# Linear Elasticity and Energy Method

Mechanics

Jul 18, 2012 (6 years ago)

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1
Linear Elasticity and Energy Method
MCEN 4173/5173
Chapter 7
Fall, 2006
2
Linear Elasticity
Linear elasticity is the most common problem in a variety of
engineering applications
Bio-MechanicsMachine Design
3
Linear Elasticity
Linear ElasticityWhat is linear elasticity about?
P
2
X
1
X
undeformed
2
X
1
X
deformed
P
Question:
If we apply a force on a
material, what are the
stresses, strains and
displacements?
Object: Linear Elastic Body (Machine elements; Human hard tissue……)
Boundary conditions
(Applied force; Applied
displacement …)
Input
Stresses, strains,
displacements, at each
material point (x1,x2,x3)
? ? ?
Output
4
Linear Elasticity
Stresses:

=
333231
232221
131211
σσσ
σσσ
σσσ
σ
2
X
1
X
3
X
22
σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ

σ
jiij
σ
σ
=
6 independent stress components
5
Linear Elasticity
Strains:
L
Δ
L
γ
L
L
x
Δ
=
ε
Normal strain
γ
γ
=
xy
Shear strain
6
Linear Elasticity
Strains:
(
)
(
)
wvuuuu,,,,
321

(
)
(
)
zyxxxx,,,,
321

x
u
x
u

=

=
1
1
11
ε
y
v
x
u

=

=
2
2
22
ε
z
w
x
u

=

=
3
3
33
ε
Normal StrainsShear Strains
x
v
y
u
x
u
x
u
xy

+

=

+

==
1
2
2
1
12
2
γε
y
w
z
v
x
u
x
u
yz

+

=

+

==
2
3
3
2
23
2
γε
x
w
z
u
x
u
x
u
xz

+

=

+

==
1
3
3
1
13
2
γε
7
Linear Elasticity

+

=
i
j
j
i
ij
x
u
x
u
2
1
ε
Linear Elasticity
Strains (kinematics) :
x
u
x
u

=

=
1
1
11
ε
y
v
x
u

=

=
2
2
22
ε
z
w
x
u

=

=
3
3
33
ε
i=j=1
i=j=2
i=j=3

+

=
1
2
2
1
12
2
1
x
u
x
u
ε

+

=
2
3
3
2
23
2
1
x
u
x
u
ε

+

=
1
3
3
1
13
2
1
x
u
x
u
ε
i=1, j=2
i=2, j=3
i=1, j=3
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.
8
Linear Elasticity
Strains (kinematics) :
21211
4321xxxxu
+
+
+
=
21212
8765xxxxu
+
+
+
=
0
3
=
u
9
Linear Elasticity
Strains (kinematics) :
1
2
3
4
5
6
7
8
9
10
11
S1
S4
S7
S10
-1
0
1
2
3
4
5
6
2
1
1
11
42x
x
u
+=

=
ε
21211
4321xxxxu
+
++=
1
2
3
4
5
6
7
8
9
10
11
S1
S4
S7
S10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2nd
order polynomial functionLinear function
10
Linear Elasticity
Constitutive equations (stress-strain relations)
()
[]
33221111
1
σσνσ
+−=
E
e
()
[]
33112222
1
σσνσ
+−=
E
e
()
[]
22113333
1
σσνσ
+−=
E
e

==
xyxy
GG
e
τγσ
1
2
1
1212

==
yzyz
GG
e
τγσ
1
2
1
2323

==
xzxz
GG
e
τγσ
1
2
1
1313
()
ijijij
EE
e
δσσσ
ν
σ
ν
332211
1
++−
+
=

=

=
ji
ji
ij
1
0
δ
ijijij
eeeGe
δ
λ
σ
)(2
332211
+
+
+
=
()()
νν
ν
λ
211−+
=
E
11
Linear Elasticity
Equilibrium Equations
2
X
1
X
1
Xd
2
Xd
f
12
Linear Elasticity
Equilibrium Equations
If we denote
0
0
0
3
3
33
2
32
1
31
2
3
23
2
22
1
21
1
3
13
2
12
1
11
=+

+

+

=+

+

+

=+

+

+

f
XXX
f
XXX
f
XXX
σσσ
σ
σσ
σ
σσ
(
)
()
i
i
X
,
•=

0
,
=
+
ijij
f
σ
2112
σ
σ
=
㌱ㄳ
σ
σ
=
㌲㈳
σ
σ
=
13
Linear Elasticity
Summation Convention:
When an index repeats itself in one term, we mean it is a summation.
332211
3
1
bababababa
i
iiii
++==

