Formula Book.pdf
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About This Presentation
Mechanical
Slide Content
Department of Technical and Vocational
Education and Training
Department of Mechanical Engineering
Tables, Data and Equations
For
ME 31031 and ME 32031 Design of Machine Elements
AGTI Third Year (Mechanical)
Education and Training
Department of Mechanical Engineering
Tables, Data and Equations
For
ME 31031 and ME 32031 Design of Machine Elements
AGTI Third Year (Mechanical)
Page 1
Stresses In Simple Machine Member
S
n(max:) =
S
x+ S
y
2
+√(
S
x−S
y
2
)
2
+ S
s
2
S
n(min:)=
S
x+ S
y
2
−√(
S
x−S
y
2
)
2
+ S
s
2
S
s(max:) =√(
S
x−S
y
2
)
2
+ S
s
2
S
s(max:)=
S
n(max:)− S
n(min:)
2
or
S
n(max:)− 0
2
or
S
n(min:)− 0
2
S
xand S
y= ±
??????
??????
±
??????×??????
??????
(+Tension),(−compression)
S
s=
T × r
J
( for circular cross section)
Where, S
n(max:) = Maximum normal Stress
S
n(min:) = Minimum normal Stress
S
s(max:) = Maximum shear Stress
P = Axial load,N
A = Area of cross section,m
2
M = Bending Moment,N.m
C = Distance from neutral axis to outer surface,m
I = Rectangular moment of inertia of cross section ,m
4
T = Torsional moment,N.m
J = Polar moment of inertia,m
4
Stresses In Simple Machine Member
S
n(max:) =
S
x+ S
y
2
+√(
S
x−S
y
2
)
2
+ S
s
2
S
n(min:)=
S
x+ S
y
2
−√(
S
x−S
y
2
)
2
+ S
s
2
S
s(max:) =√(
S
x−S
y
2
)
2
+ S
s
2
S
s(max:)=
S
n(max:)− S
n(min:)
2
or
S
n(max:)− 0
2
or
S
n(min:)− 0
2
S
xand S
y= ±
??????
??????
±
??????×??????
??????
(+Tension),(−compression)
S
s=
T × r
J
( for circular cross section)
Where, S
n(max:) = Maximum normal Stress
S
n(min:) = Minimum normal Stress
S
s(max:) = Maximum shear Stress
P = Axial load,N
A = Area of cross section,m
2
M = Bending Moment,N.m
C = Distance from neutral axis to outer surface,m
I = Rectangular moment of inertia of cross section ,m
4
T = Torsional moment,N.m
J = Polar moment of inertia,m
4
Page 2
Curved Beam hi
ho
Bending stress at the inside fiber,?????? =
� × ℎ
??????
?????? × � × �
??????
,ℎ
??????= �
�− �
??????
Bending stress at the outside fiber,?????? =
� × ℎ
�
?????? × � × �
�
,ℎ
�= �
�− �
�
Where,
ℎ
?????? = the distance from the neutral axis to the inside fiber,m
ℎ
� = the distance from the neutral axis to the outside fiber,m
�
?????? = the radius of curvature of the inside fiber,m
�
� = the radius of curvature of the outside fiber,m
C.A = Central Axis
N.A = Neutral Axis
ri
ro
rn R
e
Curved Beam hi
ho
Bending stress at the inside fiber,?????? =
� × ℎ
??????
?????? × � × �
??????
,ℎ
??????= �
�− �
??????
Bending stress at the outside fiber,?????? =
� × ℎ
�
?????? × � × �
�
,ℎ
�= �
�− �
�
Where,
ℎ
?????? = the distance from the neutral axis to the inside fiber,m
ℎ
� = the distance from the neutral axis to the outside fiber,m
�
?????? = the radius of curvature of the inside fiber,m
�
� = the radius of curvature of the outside fiber,m
C.A = Central Axis
N.A = Neutral Axis
ri
ro
rn R
e
Page 3
Figures give the location to the neutral axis, the distance from the centroidal
axis to the neutral axis, and the distance to the centrodial axis from the center
of curvature for various commonly encountered shapes C.A
N.A
rn
e
h
b
ri
R
ro ro
d ri
rn
R
C.A
N.A
e
r
n=
h
log
e(
r
o
r
i
)
e=R−r
n
R=r
i+
h
2
r
n=
[r
o
1
2
⁄
+r
i
1
2
⁄
]
2
4
e=R−r
n
R=r
i+
d
2
Figures give the location to the neutral axis, the distance from the centroidal
axis to the neutral axis, and the distance to the centrodial axis from the center
of curvature for various commonly encountered shapes C.A
N.A
rn
e
h
b
ri
R
ro ro
d ri
rn
R
C.A
N.A
e
r
n=
h
log
e(
r
o
r
i
)
e=R−r
n
R=r
i+
h
2
r
n=
[r
o
1
2
⁄
+r
i
1
2
⁄
]
2
4
e=R−r
n
R=r
i+
d
2
Page 4 ro
h ri
bi
tito
bo
t
e
rn
R
C.A
N.A ro
h
ri
ti
bi
t
e
rn
R
C.A N.A
r
n=
(b
i−t)t
i+th
(b
i−t)log
e(
r
i+t
i
r
i
)+tlog
e(
r
o
r
i
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b
i−t)
(b
i−t)t
i+th
r
n=
(b
i−t)t
i+(b
o−t)t
o+th
b
ilog
e(
r
i+t
i
r
i
)+tlog
e(
r
o−t
o
r
i+t
i
)+b
olog
e(
r
o
r
o−t
o
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b
i−t)+(b
o−t)t
o(h−
1
2
⁄t
o)
(b
i−t)t
i+(b
o−t)t
o+th
h ri
bi
tito
bo
t
e
rn
R
C.A
N.A ro
h
ri
ti
bi
t
e
rn
R
C.A N.A
r
n=
(b
i−t)t
i+th
(b
i−t)log
e(
r
i+t
i
r
i
)+tlog
e(
r
o
r
i
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b
i−t)
(b
i−t)t
i+th
r
n=
(b
i−t)t
i+(b
o−t)t
o+th
b
ilog
e(
r
i+t
i
r
i
)+tlog
e(
r
o−t
o
r
i+t
i
)+b
olog
e(
r
o
r
o−t
o
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b
i−t)+(b
o−t)t
o(h−
1
2
⁄t
o)
(b
i−t)t
i+(b
o−t)t
o+th
Page 5 ro
h ri
bo bi
rn
R
C.A N.A
e ro
ri
t/2
t/2
to
ti
e
rn
R
C.A
N.A
h
b
r
n=
(
b
i+b
o
2
)h
(
b
ir
o−b
or
i
h
)log
e(
r
o
r
i
)−(b
i−b
o)
e=R−r
n
R=r
i+
h(b
i+2b
o)
3(b
i+b
o)
r
n=
(b
i−t)(t
i+t
o)+th
b[log
e(
r
i+t
i
r
i
)+log
e(
r
o
r
o−t
o
)]+tlog
e(
r
o−t
o
r
i+t
i
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b−t)+(b−t)t
o(h−
1
2
⁄t
o)
(b−t)(t
i+t
o)+th
h ri
bo bi
rn
R
C.A N.A
e ro
ri
t/2
t/2
to
ti
e
rn
R
C.A
N.A
h
b
r
n=
(
b
i+b
o
2
)h
(
b
ir
o−b
or
i
h
)log
e(
r
o
r
i
)−(b
i−b
o)
e=R−r
n
R=r
i+
h(b
i+2b
o)
3(b
i+b
o)
r
n=
(b
i−t)(t
i+t
o)+th
b[log
e(
r
i+t
i
r
i
)+log
e(
r
o
r
o−t
o
)]+tlog
e(
r
o−t
o
r
i+t
i
)
e=R−r
n
R=r
i+
1
2
⁄h
2
t+
1
2
⁄t
i
2
(b−t)+(b−t)t
o(h−
1
2
⁄t
o)
(b−t)(t
i+t
o)+th
Page 6
Power Transmission shafting
Hollow shaft and solid shaft
Equal torsional stiffness
(
T
θ
)
S= (
T
θ
)
H
i.e (
JG
L
)
S= (
JG
L
)
H
J
S= J
H
π
32
d
4
=
π
32
(d
o
4
− d
i
4
)
∴ d
i
4
= d
o
4
− d
4
………..(1)
Equal torsional strength
(
T
S
s
)
S= (
T
S
s
)
H
i.e (
J
y
)
S= (
J
y
)
H
π
32
d
4
×
2
d
=
π
32
(d
o
4
− d
i
4
) ×
2
d
o
∴ d
i
4
= d
o
4
− d
3
d
o………..(2)
Power Transmission shafting
Hollow shaft and solid shaft
Equal torsional stiffness
(
T
θ
)
S= (
T
θ
)
H
i.e (
JG
L
)
S= (
JG
L
)
H
J
S= J
H
π
32
d
4
=
π
32
(d
o
4
− d
i
4
)
∴ d
i
4
= d
o
4
− d
4
………..(1)
Equal torsional strength
(
T
S
s
)
S= (
T
S
s
)
H
i.e (
J
y
)
S= (
J
y
)
H
π
32
d
4
×
2
d
=
π
32
(d
o
4
− d
i
4
) ×
2
d
o
∴ d
i
4
= d
o
4
− d
3
d
o………..(2)
Page 7
Reduction in weight
If N% is weight reduction
W
H= (1 − N)W
s
π
4
(d
o
2
− d
i
2
)l
hW = (1 − N)
π
4
d
2
l
sW
(d
o
2
− d
i
2
)= (1 − N)d
2
∴ d
i2
= d
o
2
− (1 − N)d
2
……………………(3)
The direction of the thrust force depends on the direction of rotation and the
hand of gear tooth. Driver
RH
Fa
Driven
RH
Fa
Driver
LH
Fa
Driven
LH
Fa
Driven
LH
Fa
Driver
Fa
Driven
Fa
Driver
Fa
LH
RH
RH
Reduction in weight
If N% is weight reduction
W
H= (1 − N)W
s
π
4
(d
o
2
− d
i
2
)l
hW = (1 − N)
π
4
d
2
l
sW
(d
o
2
− d
i
2
)= (1 − N)d
2
∴ d
i2
= d
o
2
− (1 − N)d
2
……………………(3)
The direction of the thrust force depends on the direction of rotation and the
hand of gear tooth. Driver
RH
Fa
Driven
RH
Fa
Driver
LH
Fa
Driven
LH
Fa
Driven
LH
Fa
Driver
Fa
Driven
Fa
Driver
Fa
LH
RH
RH
Page 8
Bevel Gear (straight tooth) r r
Driven
gear
pinionDriver
F
t=
M
t
r
, M
t=
9550×kW
rpm
F
r= F
t× tan∅
The separating force can be resolved into two components. Force
components along the shaft axis of the pinion is called the pinion thrust force
F
� and the force component along the shaft axis of the gear is called the gear
thrust force F
??????.
