Machine design-2

MEC2-K56- Group 15 Page 1
HANOI UNIVERSITY OF SCIENCE AND TECHNOLOGY
SCHOOL OF MECHANICAL ENGINEERING
Project 1
Designing Chain - Driven System
Supervisor: PhD Dang Bao Lam
PhD Nguyen Tuan Khoa
Class: Mechatronics 2 - K56
Group 15 Students Mai Văn Quyết (20110646)
Lã Minh Công (20110087)

Contents
A. Choosing motor and distributing transmission ratio: ............................. 7
1. Work power: ................................................................................. 7
2. Efficiency of driven system:........................................................... 7
3. Necessary power on the motor axis: ............................................... 7
4. Number of revolutions on operating axis: ....................................... 7
5. Choosing the first-aid transmission ratio:........................................ 7
6. First-aid number of revolutions on motor axis................................. 7
7. Calculating synchronous number of revolutions of motor:............... 8
8. Choosing the motor: ...................................................................... 8
9. Distributing transmission ratio: ...................................................... 8
10. Calculating the factors on the axis: ............................................... 8
11. Data Table:.................................................................................. 9
B. Calculating and designing outer transmission – chain drives: ........Error!
Bookmark notdefined.
1. Choosing the type of chain base on operating condition: .........Error!
2. Choosing number of teeth of the sprocket......Error! Bookmark not
defined.
3. Select factors...................................Error! Bookmark notdefined.
4. Select chain drive ............................Error! Bookmark notdefined.
5. Installation parameters .....................Error! Bookmark notdefined.
6. Centre distance calculation...............Error! Bookmark notdefined.
7. Selection of sprocket materials .........Error! Bookmark notdefined.
8. Data of chosen chain........................Error! Bookmark notdefined.
9. Check for chian’s durability .............Error! Bookmark notdefined.
C. Design gear transmission...........................Error! Bookmark notdefined.

1. Design Decision:.......................................Error! Bookmark notdefined.
2. Design Calculation: ...................................Error! Bookmark notdefined.
 Pitch diameters:................................Error! Bookmark notdefined.
 Bending geometry factor...................Error! Bookmark notdefined.
 Velovity factor………………………………………………………19
 Transmitted load…………………………………………………….19
 Bending stresses in pinion and gear………………………………20
 Surface stress in the pinion gear…………………………………….20
 Bending- fatigue strength…………………………………………...22
 Surface fatigue strength……………………………………………..23
 Safety factors………………………………………………………..24
3. Data table..................................................Error! Bookmark notdefined.
D. Shafts, keys and couplings calculation. ....Error! Bookmark notdefined.
I. Choosing material ...........................Error! Bookmark notdefined.
II. Shaft 1 calculations .............................Error! Bookmark notdefined.
a) Analyze the forces and moment...............................………………...29
b) Diameter calculation…………………………………………………32
c) Key and testing for shaft I…………………………………………...33
III. Shaft 2 calculations……………………………………………………38
a) Analyze the forces and moments……………………………….. ...38
b) Diameter calculation……………………………………………….40
c) Key and testing for shaft II ……………………………………….42
E. Bearing, lubrication, gear box’s cover and parts structure. ...........Error!
1. Shaft 1............................................Error! Bookmark notdefined.
2. Shaft 2............................................Error! Bookmark notdefined.
F. Gearbox's cover…………………………………………………………...49

G. Comparisionof Englishvs Vietnamese document………………………
References:................................................................................................. 66
List of tables
Table 1 Motordata table 10
Table 2 Sprocket materials 13
Table 3 Chain drive data table 14
Table 4 Geardrive data table 25
Table 5 Uncorrectedshaft I diameter data 35
Table 6 CorrectedshaftI diameter data 40
Table 7 Uncorrectedshaft II diameter data 45
Table 8 Correctedshaft II diameter data 50
Table 9 Compare English vs Vietnamese document 55-65
List of figures
Figure 1 Chain lubrication 11
Figure 2 Chain 15
Figure 3 Shaft dimensions 29
Figure 4 Force diagramin shaft I 30
Figure 5 Moment diagram in shaft I 33

F
v
1
2 3
4
5
A
Theo A (c.t.4)
@
z,p
Figure 6 Force diagram in shaft II 40
Figure 7 Moment diagram in shaft II 43
Project 1.15
1.
2. Motor
3. Elastic Clutch
4. Reduction unit
5. Chain trans. Unit
6. Conveyor chain
Fig. 1 Chain-Driven system
Input data:
1. Pulling force of conveyer chain: F = 4220 (N) Driving gear:
2. Conveyor chain speed: v = 1,36 (m/s) Spur -gear
3. Sprocketteeth number of conveyor chain: z = 14 (teeth)
4. Chain pitch of conveyor chain: p = 70 (mm)
5. Service time: lh = 11000 (hours)
6. Number of shifts: soca= 2 (shift)
7. Inclination angle of outer driven 𝛽 = 0 (degree)
8. Working characteristic: Mild impact
Shaft for calculating: Input shaft 1

A. Choosing motor and distributing transmission ratio:
1. Work power:
)(7.5
1000
36.14220
1000
.
KW
vF
Plv 


2. Efficiency of driven system:
3
. . .br OL x knh h h h h=
Where : Gear performance : ℎ 𝑏𝑟 = 0.96
Chain performance : ℎ 𝑥 = 0.9
Bearing performance: ℎ 𝑂𝐿 = 0.995
Jointing performance: ℎ 𝑘𝑛 = 1
85.019.0995.096.0... 33
 knxOLbr 
3. Necessarypoweron the motor axis:
)(75.6
85.0
7.5
KW
P
P lv
yc 

4. Number of revolutions on operating axis:
𝑛 𝑐𝑡 =
60000. 𝑣
𝑧. 𝑝
=
60001.36
1470
= 83.26 (rev/min)
5. Choosing the first-aid transmissionratio (𝒏 𝒔𝒃):
𝒖 𝒔𝒃 = 𝒖 𝒙. 𝒖 𝒃𝒓
Follow table B
2.4
21
[1]: Chain ration : 𝒖 𝒙 =2 ÷ 5
Gear ration : 𝒖 𝒃𝒓 = 3 ÷ 5
Then:
𝒖 𝒔𝒃 𝒎𝒊𝒏= 𝒖 𝒙 𝒎𝒊𝒏. 𝒖 𝒃𝒓 𝒎𝒊𝒏=2 3= 6
𝒖 𝒔𝒃 𝒎𝒂𝒙= 𝒖 𝒙 𝒎𝒂𝒙. 𝒖 𝒃𝒓 𝒎𝒂𝒙=5 5= 25
6. First-aid number of revolutions on motor axis

We have : 𝒏 𝒔𝒃 = 𝒖 𝒔𝒃. 𝒏 𝒄𝒕
𝒏 𝒔𝒃 𝒎𝒊𝒏 = 𝒖 𝒔𝒃 𝒎𝒊𝒏. 𝒏 𝒄𝒕= 500 (rev/min)
𝒏 𝒔𝒃 𝒎𝒂𝒙 = 𝒖 𝒔𝒃 𝒎𝒂𝒙. 𝒏 𝒄𝒕= 2081 (rev/min)
We choose 𝒏 𝒔𝒃 =1000 (rev/min)
7. Calculating synchronous number of revolutions of motor :
Choosing 𝒏 𝒅𝒃
𝒕
= 𝟕𝟓𝟎 (rev/min)
8. Choosing the motor:
Choosethe motor which satisfies: {
𝑛đ𝑏 ≈ 𝑛 𝑠𝑏 = 1000 (rev/min)
𝑃đ𝑐 ≥ 𝑃𝑦𝑐 = 6.75 (𝐾𝑊)
We have the motor with the details:
Siemen 1LA7 134-6AA {
𝑛 𝑑𝑐 = 1000 (rev/min)
𝑝 = 7 𝐾𝑊
9. Distributing transmission ratio:
Transmission ratio of the system:
𝒖 𝒄𝒉 =
𝒏đ𝒄
𝒏 𝒄𝒕
=
𝟏𝟎𝟎𝟎
𝟖𝟑.𝟐𝟔
=12
Choose the transmission ratio of 1st gear box: 𝑢 𝑏𝑟 =5
Transmission ratio of outer driven: 𝑢 𝑥=
𝒖 𝒄𝒉
𝒖 𝒃𝒓
=
𝟏𝟐
𝟓
≈ 𝟐. 𝟒
So, we have: {
𝑢 𝑐ℎ = 12
𝑢 𝑏𝑟 = 5
𝑢 𝑥 = 2.4
10. Calculating the factors on the axis:
Power on operating axis: 𝑃𝑐𝑡 = 𝑃𝑙𝑣 = 5.7 (Kw)

