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2D Theory of Elasticity
Ramadas Chennamsetti
Summary
Field equations
Boundary conditions
Problem formulation
Plane strain
Plane stress
Airy’s stress function
Axially loaded bar
Pure bending of beam
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Introduction
In principle all practical problems – 3D
problems – very complex – difficult to
handle
Reasonable assumptions – bring down the
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)
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Introduction
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Conservation of mass
Principle of momentum
Principle of moment of momentum
First law of thermodynamics
Second law of thermodynamics - inequality
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Field variable is independent of third
dimension => ∂∂φz = 0
Continuum approach – the following laws
hold good
Introduction
No inertia effects – quasi-static / static
Small deformation and rotations
Material => isotropic – same properties in all
the direction
Homogeneous – material properties are
uniform over the domain
Linear elastic – Hooke’s law holds good =>
Hookean material
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Field equations
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Strain-displacement equations
1  ∂ui ∂u j  1
ε ij = 
+
= (ui , j + ui , j )

2  ∂x j ∂xi  2
Six equation
ε ij ,kl + ε kl ,ij − ε ik , jl − ε jl ,ik = 0
Six equations – three independent
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Compatibility equations
Field equations
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Equilibrium equations –
σij, j + Bi = 0
Three equations
ε ij
1+υ
=
σ
E
ij
−
υ
E
σ kk δ ij
Six equations
7
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Constitutive law – Linear elastic – Hooke’s
law
σ ij = λε kk δ ij + 2 µε ij
Field equations
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ℑ(u i , ε ij , σ ij , λ , µ , Bi ) = 0
Unknowns
Lame’s constants
Body load
Not easy to solve analytically 15 equations to find out 15 unknowns
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Total – 15 equations = 3 Equilibrium +
6 strain displacement + 6 constitutive laws
Unknowns => 3 Displacements, 6 stress
components and 6 strain components
Boundary value problem casted as
Boundary conditions
Solution of a differential (partial) equations –
boundary conditions required
Boundary conditions influence the solution –
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
9
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Boundary conditions
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Tn
Traction boundary
conditions
U
Displacement boundary
conditions
Tn
U
Mixed boundary
conditions
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Various boundary conditions in elasticity –
Boundary conditions
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Tractions –
At ‘A’ – Direction cosines of
area normal => (0, 1, 0)
σyy
σxx is inside the boundary – this
need not be zero
τxy
σyy
Tx = τxy; Ty = σyy
τxy
σxx
B
y
At ‘B’ – Direction cosines of
area normal => (1, 0, 0)
Tx = σxx; Ty = τxy
x
T x = lσ
xx
+ mτ
xy
T y = lτ
xy
+ mσ
yy
σyy is inside the boundary – this
need not be zero
11
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σxx
A
Boundary conditions
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Tractions –
A
r
B
Tractions in Polar co-ordinate system
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θ
Boundary conditions
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Boundary – 1 – traction free
Tractions –
DC => (0, 1, 0)
y
Tx = 0. σxx + 1. τxy = 0
1
⇒Tx = τxy = 0
⇒Ty = 0. τxy + 1. σyy = 0 => σyy = 0
Boundary – 2 – DCs (1, 0, 0)
S
Tx = 1. σxx + 0. τxy = S = σxx
x
3
Boundary – 4
u = 0; v = 0
Ty = 1. τxy + 0. σyy = 0 = τxy
Boundary – 3 – DCs (0, -1, 0)
Tx = 0. σxx - 1. τxy = 0
Tx = τxy = 0
Ty = 0. τxy - 1. σyy = 0 => σyy = 0
13
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2
4
Boundary conditions
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Boundary - 1 – DCs (l, m, 0)
Tractions
This is a traction free boundary
y
DCs are constant on this
boundary
Tx = lσxx + m τxy = 0
Ty = mσxx + l τxy = 0
n
Boundary – 2 – Constrained
boundary – U, V = 0
σxx
3
τxy
1
σyy
o
2
Boundary – 3- Traction is acting
x
DCs (-1, 0, 0)
Tx = -1σxx + m τxy = S => σxx = - S
Individual stress components are not zero
Ty = mσxx + l τxy = 0 => τxy = 0
on boundary - 1
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S
Problem formulation
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3D elasticity 15 equations and 15 unknowns - 2D
elasticity – 8 equations and 8 unknowns
Reformulating elasticity problems – mathematically
convenient way
Two approaches –
Displacement based – express filed equation in
terms of displacements, ui – Navier equation
Stress based – express filed equation in terms of
stresses, σij
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Displacement based
Stress based formulations
Problem formulation
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In displacement based formulation
Primary variable – displacement – solution is
obtained for displacements – stresses and strains
secondary / derived variables
Strains – derivation of displacements
Stresses – use constitutive law
Primary variable – stresses – solution is obtained
for stresses – strains and displacements
secondary variables
Strain compatibility equations
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In stress based formulation
Displacement formulation
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Governing equation expressed in
displacements
Field equations –
Equilibrium σ ij , j + Bi = 0 - (1)
Constituti ve law σ ij = λε kk δ ij + 2 µε ij
- (2)
- (3)
17
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1  ∂ui ∂u j 
Strain - displacement ε ij =
+
2  ∂x j ∂xi 
Displacement formulation
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Differentiate (3) wrt ‘j’
σ ij , j = λε kk , jδ ij + 2 µε ij , j
- (4)
Plug this in (1)
Express all strain components in (5) in terms of
displacements – Use equation (2)
λ u k , kj δ ij + µ (u i , jj + u j , ij ) + B i = 0
λ u k , ki + µ u j , ji + µ u i , jj + B i = 0
18
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σ ij, j + Bi = λεkk, jδij + 2µεij, j + Bi = 0 - (5)
Displacement formulation
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λ u k , ki + µ u j , ji + µ u i , jj + B i = 0
(λ
+ µ )u j , ji + µ u i , jj + B i = 0
This gives three equations for i = 1, 2 and 3
∂
(λ + µ )
∂ x1
 ∂ u1 ∂ u 2 ∂ u 3

