The document provides equations and concepts related to stresses in mechanical elements. It includes equations for normal and shear stresses in simple members under tension and bending. It also provides locations of the neutral axis and distances to the outer surfaces for common cross sections. Additional sections cover stresses in curved beams, torsion in shafts, stresses in gears and pulleys, and equations for power screws and threaded fasteners.
1. 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)
2. Page 1
Stresses In Simple Machine Member
Sn(max: ) =
Sx + Sy
2
+ √(
Sx−Sy
2
)2 + Ss
2
Sn(min: ) =
Sx + Sy
2
− √(
Sx−Sy
2
)2 + Ss
2
Ss(max: ) = √(
Sx−Sy
2
)2 + Ss
2
Ss(max: ) =
Sn(max: ) − Sn(min: )
2
or
Sn(max: ) − 0
2
or
Sn(min: ) − 0
2
Sxand Sy = ±
𝐏
𝐀
±
𝐌×𝐂
𝐈
(+Tension), (−compression)
Ss =
T × r
J
( for circular cross section)
Where, Sn(max: ) = Maximum normal Stress
Sn(min: ) = Minimum normal Stress
Ss(max: ) = Maximum shear Stress
P = Axial load, N
A = Area of cross section, m2
M = Bending Moment, N. m
C = Distance from neutral axis to outer surface, m
I = Rectangular moment of inertia of cross section , m4
T = Torsional moment, N. m
J = Polar moment of inertia, m4
3. Page 2
Curved Beam
hi
ho
Bending stress at the inside fiber, 𝑆 =
𝑀 × ℎ𝑖
𝐴 × 𝑒 × 𝑟𝑖
, ℎ𝑖 = 𝑟𝑛 − 𝑟𝑖
Bending stress at the outside fiber, 𝑆 =
𝑀 × ℎ𝑜
𝐴 × 𝑒 × 𝑟𝑜
, ℎ𝑜 = 𝑟𝑜 − 𝑟𝑛
Where,
ℎ𝑖 = the distance from the neutral axis to the inside fiber, m
ℎ𝑜 = the distance from the neutral axis to the outside fiber, m
𝑟𝑖 = the radius of curvature of the inside fiber, m
𝑟𝑜 = the radius of curvature of the outside fiber, m
C.A = Central Axis
N.A = Neutral Axis
ri
ro
rn R
e
4. 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
rn =
h
loge (
ro
ri
)
e = R − rn
R = ri +
h
2
rn =
[ro
1
2
⁄
+ ri
1
2
⁄
]
2
4
e = R − rn
R = ri +
d
2
5. Page 4
ro
h ri
bi
ti
to
bo t
e
rn
R
C.A
N.A
ro
h ri
ti
bi
t
e
rn
R
C.A N.A
rn =
(bi − t)ti + th
(bi − t)loge(
ri + ti
ri
) + tloge(
ro
ri
)
e = R − rn
R = ri +
1
2
⁄ h2
t + 1
2
⁄ ti
2
(bi − t)
(bi − t)ti + th
rn =
(bi − t)ti + (bo − t)to + th
biloge(
ri + ti
ri
) + tloge(
ro − to
ri + ti
) + bologe(
ro
ro − to
)
e = R − rn
R = ri +
1
2
⁄ h2
t + 1
2
⁄ ti
2
(bi − t) + (bo − t)to(h − 1
2
⁄ to)
(bi − t)ti + (bo − t)to + th
6. 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
rn =
(
bi + bo
2 ) h
(
biro − bori
h
) loge (
ro
ri
) − (bi − bo)
e = R − rn
R = ri +
h(bi + 2bo)
3(bi + bo)
rn =
(bi − t)(ti + to) + th
b [loge(
ri + ti
ri
) + loge(
ro
ro − to
)] + tloge(
ro − to
ri + ti
)
e = R − rn
R = ri +
1
2
⁄ h2
t + 1
2
⁄ ti
2
(b − t) + (b − t)to(h − 1
2
⁄ to)
(b − t)(ti + to) + th
7. 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
JS = JH
π
32
d4
=
π
32
(do
4
− di
4
)
∴ di
4
= do
4
− d4
… … … . . (1)
Equal torsional strength
(
T
Ss
)S = (
T
Ss
)H
i. e (
J
y
)S = (
J
y
)H
π
32
d4
×
2
d
=
π
32
(do
4
− di
4
) ×
2
do
∴ di
4
= do
4
− d3
do … … … . . (2)
8. Page 7
Reduction in weight
If N% is weight reduction
WH = (1 − N)Ws
π
4
(do
2
− di
2
)lhW = (1 − N)
π
4
d2
lsW
(do
2
− di
2
) = (1 − N)d2
∴ di
2
= do
2
− (1 − N)d2
……………………(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
9. Page 8
Bevel Gear (straight tooth)
r r
Driven
gear
pinion
Driver
Ft =
Mt
r
, Mt =
9550 × kW
rpm
Fr = Ft × 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
F𝑝 and the force component along the shaft axis of the gear is called the gear
thrust force F𝑔.
