2D theory of elasticity - imechanica theory of elasticit.pdf · 2014-06-04 · 2D Theory of...

R&DE (Engineers), DRDO

2D Theory of Elasticity

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Ramadas Chennamsetti

2D Theory of Elasticity


Summary

� Field equations

� Boundary conditions

� Problem formulation

� Plane strain

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� Plane strain

� Plane stress

� Airy’s stress function

� Axially loaded bar

� Pure bending of beam


Introduction

� In principle all practical problems – 3D problems – very complex – difficult to handle

� Reasonable assumptions – bring down the complexity in the problem – reduces a

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complexity in the problem – reduces a dimension => 2D problem

� 2D problems – two independent variables –static problem – three independent variables – dynamic problems

� Field variables, φ = φ(x, y, t)


Introduction

� Field variable is independent of third dimension =>

� Continuum approach – the following laws hold good

0=∂∂

z

φ

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hold good� Conservation of mass

� Principle of momentum

� Principle of moment of momentum

� First law of thermodynamics

� Second law of thermodynamics - inequality


Introduction

� No inertia effects – quasi-static / static

� Small deformation and rotations

� Material => isotropic – same properties in all the direction

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the direction

� Homogeneous – material properties are uniform over the domain

� Linear elastic – Hooke’s law holds good => Hookean material


Field equations

� Strain-displacement equations

( )jijii

j

j

iij uu

x

u

x

u,,2

1

2

1 +=

∂∂

+∂∂=ε

Six equation

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� Compatibility equations

0,,,, =−−+ ikjljlikijklklij εεεε

Six equation

Six equations – three independent


Field equations

� Equilibrium equations –

� Constitutive law – Linear elastic – Hooke’s

0, =+ ijij BσThree equations

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� Constitutive law – Linear elastic – Hooke’s law

ijkkijij

ijijkkij

EEδσυσυε

µεδλεσ

−+=

+=

1

2

Six equations


Field equations

� Total – 15 equations = 3 Equilibrium + 6 strain displacement + 6 constitutive laws

� Unknowns => 3 Displacements, 6 stress components and 6 strain components

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components and 6 strain components

� Boundary value problem casted as

( ) 0 , , , , , =ℑ iijiji Bu µλσεUnknowns Lame’s constants

Body load

Not easy to solve analytically 15 equations to find out 15 unknowns


Boundary conditions

� Solution of a differential (partial) equations –boundary conditions required

� Boundary conditions influence the solution –field variable in the domain

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field variable in the domain

� BCs play a very essential role for properly formulating and solving elasticity problems

� In elasticity – boundary conditions –displacements, tractions and combination


Boundary conditions

Various boundary conditions in elasticity –Tn

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Tn

Traction boundary conditions

U

Displacement boundary conditions

U

Mixed boundary conditions


Boundary conditions

� Tractions –

A

σyy

σxx

τxy

σyy

At ‘A’ – Direction cosines of area normal => (0, 1, 0)

σxx is inside the boundary – this need not be zero

T = τ ; T = σ

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σxx

x

y

B

σyy

τxy

Tx = τxy; Ty = σyy

At ‘B’ – Direction cosines of area normal => (1, 0, 0)

Tx = σxx; Ty = τxy

σyy is inside the boundary – this need not be zero

yyxyy

xyxxx

mlT

mlT

σττσ

+=

+=


Boundary conditions

� Tractions –

A

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Br

θ

Tractions in Polar co-ordinate system


Boundary conditions

� Tractions –

y

1

Boundary – 1– traction free

DC => (0, 1, 0)

Tx = 0. σxx + 1. τxy = 0

⇒Tx = τxy = 0

⇒Ty = 0. τxy + 1. σyy = 0 => σyy = 0

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x

2

3

4 S

⇒Ty = 0. τxy + 1. σyy = 0 => σyy = 0

Boundary – 2– DCs (1, 0, 0)

