2. Syllabus
UNIT – I
INTRODUCTION: General considerations in the design of Engineering Materials and their properties – selection –
Manufacturing consideration in design, tolerances and fits –BIS codes of steels.
STRESSES IN MACHINE MEMBERS: Simple stresses – combined stresses – torsional and bending stresses – impact
stresses – stress strain relation – various theories of failure – factor of safety – design for strength and rigidity – preferred
numbers. the concept of stiffness in tension, bending, torsion and combined situations – static strength design based on
fracture toughness.
UNIT – II
STRENGTH OF MACHINE ELEMENTS: Stress concentration – theoretical stress concentration factor – fatigue stress
concentration factor notch sensitivity – design for fluctuating stresses – endurance limit – estimation of endurance strength
– Goodman’s line – Soderberg’s line – modified Goodman’s line, Gerber’s parabola.
UNIT – III
RIVETED AND WELDED JOINTS – design of joints with initial stresses – eccentric loading.
Bolted joints – design of bolts with pre-stresses – design of joints under eccentric loading – locking devices – bolts of
uniform strength.
KEYS, COTTERS AND KNUCKLE JOINTS: Design of keys-stresses in keys-cotter joints-spigot and socket, sleeve
and cotter, jib and cotter joints- knuckle joints.
UNIT – IV
SHAFTS: Design of solid and hollow shafts for strength and rigidity – design of shafts for combined bending and axial
loads – shaft sizes – BIS code. Use of internal and external circlips, gaskets and seals (stationary & rotary).
SHAFT COUPLING: Rigid couplings – muff, split muff and flange couplings; rigid flanged couplings – protected rigid
flanged, Bushed pin type flexible coupling.
UNIT – V
MECHANICAL SPRINGS:
Stresses and deflections of helical springs – extension -compression springs – springs for fatigue loading, energy storage
capacity – helical torsion springs – co-axial springs, leaf springs.
3. Contents of Design
• Engineering Mechanics
Forces, equilibrium, moments, supports,
reactions, - statics and dynamics.
• Materials science
types, techniques for improvement, properties
• Strength of materials
everything- stress, strain, theories of failure,
beams, structures, columns, springs, strain energy
• Kinematics and dynamics
links, motion generators
4. INTRODUCTION: General considerations in the design of Engineering
Materials and their properties – selection –Manufacturing consideration in
design, tolerances and fits –BIS codes of steels.
STRESSES IN MACHINE MEMBERS: Simple stresses – combined
stresses – torsional and bending stresses – impact stresses – stress strain
relation – various theories of failure – factor of safety – design for strength
and rigidity – preferred numbers. the concept of stiffness in tension, bending,
torsion and combined situations – static strength design based on fracture
toughness.
Introduction – Lecture
5. MACHINE?
• A machine is a combination of
resisting bodies with successfully
constrained relative motions
which is used to transform other
forms of energy into mechanical
energy or transmit and modify
available energy to do some
useful work.
EXAMPLE: HEAT ENGINE
Heat Energy Mechanical Energy
6. Machine Design
• It is the creation of new and better machines
and improving existing ones
• A new or a better machine is one which is
more economical in overall cost of production
and operation.
8. Purpose of design Design for what situation Material Finally the aesthetics
Adaptive Design (slight modifications in existing one) Development Design New
Design
OTHER TYPES
Rational design (Determining size based on stresses and strains)
Empirical design (Based on Empirical relations i.e. scientific principles)
Industrial design (based on market survey, low cost, production facilities, reverse
Engineering)
FACTORS TO BE CONSIDERED
Mechanism to be used Material Loading Size, shape & space
requirements Method of manufacturing Operating environment
Safety Inspect ability Maintenance, Cost & Aesthetics
Machine design
(Technical Decision making process)
9.
10. Factors influencing
Machine Design
Strength
(Internal resistance offered by a part)
Rigidity
(Resistance to deformation)
Wear resistance
(Resistance to wear)
Minimum dimensions and weight
Manufacturability
(Type of fabrication processes)
Safety
Conformance to standards
(Use of standard parts)
Reliability
(Trustworthiness)
Minimum life cycle cost
(Maintenance)
11. • Strength ability to resist external forces
• Stiffness ability to resist deformation under stress
• Elasticity property to regain its original shape
• Plasticity property which retains the deformation produced under load
• Ductility property of a material to be drawn into wire form with using tensile force
• Brittleness property of breaking a material without any deformation
• Malleability property of a material to be rolled or hammered into thin sheets
• Toughness property to resist fracture under impact load
• Machinability property of a material to be cut
• Resilience property of a material to absorb energy
• Creep material undergoes slow and permanent deformation when subjected to constant stress
with high temperature
• Fatigue failure of material due to cyclic loading
• Hardness resistant to indentation, scratch
Material selection based
on Mechanical properties
12. Failure Mechanisms
Wear Out
Overstress
Mechanical Thermal Electrical Mechanical Thermal Electrical chemical
Excess
Deformation
Yield
Fracture
EOS
ESD
Latch-up
Thermal
Overstress
Melting
Fatigue
Creep
Wear
Stress Voids
Diffusion
Related
Intermetallic
Growth
Hillock
Formation
Electro-
-migration
Corrosion
Stress
Corrosion
Dendrite
Growth
Depolymerization
Failure Mechanisms in Products
15. Important considerations in Machine Design
1. Type of LOAD and STRESSes caused by the load
Steady loads
• Dead loads
• Live loads
Variable loads
• Shock loads (suddenly)
• Impact loads (applied with
some velocity)
• Stress and strain
(Tensile, compressive, shear)
• Thermal stresses
• Torsional stresses
• Bending stress
16. Important considerations in Machine Design…..
2. KINEMATICS of the machine (Motion of the parts)
Find the simplest arrangement that would give the most efficient motion
that is required.
3. Selection of MATERIALS
Knowledge of the properties of the materials and their behaviour under
working conditions is required.
Strength, hardness, durability, flexibility, weight, resistance to heat and
corrosion, electrical conductivity, machinability, etc.
