Mechanics of structures - module1

M h i f lidMechanics of solids

Stress and strain

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

Module IStress and strain

Module I

Introduction – general concepts - definition of stress –stress tensor –stress tensor –stress analysis of axially loaded members –strength design of members –axial strains and deformations in bars axial strains and deformations in bars –stress-strain relationships –Poisson's ratio – relationships between elastic constants -h l i thermal strain –

Saint Venant's principle –elastic strain energy for uniaxial stress –statically indeterminate systems –generalised Hooke's law for isotropic materials introduction to anisotropy – orthotropy


2

py py

Introduction

• What is mechanics of solids (or strength of materials)?

– Mechanics of deformable solids

• Why to study mechanics of solids?

– Principles of mechanics of solids are applied to predict the response (forces/displacements) of a p p ( p )structure/machine part subjected to loads

– In turn helps in the design of such structures or machine parts


3

machine parts3

Loads Structure/ ResponseLoads

input

Structure/machine

Response

t tinput system output

•Support reactions•Internal force resultants Internal force resultants

(SF, BM, axial forces, twisting moments etc.)g )

•Displacements (deflections & rotations)


4Analysis, calibration, design

Geometric propertiesMaterial properties

ElasticityPlasticity

LengthArea

DuctilityMalleability

VolumeMoment of inertia

DuctilityBrittlenessToughnessHardnessStiffnessStrength


5

Types of forcesTypes of forces

• Axial forces• Tension

• CompressionCompression

• Bending

• Torsion

• Shear


6

ElasticityElasticityProperty by virtue of which a deformed body under the action of loads regains its original under the action of loads regains its original shape once the loads are removed.


7

Linear elasticity and Hooke’s lawy

Elastic limit

σess

σE σ

ε=St

re

ε

StrainStrain


8

HomogeneityHomogeneity

•Same composition throughout bodyp g y•Elastic properties are same at every point in the body•Elastic properties need not be same in all directions

Isotropy•Elastic properties are same in all directions

A i tAnisotropy•Elastic properties different in various directionsElastic properties different in various directions


9

StressStress• Internal resistance against deformation

F / f i • Force/area - average force on unit area• Axial, bending, torsional etc.• Fundamental types of stresses:

• Normal (tensile or compressive)• Tangential (shear)

• Units


1010


1111

Stresses in 3D spaceStress tensor

p

x xy xz xx xy xz

ORσ τ τ τ τ τ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

yx y yz yx yy yz

zx zy z zx zy zz

ORτ σ τ τ τ ττ τ σ τ τ τ

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


1212

Stresses in three dimensionsStresses in three dimensions

General linearly elastic material

x xσ ε⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪

yσ

11 12 19

xy xy

xz xzD D D

τ γτ γ

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

⎡ ⎤⎪ ⎪ ⎪ ⎪xyτ

yxτ

τyzτ

11 12 19

21 22 29yx yx

y y

D D Dτ γσ ε

⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

xσxzτzxτ

zyτ

91 92 99yz yz

zx zx

D D Dτ γτ γ

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪

zσ

zy zy

z z

τ γσ ε

⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭Generalised Hooke’s Law


81 independent elastic constants 13

General anisotropic material

We have, xy yxτ τ= x xσ ε⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪

p

We ave, xy yx

yz zyτ τ

τ τ

=

=

11 12 16

12 22 26

x x

y y

z z

D D DD D D

σ εσ ε

⎪ ⎪ ⎪ ⎪⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬Hence,

xz zxτ τ

Also, the constant matrix is 16 26 66

xy xy

yz yzD D Dτ γτ γ

⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦⎪ ⎪ ⎪ ⎪symmetric.zx zxτ γ

⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

21 independent elastic constants Constitutive matrix (Stiffness matrix)

{ } [ ]{ }Dσ ε=Constitutive matrix (Stiffness matrix)

