a book of machine design

MECHANICALNOTESHUB

A Book of machine Design Chapter-1

NOTES FROM SOUVIK BHATTACHARJEE

https://bhattacharjeesouvi.wixsite.com/mechanicalnoteshub

A Book of machine Design

1

INTRODUCTION:-

Machine Design or Mechanical Design can be defined as the process by which resources or

energy is converted into useful mechanical forms, or the mechanisms so as to obtain useful

output from the machines in the desired form as per the needs of the human beings.

GENERAL CONSIDERATION IN MACHINE DESIGN:-

The following consideration should be followed for designing:-

1. Type of Load and Stresses caused by the Load:-

The load on the Machine Component, may act in several ways due to which the Internal Stresses

are set up.

2. Motion of Parts:-

The successful operation of any Machine depends largely upon the simplest arrangements of the

Parts, which will give the required motion. The Motion of the Part may be Rectilinear Motion,

Curvilinear Motion, Constant Velocity, Constant or Variable Acceleration.

3. Selection of Material:-

Every Machine Design Engineer should have a thorough knowledge of the Properties of Material

and their behavior under working conditions.

4. Form and Size of the Parts:-

In order to design any Machine Part for form and size, it is necessary to know the Forces which

the Part must sustain. Any suddenly applied or impact load must be taken into consideration,

which may cause failure. The smallest Practicable Cross-Section may be used, but it may be

checked that the Stresses induced in the Designed Cross-Section are reasonably safe.

5. Frictional Resistance and Lubrication:-

There is always a Loss of Power due to Frictional Resistance. Careful attention must be given to

the matter of Lubrication of all surfaces which moves in contact with others.

6. Safety of Operator:-

A Machine Designer should always provide safety device for the safety of the operator. The

Safety Appliances should in no way interfere with the operation of the Machine.

7. Use of Standard Parts:-

The use of Standard Parts is closely related to the Cost of Machine, because the Cost of Standard

Parts is only a fraction of the cost of similar parts made to order.

8. Convenient and Economical Features:-

The operating feature of the Machine should be carefully studied. The Starting, Controlling and

Stopping Levers should be located on the basis of convenient handling.


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9. Workshop Facilities:-

A Design Engineer should be familiar with limitation of his Employer's Workshop, in order to

avoid the necessity of having work-done in some other Workshop.

10. Assembling:-

Every Machine must be assembled as a unit before it can function. The final Location of any

Machine is important and the Design Engineer must anticipate the exact location and the local

facilities for erection.

CONCEPT OF LOAD:-

It is defined as any external force acting upon a machine part. The following types of the load are

important from the subject point of view:

Based on the relation between the point of Action and direction of effect of the force, it is

classified into:

Fig.-1.1

Tension:- Tension is described as the pulling force transmitted axially by the means of a string, a

cable, chain, or similar one-dimensional continuous object, or by each end of a

rod, truss member, or similar three-dimensional object; tension might also be described as the

action-reaction pair of forces acting at each end of said elements.

Compression:- compression force is described as the pushing force transmitted axially by means

of string, cable, etc.

Shear:- Shearing forces are unaligned forces pushing one part of a body in one specific direction,

and another part of the body in the opposite direction.

Bending:- A bending force is a force that is applied to a length of material. The bending force is

applied to a point, area or volume that is some distance from a fixed portion of the component or

structure to which the force is being applied.

Torsion:- Torsion is the twisting of an object due to an applied torque. Fig. 1.1 shows how the

forces act on a particle and how its look like after applying.

Based on the nature of force applied:

https://en.wikipedia.org/wiki/Truss

https://en.wikipedia.org/wiki/Force


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Dead Load:- Constant load that is always exerted on a Body. Eg:- Weight of an apartment

without residents.

Fluctuating load:- It is also called as live load. Eg: Weight of residents of the apartment.

Inertia load:- Force that is exerted on a body to help it continue with its state of motion Eg:-

Forces acting on an accelerating car.

Point load:- Load concentrated at a point Eg: Weight of a person , when he is on a weight gauge.

Distributed load:- Load that is distributed over a small length or an area. Eg: A cantilever beam

in which load is applied at one end.

STRESS:-

When some external 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 resisting force per unit

area at any section of the body is known as 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 force or load 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 (ε).