17. Important considerations in Machine Design…..
3. Selection of MATERIALs
Physical properties: Density, Melting point, Elec/thermal properties
Mechanical properties:
• STRENGTH – resist externally applied loads without breaking
or yielding
• STIFFNESS – resist deformation under stress
• ELASTICITY – regain original shape once the force is removed
• PLASTICITY – property which retains deformation (required for
forging etc)
• DUCTILITY – ability to be drawn into a wire by a tensile force
• BRITTLENESS – sudden breaking with minimum distortion
• TOUGHNESS – resist fracture due to high impact load
• CREEP – deformation under stress and high temperature
• FATIGUE – ability to withstand cyclic stresses
• HARDNESS – resistance to wear, scratching, deformation,
machinability etc
Metal Non-metal
Ferrous Non-ferrous
18. Important considerations in Machine Design…..
4. Form and size of the parts
The size will be determined by the forces/torques applied (stresses on the
object) and the material used such that failure (fracture or deformation)
would not occur
Frictional resistance and lubrication.
Convenient and economical features.
Use of standard parts.
Safety of operation.
Workshop facilities.
Cost of construction.
Assembling
5. Other considerations
19. Why stresses are induced in a component?
Due to Energy transmitted
(kinetic energy strain energy) Example: Spring
Due to Self weight of the machine component
Due to friction
Due to Inertia (resistance to change its position) of the moving parts
Due to Change in temperature
Due to fabrication
20. Stresses vs. Resisting Area’s
For Direct loading or Axial loading
For transverse loading
For tangential loading or twisting
Where I and J Resistance properties of cross sectional area
I Area moment of inertia of the cross section about the axes lying on the section
(i.e. xx and yy)
J Polar moment of inertia about the axis perpendicular to the section
23. 1. Proportional limit
2. Elastic limit
3. Yield point
4. Ultimate stress
5. Breaking stress
Measure of ductility
Percentage reduction in area
Percentage elongation
Necking
24. In engineering practice, the machine parts are subjected to various forces which may be due to
either one or more of the following:
1. Energy transmitted,
2. Weight of machine,
3. Frictional resistances,
4. Inertia of reciprocating parts,
5. Change of temperature, and
6. Lack of balance of moving parts.
Load
It is defined as any external force acting upon a machine part.
The following four types of the load are important from the subject point of view:
1. Dead or steady load. A load is said to be a dead or steady load, when it does not change in
magnitude or direction.
2. Live or variable load.A load is said to be a live or variable load, when it changes continually.
3. Suddenly applied or shock loads. A load is said to be a suddenly applied or shock load, when
it is suddenly applied or removed.
4. Impact load. A load is said to be an impact load, when it is applied with some initial velocity.
25. Stress
When some external system of forces or loads act on a body, the internal forces (equal and
opposite) are set up at various sections of the body, which resist the external forces. This
internal force per unit area at any section of the body is known as unit stress or simply a stress.
It is denoted by a Greek letter sigma (σ). Mathematically,
Stress, σ = P/A
where P = Force or load acting on a body, and
A = Cross-sectional area of the body.
In S.I. units, the stress is usually expressed in Pascal (Pa) such that 1 Pa = 1 N/m2. In actual
practice, we use bigger units of stress i.e. megapascal (MPa) and gigapascal (GPa), such that
1 MPa = 1 × 106 N/m2 = 1 N/mm2 and 1 GPa = 1 × 109 N/m2 = 1 kN/mm2
Strain
When a system of forces or loads act on a body, it undergoes some deformation. This
deformation per unit length is known as unit strain or simply a strain. It is denoted by a Greek
letter epsilon (Ɛ). Mathematically,
Strain, Ɛ = δl / l
where δl = Change in length of the body, and
l = Original length of the body.
26. Tensile Stress and Strain
Fig. Tensile stress and strain.
When a body is subjected to two equal and opposite axial pulls P (also called tensile load) as
shown in Fig. (a), then the stress induced at any section of the body is known as tensile stress
as shown in Fig. (b).
A little consideration will show that due to the tensile load, there will be a decrease in cross-
sectional area and an increase in length of the body. The ratio of the increase in length to the
original length is known as tensile strain.
Let P = Axial tensile force acting on the body,
A = Cross-sectional area of the body,
l = Original length, and
δl = Increase in length.
∴ Tensile stress, σt = P/A and tensile strain, εt = δl / l
27. Compressive Stress and Strain
Fig. Compressive stress and strain.
When a body is subjected to two equal and opposite axial pushes P (also called compressive
load) as shown in Fig (a), then the stress induced at any section of the body is known as
compressive stress as shown in Fig (b).
A little consideration will show that due to the compressive load, there will be an increase in
cross-sectional area and a decrease in length of the body. The ratio of the decrease in length to
the original length is known as compressive strain.
Let P = Axial compressive force acting on the body,
A = Cross-sectional area of the body,
l = Original length, and
δl = Decrease in length.
∴ Compressive stress, σc = P/A and compressive strain, εc = δl /l
28. Young's Modulus or Modulus of Elasticity
Hooke's law* states that when a material is loaded within elastic limit, the stress is directly
proportional to strain, i.e.
σ ∝ ε or σ = E.ε
where E is a constant of proportionality known as Young's modulus or modulus of elasticity. In
S.I. units, it is usually expressed in GPa i.e. GN/m2 or kN/mm2. It may be noted that Hooke's
law holds good for tension as well as compression.
Shear Stress and Strain
When a body is subjected to two equal and opposite forces acting tangentially across the
resisting section, as a result of which the body tends to shear off the section, then the stress
induced is called shear stress.
The corresponding strain is known as shear strain and it is measured by the angular
deformation accompanying the shear stress. The shear stress and shear strain are denoted by the
Greek letters tau (τ) and phi (φ) respectively. Mathematically,
Shear stress, τ = Tangential force/Resisting area
29. Consider a body consisting of two
plates connected by a rivet as shown in
Fig (a). In this case, the tangential force
P tends to shear off the rivet at one
cross-section as shown in Fig (b).