{ } [ ]{ }Cε σ=


Compliance matrix (Flexibility matrix) 14

For a general anisotropic material,- Infinite number of planes of symmetry – Different properties along different directions

g p ,

Orthotropic material- Three planes of symmetry

11 12 13 0 0 0D D Dσ ε⎧ ⎫ ⎧ ⎫⎡ ⎤

p y y– Different properties along three orthogonal directions

11 12 13

12 22 23

13 23 33

0 0 00 0 00 0 0

x x

y y

z z

D D DD D DD D D

σ εσ εσ ε

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬

13 23 33

44

55

0 0 0 0 00 0 0 0 0

z z

xy xy

yz yz

DD

τ γτ γ

⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

55

660 0 0 0 0yz yz

zx zxDγ

τ γ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎣ ⎦



1 0 0 0xy xz

x y zE E Eμ μ−⎡ ⎤−

⎢ ⎥⎢ ⎥⎢ ⎥1 0 0 0xy yz

x y zxx xxE E Eμ μ

γ τ

⎢ ⎥− −⎢ ⎥

⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥1 0 0 0yy yyyzxz

zz zzx y zE E Eγ τμμγ τ

⎪ ⎪ ⎪ ⎪⎢ ⎥−−⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥10 0 0 0 0xy xy

yz yzxyGγ τγ τ

=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥

10 0 0 0 0zx zx

yzG

γ τ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎢ ⎥

⎢ ⎥⎢ ⎥

10 0 0 0 0zxG

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦


Orthotropic materialCompliance matrix16

Isotropic material

1 0 0 0E E E

μ μ− −⎡ ⎤⎢ ⎥

1 0 0 0xx xx

E E E

E E Eμ μ

γ τ

⎢ ⎥⎢ ⎥− −⎢ ⎥⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪

1 0 0 0;

yy yy

zz zz

E E E

EE E E G

γ τμ μγ τ

⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥= =⎨ ⎬ ⎨ ⎬ ( );

1 2 10 0 0 0 0

1

xy xy

yz yz

G

Gγ τ μγ τ

⎢ ⎥= =⎨ ⎬ ⎨ ⎬⎢ ⎥ +⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥10 0 0 0 0

1

zx zxG

γ τ⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎩ ⎭⎢ ⎥

⎢ ⎥⎢ ⎥10 0 0 0 0

G⎢ ⎥⎢ ⎥⎣ ⎦



Special cases of stresses in 3D space

Pure shear

τxyτ

yxτ

τyzτ 0 0

0 0xy xz xy xz

yx yz xy yz

τ τ τ ττ τ τ τ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥

xzτzxτ

zyτ0 0zx zy xz yzτ τ τ τ

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦


18

Special cases of stresses in 3D space (contd…)p p ( )

1 0 0σ⎡ ⎤⎢ ⎥

σ1σ

2

3

0 00 0

σσ

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

3σ2σ

0 0σ−⎡ ⎤⎢ ⎥

σ0 00 0

σσ

⎢ ⎥−⎢ ⎥−⎢ ⎥⎣ ⎦

σ σ Hydrostatic pressure


1919

General case of Stresses in 2D spaceGeneral case of Stresses in 2D space

yxτxyτ 0x xyσ τ⎡ ⎤

⎢ ⎥

σ

00 0 0

y

yx yτ σ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

xσ

yxτ

xyτ

yσyx

Pl t ditiPlane stress condition


2020

StrainStrain• Measure of deformation • Deformation per unit length in the direction of deformation• Fundamental types of strains:

• Normal (tensile or compressive)( p )• Tangential (shear)

• Units?


2121

Stress-strain diagrams for brittle and ductile materials

Brittle material

Stre

ss Ductile material

S

Strain


22Toughness?

Ideal stress-strain diagram – mild steelIdeal stress strain diagram mild steelActual breaking stress

El ti li itUltimate stress

Elastic limit

Upper yield pointLimit of proportionality

Stre

ss

Yield plateau

Breaking stress

S

Lower yield point

Strain


23

• Ultimate stress

W ki t• Working stress

• Factor of safety Ultimate stress=Factor of safety

Working stress


24

Normal stressN

P

L Lδ

PA

σ =Normal stress (Intensity of force)A

P -> ForceA > A f ti l t th i

( y )

A-> Area of cross-section normal to the axis

P A PLP A PLEL L A LPL

σε δ δ

= = =Modulus of elasticity (Young’s modulus)


25• St. Venant’s principlePLLAE

δ∴ =


26

• Problems• Problems• Bars of uniform section (prismatic bars)

B f i i• Bars of varying sections

• Bars of varying sections of different materials

• Uniformly tapering circular & rectangular bars

• Elongation of bar of uniform section due to self weight• Elongation of bar of uniform section due to self weight

• Elongation of uniformly tapering bar due to self weight

• Elongation of bar with a truncated cone shape

• Bar of uniform strengthg

• Compound bars - Composite section


27

10mm2

Bar of uniform section (prismatic bars)

