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Chapter: Mechanical - Design of Transmission Systems - Design of Transmission Systems for Flexible Elements

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Design of Transmission Systems for Flexible Elements

1 Selection Of V Belts And Pulleys, 2 Selection Of Flat Belts And Pulleys, 3 Wire Ropes And Pulleys, 4 Selection Of Transmission Chains And Sprockets, 5 Solved Problems.


DESIGN OF TRANSMISSION SYSTEMS FOR FLEXIBLE ELEMENTS

 

 

V- BELT

 

Selection of V belts and pulleys

 

ü Determine your drive requirements. How much power do you need to transmit and at what speed?

 

INTRODUCTION:

 

V- Belts are one type of flexible connectors for transmitting power from one pulley to another pulley. Whose center distance is approx. 3M.thier cross section is trapezoidal. The belts are operated on groove pulleys.

 

MATERIALS USED:

 

·        Cord

 

·        Fabric

 

·        Cotton

 

·        Rayon

 

 

 

 

 

POWER TRANSMISSION

 

 

Belts are the cheapest utility for power transmission between shafts that may not be axially aligned. Power transmission is achieved by specially designed belts and pulleys. The demands on a belt drive transmission system are large and this has led to many variations on the theme. They run smoothly and with little noise, and cushion motor and bearings against load changes, albeit with less strength than gears or chains. However, improvements in belt engineering allow use of belts in systems that only formerly allowed chains or gears.

 

Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity

 

P= (T1 -T2 ) v

where, T1 and T2 are tensions in the tight side and slack side of the belt respectively. They are

 

related as:




TYPES OF V BELT

 

Generally V belts are classified into various grades based on their power transmitting capacity as A, B, C, D and E. the cross sectional areas are increased order from A –E

 

SELECTION OF V BELTS AND PULLEYS

 

V belts are designed based on

 

1. Fundamental formula

2. Manufactures catalogues

FUNDAMENTAL FORMULA:


2.       Power Transmitted by belt

 


 

MANUFACTURES CATALOGUES

 

1.       At first based on amount of power to be transmitted , select the type of belt

 

2.       Calculate design power

              Design power = 

 

3.  Pitch length  = 

                                               

4.       Note inside length

 

5.       Determine Belt rating

 

6.       Design no of belts  = 

 

7.       Correct the center Length

 

8.       Also determine parameters of V groove pulleys using Manufactures data

 

 

 

 

 

 

Selection of Flat belts and pulleys

 

INTRTODUCTION

 

Flat Belts are one type of flexible connectors for transmitting power from one pulley to another pulley. Whose center distance is approx. 5-15m

 

CHARACTERISTICS OF BELT:

 


 

TYPES OF FLAT BELT DRIVES:

 

1.     Open belt drive

2.     Cross belt drive

3.     Quarter turn drive

 

4.     Belt drive with idler pulley

 

5.     Belt drive with many pulleys

 

DESIGN OF FLAT BELTS

 

1.     Using fundamental formula

 

2.     Using Manufactures catalogues

 

FUNDAMENTAL FORMULA:

 

3.        Ratio of driving tensions=


 

4.        Power Transmitted by belt


 

MANUFACTURES CATALOGUES

 

1.     How much power to be transmitted

 

2.     What may be the power transmitting capacity

 

For determining the design power and belt rating, we must consider certain factors like service, arc of contactand so on.

 

i. Arc of contact


 

ii. Load rating

                  

The load ratings have been developed for  180° of arc of contact 10m/s belt speed per mm width.

 

iii.       Length of belt                                                                      

 

           Open belt drive   Cross belt drive

 

iv.              Belt tensions

 

 

1.     Belt of 3 plies – 1.5% of L

 

2.     Belt of 4,5,6 plies – 1% of L

 

3.     Belt of 8 plies – 0.5% of L

 

 

v.                 Pulley width

 

Generally the pulleys should be slightly wider than belt width.

 

 

POWER TRANSMISSION

 

 

Belts are the cheapest utility for power transmission between shafts that may not be axially aligned. Power transmission is achieved by specially designed belts and pulleys. The demands on a belt drive transmission system are large and this has led to many variations on the theme. They run smoothly and with little noise, and cushion motor and bearings against load changes, albeit with less strength than gears or chains. However, improvements in belt engineering allow use of belts in systems that only formerly allowed chains or gears. Power transmitted between a belt and a pulley is expressed as the product of difference of tension and belt velocity

 

P = (T1- T2) v

 

where, T1 and T2 are tensions in the tight side and slack side of the belt respectively. They are

 

related as:


 

DESIGN PROCEDURE

 

1.     From the given conditions like power, type of working conditions, diameters of pulleys, speed ratio etc, determine maximum power

 

Desin power = rated power x service factor x arc of contact factor

 

Select service factor based on  nature of load and applications from PSG data book

 

2.     Decide the type of belt

 

3.     Then calculate the belt rating

 

4.     Find the reqired width by design power by belt capacity and adopt the standard available

 

5.     Determine the length of belt based on type of drive and reduce certain amount length

 

6.     Find out the pulley dimension and draw the arrangement of belt drive.

 

 

 

 

Wire ropes and pulleys

 

SELECTION PROCEDURE

 

1.     Based on the given data like nature of application, duty etc, select the type of rope

 

2.     Estimate the design load by multiplying the dead weight by three times design factor.

 

3.     3.  Determine the net cross sectional area of the rope by choosing specific strength of wire.