Mathematically,

Strain, ε = δl / l or δl = ε.l

where δl = Change in length of the body, and

l = Original length of the body.

HOOK’S LAW:-

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


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or, E =σ

ϵ

Now , σ = P/A and ε = δl / l so,

E =P. l

A. δl

∴ δl =P. l

A. E

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. Hook’s Law holds good for tension as well as

compression.

Table 1.1. Values of E for the commonly used engineering materials.

MATERIAL VALUE OF E IN GPa

Steel and Nickel 200-220

Cast Iron 100-160

Copper 90-115

Brass 75-95

Alluminium 55-85

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.

Mathematically,

Shear Stress =Tangential Force

Resisting Area

Consider a body consisting of two plates connected by a

rivet as shown in Fig. 1.2. In this case, the tangential

force P tends to shear off the rivet at one cross-section.

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

Fig.-1.2


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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 τ = PA

= 4P

πd2

But in case of double shear the area should be 2 times of single shear

Hence, A = 2 ×π

4× d2

And shear stress, τ =P

A=

2P

πd2

SHEAR MODULUS:-

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.

Table 1.2. Values of C for the commonly used engineering materials.

MATERIAL VALUE OF C IN GPa

Steel and Nickel 80-100

Cast Iron 40-50

Copper 30-40

Brass 30-50

Alluminium 20-25

BEARING STRESS:-

Bearing stress is Contact pressure between two separate bodies. It can be defined as

Compression force divided by the characteristic Area perpendicular to it. Crushing stress occurs

when the body is in contact with the other body.


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Let us consider a riveted joint subjected to a load P as shown in Fig. 1.3.

Fig.-1.3

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.

STRESS-STRAIN DIAGRAM FOR DUCTILE MATERIAL:-

The values of the stress and corresponding strain are used to draw the stress-strain diagram of the

material tested. A stress-strain diagram for a mild steel under tensile test is shown in Fig. 1.4. in

which X-axis shows strain

value and Y-axis shows

corresponding stress value.

The various properties of the

material are discussed below:

The point shows in the curve

are as follows:-

A- Limit of Proportionality

B- Elastic limit

C- Upper yield point Fig-1.4

D- Lower yield point

E-Ultimate point

F- Breaking point


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Let discuss every point below:-

Limit of Proportionality:- It is up to point A in which Hook’s Law holds good enough.

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.

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.

Ultimate Point:- Point E indicates the location of the value of the ultimate stress. The portion DE

is called the yielding of the material at constant stress. From point E onwards, the strength of the

material increases and requires more stress for deformation, until point F is reached.

Breaking Point:- A material is considered to have completely failed once it reaches the ultimate

stress. The point of fracture, or the actual tearing of the material, does not occur until point F.

The point F is fracture point and the value of stress or load at F is slightly lower than E.

STRESS STRAIN CURVE FOR BRITTLE MATERIAL:-

Brittle materials are those which can't elongate as much the

ductile materials because they don't have such a long region

b/w point yield to ultimate.

Fig1.5 shows the diagram of stress-strain for brittle materials.

In the fig point O-P is called the elastic region. If stress is

removed, the material will return to its original length. Point

U is the point of ultimate strength and after that the

material will fail.

Fig.-1.5

During elongation following changes in area and length are:-


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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

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

TORSIONAL SHEAR STRESS:-

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. The stress set up by torsion is known as torsional shear stress. It is zero at the

centroidal axis and maximum at the outer surface.

Fig.-1.6

According to fig. 1.6 the mathematical expression for torsion equation is,

T

J=

τ

r=

C. θ

l


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Where,

τ = Torsional shear stress induced at the outer surface of the shaft or maximum shear stress,

r = Radius of the shaft,

T = Torque or twisting moment,

J = Second moment of area of the section about its polar axis or polar moment of inertia,

C = Modulus of rigidity for the shaft material,

l = Length of the shaft, and

θ = Angle of twist in radians on a length l.

The equation is based on the following assumptions:

1. The material of the shaft is uniform throughout.

2. The twist along the length of the shaft is uniform.

3. The normal cross-sections of the shaft, which were plane and circular before twist, remain

plane and circular after twist.

4. All diameters of the normal cross-section which were straight before twist, remain straight

with their magnitude unchanged, after twist.