It may be noted that when the tangential force is resisted by one cross-section of the rivet (or
when shearing takes place at one cross-section of the rivet), then the rivets are said to be in
single shear. In such a case, the area resisting the shear off the rivet,
A = (π/4) × d2
and shear stress on the rivet cross-section, τ = P/A= (4P)/(π × d2)
It may be noted that when the tangential force is resisted by two cross-sections of the rivet (or
when the shearing takes place at two cross-sections of the rivet), then the rivets are said to be
in double shear.
In such a case, the area resisting the shear off the rivet, A = 2 × ((π/4) × d2)
and shear stress on the rivet cross-section, τ = P/A= (2P)/(π × d2)
Now let us consider two plates
connected by the two cover
plates as shown in Fig (a). In
this case, the tangential force P
tends to shear off the rivet at
two cross-sections as shown in
Fig (b).
30. Shear Modulus or Modulus of Rigidity
It has been found experimentally that within the elastic limit, the shear stress is directly
proportional to shear strain.
Mathematically
τ ∝ φ or τ = C . φ or τ / φ = C
where τ = Shear stress,
φ = Shear strain, and
C = Constant of proportionality, known as shear modulus or modulus
of rigidity. It is also denoted by N or G.
Bearing Stress
A localised compressive stress at the surface of contact between two members of a machine part,
that are relatively at rest is known as bearing stress or crushing stress. The bearing stress is
taken into account in the design of riveted joints, cotter joints, knuckle joints, etc. Let us
consider a riveted joint subjected to a load P as shown in Fig. 4.9. In such a case, the bearing
stress or crushing stress (stress at the surface of contact between the rivet and a plate),
σb (or σc) = P/(d*t* n)
where d = Diameter of the rivet,
t = Thickness of the plate,
d.t = Projected area of the rivet, and
n = Number of rivets per pitch length in bearing or crushing.
Average bearing pressure for a journal supported in a bearing is
given by pb =P/(l * d)
where P = Radial load on the journal, l = Length of the journal in
contact, and d = Diameter of the journal.
31. Stress-strain Diagram 1. Proportional limit. We see from the diagram that from point O to A
is a straight line, which represents that the stress is proportional to
strain. Beyond point A, the curve slightly deviates from the straight
line. It is thus obvious, that Hooke's law holds good up to point A and
it is known as proportional limit. It is defined as that stress at which
the stress-strain curve begins to deviate from the straight line.
2. Elastic limit. It may be noted that even if the load is increased
beyond point A upto the point B, the material will regain its shape and
size when the load is removed. This means that the material has
elastic properties up to the point B. This point is known as elastic limit.
It is defined as the stress developed in the material without any
permanent set.
Note: Since the above two limits are very close to each other,
therefore, for all practical purposes these are taken to be equal.
3. Yield point. If the material is stressed beyond point B, the plastic stage will reach i.e. on the removal of
the load, the material will not be able to recover its original size and shape. A little consideration will show
that beyond point B, the strain increases at a faster rate with any increase in the stress until the point C is
reached. At this point, the material yields before the load and there is an appreciable strain without any
increase in stress. In case of mild steel, it will be seen that a small load drops to D, immediately after
yielding commences. Hence there are two yield points C and D. The points C and D are called the upper and
lower yield points respectively. The stress corresponding to yield point is known as yield point stress.
4. Ultimate stress. At D, the specimen regains some strength and higher values of stresses are required for
higher strains, than those between A and D. The stress (or load) goes on increasing till the point E is
reached. The gradual increase in the strain (or length) of the specimen is followed with the uniform
reduction of its cross-sectional area. The work done, during stretching the specimen, is transformed largely
into heat and the specimen becomes hot. At E, the stress, which attains its maximum value is known as
ultimate stress. It is defined as the largest stress obtained by dividing the largest value of the load reached
in a test to the original cross-sectional area of the test piece.
32. 5. Breaking stress. After the specimen has reached the ultimate stress, a neck is formed, which
decreases the cross-sectional area of the specimen, as shown in Fig (b). A little consideration
will show that the stress (or load) necessary to break away the specimen, is less than the
maximum stress. The stress is, therefore, reduced until the specimen breaks away at point F.
The stress corresponding to point F is known as breaking stress.
6. Percentage reduction in area. It is the difference between the original cross-sectional area
and cross-sectional area at the neck (i.e. where the fracture takes place). This difference is
expressed as percentage of the original cross-sectional area.
Let A = Original cross-sectional area, and
a = Cross-sectional area at the neck.
Then reduction in area = A – a
and percentage reduction in area = ((A-a)/A)*100
7. Percentage elongation. It is the percentage increase in the standard gauge length (i.e.
original length) obtained by measuring the fractured specimen after bringing the broken parts
together.
Let l = Gauge length or original length, and
L = Length of specimen after fracture or final length.
∴ Elongation = L – l
and percentage elongation = ((L-l)/l)*100
Working Stress
When designing machine parts, it is desirable to keep the stress lower than the maximum or
ultimate stress at which failure of the material takes place. This stress is known as the working
stress or design stress. It is also known as safe or allowable stress.
33. Factor of Safety
For Ductile Materials
For Brittle Materials
For Variable loading
34. Selection of FOS
• The selection of a proper factor of safety to be used in designing any machine
component depends upon a number of considerations, such as the material, mode of
manufacture, type of stress, general service conditions and shape of the parts.
• Before selecting a proper factor of safety, a design engineer should consider the
following points :
1. The reliability of the properties of the material and change of these properties during
service
2. The reliability of test results and accuracy of application of these results to actual
machine parts
3. The reliability of applied load
4. The certainty as to exact mode of failure
5. The extent of simplifying assumptions
6. The extent of localised stresses
7. The extent of initial stresses set up during manufacture
8. The extent of loss of life if failure occurs and
9. The extent of loss of property if failure occurs.
35. Stresses in Composite Bars
• A composite bar may be defined as a bar made up of two or more different materials,
joined together, in such a manner that the system extends or contracts as one unit, equally,
when subjected to tension or compression.