+
+
 ∂ x1 ∂ x 2 ∂ x 3

 ∂2
∂2
∂2
 + µ  2 +
+
2
2
∂
x
∂
x
∂
x
2
3

 1

 u 1 + B1 = 0

Combining all three equations into a single equation
(λ
(
)
+ µ ) ∇ ∇ .U + µ ∇ 2 U + B = 0
~
~
~
19
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For i = 1 => x - direction
Displacement formulation
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Governing equations in terms of displacements
– three equations - reduced from ’15’ equations
(λ + µ )u j , ji + µui , jj + Bi = 0
ℑ(u i , λ , µ , Bi ) = 0
- (U)
Navier equation
Boundary conditions => ui = Ui on ‘S’ boundary ui = ui ( x1 , x2 , x3 )
Solve second order PDEs – get displacements
Use strain – displacement relations for calculating strains
Constitutive relations for stresses
20
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Equations (U) known as Navier’s or Lame’s equations
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
Stresses – primary variables – obtain
secondary variables displacements from
strain-displacement relations
Integrate ε-U relations – compatibility
equations come into play
21
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Stress formulation
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Boundary conditions – only tractions
From the filed equations eliminate strains and
displacements
Compatibility equation
For k = l first three compatibility equations are obtained
1+υ
υ
Constitutive law ε ij =
σ ij − σ kk δ ij
E
E
Differentiate this wrt k and l, plug in (1)
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ε ij ,kl + ε kl,ij − ε ik , jl − ε jl ,ik = 0 - (1)
Stress formulation
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Three compatibility equations for k=l = 1, 2
and 3
∂ 2 ε xx ∂ 2 ε zz
∂ 2 γ xz
+
=
2
2
∂z
∂x
∂x∂z
(i = j = 3, k = l = 1)
∂ 2 ε yy
∂ 2γ
∂ ε zz
yz
+
=
2
2
∂y
∂y∂z
∂z
(i = j = 2 , k = l = 3 )
2
23
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∂ 2 ε yy
∂ 2 γ xy
∂ 2 ε xx
+
=
2
2
∂y
∂x
∂x∂y
( i = j = 1, k = l = 2)
Stress formulation
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This gives –
σ ij , kk + σ kk ,ij − σ ik , jk − σ jk ,ik =
1+υ
(σ
mm , kk
δ ij + σ mm ,ijδ kk − σ mm , jk δ ik − σ mm ,ik δ jk )
δ kk = 3
equilibriu m eqn. σ ij , j + Bi = 0
σ ij , j = − Bi
Make use of equilibrium equation to reduce further
24
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υ
Stress formulation
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Governing equation in terms of stress –
1
υ
σ ij , kk +
σ kk ,ij =
σ mm , kk δ ij − B i , j − B j , i
1+υ
1+υ
ℑ(σ ij , λ , µ , Bi ) = 0
“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
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Compatibility equation in terms of stresses, known as
Plane strain
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Out-of-plane dimensions are large compared to
in-plane dimensions All loads and boundary conditions
independent of ‘z’ co-ordinate
y
The deformation in ‘z’ direction
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)
z
