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UNIT I

SIMPLE STRESSES

The application of laws of mechanics to find the support reactions due to applied forces isnormally covered under Mechanics.

In transferring, these forces from their point of application to supports the material of structu

develops resisting forces and undergoes deformation. The effect of these resisting forces, on thestructural element, is treated under the subject Mechanics of Solids

Though several materials are used for the benefit of mankind, the designer is most interested

in the way in which materials will respond to external loads. The characteristics of material that describ

the behaviour under the action of external loads are referred as its Mechanical Properties.

STRENGTH

It is defined as the ability of the material to resist deformation, without rupture under

the action of forces.

STIFFNESSIt is the ability of material to resist deformation (or) deflection under load.

ELASTICITY

The ability of material to deform under load and regain its original shape when the

load is removed is called Elasticity.

PLASTICITY

The ability of material to deform non elastically ie. The material will not regain its

original shape after removal of load is known as plasticity.

DUCTILITY

The ability of a material to deformed plastically without rapture under tensile load.

(or)

Ability of material to draw into wires. Ex. Gold, silver, tungsten, Ms, Cu, Al, Ni, etc.

MALLEABILITY

The ability of material to be deformed plastically without rupture under compressive load.

(or)Ability of material rolled in to thin sheets.

Ex: Rivets are made-up of malleable material.

TOUGHNESSThe ability of material to absorb energy up to fracture during plastic deformation.

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BRITTLENESS

The property of sudden fracture without any visible permanent deformation.

Thus, brittle materials does not resist to shock loads.Ex: Glass, Cast Iron

HARDNESS

The ability of material to resist Indentation (or) surface abrasion.

CREEP:The slow and progressive deformation of material with time at constant stress is called

creep.

FATIGUE

Failure of material under repeated (or) reversal stresses is called fatigue.

MACHINABILITY:Property which facilitates easy machining.

WELDABILITY

Ability of material to be jointed by welding .

STRESSES

The force of resistance offered by a body against the deformation is called the stress.

STRAIN

The ratio of change in length (dl) to the original length of member

.ldlestrain ==

TYPES OF STRESS

(i) Tensile stresses :

When the resistance offered by a section of a member is against an increase in length,.

The section is said to offer a Tensile stress.

The corresponding strain is called Tensile strain

l

dl

lengthoriginal

lengthinincreasee ==

(ii) Compressive Stress

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The resistance offered by a section of a member is against a decrease in length the

section is said to offer a compressive resistance (or) stress.

l

dl

lengthOriginal

lengthinDecreasestainecompressiv =

(iii) Shear StressConsider a rectangular block ABCD. (of area L X b ).

Let the bottom face of block be fixed to the surface. As shown in figure.

Let force P be applied tangentially along the top face of the block. Suc

force acting tangentially along a surface is called a shear force. For equilibrium of the block, The surfawill offer a tangential reaction which is equal & opposite to applied tangential force. P.

The resistance offered by a section of member (x x) is called shear resistance.

bl

P

AreaShear

cesisShearstressShear

==

tanRe

Unit of Stress

N/mm2 , N/m2

1 Pascal = 1 N /m2

1 MPa = 1 N /mm2

HOOKE S LAW

Robert Hooke, an English mathematician concluded that stress is directly proportional

to strain with in the elastic limit.

e stres=

e = strain

eE=

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Hence E is the constant of proportionality of material, known as modulus of elasticity incase

of axial loading

E is called as youngs Modulus.

Structural metal which are very stiff have high value of modulus of elasticity

For example,

Steel E = 210 GPa.Al = 73 GPa.

Plastics = 1 GPa to 14 GPa.

Youngs Modulus E =stainecompressivortensile

stresscompresiveortensile

)(

)(

ShearstrainShear

stressShear= Modulus (or) Rigidity Modulus C, N, (or) G.

