3.1 Types of Gear Trains | unit 3 gear trains

Back to Study material

TOM2

Unit -3

Gear Trains

3.1 Types of Gear Trains

Following are the different types of gear trains, depending upon the arrangement of wheels:

Simple gear train

When there is only one Gear on each shaft as shown in figure it is known as a simple gear train.

The gears are represented by their pitch circles.

When the distance between the two shafts is small, the two gears 1 and 2 are made to match with each other to transmit motion from one shaft to another, as shown in figure a.

Since the gear 1 drives the gear 2, therefore, gear one is called the driver and the gear 2 is called the driven or follower.

The motion of the driven gear is opposite to the motion of the driving gear.

Let =speed of gear 1 in r.p.m

=speed of Gear 2 in r.p.m.

=number of teeth on gear 1 and

= number of teeth on gear 2.

Since the speed ratio of the gear train is the ratio of the speed of the driver to the speed of the driven and ratio of speeds of any pair of gears in the mesh is the inverse of their numbers of teeth, therefore

Speed ratio=

The ratio of the speed of the driven or follower to the speed of the driver is known as the train value of the gear train. Mathematically,

Train value=

Sometimes, the distance between the two gears is large.

The motion from one gear to another, in such case, may be transmitted by either of the following two methods:

By providing the large size gear, or

By providing one or more intermediate gears.

It may be noted that when the number of intermediate gears is odd, the motion of both the gears is like as shown in figure b

But if the number of intermediate gears is even, the motion of the driven will be in the opposite direction of the driver as shown in figure c.

Now consider a simple train of gears with one intermediate gear as shown in figure

Let = speed of driver in RPM

=speed of intermediate gear in RPM

= speed of driven or follower in r.p.m

= number of teeth on driver

= number of teeth on the intermediate gear

= number of teeth on driven or follower

Since the driving gear 1 is in mesh with the intermediate gear 2, therefore speed ratio for these two gears is

Similarly, as the intermediate gear 2 is in mesh with the driven gear 3, therefore, the speed ratio for these two years is

The speed ratio of the gear train as shown in figure b is obtained by multiplying equations 1 and 2

speed ratio =

Train value=

2. Compound gear train

When there is more than one Gear on a shaft, as shown in figure it is called a compound train of gear.

These gears are useful in bridging over the space between the driver and the driven.

But whenever the distance between the driver and the driven or follower has to be bridged over by intermediate gears and at the same time a great speed ratio is required, then the advantage of intermediate gears is intensified by providing compound gears on intermediate shafts.

In this case, each intermediate shaft has to gears rigidly fixed to it so that they may have the same speed.

One of these two gear meshes with the driver and the other with the driven attached to the next shaft as shown in the figure.

In a compound train of gears, as shown in the figure the gear 1 is the driving gear mounted on shaft A, gears 2 and 3 are compound gears which are mounted on shaft B.

The gears 4 and 5 are also compound gears that are mounted on shaft c and the gear 6 is the driven gear mounted on shaft D.

Let =speed of driving gear 1.

= number of teeth on driving gear 1.

= speed of respective gears in RPM

= number of teeth on respective gears.

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

Similarly for Gear 3 and 4-speed ratio is

And for gears 5 and 6-speed ratio is

The speed ratio of the compound gear train is obtained by multiplying the equations 1 2 and 3

Speed ratio=

Train value=

The advantage of a compound train over a simple gear train is that a much larger speed reduction from the first shaft to the last shaft can be obtained with small gear.

3. Reverted gear train

When the axis of the first gear and the last gear is coaxial, then the gear train is known as a reverted gear train as shown in the figure.

Gear 1 drives the gear 2 in the opposite direction.

Since the gears 2 and 3 are mounted on the same shaft, therefore they form a compound gear and the gear 3 will rotate in the same direction as that of Gear 2.

The gear 3 drives the gear 4 in the same direction as that of gear 1.

Thus in a reverted gear train, the motion of the first gear and the last gear is like.

Let =number of teeth on gear 1

= pitch Circle radius of gear 1, and

= speed of gear 1 in RPM

Similarly,

=number of teeth on respective gears,

= pitch Circle radius of respective gears and

= speed of respective gears in RPM

Since the distance between the centers of shafts of gear 1 and 2 as well as Gear 3 and 4 is the same common therefore

…….. i

also, the circular pitch or module of all the gear is assumed to be the same, therefore number of teeth on each gear is directly proportional to its circumference to the radius.

……….. ii

And
speed ratio

Or ……….iii

From equations i, ii & iii we can determine the number of teeth on each gear for the given center distance, speed ratio, and module only when the number of teeth on one gear is chosen arbitrarily.