Reduce a 3D problem to
2D problem
All cross-sections have same
displacements
Out of plane strains vanish
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x
Plane strain
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In plane strain εzz = 0, γyz = 0 and γzx = 0
Linear elastic material – Hooke’s law
σ ij = λε kk δ ij + 2 µε ij
σ yy = λ (ε xx + ε yy ) + 2 µε
σ zz = λ (ε xx + ε yy )
τ xy = 2 µε xy = µγ
yy
xy
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σ xx = λ (ε xx + ε yy ) + 2 µε xx
Plane strain
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Equilibrium equations –
σ
ij , j
+ Bi = 0
∂
(λ (ε xx + ε yy ) + 2 µε xx ) + ∂ (µγ xy ) + Bi = 0
∂x
∂y
∂
∂x
  ∂u ∂v 
∂u  ∂   ∂u ∂v  
 + 2 µ
  + B x = 0
+
+
 λ 
+
 µ 
∂x  ∂y   ∂y ∂x  
  ∂x ∂y 
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∂ τ xy
∂ σ xx
+
+ B x = 0 - (1)
∂x
∂y
∂ τ xy
∂ σ xx
+
+ B y = 0 - (2)
∂x
∂y
Use constitutive law and strain-displacement equations in (1)
Plane strain
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Governing equation in x – direction
∂  ∂u ∂v 
µ∇ u + (λ + µ )  +  + Bx = 0
∂x  ∂x ∂y 
2
Use second equilibrium equation – convert to displacements
∂  ∂u ∂v 
µ∇ v + (λ + µ )  +  + B y = 0
∂y  ∂x ∂y 
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
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2
Plane strain
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Stress based formulation -
∂ 2 ε yy
∂ 2 γ xy
∂ 2 ε xx
+
=
2
2
∂y
∂x
∂x∂y
In 2D case, compatibility equation
Use constitutive law – convert into
stresses
ε yy =
ε zz = 0 =
σ zz
E
−
υ
E
E
σ yy
E
(σ
γ xy
xx
−
−
υ
E
υ
E
(σ
yy
+ σ zz )
(σ xx + σ zz )
+ σ yy ) => σ zz = υ (σ xx + σ yy )
2(1 + υ )
=
τ xy
E
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ε xx =
σ xx
Plane strain
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Compatibility equation in terms of stresses
[
]
[
∂2
∂2
(1 − υ )σ yy − υσ xx + 2 (1 − υ )σ xx − υσ
2
∂x
∂y
∂ 2τ xy
yy
] = 2 ∂x∂y
Use equilibrium equations to eliminate shear stress from above
equation

∂σ xx ∂τ xy
∂  ∂σ xx ∂τ xy
+
+ Bx = 0 =>

+
+ Bx  = 0
∂x
∂y
∂x  ∂x
∂y

∂τ xy ∂σ xx

∂  ∂τ xy ∂σ yy
+
+ B y = 0 => 
+
+ B y  = 0
∂x
∂y
∂y  ∂x
∂y

Add
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Above equation is obtained from compatibility, constitutive
equations
Plane strain
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This is obtained as –
∂ 2σ
∂ 2τ xy
∂ σ xx
∂B x ∂B y
yy
+
+2
+
+
=0
2
2
∂x
∂y
∂x∂y
∂x
∂y
∂ 2τ xy
 ∂ 2 σ xx
∂ 2σ yy
∂B x ∂B y

2
= −
+
+
+
2
2

∂x∂y
∂y
∂x
∂y
 ∂x




 ∂ 2σ xx ∂ 2σ yy ∂Bx ∂B y 
∂2
∂2

+
(1 − υ )σ yy − υσ xx + 2 (1 − υ )σ xx − υσ yy = − 2 + 2 +
2
∂y 
∂x
∂y
∂y
∂x
 ∂x
[
]
[
]
Simplifying this expression
1  ∂Bx ∂B y 


∇ (σ xx + σ yy ) = −
+
1 − υ  ∂x
∂y 
2
Single governing
equation for stress based
formulation of plane
strain problems
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2
Plane strain
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Governing equation
∇ (σ xx
2
1
+ σ yy ) = −
1−υ
 ∂B x ∂B y

