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1. DEPARTMENT OF MECHANICAL ENGINEERING
Course: Theory of Machines (21MEC503)
OTHER ASSIGNMENT / ACTIVITY – SEMINAR PRESENTATION
Torque Calculations for Epicyclic Gear
By
Mohammed Asman 4SO21ME038
Mohammed Anas 4SO21ME037
Mohammed Amaan 4SO21ME036
Mohammed Riyaz 4SO21ME033
Mohammed Afrad 4SO21ME035
Faculty
Mr Joel D’Mello
Assistant Professor
Department of Mechanical Engineering

2. Torque in Epicyclic Gear Train:
Gears in an epicyclic gear train move with uniform speeds, so they will not have angular acceleration.
The gear train is in equilibrium by the three externally applied torques. i.e.,
1) Input torque on the driving member (T1)
ii) Output torque on the driven member (T2)
iii) Holding or Fixing torque (T3)
For equilibrium, T1 + T2 + T3 = 0........ (i)
i.e., F1r1 + F2r2 + F3r3 = 0… …..(ii)
Where; F1, F2 and F3 are externally applied by forces at radii r1, r2 and r3.
If friction is neglected, the net energy dissipated by the gear train must be equal to zero.
T1ω1 + T2ω2 + T3ω3 = 0.............. (iii)

3. Where ω1, ω2 & ω3 are uniform angular velocities of driver, driven and fixed member respectively.
The equation (iii) can also be written as;
T1n1 + T2n2 + T3n3 = 0… …..(iv)
Equation (iii) or (iv) is called energy equation.

4. PROBLEM #1
In the epicyclic gear train, as shown in Fig , 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:
1. the speed and direction of rotation of gear E, and
2. the fixing torque required at C, assuming 100 % efficiency throughout and that all teeth
have the same pitch.

5. PROBLEM #1 (SOLUTION)
Given : TA=14 ; TC= 100 ; TE/TD= 98 / 41 ; PA= 1850 W ; NA= 1200 r.p.m
Let dA, dB and dC be the pitch circle diameters of gears A, B and C respectively.
From Fig, dA+ 2 dB= dC Since teeth of all gears have the same pitch and the number of teeth are
proportional to their pitch circle diameters, therefore: TA+ 2TB= TC or TB=43
The table of motions is now drawn as below:
Lets take CW as -ve and CCW as +ve
Step No. Conditions of
Motions
Arm Gear A Compound
Gear B-D
Gear C Gear E
1 Arm fixed;
-1 rev
0 -1 -1*(-TA/TB) -1*(-TA/TB)*(TB/TC) -1*(-TA/TB)*(TD/TE)
2 Arm fixed;
-x rev
0 -x x*(TA/TB) x*(TA/TB)*(TB/TC) x*(TA/TB)*(TD/TE)
3 Add -y rev -y -x-y -y+x*(TA/TB) -y+x*(TA/TB)*(TB/TC) -y+x*(TA/TB)*(TD/TE)

6. PROBLEM #1 (SOLUTION)
Since the annular gear C is fixed,
-y + x*(TA/TB)*(TB/TC) = 0;
-y + x*(14/100) = 0;
-y + 0.14x = 0; ---------(i)
Also, the gear A is rotating at 1200 r.p.m., therefore: -x -y = 1200-------(ii)
From eqns i & ii we get x = -1052.6 and y = -147.4
1. Speed and direction of rotation of gear E
From the table, speed of gear E, Ne= -y + x*(TA/TB)*(TD/TE);
= -(-147.4) - 1052.6 (14/43)*(41/98)
= 4.022 ~ 4 rpm (CCW)
2. Fixing torque required at C
We know that
Torque on A = (PA*60/2*pi*NA)=(1850*60/2*pi*1200)=14.7 N-m
Since the efficiency is 100 per cent throughout, therefore the power available at E (PE) will be equal to power
supplied at A (PA).
Torque on E = (PA*60/2*pi*NE)=(1850*60/2*pi*4)=4416 N-m
Fixing torque required at C= 4416 –14.7 = 4401.3 N-m

7. PROBLEM #2
An epicyclic gear train consists of a sun wheel S, a stationery internal gear'E' and three identical planet
wheels 'P' carried on a stat-shape planet carrier "C'. The sizes of different toothed wheels are such that
the planet carrier C rotates one revolution for every 5 revolutions of the sun wheel S.The minimum
number of teeth on any wheel (say P) is 16. The driving torque on the sun wheels of the train is 100 N-
m. Determine
(i) Number of teeth on different wheels of the train
(ii) Torque necessary to keep the internal gear stationery.

