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 VΔn QuyαΊΏt (20110646)
LΓ£ Minh CΓ΄ng (20110087)
MEC2-K56- Group 15 Page 2
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 β chain drives: ........Error!
Bookmark not defined.
1. Choosing the type of chain base on operating condition: .........Error!
Bookmark not defined.
2. Choosing number of teeth of the sprocket ...... Error! Bookmark not
defined.
3. Select factors. ..................................Error! Bookmark not defined.
4. Select chain drive ............................Error! Bookmark not defined.
5. Installation parameters .....................Error! Bookmark not defined.
6. Centre distance calculation...............Error! Bookmark not defined.
7. Selection of sprocket materials .........Error! Bookmark not defined.
8. Data of chosen chain........................Error! Bookmark not defined.
9. Check for chianβs durability .............Error! Bookmark not defined.
C. Design gear transmission ...........................Error! Bookmark not defined.
MEC2-K56- Group 15 Page 3
1. Design Decision: .......................................Error! Bookmark not defined.
2. Design Calculation: ...................................Error! Bookmark not defined.
Pitch diameters: ................................Error! Bookmark not defined.
Bending geometry factor...................Error! Bookmark not defined.
Velovity factorβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦19
Transmitted loadβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.19
Bending stresses in pinion and gearβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦20
Surface stress in the pinion gearβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.20
Bending- fatigue strengthβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...22
Surface fatigue strengthβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..23
Safety factorsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..24
3. Data table..................................................Error! Bookmark not defined.
D. Shafts, keys and couplings calculation. ....Error! Bookmark not defined.
I. Choosing material ...........................Error! Bookmark not defined.
II. Shaft 1 calculations .............................Error! Bookmark not defined.
a) Analyze the forces and moment...............................β¦β¦β¦β¦β¦β¦...29
b) Diameter calculationβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦32
c) Key and testing for shaft Iβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...33
III. Shaft 2 calculationsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦38
a) Analyze the forces and momentsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. ...38
b) Diameter calculationβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.40
c) Key and testing for shaft II β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.42
E. Bearing, lubrication, gear boxβs cover and parts structure. ...........Error!
Bookmark not defined.
1. Shaft 1 ............................................Error! Bookmark not defined.
2. Shaft 2 ............................................Error! Bookmark not defined.
F. Gear box's coverβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...49
MEC2-K56- Group 15 Page 4
G. Comparision of English vs Vietnamese documentβ¦β¦β¦β¦β¦β¦β¦β¦β¦
References: ................................................................................................. 66
List of tables
Table 1 Motor data table 10
Table 2 Sprocket materials 13
Table 3 Chain drive data table 14
Table 4 Gear drive data table 25
Table 5 Uncorrected shaft I diameter data 35
Table 6 Corrected shaft I diameter data 40
Table 7 Uncorrected shaft II diameter data 45
Table 8 Corrected shaft 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 diagram in shaft I 30
Figure 5 Moment diagram in shaft I 33
MEC2-K56- Group 15 Page 5
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. Sprocket teeth 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 π½ = 0 (degree)
8. Working characteristic: Mild impact
Shaft for calculating: Input shaft 1
MEC2-K56- Group 15 Page 6
MEC2-K56- Group 15 Page 7
A. Choosing motor and distributing transmission ratio:
1. Work power:
)(7.51000
36.14220
1000
.KW
vFPlv
2. Efficiency of driven system:
3. . .br OL x knh h h h h=
Where : Gear performance : βππ = 0.96 Chain performance : βπ₯ = 0.9
Bearing performance: βππΏ = 0.995