F
p= F
t× tan∅ sinβ
F
g= F
t× tan∅cosβ
Where β = cone angle
F
??????= tangential force
F
??????= separating force
Bevel Gear (straight tooth) r r
Driven
gear
pinionDriver
F
t=
M
t
r
, M
t=
9550×kW
rpm
F
r= F
t× tan∅
The separating force can be resolved into two components. Force
components along the shaft axis of the pinion is called the pinion thrust force
F
� and the force component along the shaft axis of the gear is called the gear
thrust force F
??????.
F
p= F
t× tan∅ sinβ
F
g= F
t× tan∅cosβ
Where β = cone angle
F
??????= tangential force
F
??????= separating force
Page 9
Pulley Driver
Driven
T1
T2
(T1+T2) sin
(T1+T2) cos
(T1+T2)
T1
T2
M
t=(T
1− T
2)× r
(T
1− T
2)=
M
t
r
Where T
1 = Tight side tension
T
2 = Slack side tension
ASME CODE Equation for Shaft Design
1.For solid shaft subjected to torsion and bending load ( without axial
load)
d
3
=
16
πS
s
√(K
bM
b)
2
+ (K
tM
t)
2
2.For solid shaft subjected to combine bending, torsion and axial load
d
3
=
16
πS
s(max:)
√[�
??????M
b+
αF
ad
8
]
2
+ (K
tM
t)
2
Pulley Driver
Driven
T1
T2
(T1+T2) sin
(T1+T2) cos
(T1+T2)
T1
T2
M
t=(T
1− T
2)× r
(T
1− T
2)=
M
t
r
Where T
1 = Tight side tension
T
2 = Slack side tension
ASME CODE Equation for Shaft Design
1.For solid shaft subjected to torsion and bending load ( without axial
load)
d
3
=
16
πS
s
√(K
bM
b)
2
+ (K
tM
t)
2
2.For solid shaft subjected to combine bending, torsion and axial load
d
3
=
16
πS
s(max:)
√[�
??????M
b+
αF
ad
8
]
2
+ (K
tM
t)
2
Page 10
3.For hollow shaft without axial load
d
o
3
=
16
πS
s(1−K
4
)
√(K
bM
b)
2
+ (K
tM
t)
2
,K =
d
i
d
o
,
4.For hollow shaft with axial load
d
o
3
=
16
πS
s(1 − K
4
)
√[K
bM
b+
αF
ad
8
(1 + K
2
)]
2
+ (K
tM
t)
2
ASME Code states for commercial steel shafting
S
s(allowable)= 8000psi(55
MN
m
2⁄)for shaft without keyway
S
s(allowable)= 6000psi(40
MN
m
2⁄)for shaft with keyway
If ultimate strength and yield strength are known,
S
s(allowable)= 0.18 S
u
S
s(allowable)= 0.3 S
y
Choose smaller value (
without keyway)
If there is a keyway or fillet, the strength of the shaft is reduced by 25%
S
s(allowable)= 0.18 S
u× 0.75
S
s(allowable)= 0.3 S
y× 0.75
Choose smaller value (
with keyway)
3.For hollow shaft without axial load
d
o
3
=
16
πS
s(1−K
4
)
√(K
bM
b)
2
+ (K
tM
t)
2
,K =
d
i
d
o
,
4.For hollow shaft with axial load
d
o
3
=
16
πS
s(1 − K
4
)
√[K
bM
b+
αF
ad
8
(1 + K
2
)]
2
+ (K
tM
t)
2
ASME Code states for commercial steel shafting
S
s(allowable)= 8000psi(55
MN
m
2⁄)for shaft without keyway
S
s(allowable)= 6000psi(40
MN
m
2⁄)for shaft with keyway
If ultimate strength and yield strength are known,
S
s(allowable)= 0.18 S
u
S
s(allowable)= 0.3 S
y
Choose smaller value (
without keyway)
If there is a keyway or fillet, the strength of the shaft is reduced by 25%
S
s(allowable)= 0.18 S
u× 0.75
S
s(allowable)= 0.3 S
y× 0.75
Choose smaller value (
with keyway)
Page 11
Combined shock and fatigue factor applied to bending
moment (Kb) and Combined shock and fatigue factor applied
to torsion moment (Kt)
For stationary load:
Kb Kt
Load gradually applied 1.0 1.0
Load suddenly applied 1.5 to 2.0 1.5 to 2.0
For Rotating shafts:
Kb Kt
Load gradually applied 1.5 1..0
Load suddenly applied
(minor shock)
1.5 to 2.0 1.0 to 1.5
Load suddenly applied
(heavy shock)
2.0 to 3.0 1.5 to 3.0
Column action Factor (α )
The Column action Factor is unity for a tensile load. For a compression
load, α may be computed by
α =
1
1 − 0.0044(
L
K
)
for
L
K
<115
α =
S
y
π
2
n E
(
L
K
)
2
for
L
K
>115
Combined shock and fatigue factor applied to bending
moment (Kb) and Combined shock and fatigue factor applied
to torsion moment (Kt)
For stationary load:
Kb Kt
Load gradually applied 1.0 1.0
Load suddenly applied 1.5 to 2.0 1.5 to 2.0
For Rotating shafts:
Kb Kt
Load gradually applied 1.5 1..0
Load suddenly applied
(minor shock)
1.5 to 2.0 1.0 to 1.5
Load suddenly applied
(heavy shock)
2.0 to 3.0 1.5 to 3.0
Column action Factor (α )
The Column action Factor is unity for a tensile load. For a compression
load, α may be computed by
α =
1
1 − 0.0044(
L
K
)
for
L
K
<115
α =
S
y
π
2
n E
(
L
K
)
2
for
L
K
>115
Page 12
K =√
I
A
=
d
4
(for circular cross section)
??????��??????�,
n = 1 for hinged ends
n = 2.25 for fixed ends
n = 1.6 for ends partly restrained
k = Radius or gyration, m
I = Rectangular moment of inertia, m
4
A = Cross section area of shaft,m
2
Sy = Yield stress in compression , N/m
2
Standard sizes of shafting
Up to 25mm in 0.5mm increments
25 to 50mm in 1mm increments
50 to 100mm in 2mm increments
100 to 200mm in 5mm increments
K =√
I
A
=
d
4
(for circular cross section)
??????��??????�,
n = 1 for hinged ends
n = 2.25 for fixed ends
n = 1.6 for ends partly restrained
k = Radius or gyration, m
I = Rectangular moment of inertia, m
4
A = Cross section area of shaft,m
2
Sy = Yield stress in compression , N/m
2
Standard sizes of shafting
Up to 25mm in 0.5mm increments
25 to 50mm in 1mm increments
50 to 100mm in 2mm increments
100 to 200mm in 5mm increments
Page 13
Power screws and Threaded Fasteners
Pitch (P)
The distance from a point on one thread to the corresponding point on
the next adjacent thread measures parallel to the axis.
P =
1
t.p.i
=
1
number of thread per inch
Lead (L)
The distance the screw would advance relative to nut in one rotation.
For single start, L= P
For double start, L= 2P
For multi start,
L= nP
Helix angle (α)
tanα =
Lead
2πr
m
tanθ
n= tanθ × cosα
Where,
do
= outer diameter
di
= inner (or) core diameter
dm = mean diameter
Power screws and Threaded Fasteners
Pitch (P)
The distance from a point on one thread to the corresponding point on
the next adjacent thread measures parallel to the axis.
P =
1
t.p.i
=
1
number of thread per inch
Lead (L)
The distance the screw would advance relative to nut in one rotation.
For single start, L= P
For double start, L= 2P
For multi start,
L= nP
Helix angle (α)
tanα =
Lead
2πr
m
tanθ
n= tanθ × cosα
Where,
do
= outer diameter
di
= inner (or) core diameter
dm = mean diameter
Page 14
Torque required to advance the screw against the load
P =
W(sinα + fcosα)
cosα − fsinα
……………..(÷ cosα)
P = W[
f + tanα
1 − ftanα
]
T
s= P × r
m
Collar Torque
If there is rubbing and axial load along the axis (ie. Between rotating
and nonrotating members), some torque is wasted in overcoming the
collar friction
T
c= W × f
c× r
c
??????ℎ���,
T
c= collar friction torque
f
c= collar friction
r
c= collar friction radius
r
c=
d
oc+ d
ic
4
( uniform wear)
r
c=
2
3
r
oc
3
− r
ic
3
r
oc
2
− r
ic
2
( uniform pressure)
T
R= T
s+ T
c
T
R= W[r
m
tanα + f
1 − ftanα
+ f
cr
c]
Torque required to advance the screw against the load
P =
W(sinα + fcosα)
cosα − fsinα
……………..(÷ cosα)
P = W[
f + tanα
1 − ftanα
]
T
s= P × r
m
Collar Torque
If there is rubbing and axial load along the axis (ie. Between rotating
and nonrotating members), some torque is wasted in overcoming the
collar friction
T
c= W × f
c× r
c
??????ℎ���,
T
c= collar friction torque
f
c= collar friction
r
c= collar friction radius
r
c=
d
oc+ d
ic
4
( uniform wear)
r
c=
2
3
r
oc
3
− r
ic
3
r
oc
2
− r
ic
2
( uniform pressure)
T
R= T
s+ T
c
T
R= W[r
m
tanα + f
1 − ftanα
+ f
cr
c]
Page 15
Torque required to lower the load (in the direction of the load)
P = W[
−tanα + f
1 + ftanα
]
T
s= W × r
m
T
s= Wr
m[
−tanα + f
1 + f tanα
]
T
L= T
s+ T
c
T
L= W[r
m
−tanα + f
1 + f tanα
+ f
cr
c]
Torque required to raise the load (against load) considering the thread angle
T = W[r
m
tanα +
f
cosθ
n
1 −
f × tanα
cosθ
n
+ f
cr
c]
Torque required to advance the screw (or nut) in the direction of load (or
lowering the load)
T
L= W[r
m
−tanα +
f
cosθ
n
1 +
f × tanα
cosθ
n
+ f
cr
c]
Torque required to lower the load (in the direction of the load)
P = W[
−tanα + f
1 + ftanα
]
T
s= W × r
m
T
s= Wr
m[
−tanα + f
1 + f tanα
]
T
L= T
s+ T
c
T
L= W[r
m
−tanα + f
1 + f tanα
+ f
cr
c]
Torque required to raise the load (against load) considering the thread angle
T = W[r
m
tanα +
f
cosθ
n
1 −
f × tanα
cosθ
n
+ f
cr
c]
Torque required to advance the screw (or nut) in the direction of load (or
lowering the load)
T
L= W[r
m
−tanα +
f
cosθ
n
1 +
f × tanα
cosθ
n
+ f
cr
c]
Page 16
??????���??????��??????????????????
