FUNDAMENTAL CONCEPTS IN DESIGN Introduction to the design process

Machine Design
 The subject Machine Deisgn is the creation of new and better
machines and improving the existing ones. A new or better
machine is one which is more economical in the overall cost
of production and operation.
 The process of design is a long and time consuming one. From
the study of existing ideas, a new idea has to be conceived.
The idea is then studied keeping in mind its commercial
success and given shape and form in the form of drawings.
 In the preparation of these drawings, care must be taken of
the availability of resources in money, in men and in
materials required for the successful completion of the new
idea into an actual reality.
 In designing a machine component, it is necessary to have a
good knowledge of many subjects such as Mathematics,
Engineering mechanics, Strength of Materials, Theory of
Machines, Workshop Processes and Engineering Drawing.

INTRODUCTION
Engineering Design:
 Engineering design is a process of applying various
scientific principles and techniques for purpose of
defining in detail a product (or) a process (or) a system to
its realisation.
 In simpler words design is formulation of a plan,
for execution towards satisfying a needs.
 Machine Design is defined as to fix dimension for
machine components.

VARIOUS STEPS IN DESIGN PROCESS
Recongnition of a need:
Identifying the customer needs through market research.
Definition of the problem:
Preparation of complete list of technical specifications.
Synthesis:
Collection of new ideas (or) modifying the existing Ideas.
Analysis:
 The forces acting on the component are determined.
 The material for the component is selected.
 The geometric dimensions of the component are
determined.

CONTINUE…
Evaluation:
The possible success of the proposal should be
verified from technical and economical stand points.
Detailed Design:
It’s the actual sizing and dimensioning all individual
components in the part.
Proto-type & Testing:
Proto-type testing may lead to some modification.
Production:
Actual component manufactured at shop floor.

TYPES OF MACHINE DESIGN
Adaptive design:
The designer only makes minor alteration (or)
modifications in the existing designs of the product.
Development design:
Modifying existing designs into a new idea by adopting a
new material or different method of manufacture.
New design:
This type of design needs lot of research, technical ability
and creative thinking.

Further classification of design
Rational design:
This type of design depends upon mathematical formulae
of principle of mechanics.
Empirical design:
This type of design based on empirical formula based on
practice and past experience.
Industrial design:
Depends upon the production aspects to manufacture.
Optimum design:
It’s the best design for the given objective function under
the specified constraints.

Continue…
System design:
To develop a system that will meet expected needs
within realistic constraints such as economical
environmental, social, political, ethical, safety and
sustainability
Element design:
Design of machine elements such as piston, crank
shaft, gear,etc.
Computer aided design:
Use of computer systems to assist in the creation,
modification, analysis and optimization of a design.

Factor influencing machine design
1. Safety
2. Reliability
3. Quality
4. Productivity
5. Cost
6. Ecological consideration
7. Availability of men, material & machines
8. Working environment
9. Energy conservation
10. Space constraints

General considerations in machine design
1. Type of load and stresses caused by the load
2. Motion of the parts or kinematics of the machine
3. Selection of materials
4. Form and size of the parts
5. Frictional resistance and lubrication
6. Convenient and economical features
7. Use of standard parts
8. Safety of operation
9. Workshop facilities
10. Number of machines to be manufactured
11. Cost of construction

Design rules
1. Design should aim at the best with the least
expenditure
2. The component should have adequate strength, wear
resistance and corrosion resistance
3. The assembly should be backlash free
4. Resonance is to be avoided
5. The design should be simple, fool proof, easy to
operate and should reduce operator’s fatigue.
6. The design should need only a minimum maintenance
7. Whenever possible, dimensions of the components
should be rounded off to standard values.

Engineering materials
a) Metals
b) Non-metals
Metals:
Ferrous – Which contains iron as the major constituent
Ex. Steel, Cast Iron
Non-ferrous – materials don't contains Iron.
Ex. Copper, Aluminium
Non-Metals:
(i) Ceramic materials – oxides, carbides and nitrides of various
metals. Ex. Glass, Brick, Concrete, Cement etc.
(ii) Organic materials – Polymeric materials composed of carbon
compounds. Ex: Paper, fuel, rubber, paints, etc.