Fp = Ft × tan ∅ sinβ
Fg = Ft × tan∅cosβ
Where β = cone angle
F𝑡 = tangential force
F𝑟 = separating force
10. Page 9
Pulley
Driver
Driven
T1
T2
(T1+T2) sin
(T1+T2) cos
(T1+T2)
T1
T2
Mt = (T1 − T2) × r
(T1 − T2) =
Mt
r
Where T1 = Tight side tension
T2 = Slack side tension
ASME CODE Equation for Shaft Design
1. For solid shaft subjected to torsion and bending load ( without axial
load)
d3
=
16
πSs
√(KbMb)2 + (KtMt)2
2. For solid shaft subjected to combine bending, torsion and axial load
d3
=
16
πSs(max: )
√[𝐾𝑏Mb +
αFad
8
]
2
+ (KtMt)2
11. Page 10
3. For hollow shaft without axial load
do
3
=
16
πSs(1−K4)
√(KbMb)2 + (KtMt)2, K =
di
do
,
4. For hollow shaft with axial load
do
3
=
16
πSs(1 − K4)
√[KbMb +
αFad
8
(1 + K2) ]
2
+ (KtMt)2
ASME Code states for commercial steel shafting
Ss(allowable) = 8000psi (55 MN
m2
⁄ ) for shaft without keyway
Ss(allowable) = 6000psi (40 MN
m2
⁄ ) for shaft with keyway
If ultimate strength and yield strength are known,
Ss(allowable) = 0.18 Su
Ss(allowable) = 0.3 Sy
Choose smaller value (without keyway)
If there is a keyway or fillet, the strength of the shaft is reduced by 25%
Ss(allowable) = 0.18 Su × 0.75
Ss(allowable) = 0.3 Sy × 0.75
Choose smaller value (with keyway)
12. 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 =
Sy
π2n E
(
L
K
)2
for
L
K
> 115
13. Page 12
K = √
I
A
=
d
4
(for circular cross section)
𝐖𝐡𝐞𝐫𝐞,
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, m4
A = Cross section area of shaft,m2
Sy = Yield stress in compression , N/m2
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
14. 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)
tan α =
Lead
2πrm
tan θn = tan θ × cosα
Where,
do = outer diameter
di = inner (or) core diameter
dm = mean diameter
15. Page 14
Torque required to advance the screw against the load
P =
W(sinα + fcosα)
cosα − fsinα
… … … … … . . (÷ cos α)
P = W [
f + tanα
1 − ftanα
]
Ts = P × rm
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
Tc = W × fc × rc
𝑊ℎ𝑒𝑟𝑒,
Tc = collar friction torque
fc = collar friction
rc = collar friction radius
rc =
doc + dic
4
( uniform wear)
rc =
2
3
roc
3
− ric
3
roc
2 − ric
2 ( uniform pressure)
TR = Ts + Tc
TR = W [rm
tanα + f
1 − ftanα
+ fcrc]
16. Page 15
Torque required to lower the load (in the direction of the load)
P = W [
−tanα + f
1 + ftanα
]
Ts = W × rm
Ts = Wrm [
−tanα + f
1 + f tanα
]
TL = Ts + Tc
TL = W [rm
−tanα + f
1 + f tanα
+ fcrc]
Torque required to raise the load (against load) considering the thread angle
T = W [rm
tanα +
f
cosθn
1 −
f × tanα
cosθn
+ fcrc]
Torque required to advance the screw (or nut) in the direction of load (or
lowering the load)
TL = W [rm
−tanα +
f
cosθn
1 +
f × tanα
cosθn
+ fcrc]
17. Page 16
𝐄𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐜𝐲
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 +
fs
cosθn
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
18. Page 17
𝑀 = 𝑊 ×
ℎ
2
𝐼 = 2𝜋𝑟𝑚𝑛 𝑏3
12
⁄ , 𝑦 =
𝑏
2
