machine design-2

MEC2-K56- Group 15

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)

MEC2-K56- Group 15

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 not defined.

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 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 chian’s durability .............Error! Bookmark not defined.

C. Design gear transmission ...........................Error! Bookmark not defined.

MEC2-K56- Group 15

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 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 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...............................………………...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 not defined.

2. Shaft 2 ............................................Error! Bookmark not defined.

F. Gear box's cover…………………………………………………………...49

MEC2-K56- Group 15

G. Comparision of English vs Vietnamese document………………………

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

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 𝛽 = 0 (degree)

8. Working characteristic: Mild impact

Shaft for calculating: Input shaft 1

MEC2-K56- Group 15

MEC2-K56- Group 15

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 : ℎ𝑏𝑟 = 0.96 Chain performance : ℎ𝑥 = 0.9

Bearing performance: ℎ𝑂𝐿 = 0.995

Jointing performance: ℎ𝑘𝑛 = 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:

𝑛𝑐𝑡 =60000.𝑣

𝑧. 𝑝=60001.36

1470= 83.26 (rev/min)

5. Choosing the first-aid transmission ratio (𝒏𝒔𝒃):

𝒖𝒔𝒃 = 𝒖𝒙 . 𝒖𝒃𝒓

Follow table B2.4

21[1]: Chain ration : 𝒖𝒙 =2 ÷ 5

Gear ration : 𝒖𝒃𝒓 = 3 ÷ 5

Then:

𝒖𝒔𝒃 𝒎𝒊𝒏= 𝒖𝒙 𝒎𝒊𝒏 . 𝒖𝒃𝒓 𝒎𝒊𝒏=23= 6

𝒖𝒔𝒃 𝒎𝒂𝒙= 𝒖𝒙 𝒎𝒂𝒙 .𝒖𝒃𝒓 𝒎𝒂𝒙=55= 25

6. First-aid number of revolutions on motor axis

MEC2-K56- Group 15

We have : 𝒏𝒔𝒃 = 𝒖𝒔𝒃. 𝒏𝒄𝒕

𝒏𝒔𝒃 𝒎𝒊𝒏 = 𝒖𝒔𝒃 𝒎𝒊𝒏. 𝒏𝒄𝒕= 500 (rev/min)

𝒏𝒔𝒃 𝒎𝒂𝒙 = 𝒖𝒔𝒃 𝒎𝒂𝒙 . 𝒏𝒄𝒕= 2081 (rev/min)

We choose 𝒏𝒔𝒃 =1000 (rev/min)

7. Calculating synchronous number of revolutions of motor :

Choosing 𝒏𝒅𝒃𝒕 = 𝟕𝟓𝟎 (rev/min)

8. Choosing the motor:

Choose the motor which satisfies: {𝑛đ𝑏 ≈ 𝑛𝑠𝑏 = 1000 (rev/min)

𝑃đ𝑐 ≥ 𝑃𝑦𝑐 = 6.75 (𝐾𝑊)

We have the motor with the details:

Siemen 1LA7 134-6AA { 𝑛𝑑𝑐 = 1000 (rev/min)

𝑝 = 7 𝐾𝑊

9. Distributing transmission ratio:

Transmission ratio of the system:

𝒖𝒄𝒉 =𝒏đ𝒄

𝒏𝒄𝒕=

𝟏𝟎𝟎𝟎

𝟖𝟑.𝟐𝟔=12

Choose the transmission ratio of 1st gear box: 𝑢𝑏𝑟 =5

Transmission ratio of outer driven: 𝑢𝑥=𝒖𝒄𝒉

𝒖𝒃𝒓=

𝟏𝟐

𝟓≈ 𝟐. 𝟒

So, we have: {

𝑢𝑐ℎ = 12𝑢𝑏𝑟 = 5𝑢𝑥 = 2.4

10. Calculating the factors on the axis:

Power on operating axis: 𝑃𝑐𝑡 = 𝑃𝑙𝑣 = 5.7 (Kw)

MEC2-K56- Group 15

Power on axis II:

