DME - Unit1.pptx

ME 8593-DESIGN OF
MACHINE ELEMENTS
Unit 1 - STEADY STRESSES AND
VARIABLE STRESSES IN MACHINE
MEMBERS
DEPARTMENT OF MECHANICAL ENGINEERING

Textbooks and Reference
TEXT BOOKS:
• 1. Bhandari V, “Design of Machine Elements”, 4th Edition, Tata McGraw-
Hill Book Co, 2016.
• 2. Joseph Shigley, Charles Mischke, Richard Budynas and Keith Nisbett “
• Mechanical Engineering Design”, 9th Edition, Tata McGraw-Hill, 2011
REFERENCES:
• 1. Alfred Hall, Halowenko, A and Laughlin, H., “Machine Design”, Tata McGraw-Hill
BookCo.(Schaum’s Outline), 2010
• 2. Ansel Ugural, “Mechanical Design – An Integral Approach", 1st Edition, Tata McGraw-Hill
Book Co, 2003.
• 3. P.C. Gope, “Machine Design – Fundamental and Application”, PHI learning private ltd, New
Delhi, 2012.
• 4. R.B. Patel, “Design of Machine Elements”, MacMillan Publishers India P Ltd., Tech-Max
Educational resources, 2011.
• 5. Robert C. Juvinall and Kurt M. Marshek, “Fundamentals of Machine Design”, 4th Edition,
Wiley, 2005
• 6. Sundararajamoorthy T. V. Shanmugam .N, “Machine Design”, Anuradha Publications,
Chennai, 2015.

UNIT I STEADY STRESSES AND VARIABLE
STRESSES IN MACHINE MEMBERS
• Introduction to the design process - factors
influencing machine design, selection of materials
based on mechanical properties - Preferred
numbers, fits and tolerances – Direct, Bending
and torsional stress equations – Impact and shock
loading – calculation of principle stresses for
various load combinations, eccentric loading –
curved beams – crane hook and ‘C’ frame- Factor
of safety - theories of failure – Design based on
strength and stiffness – stress concentration –
Design for variable loading.

STEADY STRESSES AND
VARIABLE STRESSES IN
MACHINE MEMBERS
Design
Process
fits and
tolerances
Direct,
Impact
Stress
Principle
stresses
(PSG 7.2)
Curved
Beams
(PSG 6.2,
6.3)
Theories
of Failure
(PSG 7.3)
Stress
Concentration
(PSG 7.6)
o Red colored – Most important, Orange colored – Next most important, Black – important topics

Unit 1
• Introduction to the design process

Thought bites
Is an Engine, a Machine?

Is an Engine, a Machine?
• All engines can be called machines, but not
all machines can be called engines.
• Engine is basically a prime mover which
generates power using some fuel i.e. diesel,
petrol etc. A machine needs power to do work
which must be created by hand, engine or
electric motor. Engine could be a component
of machine.

What is Machine elements?
What are the parts in an engine?

Machine elements
(Day today Life examples)

Machine Design
• Machine design is defined as the use of
scientific principles, technical information
& imagination in the description of a
machine or a mechanical system to
perform specific functions with maximum
economy & efficiency.
• Machine Design is defined as the creation
of new design (Machines) or improving the
exist one.

•Mathematics
•Engineering Mechanics
•Strength of Materials
e
e
g
• Math matics
• Engin ering Mechanics
• Stren th of Materials
• Workshop Processes
• Engineering Drawing
What is the basic knowledge required for Machine Design?
•Mathematics
•Engineering Mechanics
•Strength of Materials
•Workshop Processes
•Engineering Drawing
• Mechanics of Machines
• Mechanics of Materials
• Fluid Mechanics & Thermodynamics
16

4 C’s in Design Process
• Creativity
• Complexity
• Choice
• Compromise

Classifications of Machine Design
1. Adaptive design (Old design)
2. Development design (Modification in old design)
3. New design (Creating a new design)
a. Rational Design (Mathematical formulae)
b. Empirical design (Empirical formulae – Practice & Past
Experience)
c. Industrial design (Production aspect)
d. Optimum design (Best design)
e. System design
f. Element design
g. Computer Aided design

Basic Requirement of Machine Element
(DESIGN CONSIDERATIONS IN MACHINE DESIGN)
Factors influencing Machine Design
• Strength and Stiffness
• Type of Load and stresses
• Rigidity
• Maintenance
• Flexibility
• Size and shape
• Stiffness
• Reliability
• Kinematics of machine
• Safety of operation
• Weight
• Manufacturing considerations
• Selection of Materials
• Corrosion of Materials
• Friction and wear
• Frictional resistance and lubrication
• Life
• Assembly considerations
• Conformance to standards
• Vibrations
• Thermal considerations
• Workshop facilities
• Ergonomics
• Aesthetics
• Cost
• Noise
• Environmental factors

General procedure in Machine Design
Detailed drawing
Need or aim
Synthesis
Analysis of the FORCES
Material selection
Design of elements
Recognize and specify the problem
Select the mechanism that would give the desired
motion and form the basic model with a sketch etc
Determine the stresses and thereby the sizes of
components s.t. failure or deformation does not
occur
Modify sizes to ease construction & reduce overall cost
Modification
Production

Procedure of Design of Machine Elements

Material Selection
• The best material is one which will serve the
desired purpose at minimum costs
• Factors Considered while selecting the Material
– Availability
– Cost
– Mechanical properties: Strength, Hardness,
Toughness, Ductility, Malleabilty, etc.,
– Manufacturing considerations – Shaping, Machining,
Joinimg, surface finishing, FoS, Assembly cost

