1
Power transmission
Components used to transmit power: gears, belt, clutch and
brakes.
Gear
Objective:
Student must be able to force analysis, stress analysis using basic
formula (Lewis) and AGMA (bending stress and surface stress)
Type of gear:
Spur gear only
Types of Gear
a)
Spur Gear


b)
Teeth is parallel to axis of rotation
Can transmit power from one shaft to another
parallel shaft
Helical gear




Teeth is inclined to the axis of rotation
Smoother than spur
Develop thrust load (helix angle)
Can transmit power from one shaft to a
parallel and non-parallel shaft
2
c)
Bevel gear


d)
Teeth on conical surfaces
Transmit power between two intersecting
shafts
Worm gear
 Transmit power between
intersecting shafts
two
Terminologies
A pair of gears can represented as 2 circles
d1 = Nm
d2 = Nm
2
3
where: N : number of teeth
m : module
note: mating gear must have same m
3
Module: is the ratio if diametral pitch and number of teeth m = d/N [mm]
Face Width (F) : width of the tooth
Addendum [a] : distance between top face of the tooth to pitch circle
Dedendum [b] : distance between pitch diameter to bottom of the gear
In the following : we only concentrates on full depth gear
4
Full depth tooth
When the offset occurs between pitch. not full-depth tooth, which is called stub
Undercut: resulted from number of tooth is less than the minimum number of
tooth suggested. Results: higher stresses at the root of the tooth. (Refer to
section 13-7 interference)
Backlash : gap between mating tooth. The gap can be used for lubrication
5
Contact Ratio: the number of tooth in contact during meshing. Roughly  spur
gear (1.4 to 1.8 )
Pinion and Gear : pinion is driver and gear is driven
Gear Parameters
Metric
Module m = d/N
Imperial
Diametral Pitch P= N/d (inverse to module)
One pair of gear must have the same module
Pressure Angle: 200, 22.5o, 250
Table 13-1: relationship between addendum (a) and dedundum (b) pp
696
Table 13-2: Available pitch diametral and module.
6
Gear Train
2
V2 = V3
3
Known that
d 2 n2  d 3 n3
V
dn
60
where : n : revolution / min it
Equation 1
d 2 n2  d 3 n3
For a pair of gear m2 = m3
d2 d3

N2 N3
Equation 2
From Eq 1 and 2
d 2 N 2 n3


d 3 N 3 n2
Significance:
 d increases N increase
 d increases n reduces  to reduce rpm requires small
pinion and larger gear and vice versa.
7
Gear Train (continued)
3
2
4
V2 = V3 and V3 = V4
Therefore
V2 = V3 = V4
From previous formula:
N2 n3

N3 n2
N2 n2  N3 n3
Gear 2 and 3
n3 
N2
n2
N3
… (1)
n4 
N3
n3
N4
…(2)
n4 
N3 N2
n2
N4 N3
Gear 3 and 4
Eq (1) in eq (2)
nL 
product of driving tooth numbers
nF
product of driven tooth numbers
where:
Train value
e
nL : rotational speed of last gear (output)
nF : rotational speed of first gear (input)
product of driving tooth numbers
product of driven tooth numbers
8
Planetary Gear
5
4
2
3
Gear 2: Sun gear
Gear 3: Arm
Gear 4: Planet Gear
Gear 5: Ring Gear
Assumption Arm Fixed:
4
2
3
Train value
e
N 2 N 4 N5
N 4 N 5 N6
5
9
3 MAGIC FORMULAE FOR FBD ANALYSIS ON GEAR
Torque
d 
T   Wt
2
[ Nm ]
Wt: tangential force
Speed
V
dn
60
[ m / s]
D: pitch diameter in [m]
n: rotational speed [rpm]
Power
H  WtV
[ watts ]
10
Force Analysis (Free Body Diagram)
Input rpm direction: cw
To transfer power, T must exist.
When the pinion rotates, tooth from gear against the movement  direction W t32
must against the direction of rotation
Wt32 = H / V
T2 = Wt32. d2/2
Due to pressure angle, W r32 (radial force) is generated
Wr32 = Wt32 tan 