10kN5kN

15kN 20kN

10kN

15kN 20kN2m 3m4m 3m

B f i ti

4m 3m4m

Bar of varying sections

100kN 100kN10mm2 5mm2

20 2


28

20mm2

Bar of varying sections of different materials

4m 3m4m

Bar of varying sections of different materials

Al i i

100kN 100kN10mm2 5mm2

20 2

Steel Brass Aluminium

20mm2

Steel : ES=2x105 N/mm2S /

Brass : Eb=1x105 N/mm2

Aluminium : Ea=0.7x105 N/mm2


29

Uniformly tapering circular bar

PD1

D2DxP P

x

1

xδx

1 21x

D DD D xL−⎛ ⎞= −⎜ ⎟

⎝ ⎠1x L⎜ ⎟

⎝ ⎠

2

P xδΔ =

⎛ ⎞2

1 20

4Total elongation, L

x

P x PLLED DD E

δππ

∴ Δ = =⎛ ⎞⎜ ⎟∫

2

4xD Eπ⎛ ⎞

⎜ ⎟⎝ ⎠

1 20

44When =

x E

PLD D D L

⎜ ⎟⎝ ⎠

= Δ =


30

1 2 2When , D D D LEDπ

= Δ =

P b A i f t i t t i if l t i i di tProb: A specimen for tension test is uniformly tapering in diameterFrom D+a to D-a. Estimate the error in using the average diameter D to calculate the Young’s Modulus.g

Elongation L LΔ Δ

( )( )

1 2

1

Elongation, 4

L LPLE

L D a D aπ

Δ =Δ

=Δ + −( )( )

2 2

4L D a D aPLELD

π

π

Δ +

=Δ

21 2

21

Error 100 100E E aE D−

= × =


31

A = 800 mm2

P = 100 kNL 3

P/2 P/2

L1 = 3mL2 = 4mE = 1000N/mm2L1

L2

/

P

P


32Ans: Downward movement of pulley = 218.5 mm

A = 800 mm2

P = 100 kNL 3 P1

P2

L1

L2L1 = 3mL2 = 4mE = 1000N/mm2

P12

/

PPP


33

4

Alu Steel

Ea=7x104 MPaEs=2x105 MPaA = 3 mm2 Alu Steel

1m 1m

Aa 3 mmAs= 2 mm2

0.5md2

d1

20 kN20 kN


34

Compound bars (Composite sections)Compound bars (Composite sections)

PP1 2P P P+ =

1 2L L Lδ δ δ= =

1 2PL P L=

E

1 1 2 2A E A E

1 2 11 2

1 2 2

EE E Eσ σ σ σ= ⇒ =

Modular ratio


35

Compound tubeCompound tube

1000 kN Eb=105 MPaEs=2x105 MPaInternal dia of steel tube = 150mmThickness of steel tube = 10mmThickness of brass tube = 10mm

1000s bP P+ =

s bL L Lδ δ δ= =s b

St l t bs bP L P L

=Steel tubes s b bA E A E

Ans: Pb=360kN


36Brass tubeAns: Pb 360kNPs= 640kN

Elongation of bar of uniform section due to self weight

AX Stress at X, x

Ax xA

γσ γ= =

, ,E AγElement of length

. . .El ti f l t x

dxdx x dxσ γδ

2

Elongation of element,

. .Total elongation of bar

xx

LE E

x dx LL

γδ

γ γδ

= =

= =∫0

Total elongation of bar, 2

LE E

δ ∫Xdx


37Axγ

Bar of uniform strength • Uniform stress throughout the length of bar

( )xA dAσ +

xA dxγ

( )A dA A A dA

Ax+dA

LxAσ

( )x x xA dA A A dxσ σ γ+ = +Ax

x

L

dA dxσ γ=

AxA

γ

xA xdA dxγ=∫ ∫

P 0

log

xA

x

dxA

A x

σ

γ

=

⎛ ⎞ =⎜ ⎟

∫ ∫


38

loge xA σ

=⎜ ⎟⎝ ⎠

Temperature stress α : Coefficient of thermal expansion, T : Rise in temperature,L : Length

αTL : Change in length, if bar is free to expand.No stress in bar.