Factor of safety
• Is used to provide a design margin over the
theoretical design capacity to allow for
uncertainty in the design process.
– In the calculations,
– Material strengths,
– Manufacturing process
• FoS = Strength of the component (Max load)
Load on the component (Actual load)

AU Questions on the completed topic
• Explain various phases in Design using a flow
diagram and enumerate the factors
influencing the machine design.
(APR/MAY 2013)
• Write short notes on preferred numbers, fits
and types of fits. (APR/MAY 2012)
• What is meant by Hole basis system and Shaft
basis system? Which one is preferred and
why? (APR/MAY 2013)

STEADY STRESSES AND
MACHINE MEMBERS
Design
Process
Fits and
tolerances
Direct,
Impact
Stress
Principle
stresses
(PSG 7.2)
Curved
Beams
(PSG 6.2,
6.3)
Theories
of Failure
(PSG 7.3)
Stress
Concentration
(PSG 7.6)

Preferred Numbers (PSG 3.1 to 3.6)
• These are the kind of numbers derived from
geometric series, including the integral
powers of 10.
• Developed by French military engineer Charles
Reynard in 1877 – Standardisation.
• Basic series : R5, R10, R20, R40, R80
• R5(
5
10), R10 (
10
10), R20 (
20
10), R40
(
40
10), R80 (
80
10). (Note: This will be multiplied by ‘a’ – variable)

Applications of R Series
• R5 – Hydraulic cylinder capacities, Tolerance
grades in ISO standards.
• R10 – Hoisting – cranes, dia of wire ropes.
• R20 – Thickness of sheet metals, dia of wires
of helical springs, machine tool design
• R40 – Hydraulic cylinder diameters.

Allowance
• It is an intentional difference between the
maximum material limits of mating parts.
• It is the difference between the basic
dimensions of the mating parts.
• The allowance may be positive or negative.
When the shaft size is less than the hole size,
then the allowance is positive
• When the shaft size is greater than the hole
size, then the allowance is negative.

Tolerance
• It is the difference between the upper limit and
lower limit of a dimension. In other words, it is
the maximum permissible variation in a
dimension. The tolerance may be unilateral or
bilateral. When all the tolerance is allowed on
one side of the nominal size, e.g.,
• It is said to be unilateral system of tolerance
• The unilateral system is mostly used in industries
as it permits changing the tolerance value while
still retaining the same allowance or type of fit.

Tolerance
• When the tolerance is allowed on both sides
of the nominal size , then it is said to be
bilateral system of tolerance.
• In this case +0.002 is the upper limit and
-0.002 is the lower limit.

Lower limit =  27. 8
 0.2
Upper limit =  28 . 2
Hole
Terminology for Limits and Fits Cont.
Tolerance
zone
Tolerance :Tolerance is the difference between maximum limit of size and minimum
limit of size.
Ø28
(Basic)
Zero line

CLASSIFICATION OF FITS
Clearance fit
Interference fit
Transition fit
Fit is the relationship that exists between two mating
parts, a hole and shaft with respect to their
dimensional difference before assembly. Three types
of fit are given hereunder

Clearance Fit
Clearance Fit : In clearance Fit shaft is
always smaller than the hole. A positive
allowance exists between the largest
possible shaft and smallest possible hole.
Minimum Clearance : It is the
difference between the
maximum size of shaft and
minimum size of hole.
Maximum clearance: It is the
minimum size of the shaft and
the maximum size of hole
Hole basis
Shaft Basis

A
B
Fig. 1.2 Clearance Fit
Min hole – Max Shaft = + ve – clearance fit.
Fit with positive clearance between the hole and the shaft.
CLEARANCE FIT

Fit Cont.
Interference Fit: It is also
called Press or force fit, In
this fit shaft is always larger
than the hole
MAX interference: it is the
maximum size of hole and the
minimum size of shaft prior to
assembly
Minimum Interference: It is the
minimum size of the hole and
the maximum size of the shaft
prior to assembly.
Shaft
Hole

INTERFERENCE FIT
Fit with negative clearance between the hole and the shaft.
Fig. 1.3 Interference Fit
Max hole – Min Shaft = - ve – interferance fit

Fit Cont.
Transition Fit: it is called
sliding Fit . It occurs when the
resulting fit due to the
variations in size of the male
and female components due
to their tolerance, varies
between clearance and
interference fits. The
tolerance zones of shaft and
hole overlap
Shaft
Hole

TRANSITION FIT
Fit established when the dimensions of the hole are such that there exists
either a positive clearance or a negative clearance when the shaft is fitted
into the hole.
B

SYSTEMS OF FITS
HOLE BASIS SYSTEM: the hole is constant in diameter and various types
of fits are obtained by suitably varying the limits of the shaft.
SHAFT BASIS SYSTEM: the shaft is constant in diameter and various
types of fits are obtained by suitably varying the limits of the hole.

Fig. 1.5 Hole Basis System
Clearance Fit
B
Interference Fit
C
HOLE BASIS SYSTEM
Single hole, whose lower deviation is zero.
Minimum limit of the hole will be equal to its basic size.

HOLE BASIS SYSTEM - Design Example
Requirement
Hole basic size - 20 mm diameter.
Clearance of - 0.100 mm.
Hole tolerance - 0.025 mm.
Shaft tolerance - 0.050 mm.
Minimum limit of the hole is Ø 20 mm.
Maximum limit of the shaft = Lower limit of the hole –
Minimum clearance
= Ø 19.900 mm
Design

Minimum limit of the shaft = Maximum limit of the shaft –
Tolerance on the shaft.
= Ø 19. 850 mm.
Shaft
Max. limit = Ø 19.900 mm.
Min. limit = Ø 19. 850 mm.
Maximum limit of the hole = Maximum limit of the hole +
Tolerance on the hole.
= Ø 20. 025 mm.
Hole
Max. limit = Ø 20.025 m.
Min. limit = Ø 20. 000 mm.