Jointing performance: βππ = 1
85.019.0995.096.0... 33 knxOLbr
3. Necessary power on the motor axis:
)(75.685.0
7.5KW
PP lv
yc
4. Number of revolutions on operating axis:
πππ‘ =60000.π£
π§. π=60001.36
1470= 83.26 (rev/min)
5. Choosing the first-aid transmission ratio (πππ):
πππ = ππ . πππ
Follow table B2.4
21[1]: Chain ration : ππ =2 Γ· 5
Gear ration : πππ = 3 Γ· 5
Then:
πππ πππ= ππ πππ . πππ πππ=23= 6
πππ πππ= ππ πππ .πππ πππ=55= 25
6. First-aid number of revolutions on motor axis
MEC2-K56- Group 15 Page 8
We have : πππ = πππ. πππ
πππ πππ = πππ πππ. πππ= 500 (rev/min)
πππ πππ = πππ πππ . πππ= 2081 (rev/min)
We choose πππ =1000 (rev/min)
7. Calculating synchronous number of revolutions of motor :
Choosing ππ ππ = πππ (rev/min)
8. Choosing the motor:
Choose the motor which satisfies: {πΔπ β ππ π = 1000 (rev/min)
πΔπ β₯ ππ¦π = 6.75 (πΎπ)
We have the motor with the details:
Siemen 1LA7 134-6AA { πππ = 1000 (rev/min)
π = 7 πΎπ
9. Distributing transmission ratio:
Transmission ratio of the system:
πππ =πΔπ
πππ=
ππππ
ππ.ππ=12
Choose the transmission ratio of 1st gear box: π’ππ =5
Transmission ratio of outer driven: π’π₯=πππ
πππ=
ππ
πβ π. π
So, we have: {
π’πβ = 12π’ππ = 5π’π₯ = 2.4
10. Calculating the factors on the axis:
Power on operating axis: πππ‘ = πππ£ = 5.7 (Kw)
MEC2-K56- Group 15 Page 9
Power on axis II:
ππΌπΌ =πππ‘
πππΏ.ππ₯=
5.7
0.995Γ0.9= 6.36 (Kw) (Kw)
Power on axis I:
ππΌ =πππ‘
πππΏ .ππ₯=
6.36
0.995Γ0.96= 6.66 (Kw)
Power on the motor axis:
πΔπ =ππΌ
πππΏ .πππ=
6.66
0.995Γ1= 6.70 (Kw)
Number of revolutions on the motor axis: πΔπ= 1000 (rev/min)
Number of revolutions I:
ππΌ =πΔπ
π’ππ=
1000
1= 1000 (rev/min)
Number of revolutions II:
ππΌπΌ =ππΌ
π’ππ=
1000
5= 200 (πππ£/πππ)
Number of revolutions on the operating axis:
πππ‘ =ππΌπΌ
π’π₯=
200
2.4= 83.33 (πππ£/πππ)
Torque on the motor axis:
πΔπ = 9,55.106.πΔπ
πΔπ= 9,55.106.
6.75
1000= 64462 (π.ππ)
Torque on the axis I:
ππΌ = 9,55.106.ππΌ
ππΌ= 9,55.106.
6.66
1000= 63603 (π.ππ)
Torque on the axis II:
ππΌπΌ = 9,55.106.ππΌπΌ
ππΌπΌ= 9,55.106 .
6.36
200= 303690 (π.ππ)
Torque on the operating axis:
πππ‘ = 9,55.106 .πππ‘
πππ‘= 9,55.106 .
5.7
83.33= 653246 (π.ππ)
MEC2-K56- Group 15 Page 10
11. Data Table:
Factors\Axis Motor I II Operating
ukn = 1 ubr = 5 ux = 2.4
P(KW) PΔc= 6.75 PI = 6.66 PII = 6.36 Pct= 5.7
n(rev/min) nΔc= 1000 nI= 1000 nII= 200 nct = 83.33
T(N.mm) TΔc= 64462 TI = 63603 TII = 303690 Tct = 653246
B.Calculating and designing outer transmission β chain driver
Data requirements:
{
P = PII = 6.36(KW)T1 = TII = 303690(N.mm)
n1 = nII = 200 (rpm)u = ux = 2.4
Ξ² = 00
1. Choosing the type of chain
We choose the brushed roller chain.
MEC2-K56- Group 15 Page 11
2. Choosing number of teeth of the sprocket
The number of teeth on driver sprocket:
Z1 = 29β 2 Γu = 29β 2Γ 2.4 = 24.2 β₯ 19, choose Z1 = 24
Therefore the driven number of teeth:
π2 = π’ Γπ1 = 2 Γ24.2 = 48.4 β€ ππππ₯ = 140, πβπππ π π2 = 48
3. Select factors
Application factor: use the chart 2 (see page 102 Renold Roller Chain Catalogue)
with driver and driven sprockets smooth running we have: 1 1f
Tooth factor: π2 =19
π1=
19
24= 0.79
Selection power = Transmitted power1 2f f
= 6.36Γ1Γ0.7 =5.02 (KW)
4. Select chain 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 choose the ANSI Simplex with the chain pitch p = 19.05(mm).
5. Installation parameters
Lubrication β American Chain Rating Chart (see page 106 Renold Roller
Chain Catalogue) clearly indicates the chain need Oil Bath type 3 lubrication.
MEC2-K56- Group 15 Page 12
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 Zp
Z Z C pL
p C
Where:
C is the contemplated center distance in mm and should generally be
between 30-50 pitches. In this case:
C = 19.05Γ40 = 762 (mm)
L = Chain length (pitches)
p = Chain pitch = 19.05(mm)
Z1= 24(teeth)
Z2= 48(teeth)
πΏ =24+48
2+2Γ762
19.05+(48β24
2Γ19.05)2Γ19.05
762= 116 (pitches)
MEC2-K56- Group 15 Page 13
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 Γ 116β 48β 24+β(2Γ 116β 48β 24)2β (48β 24)2
Ο
3.88]=
C = 758.5
7. Selection of sprocket materials
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 28 3.88
pC L Z Z L Z Z Z Z
MEC2-K56- Group 15 Page 14
8. Data of chosen chain
Following this link:
http://www.renoldchainselector.com/ChainSelector
We will choose the ANSI 80 (ISO 606) Simplex chain transmission with
conditions:
Environmental condition:
Loading Classification:
Driven Machine: smooth running, Moderate Shocks.
