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DEPARTMENT OF EDUCATION AND TRAINNING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY
AND EDUCATION

THEORY OF MACHINE AND MACHINE DESIGN

CHAPTER 10: SHAFT

20-Nov-23

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Outcome

1. Getting the overview of shaft

2. Understanding the
mechanical properties of shaft

3. Calculating and design shaft

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3

Theorical contents

I. Overview

II. Basic criteria of calculation

III. Fundamentals of calculation
and design

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4 Rotating component 11/20/2023
Pulley, gear, sprocket,… 2
Overview assemble with shaft

Coupling component
- Shaft (Trục)
- Bearing (Ổ trục)
- Couplings (Khớp nối)

Most of mechanical machinery and equipment contain shaft

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Overview

1. Application

a. Supporting component (bending)
+ Shaft with a neutral wheel
+ Shaft assemble with neutral gear

b. Transmitting torque (twist)

+ Driving shaft
+ Transmitting shaft

Almost shafts are used to support and transmit torque

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Overview
2. Classification

a. Based on load
+ Axles (Trục tâm) are subjected only to bending
+ Shaft (Trục truyền) are intended not only to support revolving parts but also to
transmit torque

Axles Shaft

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Overview 11/20/2023
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2. Classification

a. Based on load
+ Parallel drive (trục truyền chung) are applied to transmit torque to multi-component
at the same time

Parallel drive

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Overview
2. Classification

b. Centerline (đường tâm)
+ Straight shaft (Trục thẳng)+ Crank shaft (Trục khuỷu)+ Flexible wire shaft (Trục mềm)

Crank shaft

Straight shaft

Flexible wire shaft
(Trục mềm)

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Overview
2. Classification

c. Structure (Kết Cấu )
+ Plain shaft: constant diameter (Trục trơn)
+ Step shaft: Changing diameter (Trục bậc)

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Overview 11/20/2023
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2. Classification

d. Cross-section (Tiết diện)

+ Solid shaft (Trục đặc)
+ Hollow shaft ( Trục rỗng)

Solid shaft Hollow shaft

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11 Overview

3. Shaft material

Technical requirement:
 Reducing the stress concentration (giảm tập trung ứng suất)

 Manufacturing and Maintenance Technology (có tính cơng nghệ)

Common material :  Carbon steel (thép carbon)
 Alloyed steel (thép hợp kim)

* Common carbon steel:
CT3, CT5, C30, C40, C45, C50
C45 has 0.45% carbon the most common steel in the world

* Common alloyed steel:
40CrNi, 40CrNi2MoA, 30CrMnTi, 30CrMnSiA
20Cr, 12CrNi3A, 18CrMnTi

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12 Overview


4. Structure

a. Journal (ngõng trục): Using to assemble bearings

b. The clamped parts (thân trục): Used to mount
the rotating component

c. Transitions location: Between 2 different diameter on
shaft

p
t

d. Another surface: Key sitting (rãnh then), fillet

(góc lượn),… d

Requirements of shaft structure

+ Strength: Satisfying the technical requirements

+ Technology: Ability of processing disassemble, assemble and
maintenance

+ Reduce stress concentration : Fillet suitable to structure

= 0.02 ÷ 0.04; ≈3( ) 20-Nov-23

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13 Overview 11/20/2023
5. Pattern failures and Design method 5
a. Break (gãy)

Overload or fatigue
Strength
validation

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14 Overview
5. Pattern failures and Design method
b. Deformation (biến dạng)

Not enough to rigidity (Không đủ độ cứng)
rigidity

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15 Overview
5. Pattern failures and Design method
c. Vibration (dao động)

Rotating Component is off centerline (lệch tâm)

Vibration

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16 11/20/2023
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Theorical contents

I. Overview

II. Basic criteria of calculation

III. Fundamentals of calculation
and design

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17 Basic criteria calculation of shaft

1. Calculating strength
Sequence of calculation and design

a. Preliminarily calculation (Tính sơ bộ)
- No shaft structure

- According to torque (twist only) - (chỉ tính momen xoắn)

- Calculating diameter preliminarily
- (Tính đường kính sơ bộ)

