6. Simple Stresses and Strains1

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PART II

Mechanics of Deformable Bodies

COURSE CONTENT IN

BRIEF

6. Simple stresses and strains

7. Statically indeterminate problems and thermal stresses

8. Stresses on inclined planes

9. Stresses due to fluid pressure in thin cylinders

The subject strength of materials deals with the relations

between externally applied loads and their internal effects on

bodies. The bodies are no longer assumed to be rigid and the

deformations, however small, are of major interest

Alternatively the subject may be called the mechanics of solids.

The subject, strength of materials or mechanics of materials

involves analytical methods for determining the strength ,

stiffness (deformation characteristics), and stability of various

load carrying members.

6. Simple stresses and strains

GENERAL CONCEPTS

STRESS

No engineering material is perfectly rigid and hence,

when a material is subjected to external load, it

undergoes deformation.

While undergoing deformation, the particles of the

material offer a resisting force (internal force). When

this resisting force equals applied load the equilibrium

condition exists and hence the deformation stops.

These internal forces maintain the externally applied

forces in equilibrium.

Stress = internal resisting force / resisting cross sectional

area

The internal force resisting the deformation per unit area is

called as stress or intensity of stress.

STRESS

A

R

gigapascal, 1GPa = 1×109 N/m2

= 1×103 MPa

= 1×103 N/mm2

SI unit for stress

N/m2 also designated as a pascal (Pa)

Pa = N/m2

kilopascal, 1kPa = 1000 N/m2

megapascal, 1 MPa = 1×106 N/m2

= 1×106 N/(106mm2) = 1N/mm2

1 MPa = 1 N/mm2

STRESS

AXIAL LOADING – NORMAL STRESS

Consider a uniform bar of cross

sectional area A, subjected to a

tensile force P.

Consider a section AB normal to

the direction of force P

Let R is the total resisting force

acting on the cross section AB.

Then for equilibrium condition,

R = P

Then from the definition of stress,

normal stress = ζ = R/A = P/A

P

P

P

R

B A

R

P

STRESS

ζ = Normal Stress Symbol:

Direct or Normal

Stress:

AXIAL LOADING – NORMAL STRESS

Intensity of resisting force perpendicular to or normal

to the section is called the normal stress.

Normal stress may be tensile or compressive

Tensile stress: stresses that cause pulling on the surface of

the section, (particles of the materials tend to pull apart

causing extension in the direction of force)

Compressive stress: stresses that cause pushing on the

surface of the section, (particles of the materials tend to push

together causing shortening in the direction of force)

STRESS

• The resultant of the internal forces for

an axially loaded member is normal

to a section cut perpendicular to the

member axis.

A

P

A

Fave

A

0lim

• The force intensity on that section is

defined as the normal stress.

STRESS

Illustrative Problems

A composite bar consists of an aluminum section

rigidly fastened between a bronze section and a steel

section as shown in figure. Axial loads are applied at

the positions indicated. Determine the stress in each

section.

Bronze

A= 120 mm2

4kN

Steel

A= 160 mm2

Aluminum

A= 180 mm2

7kN 2kN 13kN

300mm 500mm 400mm

Q 6.1

To calculate the stresses, first determine the forces in

each section.

For equilibrium condition algebraic sum of forces on

LHS of the section must be equal to that of RHS

4kN 7kN 2kN 13kN

To find the Force in bronze section,

consider a section bb1 as shown in the figure

Bronze

b

b1

Bronze

4kN 7kN 2kN 13kN

13kN 2kN 7kN

Bronze

4kN 4kN

Force acting on Bronze section is 4kN, tensile

Stress in Bronze

section = Force in Bronze section

Resisting cross sectional area of the Bronze section

= 2

22/33.33

120

10004

120

4mmN

mm

N

mm

kN

= 33.33MPa

(Tensile stress)

(= )

b1

b

4kN 7kN 2kN 13kN

2kN 7kN

Aluminum

9kN

Force in Aluminum section

Force acting on Aluminum section is 9kN,

(Compressive)

4kN 13kN

Aluminum

(= )

4kN 7kN 2kN 13kN

7kN

steel

7kN

Force in steel section

Force acting on Steel section is 7kN, ( Compressive)

4kN 2kN 13kN

steel

Stress in Steel section = Force in Steel section

Resisting cross sectional area of the Steel section

= 2

22/75.43

160

10007

160

7mmN

mm

N

mm

kN

= 43.75MPa

Stress in Aluminum

section =

Force in Al section

Resisting cross sectional area of the Al section

= 2

22/50

180

10009

180

9mmN

mm

N

mm

kN

= 50MPa

(Compressive stress)

Compressive stress

STRAIN

STRAIN :

when a load acts on the material it will undergo

deformation. Strain is a measure of deformation produced by

the application of external forces.