+
∂y
 ∂x



In this equation, equilibrium, constitutive and compatibility
equations are inbuilt.
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
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Three stress components in 2D – σxx, σyy, τxx
Plane stress
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Third dimension – much smaller than inplane dimensions
Out of plane stresses σzz, τxz and
τyz negligible
Body can have out of plane
deformation
y
εzz is non-zero, γxz and γyz = 0
Stresses are functions of in-plane
displacements
x
Thin plates – structures
Reducing a 3D problem
to a 2D problem
σ xx = σ xx ( x , y ),
σ yy = σ yy ( x , y ),
τ xy = τ xy ( x , y )
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z
Plane stress
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Formulation of plane stress problem – two
approaches
Formulation procedure – same as in plane
strain
Formulation takes care of non-zero out of
normal strain and zero normal stress
Use field equations
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Displacement based
Stress based approach
Plane stress
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Displacement based approach –
(
2
)

1
ν

0

ν
1
0
∂σ xx ∂τ xy
+
+ Bx = 0 ;
∂x
∂y

  ε x 
0  ε y 
(1 − ν ) γ 
 xy 
2 
0
Plug in stress
equilibrium
equations
∂τ xy
∂σ xx
+
+ By = 0
∂x
∂y
∂  E
Eυ
 ∂  E  ∂u ∂v 
 +  + Bx = 0
ε +
ε + 
2 xx
2 yy 

∂x 1 − υ
1−υ
 ∂y  2(1 + υ )  ∂y ∂x 
∂  E
Eυ
 ∂  E  ∂u ∂v 
 +  + B y = 0
ε +
ε + 
2 yy
2 xx 

∂y 1 − υ
1−υ
 ∂x  2(1 + υ )  ∂y ∂x 
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σ x 
E
 
σ y  =
1 −ν
τ 
 xy 
Plane stress
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Use strain displacement equations to convert
strains into displacements
Governing equations in terms of displacements
E
∂
2 (1 − υ ) ∂ x
In the above two equations – equilibrium, constitutive laws
and strain-displacements equations
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 ∂u ∂v 

 + B x = 0
+
 ∂x ∂y 
∂  ∂u ∂v 
E
2

 + B y = 0
µ∇ v +
+
2 (1 − υ ) ∂ y  ∂ x ∂ y 
µ∇ 2u +
Plane stress
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Stress based approach – field equations in plane
stress problems
Equilibrium equations
ε xx =
ε yy =
γ xy
σ xx
E
σ yy
−
−
υ
E
υ
σ yy
σ xx
E
E
2 (1 + υ )
=
τ xy
E
∂ τ xy
∂ σ xx
+
+ Bx = 0
∂x
∂y
∂ τ xy
∂ σ xx
+
+ By = 0
∂x
∂y
Compatibility equation
∂ 2ε yy
∂ 2γ xy
∂ ε xx
+
=
2
2
∂y
∂x
∂x∂y
2
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Constitutive laws
Plane stress
Use constitutive laws in compatibility equation –
eliminate strains in compatibility using constitutive
law
Compatibility equation in terms of stress – stress
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
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Plane stress
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The governing equation –
∇ (σ xx
2
 ∂B x ∂B x 
+ σ yy ) = − (1 + υ )
+

∂x 
 ∂x
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This is derived making use of strain compatibility equation,
constitutive laws and equilibrium equations
Summary – Plane stress and strain
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Plane strain
µ∇ 2u + (λ + µ )
Displacement
∂  ∂u ∂v 
 +  + Bx = 0
∂x  ∂x ∂y 
∂  ∂u ∂v 
µ∇ 2 v + (λ + µ )  +  + By = 0
∂y  ∂x ∂y 
Stress
∇ 2 (σ xx + σ
yy
)= −
1
1−υ
 ∂B x ∂B y

+
∂
x
∂y

Plane stress
µ∇ 2u +
E
∂  ∂u ∂v 
 +  + Bx = 0
2(1 − υ ) ∂x  ∂x ∂y 
µ∇ 2 v +
E
∂  ∂u ∂v 
 +  + By = 0
2(1 − υ ) ∂y  ∂x ∂y 
 2
 ∂B x ∂B x 
 ∇ σ xx + σ yy = − (1 + υ )
+