Why the Hole Basis System is Preferred?
Holes are produced by drilling, boring, reaming, broaching, etc.,
Shafts are either turned or ground.
Shaft basis system - Holes of different sizes are required,
(requires tools of different types and
sizes).
Hole basis system - Only one tool is required, to
produce the hole and the shaft can
be machined to any desired size.

Hole Basis System Shaft Basis System
 Hole is keep constant and the
shaft diameter is varied
 The basic size of the hole is taken
as the low limit
 The high limit of the size of the
hole and the two limits of size of
the shaft are selected to give the
desired fit
 The actual size of the hole is
within the tolerance limit.
 In this system Hole gets the letter
H and the shaft gets letter o decide
the position of tolerance
 Shaft is kept constant and the hole
diameter is varied.
 The basic size of the shaft is taken as one
of the limits(maximum) of size of shaft
 The other limit of size of the shaft and the
two limits of hole are then selected to give
the desired fit
 The actual size of a hole that is within the
tolerance limits is always less than the
basic size.
 In this system Shaft gets the letter h and
the hole gets different letter o decide the
position of the tolerance zone to obtain
desired fit

Fundamental deviation for Shafts
(IT – Std tolerance)
For holes
• Upper deviation  ES
• Lower deviation  EI
• EI = ES – IT (OR) ES = EI + IT
For Shaft
• Upper deviation  es
• Lower deviation  ei
• ei = es – IT (or) es = ei + IT

POINTS SHOULD REMEMBER Before
SOLVING PROBLEMS (For the whole Unit)
• All the parameters should convert to mm.
• The answer value should be in the unit of
mm.

Problem on Tolerance
• A shaft of 40mm dia having tolerance grade of
5. Calculate the IT tolerance value by using the
equation and Table.

A shaft of 40mm dia having tolerance grade of 5. Calculate the IT
tolerance value by using the equation and Table.
• Given:
– D = 40 mm
– Tolerance Grade = 5
• To Find:
– IT tolerance by both table and equation

PSG data book page no: 3.3 (For Table)

Result
• Using Table:
– IT tolerance value = 11 microns =0.001 mm

A shaft of 40mm dia having tolerance grade of 5.
PSG data book page no: 3.6 (For Equation)
• i = 0.45 3√D + 0.001D
• Where D = √D1D2
• D1 = Max dia from PSG data book Pg no.3.3
• D2 = Min dia from PSG data book Pg no.3.3
• D1 = 50
• D2 = 30
• Hence D = √D1D2 = √30 * 50 = 38.73mm

A shaft of 40mm dia having tolerance grade of 5.
PSG data book page no: 3.6 (For Equation)
• i = 0.45 3√D + 0.001D
• i = 0.45 3√ 38.73 + 0.001 (38.73)
• i = 1.5612 microns
• From PSG data book page no: 3.6
• For IT grade value 5 the tolerance value is 7i
• Hence 7 I = 7 x 1.5612= 10.9285 microns ≈ 11
microns

Geometric Dimensioning and
Tolerancing (GD&T)

AU Questions on the completed topic
• (i) Discuss in detail about the factors
influencing machine design. (OR) What are the
factors influencing machine design? Explain it.
• (ii) Write short notes on the following:
• Interchangeability
• Tolerance
• Allowance
(APR/MAY 2012)(MAY/JUNE 2014)

numbers, fits and tolerances
• Direct, Bending and torsional stress equations –
Impact and shock loading – calculation of
principle stresses for various load combinations,
eccentric loading – curved beams – crane hook
and ‘C’ frame- Factor of safety - theories of failure
– Design based on strength and stiffness – stress
concentration – Design for variable loading.

TYPE OF LOADS AND STRESSES IN
REAL- TIME APPLICATIONS
Steady loads
• Dead loads
• Live loads
Variable loads
• Shock loads (suddenly)
• Impact loads (applied
with some velocity)
 axial (tension-
compression) ,
flexural (bending)
torsion(twisting) in nature.
 Thermal Loading

Static Load
Time
Stress
F and P are applied and remain constant
Stress Ratio, R = 1.0
A static (steady) load is a
stationary force or couple
(applied to a member) that
does not change in magnitude,
point of application, and
direction.

• An element subjected to Dynamic loads
(tensile and compressive stresses involved)
Continuous total load reversal over time
Dynamic Load:

Stress is dependent on the load characteristics.
Strength is an inherent property of the material.
•.
Factor of safety depends on
Type of material
How controllable are environment conditions
Type of loading and the degree of certainty
with which the stresses are calculated
Type of application
Failure can mean a part
•has separated into two or more pieces; (brittle)
•has become permanently distorted, thus ruining its geometry; (ductile)
•has had its function compromised
Strength Vs Stress
Factor of safety = Strength / Stress = S / σ

Factor of Safety
• FOS (Ductile) =
𝑌𝑖𝑒𝑙𝑑 𝑆𝑡𝑟𝑒𝑠𝑠
𝑊𝑜𝑟𝑖𝑛𝑔 𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑡𝑟𝑒𝑠𝑠
• FOS (Brittle) =
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑠𝑠
𝑊𝑜𝑟𝑖𝑛𝑔 𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑆𝑡𝑟𝑒𝑠𝑠

Allowable stress
. . . While designing a component, it must be
ensured that the maximum stress that may be induced
during working life, do not exceed a certain safe limit . .
Such a safe limiting stress is known as allowable,
permissible or design stress

TENSILE STRESS:
When a body is subjected to two equal and
opposite axial pulling forces, F - F as shown in
Fig.
the stress induced at any section A – A of the body
is known as tensile stress and the corresponding
strain, the tensile strain (the ratio of total
elongation, 8 to the original length, l).

