1. The document discusses the 2-D formulation of plane theory of elasticity, including plane strain and plane stress. It introduces the field equations, compatibility equations, and equilibrium equations for both plane strain and plane stress. 2. It describes the Airy stress function method, which reduces the general plane problem to a single biharmonic equation in terms of the stress function. 3. It provides examples of using polynomial solutions to solve 2-D problems, including using a quadratic Airy stress function to model pure bending of a beam and comparing to the beam theory solution.

1. Lecture 7
2-D formulation
Plane theory of elasticity
Print version Lecture on Theory of Elasticity and Plasticity of
Dr. D. Dinev, Department of Structural Mechanics, UACEG
7.1
Contents
1 Plane strain 1
2 Plane stress 3
3 Plane strain vs. plane stress 5
4 Airy stress function 5
5 Polynomial solution of 2-D problem 7
6 General solution strategy 14 7.2
1 Plane strain
Plane strain
Introduction
• Because of the complexity of the field equations analytical closed-form solutions to full
3-D problems are very difficult to accomplish
• A lot of problems into the area of engineering can be approximated by 1-D or 2-D strain
or stress state
– rods, beams, columns, shafts etc.
– Retaining walls, disks, plates, shells
7.3
Plane strain
1

2. Problem definition
• Consider an infinitely long prismatic body
• If the body forces and surface tractions have no components on z-direction the deformation
field can be reduced into
u = u(x,y)
v = v(x,y)
w = 0
• This deformation is called as a state of plane strain in the (x,y)-plane
• Thus all cross-sections will have same displacements
7.4
Plane strain
Field equations
• The strain-displacement relations become
εxx =
∂u
∂x
, εyy =
∂v
∂y
, εxy =
1
2
∂u
∂y
+
∂v
∂x
εzz = εxz = εyz = 0
• In matrix form


εxx
εyy
2εxy

 =



∂
∂x 0
0 ∂
∂y
∂
∂y
∂
∂x



u
v
• The St.-Venant’s compatibility equation is
∂2εxx
∂y2
+
∂2εyy
∂x2
= 2
∂2εxy
∂x∂y
7.5
Plane strain
Field equations
• The stress-strain relations are
σxx = (λ +2µ)εxx +λεyy
σyy = λεxx +(λ +2µ)εyy
σzz = λεxx +λεyy
σxy = µ2εxy
σxz = σyz = 0
2

3. • In matrix form




σxx
σyy
σzz
σxy



 =




λ +2µ λ 0
λ λ +2µ 0
λ λ 0
0 0 2µ






εxx
εyy
εxy


7.6
Plane strain
Field equations
• The equilibrium equations are reduced to
∂σxx
∂x
+
∂σxy
∂y
+ fx = 0
∂σxy
∂x
+
∂σyy
∂y
+ fy = 0
• In matrix form
σxx σxy
σxy σyy
∂
∂x
∂
∂y
+
fx
fy
=
0
0
7.7
Plane strain
Field equations
• The Navier’s displacement equilibrium equations are
µ∇2
u+(λ + µ)
∂
∂x
∂u
∂x
+
∂v
∂y
+ fx = 0
µ∇2
v+(λ + µ)
∂
∂y
∂u
∂x
+
∂v
∂y
+ fy = 0
where ∇2 = ∂2
∂x2 + ∂2
∂y2 - Laplacian operator
• The Beltrami-Michell stress equation is
∇2
(σxx +σyy) = −
1
1−ν
∂ fx
∂x
+
∂ fy
∂y
• The surface tractions (stress BCs)are
tx
ty
=
σxx σxy
σxy σyy
nx
ny
7.8
2 Plane stress
Plane stress
Problem definition
3

4. • Consider an arbitrary disc which thickness is small in comparison to other dimensions
• Assume that there is no body forces and surface tractions in z-directions and the surface of
the disc is stress free
• Thus imply a stress field
σxx = σxx(x,y)
σyy = σyy(x,y)
σxy = σxy(x,y)
σzz = σxz = σyz = 0
7.9
Plane stress
Field equations
• The Hooke’s law




εxx
εyy
εzz
εxy



 =
1
E




1 −ν 0
−ν 1 0
−ν −ν 0
0 0 1+ν






σxx
σyy
σxy


• Relation between normal strains
εzz = −
ν
1−ν
(εxx +εyy)
7.10
Plane stress
Field equations
• Strain-displacement equations