∂
x
∂x 


(
)
41
Ramadas Chennamsetti
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Formulation
Summary – Plane stress and strain
R&DE (Engineers), DRDO
No elastic constants appear in governing equation
Solving these equations – difficult – some text book
problems
42
Ramadas Chennamsetti
rd_mech@yahoo.co.in
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
In absence of body loads – in stress formulation –
governing equations of plane stress and strain are
same
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
function’ (φ)
Aim – reduce the governing equations from three to
one
Find the distribution of single variable ‘φ’ – get
stresses from that
43
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Airy’s stress function
R&DE (Engineers), DRDO
Airy’s stress function (ASF) defined for stress
based formulations – both plane stress and
plane strain
Basic stress equilibrium equations –
Body loads can be derived
from potential function ‘V’
Defining potential
B = −∇ V => Bi = V,i
~
∂V
∂V
Bx = − , By = −
∂x
∂y
Substitute in equilibrium
equations
Ramadas Chennamsetti
44
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∂ τ xy
∂ σ xx
+
+ Bx = 0
∂x
∂y
∂ τ xy
∂ σ xx
+
+ By = 0
∂x
∂y
Airy’s stress function
R&DE (Engineers), DRDO
Equilibrium equations become
∂τ xy
∂
(σ xx − V ) +
= 0;
∂x
∂y
∂τ xy
∂
+ (σ yy − V ) = 0
∂x ∂y
Defining ASF as
τ xy
∂ 2φ
=−
;
∂x∂y
σ yy
∂ 2φ
−V =
∂x 2
ASF, φ = φ(x, y)
ASF satisfies both equilibrium equations.
In stress based formulation – it has to satisfy one more
governing equation
Ramadas Chennamsetti
45
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σ xx
∂ 2φ
−V =
;
2
∂y
Airy’s stress function
R&DE (Engineers), DRDO
In plane strain – stress based formulation –
governing equation
1  ∂Bx ∂B y 


∇ (σ xx + σ yy ) = −
+
1 − υ  ∂x
∂y 
∂ 2φ  
∂ 2φ  
1  ∂  ∂V  ∂  ∂V

 −
 + V +
 = −
+
 +  −


∂y  
∂x  
1 − υ  ∂x  ∂x  ∂y  ∂y
2

∇  V

2

 

 ∂ 2φ ∂ 2φ 
1
 + 2∇ 2V =
∇ 
+
∇ 2V
∂y 
1−υ
 ∂x
 1

2
2
∇ .∇ φ = 
− 2 ∇ 2V
 1 −υ

 1 − 2υ  2
Bi-harmonic operator ∇ 4φ = −
∇ V
 1−υ 
Ramadas Chennamsetti
46
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2
Airy’s stress function
R&DE (Engineers), DRDO
In plane stress – stress based formulation –
governing equation
 ∂Bx ∂By 

∇ (σ xx + σ yy ) = −(1 + υ )
+
∂y 
 ∂x
2
2
2

φ
φ
∂
∂
2
 + 2∇ 2V = (1 + υ )∇ 2V
∇ 
+
 ∂x ∂y 
∇ 2 .∇ 2φ = (1 + υ − 2)∇ 2V
∇ 4φ = −(1 − υ )∇ 2V
Ramadas Chennamsetti
47
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2
2