Factors to be considered for the selection of
materials
1. Availability
2. Cost
3. Physical properties
4. Mechanical Properties
5. Manufacturing process
Physical properties:
 Colour Electical conductivity
 Shape Thermal conductivity
 Size
 Density

 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

Direct Stress
Load:
Any external force acting upon a machine member
Types of load:
(i) Dead (or) Steady (or) Static load:
The load which does not change in magnitude
and direction.
Ex. Self weight
(ii) Live (or) Varying load:
The load which is continuously changing.
Ex. Vehicle pass over a bridge

Continue…
(iii) Suddenly applied load (or) shock load:
The load which is applied suddenly
Ex: Blows of a hammer
(iv) Impact load:
The load which is applied with some initial velocity
(or) The load which is dropped from certain height.
Ex: forging

Continue…
Stress:
The internal resistance of force per unit area is called stress.
σ =P/A
Where P = Load or force acting on the body
A = Cross- sectional area of the body
Strain:
The rate of change of deformation (or) It’s the ratio of change in
dimension to the original dimension.
e =δl/l
Relationship between σ,e,E, δl
Deformation, ( δl ) =pl/AE

TYPES OF STRESSES AND
STRAINS
Tensile Stress: The stress induced in a body, when subjected to two
equal and opposite pulls as result of which there is an increase in
length, is known as tensile stress. The ratio of increase in length to
the original length is known as tensile strain.
Tensile stress σ = Resisting force =
Cross sectional area
and tensile strain, ε = Increase length =
Original length
A
P
l
l


Compressive Stress
The stress induced in a body, when subjected to two equal
and opposite pushes as a result of which there is a
decrease in length of the body, is known as compressive
stress. The ratio of decrease in length to the original length
is known as compressive strain.
Compressive stress, σ = Resisting force =
Cross sectional area
Compressive strain, ε = Decrease in length =
Original length
A
P
l
l


Shear Stress:
The stress induced in a body, when subjected to two equal
and opposite forces which are acting tangentially across
the resisting section as a result of which the body tends to
shear off across the section, is known as shear stress. The
corresponding strain is known as shear strain.
Shear stress, q = Shear resistance =
Shear area
Shear strain Φ = Transverse displacement =
Distance
A
P
l
l


FACTOR OF SAFETY AND
LATERAL STRAIN
FACTOR OF SAFETY:
It is defined as the ratio of the ultimate stress to the working stress
of the material.
Factor of Safety = Ultimate Stress
Working Stress
LATERAL STRAIN:
The strain at right angles to the direction of applied load is known as
lateral strain.
Lateral Strain = Increase or Decrease in Lateral Dimension
Original Lateral Dimension

Modulus of Elasticity (Young’s Modulus (E):
According to Hooke’s law, the stress in a material is proportional to the
strain upto the elastic limit. Therefore within the elastic limit, the ratio of
the axial stress to the corresponding axial strain is found to be a constant.
This constant is called Modulus of Elasticity or Young’s Modulus. It is
denoted by E.
Modulus of Elasticity or Young’s Modulus = Axial Stress
Axial Strain
E =
Modulus of Rigidity or Shear Modulus (G):
The ratio of shear stress to the corresponding shear strain is found to be a
constant upto the elastic limit of the material. This constant is called
Modulus of Rigidity or Shear Modulus of the material. It is denoted by G.
Modulus of Rigidity or Shear Modulus = Shear Stress = G =
Shear Strain



q

Poisson’s Ratio (μ) or (1/m):
It has been experimentally found, that if a body is stressed within its
elastic limit, the lateral strain bears a constant ratio to the linear
strain. Mathematically,
Lateral strain = A constant
Linear strain
This constant is known as poisson’s ratio and is denoted by 1/m or μ.
Strain Energy (or) Resilence:
When a body is loaded with in the elastic limit the work done on the
body is stored in the form of energy . The strained body is now
capable of doing some external work on removal of the load. The
energy stored in the body due to internal strain is called strain energy
or resilience.
Strain energy = (σ2
/2E) ×A×l------------- N.mm

It states that, “ Within elastic limit the stress induced in the
material is directly proportional to strain”.
σ e
σ = E e.
E = σ
e
Where, σ – Stress; e – Strain; E – Young’s Modulus

Stress Strain Diagram
STRESS/LOAD
EXTENSION/STRAIN
Proportional Limit
Elastic Limit
Yield Point
Ultimate Strength
Breaking Limit

•Point P: Proportional Limit: Within Proportional Limit stress is directly
proportional to strain. Hence the material will regain its original shape after
unloading.
The stress corresponding to the load is known as Limit of Proportionality.
•Point E represents the elastic limit. In the region PE, the stress is not
proportional to strain. It means the stress strain diagram is not a straight line.
•Any loading beyond point E, will cause permanent deformation. The stress
corresponding to the load at E is called at Elastic Limit.
•Yield Stress (Point Y) :The loading beyond E causes extension much larger
than the extensions observed earlier. The material yields to a greater extent
and the stress corresponding to the load at Y is termed as Yield Stress
•Beyond Y, a much smaller increase in the load causes considerable
extension and the materials is said to be semi plastic mode.

