Unit VGear TrainsTable of Contents

5.1 Introduction25.2 Types of Gear Trains35.2.1 Simple Gear Train35.2.2 Torque and Efficiency55.2.3 Compound Gear Train85.3 Speed Ratio of the Compound Gear Train95.4 Design of Spur Gears95.4.1 Reverted Gear Train105.4.2 Planetary Gear Train (Epicyclic Gear Train)135.4.3 Analysis of Epicyclic Gear Trains155.4.4 Compound Epicyclic Gear TrainSun and Planet Gear205.4.5 Epicyclic Gear Train with Bevel Gears245.5 Differential Gear of an Automobile255.5.1 Application of Gear Trains in Automobiles285.6 Questions315.6.1 Short Questions325.6.2 Essay type questions33

5.1 IntroductionA gear train consists of two or more gears working together by meshing their teeth and turning each other in a system to generate power and speed. It reduces speed and increases torque. To create large gear ratio, gears are connected together to form gear trains. They often consist of multiple gears in the train.Comment by logeshwaran: Fig.5.1 Title?

Fig. 5.1 Gear TrainThe most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type can be spur, helical or herringbone. The angular velocity is simply the reverse of the tooth ratio.Any combination of gear wheels employed to transmit motion from one shaft to the other is called a gear train. The meshing of two gears may be idealized as two smooth discs with their edges touching and no slip between them. This ideal diameter is called the Pitch Circle Diameter (PCD) of the gear.Electric motors are used with the gear systems to reduce the speed and increase the torque. 5.2 Types of Gear Trains Simple Gear Train Compound Gear Train Reverted Gear Train Planetary Gear Train5.2.1 Simple Gear Train The most common of the gear train is the gear pair connecting parallel shafts. The teeth of this type can be spur, helical or herringbone. Only one gear may rotate about a single axis

Fig. 5.2 Simple Gear Train Speed ratio (or velocity ratio) of gear train, is the ratio of the speed of the driver to the speed of the driven or follower. The ratio of speeds of any pair of gears in mesh is the inverse of their number of teeth.

Speed ratio =

Train value = It has no effect on the gear ratio. The teeth on the gears must all be the same size; so if gear A advances one tooth, so does B and C.

Fig.5.3 Simple Gear Train

Speed ratio =

Train Value = Applicationa) To connect gears where a large center distance is required.b) To obtain desired direction of motion of the driven gear (CW or CCW).c) To obtain high speed ratio.5.2.2 Torque and Efficiency The power transmitted by a torque T N-m applied to a shaft rotating at N rev/min is given by:

In an ideal gear box, the input and output powers are the same so;

It follows that if the speed is reduced, the torque is increased and vice versa. In a real gear box, power is lost through friction and the power output is smaller than the power input. The efficiency is defined as:

Since the torque in and out is different, a gear box has to be clamped in order to stop the case or body rotating. A holding torque T3 must be applied to the body through the clamps.

The total torque must add up to zero. T1 + T2 + T3 = 0If we use a convention that anti-clockwise is positive and clockwise is negative, we can determine the holding torque. The direction of rotation of the output shaft depends on the design of the gear box.

Fig.5.4 Gear Box Problem 5.1:A gear box has an input speed of 1500 rpm clockwise and an output speed of 300 rpm anticlockwise. The input power is 20 kW and the efficiency is 70%. Determine the following:i. The gear ratio, ii. The input torque, iii. The output power, iv. The output torque, v. The holding torque.Solution:

(Negative Clockwise)

(Positive anticlockwise)

Clockwise5.2.3 Compound Gear Train For large velocities, compound arrangement is preferred. Two or more gears may rotate about a single axis.

