This document contains notes from a lecture on machine dynamics and design taught by Dr. Muhammad Wasif. The notes cover topics including spur gear tooth profiles, standard spur gears, gear trains, and gear train analysis problems. Specifically, it discusses the involute and cycloidal tooth profiles, AGMA gear standards, simple, compound, and epicyclic gear trains, and provides examples of calculating speed ratios and tooth counts for different gear train configurations.

1. 3/16/2013
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ME‐205 : Element of machine dynamics and design
Dr. Muhammad Wasif
Assistant Professor – I.M.D.
Ph.D. (CAD/CAM – Canada), M.Engg. (Mfg. Engg. – NEDUET),
B.E. (Mech. Engg. – NED UET). Member ASME and PEC
Room : 1st on LHS of main corridor, ground floor – IM Building
Machine dynamics : Gear Trains
1
Spur Gear Tooth Profile
2ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• Two types of spur gear tooth‐profiles are used;
– Cycloidal profile,
– Involute profile
• Velocity ratio in involute gear does not change
with slight misalignments (center distance).
• The pressure angle, from the start of the
engagement of teeth to the end of the
engagement, remains constant which results in
smooth running and less wear of gears.
• The face and flank of involute teeth are generated
by a single curve whereas in cycloidal gears,
double curves (i.e. epicycloid and hypocycloid).
• Involute profile gears are Easy to machine.
Cycloidal profile
Involute profile

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Spur Gear Tooth Profile
3ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• An involute is one of a class of curves
called ‘conjugate curves’. Involute
profile gears are easy to machine.
• A cord is wound clockwise around the
base circle of gear 1, pulled tight
between points a and b, and wound
counterclockwise around the base
circle of gear 2. If, now, the base circles
are rotated in different directions so as
to keep the cord tight, a point g on the
cord will trace out the involutes cd on
gear 1 and ef on gear 2.
Standard Spur Gears
4ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• Standards are designed by AGMA : American gear manufacturing Association.
• 14.5 composite system : cycloidal curve at face, involute at flank.
• 14.5 full depth involute system : to cut by hobs
• 20 full depth involute system : to cut by hobs
• 20 stub involute system : stronger.

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Standard Spur Gears
5ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• The phenomenon when the tip of a tooth undercuts the root on
its mating gear is known as interference.
Standard Spur Gears
6ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
A 25 pressure angle, 43 tooth spur gear has a module 5, find the
pitch dia, addendum, dedendum, working depth, total depth, dia of
addedndum circle, circular pitch.
 = 25 ; N=43; m=5
Sol :
Pitch dia (dp) = Nm = (43) (5)= 215mm
Addendum (a) = 1 m = (1)  (5) = 5mm
Dedendum (b) = 1.25m = 1.25(5) = 6.25mm
Working depth (h) = 2 m = 2(5) = 10mm
Total depth = 2.25 m = 2.25(5) = 11.25mm
Dia. Of addendum = dp+2a = 215+2(5)= 225mm
Circular pitch =  m = 15.71mm

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Standard Spur Gears
7ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A 57‐tooth spur gear is mesh with a 23‐tooth pinion, The
module is 5 and pressure angle is 25, what will be the pressure
angle, if the center distance is increased by 5%.
• Sol :  = 25; Ng=57; Np=23 ; m=5
• rb= rp cos 
• dp= Ng m = (57)(5) = 285mm
• rp=0.5  dp = 142.5mm
• new =cos‐1(
rb
rp,new
)=cos‐1(
rp cos 
rp,new
)= cos‐1(
rp cos 25
. rp
)=30.32
Gear Trains
8ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A gear train is any collection of two
or more meshing gears.
• A pair of gear is the simplest form of
gear train and limited to a ratio
about 10:1.
• Gear trains have following three
types;
– Simple
– Compound
– Epi‐cyclic

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Simple Gear Train
9ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A gear train in which each shaft carries one gear.
• A single idle gear may be used to change the direction of
rotation.
Velocity ratio =
∙ ∙ =
∙ ∙
Velocity ratio =
T=No. of teeth, N=rpm
Intermediate gears
Driver
Driven
Driver Driven
Simple Gear Train
10ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
From above equation
• For the odd number of intermediate gears, directions of rotation of driver
and driven are same.
• For even number of intermediate gear, direction of rotation of driver and
driven are opposite.
• Intermediate gears act as ideal gear and do not effect the velocity ratio.
• Intermediate gears are used to connect the driver and driven at large center
distance and to attain the desired direction of motion of the driven gear.
Velocity ratio =
∙ ∙
Intermediate gears

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Compound Gear Trains
11ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A gear train in which at least one shaft carries more than one gear.
• Usually used to get the reduction ratio more than 10:1.
• This train may contain, parallel or series‐parallel combination of
gears.
Gear 1 is driver mounted on shaft A.
Gear 2,3 and 4,5 are compound gears mounted
on shaft B and C respectively.
Gear 6 is driven mounted on shaft D.
V.R. =
∙ ∙
= ∙ ∙
. .
. .
Reverted Gear Trains
12ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A gear compound gear train in which axis of driver and driven
gears are co‐axial.
Since the center distance b/w gear 1,2 and as
well as 3,4 is same, therefore,
∙
+
∙
=
∙
+
∙
.
.
c

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Epi‐cyclic Gear Train
13ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• A gear train having at least one planetary gear, connected with a
link to a sun gear is called epi‐cyclic gear train.
• Used to transmit high velocity ratio, with gears of moderate size and
lesser space.
Velocity ratio of epi‐cyclic gear train can be
determined by two methods;
• Algebraic method : Motion of each
element in gear train is related to the
arm. No. of equations depends upon the
number of elements in the train.
/ ; /
Since, gear A and B are meshed directly.
/
/
Epi‐cyclic Gear Train
14ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
• Tabular Method :
Step Condition of meshing Revolution of elements
Arm C
(
Gear A
( )
Gear B
( . )
1. Arm fixed, gear ‘A’ rotates through 1 rev. 0 1
2. Arm fixed, gear ‘A’ rotates through ‘ ’ rev. 0
.
3. Add ‘ ’ motion to all elements
.