If a bar is subjected to a direct load, and hence a stress, the

bar will changes in length. If the bar has an original length L

and change in length by an amount δL, the linear strain

produced is defined as,

L

L

Original length

Change in length =

Strain is a dimensionless quantity.

Linear strain,

Linear Strain

strain normal

stress

L

A

P

L

A

P

A

P

2

2

LL

A

P

2

2

STRESS-STRAIN DIAGRAM

In order to compare the strength of various materials it is

necessary to carry out some standard form of test to establish

their relative properties.

One such test is the standard tensile test in which a circular

bar of uniform cross section is subjected to a gradually

increasing tensile load until failure occurs.

Measurement of change in length over a selected gauge

length of the bar are recorded throughout the loading

operation by means of extensometers.

A graph of load verses extension or stress against strain is

drawn as shown in figure.


Typical tensile test curve for mild steel

Proportionality limit


Typical tensile test curve for mild steel showing upper yield point

and lower yield point and also the elastic range and plastic range

Limit of Proportionality :

From the origin O to a point called proportionality limit the

stress strain diagram is a straight line. That is stress is

proportional to strain. Hence proportional limit is the maximum

stress up to which the stress – strain relationship is a straight

line and material behaves elastically.

From this we deduce the well known relation, first postulated

by Robert Hooke, that stress is proportional to strain.

Beyond this point, the stress is no longer proportional to strain

A

PPP Load at proportionality limit

Original cross sectional area =

Stress-strain Diagram

Elastic limit:

It is the stress beyond which the material will not return to its

original shape when unloaded but will retain a permanent

deformation called permanent set. For most practical purposes

it can often be assumed that points corresponding proportional

limit and elastic limit coincide.

Beyond the elastic limit plastic deformation occurs and strains

are not totally recoverable. There will be thus some permanent

deformation when load is removed.

A

PEE Load at proportional limit



Yield point:

It is the point at which there is an appreciable elongation or

yielding of the material without any corresponding increase of

load.

A

PYY

Load at yield point



Ultimate strength:

It is the stress corresponding to

maximum load recorded during

the test. It is stress corresponding

to maximum ordinate in the

stress-strain graph.

A

PUU Maximum load taken by the material


Rupture strength (Nominal Breaking stress):

It is the stress at failure.

For most ductile material including structural steel breaking

stress is somewhat lower than ultimate strength because the

rupture strength is computed by dividing the rupture load

(Breaking load) by the original cross sectional area.

A

PBB load at breaking (failure)


True breaking stress = load at breaking (failure)

Actual cross sectional area


The capacity of a material to allow these large plastic

deformations is a measure of ductility of the material

After yield point the graph becomes much more shallow and

covers a much greater portion of the strain axis than the

elastic range.

Ductile Materials:

The capacity of a material to allow large extension i.e. the

ability to be drawn out plastically is termed as its ductility.

Material with high ductility are termed ductile material.

Example: Low carbon steel, mild steel, gold, silver, aluminum



Percentage elongation

A measure of ductility is obtained by measurements of the

percentage elongation or percentage reduction in area,

defined as, increase in gauge length (up to fracture)

original gauge length ×100

Percentage reduction in

area original area ×100

=

=

Reduction in cross sectional area of necked portion (at fracture)

Cup and cone fracture for a Ductile

Material


Brittle Materials :

A brittle material is one which exhibits relatively small

extensions before fracture so that plastic region of the tensile

test graph is much reduced.

Example: steel with higher carbon content, cast iron,

concrete, brick

Stress-strain diagram for a typical brittle material

HOOKE‟S LAW

Hooke‟s Law

For all practical purposes, up to certain limit the relationship

between normal stress and linear strain may be said to be

linear for all materials

Thomas Young introduced a constant of proportionality that

came to be known as Young‟s modulus.

stress (ζ) α strain (ε)

stress (ζ)

strain (ε) = constant

stress (ζ)

strain (ε) = E

Modulus of Elasticity

Young‟s Modulus = or

HOOKE’S LAW

Young‟s Modulus is defined as the ratio of normal stress to

linear strain within the proportionality limit.

From the experiments, it is known that strain is always a very

small quantity, hence E must be large.