=G

STRESS - STRAIN RELATION :

The stress strain relation of any material is obtained by conducting tension test o

standard specimen.Different materials behave differently

Mild steel

The standard specimen of mild steel is set for tensile test. The specimen is gripped it end

in Universal Testing Machine (UTM)

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The following salient points are observed on stress-strain curve

A : LIMIT OF PROPORTIONALITY - It is the limiting value of stress up to which stress is

proportional to strain

1A : Elastic Limit This is the limiting value of stress up to which the material can regain it

original shape & size. This point is slightly beyond the proportionality limit.

B : Upper Yield Point -

This is the stress at which, the load starts reducing and the extension increases. This

phenomenon is called yielding of material.

C : Lower yield point

At this stage stress remains same but strain increase for sometime.

D : Ultimate Stress -

This is the maximum stress the material can resist.

This stress is about 370 - 400 N/mm2.

At this stage Cross sectional area at particular section starts reducing very fast. This is calle

Necking.

E : Breaking PointThe stress at which finally the specimen fails is called breaking point.

Note: For Brittle materials like cast Iron. There is no yield point & no necking takes place. Ultimate stre& breaking stress are same.

Percentage Elongation = 1001001

=L

LL

lengthOriginal

ruptureatExtensionFinal

Where L1 = length of member at rupture

L = Original Length.

Working stress,

The maximum stress to which any member is designed is much less than ultimate stress, an

this stress is called working stress.

Factor of Safety:

It is not possible to design a mechanical component (or) structural component

permitting stressing up to ultimate stress. Due to following reasons.

Reliability of material may not be 100 percent. There may be small spots of flaws.

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The resulting deformation may obstruct the functional performance of the component.

The loads taken by designer are only estimated loads. Occasionally there can be overloading.

Unexpected impact and temperature loading may act in the life time of member.

There are certain ideal conditions assumed in analysis.

Therefore, the calculated stresses will not be 100 percent real stresses.

Hence, the maximum stress to which any member is designed is much less than ultimatestress, and this stress is called working stress.

StressWorking

StressUltimatesafetyofFactor =

For steel - 1.85

For concrete - 3

For timber - 4 to 6

PROBLEMS

1. An elastic rod 25mm in diameter , 200mm long extends by 0.25mm under tensile load of 40KN.

Find

stress, strain & elastic Modulus of material of rod.

Sol:- Dia of rod = d = 25 mm

Length = l = 200 mm

Tensile load = P = 40KN

A

PStress =

222

87.4904

25

4sec mm

dAtionrodofArea =

==

Stress =2/49.81

87.490

100040mmN=

00125.0200

25.0===

l

lsestrain

Elastic Modulus E =2

/6519200125.0

49.81mmNstrain

Stress

==

= 65.192 KN/mm2

2. A circular rod of diameter 20mm and 500mm long is subjected to a tensile force 45 KN. The

modulus of elasticity for steel may be taken as 200 KN/mm2 . find elongation of bar due toapplied load.

Sol : Dia of rod = d = 20 mm

Length = l = 500 mm

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Load = P = 45 KN.

Modulus of elasticity E = 200KN/mm2

222

159.3142044

mmdArodofArea ===

Stress =2

/24.143159.314

100045mmN

A

P=

=

= E e

L

EA

P =

Elongation =1000200159.314

500100045

=AE

PL

= 0.358mm*3. The following data refer to a mild steel specimen tested in a lab

(i) dia of specimen = 25mm

(ii) Length of specimen = 300 mm

(iii) Extension under load of 15KN = 0.045mm(iv) Load at yield point = 127.65KN

(v) Max. Load = 208.60 KN(vi) Length of specimen after failure = 375 mm

(vii) Neck dia = 17.75mm

Determine (a) Youngs Modulus (b) yield point (c ) Ultimate stress (d ) percentage of elongation

(e ) safe stress adopting FOS of 2

Sol:-

Area of specimen A

22287.49025

44mmd

===

Stress at 15KN load ( ) 23

/56.3087.490

1015mmN

A

P=

==

Strain at 15 KN load (e) =4105.1

300

045.0 ==l

ls

(a) youngs Modulus E =e

=25

4/10037.2

105.1

56.30mmN=

(b) Yield Point =87.490

1065.127int3

=A

poYieldatLoad= 260 N/mm2

( c) Ultimate stress =87.490

1060.208.3

=A

loadMax= 425 N/mm2.