Tx = 1. σxx + 0. τxy = S = σxx

Ty = 1. τxy + 0. σyy = 0 = τxy

Boundary – 3– DCs (0, -1, 0) Tx = 0. σxx - 1. τxy = 0Tx = τxy = 0Ty = 0. τxy - 1. σyy = 0 => σyy = 0

Boundary – 4

u = 0; v = 0


Boundary conditions

� Tractions Boundary - 1– DCs (l, m, 0)

This is a traction free boundary

Tx = lσxx + m τxy = 0

Ty = mσxx + l τxy = 0

y

n

DCs are constant on this boundary

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Boundary – 2– Constrained boundary – U, V = 0

Boundary – 3- Traction is acting

DCs (-1, 0, 0)Tx = -1σxx + m τxy = S => σxx = - S

Ty = mσxx + l τxy = 0 => τxy = 0

x

n

σyy

τxy

σxx

o

1

2

3S

Individual stress components are not zero on boundary - 1


Problem formulation

� 3D elasticity 15 equations and 15 unknowns - 2D elasticity – 8 equations and 8 unknowns

� Reformulating elasticity problems – mathematically convenient way

� Two approaches –

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� Two approaches –� Displacement based � Stress based formulations

� Displacement based – express filed equation in terms of displacements, ui – Navier equation

� Stress based – express filed equation in terms of stresses, σij


Problem formulation

� In displacement based formulation� Primary variable – displacement – solution is

obtained for displacements – stresses and strains secondary / derived variables

� Strains – derivation of displacements

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� Strains – derivation of displacements� Stresses – use constitutive law

� In stress based formulation� Primary variable – stresses – solution is obtained

for stresses – strains and displacements secondary variables

� Strain compatibility equations


Displacement formulation

� Governing equation expressed in displacements

� Field equations –

(1) - 0 mEquilibriu , =+ ijij Bσ

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(1) - 0 mEquilibriu , =+ ijij Bσ

(2) - 2

1nt displaceme-Strain

∂∂

+∂∂=

i

j

j

iij x

u

x

uε

(3) - 2 law veConstituti ijijkkij µεδλεσ +=



� Differentiate (3) wrt ‘j’

(4) - 2 ,,, jijijjkkjij µεδλεσ +=Plug this in (1)

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(5) - 0 2 ,,, =++=+ ijijijjkkijij BB µεδλεσExpress all strain components in (5) in terms of displacements – Use equation (2)

( )0

0

,,,

,,,

=+++

=+++

ijjijijkik

iijjjjiijkjk

Buuu

Buuu

µµλµδλ



( ) 0

0

,,

,,,

=+++

=+++

ijjijij

ijjijijkik

Buu

Buuu

µµλµµλ

This gives three equations for i = 1, 2 and 3

For i = 1 => x - direction

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For i = 1 => x - direction

( ) 01123

2

22

2

21

2

3

3

2

2

1

1

1

=+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂+

∂∂

∂∂+ Bu

xxxx

u

x

u

x

u

xµµλ

Combining all three equations into a single equation

( ) ( ) 0. ~~

2

~=+∇+∇∇+ BUU µµλ



� Governing equations in terms of displacements – three equations - reduced from ’15’ equations

( ) (U) - 0,, =+++ ijjijij Buu µµλ( ) 0 , , , =ℑ Bu µλ Navier equation

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Equations (U) known as Navier’s or Lame’s equations

Boundary conditions => ui = U i on ‘S’ boundary

Solve second order PDEs – get displacements

Use strain – displacement relations for calculating strains

Constitutive relations for stresses

( )321 , , xxxuu ii =

( ) 0 , , , =ℑ ii Bu µλ Navier equation


Stress formulation

� In displacement formulation compatibility equations play no role – displacement obtained are single valued and continuous