Service Conditions: Recommended.
Environnent Condition: Normal environnement, Indoor application.
Expected Working Life of the Chain
πππ§π¨π₯π ππ²π§ππ«π π²π»π΄
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/mmΒ²
Torque: T = 303.69 Nm Bearing Area: f = 1.78 cmΒ²
Static Force: F = 3129.9 N Weight: q = 2.8 kg/m
MEC2-K56- Group 15 Page 15
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.
MEC2-K56- Group 15 Page 16
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 chainβs durability
Durability factor:
π =πΈ
πΔππ +ππ +ππ
Where:
Q β ISO Breaking Load:
Q =31.8(KN)
Chainβs mass:
q = 2.8(kg/m).
MEC2-K56- Group 15 Page 17
F - Static force:
πΉπ‘ =1000Γπ
π£ , π€ππ‘β π£ = 2π/π
πΉπ‘ =1000Γπ
π£ =
1000Γ6.36
2= 3180(N)
Ff - Centrifugal Force:
πΉπ = π. π£2 = 2.8 Γ 22 = 11.2 (π)
F0- The tension caused by the weight of chain:
aqkFf
...81,90 ,
with:
a - Centre distance: a=762(mm)
kf - Deflection coefficient of chain: because of = 0o kf = 1
πΉ0 = 9.81Γ 1 Γ2.8 Γ 762Γ 10β3 = 20.93 (π)
Therefore:
π =Q
kΔFt+F0+Fv with kd is dynamic loading coefficient with
characteristic is Mild: kd=1
π =π
πΔπΉπ‘+πΉ0+πΉπ£=
31800
1Γ3180+20.93+11.2 = 9.9
[s] β 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.
MEC2-K56- Group 15 Page 18
MEC2-K56- Group 15 Page 19
C. Design gear transmission .
Data requirement:
{
π· = π·π° = π. ππ (π²πΎ)π» = π»π° = πππππ (π΅.ππ)
ππ = ππ° = ππππ (πππ)
π = πππ = ππ³π = ππππππ
1. Design decision
Material selection
Material: Steel C45
Select diameter pitch
From Standard Diameter Pitches (Robert L. Norton, Machine Design,
p.645, table 11-2):
Choose p = 8
Select number of teeth (with u=5)
Driving pinion: ππ = 20 π‘πππ‘β
Driven gear: 100 teeth
2. Design calculation
Pitch diameter
Pitch diameter of driving pinion: ππ =ππ
π=
20
8= 2.5 (ππ)
Pitch diameter of driven gear: dg = 100
8 = 12.5 (in)
Bending geometry factor
MEC2-K56- Group 15 Page 20
The bending geometry factors J for this combination are found in the
Robert L. Norton, Machine Design, p.666, table 11-9, for the highest point
of the single-tooth contact (HPSTC) and are approximately : {π½π = 0.35
π½πΊ = 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 :
ππ‘ =ππ
2Γ ππ =
2.5
2Γ12Γ (1000πππ)(2π) = 654.5 (ft/min)
825.04
)612(
4
)12( 3/23/2
vQ
B
8.59)825.01(5650)1(5650 BA
πΎπ£ = πΆπ£ = (π΄
π΄+ βππ‘)
π΅
= (59.8
59.8+ β654.5)
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)/(84.394332
maxftQvAVt
We can see that maxt tV V , so tV is acceptable.
Transmitted load
( )v vK C
6vQ tV
tV
MEC2-K56- Group 15 Page 21
ππ‘ =33000π»
π=33000Γ 10.32
654.5= 520.3(ππ)
Bending stresses in 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):
Fβ 12
ππ=
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:
πππ =ππ‘ππ
πΉπ½πΓπΎππΎπ
πΎπ£ΓπΎπ πΎπ΅πΎπΌ =
520.3Γ8
1.5Γ0.35Γ
1Γ1.6
0.745Γ 1Γ 1Γ 1 = 17027.40 psi
πππ =ππ‘ππ
πΉπ½πΓ
πΎππΎπ
πΎπ£Γ πΎπ πΎπ΅πΎπΌ =
520.3Γ8
1.5Γ0.43Γ
1Γ1.6
0 .745Γ 1Γ 1Γ 1 = 13859.51 psi
p
t a mb s B I
v
W p K KK K K
FJ K
F
( )a aK C 1a aK C
( )m mK C 1.6mK
2F
sK BK
MEC2-K56- Group 15 Page 22
Surface stress in the pinion gear
The surface stress in the pinion gear can be estimated by
equation 11.21 (Robert L. Norton, Machine Design, p.672):
t a mcpg p s f
p v
W C CC 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): 0.52300pC psi for steel on steel.
I is geometry factor can estimated by equation 11.22a (Robert L.