8. PROBLEM #2 (SOLUTION)
Data: NC = 1/5 NS, NE = 0 rpm
From the geometric of the gear shown; 𝐷𝐸= 𝐷𝑆+ 2𝐷𝑃 ∴ 𝑇𝐸= 𝑇𝑆+ 2𝑇𝑃
NC = 1/5 NS ∴ 𝑋 = 1/5 (𝑋 + 𝑌) ∴ 𝑌 = 4𝑋 --------------(1)
𝑁𝐸= 0 = 𝑋 − 𝑌 (𝑇𝑆/𝑇𝐸 ) =-----------------------------------------(2)
∴ 𝑇𝐸> 𝑇𝑃> 𝑇𝑆
From 1 , 2 then 𝑇𝐸= 4𝑇𝑆, 𝑇𝐸=(8/3)𝑇P and 𝑇𝑃=(3/2)𝑇𝑆
Action Rev of Arm Rev of S Rev of P Rev of E
Give whole gears +x
rpm
x x x x
Arm is fixed +y rev 0 y -y(Ts/Tp) -y(Ts/Te)
Total motion x x+y x-y(Ts/Tp) x-y(Ts/Te)

9. PROBLEM #2 (SOLUTION)
∴ 𝑇𝐸> 𝑇𝑃> 𝑇𝑆
∴ 𝑇𝑆=16 𝑡𝑒𝑒𝑡ℎ , 𝑇𝑃=(3/2)*16 =24 𝑡𝑒𝑒𝑡ℎ and
𝑇𝐸 = 4 × 16 =64 𝑡𝑒𝑒𝑡ℎ
The torque necessary to keep the internal gear stationary
𝑀𝑆×𝜔𝑆 = 𝑀𝐶× 𝜔𝐶
∴ 𝑀𝐶= 𝑀𝑆(𝜔𝑆/𝜔𝐶)=10 (𝑁𝑆/𝑁𝐶)
=10 × 5 = 50 𝐾𝑔. 𝑚

10. PROBLEM #3
An epicyclic gear is constructed as follows. A fixed annular wheel A and a smaller concentric wheel
B are connected by a compound wheel A1-B1. A1 gearing with A.
B1 gearing with B. The compound wheel revolves on a stud which is carried around an arm which
revolves about the axis A and B. 'A' has 130 teeth, B = 20 teeth, B1 =80 teeth, pitch of A and A1
being twice that of pitch of B1 and B. How many revolutions B will make for one revolution of
arm?

11. PROBLEM #3 (SOLUTION)
As pitch circle radius is proportional to number of teeth and pitch of A and A, being twice that of B and B,
rA = rB + rB1 + rA1
i.e., 2TA= TB+ TB1 + 2TA1
2*130 = 20 + 80 + 2TA1
Therefore the number of teeth on Gear A1 i.e T1 = 80
Condition of Motion Arm Rev of Gear B Rev of Compound
Gear B1&A1
Rev of Gear A
Fix the arm and give
+1 revolutions to gear
B
0 +1 -TB/TB1 -(TB/TB1)*(TA1/TA)
Multiply ‘x’ revolutions 0 x -(TB/TB1)*x -(TB/TB1)*(TA1/TA)*x
Add ‘y’ y x+y y-(TB/TB1)*x y-(TB/TB1)*(TA1/TA)*x

12. PROBLEM #3 (SOLUTION)
Condition of Motion Arm Rev of Gear B Rev of Compound Gear
B1&A1
Rev of Gear A
Fix the arm and give +1
revolutions to gear B
0 +1 -20/80 -(20/80)*(80/130)
Multiply ‘x’ revolutions 0 x -(20/80)*x -(20/80)*(80/130)*x
Add ‘y’ y x+y y-(20/80)*x y-(20/80)*(80/130)*x
Arm makes one revolution, i.e., y = 1
Annular wheel A is fixed
i.e, y-(TB/TB1)*(TA1/TA)=0
1-(20/80)*(80/130)*x=0
x=6.5 rev
Speed of Gear B= y+x =1+6.5=7.5
i.e. for one revolution of arm, gear B will make 7.5 revolutions, in the same direction of arm.

13. Thank you.
By
Mohammed Asman 4SO21ME038
Mohammed Anas 4SO21ME037
Mohammed Amaan 4SO21ME036
Mohammed Riyaz 4SO21ME033
Mohammed Afrad 4SO21ME035