� In stress formulation express governing equation in terms of stresses

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equation in terms of stresses� Stresses – primary variables – obtain

secondary variables displacements from strain-displacement relations

� Integrate ε-U relations – compatibility equations come into play


Stress formulation

� Boundary conditions – only tractions

� From the filed equations eliminate strains and displacements

� Compatibility equation

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� Compatibility equation

ijkkijij EEδσυσυε −+= 1

(1) - 0,,,, =−−+ ikjljlikijklklij εεεε

Constitutive law

Differentiate this wrt k and l, plug in (1)

For k = l first three compatibility equations are obtained


Stress formulation

� Three compatibility equations for k=l = 1, 2 and 3

2

2

2

2

2

∂∂∂

=∂

∂+

∂∂

yxxyxyyyxx

γεε

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2) ,1( ====∂∂∂∂

lkji

yxxy

( )1 ,3

2

2

2

2

2

====∂∂

∂=∂

∂+∂

∂

lkjizxxzxzzzxx γεε ( )3 ,2

2

2

2

2

2

====∂∂

∂=

∂∂+

∂∂

lkji

zyyzyzzzyy γεε


Stress formulation

This gives –

( )jkikmmikjkmmkkijmmijkkmm

ikjkjkikijkkkkij

−−++

=−−+

,,,,

,,,,

1δσδσδσδσ

υυ

σσσσ

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ijij

ijij

kk

B

B

−=

=+=

+

,

, 0 eqn. mequilibriu

31

σσ

δυ

Make use of equilibrium equation to reduce further


Stress formulation

Governing equation in terms of stress –

ijjiijkkmmijkkkkij BB ,,,,, 11

1 −−+

=+

+ δσυ

υσυ

σ

( ) 0 , , , =ℑ iij Bµλσ

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Compatibility equation in terms of stresses, known as

“Beltrami-Michell” compatibility equation

Six equations can be obtained for i, j = 1, 2, 3

Only three are independent – similar to strain compatibility equations

Three more equations are required – complemented by three equilibrium equations - Six unknowns – six equations

( ) 0 , , , =ℑ iij Bµλσ


Plane strain

� Out-of-plane dimensions are large compared to in-plane dimensions

y

All loads and boundary conditions independent of ‘z’ co-ordinate

The deformation in ‘z’ direction assumed to be zero. w = 0

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x

z

assumed to be zero. w = 0

Deformations in ‘x’ and ‘y’ directions depend on (x, y)only

u = u(x, y)and v = v(x, y)

All cross-sections have same displacements

Out of plane strains vanishReduce a 3D problem to 2D problem


Plane strain

� In plane strain εzz= 0, γyz = 0 and γzx = 0

� Linear elastic material – Hooke’s law

( )ijijkkij

µεεελσµεδλεσ

++=

+= 2

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( )( )

( )xyxyxy

yyxxzz

yyyyxxyy

xxyyxxxx

µγµετεελσ

µεεελσµεεελσ

==

+=

++=

++=

2

2

2


Plane strain

� Equilibrium equations –

(1) - 0

0,

=+∂+∂

=+∂

∂+

∂∂

=+

xy

xxyxx

ijij

Byx

B

στ

τσ

σ

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(2) - 0=+∂

∂+∂

∂y

xxxy Byx

στ

Use constitutive law and strain-displacement equations in (1)

( )( ) ( )

02

02

=+

∂∂+

∂∂

∂∂+

∂∂+

∂∂+

∂∂

∂∂

=+∂∂+++

∂∂

x

ixyxxyyxx

Bx

v

y

u

yx

u

y

v

x

u

x

Byx

µµλ

µγµεεελ


Plane strain

( ) 02 =+

∂∂+

∂∂

∂∂++∇ xB

y

v

x

u

xu µλµ

Governing equation in x – direction

Use second equilibrium equation – convert to displacements

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( ) 02 =+

∂∂+

∂∂

∂∂++∇ yB

y

v

x

u

yv µλµ

In each of the above two equations, equilibrium, constitutive law and displacement – strain equations are inbuilt -Displacement formulation