Power on axis II:
𝑃𝐼𝐼 =
𝑃𝑐𝑡
𝑛 𝑂𝐿.𝑛 𝑥
=
5.7
0.995×0.9
= 6.36(Kw) (Kw)
Power on axis I:
𝑃𝐼 =
𝑃𝑐𝑡
𝑛 𝑂𝐿.𝑛 𝑥
=
6.36
0.995×0.96
= 6.66 (Kw)
Power on the motor axis:
𝑃đ𝑐 =
𝑃 𝐼
𝑛 𝑂𝐿.𝑛 𝑘𝑛
=
6.66
0.995×1
= 6.70 (Kw)
Number of revolutions on the motor axis: 𝑛đ𝑐= 1000 (rev/min)
Number of revolutions I:
𝑛𝐼 =
𝑛đ𝑐
𝑢 𝑘𝑛
=
1000
1
= 1000 (rev/min)
Number of revolutions II:
𝑛𝐼𝐼 =
𝑛 𝐼
𝑢 𝑏𝑟
=
1000
5
= 200 (𝑟𝑒𝑣/𝑚𝑖𝑛)
Number of revolutions on the operating axis:
𝑛 𝑐𝑡 =
𝑛𝐼𝐼
𝑢 𝑥
=
200
2.4
= 83.33 (𝑟𝑒𝑣/𝑚𝑖𝑛)
Torque on the motor axis:
𝑇đ𝑐 = 9,55.106
.
𝑃đ𝑐
𝑛đ𝑐
= 9,55.106
.
6.75
1000
= 64462 (𝑁. 𝑚𝑚)
Torque on the axis I:
𝑇𝐼 = 9,55.106
.
𝑃 𝐼
𝑛 𝐼
= 9,55.106
.
6.66
1000
= 63603 ( 𝑁. 𝑚𝑚)
Torque on the axis II:
𝑇𝐼𝐼 = 9,55.106
.
𝑃 𝐼𝐼
𝑛𝐼𝐼
= 9,55.106
.
6.36
200
= 303690 ( 𝑁. 𝑚𝑚)
Torque on the operating axis:
𝑇𝑐𝑡 = 9,55.106
.
𝑃𝑐𝑡
𝑛 𝑐𝑡
= 9,55.106
.
5.7
83.33
= 653246 (𝑁. 𝑚𝑚)

11. Data Table:
FactorsAxis Motor I II Operating
ukn = 1 ubr = 5 ux = 2.4
P(KW) Pđc= 6.75 PI = 6.66 PII = 6.36 Pct= 5.7
n(rev/min) nđc= 1000 nI= 1000 nII= 200 nct = 83.33
T(N.mm) Tđc= 64462 TI = 63603 TII = 303690 Tct = 653246
B.Calculating and designing outer transmission – chain driver
Data requirements:
{
P = PII = 6.36(KW)
T1 = TII = 303690(N.mm)
n1 = nII = 200 (rpm)
u = ux = 2.4
β = 00
1. Choosing the type of chain
We choosethe brushed roller chain.

2. Choosing number of teeth of the sprocket
The number of teeth on driver sprocket:
Z1 = 29 − 2 × u = 29 − 2 × 2.4 = 24.2 ≥ 19, choose Z1 = 24
Therefore the driven number of teeth:
𝑍2 = 𝑢 × 𝑍1 = 2 × 24.2 = 48.4 ≤ 𝑍 𝑚𝑎𝑥 = 140, 𝑐ℎ𝑜𝑜𝑠𝑒 𝑍2 = 48
3. Selectfactors
Application factor: use the chart 2 (see page 102 Renold Roller Chain Catalogue)
with driver and driven sprockets smoothrunning we have: 1 1f 
Toothfactor: 𝑓2 =
19
𝑍1
=
19
24
= 0.79
Selection power = Transmitted power 1 2f f 
= 6.36×1×0.7 =5.02 (KW)
4. Selectchain drive
According to American Chain Rating Chart (see page 106 Renold Roller
Chain Catalogue) by cross reference power 5.02(KW) and speed 200(rpm),
we will choosethe ANSI Simplex with the chain pitch p = 19.05(mm).
5. Installation parameters
Lubrication – American Chain Rating Chart (see page 106 Renold Roller
Chain Catalogue) clearly indicates the chain need Oil Bath type 3 lubrication.

Figure 2.1.Chain lubrication.
Now we calculate the chain length (see page 105 Renold Roller Chain Catalogue):
22 1
1 2
( )
2 2
2
Z Z
p
Z Z C p
L
p C


  
  
Where:
C is the contemplated center distance in mm and should generally be
between 30-50 pitches. In this case:
C = 19.05×40 = 762 (mm)
L = Chain length (pitches)
p = Chain pitch = 19.05(mm)
Z1= 24(teeth)
Z2= 48(teeth)
𝐿 =
24+48
2
+
2×762
19.05
+
(
48−24
2×19.05
)2×19.05
762
= 116 (pitches)

6. Centre distance calculation
The center distance of the drive can now be calculated using the
formula below (see page 105 Renold Roller Chain Catalogue):
C =
19.05
8
[2 × 116− 48 − 24 + √(2 × 116− 48 − 24)2 − (48 − 24)2 π
3.88
]=
C = 758.5
7. Selectionofsprocketmaterials
Depend on the table (p.104 Renold Roller Chain Catalogue):
Table 2.1.Sprocket materials.
Driver sprocket material is cast iron
Driven sprocket material is EN8 or EN9
   
2 2
2 1 2 1 2 12 2
8 3.88
p
C L Z Z L Z Z Z Z
 
          
 

8. Data of chosenchain
Following this link:
http://www.renoldchainselector.com/ChainSelector
We will choosethe ANSI 80 (ISO 606) Simplex chain transmission with
conditions:
Environmental condition:
Loading Classification:
Driven Machine: smoothrunning, Moderate Shocks.
Service Conditions: Recommended.
Environnent Condition: Normal environnement, Indoor application.
ExpectedWorking Life of the Chain
𝐑𝐞𝐧𝐨𝐥𝐝 𝐒𝐲𝐧𝐞𝐫𝐠𝐲 𝑻𝑴
Chain: ANSI 80 (ISO 606)Simplex
The working life of the chain is greater than 3000 hours.
After this time:
The chain will reach 3% elongation.
Serial Number: GY80A1
Input Power: P = 6.36 kW Pitch: p = 19.05 mm
Input Speed: n = 200 rpm ISO Breaking Load: Fb = 55600 N
Chain Linear Velocity: v = 2 m/s Bearing Pressure: pr = 24.831 N/mm²
Torque: T = 303.69 Nm Bearing Area: f = 1.78 cm²
Static Force: F = 3129.9 N Weight: q = 2.8 kg/m

Dynamic Force: Fd = 4408.3 N Chain Length: l = 2489.2 mm
Centrifugal Force: Ff = 11.56 N Centre Distance: a = 762mm
Total Force: Fg = 4419.9 N Number of Links: X = 98
Chain Tensioner required
static = 17.8
dynamic = 12.6
Figure 2.2.Chain.

Chain Drive:
Sprocket Driving (Z1) Driven (Z2)
Number of Teeth: 24 48 Ratio: i = 2
Pitch Circle Diameter: 194.597 mm 388.361 mm
Loading
Classification:
Smooth running Moderate
Shocks
Environment Conditions:
Environment
Conditions:
Indoor, Normal
Service Conditions: Recommended
Recommended
Lubrication:
Drip Lubrication
.
9. Check for chain’s durability
Durability factor:
𝒔 =
𝑸
𝒌đ 𝑭𝒕 + 𝑭 𝟎 + 𝑭 𝒗
Where:
 Q – ISO Breaking Load:
Q =31.8(KN)
 Chain’s mass:
q = 2.8(kg/m).

 F - Static force:
𝐹𝑡 =
1000×𝑃
𝑣
, 𝑤𝑖𝑡ℎ 𝑣 = 2𝑚/𝑠
𝐹𝑡 =
1000×𝑃
𝑣
=
1000×6.36
2
= 3180(N)
 Ff - Centrifugal Force:
𝐹𝑓 = 𝑞. 𝑣2
= 2.8 × 22
= 11.2 (𝑁)
 F0- The tension caused by the weight of chain:
aqkF f
...81,90
 ,
with:
a - Centre distance: a=762(mm)
kf - Deflection coefficient of chain: because of  = 0o  kf = 1
𝐹0 = 9.81 × 1 × 2.8 × 762× 10−3
= 20.93 (𝑁)
Therefore:
𝒔 =
Q
kđFt+F0+Fv
with kd is dynamic loading coefficient with
characteristic is Mild: kd=1
𝒔 =
𝑄
𝑘đ 𝐹𝑡+𝐹0+𝐹𝑣
=
31800
1×3180+20.93+11.2
= 9.9
[s] – Allowed safe factor:
Find in table
5.10
(1)
86
B with p = 19.05(mm); n1 = 200(rpm) we get
[s]=8.2. We find that s>[s] therefore the chain has satisfied safe factor.