The reverted gear trains are used in an automotive transmission, lathe back gears, industrial speed reducers, and clocks.

4. Epicyclic gear train

In an epicyclic gear train, the axes of the shafts over which the gears are mounted may move relative to a Fixed Axis.

A simple epicyclic gear train is shown in the figure where a gear A and the arm C have common access at

about which they can rotate.

The gear B meshes with gear A and has its axis on the arm at

about which the gear B can rotate.

If the arm is fixed, the gear train is simple and gear A can drive gear B or vice versa, but if gear A is fixed and the arm is rotated about the axis of gear A, then the gear B is forced to rotate upon and around gear A.

Such a motion is called epicyclic and the gear trains arranged in such a manner that one or more of their members move upon and around another member are known as epicyclic gear trains.

The epicyclic gear trains may be simple or compound.

The epicyclic gear train is useful for transmitting high-velocity ratios with gears of moderate size is comparatively less space.

The epicyclic gear trains are used in the back gear of lathe, differential gears of automobiles, hoists, Pulley blocks, wristwatches, etc.

3.2 Analysis of epicyclic gear trains

The following two methods may be used for finding out the velocity ratio of an epicyclic gear train.

Tabular method:-

Consider an epicyclic gear train as shown in figure

Let = number of teeth on gear A, and

= number of teeth on gear B.

Let us suppose that the army is fixed.

Therefore the axes of both the gears are also fixed relative to each other.

When the gear A makes one revolution anticlockwise, the gear B will make

revolutions, clockwise.

Assuming the anticlockwise rotation as positive and clockwise as negative, we may say that when gear A makes plus one revolution then the gear B will take (

revolutions.

This statement of relative motion is entered in the first row of the table.

Secondly, if the gear A makes +x revolutions, then the gear B will make -x ×

revolutions.

This statement is entered in the second row of the table.

Thirdly, each element of an epicyclic gear train is given +y revolutions and entered in the third row.

Finally, the motion of each element of the gear train is added up and entered in the fourth row.

Step No.	Conditions of motion	Arm C	Gear A	Gear B
1	Arm fixed gear A rotates through plus one revolution	0	+1
2	Arm fixed gear A rotates through +x revolutions	0	+x
3	Add +y revolutions to all elements	+y	+y	+y
4	Total motion	+y	x+y

A little consideration will show that when two conditions about the motion of rotation of any two elements are known, then the unknown speed of the third element may be obtained by substituting the given data in the third column of the fourth row.

2. Algebraic method:-

In this method, the motion of each element of the epicyclic gear 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 applied in any epicycle Gear train are some element is fixed and the other has specified motion.

These two conditions are sufficient to solve all the equations; hence determining the motion of any element in the epicyclic gear train.

Let the arm C be fixed in an epicyclic gear train as shown in the figure.

Therefore speed of the gear A relative to the arm C.

And speed off the gear B relative to the arm C,

Since the gears A and B are meshing directly, therefore they will revolve in opposite directions.

Since the arm c is fixed, therefore its speed.

If the gear A is fixed, then

3.3 Holding torque – Simple, compound and epicyclic gear trains

Torques are transmitted from one element to another when a geared system transmits power.

When the rotating parts of an epicyclic gear train, as shown in the figure have no angular acceleration, the gear train is kept in equilibrium by three externally applied torques

Input torque on the driving member

Output torque for resisting or load torque on the driving member.

Holding or breaking or fixing torque on the fixed member

The net torque applied to the gear train must be zero.

In other words,

Where are the corresponding externally applied forces at radii

Further if

are the angular speed of the driving, driven, and fixed members respectively, and the friction is neglected, then the net kinetic energy dissipated by the gear train must be zero.

But for a fixed member

3.4 Torque on sun and planetary gear train

A compound epicyclic gear train is shown in the figure.

It consists of two coaxial shafts

an annulus gear A which is fixed, the compound gear B-C, the sun gear D, and the arm H.

The annulus gear has internal teeth and the compound gear is carried by the arm and revolves freely on a pin of the arm H.

The sun gear is coaxial with the annulus gear and the arm but independent of them.

The annulus gear A meshes with the gear B.

The sun gear D meshes with the gear C.

When the annulus gear is fixed, the sun gear provides the drive and when the sun gear is fixed, the annulus gear provides the drive.

In both cases, the arm acts as a follower.

The gear at the center is called the Sun gear and the gears whose axes move are called planet gear.

Let

be the teeth and.

be the speeds for the gear A, B, C, and D respectively.

