Chapter 5
Design for Different
Type of Loading
Lecture Notes
Dr. Rakhmad Arief Siregar
Kolej Universiti Kejuruteraan Utara Malaysia
Machine Element in Mechanical Design
Fourth Edition in SI Unit
Robert L. Mott
1
Chapter 5
Design for Different Types of Loading
Objectives:





Identify various kinds of loading commonly encountered by
machine parts, including static, repeated and reversed,
fluctuating, shock or impact and random
Define the term stress ratio and compute its value for the
various kinds of loading
Define the concept of fatigue
Define the material property of endurance strength and
determine estimates of its magnitude for different materials
Recognize the factors that affect the magnitude of
endurance strength
2
Chapter 5
Design for Different Types of Loading
Objectives:







Define the term design factor
Specify a suitable value for the design factor
Define the maximum normal stress theory of failure and the
modified Mohr method for design with brittle materials
Define maximum shear stress theory of failure
Define he distortion energy theory, also called the von Mises
theory or the Mises-Hencky theory
Describe the Goodman method and apply it to the design of
parts subjected to fluctuating stresses
Consider statistical approaches, finite life and damage
accumulation method for design
3
Types of Loading &
Stress Ratio

Types of loading:





Static: when a part is subjected to a load that is applied
slowly, without shock, and is held at constant value.
Repeated and Reversed: when a part is subjected to a
certain level of tensile stress followed by the same level
of compressive stress
Fluctuating stress: when a load-carrying member is
subjected to an alternating stress with a nonzero mean.
Shock or impact: loads applied suddenly and rapidly
cause shock or impact, i.e., hammer blow, weight falling
Random: when varying loads are applied that are not
regular in their amplitude
4
Figures



See Fig. 5-1 for static stress
See Fig. 5-2 for repeated, reversed stress
See Fig. 5-4 for Fluctuating stress
5
Impact Load
Strain Gage A
Striker Bar
V
Strain Gage B
Input Bar
Output Bar
Signal Conditioner
Photodiode velocity sensor
(a)
Digital Oscilloscope
6
Impact Load
60
Incident wave
Transmitted wave x 20%
40
Stress [ MPa ]
20
0
-20
-40
-60
0
200
400
600
800
1000
Time [ sec ]
7
Stress ratio

Stress ratio is one of method to characterize variation of
stresses.
 min
Stress ration R 
 max
a
Stress ration A 
m




Maximum stress, max
Minimum stress, min
Mean stress, m
Alternating stress, a (stress amplitude)
a 
m 
 max   min
2
 max   min
2
8
Photographs of failed parts
Failure of a truck drive shaft spline due to corrosion fatigue
9
Photographs of failed parts
Failure of a stamped steel bracket due to residual stresses
10
Photographs of failed parts
Failure of an automotive drag link (steering wheel)
11
Photographs of failed parts
Failure of bolt in the overhead-pulley
12
Photographs of failed parts
Automotive rocker-arm articulation-joint fatigue failure
13
Photographs of failed parts
Valve-spring failure caused by spring surge in
an over speed engine
14
Photographs of failed parts
Brittle facture of a lock washer in one-half cycle
15
Failure resulting from static
loading

Static loading







Direct tension and compression
Direct shear
Torsional shear
Vertical shearing stresses
Bending
Buckling
How to predict failure if the component is
subjected to combine loading?
16
Ductile materials
Maximum shear stress


Also known as Tresca theory
The maximum shear stress hypothesis states that
yielding begin “whenever the maximum shear
stress in any element becomes equal to the
maximum shear stress in a tension test specimen of
the same material when specimen begins to yield”
 max 
Sy
2
or
1   2  S y
S sy  0.5S y
17
Triaxial shear stresses
 1/ 2 
1   2
2
 2/3 
 2 3
2
 1/ 3 
1   3
2
The maximum shear
stress graphically
represented in three
dimensions
18
Biaxial stress
1
 max 
Sy
2
or
1   2  S y
Sy
S sy  0.5S y
Sy
-Sy
-Sy
2
19
Ductile materials
Distortion energy