C. Design gear transmission .
Data requirement:
{
𝑷 = 𝑷𝑰 = 𝟔. 𝟔𝟔 (𝑲𝑾)
𝑻 = 𝑻 𝑰 = 𝟔𝟑𝟔𝟎𝟑 (𝑵. 𝒎𝒎)
𝒏 𝟏 = 𝒏𝑰 = 𝟏𝟎𝟎𝟎 ( 𝒓𝒑𝒎)
𝒖 = 𝒖 𝒃𝒓 = 𝟓
𝑳 𝒉 = 𝟏𝟏𝟎𝟎𝟎𝒉
1. Designdecision
 Materialselection
Material: Steel C45
 Selectdiameter pitch
From Standard Diameter Pitches (Robert L. Norton, Machine Design,
p.645, table 11-2):
Choosep = 8
 Selectnumber of teeth (with u=5)
Driving pinion: 𝑁𝑝 = 20 𝑡𝑒𝑒𝑡ℎ
Driven gear: 100 teeth
2. Designcalculation
 Pitch diameter
Pitch diameter of driving pinion: 𝑑 𝑝 =
𝑁 𝑝
𝑝
=
20
8
= 2.5 ( 𝑖𝑛)
Pitch diameter of driven gear: dg =
100
8
= 12.5 (in)
 Bending geometryfactor

The bending geometry factors J for this combination are found in the
RobertL. Norton, Machine Design, p.666, table 11-9, for the highest point
of the single-tooth contact(HPSTC)and are approximately : {
𝐽𝑝 = 0.35
𝐽 𝐺 = 0.43
 Velocity factor
The velocity factor is calculated from equations 11.16
and 11.17 (Robert L. Norton, Machine Design, p.665) base on the assumed
gear quality index and pitch-line velocity :
𝑉𝑡 =
𝑑 𝑝
2
× 𝜔 𝑝 =
2.5
2×12
× (1000𝑟𝑝𝑚)(2𝜋) = 654.5(ft/min)
825.0
4
)612(
4
)12( 3/23/2




 v
Q
B
8.59)825.01(5650)1(5650  BA
𝐾𝑣 = 𝐶𝑣 = (
𝐴
𝐴 + √ 𝑉𝑡
)
𝐵
= (
59.8
59.8+ √654.5
)
0.825
= 0.745
Maximum velocity checking
should be checked against the maximum allowable pitch-line
velocity for this quantity gear using equation 11.18 (Robert L. Norton,
Machine Design, p.668):
   min)/(84.39433
2
max
ftQvAVt 
We can see that maxt tV V , so tV is acceptable.
 Transmitted load
( )v vK C
6vQ  tV
tV

𝑊𝑡 =
33000𝐻
𝑉
=
33000× 10.32
654.5
= 520.3(𝑙𝑏)
 Bending stressesin pinion and gear
Bending stress can be estimated using equation 11.15 (Robert
L. Norton, Machine Design, p.664):
Where:
is face width can be estimated by this equation (Robert L. Norton,
Machine Design, p.669):
F≅
12
𝑃 𝑑
=
12
8
= 1.5 in
is application factor. We choose from table
11.17(Robert L. Norton, Machine Design, p.669):
is load distribution factor. We choose from table
11.16 with (Robert L. Norton, Machine Design, p.669).
is size factor and rim bending factor are equal 1 for these small
gears.
Apply all these factors; we can estimate bending stress in:
𝜎𝑏 𝑝
=
𝑊𝑡 𝑝 𝑑
𝐹𝐽 𝑝
×
𝐾 𝑎 𝐾 𝑚
𝐾𝑣
× 𝐾𝑠 𝐾 𝐵 𝐾𝐼 =
520.3×8
1.5×0.35
×
1×1.6
0.745
× 1 × 1 × 1 = 17027.40 psi
𝜎𝑏 𝑔
=
𝑊𝑡 𝑝 𝑑
𝐹𝐽 𝑔
×
𝐾 𝑎 𝐾 𝑚
𝐾𝑣
× 𝐾𝑠 𝐾 𝐵 𝐾𝐼 =
520.3×8
1.5×0.43
×
1×1.6
0.745
× 1 × 1 × 1 = 13859.51 psi
p
t a m
b s B I
v
W p K K
K K K
FJ K
 
F
( )a aK C 1a aK C 
( )m mK C 1.6mK 
2F 
sK BK

 Surface stress in the pinion gear
The surface stress in the pinion gear can be estimated by
equation 11.21 (RobertL. Norton, Machine Design, p.672):
t a m
cpg p s f
p v
W C C
C C C
FId C
 
Where:
pC is elastic coefficient accounts for differences in tooth materials
and can be found in table 11.18 (Robert L. Norton, Machine Design,
p.674): 0.5
2300pC psi for steel on steel.
I is geometry factor can estimated by equation 11.22a (Robert L.
Norton, Machine Design, p.673):
cos
1 1
p
p g
I
d

 

 
  
 
Where :
+  is pressure angle of pinion teeth. We choose 25  
+ p and g are the radius of curvature of the pinion and gear teeth,
respectively. These factors are calculated from the geometry (Robert L.
Norton, Machine Design, p.673, equation 11.22b):
𝜌 𝑝 = √( 𝑟𝑝 +
1
𝑃 𝑑
)
2
− (𝑟𝑝 𝑐𝑜𝑠∅)2 -
𝜋
𝑃 𝑑
𝑐𝑜𝑠∅

𝜌 𝑝 = √(1.25 +
1
8
)
2
− (1.25𝑐𝑜𝑠250)2 −
𝜋
8
𝑐𝑜𝑠250
 𝜌 𝑝 = 0.33 𝑖𝑛
𝜌 𝑔 = 𝐶𝑠𝑖𝑛∅ − 𝜌1 = (𝑟𝑝 + 𝑟𝑔)𝑠𝑖𝑛∅− 𝜌1 = (1.25 +6.25)× sin 250
− 0.33 =2.84
in
Therefore, geometry factor now can be estimated:
I =
𝑐𝑜𝑠∅
(
1
𝜌 𝑝
±
1
𝑝 𝑔
)
=
𝑐𝑜𝑠250
(
1
0.33
−
1
2.84
)×2.5
= 0.14
fC is surface factor and can be set to 1 for well-finished gears made
by conventional methods.
Apply all these factors, now we can estimate surface stress in the gear
mesh:
t a m
cpg p s f
p v
W C C
C C C
FId C
 
=2300√
520.3
1.5×0.14×2.5
×
1×1.6
0.745
× 1 × 1 = 106079.6 psi
 Bending-fatigue strength
The correction (or corrected) bending -fatigue strength of gears can
be computed using formula 11.24 (Robert L. Norton, Machine Design,
p.678):

'
L
fb fb
T R
K
S S
K K

Where:
lK is life factor which is found from the appropriate equation in
Figure 11.24(Robert L. Norton, Machine Design, p.679) based on the
required number of cycles in the life of the gears. The pinion sees the
largest number of repeat tooth-loadings, so we calculate the life based
on it. First, calculate the number of cycles N for the required life of
16000h, one shift:
N = 1000rpm (
60𝑚𝑖𝑛
ℎ𝑟
) × 11000ℎ𝑟 = 0.66 × 109
(𝑐𝑦𝑐𝑙es)
then calculate the life factor:
𝐾𝐿 = 1.3558× 𝑁−0.0178
= 1.3558× (0.66× 109)−0.0178
= 0.906
TK is temperature factor. At the normal condition, 1TK  .
The gear-material data are all taken at a reliability level of 99%. This
is satisfactory in this case, making 1RK 
'fbS is uncorrected bending-fatigue strength can be made from the
curves of Figure 11.25(Robert L. Norton, Machine Design, p.681). We
will try an AGMA Grade 1 steel, through hardened to 250 HB. The
uncorrected bending-fatigue strength is found from the lower curve of
the figure:
2
'
2
274 167 0.152
274 167 240 0.152 240 31051
fbS HB HB
psi
   
      

Apply all these factors, now we can calculate the corrected bending-
fatigue strength:
𝑆𝑓𝑏 =
𝐾𝐿
𝐾 𝑇 𝐾 𝑅
𝑆𝑓𝑏′ =
0.906
1
× 31051 = 28132𝑝𝑠𝑖
 Surface fatigue strength
The correction (or corrected) surface-fatigue strength of gears can be
computed using formula 11.25 (Robert L. Norton, Machine Design,
p.679):
'
L H
fc fc
T R
C C
S S
C C

Where:
lC is life factor which is found from the appropriate equation in
Figure 11.25(Robert L. Norton, Machine Design, p.679) based on the
required number of cycles in the life of the gears. The pinion sees the
largest number of repeat tooth-loadings, so we calculate the life based
on it. First, calculate the number of cycles N found above:
𝑆 𝑓𝑐′ = 2600+ 327𝐻𝐵 = 2600+ 327(240) = 104480 psi
𝐶𝐿 = 1.4488𝑁−0.023
= 1.4488× (0.66 × 109
)−.0023
= 0.908
1 TT KC and 1 RR KC
Since the gears and pinion are of the same hardness material in this
case, 1HC  .
Apply all these factors, now we can calculate the corrected surface-
fatigue strength:

𝑆𝑓𝑐 =
𝐶𝐿 𝐶 𝐻
𝐶 𝑇 𝐶 𝑅
𝑆𝑓𝑐′ =
0.908 × 1
1 × 1
× 104480 = 94867.84𝑝𝑠𝑖
 Safety factors
The safety factors against bending failure are found by comparing the
corrected bending strength to the bending stress for each gear in the
mesh:
𝑁𝑏 𝑝
=
𝑆𝑓𝑏
𝜎𝑏 𝑝
=
28132
17027.40
= 1.65
𝑁𝑏 𝑔
=
𝑆𝑓𝑏
𝜎𝑏 𝑔
=
28132
13859.51
= 2.03
The safety factor against surface failure is found by comparing the
actual load to the load that would produce a stress equal to the
material’s corrected surface strength. Because surface stress is related to
the squarer root of the load, the surface-fatigue safety factor can be
calculated as the quotient of the square of the corrected surface strength
divided by the square of the surface stress for each gear in the mesh:
𝑁𝐶 𝑝−𝑔
= (
𝑆 𝑓𝑐
𝜎 𝑐 𝑝𝑖𝑛𝑖𝑜𝑛
)
2
= (
94867.84
106079.6
)
2
=0.8

3. Data table
Gear details Symbol Value
Diametral pitch dp 8
Pressure angle  25
Tangential force tW 520.3(lb)
Number of teeth on pinion pN
20
Number of teeth on gear gN
100
Pitch diameter of pinion pd
2.5 (in)
Pitch diameter of gear gd
12.5 (in)
Pitch-line velocity tV 654.5(ft/min)
Bending stress-pinion tooth pb 17027.40psi
Bending stress-gear tooth gb 13859.51psi
Surface stress in pinion and gear cpg
106079.6psi
Uncorrected bending strength '
fb
S
31051psi
Corrected bending strength fbS
28132psi
Uncorrected surface strength '
fc
S
104480psi
Corrected surface strength fc
S
94867.84psi
Bending safety factor for pinion bpN
1.65
Bending safety factor for gear bgN
2.03
Surface safety factor for mesh p gcN  0.8
Table 3.1.Gear data table.

D. Shafts, keys and coupling calculation
I. Choosing material
In order to minimize the deflections, we try an inexpensive, low-carbon,
cold-rolled steel such as SAE 1040 with 86utS kpsi and 54yS kpsi . Though
not exceptionally strong, this material has low notch sensitivity, which will
be an advantage given the large stress concentration.
The endurance limit strength can be estimated using equation 6.6 (Robert L.
Norton, Machine Design, p326):
'e load size surf temp reliab e
S C C C C C S     
Where:
1loadC  because the loading is bending and torsion.
1sizeC  because we do not know the part size, therefore sizeC is temporarily
assumed and will be adjusted later.
surfC is approximate surface factor and can be estimated using
equation 6.7e (Robert L. Norton, Machine Design, p329): ( )b
surf utC A S 
with A= 2.7 and b= -0.265 are found in table 6.3 (Robert L. Norton,
Machine Design, p329). Calculate above equation with given factors, we
have:
𝑪 𝒔𝒖𝒓𝒇 = 𝑨 × (𝑺 𝒖𝒕) 𝒃
= 𝟐. 𝟕 × (𝟖𝟔)−𝟎.𝟐𝟔𝟓
=0.83<1 (satisfied)
1tempC  since temperature is not evaluated.
1reliabC  while we assume 50% reliability at this preliminary design
stage.

'
e
S is uncorrected endurance strength and calculated using equation 6.5a
(Robert L. Norton, Machine Design, p324) for 200utS kpsi :
' 0.5 0.5 86000 43000ute
S S psi    
Apply all above factor, now we can calculate the corrected endurance
strength:
'e load size surf temp reliab e
S C C C C C S     
= 1× 𝟏 × 𝟎. 𝟖𝟑 × 𝟏 × 𝟏 × 𝟒𝟑𝟎𝟎𝟎 = 𝟑𝟓𝟔𝟗𝟎 psi
Another factor of material is notch sensitivity which is found from equation
6.13 (Robert L. Norton, Machine Design, p339) or figure 6.36 (Robert L.
Norton, Machine Design, p340-341):
𝒒 𝒃𝒆𝒏𝒅𝒊𝒏𝒈 = 𝟎. 𝟔
𝒒𝒕𝒐𝒓𝒔𝒊𝒐𝒏 = 𝟎. 𝟖𝟓
The fatigue stress-concentration factor is found from equation 6.11b
(Robert L. Norton, Machine Design, p339) using the assumed geometric
stress-concentration factor noted above: {
𝑲 𝒕 = 𝟐. 𝟓
𝑲 𝒇𝒔 = 𝟏. 𝟖
The fatigue stress concentration factor for bending stress at section 1:
𝑲 𝒇 = 𝟏 + 𝒒 × ( 𝑲 𝒕 − 𝟏) = 𝟏 + 𝟎. 𝟔 × ( 𝟐. 𝟓 − 𝟏) = 𝟏. 𝟗
The fatigue stress concentration factor for torsion at section 1:
𝑲 𝒇𝒔 = 𝟏 + 𝒒 × ( 𝑲 𝒕𝒔 − 𝟏) = 𝟏 + 𝟎. 𝟖𝟓 × ( 𝟏. 𝟖 − 𝟏) = 𝟏. 𝟔𝟖

From equation 6.17 (Robert L. Norton, Machine Design, p360), we find
that in this case, the same factor should be used on the mean torsional stress
component: 𝑲 𝒇𝒎𝒔 = 𝑲 𝒇𝒔= 1.68
II. Shaft 1 calculations
Given 𝐹𝑔 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙
= 520.3 ( 𝑙𝑏)
𝑇1 = 63603 (𝑁. 𝑚𝑚)
Figure 4.1.Shaft dimensions.
We choosediameter of shaft I is 1 25( )ed mm , according to the standard of
the SKF bearing manufacture, the width of roller-bearing 01 17( )b mm . And
1 2 310, 15, 20nk k k h    .
 Length of the gear hub: 13 11.2 1.2 25 37.5( )m el d mm    
 Length of the hub of a haft coupling: 12 12 2 25 50( )m el d mm    
 12 12 01 30.5 ( ) 0.5 (50 17) 15 20 68.5( )c m nl l b k h mm          
 13 13 01 1 20.5 ( ) 0.5 (30 17) 10 10 43.5( )ml l b k k mm          
 12 12 68.5( )cl l mm   

 11 132 2 43.5 87( )l l mm    
a) Analyze the forces and moments
We assume that 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 = 0.2 × 𝑊𝑡 = 0.2 × 520.3 = 104.06 (𝑙𝑏), according
Figure 4.2.Force diagram of shaft 1.
The tangential force on gear is found from the torque and its radius. The
tangential force at the spur-gear tooth as shown in part C:
𝐹𝑔 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙
= 520.3 (𝑙𝑏)
The spur gear has a 250 pressure angle, which means that there will also be
a radial component of force that gear tooth of:
𝐹𝑔 𝑟𝑎𝑑𝑖𝑎 𝑙
= 𝐹𝑔 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙
× tan(25 𝑜) = 520.3 × tan(25 𝑜) = 242.6 (𝑙𝑏)
Solve the reaction forces in the xz and yz planes using 0, 0x xF M  
and 0, 0y yF M   with the beam dimensions as above:

0 2 1 2 1 1 2 3
2
2
`0 2
0 2
( ) F F ( ) 0
(87 68.5) R 87 (87 68.5 43.5) 0
(87 199 )
0.56 1.28
155.5
0
g coupling
g coupling
g coupling
g coupling
g coupling
g coupling g coupli
M R l l l l l l
F F
F F
R F F
F R F F R
R F F R F F
          
         
   
      
     
        ( 0.56 1.28 )
0.44 2.28
ng g coupling
g coupling
F F
F F
    
    
Above equations can be solved for 𝑅0 and 2R in each plane, using the
approximate components of the apply load gF and couplingF :
𝑅0 𝑥
= −0.44 × 𝐹𝑔 𝑥
+ 2.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 𝑥
= −0.44 × 242.6 + 2.28 × 104.06
= 130.5 ( 𝑙𝑏)
𝑅0 𝑦
= −0.44 × 𝐹𝑔 𝑦
+ 2.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 𝑦
= −0.44 × (−520.3)+ 2.28 × 0 = 228.93 ( 𝑙𝑏)
𝑅2 𝑥
= −0.56 × 𝐹𝑔 𝑥
− 1.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 𝑥
= −0.56 × 242.6 − 1.28 × 104.06
= −269.05( 𝑙𝑏)
𝑅2 𝑦
= −0.56 × 𝐹𝑔 𝑦
− 1.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 𝑦
= −0.56 × (−520.3)− 1.28 × 0 = 291.37 (𝑙𝑏)
Moment calculations:
At point 0:
𝑀 𝑥
(0)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(0)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(1)
= 556.5(𝑙𝑏. 𝑖𝑛)
At point 1:
𝑀 𝑥
(1)
= 𝑅0 𝑥
× 𝑙1 = 130.5 × 3.4 = 443.7( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(1)
= 𝑅0 𝑦
× 𝑙1 = 228.93 × 3.4 = 778.36( 𝑙𝑏. 𝑖𝑛)
𝑇(1)
= 556.5(𝑙𝑏. 𝑖𝑛)

