MCEN 4173/5173
Chapter 7
Linear Elasticity and Energy Method
Fall, 2006
1
Linear Elasticity
Linear elasticity is the most common problem in a variety of
engineering applications
Machine Design
Bio-Mechanics
2
Linear Elasticity
Linear Elasticity
What is linear elasticity about?
P
P
deformed
undeformed
X2
Question:
If we apply a force on a
material, what are the
stresses, strains and
displacements?
X2
X1
X1
Object: Linear Elastic Body (Machine elements; Human hard tissue……)
Input
Boundary conditions
(Applied force; Applied
displacement …)
Output
Stresses, strains,
displacements, at each
material point (x1,x2,x3)
???
3
Linear Elasticity
σ 33
Stresses:
σ 32
σ 31
⎡σ 11 σ 12 σ 13 ⎤
σ = ⎢⎢σ 21 σ 22 σ 23 ⎥⎥
⎢⎣σ 31 σ 32 σ 33 ⎥⎦
σ 23
σ 13
σ 12
X3
σ 21
σ 22
σ 11
X2
X1
Normal Stresses:
σ 11 σ 22 σ 33
Shear Stresses:
σ 12 σ 21 σ 23 σ 32 σ 13 σ 31
σ ij = σ ji
6 independent stress components
4
Linear Elasticity
Strains:
γ
L
ΔL
εx =
L
Normal strain
ΔL
γ xy = γ
Shear strain
5
Linear Elasticity
(x1 , x2 , x3 ) ⇔ (x, y, z )
(u1 , u2 , u3 ) ⇔ (u, v, w)
Strains:
Normal Strains
∂u1 ∂u
ε 11 =
=
∂x1 ∂x
ε 22
∂u 2 ∂v
=
=
∂x2 ∂y
∂u3 ∂w
ε 33 =
=
∂x3 ∂z
Shear Strains
∂u1 ∂u2 ∂u ∂v
2ε 12 = γ xy =
+
=
+
∂x2 ∂x1 ∂y ∂x
∂u2 ∂u3 ∂v ∂w
2ε 23 = γ yz =
+
= +
∂x3 ∂x2 ∂z ∂y
∂u1 ∂u3 ∂u ∂w
2ε 13 = γ xz =
+
=
+
∂x3 ∂x1 ∂z ∂x
6
Linear Elasticity
Strains (kinematics) :
1 ⎛⎜ ∂ui ∂u j ⎞⎟
ε ij = ⎜
+
2 ⎝ ∂x j ∂xi ⎟⎠
∂u1 ∂u
=
i=j=1 ε 11 =
∂x1 ∂x
i=1, j=2
∂u 2 ∂v
=
=
∂x2 ∂y
i=2, j=3
∂u3 ∂w
=
i=j=3 ε 33 =
∂x3 ∂z
i=1, j=3
i=j=2 ε 22
A letter in a subscript is an index,
such as i, j, k, …
In 3D solid mechanics problems,
an index runs from 1 to 3.
In 2D solid mechanics problems,
an index runs from 1 to 2.
