PrePared by

AjAy kumar
AssociAte professor
Mechanical engineering departMent
Kiet, ghaziabad
 Introduction
Definition, Design requirements of machine elements, Design
procedure, Standards in design, Selection of preferred sizes,
Indian Standards designation of carbon & alloy steels,
Selection of materials for static and fatigue loads.
 Machine Design
The subject Machine Design is the creation of new and better machines
and improving the existing ones.
General Considerations/requirements in Machine Design
 Type of load and stresses caused by the load
 Motion of the parts or kinematics of the machine
 Selection of materials
 Form and size of the parts
 Frictional resistance and lubrication
 Convenient and economical features
 Use of standard parts
 Safety of operation
 Workshop facilities
 Number of machines to be manufactured
 Cost of construction
 Assembling
General Procedure in Machine Design
In designing a machine component, there is no rigid rule.
However, the general procedure to solve a design problem is as follows :
1. Recognition of need.
2. Synthesis (Mechanisms).
3. Analysis of forces.
4. Material selection.
5. Design of elements (Size and Stresses).
6. Modification.
7. Detailed drawing.
8. Production.

Selection of Material
The following factors should be considered in design:
1. Availability of Material
2. Suitability of Material for Working Condition
3. The Cost of Material
4. Suitability of material for Manufacturing
 Preferred Numbers
When a machine is to be made in several sizes with different powers or capacities, it is
necessary to decide what capacities will cover a certain range efficiently with minimum
number of sizes. These numbers are called preferred number.

It has been found by experience that a certain range can be covered efficiently when it
follows a geometrical progression with a constant ratio.

The preferred numbers are the conventionally rounded off values derived from
geometric series including the integral powers of 10 and having as common ratio of the
following factors:
5√10, 10√10, 20√10, 40√10,

These ratios are approximately equal to 1.58, 1.26, 1.12 and 1.06. The series of
preferrednumbers are designated as *R5, R10, R20 and R40 respectively.
The knowledge of materials and their properties is of great significance for a design engineer.
Classification of Engineering Materials
1. Metals and their alloys, such as iron, steel, copper, aluminium, etc.
2. Non-metals, such as glass, rubber, plastic, etc.
The metals may be further classified as :
(a) Ferrous metals: those which have the iron as their main constituent, such as cast iron,
wrought iron and steel.
(b) Non-ferrous metals: those which have a metal other than iron as their main constituent,
such as copper, aluminium, brass, tin, zinc, etc.
Wrought Iron
It is the purest iron which contains at least 99.5% iron but may contain upto 99.9% iron. The
typical composition of a wrought iron is
Carbon = 0.020%, Silicon = 0.120%, Sulphur = 0.018%, Phosphorus = 0.020%, Slag = 0.070%,
and the remaining is iron.

The wrought iron is a tough, malleable and ductile material. It cannot stand sudden and
excessive shocks. Its ultimate tensile strength is 250 MPa to 500 MPa and the ultimate
compressive strength is 300 MPa.
It can be easily forged or welded. It is used for chains, crane hooks, railway couplings, water
and steam pipes.
Cast Iron
It is obtained by re-melting pig iron with coke and limestone in a furnace known as cupola.
It is primarily an alloy of iron and carbon. The carbon contents in cast iron varies from 1.7
percent to 4.5 percent.
brittle material, therefore, it can’t be used in machines parts which are subjected to shocks.
low cost, good casting characteristics, high compressive strength, wear resistance and excellent
machinability. The compressive strength of cast iron is much greater than the tensile strength.
Following are the values of ultimate strength of cast iron :
Tensile strength = 100 to 200 MPa*
Compressive strength = 400 to 1000 MPa
Shear strength = 120 MPa
1. Grey cast iron.
Types of Cast Iron
2. White cast iron
3. Chilled cast iron
4. Mottled cast iron
5. Malleable cast iron
(a) white heart malleable iron casting,
(b) black heart malleable iron casting, and
(c) pearlitic malleable iron casting.
6. Nodular or spheroidal graphite cast iron.
Types of Cast Iron
1. Grey cast iron. It is an ordinary commercial iron having the following compositions :
Carbon = 3 to 3.5%; Silicon = 1 to 2.75%; Manganese = 0.40 to 1.0%; Phosphorous = 0.15 to 1% ;
Sulphur = 0.02 to 0.15% ; and the remaining is iron.
The grey colour is due to the fact that the carbon is present in the form of *free graphite. It has
a low tensile strength, high compressive strength and no ductility.