At point 2:
𝑀𝑥
(2)
= 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 × 𝑙3 = 104.06× 1.71 = 177.94( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(2)
= 𝑅0 𝑦
× (𝑙1 + 𝑙2) − 𝐹𝑔 𝑦
× 𝑙2 = 228.93× 6.1− 520.3 × 2.69
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(2)
= 556.5(𝑙𝑏. 𝑖𝑛)
At point 3:
𝑀𝑥
(3)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(3)
= 556.5(𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(3)
= 0( 𝑙𝑏. 𝑖𝑛)

Figure 4.3.Moment diagram of shaft 1.

b) Diameter calculation
Firstly, we choosethe safety factor equal to 2.5
Apply the equation 9.6a (Robert L. Norton, Machine Design, p.512) to
determine the diameter of the shaft’s section:
1/31/22 2
32 3
4
f a m
f fsm
f y
N M T
d K K
S S
                         
Consider the section I-0 at the point 0 of the 1st bearing:
0( . ),T 891.60(lb.in)aM lbin  , the minimum recommended diameter 0 :d
𝑑0 = {
32 × 2.5
𝜋
[(1.9 ×
0
35690
)
2
+
3
4
(1.68 ×
556.5
54000
)
2
]
1
2
}
1
3
= 0.73(𝑖𝑛)
Consider the section I-1 at the point 1 of the gear:
𝑀 𝑎 = 895.94( 𝑙𝑏. 𝑖𝑛), 𝑇 = 556.5(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter 1 :d
𝑑1 = {
32 × 2.5
𝜋
[(1.9 ×
895.94
35690
)
2
+
3
4
(1.68 ×
556.5
54000
)
2
]
1
2
}
1
3
= 1.08(𝑖𝑛)
Consider the section I-2 at the point 2 of the 2nd bearing position:
𝑀 𝑎 = 177.96( 𝑙𝑏. 𝑖𝑛), 𝑇 = 556.5(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter 2 :d
𝑑2 = {
32 × 2.5
𝜋
[(1.9 ×
177.96
35690
)
2
+
3
4
(1.68 ×
556.5
54000
)
2
]
1
2
}
1
3
= 0.77(𝑖𝑛)
Consider the section I-3 at the point 3 of the coupling position:
𝑀 𝑎 = 0( 𝑙𝑏. 𝑖𝑛), 𝑇 = 556.5(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter 3 :d

𝑑3 = {
32 × 2.5
𝜋
[(1.9 ×
0
35690
)
2
+
3
4
(1.68 ×
556.5
54000
)
2
]
1
2
}
1
3
= 0.73(𝑖𝑛)
From these preliminary calculations, we can determine reasonable sizes for
the four step diameters d0, d1, d2, d3:
Position Symbol Minimum(in) Standard(in) Standard(mm)
1stbearing d0 0.73 0.98 25
Gearing d1 1.08 1.18 30
2ndbearing d2 0.77 0.98 25
Coupling d3 0.73 0.89 22
Table 4.1.Uncorrected shaft 1 data.
c) Key and testing for shaft I
Assuming use square, parallel keys with end-milked keyways. Low-carbon
steel, SAE 1040 with 86 , 54ut yS kpsi S kpsi  will be used. 35690eS psi as
calculated above, stress-concentration factor can be seen in figure 9.16
(Robert L. Norton, Machine Design, p.529).
These are 2 locations with keys on this shaft, at point 1 and 3. The design
diameters chosen for these sections were 1 30 (1.18 )d mm in and 𝑑3 =
22𝑚𝑚 (0.89𝑖𝑛). Table 9.2 (RobertL. Norton, Machine Design, p.525) shows
that the standard key width for d1 is 0.250(in) and for d3 is 0.187(in).
At point 1, the mean and alternating components of force on the key are
found from the torque component divided by the shaft radius at that point:
𝐹𝑎 =
𝑇𝑎
𝑟
=
556.5
0.59
= 943.22𝑙𝑏

𝐹𝑚 =
𝑇 𝑚
𝑟
=
556.5
0.59
= 943.22𝑙𝑏
Assuming a key length of 1 in and calculate the alternating and mean shear
stress components from equation 4.9 (Robert L. Norton, Machine Design,
p.151):
𝜏 𝑎 = 𝜏 𝑚 =
𝐹𝑎
𝐴 𝑠ℎ𝑒𝑎𝑟
=
943.22
1 × 0.25
= 3772.88 𝑝𝑠𝑖
To find the safety factor for shear fatigue of the key, compute the Von
Mises equivalent stresses for each these components from equation 5.7d
(Robert L. Norton, Machine Design, p.245):
𝜎𝑎
′
= 𝜎 𝑚
′
= √ 𝜎𝑥
2 + 𝜎𝑦
2 − 𝜎𝑥. 𝜎𝑦 + 3𝜏 𝑥𝑦
2 = √3 × 3772.882 = 6534.8 𝑝𝑠𝑖
Then use them in equation 6.18e (Robert L. Norton, Machine Design,
p.364):
𝑁𝑓 =
1
𝜎 𝑎
′
𝑆 𝑒
+
𝜎 𝑚
′
𝑆 𝑢𝑡
=
1
6534.8
35690
+
6534.8
86000
= 3.86
𝐹𝑎 = 𝐹𝑚 =
𝑇𝑎
𝑟
=
556.5
0.445
= 1250.56𝑙𝑏
Assuming a key length of 0.5 in and calculate the alternating and mean
shear stress components from:
𝜏 𝑎 = 𝜏 𝑚 =
𝐹𝑎
=
1250.56
0.5 × 0.187
= 13375 𝑝𝑠𝑖

𝜎𝑎
′
= 𝜎 𝑚
′
= √ 𝜎𝑥
2 + 𝜎𝑦
2 − 𝜎𝑥. 𝜎𝑦 + 3𝜏 𝑥𝑦
2 = √3 × 133752 = 23166.2 𝑝𝑠𝑖
p.364):
𝑁𝑓 =
1
𝜎 𝑎
′
𝑆 𝑒
+
𝜎 𝑚
′
𝑆 𝑢𝑡
=
1
23166.2
35690
+
23166.2
86000
= 1.09
End-mill radius versus shaft diameter ratio:
 At point 1:
1
0.02
0.017
1.18
r
d
 
 At point 3:
3
0.02
0.023
0.89
r
d
 
The corresponding stress-concentration factors are read from figure 9.16
 At point 1:
2.25
3.15
t
ts
K
K



 At point 3:
2.10
2.85
t
ts
K
K



These are used to obtain the fatigue stress concentration factors, which for
materials notch sensitivity q=0.65 are:
 At point 1:
1 ( 1) 1 0.65 (2.25 1) 1.81
1 ( 1) 1 0.65 (3.15 1) 2.39
f t
fs ts
K q K
K q K
        

        
 At point 3:
1 ( 1) 1 0.65 (2.10 1) 1.71
1 ( 1) 1 0.65 (2.85 1) 2.20
f t
fs ts
K q K
K q K
        

        

For both point f fm
fsm fs
K K
K K



The new safety factors are then calculated using equation 9.8 (Robert L.
Norton, Machine Design, p.514) with the data from equations (b) and (c)
with the design values for shaft diameter and above stress-concentration
values inserted:
 At point 1:
1
3
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SN
d
k M K T
S

  
    
  
    
   
      
  
   
1.18 =
{
32 × 𝑁𝑓
𝜋
[
√(1.81× 895.94)2 +
3
4
(2.39 × 556.5)2
35690
+
√(1.81 × 895.94)2 +
3
4
(2.39 × 556.5)2
86000
]}
1
3
→ 𝑁𝑓 = 2.05
 At point 3:

1
3
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SN
d
k M K T
S

  
    
  
    
   
      
  
   
0.89 =
{
32 × 𝑁𝑓
𝜋
[
√(1.71 × 0)2 +
3
4
(2.20 × 556.5)2
35690
+
√(1.71 × 0)2 +
3
4
(2.20 × 556.5)2
86000
]}
1
3
→ 𝑁𝑓 = 1.65
At both point, the safety factor are smaller than specified valued of 2.5. So
we retry the above step with different diameter we find:
Increasing the diameter at point 1 to 1.57 in (35mm) gives a safety factor of
4.33.
Increasing the diameter at point 3 to 1.18 in (25mm) gives a safety factor of
2.53.
According to this change, all diameters of shaft I now become:
Position Symbol
Minimum
(in)
Standard
(in)
Standard
(mm)
1stbearing d0 0.720 1.180 30
Gearing d1 0.865 1.181 35