When a body is subjected to two equal and opposite
axial pushing forces, F
COMPRESSIVE STRESS:

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.
SHEAR STRESS:
Single shearing of a riveted joint.

the area resisting the shear off the rivet
Double shearing of a riveted joint.

Bending Stress
• 𝜎𝑏 =
𝑀𝑏
𝑍
• 𝑀𝑏 or M - Bending Moment
• Z =
I
𝑦
– Sectional Modulus (PSG 6.1)

Problem
• The piston of a reciprocating compressor has
a dia of 60mm. The max pressure on the
piston tall is 1.25MN/m2 (12.5 bar). Assuming
the gudgeon pin passing through the small
end of the connecting rod can be safely
loaded in shear upto 10 MN/m2. Cal the min
dia of the gudgeon pin.

The piston of a reciprocating compressor has a dia of 60mm. The max
pressure on the piston tall is 1.25MN/m2 (12.5 bar). Assuming the gudgeon
pin passing through the small end of the connecting rod can be safely loaded
in shear upto 10 MN/m2. Cal the min dia of the gudgeon pin.
Given:
• D = 60mm
• P = 1.25MN/m2
• = 1.25 N/mm2
• 𝜏 = 10 MN/m2
• = 10 N/mm2
To Find:
• Gudgeon pin dia, d
Hint:
1 MN/m2 = 1 N/mm2
(1 MN/m2 = 1 x 106 N/ 106mm)

D = 60mm; P = 1.25 N/mm2;
𝜏 = 10 N/mm2
• Solution:
• Piston area = D2/4 = 2,826 mm2
• Piston force = Pressure x Area
= 1.25 N/mm2 x 2,826 mm2
= 3,560.76 N

D = 60mm; P = 1.25 N/mm2;
𝜏 = 10 N/mm2
Solution:
• Gudgeon pin has double shear
• Hence, P = 2 x 𝜏 x Area
• 3,560.76 = 2 x 10 x ∏d2/4
• d = 15.06 mm
Result:
• Diameter of Gudgeon pin = 15.06mm

Consider a straight beam subjected to a bending moment M
as shown in Fig.
The bending equation
is given by

Problem
• A cantilever of span 500mm carries a vertical
downward load of 6kN at free end. Assume
the yield value is 350MPa and FoS is 3. Find
the economical section for the cantilever
among
• a) Circular cross section of dia “d”
• b) Rectangular section of depth “h” and width
“t” where h/t = 2
• c) ɪ section of depth 7t and flange width 5t
where t is thickness . Specify the dimension
and area.

A cantilever of span 500mm carries a vertical downward load of 6kN at free
end. Assume the yield value is 350MPa and FoS is 3. Find the economical
section for the cantilever among
a) Circular b) Rectangular c) ɪ section
Given data:
• Length, L = 500 mm
• Load, P = 6000 N
• σy = 350 N/mm2
• FoS, n = 3
To Find:
• Economical section among Circular,
rectangular and I section

L = 500 mm; P = 6000 N; σy = 350 N/mm2; FoS, n = 3
• Mb = P x L = 6000 x 500 = 3x106 N-mm
• Allowable stress, σb = σy / n = 350/3 = 116.67
N/mm2.
• a) Circular Section:
• Bending moment equation:
•
σb
𝑦
=
Mb
ɪ

σb
𝑑
=
Mb
ɪ
σ =
32Mb
πd3
• W.K.T, ɪ = πd4 / 64 (PSG 6.1) & y = d/2

Area for Circular rod
• d = 63.98 ≈ 64 mm
• Cross sectional area , A = πd2 / 4 = 3215.63 mm2

L = 500 mm; P = 6000 N; σy = 350
N/mm2; FoS, n = 3; σb = 116.67 N/mm2
b) Rectangular section:
• Given that h = 2t
• Mb = P x L = 6000 x 500 = 3x106 N-mm
•
σb
𝒚
=
Mb
ɪ
• Where y = h/2, ɪ = th3/12
(PSG 6.1)
t

Rectangular Section
• Now, σb = Mb x (h/2) / th3/12
• σb = Mb x (1/2) / th2/12
• σb = Mb / 2 t(2t)2/12
• σb = Mb x 6 / t(2t)2
• t3 = Mb x 6 / 4 x σb
• t3 = 3x 106 x 6 / 4 x 116.67  t = 33.78 ≈ 34
mm

Rectangular Section
• Area = h x t = 2 x 34 x 34 = 2313 mm2

ɪ section
• Given that depth is 7t and width is 5t
• ɪ = ɪ1 - ɪ2
• = b1 h1
3 / 12 - 2 x b2 h2
3 / 12
• = [5t (7t)3 – 2 x 2t (5t)3 ]/12
• ɪ = 1125t4 / 12
•
σb
𝑦
=
Mb
ɪ
• Where y = 3.5t (i.e., h/2)
7t
5t
1t
1t
5t
t

l = 500 mm; P = 6000 N; σy = 350
N/mm2; FoS, n = 3; σb = 116.67 N/mm2
• σb =
Y . Mb
ɪ
• 116.67 = 3.5t x 3x106 / 1125t4 /12
• t = 9.8 mm
• h = 7t = 68mm
• b = 5t = 49mm
• Area = b h – 2 * (b2 h2)
= (5t * 7t) – 2 * (2t * 5t)
Area = 1,593 mm2

Result
• Area of ɪ section = 1593 mm2
• Area of Rectangular section = 2313 mm2
• Area of Circular section = 3215 mm2
• Hence the economical section for cantilever
beam based on area is ɪ section.
• ɪ section is chosen.