Using bending equation
Sb
y
=
M
I
Sb =
3Wh
2πrmnb2
(2) Shear stress at the root of thread
𝑠 =
𝑊
𝑟𝑜𝑜𝑡 𝑎𝑟𝑒𝑎
=
𝑊
2𝜋𝑟𝑚𝑛𝑏
(3)Bending stress (or) Bearing pressure between the contacting surface
of the screw and nut thread
p =
W
projected area
=
W
πdmhn
, 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 , Ss =
Tri
J
(𝑖𝑖)direct stress , Sc =
4W
πdi
2
19. Page 18
(𝑖𝑖𝑖)𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑆𝑏 =
32𝑀
𝜋𝑑𝑖
3
(iv)Maximum shear stress, Ss(max) = √(
Sx
2
)
2
+ Ss
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 𝑆 =
𝑃
𝐴
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-
Fcr =
Cπ2
EA
(
L
K
)2
The J.B Johnson formula for the critical load for moderate length
columns of uniform cross section is
Fcr = SyA [1 −
Sy(
L
K
)2
4Cπ2E
]
20. 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/m2
(E=200× 209
N/m2
)
A = Area of transverse section, m2
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/m2
21. 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
22. Page 21
The safe load is obtained by dividing the critical load by a factor of safety N
For Euler equation,
F =
Fcr
N
=
Cπ2
EA
N(
L
K
)2
For J.B Johnson equation,
F =
SyA
N
[1 −
Sy(
L
K
)2
4Cπ2E
]
The value of
𝐿
𝐾
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.
Cπ2EA
N(
L
K
)2
=
SyA
N
[1 −
Sy(
L
K
)2
4Cπ2E
] From which
L
K
= √
2Cπ2E
Sy
The value of
𝐿
𝐾
above which the Euler equation should be used and below
which the J.B Johnson formula should be used
24. 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-
Seq =
F
A
[
Sy(
L
K
)2
Cπ2E
]
The equivalent column stress for an actual load F, derived from J.B Johnson
equation is
Seq =
F
A
[
1
1 −
Sy(
L
K
)2
4Cπ2E]
In the equivalent stress equations, the factor of safety is
N =
Fcr
F
=
Sy
Seq
25. 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
26. Page 2
Case I Known diameter Case
(
1
m2 × y
)
all
=
s k π2
Ft
. … … … . . (1)
s = so (
3
3 + V
) for V less than 10 m
s
⁄ , V. F = (
3
3 + V
)
s = so (
6
6 + V
) for V 10 m
s
⁄ to 20 m
s
⁄ , V. F = (
6
6 + V
)
s = so (
5.6
5.6 + √V
) for V greater than 20 m
s
⁄ , V. F = (
5.6
5.6 + √V
)
k = 4(maximum)
Ft =
9550 × kW
rpm × D
2
⁄
, Ft =
𝑀𝑡
D
2
⁄
Substitute these value in equation (1) and we get(
1
m2×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
m2 × y
)
ind
= − − − − −
Compare (
1
m2×y
)
ind
and (
1
m2×y
)
all
value
27. Page 3
If (
1
m2 × y
)
ind
< (
1
m2 × y
)
all
design is satisfied, decrease the module and try again
If (
1
m2 × y
)
ind
> (
1
m2 × y
)
all
design is not satisfied, increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m2 × y
)
ind
≤ (
1
m2 × y
)
all
The reduce k value
kred = kmax ×
(
1
m2 × y
)
ind
(
1
m2 × y
)
all
Face width, b
b = kred × π × m
Case II Unknown Diameter Case
Sind =
2Mt
kyπ2Nm3
− − − − − (2)
Where Mt =
9550 × kW
rpm
k = 4(maximum)
28. Page 4
N = Number of teeth of weaker, y from table of appendix I
Substituting these value in equation 2 and we get Sind =
…….