η =
Work output
Work input
=
W × lead
2πT
Self lock the load should not descend under its own weight
Overhaul the load descend under its own weight
App torque
(+) Self Lock, (-) overhaul
Check −tanα +
f
s
cosθ
n
Stress in screw members
(1) Bending stress are estimated by considering the thread to be a short
cantilever beam projecting from the root cylinder h/2
b
h
2πrmn
??????���??????��??????????????????
η =
Work output
Work input
=
W × lead
2πT
Self lock the load should not descend under its own weight
Overhaul the load descend under its own weight
App torque
(+) Self Lock, (-) overhaul
Check −tanα +
f
s
cosθ
n
Stress in screw members
(1) Bending stress are estimated by considering the thread to be a short
cantilever beam projecting from the root cylinder h/2
b
h
2πrmn
Page 17
� = ?????? ×
ℎ
2
?????? = 2??????�
��
�
3
12
⁄,?????? =
�
2
Using bending equation
S
b
y
=
M
I
S
b=
3Wh
2πr
mnb
2
(2) Shear stress at the root of thread
� =
??????
���� ����
=
??????
2??????�
���
(3) Bending stress (or) Bearing pressure between the contacting surface
of the screw and nut thread
p =
W
projected area
=
W
πd
mhn
,n =
nut length
Pitch
(4) Stresses in the root cylinder of the screw may be estimated by
considering loads and torques carried by the bore cylinder
(neglecting strengthening effect of thread)
(i)torsional shear stress ,S
s=
Tr
i
J
(????????????)direct stress ,S
c=
4W
πd
i
2
� = ?????? ×
ℎ
2
?????? = 2??????�
��
�
3
12
⁄,?????? =
�
2
Using bending equation
S
b
y
=
M
I
S
b=
3Wh
2πr
mnb
2
(2) Shear stress at the root of thread
� =
??????
���� ����
=
??????
2??????�
���
(3) Bending stress (or) Bearing pressure between the contacting surface
of the screw and nut thread
p =
W
projected area
=
W
πd
mhn
,n =
nut length
Pitch
(4) Stresses in the root cylinder of the screw may be estimated by
considering loads and torques carried by the bore cylinder
(neglecting strengthening effect of thread)
(i)torsional shear stress ,S
s=
Tr
i
J
(????????????)direct stress ,S
c=
4W
πd
i
2
Page 18
(??????????????????)����??????�?????? ������,??????
??????=
32�
??????�
??????
3
(iv)Maximum shear stress,S
s(max) =√(
S
x
2
)
2
+ S
s
2
Column action
Column action due to axial loading of machine parts occurs very
frequently if the axial load is tensile load, then the application of ?????? =
??????
??????
is in order. If the axial load is compression load, then an appropriate
column equation should be used
The Euler equation for the critical load for slender columns of uniform cross section is-
F
cr=
Cπ
2
EA
(
L
K
)
2
The J.B Johnson formula for the critical load for moderate length
columns of uniform cross section is
F
cr= S
yA[1 −
S
y(
L
K
)
2
4Cπ
2
E
]
(??????????????????)����??????�?????? ������,??????
??????=
32�
??????�
??????
3
(iv)Maximum shear stress,S
s(max) =√(
S
x
2
)
2
+ S
s
2
Column action
Column action due to axial loading of machine parts occurs very
frequently if the axial load is tensile load, then the application of ?????? =
??????
??????
is in order. If the axial load is compression load, then an appropriate
column equation should be used
The Euler equation for the critical load for slender columns of uniform cross section is-
F
cr=
Cπ
2
EA
(
L
K
)
2
The J.B Johnson formula for the critical load for moderate length
columns of uniform cross section is
F
cr= S
yA[1 −
S
y(
L
K
)
2
4Cπ
2
E
]
Page 19
K =√
I
A
=
d
4
( for circular cross section)
Where,
Fcr = Critical load to cause buckling, N
C = Constant depending upon the end condition
E = Modulus of Elasticity, N/m
2
(E=200× 20
9
N/m
2
)
A = Area of transverse section, m
2
L = Length of column, m
K =
Minimum radius of gyration,m
Where, I is the minimum moment of inertia about the axis
of bending
Sy = Yield stress in compression , N/m
2
K =√
I
A
=
d
4
( for circular cross section)
Where,
Fcr = Critical load to cause buckling, N
C = Constant depending upon the end condition
E = Modulus of Elasticity, N/m
2
(E=200× 20
9
N/m
2
)
A = Area of transverse section, m
2
L = Length of column, m
K =
Minimum radius of gyration,m
Where, I is the minimum moment of inertia about the axis
of bending
Sy = Yield stress in compression , N/m
2
Page 20
The value of C depends on the end condition in figure C=¼ C=1 C=2 C=4
C = 4, both ends fixed so that the tangent to the elastic curve at
each end is parallel to the original axis of the column
C =
2, one end fixed and one end free to rotate but not free to
move laterally
C =
1, both ends free to rotate , but not free to move laterally (
so called round, or pivot, or hinged end columns)
C = ¼ , one end fixed and the other end free of all restraint
The value of C depends on the end condition in figure C=¼ C=1 C=2 C=4
C = 4, both ends fixed so that the tangent to the elastic curve at
each end is parallel to the original axis of the column
C =
2, one end fixed and one end free to rotate but not free to
move laterally
C =
1, both ends free to rotate , but not free to move laterally (
so called round, or pivot, or hinged end columns)
C = ¼ , one end fixed and the other end free of all restraint
Page 21
The safe load is obtained by dividing the critical load by a factor of safety N
For Euler equation,
F =
F
cr
N
=
Cπ
2
EA
N(
L
K
)
2
For J.B Johnson equation,
F =
S
yA
N
[1 −
S
y(
L
K
)
2
4Cπ
2
E
]
The value of
�
�
which determines whether the Euler equation or J.B Johnson
equation should be used is found by equating the critical load from the Euler
equation to the critical load from the J.B Johnson formula.
Cπ
2
EA
N(
L
K
)
2
=
S
yA
N
[1 −
S
y(
L
K
)
2
4Cπ
2
E
] From which
L
K
=√
2Cπ
2
E
S
y
The value of
�
�
above which the Euler equation should be used and below
which the J.B Johnson formula should be used
The safe load is obtained by dividing the critical load by a factor of safety N
For Euler equation,
F =
F
cr
N
=
Cπ
2
EA
N(
L
K
)
2
For J.B Johnson equation,
F =
S
yA
N
[1 −
S
y(
L
K
)
2
4Cπ
2
E
]
The value of
�
�
which determines whether the Euler equation or J.B Johnson
equation should be used is found by equating the critical load from the Euler
equation to the critical load from the J.B Johnson formula.
Cπ
2
EA
N(
L
K
)
2
=
S
yA
N
[1 −
S
y(
L
K
)
2
4Cπ
2
E
] From which
L
K
=√
2Cπ
2
E
S
y
The value of
�
�
above which the Euler equation should be used and below
which the J.B Johnson formula should be used
Page 22
For different representative data are.
C E, Gpa ??????
??????, Mpa
(
�
�
)
2
�
�
¼ 207
550
480
415
345
275
1849
2113
2465
2958
3697
43
46
50
54
61
1 207
550
480
415
345
275
7394
8451
9860
11832
1488789
86
92
99
109
121
2 207
550
480
415
345
275
14789
16902
19719
23663
29579
121
130
140
154
172
For different representative data are.
C E, Gpa ??????