1 ⎛ ∂u1 ∂u2 ⎞
⎟⎟
ε 12 = ⎜⎜
+
2 ⎝ ∂x2 ∂x1 ⎠
1 ⎛ ∂u2 ∂u3 ⎞
⎟⎟
+
ε 23 = ⎜⎜
2 ⎝ ∂x3 ∂x2 ⎠
1 ⎛ ∂u1 ∂u3 ⎞
⎟⎟
+
ε 13 = ⎜⎜
2 ⎝ ∂x3 ∂x1 ⎠
7
Linear Elasticity
Strains (kinematics) :
u1 = 1 + 2 x1 + 3 x2 + 4 x1 x2
u2 = 5 + 6 x1 + 7 x2 + 8 x1 x2
u3 = 0
8
Linear Elasticity
Strains (kinematics) :
∂u1
= 2 + 4 x2
ε 11 =
∂x1
u1 = 1 + 2 x1 + 3 x2 + 4 x1 x2
6
2
1.8
5
1.6
4
1.4
3
1.2
1
2
0.8
0.6
1
S10
0
S10
0.4
S7
S7
0.2
S4
-1
1
2
3
4
5
6
7
8
9
S1
10 11
2nd order polynomial function
S4
0
1
2
3
4
5
6
7
8
S1
9
10 11
Linear function
9
Linear Elasticity
Constitutive equations (stress-strain relations)
1
e11 = [σ 11 −ν (σ 22 + σ 33 )]
E
1
e22 = [σ 22 − ν (σ 11 + σ 33 )]
E
1
e33 = [σ 33 − ν (σ 11 + σ 22 )]
E
1
e12 =
σ 12
2G
1
e23 =
σ 23
2G
1
e13 =
σ 13
2G
1
⎞
⎛
⎜ γ xy = τ xy ⎟
G ⎠
⎝
1
⎞
⎛
⎜ γ yz = τ yz ⎟
G ⎠
⎝
1 ⎞
⎛
⎜ γ xz = τ xz ⎟
G ⎠
⎝
1 +ν
ν
σ ij − (σ 11 + σ 22 + σ 33 )δ ij
eij =
E
E
⎧0 i ≠ j
δ ij = ⎨
⎩1 i = j
σ ij = 2Geij + λ (e11 + e22 + e33 )δ ij
λ=
νE
(1 +ν )(1 − 2ν )
10
Linear Elasticity
Equilibrium Equations
f
d X2
d X1
X2
X1
11
Linear Elasticity
Equilibrium Equations
∂σ 11 ∂σ 12 ∂σ 13
+
+
+ f1 = 0
∂X 1 ∂X 2 ∂X 3
∂σ 21 ∂σ 22 ∂σ 23
+
+ f2 = 0
+
∂X 1 ∂X 2 ∂X 3
∂σ 31 ∂σ 32 ∂σ 33
+
+
+ f3 = 0
∂X 1 ∂X 2 ∂X 3
σ 12 = σ 21 σ 13 = σ 31 σ 23 = σ 32
If we denote ∂ (•) = (•) ,i
∂X i
σ ij , j + f i = 0
12
Linear Elasticity
Summation Convention:
When an index repeats itself in one term, we mean it is a summation.
3
ai bi = ∑ ai bi = a1b1 + a2b2 + a3b3
i =1
3
σ ij , j = ∑ σ ij , j = σ i1,1 + σ i 2, 2 + σ i 3,3
j =1
13
Linear Elasticity
Summary of Linear Elasticity
Kinematics:
Constitutive:
eij =
eij =
1
(ui, j + u j ,i )
2
ν
1 +ν
σ ij − σ kk δ ij
E
E
6 equations
σ ij = 2Geij + λekk δ ij 6 equations
σ ij , j + f i = 0
Equilibrium:
3 equations
Totally: 15 equations
Unknowns:
Stresses:
σ 11 σ 22 σ 33 σ 12 σ 21 σ 23
Strains:
ε 11 ε 22 ε 33 ε 12 ε 21 ε 23
Displacements:
u1 u 2 u3
Totally: 15 unknowns
14
Linear Elasticity
Good news:
We have 15 unknowns, and we have 15 equations. So with proper
boundary conditions, we have a unique (one, and only one) solution
to this problem.
Bad news:
How to get the solution?
15
Linear Elasticity
u1 u2 u3
Start with displacements,
eij =
1
(ui, j + u j ,i )
2
Strains in terms of displacements
σ ij = 2Geij + λekk δ ij
Stresses in terms of displacements
σ ij , j + f i = 0
Equilibrium equations in terms of displacements
σ (u1 , u2 , u3 )ij , j + f i = 0
These equations are second order partial differential equations!!
16
Linear Elasticity
P
deformed
eij =
1
(ui, j + u j ,i )
2
σ ij = 2Geij + λekk δ ij
σ ij , j + f i = 0
X2
Boundary conditions
X1
What we mean by saying a solution to a linear elasticity problem?
A solution that satisfies above equations and boundary conditions at
EACH material point.
This is a very strong requirement!!
17
Linear Elasticity
Since finding a precise solution is too difficult, we step back a little.
σ (u1 , u2 , u3 )ij , j + f i = 0
P
deformed
In stead of asking above equations are
satisfied at each material point, we ask the
above equations to be satisfied in the sense
of integral, i.e.