5. The maximum shear stress induced in the shaft due to the twisting moment does not exceed its

elastic limit value.

[Note:- 1. The derivation of this equation is in the Strength Of Materials portion.

2. Strength of shaft for solid T= π

16× τ × d3

For hollow T= π

16× τ ×

D4−d4

D

3. The term torsional rigidity = C×J

4. Power P = T×ω]

BENDING EQUATION:-

In engineering practice, the machine parts of structural members may be subjected to static or

dynamic loads which cause bending stress in the sections besides other types of stresses such as

tensile, compressive and shearing stresses.

The following assumptions are usually made while deriving the bending formula.

1. The material of the beam is perfectly homogeneous (i.e. of the same material throughout) and

isotropic (i.e. of equal elastic properties in all directions).

2. The material of the beam obeys Hooke’s law.

3. The transverse sections which were plane before bending, remain plane after bending also.


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4. Each layer of the beam is free to expand or contract, independently, of the layer, above or

below it.

5. The Young’s modulus (E) is the same in tension and compression.

6. The loads are applied in the plane of bending.

Fig.-1.7

According to fig. 1.7 the mathematical expression for bending equation is

M

I=

σb

y=

E

R

Where,

M = Bending moment acting at the given section,

σ = Bending stress,

I = Moment of inertia of the cross-section about the neutral axis,

y = Distance from the neutral axis to the extreme fibre,

E = Young’s modulus of the material of the beam, and

R = Radius of curvature of the beam.

[ Note:- 1. The ratio of I/y is called Section Modulus (Z).

2. The product of E and I is called Flexural rigidity.

3. the ratio of bending moment to the section modulus is called Strength of beam. ]

CREEP STRAIN AND CURVE:-Deformation of material at higher temperature is far more than

that at normal temperature known as creep. Subsequently strain produced in heated material is

called creep strain. Creep strain is function of temperature.


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Fig.:-1.8

At elevated temperatures and stresses, much less than the high-temperature yield stress, metals

undergo permanent plastic deformation called creep. Figure 1.8 shows a schematic creep curve

for a constant load; a plot of the change in length verses time. The weight or load on the

specimen is held constant for the duration of the test. There are four portions of the curve that are

of interest:

An initial steep rate that is at least partly of elastic origin, from point "0" to point "A" in Figure

1.8.

This is followed by a region in which the elongation or deformation rate decreases with time,

the so-called transient or primary creep, from region "A" to "B" of Figure 1.8. The portion from

point "0" to point "B" occurs fairly quickly.

The next portion of the creep curve is the area of engineering interest, where the creep rate is

almost constant. The portion from "B" to "C" is nearly linear and predictable. Depending on the

load or stress, the time can be very long; two years in a test and several decades in service.

The fourth portion of the creep curve, beyond the constant-creep-rate or linear region, shows a

rapidly increasing creep rate which culminates in failure. Even under constant-load test

conditions, the effective stress may actually increase due to the damage that forms within the

microstructure.


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FATIGUE:-

Fatigue is defined as the weakening of a material caused by cyclic loading that results in

progressive and localised structural damage and the growth of cracks. Once a fatigue crack has

initiated, each loading cycle will grow the crack a small amount, typically

producing striations on some parts of the fracture surface. The crack will continue to grow until

it reaches a critical size, which occurs when the stress intensity factor of the crack exceeds

the fracture toughness of the material, producing rapid propagation and typically complete

fracture of the structure.

ENDURANCE LIMIT:-

The endurance limit , also known as the fatigue limit or fatigue strength, is the stress level below

which an infinite number of loading cycles can be applied to a material without

causing fatigue failure.