For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2

stress (ζ)

strain (ε) = E =

LA

PL

L

L

A

P

The value of the Young‟s modulus is a definite property of a

material

Deformations Under Axial Loading

AE

P

EE

• From Hooke‟s Law:

• From the definition of strain:

L

• Equating and solving for the

deformation,

AE

PL

• With variations in loading, cross-

section or material properties,

i ii

ii

EA

LP

A specimen of steel 20mm diameter with a gauge length of

200mm was tested to failure. It undergoes an extension of

0.20mm under a load of 60kN. Load at elastic limit is

120kN. The maximum load is 180kN. The breaking load is

160kN. Total extension is 50mm and the diameter at

fracture is 16mm. Find:

a) Stress at elastic limit

b) Young‟s modulus

c) % elongation

d) % reduction in area

e) Ultimate strength

f) Nominal breaking stress

g) True breaking stress

Q.6.2

Solution:

a) Stress at elastic limit,

ζE =

Load at elastic limit

Original c/s area

MPamm

Nmm

kN

A

PE 97.38197.38116.314

12022

b) Young‟s Modulus,

GPa

MPa

mmN

mmmm

mmkN

LL

AP

E

98.190

190980

190980101

98.190

20020.0

16.31460

23

2

(consider a load which is within the elastic limit)

c) % elongation,

% elongation = Final length at fracture – original length

Original length

%25100200

50

d) % reduction in area =

%3610016.314

41616.314

2

Original c/s area -Final c/s area at fracture

Original c/s area

e) Ultimate strength,

Ultimate strength = Maximum load

Original c/s area )(

/96.57216.314

180 2

2

MPa

mmNmm

kN

f) Nominal breaking

Strength = MPa

kN29.509

16.314

160

Breaking load

Original c/s area

g) True breaking

Strength =

MPamm

kN38.795

06.201

1602

Breaking load

c/s area at fracture

A composite bar consists of an aluminum section rigidly

fastened between a bronze section and a steel section as

shown in figure. Axial loads are applied at the positions

indicated. Determine the change in each section and the

change in total length. Given

Ebr = 100GPa, Eal = 70GPa, Est = 200GPa

Bronze

A= 120 mm2

4kN

Steel

A= 160 mm2

Aluminum

A= 180 mm2

7kN 2kN 13kN

300mm 500mm 400mm

Q.6.3

From the Example 1, we know that,

Pbr = +4kN (Tension)

Pal = -9kN (Compression)

Pst = -7kN (Compression)

stress (ζ)

strain (ε) = E = LA

PL

AE

PLL Change in length =

Change in length of

bronze = )/(10100120

3004000232 mmNmm

mmNLbr

= 0.1mm

Deformation due to

compressive force is

shortening in length, and is

considered as -ve

stalbr LLL Change in total

length =

Change in length of

steel section = )/(10200160

5007000232 mmNmm

mmNLst

= -0.109mm

Change in length of

aluminum section = )/(1070180

4009000232 mmNmm

mmNLal

= -0.286mm

+0.1 – 0.286 - 0.109

= -0.295mm

An aluminum rod is fastened to a steel rod as

shown. Axial loads are applied at the positions

shown. The area of cross section of aluminum and

steel rods are 600mm2 and 300mm2 respectively.

Find maximum value of P that will satisfy the

following conditions.

a)ζst ≤ 140 MPa

b)ζal ≤ 80 MPa

c)Total elongation ≤ 1mm,

2P Steel

Aluminum 2P 4P

2.8m 0.8m

Q.6.4

Take Eal = 70GPa, Est = 200GPa

To find P, based on the condition, ζst ≤ 140 MPa

Stress in steel must be less than or equal to 140MPa.

Hence, ζst =

= 140MPa

st

st

A

P

2P Steel

Aluminum 2P 4P

4P 2P 2P

2P 2P

2/1402

mmNA

P

st

kNNA

P st 21210002

140

Tensile

To find P, based on the condition, ζal ≤ 80 MPa

Stress in aluminum must be less than or equal to 80MPa.

