MAE 5100 - Continuum Mechanics
Course Notes
Prof. Brandon Runnels, University of Colorado Colorado Springs
Contents
L ECTURE 1
0
1
L ECTURE 2
L ECTURE 3
L ECTURE 4
Introduction
0.1
Motivation . . . . . . . . . . . . . . . . . . . . . . . . . .
0.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . .
0.2.1 Sets . . . . . . . . . . . . . . . . . . . . . . . .
0.2.2 Proof notation . . . . . . . . . . . . . . . . . . .
Tensor Analysis
1.1
Index notation and the Einstein summation convention
1.1.1 Vector equality . . . . . . . . . . . . . . . . . .
1.1.2 Inner product . . . . . . . . . . . . . . . . . . .
1.1.3 Kronecker delta . . . . . . . . . . . . . . . . . .
1.1.4 Components of a vector . . . . . . . . . . . . .
1.1.5 Norm of a vector . . . . . . . . . . . . . . . . .
1.1.6 Permutation tensor . . . . . . . . . . . . . . . .
1.1.7 Cross product . . . . . . . . . . . . . . . . . . .
1.1.8 Notation . . . . . . . . . . . . . . . . . . . . . .
1.2
Mappings and tensors . . . . . . . . . . . . . . . . . . .
1.2.1 Second order tensors . . . . . . . . . . . . . . .
1.2.2 Index notation . . . . . . . . . . . . . . . . . . .
1.2.3 Dyadic product . . . . . . . . . . . . . . . . . .
1.2.4 Tensor components . . . . . . . . . . . . . . .
1.2.5 Higher order tensors . . . . . . . . . . . . . . .
1.2.6 Transpose . . . . . . . . . . . . . . . . . . . . .
1.2.7 Trace (first invariant) . . . . . . . . . . . . . . .
1.2.8 Determinant (third invariant) . . . . . . . . . . .
1.2.9 Inverse . . . . . . . . . . . . . . . . . . . . . . .
1.2.10 The special orthogonal group . . . . . . . . . .
1.3
Tensor calculus . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Gradient . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Divergence . . . . . . . . . . . . . . . . . . . . .
1.3.3 Laplacian . . . . . . . . . . . . . . . . . . . . .
1.3.4 Curl . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.5 Gateaux derivatives . . . . . . . . . . . . . . . .
1.3.6 Notation . . . . . . . . . . . . . . . . . . . . . .
1.3.7 Evaluating derivatives . . . . . . . . . . . . . .
1.4
The divergence theorem . . . . . . . . . . . . . . . . . .
1.5
Curvilinear coordinates . . . . . . . . . . . . . . . . . .
All content © 2016-2018, Brandon Runnels
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1.1
1.1
1.2
1.2
1.2
1.2
1.3
1.4
1.4
1.4
1.5
1.5
1.5
1.5
2.1
2.1
2.1
2.2
2.2
2.3
2.3
2.3
2.3
2.4
3.1
3.1
3.1
3.2
3.2
3.2
3.3
3.3
3.3
3.4
4.1
4.2
1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
L ECTURE 5
2
L ECTURE 6
L ECTURE 7
L ECTURE 8
L ECTURE 9
L ECTURE 10
L ECTURE 11
L ECTURE 12
3
L ECTURE 13
L ECTURE 14
L ECTURE 15
L ECTURE 16
1.5.1 The metric tensor . . . . . . . . . . . . .
1.5.2 Orthonormalized basis . . . . . . . . . .
1.5.3 Change of basis . . . . . . . . . . . . . .
1.6
Calculus in curvilinear coordinates . . . . . . . .
1.6.1 Gradient . . . . . . . . . . . . . . . . . .
1.6.2 Divergence . . . . . . . . . . . . . . . . .
1.6.3 Curl . . . . . . . . . . . . . . . . . . . . .
1.7
Tensor transformation rules . . . . . . . . . . . .
Kinematics of Deformation
2.1
Eulerian and Lagrangian frames . . . . . . . . .