If deformation of bar is prevented, stress develops in barIf elongation is prevented, stress is compressive If shortening is prevented, stress is tensileIf shortening is prevented, stress is tensile

Compressive strain, TL TL TL TL Lα αε αα

= ≅ =+

LCompressive str ,e ss

L TL LTEα

σ α+

=

If expansion is allowed partially,

Compressive strain TLα δε −=


39

Compressive st , rainL

ε =L δ

Example 1: What is the maximum stress in the steel rod if temperature i th h 30oC?

7 2 62 10 N/cm 12 10 cm/cm/oE Cα −= × = ×

212 cm218 cm

rises through 30oC? 2 10 N/cm , 12 10 cm/cm/E Cα= × = ×

30 cm 45 cm

( ) PL PL( ) 1 21 2

1 2

Total elongation, PL PLT L LAE A E

α + = +

67

30 45i.e., 12 10 30 752 10 12 18

P− ⎛ ⎞× × × = +⎜ ⎟× ⎝ ⎠108000NP∴ =

21 2

108000N 9000N cm12

σ∴ = =


1 212cm40


41Problems: Sc P 90-96 Pun


42

7

8


43

STRAIN ENERGY

due to axial loadaxial loadbending

shear torsion


44

– When an elastic body is loaded it undergoes deformation

STRAIN ENERGY

– When an elastic body is loaded, it undergoes deformation– Energy is stored during this process, and released when the

load is removed This energy is strain energy– This energy is strain energy

• Within elastic limit energy stored is RESILIENCE• Within elastic limit, energy stored is RESILIENCE

• Maximum energy stored upto elastic limit is PROOF RESILIENCERESILIENCE

• Proof resilience is the capacity to bear shocks. Proof resilience per l funit volume of piece is MODULUS OF RESILIENCE


45

Strain energy due to axial load – Variable force St a e e gy due to ax a oad Va ab e o ce along length

Axial strain energy in a member of length L and axial rigidity EA, subject to tensile force P(x) and corresponding axial strain ε(x) = P(x)/EA is obtained as below:

( )1 1 P d

P(x)/EA is obtained as below:

For an elemental length dx, ( ) ( ) ( )1 1

2 2axial

P x dxlu P x x

EAPδ= =strain energy

2( ) ( )2

2

0 0

1 12 2

L L

axial

P xU dx EA x dx

EAε= =∫ ∫strain energy

*axial axialU U=Complementary strain energy


46

axial axial

for linear elastic material

Strain energy due to axial load–gyUniform force throughout length

Strain energy stored in the bar, 12

U W lδ= W2

2 21 l Al Vσ σ σ⎛ ⎞

W

lδ( )1

2 2 2l Al VU A

E E Eσ σ σσ ⎛ ⎞= = =⎜ ⎟⎝ ⎠

If σP is proof stress (stress upto elastic limit), then proof resilience =

2P Vσp

2E

Modulus of resilience = 2

2P

Eσ


47

2E

Example 1: A bar 100 cm length is subjected to an axial pull, such that the maximum stress is 150MN/m2. Its area of cross-section is 2 cm2 over a length of 95 cm and for the middle 5 cm it is only 1 g ycm2. If E= 200 GN/m2 , calculate the strain energy stored in bar.


48


49

Example 2: Two elastic bars, whose proportions are shown in figure, areto absorb the same amount of energy delivered by axial forces at theto absorb the same amount of energy delivered by axial forces at thefree end. Compare the stresses in the bars at the free end. Cross-sectional areas are shown.

1 2U U=

( ) ( )222 2

1 2 2 2 .3AL AL A Lσ

σ σ ⎛ ⎞ ⎛ ⎞ L34L 2A

( ) ( )1 2 2 2 .3

2 2 4 2 4AL A L

E E Eσ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠A L

A42 11.265σ σ=


50

Strain energy due to axial load for multiple membersSt a e e gy due to ax a oad o u t p e e be s– Uniform force throughout length

If the axial force is constant, and the ith element has an axial stiffness EA/L, (i.e., axial flexibility L/EA), and a tensile force Pi is acting on the ith element causing an elongation ei, axial strain energy is given by

2 * 21 12 2truss i truss i

i i

EA LU e U PL EA

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ ∑2 2i ii iL EA⎝ ⎠ ⎝ ⎠


51

Elastic constantsElastic constants

M d l f l ti it (Y ’ d l ) E• Modulus of elasticity (Young’s modulus) E• Modulus of rigidity (Shear modulus) G

B lk d l K• Bulk modulus K• Poisson’s ratio µ


52

Poisson’s ratioLinear and lateral strain

LC

B Bδ−B 1σ

L Lδ+

A normal stress in a direction causes linear strain in its direction and

Lδ Lateral strain Poisson's ratio = = B Bδμ

causes lateral strains in directions perpendicular to it.