• In case of composite bars, the following points should be kept in view:
1. The extension or contraction of the bar being equal, the strain i.e. deformation per unit
length is also equal.
2. The total external load on the bar is equal to the sum of the loads carried by different
materials.
Consider a composite bar made up of two different materials as
shown in Fig.
Let P1 = Load carried by bar 1,
A1 = Cross-sectional area of bar 1,
σ1 = Stress produced in bar 1,
E1 = Young's modulus of bar 1,
P2, A2, σ2, E2 = Corresponding values of bar 2,
P = Total load on the composite bar,
l = Length of the composite bar, and
δl = Elongation of the composite bar.
Fig. Stresses in composite bars.
(i)
36. From the above equations, we can find out the stresses produced in the different bars. We also know that
P = P1 + P2 = σ1.A1 + σ2.A2
From this equation, we can also find out the stresses produced in different bars.
37. Stresses due to Change in Temperature—Thermal Stresses
• Whenever there is some increase or decrease in the temperature of a body, it causes the
body to expand or contract.
• A little consideration will show that if the body is allowed to expand or contract freely, with
the rise or fall of the temperature, no stresses are induced in the body. But, if the
deformation of the body is prevented, some stresses are induced in the body. Such
stresses are known as thermal stresses.
38. Linear and Lateral Strain
Consider a circular bar of diameter d and length l, subjected to a tensile force P as shown in
Fig (a).
A little consideration will show that due to tensile force, the length of the bar increases by an
amount δl and the diameter decreases by an amount δd, as shown in Fig (b). Similarly, if the
bar is subjected to a compressive force, the length of bar will decrease which will be followed
by increase in diameter.
It is thus obvious, that every direct stress is accompanied by a strain in its own direction which
is known as linear strain and an opposite kind of strain in every direction, at right angles to it,
is known as lateral strain.
Poisson's Ratio
It has been found experimentally that when a body is stressed within elastic limit, the lateral
strain bears a constant ratio to the linear strain, Mathematically,
Lateral strain/Linear strain= Constant
This constant is known as Poisson's ratio and is denoted by 1/m or μ.
39. Volumetric Strain
• When a body is subjected to a system of forces, it undergoes some changes in its
dimensions.
• In other words, the volume of the body is changed. The ratio of the change in volume to
the original volume is known as volumetric strain. Mathematically,
volumetric strain, εv = δV / V
where δV = Change in volume,
and V = Original volume.
Bulk Modulus
• When a body is subjected to three mutually perpendicular stresses, of equal intensity, then
the ratio of the direct stress to the corresponding volumetric strain is known as bulk
modulus. It is usually denoted by K. Mathematically,
bulk modulus, K = Direct stress/Volumetric strain = σ/(Δv/V)
Relation Between Bulk Modulus and Young’s Modulus
• The bulk modulus (K) and Young's modulus (E) are related by the following relation,
Relation Between Young’s Modulus and Modulus of Rigidity
• The Young's modulus (E) and modulus of rigidity (G) are related by the following relation,
40. Impact stress
• Sometimes, machine
members are subjected to
the load with impact. The
stress produced due to
falling load is called impact
stress.
2
{
1 1 }
2
[1 1
W hAE
i A Wl
W
s
A
Wl
AE
then
h
s
i
41. Find the safe diameter of the pin A, if the allowable shear and tensile stresses
are 170 MPa and 350 MPa respectively?
Force at B = 5x0.1 / 0.15 = 3.33KN
Resultant force at A= = 6 kN.
Stresses developed in pin A: (a) shear stress (b) bearing stress
Considering double shear at A,
pin diameter d = d= 4.7 mm
Considering bearing stress at A, pin diameter d =
d =6.70 mm. A safe pin diameter is 10 mm.
2 2
5 3.33
3
6
2 6 10
170 10
3
6
4 6 10
170 10
42. A mass of 50 kg drops through 25 mm at the centre of a 250 mm long simply
supported rod having square cross section. The bar is made of steel having yield
strength of 400 N/mm2. Determine the dimension of the cross section.
Soln: 50 9.81 250 30656.25
4 4
3 4 4
12 12
490.5 250 30656.25
4
30656.25 ( )
2
4/12
2
400 200 /
2
4
2(25)
183973.5
200 [1 1
3 9526.11
3 1
846160.82 459.84 185.12
Neglecting second t
Wl
M N mm
b
bd a
I mm
Wl
AE
a
M y
b
s I a
y
N mm
i FOS
a
a
a a
erm
we get a 67.6 .
mm
43. A unknown weight falls through 10 mm on a collar rigidly attached to the lower end
of a vertical bar 3 m long and 600 mm2 in section. If the maximum instantaneous
extension is known to 2 mm. what is the corresponding stress and the value of
unknown weight? Take E = 200 kN/mm2.. (Nov. 2014)
Young’s modulus =
W= 6666.7 N
.
3 2
. 200 10 2 133.33 /
3000
stress l
strain l
E l N mm
l
2
{1 1 }
W hAE
i A Wl
45. Bending stress in straight beams
In a machine part structural members may be subjected to static and
dynamic loads which causes bending stress which is a combination of
tensile compressive and shear stresses.
The fibers on the upper side of the beam will be subjected to
compression and those on the lower side will be subjected to tension.
Some where in between there will be a surface which is subjected to
neither tension nor compression called neutral surface.
46.
47. Where Z = Section modulus
3
32
32
3
M M
y
I Z
d
Z
M
d
A pump lever shaft as shown in figure. It exerts force of 25 kN and 35 kN
concentrated at 150 mm and 200 mm from the left and right hand bearing
respectively. Find the diameter of the central portion of the shaft, if the stress is not
to exceed 100 MPa.
