Machine Design I
ME3011D
Dr. Ummen Sabu
Assistant Professor
Department of Mechanical Engineering
Department of Mechanical Engineering, National Institute of Technology Calicut
1
Introduction to Design
Problem Solving
▪ Fundamentally, design is the process
of problem solving.
▪ Within the domain of engineering
design, there are many sub-domains.
▪ Machine design is a subset of
mechanical design with a focus on
structures and motion.
Design
Engineering
Design
Mechanical
Design
Machine
Design
The hierarchy of problem solving
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Consider the case of design of IC Engine and Gear-Box
▪ The design of IC engine mainly integrates principles from heat transfer,
thermodynamics and combustion (Mechanical Design)
▪ The design of gear-box integrates principles from strength of materials, solid body
mechanics, kinematics and dynamics (Machine Design)
▪ Machine design involves proper sizing of a machine member to safely withstand
max. stress-induced when it is subjected to an external load
(Axial/Transverse/Torsional/Bending)
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Design Process
Recognize
the Need
Create a
Design
Prepare a
Model
Test and
evaluate
(Physical/mathematical or
use of software packages)
Improvement
of design
The process of design
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Presentation
Design Factors
▪ A number of factors must be considered in a given design situation involving an
element or the configuration of the total system.
Many of the important ones are:
➢ Strength
➢ Stiffness
➢ Wear resistance and friction
➢ Corrosion
➢ Dimension and weight
➢ Safety
➢ Reliability
➢ Thermal properties
➢ Life time
➢ Conformance to standards
➢ Maintainability
➢ Manufacturability
➢ Economic considerations
➢ Aesthetics
➢ Ergonomics
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Principles of Standardization
▪ A standard is a set of specifications for parts, materials or processes intended to
achieve uniformity and provide a reasonable inventory of tooling, sizes, shapes and
varieties.
▪ In design, the aim is to use only standardized components as possible. The standards
of specifications and testing procedures of machine elements improve their quality
and reliability.
For e.g., SKF bearings and Dunlop belts have a good reputation.
▪ Organizations and societies establish specifications for standards. Some of them are:
• Bureau of Indian Standards (BIS)
• American Society of Testing and Materials (ASTM)
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Selection of Materials
▪ Selection of a proper material for the component is an important step in the process
of machine design.
▪ The material should serve the desired purpose at a minimum cost.
▪ The important factors considered for material selection are:
➢ Availability
➢ Cost
➢ Mechanical properties (Strength, stiffness, hardness, toughness, plasticity etc.)
➢ Manufacturing considerations (Casting, forging, machining, welding etc.)
▪ Cast iron, steel, aluminium, copper, ceramics, plastics, rubber etc. are common
examples of materials selected for components
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Statistical Considerations
▪ An uncertainty exists as to whether a machine component will actually perform
satisfactorily, and the statistical measure of the probability that it will not fail in
use is called reliability (0 ≤ R < 1)
▪ A design engineer has to judiciously select materials, processes, and dimensions
to achieve the desired reliability.
▪ To make a sound analysis of the design using statistical techniques, all the
needed stochastic data should be available in sufficient quantities.
▪ The statistical approach to design is relatively new, whereas the factor-of-safety
method of design is time-proven.
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Factor of Safety
▪ A sufficient reserve strength is ensured during machine design considering the
chance of an accident.
▪ The factor of safety (fs) is defined as:
fs =
Failure stress
Allowable stress
▪ The allowable stress is the stress value used in design to determine the dimensions
of the component.
▪ The designer expects that the stress under normal operating conditions does not
exceed the allowable stress.
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Simple Stresses in Machine Elements
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Normal stress due to direct axial load
▪ The normal stress, strain and deformation are given by:
𝑃
𝜎𝑡 =
𝐴
𝜕
𝜀=
𝑙
𝑃𝐿
𝛿=
𝐴𝐸
Using the Hooke′ s law
𝜎𝑡 𝛼 𝜀, 𝜎𝑡 = 𝐸 𝜀
𝜎𝑡 = Tensile stress, 𝑃 = External Force, A= cross-sectional area
𝜀 = Strain, 𝜕 = deformation, 𝑙= original length, E = Elastic modulus
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Shear stress due to direct shear load
▪ The shear stress in the rivet is given by:
𝑃
𝜏=
𝐴
▪ Shear strain is given by:
𝜏 = 𝐺𝛾
𝐸 = 2𝐺 1 + 𝜇
𝜏= Shear stress, 𝑃 = External Force, A= cross-sectional area of rivet
𝛾= Shear strain, E = Elastic modulus, G = Shear modulus,
𝜇 = Poissons ratio
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Stresses due to bending moment
▪ Due to the bending moment, the beam is subjected to a combination of tensile
stress on one side and compressive stress on the other side of the nuetral axis
▪ The bending stress at a distance y from neutral axis is given by:
𝑀𝑏𝑦
𝜎𝑏 =
𝐼
𝑀𝑏 = Applied bending moment, I = Moment of inertia of the cross-section about the neutral axis.
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Stresses due to torsional moment
▪ The torsional shear stress is given by:
𝑀𝑡 𝑟
𝜏=
𝐽
▪ The angle of twist is given by:
𝑀𝑡𝑙
𝜃=
𝐽𝐺
Mt = Applied torque, r = radial distance from axis of rotation,
J = Polar moment of inertia of the crosssection about the axis of rotation,
𝑙 = length of shaft , G = Shear modulus
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▪ The formulas reviewed for computing simple stresses due to direct tensile and
compressive forces, shear force, bending moments, and torsional moments are
applicable under certain conditions.
▪ One condition is that the geometry of the member is uniform throughout the
section of interest.
▪ In many typical machine design situations, inherent geometric discontinuities
are necessary for the parts to perform their desired functions.
▪ Any of these geometric discontinuities will cause the actual maximum stress in
the part to be higher than the simple formulas predict.
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Stress concentration factor
▪ Defined as a factor by which the actual maximum stress exceeds the nominal
stress.
▪ Usually denoted as Kt
▪ The value of Kt depends on the shape of the discontinuity, the specific
geometry, and the type of stress.
▪ A good design reduces stress concentration by limiting abrupt changes in
geometry.
▪ Stress concentration factor curves are used for design calculations.
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Stress concentration factor
Stress concentration factor can be estimated through:
▪ Mathematical calculations based on elasticity theory.
▪ Experimentally through photoelastic methods.
▪ Stress concentration factors for various shapes and loading conditions can
be obtained from charts in design data book.
▪ Finite Element Analysis (FEA) packages can be used for any shape.
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Stress concentration factor curves
H
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Problem 1:
A round steel rod is subjected to a tensile load of 90 KN. Taking the yield stress for
the steel as 328.6 Mpa and the factor of safety as 1.8, Determine the suitable
diameter for the rod.
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Solution:
Load, F = 90 KN = 9000 N
Yield Stress, 𝜎𝑦 = 328.6 Mpa
FOS = 1.8
∴ Allowable stress, 𝜎 = 𝜎𝑦 / FOS = 328.6/1.8 = 182.56 Mpa
For the given axial load, we can calculate the stress developed by the relation, 𝜎 = F/A
∴ 𝜎 = 9000 / A
Equating the stress developed to the maximum allowable stress, we get
182.56 = 9000 / A, or A = 9000/182.56, ∴ A = 493 mm2
𝜋 2
𝑑
4
= 493, diameter, d = 25.054 mm ~ 25 mm
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Problem 2:
Calculate the maximum stress in a 6 mm thick stepped flat plate subjected to an axial
tensile force of 9800 N.