Eight equations => reduced to two equations in displacements


Plane strain

� Stress based formulation -In 2D case, compatibility equation

Use constitutive law – convert into stresses

yxxyxyyyxx

∂∂∂

=∂

∂+

∂∂ γεε 2

2

2

2

2

( )xx σσυσε +−=

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( )

( )

( ) ( )( )

xyxy

yyxxzzyyxxzz

zz

zzxxyy

yy

zzyyxx

xx

E

EE

EE

EE

τυγ

σσυσσσυσε

σσυσε

σσυσε

+=

+==>+−==

+−=

+−=

12

0


Plane strain

� Compatibility equation in terms of stresses

( )[ ] ( )[ ]yxyxxy

yyxxxxyy ∂∂∂

=−−∂∂+−−

∂∂ τ

υσσυυσσυ2

2

2

2

2

211

Above equation is obtained from compatibility, constitutive

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Above equation is obtained from compatibility, constitutive equations

Use equilibrium equations to eliminate shear stress from above equation

0y

0

0 0

=

+

∂∂

+∂

∂∂∂=>=+

∂∂+

∂∂

=

+

∂∂

+∂

∂∂∂=>=+

∂∂

+∂

∂

yyyxy

yxxxy

xxyxx

xxyxx

Byx

Byx

Byxx

Byx

στστ

τστσ

Add


Plane strain

This is obtained as –

∂∂

+∂

∂+∂

∂+

∂∂−=

∂∂∂

=∂

∂+

∂∂+

∂∂∂

+∂

∂+

∂∂

y

B

x

B

yxyx

y

B

x

B

yxyx

yxyyxxxy

yxxyyyxx

2

2

2

22

2

2

2

2

2

2

02

σστ

τσσ

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∂+

∂+

∂+

∂−=

∂∂ yxyxyx 222

( )[ ] ( )[ ]

∂∂

+∂

∂+∂

∂+

∂∂−=−−

∂∂+−−

∂∂

y

B

x

B

yxyxyxyyxx

yyxxxxyy 2

2

2

2

2

2

2

2

11σσυσσυυσσυ

Simplifying this expression

( )

∂∂

+∂

∂−

−=+∇y

B

x

B yxyyxx υ

σσ1

12

Single governing equation for stress based formulation of plane strain problems


Plane strain

� Governing equation

( )

∂∂

+∂

∂−

−=+∇y

B

x

B yxyyxx υ

σσ1

12

In this equation, equilibrium, constitutive and compatibility equations are inbuilt.

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equations are inbuilt.

Three stress components in 2D –σxx, σyy, τxx

Only one governing equation – three unknowns

In 2D two stress equilibrium equations exist – use these two with above governing equation

Boundary conditions given in terms of tractions. Relate them to stresses


Plane stress

� Third dimension – much smaller than in-plane dimensions Out of plane stresses σzz, τxz and

τyz negligible

Body can have out of plane deformation

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x

yz

deformation

εzz is non-zero, γxz and γyz = 0

Stresses are functions of in-plane displacements

Thin plates – structures

Reducing a 3D problem to a 2D problem

( )( )( )yx

yx

yx

xyxy

yyyy

xxxx

,

, ,

, ,

ττσσσσ

=

==


Plane stress

� Formulation of plane stress problem – two approaches � Displacement based

� Stress based approach

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� Formulation procedure – same as in plane strain

� Formulation takes care of non-zero out of normal strain and zero normal stress

� Use field equations


Plane stress

� Displacement based approach –

( ) ( )