S-N CURVE FOR FATIGUE STRENGTH:-

Fig.-1.9

Fatigue properties of materials are often described using the fatigue limit or the S-N curve

(fatigue curve, Wöhler curve). The S-N curve describes the relation between cyclic stress

amplitude and number of cycles to failure. The figure below shows a typical S-N curve. On the

horizontal axis the number of cycles to failure is given on logarithmic scale. On the vertical axis

(either linear or logarithmic) the stress amplitude (sometimes the maximum stress) of the cycle is

https://en.wikipedia.org/wiki/Striation_(fatigue)

https://en.wikipedia.org/wiki/Stress_intensity_factor

https://en.wikipedia.org/wiki/Fracture_toughness

https://en.wikipedia.org/wiki/Fatigue_(material)


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given. S-N curves are derived from fatigue tests. Tests are performed by applying a cyclic stress

with constant amplitude (CA) on specimens until failure of the specimen. In some cases the test

is stopped after a very large number of cycles (N>106). The results is then interpreted as infinite

life. Fatigue curves are often given for Kt=1 (unnotched specimens). Those curves describe the

fatigue properties of a material. Actual structures are better described with S-N curves for Kt>1

(notched specimens).

FACTOR OF SAFETY:-

When you design a component the load that component has to carry during its service is

calculated taking in to consideration of all the possible loads that might act on it. This calculated

load is known as design load. When this design load acts, we expect the component to be safe.

That means the stress induced in the member when this design load acts on it has to be less than

yield point stress. This is because if the stress reaches yield point stress, the component is

considered to be failed Hence we need to keep the stress during its service well below the yield

point stress. By what factor you reduce this stress is left to the design engineer. He select a

suitable number (Factor of safety-FOS) by which he divide the yield stress to get the safe value

of stress which is known as design stress or working stress.

For ductile materials which has a well defined yield point factor of safety is defined as, the ratio

of yield stress to working stress for ductile materials.

FOS = yield stress

working stress

Or, Design stress = yield stress

FOS

For brittle materials which has no well defined yield point the factor of safety is calculated based

on ultimate stress.

FOS =Ultimate stress

working stress

When factor of safety is defined based on the ultimate stress, higher values are generally used.

STRESS CONCENTRATION:-

Whenever a machine component changes the shape of its cross-section, the simple stress

distribution no longer holds good and the neighbourhood of the discontinuity is different. This

irregularity in the stress distribution caused by abrupt changes of form is called stress


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concentration. It occurs for all kinds of stresses in the presence of fillets, notches, holes,

keyways, splines, surface roughness or scratches etc.

Fig.-1.10

In order to understand fully the idea of stress concentration, consider a member with different

cross-section under a tensile load as shown in Fig. 1.9. A little consideration will show that the

nominal stress in the right and left hand sides will be uniform but in the region where the

crosssection is changing, a re-distribution of the force within the member must take place. The

material near the edges is stressed considerably higher than the average value. The maximum

stress occurs at some point on the fillet and is directed parallel to the boundary at that point.

Cause:-

Various causes of stress concentration are as follows:

(i) Abrupt change of cross section.

(ii) Poor surface finish.

(iii) Localized loading.

(iv) Variation in the material properties.

Remedies:-

The presence of stresses concentration cannot be totally eliminated but it can be reduced, so

following are the remedial measures to control the effects of stress concentration.

(i) Provide additional notches and holes in tension members.

(ii) Fillet radius, undercutting and notch for member in bending.

(iii) Reduction of stress concentration in threaded member.

(iv) Provide taper cross-section to the sharp corner of member.

PROPERTIES OF ENGINEERING MATERIALS:-

The following properties are

Strength:-

It is the property of a material which opposes the deformation or breakdown of material in

presence of external forces or load. Materials which we finalize for our engineering products,


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must have suitable mechanical strength to be capable to work under different mechanical forces

or loads.

Toughness:-

It is the ability of a material to absorb the energy and gets plastically deformed without

fracturing. Its numerical value is determined by the amount of energy per unit volume. Its unit is

Joule/ m3. Value of toughness of a material can be determined by stress-strain characteristics of a

material. For good toughness, materials should have good strength as well as ductility.

For example: brittle materials, having good strength but limited ductility are not tough enough.

Conversely, materials having good ductility but low strength are also not tough enough.

Therefore, to be tough, a material should be capable to withstand both high stress and strain.

Hardness:-

It is the ability of a material to resist to permanent shape change due to external stress. There are

various measure of hardness – Scratch Hardness, Indentation Hardness and Rebound Hardness.

• Scratch Hardness Scratch Hardness is the ability of materials to the oppose the scratches to

outer surface layer due to external force.

• Indentation Hardness It is the ability of materials to oppose the dent due to punch of external

hard and sharp objects.