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Solution:
(From design data book)
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Combined Stresses in Machine Elements
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Complex Stresses
▪ In practical applications, machine members are subjected to combined stresses due
to simultaneous action of direct stress ( either tensile or compression) combined with
shear stresses (torsional) and/or bending stress.
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Complex Stresses
(Stress at a point)
𝜎=
𝜎xx 𝜏xy 𝜏xz
𝜏yx 𝜎yy 𝜏yz
𝜏zx 𝜏zy 𝜎zz
▪ If the stress vector acting on 3
mutually Ʇr planes are known, then
we can determine stress vector on
any other arbitrary plane.
▪ Six stress components are required to
define stress at a point.
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State of Stress in 2 Dimensions
(Plane Stress)
𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
𝜎=
+
cos 2𝜃 + 𝜏xy sin 2𝜃
2
2
𝜎𝑥 − 𝜎𝑦
𝜏 =
sin 2𝜃 − 𝜏xy cos 2𝜃
2
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State of Stress in 2 Dimensions
(Principal Stresses)
▪ 𝑇𝑎𝑛2𝜃 =
𝜎𝑥 + 𝜎𝑦
𝜎1 =
+
2
𝜎𝑥 − 𝜎𝑦
2
2
𝜎𝑥 + 𝜎𝑦
𝜎2 =
−
2
𝜎𝑥 − 𝜎𝑦
2
2
+ 𝜏𝑥𝑦 2
▪ 𝜃 + 90o
+ 𝜏𝑥𝑦 2
▪ On the Principal plane, shear stress is ZERO
▪ In 2D system, there are two principal planes separated by 90o
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2𝜏𝑥𝑦
𝜎𝑥− 𝜎𝑦
State of Stress in 2 Dimensions
(Maximum shear stress)
𝜏𝑚𝑎𝑥
(𝜎1 −𝜎2 )
=±
=±
2
𝜎𝑥 − 𝜎𝑦
2
2
+ 𝜏𝑥𝑦 2
▪ 𝑇𝑎𝑛2𝜃 = −
▪ 𝜃 + 90o
▪ On the 𝝉𝒎𝒂𝒙 plane normal stresses also act, 𝜎′ =
𝜎𝑥 +𝜎𝑦
2
=
𝜎1 +𝜎2
2
▪ In 2D system, there are two 𝝉𝒎𝒂𝒙 planes separated by 90o.
▪ The angle between any principal plane and nearest 𝝉𝒎𝒂𝒙 plane is 45o.
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𝜎𝑥− 𝜎𝑦
2𝜏𝑥𝑦
State of Stress in 2 Dimensions
𝜎′
𝜏𝑚𝑎𝑥
𝜎2
𝜏=0
Second 𝜏𝑚𝑎𝑥 Plane
Minor Principal Plane
45𝑜
𝜎′
𝑜
𝜃 45
𝜏𝑚𝑎𝑥
First 𝜏𝑚𝑎𝑥 Plane
𝜎1
𝜏=0
Major Principal Plane
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Problem 3
▪ A cantilever beam of circular cross-section is loaded as shown. Determine the
maximum and minimum normal stresses and maximum shear stress at points A
and B.
3 kN
250 mm
A
50 mm
1 kN.m
B
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15 kN
3 kN
250 mm
Solution:
A
50 mm
1 kN.m
B
𝜋
4
𝜋
4
Axial stress, 𝛔d = F/ 𝑑2 = 15 X 103 / 502 = 𝟕. 𝟒 𝐌𝐏𝐚
Bending stress, 𝛔b = M x y /
𝜋 4
𝑑
64
= 3x
Torsional shear stress, 𝝉 = T x r /
103
𝜋 4
𝑑 =
32
x 250 x 25 /
1 x106x25/
𝜋
504
64
𝜋
504
32
= 𝟔𝟏. 𝟏𝟐 𝐌𝐏𝐚
= 𝟒𝟎 . 𝟕𝟒 𝐌𝐏𝐚
Resultant normal stress at A, 𝛔A = 𝛔d + 𝛔b = 7.4 + 61.12 = 68.76 Mpa
Resultant normal stress at B, 𝛔B = 𝛔d - 𝛔b = 7.4 - 61.12 = -53.48 Mpa
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15 kN
For Principal stress at A, 𝛔1A,𝛔X = 𝛔A = 68.76 MPa , 𝛔Y = 0, 𝝉 = 40.74 MPa
𝛔1A =
𝛔2A =
𝜎𝑥 +𝜎𝑦
2
𝜎𝑥 +𝜎𝑦
𝜏𝑚𝑎𝑥𝐀 =
2
+
−
(𝜎1 −𝜎2 )
2
𝜎𝑥 −𝜎𝑦 2
2
𝜎𝑥 −𝜎𝑦 2
2
=
+ 𝜏𝑥𝑦 2 = 87.69 MPa
+ 𝜏𝑥𝑦 2 = -18.93 MPa
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2 = 53.31 MPa
For Principal stress at B, 𝛔1B,𝛔X = 𝛔B = -53.48 MPa , 𝛔Y = 0, 𝝉 = 40.74 MPa
𝛔1B = 22 MPa
𝛔2B = -75.47 MPa
𝜏𝑚𝑎𝑥𝐁 = 48.73 MPa
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Problem 4 (Assignment)
▪ A hollow shaft of 40 mm external diameter and 25 mm inner diameter is
subjected to a twisting moment of 118 Nm, an axial thrust of 9806 N and a
bending moment of 79 Nm. Calculate the maximum compressive and shear
stresses at A and B.
79 N.m
A
25 mm
40 mm 118 N.m
B
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9806 N
Problem 5
▪ Determine the maximum normal stress and maximum shear stress at a section A-A
for the crank shown in Figure, when a load of 10 kN is assumed to be concentrated
at the center of the crank pin.
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Bending stress, 𝛔b = M x y /
𝜋 4
𝑑
64
Torsional shear stress, 𝝉 = T x
= 10x103x90 x(75/2) /
𝜋 4
r/ 𝑑
32
𝜋
754
64
= 10 x103x125x(75/2) /
= 21.73 MPa
𝜋
754
32
= 15.10 MPa
Resultant normal stress at A, 𝛔X = 𝛔d + 𝛔b = 0 + 21.73 = 21.73 MPa
Principal stress:
𝛔1A =
𝜎𝑥 +𝜎𝑦
2
+
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2 =
21.73+0
2
+
21.73−0 2
2
+ 15.102 = 29.47 Mpa
Maximum shear stress:
𝜏𝑚𝑎𝑥𝐀 =
(𝜎1 −𝜎2 )
2
=
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2 =
21.73−0 2
2
+ 15.102 = 18.60 MPa
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Problem 6 (Assignment)
▪ Determine the principal stresses and the maximum shear stress at A-A for the
crankshaft bearing shown.
