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DEPARTMENT OF EDUCATION AND TRAINNING

HOCHIMINH CITY UNIVERSITY OF TECHNOLOGY
AND EDUCATION

THEORY OF MACHINE AND MACHINE

Chapter 4: DESIGN

FUNDAMENTAL OF COMPONENT DESIGN

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Contents

I. Loads and stresses in
machine parts

II. Basic criteria of machine
part

III. Mechanical transmission
system and their principal
feature

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I. Loads and stresses in machine parts
1. Loads

- Load in mechanisms is the physical unit on mechanical system or component.
- Classification: Static load and dynamic load

Static load Dynamic load

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I. Loads and stresses in machine parts

1. Loads

• 3 kinds of load:
 Rated load: The longest or maximum effective load
 Equivalent load: a constant value represented for dynamic load

Qtd  QdnK E
 Assumed load: applied to design machine parts


Ke: Coefficient depends on load mode
Qtt  Qtd K tt K d K dk

Ktt: Coefficient considers nonuniform distribution
Kd: Coefficient of dynamic load
Kdk: Coefficient depends on working condition

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I. Loads and stresses in machine parts
2. Stress

- Stress is internal resistance due to external forces

2 Types of stress:

Static stress Dynamic stress 15-Sep-21

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6 I. Loads and stresses in machine parts

2. Stress


- Stress cycle: The minimum time for the stress to return to original state
- 2 types of cycle stress:

 Cycle stress stable over time
 Cycle stress oscillated over time

- Amplitude stress:  a   max   min
- Mean stress: 2
- Stress ratio:
 m   max  min
2

r   min
 max

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7 I. Loads and stresses in machine parts

2. Stress

σ

σa σmax σa σmax σmax = σmin
σa σm σa


σmax σm
σmax

σmin r = 0 r = +C σmin n
σm
r = -C
σmin
r = -1

a) b) c) d) e)
Symmetrical period asymmetrical period Closed circuit period asymmetrical period Constant stress

opposite sign same sign

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8 I. Loads and stresses in machine parts

2. Stress

 k(n)  F  [ ] c  F  [ c ]   F  [ ]  F  F  [ F ]  H  [ H ]
A A Wo W

F: Force (N) and A: Cross-section (mm2)
W, Wo: The resistance moment (mm4)

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9 I. Loads and stresses in machine parts

2. Stress

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2. Stress

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Contents

I. Loads and stresses in
machine parts

II. Basic criteria of machine

part

III. Mechanical drive
transmission system

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II. Basic criteria of machine part

1. Strength

 Definition: The ability of material to resist breaking .

 2 types of destruction:

 Destruction caused by exceeding working stress (overload)

 Destruction caused by long-tern effects of changing stress value is

over the limit of ultimate strength

 Calculate method: Static stress of element

 Plasticity material:   [ ]  ch ;  [ ]   ch
[s] [s]

 Brittle material:   [ ]   b ;  [ ]   b
ε: Coefficient of measurement [s]Ks [s]Ks


Ks: Coefficient of concentration static load

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1. Strength

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II. Basic criteria of machine part

1. Strength

 Element has dynamic stress σa, σm= const. Calculation stress:   [ ]  lim KL
[s]K

KL: Life coefficient depends on N
β: coefficient of durable increase

 Calculate cycle of operations element:
 N > N0: Limited stress is limited long fatigue KL=1 N  60Lhn  NLE  Nk


  [ ]   r [s]: 1.5-2 safety coefficient
[s]K

 N < N0: Limited stress is limited short fatigue

  [ ]   r KL,KL  m N0
[s]K NLE

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1. Strength

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1. Strength

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II. Basic criteria of machine part

2. Rigidity

 Definition: The ability of material to resist deformation

 Calculate: l  Fal  [l ]
 Axial force: Long transposed EA
E: Modulus of elasticity (MPa)

A: Axial cross-section

L: Length of axis

 Moment of deflection y  [y];  [ ]
y: displacement

θ: Rotate angle   Tl  []
 Moment of torsion GJ0

G: Modulus of elasticity

J0: Inertia moment

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II. Basic criteria of machine part

3. Wear strength

 Definition: The ability of element that working under time of limited wear
 Calculate:

 Limited pressure: Relationship between pressure and distance
Pm.S  const

 Condition: p ≤ [p]

4. Heating

 During the working process, the machine will generate heat due to friction,
engine, ….