Norton, Machine Design, p.673):
cos
1 1p
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):
ππ = β(ππ +1
ππ)2
β (πππππ β )2 -
π
πππππ β
MEC2-K56- Group 15 Page 23
ππ = β(1.25 +1
8)2
β (1.25πππ 250)2 βπ
8πππ 250
ππ = 0.33 ππ
ππ = πΆπ ππβ β π1 = (ππ + ππ)π ππβ β π1 = (1.25+6.25)Γ sin 250 β 0.33=2.84
in
Therefore, geometry factor now can be estimated:
I = πππ β
(1
ππΒ±1
ππ) =
πππ 250
(1
0.33β
1
2.84)Γ2.5
= 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 mcpg p s f
p v
W C CC C C
FId C
=2300β520.3
1.5Γ0.14Γ2.5Γ1Γ1.6
0.745Γ 1 Γ 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):
MEC2-K56- Group 15 Page 24
'L
fb fb
T R
KS 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 (60πππ
βπ) Γ 11000βπ = 0.66 Γ 109(ππ¦ππes)
then calculate the life factor:
πΎπΏ = 1.3558Γ πβ0.0178 = 1.3558Γ (0.66Γ 109)β0.0178 = 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
MEC2-K56- Group 15 Page 25
Apply all these factors, now we can calculate the corrected bending-
fatigue strength:
πππ =πΎπΏπΎππΎπ
πππβ² =0.906
1Γ 31051= 28132ππ π
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 CS 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:
πππβ² = 2600+ 327π»π΅ = 2600+ 327(240) = 104480 psi
πΆπΏ = 1.4488πβ0.023 = 1.4488Γ (0.66Γ 109)β.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:
MEC2-K56- Group 15 Page 26
πππ =πΆπΏπΆπ»πΆππΆπ
πππβ² =0.908Γ 1
1 Γ 1Γ104480= 94867.84ππ π
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:
πππ =ππππππ
=28132
17027.40= 1.65
πππ =ππππππ
=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
materialβs 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:
ππΆπβπ = (πππ
ππππππππ
)
2
= (94867.84
106079.6 )2
=0.8
MEC2-K56- Group 15 Page 27
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 'fbS
31051psi
Corrected bending strength fbS
28132psi
Uncorrected surface strength 'fcS
104480psi
Corrected surface strength fcS
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.
MEC2-K56- Group 15 Page 28
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:
πͺππππ = π¨Γ (πΊππ)π = π. π Γ (ππ)βπ.πππ =0.83<1 (satisfied)
1tempC since temperature is not evaluated.
1reliabC while we assume 50% reliability at this preliminary design
stage.
MEC2-K56- Group 15 Page 29
'eS is uncorrected endurance strength and calculated using equation 6.5a
(Robert L. Norton, Machine Design, p324) for 200utS kpsi :
' 0.5 0.5 86000 43000uteS S psi
Apply all above factor, now we can calculate the corrected endurance
strength:
'e load size surf temp reliab eS C C C C C S
= 1Γ πΓπ. ππ Γπ Γπ Γπππππ = πππππ 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):
πππππ πππ = π. π
ππππππππ = π. ππ
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: {π²π = π. π π²ππ = π. π
The fatigue stress concentration factor for bending stress at section 1:
π²π = π+ π Γ (π²π βπ) = π+ π. π Γ (π. π β π) = π. π
The fatigue stress concentration factor for torsion at section 1:
π²ππ = π+ π Γ (π²ππ βπ) = π+ π. ππ Γ (π. π β π) = π. ππ
MEC2-K56- Group 15 Page 30
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: π²πππ = π²ππ= 1.68
II. Shaft 1 calculations
Given πΉππ‘ππππππ‘ππ = 520.3 (ππ)
π1 = 63603 (π.ππ)
Figure 4.1.Shaft dimensions.
We choose diameter 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
MEC2-K56- Group 15 Page 31
11 132 2 43.5 87( )l l mm
a) Analyze the forces and moments
We assume that πΉπππ’πππππ = 0.2 Γππ‘ = 0.2 Γ 520.3 = 104.06 (ππ), 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:
πΉππ‘ππππππ‘ππ = 520.3 (ππ)
The spur gear has a 250 pressure angle, which means that there will also be
a radial component of force that gear tooth of:
πΉπππππππ = πΉππ‘ππππππ‘ππ Γ tan(25π) = 520.3 Γ tan(25π) = 242.6 (ππ)
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:
MEC2-K56- Group 15 Page 32
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 FR 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 π 0 and 2R in each plane, using the
approximate components of the apply load gF and couplingF :
π 0π₯ = β0.44 Γ πΉππ₯ + 2.28 Γ πΉπππ’ππππππ₯ = β0.44 Γ 242.6 + 2.28 Γ 104.06
= 130.5 (ππ)
π 0π¦ = β0.44 Γ πΉππ¦ +2.28 Γ πΉπππ’ππππππ¦ = β0.44 Γ (β520.3)+ 2.28 Γ 0 = 228.93 (ππ)
π 2π₯ = β0.56 Γ πΉππ₯ β1.28 Γ πΉπππ’ππππππ₯ = β0.56 Γ 242.6 β 1.28 Γ 104.06
= β269.05(ππ)
π 2π¦ = β0.56 Γ πΉππ¦ β 1.28 Γ πΉπππ’ππππππ¦ = β0.56 Γ (β520.3)β 1.28 Γ 0 = 291.37 (ππ)
Moment calculations:
At point 0:
ππ₯(0) = 0(ππ. ππ)
ππ¦(0) = 0(ππ. ππ)
π (1) = 556.5(ππ. ππ)
At point 1:
ππ₯(1) = π 0π₯ Γ π1 = 130.5 Γ 3.4 = 443.7(ππ. ππ)
ππ¦(1) = π 0π¦ Γ π1 = 228.93 Γ 3.4 = 778.36(ππ. ππ)
π (1) = 556.5(ππ. ππ)
MEC2-K56- Group 15 Page 33
At point 2:
ππ₯(2) = πΉπππ’πππππ Γ π3 = 104.06Γ 1.71 = 177.94(ππ. ππ)
ππ¦(2) = π 0π¦ Γ (π1 + π2)β πΉππ¦ Γ π2 = 228.93Γ 6.1β 520.3 Γ 2.69
= 0(ππ. ππ)
π(2) = 556.5(ππ. ππ)
At point 3:
ππ₯(3) = 0(ππ. ππ)
π(3) = 556.5(ππ. ππ)
ππ¦(3) = 0(ππ. ππ)
MEC2-K56- Group 15 Page 34
Figure 4.3.Moment diagram of shaft 1.