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Mohr’s Circle (2D)
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Mohr’s Circle (2D)
𝝉
(σy ,τxy )
𝛔
(σx , - τxy )
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Mohr’s Circle (2D)
𝝉
(σy ,τxy )
O
B
(σy ,0 )
A
(σx , 0)
(σx , - τxy )
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𝛔
Mohr’s Circle (2D)
𝝉
R = 𝜏𝑚𝑎𝑥
(σy ,τxy )
=±
𝜎𝑥 − 𝜎𝑦
2
O
(σy ,0 )
2𝜃
(σx , 0) (σ1 , 0)
(σx , - τxy )
𝜎𝑥 + 𝜎𝑦
2
𝜎𝑥 − 𝜎𝑦
2
2
+ 𝜏𝑥𝑦 2
A
B
(σ2 , 0)
(𝜎1 −𝜎2 )
=±
2
𝛔
▪ In Mohr’s circle angles are twice of
their actual value
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State of Stress in 3 Dimensions
▪ The general three-dimensional state of stress consists of three unequal principal
stresses acting at a point
▪ 𝜎3 − 𝜎x+𝜎y+𝜎z 𝜎2 + (𝜎x𝜎y + 𝜎y𝜎z + 𝜎x𝜎z - 𝜏xy2 - 𝜏yz2 - 𝜏xz2)𝜎 −
(𝜎x𝜎y 𝜎z + 2 𝜏xy 𝜏yz 𝜏xz - 𝜎x 𝜏yz2 - 𝜎y 𝜏xz2 - 𝜎z 𝜏xy2) = 0
▪ The three roots of the above equation gives the three principal stresses: 𝜎1 , 𝜎2 , 𝜎3
▪ The maximum shear stress is given by:
𝝉𝒎𝒂𝒙 =
𝝈𝟏 −𝝈𝟑
𝟐
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Mohr’s Circle (3D)
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Mohr’s Circle
▪ Draw Mohr’s circle for:
o Uniaxial tension
o Pure shear
o Hydrostatic pressure
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Problem 7:
The state of stress at a point is given by the following stress tensor. All the
components are in Mpa. Calculate the maximum shear stress at the point.
𝜎=
30
40
0
40 0
90 0
0 −10
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Solution:
σzz = - 10 Mpa, acts on a principal plane (No shear stress acting)
∴ Convert the matrix into 2D
𝜎𝑥 + 𝜎𝑦
𝜎1 =
+
2
𝜎1 =
30+90
2
+
30−90 2
2
𝜎𝑥 − 𝜎𝑦
2
2
+ 𝜏𝑥𝑦 2
+ 402 = 110 MPa, 𝜎2 = 60 − 50 = 10 MPa
𝜎1 − 𝜎3
110 − −10
𝝉𝒎𝒂𝒙 =
=
= 𝟔𝟎 𝐌𝐏𝐚
2
2
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Theories of Failure
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Engineering Stress/Strain &
True Stress/Strain
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Theories of Failure
▪ When a material is subjected to a single type of stress, it is easy to predict when the
failure is likely to occur.
▪ However, if the material is subjected to a complex stress system, then it is difficult to
predict the failure.
▪ Failure by yielding or fracture is possible.
▪ Ductile materials, yield stress is taken as limiting strength (Fails by shear)
▪ Brittle materials, ultimate stress is taken as limiting strength (Fails by normal stress)
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Max Normal Stress Theory
(Principal Stress Theory/Rankine’s Theory)
▪ According to this theory, failure of an element subjected to complex state of
stress occurs when the maximum principal stresses of the component exceeds the
yield strength of the material in simple tension test.
▪ Failure occurs when, 𝛔1 > 𝑺yt
▪ To avoid failure, 𝛔1 ≤ 𝑺yt
▪ The dimensions of the component are determined using a factor of safety:
𝛔𝟏 =
𝑺yt
𝐅𝐒
or 𝛔𝟏 =
𝑺yc
𝐅𝐒
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𝛔2
𝑺yt
𝑺yt
−𝛔1 − 𝑺yc
−𝛔2
𝛔1
− 𝑺yc
▪ Square represents region of safety
▪ If a point with coordinates fall outside the square, it indicates failure condition
▪ Applicable only for brittle material
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Max. Shear Stress Theory
(Guest’s/Tresca’s Theory)
▪ According to this theory, failure of an element subjected to complex state of
stress occurs when the maximum shear stress of the component at any point
exceeds the shear strength at yield point (𝑺SY) in simple tension test.
▪ Failure occurs when, 𝝉𝒎𝒂𝒙 > 𝑺SY
𝑺yt
𝝈𝟏 −𝝈𝟐
▪ From a simple tension test, 𝝉𝒎𝒂𝒙 = 𝑺SY =
, we know 𝝉𝒎𝒂𝒙 =
𝟐
𝝈𝟏 −𝝈𝟐
𝟐
≤ 𝑺SY or
𝝈𝟏 −𝝈𝟐
𝟐
=
𝑺yt
𝟐𝐅𝐒
𝝈𝟏 −𝝈𝟐
𝟐
𝟐
≤
𝑺yt
𝟐
or 𝝈𝟏 −𝝈𝟐 =
𝑺yt
𝐅𝐒
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𝛔2
𝑺yt
- 𝑺yt
𝑺yt
−𝛔1
𝛔1
−𝑺yt
−𝛔2
▪ Hexagon represents region of safety
▪ If a point with coordinates fall outside the hexagon, it indicates failure condition
▪ Applicable for ductile material
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Distortion Energy Theory
(Vonmises Theory)
According to this theory, failure of an element subjected to complex state of stress
occurs when the maximum distortion energy stored at a point in the material exceeds
the distortion energy at yield point under simple tension.
▪ The total strain energy has components corresponding to strain energy due to
change of volume and strain energy due to distortion.
▪ According to this theory, criterion of failure is given by:
𝑠𝑦𝑡
=
𝐹𝑆
1
⋅ [(𝜎1 −𝜎2 )2 + (𝜎1 −𝜎2 )2 + (𝜎1 − 𝜎2 )2 ]
2
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For bi − axial stress,
𝑠𝑦𝑡
=
𝐹𝑆
𝛔2 𝑺
yt
𝜎1 2 − 𝜎1 𝜎2 + 𝜎2 2
For a component in pure shear,
𝜎1 = 𝜏, 𝜎2 = − 𝜏
−𝑺yt
𝑺yt
−𝛔1
syt = 3 𝜏 𝑜𝑟 𝒔𝒚𝒕 = 𝟑 𝒔𝐬𝐲 ,
𝑠sy = 0. 577 syt
−𝛔2
𝛔1
−𝑺yt
▪ According to this theory, yield strength in shear is 0.577 times the yield strength in
tension.
▪ The region of safety is an ellipse
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Maximum Normal Strain Theory
(Principal Strain Theory/Saint Venant Theory)
According to this theory, failure of an element subjected to complex state of stress
occurs when the maximum principal strain of the component exceeds the maximum
strain of the material in simple tension.
ε1 =
σ1
E
−
μσ2
E
𝑠𝑦𝑡
𝐹𝑆
−
μσ3
E
, ε2 =
σ2
E
−
= 𝜎1 − 𝜇𝜎2 − 𝜇𝜎3
μσ3
E
or
−
μσ1
E
𝑠𝑦𝑡
𝐹𝑆
, ε3 =
σ3
E
−
= 𝜎1 − 𝜇𝜎2
▪ This theory is not used in general.
Department of Mechanical Engineering, National Institute of Technology Calicut
μσ1
E
−
μσ2
E
Maximum Strain Energy Theory
(Haigh’s Theory)
According to this theory, failure of an element subjected to complex state of stress
occurs when the maximum strain energy stored at a point in the material exceeds the
strain energy at yield point under simple tension.
▪ According to this theory, criterion of failure is given by:
𝑠𝑦𝑡
=
𝐹𝑆
𝜎1 2 + 𝜎2 2 − 2μ𝜎1 𝜎2
▪ This theory is applicable to ductile materials and yields good approximation.
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Selection of Theory
▪ Maximum
shear
stress
theory gives results on the
conservative side.
▪ On
the
other
hand,
distortion energy theory is
slightly liberal.