48. Taking moments about A,
35 * 750 + 25 *150 - RB * 950 = 0
RB = 30 000 / 950 = 31.58 kN = 31.58 × 103 N
RB = 31.58 kN. and RA = (35+25)-31.58=28.42 kN.
Bending moment at C = RA X 150 = 4.263 X 106 N-mm
Bending moment at D = RB X 200 = 6.316 X 106 N-mm
It is clear that maximum bending moment is at D, therefore maximum bending
moment,
M = 6.316 × 106 N-mm.
Z = (π*d3)/32= 0.0982 d3
Bending stress (σb) given as 100 MPa and it is equal to M/Z
100=M/Z⇒100=(6.316*106)/(0.0982 d3 ) ⇒d=86.3 mm say 90 mm.
51. R = Radius of shaft, L = Length of the shaft
T = Torque applied at the free end
C = Modulus of Rigidity of a shaft material
τ = torsional shear stress induced at the cross section
Ø = shear strain, θ = Angle of twist
Torsional Equation
52. Design for Bending
Design for Bending & Twisting
When a shaft is subjected to pure rotation, then it has to be designed for
bending stress which is induced due to bending moment caused by self
weight of the shaft.
Example: Rotating shaft between two bearings.
When a gear or pulley is mounted on a shaft by means of a key, then it
has to be designed for bending stress (induced due to bending moment)
and also torsional shear stress which is caused due to torque induced
by the resistance offered by the key .
Example: gearbox shaft (splines)
54. Torsional Equation :
C = Modulus of rigidity for the shaft material,
For a solid circular shaft polar moment of inertia is
r =d/2 (Distance of neutral axis from the outermost fiber).
C
T
r J l
4
32
d
J
4
2
32
3
16
T
d
d
T d
A shaft is transmitting 100 kW at 160 rpm. Find a suitable
diameter for the shaft, if the maximum torque transmitted
exceeds the mean by 25%. Take allowable shear stress as 70
MPa.
55. 3
2
60
60 100 10 60
2 2
3
max
3
16
1.25 7458 10
81.5
NT
P
mean N N
mean
P
T T
T T N mm
T d
d mm
A shaft transmitting 97.5 kW at 180 rpm. If the allowable shear stress in
the material is 60 MPa. Find the suitable diameter of the shaft if the twist
is not to exceed 1° in a length of 3 meters. Take C = 80 GPa.
56. 2
60
3 2 180
97.5 10
60
3
5172 10
3
16
76
NT
P
T
T N mm
T d
d mm
4 4
0.0982
32
3 3
5172 10 80 10 0.0174
4 3000
0.0982
103
C
T
J l
J d d
d
d mm
Considering the strength of the shaft
Considering the rigidity of the shaft
Taking maximum of these two d = 103 ≈105 mm
57. A hollow shaft is required to transmit 600 kW at 110 r.p.m., the maximum torque being 20% greater than the
mean. The shear stress is not to exceed 63 MPa and twist in a length of 3 metres not to exceed 1.4 degrees.
Find the external diameter of the shaft, if the internal diameter to the external diameter is 3/8. Take modulus of
rigidity as 84 GPa.
Home work problem
58. Principal Stresses for
various load combinations
In a machine component several types of loads are acting
simultaneously, then it is very important to find the maximum stress
induced plane in a component to avoid failure.
This max value of normal stress to the plane is known as principal stress
and the plane is known as principal plane. The value of shear stress
induced in that plane is zero.
Bending stresses in
two directions
Torsional shear stress
due to twisting
59. Determination of Principal Stresses for a Member Subjected to Bi-axial Stress
• When a member is subjected to bi-axial stress (i.e.
direct stress in two mutually perpendicular planes
accompanied by a simple shear stress), then the
normal and shear stresses are obtained.
• Consider a rectangular body ABCD of uniform cross-
sectional area and unit thickness subjected to normal
stresses σ1 and σ2 as shown in Fig. (a). In addition to
these normal stresses, a shear stress τ also acts.
(a) Direct stress in two mutually
prependicular planes accompanied by a
simple shear stress.
• WKT (from Strength of Materials) that the normal
stress across any oblique section such as EF inclined at
an angle θ with the direction of σ2, as shown in Fig.
(a), is given by
σt =
σ1+ σ2
2
+
σ1+ σ2
2
cos 2θ + τ sin 2θ ...(i)
and tangential stress (i.e. shear stress) across the section
EF,
τ1 =1/2(σ1 – σ2) sin 2θ – τ cos 2θ ...(ii)
• Since the planes of maximum and minimum normal
stress (i.e. principal planes) have no shear stress,
therefore the inclination of principal planes is
obtained by equating τ1 = 0 in the above equation (ii),
i.e.
1/2(σ1 – σ2) sin 2θ – τ cos 2θ ∴ tan 2θ =
2τ
σ1−σ2
(b) Direct stress in one plane accompanied
by a simple shear stress.
Fig. Principal stresses for a member subjected
to bi-axial stress.
60. We know that there are two principal planes at right angles to
each other. Let θ1 and θ2 be the inclinations of these planes with
the normal cross-section.
From Fig. shown here, we find that
sin 2𝜃 = ±
2𝜏
𝜎1−𝜎2
2+4𝜏2
∴ sin 2𝜃1 = +
2𝜏
𝜎1−𝜎2
2+4𝜏2
and sin 2𝜃2 = −
2𝜏
𝜎1−𝜎2
2+4𝜏2
also cos 2𝜃 = ±
𝜎1−𝜎2
𝜎1−𝜎2
2+4𝜏2
∴ cos 2𝜃1 = +
𝜎1−𝜎2
𝜎1−𝜎2
2+4𝜏2
and cos 2𝜃2 = −
𝜎1−𝜎2
𝜎1−𝜎2
2+4𝜏2
The maximum and minimum principal stresses may now be obtained by substituting the
values of sin 2θ and cos 2θ in equation (i).
∴ Maximum principal (or normal) stress, σt1 =
𝜎1+ 𝜎2
2
+
1
2
𝜎1 − 𝜎2
2 + 4𝜏2 …. (iv).