MEC2-K56- Group 15 Page 35
b) Diameter calculation
Firstly, we choose the safety factor equal to 2.5
Apply the equation 9.6a (Robert L. Norton, Machine Design, p.512) to
determine the diameter of the shaftβs section:
1/31/2
2 2
32 3
4
f a mf fsm
f y
N M Td K K
S S
Consider the section I-0 at the point 0 of the 1st bearing:
0( . ),T 891.60(lb.in)aM lb in , the minimum recommended diameter 0 :d
π0 = {32 Γ 2.5
π[(1.9 Γ
0
35690)2
+3
4(1.68 Γ
556.5
54000)2
]
12
}
13
= 0.73(ππ)
Consider the section I-1 at the point 1 of the gear:
ππ = 895.94(ππ. ππ), π = 556.5(ππ. ππ), the minimum recommended diameter 1 :d
π1 = {32 Γ 2.5
π[(1.9 Γ
895.94
35690)2
+3
4(1.68 Γ
556.5
54000)2
]
12
}
13
= 1.08(ππ)
Consider the section I-2 at the point 2 of the 2nd bearing position:
ππ = 177.96(ππ. ππ), π = 556.5(ππ. ππ), the minimum recommended diameter 2 :d
π2 = {32 Γ 2.5
π[(1.9 Γ
177.96
35690)2
+3
4(1.68 Γ
556.5
54000)2
]
12
}
13
= 0.77(ππ)
Consider the section I-3 at the point 3 of the coupling position:
ππ = 0(ππ. ππ), π = 556.5(ππ. ππ), the minimum recommended diameter 3 :d
MEC2-K56- Group 15 Page 36
π3 = {32 Γ 2.5
π[(1.9 Γ
0
35690)2
+3
4(1.68 Γ
556.5
54000)2
]
12
}
13
= 0.73(ππ)
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 π3 =
22ππ (0.89ππ). Table 9.2 (Robert L. 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:
πΉπ =πππ=556.5
0.59= 943.22ππ
MEC2-K56- Group 15 Page 37
πΉπ =πππ=556.5
0.59= 943.22ππ
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):
ππ = ππ =πΉπ
π΄π βπππ=943.22
1 Γ 0.25= 3772.88 ππ π
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):
ππβ² = ππ
β² = βππ₯2 + ππ¦
2 βππ₯ . ππ¦ + 3ππ₯π¦2 = β3 Γ 3772.882 = 6534.8 ππ π
Then use them in equation 6.18e (Robert L. Norton, Machine Design,
p.364):
ππ =1
ππβ²
ππ+
ππβ²
ππ’π‘
=1
6534.835690 +
6534.886000
= 3.86
At point 3, the mean and alternating components of force on the key are
found from the torque component divided by the shaft radius at that point:
πΉπ = πΉπ =πππ=556.5
0.445= 1250.56ππ
Assuming a key length of 0.5 in and calculate the alternating and mean
shear stress components from:
ππ = ππ =πΉπ
π΄π βπππ=
1250.56
0.5 Γ 0.187= 13375 ππ π
MEC2-K56- Group 15 Page 38
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):
ππβ² = ππ
β² = βππ₯2 +ππ¦
2 β ππ₯ . ππ¦ +3ππ₯π¦2 = β3Γ 133752 = 23166.2 ππ π
Then use them in equation 6.18e (Robert L. Norton, Machine Design,
p.364):
ππ =1
ππβ²
ππ+
ππβ²
ππ’π‘
=1
23166.235690 +
23166.286000
= 1.09
End-mill radius versus shaft diameter ratio:
At point 1:1
0.020.017
1.18
r
d
At point 3:3
0.020.023
0.89
r
d
The corresponding stress-concentration factors are read from figure 9.16
(Robert L. Norton, Machine Design, p.529):
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
MEC2-K56- Group 15 Page 39
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
32 2
2 2
3( ) ( )
4
32
3( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SNd
k M K T
S
1.18 =
{
32 Γ πππ
[ β(1.81Γ 895.94)2 +
34(2.39 Γ 556.5)2
35690
+
β(1.81 Γ895.94)2 +34(2.39 Γ 556.5)2
86000
]
}
13
β ππ = 2.05
At point 3:
MEC2-K56- Group 15 Page 40
1
32 2
2 2
3( ) ( )
4
32
3( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SNd
k M K T
S
0.89 =
{
32 Γ πππ
[ β(1.71 Γ 0)2 +
34(2.20 Γ 556.5)2
35690
+β(1.71 Γ 0)2 +
34(2.20 Γ 556.5)2
86000]
}
13
β ππ = 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
MEC2-K56- Group 15 Page 41
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.