AU Question on the completed topic
• 7) A cantilever beam of rectangular cross-section is used to support a
pulley as shown in Fig. 11a. The tension in the wire rope is 10 kN. The
beam is made of cast iron whose ultimate strength σut= 240 MPa and the
factor of safety is 3. The ratio of depth to width of cross-section is 2.
Determine the dimensions of the cross-section of the beam. (NOV/DEC
2020 AND April/May 2021)

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 or crushing stress (stress at the surface of contact
between the rivet and a plate

Impact Stress
• Stress produced in the member due
to load with impact or falling load is
known as Impact stress.
Where,
E – Young’s Mod
l – Bar length
l – Bar deformation
h – height through
which the load falls

Problem
• An unknown weight falls through 10mm on a
collar rigidly attached to the lower end of a
vertical bar 3m long and 600mm2 in section. If
the max instantaneous extension is known to
be 2mm. What is the corresponding stress and
the value of unknown weight. Take
E =200kN/mm2.

An unknown weight falls through 10mm on a collar rigidly attached to the
lower end of a vertical bar 3m long and 600mm2 in section. If the max
instantaneous extension is known to be 2mm. What is the corresponding
stress and the value of unknown weight. Take E =200kN/mm2.
Given:
• h = 10mm
• l = 3m = 3000mm
• A = 600mm2
• l = 2 mm
• E =200 x 103 N/mm2
To Find :
W

h = 10mm; l = 3m = 3000mm; A = 600mm2
; l = 2 mm; E =200 x 103 N/mm2
Solution:
• E = l/ l  E l/l = 
•  = 200 x 103 * 2 / 3000
•  = 133.3 N/mm2
•  = W/A [1 + 1 +
2ℎ𝐴𝐸
𝑊𝑙
]
• W = 6666.7 N

When a machine member is subjected to the action of
two equal and opposite couples acting in parallel planes
(or torque or twisting moment), then the machine
member is said to be subjected to torsion.
Torsional Shear Stress

Torsional Stresses
• Shear Stress, 𝜏 = Mt . r/ J
• 𝜏max = Mt . r/ (πd4 / 32)
= Mt . (d/2) / (πd4 / 32)
𝜏max = 16 Mt / πd3

AU Problem on Torsional equation
18) A hollow shaft is required to transmit 500 W
at 100 rpm. The maximum torque being 20%
greater than the mean. The shear stress is not to
exceed 65 MPa and twist in a length of 2 metres
not to exceed 1.2 degrees. Find the external and
internal diameter of the shaft, if the ratio of
internal to external diameter is 3/8. Take
modulus of rigidity as 84 GPa. (APR/MAY 2019)

STEADY STRESSES AND
MACHINE MEMBERS
Design
Process
Fits and
tolerances
Direct
(CSA),
Impact
Stress
Principle
stresses
(PSG 7.2)
Curved
Beams
(PSG 6.2,
6.3)
Theories
of Failure
(PSG 7.3)
Stress
Concentration
(PSG 7.6)

Principal stresses for various load
consideration
• Practically mechanical components are
subjected to several types of external loads
simultaneously.
• Principal stresses are the maximum and
minimum normal stresses on a particular
plane, well we can also determine extreme
values of normal stresses possible in the
material.

Principal stresses for various load
consideration
• From PSG data book 7.2

Approach to Principal Stresses
problems
1. Calculate Direct stress, 𝜎𝑑 =
𝑃𝑎
𝐴
2. Calculate Bending stress, 𝜎𝑏 =
32𝑀𝑏
𝜋𝑑3
3. Calculate Torsional Shear stress, 𝜏𝑥𝑦 =
16𝑀𝑡
𝜋𝑑3
4. If it is bending, 𝜎𝑥 = 𝜎𝑑 + 𝜎𝑏
If it is Compression, 𝜎𝑥 = 𝜎𝑑 − 𝜎𝑏
5. From PSG DB 7.2, find Max (use +ve) & Min (-ve)
Principal Stress and Shear Stress (Assume 𝜎𝑦 = 0)

Problem on Principal stresses
16) A circular shaft of 30mm dia is subjected to
an axial load, bending moment and twisting
moment as shown. Det the max principal stress,
min principal stress and maximum shear stress
at point A and B.
Axial/Direct Load
Bending Load
Torsion

Hint: Bending moment equation:
σb
𝑦
=
Mb
ɪ

σb
𝑑
=
Mb
ɪ
σb =
32Mb
πd3

• Now, PSG Data book page no 7.2 to find Max
principal stress, min principal stress and Shear
stress

16) A shaft, as shown in Figureis subjectedto a bendingload of 3kN, pure
torque of 1000 N-m and an axial pullingforce of 15 kN. Calculate the
stressesat A and B
Given :
W = 3 kN = 3000 N ;
T = 1000 N-m = 1 × 106 N-mm ;
P = 15 kN= 15 × 103 N ;
d = 50 mm;
Length (l) x= 250 mm
(APR/MAY 2010) (APR/MAY 2016)

This bending stress is tensile at point A and compressive at point B.
∴ Resultant tensile stress at point A

Resultant compressive stress at point B,

Problem
17) A crank shaft of 20mm dia is subjected to a
load of 10kN as shown. Det the max and min
principal stresses and max shear stress at the
crank shaft bearing.