Hence, ζal =

= 80MPa al

al

A

P

2P Steel

Aluminum 2P 4P

4P 2P 2P

2P 2P

2/802

mmNA

P

al

kNNA

P al 24240002

80

Compressive

To find P, based on the condition, total elongation ≤ 1mm

Total elongation = elongation in aluminum + elongation in steel.

stal AE

PL

AE

PL

stst

st

alal

al

EA

PL

EA

PL 22

33 10200300

28002

1070600

8002 PP

1mm

1mm

1mm

P = 18.1kN

Ans: P = 18.1kN (minimum of the three values)

Q.6.5

Derive an expression for the total extension of the tapered bar

of circular cross section shown in the figure, when subjected to

an axial tensile load , W

W W

A B

L Diameter

d1 Diameter

d2

Consider an element of length, δx at a distance x from A

B

W W

A

x d1 d2 dx

Diameter at x,

xL

ddd

12

1c/s area at x, 21

2

1

44kxd

d

xkd 1

Change in length over a

length dx is

Ekxd

Wdx

AE

PL

dx2

14


length L is

L

Ekxd

Wdx

0 2

14


Put d1+kx = t,

Then k dx = dt


length L is

L

Ekxd

Wdx

0 2

14

L

Et

k

dtW

0 2

4

LLL

kxdEk

W

tEk

Wt

Ek

W

0100

12

)(

1414

1

4

Edd

WL

dEd

WL

4

4

2121

Q.6.6

A two meter long steel bar is having uniform diameter of 40mm

for a length of 1m, in the next 0.5m its diameter gradually

reduces to 20mm and for remaining 0.5m length diameter

remains 20mm uniform as shown in the figure. If a load of

150kN is applied at the ends, find the stresses in each section

of the bar and total extension of the bar. Take E = 200GPa.

500mm

Ф = 40mm Ф = 20mm

150kN 150kN

500mm 1000mm

500mm

Ф = 40mm Ф = 20mm

150kN 150kN

500mm 1000mm

If we take a section any where along the length of the bar, it is

subjected to a load of 150kN.

2

1

3

MPakN

MPakN

MPakN

d

kN

MPakN

46.477

420

150

46.477

420

150

37.119

440

150

4

150

37.119

440

150

23

2min.2,

2.max,222

21

500mm

Ф = 40mm Ф = 20mm

150kN 150kN

500mm 1000mm

If we take a section any where along the length of the bar, it is

subjected to a load of 150kN.

2

1

3

mm

E

kNl

mmE

kN

dEd

PLl

mmE

kNl

194.1

420

500150

597.02040

50015044

597.0

440

1000150

23

21

2

21

mml 388.2 total,

Q.6.7

Derive an expression for the total extension of the tapered bar

AB of rectangular cross section and uniform thickness, as

shown in the figure, when subjected to an axial tensile load ,W.

W W

A B

L

d1 d2

b

b

W W

A B

x

d1 d2

b

b dx


depth at x,

xL

ddd

12

1c/s area at x, bkxd 1

xkd 1


length dx is

Ebkxd

Wdx

AE

PL

dx 1


length L is

L

Ebkxd

Wdx

01

12 loglog ddkEb

Pee

12

12

loglog302.2

ddddEb

LP

Q.6.8

Derive an expression for the total extension produced by self

weight of a uniform bar, when the bar is suspended vertically.

L

Diameter

d

P1 x

P1 = weight of the bar below

the section,

= volume × specific weight

= (π d2/4)× x ×

= A× x ×

Diameter

d

dx dx

element

Extension of the element due to weight of the bar below that,

AE

dxxA

AE

dxP

AE

PL

dx

)(1

The above expression can also be written as

Hence the total extension entire bar

E

L

E

x

AE

dxxAL

L

22

)( 2

0

2

0

AE

PL

AE

LAL

A

A

E

L

2

1

2

)(

2

2

Where, P = (AL)×

= total weight of the bar

SHEAR STRESS

Consider a block or portion of a material shown in Fig.(a)

subjected to a set of equal and opposite forces P. then there is a

tendency for one layer of the material to slide over another to

produce the form failure as shown in Fig.(b)

P

The resisting force developed by any plane ( or section) of the

block will be parallel to the surface as shown in Fig.(c).

P Fig. a Fig. b Fig. c

P

P

R R

The resisting forces acting parallel to the surface per unit area is

called as shear stress.

Shear stress (η)

=

Shear resistance

Area resisting shear

η

If block ABCD subjected to shearing stress as shown in

Fig.(d), then it undergoes deformation. The shape will not

remain rectangular, it changes into the form shown in Fig.(e),

as AB'C'D.

B

Fig. d

Shear strain

A

P

This shear stress will always be tangential to the area on which

it acts

η D

C

A

η B'

D

C'

A

η

B C

Fig. e

The angle of deformation is measured in radians and hence

is non-dimensional.