𝑃𝐼𝐼 =𝑃𝑐𝑡

𝑛𝑂𝐿.𝑛𝑥=

5.7

0.995×0.9= 6.36 (Kw) (Kw)

Power on axis I:

𝑃𝐼 =𝑃𝑐𝑡

𝑛𝑂𝐿 .𝑛𝑥=

6.36

0.995×0.96= 6.66 (Kw)

Power on the motor axis:

𝑃đ𝑐 =𝑃𝐼

𝑛𝑂𝐿 .𝑛𝑘𝑛=

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

𝑢𝑘𝑛=

1000

1= 1000 (rev/min)

Number of revolutions II:

𝑛𝐼𝐼 =𝑛𝐼

𝑢𝑏𝑟=

1000

5= 200 (𝑟𝑒𝑣/𝑚𝑖𝑛)

Number of revolutions on the operating axis:

𝑛𝑐𝑡 =𝑛𝐼𝐼

𝑢𝑥=

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.𝑃𝐼

𝑛𝐼= 9,55.106.

6.66

1000= 63603 (𝑁.𝑚𝑚)

Torque on the axis II:

𝑇𝐼𝐼 = 9,55.106.𝑃𝐼𝐼

𝑛𝐼𝐼= 9,55.106 .

6.36

200= 303690 (𝑁.𝑚𝑚)

Torque on the operating axis:

𝑇𝑐𝑡 = 9,55.106 .𝑃𝑐𝑡

𝑛𝑐𝑡= 9,55.106 .

5.7

83.33= 653246 (𝑁.𝑚𝑚)

MEC2-K56- Group 15

11. Data Table:

Factors\Axis 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 choose the brushed roller chain.

MEC2-K56- Group 15

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 ≤ 𝑍𝑚𝑎𝑥 = 140, 𝑐ℎ𝑜𝑜𝑠𝑒 𝑍2 = 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: 𝑓2 =19

𝑍1=

19

24= 0.79

Selection power = Transmitted power1 2f f

= 6.36×1×0.7 =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 – 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

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 = 19.05×40 = 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)

MEC2-K56- Group 15

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. 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

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

𝐑𝐞𝐧𝐨𝐥𝐝 𝐒𝐲𝐧𝐞𝐫𝐠𝐲𝑻𝑴

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


MEC2-K56- Group 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

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

𝒌đ𝑭𝒕 +𝑭𝟎 +𝑭𝒗

Where:

Q – ISO Breaking Load:

Q =31.8(KN)

Chain’s mass:

q = 2.8(kg/m).

MEC2-K56- Group 15

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:

aqkFf

...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

𝑘đ𝐹𝑡+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.

MEC2-K56- Group 15

MEC2-K56- Group 15

C. Design gear transmission .

Data requirement:

{

𝑷 = 𝑷𝑰 = 𝟔. 𝟔𝟔 (𝑲𝑾)𝑻 = 𝑻𝑰 = 𝟔𝟑𝟔𝟎𝟑 (𝑵.𝒎𝒎)

𝒏𝟏 = 𝒏𝑰 = 𝟏𝟎𝟎𝟎 (𝒓𝒑𝒎)

𝒖 = 𝒖𝒃𝒓 = 𝟓𝑳𝒉 = 𝟏𝟏𝟎𝟎𝟎𝒉

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: 𝑁𝑝 = 20 𝑡𝑒𝑒𝑡ℎ

Driven gear: 100 teeth

2. Design calculation

Pitch diameter

Pitch diameter of driving pinion: 𝑑𝑝 =𝑁𝑝

𝑝=

20

8= 2.5 (𝑖𝑛)

Pitch diameter of driven gear: dg = 100

8 = 12.5 (in)

Bending geometry factor

MEC2-K56- Group 15

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 : {𝐽𝑝 = 0.35

𝐽𝐺 = 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 =𝑑𝑝

2× 𝜔𝑝 =

2.5

2×12× (1000𝑟𝑝𝑚)(2𝜋) = 654.5 (ft/min)

825.04

)612(

4

)12( 3/23/2

vQ

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)/(84.394332

maxftQvAVt

We can see that maxt tV V , so tV is acceptable.