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Theories of Failure
Theory
Mathematical relation
1 Rankine’s theory
(Max. normal stress theory)
𝛔𝟏 =
𝑺yc
𝐅𝐒
𝝈𝟏 −𝝈𝟐 = 𝐅𝐒yt
𝒔𝒚𝒕
=
𝑭𝑺
𝟏
]
⋅ [(𝝈𝟏 −𝝈𝟐 )𝟐 + (𝝈𝟏 −𝝈𝟐 )𝟐 + (𝝈𝟏 − 𝝈𝟐 )𝟐
𝟐
𝒔𝒚𝒕
4 Saint Venants theory
(Max. principal strain
theory)
5 Haigh’s theory
(Max. strain energy theory)
or 𝛔𝟏 =
𝑺
2 Trescas theory
(Max. shear stress theory)
3 Vonmises theory
(Distortion energy theory)
𝑺yt
𝐅𝐒
𝑭𝑺
𝒔𝒚𝒕
=
𝑭𝑺
= 𝝈𝟏 − 𝝁𝝈𝟐
𝝈𝟏 𝟐 + 𝝈𝟐 𝟐
−
𝟐𝛍𝝈𝟏 𝝈𝟐
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𝝉 /𝑺yt
Applicability
1
Brittle Material
0.5
Ductile Material
0.57
Ductile Material
0.77
-
0.62
Ductile Material
Problem 8:
A component is subjected to torsion. Estimate the maximum shear stress permitted,
according to the five different theories of failure, in terms of yield strength in
tension Syt. (Take μ =0.3, FS = 1)
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Solution:
𝜎𝑥 = 0, 𝜎𝑦 = 0, 𝜏𝑥𝑦 = 𝜏, 𝜎1 = 𝜏, 𝜎2 = - 𝜏
𝝉
𝜏
1. Rankine’s Theory: 𝛔𝟏 =
𝑺yt
𝐅𝐒
𝜏 = 𝑆yt
𝛔
2. Trescas Theory: 𝛔𝟏 − 𝛔𝟐 =
𝑺yt
𝐅𝐒
𝜏 − (−𝜏) = 𝑆yt
𝜏=
𝑆yt
2
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-𝜏
3. Distortion Energy Theory
𝜏𝟐
𝒔𝒚𝒕
𝑭𝑺
− 𝜏 x(−𝜏) +
= 𝝈𝟏 𝟐 − 𝝈𝟏 𝝈𝟐 + 𝝈𝟐 𝟐
(−𝜏)𝟐
= 𝑆yt
3𝜏 = 𝑆yt , 𝜏
𝑆yt
=
3
𝜏 = 0.577 𝑆yt
4. Saint Venants Theory
𝒔𝒚𝒕
𝑭𝑺
= 𝜎1 − 𝜇𝜎2
𝜏 − 0.3 x (−𝜏) = 𝑆yt
5. Haigh’s Theory
𝜏𝟐 + −𝜏
𝟐
𝒔𝒚𝒕
𝑭𝑺
=
𝜏 = 0.77 𝑆yt
𝜎1 2 + 𝜎2 2 − 2μ𝜎1 𝜎2
− 2x0.3(𝜏x − 𝜏) = 𝑆yt
𝜏 = 0.62 𝑆yt
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 9:
A bolt is subjected to an axial force of 1000N with a transverse force of 5000 N. Find
the diameter of the bolt required according to the different theories of failure. It is
assumed that the permissible tensile stress at elastic limit a 100 MPa and Poisson’s
ratio as 0.3.
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Solution:
𝑃
𝐴
𝜎𝑥 = 0, 𝜎𝑦 = =
12732
𝑑2
𝑃
𝐴
, 𝜏𝑥𝑦 = =
1. Rankine’s Theory: 𝛔𝟏 =
15369
𝑑2
−
15369
,
𝑑2
𝜎2 =
2637
𝑑2
d = 12.4 mm
= 100
15369
𝑑2
𝜎1 = -
𝑺yt
𝐅𝐒
2. Trescas Theory: 𝛔𝟏 − 𝛔𝟐 =
−
6366
,
𝑑2
𝑺yt
𝐅𝐒
2637
𝑑2
15369
𝑑2
= 100
= 100
d = 13.5 mm
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3. Distortion Energy Theory
15369
−
𝑑2
2
−
− 0.3 x
2637
( 2 )
𝑑
5. Haigh’s Theory
−
15369
𝑑2
2
𝑭𝑺
= 𝝈𝟏 𝟐 − 𝝈𝟏 𝝈𝟐 + 𝝈𝟐 𝟐
15369 2637
2637
− (−
x 2 )+
𝑑2
𝑑
𝑑2
4. Saint Venants Theory
15369
𝑑2
𝒔𝒚𝒕
+
2637
𝑑2
𝒔𝒚𝒕
𝑭𝑺
2
𝒔𝒚𝒕
𝑭𝑺
= 100
d = 13 mm
= 𝜎1 − 𝜇𝜎2
= 100
=
2
−16160
𝑑2
= 100
d = 13 mm
𝜎1 2 + 𝜎2 2 − 2μ𝜎1 𝜎2
− 2x0.3(−
15369 2637
x 2 ) = 100
2
𝑑
𝑑
d = 13 mm
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 10:
An element is subjected to 𝜎X = 60 Mpa and 𝜏 = 40 Mpa. If the material has 𝑆yt = 330
MPa. Calculate the factor of safety according to Rankine’s theory.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
𝜎X = 60 MPa
𝜏 = 40 MPa
𝑆yt = 330 MPa
𝜎1 =
𝜎𝑥 +𝜎𝑦
2
+
𝜎𝑥 −𝜎𝑦 2
2
+ 𝜏𝑥𝑦 2 = 80 Mpa
According to Rankine’s theory, 𝜎1 =
80 =
330
FS
𝑆yt
FS
, 𝐅𝐒 = 𝟒. 𝟏𝟐
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Problem 11:
An machine member 50 mm in diameter, 250 mm long and supported one end as a
cantilever is used for axial tensile load of 235 KN. 𝑆yt = 480 MPa. Calculate the
maximum shear stress and factor of safety according to Guest’s theory.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
𝑑 = 50 𝑚𝑚
L= 250 mm
P = 235 kN
𝑆yt = 480 MPa
σX = F/
𝜎1 =
π 2
d
4
𝜎𝑥 +𝜎𝑦
2
= 120 Mpa
𝜎𝑥 −𝜎𝑦 2
+
2
+ 𝜏𝑥𝑦 2 = 120 Mpa
𝜎2 = 0
𝝉𝒎𝒂𝒙 =
𝜎1 − 𝜎2
2
= 60 Mpa
According to Guest’s theory, 𝝈𝟏 −𝝈𝟐 =
FS = 480/120 = 4
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𝑺yt
𝐅𝐒
Problem 12:
A load (P) of 45 kN is applied to a crankshaft of diameter 90 mm at a distance of 200
mm, as shown in the figure. The material is 30C4 with 𝑆yt = 315 Mpa. The factor of
safety according to Guest’s theory.