D

η B' C'

A η

Fig. e

B C

tanstrain shear AB

BB

The angle of deformation is then termed as shear strain

Shear strain is defined as

the change in angle

between two line element

which are originally right

angles to one another.

SHEAR MODULUS

For materials within the proportionality limit the shear strain is

proportional to the shear stress. Hence the ratio of shear stress

to shear strain is a constant within the proportionality limit.

For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2

Shear stress (η)

Shear strain (θ) = constant =

The value of the modulus of rigidity is a definite property

of a material

G

Shear Modulus

or

Modulus of Rigidity

=

example: Shearing Stress

• Forces P and P‘ are applied

transversely to the member AB.

A

Pave

• The corresponding average shear stress is,

• The resultant of the internal shear

force distribution is defined as the

shear of the section and is equal to

the load P.

• Corresponding internal forces act in

the plane of section C and are

called shearing forces.

• The shear stress distribution cannot be

assumed to be uniform.

η

State of simple shear

Force on the face AB = P = η × AB × t

Consider an element ABCD in a strained material subjected to shear stress, η as shown in the figure

Where, t is the thickness of the element.

η

A B

C D

Force on the face DC is also equal to P

P


The element is subjected to a clockwise moment

Now consider the equilibrium of the element.

(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)

P × AD = (η × AB × t) × AD

P

A B

C D

But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )

For the force diagram shown,

ΣFx = 0, & ΣFy = 0,

But ΣM = 0

force


Equn.(1) can be written as

If η1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as,

P ' = η ' × AD × t

P ' × AB = P × AD

= (η × AB × t)× AD ----- (1)

P

P

A B

C D

P ' P '

(η ' × AD× t ) × AB = (η × AB × t) × AD ----- (1)

η ' = η

η

η

A B

C D

η ' η '


Thus in a strained material a shear stress is always

accompanied by a balancing shear of same intensity at

right angles to itself. This balancing shear is called

“complementary shear”.

The shear and the

complementary shear together

constitute a state of simple

shear

A B

C D

η'= η

η

η

η'= η

Direct stress due to pure shear

Consider a square element of side „a‟ subjected to shear

stress as shown in the Fig.(a). Let the thickness of the

square be unity.

Fig.(b) shows the deformed shape of the element. The length of

diagonal DB increases, indicating that it is subjected to tensile

stress. Similarly the length of diagonal AC decreases indicating

that compressive stress.

a

A B

C D

η

η

η

η

a

A B

C D

η

η

η

η a

a

Fig.(a). Fig.(b).


Now consider the section, ADC of the element, Fig.(c).

Resolving the forces in ζn direction, i.e., in the X-direction

shown

a

Fig.(c).

a

a

A

C D

a2

For equilibrium

A ζn

C D

η

η

a

X

n

n aa

Fx

45cos212

0


Therefore the intensity of normal tensile stress

developed on plane BD is numerically equal to the

intensity of shear stress.

Similarly it can be proved that the intensity of compressive

stress developed on plane AC is numerically equal to the

intensity of shear stress.

Poisson‟s Ratio:

Consider the rectangular bar shown in Fig.(a) subjected to a

tensile load. Under the action of this load the bar will increase

in length by an amount δL giving a longitudinal strain in the

bar of

POISSON‟S RATIO

l

ll

Fig.(a)

The associated lateral strains will be equal and are of

opposite sense to the longitudinal strain.

POISSON‟S RATIO

The bar will also exhibit, reduction in dimension laterally, i.e.

its breadth and depth will both reduce. These change in

lateral dimension is measured as strains in the lateral

direction as given below.

d

d

b

blat

Provided the load on the material is retained within the elastic

range the ratio of the lateral and longitudinal strains will

always be constant. This ratio is termed Poisson’s ratio (µ)