Q3.An overhang crank with pin and shaft is shown in Fig. 5.18. A tangential
load of 15 kN acts on the crank pin. Determine the maximum principal stress
and the maximum shear stress at the centre of the crankshaft bearing.
Given :
W = 15 kN = 15 × 103 N ;
d = 80 mm ;
y = 140 mm ;
l = 120 mm

Bending moment at the centre of the
crankshaft bearing,
M = W * l =
Torque transmitted at the axis of the shaft,
T = W × y =

Shear stress due to the torque transmitted,

AU Problems
• A hollow circular column of
external diameter 250 mm and
internal diameter 200 mm carries
a projecting bracket on which a
load of 20 kN rests as shown in
Fig. The centre of the load from
the centre of the column is 500
mm. Find the stresses at the
sides of the column. All
dimensions in mm. (NOV DEC
2016)

Approach to this problems
1. Calculate Direct stress, 𝜎𝑑 =
𝑃𝑎
𝐴
Where A = /4 (D2 – d2)
2. Calculate Bending stress, 𝜎𝑏 =
𝑀𝑏𝑦
𝐼
(Take I
from PSG 6.2)
3. If it is bending, 𝜎𝑥 = 𝜎𝑑 + 𝜎𝑏
If it is Compression, 𝜎𝑥 = 𝜎𝑑 − 𝜎𝑏

Short Solution
• A = 0.018 mm2
• 𝜎𝑑 =
𝑃𝑎
𝐴
= 11,11,000 N/mm2
• I = 0.000113 mm
• Y = 250/2 = 125 mm
• 𝜎𝑏 =
𝑀𝑏𝑦
𝐼
= 11,00,000 N/m2
• 𝐴𝑡 𝑆𝑖𝑑𝑒 1, 𝜎1𝑥 = 𝜎𝑑 + 𝜎𝑏 = 13 N/m2
• 𝐴𝑡 𝑆𝑖𝑑𝑒 2 , 𝜎2𝑥 = 𝜎𝑑 − 𝜎𝑏 = -9 N/m2

and ‘C’ frame- Factor of safety - theories of
failure – Design based on strength and stiffness –
stress concentration – Design for variable loading.

Curved Beams
• ro – Outer Radius
• ri - Inner radius
• rn - Neutral axis radius
• R – Centroidal axis radius
• hi - Distance between neutral axis and
inner fibre = rn - ri
• ho - Distance between neutral axis and
outer fibre = ro - rn
• e- distance between centriodal axis and
neutral axis
• Mb - Bending moment

Approach to Curved beam problems
• Step 1: Find Direct Stress d = P/A
• Step 2: Find the bending stress bi = M .h/a.e.r
(Mostly for Inner fibre) – Find out other
parameters [PSG DB – 6.2]
– Step 2.1 : Various Radius: R, ri, ro, rn,
– Step 2.2 : Eccentricity : e, hi
– Step 2.3 : Moment Mb
• Either Load, diameter or max stress has to be
found

Problems on Curved Beams
• A C-shaped link of circular
section having dia 20mm is
loaded as shown. Det the
max stress in the link.

A C-shaped link of circular section having dia 20mm is
loaded as shown. Det the max stress in the link.
Given Data:
• Load, P = 1 kN
• Dia of circular rod, d = 20 mm
• Centroidal axis radius, R = 50mm
To find:
• Max Stress, max

Problems on Curved Beam (Crane
Hooks)
• A trapezoidal cross section crane hook having
a yield strength of 380 N/mm2 is loaded as
shown below. Assume the FoS for hook
material is 3.5. Det the load carrying capacity.

A trapezoidal cross section crane hook having a yield strength of
380 N/mm2 is loaded as shown below. Assume the FoS for hook
material is 3.5. Det the load carrying capacity.
Given Data:
• Yield strength,
y = 380 N/mm2
• fs = 3.5
• Inner radius, ri = 45 mm
• For trapezoidal cross
section:
– bi = 80 mm
– bo = 25 mm
– h = 110 mm
To Find:
• Load carrying capacity, P

• Factor of safety is given in the problem
• WKT, FoS (n) = Yield stress/ allowable stress

Problems on Curved Beams
• A S-link having circular cross
section is subjected to a load
of 2 KN shown. Dia of link is
30mm. Det the max tensile
stress and max shear stress of
the S-link.

A S-link having circular cross section is subjected to a
load of 2 KN shown. Dia of link is 30mm. Det the max
tensile stress and max shear stress of the S-link.

AU Problems on Curved Beams
• The C-frame of a 100 kN capacity press is
shown in Fig. The material of the frame is grey
cast iron FG 200 and the factor of safety is 3.
Determine the dimensions of the frame.
• (APR/MAY 2010) (APR/MAY 2018) (NOV/DEC
2018)

• A punch press of capacity 50 KN has a c-frame
of ‘T’ cross section as shown in the fig. The
Tensile strength of material is 350 MPa. Take
FoS as 3.5. Determine the dimensions of C-
frame. (NOV 2021)

• A link shaped in the form of a letter S is made
up of 30 mm diameter bar, as shown in figure
Determine the maximum tensile stress and
maximum shear stress in the link.
• (APR MAY 2017)

• (i) The frame of a punch press is shown in fig. Find the
stresses at the inner and outer surface at section X-X of
the frame, if W = 5000 N.
(MAY/JUNE 2014)
• (ii) What is factor of safety? List the factors to be
considered while deciding the factor of safety.
(MAY/JUNE 2014)

• Determine the stress at point A and B split ring
shown in fig. If a compressive force = 20 kN is
applied point ‘C’. (APR/MAY 2018) (PART C)

• A wall bracket with a rectangular cross-section is
shown in Fig. The depth of the cross-section is
twice of the width. The force P acting on the
bracket at 600 to the vertical is 5 kN. The material
of the bracket is grey cast iron FG 200 and the
factor of safety is 3.5. Determine the dimensions
of the cross-section of the bracket. Assume
maximum normal stress theory of failure.
(APR/MAY 2018)

and ‘C’ frame- Factor of safety - theories of
failure – Design based on strength and stiffness
– stress concentration – Design for variable
loading.