m3
Assume V. F =
1
2
and equation 2 becomes (
so
2
) = (
… … .
m3
) , 𝑚 = − − −
and we get m value . Then with std: module series
Try m = − − − −
D = Nm, V =
πDrpm
60
sall = so × V. F
sind = (
− − −
m3
)
Compare the value of Sind and Sall
If Sind < sall
design is satisfied, decrease module and try again
If Sind > sall
design is not satisfied, increase module and try again
Try these procedures until Sindjust ≤ Sall and
take the smallest module that satisfies the condition
30. Page 6
Where,
Dp = Pitch diameter of the pinion, m
B = Face width, m
K = Stress fatigue factor, N/m2
Ses = Surface endurance limit of a gear pair, N/m2
BHNavg = Average brinell hardness number of pinion and gear material
Ep = Modulus of elasticity of the pinion material, N/m2
Eg = Modulus of elasticity of the gear material, N/m2
𝐃𝐲𝐧𝐚𝐦𝐢𝐜 𝐅𝐨𝐫𝐜𝐞, 𝐅𝐝
𝐹𝑑 = 𝐹𝑡 +
21𝑉 (𝑏𝐶 + 𝐹𝑡)
21𝑉 + √𝑏𝐶 + 𝐹𝑡
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.
31. 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/m2
)
Stress Fatigue Facto,K
(kN/m2
)
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
32. 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
33. Page 9
Helical Gears
Formative or virtual number of teeth
Nf =
N
cos3
Strength Design
Case I Known diameter Case
(
1
m2 × y
)all =
skπ2
Ft
cos − − − −(1)
s = so (
5.6
5.6 + √V
) , V =
π D rpm
60
, V. F = (
5.6
5.6 + √V
)
k = 6(maximum)
Ft =
9550 × kW
rpm × D
2
⁄
Substitute these value in equation (1) and we get(
1
m2×y
)
all
value
Assume y=0.15…….m=…., and write std: module series
Try m=
N =
D
m
Nf =
N
cos3
, =……..,y=………..(from the table of Appendix I)
34. Page 10
(
1
m2 × y
)
ind
= − − − − −
Compare (
1
m2×y
)
ind
and (
1
m2×y
)
all
value
If (
1
m2 × y
)
ind
< (
1
m2 × y
)
all
design is satisfied, decrease the module and try again
If (
1
m2 × y
)
ind
> (
1
m2 × y
)
all
design is not satisfied, increase the module and try again
And take the smallest module that satisfies the condition of
(
1
m2 × y
)
ind
≤ (
1
m2 × y
)
all
The reduce k value
kred = kmax ×
(
1
m2 × y
)
ind
(
1
m2 × y
)
all
Face width, b
b = kred × π × m
35. Page 11
Case II Unknown diameter Case
Sind =
2Mt
kyπ2Nm3cos
− − − − − (2)
Where Mt =
9550 × kW
rpm
k = 6(maximum)
N = − − − −
Nf =
N
cos3
, =……..,y=………..(from the table of Appendix 1)
Substituting these values in equation 2 and we get Sind =
…….
m3
Assume V. F =
1
2
and equation 2 becomes (
so
2
) = (
… … .