150 mm
200 mm
P
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Solution:
Bending stress, 𝛔b = M x y /
𝜋 4
𝑑
64
Torsional shear stress, 𝝉 = T x
𝜎𝑥 + 𝜎𝑦
𝜎1 =
+
2
= (𝑃x200) x 45 /
𝜋 4
r/ 𝑑 =
32
𝜎𝑥 − 𝜎𝑦
2
𝜋
904
64
(Px150) x 45/
2
+ 𝜏𝑥𝑦
2
= 𝟏𝟐𝟓 𝐌𝐏𝐚
𝜋
904
32
125
=
+
2
125
2
𝜎1 = 140.79, 𝜎2 = −15.79
𝝉𝒎𝒂𝒙 =
𝝈𝟏 −𝝈𝟐
𝟐
= 78.3 Mpa
According to Guest’s theory, 𝝉𝒎𝒂𝒙 =
78.3 =
𝟑𝟏𝟓
,
𝟐𝐱𝐅𝐒
= 𝟒𝟕. 𝟏 𝐌𝐏𝐚
𝑺yt
𝟐𝐅𝐒
FS = 2
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2
+ 47.12
Problem 13:
The stresses induced at a critical point in a machine component made of 45C8 with
yield strength of 380 Mpa are shown below. Calculate the Factor of Safety by
(i) Maximum Normal Stress Theory
40 MPa
(ii) Maximum Shear Stress Theory
(iii) Distortion Energy Theory
80 MPa
100 MPa
100 MPa
80 MPa
40 MPa
Solution:
(i) FS = 2.44
(ii) FS = 2.22
(iii)FS = 2.32
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Subjected to Torsion
𝜃
Brittle
45𝑜
𝜎1
Ductile
𝜏𝑚𝑎𝑥
(Recall chalk piece experiment in class)
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Subjected to Uni-axial Tension
Ductile
(Recall chalk piece experiment in class)
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Brittle
Design for Impact and Fatigue Load
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Operational Loads
▪ Machine components are subjected to external force or load, which can be either
static or dynamic.
▪ The dynamic load is further classified into cyclic and impact loads.
▪ The various types of forces are:
• Useful loads due to the energy transmitted by the element
• Dead weight
• Inertial forces
• Thermal stresses
• Frictional forces
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Design for Impact Load
▪ A load applied with some initial velocity is said to be an impact load.
Impact stress, 𝜎 ′ = 𝜎
1+ 1+
2𝐴𝐸ℎ
WL
=
𝑊
A
1+ 1+
Deformation under impact action, 𝛿′ = 𝛿 { 1 + 1 +
Impact or shock factor,
𝜎′
𝜎
={1+ 1+
2ℎ
𝛿
2ℎ
𝛿
2𝐴𝐸ℎ
WL
}
}
W = load applied with impact; h = height through which load falls; A = Cross-sectional area;
L= Length of bar; E = Young’s modulus of bar material; 𝜎 = static stress; 𝛿′ = deformation due to impact;
𝛿 = deformation due to static action.
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Design for Impact Load
Bending:
Impact stress due to bending, 𝜎𝑏 ′ = 𝜎𝑏 { 1 + 1 +
Deformation under impact action, 𝑦 ′ = 𝑦
2ℎ
𝑦
1+ 1+
}
2ℎ
𝑦
Torsion:
Impact shear stress,
𝜏′
= 𝜏{ 1 + 1 +
2ℎ
𝑟Ɵ
}
′
Angular deformation under impact action, Ɵ = Ɵ 1 + 1 +
2ℎ
𝑟Ɵ
𝜎𝑏 = Bending stress due to static weight; h = height through which load falls; y= static deflection of the
beam due to weight W.
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Problem 14:
A rectangular bar 200 mm long is subjected to an impact load of 2 kN that falls from
a height of 20 mm. Determine the dimensions of the bar if the allowable stress is
125 Mpa. Assume the thickness as twice the width. Take E= 200 GPa
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Solution:
𝜎′ = 𝜎
1+ 1+
125 =
2𝐴𝐸ℎ
WL
or 𝜎 ′ =
𝑊
A
1+ 1+
2𝐴𝐸ℎ
WL
2000
2𝐴 𝑋 2𝑋 105 𝑋 20
1+ 1+
A
2000 X 200
0.0625 A = [ 1 + 1 + 20 𝐴 ]
A = 5147 mm2
A = b X t = b X (2b)
5147 = 2b2
b = 52 mm, t = 104 mm
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 15:
A bar of 16 mm diameter gets stretched by 4 mm under a steady load of 10 kN.
What stress would be produced in the bar by a weight of 1000 N, which falls through
100 mm before commencing the stretch of rod which is initially unstressed. Also
calculate the impact factor for stress.
Take E = 200 GPa.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
𝜎′ =
𝛿=
𝜎′
=
1000
201.6
𝑃𝐿
,
𝐴𝐸
4=
1+ 1+
Impact factor = 1 + 1 +
𝑊
A
1+ 1+
2𝐴𝐸ℎ
WL
10000 𝑋𝐿
201.6 𝑋 (2 𝑋 105)
L = 16128 mm
2 𝑋 201.6 𝑋 2 𝑋105𝑋100
1000 X 16084
2ℎ
𝛿
== 1 + 1 +
, 𝜎 ′ = 116 MPa
2 𝑋 100
4
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= 𝟖. 𝟏𝟒
Problem 16:
A beam of rectangular cross-section has a width of 30 mm and 40 mm depth. The
beam is simply supported and the length of beam is 0.9 m. If it is struck by a mass of
10 kg falling through a height of 80 mm, find the instantaneous stress developed.
Take E = 210 GPa.
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Solution:
′
Impact stress due to bending, 𝜎𝑏 = 𝜎𝑏
𝛔𝐛 =
𝐌
,
𝐙
I
y
Z= =
bd3
12 X d/2
=
bd2
6
=
30 X 402
6
1+ 1+
2ℎ
𝑦
= 8000 mm3
For a simply supported beam with point load at mid span, bending moment is given by
(98 X 900)
𝐖𝐋
M=
=
= 22 x 103 N.mm, 𝛔 = M = 22 x 103 = 2.76 MPa
𝐛
Z
8000
𝟒
4
𝟑
𝐖𝐋
Deflection, y =
𝟒𝟖 𝐄𝐈
′
𝜎𝑏 = 𝜎𝑏
9003
= 48 X21098xx103
= 0.0433 mm
x160000
1+ 1+
2ℎ
𝑦
= 2.76
1+ 1+
2 𝑋 80
0.0433
= 168.57
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MPa
Problem 17:
A cantilever beam 12 mm deep, 8 mm wide and 300 mm long is loaded as follows, at
the free end. Determine the maximum bending stress in each case.
(i) A load of 50 N applied gradually.
(ii) A load of 50 N dropped through a distance of 5 mm.
Take E = 210 GPa.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
Impact stress due to bending, 𝜎𝑏 ′ = 𝜎𝑏
𝛔𝐛 =
𝐌
,
𝐙
I
y
Z= =
bd3
12 X d/2
=
bd2
6
=
8 X 122
6
1+ 1+
2ℎ
𝑦
= 192 mm3
For a simply supported beam with point load at mid-span, bending moment is given by
M = WL = 50 x 300 = 15000 N.mm, σb =
Deflection, y =
WL3
3EI
𝜎 𝑏 ′ = 𝜎𝑏
15000
192
= 𝟕𝟖. 𝟏𝟐 𝐌𝐏𝐚
50 x 3003
= 3 x 210
= 1.86 mm
x103 x1152
1+ 1+
2ℎ
𝑦
= 78.12
1+ 1+
2𝑋5
1.86
= 275.40
Department of Mechanical Engineering, National Institute of Technology Calicut
MPa
Problem 18:
A mass of 500 kg is lowered by means of a steel wire rope having cross-sectional
area of 250 mm2. The velocity of weight is 0.5 m/sec. When the length of the
extended rope is 20 m, the sheave gets stuck up. Determine the stresses induced in
the rope due to sudden stoppage of the sheave. Neglect friction and take E=190
Gpa.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
𝛔𝟐
𝟐𝐄
Strain energy per unit volume, 𝐔 =
Volume, V = A x L = 250 x 20 x 103 = 5 x 106 mm3
Total strain energy, U =
σ2
2E
.V
2
σ
=
x 5 x 106 = 0.01357 σ2 N.m
2 x 190 x 103
Kinetic energy of the system, Ek =
mv2
2
= 500 x2 0.52 = 62.5 N.m
0.01357 σ2 = 62.5
𝛔 = 𝟔𝟖. 𝟗𝟐 𝐌𝐏𝐚
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Design for Fatigue Load
▪ When a material is subjected to repeated stresses (due to fluctuating/repeated
/alternating loads), it fails at a stress far below the yield point stress. Such type of
failure is referred to as fatigue.
o Fatigue is a progressive localized structural failure.
o Fatigue failure is always brittle and catastrophic.
o Accounts for 90% of all failures in metals.
o Fatigue life is influenced by various factors.
o Size variation, holes, notches and other stress raisers contribute to fatigue failure.
o Smooth surface finish and residual compressive stresses increases fatigue
strength.