−−=

xy

y

x

xy

y

xE

γεε

νν

ν

ντσσ

2

100

01

01

1 2

Plug in stress equilibrium equations

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xyxy

2

0 ; 0 =+∂

∂+∂

∂=+

∂∂

+∂

∂y

xxxyx

xyxx Byx

Byx

σττσ

( ) 01211 22

=+

∂∂+

∂∂

+∂∂+

−+

−∂∂

xyyxx Bx

v

y

uE

y

EE

x υε

υυε

υ

( ) 01211 22

=+

∂∂+

∂∂

+∂∂+

−+

−∂∂

yxxyy Bx

v

y

uE

x

EE

y υε

υυε

υ


Plane stress

� Use strain displacement equations to convert strains into displacements

� Governing equations in terms of displacements

02 =+

∂+∂∂+∇ B

vuEuµ

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( )

( ) 012

012

2

2

=+

∂∂+

∂∂

∂∂

−+∇

=+

∂∂+

∂∂

∂∂

−+∇

y

x

By

v

x

u

y

Ev

By

v

x

u

x

Eu

υµ

υµ

In the above two equations – equilibrium, constitutive laws and strain-displacements equations


Plane stress

� Stress based approach – field equations in plane stress problems

0=+∂

∂+

∂∂

xxyxx Byx

τσ

συσε −=

Constitutive laws

Equilibrium equations

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0=+∂

∂+∂

∂y

xxxy Byx

στ

( )xyxy

xxyy

yy

yyxx

xx

E

EE

EE

τυγ

συσε

συσε

+=

−=

−=

12

yxxyxyyyxx

∂∂∂

=∂

∂+

∂∂ γεε 2

2

2

2

2

Compatibility equation


Plane stress

� Use constitutive laws in compatibility equation –eliminate strains in compatibility using constitutive law

� Compatibility equation in terms of stress – stress compatibility equation

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compatibility equation� Eliminate shear stress in stress compatibility

equation using two equilibrium equations� Differentiate x – direction equilibrium equation wrt

‘x’ – y – direction eqn. wrt ‘y’, add these two get shear stress

� Plug this in stress compatibility equation


Plane stress

� The governing equation –

( ) ( )

∂∂+

∂∂+−=+∇

x

B

x

B xxyyxx υσσ 12

This is derived making use of strain compatibility equation,

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This is derived making use of strain compatibility equation, constitutive laws and equilibrium equations


Summary – Plane stress and strain

Formulation Plane strain Plane stress

Displacement( ) 02 =+

∂∂+

∂∂

∂∂++∇ xB

y

v

x

u

xu µλµ

( ) =+

∂+∂∂++∇ vuµλµ

( )

( ) 012

012

2

2

=+

∂∂+

∂∂

∂∂

−+∇

=+

∂∂+

∂∂

∂∂

−+∇

y

x

By

v

x

u

y

Ev

By

v

x

u

x

Eu

υµ

υµ

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Stress

( ) 02 =+

∂∂+

∂∂

∂∂++∇ yB

y

v

x

u

yv µλµ ( ) 0

12=+

∂

+∂∂−

+∇ yByxy

vυ

µ

( )

∂∂

+∂

∂−

−=+∇y

B

x

B yxyyxx υ

σσ1

12 ( ) ( )

∂∂+

∂∂+−=+∇

x

B

x

B xxyyxx υσσ 12


Summary – Plane stress and strain

� Basic difference – coefficients involving elastic material constants

� Solving one type of plane problem gives solution to other type also using simple transformation of elastic coefficients

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elastic coefficients� In absence of body loads – in stress formulation –

governing equations of plane stress and strain are same� No elastic constants appear in governing equation

� Solving these equations – difficult – some text book problems


Airy’s stress function

� In stress based formulation – one equation in terms of normal stresses and two equilibrium equations for finding out the complete stress state

� Reduce the complexity of equations

� Introduce a new filed variable ‘Airy’s stress

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� Introduce a new filed variable ‘Airy’s stress function’ (φ)

� Aim – reduce the governing equations from three to one

� Find the distribution of single variable ‘φ’ – get stresses from that



� Airy’s stress function (ASF) defined for stress based formulations – both plane stress and plane strain