- Selecting bearing preliminarily, sketch out (phát thảo) shaft structure

b. Calculating strength (Tính sức bền tĩnh)
- Determining the acting force (torque and bending) and Reactive force

- Calculating exactly shaft diameters at critical cross sections and selecting

parameter based on standard)
- Design shaft structure

c. Examining (Tính kiểm nghiệm)

- Examining overload (kiểm nghiệm quá tải)

- Examining strength and fatigue (kiểm nghiệm độ bền mỏi)

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18 Basic criteria calculation of shaft

1. Calculating strength

a. Preliminary calculation (Tính sơ bộ)
Selecting material and allowable torque [ ′], MPa

= = ≤ [ ′]

.

where, T is a torque acting on shaft, N.mm

=> The preliminary shaft d, mm

5 Vị trí xác định đường kính:
≥ + Đầu trục đối với trục vào và ra

+ Thân trục lắp bánh dẫn với trục trung gian
′

Selecting preliminary bearing and sketch out shaft structure

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1. Calculating strength

b. Calculating strength

- Determining all forces acting on shaft
- Calculating the support reactions (Phản lực) at pins (Gối đỡ)
- Drawing the diagram of bending or torque moment
- Obtaining exactly diameter of shaft by using the IV strength theory
- Completing the shaft structure

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20 Basic criteria calculation of shaft

1. Calculating strength

b. Calculating strength

Based on IV strength theory (Thuyết bền 4)

= =

0.1
đ = + 3 ≤ , where,
=
= 0.2

The value Mtđ determining by formula đ= + 0.75

đ đ

đ = 0.1 ≤ ≥
0.1

Note: - Key seat (Rãnh then) increases d about 6-10%
- Diameter of journal must be standardized in (TLTK)

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21 Basic criteria calculation of shaft

1. Calculating strength

Examble:
Design shaft I with the parameter in picture. Material C35 has allowable twist

torque [ ′]=20 Mpa and =20 Mpa .

Belt force acting on shaft Fđ = 791.8N and

Force acting on gearing Fr1 = 1178.9N , Ft1 = 3239N


Fđ=791.8N Fr1=1178.9N

P1=5.66kW Ft1=3239N Oz
T1=181385Nmm B

A

C D
n1=298v/p x
y

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22 Basic criteria calculation of shaft 11/20/2023
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1. Calculating strength

+ Calculating diameter preliminary

≥ = 35.6 mm;

Selecting diameter based on standard d = 36 mm,

+ Sketch out shaft structure

- Selecting diameter of shaft based on standard

f=90 l=145

d=36 d=40 d=40

d=36
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23 Basic criteria calculation of shaft

1. Calculating strength
+ Determining the equivalent moment at cross-section j Mtđ

đ= + + 0.75

+ Calculating shaft diameter at cross-section

≥đ f=90 l=145
0.1

Selecting dj according standard d=40
d=45
d=40

+ Completing structure

Ability of processing disassemble, assemble and maintenance

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24 Basic criteria calculation of shaft

1. Calculating strength

c. Examining


• Examining over load

= +3 ≤ = 0.8 ℎ

where,
, bending and torque at danger cross − section
, ℎ allowable stress when overload and elastic (chảy)

• Examining strength and fatigue
safety factor 1.5~2.5

= + ≥[ ] , safety factor of bending and torque

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25 Basic criteria calculation of shaft 11/20/2023
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2. Calculating rigidity. Condition of rigidity: f≤[ ]
a. Rigidity ≤ 

F

f
+ [f] = 0,01m – Shaft assemble with cylindrical gears
+ [f] = 0,005m – Shaft assemble with bevel gears
+ [] = 0,01rad - Thrust bearing (ổ bi đỡ)
+ [] = 0,05rad- Ball bearings (ổ bi lòng cầu)
+ [] = 0,001rad – Friction bearing (ổ bi trượt)
In machine manufacturing, we can be selected : [f] =(0,0002 ÷ 0,0003)I

where, l: Pin distance.