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▪ The three stages involved in fatigue failure are:
• Crack initiation (Small cracks in areas of localized stress concentration)
• Crack propagation (opening and closing of crack during each cycle continues
and results in crack propagation)
• Fracture (As crack size increase, size of cross-sectional area resisting applied
stress decrease and finally becomes insufficient to resist applied stress and
results in a sudden fracture)
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Varying Stresses
▪ Fluctuating stress - varies in sinusoidal manner from a max value to a min value of
same nature (tensile or compressive)
▪ Repeated stress – varies in sinusoidal manner from zero to a max. value (tensile or
compressive)
▪ Reversed stress – varies in sinusoidal manner from one value in compression to the
same value in tension
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Terminology
▪ Range of stress: 𝚫𝛔 = 𝝈𝒎𝒂𝒙 − 𝝈𝒎𝒊𝒏
▪ Mean/average stress: 𝝈𝐦 =
▪ Stress amplitude: 𝝈𝒂 =
▪ Stress ratio: R =
𝝈𝒎𝒂𝒙 + 𝝈𝒎𝒊𝒏
𝟐
𝝈𝒎𝒂𝒙 − 𝝈𝒎𝒊𝒏
𝟐
𝝈
𝝈𝐦 , 𝝈𝒂 = 𝒎𝒂𝒙
𝟐
R=0, A=1
𝝈𝒎𝒊𝒏
𝝈𝒎𝒂𝒙
▪ Amplitude ratio: A =
𝝈𝒂
𝝈𝒎
Department of Mechanical Engineering, National Institute of Technology Calicut
𝝈𝐦𝒂𝒙 = - 𝛔𝐦𝐢𝐧
R=-1, A=∞
S-N Curve/Endurance Strength
▪ S-N curve (Strength-Life curve): A plot of fatigue alternating stress versus the
number of cycles to failure plotted on semi-log paper.
▪ S-N curve for a material is generated in laboratory by RR Moore test.
▪
Schematic of RR Moore rotating beam fatigue machine
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RR Moore Test
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RR Moore Test
▪ The beam specimen is subjected to completely reversed stress with tensile stress
in the first half and compressive in the second half.
▪ One stress cycle is completed in one revolution.
▪ The amplitude of the cycle is given by:
𝑴 𝒃𝒚
𝝈𝒕 𝒐𝒓 𝝈𝒄 =
𝑰
▪ First test is selected such that it produces a stress equal to the static strength Sut
(Generally 0.9 Sut)
▪ Second test is conducted at a stress less than that of the first test.
▪ Each test on the fatigue testing machine gives one failure point on the S–N
diagram.
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RR Moore Test
▪ In each test, two readings are taken:
Stress amplitude (Sf) and number of cycles (N)
▪ These readings are used as two co-ordinates of the failure point in S-N curve.
▪ Large number of tests are performed to determine the endurance limit of a
material.
▪ The fatigue or endurance limit of a material is defined as the maximum
amplitude of completely reversed stress that the standard specimen can
sustain for an unlimited number of cycles without fatigue failure.
▪ The results of the tests are plotted to get the S-N curve.
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S-N curve for steel
▪ The magnitude of stress
amplitude corresponding to
an infinite number of stress
cycles (106 cycles) represents
the endurance limit of the
material.
▪ The endurance limit is not a
material property.
▪ It depends on size, shape,
finish, temperature etc.
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Department of Mechanical Engineering, National Institute of Technology Calicut
Design for Fatigue Load
Low Cycle Fatigue (LCF)
High Cycle Fatigue (HCF)
▪ Strain controlled – Plastic
strain occurs
▪ Stress controlled – Strain
confined to elastic region
▪ High loads & short lives
▪ Low loads & longer lives
▪ Low number of cycles to
produce fracture < 103
▪ High number of cycles to
produce fracture > 103
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Endurance/fatigue limit
Endurance/fatigue strength
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Endurance limit modifying factors
▪ The endurance limit of a component is different from the endurance limit of a
rotating beam specimen due to a number of factors.
▪ The difference arises due to the fact that there are standard specifications and
working conditions for the rotating beam specimen, while the actual
components have different specifications and work under different conditions.
▪ Different modifying factors are used in practice to account for this difference.
▪ The purpose of derating factors is to reduce the endurance limit of a rotating
beam specimen to suit the actual component.
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Endurance limit modifying factors
▪ Se’ = endurance limit stress of a rotating beam specimen subjected to reversed
bending stress
▪ Se = endurance limit stress of a particular mechanical component subjected to
reversed bending stress
Se = Ka Kb Kc Kd Se’
Ka = surface finish factor
Kb = size factor
Kc = reliability factor
Kd = modifying factor to account for stress concentration.
Department of Mechanical Engineering, National Institute of Technology Calicut
Surface finish factor, Ka
▪ The surface of the rotating beam specimen is polished to mirror finish.
▪ The actual component will have scratches and geometric irregularities on the
surface. These surface scratches serve as stress raisers and result in stress
concentration.
▪ The endurance limit is reduced due to introduction of stress concentration at
these scratches.
The values of Ka based on ultimate stress are given in Pg 32, and Pg 46, Databook
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Size factor, Kb
▪ The larger the machine part, the greater the probability that a flaw exists in the
component. A larger part is thus more likely to fail at lower stress
▪ Size factor Kb takes into account the reduction in endurance limit due to increase
in the size of the component.
▪ Kb⍺
1
d
▪ Empirical relation:
o Bending and torsion:
Only for circular crossFor 2.79 mm ≤ d ≤ 51 mm: Kb = 1.24 d–0.107
section with rotation
For 51 mm ≤ d ≤ 254 mm: Kb = 0.859 – 0.000 873 d
o Axial load:
Kb =1
Department of Mechanical Engineering, National Institute of Technology Calicut
Size factor, Kb
▪ An effective diameter is calculated for non-circular cross-section based on an
equivalent circular cross-section.
The effective diameter is obtained by equating the volume of the material stressed at and above 95% of the
maximum stress to the equivalent volume in the rotating beam specimen. When these two volumes are equated, the
lengths
▪ For bending and torsion, the values of size factor (Kb):
The values of Kb are given in Pg 22, and Pg 33, Databook
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Reliability factor, Kc
▪ There is considerable dispersion in fatigue
test results.
▪ The reliability factor Kc depends upon the
reliability that is used in the design of the
component.
▪ The more is the reliability and lower is the
reliability factor.
To ensure that more parts will survive, the stress amplitude on the component should
be lower than the tabulated value of the endurance limit. The reliability factor is used
to achieve this reduction.
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Stress concentration factor, Kd
▪ The endurance limit is reduced due to stress concentration.
