

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Tangential force, F =

hp × 63,000 ω

r 

 =

25 × 63000 1750 × 1.8

= 500 lb

Hence wear strength is satisfactory Pitch line velocity, V = π × 3.6 × 1750 / 12 = 1649.336 fpm 600  = 0.2667 Velocity factor, k = 600 + 1649.336 From Table 7.2, Y = 0.308 S W Y k  20000 × 0.5 ×  0.308 × 0.2667  = 164 lb  = Allowable bending strength, F bd = 5 P This is less than the force, hence W needs to be increased. Make, W = 2 in Allowable bending strength = 4 × 164 = 656 lb Base circle diameter of pinion, = d cos φ = 3.6 × cos 20 = 3.383 in Base circle diameter of gear, = d cos φ = 12.6 × cos 20 = 11.84 in Addendum, a = 1/5 = 0.2 To calculate the contact ratio, we need Z which is given by 2 2 Z = (R + a) 2 − R  b + (r  + a) 2 − r  b − c sinφ   2

=

(1.8 + 0.2)

2

2

2

3.383   12.6 + 0.2  −  11.84  − 8.1 sin 20   −       +       2     2   2  

= 0.9811 in Circular pitch, p = π / P = π / 5 = 0.6283 in 0.9811 Z  =  = 1.6616 in Contact ratio = 0.6283 × cos 20  p cos φ Since contact ratio is greater than 1.4, it is satisfactory Final parameters as designed: The actual diameter of Gear, Dg = 12.6 in The actual diameter of pinion, D p = 3.6 in The actual centre distance, C = 8.1 in Face width, W = 2 in Diametral pitch, P = 5 Design of Shafts Input shaft:

Torque transmitted, T = 63000 × 25  = 900 lb-in 1750  

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  In addition to this there is a force of 500 lb coming from the gear set acting in the  pressure angle direction, which needs to be taken into account while calculating the shaft diameter. Assuming a shaft length of 8 inches to accommodate the gear, keyways, and bearings at  both ends, Horizontal force component = 500 × cos 20 = 469.846 lb The shear force acting at the bearing points, R = 234.9 lb Maximum bending moment, M = 234.9 × (8 / 2) = 939.69 lb-in Vertical force component = 500 × sin 20 = 171 lb The shear force acting at the bearing points, R = 85.5 lb Maximum bending moment, M = 85.5 × (8 / 2) = 342 lb-in 939.69  2

Total BM acting =

+ 342 2  = 1000 lb-in

Step 1: Equivalent torque method

900 2  + 1000 2 = 1345 lb-in

Equivalent torque, TE = Shaft dia, d =

3

16 T  E 

 =

3

16 × 1345

π   τ  

π  

× 6000

 = 1.045 in

Step 2: Equivalent bending moment method Equivalent bending moment, ME = 0.5 × (1000 + 1345) = 1172.5 lb-in

Shaft dia, d =

3

32  M  E  π  

S 

 =

3

32 × 1172.5 π  

×12000

 = 0.9984 in

Take the larger value, d = 1.045 in Output shaft:

Torque transmitted, T =

63000 × 25 500

 = 3150 lb-in

Total BM acting = 1000 lb-in (same as in Input shaft) Step 1: Equivalent torque method 2

Equivalent torque, TE =

 

2

3150   + 1000 = 3305 lb-in

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Shaft dia, d =

3

16 T  E 

 =

3

16 × 3305

π   τ  

π  

× 6000

 = 1.41 in

Step 2: Equivalent bending moment method Equivalent bending moment, ME = 0.5 × (1000 + 3305) = 2152.5 lb-in

Shaft dia, d =

3

32  M   E  E   = π   S 

32 × 2152.5  = 1.2225 in π   ×12000

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Take the larger value, d = 1.41 in Shaft diameter, therefore can be taken as 1.5 in. Keys

Keyway in the motor shaft is 0.5 length of the key Input shaft Key length, l =

× 0.5 in. Taking the same cross section, calculate the

T   = 900  = 0.2 in 0.5 × 1.8 × 5000 w r  τ   T 

Output shaft Key length, l =

w r  τ  

 =

3150 0.5 × 1.8 × 5000

 = 0.7 in

Bearings

Calculate for the output shaft: Since there is no axial load, we use single row deep groove ball bearings. The radial load = 500 lb 6 The life required = 20000 × 500 × 60 = 600 × 10  revolutions  b

3

600   C   C  =   =   ;   90 L 2  R    500  L1

Radial capacity, C = 941 lb From the tables NSK, we select 6208 bearings to suit the purpose

 

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Couplings

From Love joy S-flex inc.com/catalog/sf.pdf):

coupling

selection

procedure

(http://www.lovejoy-

Application service factor = 1.5 Design torque = 1.5 × 3150 = 4725 lb-in Size 12 coupling can be selected.

Housing Lubrication Seals

Final layout drawing of the assembly of individual component components of the designed power transmission showing all the necessary dimensions.

 

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9.5

Ø1.5

INPUT 8.1 18

OUTPUT

 

 

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