Percentage elongation = %25100300

75100

300

300375100

1

==

=L

LL

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Safe stress =2/130

2

260mmN

safetyoffactor

stressYield==

*Bars of Varying Sections:

Consider a bar consists of three lengths L1,L2, L3 with sectional areas A1, A2, A3, and

subjected to an axial load P.

Even though the total force as each section is same, the intensities of stress will bedifferent for three section.

Stress of section1A

PAB =

2A

PBC =

3A

PCD =

Strain of section AB ;E

e ABAB

= where E = youngs Modulus

;E

eBC

BC

= E

eCD

CD

=

Change in length of section AB = dLAB = eAB L1

dLBC = eBC L2dLCD = eCD L3

Total change in length of bar = dLAB + dLBC + dLCD.

= eAB L1 + eBC L2 + eCD L3

=1L

E

AB +2L

E

BC +

3LE

CD

=EA

PL

EA

PL

EA

PL

3

3

2

2

1

1 ++

++=3

3

2

2

1

1

A

L

A

L

A

L

E

Pdl

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4. Determine the stresses in three parts and the total extension of bar for a axial pull of 40KNas shown below fig.

Take E = 2 25 /10 mmN

Sol:

Sol: Given data :

Load P = 40KN = 40 x 103 N

E = 2x10

5

N/mm

2

Stress in section 1 =2

221

/58.56

)30(4

4000

4

4000mmN

dA

P===

Similarly

2 =2

222

/4.127

)20(4

4000

4

4000mmN

dA

P===

3 =2

2

/21.48

)5.32(4

4000 mmN=

Total extension dl = dl1 + dl2 + dl3

=

++=3

3

2

2

1

1

A

L

A

L

A

L

E

P(or) [ ]3322111

lllE

++

++=

32

3

22

2

12

14

d

L

d

L

d

L

E

P

++

=

2222

)5.32(

160

)20(

260

)30(

180

102

400004

= 0.255 mm5. Find the decrease in length of steel bar loaded as shown in fig.

Take E = 2 x 105 N/mm2

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Sol: Stress in upper part

21

1

1

)10(4

2000

=A

P

= 25.46N/mm2

Stress in lower part

22

22

154

50002000

+==

A

P

= 39.6 N/mm2

Decrease in length of bar

E

l

E

ll

2211 +=

= [ ]22111

llE

+

= ]2006.3918046.25[102

15

+

= 0.0625mm

6. A steel tie rod 50mm in diameter and 2.5m long is subjected to a pull of 100KN. To whatlength of rod should be bored centrally so that the total extension will increase by 15%

under same pull, the bore is being 25mm dia? Take E = 200 GN/m 2Sol:-

Given data

Dia of rod d = 50mmLength of rod = 2.5m = 2500mm

Load P = 100KN

Dia of bore d1 = 25mm

E = 200 x 109 x 10-6N\mm2

= 2 x 105N\mm2

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Let us assume length of bore = x mm.

Stress in solid rod

2

2

3

\93.50

)50(4

10100mmN

A

P=

==

mmE

Ll 64.0

102

250093.505

=

==

Elongation of the rod is bored l1 = 1.15 x l = 1.15 x 0.64

= 0.732mm

Area of bored rod A1 =222 62.1472)2550(

4mm=

Stress in bored rod2

3

1

1/9.67

62.1472

10100mmN

A

P=

==

Elongation of bored rod 1ls =( )

73.0102

9.67

1052

)2500(93.5025005

1

=

+

+

XxE

x

E

x

0.732 = 510297.16127325

x

x+

X = 1124 mm (or) 1.124m

*BARS WITH CONTINUOUSLY VARYING CROSS SECTIONS

*FOR A PLATE MATERIAL

A bar of uniform thickness t tapers uniformly from a width of b1 at one end to b2 at otherend in a length L as shown in fig below. The expression for the change in length of bar whe

subjected to an axial force P

Consider an elemental length dx at a distance x from larger end.