� Basic stress equilibrium equations –

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0

0

=+∂

∂+∂

∂

=+∂

∂+

∂∂

yxxxy

xxyxx

Byx

Byx

στ

τσ

Body loads can be derived from potential function ‘V’

Defining potential

y

VB

x

VB

VBVB

yx

ii

∂∂−=

∂∂−=

==>−∇=

,

,~

Substitute in equilibrium equations



Equilibrium equations become

( ) ( ) 0 ;0 =−∂∂+

∂∂

=∂

∂+−

∂∂

Vyxy

Vx yy

xyxyxx σ

ττσ

Defining ASF as

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Defining ASF as

2

22

2

2

; ;x

Vyxy

V yyxyxx ∂∂=−

∂∂∂−=

∂∂=− φσφτφσ

ASF, φ = φ(x, y)

ASF satisfies both equilibrium equations.

In stress based formulation – it has to satisfy one more governing equation



� In plane strain – stress based formulation –governing equation

( )

VV

y

B

x

B yxyyxx

22

2

1

1

1

∂−∂+ ∂−∂−=

∂++

∂+∇

∂∂

+∂

∂−

−=+∇

φφ

υσσ

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V

V

VVyx

y

V

yx

V

xxV

yV

24

222

2222

2

222

1

21

21

1.

1

12

1

1

∇

−−−=∇

∇

−−

=∇∇

∇−

=∇+

∂∂+

∂∂∇

∂∂−

∂∂+

∂∂−

∂∂

−−=

∂∂++

∂∂+∇

υυφ

υφ

υφφ

υφφ

Bi-harmonic operator



� In plane stress – stress based formulation –governing equation

( ) ( )y

B

x

B yxyyxx

22

2 1

∂∂ ∂∂ ∂ ∂

∂∂

+∂

∂+−=+∇

φφ

υσσ

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( )

( )

( )( ) V

V

VVyx

y

V

yx

V

xxV

yV

24

222

2222

2

222

1

21.

12

1

∇−−=∇∇−+=∇∇

∇+=∇+

∂∂+

∂∂∇

∂∂−

∂∂+

∂∂−

∂∂+−=

∂∂++

∂∂+∇

υφυφ

υφφ

υφφ



� Governing equations in terms of ASF

( ) stress Plane 1

strain Plane 1

21

24

24

V

V

∇−−=∇

∇

−−−=∇

υφυυφ

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If body forces are neglected – V = 0 – In both cases the governing equation is same

0204

4

22

4

4

44 =

∂∂+

∂∂∂+

∂∂=>=∇

yyxx

φφφφ

Bi-harmonic equation – solution to this equation => ASF

Problem of elasticity reduced to a single equation – find φ in solution domain and boundaries – BCs - Tractions


Stress formulation

� This formulation is appropriate for use with traction boundary conditions

� Use Hooke’s law for calculating strains from stresses

� Compute displacements from strain-displacement relations

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relations� Since compatibility equations are used in

formulating governing equations, the displacements obtained from integration of strain-displacement equations yield single valued, continuous filed

� All equations are used for solving the problem� Closed form analytical solutions for elasticity

problems – difficult – rely on numerical methods


Uniaxial tension – ASF approach

� A bar subjected to tension

x

y

2l

2c TT

12

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Boundary – 1

DCs => (1, 0, 0)

Tx = T = lσxx + m τxy = σxx = T

Ty = 0 = mσyy + l τxy => τxy = 0

Boundary – 2

DCs => (-1, 0, 0)

Tx = -T = lσxx + m τxy = σxx = T

Ty = 0 = mσyy + l τxy => τxy = 0

Remaining boundaries – stress free.