Fig.5.5 Compound Gear TrainLet,N1 = Speed of driving gear 1,T1 = Number of teeth on driving gear 1,N2, N3, N6 = Speed of respective gears in r.p.m., andT2, T3, . T6 = Number of teeth on respective gears,

Since gear 1 is in mesh with gear 2, therefore, its speed ratio is

Similarly, for gears 3 and 4, speed ratio is

And for gears 5 and 6 speed ratio is 5.3 Speed Ratio of the Compound Gear Train

or

Speed Ratio =

=

Train Value =

= 5.4 Design of Spur Gearsx = Distance between the centres of two shafts,N1= Speed of the driver,T1= Number of teeth on the driver,d1= Pitch circle diameter of the driver,N2 , T2 and d2= Corresponding values for the driven or follower,pc= Circular pitch.

5.4.1 Reverted Gear TrainWhen the axes of the first gear (i.e. first driver) and the last gear (i.e. last driven or follower) are co-axial, then the gear train is known as Reverted Gear Train.

Fig.5.6 Reverted Gear TrainThe driver and driven axes lies on the same line. These are used in speed reducers, clocks and machine tools.

If R and T=Pitch circle radius and number of teeth of the gear respectively,RA + RB = RC + RDtA + tB = tC + tD

Problem 5.2:The speed ratio of the reverted gear train, as shown in Fig.5.7 is to be 12. The module pitch of gears A and B is 3.125 mm and of gears C and D is 2.5 mm. Calculate the suitable numbers of teeth for the gears. No gear is to have less than 24 teeth.Let NA = Speed of gear A,TA = Number of teeth on gear A,rA = Pitch circle radius of gear A,NB , NC , ND = Speed of respective gears,TB , TC, TD = Number of teeth on respective gears, andrB, rC , rD = Pitch circle radii of respective gears.

Fig.5.7 Reverted Gear TrainSince the speed ratio between the gears A and B and between the gears C and D are to be same, therefore,

Also the speed ratio of any pair of gears in mesh is the inverse of their number of teeth, therefore,

------ (i)We know that the distance between the shafts,X = rA + rB = rC + rD = 200mmOr

3.125(TA + TB) = 2.5(TC+TD) = 400(mA = mB, and mC = mD)TA + TB = 400 / 3.125 = 128------- (ii)And, TC + TD = 400 / 2.5 = 160------- (iii)From equation (i), TB = 3.464 TA . Substituting this value of TB in equation (ii),TA + 3.464 TA = 128 or TA = 128 / 4.464 = 28.67 And, TB = 128 28 = 100 Again from equation (i), TD = 3.464 TC. Substituting this value of TD in equation (iii),TC + 3.464 TC = 160 or TC = 160 / 4.464 = 35.84 say 36And, TD = 160 36 = 124Note: The speed ratio of the reverted gear train with the calculated values of number of teeth on each gear is,

5.4.2 Planetary Gear Train (Epicyclic Gear Train)Epicyclic would mean that one gear revolves upon and around another. The design involves planet and sun gears; as one orbits the other like a planet orbiting around the sun.

Fig. 5.8 Planetary Gear TrainThis design can produce large gear ratios in a small space and are used on a wide range of applications from marine gear boxes to electric screw drivers.

Fig.5.9 Epicyclic Gear TrainIt consists of a small gear at the center called the sun, several medium sized gears called the planets, and a large external gear called the ring gear.

Fig.5.10 Epicyclic Gear TrainAdvantages They have higher gear ratios. They are popular for automatic transmissions in automobiles. They are also used in bicycles for controlling power of pedaling automatically or manually. They are also used for power train between internal combustion engine and an electric motor.5.4.3 Analysis of Epicyclic Gear Trains

Fig.5.11 Analysis of Epicyclic Gear TrainsThe analysis of epicyclic gear trains can be done by Tabular and Analytical methods. 1 Tabular methodObserve point p, and you will see that gear B also revolves once on its own axis. Any object orbiting around a center must rotate once. Now consider that B is free to rotate on its shaft and meshes with C.

Fig.5.12 Tabular Method

Suppose the arm is held stationary and gear C is rotated once. B spins about its own center and the number of revolutions it makes is the ratio: B will rotate by this number for every complete revolution of C.