??????, Mpa
(
�
�
)
2
�
�
¼ 207
550
480
415
345
275
1849
2113
2465
2958
3697
43
46
50
54
61
1 207
550
480
415
345
275
7394
8451
9860
11832
1488789
86
92
99
109
121
2 207
550
480
415
345
275
14789
16902
19719
23663
29579
121
130
140
154
172
Page 23
Equivalent column stresses are used where column action is to be combined
with other effects as torsion and bending. The equivalent column stress for an
actual load,F derived from Euler’s equation is-
S
eq=
F
A
[
S
y(
L
K
)
2
Cπ
2
E
]
The equivalent column stress for an actual load F, derived from J.B Johnson
equation is
S
eq=
F
A
[
1
1 −
S
y(
L
K
)
2
4Cπ
2
E]
In the equivalent stress equations, the factor of safety is
N =
F
cr
F
=
S
y
S
eq
Equivalent column stresses are used where column action is to be combined
with other effects as torsion and bending. The equivalent column stress for an
actual load,F derived from Euler’s equation is-
S
eq=
F
A
[
S
y(
L
K
)
2
Cπ
2
E
]
The equivalent column stress for an actual load F, derived from J.B Johnson
equation is
S
eq=
F
A
[
1
1 −
S
y(
L
K
)
2
4Cπ
2
E]
In the equivalent stress equations, the factor of safety is
N =
F
cr
F
=
S
y
S
eq
Page 1
Super Gear Design
Proportions of standard gear teeth
14 ½
Composite
14 ½
Full depth
Involute
20
Full depth
Involute
20
Stub
Involute
Addendum m m m 0.8m
Minimum
Dedendum
1.157m 1.157m 1.157m m
Whole depth 2.157m 2.157m 2.157m 1.8m
Clearance 0.157m 0.157m 0.157m 0.2m
Standard module series
Preferred:
1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50
Second choice:
1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45
Super Gear Design
Proportions of standard gear teeth
14 ½
Composite
14 ½
Full depth
Involute
20
Full depth
Involute
20
Stub
Involute
Addendum m m m 0.8m
Minimum
Dedendum
1.157m 1.157m 1.157m m
Whole depth 2.157m 2.157m 2.157m 1.8m
Clearance 0.157m 0.157m 0.157m 0.2m
Standard module series
Preferred:
1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 40, 50
Second choice:
1.125, 1.375, 1.75, 2.25, 2.75, 3.5, 4.5, 5.5, 7, 9, 11, 14, 18, 22, 28, 36, 45
Page 2
Case I Known diameter Case
(
1
m
2
× y
)
all
=
s k π
2
F
t
.………..(1)
s = s
o(
3
3 + V
)for V less than 10
m
s
⁄,V.F =(
3
3 + V
)
s = s
o(
6
6 + V
)for V 10
m
s
⁄to 20
m
s
⁄,V.F =(
6
6 + V
)
s = s
o(
5.6
5.6 +√V
)for V greater than 20
m
s
⁄,V.F =(
5.6
5.6 +√V
)
k = 4(maximum)
F
t=
9550×kW
rpm ×
D
2
⁄
,F
t=
�
�
D
2
⁄
Substitute these value in equation (1) and we get(
1
m
2
×y
)
all
value
Assume y=0.1…….m=…., and write std: module series
Try m=
N =
D
m
, =…….., y=………..(from the table of Appendix I)
(
1
m
2
× y
)
ind
= − − − − −
Compare (
1
m
2
×y
)
ind
and (
1
m
2
×y
)
all
value
Case I Known diameter Case
(
1
m
2
× y
)
all
=
s k π
2
F
t
.………..(1)
s = s
o(
3
3 + V
)for V less than 10
m
s
⁄,V.F =(
3
3 + V
)
s = s
o(
6
6 + V
)for V 10
m
s
⁄to 20
m
s
⁄,V.F =(
6
6 + V
)
s = s
o(
5.6
5.6 +√V
)for V greater than 20
m
s
⁄,V.F =(
5.6
5.6 +√V
)
k = 4(maximum)
F
t=
9550×kW
rpm ×
D
2
⁄
,F
t=
�
�
D
2
⁄
Substitute these value in equation (1) and we get(
1
m
2
×y
)
all
value
Assume y=0.1…….m=…., and write std: module series
Try m=
N =
D
m
, =…….., y=………..(from the table of Appendix I)
(
1
m
2
× y
)
ind
= − − − − −
Compare (
1
m
2
×y
)
ind
and (
1
m
2
×y
)
all
value
Page 3
If (
1
m
2
× y
)
ind
<(
1
m
2
× y
)
all
design is satisfied,decrease the module and try again
If (
1
m
2
× y
)
ind
>(
1
m
2
× y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m
2
× y
)
ind
≤(
1
m
2
× y
)
all
The reduce k value
k
red= k
max×
(
1
m
2
× y
)
ind
(
1
m
2
× y
)
all
Face width, b
b = k
red× π × m
Case II Unknown Diameter Case
S
ind=
2M
t
kyπ
2
Nm
3
− − − − − (2)
Where M
t=
9550×kW
rpm
k = 4(maximum)
If (
1
m
2
× y
)
ind
<(
1
m
2
× y
)
all
design is satisfied,decrease the module and try again
If (
1
m
2
× y
)
ind
>(
1
m
2
× y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m
2
× y
)
ind
≤(
1
m
2
× y
)
all
The reduce k value
k
red= k
max×
(
1
m
2
× y
)
ind
(
1
m
2
× y
)
all
Face width, b
b = k
red× π × m
Case II Unknown Diameter Case
S
ind=
2M
t
kyπ
2
Nm
3
− − − − − (2)
Where M
t=
9550×kW
rpm
k = 4(maximum)
Page 4
N = Number of teeth of weaker,y from table of appendix I
Substituting these value in equation 2 and we get S
ind=
…….
m
3
Assume V.F =
1
2
and equation 2 becomes(
s
o
2
)=(
…….
m
3
),� = − − −
and we get m value .Then with std:module series
Try m = − − − −
D =Nm,V =
πDrpm
60
s
all= s
o× V.F
s
ind=(
− − −
m
3
)
Compare the value of Sind and Sall
If S
ind< s
all
design is satisfied,decrease module and try again
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition
N = Number of teeth of weaker,y from table of appendix I
Substituting these value in equation 2 and we get S
ind=
…….
m
3
Assume V.F =
1
2
and equation 2 becomes(
s
o
2
)=(
…….
m
3
),� = − − −
and we get m value .Then with std:module series
Try m = − − − −
D =Nm,V =
πDrpm
60
s
all= s
o× V.F
s
ind=(
− − −
m
3
)
Compare the value of Sind and Sall
If S
ind< s
all
design is satisfied,decrease module and try again
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition
Page 5
Reduce k
�
���= �
�??????�×
�
??????��
�
??????��
Face width, b
� = �
���× ?????? × �
Dynamic Check
����????????????��� ��??????��,�
�
�
�= �
��????????????�
??????�???????????? ��??????��,�
??????(??????�???????????? ����� ��??????�)
����??????��ℎ�� �����??????��
�
�= �
����
� =
2�
??????
�
�+ �
??????
=
2�
??????
�
�+ �
??????
� =
�
��×�??????�∅ (
1
??????
??????
⁄+
1
??????
??????)
⁄
2
1.4
�
��= (2.75 ���
??????�??????−70)
��
�
2⁄
Reduce k
�
���= �
�??????�×
�
??????��
�
??????��
Face width, b
� = �
���× ?????? × �
Dynamic Check
����????????????��� ��??????��,�
�
�
�= �
��????????????�
??????�???????????? ��??????��,�
??????(??????�???????????? ����� ��??????�)
����??????��ℎ�� �����??????��
�
�= �
����
� =
2�
??????
�
�+ �
??????
=
2�
??????
�
�+ �
??????
� =
�
��×�??????�∅ (
1
??????
??????
⁄+
1
??????
??????)
⁄
2
1.4
�
��= (2.75 ���
??????�??????−70)
��
�
2⁄
Page 6
Where,
Dp = Pitch diameter of the pinion, m
B = Face width, m
K = Stress fatigue factor, N/m
2
Ses = Surface endurance limit of a gear pair, N/m
2
BHNavg = Average brinell hardness number of pinion and gear material
Ep = Modulus of elasticity of the pinion material, N/m
2
Eg = Modulus of elasticity of the gear material, N/m
2
�??????�??????��� ��??????��,�
�
�
�= �
�+
21� (��+ �
�)
21� +√��+ �
�
�ℎ��� � = ��������??????�� ������,
�
�
⁄
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
Where,
Dp = Pitch diameter of the pinion, m
B = Face width, m
K = Stress fatigue factor, N/m
2
Ses = Surface endurance limit of a gear pair, N/m
2
BHNavg = Average brinell hardness number of pinion and gear material
Ep = Modulus of elasticity of the pinion material, N/m
2
Eg = Modulus of elasticity of the gear material, N/m
2
�??????�??????��� ��??????��,�
�
�
�= �
�+
21� (��+ �
�)
21� +√��+ �
�
�ℎ��� � = ��������??????�� ������,
�
�
⁄
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
Page 7
Value for Ses as used in the wear load equation depend upon
a combination of the gear and pinion materials. Some values
for various materials for both Ses and K are tabulated
Average Brinell Hardness
Number of steel pinion and
gear
Surface
Endurance
Limit
Ses(MN/m
2
)
Stress Fatigue Facto,K
(kN/m
2
)
14 ½ 20
150
200
250
300
400
342
480
618
755
1030
206
405
673
1004
1869
282
555
919
1372
2553
Brinell Hardness Number,BHN
Steel
pinion
Gear
150
200
250
150
200
C.I
C.I
C.I
Phosphor Bronze
Phosphor Bronze
342
480
618
342
445
303
600
1000
317
503
414
820
1310
427
689
C.I Pinion
C.I Pinion
C.I Gear
C.I Gear
549
618
1050
1330
1420
1960
Value for Ses as used in the wear load equation depend upon
a combination of the gear and pinion materials. Some values
for various materials for both Ses and K are tabulated
Average Brinell Hardness
Number of steel pinion and
gear
Surface
Endurance
Limit
Ses(MN/m
2
)
Stress Fatigue Facto,K
(kN/m
2
)
14 ½ 20
150
200
250
300
400
342
480
618
755
1030
206
405
673
1004
1869
282
555
919
1372
2553
Brinell Hardness Number,BHN
Steel
pinion
Gear
150
200
250
150
200
C.I
C.I
C.I
Phosphor Bronze
Phosphor Bronze
342
480
618
342
445
303
600
1000
317
503
414
820
1310
427
689
C.I Pinion
C.I Pinion
C.I Gear
C.I Gear
549
618
1050
1330
1420
1960
Page 8
Value of Deformation factor C in kN/m for dynamic load check
Materials Involute
tooth form
Tooth error,mm
Pinion Gear 0.01 0.02 0.04 0.06 0.06
Cast iron
Steel
Steel
Cast iron
Steel
Steel
Cast iron
Steel
steel
Cast iron
Cast iron
Steel
Cast iron
Cast iron
Steel
Cast iron
Cast iron
steel
14 ½
14 ½
14 ½
20 full
depth
20 full
depth
20 full
depth
20 stub
20 stub
20 stub
55