π1 = π2 = 43.5(ππ) = 1.71(ππ)
π3 = 68.5(ππ) = 2.69(ππ)
πΉππ‘ππππππ‘ππ = 520.3(ππ)
πΉπππππππ = 242.6(ππ)
MEC2-K56- Group 15 Page 42
πΉ = 4419.9(π) πΉπ₯ = πΉ Γ πππ 60π = 4419.9 Γ πππ 60π = 2209.95(π) = 491.1(ππ)
πΉπ¦ = πΉ Γ π ππ60π = 4419.9 Γ π ππ60π = 3827.75(π) = 850.6(ππ)
ππΌπΌ = 303690(π.ππ) = 2656.95(ππ. ππ)
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 ) 155.50.5
87 87
0
155.5( 0.5 )
87
68.50.5
87
g
g
g
g
g
g g g
g
M R l l l l l l
F F
F FR 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 :
π 4π₯ = β0.5 Γ πΉππ₯ +68.5
87Γ πΉπ₯ = β0.5 Γ 242.6 +
68.5
87Γ 491.1 = 265.37(ππ)
π 4π¦ = β0.5 Γ πΉππ¦ +68.5
87Γ πΉπ¦ = β0.5 Γ (β520.3)+
68.5
87Γ 850.6 = 929.87(ππ)
π 6π₯ = β0.5 Γ πΉππ₯ +68.5
87Γ πΉπ₯ = β0.5 Γ 242.6 β
155.5
87Γ 491.1 = β999.07(ππ)
π 6π¦ = β0.5 Γ πΉππ¦ +68.5
87Γ πΉπ¦ = β0.5 Γ (β520.3)β
155.5
87Γ 850.6 = β1260.2(ππ)
Moment calculations:
At point 4:
ππ₯(4) = 0(ππ. ππ)
ππ¦(4) = 0(ππ. ππ)
π (4) = 2656.95(ππ. ππ)
MEC2-K56- Group 15 Page 43
At point 5:
ππ₯(5) = π 4π₯ Γ π1 = 491.1 Γ 1.71 = 839.8(ππ. ππ)
ππ¦(6) = π 4π¦ Γ π1 = 850.6 Γ 1.71 = 1454.5(ππ. ππ)
π (6) = 2656.95(ππ. ππ)
At point 6:
ππ₯(6) = πΉπ₯ Γ π3 = 491.1 Γ 2.69 = 1321.1(ππ. ππ)
ππ¦(6) = 0(ππ. ππ)
π (6) = 2656.95(ππ. ππ)
At point 7:
ππ₯(7) = 0(ππ. ππ)
ππ¦(7) = 0(ππ. ππ)
π (7) = 2656.95(ππ. ππ)
MEC2-K56- Group 15 Page 44
Figure 4.5.Moment diagram of shaft 2.