POISSON’S RATIO Lateral strain

Longitudinal strain =

ll

dd

)(

ll

bb

)(

OR

Poisson‟s Ratio = µ

For most engineering metals the value of µ lies between 0.25 and

0.33

In general

z

y

x P P

Poisson‟s

Ratio

Lateral strain

Strain in the direction of

load applied

=

x

x

y

y

ll

l

l

OR

x

x

z

z

ll

ll

Lx

Ly Lz

Poisson‟s Ratio = µ

In general

Strain in X-direction = εx

z

y

x Px Px

Lx

Ly Lz

x

x

l

l

Strain in Y-direction = εy

Strain in Z-direction = εz

x

x

y

y

l

l

l

l

x

x

z

z

l

l

l

l

Load applied in Y-direction

Poisson‟s

Ratio

Lateral strain


load applied

=

y

y

x

x

l

l

ll

OR

y

y

z

z

l

l

ll

z

y

x

Py

Lx

Ly Lz

Py


y

y

x

x

l

l

l

l

Load applied in Z-direction

Poisson‟s

Ratio

Lateral strain


load applied

=

z

z

x

x

ll

ll

OR

z

z

y

y

ll

l

l

y

z

x

Pz

Lx

Ly Lz

Pz


z

z

x

x

l

l

l

l

Load applied in X & Y direction


z

y

x Px Px

Lx

Ly Lz

Py

Py


EE

xy

Strain in Z-direction = εz EE

xy

EE

yx

General

case:



Strain in Z-direction = εz

z

y

x Px Px

Py

Py Pz

Pz

EEE

zyxx

EEE

zxy

y

EEE

xyzz

ζx

ζz

ζy

ζx

ζz ζy

Bulk Modulus

Bulk Modulus

A body subjected to three mutually perpendicular equal direct

stresses undergoes volumetric change without distortion of

shape.

If V is the original volume and dV is the change in volume,

then dV/V is called volumetric strain.

Bulk modulus, K

A body subjected to three mutually perpendicular equal direct

stresses then the ratio of stress to volumetric strain is called

Bulk Modulus.

V

dV

Relationship between volumetric strain and linear strain

Relative to the unstressed state, the change

in volume per unit volume is

eunit volumper in volume change

1111111

zyx

zyxzyx

dV

Consider a cube of side 1unit, subjected to

three mutually perpendicular direct

stresses as shown in the figure.

Relationship between volumetric strain and linear strain

EEE

zyx

EEE

zxy

EEE

xyz

zyxV

dV

Volumetric strain

zyx

E

21

For element subjected to uniform hydrostatic pressure,

2-13KE

or

modulusbulk 213

E

K

zyx

321

21

EV

dV

EV

dVzyx

V

dVK

Relationship between young’s modulus of elasticity (E)

and modulus of rigidity (G) :-

A

D η

B

η

a

a 45˚

A1

θ θ

B1

Consider a square element ABCD of side ‘a’ subjected to pure shear

‘η’. DA'B'C is the deformed shape due to shear η. Drop a perpendicular

AH to diagonal A'C.

Strain in the diagonal AC = η /E – μ (- η /E) [ ζn= η ]

= η /E [ 1 + μ ] -----------(1)

Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC

C

H

In Δle AA'H

Cos 45˚ = A'H/AA'

A'H= AA' × 1/√2

AC = √2 × AD ( AC = √ AD2 +AD2)

Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=θ/2 ----(2)

Modulus of rigidity = G = η /θ

θ = η /G

Substituting in (2)

Strain along the diagonal AC = η /2G -----------(3)

Equating (1) & (3)

η /2G = η /E[1+μ]

E=2G(1+ μ)

Substituting in (1)

E = 2G[ 1+(3K – 2G)/ (2G+6K)]

E = 18GK/( 2G+6K)

E = 9GK/(G+3K)

Relationship between E, G, and K:-

We have

E = 2G( 1+ μ) -----------(1)

E = 3K( 1-2μ) -----------(2)

Equating (1) & (2)

2G( 1+ μ) =3K( 1- 2μ)

2G + 2Gμ=3K- 6Kμ

μ= (3K- 2G) /(2G +6K)

(1) A bar of certain material 50 mm square is subjected to an axial

pull of 150KN. The extension over a length of 100mm is 0.05mm

and decrease in each side is 0.0065mm. Calculate (i) E (ii) μ (iii) G

(iv) K

Solution:

(i) E = Stress/ Strain = (P/A)/ (dL/L) = (150×103 × 100)/(50 × 50 × 0.05)

E = 1.2 x 105N/mm2

(ii) µ = Lateral strain/ Longitudinal strain = (0.0065/50)/(0.05/100) = 0.26

(iii) E = 2G(1+ μ)

G= E/(2 × (1+ μ)) = (1.2 × 105)/ (2 × (1+ 0.26)) = 0.47 ×105N/mm2

(iv) E = 3K(1-2μ)

K= E/(1-2μ) = (1.2 × 105)/ (3 × (1- 2 × 0.26)) = 8.3 × 104N/mm2

(2) A tension test is subjected on a mild steel tube of external

diameter 18mm and internal diameter 12mm acted upon by

an axial load of 2KN produces an extension of 3.36 x 10-

3mm on a length of 50mm and a lateral contraction of 3.62

x 10-4mm of outer diameter. Determine E, μ,G and K.