And normal principal (or normal) stress, σt2 =
𝜎1+ 𝜎2
2
−
1
2
𝜎1 − 𝜎2
2 + 4𝜏2 …. (v).
The planes of maximum shear stress are at right angles to each other and are inclined at 45°
to the principal planes. The maximum shear stress is given by one-half the algebraic
difference between the principal stresses, i.e.
τmax =
𝜎𝑡1+ 𝜎𝑡2
2
+
1
2
𝜎1 − 𝜎2
2 + 4𝜏2 …. (vi)
Notes: 1. When a member is subjected to direct stress in one plane accompanied by a simple shear stress as shown in
Fig. (previous slide), then the principal stresses are obtained by substituting σ2 = 0 in equation (iv), (v) and (vi).
2. In the above expression of σt2, the value of
1
2
𝜎1 − 𝜎2
2 + 4𝜏2 is more than
𝜎1
2
. Therefore the nature of σt2 will be
opposite to that of σt1, i.e. if σt1 is tensile then σt2 will be compressive and vice-versa.
61. Application of Principal Stresses in Designing Machine Members
• There are many cases in practice, in which machine members are subjected to combined
stresses due to simultaneous action of either tensile or compressive stresses combined with
shear stresses.
• In many shafts such as propeller shafts, C-frames etc., there are direct tensile or
compressive stresses due to the external force and shear stress due to torsion, which acts
normal to direct tensile or compressive stresses.
• The shafts like crank shafts, are subjected simultaneously to torsion and bending. In such
cases, the maximum principal stresses, due to the combination of tensile or compressive
stresses with shear stresses may be obtained.
1. Maximum tensile stress, σt(max) =
𝜎𝑡
2
+
1
2
𝜎𝑡
2 + 4𝜏2
2. Maximum compressive stress, σc(max) =
𝜎𝑐
2
+
1
2
𝜎𝑐
2 + 4𝜏2
3. Maximum shear stress, τmax =
1
2
𝜎𝑡
2 + 4𝜏2
where σt = Tensile stress due to direct load and bending,
σc = Compressive stress, and
τ = Shear stress due to torsion.
Notes : 1. When τ = 0 as in the case of thin cylindrical shell subjected in internal fluid pressure,
then σt (max) = σt
2. When the shaft is subjected to an axial load (P) in addition to bending and twisting
moments as in the propeller shafts of ship and shafts for driving worm gears, then the stress
due to axial load must be added to the bending stress (σb). This will give the resultant tensile
stress or compressive stress (σt or σc) depending upon the type of axial load (i.e. pull or push).
62. A hollow shaft of 40 mm outer diameter and 25 mm inner diameter is
subjected to a twisting moment of 120 Nm simultaneously it is
subjected to an axial thrust of 10 kN and a bending moment of
80 N-m. Calculate the maximum compressive and shear stress.
Soln: 2 2 2
0
4
( ) 766
i
A d d mm
Direct compressive stress due to axial thrust
3 2
10 10 1305 /
766
P N mm
c A
4 4
3
0
[ ] 5325
32
d di
Z mm
do
Bending stress
2
15.02 /
M N mm
b Z
63. 4 4
[ ]
16
2
11.27 /
d d
o i
T
do
N mm
(max)
2 2
1 4
1 2 2
c
c
(max)
2 2
1 4
2 c
σc(max) = 32.035 N/mm2
τ(max)=18 N/mm2
Resultant compressive stress
σc = σb + σo = 15.02 + 13.05 = 28.07 N/mm2 = 28.07 MPa
We know that twisting moment (T)
Maximum compressive stress
We know that maximum compressive stress,
Maximum shear stress
We know that maximum shear stress,
64. Design of curved
beams
In curved beam neutral axis of beam in unloaded condition is curve
instead of straight.
Neutral axis is shifted towards the centre of curvature. Centroidal axis
and neutral axis are not same.
Bending stress distribution is hyperbolic and not varying in a linear
manner. The beam is initially curved before unloading. Thus the length
of the inner fiber is less than the outer fiber. So the bending stress is high
at the inner portion of a beam.
In case of symmetrical cross section always stress is
high at the inner fiber. But in case of unsymmetrical
Cross section, both inner and outer bending stresses
have to be calculated.
65. Theories of Failure Under Static Load
• It has already been discussed in the previous chapter that strength of machine members is
based upon the mechanical properties of the materials used.
• Since these properties are usually determined from simple tension or compression tests,
therefore, predicting failure in members subjected to uniaxial stress is both simple and
straight-forward.
• But the problem of predicting the failure stresses for members subjected to bi-axial or tri-
axial stresses is much more complicated. In fact, the problem is so complicated that a large
number of different theories have been formulated.
The principal theories of failure for a member subjected to bi-axial stress are as follows:
1. Maximum principal (or normal) stress theory (also known as Rankine’s theory).
2. Maximum shear stress theory (also known as Guest’s or Tresca’s theory).
3. Maximum principal (or normal) strain theory (also known as Saint Venant theory).
4. Maximum strain energy theory (also known as Haigh’s theory).
5. Maximum distortion energy theory (also known as Hencky and Von Mises theory).
• Since ductile materials usually fail by yielding i.e. when permanent deformations occur in
the material and brittle materials fail by fracture, therefore the limiting strength for these
two classes of materials is normally measured by different mechanical properties.
• For ductile materials, the limiting strength is the stress at yield point as determined from
simple tension test and it is, assumed to be equal in tension or compression. For brittle
materials, the limiting strength is the ultimate stress in tension or compression.
66. Theories of Failure
Predicting failure in the members subjected to uniaxial stress is very
simple and straightforward. Because all failure criterions are reaching
the critical limit at an instant.
But in multi axial loading the prediction of failure is much
complicated. Because predicting the cause of failure i.e. which quantity
of failure criterion is causing failure is difficult to find.
Thus, theories were formulated to predict this issue, which are known
as failure theories.
67. Simple Tension Test
In simple tension test, all six quantities reaches its critical
values simultaneously (at a single instant).