Transmitted load

( )v vK C

6vQ tV

tV

MEC2-K56- Group 15

𝑊𝑡 =33000𝐻

𝑉=33000× 10.32

654.5= 520.3(𝑙𝑏)

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):

F≅ 12

𝑃𝑑=

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:

𝜎𝑏𝑝 =𝑊𝑡𝑝𝑑

𝐹𝐽𝑝×𝐾𝑎𝐾𝑚

𝐾𝑣×𝐾𝑠𝐾𝐵𝐾𝐼 =

520.3×8

1.5×0.35×

1×1.6

0.745× 1× 1× 1 = 17027.40 psi

𝜎𝑏𝑔 =𝑊𝑡𝑝𝑑

𝐹𝐽𝑔×

𝐾𝑎𝐾𝑚

𝐾𝑣× 𝐾𝑠𝐾𝐵𝐾𝐼 =

520.3×8

1.5×0.43×

1×1.6

0 .745× 1× 1× 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

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):

𝜌𝑝 = √(𝑟𝑝 +1

𝑃𝑑)2

− (𝑟𝑝𝑐𝑜𝑠∅)2 -

𝜋

𝑃𝑑𝑐𝑜𝑠∅

MEC2-K56- Group 15

𝜌𝑝 = √(1.25 +1

8)2

− (1.25𝑐𝑜𝑠250)2 −𝜋

8𝑐𝑜𝑠250

𝜌𝑝 = 0.33 𝑖𝑛

𝜌𝑔 = 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

𝜌𝑝±1

𝑝𝑔) =

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 mcpg p s f

p v

W C CC 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):

MEC2-K56- Group 15

'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 (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

MEC2-K56- Group 15

Apply all these factors, now we can calculate the corrected bending-

fatigue strength:

𝑆𝑓𝑏 =𝐾𝐿𝐾𝑇𝐾𝑅

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 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:

𝑆𝑓𝑐′ = 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:

MEC2-K56- Group 15

𝑆𝑓𝑐 =𝐶𝐿𝐶𝐻𝐶𝑇𝐶𝑅

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:

𝑁𝑏𝑝 =𝑆𝑓𝑏𝜎𝑏𝑝

=28132

17027.40= 1.65

𝑁𝑏𝑔 =𝑆𝑓𝑏𝜎𝑏𝑔

=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:

𝑁𝐶𝑝−𝑔 = (𝑆𝑓𝑐

𝜎𝑐𝑝𝑖𝑛𝑖𝑜𝑛

)

2

= (94867.84

106079.6 )2

=0.8

MEC2-K56- Group 15

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

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:

𝑪𝒔𝒖𝒓𝒇 = 𝑨× (𝑺𝒖𝒕)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.

MEC2-K56- Group 15

'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

= 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):

𝒒𝒃𝒆𝒏𝒅𝒊𝒏𝒈 = 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: {𝑲𝒕 = 𝟐. 𝟓 𝑲𝒇𝒔 = 𝟏. 𝟖

The fatigue stress concentration factor for bending stress at section 1:

𝑲𝒇 = 𝟏+ 𝒒 × (𝑲𝒕 −𝟏) = 𝟏+ 𝟎. 𝟔 × (𝟐. 𝟓 − 𝟏) = 𝟏. 𝟗

The fatigue stress concentration factor for torsion at section 1:

𝑲𝒇𝒔 = 𝟏+ 𝒒 × (𝑲𝒕𝒔 −𝟏) = 𝟏+ 𝟎. 𝟖𝟓 × (𝟏. 𝟖 − 𝟏) = 𝟏. 𝟔𝟖

MEC2-K56- Group 15

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: 𝑲𝒇𝒎𝒔 = 𝑲𝒇𝒔= 1.68

II. Shaft 1 calculations

Given 𝐹𝑔𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙 = 520.3 (𝑙𝑏)

𝑇1 = 63603 (𝑁.𝑚𝑚)

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

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:

𝐹𝑔𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙 = 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:

𝐹𝑔𝑟𝑎𝑑𝑖𝑎𝑙 = 𝐹𝑔𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙 × tan(25𝑜) = 520.3 × tan(25𝑜) = 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:

MEC2-K56- Group 15

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 𝑅0 and 2R in each plane, using the

approximate components of the apply load gF and couplingF :

𝑅0𝑥 = −0.44 × 𝐹𝑔𝑥 + 2.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔𝑥 = −0.44 × 242.6 + 2.28 × 104.06

= 130.5 (𝑙𝑏)

𝑅0𝑦 = −0.44 × 𝐹𝑔𝑦 +2.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔𝑦 = −0.44 × (−520.3)+ 2.28 × 0 = 228.93 (𝑙𝑏)

𝑅2𝑥 = −0.56 × 𝐹𝑔𝑥 −1.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔𝑥 = −0.56 × 242.6 − 1.28 × 104.06

= −269.05(𝑙𝑏)

𝑅2𝑦 = −0.56 × 𝐹𝑔𝑦 − 1.28 × 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔𝑦 = −0.56 × (−520.3)− 1.28 × 0 = 291.37 (𝑙𝑏)

Moment calculations:

At point 0:

𝑀𝑥(0) = 0(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(0) = 0(𝑙𝑏. 𝑖𝑛)

𝑇 (1) = 556.5(𝑙𝑏. 𝑖𝑛)

At point 1:

𝑀𝑥(1) = 𝑅0𝑥 × 𝑙1 = 130.5 × 3.4 = 443.7(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(1) = 𝑅0𝑦 × 𝑙1 = 228.93 × 3.4 = 778.36(𝑙𝑏. 𝑖𝑛)

𝑇 (1) = 556.5(𝑙𝑏. 𝑖𝑛)

MEC2-K56- Group 15

At point 2:

𝑀𝑥(2) = 𝐹𝑐𝑜𝑢𝑝𝑙𝑖𝑛𝑔 × 𝑙3 = 104.06× 1.71 = 177.94(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(2) = 𝑅0𝑦 × (𝑙1 + 𝑙2)− 𝐹𝑔𝑦 × 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)

𝑀𝑦(3) = 0(𝑙𝑏. 𝑖𝑛)

MEC2-K56- Group 15

Figure 4.3.Moment diagram of shaft 1.

MEC2-K56- Group 15

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 shaft’s 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

𝑑0 = {32 × 2.5

𝜋[(1.9 ×

0

35690)2

+3

4(1.68 ×

556.5

54000)2

]

12

}

13

= 0.73(𝑖𝑛)

Consider the section I-1 at the point 1 of the gear:

𝑀𝑎 = 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

]

12

}

13

= 1.08(𝑖𝑛)

Consider the section I-2 at the point 2 of the 2nd bearing position:

𝑀𝑎 = 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

]

12

}

13

= 0.77(𝑖𝑛)

Consider the section I-3 at the point 3 of the coupling position:

𝑀𝑎 = 0(𝑙𝑏. 𝑖𝑛), 𝑇 = 556.5(𝑙𝑏. 𝑖𝑛), the minimum recommended diameter 3 :d

MEC2-K56- Group 15

𝑑3 = {32 × 2.5

𝜋[(1.9 ×

0

35690)2

+3

4(1.68 ×

556.5

54000)2

]

12

}

13

= 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 (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:

𝐹𝑎 =𝑇𝑎𝑟=556.5

0.59= 943.22𝑙𝑏

MEC2-K56- Group 15

𝐹𝑚 =𝑇𝑚𝑟=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):

𝜏𝑎 = 𝜏𝑚 =𝐹𝑎

𝐴𝑠ℎ𝑒𝑎𝑟=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′ = 𝜎𝑚

′ = √𝜎𝑥2 + 𝜎𝑦

2 −𝜎𝑥 . 𝜎𝑦 + 3𝜏𝑥𝑦2 = √3 × 3772.882 = 6534.8 𝑝𝑠𝑖

Then use them in equation 6.18e (Robert L. Norton, Machine Design,

p.364):

𝑁𝑓 =1

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑢𝑡

=1

6534.835690 +

6534.886000

= 3.86



𝐹𝑎 = 𝐹𝑚 =𝑇𝑎𝑟=556.5

0.445= 1250.56𝑙𝑏

Assuming a key length of 0.5 in and calculate the alternating and mean

shear stress components from:


𝐴𝑠ℎ𝑒𝑎𝑟=

1250.56

0.5 × 0.187= 13375 𝑝𝑠𝑖

MEC2-K56- Group 15





′ = √𝜎𝑥2 +𝜎𝑦

2 − 𝜎𝑥 . 𝜎𝑦 +3𝜏𝑥𝑦2 = √3× 133752 = 23166.2 𝑝𝑠𝑖


p.364):

𝑁𝑓 =1

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑢𝑡

=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


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

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 × 𝑁𝑓𝜋

[ √(1.81× 895.94)2 +

34(2.39 × 556.5)2

35690

+

√(1.81 ×895.94)2 +34(2.39 × 556.5)2

86000

]

}

13

→ 𝑁𝑓 = 2.05

At point 3:

MEC2-K56- Group 15

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 × 𝑁𝑓𝜋

[ √(1.71 × 0)2 +

34(2.20 × 556.5)2

35690

+√(1.71 × 0)2 +

34(2.20 × 556.5)2

86000]

}

13

→ 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

MEC2-K56- Group 15

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)

𝐹𝑔𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑎𝑙 = 520.3(𝑙𝑏)

𝐹𝑔𝑟𝑎𝑑𝑖𝑎𝑙 = 242.6(𝑙𝑏)

MEC2-K56- Group 15

𝐹 = 4419.9(𝑁) 𝐹𝑥 = 𝐹 × 𝑐𝑜𝑠60𝑜 = 4419.9 × 𝑐𝑜𝑠60𝑜 = 2209.95(𝑁) = 491.1(𝑙𝑏)

𝐹𝑦 = 𝐹 × 𝑠𝑖𝑛60𝑜 = 4419.9 × 𝑠𝑖𝑛60𝑜 = 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 ) 155.50.5

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 :

𝑅4𝑥 = −0.5 × 𝐹𝑔𝑥 +68.5

87× 𝐹𝑥 = −0.5 × 242.6 +

68.5

87× 491.1 = 265.37(𝑙𝑏)

𝑅4𝑦 = −0.5 × 𝐹𝑔𝑦 +68.5

87× 𝐹𝑦 = −0.5 × (−520.3)+

68.5

87× 850.6 = 929.87(𝑙𝑏)

𝑅6𝑥 = −0.5 × 𝐹𝑔𝑥 +68.5

87× 𝐹𝑥 = −0.5 × 242.6 −

155.5

87× 491.1 = −999.07(𝑙𝑏)

𝑅6𝑦 = −0.5 × 𝐹𝑔𝑦 +68.5

87× 𝐹𝑦 = −0.5 × (−520.3)−

155.5

87× 850.6 = −1260.2(𝑙𝑏)

Moment calculations:

At point 4:

𝑀𝑥(4) = 0(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(4) = 0(𝑙𝑏. 𝑖𝑛)

𝑇 (4) = 2656.95(𝑙𝑏. 𝑖𝑛)

MEC2-K56- Group 15

At point 5:

𝑀𝑥(5) = 𝑅4𝑥 × 𝑙1 = 491.1 × 1.71 = 839.8(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(6) = 𝑅4𝑦 × 𝑙1 = 850.6 × 1.71 = 1454.5(𝑙𝑏. 𝑖𝑛)

𝑇 (6) = 2656.95(𝑙𝑏. 𝑖𝑛)

At point 6:

𝑀𝑥(6) = 𝐹𝑥 × 𝑙3 = 491.1 × 2.69 = 1321.1(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(6) = 0(𝑙𝑏. 𝑖𝑛)

𝑇 (6) = 2656.95(𝑙𝑏. 𝑖𝑛)

At point 7:

𝑀𝑥(7) = 0(𝑙𝑏. 𝑖𝑛)

𝑀𝑦(7) = 0(𝑙𝑏. 𝑖𝑛)

𝑇 (7) = 2656.95(𝑙𝑏. 𝑖𝑛)

MEC2-K56- Group 15

Figure 4.5.Moment diagram of shaft 2.

