2-D formulation Plane theory of elasticity

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