2ndbearing d2 0.680 1.180 30
Coupling d3 0.720 0.984 25
Width(in) Length(in)
Key
Point 1 0.375 1.0
Point 3 0.250 0.5
Table 4.2.Corrected shaft 1 data.
III. Shaft 2 calculations
a) Analyze the forces and moments
Figure 4.4.Force diagram of shaft 2.
Choose diameter of shaft 2 is de2=30(mm), according to the standard of the
SKF bearing manufacture, the width of roller bearing is b02= 19mm.
𝑙1 = 𝑙2 = 43.5( 𝑚𝑚) = 1.71( 𝑖𝑛)
𝑙3 = 68.5( 𝑚𝑚) = 2.69( 𝑖𝑛)
𝐹𝑔 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙
= 520.3( 𝑙𝑏)
𝐹𝑔 𝑟𝑎𝑑𝑖𝑎𝑙
= 242.6(𝑙𝑏)

𝐹 = 4419.9(𝑁) 𝐹𝑥 = 𝐹 × 𝑐𝑜𝑠60 𝑜
= 4419.9 × 𝑐𝑜𝑠60 𝑜
= 2209.95( 𝑁) = 491.1(𝑙𝑏)
𝐹𝑦 = 𝐹 × 𝑠𝑖𝑛60 𝑜
= 4419.9 × 𝑠𝑖𝑛60 𝑜
= 3827.75( 𝑁) = 850.6(𝑙𝑏)
𝑇𝐼𝐼 = 303690( 𝑁. 𝑚𝑚) = 2656.95(𝑙𝑏. 𝑖𝑛)
Solve the reaction forces in the xz and yz planes using 0, 0x xF M  
and 0, 0y yF M   with the beam dimensions as above:
0 6 1 2 1 1 2 3
6
6
`4 6
4 6
( ) F F ( ) 0
(43.5 43.5) R 43.5 (43.5 43.5. 68.5) 0
(43.5 155.5 ) 155.5
0.5
87 87
0
155.5
( 0.5 )
87
68.5
0.5
87
g
g
g
g
g
g g g
g
M R l l l l l l
F F
F F
R F F
F R F F R
R F F R F F F F
F F
          
         
   
      
     
            
    
Above equations can be solved for 1R and 2R in each plane, using the
approximate components of the apply load gF and couplingF :
𝑅4 𝑥
= −0.5 × 𝐹𝑔 𝑥
+
68.5
87
× 𝐹𝑥 = −0.5 × 242.6 +
68.5
87
× 491.1 = 265.37(𝑙𝑏)
𝑅4 𝑦
= −0.5 × 𝐹𝑔 𝑦
+
68.5
87
× 𝐹𝑦 = −0.5 × (−520.3)+
68.5
87
× 850.6 = 929.87(𝑙𝑏)
𝑅6 𝑥
= −0.5 × 𝐹𝑔 𝑥
+
68.5
87
× 𝐹𝑥 = −0.5 × 242.6 −
155.5
87
× 491.1 = −999.07(𝑙𝑏)
𝑅6 𝑦
= −0.5 × 𝐹𝑔 𝑦
+
68.5
87
× 𝐹𝑦 = −0.5 × (−520.3)−
155.5
87
× 850.6 = −1260.2(𝑙𝑏)
Moment calculations:
 At point 4:
𝑀 𝑥
(4)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(4)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(4)
= 2656.95(𝑙𝑏. 𝑖𝑛)

 At point 5:
𝑀 𝑥
(5)
= 𝑅4 𝑥
× 𝑙1 = 491.1 × 1.71 = 839.8( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(6)
= 𝑅4 𝑦
× 𝑙1 = 850.6 × 1.71 = 1454.5( 𝑙𝑏. 𝑖𝑛)
𝑇(6)
= 2656.95(𝑙𝑏. 𝑖𝑛)
At point 6:
𝑀 𝑥
(6)
= 𝐹𝑥 × 𝑙3 = 491.1 × 2.69 = 1321.1( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(6)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(6)
= 2656.95(𝑙𝑏. 𝑖𝑛)
 At point 7:
𝑀 𝑥
(7)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑀 𝑦
(7)
= 0( 𝑙𝑏. 𝑖𝑛)
𝑇(7)
= 2656.95(𝑙𝑏. 𝑖𝑛)

Figure 4.5.Moment diagram of shaft 2.

b) Diameter calculation
Firstly, we choosethe safety factor equal to 2.5
Apply the equation 9.56a (Robert L. Norton, Machine Design, p.512)
to determine the diameter of the shaft’s section:
Consider the section II-4 at the point 4 of the 1st bearing:
𝑀 𝑎 = 0( 𝑙𝑏. 𝑖𝑛), 𝑇 = 2656.95( 𝑙𝑏. 𝑖𝑛), the minimum recommended diameter
𝑑0 = {
32× 2.5
𝜋
[(1.9 ×
0
35690
)
2
+
3
4
(1.68 ×
2656.95
54000
)
2
]
1
2
}
1
3
= 1.22(𝑖𝑛)
Consider the section II-5 at the point 5 of the gear:
𝑀 𝑎 = 1679.5(𝑙𝑏. 𝑖𝑛), 𝑇 = 2656.95(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter
𝑑1 = {
32 × 2.5
𝜋
[(1.9 ×
1679.5
35690
)
2
+
3
4
(1.68 ×
2656.95
54000
)
2
]
1
2
}
1
3
= 1.43(𝑖𝑛)
Consider the section II-6 at the point 6 of the 2nd bearing position:
𝑀 𝑎 = 1321.1(𝑙𝑏. 𝑖𝑛), 𝑇 = 2656.95(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter
𝑑2 = {
32 × 2.5
𝜋
[(1.9 ×
1321.1
35690
)
2
+
3
4
(1.68 ×
2656.95
54000
)
2
]
1
2
}
1
3
= 1.37(𝑖𝑛)
Consider the section II-7 at the point 7 position:
0( . ),T 3385.74(lb.in)aM lbin  , the minimum recommended diameter
1/31/22 2
32 3
4
f a m
f fsm
f y
N M T
d K K
S S
                         
0 :d
1 :d
2 :d
3 :d

𝑑3 = {
32× 2.5
𝜋
[(1.9 ×
0
35690
)
2
+
3
4
(1.68 ×
2656.95
54000
)
2
]
1
2
}
1
3
= 1.22(𝑖𝑛)
From these preliminary calculations, we can determine reasonable sizes for
the four step diameters d0, d1, d2, d3:
Position Symbol Minimum(in) Standard(in) Standard(mm)
1stbearing d0 1.22 1.57 40
Gearing d1 1.43 1.77 45
2ndbearing d2 1.37 1.57 40
Chain d3 1.22 1.37 35
Table 4.3.Uncorreted shaft II data.
c) Key and testing for shaft II
Assuming use square, parallel keys with end-milked keyways. Low-
carbonsteel, SAE 1040 with will be used.
as calculated above, stress-concentration factor can be seen in figure 9.16
(Robert L. Norton, Machine Design, p.529).
These are 2 locations with keys on this shaft, at point 5 and 7. The
design diameters chosenfor these sections were 1 30 (1.5 )d mm in and
3 20 (0.72 )d mm in . Table 9.2 (RobertL. Norton, Machine Design, p.525)
shows that the standard key width for d1 is 0.500(in) and for d3 is 0.312(in).
At point 5, the mean and alternating components of force on the key
are found from the torque componentdivided by the shaft radius at that point:
86 , 54ut yS kpsi S kpsi  35690eS psi

𝐹𝑎 =
𝑇𝑎
𝑟
=
2656.95
0.885
= 3002.2( 𝑙𝑏)
𝐹𝑚 =
𝑇 𝑚
𝑟
=
2656.95
0.885
= 3002.2(𝑙𝑏)
Assuming a key length of 1in and calculate the alternating and mean shear
stress components from equation 4.9 (Robert L. Norton, Machine Design,
p.151):
𝜏 𝑎 = 𝜏 𝑚 =
𝐹𝑎
=
2656.95
1 × 0.5
= 5313.9 𝑝𝑠𝑖
𝜎𝑎
′
= 𝜎 𝑚
′
= √ 𝜎𝑥
2 + 𝜎𝑦
2 − 𝜎𝑥. 𝜎𝑦 + 3𝜏 𝑥𝑦
2 = √3 × 5313.92 = 9204 𝑝𝑠𝑖
p.364):
𝑁𝑓 =
1
𝜎 𝑎
′
𝑆 𝑒
+
𝜎 𝑚
′
𝑆 𝑢𝑡
=
1
9204
35690
+
9204
86000
= 2.74
𝐹𝑎 = 𝐹𝑚 =
𝑇𝑎
𝑟
=
2656.95
0.685
= 3878.76𝑙𝑏
Assuming a key length of 1 in and calculate the alternating and mean shear
stress components from:
𝜏 𝑎 = 𝜏 𝑚 =
𝐹𝑎
=
3878.76
1 × 0.312
= 12431.92 𝑝𝑠𝑖