MEC2-K56- Group 15 Page 45
b) Diameter calculation
Firstly, we choose the safety factor equal to 2.5
Apply the equation 9.56a (Robert L. Norton, Machine Design, p.512)
to determine the diameter of the shaftβs section:
Consider the section II-4 at the point 4 of the 1st bearing:
ππ = 0(ππ. ππ), π = 2656.95(ππ. ππ), the minimum recommended diameter
π0 = {32Γ 2.5
π[(1.9 Γ
0
35690)2
+3
4(1.68 Γ
2656.95
54000)2
]
12
}
13
= 1.22(ππ)
Consider the section II-5 at the point 5 of the gear:
ππ = 1679.5(ππ. ππ),π = 2656.95(ππ. ππ), the minimum recommended diameter
π1 = {32 Γ 2.5
π[(1.9 Γ
1679.5
35690)2
+3
4(1.68 Γ
2656.95
54000)2
]
12
}
13
= 1.43(ππ)
Consider the section II-6 at the point 6 of the 2nd bearing position:
ππ = 1321.1(ππ. ππ),π = 2656.95(ππ. ππ), the minimum recommended diameter
π2 = {32 Γ 2.5
π[(1.9 Γ
1321.1
35690)2
+3
4(1.68 Γ
2656.95
54000)2
]
12
}
13
= 1.37(ππ)
Consider the section II-7 at the point 7 position:
0( . ),T 3385.74(lb.in)aM lb in , the minimum recommended diameter
1/31/2
2 2
32 3
4
f a mf fsm
f y
N M Td K K
S S
0 :d
1 :d
2 :d
3 :d
MEC2-K56- Group 15 Page 46
π3 = {32Γ 2.5
π[(1.9 Γ
0
35690)2
+3
4(1.68 Γ
2656.95
54000)2
]
12
}
13
= 1.22(ππ)
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-
carbon steel, 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 chosen for these sections were 1 30 (1.5 )d mm in and
3 20 (0.72 )d mm in . Table 9.2 (Robert L. 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 component divided by the shaft radius at that point:
86 , 54ut yS kpsi S kpsi 35690eS psi
MEC2-K56- Group 15 Page 47
πΉπ =πππ=2656.95
0.885= 3002.2(ππ)
πΉπ =πππ=2656.95
0.885= 3002.2(ππ)
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):
ππ = ππ =πΉπ
π΄π βπππ=2656.95
1 Γ 0.5= 5313.9 ππ π
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):
ππβ² = ππ
β² = βππ₯2 +ππ¦
2 β ππ₯ . ππ¦ +3ππ₯π¦2 = β3Γ 5313.92 = 9204 ππ π
Then use them in equation 6.18e (Robert L. Norton, Machine Design,
p.364):
ππ =1
ππβ²
ππ+
ππβ²
ππ’π‘
=1
920435690+
920486000
= 2.74
At point 7, the mean and alternating components of force on the key are
found from the torque component divided by the shaft radius at that point:
πΉπ = πΉπ =πππ=2656.95
0.685= 3878.76ππ
Assuming a key length of 1 in and calculate the alternating and mean shear
stress components from:
ππ = ππ =πΉπ
π΄π βπππ=
3878.76
1 Γ 0.312= 12431.92 ππ π
MEC2-K56- Group 15 Page 48
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):
ππβ² = ππ
β² = βππ₯2 +ππ¦
2 β ππ₯ . ππ¦ +3ππ₯π¦2 = β3Γ 12431.922 = 21532.72 ππ π
Then use them in equation 6.18e (Robert L. Norton, Machine Design,
p.364):
ππ =1
ππβ²
ππ+
ππβ²
ππ’π‘
=1
21532.7235690 +
21532.7286000
= 1.17
End-mill radius versus shaft diameter ratio:
At point 5:1
0.020.005
3.93
r
d
At point 7:3
0.020.005
3.54
r
d
The corresponding stress-concentration factors are read from figure 9.16
(Robert L. Norton, Machine Design, p.529):
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
MEC2-K56- Group 15 Page 49
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
32 2
2 2
2 2
2 2
3( ) ( )
4
32
3( ) ( )
4
3(2.95 2150.7) (2.95 3385.74)
432 35690
1.773
(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
SNd
k M K T
S
N
1
3
3.32fN
1
32 2
2 2
3( ) ( )
4
32
3( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SNd
k M K T
S
f fm
fsm fs
K K
K K
MEC2-K56- Group 15 Page 50
1.77 =
{
32 Γ πππ
[ β(2.95 Γ 1679.5)2 +
34(2.95 Γ 2656.95)2
35690
+
β(2.95 Γ 1679.5)2 +34(2.95 Γ 2656.95)2
86000
]
}
13
β ππ = 1.63
At point 7:
1
32 2
2 2
3( ) ( )
4
32
3( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SNd
k M K T
S
1
32 2
2 2
3(2.95 0) (2.95 3385.74)
432 35690
1.373
(2.95 0) (2.95 3385.74)4
86000
fN
2.52fN
1
32 2
2 2
3( ) ( )
4
32
3( ) ( )
4
f a fs a
ff
fm m fsm m
ut
k M k T
SNd
k M K T
S
MEC2-K56- Group 15 Page 51
1.37 =
{
32 Γ πππ
[ β(2.95Γ 0)2 +
34(2.95 Γ 2656.95)2
35690
+
β(2.95 Γ 0)2 +34(2.95 Γ 2656.95)2
86000
]
}
13
β ππ = 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
MEC2-K56- Group 15 Page 52
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 bearingβs 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 357000.62 704545( )
86.94p p
CL K revs
P
For the reaction load of 70.61 lb at 2nd bearing:
10 10
3 357000.62 1409571( )
70.61p p
CL 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
bearingβs 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
MEC2-K56- Group 15 Page 53
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 bearingβs 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 3106000.62 228772( )
226.57p p
CL K revs
P
For the reaction load of 358.31 lb at 2nd bearing:
10 10
3 310600
358.310.62 49645( )p p
CL 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
bearingβs 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.