Rate of change of breadth isL

bb 21

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Width at distance x is b = b1 -L

bb 21 x

b = b1 Kx

=

L

bbK 21

btAelementofareationalCross = sec

= t (b1 Kx)Extension of element =

AE

Pdx

=EKxbt

Pdx

)( 1

Total extension of bar is = =L

O

L

O Kxb

dx

tE

P

EKxbt

Pdx

11 )(

= [ ]L

OKxb

tE

P)1(log

K

1

=

L

O

xL

bbb

tEK

P

)(log 211

= )log(log 12 bbtEK

P

= )log(log 21 bbtEK

P

=2

1logb

b

tEK

P

*FOR A ROD MATERIAL

A tapering rod has diameter d1 at one end & it tapers uniformly to a diameter d2 at other en

in a length L as shown in fig. if modulus of elasticity of material is E when it subjected to axi

force PThe change in length is

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Change in dia of rod throughout its length L is d1 d2

L

ddKdiaofchangeofRate

21 =

Consider a element of bar of length dx at a distance x from larger end.

The dia of element d = d1 - Kx

Cross sectional area A =2

1

2)(

44Kxdd =

Extension of element =EKxd

Pdx

2

1 )(4

Total extension of bar = L

O EKxd

Pdx

2

1 )(4

= EP4 L

O2

1 )( Kxd

Pdx

=E

P

4L

OKxd

1

1

=E

P

4

L

O

xL

ddd

211

1

k

1

=E

P

4

1221

11

dddd

L

= ( )

21

21

21 )(4

dddd

ddEPL

=21

4

dEd

PL

7. A 1.5m long steel bar is having uniform dia of 40mm for a length of 1m and in next 0.5m i

diameter gradually reduces from 40mm to 20mm as shown in fig below. Determine the elongatio

of this bar when subjected to an axial tensile load of 160KN.takeE = 200GN/m2

Sol: Given Data

Load P = 160 KN

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Consider the bars in two section (1) & (2)

Extension of uniform portion 52

3

1

11

102)40(4

100010160

==EA

PL

= 0.6366mm

Extension of tapering portion 5

3

21

221022040

5001016044

== EddPL

= 0.2122mm.

Total Extension of bar 21 +=

= 0.6366 + 0.2122

= 0.8488mm.BARS OF VARYING LOADS

8. A Brass bar having a cross Sectional area of 1000mm2 is subjected to axial forces asshown below fig. find the total change in length of bar.

Take E = 1.05 x 105

N/mm2

Sol: Given data

Cross sectional area A = 1000mm2

E = 1.05 x 105 N/mm2

Split the bar in to 3 sections

Extension of AB AB =AE

lP11

= 5

3

1005.11000

6001050

= 0.2857 (extension)mm

Charge in length of BC

AE

lPBC

22=

= 5

3

1005.11000

10001030

= 0.2857m(contraction)

Change in length of CD

AE

lPCD

33= = 53

1005.11000

12001010

= 0.1143mm (contraction)

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Total change in length bar CDBCAB ++=

= 0.2857 0.2857 0.1143

= -0.1143mm

Decrease in length of bar = 0.1143mm

*ELONGATION DUE TO SELF WEIGHT

9. A bar of uniform cross section A and length L is suspended from top. Find the expression forextension of bar due to self weight only if youngs modulus is E & unit weight material is

Sol : Consider an element of length dx at a distance x from free and

The load acting on element P = Ax

Extension of element =AE

lP

=AE

dxxA

=E

dxx

Extension of bar =

=

L

O

L

O

x

Edx

E

x

2

2

=2

2L

E

Thus this extension is half the extension of bar if the load equal to self weight is applied at end.

self wt =ALP=

AE

lP=

E

L

AE

LAL 2)( =

BARS OF COMPOSITE SECTIONS

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Suppose the cross section of member consists of different Materials, the load applied on the memb

will be shared by various components of section.