• At U, the material yields at a particular point and a neck is
formed there. The stress corresponding to that load at U is
called maximum stress (Ultimate stress)
• Beyond U, the extension governed by the time of loading.
The load required to cause extension is smaller than the load
at M.
•The area of cross section is considerable reduced.
• The elongation continues till the material breaks at B.
•The stress corresponding to the load at B is called Breaking
stress.

Continue…
 Percentage reduction in area:
% reduction in area = (A-A1)/A ×100
A – Original area of cross-section.m2
A1 – Cross- sectional area after fructure at the neck,m2
Percentage elongation:
% elongation in length= (l1-l/l × 100
l1 – Length of specimen after fracture, mm
l – original length, mm

Torsional shear stress
 When a machine member is subjected to the action
of two equal and opposite couples acting in parallel
planes, then the machine member is said to be subjected to
torsion.
 The stress set up by torsion is known as torsional
shear stress.
 Consider a shaft fixed at one end and subjected to a torque
(T) at the other end.
 As a result of this torque every cross-section of the shaft is
subjected to torsional shear stress.

Continue…
T/J = τ/R = Cθ/l
θ= angle of twist, radian
T=Torque = twisting moment = turning moment, N-
mm
J= Polar movement of Inertia, mm4
τ= Torsional shear stress , N/mm2
C= Modulus of rigidity (or) shear modulus, N/mm2
l= Length of the shaft, mm

Continue…
Polar modulus zp:
T/J = τ/R
T = τ× J/R
J/R is called the polar modulus (zp)
T= τ. zp
For solid shaft of diameter (d)
J=πd4
/32
R = d/2
Z p = J/R = πd3
/16
T = τ× (πd3
/16) ---- N-mm

Continue…
Torsional rigidity:
T/J =cθ/l
θ= T×l/c×J
The quantity c×J is called torsional rigidity
Power transmitted by shaft (p):
P = 2πNT/60 ---- Watts (or) N.m/s
T= torque-----N-m
N=speed------rpm

Impact and shock loading
The load acting on any machine component can be of either of
these two types
1. Gradual load
2. Suddenly applied (or) Impact (or) Shock load.
Gradual load:
Gradual load is one which, goes on increasing over a
period of time till the maximum value is reached.
Suddenly applied (or) Impact (or) Shock load:
Impact (or) shock load which is applied suddenly (or)
with some initial velocity.
Ex: Punching presses, Hammer

Principal Stress
 Principal plane is a plane in which the shear
stress is zero, and the direct stresses acting
along these planes are known as principal
stress.
 When shear stress is also acting in addition we
have to find out maximum and minimum
principal stresses.

Eccentric loading
 An external load, whose line of action is
parallel but does not coinside with the
centroidal axis of the machine component, is
known as an eccentric load (P).
 The distance between the centroidal axis of the
machine component and the eccentric load is
called eccentricity.
 It is generally denoted by e.

Problems
1. A cast-iron link, as shown in fig., is to carry a load of 20
kN. If the tensile and compressive stresses in the link are
not to exceed 25 MPa and 80 MPa respectively, obtain
the dimensions of the cross-section of the link at the
middle of its length.

Continue…
A mild steel bracket as shown in fig., is subjected to a pull
of 6000 N acting at 45º to its horizontal axis. The bracket
has a rectangular section whose depth is twice the
thickness. Find the cross- sectional dimensions of the
bracket, if the permissible stress in the material of the
bracket is limited to 60 MPa.

Continue…
Ex:-
4. A wall bracket of rectangular cross-section as shown in
fig. It is subjected to a pull of 6 kN acting at 30º to the
horizontal. If the maximum stress induced in the bracket
material is not to exceed 25 N/mm2
in both tension and
compression. Design the cross-section of the bracket.