Now consider that the sun gear C is restricted to rotate, and the arm A is revolved once. Gear B will revolve because of the orbit. It is this extra rotation that causes confusion. One way to get round this is to imagine that the whole system is revolved once. Then, identify the gear that is fixed and revolve it back for one revolution. Work out the revolutions of the other gears and add them up. Consider the following example. Suppose gear C is fixed and the arm A makes one revolution. Determine how many revolutions the planet gear B makes.Step 1: Revolve all elements once about the center.Step 2: Identify that C should be fixed and rotate it backwards one revolution keeping the arm fixed as it should only do one revolution in total. Work out the revolutions of B.Step 3: Add them up and we find the total revolutions of C is zero and for the arm is1.StepActionABC

1 Revolve all once1 11

2 Revolve C by 1 revolution, keeping the arm fixed0

-1

3 Add 1

0

A simple epicyclic gear has a fixed sun gear with 100 teeth and a planet gear with 50 teeth. If the arm is revolved once, how many times does the planet gear revolve? StepActionABC

1 Revolve all once1 11

2 Revolve C by 1 revolution, keeping the arm fixed0

-1

3 Add1 30

This can be generalized by the following table. Prepare the following table for any such problem and solve for the given conditions. Sl. No.Conditions of motionRevolutions

Arm CGear AGear B

1.Arm fixed-gear A rotates through +1 revolution i.e., 1 rev. anticlockwise0+1

2.Arm fixed-gear A rotates through +x revolutions0+x

3.Add +y revolutions to all elements+y+y+y

4.Total motion+yx+y

2.Algebraic MethodIn this method, the motion of each element of the epicyclic train relative to the arm is set down in the form of equations. The number of equations depends upon the number of elements in the gear train. But the two conditions are, usually, supplied in any epicyclic train viz. some element is fixed and the other has specified motion.These two conditions are sufficient to solve all the equations, and hence, to determine the motion of any element in the epicyclic gear train.Speed of the gear A relative to the arm C= NA - NCThe speed of the gear B relative to the arm C, = NB - NCSince the gears A and B are meshing directly, therefore, they will revolve in opposite directions.

Since the arm C is fixed, therefore, its speed, NC = 0.

If the gear A is fixed, then NA = 0.

orProblem 5.3:In a reverted epicyclic gear train, the arm A carries two gears B and C, and a compound gear D and E. The gear B meshes with gear E and the gear C meshes with gear D. The number of teeth on gears B, C and D are 75, 30 and 90 respectively. Find the speed and direction of gear C, when gear B is fixed and the arm A makes 100 r.p.m. clockwise.TB = 75 ; TC = 30 ; TD = 90 ;NA = 100 r.p.m. (clockwise)Let, dB ,dC , dD and dE be the pitch circle diameters of gears B,C, D and EdB + dE = dC + dDSince the number of teeth on each gear, for the same module, are proportional to their pitch circle diameters, therefore, TB + TE = TC + TDTE = TC + TD TB = 30 + 90 75 = 45

Fig.5.13 Reverted Epicyclic Gear TrainSl. No.Conditions of motionRevolutions

Arm ACompound Gear D - EGear BGear C

1.Arm fixed compound gear D E rotated thorough +1 revolution (i.e., 1 rev. anticlockwise)0+1

2.Arm fixed-gear D-E rotates through +x revolutions0+x

3.Add +y revolutions to all elements+y+y+y+y

4.Total motion+yx+y

Since the gear B is fixed, therefore, from the fourth row of the table,

= 0orY 0.6 x = 0Also, the arm A makes 100 r.p.m. clockwise, therefore, y = - 100Substituting, y = -100 in equation We get, -100 0.6x = 0or x = -100 / 0.6 = 166.67From the fourth row of the table, speed of gear C,

NC = y x r.p.m = 400r.p.m (anticlockwise)5.4.4 Compound Epicyclic Gear TrainSun and Planet Gear