For instance, suppose a column consists of outer tube of area A1 and youngs Modulus E1 and a

Inner tube of area A2 and youngs Modulus E2. Let the length of column be L. Suppose a load P beapplied on the column. Let the unit stresses on the outer and inner tube sections be P1 and P2Total load on the column = Load on outer tube + Load on inner tube

P = 11 A + 22 A _________

Let dl be the change in length of column

Strain of each tubeL

dle =

2

2

1

1

EEe

== _____________

From &

1 , 2 may be computed

From

1 =2

1

E

E 2

* 10. A compound bar consists of circular rod of steel of dia 20mm rigidly fitted into a coppertube of internal dia 20mm and thickness 5mm as shown in fig. if the bar, is subjected to a load of

100KN, find the stresses developed in two materials

Take E S = 2 x 105 N/mm2 , EC = 1.2 x 10

5 N/mm2

AS =220

4

= 100 2mm

External dia of copper tube = 20 + (2x 5)

= 30 mm

AC =222

125)2030(4

mm

=

From static equilibrium condition

Load on steel rod + load on copper tube = Total load on compound bar.

Given total loadP = 100KN

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Scon 067.0= ----------------

Total load on column = load on concrete + Load on steel

P = Pcon + PS

360 x 103 = SSconcon AA + = 0.067 x 87486 S + 2514 S

360 x 103 = 8346 S

S = 43.13 N/mm2

Then

con = 0.067 S

con = 2.87 N/mm2

*12. A steel bolt of 16mm diameter passes centrally through a copper tube of internal diamet20mm and external dia 30mm. the length of the whole assembly is 500mm. After tight fitting o

assembly, the nut is over tightened by quarter of a turn. What are the stresses in bolt and tube if pitc

of nut is 2mm

Take

ES = 2x 105 N/mm2

EC = 1.2 x 105 N/mm2

Sol : Since there is no external force (or) load on the assembly

Load on copper tube + Load on steel bolt = 0

PC + PS = 0

PC = - PS -------------

i.e from equal & opposite forces are introduced.

As the Nut was tightened then due to this tightening the copper tube will shortened & steebolt gets extended

Movement of nut is to overcome extension of bolt & shortening of tube

CSN+=

Movement of nut N = mmpitch 5.024

1

4

1==

0.5 =CC

C

SS

S

EA

LP

EA

LP+

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= PL

+

CCSS EAEA

11 ( )CS PP =

0.5 =

+

1052.17.392

1

10526.201

1500P

AS =

216

4

= 201.6mm2

AC =)2030(

4

22

= 392.7mm2

P = 21728.64N

2

/78.1076.201

64.21728mmN

A

P

S

S ===

2/3.55

7.392

64.21728mmN

A

P

CC

===

TEMPERATURE STRESSES

Every Material expands when temperature rises and contracts when temp falls. The change

in length due to change is temp is found to be directly proportional to length of the member & also tocharge in temp.

Lt

Where Coefficient of thermal expansion [ change in unit length of material due to unit change in tem

t = change intemp..