76
110
57
79
114
59
81
119
110
152
220
114
158
228
118
162
238
220
304
440
228
316
456
236
324
476
330
456
660
342
474
684
354
486
714
440
608
880
456
632
912
472
648
952
Value of Deformation factor C in kN/m for dynamic load check
Materials Involute
tooth form
Tooth error,mm
Pinion Gear 0.01 0.02 0.04 0.06 0.06
Cast iron
Steel
Steel
Cast iron
Steel
Steel
Cast iron
Steel
steel
Cast iron
Cast iron
Steel
Cast iron
Cast iron
Steel
Cast iron
Cast iron
steel
14 ½
14 ½
14 ½
20 full
depth
20 full
depth
20 full
depth
20 stub
20 stub
20 stub
55
76
110
57
79
114
59
81
119
110
152
220
114
158
228
118
162
238
220
304
440
228
316
456
236
324
476
330
456
660
342
474
684
354
486
714
440
608
880
456
632
912
472
648
952
Page 9
Helical Gears
Formative or virtual number of teeth
N
f=
N
cos
3
Strength Design
Case I Known diameter Case
(
1
m
2
× y
)
all=
skπ
2
F
t
cos− − − −(1)
s = s
o(
5.6
5.6 +√V
),V =
π D rpm
60
,V.F =(
5.6
5.6 +√V
)
k = 6(maximum)
F
t=
9550×kW
rpm ×
D
2
⁄
Substitute these value in equation (1) and we get(
1
m
2
×y
)
all
value
Assume y=0.15…….m=…., and write std: module series
Try m=
N =
D
m
N
f=
N
cos
3
, =……..,y=………..(from the table of Appendix I)
Helical Gears
Formative or virtual number of teeth
N
f=
N
cos
3
Strength Design
Case I Known diameter Case
(
1
m
2
× y
)
all=
skπ
2
F
t
cos− − − −(1)
s = s
o(
5.6
5.6 +√V
),V =
π D rpm
60
,V.F =(
5.6
5.6 +√V
)
k = 6(maximum)
F
t=
9550×kW
rpm ×
D
2
⁄
Substitute these value in equation (1) and we get(
1
m
2
×y
)
all
value
Assume y=0.15…….m=…., and write std: module series
Try m=
N =
D
m
N
f=
N
cos
3
, =……..,y=………..(from the table of Appendix I)
Page 10
(
1
m
2
× y
)
ind
= − − − − −
Compare (
1
m
2
×y
)
ind
and (
1
m
2
×y
)
all
value
If (
1
m
2
× y
)
ind
<(
1
m
2
× y
)
all
design is satisfied,decrease the module and try again
If (
1
m
2
× y
)
ind
>(
1
m
2
× y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m
2
× y
)
ind
≤(
1
m
2
× y
)
all
The reduce k value
k
red= k
max×
(
1
m
2
× y
)
ind
(
1
m
2
× y
)
all
Face width, b
b = k
red× π × m
(
1
m
2
× y
)
ind
= − − − − −
Compare (
1
m
2
×y
)
ind
and (
1
m
2
×y
)
all
value
If (
1
m
2
× y
)
ind
<(
1
m
2
× y
)
all
design is satisfied,decrease the module and try again
If (
1
m
2
× y
)
ind
>(
1
m
2
× y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m
2
× y
)
ind
≤(
1
m
2
× y
)
all
The reduce k value
k
red= k
max×
(
1
m
2
× y
)
ind
(
1
m
2
× y
)
all
Face width, b
b = k
red× π × m
Page 11
Case II Unknown diameter Case
S
ind=
2M
t
kyπ
2
Nm
3
cos
− − − − − (2)
Where M
t=
9550×kW
rpm
k = 6(maximum)
N = − − − −
N
f=
N
cos
3
, =……..,y=………..(from the table of Appendix 1)
Substituting these values in equation 2 and we get S
ind=
…….
m
3
Assume V.F =
1
2
and equation 2 becomes(
s
o
2
)=(
…….
m
3
)
and we get m value .Then with std:module series
Try m = − − − −
D =Nm,V =
πDrpm
60
s
all= s
o× V.F, V.F =(
5.6
5.6 +√V
)
s
ind=(
− − −
m
3
)
Compare the value of Sind and Sall
Case II Unknown diameter Case
S
ind=
2M
t
kyπ
2
Nm
3
cos
− − − − − (2)
Where M
t=
9550×kW
rpm
k = 6(maximum)
N = − − − −
N
f=
N
cos
3
, =……..,y=………..(from the table of Appendix 1)
Substituting these values in equation 2 and we get S
ind=
…….
m
3
Assume V.F =
1
2
and equation 2 becomes(
s
o
2
)=(
…….
m
3
)
and we get m value .Then with std:module series
Try m = − − − −
D =Nm,V =
πDrpm
60
s
all= s
o× V.F, V.F =(
5.6
5.6 +√V
)
s
ind=(
− − −
m
3
)
Compare the value of Sind and Sall
Page 12
If S
ind< s
all
design is satisfied,decrease module and try again
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition
Reduce k
k
red= k
max×
S
ind
S
all
Face width, b
b = k
red× π × m
End thrust
F
e= F
t× tan
Dynamic Check
����????????????��� ��??????��,�
�
F
o= S
obyπmcos
If S
ind< s
all
design is satisfied,decrease module and try again
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition
Reduce k
k
red= k
max×
S
ind
S
all
Face width, b
b = k
red× π × m
End thrust
F
e= F
t× tan
Dynamic Check
����????????????��� ��??????��,�
�
F
o= S
obyπmcos
Page 13
??????�???????????? ��??????��,�
??????(??????�???????????? ����� ��??????�)
����??????��ℎ�� �����??????��
�
�=
D
pbKQ
cos
2
Q =
2N
g
N
p+ N
g
=
2D
g
D
p+ D
g
K =
S
es×sin∅
n (
1
E
p
⁄+
1
Eg)
⁄
2
1.4
tan∅
n= tan∅ × cos
s
es= (2.75 BHN
avg−70)
MN
m
2⁄
�??????�??????��� ��??????��,�
�
F
d= F
t+
21V (bCcos
2
+ F
t)cos
21V+√bCcos
2
+ F
t
Where C = deformation factor,
N
m
⁄
Values for deformation factor, C
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
??????�???????????? ��??????��,�
??????(??????�???????????? ����� ��??????�)
����??????��ℎ�� �����??????��
�
�=
D
pbKQ
cos
2
Q =
2N
g
N
p+ N
g
=
2D
g
D
p+ D
g
K =
S
es×sin∅
n (
1
E
p
⁄+
1
Eg)
⁄
2
1.4
tan∅
n= tan∅ × cos
s
es= (2.75 BHN
avg−70)
MN
m
2⁄
�??????�??????��� ��??????��,�
�
F
d= F
t+
21V (bCcos
2
+ F
t)cos
21V+√bCcos
2
+ F
t
Where C = deformation factor,
N
m
⁄
Values for deformation factor, C
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
Page 14
Bevel Gear
= cone angle,p= pitch angle(or) ½ cone angle (pinion)
g= pitch angle (or) ½ cone angle (gear)
L = cone length =
1
2
√D
p
2
+ D
g
2
=
N
pm
2
√1 + V.R
2
,(m)
b = face width(close to never greater than
L
3
⁄ ),(m)
The formative or virtual number of teeth
Nf, for a bevel gear is the number of teeth having the same pitch as the actual
gear, that could be cut on a gear having a pitch radius equal to the radius of
back cone
N
fp=
N
p
cosα
p
=
N
p
D
g
2L
⁄
=
N
p
N
g
√N
p
2
+ N
g
2
N
fg=
N
g
cosα
g
=
N
g
D
p
2L
⁄
=
N
g
N
p
√N
p
2
+ N
g
2
Bevel Gear
= cone angle,p= pitch angle(or) ½ cone angle (pinion)
g= pitch angle (or) ½ cone angle (gear)
L = cone length =
1
2
√D
p
2
+ D
g
2
=
N
pm
2
√1 + V.R
2
,(m)
b = face width(close to never greater than
L
3
⁄ ),(m)
The formative or virtual number of teeth
Nf, for a bevel gear is the number of teeth having the same pitch as the actual
gear, that could be cut on a gear having a pitch radius equal to the radius of
back cone
N
fp=
N
p
cosα
p
=
N
p
D
g
2L
⁄
=
N
p
N
g
√N
p
2
+ N
g
2
N
fg=
N
g
cosα
g
=
N
g
D
p
2L
⁄
=
N
g
N
p
√N
p
2
+ N
g
2
Page 15
Strength Check
Case I Known Diameter Case
(
1
m × y
)
??????��=
s b π
F
t
(
L − b
L
)− − − (1)
s = s
o× V.F
V.F =
6
6 + V
(cut teeth)
V.F =
5.6
5.6 +√V
(generate teeth)
V =
π D rpm
60
L =
1
2
√D
p
2
+ D
g
2
,b close tonot greater than
L
3
⁄
F
t=
9550×kW
rpm ×
D
2
⁄
(
1
m × y
)
all
= − − − − −
Assume y=0.1, m=------
Write down Std module series
Try m=
Strength Check
Case I Known Diameter Case
(
1
m × y
)
??????��=
s b π
F
t
(
L − b
L
)− − − (1)
s = s
o× V.F
V.F =
6
6 + V
(cut teeth)
V.F =
5.6
5.6 +√V
(generate teeth)
V =
π D rpm
60
L =
1
2
√D
p
2
+ D
g
2
,b close tonot greater than
L
3
⁄
F
t=
9550×kW
rpm ×
D
2
⁄
(
1
m × y
)
all
= − − − − −
Assume y=0.1, m=------
Write down Std module series
Try m=
Page 16
N =
D
m
N
f= − − − , =……..,y=………..(from the table of Appendix 1)
(
1
m × y
)
ind
= − − − − −
Compare (
1
m×y
)
ind
and (
1
m×y
)
all
value
If (
1
m × y
)
ind
<(
1
m × y
)
all
design is satisfied,decrease the module and try again
If (
1
m × y
)
ind
>(
1
m × y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m × y
)
ind
≤(
1
m × y
)
all
Case II Unknown Diameter Case
S
ind=
2M
t
m
2
byπN
(
L
L − b
)− − − −(2)
M
t=
9550 ×kW
rpm
L =
mN
p
2
√1 + (V.R)
2
N =
D
m
N
f= − − − , =……..,y=………..(from the table of Appendix 1)
(
1
m × y
)
ind
= − − − − −
Compare (
1
m×y
)
ind
and (
1
m×y
)
all
value
If (
1
m × y
)
ind
<(
1
m × y
)
all
design is satisfied,decrease the module and try again
If (
1
m × y
)
ind
>(
1
m × y
)
all
design is not satisfied,increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m × y
)
ind
≤(
1
m × y
)
all
Case II Unknown Diameter Case
S
ind=
2M
t
m
2
byπN
(
L
L − b
)− − − −(2)
M
t=
9550 ×kW
rpm
L =
mN
p
2
√1 + (V.R)
2
Page 17
L
L − b
=
3
2
S
ind=
− − −
m
3
Assume V.F =
1
2
⁄ and S
ind becomes
S
o
2
∴
S
o
2
=
− − −
m
3
,m = − − −
Write down standard module series
Try with module from the standard module series and find the induce and
allowable value of s
Try m = − − −
D = N m,V =
π D rpm
60
s
all= s
o× V.F
s
ind=
− − −
m
3
Compare the value of Sind and Sall
If S
ind< s
all
design is satisfied,decrease module and try again
L
L − b
=
3
2
S
ind=
− − −
m
3
Assume V.F =
1
2
⁄ and S
ind becomes
S
o
2
∴
S
o
2
=
− − −
m
3
,m = − − −
Write down standard module series
Try with module from the standard module series and find the induce and
allowable value of s
Try m = − − −
D = N m,V =
π D rpm
60
s
all= s
o× V.F
s
ind=
− − −
m
3
Compare the value of Sind and Sall
If S
ind< s
all
design is satisfied,decrease module and try again
Page 18
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition.