εxx
εyy
εzz
2εxy



 =





∂
∂x 0 0
0 ∂
∂y 0
0 0 ∂
∂z
∂
∂y
∂
∂x 0







u
v
w


εyz = εzx = 0
• St.-Venant’s compatibility equation is
∂2εxx
∂y2
+
∂2εyy
∂x2
= 2
∂2εxy
∂x∂y
7.11
Plane stress
Field equations
• Equilibrium equations - same as in plane strain
σxx σxy
σxy σyy
∂
∂x
∂
∂y
+
fx
fy
=
0
0
7.12
4

5. E ν
Plane
stress to strain E
1−ν2
ν
1−ν
Plane
strain to stress E(1+2ν)
(1+ν)2
ν
1+ν
Plane stress
Field equations
• The Navier’s displacement equilibrium equations are
µ∇2
u+
E
2(1−ν)
∂
∂x
∂u
∂x
+
∂v
∂y
+ fx = 0
µ∇2
v+
E
2(1−ν)
∂
∂y
∂u
∂x
+
∂v
∂y
+ fy = 0
• The Beltrami-Michell stress equation is
∇2
(σxx +σyy) = −(1+ν)
∂ fx
∂x
+
∂ fy
∂y
• The surface tractions (stress BCs)are
tx
ty
=
σxx σxy
σxy σyy
nx
ny
7.13
3 Plane strain vs. plane stress
Plane strain vs. plane stress
Summary
• The plane problems have identical equilibrium equations, BCs and compatibility equa-
tions
• The similar equations show that the differences are due to different constants involving
different material constants
• The field equations of plane stress can be obtained from equations of plane strain by fol-
lowing substitution
• When ν = 0 plane strain ≡ plane stress
7.14
4 Airy stress function
Airy stress function
The Method
• A popular method for the solution of the plane problem is using the so called Stress func-
tions
• It employs the Airy stress function and reduce the general formulation to a single equation
in terms of a single unknown
• The general idea is to develop a stress field that satisfies equilibrium and yields a single
governing equation from the compatibility equations.
• The obtained equilibrium equation ca be solved analytically in closed-form
7.15
5

6. Airy stress function
The Method
• Assume that the body forces are zero
• The Beltrami-Michell stress compatibility equations are
∇2
(σxx +σyy) = 0
• Equilibrium equations are
∂σxx
∂x
+
∂σxy
∂y
= 0
∂σxy
∂x
+
∂σyy
∂y
= 0
• The stress BCs are
tx
ty
=
σxx σxy
σxy σyy
nx
ny
7.16
Airy stress function
The Method
• The Beltrami-Michell equation can be expanded as
∂2σxx
∂x2
+
∂2σyy
∂x2
+
∂2σxx
∂y2
+
∂2σyy
∂y2
= 0
• The equilibrium equations are satisfied if we choose the representation
σxx =
∂2φ
∂y2
, σyy =
∂2φ
∂x2
, σxy = −
∂2φ
∂x∂y
where φ = φ(x,y) is an arbitrary form called an Airy stress function
• Substitution of the above expressions into the Beltrami-Michell equations lied to
∂4φ
∂x4
+2
∂4φ
∂x2∂y2
+
∂4φ
∂y4
= 0
7.17
Airy stress function
The Method
• George Biddell Airy (1801-1892)
7.18
6

7. Airy stress function
The Method
• The previous expression is a biharmonic equation. In short notation
∇2
∇2
φ(x,y) = 0
• Thus all equations of the plane problem has been reduced to a single equation in terms of
the Airy stress function φ(x,y).
• This function is to be determined in the 2-D region R bounded by the boundary S
• Appropriate BCs are necessary to complete a solution
7.19
5 Polynomial solution of 2-D problem
Polynomial solution of 2-D problem
The Method
• The solution with polynomials is applicable in Cartesian coordinates and useful for prob-
lems with rectangular domains
• Based on the inverse solution concepts - we assume a form of the solution of the equation
∇2∇2φ(x,y) = 0 and then try to determine which problem may be solved by this solution
7.20
Polynomial solution of 2-D problem
The Method
• The assumed solution is taken to be a general polynomial and can be expressed in the
power series
φ(x,y) =
∞
∑
m=0
∞
∑
n=0
Cmnxm
yn
where Cmn are constants to be determined
• The method produces a polynomial stress distribution and not satisfies the general BCs
• We need to modify the BCs using St.-Venant principle- with statically equivalent BCs
• The solution would be accurate at points sufficiently far away from the modified boundary
7.21
Polynomial solution of 2-D problem
Example 1
• Let’s use a trial solution- first order polynomial
φ(x,y) = C1x+C2y+C3
• The solution satisfies the biharmonic equation
• Go to the stress field
σxx =
∂2φ
∂y2
= 0, σyy =
∂2φ
∂x2
= 0, σxy = −
∂2φ
∂x∂y
= 0
Question
• What this solution mean?
7.22
7