Fig.5.14 Sun and Planet GearLet TA , TB , TC , and TD be the teeth, and NA, NB, NC and ND be the speeds for the gears A, B,C and D respectively. When the arm is fixed and the sun gear D is turned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwise direction.Sl. No.Conditions of motionRevolutions of elements

Arm Gear DCompound gear B-CGear A

1.Arm fixed-gear D rotates through +1 revolution0+1

2.Arm fixed-gear D rotates thorough +x revolutions0+x

3.Add +y revolutions to all elements+y+y+y+y

4.Total motion+yx+y

Problem 5.4:An epicyclic train of gears is arranged as shown in figure. How many revolutions does the arm, to which the pinions B and C are attached, make:1. When A makes one revolution clockwise and D makes half a revolution anticlockwise, and 2. When A makes one revolution clockwise and D is stationary?The number of teeth on the gears A and D are 40 and 90 respectively.

Fig.5.15 Epicyclic Train of GearsTA = 40; TD = 90

dA + dB = dC = dDordA + 2dB = dD------- (dB = dC)Since the number of teeth are proportional to their pitch circle diameters, therefore,TA = 2TB = TDor40 + 2TB = 90

TB = 25, and TC = 25 -------- (TB = TC)Sl. No.Conditions of motionRevolutions of elements

Arm Gear ACompound gear B-CGear D

1.Arm fixed-gear A rotates through -1 revolution0-1

2.Arm fixed-gear A rotates thorough -x revolutions0-x

3.Add -y revolutions to all elements -y-y-y-y

4.Total motion-y-x-y

Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise. Since the gear A makes 1 revolution clockwise, therefore, from the fourth row of the table, x y = 1 or x + y = 1------ (i)Also, the gear D makes half revolution anticlockwise, therefore,

or40 x 90 y = 45orx 2.25 y = 1.125------- (ii)From equations (i) and (ii), x = 1.04and y = -0.04.Speed of arm = -y = -(-0.04) = + 0.04= 0.04 revolution anticlockwiseSpeed of arm when A makes 1 revolution clockwise and D is stationary.Since the gear A makes 1 revolution clockwise, therefore from the fourth row of the table,- x y = - 1orx + y = 1------- (iii)Also the gear D is stationary, therefore,

or40 x 90 y = 0andx 2.25 y = 0-------- (iv)From equations (iii) and (iv)X = 0.692andy = 0.308Speed of arm = -y = -0.308 = 0.308 revolution clockwiseProblem 5.5:In an epicyclic gear train shown in Fig.5.16, the arm A is fixed to the shaft S. The wheel B is having 100 teeth and rotates freely on the shaft S. The wheel F having 150 teeth is driven separately. If the arm rotates at 200 rpm and wheel F at 100 rpm in the same direction; find (a) number of teeth on the gear C and (b) speed of wheel B.

Fig.5.16 Epicyclic Gear TrainGiven: TB=100;TF=150;NA=200rpm;NF=100rpm:

The gear B and gear F rotates in the opposite directions:

The Gear B rotates at 350 rpm in the same direction of gears F and Arm A.5.4.5 Epicyclic Gear Train with Bevel Gears The bevel gears are used to make a more compact epicyclic system and they permit a very high speed reduction with few gears. The useful application of the epicyclic gear train with bevel gears is found in Humpages speed reduction gear and differential gear of an automobile.

Fig 5.17 Epicyclic Gear Train with Bevel Gears5.5 Differential Gear of an AutomobileThe differential gear used in the rear drive of an automobile,(a) to transmit motion from the engine shaft to the rear driving wheels, and(b) to rotate the rear wheels at different speeds while the automobile is taking a turn.Sl. No.Conditions of motionRevolutions of elements

Gear BGear DGear EGear D

1.Gear B fixed Gear C rotated through +1 revolution (i.e., 1 revolution anticlockwise)0+1

2.Gear B fixed-Gear C rotated through + x revolutions0+x

-x

3.Add + y revolutions to all elements+y+y+y+y

4.Total motion+yx+y

y - x

Problem 5.6:In a gear train, as shown in following Fig.5.18, gear B is connected to the input shaft and gear F is connected to the output shaft. The arm A carrying the compound wheels D and E, turns freely on the output shaft. If the input speed is 1000 rpm counter- clockwise when seen from the right, determine the speed of the output shaft under the following conditions:1. When gear C is fixed, and 2. When gear C is rotated at 10 rpm counter clockwise.