1

2

Linear strain =

Lateral strain =

LL

BB

δε

δε −

Poisson s ratio Linear strain L L

μδ

2 12 1= ε σμ ε με μ−

⇒ = − = −


53

B 2 11 E

μ ε με με

⇒

• Poisson’s ratio is a material propertyPoisson s ratio is a material property


54

Volumetric strain

• A single normal stress – rectangular cross section

BPP

Hσ σ

Linear (longitudinal) strain 1 (Tensile)Lδ σε = =Linear (longitudinal) strain 1 (Tensile)L E

ε

2 (Compressive)BB Eδ σε μ−

= = −Lateral strainB E

3 (Compressive)HH Eδ σε μ−

= = −Lateral strain


55

BPP

HP

σ σ

V LBH=

( ) ( ) ( )V V L L B B H Hδ δ δ δ+ = + + +

L B Hδ δ δ⎛ ⎞⎛ ⎞⎛ ⎞1 1 1L B HLBHL B Hδ δ δ⎛ ⎞⎛ ⎞⎛ ⎞= + + +⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

1 L B HV δ δ δ⎛ ⎞≅ + + +⎜ ⎟

L B HV V δ δ δδ ⎛ ⎞⇒ ≅ + +⎜ ⎟

1VL B H

≅ + + +⎜ ⎟⎝ ⎠

V VL B H

δ⇒ ≅ + +⎜ ⎟⎝ ⎠

1 2 3Volumetric strain vV

V E E Eδ σ σ σε ε ε ε μ μ= ≅ + + = − −V E E E

( )= 1 2v Eσε μ−


56

E

A l l lVolumetric strain

P

• A single normal stress – circular cross section

P P

2

4D LV π

=

( )2

2

2 . .4

2

V D L D D L

D L D L

πδ δ δ

π δ δ

= +

⎛ ⎞+⎜ ⎟ Lδ24 D L

= +⎜ ⎟⎝ ⎠

2V D LV D Lδ δ δ

= +

1Here,

and

LL ED

δ σε

δ σε μ

= =

= =V D L 2and, D E

ε μ= = −

( )Volumetric strain, 1 2vV

V Eδ σε μ= = −


57

V E

Volumetric strain

• Three mutually perpendicular normal stresses

1 2 3Volumetric strain vV

Vδε ε ε ε= ≅ + +

2P 2σ2ε

1 2 31 E E E

σ σ σε μ μ= − −1P

3P1σ1ε 2 3 1

2 E E Eσ σ σε μ μ= − −

3P3σ

3ε3 1 2

3 E E Eσ σ σε μ μ= − −

( ) 1 2 31 2 3 1 2v

Vδ σ σ σε ε ε ε μ + +⎛ ⎞∴ = ≅ + + = − ⎜ ⎟⎝ ⎠


58

( )1 2 3v V Eμ ⎜ ⎟

⎝ ⎠

Hydrostatic pressure: Equal pressure on all sidesHydrostatic pressure: Equal pressure on all sides

1 2 3σ σ σ σ= = = 1 2 3ε ε ε ε= = =

( )1 2 333 1 2v

VV Eδ σε ε ε ε ε μ ⎛ ⎞∴ = ≅ + + = = − ⎜ ⎟

⎝ ⎠( )1 2 3v V E

μ ⎜ ⎟⎝ ⎠

( )1 2Eσε μ ⎛ ⎞= − ⎜ ⎟⎝ ⎠σ σ

σ

Maximum possible value of Poisson’s ratio

E⎝ ⎠σ

( )1 2 σε μ ⎛ ⎞⎜ ⎟

Maximum possible value of Poisson s ratio

( ) a +ve value1 2Eε μ= −( )1 2E

ε μ= − ⎜ ⎟⎝ ⎠

( )

( )

a +ve value1 2 ,

11 2 02

μσ

μ μ∴ − > ⇒ <


59

( )2

Bulk modulus

• When a body is subjected to mutually perpendicular likestresses of equal intensity the ratio of normal stress to the stresses of equal intensity σ, the ratio of normal stress σ to the corresponding volumetric strain εv is defined as bulk modulus of the material