• Rebound Hardness Rebound hardness is also called as dynamic hardness. It is determined by

the height of “bounce” of a diamond tipped hammer dropped from a fixed height on the material.

Hardenability:-

It is the ability of a material to attain the hardness by heat treatment processing. It is determined

by the depth up to which the material becomes hard. The SI unit of hardenability is meter

(similar to length). Hardenability of material is inversely proportional to the weld-ability of

material.

Brittleness:-

https://www.electrical4u.com/si-system-of-units/


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Brittleness of a material indicates that how easily it gets fractured when it is subjected to a force

or load. When a brittle material is subjected to a stress it observes very less energy and gets

fractures without significant strain. Brittleness is converse to ductility of material. Brittleness of

material is temperature dependent. Some metals which are ductile at normal temperature become

brittle at low temperature.

Malleability:-

Malleability is a property of solid materials which indicates that how easily a material gets

deformed under compressive stress. Malleability is often categorized by the ability of material to

be formed in the form of a thin sheet by hammering or rolling. This mechanical property is an

aspect of plasticity of material. Malleability of material is temperature dependent. With rise in

temperature, the malleability of material increases.

Ductility:-

Ductility is a property of a solid material which indicates that how easily a material gets

deformed under tensile stress. Ductility is often categorized by the ability of material to get

stretched into a wire by pulling or drawing. This mechanical property is also an aspect of

plasticity of material and is temperature dependent. With rise in temperature, the ductility of

material increases.

Resilience:-

Resilience is the ability of material to absorb the energy when it is deformed elastically by

applying stress and release the energy when stress is removed. Proof resilience is defined as the

maximum energy that can be absorbed without permanent deformation. The modulus of

resilience is defined as the maximum energy that can be absorbed per unit volume without

permanent deformation. It can be determined by integrating the stress-strain cure from zero to

elastic limit. Its unit is joule/m3.

THEORIES OF FALIURES:-

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).


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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).

1. Maximum principal 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.

the maximum principal or normal stress (σt1) in a bi-axial stress system is given by

σ1 =σyt

FOS… … … … … . For ductitle materials

σ1 =σu

FOS… … … … … … 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.

2. Maximum shear stress theory (Guest’s or Tresca’s theory):-


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 /FOS

Where, τmax = Maximum shear stress in a bi-axial stress system,

τyt = Shear stress at yield point as determined from simple tension test,

and FOS = Factor of safety.

3. Maximum principle strain theory (Saint Venant theory):-


maximum principal (or normal) strain in a bi-axial stress system reaches the limiting value of

strain (i.e. strain at yield point) as determined from a simple tensile test. The maximum principal

(or normal) strain in a bi-axial stress system is given by

σt1 −σt2

m=

σyt

FOS


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Where,

σt1 and σt2 = Maximum and minimum principal stresses in a bi-axial stress system,

1/m = Poisson’s ratio,

E = Young’s modulus, and

FOS = Factor of safety and σyt= yield point stress.

This theory is not used, in general, because it only gives reliable results in particular cases.

4. Maximum strain energy theory (Haigh’s theory):-

According to this theory, the failure or yielding occurs at a point in a member when the strain

energy per unit volume in a bi-axial stress system reaches the limiting strain energy (i.e. strain

energy at the yield point) per unit volume as determined from simple tension test.

According to above theory,

(σt1)2 + (σt2)2 −2σt1 × σt2

m= (

σyt

FOS)2

Where,





This theory may be used for ductile materials.

5. 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 (also called shear strain energy) per unit volume in a bi-axial stress system reaches

the limiting distortion energy (i.e. distortion energy at yield point) 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 = (𝜎𝑦𝑡

𝐹𝑂𝑆)2

Where,




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This theory is mostly used for ductile materials in place of maximum strain energy theory. The

maximum distortion energy is the difference between the total strain energy and the strain energy

due to uniform stress.

SOME EXAMPLES OF NUMERICALS:-

N-1.1 A hydraulic press exerts a total load of 3.5 MN. This load is carried by two steel rods,

supporting the upper head of the press. If the safe stress is 85 MPa and E = 210 kN/mm2, find :

1. diameter of the rods, and 2. extension in each rod in a length of 2.5 m.