(i) E = Stress/Strain = (2 ×103 × 50)/ (π/4(182 – 122)× 3.36× 10-3)

= 2.11× 105N/mm2

ii) μ=lateral strain/longitudinal strain = [(3.62 ×10-4)/18]/[(3.36 × 10-3)/50]

= 0.3 iii) E = 2G (1 + μ)

G = E / 2(1+ μ) = (2.11 × 105)/(2 × 1.3) = 81.15 × 103N/mm2

iv) E = 3K(1 -2 μ)

K = E/ [3×(1-2 μ)] = (2.11×105)/{3×[1-(2 × 0.3)]} = 175.42 ×103N/mm2

Working stress: It is obvious that one cannot take risk of

loading a member to its ultimate strength, in practice. The

maximum stress to which the material of a member is

subjected to in practice is called working stress.

This value should be well within the elastic limit in elastic

design method.

Factor of safety: Because of uncertainty of loading

conditions, design procedure, production methods, etc.,

designers generally introduce a factor of safety into their

design, defined as follows

Factor of safety = Allowable working stress

Maximum stress

Allowable working stress

Yield stress (or proof stress) or

Homogeneous: A material which has a uniform structure

throughout without any flaws or discontinuities.

Malleability: A property closely related to ductility, which

defines a material‟s ability to be hammered out in to thin

sheets

Isotropic: If a material exhibits uniform properties throughout

in all directions ,it is said to be isotropic.

Anisotropic: If a material does not exhibit uniform properties

throughout in all directions ,it is said to be anisotropic or

nonisotropic.

Q.6.9

A metallic bar 250mm×100mm×50mm is loaded as shown in

the figure. Find the change in each dimension and total

volume. Take E = 200GPa, Poisson's ratio, µ = 0.25

250

400kN 50

100

2000kN

4000kN

4000kN

400kN

2000kN

Stresses in different

directions 100

250

400kN 50

2000kN

4000kN

100

250

50 MPa

mm

Nx 80

50100

10004002

MPamm

Ny 160

100250

100040002

MPamm

Nz 160

50250

100020002

Stresses in different direction

MPa80

MPa160

MPa160 EEE

zyxx

410416016080

EEEx

mml

l

l

l

x

x

x

x

1.0

104250

4

EEE

zxy

y

3101.116080160

EEEy

mml

l

l

l

y

y

y

y

005.0

101.150

3

MPa80

MPa160

MPa160

mml

l

l

l

z

z

z

z

09.0

109250

4

EEE

xyzz

410980160160

EEEz

MPa80

MPa160

MPa160

3

44

44

250

50100250102102

102109114

mmdV

VdV

V

dV

zyxV

dV

To find change in volume

410280E

2-1

1601608021

21

EV

dV

EV

dVzyx

Alternatively,

MPa80

MPa160

MPa160

Q.6.10

A metallic bar 250mm×100mm×50mm is loaded as shown in

the Fig. shown below. Find the change in value that should

be made in 4000kN load, in order that there should be no

change in the volume of the bar. Take E = 200GPa, Poisson's

ratio, µ = 0.25

250

400kN 50

100

2000kN

4000kN

We know that

zyx

EV

dV

21

In order that change in volume to be

zero

0

210

zyx

zyxE

kNP

P

MPa

y

y

y

y

6000

100250240

240

016080

MPa80

MPa160

MPa160

The change in value should be an

addition of 2000kN compressive force

in Y-direction

Exercise Problems

Q1. An aluminum tube is rigidly fastened between a brass

rod and steel rod. Axial loads are applied as indicated in the

figure. Determine the stresses in each material and total

deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa

500mm 700mm 600mm

steel aluminum

brass 20kN 15kN 15kN 10kN

Ab=700mm2

Aa=1000mm2

As=800mm2

Ans: ζb=28.57MPa, ζa=5MPa, ζs=12.5MPa, δl = - 0.142mm

Q2. A 2.4m long steel bar has uniform diameter of 40mm for

a length of 1.2m and in the next 0.6m of its length its

diameter gradually reduces to „D‟ mm and for remaining

0.6m of its length diameter remains the same as shown in

the figure. When a load of 200kN is applied to this bar

extension observed is equal to 2.59mm. Determine the

diameter „D‟ of the bar. Take E =200GPa

Ф = 40mm

Ф = D mm

200kN 200kN

500mm 500mm 1000mm

Q3. The diameter of a specimen is found to reduce by

0.004mm when it is subjected to a tensile force of 19kN.