2.2 Time-dependent deformation . . . . . . . . . . .
2.2.1 The material derivative . . . . . . . . . .
2.3 Kinematics of local deformation . . . . . . . . .
2.4 Metric changes . . . . . . . . . . . . . . . . . . .
2.4.1 Change of length . . . . . . . . . . . . .
2.4.2 Change of angle . . . . . . . . . . . . . .
2.4.3 Determinant identities . . . . . . . . . .
2.4.4 Change of volume . . . . . . . . . . . .
2.4.5 Change of area . . . . . . . . . . . . . .
2.5 Tensor decomposition . . . . . . . . . . . . . . .
2.5.1 Eigenvalues and Eigenvectors . . . . . .
2.5.2 Symmetric and positive definite tensors
2.5.3 Spectral theorem (symmetric tensors) .
2.5.4 Spectral theorem (general case) . . . .
2.5.5 Functions of tensors . . . . . . . . . . .
2.5.6 Polar decomposition . . . . . . . . . . .
2.6 Principal deformations . . . . . . . . . . . . . . .
2.7
Compatibility . . . . . . . . . . . . . . . . . . . .
2.7.1 Continuous case . . . . . . . . . . . . .
2.7.2 Discontinuous case (Hadamard) . . . .
2.8 Other deformation measures . . . . . . . . . . .
2.9 Linearized kinematics . . . . . . . . . . . . . . .
2.9.1 Linearized metric changes . . . . . . . .
2.9.2 Small strain compatibility . . . . . . . .
2.10 The spatial/Eulerian picture . . . . . . . . . . . .
Conservation Laws
3.1
Conservation of Mass . . . . . . . . . . . . . . .
3.1.1 Control volume . . . . . . . . . . . . . .
3.2 Conservation of linear momentum . . . . . . . .
3.2.1 Forces, tractions, and stress tensors . .
3.2.2 Balance laws . . . . . . . . . . . . . . .
3.2.3 Navier-Stokes momentum equations . .
3.3 Conservation of angular momentum . . . . . . .
3.4 Conservation of energy . . . . . . . . . . . . . .
3.4.1 Energetic quantities . . . . . . . . . . .
All content © 2016-2018, Brandon Runnels
Course Notes
https://canvas.uccs.edu/courses/22031
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4.2
4.3
4.3
4.4
4.4
5.1
5.2
5.2
5.3
6.1
6.2
6.3
6.4
7.1
7.1
7.1
7.2
7.2
7.3
8.2
8.2
8.3
8.3
9.2
9.2
9.3
9.3
10.1
10.1
10.3
10.4
11.1
11.2
11.4
12.1
12.2
13.1
13.2
13.2
13.2
14.1
14.2
15.1
16.1
16.1
2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
L ECTURE 17
L ECTURE 18
L ECTURE 19
4
L ECTURE 20
L ECTURE 21
L ECTURE 22
L ECTURE 23
L ECTURE 24
L ECTURE 25
L ECTURE 26
5
Course Notes
https://canvas.uccs.edu/courses/22031
3.4.2 Balance laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.3 Power-conjugate pairs . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Introduction to statistical thermodynamics and entropy . . . . . .
3.5.2 Internal entropy generation . . . . . . . . . . . . . . . . . . . . . .
3.5.3 Continuum formulation . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Review and summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Constitutive Theory
4.1
Introduction to the calculus of variations . . . . . . . . . . . . . . . . . . .
4.1.1 Stationarity condition . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Variational formulation of linear momentum balance . . . . . . . . . . . .
4.3 Material frame indifference . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Elastic modulus tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Elastic material models . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.1 Useful identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.2 Pseudo-Linear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.3 Compressible neo-Hookean . . . . . . . . . . . . . . . . . . . . . .
4.6 Internal constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6.1 Review of Lagrange multipliers . . . . . . . . . . . . . . . . . . . .