Solution. Given : P = 3.5 MN = 3.5 × 106 N ; σt = 85 MPa = 85 N/mm2 ; E = 210 kN/mm2 = 210

× 103 N/mm2 , l = 2.5 m = 2.5 × 103 mm

1. Diameter of the rods:-

Let d = Diameter of the rods in mm.

∴ Area, A = 𝜋

4× 𝑑2 = 0.7854𝑑2

Since the load P is carried by two rods, therefore load carried by each rod,

P1 = P

2=

3.5 × 106

2= 1.75 × 106N

We know that load carried by each rod (P1),

1.75 × 106 = σt . A = 85 × 0.7854 d2 = 66.76 d2

∴ d 2 = 1.75 × 106/66.76 = 26 213 or d = 162 mm Ans.

2. Extension in each rod:-

Let δl = Extension in each rod.

We know that Young's modulus (E),

210 × 103 =P1 × l

A × δl

Or, δl =σt×l

E=

85×2.5×103

210×103= 1.012mm

N-1.2 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. Find,

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?


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Solution. Given d = 35 mm or r = 17.5 mm; l = 1.2m = 1200 mm; D = 500

R = 250 mm; C = 80 GPa = 80kN/mm2 = 80 × 103N/mm2; τ = 60 MPa = 60N/mm2

1. Load applied to the tangent to the rim of the wheel:-

Let W = Load applied (in newton) to tangent to the rim of the wheel.

We know that torque applied to the hand wheel,

T = WR = W × 250 = 250 WN‐mm

and polar moment of inertia of the shaft,

J =π

32× d4 =

π

32 (35)4 = 147.34 × 103mm4

We know that T

J=

τ

r

250W

147.34×103=

60

17.5 or W =

60×147.34×103

17.5×250= 2020N Ans.

2. Number of degrees which the wheel 𝐰𝐢𝐓𝐓 turn when load 𝐖 = 𝟐𝟎𝟐𝟎𝐍 𝐢𝐬 𝐚pplied:-

Let θ = Required number of degrees.

We know that T

J=

C.θ

l

θ =T.⋅l

CJ=

250×2020×1200

80×103×147.34×103= 0.05∘ Ans.

N-1.3 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.

Solution. Given : P = 97.5 kW = 97.5 × 103 W ; N = 180 r.p.m. ; τ = 60 MPa = 60 N/mm2 ;

θ = 1° = π / 180 = 0.0174 rad ; l = 3 m = 3000 mm ; C = 80 GPa = 80 × 109 N/m2 = 80 × 103

N/mm2

Let T = Torque transmitted by the shaft in N-m, and

d = Diameter of the shaft in mm.

We know that the power transmitted by the shaft (P),

97.5 × 103 =2𝜋 × 180 × 𝑇

60

or, T = 5172 N − m = 5172 × 103N − mm

Now let us find the diameter of the shaft based on the strength and stiffness.

1. Considering strength of the shaft:-

We know that the torque transmitted (T) ,


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5172 × 103 =π

16× τ × d3 =

π

16× 60 × d3 = 11.78d3

d3 = 5172 × 103/11.78 = 439 × 103 or d = 76 mm

2. Considering stiffness of the shaft:-

Polar moment of inertia of the shaft,

J =π

32× d4 = 0.0982d4

We know that T

J=

C.θ

l

5172×103

0.0982d4=

80×103×0.0174

3000

Or, 52.7×106

d4= 0.464

Or, d4 = 52.7 × 106/0.464 = 113.6 × 106

Or, d = 103 mm

Taking larger of the two values, we shall provide d = 103 say 105 mm Ans.

N-1.4 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.

Given: W = 400 N ; L = 300 mm ; σb = 40 MPa = 40

N/mm2 ; h = 2b The beam is shown in Fig. 1.11.

Let b = Width of the beam in mm, and h = Depth of the

beam in mm.

∴ Section modulus,

Fig.-1.11

𝑍 =𝑏.ℎ2

6=

𝑏(2𝑏)2

6=

2𝑏3

3mm3

Maximum bending moment (at the fixed end),

𝑀 = 𝑊𝐿 = 400 × 300 = 120 × 103 N‐mm

We know that bending stress (σ𝑏) ,

40 =𝑀

𝑍=

120 × 103 × 3

2𝑏3=

180 × 103

𝑏3

𝑏3 = 180 × 103/40 = 4.5 × 103 or 𝑏 = 16.5 mm Ans.

and ℎ = 2𝑏 = 2 × 16.5 = 33 mm Ans.