K σ=

vε


60

Relation between Young’s modulus E and bulk modulus KRelation between Young s modulus E and bulk modulus K

Consider equal pressure on all sides of a cube

σ

Consider equal pressure on all sides of a cubeof side L

σ σ( ) 31 2vV

V Eδ σε μ ⎛ ⎞= = − ⎜ ⎟

⎝ ⎠

( ) 31 2K Eσ σμ ⎛ ⎞= − ⎜ ⎟

⎝ ⎠

( )3 1 2E K μ= −


61

Problem


62


63

Shear modulus (modulus of rigidity)

Shear stressShear modulusSh t

, = i

G τ=

PA

Shear strain γ

Sl N M t i l M d l f Sl. No. Material Modulus of rigidity

1 Aluminium (pure) 26

γ γ

P Aτ =2 Aluminium alloys 26-30

3 Brass 36-41

4 B 36 44 γ4 Bronze 36-44

5 Cast iron 32-69

6 Copper (pure) 40-476 Copper (pure) 40-47

7 Rubber 0.0002-0.001

8 Steel 75-80


649 Wrought iron 75

•Complimentary shear stress

AEτ ′F

AB τ

τAD AF = BC BEτ τ× × × ×

CD τ ′

AD AF BC BEτ τ× × × ×

( )( )

Moment of the couple

Moment of the balancing couple

= AD AF AB

= AB BE AD

τ

τ

× × ×

′× × ×( )g pEquating the above two moments, τ τ ′=

Principle of complimentary shear stresses: Shear stress in a direction cannot exist without a complimentary shear stress

di l t it


65

perpendicular to it

τ τRelation between Young’s modulus E and shear modulus G

τ τ

τ045

A B

ττ τ nτ

045

C iτ

Square element cos45 cos45 0BC AB ACτ τ τ= − =

nσC Compressive

Square elementUnder pure shear

. . .cos45 . .cos45 00

n

n

BC AB ACτ τ ττ

= =⇒ =

. . .sin 45 . .sin 45

sin 45 sin 45

n BC AC ABAC AC AC

σ τ τ

σ τ τ

= +

= + τ045

nσ τ=Tensile

. . .sin 45 . .sin 45sin 45n AC ACσ τ τ= +

22 sin 45nσ τ=045


66nσ τ= Compressiveτ

τC 045≅

ττ2γSquare

element γ γ

C C’

C’’

045

45≅

τ2γ

γ

A D

AC AC C CAC AC

ε′ ′′ ′ ′′−

= =′′

Tensile strain in diagonal

A D

( )cos45cos45 2 2 2

CC CCCD CD G

γ τ′ ′= = = =( )

Tensile strain in diagonal

( )1τ τ τε μ μ= + = +

τ

Tensileτ( )1E E E

ε μ μ+ +ττ

Tensile

Compressive

τ

τ( )1τ τ μ= + ( )2 1E G μ+


67τ

Compressive( )12G E

μ= + ( )2 1E G μ= +


68


69

Prob: P1= 480 kN, P2= 900 kN, P3= 1000 kN P3P

1 , 2 , 3AB = 250 mm, AC = 80 mm, CD = 100 mm,E = 2x105 N/mm2, Poisson’s ratio = 0.25.Find a) shear modulus, b) change in volume.

1P

2P

C

D

) ) g1P

A B( )( )2 1 GE G μ= + ⇒ = 0.8x105 N/mm2

1 2 31

1 2 3

P P PA E A E A E

ε μ μ= − + = 2.9375x10-4

2 1 32

2 1 3

P P PA E A E A E

ε μ μ= − +1 2 3v

VVδε ε ε ε∴ = = + +

= 2x10-4

1 625 10 42 1 3

3 1 23

P P PA E A E A E

ε μ μ= − − − = -3.3125x10-4

= 1.625x10-4

vV Vδ ε=


70

33 1 2A E A E A E = 325 mm3

SummarySummaryStress and strain

Introduction - general concepts - definition of stress - stress tensor -stress analysis of axially loaded members - strength design of members - axial strains and deformations in bars - stress-strain relationships - Poisson's ratio - thermal strain - Saint Venant's pprinciple - elastic strain energy for uniaxial stress - statically indeterminate systems - generalised Hooke's law for isotropic materials - relationships between elastic constants - introduction to materials relationships between elastic constants introduction to anisotropy – orthotropy


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