m3
)
and we get m value . Then with std: module series
Try m = − − − −
D = Nm, V =
πDrpm
60
sall = so × V. F, V. F = (
5.6
5.6 + √V
)
sind = (
− − −
m3
)
Compare the value of Sind and Sall
36. Page 12
If Sind < sall
design is satisfied, decrease module and try again
If Sind > sall
design is not satisfied, increase module and try again
Try these procedures until Sindjust ≤ Sall and
take the smallest module that satisfies the condition
Reduce k
kred = kmax ×
Sind
Sall
Face width, b
b = kred × π × m
End thrust
Fe = Ft × tan
Dynamic Check
𝐄𝐧𝐝𝐮𝐫𝐚𝐧𝐜𝐞 𝐅𝐨𝐫𝐜𝐞, 𝐅𝐨
Fo = Sobyπmcos
37. Page 13
𝐖𝐞𝐚𝐫 𝐅𝐨𝐫𝐜𝐞, 𝐅𝐰(𝐖𝐞𝐚𝐫 𝐭𝐨𝐨𝐭𝐡 𝐥𝐨𝐚𝐝)
𝐵𝑢𝑐𝑘𝑖𝑛𝑔ℎ𝑎𝑚 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝐹
𝑤 =
DpbKQ
cos2
Q =
2Ng
Np + Ng
=
2Dg
Dp + Dg
K =
Ses×sin∅n (1
Ep
⁄ +1
Eg)
⁄
2
1.4
tan ∅n = tan ∅ × cos
ses = (2.75 BHNavg − 70) MN
m2
⁄
𝐃𝐲𝐧𝐚𝐦𝐢𝐜 𝐅𝐨𝐫𝐜𝐞, 𝐅𝐝
Fd = Ft +
21V (bCcos2
+ Ft)cos
21V + √bCcos2 + Ft
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.
38. 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
√Dp
2
+ Dg
2
=
Npm
2
√1 + V. R2, (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
Nfp =
Np
cosαp
=
Np
Dg
2L
⁄
=
Np
Ng
√Np
2
+ Ng
2
Nfg =
Ng
cosαg
=
Ng
Dp
2L
⁄
=
Ng
Np
√Np
2
+ Ng
2
39. Page 15
Strength Check
Case I Known Diameter Case
(
1
m × y
)𝑎𝑙𝑙 =
s b π
Ft
(
L − b
L
) − − − (1)
s = so × 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
√Dp
2
+ Dg
2
, b close tonot greater than L
3
⁄
Ft =
9550 × kW
rpm × D
2
⁄
(
1
m × y
)
all
= − − − − −
Assume y=0.1, m=------
Write down Std module series
Try m=
40. Page 16
N =
D
m
Nf = − − − , =……..,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
Sind =
2Mt
m2byπN
(
L
L − b
) − − − −(2)
Mt =
9550 × kW
rpm
L =
mNp
2
√1 + (V. R)2
41. Page 17
L
L − b
=
3
2
Sind =
− − −
m3
Assume V. F = 1
2
⁄ and Sind becomes
So
2
∴
So
2
=
− − −
m3
, 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
sall = so × V. F
sind =
− − −
m3
Compare the value of Sind and Sall
If Sind < sall
design is satisfied, decrease module and try again
42. Page 18
If Sind > sall
design is not satisfied, increase module and try again
Try these procedures until Sindjust ≤ Sall and
take the smallest module that satisfies the condition.
Then find the face width.
Dynamic Check
The limiting endurance load
Fo = Sobyπm(
L − b
L
)
The limiting wear load
Fw =
0.75 DpbKQ
cosαp
Q =
2Nfg
Nfp + Nfg
K =
Ses×sin∅ (1
Ep
⁄ +1
Eg)
⁄
2
1.4
ses = (2.75 BHNavg − 70) MN
m2
⁄
43. Page 19
𝐃𝐲𝐧𝐚𝐦𝐢𝐜 𝐅𝐨𝐫𝐜𝐞, 𝐅𝐝