Any one of the following will cause failure.
• Principal normal stress yield stress σmax = σy or σu]
• Principal shear stress yield shear stress τmax = σy /2
• principal strain energy strain energy at yield point Utotal = ½ [σy εy]
• Principal strain strain at yield point εmax = σy /E (or) σu /E
• Distortion energy distortion energy at yield point
Udistortion = (1+µ)/6E * [2σy
2]
68. Maximum Principal or Normal Stress Theory
(Rankine’s Theory)
According to this theory, the failure or yielding occurs at a point in a member when the
maximum principal or normal stress in a bi-axial stress system reaches the limiting
strength of the material in a simple tension test.
This theory is based on failure in tension or compression and ignores the possibility of
failure due to shearing stress, therefore it is not used for ductile materials.
Max principal stress [σ1] ≥ [σy] yield stress
(In a multi axial loading) (In a simple tension test)
Since the limiting strength for ductile materials is yield point stress and for brittle materials
(which do not have well defined yield point) the limiting strength is ultimate stress.
Therefore according to the above theory, taking factor of safety (F.S.) into consideration, the
maximum principal or normal stress (σt1) in a bi-axial stress system is given by
σt1 = σyt/F S for ductile materials
= σu/F S for brittle materials
where σyt = Yield point stress in tension as determined from simple tension test, and
σu = Ultimate stress.
Since the maximum principal or normal stress theory is based on failure in tension or
compression and ignores the possibility of failure due to shearing stress, therefore it is not used
for ductile materials.
However, for brittle materials which are relatively strong in shear but weak in tension or
compression, this theory is generally used.
69. Maximum Shear Stress Theory (Guest’s or Tresca’s Theory)
According to this theory, the failure or yielding occurs at a point in a member when the
maximum shear stress in a bi-axial stress system reaches a value equal to the shear stress at
yield point in a simple tension test. Mathematically,
τmax = τyt /F.S. ...(i)
where τmax = Maximum shear stress in a bi-axial stress system, τyt = Shear stress at yield point
as determined from simple tension test, and F.S. = Factor of safety.
Since the shear stress at yield point in a simple tension test is equal to one-half the yield
stress in tension, therefore the equation (i) may be written as
τmax = σyt/(2 × F S)
This theory is mostly used for designing members of ductile materials.
70. Maximum Principal Strain Theory (Saint Venant’s Theory)
According to this theory, the failure or yielding occurs at a point in a member
when the maximum principal strain in a bi-axial stress system reaches the
limiting value of strain (strain at yield point) as determined from a simple
tension test.
The maximum principal (or normal) strain in a
bi-axial stress system is given by
εmax = (σt1/E)-(σt2/m E)
∴ According to the above theory,
εmax = (σt1/E) - (σt2/m E)= ε= σyt / E x F.S
This theory is not applicable if the failure in elastic behavior is by yielding. It is
applicable when the conditions are such that failure occurs by brittle fracture.
where σt1 and σt2 = Maximum and minimum principal stresses in a bi-axial stress system,
ε = Strain at yield point as determined from simple tension test,
1/m = Poisson’s ratio,
E = Young’s modulus, and
F.S. = Factor of safety.
71. According to this theory, the failure or yielding occurs at a point in a member
when the strain energy per unit volume in a biaxial stress system reaches the
limiting strain energy (strain energy at yield point) per unit volume as
determined from the simple tension test.
Maximum Strain Energy Theory (Haigh’s Theory)
We know that strain energy per unit volume in a bi-axial stress system,
𝑈1 =
1
2𝐸
𝜎𝑡1
2 + 𝜎𝑡2
2 −
2𝜎𝑡1 ∗ 𝜎𝑡2
𝑚
and limiting strain energy per unit volume for yielding as determined from simple tension test,
𝑈2 =
1
2𝐸
𝜎𝑦𝑡
𝐹. 𝑆
2
According to this theory, U1 = U2
∴
1
2𝐸
𝜎𝑡1
2 + 𝜎𝑡2
2 −
2𝜎𝑡1∗𝜎𝑡2
𝑚 =
1
2𝐸
𝜎𝑦𝑡
𝐹.𝑆
2
or 𝜎𝑡1
2 + 𝜎𝑡2
2 −
2𝜎𝑡1∗𝜎𝑡2
𝑚
=
𝜎𝑦𝑡
𝐹.𝑆
2
This theory may be used for ductile materials.
72. Maximum Distortion Energy Theory (Hencky and Von Mises
Theory)
According to this theory, the failure or yielding occurs at a point in a member when the
distortion strain energy (shear strain energy) per unit volume in a biaxial stress system
reaches the limiting distortion energy (distortion energy per unit volume) as determined
from a simple tension test. Mathematically, the maximum distortion energy theory for
yielding is expressed as
𝝈𝒕𝟏
𝟐
+ 𝝈𝒕𝟐
𝟐
− 𝟐𝝈𝒕𝟏 ∗ 𝝈𝒕𝟐 =
𝝈𝒚𝒕
𝑭. 𝑺
𝟐
This theory is mostly used for ductile materials in place of maximum strain energy theory.
73. The load on a bolt consists of an axial pull of 10 kN together with a transverse shear
force of 5 kN. Find the diameter of the bolt using various theories of failure.
Take permissible tensile stress = 100 MPa and Poisson's ratio= 0.3.
Given : Pt1 = 10 kN ; Ps = 5 kN ; σt(el) = 100 MPa = 100 N/mm2 ; 1/m = 0.3
1. According to maximum principal stress theory
76. 4. According to maximum strain energy theory
5. According to maximum distortion energy theory
77. Home work Problem
A mild steel shaft of 60 mm diameter is subjected to a bending moment of 3000 N-m and a
torque T. If the yield point of the steel in tension is 300 MPa, find the maximum value of this
torque without causing yielding of the shaft according to
1. the maximum principal stress;
2. the maximum shear stress; and
3. the maximum distortion strain energy theory of yielding.