 The harmful effects of heat:
 Changing characteristic of material
 Machinery parts will stick together
 Reduce accuracy of machine because of thermal deformation

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II. Basic criteria of operating capacity of machine part
4. Heating

Cutting temperature generation

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II. Basic criteria of operating capacity of machine part

5. Resistance vibration

𝐹𝑙𝑡 = 𝑐𝑦 e
ey

where,

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

𝐹𝑙𝑡 = 𝑚𝑟𝜔 = 𝑚(𝑦 + 𝑒)𝜔

𝑚𝜔 𝑒 𝑒
𝑦(𝜔) = 𝑐 − 𝑚𝜔 − 1 = 𝑐
𝑚𝜔

at 𝜔 = 𝜔𝑛 = (natural frequency)


then 𝑦 → ∞ (𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑐𝑒)

𝑐 = 48𝐸𝐽 stiffness
𝑙
𝝎
Note: r𝐞𝐬𝐨𝐧𝐚𝐧𝐜𝐞 𝐟𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲 ≈ (𝟎. 𝟕 ÷ 𝟏. 𝟐)
𝝎𝒏

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II. Basic criteria of operating capacity of machine part
5. Resistance vibration

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Contents

I. Loads and stresses in
machine parts

II. Basic criteria of machine
part


III. Mechanical drive
transmission system

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1. Power III. Mechanical drive transmission

Ft

P  Ftv (kW ) Ft: Tangential force (N) v
1000 v: Velocity (m/s)

2. Velocity n

• Velocity of driving pulley: v   Dn (m / s) D T
60000

• Velocity of chain: v  pzn (m / s) P
60000

• Rotating speed of pulley: n  60000v (rpm) D: Diameter (mm)
D


• Rotating speed of chain: n  60000v (rpm) n: Rotating speed (rpm)
zp p: Pitch of chain

• Angular velocity:   2 n (rad / s) z: Number of teeth in
60 sprockets

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III. Mechanical drive transmission

3. Torque In T1 P1
Drive
• Given Power and rotation speed I x
n1
T1  9.55x106 xP1 (Nmm) 12  PII
u12  nI PI
n1 nII

• Given tangential Force and diameter T2

T1  Ft d1 (Nmm) II P2 Out
2
n2 x X

driven

4. Transmission Ratio


• Belt drives, friction drives:

u12  d2
(1  )d1

• Gear drives, worm-gear drives, chain drives:

u12  z2
z1

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III. Mechanical drive transmission

5. Efficiency P2 P1  Pm
 
P1 P1

• Series:   1.2.3...n

• Parallel:   1  2  ...  n

PI  PII P  Ftv (kW )
12 1000


u12  n1  n2  n1 n  6 0 0 0 0 v ( r p m )
n2 u12 D

TII=uɳTI T  9,55 106  P (Nmm)
n

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ESSAY

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III. Mechanical drive transmission

Pct nI x PI x Làm việc Plv (Ptđ)
Fv Tn nlv (III)
M I

nđc

nII x II x PII


ht  d1 12 23 uht  ud1  u12  u23
 nt,o br ,o x,o  unt  ubr  ux

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III. Mechanical drive transmission

1. Determine the required power on the motor’s shaft
 Based on diagram, the system includes 1 clutch, 1 pair of helical gear, 3
bearing roller gear and 1 chain drive

η  ηnt  ηbr  ηol 3  ηx

where, the reference efficiencies are shown on the given Table

Pct  Plv
η

2. Determine the preliminary speed of motor’s shaft
Based on diagram, the system includes 1 pair of helical gear and 1 chain drive
usb = uh x ux

where, the reference ratios are shown on the given Table (minimum values)

nsb = usb x nlv compared to 750; 1000; 1500 & 3000 (synchronous speed)

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III. Mechanical drive transmission

3. Transmission Ratios

utt  nm  ubr  ux where, ubr is chosen in the givenTable  ux  utt
nlv ubr

1.0 1.25 1.6 2.0 2.5 3.15 4.0 5.0 6.3 8.0

Try to select ux based on the rated range
 Nên chọn un (ud hoặc ux) theo tiêu chuẩn rồi kiểm tra sai số tỉ số truyền tổng thể

Check the tolerance of total ratio
u  utt  u compared to the required tolerance (4  5%)

utt

 Nếu chọn un theo tiêu chuẩn mà khơng thỏa sai số tỉ số truyền thì dùng luôn
giá trị đã tính được sau khi chọn uh

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Example

I

x

II x Làm việcx
III Tn

M

đc

Giving the parameter on the working shaft:
T = 425000 Nmm, n = 120 rpm

1. Motor specification

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Motor selection


1. Determine the power of working shaft

6 Tlv  nlv 425000 120
Tlv   Plv  6 6  5,34(kW )
9,55.10 .Pt

nlv 9,55 10 9,55 10

The load is static: Pt = Plv = 5,34 kW

 Based on diagram: 1 belt drive, 3 bearing roller, 1 pair of helical gear, 1 clutch