Fd = Ft +
21V (bC + Ft)
21V + √bC + Ft
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 So(rpm)pDpbyπ
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/m2
44. Page 20
2. AGMA standard for wear (durability)
Power in kW = 0.8 CmCBb for straight bevel gear
Power in kW = CmCBb for spiral bevel gear
Where,
Cm = material factor as listed in following Table
CB =
Dp
1.5(rpm)p
0.032
(
5.6
5.6 + √V
)
45. 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
𝐼 = 𝐴𝑛𝑛𝑒𝑙𝑒𝑑 𝑠𝑡𝑒𝑒𝑙
𝐼𝐼 = 𝐻𝑒𝑎𝑡 𝑡𝑟𝑒𝑎𝑡𝑒𝑑 𝑠𝑡𝑒𝑒𝑙
𝐼𝐼𝐼 = 𝑂𝑖𝑙 𝑜𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 − ℎ𝑎𝑟𝑑𝑒𝑛𝑒𝑑 𝑠𝑡𝑒𝑒𝑙
𝐼𝑉 = 𝐶𝑎𝑠𝑒 − ℎ𝑎𝑟𝑑𝑒𝑛𝑒𝑑 𝑠𝑡𝑒𝑒l
46. Page 22
Worm Gear
𝐶 =
𝐷𝑔 + 𝐷𝑤
2
𝑃𝑐 = 𝑃𝑎 𝑓𝑜𝑟 𝑠𝑚𝑜𝑜𝑡ℎ 𝑒𝑛𝑔𝑎𝑔𝑒𝑚𝑒𝑛𝑡
𝑉. 𝑅 =
𝑁𝑔
𝑁𝑤
=
𝐷𝑔
𝐷𝑤𝑡𝑎𝑛𝛼
tan 𝛼 =
𝐿𝑒𝑎𝑑
𝜋𝐷𝑤
=
𝑃𝑐𝑁𝑤
𝜋𝐷𝑤
=
𝑚𝑎𝑁𝑤
𝐷𝑤
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
47. Page 23
AGMA equation include the following design equation
𝐷𝑤 ≈
𝐶0.875
3.48
≈ 3𝑃𝑐, 𝑏 ≈ 0.73 × 𝐷𝑤, 𝐿 ≈ 𝑃𝑐(4.5 +
𝑁𝑔
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
𝑚𝑎 =
𝑃𝑐
𝜋
≈
𝐷𝑤
3𝜋
, 𝐷𝑔(𝑎𝑝𝑝𝑟𝑜𝑥:) = 2𝐶 − 𝐷𝑤(𝑎𝑝𝑝𝑟𝑜𝑥:), 𝑉. 𝑅 =
𝐷𝑔(𝑒𝑥)
𝑚𝑎𝑁𝑤
From the above equation we get,
Nw
Dg
𝐷𝑔(𝑒𝑥𝑎𝑐𝑡) = 𝑉. 𝑅 𝑚𝑎𝑁𝑤
48. 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,
𝐹 = 𝑠 𝑏 𝑦 𝑃𝑛𝑐 = 𝑠 𝑏 𝑦 𝜋 𝑚𝑛
𝑠 = 𝑠𝑜(
6
6 + 𝑉
𝑔
)
𝑚𝑛 = 𝑚𝑎 𝑐𝑜𝑠𝛼
Where,
F = Permissible tangential load, N
s = Allowable stress, MN/m2
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
49. Page 25
Dynamic Check
Endurance load, Fo
𝐹
𝑜 = 𝑠𝑜 𝑏 𝑦 𝜋𝑚𝑛
𝑊𝑒𝑎𝑟 𝐿𝑜𝑎𝑑, 𝐹
𝑤
𝐹
𝑤 = 𝐷𝑔 𝑏 𝐵
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
50. Page 26
Constant B
Worm Gear B(kN/m2
)
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
𝐹𝑑 = (
6 + 𝑉
𝑔
6
) 𝐹
Where,
𝐹 = 𝑎𝑐𝑡𝑢𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑙𝑜𝑎𝑑
51. 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
𝑃 =
𝑟𝑝𝑚𝑤 𝐾 𝑄 𝑚
𝑉. 𝑅
𝑄 =
𝑉. 𝑅
𝑉. 𝑅 + 2.5
𝑚 =
2.3
2.3 + 𝑉
𝑤 +
3𝑉
𝑤
𝑉. 𝑅
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
53. 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
𝑃 =
3650 𝐶1.7
𝑉. 𝑅 + 5
(ℎ𝑒𝑎𝑡 𝑐ℎ𝑒𝑐𝑘)
Where,
P = Permissible power input in kW
C = Center distance in meters
V.R = Transmission ratio
Allowable Power
𝑃 = 𝐹 𝑉
𝑔
The efficiency of a worm gear unit, assuming square threads, may be
approximated by
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
1 − 𝑓 𝑡𝑎𝑛𝛼
1 + 𝑓 𝑡𝑎𝑛𝛼
54. 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.
55. 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
56. 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