78.
79. Stress Concentration
Reasons for stress
concentration
Variation in properties of
materials
Load application
Abrupt changes in cross
section
Discontinuities in the
component
Machining scratches
Stress concentration: Localization of
high stresses due to the irregularities
present in the component and abrupt
changes of the cross section
80. Stress concentration in brittle materials
Brittle materials do not yield locally and there is no readjustment of stresses at
the discontinuities. (due to inability of plastic deformation)
When the magnitude of stress reaches the ultimate strength of the material, a
crack will nucleate and increases the stress concentration at the crack.
Therefore, stress concentration factors have to be used in the design of brittle
materials.
Stress concentration in ductile materials (static load)
When the stress reaches the yield point, then there will be a local plastic
deformation near the discontinuity which will lead to redistribution of stresses
near the stress concentration zone.
There is no remarkable damage to the machine component. This redistribution
of stresses will be restricted to very small area.
81. Stress concentration in ductile materials
(fluctuating load)
Due to fluctuating load the component may fail due to fatigue. stress
concentration will leads to the reduction in endurance limit of the
ductile materials.
Therefore stress concentration factors have to be used in the design of
machine components made of ductile materials.
82. Eccentric Loading
If the line of action of a load is not passing through the Centroid of
the machine component, then that is knows as eccentric load.
There are different kinds of stresses will be induced during
eccentric loading
For eccentric axial load,
Direct stress and bending stress
To find out the magnitude of resultant stress, these combination of
stresses have to be super imposed.
85. A rectangular strut 150 mm wide and 120 mm thick carries a load of
180 kN at an eccentricity of 10 mm in a plane intersecting the
thickness as shown in figure. Find the maximum and minimum
intensities of stress in the section.
Cross sectional area of the part = b.d
= 150 X 120 = 18 X 103 mm2.
Direct compressive stress
= 10 N/mm2.
Section modulus Z = I/y = db2/6
= 450 X 103 mm3
Bending moment M = P. e = 180 X 103 X 10
= 1.8 X 106 N-mm
P
c A
86. Bending stress
= 4 N/ mm2.
As the compressive stress is more than bending stress the strut will be subjected to
more compressive stress
Maximum intensity of compressive stress = 10+4 = 14 N/mm2
Minimum intensity of compressive stress = 10-4 = 6 N/mm2.
M
b z
87. Fluctuating stresses
σmax = max stress ; σmin = min stress ; σa = stress amplitude
σmean = mean stress
The stresses induced in a machine component due to dynamic load
(change in magnitude with respect to time) is known as fluctuating
stresses.
88. Variable loading
• Change in magnitude of the
applied load
Example: Punching machine
• Change in direction of the load
Example: Connecting rod
• Change in point of application
Example: Rotating shaft
Types of loading
• Fully Reversed loading
• Repeated loading
89. Design of machine components for fluctuating load
Fatigue
Mean stress
Number of
cycles
Stress
amplitude
Stress
concentration
Residual
stresses
Corrosion
& creep
90. Endurance limit or fatigue 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.
106 cycles are considered as a sufficient number of cycles to define the endurance limit.
Fatigue life: the total number of stress cycles that the standard specimen can complete
during the test before appearance of the first fatigue crack.
91. Fatigue failure
( time delayed fracture under cyclic loading)
Fatigue failure begins with a crack at some point in the material .
Regions of discontinuities (oil holes, keyways and screw threads)
Regions of irregularities in machining operations (scratches on the
surface, stamp mark, inspection marks)
Internal cracks due to defects in materials like blow holes
These regions are subjected to stress concentration due to crack,
then due to fluctuating load the crack spreads.
92. Region indicating slow growth of
crack with a fine fibrous
appearance
Region of sudden fracture with
a coarse granular appearance
93. Notch sensitivity factor (q)
In case of dynamic loading, if stress concentration present in the material, then it will
reduce the endurance limit.
The actual reduction in the endurance limit of a material due to stress concentration
under dynamic loading is varied by the theoretical values predicted using theoretical stress
concentration factor.
Therefore two separate stress concentration factors are used . i.e. Kt and Kf.
kf is the fatigue stress concentration factor
kf = Endurance limit of the notch free specimen / Endurance limit of the notched
specimen
Notch sensitivity [q] : Susceptibility of a material to succumb to the damaging effects of
stress raising notches in fatigue loading.
q = Increase of actual stress over nominal stress / Increase of theoretical stress over
nominal stress
94. σo = nominal stress obtained by the elementary equations
Actual stress due to fatigue loading = Kf σ0
Theoretical stress = Kt σ0
Increase of actual stress over nominal stress = (Kf σ0 - σ0)
Increase of theoretical stress over nominal stress = (Kt σ0 - σ0)
q =
Kf = 1 + q (Kt – 1)
When the material has no sensitivity to notches,
q = 0 and Kf = 1
When the material is fully sensitive to notches,
q = 1 and Kf = Kt
97. Combined variable normal and variable shear stress
When and machine member is subjected to both variable normal stress and
variable shear stress the Maximum shear stress theory and Maximum normal stress
theory are used.
According to Soderberg’s equation
equivalent normal stress
similarly
1
Multiplying by
term
y
ne
is known as
m v
n y e
y v y
m
n e
y
The n
y v y
m
n e
y
es
n
98. 2
(max)
2
1 4
2
es ne es
2
(max)
2
1 4
2 2
ne
ne ne es
According to maximum shear stress theory
According to maximum normal stress theory
A hot rolled steel shaft is subjected to a torsional moment that varies from
330 N-m clockwise to 110 N-m counterclockwise and an applied bending
moment at a critical section varies from 440 N-m to – 220 N-m. The shaft is of
uniform cross-section and no keyway is present at the critical section.
Determine the required shaft diameter. The material has an ultimate strength
of 550 MN/m2 and a yield strength of 410 MN/m2. Take the endurance limit as
half the ultimate strength, factor of safety of 2, size factor of 0.85 and a
surface finish factor of 0.62.