MEC2-K56- Group 15 Page 54
F. Gear boxβs 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 shaftβs 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
MEC2-K56- Group 15 Page 55
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
/ 2 with 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 gearβs 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
MEC2-K56- Group 15 Page 56
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 transmission ratio are
nearly similar to English
Document
-Vietnamese and English
documents use the same input data
-work power
.
1000lv
F vP
-The same Efficiency of gear
train,chain gear,bearing,coupling
-the similar way to calculate the
necessary power or the revolutions
-According to Vietnamese
Document,Choosing motor can be
from 200rpm to 4000rpm
-the distributing transmission ratio
and torque on the axis is calculated
in the same way for Vietnamese
and English document
- Using βTinh Toan He Dan Dong
Co Khiβ to choose the motor.
In part A: Choosing motor and
distributing transmission ratio are
nearly similar to the Vietnamese
Document.
-Input data: Pulling force of
conveyor chain, conveyor speed
chainβ¦
-work power
.
1000lv
F vP
-The necessary power lvyc
PP
-the revolutions 60000.
.ct
vn
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 choose motor.
In part B: Calculating and designing
outer transmission β chain drives
In part B: Calculating and designing
MEC2-K56- Group 15 Page 57
- 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 β chain drives
-English version: Ux=2.4 =>
choose:
Z1=24
Z2=48 -kz βTooth factor
-kn β Rotating factor
-Using selection factor
F1,F2,F3,F4,F5,F6
-English document use Tooth factor
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 both torque 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
ratioβ¦
-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
MEC2-K56- Group 15 Page 58
power to calculate
-Vietnamese have testing for gear
transmission
-in Vietnamese document,we
calculate for diameter of teeth
distinctly with gear
factor,overload factorβ¦
-using only transmission power to
calculate
-in English document,we adjust the
contact stress to check
-Englisht documents donβt calculate
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 motorβs 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 shaftβ
s section.
13
0,1[ ]
tdMd
-in Vietnamese document,the shaft
In part D. Shafts, keys and couplings
calculation
- dcT , T , IIT β¦
-At first,the material is choosen by
using working charact
-
300
10,2
ITd
- 13ml , 12l , 11l β¦
- couplingF
-according to norton book
3
1
2
122
..4
3.
32
y
mfsm
f
af
f
S
Tk
S
MK
Nd
-in English Document,the shaft
diameter is smaller because we chose
material which has higher hardness
than Vietnamese Document
MEC2-K56- Group 15 Page 59
diameter is greater than English
Document
-in Vietnamese document using
safety factor and stress
-in english document,Testing for
shaft using Torque,force and safety
factor
E.Bearing and lubricant,Designβs
box cover,the oil diptick
E.Bearing and lubricant,Designβs
box cover,the oil diptick
I. Data Table
a. English document
Factors\Axis Motor I II Operating
ukn = 1 ubr = 5 ux = 2.4
P(KW) PΔc= 6.75 PI = 6.66 PII = 6.36 Pct= 5.7
n(rev/min) nΔc= 1000 nI= 1000 nII= 200 nct = 83.33
T(N.mm) TΔc= 64462 TI = 63603 TII = 303690 Tct = 653246
b. Vietnamese document
ThΓ΄ng
sα»\Trα»₯c
Δα»ng cΖ‘ I II CΓ΄ng tΓ‘c
ukn = 0.99 ubr = 4 ux = 2.19
P(KW) PΔc=4.4 PI=4.33 PII=4.09 Pct=3.74
n(v/ph) nΔc=716 nI=716 nII=179 nct=327
MEC2-K56- Group 15 Page 60
T(N.mm) TΔc=58687 TI=57753 TII=218209 Tct=109225
we can see the different about the way we choose ubr and ux, that make the
different design.
II. Calculating and designing outer transmission β 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 ππ 8
Pressure angle 25o
MEC2-K56- Group 15 Page 61
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 'fbS 31051psi
Corrected bending strength 'fbS 28132psi
Uncorrected surface strength 'fcS 104480psi
Corrected surface strength fc
S 94867.84psi
MEC2-K56- Group 15 Page 62
Bending safety factor for
pinion bpN 1.65
Bending safety factor for gear bgN 2.03
Surface safety factor for mesh cpN 0.8
MEC2-K56- Group 15 Page 63
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)
MEC2-K56- Group 15 Page 64
db2 223.18(mm)
Profin angle
x1 0
Profin Gearβs 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)
MEC2-K56- Group 15 Page 65
Diameter of the shaftβ s section.
-English version : Using this equation to calculate
3
1
2
122
..4
3.
32
y
mfsm
f
af
f
S
Tk
S
MK
Nd
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
MEC2-K56- Group 15 Page 66
-Vietnamese version : Using this equation to calculate
3
0.1[ ]
tdj
j
Md
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
MEC2-K56- Group 15 Page 67
References:
1. Trα»nhChαΊ₯t, LΓͺVΔnUyα»n β TΓnhtoΓ‘nthiαΊΏtkαΊΏhα»dαΊ«nΔα»ngcΖ‘khΓ, tαΊp 1. NXB GiΓ‘odα»₯c,
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.