DESIGN OF CURVED BEAMS –
CRANE HOOK AND C- FRAME
 In the straight beam, the neutral axis of the section
coincides with its centroidal axis and stress distribution in
the beam is linear.
But in the case of curved beams, the neutral axis of the
section not coincides with its centroidal axis and stress
distribution in the beam is non-linear.
Bending stress general expression:
σb =
M – Bending moment
A – Area of cross-section







 y
R
y
e
A
M
n

Bending stress in a curved beam

Continue…
e- Distance from centroidal axis to the neutral axis (R-Rn)
R- Radius of curvature of the centroidal axis
Rn -Radius of the curvature of the neutral axis
y- Distance from the neutral axis to the fibre under
consideration
 If the section is symmetrical such as a circle, rectangle, I-
beam with equal flange, then the maximum bending stress
will always occur at the inside fibre.
If the section is unsymmetrical, then the maximum
bending stress may occur at either the inside fibre (or) the
outside fibre.

Continue…
Maximum Bending stress at the inside fibre:
σbi =
yi= Distance from the neutral axis to the inside fibre = Rn –Ri
Ri = Radius of curvature of the inside fibre.
Maximum Bending stress at the outside fibre:
σbo =
yo = Distance from the neutral axis to the outside fibre = Ro –Rn
Ro = Radius of curvature of the outside fibre.
Resultant stress σ =









i
i
R
e
A
y
M
b
d 
 









o
o
R
e
A
y
M

Problems
1. The crane hook carries a load of 20 kN as shown in fig.
The section at X-X is rectanglar whose horizontal side is
100 mmm. Find the stresses in the inner and outer fibres
at the given section.

Continue…
2. 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= 5000N. (NOV/DEC 2013)

Continue…
3. A C- clamp is subjected to a maximum load of W, as
shown in fig. If the maximum tensile stress in the clamp is
limited to 140 Mpa, find the value of load W. (M/J 2012)

Continue…
Exercises:
1. A punch press, used for stamping sheet metal, has a punching
capacity 50 kN. The section of the frame is as shown in fig.
Find the resultant stress at the inner and outer fibre of the
section.

Stress concentration
Stress concentration:
 In most of the engineering components stress distribution
is not uniform, stress distribution will be uniform only
when there is no change in cross section.
Irregularity of the stress distribution due to abrupt changes
of form is called stress concentration.
Stress raisers:
Sudden changes in cross section and material
discontinuity is referred as stress raisers.
Ex: Holes, Notches, steps, threads, etc.

How to minimize stress
concentration
Suitably modifying the shape of the components, stress
concentration near the stress can be minimized.

Continue…
Stress concentration factor (kt):
max. stress induced due to stress concentration
kt = --------------------------------------------------------------------
stress caused without considering the stress concentration
maximum stress (σmax)
kt = ----------------------------------
Nominal stress (σ)
σmax = kt × stress at net section
Notch Sensitivity (q):
This is defined as the degree to which the actual stress
concentration effect compares with theoretical stress concentration
effect.

Continue…
Endurance limit:
It is defined as maximum value of completely reversed
bending stress which a polished specimen can
withstand without failure for infinite number of cycles.
Fatigue stress concentration factor (kf):
Endurance limit without stress concentration
kf = --------------------------------------------------------
Endurance limit with stress concentration
Kt, kf and q related as kf = 1+q (Kt -1)

Continue…
Taking stress concentration into account
find the maximum stress induced when a
tensile load of 20 kN is applied to a
stepped shaft of diameters 60 mm and 30
mm with fillet radius of 6 mm.

Design for variable loading
Variable load:
Variable load is a load which varies with respect to
time either in magnitude or in direction, sometimes both.
design of variable loading:
 Consider a rotating beam of circular cross section and
carrying a load of W, this load induces stresses in the
beam which are cyclic in nature.

Continue…
 Upper fibers of beam(A) under compression and lower
fiber (B) tensile after half revolution the point B occupies
the position of point A and point A occupies the point of
B. thus point B is now compression and point A is tensile.
 The stresses which vary from one value of compressive to
same value of tensile or vice versa are known as
completely reversed or cyclic stresses.
 The stresses which vary from a minimum value to a
maximum value of same nature is called fluctuating
stresses.
 The stresses which vary from zero to a certain maximum
value are called repeated stresses.

Continue…
 The stresses which vary from a minimum value to a maximum
value of the opposite nature is called alternative stresses (from
a certain minimum compressive to a maximum tensile or
minimum tensile to a certain maximum compressive).

problems
1. A circular bar of 500 mm length is supported
freely at its two ends. It is acted upon by a
central concentrated cylic load having a
minimum value of 20 kN and a maximum
value of 50 kN. Determine the diameter of bar
by taking a factor of safety of 1.5, size effect
of 0.85, surface finish factor of 0.9. The
material properties of bar are given by:
ultimate strength of 650 MPa, yield strength of
500 MPa and endurance strength of 350 MPa.