2. White cast iron. The white cast iron shows a white fracture and has the following
approximate
compositions :
Carbon = 1.75 to 2.3% ; Silicon = 0.85 to 1.2% ; Manganese = less than 0.4% ; Phosphorus
= less than 0.2% ; Sulphur = less than 0.12%, and the remaining is iron.
The white colour is due to fact that it has no graphite and whole of the carbon is in the form of
carbide (known as cementite) which is the hardest constituent of iron. The white cast iron has a
high
tensile strength and a low compressive strength. Since it is hard, therefore, it cannot be
machined with
ordinary cutting tools but requires grinding as shaping process.
3. Chilled cast iron. It is a white cast iron produced by quick cooling of molten iron.
According to Indian standard specifications (IS: 210 – 1993), the grey cast iron is designated by
the alphabets ‘FG’ followed by a figure indicating the minimum tensile strength in MPa or
N/mm2. For example, ‘FG 150’ means grey cast iron with 150 MPa or N/mm2 as minimum
tensile strength.
According to Indian standard specification (IS : 1865-1991), the nodular or spheroidal graphite
cast iron is designated by the alphabets ‘SG’ followed by the figures indicating the minimum
tensile strength in MPa or N/mm2 and the percentage elongation. For example, SG 400/15
means spheroidal graphite cast iron with 400 MPa as minimum tensile strength and 15 percent
elongation.
According to IS: 14329 – 1995, the whiteheart, blackheart and pearlitic malleable cast irons are
designated by the alphabets WM, BM and PM respectively. These designations are followed by a
figure indicating the minimum tensile strength in MPa or N/mm2. For example ‘WM 350’
denotes whiteheart malleable cast iron with 350 MPa as minimum tensile strength.
Steel
It is an alloy of iron and carbon, with carbon content up to a maximum of 1.5%. The carbon
occurs in the form of iron carbide, because of its ability to increase the hardness and strength of
the steel. Other elements e.g. silicon, sulphur, phosphorus and manganese are also present to
greater or lesser amount to impart certain desired properties to it. Most of the steel produced
now-a-days is plain carbon steel or simply carbon steel. A carbon steel is defined as a steel
which has its properties mainly due to its carbon content and does not contain more than 0.5%
of silicon and 1.5% of manganese. The plain carbon steels varying from 0.06% carbon to 1.5%
carbon are divided into the following
types depending upon the carbon content.
1. Dead mild steel — up to 0.15% carbon
2. Low carbon or mild steel — 0.15% to 0.45% carbon
3. Medium carbon steel — 0.45% to 0.8% carbon
4. High carbon steel — 0.8% to 1.5% carbon
According to Indian standard *[IS : 1762 (Part-I)–1974], a new system of designating the
steel is recommended. According to this standard, steels are designated on the following two
basis :
(a) On the basis of mechanical properties, and (b) On the basis of chemical composition.
Steels Designated on the Basis of Mechanical Properties
These steels are carbon and low alloy steels where the main criterion in the selection and
inspection of steel is the tensile strength or yield stress. According to Indian standard **IS: 1570
(Part–I)- 1978 (Reaffirmed 1993), these steels are designated by a symbol ‘Fe’ or ‘Fe E’
depending on whether the steel has been specified on the basis of minimum tensile strength or
yield strength, followed by the figure indicating the minimum tensile strength or yield stress in
N/mm2. For example ‘Fe 290’ means a steel having minimum tensile strength of 290 N/mm2
and ‘Fe E 220’ means a steel having yield strength of 220 N/mm2.

Steels Designated on the Basis of Chemical Composition

According to Indian standard, IS : 1570 (Part II/Sec I)-1979 (Reaffirmed 1991), the carbon
steels are designated in the following order :
(a) Figure indicating 100 times the average percentage of carbon content,
(b) Letter ‘C’, and
(c) Figure indicating 10 times the average percentage of manganese content. The figure after
multiplying shall be rounded off to the nearest integer.
For example 20C8 means a carbon steel containing 0.15 to 0.25 per cent (0.2 per cent on
an average) carbon and 0.60 to 0.90 per cent (0.75 per cent rounded off to 0.8 per cent on an
average)
manganese.
Free Cutting Steels
The free cutting steels contain sulphur and phosphorus. These steels have higher sulphur
content than other carbon steels. In general, the carbon content of such steels vary from 0.1 to
0.45 per cent and sulphur from 0.08 to 0.3 per cent.
Indian Standard Designation of Low and Medium Alloy Steels

1. Figure indicating 100 times the average percentage carbon.

2. Chemical symbol for alloying elements each followed by the figure for its average
percentage content multiplied by a factor as given below :
Element Multiplying factor
Cr, Co, Ni, Mn, Si and W 4
Al, Be, V, Pb, Cu, Nb, Ti, Ta, Zr and Mo 10
P, S and N 100
For example 40 Cr 4 Mo 2 means alloy steel having average 0.4% carbon, 1% chromium
and 0.25% molybdenum.
Notes : 1. The figure after multiplying shall be rounded off to the nearest integer.
2. Symbol ‘Mn’ for manganese shall be included in case manganese content is equal to or
greater than 1 per cent.
3. The chemical symbols and their figures shall be listed in the designation in the order of
decreasing content.
Indian Standard Designation of High Alloy Steels (Stainless Steel and
Heat Resisting Steel)
According to Indian standard, IS : 1762 (Part I)-1974 (Reaffirmed 1993), the high alloy steels
(i.e. stainless steel and heat resisting steel) are designated in the following order:
1. Letter ‘X’.
2. Figure indicating 100 times the percentage of carbon content.
3. Chemical symbol for alloying elements each followed by a figure for its average percentage
content rounded off to the nearest integer.
4. Chemical symbol to indicate specially added element to allow the desired properties.
For example, X 10 Cr 18 Ni 9 means alloy steel with average carbon 0.10 per cent, chromium
18 per cent and nickel 9 per cent.
 Design for Static Load
Modes of failure, Factor of safety, Principal stresses, Stresses
due to bending and torsion, Theory of failure.
 Modes of Failure