Fig.5.18 Gear TrainTB = 20 TC = 80 TD = 60 TE = 30 TF = 32 NB = 1000 rpm.(counter-clockwise)Step No.Conditions of motionRevolutions of elements

Arm AGear B (or input shaft)Compound wheel D-EGear CGear F (or output shaft)

1Arm fixed, gear B rotated through +1 revolution (i.e., 1 revolution anticlockwise)0+ 1

2Arm fixed, gear B rotated through +x revolutions0+ x

3Add +y revolutions to all elements+y+ y+y+y+y

4Total motion+yx + y

Speed of the output shaft when gear C is fixedSince, the gear C is fixed, therefore, from the fourth row of the table,

orY 0.25 x = 0---- (i)We know that the input speed (or the speed of gear B) is 1000 r.p.m. counter clockwise, therefore, form the fourth row of the table, x + y = +1000.From equation (i) and (ii), x = + 800, and y = + 200----- (ii)

Speed of output shaft = Speed of gear

= r.p.m. = 12.5 r.p.m (counter clockwise)

Speed of the output shaft when gear C is rotated at 10 r.p.m. counter clockwiseSince the gear C is rotated at 10 r.p.m. counter clockwise, therefore, from the fourth row of the table,

orY 0.25 x = 10------- (iii)From equations (ii) and (iii),X = 792andy = 208Speed of output shaft

= Speed of gear F = = 208 185.6 = 22.4 r.p.m = 22.4 r.p.m (counter clockwise)5.5.1 Application of Gear Trains in Automobiles Fig. 5.19 gives a clear picture of the way gears are connected in the gear train in different gear modes in automobile.

Fig. 5.19 (a) Automobile - Neutral

Fig. 5.19 (b) Automobile First Gear

Fig. 5.19 (c) Automobile Second Gear

Fig. 5.19 (d) Automobile Third Gear

Fig. 5.19 (e) Automobile Fourth Gear

Fig. 5.19 (f) Automobile Reverse Gear

Fig. 5.19 (g) Automobile Gear Mechanism Below detailed, is a very simple example to understand the gear train.

A. Compute the gear ratio of a single pair of gears.

Fig.5.20 Single Pair of GearsGear ratio = 40 to 8 or, simplifying, 5 to 1. B. Compute the gear ratio of a compound pair of gears

Fig.5.21 Compound Pair of Gears5.6 Questions1. What is the gear ratio of this pair of gears?Ans: 3:1

2. If axle 1 were to make90 rotations, how many rotations would 2 axles make?Ans: 303. If axle 2 were to make 30 rotations, how many rotations would axle 3 make?Ans: 104. What does that imply about what the overall gear ratio is?To compute the compound gear ratio, we multiply the gear ratios of each pair of gears in the gear box. (3 to 1) x (3 to 1) = 9 to 1Answer: 9 to 15. What is the compound gear ratio of this gear box going from bottom to top?

Fig.5.22 Compound Gear RatioAns: 45 to 15.6.1 Short Questions What is power transmission? Why gear drives are called positively driven? What is backlash in gears? What are the types of gears available? What is gear train? Why gear trains are used? Why intermediate gear in simple gear train is called idler? What is the advantage of using helical gear over spur gear? List out the applications of gears. Define the term module in gear tooth. What is herringbone gear?5.6.2 Essay type questions With sketch, explain various types of gears. With sketch, explain three types of gear trains. With neat sketch, explain the nomenclature of spur gear. Write the applications, advantages and disadvantages of gear drives.

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