The initial diameter of the specimen was 20mm. Taking

modulus of rigidity as 40GPa determine the value of E and

µ

Ans: E=110GPa, µ=0.36

Q4. A circular bar of brass is to be loaded by a shear load of

30kN. Determine the necessary diameter of the bars (a) in

single shear (b) in double shear, if the shear stress in

material must not exceed 50MPa.

Ans: 27.6, 19.5mm

Q5. Determine the largest weight W that can be supported

by the two wires shown. Stresses in wires AB and AC are

not to exceed 100MPa and 150MPa respectively. The cross

sectional areas of the two wires are 400mm2 for AB and

200mm2 for AC.

Ans: 33.4kN

W A

C B

300 450

Q6. A homogeneous rigid bar of weight 1500N carries a

2000N load as shown. The bar is supported by a pin at B

and a 10mm diameter cable CD. Determine the stress in

the cable

Ans: 87.53MPa

3m

A C B

2000 N

3m

D

Q7. A stepped bar with three different cross-sectional

areas, is fixed at one end and loaded as shown in the

figure. Determine the stress and deformation in each

portions. Also find the net change in the length of the

bar. Take E = 200GPa

250mm 270mm 320mm

300mm2 450mm2

250mm2

10kN 40kN

20kN

Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm

Q8. The coupling shown in figure is constructed from steel of

rectangular cross-section and is designed to transmit a

tensile force of 50kN. If the bolt is of 15mm diameter

calculate:

a) The shear stress in the bolt;

b) The direct stress in the plate;

c) The direct stress in the forked end of the coupling.

Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa

Q9. The maximum safe compressive stress in a hardened

steel punch is limited to 1000MPa, and the punch is used to

pierce circular holes in mild steel plate 20mm thick. If the

ultimate shearing stress is 312.5MPa, calculate the

smallest diameter of hole that can be pierced.

Ans: 25mm

Q10. A rectangular bar of 250mm long is 75mm wide and

25mm thick. It is loaded with an axial tensile load of 200kN,

together with a normal compressive force of 2000kN on

face 75mm×250mm and a tensile force 400kN on face

25mm×250mm. Calculate the change in length, breadth,

thickness and volume. Take E = 200GPa & µ=0.3

Ans: 0.15,0.024,0.0197mm, 60mm3

Q11. A piece of 180mm long by 30mm square is in

compression under a load of 90kN as shown in the figure. If

the modulus of elasticity of the material is 120GPa and

Poisson‟s ratio is 0.25, find the change in the length if all

lateral strain is prevented by the application of uniform

lateral external pressure of suitable intensity.

180

90kN 30

30

Ans: 0.125mm

Q12. Define the terms: stress, strain, elastic limit,

proportionality limit, yield stress, ultimate stress, proof

stress, true stress, factor of safety, Young‟s modulus,

modulus of rigidity, bulk modulus, Poisson's ratio,

Q13. Draw a typical stress-strain diagram for mild steel rod

under tension and mark the salient points.

Q14 Diameter of a bar of length „L‟ varies from D1 at one end

to D2 at the other end. Find the extension of the bar under

the axial load P

Q15. Derive the relationship between Young‟s modulus and

modulus of rigidity.

Q17 A flat plate of thickness „t‟ tapers uniformly from a width

b1at one end to b2 at the other end, in a length of L units.

Determine the extension of the plate due to a pull P.

Q18 Find the extension of a conical rod due to its own weight

when suspended vertically with its base at the top.

Q19. Prove that a material subjected to pure shear in two

perpendicular planes has a diagonal tension and

compression of same magnitude at 45o to the planes of

shear.

Q16 Derive the relationship between Young‟s modulus and

Bulk modulus.

Q20. For a given material E=1.1×105N/mm2&

G=0.43×105N/mm2 .Find bulk modulus & lateral

contraction of round bar of 40mm diameter & 2.5m

length when stretched by 2.5mm. ANS:

K=83.33Gpa, Lateral contraction=0.011mm

Q21. The modulus of rigidity of a material is 0.8×105N/mm2 ,

when 6mm×6mm bar of this material subjected to an axial

pull of 3600N.It was found that the lateral dimension of the

bar is changed to 5.9991mm×5.9991mm. Find µ & E.

ANS: µ=0.31, E= 210Gpa.

For more slides,

assignments, data related

to the subjects please visit

www.techahoy.in

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