Kd =
1
,
Kf
K f = 1 + q(K t -1)
K f = Fatigue stress concentration factor
K t = Theoretical stress concentration factor
▪ The stress concentration factor used for cyclic loading is less than the theoretical
stress concentration factor, due to the notch sensitivity of the material.
▪ Notch sensitivity factor, q =
Increase of actual stress over nominal
Increase of theoretical stress over nominal
▪ Values of q are given in Pg 21, 32, and 46, Databook
(q varies from 0 to 1)
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Se’ and Sut
▪ Approximate relationship between the endurance limit and the ultimate tensile
strength (Sut) of the material.
Department of Mechanical Engineering, National Institute of Technology Calicut
Fatigue Failure Theories
𝝈𝒂
𝑺𝒆
Tension zone
Compression zone
0
𝝈𝒎
𝝈𝒖𝒕
▪ If the load is compressive in nature, we can ignore fatigue for design
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Effect of mean stress on endurance strength
Department of Mechanical Engineering, National Institute of Technology Calicut
Fatigue Failure Theories
▪ Subjected to fluctuating axial or bending stress
o Soderberg criteria
𝜎𝑎 𝜎𝑚
+
=1
𝑠𝑒 𝑠𝑦𝑡
𝜎𝑎 𝜎𝑚
1
+
=
𝑠𝑒 𝑠𝑦𝑡 𝐹𝑆
▪ Most conservative
safest design model
𝜎𝑎 𝜎𝑚
1
+
=
𝑠𝑒 𝑠𝑢𝑡 𝐹𝑆
▪ Safe design
o Goodman criteria
𝜎𝑎 𝜎𝑚
+
=1
𝑠𝑒 𝑠𝑢𝑡
Department of Mechanical Engineering, National Institute of Technology Calicut
and
Fatigue Failure Theories
o Gerber’s criteria
𝜎𝑎
𝜎𝑚
+
𝑠𝑒
𝑠𝑢𝑡
2
𝜎𝑎
𝜎𝑚
𝐹𝑆 +
. 𝐹𝑆
𝑠𝑒
𝑠𝑢𝑡
=1
▪ Accurately predicts failure points
▪ Not commonly used
2
=1
Department of Mechanical Engineering, National Institute of Technology Calicut
Fatigue Failure Theories
▪ Considering the stress concentration factors, the criterions can be rewritten as:
o For ductile materials
𝐾𝑓 𝜎𝑎 𝜎𝑚
1
+
=
𝑠𝑒
𝑠𝑦𝑡 𝐹𝑆
▪ Local stress relief through
plastic deformation
o For brittle materials
𝐾𝑓 𝜎𝑎 𝐾𝑡 𝜎𝑚
1
+
=
𝑠𝑒
𝑠𝑢𝑡
𝐹𝑆
▪ 𝐾𝑡 is also used
Department of Mechanical Engineering, National Institute of Technology Calicut
Modified Goodman line
Subjected to fluctuating axial or bending stress
▪ Goodman line is ‘modified’ by
combining fatigue failure line
with yield line.
𝜎𝑎
𝑃𝑎
𝑇𝑎𝑛𝜃 =
=
𝜎𝑚 𝑃𝑚
(𝑀𝑏)𝑎
𝑇𝑎𝑛𝜃 =
(𝑀𝑏)𝑚
▪ (Sm, Sa) represent the limiting
values of stresses, used to
calculate dimensions of the
component.
Department of Mechanical Engineering, National Institute of Technology Calicut
Endurance limit
Subjected to fluctuating torsional shear stress
The endurance limit (Sse) of a component subjected to fluctuating torsional shear
stresses is obtained from the endurance limit in completely reversed bending (Se)
using theories of failures.
▪ According to the maximum shear stress theory,
Sse = 0.5 Se
▪ According to distortion energy theory,
Sse = 0.577 Se
Endurance limit in axial loading
▪ lower than that of the rotating beam test.
▪ (Se)a = 0.8 Se
Department of Mechanical Engineering, National Institute of Technology Calicut
Fatigue Failure Theories
Subjected to fluctuating torsional shear stress
o Soderberg criteria
𝜏𝑎 𝜏𝑚
+
=1
𝑠𝑠𝑒 𝑠𝑠𝑦
𝜏𝑎 𝜏𝑚
1
+
=
𝑠𝑠𝑒 𝑠𝑠𝑦 𝐹𝑆
𝐾𝑠𝑓 𝜏𝑎 𝜏𝑚
1
+
=
𝑠𝑠𝑒
𝑠𝑠𝑦 𝐹𝑆
𝜏 𝑎 𝜏𝑚
1
+
=
𝑠𝑠𝑒 𝑠𝑠𝑢 𝐹𝑆
𝐾𝑠𝑓 𝜏𝑎 𝜏𝑚
1
+
=
𝑠𝑠𝑒
𝑠𝑠𝑢 𝐹𝑆
o Goodman criteria
𝜏𝑎 𝜏𝑚
+
=1
𝑠𝑠𝑒 𝑠𝑠𝑢
Department of Mechanical Engineering, National Institute of Technology Calicut
Modified Goodman line
Subjected to fluctuating torsional shear stress
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Cumulative Fatigue
Miners equation
▪ Sometimes, a machine element is subjected to different stress levels for different
parts of the work cycle. The life of such a component is determined by Miner’s
equation.
▪
𝑛1
𝑁1
▪
⍺1𝑁
𝑁1
+
𝑛2
𝑁2
+
+
⍺2𝑁
𝑁2
𝑛3
𝑁3
+⋯+
+
⍺3𝑁
𝑁3
𝑛𝑥
𝑁𝑥
+⋯+
=1
⍺𝑥𝑁
𝑁𝑥
=1,
⍺1
𝑁1
+
⍺2
𝑁2
+
⍺3
𝑁3
+⋯+
⍺𝑥
𝑁𝑥
=
1
𝑁
N1: Total life if only 𝜎1 is acting; n1: no of cycles of 𝜎1 ; N = Total life of component;
⍺ = Proportion of total life
▪ Total life of the component subjected to different stress levels can be determined.
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 19:
A transmission shaft of cold drawn steel 27Mn2 (Sut = 500 N/mm2 and Syt = 300
N/mm2) is subjected to a fluctuating torque which varies from –100 N-m to + 400
Nm. The factor of safety is 2 and the expected reliability is 90%. Neglecting the
effect of stress concentration, determine the diameter of the shaft.
Assume the distortion energy theory of failure
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
Cold drawn steel
Sut = 500 N/mm2, Syt = 300 N/mm2, Tmax = 400 Nm, T min = -100 Nm, FS=2, Reliability = 90% , Neglect Kf
Endurance limit
Se’ = 0.5Sut = 0.5 (500) = 250 N/mm2
For cold drawn steel and Sut = 500 N/mm2 , Ka = 0.79
Assuming 7.5 < d <50 mm, Kb = 0.85
For 90% reliability, Kc = 0.897
Se = Ka Kb Kc Kd Se’ = 0.79 (0.85) (0.897) (250) = 150.58 N/mm2
Department of Mechanical Engineering, National Institute of Technology Calicut
Sse = 0.577Se = 0.577 (150.58) = 86.88 N/mm2
Ssy = 0.577Syt = 0.577 (300) = 173.1
𝑇𝑚𝑎𝑥 − 𝑇𝑚𝑖𝑛
2
𝑇𝑎 =
𝑇m =
𝑇𝑚𝑎𝑥 + 𝑇𝑚𝑖𝑛
2
𝑇𝑎𝑛𝜃 =
𝑇𝑎
𝑇𝑚
=
N/mm2
According to distortion energy theory
400 −(−100)
= 250 N m
2
= 400 + −100 = 150 N m
250
= 150 = 𝟏. 𝟔𝟕, 𝜃 = 59.04o
The ordinate of the point X is Sse or 86.88 N/mm2
Ssa = 86.88 N/mm2
𝜏𝑎 =
Ssa
𝐹𝑆
16𝑇𝑎
𝜋𝑑3
=
Ssa
𝐹𝑆
16 x 250 x 1000
𝜋𝑑3
=
86.88
2
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𝐝 = 𝟑𝟎. 𝟖 𝐦𝐦
Problem 20:
A steel rod(Sut = 1090 N/mm2 and Syt = 690 N/mm2, Se = 428 N/mm2) is subjected to
a tensile load, which varies from 120 KN to 40 KN. Design the safe diameter of the
rod using Soderberg diagram. Take FS=2, stress concentration factor = 1, correction
for size and surface as 0.85 and 0.91 respectively.