1. Yielding
2. Deflection Beyond a certain range
3. Buckling
4. Fatigue
5. Fracture
6. Creep
7. Environmental Degradation
8. Vibration
9. Impact
10. Wear
 Factor of Safety
In case of ductile materials

In case of brittle materials

Selection of Factor of Safety
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.
Each of the above factors must be carefully considered and evaluated. The high factor of safety
results in unnecessary risk of failure. The values of factor of safety based on ultimate strength
for
different materials and type of load are given in the following table:
Principle stress and Principle Plane
 Problems on simple stress
Example 4.5. The piston rod of a steam engine is 50 mm in diameter and 600 mm long. The
diameter of the piston is 400 mm and the maximum steam pressure is 0.9 N/mm2. Find the
compression of the piston rod if the Young's modulus for the material of the piston rod is 210
kN/mm2. 0.165 mm Ans.

Example 4.7. A pull of 80 kN is transmitted from a bar X to the bar Y through a pin as shown
in Fig. If the maximum permissible tensile stress in the bars is 100 N/mm2 and the permissible
shear stress in the pin is 80 N/mm2, find the diameter of bars and of the pin.

Diameter of the bars Db = 32 mm Diameter of the pin Dp = 25.2 mm

Example 4.16. A mild steel rod supports a tensile load of 50 kN. If the stress in the rod is limited
to 100 MPa, find the size of the rod when the cross-section is 1. circular, 2. square, and
3. rectangular with width = 3 × thickness. 25.23 mm Ans. 22.4 mm Ans , 12.9 mm 38.7 mm Ans
Example 4.17. A steel bar 2.4 m long and 30 mm square is elongated by a load of 500 kN. If
poisson's ratio is 0.25, find the increase in volume. Take E = 0.2 × 106 N/mm2. 3024 mm3 Ans
Simple Bending
 Problems on Bending
Example 5.6. A shaft is shown in fig. exerts forces of 25kN and 35kN concentrated at 150mm
and 200mm from the left and right hand bearing respectively. Find the diameter of shaft, if σy
=300MPa & FOS = 3

86.3 say 90 mm Ans

Example 5.8. A beam of uniform rectangular cross-section is fixed at one end and carries an
electric motor weighing 400 N at a distance of 300 mm from the fixed end. The maximum
bending stress in the beam is 40 MPa. Find the width and depth of the beam, if depth is twice
that of width.

b = 16.5 mm h = 33 mm
Torsion
 Problems on Torsion

Example 5.2. A steel shaft 35 mm in diameter and 1.2 m long held rigidly at one end has a
hand wheel 500 mm in diameter keyed to the other end. The modulus of rigidity of steel is 80
GPa.
1. What load applied to tangent to the rim of the wheel produce a torsional shear of 60 MPa?
2. How many degrees will the wheel turn when this load is applied? 2020 N 0.050
Example 5.3. A shaft is transmitting 97.5 kW at 180 r.p.m. If the allowable shear stress in the
material is 60 MPa, find the suitable diameter for the shaft. The shaft is not to twist more that 1°
in a length of 3 metres. Take C = 80 GPa. d = 76 mm d = 103 say 105 mm

Example 5.4. 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.
do = 172.7 mm do = 176.2 say 180 mm
 Combined Direct, Bending and Torsion
Example 5.13. A hollow shaft of 40 mm outer diameter and 25 mm inner diameter is subjected to
a twisting moment of 120 N-m, 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 stresses.
BENDING STRESS =15.02 MPa DIRECT STRESS =13.05 MPa SHEAR STRESS 11.27 MPa
COMBINED EFFECT BENDING AND DIRECT =28.07 MPa
MAX PRINCIPLE STRESS = 32.035 MPa Ans.
MAXIMUM SHEAR STRESS = 18 MPa Ans.
Theories of Failure Under Static Load
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).
Example 5.16. 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 bolt required according to
1. Maximum principal stress theory; 2. Maximum shear stress theory; 3. Maximum principal
strain theory; 4. Maximum strain energy theory; and 5. Maximum distortion energy theory.
Take permissible tensile stress at elastic limit = 100 MPa and poisson’s ratio = 0.3.
Example 5.17. A cylindrical shaft made of steel of yield strength 700 MPa is subjected to static
loads consisting of bending moment 10 kN-m and a torsional moment 30 kN-m. Determine the
diameter of the shaft using two different theories of failure, and assuming a factor of safety of 2.
Take E = 210 GPa and poisson's ratio = 0.25.

Example 5.18. A mild steel shaft of 50 mm diameter is subjected to a bending moment of 2000
N-m and a torque T. If the yield point of the steel in tension is 200 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.