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26 Basic criteria calculation of shaft

2. Calculating rigidity.

b. Torsional deflection

Basic criterial of Torsional deflection of shafts:

= ≤[ ]

where,
[ ]: allowable angel of twist (góc xoắn cho phép), rad
G: shear modulus module (trượt đàn hồi) G=8,3.104,MPa
J0: moment of inertia in torsional (moment quán tính) (J0= /32)
l: length of twist (chiều dài đoạn trục xoắn), mm

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3. Calculating oscillation.

a. Cause

- Off centerline generate external
forced generate vibration (dao

động)

b. Damage

- Making additional stress (ứng suất phụ)
=> Effecting on strength

- Resonance (vùng cộng hưởng)
=> Breaking shaft

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3. Calculating oscillation.

c. Criteria of calculation

- Calculating amplitude ≤ [ ]

- Determining resonance (vùng cộng hưởng).

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29 Basic criteria calculation of shaft

3. Calculating oscillation.

d. Oscillation problems


- According the off-centerline m

= e
where, ey

l/2 l/2 l/2 Flt l/2

= = (+)

( )= − =

−1

f= = (natural frequency) → ∞ ( )

48

= The rigidly of shaft (TLTK)

Notice: Danger working conditions ≈ ( . ÷ . )

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30

Theorical contents

I. Overview

II. Basic criteria calculation of

shaft

III. Fundamentals of calculation
and design

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Fundamentals of calculation and design

Notice:

Moment

Bending Fa =

Torque Ft = ∗ = ∗

Sequence of calculation

1. Analyzing acting force in 2 plane (coordinates) Oz

• Plane: yoz (Fr & Ma ) x
• Plane: xoz (Only Ft) y
2. Calculating support reaction at pins on shaft

3. Drawing moment diagram (From left to right)

4. Determine diameter at critical cross-section


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Fundamentals of calculation and design
1. Calculating force acting

a. Specifying gears parameter

- Determining the pitch diameter by formula :
+ Spur gear : d = mz1 (mm)
+ Helical gear: d = mnz1/cosβ (mm)
+ Bevel gear: d = mmz1 (mm)

where,
m – module
z – Number of teeth on gear
β – Helix angle

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Fundamentals of calculation and design
1. Calculation acting force

b. Obtaining the acting Force

- Radial force Fr : Fr  2F0 sin(α1 / 2) (N)

+ V belt:
+ Chain: 7

Fr  kxFt  6.10 kxP / zpn (N)

- Tangential force on coupling Ft : Ft = 2T/Dkn (N)
where,
+ Dkn – coupling diameter (đường kính khớp nối) (mm)
+ T – twist torque (N.mm)
The radial acting force on shaft Fr due to misalignment can be calculated by
Fr  (0.2  0.3)Ft

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Fundamentals of calculation and design
2. Analysis the acting force

a. Force acting diagram
- The Figure describes force acting diagram of transmission

+ Fa2 x X
+Ft2 Fr2
Fxx
Ft1 Fxy
Fa1
z
Fr1
x


Fkn xy

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Fundamentals of calculation and design
3. Determining diameter of shaft

b. Drawing moment diagram

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+ L1 = L2 = L3 = 125mm
+ Ft1 = 500N, Fr1 = 182N
+ Ft2 = 800N, Fr2 = 291N
+ T = 50000 N.mm

Reaction force at pin A and D:

+ RAY = 24.3 N, RDY = 133.3 N

+ RDX = 700 N, RAX = 600 N 20-Nov-23

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Exercise 11/20/2023

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Q 1 Given the shaft of transmission system, Torque on shaft T = 100000 Nmm.

Helical gears, mn = 2 mm and helix gear β = 12o. The number of teeth on gearing:
Z1 = 22 and Z2 = 67. The yield stress of shaft material is [σF] = 60 MPa.
Calculating the force acting on shaft?

Helical gears

d1  mnZ1 2T Fr 1  Ft1 tan α  1655N Fa1  Ft1 tan β  945N
cos β Ft1   4446N
cos β
 44,98mm d1

d2  mnZ2 2T Fr 2  Ft 2 tan α  543N Fa2  Ft2 tan β  310,32N
cos β Ft2   1459,9N
cos β
d2

Spur gears 2T Fr 2  Ft 2 tan α Fa2  0
d2  mZ2 Ft 2 

d2

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Exercise


Q 2 Gear 1 = 400 ; Gear 2 = 160 .