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 20:
Sut = 1090 N/mm2, Syt = 690 N/mm2, Se’ = 428 N/mm2
F = 120 KN to 40 KN. FS=2, Kt= 1, Kb = 0.85, Ka = 0.91
Soderberg’s criteria:
𝑲𝒇 𝝈𝒂
𝒔𝒆
+
𝐹
−𝐹
120 − 40
𝑭𝒂 = 𝑚𝑎𝑥 𝑚𝑖𝑛 =
2
2
𝐹
+𝐹
120+ 40
𝑭𝒎 = 𝑚𝑎𝑥 𝑚𝑖𝑛 =
2
2
𝝈𝒎
𝒔𝒚𝒕
=
𝟏
𝑭𝑺
= 40 KN,
= 80 KN,
𝑭
𝝈𝒂 = 𝒂
𝑨
𝑭
𝝈𝒎 = 𝒂
𝑨
50.91 x 103
=𝜋 2=
𝑑2
𝑑
4
𝑭𝒂
101.86 x 103
=𝜋 2=
𝑑2
𝑑
𝑭𝒂
4
For uniform cross-section, Kt= 1, q = 0,
K f = 1 + q(K t -1) Kf= 1
Kd= 1
Se = Ka Kb Kc Kd Se’ = 0.91 x 0.85 x 1 x 428 = 331 N/mm2
Department of Mechanical Engineering, National Institute of Technology Calicut
Se = 331 N/mm2
𝑲𝒇 𝝈𝒂 𝝈𝒎
𝟏
+
=
𝒔𝒆
𝒔𝒚𝒕 𝑭𝑺
50.91 x 103 101.86 x 103
1x
1
2
𝑑2
𝑑
+
=
331
690
2
𝐝 = 𝟐𝟓 𝐦𝐦
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 21:
A plate made of steel 20C8 (Sut = 440 N/mm2) in hot rolled and normalised
condition is shown in Fig. It is subjected to a completely reversed axial load of 30 kN.
The notch sensitivity factor q can be taken as 0.8 and the expected reliability is 90%.
The size factor is 0.85. The factor of safety is 2. Determine the plate thickness for
infinite life.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
P = ± 30 kN Sut = 440 N/mm2
FS= 2
R = 90%
q = 0.8
Kb = 0.85
Se’ = 0.5 Sut = 0.5 x 440 = 220 N/mm2
For hot rolled steel and Sut = 440 N/mm2 , Ka = 0.67
For 90% reliability, Kc = 0.897
For Kt ,
d
W
=
10
50
= 0.2,
Kt = 2.51, K f = 1 + q(K t -1) = 1 + 0.8(2.51-1) = 2.208
Kd = 0.4529
Se = Ka Kb Kc Kd Se’ = 0.67 x 0.85 x 0.897 x 0.4529 x 220 = 50.9 N/mm2
For axial load, (Se)a = 0.8 Se = 0.8 x 50.9 = 40.72 N/mm2
Department of Mechanical Engineering, National Institute of Technology Calicut
𝜎𝑎 =
(Se)a 40.72
=
= 20.36 N/mm2
FS
Also, 𝜎𝑎 =
2
Pa
w−d t
30 x 1000
=
50−10 t
30 x 1000
= 20.36
50 − 10 t
t = 36.84 mm
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 22:
For a machine element made of steel with geometry as shown in figure is subjected
to a complete reverse load P. Given Sut=1400 Mpa, Ka = 0.89, Kb = 1, Kc = 0.897, Kt
=2.3, q= 0.95. Thickness of the element is 10 mm. Determine the value of load P for
an infinite life.
r=1
H = 20
H = 30
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
Sut = 1400 N/mm2 Ka = 0.89, Kb = 1, Kc = 0.897, Kt =2.3, q= 0.95.
K f = 1 + q(K t -1) = 1 + 0.95 (2.3-1) = 2.235
K 𝑑 = 1/K f = 0.447
Se’ = 0.5 Sut = 0.5 x 1400 = 700 N/mm2
Se = Ka Kb Kc Kd Se’ = 0.89 x 1 x 0.897 x 0.447 x 700 = 249.79 N/mm2
Considering the bending moment acting on the critical section
𝑀
− 𝑀𝑚𝑖𝑛
𝑀𝑎 = 𝑚𝑎𝑥
2
𝑀𝑚 = 0
= P x 100 Nmm
Department of Mechanical Engineering, National Institute of Technology Calicut
σa =
Ma y
I
x (20/2)
= Px100
= 0.15 P
(10 x 203)/12
𝐾𝑓 𝜎𝑎 𝜎𝑚
1
+
=
𝑠𝑒
𝑠𝑢𝑡 𝐹𝑆
𝐾𝑓 𝜎𝑎
1
=
𝑠𝑒
𝐹𝑆
2.235 x 0.15 P = 1 x 249.7
P = 745 N
Department of Mechanical Engineering, National Institute of Technology Calicut
Problem 23:
The work cycle of a mechanical component subjected to complete ly reversed
bending stresses consists of the following three elements:
(i) ± 350 N/mm2 for 85% of time
(ii) ± 400 N/mm2 for 12% of time
(iii) ± 500 N/mm2 for 3% of time
The material for the component is 50C4 (Sut = 660 N/mm2) and the corrected
endurance limit of the component is 280 N/mm2.
Determine the life of the component.
Department of Mechanical Engineering, National Institute of Technology Calicut
Solution:
Sut = 660 N/mm2 Se = 280 N/mm2
Step I Construction of S–N diagram
0.9Sut = 0.9(660) = 594 N/mm2
log10 (0.9Sut) = log10 (594) = 2.7738
log10 (Se) = log10 (280) = 2.4472
log10 (s1) = log10 (350) = 2.5441
log10 (s2) = log10 (400) = 2.6021
log10 (s3) = log10 (500) = 2.6990
The S–N curve for this problem is
Department of Mechanical Engineering, National Institute of Technology Calicut
log10 N = 3 + 9.1855 (2.7738 – log10 𝜎)
log10 (N1) = 3 + 9.1855 (2.7738 – 2.5441) or N1 = 128 798
log10 (N2) = 3 + 9.1855 (2.7738 – 2.6021) or N2 = 37 770
log10 (N3) = 3 + 9.1855 (2.7738 – 2.6990) or N3 = 4865
⍺1
We know,
𝑁1
+
⍺2
𝑁2
0.85
128798
⍺3
+
𝑁3
+
⍺𝑥
𝑁𝑥
=
1
𝑁
0.03
4865
=
1
𝑁
+⋯+
0.12
37770
+
N = 62,723 cycles
Department of Mechanical Engineering, National Institute of Technology Calicut