X2
Ω
X1
This is a big release, because
18
Linear Elasticity
Weighted Residual Method
[∫ σ (u~ , u~ , u~ )
1
2
3 ij , j
]
+ f i dV = 0
V
19
Linear Elasticity
Weighted Residual Method
By choosing different weight functions, we obtained different numerical
approximation methods.
Point-Collocation Method
Subdomain Collocation Method
Method of Moment
Least Squares Methods Method
Galerkin Method
20
Linear Elasticity
Weighted Residual Method
Example:
d 2u
+u + x = 0
2
dx
0 ≤ x ≤1
Boundary conditions
x=0
u=0
Accurate solution:
x=1
u=0
sin x
u=
−x
sin 1
Assuming an approximate solution is:
u = x(1 − x)(a1 + a2 x + a3 x 2 + L)
1
Weighted Residual is:
∫ W Rdx = 0
0
i
21
Linear Elasticity
Weighted Residual Method: example
d 2u
+u + x = 0
2
dx
0 ≤ x ≤1
First order approximation:
n=1
u = a1 x(1 − x)
Second order approximation
n=2
u = x(1 − x)(a1 + a2 x)
22
Linear Elasticity
Weighted Residual Method: example
Point collocation method:
Allocate several points within the domain
(0 ≤ x ≤ 1)
R( xi ) = 0
n=1, we choose the mid-point
x = 1/ 2
(
)
R1 ( x = 1 / 2 ) = x + a1 − 2 + x − x 2 = 0
2
a1 =
7
1 7
− a1 = 0
2 4
n=2, we choose
x = 1/ 3
2
u1 = x(1 − x)
7
x = 2/3
1 16
2
R2 ( x = 1 / 3) = − a1 + a2 = 0
3 9
27
2 16
50
R2 ( x = 2 / 3) = − a1 − a2 = 0
3 9
27
a1 = 0.1948
a2 = 0.1731
u 2 = x(1 − x)(0.1948 + 0.1731x )
23
Linear Elasticity
Weighted Residual Method: example
n
Galerkin method
u = ∑ N i ai
Wi = N i
i =1
First order approximation:
W1 = N1 = x(1 − x )
u = a1 x(1 − x)
n=1
(
R1 = x + a1 − 2 + x − x 2
(
(
)
))
2
(
)
W
Rdx
=
0
=
x
1
−
x
x
+
a
−
2
+
x
−
x
dx
1
∫ 1
∫
1
1
0
0
a1 =
5
= 0.2727
18
u = 0.2727 x(1 − x)
24
Linear Elasticity
Weighted Residual Method: example
n
Galerkin method
u = ∑ N i ai
Wi = N i
i =1
Second order approximation:
W1 = N1 = x(1 − x )
u = x(1 − x)(a1 + a2 x)
W2 = N 2 = x 2 (1 − x )
(
)
(
R = x + a1 − 2 + x − x 2 + a2 2 − 6 x + x 2 − x 3
)
[ (
) (
∫ W Rdx = ∫ x (1 − x )[x + a (− 2 + x − x ) + a (2 − 6 x + x
)]
− x )]dx = 0
2
2
3
(
)
W
Rdx
=
x
1
−
x
x
+
a
−
2
+
x
−
x
+
a
2
−
6
x
+
x
−
x
dx = 0
1
2
∫ 1
∫
1
1
0
0
1
0
1
2
0
2
2
1
2
2
3
u = x(1 − x)(0.1924 + 1707 x)
25
Linear Elasticity
Weighted Residual Method: example
Errors
Solution
u=
x=0.25
sin x
−x
sin 1
Point coll. n=1
2
u1 = x(1 − x)
7
Galerkin, n=1
u1 = 0.2727 x(1 − x)
Error
%
0.04401
x=0.5
Error
%
0.06975
x=0.75
Error
%
0.06006
0.05355
21.7
0.07143
2.4
0.05357
-10.8
0.05208
18.3
0.06944
-0.4
0.05208
-13.3
0.04464
1.4
0.07034
0.8
0.06087
1.3
0.04408
0.2
0.06944
-0.4
0.06008
0.03
Point coll. n=2
u 2 = x(1 − x)(0.1948 + 0.1731x )
Galerkin, n=1
u = x(1 − x)(0.1924 + 1707 x)
26
Linear Elasticity
Principle of Virtual Displacements
Concepts
Admissible
Admissible displacements
P
A set of displacements that satisfy the
geometric constraints.