Then find the face width.
Dynamic Check
The limiting endurance load
F
o= S
obyπm(
L − b
L
)
The limiting wear load
F
w=
0.75 D
pbKQ
cosα
p
Q =
2N
fg
N
fp+ N
fg
K =
S
es×sin∅ (
1
E
p
⁄+
1
E
g)
⁄
2
1.4
s
es= (2.75 BHN
avg−70)
MN
m
2⁄
If S
ind> sall
design is not satisfied,increase module and try again
Try these procedures until S
indjust ≤ Sall and
take the smallest module that satisfies the condition.
Then find the face width.
Dynamic Check
The limiting endurance load
F
o= S
obyπm(
L − b
L
)
The limiting wear load
F
w=
0.75 D
pbKQ
cosα
p
Q =
2N
fg
N
fp+ N
fg
K =
S
es×sin∅ (
1
E
p
⁄+
1
E
g)
⁄
2
1.4
s
es= (2.75 BHN
avg−70)
MN
m
2⁄
Page 19
�??????�??????��� ��??????��,�
�
F
d= F
t+
21V (bC+ F
t)
21V+√bC+ F
t
Where C = deformation factor,
N
m
⁄
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
The American gear manufacturing association (AGMA) standards recommend
the following power ratings
1. Power rating based on peak or full load for both straight and spiral bevel
gear
Power in kW=
m S
o(rpm)
pD
pbyπ
19100
(
L − 0.5b
L
)(
5.6
5.6 +√V
)
Where,
s = 1.7 times the BHN of the weaker gear for gears hardened
after cutting
s = 2 times the BHN of the weaker gear if the gear is case
hardened,N/m
2
�??????�??????��� ��??????��,�
�
F
d= F
t+
21V (bC+ F
t)
21V+√bC+ F
t
Where C = deformation factor,
N
m
⁄
C is the function of material of the gear pair, pressure angle , tooth error e.
The required condition to satisfy the dynamic check is Fo,Fw Fd.
The American gear manufacturing association (AGMA) standards recommend
the following power ratings
1. Power rating based on peak or full load for both straight and spiral bevel
gear
Power in kW=
m S
o(rpm)
pD
pbyπ
19100
(
L − 0.5b
L
)(
5.6
5.6 +√V
)
Where,
s = 1.7 times the BHN of the weaker gear for gears hardened
after cutting
s = 2 times the BHN of the weaker gear if the gear is case
hardened,N/m
2
Page 20
2. AGMA standard for wear (durability)
Power in kW= 0.8 C
mC
Bb for straight bevel gear
Power in kW= C
mC
Bb for spiral bevel gear
Where, C
m= material factor as listed in following Table
C
B=
D
p
1.5
(rpm)
p
0.032
(
5.6
5.6 +√V
)
2. AGMA standard for wear (durability)
Power in kW= 0.8 C
mC
Bb for straight bevel gear
Power in kW= C
mC
Bb for spiral bevel gear
Where, C
m= material factor as listed in following Table
C
B=
D
p
1.5
(rpm)
p
0.032
(
5.6
5.6 +√V
)
Page 21
Material Factor, Cm
Gear Pinion
Material Brinell Material Brinell Cm
I 160_200 II 210_245 0.3
II 245_280 II 285_325 0.4
II 285_325 II 335_360 0.5
II 210_245 III 500 0.4
II 285_325 IV 550 0.6
III 500 IV 550 0.9
IV 500 IV 550 1.0
� = ������� �����
��= ���� ������� �����
���= �??????� �� ������� − ℎ������� �����
��= ���� − ℎ������� ����l
Material Factor, Cm
Gear Pinion
Material Brinell Material Brinell Cm
I 160_200 II 210_245 0.3
II 245_280 II 285_325 0.4
II 285_325 II 335_360 0.5
II 210_245 III 500 0.4
II 285_325 IV 550 0.6
III 500 IV 550 0.9
IV 500 IV 550 1.0
� = ������� �����
��= ���� ������� �����
���= �??????� �� ������� − ℎ������� �����
��= ���� − ℎ������� ����l
Page 22
Worm Gear
� =
�
??????+ �
�
2
�
�= �
?????? ��� �����ℎ ����������
�.� =
�
??????
�
�
=
�
??????
�
����??????
tan?????? =
����
??????�
�
=
�
��
�
??????�
�
=
�
??????�
�
�
�
Where,
Dg = Diameter of gear
Dw = Diameter of worm
Ng = Number of teeth of gear
Nw = Number of start
Pc = Circular Pitch
Pa = Axial Pitch
ma = Axial module
mn = Normal module
V.R = Velocity ratio
Worm Gear
� =
�
??????+ �
�
2
�
�= �
?????? ��� �����ℎ ����������
�.� =
�
??????
�
�
=
�
??????
�
����??????
tan?????? =
����
??????�
�
=
�
��
�
??????�
�
=
�
??????�
�
�
�
Where,
Dg = Diameter of gear
Dw = Diameter of worm
Ng = Number of teeth of gear
Nw = Number of start
Pc = Circular Pitch
Pa = Axial Pitch
ma = Axial module
mn = Normal module
V.R = Velocity ratio
Page 23
AGMA equation include the following design equation
�
�≈
�
0.875
3.48
≈ 3�
�,� ≈ 0.73 × �
�,� ≈ �
�(4.5 +
�
??????
50
)
Where,
Dw = Pitch diameter of the worm,m
C = Center distance between axis of worm and axis of gear,m
b = Face width of gear, m
Pc = Circular pitch of gear, m
L = Axial length of worm, m
The equations are for estimating the approximate proper proportions of the
gear unit
�
??????=
�
�
??????
≈
�
�
3??????
,�
??????(??????�����:)= 2� − �
�(??????�����:),�.� =
�
??????(��)
�
??????�
�
From the above equation we get,
Nw
Dg
�
??????(��??????��)= �.� �
??????�
�
AGMA equation include the following design equation
�
�≈
�
0.875
3.48
≈ 3�
�,� ≈ 0.73 × �
�,� ≈ �
�(4.5 +
�
??????
50
)
Where,
Dw = Pitch diameter of the worm,m
C = Center distance between axis of worm and axis of gear,m
b = Face width of gear, m
Pc = Circular pitch of gear, m
L = Axial length of worm, m
The equations are for estimating the approximate proper proportions of the
gear unit
�
??????=
�
�
??????
≈
�
�
3??????
,�
??????(??????�����:)= 2� − �
�(??????�����:),�.� =
�
??????(��)
�
??????�
�
From the above equation we get,
Nw
Dg
�
??????(��??????��)= �.� �
??????�
�
Page 24
Compare approximate Dg and exact value of Dg from table and choose exact
Dg close to approximate Dg. Then find exact Dw
Dw(exact)= 2C-Dg(exact)
Strength design of the worm wheel is based on the Lewis equation,
� = � � ?????? �
��= � � ?????? ?????? �
�
� = �
�(
6
6 + �
??????
)
�
�= �
?????? ���??????
Where,
F = Permissible tangential load, N
s = Allowable stress, MN/m
2
so =
About 1/3 of the ultimate strength, based on an average value for
stress concentration
Vg = Pitch line velocity of the gear, m/s
Pnc = Normal circular pitch, m
mn = Normal module
Compare approximate Dg and exact value of Dg from table and choose exact
Dg close to approximate Dg. Then find exact Dw
Dw(exact)= 2C-Dg(exact)
Strength design of the worm wheel is based on the Lewis equation,
� = � � ?????? �
��= � � ?????? ?????? �
�
� = �
�(
6
6 + �
??????
)
�
�= �
?????? ���??????