=
3,32,21,1
3
1
,,iii
j
jijjij
σσσσσ
++==

=
14
Linear Elasticity
Summary of Linear Elasticity
Kinematics:
(
)
ijjiij
uue
,,
2
1
+=
Constitutive:
ijkkijij
EE
e
δσ
ν
σ
ν

+
=
1
ijkkijij
eGe
δ
λ
σ
+
=2
Equilibrium:
0
,
=
+
ijij
f
σ
Totally: 15 equations
Unknowns:
Stresses:
11
σ

σ

σ

σ

σ

σ

ε

ε

ε

ε

ε

ε
䑩獰污捥浥湴猺
1
u
2
u
3
u
Totally: 15 unknowns
6 equations
6 equations
3 equations
15
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?
16
Linear Elasticity
Start with displacements,
1
u
2
u
3
u
(
)
ijjiij
uue
,,
2
1
+=
Strains in terms of displacements
ijkkijij
eGe
δ
λ
σ
+
=
2
Stresses in terms of displacements
0
,
=
+
ijij
f
σ
These equations are second order partial differential equations!!
Equilibrium equations in terms of displacements
(
)
0,,
,
321
=
+
i
jij
fuuu
σ
17
2
X
1
X
deformed
P
Linear Elasticity
()
ijjiij
uue
,,
2
1
+=
ijkkijij
eGe
δ
λ
σ
+
=2
0
,
=
+
ijij
f
σ
What we mean by saying a solution to a linear elasticity problem?
Boundary conditions
A solution that satisfies above equations and boundary conditions at
EACHmaterial point.
This is a very strong requirement!!
18
Linear Elasticity
Since finding a precise solution is too difficult, we step back a little.
2
X
1
X
deformed
P
(
)
0,,
,
321
=
+
i
jij
fuuu
σ
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.
Ω
This is a big release, because
19
Linear Elasticity
Weighted Residual Method
(
)
[
]
0
~
,
~
,
~
,
321
=+

dVfuuu
V
i
jij
σ
20
Linear Elasticity
Weighted Residual Method
By choosing different weight functions, we obtained different numerical
approximation methods.
Point-Collocation Method
SubdomainCollocation Method
Method of Moment
Least Squares Methods Method
GalerkinMethod
21
Linear Elasticity
Weighted Residual Method
Example:
100
2
2
≤≤=++xxu
dx
ud
Boundary conditions
x=0u=0x=1u=0
Assuming an approximate solution is:
))(1(
2
321
L+++−=xaxaaxxu
Weighted Residual is:

=
1
0
0RdxW
i
Accurate solution:
x
x
u−=
1sin
sin
22
Linear Elasticity
Weighted Residual Method: example
First order approximation:
n=1
)1(
1
xxau

=
Second order approximation
n=2
))(1(
21
xaaxxu+

=
100
2
2
≤≤=++xxu
dx
ud
23
Linear Elasticity
Weighted Residual Method: example
Point collocation method:
Allocate several points within the domain
)10(

x
(
)
0
=
i
xR
n=1, we choose the mid-point
2/1
=
x
()
(
)
022/1
2
11
=−+−+==xxaxxR
0
4
7
2
1
1
=−a
7
2
1
=a
)1(
7
2
1
xxu−=
n=2, we choose
3/1
=
x
3/2
=
x
()
0
27
2
9
16
3
1
3/1
212
=+−==aaxR
()
0
27
50
9
16
3
2
3/2
212
=−−==aaxR
1948.0
1
=
a1731.0
2
=
a
(
)
xxxu1731.01948.0)1(
2
+

=
24
Linear Elasticity
Weighted Residual Method: example
Galerkinmethod

=
=
n
i
ii
aNu
1
ii
NW
=
First order approximation:
n=1
)1(
1
xxau

=
(
)
xxNW

=
=
1
11
()
(
)
(
)
∫∫
−+−+−==
1
0
2
1
1
0
1
210dxxxaxxxRdxW
(
)
2
11
2xxaxR−+−+=
272
7
.0
18
5
1
==a
)1(2727.0xxu

=
25
Weighted Residual Method: example
Galerkinmethod

=
=
n
i
ii
aNu
1
ii
NW
=
Linear Elasticity
Second order approximation:
))(1(
21
xaaxxu
+
−=
(
)
xxNW