Continue…
2. A simply supported beam has a concentrated load
at the centre which fluctuates from a valve of P to
4P. The span of the beam is 500 mm and its cross-
sectional is circular with a diameter of 60mm.
Taking for the beam material an ultimate stress of
700 MPa, yield stress of 500 MPa, endurance limit
of 330 MPa for reversed bending and a factory of
safety of 1.3, calculate the maximum valve of P.
Take a size factor of 0.85 and a surface finish
factor of 0.9

Continue…
3. 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.

Continue…
A pulley is keyed to a shaft midway between two bearings. The
shaft is made of cold drawn steel for which the ultimate strength
is 550 MPa and the yield strength is 400 MPa. The bending
moment at the pulley varies from -150 N-m to 400 N-m as the
torque on the shaft varies from -50 N-m to 150 N-m. Obtain the
diameter of the shaft for an indefinite life. The stress
concentration factors for the keyway at the pulley in bending and
in torsion are 1.6 and 1.3 respectively. Take the following values:
Factor of safety = 1.5
Load correction factors = 1.0 in bending, 0.6 in torsion
Size effect factor = 0.85
Surface effect factor =0.88

Continue…
1. Nominal size:
It is the size of a part specified in the drawing.
2. Basic size:
It is the size of a part to which all limits of variation (i.e
tolerances) are applied to arrive at final dimensioning of
the mating parts. The nominal or basic size of a part is
often the same.
3. Actual size:
It is the actual measured dimension of the part.
4. Upper limit:
The largest size for a dimension of the part is called
upper limit.

Continue…
5. Lower limit:
The smallest size for a dimension of the part is called lower
limit.
6. Allowance:
It is the difference between the basic dimensions of the mating
parts.
Tolerance
It is the difference between the upper limit and lower limit of a
dimension.

Continue…
Unilateral tolerance:
When all the tolerance is allowed on one side of the
nominal size.
Ex:
Bilateral tolerance:
When the tolerance is allowed on both sides of the
nominal size.
Ex:
20
00
.
0
004
.
0


20
002
.
0
002
.
0



Continue…
Zero line:
It is a straight line corresponding to the basic size. The deviations
are measured from this line.
Upper deviation:
It is the algebraic difference between the maximum size and
basic size. The upper deviation of a hole is represented by a
symbol ES (Ecart Superior) and of a shaft, it is represented by es.
Lower deviation:
It is the algebraic difference between the minimum size and basic
size. The lower deviation of a hole is represented by a symbol EI
(Ecart Inferior) and of a shaft, it is represented by ei.

Continue…
Actual deviation:
It is the algebraic difference between in actual
size and the corresponding basic size.
Mean deviation:
It is the arithmetical mean between the upper
and lower deviations.

Fits
The degree of tightness or looseness between the two mating parts
is known as a fit of the parts.
Types of fits:
1. Clearance fit
2. Interference fit
3. Transition fit
Clearance fit:
The difference between the sizes of the hole and the shaft before
assembly. The difference must be positive.
Interference fit:
The arithmetical difference the sizes of the hole and shaft, before
assembly. The difference must be negative.

Continue…
Clearance fit:
 In a clearance fit, the
difference between the minimum
size of a hole and the maximum
size of the shaft is known as
minimum clearance.
 The difference between the
maximum size of a hole and the
maximum size of the shaft is
called maximum clearance.
 The clearance fits may be slide fit,
easy sliding fit, running fit, slack
running fit and loose running fit.

Continue..
Interference fit:
In an Interference fit, the
tolerance zone of the hole is
entirely below the tolerance
zone of the shaft.
Transition fit:
In a transition fit, the
tolerance zones of hole and
shaft over lap.

Basis of limit system
Hole basis system:
When the hole is kept as a
constant member and
different fits are obtained by
varying the shaft size.
Shaft basis system:
When the shaft is kept as a
constant member and
different fits are obtained by
varying the hole size.

FUNDAMENTAL CONCEPTS IN DESIGN Introduction to the design process

More Related Content

Similar to FUNDAMENTAL CONCEPTS IN DESIGN Introduction to the design process (20)

Recently uploaded (20)

FUNDAMENTAL CONCEPTS IN DESIGN Introduction to the design process