MEC2-K56- Group 15

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 shaft’s section:

Consider the section II-4 at the point 4 of the 1st bearing:

𝑀𝑎 = 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

]

12

}

13

= 1.22(𝑖𝑛)

Consider the section II-5 at the point 5 of the gear:

𝑀𝑎 = 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

]

12

}

13

= 1.43(𝑖𝑛)

Consider the section II-6 at the point 6 of the 2nd bearing position:

𝑀𝑎 = 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

]

12

}

13

= 1.37(𝑖𝑛)

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

𝑑3 = {32× 2.5

𝜋[(1.9 ×

0

35690)2

+3

4(1.68 ×

2656.95

54000)2

]

12

}

13

= 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-

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

𝐹𝑎 =𝑇𝑎𝑟=2656.95

0.885= 3002.2(𝑙𝑏)

𝐹𝑚 =𝑇𝑚𝑟=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):


𝐴𝑠ℎ𝑒𝑎𝑟=2656.95

1 × 0.5= 5313.9 𝑝𝑠𝑖





′ = √𝜎𝑥2 +𝜎𝑦

2 − 𝜎𝑥 . 𝜎𝑦 +3𝜏𝑥𝑦2 = √3× 5313.92 = 9204 𝑝𝑠𝑖


p.364):

𝑁𝑓 =1

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑢𝑡

=1

920435690+

920486000

= 2.74



𝐹𝑎 = 𝐹𝑚 =𝑇𝑎𝑟=2656.95

0.685= 3878.76𝑙𝑏

Assuming a key length of 1 in and calculate the alternating and mean shear

stress components from:


𝐴𝑠ℎ𝑒𝑎𝑟=

3878.76

1 × 0.312= 12431.92 𝑝𝑠𝑖

MEC2-K56- Group 15





′ = √𝜎𝑥2 +𝜎𝑦

2 − 𝜎𝑥 . 𝜎𝑦 +3𝜏𝑥𝑦2 = √3× 12431.922 = 21532.72 𝑝𝑠𝑖


p.364):

𝑁𝑓 =1

𝜎𝑎′

𝑆𝑒+

𝜎𝑚′

𝑆𝑢𝑡

=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


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

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

1.77 =

{

32 × 𝑁𝑓𝜋

[ √(2.95 × 1679.5)2 +

34(2.95 × 2656.95)2

35690

+

√(2.95 × 1679.5)2 +34(2.95 × 2656.95)2

86000

]

}

13

→ 𝑁𝑓 = 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

1.37 =

{

32 × 𝑁𝑓𝜋

[ √(2.95× 0)2 +

34(2.95 × 2656.95)2

35690

+

√(2.95 × 0)2 +34(2.95 × 2656.95)2

86000

]

}

13

→ 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

MEC2-K56- Group 15

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 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

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

MEC2-K56- Group 15

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 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

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.

MEC2-K56- Group 15

F. Gear box’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

MEC2-K56- Group 15

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 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

MEC2-K56- Group 15

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 “Tinh Toan He Dan Dong

Co Khi” 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

chain…

-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 – chain drives

In part B: Calculating and designing

MEC2-K56- Group 15

- 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 –Tooth factor

-kn – 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

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

MEC2-K56- Group 15

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’t 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 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.

13

0,1[ ]

tdMd

-in Vietnamese document,the shaft

In part D. Shafts, keys and couplings

calculation

- dcT , T , IIT …

-At first,the material is choosen by

using working charact

-

300

10,2

ITd

- 13ml , 12l , 11l …

- 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

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,Design’s

box cover,the oil diptick

E.Bearing and lubricant,Design’s

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) 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

số\Trục

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

MEC2-K56- Group 15

T(N.mm) Tđc=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 – 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 𝑑𝑝 8

Pressure angle 25o

MEC2-K56- Group 15

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

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

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

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)

MEC2-K56- Group 15

Diameter of the shaft’ 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

-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

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

Design

uncorrected shaft

shaft dimensions29figure

shaft ii40figure

shaft i30figure

shaft i33figure

input shaft

shaft ii43project

chain performance