η  ηd .ηol 3.ηbr .ηnt As shown on Table 2.3
ηd = 0,96 belt drive efficiency
 0,96  0,993  0,97  0,9  0,81

Pct  Pt  5,34  6.59 (kW ) ηol = 0,99 bearing efficiency
η 0,81 ηbr = 0,97 1 pair spur efficiency
ηnt = 0.9 clutch efficiency

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Table 2.3

Name Efficiency η

Closed Open


Spur gear drive 0,96 – 0,98 0,93 – 0,95
Bevel gear drive 0,95 – 0,97 0,92 – 0,94
Worm gear drive
- Self – locking 0,3 – 0,4 0,2 – 0,3
- Unself – locking with
0,7 – 0,75 0,9 – 0,93
z1 =1 0,75 – 0,82 0,7 – 0,88
z1 = 2 0,87 – 0,92 0,95 – 0,96
z1 = 4 0,95 – 0,96
Chain drives 0,9 – 0,96
Friction drives
Belt drives 0,99 – 0,995
Pair of rolling bearing 0,98 – 0,99
Pair of plain bearing

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Motor selection

4. Calculating the preliminary speed of motor

 Preliminary ratio (based on Table 3.2) Gear Ratio
usb = uh x uđ Spur gear

= 1.6 x 2 (select minimum value) - 1 level gear box 1,6 – 8
- 2 level gear box 8 – 40

nsb = usb,.nlv,= 120x3.2 = 384 rpm Bevel gear box 1 – 6,3
nsb < 750 selection Motor~ 750 rpm - 1 speed gear box 8 – 30
- 2 speed gear box

Flat belt drive
 Selection of motor (3 phases 380V-50Hz)

is satisfied by: - Simple 2 –5
4 –6
Pđc ≥ Pct = 6.59 kW - Tensioner 2–5
nđc ≥ nsb (~ 750 rpm) V belt drive 2 –5
 Selection of electric power in Catalogue Chain drive 2 –4

Friction belt drive

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Motor selection

Spec Motor M2QA16L8A
Power (kW) 7,5
Rotation speed(rpm) 720
TS/TN 2,1

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Distribution the transmission ratio

1. Ratio

utt  nm  720  6  ubr  ud Select ubr  2,5  ud  utt  6  2, 4
nlv 120 ubr 2,5

1.0 1.25 1.6 2.0 2.5 3.15 4.0 5.0 6.3 8.0

Try to select ud =2,5 (based on the given rate range)

𝑢đ= 2; 2,24; 2,5; 2,8; 3,15; 3,56; 4; 4,5 & 5

Then, check the tolerance of total ratio

u  utt  u  utt  ubr  ud  6  (2,5  2,5)  4,1%  4% not satisfied
utt u tt 6

 Chọn ud theo tiêu chuẩn không thỏa thì thì phải thiết kế theo số liệu đã tính

ud=2,4

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Distribution the transmission ratio

2. Rotation speed of shafts n1  ndc  720  300 rpm
 Rotation speed of shaft I: ud 2,4

 Rotation speed of shaft II: n2  n1  300  120rpm
ubr 2,5

 Rotation speed of working shaft: nlv  n2  120rpm

3. Calculating the power on shafts

 Power on shaft II: P2  Plv  5,34  6,0 kW
ηnt .ηol 0,9x0,99

 Power on shaft I: P1  P2  6  6,18 kW
ηbr .ηol 0,98x0,99

 Power on motor’s shaft : Pdc  P1  6,18  6,5 kW
ηd .ηol 0,96x0,99

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Distribution the transmission ratio

4. Calculation torque on shafts

9,55 10 .Pdc 9,55 10  6,56 6

 Motor’s shaft: Tdc    86215 N.mm

ndc 720

 Shaft I: T1  6  9,55 106  6,18  196730 N.mm

9,55 10 .P1

n1 300

 Shaft II: T2  6  9,55 106  6,0  477500 N.mm
 Working shaft :
9,55 10 .P2

n2 120

Tlv  6  9,55 106  5,34  424975 N.mm

9,55 10 .Plv

nlv 120


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5. Results

Shaft Motor I II Working

Parameters 6,5 6,18 6,0 5,34
Motor (kW) 2,4 1
Ratio 2,5
Rotation speed (rpm) 720 120
Moment (N.mm) 86215 300 120 424975

196730 477500

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