Maximum Principal Stress Theory
• When the maximum
principal stress induced in
a material under complex
load condition exceeds the
maximum normal strength
in a simple tension test the
material fails.
• Good for brittle materials.

Maximum Shear Stress Theory
shear strength in actual
case exceeds maximum
allowable shear stress in
simple tension test the
material case.
• Good for ductile materials

Maximum Principal Strain Theory
normal strain in actual
case is more than
maximum normal strain
occurred in simple
tension test case the
material fails.
• Not recommended

Maximum Strain Energy Density Theory
• When the total strain
energy in actual case
exceeds the total strain
energy in simple tension
test at the time of
failure, the material
fails.
• Good for ductile
material

Maximum Distortion Energy Density
Theory
• When the shear strain
energy in the actual case
exceeds shear strain
energy in simple tension
test at the time of failure
the material fails.
• Highly recommended

Approaches for
Problem on Theories of Failure
1. Calculate Bending moment and Torsional
Moment
2. Calculate Stresses: 𝜎𝑥, 𝜏𝑥𝑦
3. Calculate Principle stresses (PSG 7.2)
4. Use any theory of failure to the required
dimension. (PSG 7.3)

Problem on Theories of Failure

Part C AU Problems
• A shaft which has a diameter of 30 mm is
subjected to an axial tension load of 15kN and a
torque of 400 Nm. In addition to these, there is a
bending moment of 300 Nm on the shaft. The
Shaft is made of steel having the properties of Su
= 780MPa ad Sy = 600 MPa.
• Neglecting the column action, determine the FOS
using
• Distortion energy theory of failure
• Shear Stress theory of failure (NOV/DEC 2019)
Note: Assume Poisson’s Ratio = 0.25 to 0.3

AU Problems
• A rod is subjected to axial tensile load of 20
KN and torsional load of 10 KN-m. Determine
the diameter of rod according to
• (1) Rankine’s theory
• (2) St. Venant’s theory
• (3) Trasca theory.
• Take factor of safety = 2.5, Poisson’s ratio =
0.25, s y =300N /mm2 . (7) (NOV 2021)

AU Problems
• A solid circular shaft of diameter 45 mm
is loaded by bending moment 650 Nm,
torque 900 Nm and an axial tensile force
of 30 kN. The shaft material is ductile
with yield strength of 280 MPa.
Determine the factor of safety according
to Maximum principal stress, Tresca and
Von misses theories of failure. (APR MAY
2017)

AU Problems
• A bolt is subjected to a direct load of 25 kN and
shear load of 15 kN. Considering theories of
failure. Determine a suitable size of the bolt (PSG
5.49) if the material of the bolt is C15 having 200
N/mm2 yield strength. Assume F.O.S. as 2 and
also give your comments.
• i) Maximum normal stress theory
• ii) Maximum shear stress theory
• iii) Von mises theory. (NOV/DEC 2017)

Stress Concentration
• A stress concentration (often called
stress raisers or stress risers) is a location
in an object where stress is concentrated.
• An object is strongest when force is
evenly distributed over its area, so a
reduction in area, e.g., caused by a crack,
results in a localized increase in stress.

Stress Concentration
• The existence of irregularities or discontinuities,
such as holes, grooves, or notches, in a part
increase the magnitude of stresses significantly in
the immediate vicinity of the discontinuity.
Fatigue failure mostly originates from such
places.
• Stress concentration factor need not be used with
ductile materials when they are subjected to only
static loads, because (local) yielding will relieve
the stress concentration.

Techniques to reduce stress
concentration
• Avoiding sharp corners and only using rounded
corners with maximum radii.
• Sanding and polishing surfaces to remove any notches
or defects that occur during forming and processing.
• Lowering the stiffness of straight load-bearing
segments.
• Placing notches and threads in low-stress areas.
• Provide fillets
• Use of multiple holes instead of single holes
• Undercutting the shoulder parts.

concentration
• Additional Notches and Holes in Tension Member
• Fillet Radius, Undercutting and Notch for Member in Bending

concentration
• Drilling Additional Holes for Shaft
• Reduction of Stress Concentration in Threaded Members

Notch Sensitivity
• The degree to which actual stress
concentration effect compares with the
theoretical stress concentration effect.
• The values of q are between zero and unity.
It is evident that if q=0, then Kf =1, and the
material has no sensitivity to notches at all.
On the other hand if q=1, then Kf = Kt, and
the material has full notch sensitivity.

Factors for Notch sensitivity
• Notch radius
• Material
• Size of the component
• Type of loading

Notch sensitivity of common
materials

Design Consideration &
Calculation of K
• For design consideration, the result obtained
depends on the following: In case of a circular
hole in a bar we consider the ratio 𝒓 / 𝒅. For
the fillet the ratio is 𝒓 / 𝒅 𝑎𝑛𝑑 𝑫 / 𝒅.
• To find the value of K we need to find use
graph.