L = length of member .Lt=

LtAE

PL=

tEtEA

PT ==

here T = Temp stress

Material Coefficient of thermal expansion

steel 12 x 10

-6

/

0

CCopper 17.5 x 10-6/0CStainless steel 18 x10-6/0C

Brass, Bronze 19x 10-6/0C

13. A composite bar is rigidly fitted at the supports A and B as shown in fig. Determine the reaction

at the supports when temp rises by 200C

Talke Ea = 70GN/m2

ES = 200GN/m2

a = 11 x10-6/0C

S = 12 x10-6/0C

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Sol : GivenTemp rise t = 200

The free expansion

SSaa LtLt +=

= 11 x 10-6 x 20 x1000 + 12x10-6 x 20 x 3000

= 0.22 + 0.72

= 0.94mmIf P is support reaction at expanding end . then

=SS

S

aa

a

EA

LP

EA

LP+

0.94 = 254 10102300

3000

107600

1000

+

PP

0.94 = 43 1021042 x

PP+

0. 94=P

+

000,20

1

42000

1

P = 12735.48N.

Temperature Stresses In Compound Bars:

When temp rises the two materials of compound bar experience different free

expansions.

Consider compound bar of length L, The rise in temp be t. Let 2,1 be coeff icient of

thermal expansion & E 1, E2 be moduli of two material.

Let 21

Free expansion of bar 1 = Lt1

Bar 2 = Lt2

If free expansions are permitted, are at AA and BB as show in fig.

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Sol :

Given data

Temp rise t = 800C

Length of bar at normal temp L = 1 meter = 1000mm

ES = 200GN/m2 = 2 x 105 N/mm2

EC = 100GN/m2 = 1 x 105 N/mm2

2

2

2 0 054 0

6 0 01 06 0s e c

m mA

m mAp l a tes t e e l

t i o nc r o s so fA r e a

C

S

==

==

From above fig copper plate has high free expansion compared to steel plate, but those two plates are joined rigidly so tensile forces acts on steel plate & compressive forces acts on copper

plate

So tensile forces gets equilibrium with compressive forces. As. Force on each copper plat

is taken equal since area are same

The equilibrium forces give PS = 2PC ----------

Change in length tLLt SCCS =+

SS

S

EA

LP+ LttL

EA

LPSC

CC

C =

Substitute in above

LtLtEA

LP

EA

LPSC

CC

C

SC

C =+

2

LttLEAEA

LPSC

CCSS

C =

+

12

( ) 10008010121017101200

1

102600

21000 66

55=

+

CP

PC = 6000 N

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PS = 2PC = 12KN

Change in length of compound bar = tLSS +

=SS

S

EA

LPtLS+

= 1000801012102600

100012000 65

+

= 1.06mm

15. A rigid bar ABC is pinned at A and is connected by a steel bar CE and copper bar BD as

shown is fig. if temp of whole assembly is raised by 400C, find the stresses induced in steel and

copper rods given.

For Steel bar For copper bar

Area 400mm2 600mm2

Modulus of elasticity 2 x 105 N/mm2 1 x 105 N/mm5

Coefficient of thermal expansion 12 x 10-6 /0C 18 x 10-6/0C

Since SC > , expansion of Cu is more than steel. But bar ABC is rigid & is hinged.

For equilibrium take moment about A is

PS x 2 = PC x 1

PC = 2 PS ----------------------------------As le ABD le ACE

=1

CCc Lt

2

SSS Lt

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) tLLc SSCCS =+ 22

+SS

SS

EA

LP

CC

CC

EA

LP2) tLL SSCC = 2 -------------------

Substitute in

+

SS

SS

EA

LP

CC

CS

EA

LP4( ) tLL SSCC = 2

40)101280010182(101600

8004

102400

600 6655

=

+

S

P

PS = 14202.74 N

PC = 2 PS = 28405.48N.

Stress in steel rod =2

/5.35400

74.14202mmN

A

P

S

S ==

Stress in copper rod =2

/34.47600

48.28405mmN

A

P

C

C ==

POISSONS RATIO :

When a material undergoes changes in length, it undergoes changes ofopposite nature in lateral directions. For example, if a bar is subjected to direct tension in its axial

direction it elongates and at the same times its sides contract.