Where,
F = Permissible tangential load, N
s = Allowable stress, MN/m
2
so =
About 1/3 of the ultimate strength, based on an average value for
stress concentration
Vg = Pitch line velocity of the gear, m/s
Pnc = Normal circular pitch, m
mn = Normal module
Page 25
Dynamic Check
Endurance load, Fo
�
�= �
� � ?????? ??????�
�
���� ����,�
�
�
�= �
?????? � �
Where,
Dg = Pitch diameter of the gear, m
b = Gear face width, m
B =
A constant depending upon the combination of the materials used
for the worm and gear, as listed in Table
Dynamic Check
Endurance load, Fo
�
�= �
� � ?????? ??????�
�
���� ����,�
�
�
�= �
?????? � �
Where,
Dg = Pitch diameter of the gear, m
b = Gear face width, m
B =
A constant depending upon the combination of the materials used
for the worm and gear, as listed in Table
Page 26
Constant B
Worm Gear B(kN/m
2
)
Hardened steel Cast iron 345
Steel, 250 BHN Phosphor bronze 415
Hardened steel Phosphor bronze 550
Hardened steel Chilled phosphor bronze 830
Hardened steel Antimony bronze 830
Cast iron Phosphor bronze 1035
The above values for B are suitable for lead angles up to 10 For
angles between 10 and 25, increase B by 25% For lead angles
greater than 25, increase B by 50%
Dynamic load, Fd
�
�=(
6 + �
??????
6
)�
Where,
� = ������ ������??????���� �������??????�� ����
Constant B
Worm Gear B(kN/m
2
)
Hardened steel Cast iron 345
Steel, 250 BHN Phosphor bronze 415
Hardened steel Phosphor bronze 550
Hardened steel Chilled phosphor bronze 830
Hardened steel Antimony bronze 830
Cast iron Phosphor bronze 1035
The above values for B are suitable for lead angles up to 10 For
angles between 10 and 25, increase B by 25% For lead angles
greater than 25, increase B by 50%
Dynamic load, Fd
�
�=(
6 + �
??????
6
)�
Where,
� = ������ ������??????���� �������??????�� ����
Page 27
As for spur, helical and bevel gears, Fo and Fw are allowable values which
must be greater than the dynamic load Fd (Fo,Fw Fd)
AGMA power rating equations are based on wear and the heat dissipation
capacity of the worm gear unit.
From the stand point of wear
� =
���
� � � �
�.�
� =
�.�
�.� + 2.5
� =
2.3
2.3 + �
�+
3�
�
�.�
Where,
P = Input power in kW
rpmw = Speed of worm in rev/min
V.R = Transmission ratio
K =
A pressure constant depending upon center distances, as listed
in Table
m =
Velocity factor depending upon the center distance transmission
ratio and worm speed
Vw = Pitch line velocity of worm, in m/s
As for spur, helical and bevel gears, Fo and Fw are allowable values which
must be greater than the dynamic load Fd (Fo,Fw Fd)
AGMA power rating equations are based on wear and the heat dissipation
capacity of the worm gear unit.
From the stand point of wear
� =
���
� � � �
�.�
� =
�.�
�.� + 2.5
� =
2.3
2.3 + �
�+
3�
�
�.�
Where,
P = Input power in kW
rpmw = Speed of worm in rev/min
V.R = Transmission ratio
K =
A pressure constant depending upon center distances, as listed
in Table
m =
Velocity factor depending upon the center distance transmission
ratio and worm speed
Vw = Pitch line velocity of worm, in m/s
Page 28
Pressure Constant, K
Center Distance
C(mm)
K
(kW/rpm)
Center Distance
C(mm)
K
(kW/rpm)
25 0.0092 250 0.881
50 0.0184 375 2.94
75 0.0294 500 5.87
100 0.0661 750 21.3
125 0.125 1000 48.5
150 0.213 1250 88.1
175 0.330 1500 147
200 0.485 1750 235
225 0.727 2000 235
Pressure Constant, K
Center Distance
C(mm)
K
(kW/rpm)
Center Distance
C(mm)
K
(kW/rpm)
25 0.0092 250 0.881
50 0.0184 375 2.94
75 0.0294 500 5.87
100 0.0661 750 21.3
125 0.125 1000 48.5
150 0.213 1250 88.1
175 0.330 1500 147
200 0.485 1750 235
225 0.727 2000 235
Page 29
From the standpoint of heat dissipation
Based on AGMA recommendations, the limiting input power rating of a plain
worm gear unit from the standpoint of heat dissipation, for worm gear speeds
up to 2000 rev/min, may be estimated by
� =
3650 �
1.7
�.� + 5
(ℎ��� �ℎ���)
Where,
P = Permissible power input in kW
C = Center distance in meters
V.R = Transmission ratio
Allowable Power
� = � �
??????
The efficiency of a worm gear unit, assuming square threads, may be
approximated by
���??????�??????���?????? =
1 − � ���??????
1 + � ���??????
From the standpoint of heat dissipation
Based on AGMA recommendations, the limiting input power rating of a plain
worm gear unit from the standpoint of heat dissipation, for worm gear speeds
up to 2000 rev/min, may be estimated by
� =
3650 �
1.7
�.� + 5
(ℎ��� �ℎ���)
Where,
P = Permissible power input in kW
C = Center distance in meters
V.R = Transmission ratio
Allowable Power
� = � �
??????
The efficiency of a worm gear unit, assuming square threads, may be
approximated by
���??????�??????���?????? =
1 − � ���??????
1 + � ���??????
Page 30
Where,
f = Coefficient of friction
= Lead angle
If the efficiency is less than 50% than the device will be self locking.
That is, it cannot be driven by applying a torque to the wheel. This
characteristic can be a useful safety feature in some applications.
Where,
f = Coefficient of friction
= Lead angle
If the efficiency is less than 50% than the device will be self locking.
That is, it cannot be driven by applying a torque to the wheel. This
characteristic can be a useful safety feature in some applications.
Page 31
Module of Spur Gears, mm
Error (e),
mm
Permissible Error, mm
Pitch line velocity, m/s (Spur Gear)
First Class Commercial Gears
Precision Gears
Module of Spur Gears, mm
Error (e),
mm
Permissible Error, mm
Pitch line velocity, m/s (Spur Gear)
First Class Commercial Gears
Precision Gears
Page 1
Table I Form or Lewis factory Y for spur gears with load at tip of tooth
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0.067
0.071
0.075
0.078
0.081
0.084
0.086
0.088
0.090
0.092
0.093
0.094
0.0955
0.097
0.098
0.099
0.09966
0.10033
0.101
0.078
0.083
0.088
0.092
0.094
0.096
0.092
0.100
0.102
0.104
0.105
0.106
0.107
0.108
0.1095
0.111
0.112
0.113
0.114
0.099
0.103
0.108
0.111
0.115
0.117
0.120
0.123
0.125
0.127
0.1285
0.130
0.1315
0.133
0.1345
0.136
0.137
0.138
0.139
Table I Form or Lewis factory Y for spur gears with load at tip of tooth
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0.067
0.071
0.075
0.078
0.081
0.084
0.086
0.088
0.090
0.092
0.093
0.094
0.0955
0.097
0.098
0.099
0.09966
0.10033
0.101
0.078
0.083
0.088
0.092
0.094
0.096
0.092
0.100
0.102
0.104
0.105
0.106
0.107
0.108
0.1095
0.111
0.112
0.113
0.114
0.099
0.103
0.108
0.111
0.115
0.117
0.120
0.123
0.125
0.127
0.1285
0.130
0.1315
0.133
0.1345
0.136
0.137
0.138
0.139
Page 2
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
0.10175
0.1025
0.10325
0.104
0.1045
0.105
0.1055
0.106
0.1064
0.1064
0.1072
0.1076
0.108
0.10828
0.10857
0.10885
0.10914
0.10942
0.10971
0.110
0.1103
0.1106
0.1109
0.115
0.116
0.117
0.118
0.119
0.120
0.121
0.122
0.1228
0.1236
0.1244
0.1252
0.126
0.12657
0.12714
0.12771
0.12828
0.12885
0.12942
0.130
0.1304
0.1308
0.1312
0.13975
0.1405
0.1425
0.142
0.14275
0.1435
0.14425
0.145
0.1454
0.1458
0.1462
0.1466
0.147
0.14757
0.14814
0.14871
0.14928
0.14985
0.15042
0.151
0.1513
0.1516
0.1519
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
0.10175
0.1025
0.10325
0.104
0.1045
0.105
0.1055
0.106
0.1064
0.1064
0.1072
0.1076
0.108
0.10828
0.10857
0.10885
0.10914
0.10942
0.10971
0.110
0.1103
0.1106
0.1109
0.115
0.116
0.117
0.118
0.119
0.120
0.121
0.122
0.1228
0.1236
0.1244
0.1252
0.126
0.12657
0.12714
0.12771
0.12828
0.12885
0.12942
0.130
0.1304
0.1308
0.1312
0.13975
0.1405
0.1425
0.142
0.14275
0.1435
0.14425
0.145
0.1454
0.1458
0.1462
0.1466
0.147
0.14757
0.14814
0.14871
0.14928
0.14985
0.15042
0.151
0.1513
0.1516
0.1519
Page 3
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
0.1112
0.1115
0.1118
0.1121
0.1124
0.1127
0.113
0.11313
0.11326
0.11339
0.11353
0.11366
0.11379
0.11393
0.11406
0.11419
0.11433
0.11446
0.11459
0.11473
0.011483
0.115
0.11508
0.1316
0.1320
0.1324
0.1323
0.1332
0.1336