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N-1.5 A cast iron pulley transmits 10 kW at 400 r.p.m. The diameter of the pulley is 1.2 metre

and it has four straight arms of elliptical cross-section, in which the major axis is twice the minor

axis. Determine the dimensions of the arm if the allowable bending stress is 15 MPa.

Solution. Given : P = 10 kW = 10 × 103 W ; N = 400 r.p.m ; D = 1.2 m = 1200 mm or R = 600

mm ; σb = 15 MPa = 15 N/mm2

Let T = Torque transmitted by the pulley.

We know that the power transmitted by the pulley (P),

mmN10238m238NTor,

60

T4002π

60

2ππN1010

3

3

−=−=

==

Since the torque transmitted is the product of the tangential load and

therefore tangential load acting on the pulley

=𝑇

𝑅=

238 × 103

600= 396.7N

Since the pulley has four arms, therefore tangential load on each arm,

𝑊 = 396.7/4 = 99.2N

and maximum bending moment on the arm,

𝑀 = 𝑊 × 𝑅 = 99.2 × 600 = 59520 N‐mm

Let 2𝑏 = Minor axis in mm, and

2𝑎 = Major axis in mm = 2 × 2𝑏 = 4𝑏

Section modulus for an elliptical cross‐section,

𝑍 =𝜋

4× 𝑎2𝑏 =

𝜋

4(2𝑏)2 × 𝑏 = 𝜋𝑏3mm3

We know that bending stress (0𝑏) ,

15 =𝑀

𝑍=

59520

𝜋𝑏3=

18943

𝑏3

or 𝑏3 = 18943/15 = 1263 or 𝑏 = 10.8 mm

Minor axis, 2𝑏 = 2 × 10.8 = 21.6 mm Ans.

and major axis, 2𝑎 = 2 × 2𝑏 = 4 × 10.8 = 43.2 mm Ans.

EXERCISES

NUMERICALS:-

1. A wrought iron bar 50 mm in diameter and 2.5 m long transmits a shock energy of 100 N-m.

Find the maximum instantaneous stress and the elongation. Take E = 200 GN/m2.


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2. An axle 1 metre long supported in bearings at its ends carries a fly wheel weighing 30 kN at

the centre. If the stress (bending) is not to exceed 60 MPa, find the diameter of the axle.

3. A shaft is transmitting 100 kW at 180 r.p.m. If the allowable stress in the material is 60 MPa,

find the suitable diameter for the shaft. The shaft is not to twist more than 1° in a length of 3

metres. Take C = 80 GPa.

4. Design a suitable diameter for a circular shaft required to transmit 90 kW at 180 r.p.m. The

shear stress in the shaft is not to exceed 70 MPa and the maximum torque exceeds the mean by

40%. Also find the angle of twist in a length of 2 metres. Take C = 90 GPa.

5. Compare the weights of equal lengths of hollow shaft and solid shaft to transmit a given

torque for the same maximum shear stress. The material for both the shafts is same and inside

diameter is 2/3 of outside diameter in case of hollow shaft.

6. A shaft is supported in bearings, the distance between their centres being 1 metre. It carries a

pulley in the centre and it weighs 1 kN. Find the diameter of the shaft, if the permissible bending

stress for the shaft material is 40 MPa.

THEORY:-

I.Define the terms load , stress and strain. Discuss the various types of stresses and strain.

II.Draw stress-strain diagram for mild steel and cast iron and discuss each points.

III.Define the following terms:- Ductility, Malleability, Toughness, Resilience, Creep, Fatigue

Endurance Limit.

IV.Draw and explain creep curve. What is creep strain?

V.Draw S-N curve fatigue life and discuss it.

VI.What is stress concentration? Write down it causes and remedies.

VII.Discuss in brief the following theories:-

a. Maximum principal (or normal) stress theory.

b. Maximum shear stress theory.

c. Maximum principal (or normal) strain theory.

d. Maximum strain energy theory.

e. Maximum distortion energy theory.

MCQ’S:-

01. Stress concentration is caused due to :

a) Variation in properties of material from point to point in a member b) Pitting at points or areas

at which loads on a members are applied. c) Abrupt change of section d) All of the above.