Linear Strain: the ratio of change in axial direction to original length of member in called asLinear Stain

Lateral strain : The ratio of change in lateral direction to the original lateral dimension called as

lateral strain

Within elastic limit there is a constant ratio between lateral strain and linear strain. This

constant ratio is called Poissons ratio.

)(1

' ormstrainLinear

strainLateralratiosPoisson ==

For most of metals its value is 0.25 0.33

Steel 3.01 =m

Concrete 15.01=

m

VOLUMETRIC STRAIN :

When a member is subjected to stresses, it undergoes deformation in alldirections.

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Hence there will be change in volume.

The ratio of change in volume to original volume is called volumetric. Strain.

v

sVeV =

Volumetric strain is sum of strains in three mutually perpendicular directions.

eV = ex + ey + ezELASTIC CONSTANTS :

strainLinear

stressLinearElasticityofModulus =

q

strainShearing

stressShearingGRigidityofModulus ==

strainVolumetric

directionslarperpendicuinpressureidenticalKModulusBulk

3=

K =Ve

P

Ve = strainVolumetricV

V

RELATION BETWEEN E AND G: A solid cube LMST subjected to a shearing force F.

Let be the shear stress produced in the face MS and LT due to shearing force.

Due to shearing load, the tube is distorted to LMST, and as such, the edge M movesto M S to s and the diagonal LS to LS1 .

Shear strainST

SS1=

Also shear strain =STSS

G

1= ---------

On the diagonal LS , draw a perpendicular SN from S.

Now diagonal strain =LS

NS

LN

NS11

= -------------

From2

45cos1

111 SSSSNSNSSel

O ==

(LL S1T 2)sin 1 STLSsmallveryisSSeSTLL =

Substitute value of LS in ------------------

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Diagonal strain =ST

SS

ST

SS

222

11

=

But from,GST

SS =1

Diagonal strain = GGn

22

= -------------------

Where n is normal stress due to shear stress

The net strain in diagonal LS =mEE

nn

+ LS and MT have normal tensile & compressive stress n=

+mE

n 11

-----------------

From &

+=mEG

nn1

12

+=m

GE112 -------------------- ( I )

RELATION BETWEEN E AND K :

If solid cube is subjected to n (normal compressive stress) on all faces. The direct strain

in each axis =E

n

(comp)

Lateral strain in other axis = )(tensileE

n

net strain in each axis = mEmEEnnn

=

mE

n 21

Volumetric strain ev = 3 x linear strain

= 3

mE

n2

1

but ev =K

n

=mEK

nn2

13

=

mKE

213 --------------- II

From equation I and II by eliminating m we can get relation between E,G,K.

From I

GE

Gm

2

2

=

Substitute m in II

E = 3K

)2/(2

21

GEG

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3K

G

GE 21

G

GEG

K

E 2

3

+=

=G

EG 3

G

E

K

E

= 33

33

=+G

E

K

E

EG +3KE = 9KGE (3K+G)= 9KG

GK

KGE

+=3

9

(or)

KGE

139+=

HOOP STRESS:

If a thin steel type of internal diameter d. such a tyre can be shrunk on to a wheel of slight

bigger diameter D. The steel tyre is heated so that its dia exceeds from d to D. In this stage t

steel tyre is slipped on to the wheel. If now the tyre be cooled it will grip the wheel.

Hence a tensile stress is induced circumferentially along the tyre. Such stress is called a Hoo

stress

Temp straind

dD

original

preventedncontractioe

==

d

dDe

=

Hoop stress due to fall of temp = .Ed

dDeE

==

STRAIN ENERGY:

Energy absorbed (or) stored by a member when work is done on it to deform it.

Consider a bar of length L, cross sectional area A and subjected to axial load P. Let resistancedeveloped is R.