Tangential force = 1000 and radial force = 364
Tangential force = 2500 and radial force = 910
Lengths = 150 , = 250 , = 150 .

Allowable stress = 50

a.Determining the reaction support at pin B and C
b.Drawing moment diagram , , T
c. Determining diameter at cross-section

Oz

x
y

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39 Exercise

Q 2 2.a The reactive force at pin B and C:

+ The moment equation at B in Y direction (at point B)

=− ∗ + − ∗ + =0

 = ∗ ∗ = ∗ ∗( ) = ,

+ The force equation: (at point C)

=− + + − =0

 = + − = 1674,3 + 364 − 910 = ,

+ The moment equation at B in X direction (at point B)

∑ =− ∗ − ∗ + ∗ + =0

 = ∗ ∗ ∗ ∗( ) =

=

+ The force equation (at point C)

= − − + =0

 = − + + = −3400 + 1000 + 2500 =

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40 11/20/2023
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Exercise

Q 2

2.b

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41

Exercise

Q 2

2.b

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42

Exercise
Q 2

2c Diameter at danger cross-section C:

Equivalent moment at C:

đ= + + 0,75

= 136489 + 375000 + 0,75 ∗ 200000
=
Diameter of shaft at cross-section C:

≥ đ =,

0,1 ∗

For assembling bearing at C we choose dc= 45 mm


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43 11/20/2023
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Exercise

Q 3 a. Determining the reaction
support at pin B and C
Given the working shaft in picture 3. b.Drawing moment diagram ,
The helical gear parameter
,T
= 240 , = 5000 , = 1885 , c. Determining the diameter of
= 1340 . shaft at danger cross-section C
The spur gear parameter

= 120 , = 10000 , = 3640 .
Length

= 120 , = 150 , = 150 .
Yield stress = 60 .

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44

Exercise

Q 3


Determining the reaction support at pin B and C

= = 1340 ∗ 120 = ;

= ∗= ∗=
2 2

+ The moment equation at B in Y direction : ∗( + )=0
= ∗− + ∗−

 = ∗ ∗ = ()

+ The force equation: ∑ = − − + = 0

 =− + + = ()

+ The moment equation at B in X direction :

∑ =− ∗ + ∗+ ∗( + )=0
()
 = ∗ ∗ =

( )

+ The force equation: ∑ = − + − − =0

=> = + + = ()

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45

Exercise

Q 3

4.b

In yoz plane: Fr or Ma
Ma1

In xoz plane: only Ft

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46 Exercise 11/20/2023
16
Q 3 + Equivalent moment at C:

đ= + + 0.75

= 440100 + 225000 + 0,75 ∗ 600000 =
Diameter of shaft at cross-section C:

≥ đ ≥.

0.1

For assembling gear at C, we choose: = ( )


Bonus:
+ Equivalent moment at B:

đ= + + 0.75

= 436764 + 1200000 + 0.75 ∗ 600000 =
Diameter of shaft at cross-section C::

≥ đ =,

0.1

For assembling bearing at B, we choose: = ( )

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Exercise 1 = 325 ,
2 = 486 .
Q 4 The bevel gear 1 : = 200 , 1 = 1000 , 1 = 163 ,

and helical gear Z2 2 = 160 , 2 = 1250 , 2 = 465 ,
Distance 1 = 100 , 2 = 200 , 3 = 100 .
The allowable bending stress [ ] = 60 .
a. Determine the reaction support at pin A and D.
b. Drawing the moment diagram Mx, My and T.
c. Determining the diameter of shaft at danger cross-section B.

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Exercise

Q 4

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Exercise 11/20/2023
Q 4 17

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Exercise
Q 4

In yoz plane: Fr or Ma Ma1

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Exercise
Q 4

In xoz plane: only Ft

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11/20/2023

Exercise
Q 4


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