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Problems
15ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
In a compound gear train, the number of teeth on gear 1, 2, 3, 4, 5
and 6 are 80, 40, 50, 25, 30 and 12 respectively. If gear 1 runs at 200
rpm, find the speed and direction of gear 6.
Sol:
T1=80, T2=40, T3=50, T4=25, T5=30, T6=12; N1=200rpm; N6=?
. .
. .
N6=2000rpm, Direction = opposite
Problems
16ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
The speed ratio of reverted gear train is 12(constant). The module of
gear ‘1’ and ‘2’ is 3.125mm, and gear ‘3’ and ‘4’ is 2.5mm. Find the
number of teeth on each gear. (center distance =200mm)
Sol:
c
NA/ND=12
(NA/NB).(NC/ND)= 120.5 . 120.5 =12
NA/NB =120.5 ; NC/ND= 120.5
rA+rB=rC+rD=c
mA.TA/2+ mB.TB/2= mC.TC/2+ mD.TD/2=c
0.5mA. (TA+ TB)=0.5mc. (Tc+ TD)=c
0.5.(3.125). (TA+ TB)=200

9. 3/16/2013
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Problems
17ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
The speed ratio of reverted gear train is 12(constant). The module of
gear ‘1’ and ‘2’ is 3.125mm, and gear ‘3’ and ‘4’ is 2.5mm. Find the
number of teeth on each gear. (center distance =200mm)
Sol:
c
TA+ TB=200x2/3.125 ‐‐‐‐‐(1)
TC+ TD=200x2/2. 5 ‐‐‐‐‐(2)
NA/NB = TA/TB = 120.5 ‐‐‐‐ (3)
NC/ND = TC/TD = 120.5 ‐‐‐‐ (3)
Solving equations 1 with 3 and 2 with 4
TA=29; TB = 100
TC= 36; TD = 124
Problems
18ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
In an epi‐cyclic gear train, an arm carries two gears ‘A’ and ‘B’ having
36 and 45 teeth respectively. If the arm rotates 150rpm in an
anticlockwise direction about a centre of gear A, which is fixed.
Determine the speed of ‘B’. If gear ‘A’ instead of being fixed makes
300 rpm in clockwise direction. What will be the speed of gear ‘B’.
TA = 36 ; TB = 45 ; NC = 150rpm (A‐CW);
(a) NB =? When NA = 0 ; (b) NB =? When NA = 300rpm
Algebraic method
(a)
/
/

=270 rpm
(b)
/
/

=510 rpm

10. 3/16/2013
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Problems
19ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
In an epi‐cyclic gear train, an arm carries two gears ‘A’ and ‘B’ having 36 and 45 teeth
respectively. If the arm rotates 150rpm in an anticlockwise direction about a center
of gear A, which is fixed. Determine the speed of ‘B’. If gear ‘A’ instead of being fixed
makes 300 rpm in clockwise direction. What will be the speed of gear ‘B’.
St
ep
Condition of meshing Revolution of elements
Arm
C
)
Gear
A
( )
Gear B
(
. )
1. Arm fixed, gear ‘A’ rotates
through 1 rev.
0 1
2. Arm fixed, gear ‘A’ rotates
through ‘ ’ rev.
0
.
3. Add ‘ ’ motion to all elements
.
(a)
0 150
150
.
150
150 150 =270rpm
(b)
300 150
450
.
150 450 =510rpm
Problems
20ME‐205, Elements of Machine design and dynamics, conducted by Dr. Muhammad Wasif (Asst. Professor ‐ IMD, NEDUET)
In an epi‐cyclic gear train, as shown in the fig. The no. of teeth on gear A, B and C are
48, 24 and 50 respectively. If the arm rotates at 400rpm CW. Find
(a) Speed of gear C when A is fixed, (b) speed of gear A when C is fixed.
(a)
0 400
400
.
400 400 =16rpm
b .
0 400  416.667
416.667 400 16.667
A B C
Arm D

11. 3/16/2013
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Refernces
21
• R.S. Khurmi, G.K. Gupta, 2005, A text book of machine design, New Dehli‐ India,
EURASIA PUBLISHING HOUSE.
• R. G. Budynass, J.K. Nisbett, 2005, Shigley’s Mechanical Engineering Deisgn, New
York –USA, McGraw Hill.
• R.C. Juvinall, K.M. Marshek, 2012, Fundamental of Machine Componenet Design,
NJ, John Willey and Sons.
• R.L. Norton, 1996, Machine design and integrated approach, USA, Prentice Hall.
• http://www.mae.ncsu.edu/eischen/
• http://www.khkgears.co.jp/en/gear_technology/pdf/gearabc_b.pdfB. J. Hamrock
• http://www.xtek.com/pdf/wp‐gear‐terminology.pdf
• http://www.khkgears.co.jp/en/gear_technology/pdf/gearabc_b.pdf