𝜎𝑎
′
= 𝜎 𝑚
′
= √ 𝜎𝑥
2 + 𝜎𝑦
2 − 𝜎𝑥. 𝜎𝑦 + 3𝜏 𝑥𝑦
2 = √3 × 12431.922 = 21532.72 𝑝𝑠𝑖
p.364):
𝑁𝑓 =
1
𝜎 𝑎
′
𝑆 𝑒
+
𝜎 𝑚
′
𝑆 𝑢𝑡
=
1
21532.72
35690
+
21532.72
86000
= 1.17
End-mill radius versus shaft diameter ratio:
 At point 5:
1
0.02
0.005
3.93
r
d
 
 At point 7:
3
0.02
0.005
3.54
r
d
 
The corresponding stress-concentration factors are read from figure 9.16
 At point 5:
4.00
4.00
t
ts
K
K



 At point 7:
4.00
4.00
t
ts
K
K



These are used to obtain the fatigue stress concentration factors, which for
materials notch sensitivity q=0.65 are:
 At point5:
1 ( 1) 1 0.65 (4.00 1) 2.95
1 ( 1) 1 0.65 (4.00 1) 2.95
f t
fs ts
K q K
K q K
        

        

 At point 7:
1 ( 1) 1 0.65 (4.00 1) 2.95
1 ( 1) 1 0.65 (4.00 1) 2.95
f t
fs ts
K q K
K q K
        

        
For both point
The new safety factors are then calculated using equation 9.8 (Robert L.
Norton, Machine Design, p.514) with the data from equations (b) and (c)
with the design values for shaft diameter and above stress-concentration
values inserted:
 At point 5:
1
3
2 2
2 2
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
3
(2.95 2150.7) (2.95 3385.74)
4
32 35690
1.77
3
(2.95 2150.7) (2.95 3385.74)
4
86000
f a fs a
ff
fm m fsm m
ut
f
k M k T
SN
d
k M K T
S
N


  
    
  
    
   
      
  
   
 
   
 
  
 
   

 
1
3
3.32fN
 
 
 
 
 
 
 
  
 
1
3
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SN
d
k M K T
S

  
    
  
    
   
      
  
   
f fm
fsm fs
K K
K K




1.77 =
{
32 × 𝑁𝑓
𝜋
[
√(2.95 × 1679.5)2 +
3
4
(2.95 × 2656.95)2
35690
+
√(2.95 × 1679.5)2 +
3
4
(2.95 × 2656.95)2
86000
]}
1
3
→ 𝑁𝑓 = 1.63
 At point 7:
1
3
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SN
d
k M K T
S

  
    
  
    
   
      
  
   
1
3
2 2
2 2
3
(2.95 0) (2.95 3385.74)
4
32 35690
1.37
3
(2.95 0) (2.95 3385.74)
4
86000
fN

  
    
  
     
          
2.52fN 
1
3
2 2
2 2
3
( ) ( )
4
32
3
( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SN
d
k M K T
S

  
    
  
    
   
      
  
   

1.37 =
{
32 × 𝑁𝑓
𝜋
[
√(2.95× 0)2 +
3
4
(2.95 × 2656.95)2
35690
+
√(2.95 × 0)2 +
3
4
(2.95 × 2656.95)2
86000
]}
1
3
→ 𝑁𝑓 = 0.94
Position Symbol
Minimum
(in)
Standard
(in)
Standard
(mm)
1stbearing d0 1.22 1.57 40
Gearing d1 1.43 1.77 45
2ndbearing d2 1.37 1.57 40
Chain d3 1.22 1.37 35
Width(in) Length(in)
Key
Point 5 0.500 1.0
Point 7 0.500 1.0
Table 4.4.Corrected shaft II data.
E. Bearing and lubricant
We select roller bearings for the designed shafts.
1. Shaft 1: d0 =35 mm = 1.37 in

From figure 10.23(Robert L. Norton, Machine Design, p615), choose
a #6306 bearing with 35 mm inside diameter. Its dynamic load rating factor is
C=5700 lb, the static load rating C0 = 4000 lb. The static applied load of
86.94 lb (at 1st bearing) and 70.61 lb (at 2nd bearing) is obviously well below
the bearing’s static rating.
From table 10.5(Robert L. Norton, Machine Design, p614), choose
the factor for a 5% failure rate: KR =0.62
Calculate the projected life with equation 10.20a and 10.19 or their
combination, equation 10.20d (Robert L. Norton, Machine Design, p615).
Note that the equivalent load in this case is simply the applied radial load due
to the absence of any thrust load. For the reaction load of 86.94 lb at 1st
bearing:
10 10
3 35700
0.62 704545( )
86.94
p p
C
L K revs
P
   
       
   
For the reaction load of 70.61 lb at 2nd bearing:
10 10
3 35700
0.62 1409571( )
70.61
p p
C
L K revs
P
   
       
   
There bearing are obviously very light loaded, but their size was
dictated by considerations of stresses in the shaft that defined the shaft
diameter.
From figure 10.23 (Robert L. Norton, Machine Design, p615), this
bearing’s limiting speed is 8500 rpm, well above the operating speed of
shaft 960 rpm.
2. Shaft 2: d0 =50 mm = 1.96 in
From figure 10.23(Robert L. Norton, Machine Design, p615), choose
a #6308 bearing with 50 mm inside diameter. Its dynamic load rating factor is

C=10600 lb, the static load rating C0 = 8150 lb. The static applied load of
226.57 lb (at 1st bearing) and 358.31 lb (at 2nd bearing) is obviously well
below the bearing’s static rating.
From table 10.5(Robert L. Norton, Machine Design, p614), choose
the factor for a 5% failure rate: KR =0.62
Calculate the projected life with equation 10.20a and 10.19 or their
combination, equation 10.20d (Robert L. Norton, Machine Design, p615).
Note that the equivalent load in this case is simply the applied radial load due
to the absence of any thrust load. For the reaction load of 226.57 lb at 1st
bearing:
10 10
3 310600
0.62 228772( )
226.57
p p
C
L K revs
P
   
       
   
For the reaction load of 358.31 lb at 2nd bearing:
10 10
3 310600
358.31
0.62 49645( )p p
C
L K revs
P
   
       
   
There bearing are obviously very light loaded, but their size was
dictated by considerations of stresses in the shaft that defined the shaft
diameter.
From figure 10.23 (Robert L. Norton, Machine Design, p615), this
bearing’s limiting speed is 6000 rpm, well above the operating speed of shaft
320 rpm.
We choose the petroleum oils for lubricant for the work characteristic
very mild impact of system.
We choose Rzeppar coupling for its characteristic constant velocity.

F. Gearbox’s cover
 Box cover has responsibility of maintaining the relative position among
machine parts and machine elements. It receives loads from part that attached
in box cover, contains lubricating oil, protects inside element from dirt and
dust.
 Material: gray cast iron GX 15-32
 Assembling surface go through shaft’s center line in order assemble
machine element conveniently.
 The cap surface and the black rubber or grinded in order to make
interference fit, when fitting there are a layer of liquid coat or special coat.
 The bottom surface inclines to the oil outlet about 1degree.
 We have table of structure and dimensions of the gear box cover:
Name Value
Thickness:
1
( )
( )
Body
Cap


1
1
0.03 3 0.03 120.47 3 6.61( )
7 6( )
0.9 0.9 7 6.3( )
7 6( )
a mm
mm
mm
mm


 

      
  
    
  
Reinforcing rib:
( )
( )
Depth e
Height h
Slope
(0.8 1) (0.8 1) 7 5.6 7( ) 7
5 5 7 35( ) 30
2
e mm e mm
h mm h mm
About


        
      

Diameter:
Foundation bolt 1( )d
Side bolt 2( )d
Assembly cap and body
bolt 3( )d
Cap connecting screw
4( )d
Oil outlet cap
connecting screw 5( )d
1
1
2 1 2
3 2 3
4 2 4
5 2 5
0.04 10 0.04 120.47 10 14.81( )
16
(0.7 0.8)d 12
(0.8 0.9)d 10
(0.6 0.7)d 6
(0.5 0.6)d 6
d a mm
d mm
d d mm
d d mm
d d mm
d d mm
      
 
   
   
   
   

Bearing dimension:
Width of the assembly
surface of drive side
bolt: K2
Center of drive side
bolt: E2
Distance between bolt
center and hole side (k)
2 2
2 2
2 2 2
2
3 2
1.6 1.6 12 19.2( )
1.3 1.3 12 15.6( )
(3 5) 19.2 15.6 3 37.8( )
38
/ 2with k 1,2 d 7.2( )
E d mm
R d mm
K E R mm
K mm
C D mm
    
    
       
 
   
Cap and body connecting
flange:
Body depth 3( )S
Cap depth 4( )S
Flange width 3(K )
3 3 3
4 3 4
3 2 3
(1.4 1.5) 14
(0.9 1)S 13
(3 5) 35
S d S mm
S S mm
K K K mm
   
   
    
Cover box foundation
surface:
Depth: without stub S1
Width of box
foundation: K1 and q
1 1 1
1
1
(1.3 1.5) 22
3 1 3 15 45
2 45 2 7 59
S d S mm
K d mm
q K mm
   
    
     