When deformation is zero, resistance R= 0

When deformation is = e L, R =PWork done by the resisting force = average resistance x

= eLPo

+2

= eLP2

1

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= eLA2

1

= Ve2

1( AL = Volume)

But strain energy = work done by internal force R

SE = Ve2

1

= volumestrainstress 2

1

SE = VE

2

1

Strain energy = VE2

2

Strain energy per unit volume is defined as Resilience

Resilience =E2

2

The maximum strain energy which can be stored by a body without undergoing permanent

deformation is called Proof Resilience. Hence proof resilience is equal to strain energy in the body

corresponding to stress at elastic limit ( Y )

i.e. proof Resilience =E

Y

2

2

Stress analysis due to various types of loads can be done by strain energy method. In this method

strain energy is equated to work done by the loads. This procedure illustrated by various load cases

below.

(i) Gradually Applied Load

Consider a bar of length L, cross sectional area A subjected to load.If load P is gradually applied, the load increases from O to P as extension increases from O to gradually. Hence work done by load

= averageP

= eLPPO

22=

+

Equating it to strain energy we get,

eLL

PVe

22

1=

PLLA =

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)(2

2

LE

hWALE

+=

)(22 L

Eh

AL

EW +=

A

W

AL

EWh 222 +=

(or)

0222 =

AL

EWh

A

W

+

+=

AL

EWh

A

W

A

W 222

2

12

++=

2

2

4

241

22

2

1

W

A

AL

EWh

A

W

A

W

++= WL

EWh

A

W 211

-----------------------

Thus in this stress produced by W is more than suddenly applied case. In most of cases, compare

to L, is too small. Hence if, is neglected in comparison to hWork done by load = Wh

Equating to strain energyAL

ehWWhAL

E

2

2

2

==

The same result is obtained by neglecting small quantity , compared toWL

AEh2in equation

(iv) Shock Load :

Shock load is measured in terms of energy. It is an externally applied energy. If U units

shock load is applied to bar, the stress produced in the bar can be obtained by equating strainenergy to shock energy.

ALE2

2

=U

AL

UE2=

**16. A 100N load falls from a height of 60mm on a collar attached to a bar of 300mm

diameter and 40mm long find the instantaneous stress and extension produced in the bar. Take E =

2x 105

N/mm2

. what is the percentage error, if extension of bar is neglected in final work done byload.Sol :

Given data

Area of cross section A =230

4

= 225 2mm

W = 100 N

E = 2x105N/mm2

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W = 100N

E = 2x 105 N/mm2

h = 60mmL = 400mm

Instantaneous stress

++=WL

EWh

A

W 211

=

++400100

60102225211

225

1005

= 92.273 N/mm2

If extension of bar is neglected in calculating work done by load

WhALE =2

2

AL

EWh22 =

400225

1006010225

=

= 92.132 N/mm2

Hence, percentage error in approximating is

= 100273.92

132.92273.92

= 0.153

Instantaneous extension produced = LE

= 400102

132.925

= 0.1842mm

17. A steel bar 20mm diameter and 1m long is freely suspended from a roof and is provided with a

collar at other end . If modulus of elasticity is 2x105 and Max. Permissible stress is 300N/mm2, find(a) the Maximum load which can fall from a height of 500mm on collar

(b) The maximum height from which a 600N load can fall on collar.

Sol: Given

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A =22

100204

mm

=

L = 1m = 1000 mm

E = 2x 105 N/mm2

h = 50mm(a) Maximum stress = 300N/mm2

Instantaneous extension permitted

LE=

=5102

1000300

= 1.5mm

work done by load W = W (50+1.5) = 51.5 W N-mm.

Strain energy = VolumeE

2

2

= ALE2

2

= 1000100

1022

3003005

= 45000 N- mmEquating work done by load to strain energy

51. 5W = 45000

W = 2745.08N(b) When W = 600N

let h be the maximum height

work done by the load = W ( h+ )= 600 (h +)

Equating work done by load to strain energy

600 ( h + 1.5) = 45000

h = 234.12mm