8. Polynomial solution of 2-D problem
Example 2
• Use a higher order polynomial
φ(x,y) = C1y2
• The solution also satisfies the biharmonic equation
• The stress field
σxx =
∂2φ
∂y2
= 2C1, σyy =
∂2φ
∂x2
= 0, σxy = −
∂2φ
∂x∂y
= 0
7.23
Polynomial solution of 2-D problem
Example 2
• The solution fits with the uniaxial tension of a disc
• The boundary conditions are
σxx(± ,y) = t
σyy(x,±h) = 0
σxy(± ,y) = σxy(x,±h) = 0
• The constant C1 can be obtained from the BCs
7.24
Polynomial solution of 2-D problem
Example 3
• Pure bending of a beam - a comparison with the MoM solution
7.25
Polynomial solution of 2-D problem
Review of the beam theory
• á la Speedy Gonzales
• Assumptions
– Long beam- h
– Small displacements- u h and v h
– Small strains- εxx 1
– Bernoulli hypothesis- εyy ≈ 0 and σxy ≈ 0
7.26
8

9. Polynomial solution of 2-D problem
Review of the beam theory
• Displacement field
u(x,y,z) = yθ
v(x,y,z) = v(x,y)
• Because γxy = 0, thus
θ = −
∂v
∂x
• The final displacements are
u(x,y,z) = −y
∂v
∂x
v(x,y,z) = v(x,y)
7.27
Polynomial solution of 2-D problem
Review of the beam theory
• Strain field
εxx = −y
∂2v
∂x2
• Compatibility equation
1
r
= κ =
dθ
ds
≈
dθ
dx
=
d2v
dx2
• The strain field can be expressed
εxx = −yκ
• Hooke’s law
σxx = Eεxx = −yEκ
7.28
Polynomial solution of 2-D problem
Review of the beam theory
9

10. • Bending moment
M =
A
σxxydA = −EIκ
• Equilibrium equations
dV
dx
= −q
dM
dx
= −V
7.29
Polynomial solution of 2-D problem
Review of the beam theory
• Differential equation
EI
d4v
dx4
= q
• 4-th order ODE- needs of four BCs
7.30
Polynomial solution of 2-D problem
Example 3- MoM solution
• MoM solution
7.31
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Elasticity solution
7.32
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Strong BCs
σyy(x,±c) = 0, σxy(x,±c) = 0, σxy(± ,y) = 0
7.33
10

11. Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Weak BCs- imposed in a weak form (using the St.-Venant principle)
c
−c
σxx(± ,y)dy = 0,
c
−c
σxx(± ,y)ydy = −M
7.34
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Based on the MoM solution (linear σxx distribution) we try a following solution
φ(x,y) = A1y3
• The function satisfies ∇4φ(x,y) = 0
• The stress functions are
σxx = 6A1y
σyy = 0 → satisfies σyy(x,±c) = 0
σxy = 0 → satisfies σxy(x,±c) = 0, σxy(± ,y) = 0
• This trial solution fits with the BCs
7.35
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The constant A1 is obtained from the weak BC at x = ±
c
−c
σxx(± ,y)dy ≡ 0
c
−c
σxx(± ,y)ydy = 4c3
A1 = −M, → A1 = −
M
4c3
• Thus the Airy stress function is
φ(x,y) = −
M
4c3
y3
• Corresponding stresses are
σxx = −
3M
2c3
y
σyy = 0
σxy = 0
7.36
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Strain field-by the Hooke’s law
εxx = −
3M
2Ec3
y
εyy =
3Mν
2Ec3
y
εxy = 0
• Displacement field- by strain-displacement equations
u = −
3M
2Ec3
xy+ f(y)
v =
3Mν
4Ec3
y2
+g(x)
• The functions f(y) and g(x) have to be determined from the definition of the shear strain
7.37
11

12. Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The definition of the shear strain gives
εxy = −
3M
4Ec3
x+
1
2
∂ f(y)
∂y
+
1
2
∂g(x)
∂x
• This result can be compared with the shear strain obtained from the constitutive relations
εxy = 0
−
3M
4Ec3
x+
1
2
∂g(x)
∂x
+
1
2
∂ f(y)
∂y
= 0
7.38
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The equation can be partitioned into
∂ f(y)
∂y
= ω0
∂g(x)
∂x
=
3M
2Ec3
x+ω0
where ω0 is an arbitrary constant
• Integration of the above equation gives
f(y) = uo +yω0
g(x) =
3M
4c3E
x2
+xω0 +v0
• The constants u0, v0 and ω0 express the rigid-body motion of the beam
7.39
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The back substitution into the displacement field gives
u(x,y) = u0 +yω0 −
3M
2c3E
xy
v(x,y) = v0 +xω0 +
3M
4c3E
x2
+
3Mν
4c3E
y2
• The constants can be found from the essential BCs
7.40
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The essential BCs are concentrated at points of beam ends
u(− ,0) = 0 → u0 = 0
v(− ,0) = 0 →
3M 2
4c3E
+v0 − ω0 = 0
v( ,0) = 0 →
3M 2
4c3E
+v0 + ω0 = 0
7.41
12

13. Polynomial solution of 2-D problem
Example 3- Elasticity solution
• The constants are
u0 = 0
ω0 = 0
v0 = −
3M 2
4c3E
• Displacement field can be completed as
u(x,y) = −
3M
2c3E
xy
v(x,y) =
3M
4c3E
(x2
+νy2
− 2
)
7.42
Polynomial solution of 2-D problem
3 2 1 0 1 2 3
2
1
0
1
2
Example 3- Elasticity solution
• Vector plot of the displacement field
7.43
Polynomial solution of 2-D problem
Example 3- Elasticity solution
• Elasticity solution
u(x,y) = −
M
EI
xy
v(x,y) =
M
2EI
(x2
+νy2
− 2
)
• MoM solution
u(x) = −
M
EI
xy
v(x) =
M
2EI
(x2
− 2
)
where I = 2
3 c3
Note
It is convenient to use a computer algebra system for the mathematics (Maple, Mathematica etc.)
7.44
13

14. Polynomial solution of 2-D problem
Example 3
• General conclusion
7.45
6 General solution strategy
General solution strategy
Selection of the polynomial order
• Step 1- Determine the maximum order of polynomial using MoM arguments
Example 1
• Normal loading-q(x) → xn
• Shear force- V(x) → xn+1
• Bending moment- M(x) → xn+2
• Stress- σxx → xn+2y
• Airy function- xn+2y3
• Maximum order= n+5
7.46
General solution strategy
Example 2
• Shear loading-n(x) → xm
• Shear force- V(x) → xm
• Bending moment- M(x) → xm+1
• Stress- σxx → xm+1y
• Airy function- xm+1y3
• Maximum order= m+4
7.47
14

15. General solution strategy
Selection of the polynomial order
• Step 2- Write down a polynomial function φ(x,y) that contains all terms up to order
max(m+4,n+5)
φ(x,y) = C1x2
+C2xy+C3y2
+C4x3
+...
7.48
General solution strategy
Selection of the polynomial order
• May use the Pascal’s triangle for the polynomial
1
x y
x2 xy y2
x3 x2y xy2 y3
x4 x3y x2y2 xy3 y4
x5 x4y x3y2 x2y3 xy4 y5
• And constants
C1 C2 C3
C4 C5 C6 C7
C8 C9 C10 C11 C12
C13 C14 C15 C16 C17 C18
• The first three terms have no physical meaning (zero stress field)
7.49
General solution strategy
Selection of the polynomial order
• Step 3 Compatibility condition
∇2
∇2
φ(x,y) = 0
• Step 4 Boundary conditions- strong and weak. Lead to a set of equations for Ci
• Step 5 Solve all equations and determine Ci
Other types of solution
• Fourier series method
• ......
7.50
General solution strategy
The End
• Any questions, opinions, discussions?
7.51
15