S P1
Virtual displacements
S P2
A set of small displacement variations
upon which the geometric constraints are
satisfied.
δui
dui
t
u
Su
X2
Β
X1
Not Admissible
27
Linear Elasticity
Principle of Virtual Displacements
Some math:
a, j b = (ab ), j − ab, j
(ab ), j = a, j b + ab, j
Green Theorems:
∫ a dB =∫
B
,i
∂B
a ni dS
28
Linear Elasticity
Principle of Virtual Displacements
σ ij , j + f i = 0
B.C.:
σ ij n j − Ti = 0
P
in B
S P1
on ∂B
T
Now, we want to use weighted residual
method to obtain the integral form of
equilibrium equations.
We use the virtual displacements δ ui as
weight function
∫ (σ
B
ij , j
u
Su
X2
Β
X1
+ f i )δui dV − ∫ (σ ij n j − Ti )δui dS = 0
∂B
29
Linear Elasticity
Principle of Virtual Displacements
∫ (σ
B
ij , j
+ f i )δui dV − ∫ (σ ij n j − Ti )δui dS = 0
∂B
30
Linear Elasticity
Principle of Virtual Displacements
∫ σ δu
B
ij
i, j
dV
31
Linear Elasticity
Principle of Virtual Displacements
∫B σ ij , jδui dV = ∫∂B (σ ijδui )n j dS − ∫B σ ijδeij dV
∫ (σ
B
ij , j
+ f i )δui dV − ∫ (σ ij n j − Ti )δui dS = 0
∂B
∫B (σ ijδeij − f iδui )dV = ∫∂B Tiδui dS
Internal work on virtual
displacements
External work on virtual
displacements
Principle of virtual displacement:
The work done by external force on virtual displacements is equal
to the work done by the internal force on virtual displacements.
32
Linear Elasticity
Principle of Minimum Potential Energy
General form of linear elastic constitutive equations
σ ij = 2Geij + λekk δ ij
σ ij = Dijkl ekl
Dijkl = Dklij
From principle of virtual displacements
∫B (σ ijδeij − f iδui )dV − ∫∂B Tiδui dS = 0
33
Linear Elasticity
Principle of Minimum Potential Energy
⎡ ⎛1
⎤
⎞
∫B ⎢⎣δ ⎜⎝ 2 Dijkl ekl eij ⎟⎠ − f iδui ⎥⎦ dV − ∫∂B Tiδui dS = 0
( )
1
Dijkl ekl eij = U eij = U (ui )
2
Strain energy density
Stored energy density
Stress
Strain
34
Linear Elasticity
Principle of Minimum Potential Energy
Conservative force: If a force acting on an object is a function of position
only, it is said to be a conservative force, and it can be
represented by a potential energy function.
dV
Fi = −
dui
V is a potential
Fi dui = −dV
Examples of conservative forces:
Examples of non-conservative forces:
In linear elasticity, we only consider conservative forces.
35
Linear Elasticity
Principle of Minimum Potential Energy
∫ [δU (u ) − f δu ]dV − ∫
B
We say both fi and
i
i
i
∂B
Tiδui dS = 0
Ti are conservative forces.
− f iδui = δΦ (ui )
− Tiδui = δΨ (ui )
36
Linear Elasticity
Principle of Minimum Potential Energy
System potential energy
Π p = ∫ [U (ui ) + Φ(ui )]dV + ∫ Ψ (ui )dS
∂B
B
δΠ p = 0
Among all the possible displacements, the true solution (real
displacements) results in the smallest system potential energy.
In other words, any approximate solutions give larger
system potential energy than the real solution does.
⎡1
⎤
Π p = ∫ ⎢ Dijkl eij ekl − f i ui ⎥ dV − ∫ Ti ui dS
B 2
∂B
⎣
⎦
37