=
=
1
11
(
)
xxNW−==1
2
22
(
)
(
)
Π2
2
2
1
622xxxaxxaxR−+−+−+−+=
()
(
)()
[
]
06221
1
0
32
2
2
1
1
0
1
=−+−+−+−+−=
∫∫
dxxxxaxxaxxxRdxW
()
(
)()
[
]
06221
1
0
32
2
2
1
2
1
0
2
=−+−+−+−+−=
∫∫
dxxxxaxxaxxxRdxW
)17071924.0)(1(xxxu
+

=
26
Weighted Residual Method: example
Linear Elasticity
Errors
0.030.06008-0.40.069440.20.04408
Galerkin, n=1
1.30.060870.80.070341.40.04464
Point coll. n=2
-13.30.05208-0.40.0694418.30.05208
Galerkin, n=1
-10.80.053572.40.0714321.70.05355
Point coll. n=1
0.060060.069750.04401
Error
%
x=0.75
Error
%
x=0.5
Error
%
x=0.25Solution
x
x
u−=
1sin
sin
)1(
7
2
1
xxu−=
)1(2727.0
1
xxu−=
()
xxxu1731.01948.0)1(
2
+−=
)17071924.0)(1(xxxu+−=
27
Linear Elasticity
Principle of Virtual Displacements
Concepts
Admissible displacements
A set of displacements that satisfy the
geometric constraints.
Virtual displacements
A set of small displacement variations
upon which the geometric constraints are
satisfied.
2
X
1
X
Β
u
S
1
P
S
2
P
S
P
t
u
Admissible
Not Admissible
i
u
δ
i

28
Linear Elasticity
Principle of Virtual Displacements
Some math:
Green Theorems:
∫∫

=
B
i
B
i
dSnadBa,
()
jj
j
abbaab
,,
,
+
=
(
)
j
j
j
ababba
,
,
,

=
29
Linear Elasticity
Principle of Virtual Displacements
0
,
=+
ijij
f
σ
B.C.:
in B
0=−
ijij
Tn
σ
on
B

2
X
1
X
Β
u
S
1
P
S
P
u
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
()
(
)
0
,
=−−+
∫∫
∂B
iijij
B
iijij
dSuTndVuf
δσδσ
30
Linear Elasticity
Principle of Virtual Displacements
()
(
)
0
,
=−−+
∫∫
∂B
iijij
B
iijij
dSuTndVuf
δσδσ
31
Principle of Virtual Displacements
Linear Elasticity

B
jiij
dVu
,
δσ
32
(
)
∫∫∫
−=
∂B
ijij
B
jiij
B
ijij
dVedSnudVu
δσδσδσ
,
Principle of Virtual Displacements
Linear Elasticity
()
(
)
0
,
=−−+
∫∫
∂B
iijij
B
iijij
dSuTndVuf
δσδσ
()
∫∫

=−
B
ii
B
iiijij
dSuTdVufe
δδδσ
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.
33
Linear Elasticity
Principle of Minimum Potential Energy
General form of linear elastic constitutive equations
ijkkijij
eGe
δ
λ
σ
+
=2
klijklij
eD
=
σ
From principle of virtual displacements
()
0=−−
∫∫
∂B
ii
B
iiijij
dSuTdVufe
δδδσ
klijijkl
DD
=
34
Linear Elasticity
Principle of Minimum Potential Energy
0
2
1
=−

∫∫
∂B
ii
B
iiijklijkl
dSuTdVufeeD
δδδ
()
()
iijijklijkl
uUeUeeD==
2
1
Strain energy density
Stored energy density
Stress
Strain
35
Linear Elasticity
Principle of Minimum Potential Energy
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.
Conservative force:
i
i
du
dV
F−=
Examples of conservative forces:
Examples of non-conservative forces:
In linear elasticity, we only consider conservative forces.
dVduF
ii

=

36
Linear Elasticity
Principle of Minimum Potential Energy
()
[]
0=−−
∫∫
∂B
ii
B
iii
dSuTdVufuU
δδδ
We say both fi
and are conservative forces.
i
T
(
)
iii
uuf
Φ
=−
δ
δ
(
)
iii
uuTΨ=−
δδ
37
(
)
(
)
[]
(
)
∫∫

Ψ+Φ+=Π
B
i
B
iip
dSudVuuU
Linear Elasticity
Principle of Minimum Potential Energy
System potential energy
0
=
Π
p
δ
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.
∫∫

−=Π
B
ii
B
iiklijijklp
dSuTdVufeeD
2
1