 ∂  ∂V  ∂  ∂V  




∂
φ
∂
φ
2

  = −(1 + υ )  −
 + V +


∇  V +
+
−





∂
∂
∂
∂
∂
∂
y
x
x
x
y
y












Airy’s stress function
R&DE (Engineers), DRDO
Governing equations in terms of ASF
 1 − 2υ  2
∇ φ = −
∇ V
 1−υ 
∇ 4φ = − (1 − υ )∇ 2V
4
Plane strain
Plane stress
4
4
4
∂
φ
φ
φ
∂
∂
4
∇ φ = 0 => 4 + 2 2 2 + 4 = 0
∂x
∂x ∂y
∂y
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
Ramadas Chennamsetti
48
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If body forces are neglected – V = 0 – In both cases the governing
equation is same
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
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
49
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Uniaxial tension – ASF approach
R&DE (Engineers), DRDO
A bar subjected to tension
y
2
2l
T
x
Boundary – 1
Boundary – 2
DCs => (1, 0, 0)
DCs => (-1, 0, 0)
Tx = T = lσxx + m τxy = σxx = T
Tx = -T = lσxx + m τxy = σxx = T
Ty = 0 = mσyy + l τxy => τxy = 0
Ty = 0 = mσyy + l τxy => τxy = 0
Remaining boundaries – stress free.
Stress in x – direction = T = constant
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50
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T
2c
1
Uniaxial tension – ASF approach
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Stress in x – direction given by ASF
At x = ± l,
σ
σyy and τxy = 0
υ
∂u
T
εx =
− σy =
=
- (1)
E
E
∂x
E
σy υ
∂v
υT
− σx =
= −
- (2)
εy =
E
E
∂y
E
x
Integrate (1) and (2)
T
x + f (y)
E
υT
v = −
y + g (x)
RamadasEChennamsetti
u =
f(y) and g(x) –
arbitrary functions
51
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σ xx
∂ 2φ
1 2
= constant = 2 = T => φ = Ty + c1 y + c2
∂y
2
Uniaxial tension – ASF approach
R&DE (Engineers), DRDO
Shear strain,
γ
xy
τ xy
∂u
∂v
=
+
=
= 0
∂y
∂x
G
=> f ' ( y ) + g ' ( x ) = 0
∂f
=>
= − ω o => f = − ω o y + u o
∂y
∂g
= ω o => g = ω o x + v o
∂x
ωo, uo, vo – arbitrary constants of integration.
Ramadas Chennamsetti
52
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=> − f ' ( y ) = g ' ( x ) = ω o
Uniaxial tension – ASF approach
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 straindisplacement 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 53
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Pure bending of beam – ASF approach
R&DE (Engineers), DRDO
Beam subjected to pure bending –
y
M
M
2c
x
2l
(x, ± c) = 0
τ
xy
(x, ± c) = 0
τ
xy
(± l, y ) = 0
Boundary
Conditions
yy
Fx =
+c
∫σ
xx
( ± l , y ) dy = 0
−c
Statically equivalent
boundary condition
+c
σ
∫
Ramadas Chennamsetti
− M
=
xx
−c
( ± l , y ) ydy
54
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σ
Pure bending of beam – ASF approach
R&DE (Engineers), DRDO
In bending, variation of bending stress is linear
along ‘y’ co-ordinate
‘φ’ should be a cubic polynomial
55
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Key point in ASF approach – selection of an
appropriate stress function - φ
It has to satisfy all boundary conditions –
capture physics of the problem
Some knowledge about structure’s behavior
helps in selection of ‘φ’
Pure bending of beam – ASF approach
R&DE (Engineers), DRDO
Linear variation of bending stress
∂ 2φ
σ xx = Ay = 2
Integrating the function
∂y
A 3
φ = y + C1 y + C 2 ASF satisfies all BCs
6
+c
∫ σ (± l , y ) ydy = − M
xx
+c
+c
y 
3M
∫− c Ay dy = − M => − A 3  = M => A = − 2 c 3
−c
3
2
3 M y3
M 3
φ=−
+ C1 y + C 2 = − 3 y + C1 y + C 2
3
2 c 6
4c
Ramadas Chennamsetti
56
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−c
Pure bending of beam – ASF approach
R&DE (Engineers), DRDO
Stresses and strains
M 3
∂ 2φ
3M
φ = − 3 y + C1 y + C2 => σ xx = 2 = − 3 y
4c
∂y
2c
σ yy = 0, τ xy = 0
σ yy
∂u
3 M
ε xx =
y
−υ
=>
=−
3
E
E
∂x
2 Ec
Integrate strain –
displacement
σ yy
σ xx 3 Mυ
ε yy =
−υ
=
y
equations
3
E
E 2 Ec
u=−
3M
xy + f ( y );
3
2 Ec
v=
3M υ 2
y + g ( x)
3
4 Ec
Ramadas Chennamsetti
57
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σ xx
Pure bending of beam – ASF approach
R&DE (Engineers), DRDO
∂u ∂v
3M
'
'
γ xy =
+
= 0 => −
x
+
f
(
y
)
+
g
( x) = 0
3
∂y ∂x
2 Ec
3M
'
'
=> −
x
+
g
(
x
)
=
−
f
( y ) = ω o Integrate
3
2 Ec
3M x 2
g ( x) =
+ ω o x + vo
3
2 Ec 2
f ( y ) = −ω o y + u o
Functions g(x) and f(y) represent rigid body motions
To evaluate constants, ωo, uo, vo – constraints required
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58
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Shear strain
59
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R&DE (Engineers), DRDO