Also known as von Misses – Hencky theory
The maximum strain energy hypothesis predicts
failure by yielding occurs “when distortion energy in
a unit volume equals the distortion energy in the
same when uniaxially stressed to the yield strength”
   Sy
 
 1   2 2   2   3 2   3   1 2
2
20
Ductile materials
Distortion energy

Under the name of octahedral shear stress this
theory predicts failure occurs “whenever the
octahedral shear stress for any stress state equal or
exceeds the octahedral shear stress for the simple
tension test at failure”
 oct  S y
 oct 
1
3
 1   2 2   2   3 2   3   1 2
21
Triaxial stress
3
The distortion
energy theory
graphically
represented in
three dimensions
2
1
22
Biaxial stress
1
   Sy
Sy
    12   1 2   2 2
Sy
-Sy
2
-Sy
23
Problem 1
A hot-rolled steel is subjected to principle
stress 1 = 210 MPa, 1 = 480 MPa and 3 = 0
MPa. By utilizing UTM the hot-rolled steel has a
yield strength of Syt=Syc = 690 MPa and a true
strain at fracture of f = 0.55. Estimate the
factor of safety.
24
Solution

Maximum shear stress theory
 max 
SF 

1   2
2
Sy
2 max

480  0

 240
2
690
 1.44
2240
Distortion energy theory
 
480  2102  210  02  0  4802
2
 416.8 MPa
Sy
690
SF 

 1.66
  416.8
25
Brittle materials
Maximum normal stress



Also known as Rankine theory
The maximum normal stress hypothesis predicts
failure occurs “whenever one of the three principle
stresses equals or exceeds the strength”
Suppose we arrange: 1 > 2 > 3
n 1  S yt
n 3  S yc
Note: n = safety factor
26
Brittle materials
Modification of Mohr

Coulomb-Mohr
Sut
1 
n
1  B 1


Sut Suc n

0   1  Sut
0   2  Sut
0   1  Sut
 Suc   2  Sut
Mod. I-Mohr
Sut
1 
0   1  Sut  Sut   2  Sut
n
S 
Suc Sut
 1  ut 2 
0   1  Sut
Suc  Sut n( Suc  Sut )
 Suc   2   Sut
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Brittle materials
Modification of Mohr

Mod. II-Mohr
Sut
1 
0   1  Sut
n
2
 n 1  Sut 
n1

 1  
Sut
  Suc  Sut 
 Sut   2  Sut
0   1  Sut
 Suc   2   Sut
28
Problem 2
A cast iron is subjected to principle stress
1 = 210 MPa, 1 = 480 MPa and 3 = 0 MPa.
By utilizing UTM the cast iron has a yield strength of
Sut=215 MPa and Suc = 750 MPa. Estimate the factor
of safety by using:
(1) Coulomb-Mohr failure model
(2) Mod. I-Mohr failure model
(3) Mod. II-Mohr failure model
29
Solution

Maximum shear stress theory
 max 
SF 

1   2
2
Sy
2 max

480  0

 240
2
690
 1.44
2240
Distortion energy theory
 
480  2102  210  02  0  4802
2
 416.8 MPa
Sy
690
SF 

 1.66
  416.8
30
Fatigue

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
A machine components often subjected to dynamic
loading such as: variable, repeated alternating or
fluctuating stresses.
In most cases machine members are found to have
failed under the action of repeated or fluctuating
stresses.
The analysis reveals that the actual maximum
stresses were below the ultimate strength of the
material and quite frequently even below the yield
stress
31
Fatigue



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This kind failure CAN NOT be detected by naked
eye and even quite difficult to locate in a Magnaflux
or X-ray inspection
This failure called as fatigue failure.
Begins with a small crack and develops a point of
discontinuity in materials such as change in cross
section, a keyway or a hole.
Once developed, the stress-concentration effect
becomes greater and crack progresses more rapidly
32
Endurance strength



Endurance strength of material is its ability to
withstand fatigue loads.
Endurance strengths are usually charted on a graph
like shown in Fig. 5-7, called as S-N diagram.
Factors affecting endurance strength:





Surface finish
Material factor
Type of stress factor
Reliability Factor
Size Factor
33