0.134
0.13426
0.13453
0.1348
0.13506
0.13533
0.1356
0.13586
0.12613
0.1364
0.13666
0.13639
0.13712
0.13746
0.13773
0.138
0.13816
0.1522
0.1525
0.1528
0.1531
0.1534
0.1537
0.154
0.15426
0.15453
0.1548
0.15506
0.15533
0.1556
0.15586
0.15613
0.1561
0.15666
0.15693
0.15712
0.15746
0.15773
0.158
0.15812
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
0.1112
0.1115
0.1118
0.1121
0.1124
0.1127
0.113
0.11313
0.11326
0.11339
0.11353
0.11366
0.11379
0.11393
0.11406
0.11419
0.11433
0.11446
0.11459
0.11473
0.011483
0.115
0.11508
0.1316
0.1320
0.1324
0.1323
0.1332
0.1336
0.134
0.13426
0.13453
0.1348
0.13506
0.13533
0.1356
0.13586
0.12613
0.1364
0.13666
0.13639
0.13712
0.13746
0.13773
0.138
0.13816
0.1522
0.1525
0.1528
0.1531
0.1534
0.1537
0.154
0.15426
0.15453
0.1548
0.15506
0.15533
0.1556
0.15586
0.15613
0.1561
0.15666
0.15693
0.15712
0.15746
0.15773
0.158
0.15812
Page 4
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
0.11516
0.11524
0.11535
0.11540
0.11548
0.11556
0.11564
0.11572
0.11580
0.11588
0.11596
0.11604
0.11612
0.11620
0.11628
0.11636
0.11644
0.11652
0.11660
0.11668
0.11676
0.11684
0.11692
0.117
0.13832
0.13848
0.13864
0.1388
0.13896
0.13912
0.13928
0.13944
0.13960
0.13976
0.13992
0.14008
0.14024
0.1404
0.14056
0.14072
0.14088
0.14104
0.14120
0.14136
0.14152
0.14168
0.14184
0.142
0.15824
0.15836
0.15848
0.1586
0.15872
0.15884
0.15896
0.15908
0.15920
0.15932
0.15944
0.15956
0.15968
0.15980
0.15992
0.16004
0.16016
0.16028
0.16040
0.16052
0.16064
0.16076
0.16088
0.161
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
0.11516
0.11524
0.11535
0.11540
0.11548
0.11556
0.11564
0.11572
0.11580
0.11588
0.11596
0.11604
0.11612
0.11620
0.11628
0.11636
0.11644
0.11652
0.11660
0.11668
0.11676
0.11684
0.11692
0.117
0.13832
0.13848
0.13864
0.1388
0.13896
0.13912
0.13928
0.13944
0.13960
0.13976
0.13992
0.14008
0.14024
0.1404
0.14056
0.14072
0.14088
0.14104
0.14120
0.14136
0.14152
0.14168
0.14184
0.142
0.15824
0.15836
0.15848
0.1586
0.15872
0.15884
0.15896
0.15908
0.15920
0.15932
0.15944
0.15956
0.15968
0.15980
0.15992
0.16004
0.16016
0.16028
0.16040
0.16052
0.16064
0.16076
0.16088
0.161
Page 5
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
0.11704
0.11708
0.11712
0.11716
0.11720
0.11724
0.11728
0.11732
0.11736
0.11740
0.11744
0.11748
0.11724
0.11756
0.11760
0.11764
0.11768
0.11772
0.11776
0.11780
0.11784
0.11788
0.11792
0.11796
0.14203
0.14216
0.14224
0.14232
0.14240
0.14248
0.14256
0.14262
0.14272
0.14280
0.14288
0.14296
0.14304
0.14132
0.14320
0.14328
0.14336
0.14344
0.14352
0.14360
0.14368
0.14376
0.14384
0.14392
0.16108
0.16118
0.16124
0.16132
0.16140
0.16148
0.16156
0.16164
0.16172
0.16180
0.16188
0.16196
0.16204
0.16212
0.16220
0.16228
0.16236
0.16244
0.16252
0.16260
0.16268
0.16272
0.16284
0.16292
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
0.11704
0.11708
0.11712
0.11716
0.11720
0.11724
0.11728
0.11732
0.11736
0.11740
0.11744
0.11748
0.11724
0.11756
0.11760
0.11764
0.11768
0.11772
0.11776
0.11780
0.11784
0.11788
0.11792
0.11796
0.14203
0.14216
0.14224
0.14232
0.14240
0.14248
0.14256
0.14262
0.14272
0.14280
0.14288
0.14296
0.14304
0.14132
0.14320
0.14328
0.14336
0.14344
0.14352
0.14360
0.14368
0.14376
0.14384
0.14392
0.16108
0.16118
0.16124
0.16132
0.16140
0.16148
0.16156
0.16164
0.16172
0.16180
0.16188
0.16196
0.16204
0.16212
0.16220
0.16228
0.16236
0.16244
0.16252
0.16260
0.16268
0.16272
0.16284
0.16292
Page 6
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
0.118
0.11804
0.11808
0.11812
0.11816
0.11820
0.11824
0.11828
0.11832
0.11836
0.11840
0.11844
0.11848
0.11852
0.11856
0.11860
0.11864
0.11868
0.11872
0.11876
0.11880
0.11884
0.11888
0.144
0.14408
0.14416
0.14424
0.14432
0.14440
0.14448
0.14456
0.14464
0.14472
0.14480
0.14488
0.14496
0.14504
0.14512
0.14520
0.14528
0.14536
0.14544
0.14552
0.14560
0.14568
0.14576
0.163
0.16308
0.16316
0.16324
0.16332
0.16340
0.16348
0.16356
0.16364
0.16372
0.16380
0.16388
0.16398
0.16404
0.16421
0.16420
0.16428
0.16436
0.16444
0.16452
0.16460
0.16468
0.16476
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
0.118
0.11804
0.11808
0.11812
0.11816
0.11820
0.11824
0.11828
0.11832
0.11836
0.11840
0.11844
0.11848
0.11852
0.11856
0.11860
0.11864
0.11868
0.11872
0.11876
0.11880
0.11884
0.11888
0.144
0.14408
0.14416
0.14424
0.14432
0.14440
0.14448
0.14456
0.14464
0.14472
0.14480
0.14488
0.14496
0.14504
0.14512
0.14520
0.14528
0.14536
0.14544
0.14552
0.14560
0.14568
0.14576
0.163
0.16308
0.16316
0.16324
0.16332
0.16340
0.16348
0.16356
0.16364
0.16372
0.16380
0.16388
0.16398
0.16404
0.16421
0.16420
0.16428
0.16436
0.16444
0.16452
0.16460
0.16468
0.16476
Page 7
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
0.11892
0.11896
0.119
0.11902
0.11904
0.11906
0.11908
0.11910
0.11912
0.11914
0.11916
0.11918
0.1192
0.11922
0.11924
0.11926
0.11928
0.11930
0.11932
0.11934
0.11936
0.11938
0.11940
0.11942
0.14584
0.14592
0.146
0.146026
0.146053
0.14608
0.146106
0.146133
0.14616
0.146186
0.146213
0.14624
0.146266
0.146293
0.146312
0.146346
0.146374
0.1464
0.146426
0.146453
0.146480
0.146506
0.146533
0.14656
0.16484
0.16498
0.165
0.165033
0.165066
0.1651
0.165133
0.165166
0.1652
0.165233
0.165266
0.1653
0.165333
0.165366
0.1654
0.165433
0.165466
0.1655
0.165533
0.165566
0.1656
0.165633
0.165666
0.1657
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
0.11892
0.11896
0.119
0.11902
0.11904
0.11906
0.11908
0.11910
0.11912
0.11914
0.11916
0.11918
0.1192
0.11922
0.11924
0.11926
0.11928
0.11930
0.11932
0.11934
0.11936
0.11938
0.11940
0.11942
0.14584
0.14592
0.146
0.146026
0.146053
0.14608
0.146106
0.146133
0.14616
0.146186
0.146213
0.14624
0.146266
0.146293
0.146312
0.146346
0.146374
0.1464
0.146426
0.146453
0.146480
0.146506
0.146533
0.14656
0.16484
0.16498
0.165
0.165033
0.165066
0.1651
0.165133
0.165166
0.1652
0.165233
0.165266
0.1653
0.165333
0.165366
0.1654
0.165433
0.165466
0.1655
0.165533
0.165566
0.1656
0.165633
0.165666
0.1657
Page 8
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
0.11944
0.11946
0.11948
0.11950
0.11952
0.11954
0.11956
0.11953
0.11960
0.11962
0.11964
0.11966
0.11963
0.11970
0.11972
0.11974
0.11976
0.11978
0.11980
0.11982
0.11984
0.11986
0.11988
0.146586
0.146613
0.14664
0.146666
0.146693
0.146712
0.146746
0.146773
0.1468
0.146826
0.146853
0.146880
0.146906
0.146933
0.14696
0.146985
0.147013
0.14704
0.147066
0.147093
0.147112
0.147146
0.147173
0.165733
0.165766
0.1658
0.165833
0.165866
0.1659
0.165933
0.165966
0.1660
0.166033
0.166066
0.1661
0.166133
0.166166
0.1662
0.166233
0.166266
0.1663
0.166333
0.166366
0.1664
0.166433
0.166466
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
0.11944
0.11946
0.11948
0.11950
0.11952
0.11954
0.11956
0.11953
0.11960
0.11962
0.11964
0.11966
0.11963
0.11970
0.11972
0.11974
0.11976
0.11978
0.11980
0.11982
0.11984
0.11986
0.11988
0.146586
0.146613
0.14664
0.146666
0.146693
0.146712
0.146746
0.146773
0.1468
0.146826
0.146853
0.146880
0.146906
0.146933
0.14696
0.146985
0.147013
0.14704
0.147066
0.147093
0.147112
0.147146
0.147173
0.165733
0.165766
0.1658
0.165833
0.165866
0.1659
0.165933
0.165966
0.1660
0.166033
0.166066
0.1661
0.166133
0.166166
0.1662
0.166233
0.166266
0.1663
0.166333
0.166366
0.1664
0.166433
0.166466
Page 9
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
195
196
197
198
199
200
300
RACK
0.11990
0.11992
0.11994
0.1196
0.1998
0.120
0.122
0.124
0.1472
0.147226
0.147253
0.147280
0.147306
0.147333
0.150
0.154
0.1665
0.166533
0.166566
0.1666
0.166633
0.166666
0.170
0.175
No. of
teeth
14 ½ full Depth
Involute or Composite
20 Full Depth
Involute
20 Stub
Involute
195
196
197
198
199
200
300
RACK
0.11990
0.11992
0.11994
0.1196
0.1998
0.120
0.122
0.124
0.1472
0.147226
0.147253
0.147280
0.147306
0.147333
0.150
0.154
0.1665
0.166533
0.166566
0.1666
0.166633
0.166666
0.170
0.175
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