11
On Gear 3,
Wt23 and W r23 must in the opposite direction.
To be statically analytical, T3 is against Wt23
T3 = Wt23. d3/2
Note: Wt23 can be calculated using Wt32= H/ V, please remember that all the
parameters must be based on gear 3.
Discuss example 13-7
12
Example
You are responsible to design a gear system for speed reducer. The
speed reducer is a two stage reduction which each pinion has 18T
(Gear 2 and 4 in Figure 1). One of the constraints is that the
maximum allowable reduction is 10 at each stage.
Based on this, answer the following questions.
a.
Suggest the two possible number of teeth for Gear 3 and 5 if
the speed has to be reduced by 24 times. Note: if Gear 3 has X teeth
and Gear 5 has Y teeth and vice versa, the answer is considered as
one)
Number of Tooth
Gear 3
Gear 5
Combination 1
________
________
Combination 2
________
________
Assume gear 3 and 5 have 72 and 90 teeth respectively and
comprises of m = 4 and 200 pressure angle gears. The motor is 8kW
at 1000 rpm clockwise.
b. Calculate the rpm of the output shaft.
c. Draw and calculate all the resultant forces on all of the gears.
d. Based on the above calculation, discuss the relationship
between torque and gear ratio.
18T
18T
2
Syaf masukan
dari motor
4
Syaf Keluaran
3
5
Figure 1
13
Example
The figure below shows a dual output power transmission system. A
8kW motor with 1000 rpm in clockwise direction is attached to shaft
1 at A. 40% of the power is delivered to shaft 2 using gear 2 and 4
and the remaining 60% of the power is delivered to shaft 3 via gear
3 and 5. All the gear module is 2 mm (m = 2mm) with pressure
angle of 200.
Based on this information, answer the following questions
a)
Draw the FBD for every gear and also calculate all the forces
and torques if the speed of both output shafts have been
reduced by 3.
b)
Calculate N5 if torque on shaft 2 and 3 is equal and N4 is 72.
14
Failure Types
Bending:
resulted from bending stress.
Wt act on the tooth
Lewis formula and AGMA
Pitting:
resulted from surface stress
Repetition of high contact stresses
Scoring:
resulted from insufficiency of lubrication
Bending Stresses
Basic Formula:
Wt

FmY
*take note that there are two formulae in pair in the textbook (imperial and metric)
15
Assumption Basic Formula
 Cantilever Beam Problem
 Ft is carried by one tooth only.
However, dynamic effects are present when a pair of gear at
moderate and high speed. ( KV’)
 
KV' W t
FmY
… eq 14-1 pp738
Y: Lewis form factor Table 14-2 pp 718. (interpolation if it is required)
Dynamic Effect
KV' 
3.05  V
3.05
(cast iron, cast profile)
KV' 
6.1  V
6.1
(cut or milled profile)
16
KV' 
3.56  V
3.56
K 'V 
5.56  V
5.56
(hobbed or shaped profile)
(finishing process on gears:
shaved or ground profile)
17
Fatigue Strength
Endurance Limit
'
 Se  0.5Sut
For steel
For cast iron
AGMA
Se  ka kb kc kd keS'e
18
Surface Factor ka
 k a  aSut
b
0.265
Machining process k a  4.51S ut
19
Size Factor kb
1.24 d e 0.107
 kb  
0.157
 1.51d e
2.79  d  51mm
51  d  254mm
In the case of gear tooth d = de
d e  0.808 Ft
3Ym
2 (Eq 14-3)
t  4 xl (Eq b)
x
l = add + dedd (refer to Table 13-1)
Stress Concentration Factor
 K f  K f 1K f 2
Stress concentration factor due to load (Kf1)
kf1 = 1.66 (one way bending)
kf1 = 1 (two way bending)
K f1 
1
k f1
Stress concentration 2 (Kf2)
20
Refer to Table A-15-6
r = 0.3m
d=t
D = infinite
Based on r/d and D/d = infinite, uses the largest D/d,
get the value of Kt
Kf2= 1 + q (Kt – 1)
Definition of one and two way bending
One way bending: tooth subjected one direction rev.
Two way: tooth subjected to both direction rev (forward
and reverse)
Fatigue failure theories
K fm
Goodman Line
Sut

K fa
Se

1
n
 nK f  m  nK f  a

 
1
Se
 Sut 
2
Gerber Failure
Determine the mean and alternating stress
One way bending
m  a 

2
Two way bending
a  
m  0
Where
KV W t

FmY
Project
You have to set m, F, NP, NG
The constraint
Min no of tooth for pinion: Eq 13-11
Max no of tooth for gear: Eq 13-12
Face width 2p < F < 5p
Material : Figure 14-2, 14-3
21
22
SURFACE DURABILITY
Surface Stresses (compressive –ve)
1
 K W  1 1  2
 C  CP  V t   
 F cos   r1 r 2 
= pressure angle
P = pinion
G = gear
Cp = elastic coefficient
1

2


1


CP 
  1   P2 1   G2  


  
EG  
  EP
 = Poisson Ration (Table A-5 )
E = Modulus of Elasticity (Table A-5)
Radius of curvature of the tooth profile
r1 
dp sin 
2
d sin 
r2  G
2
23
Example
A 19-TOOTH 300 Bhn HOBBED STEEL SPUR GEAR PINION
TRANSMITS 15 Kw AT A PINION SPEED OF 360 rev/min TO A 77
TOOTH OF THE SAME MATERIAL GEAR. THE FACE WIDTH IS 75
mm, = 200 AND m = 6mm.
a) USING LEWIS FORMULA CALCULATE THE STRESSES DUE
TO BENDING AND THE CONTACT STRESSES?
b) CALCULATE THE F.S OF THE BENDING STRESS AGAINST
ITS FATIGUE STRENGTH USING GERBER THEORY?
c) CALCULATE THE F.S OF THE CONTACT STRESSES
AGAINST CONTACT STRESS ENDUCRANCE LIMIT (Sc)?
24
Soln
a)
Bending Stresses
Pinion
KvWt
 