When the arm is fixed and the sun gear D it is turned anticlockwise, then the compound gear B-C and the annulus gear A will rotate in the clockwise direction

The motion of rotations of the various elements are shown in the table below,

Step No.	Conditions of motion	Arm	Gear D	Compound gear B-C	Gear A
1	Arm fixed- gear D rotates through plus one revolution	0	+1
2	Arm fixed -gear D rotates through +x revolutions	0	+x
3	Add +y revolutions to all elements	+y	+y	+y
4	Total motion	+y	x+y

3.5 Compound epicyclic gear train

When an epicyclic gear consists of many epicyclic gears in series such that the pin of the arm of the first epicyclic gear drives an element of another epicyclic gear, it is known as a compound epicyclic gear.

The figure shows a compound epicyclic gear.

It consists of three simple epicyclic gear namely

The planet

of the first epicyclic gear rotates freely on the pin carried by the arm

As the arm

is integral with the annulus

of 2nd epicyclic gear and the sun wheels of the third, the pin of the arm

also drives

The annulus and the sun wheel of the second and the third epicyclic gear respectively at the same speed and direction as they own about the axis of the arm.

Also, the sun wheel of the first is integral with that of second and the arm of the second with that of the third.

To analyze a compound epicyclic gear, each epicyclic gear or Sun and planet gear is treated separately. Thus for the first epicyclic gear.

Speed of arm

Thus if the speed

are known

can be calculated.

Similarly for the second Sun and planet gear

As is integral with

Thus can be known.

In the same way,

is integral with

So can be calculated.

3.6 Bevel epicyclic Gear train

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 Humpage's speed reduction gear and differential gear of an automobile as discussed below:

Humpage's speed reduction gear.

The Humpage’s speed reduction gear was originally designed as a substitute for back gearing of a lathe, but its use is now considerably extended to all kinds of workshop machines and also in electrical machinery.

In Humpage’s speed reduction gear, as shown in figure the driving shaft X and the driven shaft Y are co-axial.

The driving shaft carries a bevel gear A and driven shaft carries a bevel gear E.

The bevel gear B meshes with gear A(also known as pinion) and a fixed gear C.

The gear E meshes with gear D which is compound with gear B.

This compound gear B-D is mounted on the arm or spindle F which is rigidly connected with the hollow sleeve G.

The sleeve revolves freely loose on the axes of the driving or driven shafts.

Numericals

In an epicyclic gear train, an arm carries two gears A and B having 36 and 45 teeth respectively. If the arm rotates at 150 r.p.m. in the anticlockwise direction about the center of the gear A which is fixed, Determine the speed of gear B. If the gear A instead of being fixed, makes 300r.p.m in the clockwise direction what will be the speed of gear B?

Given

Step No.	Conditions of motion	Arm C	Gear A	Gear B
1	Arm fixed gear A rotates through plus one revolution	0	+1
2	Arm fixed gear A rotates through +x revolutions	0	+x
3	Add +y revolutions 2 all elements	+y	+y	+y
4	Total motion	+y	x+y

Speed of gear B when gear A is fixed

Since the speed of arm is 150 r.p.m. anticlockwise, therefore from the fourth row of the table,

Y=+150 r.p.m

Also, the gear is fixed, therefore

x+y=0 or x=-y= -150r.p.m.

Speed of gear B

Speed of gear B when gear A makes 300 rpm clockwise

Since the gear A makes 3000r.p.m. clockwise, therefore from the fourth row of the table,

x+y=-300. Or x =-300 - y = -300-150= - 450r.p.m

Speed of gear B,

2. An internal wheel B with 80 teeth is keyed to a shaft F. A fixed internal wheel C with 82 teeth is concentric with B. A compound wheel D-E gears with the two internal wheels: D has 28 teeth and gears with C while E gears with B. The compound wheels revolve freely on a pin which projects from a disc keyed to a shaft. A coaxial with F if the wheels have the same pitch and the shaft A makes 800 r.p.m. what is the speed of the shaft F? Sketch the arrangement.

Solution

The arrangement is shown in figure

Let be the pitch circle diameter of wheels B, C, D, and E respectively. From the geometry of the figure,

Since the number of teeth is proportional to their pitch circle diameter for the same pitch, therefore

The table of motions is given below

Step No.	Conditions of motion	Arm (or shaft A)	Wheel B (or shaft F)	Compound gear D-E	Wheel C
1	Arm fixed – wheel B rotated through +1 revolution anticlockwise	0	+1
2	Arm fixed-wheel B rotated through +x revolutions	0	+x
3	Add +y revolutions to all elements	+y	+y	+y	+y
4	Total motion	+y	x+y

Since the wheel, C is fixed, therefore from the fourth row of the table,

the shaft A makes 800 r.p.m. therefore from the fourth row of the table,

y=800----(2)

From equations 1 and 2

x=-762

Speed of shaft F= speed of wheel B=x+y =-762+800=+38r.p.m.