  


4.     Find out the diameter of rope

 

5.     Select the next standard dia of rope and note down the max breaking strength

 

6.     Compute the load applied at normal working acceleration and starting etc. find out the actual factor of safety by dividing the breaking strength by above loads

 

7.     For safe design the actual factor of safety should not be less than 5 at any circumstances

 

8.     Then calculate the drum and pulley dimensions.

 

 

 

 

Selection of Transmission chains and Sprockets

 

 

SELECTION PROCEDURE

 

1.           Depending upon the amount amount of power to be transmitted and another working conditions such as available space, chain speed, position of chain drive etc

 

2.           Assuming the center distance between the chaun sprockets interms of pitches

 

3.           Calculate the developed load for breaking the chain using expression as

 

4.           For determining pitch, choose suitable chain from PSG Data Book

 

5.           Find out the actual factor of safety

 

6.           Determine the induced stress over the projected area of the chain using the relation as                            

                                      

                                     

7.           Find the length of chain and         provide allowance for initial sagging.

                   =                

8.           Evaluate the pitch diameter of pinion sprocket (d1) and wheel sprocket (d2)

                 

 

9.           Draw a neat sketch of chain drive with calculated specifications.

 

 

 

 

SOLVED PROBLEMS

 

1. Design a V – Belt drive to the following specifications Power transmitted = 75kw

Speed of driving wheel = 1440rpm

 

Speed of driven wheel = 400rpm

 

Diameter of driving wheel = 300mm

 

Center distance = 2500mm

 

Service = 16hrs/day

 

 

 

Solution

 

For the given power of 75 kw D type or E type belts are suited. Let us selected D type

 

belt.


Service factor = 1.5 (for heavy duty and 16 hrs/ day with ac motor high torque)

 

Pitch length


Now = 300 mm

The next standard pitch length = 7648mm

Corresponding inside length = 7569mm

 

Length factor = 1.05

 

Arc of contact


 

Initial Tension = 0.75%L

 

Final center distance = 2788 mm

 

Specification

 

Type of belt = D7569 50 IS294

 

Number of belts required  = 8

 

Pitch diameter of small pulley = 1080mm

 

Center distance = 2788mm

 

 

 

2. Design a Flat belt drive to transmit 25 kw at 720 rpm to an aluminium rolling machine the speed reduction being 3.0. The distance between the shaft is 3m.Diameter of rolling machine pulley is 1.2m.

 

Solution:

 

Given

 


Specification of the belt drive are

 

Dia of motor pulley are = 400mm

 

Dia of rolling machine pulley = 1200mm

 

Center distance = 30000 mm

 

Width of belt  = 305mm

 

Width of the belt = 345 mm

 

Width of pulleys = 345mm


 

3.A roller chain is to be used on a paving machine to transmit 30 hp from the 4-cylinder Diesel engine to a counter-shaft; engine speed 1000 rpm, counter-shaft speed 500 rpm. The center distance is fixed at 24 in. The cain will be subjected to intermittent overloads of 100 %. (a) Determine the pitch and the number of chains required to transmit this power. (b) What is the length of the chain required? How much slack must be allowed in order to have a whole number of pitches? A chain drive with significant slack and subjected to impulsive loading should have an idler sprocket against the slack strand. If it were possible to change the speed ratio slightly, it might be possible to have a chain with no appreciable slack. (c) How much is the bearing pressure between the roller and pin?

 

Solution:

 

(a) design hp = 2(30) = 60 hp intermittent

 





 

4.In a coal-mine hoist, the weight of the cage and load is 20 kips; the shaft is 400 ft. deep. The cage is accelerated from rest to 1600 fpm in 6 sec. A single 6 x 19, IPS, 1 ¾ -in. rope is used, wound on an 8-ft. drum. (a) Include the inertia force but take the static view and compute the factor of safety with and without allowances for the bending load. (b) If

 

N =1.35 , based on fatigue, what is the expected life? (c) Let the cage be at the bottom of the shaft and ignore the effect of the rope’s weight. A load of 14 kips is gradually applied on the

 

6-kip cage. How much is the deflection of the cable due to the load and the additional energy absorbed? (d) For educational purposes and for a load of 0.2Fu , compute the energy that this

 

400-ft rope can absorb and compare it with that for a 400-ft., 1 ¾ -in., as-rolled-1045 steel rod. Omit the weights of the rope and rod. What is the energy per pound of material in each case?


 


 


 

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