FmY
d = Nm = (19)(6) = 114 mm = 0.114 m
V
dn
Wt 
60

 (0.114)(360)
60
[m / s] = 2.149 m/s
H 15,000

= 6980 N
V
2.149
Hobbed KV' 
3.56  V
= 1.402
3.56
Y = 0.314 (Table 14-2)
KvWt
 
FmY
=
69.8 MPa
GEAR
K v W t (1.412)( 6980)


FmY
(75)( 6)Y
Y interpolation (77 tooth)Y = 0.436
K v W t (1.412)( 6980)


FmY
(75)( 6)Y = 50.23 MPa
25
b)
Surface Stresses
1
 K W  1 1  2
 C  CP  V t   
 F cos   r1 r 2 
Kv = 1.40
Wt = 6980 N
F = 75mm
dp = Nm = (19)(6) = 114 mm
dg = Nm = (77)(6) = 462 mm
r1 
dp sin 
= 19.5 mm
2
dG sin 
r2 
2
= 79 mm
1

2


1


CP 
  1   P2 1   G2  


  
EG  
  EP
Carbon Steel:
V = 0.292
E = 207 GPa = 207 x 103 MPa
Cp = 189.780 MPa
1
 K W  1 1  2
 C  CP  V t   
 F cos   r1 r 2 
 (1.402)( 6980)  1
1 
 (189,780)



(
75
)
cos(
20
)
19
.
5
79



= -565.5 MPa
1
2
26
Safety factor
S 
n   c 
 c 
Sc  6.89(0.4HB  10) MPa
Sc = 757.9
n = 1.15
27
Limits
Se
k a k b k c k d k e Se'


F .S
F .S
 all
It is advisable to have F.S > 3
Contact Stress
 K vW t  1 1 
  
 c  C p 
 F cos   r1 r2 

1
2
… Eq 14-14 pp 732
Negative: because it is always in compression
Cp …. Eq 14.13
E : Modulus of elasticity
u : Poisson’s Ratio
r1 , r2 : Eq 14-12
Table A-5
Table A-5
28
Example
You are responsible to design a gear system for speed reducer. The
speed reducer is a two stage reduction which each pinion has 18T
(Gear 2 and 4 in Figure 1). One of the constraints is that the
maximum allowable reduction is 10 at each stage.
Based on this, answer the following questions.
a.
Suggest the two possible number of teeth for Gear 3 and 5 if
the speed has to be reduced by 24 times. Note: if Gear 3 has X teeth
and Gear 5 has Y teeth and vice versa, the answer is considered as
one)
Number of Tooth
Gear 3
Gear 5
Combination 1
________
________
Combination 2
________
________
Assume gear 3 and 5 have 72 and 90 teeth respectively and
comprises of m = 4 and 200 pressure angle gears. The motor is 8kW
at 1000 rpm clockwise.
b.
Calculate the rpm of the output shaft.
c.
Draw and calculate all the resultant
forces on all of the gears.
d.
Based on the above calculation,
discuss the relationship between
torque and gear ratio.
18T
18T
2
Syaf masukan
dari motor
4
Syaf Keluaran
3
5
Figure 1
29
a)
G3
G5
Comb 1
144 (8 x 18)
54 ( 3 x 18)
Comb 2
108 (6 x 18)
72 ( 4 x 18)
N 2N 4
N 4N5
e
18 x18
72x90
 0.05

b)
nL
= e.nF
= (0.05)(1000)
= 50 rpm
c) FBD at Gear 2
Diameter pinion d = Nm
= (18)(4x10-3)
d
= 0.072 m
Calculating the tangential for
60H
dn
60(8,000)

 (0.072)(1000)
 2,122N
 2.122kN
Ft 
Direction of Ft: when gear 2 pushes gear 3, gear 3 will resist and therefore
the direction is UP
Fr= Ft tan 200 = 0.773kN
T2= Ft d/2 = (2,122)(0.072/2) = 76.4 Nm
30
FBD at Gear 3
Reaction at gear 3 is equal in magnitude and opposite direction to gear 2
Diameter d3 = Nm
= (72) (4x10-3) m
= 0.288 m
T3
= Fr d3/2
= 2122 (0.288/2)
= 305.6 Nm
Rotation direction ccw
FBD at Gear 4
T3 must be equal to T4 with opposite direction. This is because the G3 and G4 is
on the same shaft, therefore, the total Torque = 0. Rotation direction ccw
T3 = T4 = 305.6 Nm
Gear 4 (ccw) pushes gear 3: dirction of Ft is DOWN.
Ft = 2 T/d
d = 0,072 m
Ft = 8.48 kN
Fr = 3.09 kN
60H
dn
60(8,000)

 (0.072)( 250)
 8.48kN
Ft 
Or
31
Gear 5