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02. The yield point in static loading is -------------as compared to fatigue loading

a) Higher b) Lower c) Same d) unpredicted

03. Factor of safety is the ratio of :

a) breaking stress to working stress b) endurance limit to yield stress c) elastic limit to ultimate

stress d) yield stress to working stress.

04. A mechanical component may fail as a result of which of the following:

a) elastic deflection b) general yielding c)fracture d)each of the mentioned.

05. Which one of the following materials is highly elastic?

a) Rubber b) Brass c) Steel d) Glass

06. Cast iron is used for machine beds because of its high

a) tensile strength b) endurance strength c) damping capacity d) compressive strength.

07. Which of the following material has the maximum ductility? a) Mild steel b) Copper c) Zinc

d) Aluminium

08. The main objective of design synthesis is a) Maximization b) Minimization c) Optimization

d) None of the above

09. When a machine member is subjected to torsion, the torsional shear stress set up in the

member is :

a) Zero at both the centroidal axis and outer surface of the member b) Maximum at both the

centroidal axis and outer surface of the member c) Zero at the centroidal axis and maximum at

the outer surface of the member d) Maximum at the centroidal axis and zero at the outer surface

of the member

10. The endurance limit, i.e., resistance to fatigue of a machine element can be improved by:

a) heat treatment b) grinding and lapping of surface c) electroplating d) polishing.

11. The selection of the factor of safety in design is not dependent upon:

a) service condition b) nature of load c) size of the component d) material of the component.

12. The measure of stiffness is:

a) Modulus of elasticity b) Modulus of plasticity c)Resilience d) toughness

13. The basic hole system is one whose

a) upper deviation is zero b) lower deviation is zero c) both the lower and upper deviations are

zero

d) lower and upper deviations are within prescribed limits.

14. The highest stress that a material can withstand for a specified length of time without

excessive deformation is called


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a) fatigue strength b) endurance strength c) creep strength d) creep rupture strength.

15. The material in which large deformation is possible before the absolute failure or fracture is

called

a) elastic b) Plastic c) ductile d) isotropic 16. The endurance limit for a mirror polished material

will be __________ as compared to unpolished material. a) same b) less c) more d) Cann’t be

predicted

17. When screw threads are to be used in a situation where power is being transmitted in one

direction only, then the screw threads suitable for this will be:

a) Knuckle threads b) Acme threads c) Square threads d) Buttress threads.

18. For the circumferential joint in boilers, the type of joint used is:

a)butt joint with single cover plate b) butt joint with double cover plate c) lap joint with one ring

overlapping the other d) any one of the above.

19. The total area under the stress strain curve of a mild steel specimen tested upto failure under

tension is a measure of

a) ductility b) ultimate strength c) stiffness d) toughness

20. A test specimen is stressed slightly beyond the yield point and then unloaded. Its yield

strength will a) decrease b) increase c) remain same d) become equal to ultimate tensile strength.

21. The most suitable theory of failure for brittle material is

a) maximum normal stress theory b) maximum strain energy theory c) maximum distortion

energy theory d) maximum shear stress theory.

22. During a tensile testing of a specimen using a UTM, the parameters actually measured

include

a) true stress and true strain b) Poisson’s ratio and Young’s modulus c) Engineering stress and

engineering strain d) Load and elongation.

23. Interchangeability can be achieved by

a) standardization b) better process planning c) simplification d) better product planning.

24. In design process, which process is followed after selecting the material?

a. Selecting factor of safety b. Synthesis c. Analysis of forces d. Determining mode of failure

25. The component deforming progressively under load at high temperatures is called as

a). Resilience b). Creep c) Fatigue d) Damping

26. When a nut is tightened by placing a washer below it, the bolt will be subjected to

a). tensile stress b). compressive stress c). shear stress d). none of these


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MCQ’S ANSWER

1. d) 2. a) 3.d) 4.a) 5. c) 6.c) 7. a) 8.c) 9. c) 10.b)

11.c) 12.a) 13.b) 14. c) 15.c) 16. c) 17.d) 18.c) 19. d) 20. b)

21.a) 22.d) 23.a) 24.d) 25. b) 26.a)

a book of machine design

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