=38 r.p.m.anticlockwise

3. In a reverted epicyclic gear train, the arm A carries two gears B and C and compound gear D-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.

Given

Let the pitch circle diameter of gear B, C, D, and E respectively. From the geometry of the figure

Since the number of teeth on each gear, for the same module, are proportional to their pitch circle diameter therefore

The table of motions is drawn as follows:

Step No.	Conditions of motion	Arm (or shaft A)	Wheel B (or shaft F)	Compound gear D-E	Wheel C
1	Arm fixed – compound gear D-E rotated through +1 revolution anticlockwise	0	+1
2	Arm fixed-compound gear D-E rotated through +x revolutions	0	+x
3	Add +y revolutions to all elements	+y	+y	+y	+y
4	Total motion	+y	x+y

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

y - 0.6=0---(1)

Also, the arm A makes 100 r.p.m. clockwise, therefore

y= -100---(2)

Substituting above value in equation 1 we get

- 100 - 0.6x = 0. Or x=-100/0.6=-166.67

From the fourth row of the table, speed of gear C

=400 r.p.m. anticlockwise.

4. The annulus A in the gear shown in figure rotates at 300 r.p.m. about the axis of the fixed wheel S which has 80 teeth. The three-armed spider is driven at 180 rpm. Determine the number of teeth required on the wheel P.

Solution

Step No.	Conditions of motion	Arm a	Gear S	Gear P	Gear A
1	Arm a fixed – gear S rotated through +1 revolution anticlockwise	0	+1
2	Arm a fixed – gear S rotated through +x revolution anticlockwise	0	+x
3	Add +y revolutions to all elements	+y	+y	+y	+y
4	Total motion	+y	x+y

From given conditions,

From 1 and 2 x= -y= -180

Solving 3

The pitch diameters of the wheels are proportional to the number of teeth on them.

Or or

5. An epicyclic gear train consists of sun wheel S, a stationary internal gear E and three identical planet wheels P carried on a star-shaped planet carrier C. The size of different tooth wheels are such that the planet carrier C rotates at 1/5th of the speed of the sun wheel S. The minimum number of teeth on any wheel is 16. The driving torque on the sun wheel is 100N-m. Determine

Number of teeth on different wheels of the train, and

Torque necessary to keep the internal gear stationary.

Solution Given

Number of teeth on different wheels

The arrangement of the epicyclic gear train is shown in the figure.

Let be the number of teeth on the sun wheel S and the internal gear E respectively. The table of motions is given below:

Step No.	Conditions of motion	Arm (or shaft A)	Wheel B (or shaft F)	Compound gear D-E	Wheel C
1	Planet carrier C fixed, sun wheel S rotates through +1 revolution anticlockwise	0	+1
2	Planet carrier C fixed, sun wheel S rotated through +x revolutions	0	+x
3	Add +y revolutions to all elements	+y	+y	+y	+y
4	Total motion	+y	x+y

We know that when the sun wheel as makes 5 revolutions, the planet carrier C makes one revolution therefore from the fourth row of the table

y=1 and. x+y =5 or x= 5 – y = 5 – 1 = 4

Since the gear E is stationary, therefore from the fourth row of the table.

Since the minimum number of teeth on any wheel is 16, therefore, let us take the number of teeth on the sun wheel,

Let be the pitch circle diameter of the wheel S, P, and E respectively. Now from the geometry of the figure

assuming the module of all the gears to be same, the number of teeth are proportional to their pitch circle diameter,

2. Torque necessary to keep the internal gear stationary

Torque on S ×Angular speed of S = torque on C ×Angular speed of C

Torque necessary to keep the internal gear stationary

=500-100=400N-m

6. In the epicyclic gear train, as shown in the figure, the driving gear A rotating in clockwise direction has 14 teeth and the fixed annular gear C has 100 teeth. The ratio of teeth in gears E and D is 98: 41. If 1.85 kW is supplied to the gear A rotating at 1200 r.p.m. find

The speed and direction of rotation of gear E

The fixing torque required at C, assuming 100% efficiency throughout and that all teeth have the same pitch.

Solution Given.

Let the pitch circle diameter of gear A, B, and C respectively. From figure

Since teeth of all gears have the same pitch and the number of teeth is proportional to their pitch circle diameters, therefore

The table of motions is not drawn as below:

Step No.	Conditions of motion	Arm	Gear A	Compound gear B-D	Gear C	Gear E
1	Arm fixed-Gear A rotates through -1 revolution clockwise	0	-1
2	Arm fixed-Gear A rotated through +x revolutions	0	-x
3	Add -y revolutions to all elements	-y	-y	-y	-y	-y
4	Total motion	-y	-y-x	-