Stress in x – direction = T = constant



Stress in x – direction given by ASF

212

2

2

2

1 constant cycTyT

yxx ++==>=∂∂== φφσ

At x = ± l, σyy and τxy = 0

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(2) -

(1) -

E

T

y

v

EE

E

T

x

u

EE

xy

y

yx

x

υσυσε

συσε

−=∂∂=−=

=∂∂=−=

Integrate (1) and (2)

)(

)(

xgyE

Tv

yfxE

Tu

+−=

+=

υ

f(y) and g(x) –arbitrary functions



Shear strain, xy

xy

xgyf

xgyf

Gx

v

y

u

==−=>=+=>

==∂∂+

∂∂=

ω

τγ

)()(

0)()(

0

''

''

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ooo

ooo

o

vxgx

g

uyfy

f

xgyf

+==>=∂∂

+−==>−=∂∂=>

==−=>

ωω

ωω

ω)()( ''

ωo, uo, vo – arbitrary constants of integration.



� Functions ‘f’ and ‘g’ represent rigid body motions

� Rigid body rotation -ωo and translations in ‘x’ and ‘y’ directions => uo and vo

� Terms related to rigid body motion result from strain-displacement relations

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displacement relations

� Displacements are determined from strain fields only up to an arbitrary rigid motion

� For complete determination of displacement filed –additional boundary conditions required – evaluate ωo, uo and vo

� If rod has no rigid body motion => ωo, uo and vo = 0


Pure bending of beam – ASF approach

� Beam subjected to pure bending –

x

y

2l

2c

M M

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∫+

−

+

−

±=−

=±=

=±

=±

=±

c

c

xx

c

c

xxx

xy

xy

yy

ydyylM

dyylF

yl

cx

cx

),(

0),(

0) ,(

0) ,(

0) ,(

σ

σ

ττσ

Boundary

Conditions

Statically equivalent boundary condition



� Key point in ASF approach – selection of an appropriate stress function -φ

� It has to satisfy all boundary conditions –capture physics of the problem

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capture physics of the problem

� Some knowledge about structure’s behavior helps in selection of ‘φ’� In bending, variation of bending stress is linear

along ‘y’ co-ordinate

� ‘φ’ should be a cubic polynomial



Linear variation of bending stress

213

2

2

6

CyC yA

yAyxx

++=

∂∂==

φ

φσ Integrating the function

ASF satisfies all BCs

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216CyC y ++=φ ASF satisfies all BCs

( )

213

321

3

3

3

32

462

3

2

3

3

,

CyCyc

MCyC

y

c

M

c

MAM

yAMdyAy

Mydyyl

c

c

c

c

c

c

xx

++−=++−=

−==>=

−=>−=

−=±

+

−

+

−

+

−

∫

∫

φ

σ



� Stresses and strains

yc

M

yCyCy

c

M

xyyy

xx 32

2

213

3

0 ,0

2

3

4

τσ

φσφ

==

−=∂∂==>++−=

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yEc

M

EE

yEc

M

x

u

EE

xxyyyy

yyxxxx

xyyy

3

3

2

3

2

3

0 ,0

υσυσ

ε

συσε

τσ

=−=

−=∂∂=>−=

==

Integrate strain –displacement equations

)(4

3 );(

2

3 233

xgyEc

Mvyfxy

Ec

Mu +=+−= υ



Shear strain

o

xy

yfxgxEc

M

xgyfxEc

M

x

v

y

u

=−=+−=>

=++−=>=∂∂+

∂∂=

ω

γ

Integrate )()(2

3

0)()(2

30

''3

''3

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oo

oo

o

uyyf

vxx

Ec

Mxg

yfxgxEc

+−=

++=

=−=+−=>

ω

ω

ω

)(22

3)(

Integrate )()(2

2

3

3

Functions g(x) and f(y) represent rigid body motions

To evaluate constants, ωo, uo, vo – constraints required


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2D theory of elasticity - imechanica theory of elasticit.pdf · 2014-06-04 · 2D Theory of...

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Transcript of 2D theory of elasticity - imechanica theory of elasticit.pdf · 2014-06-04 · 2D Theory of...