Clearance among
elements:
Between gear and inside
wall of box.
Between gear top and
box bottom.
Between gear’s sides.
1 1
2 2
(1 2) 10
(3 5) 30
8
mm
mm
mm



     
     
    
Number of foundation bolt ( ) / (200 300) Z 4Z L B    

G. Comparisons of Designing chain - Driven system between Using
Vietnamese Document and English Document
Vietnamese Document English Document
In part A: Choosing motor and
distributing transmissionratio are
nearly similar to English
Document
-Vietnamese and English
documents use the same input data
-work power
.
1000
lv
F v
P 
-The same Efficiency of gear
train,chain gear,bearing,coupling
-the similar way to calculate the
necessarypoweror the revolutions
-According to Vietnamese
Document,Choosing motor can be
from 200rpm to 4000rpm
-the distributing transmissionratio
and torque on the axis is calculated
in the same way for Vietnamese
and English document
- Using “Tinh Toan He Dan Dong
Co Khi” to choosethe motor.
In part A: Choosing motor and
distributing transmissionratio are
nearly similar to the Vietnamese
Document.
-Input data: Pulling force of
conveyor chain, conveyor speed
chain…
-work power
.
1000
lv
F v
P 
-The necessary power lv
yc
P
P


-the revolutions
60000.
.
ct
v
n
z p

-in English Document, manufacturer
only provide motors from 500rpm to
2081rpm which are very commom in
industry
-Using Siemens or Toshiba catalogue
to choosemotor.
In part B: Calculating and designing
outer transmission – chain drives
In part B: Calculating and designing

- Vietnamese version: Ux=2.19 =>
choose:
Z1 = 25
Z2 = 55
-Both of document have choosing
number of teeth of the sprocket
-Vietnamese and English document
use the same factor to determine the
chain step
-Selecting the chain drive by
calculating center distance and force
-Vietnamese and English also have
testing chain durability
outer transmission – chain drives
-English version: Ux=2.4 =>
choose:
Z1=24
Z2=48
-kz –Toothfactor
-kn – Rotating factor
-Using selection factor
F1,F2,F3,F4,F5,F6
-English document use Toothfactor
and speed to select the chain drive in
catalogue
-Using wrecking load
-Testing chain is more difficult in
English Document.
In part C: Design gear transmission
-Vietnamese version:Ubr=4 =>
choose : Ppinion=60.08(mm)
PGear = 237.92(mm)
-The same input data
-in Vietnamese Document,we choose
material for gear before we
calculated pitch
-both of documents calculated pitch
diameter,number of teeth,pitch
diameter,distance,but using different
formula and factor
-Using bothtorque and transmission
In part C: Design gear transmission
- English version: Ubr=5 => choose:
Ppinion=8.4(mm)
PGear = 72.2(mm)
-power,number of shift,transmission
ratio…
-we finished for calculating
pitch,choosing material is at the end
of the process
-using look up table for choosing
quality,the tooth form,bending
geometry factor,size factor,rim
thickness factor,velocity

power to calculate
-Vietnamese have testing for gear
transmission
-in Vietnamese document,we
calculate for diameter of teeth
distinctly with gear
factor,overload factor…
-using only transmission power to
calculate
-in English document,we adjust the
contact stress to check
-Englisht documents don’tcalculate
for teeth,we use look up table for
choosing gear and teeth
In part D. Shafts, keys and couplings
calculation
-the same input Data
-At first,coupling is calculated by
using transmission momment and
diameter of motor’s shaft
-The same way to calculate shaft
diameter
-the similar way to calculate length
of the gear hub
-Vietnamese documents name KnF
instead of couplingF in English document
-the different formular for
calculating the diameter of the shaft’
s section.
1
3
0,1[ ]
tdM
d


-in Vietnamese document,the shaft
In part D. Shafts, keys and couplings
calculation
- dcT , T , IIT …
-At first,the material is choosenby
using working charact
-
 
300
10,2
IT
d


- 13ml , 12l , 11l …
- couplingF
-according to norton book
3
1
2
1
22
..
4
3
.
32






































y
m
fsm
f
a
f
f
S
T
k
S
M
K
N
d

-in English Document,the shaft
diameter is smaller because we chose
material which has higher hardness
than Vietnamese Document

diameter is greater than English
Document
-in Vietnamese document using
safety factor and stress
-in english document,Testing for
shaft using Torque,forceand safety
factor
E.Bearing and lubricant,Design’s
box cover,the oil diptick
E.Bearing and lubricant,Design’s
box cover,the oil diptick
I. Data Table
a. English document
FactorsAxis Motor I II Operating
ukn = 1 ubr = 5 ux = 2.4
P(KW) Pđc= 6.75 PI = 6.66 PII = 6.36 Pct= 5.7
n(rev/min) nđc= 1000 nI= 1000 nII= 200 nct = 83.33
T(N.mm) Tđc= 64462 TI = 63603 TII = 303690 Tct = 653246
b. Vietnamese document
Thông
sốTrục
Động cơ I II Công tác
ukn = 0.99 ubr = 4 ux = 2.19
P(KW) Pđc=4.4 PI=4.33 PII=4.09 Pct=3.74
n(v/ph) nđc=716 nI=716 nII=179 nct=327

T(N.mm) Tđc=58687 TI=57753 TII=218209 Tct=109225
 we can see the different about the way we chooseubr and ux, that make the
different design.
II. Calculating and designing outer transmission – chain drives:
- English document: Ux=2.4 => choose: Z1 = 24
Z2 = 48
- Vietnamese document: Ux=2.19 => choose: Z1 = 25
Z2 = 55
III. Design gear transmission.
- English document: Ubr= 5 => choose: Ppinion = 8.4(mm)
PGear = 72.2(mm)
- English document: Ubr=4 => choose: Ppinion = 60.08(mm)
PGear = 237.92(mm)
We have Data of gear transmission:
a. English document
Gear details Symbol Value
Diametral pitch 𝑑 𝑝 8
Pressure angle  25o

Tangential force tW 520.3(lb)
No. of teeth on pinion pinionN 20
No. of teeth on gear gearN 100
Pitch diameter of pinion piniond 2.5(in)
Pitch diameter of gear geard 12.5(in)
Pitch-line velocity tV 654.5(ft/min)
Bending stress-pinion tooth bpinion 17027.40psi
Bending stress-gear tooth bgear 13859.51psi
Surface stress in pinion and
gear
cpinion 106079.6psi
Uncorrected bending strength '
fb
S 31051psi
Corrected bending strength '
fb
S 28132psi
Uncorrected surface strength '
fc
S 104480psi
Corrected surface strength fc
S 94867.84psi

Bending safety factor for
pinion
bpN 1.65
Bending safety factor for gear bgN 2.03
Surface safety factor for mesh cpN 0.8

b. Vietnamese document
Information Symbol Value
Spindle diameter aw 149(mm)
Z1 24
Z2 95
Rolling diameter d1 60(mm)
d2 237.5(mm)
Peak diameter dw1 60.08(mm)
dw2 237.92(mm)
Bottom diameter da1 65(mm)
da2 232.5(mm)
Base diameter df1 53.75(mm)
df2 231.25(mm)
Moving ratio db1 56.38(mm)

db2 223.18(mm)
Profin angle
x1 0
Profin Gear’s angle x2 0
Relative angle  25o
Coincidental ratio w 35 o
Mode  1.7
Band wide (mm) m 2.5(mm)
Coincidental force bw1 50
bw2 45
Dividing diameter Ft 1725.73(N)
Fr 1812.02 (N)

Diameter of the shaft’ s section.
-English version : Using this equation to calculate
3
1
2
1
22
..
4
3
.
32






































y
m
fsm
f
a
f
f
S
T
k
S
M
K
N
d

Position Notion Minimum
diameter
( in)
Standard
diameter is
choosen (mm)
Standard
diameter is
choosen (inch)
1st bearing d0 0.73 25 0.98
2nd bearing d1 1.08 30 1.18
Coupling d2 0.77 25 0.98
Gearing d3 0.73 22 0.89

-Vietnamese version : Using this equation to calculate
3
0.1[ ]
tdj
j
M
d

 2 2
j0.75tdj jM M T  2 2
yjj xjM M M 
Position Notion Minimum
diameter
( calculated)
Standard
diameter is
choosen (mm)
Standard
diameter is
choosen (inch)
1st bearing d0 19.5 25 0.98
2nd bearing d1 22.5 25 0.98
Coupling d2 19.95 20 0.79
Gearing d3 20.27 30 1.18

References:
1. TrịnhChất, LêVănUyển – Tínhtoánthiếtkếhệdẫnđộngcơkhí, tập 1. NXB Giáodục,
2005
2. http://www.renoldchainselector.com/ChainSelector
3. Siemens Cooperation, Siemens motor choosing document.
4. Dudley's Handbook of Practical Gear Design and Manufacture
5. Robert L. Norton, Machine design.

Machine design-2

Recommended

Recommended

More Related Content

What's hot

What's hot (9)

Similar to Machine design-2

Similar to Machine design-2 (20)

Recently uploaded

Recently uploaded (20)

Machine design-2