PSG data book page no is 7.8 to 7.16

Classification of Loads
• Static Stress
• Variable Stress
– Repeated and reversed
– Fluctuating
– Shock or impact

Reversed or Repeated or Cyclic
Loading

Endurance Limit
• Max value of stress that the standard
specimen can sustain for a infinite number of
cycles (10^6 cycles) without failure.
• Stress ratio =
𝜎𝑚𝑎𝑥
𝜎𝑚𝑖𝑛

Factors affecting Endurance limit
• Size factor (KS)
• Load Factor (KL)
• Surface Factor (Ksf)
• Reliability Factor (KR)
• Other factors/Miscellaneous factor such as
Temperature factor, impact factor, etc
• Fatigue stress concentration factor,
Kf = 1+q(Kt- 1) [Kt - Theoretical stress
concentration factor, q- notch sensitivity factor]

• For goodman’s equation:
1
𝑛
=
𝐾𝑡
𝐾𝑆𝑍. 𝐾𝑆𝐹
[
𝜎𝑚
𝜎𝑦
+
𝜎𝑎
𝜎−1
]

AU Problems
• A component machined from a plate made of 45C8
(σu= 650 MPa) as shown in Fig. 11b. It is subjected to a
completely reversed axial force of 100 kN. The
reliability factor, kc = 0.897; factor of safety = 2. The
size factor, kb = 0.8, surface finish factor, ka = 0.76.
determine the thickness of the plate, for infinite life, if
the notch sensitivity factor, q = 0.8. (NOV/DEC 2020
AND April/May 2021)

AU Problems
• A shaft of diameter 'd' is subjected to a torque
varying between 900 Nm to 1800 Nm.
Assuming a factor of safety 2 and a stress
concentration factor of 1.2, find the diameter
of the shaft. Take au = 650 N/mm2, ay = 480
N/mm2, Size factor B = 0.85 and surface finish
factor C = 0.5. (NOV/DEC 2014)

AU Problems
• A cantilever beam made of cold drawn carbon steel of
circular cross section as shown in fig., is subjected to a load
which varies from –F to 3 F. Determine the maximum load
that this member can withstand for an indefinite life using a
factor of safety as 2. The theoretical stress concentration
factor is 1.42 and the notch sensitivity is 0.9. Assume the
following values:
– Ultimate stress = 550 MPa
– Yield stress = 470 MPa
– Endurance limit = 275 MPa
– Size factor = 0.85
• Surface finish factor = 0.89.
• (APR/MAY 2011)

PART C AU Problems
• A machine component is subjected to a flexural
stress which fluctuates between +300 MN/m2
and -150MN/m2. Determine the value of
minimum ultimate strength according to
• 1) Gerber relation
• 2) Modified Goodman relation and
• 3) Soderberg relation.
• Take yield strength =0.55 Ultimate strength;
Endurance strength =0.5 Ultimate strength; and
factor of safety =2. (NOV/DEC 2017)

AU Problems
• A 40 mm diameter shaft is made from carbon steel
having ultimate tensile strength of 600 MPa. It is
subjected to a torque which fluctuates between 1500 Nm
to -900 Nm. Using Soderberg method, calculate the
factor of safety. Assume suitable values for any other
data needed. (APR/MAY 2019)

AU Problems
• A steel cantilever is 200 mm long. It is subjected to an axial load which
varies from 150 N (compression) to 450 N (tension) and also a transverse
load at its free end which varies from 80 N up to 120 N down. The
cantilever is of circular cross-section. It is of diameter 2d for the first 50
mm and of diameter d for the remaining length. Determine its diameter
taking a factor of safety of 2. Assume the following values :
– Yield stress = 330 MPa
– Endurance limit in reversed loading = 300 MPa
– Correction factors = 0.7 in reversed axial loading = 1.0 in reversed
bending
– Stress concentration factor = 1.44 for bending = 1.64 for axial loading
– Size effect factor = 0.85
– Surface effect factor = 0.90
– Notch sensitivity index = 0.90 (APR/MAY 2016)

AU Problems
• A cantilever rod of length 120mm with circular
section is subjected to a cyclic transverse load;
varying from -100 N to 300N at its free end.
Determine the diameter “d” of the rod, by
(i) Goodman method and (ii) Soderberg method
using the following data.
Factor of safety =2; Theoretical stress
concentration factor=1.4; Notch sensitivity
factor=0.9; ultimate strength =550MPa; Yield
Strength =320MPa; Endurance limit =275MPa;
size correction factor=0.85; Surface correction
factor=0.9. (NOV/DEC 2015)

AU Problems
• A cantilever rod of length 120mm with circular
section is subjected to a cyclic transverse load;
varying from -100 N to 300N at its free end.
Determine the diameter “d” of the rod, by (i)
Goodman method and (ii) Soderberg method
using the following data. Factor of safety =2;
Theoretical stress concentration factor=1.4;
Notch sensitivity factor=0.9; ultimate strength
=550MPa; Yield Strength =320MPa; Endurance
limit =275MPa; size correction factor=0.85;
Surface correction factor=0.9. (NOV/DEC 2015)

AU Problems
• A shaft of diameter 'd' is subjected to a
torque varying between 900 Nm to 1800
Nm. Assuming a factor of safety 2 and a
stress concentration factor of 1.2, find
the diameter of the shaft. Take au = 650
N/mm2, ay = 480 N/mm2, Size factor B =
0.85 and surface finish factor C = 0.5.
(NOV/DEC 2014)

AU Problems
• Define Stress concentration, Give some
methods of reducing stress concentration.
(NOV/DEC 2011)
• What is the difference betwaeen Gerber curve
and Soderberg and Goodman lines?
(APR/MAY 2013)

STEADY STRESSES AND
MACHINE MEMBERS
Design
Process
fits and
tolerances
Impact
Stress
Principle
stresses
(PSG 7.2)
Curved
Beams
(PSG 6.2,
6.3)
Theories
of Failure
(PSG 7.3)
Stress
Concentration
(PSG 7.6)

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