4.6.2 Examples of internal constraints . . . . . . . . . . . . . . . . . . .
4.6.3 Lagrange multipliers in the variational formulation of balance laws
4.7
Linearized constitutive theory . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7.1 Major & minor symmetry and Voigt notation . . . . . . . . . . . . .
4.7.2 Material symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7.3 The Cauchy-Navier equation and linear elastodynamics . . . . . .
4.8 Thermodynamics of solids and the Coleman-Noll framework . . . . . . . .
4.8.1 Other thermodynamic potentials . . . . . . . . . . . . . . . . . . .
4.8.2 Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8.3 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . .
4.8.4 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8.5 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9 Inelastic constitutive modeling . . . . . . . . . . . . . . . . . . . . . . . . .
4.9.1 Crystal plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Computational mechanics
5.1
The finite element method . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Shape functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Weak formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.3 Numerical quadrature . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Linearized kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 3D linearized elasticity . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Finite kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Computational fluid dynamics . . . . . . . . . . . . . . . . . . . . . . . . .
All content © 2016-2018, Brandon Runnels
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
16.3
17.1
17.2
17.2
18.2
18.3
18.4
19.1
19.1
20.1
20.2
21.1
21.1
21.2
21.2
22.1
22.1
23.1
23.1
23.2
23.2
23.3
23.4
23.5
24.1
24.1
25.1
25.3
25.3
25.4
25.4
25.5
25.5
26.1
26.1
26.1
26.3
26.4
26.4
26.5
26.6
26.7
26.7
3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes
https://canvas.uccs.edu/courses/22031
About
These notes are for the personal use of students who are enrolled in or have taken MAE 5100 - Continuum Mechanics at the University of Colorado Colorado Springs, Continuum Mechanics at the Missouri University of Science and
Technology, or Continuum Mechanics at Auburn University. Please do not share or redistribute these notes without
permission.
All content © 2016-2018, Brandon Runnels
4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes
https://canvas.uccs.edu/courses/22031
Nomenclature
A
Lagrangian acceleration
{NI }
Material principal directions
a
Eulerian acceleration
{ni }
Spatial principal directions
ai
Covariant basis vectors
O(n)
Orthogonal group
α
The rotation vector
Ω
A set, typically ⊂ R3 , denoting a body
∀
“For all”
Ω
The microcanonical partition function
B
The left Cauchy-Green deformation tensor
∂Ω
The boundary of a body
B
Material body force
P
The first Piola-Kirchhoff stress tensor
Spatial body force
P (Ω)
Deformation power
β
The displacement gradient tensor
P E (Ω)
External power
β
Reciprocal temperature
φ
Deformation mapping
C
The right Cauchy-Green deformation tensor
Q
Heat
R
Lagrangian mass density
C
The set of complex numbers
Rn
The set of n-dimensional vectors
⊂
“subset of”
r
The infinitesimal rotation tensor
d
The rate of strain tensor
ρ
Eulerian mass density
δij
The Kronecker delta
S
The second Piola-Kirchoff stress tensor
E
The Green-Lagrange strain tensor
S
Entropy
E (Ω)
Internal Energy
Sn , sn
Internal heat generation
ei
Standard Cartesian unit vectors
SO(n)
ε
The small strain tensor
Special orthogonal group (rotation tensors)
ijk
The Levi-Civita alternator
σ
The Cauchy stress tensor
∃
“there exists”
T
Temperature (Lagrangian frame)
∈
“in”
Θ
Temperature (Eulerian frame)
g
Metric tensor
U, u
GL(n)
The general linear group in n dimensions.
Material and spatial intensive internal energy
F
The deformation gradient
u
Displacement
G
Angular momentum vector
V
Lagrangian velocity
Gi
Unit vectors for undeformed configuration
v
Eulerian velocity
Unit vectors for deformed configuration
w
gi
The spin tensor
Outward heat flux vectors
ω
H, h
The vorticity vector
The Jacobian
X
J
Material position
Kinetic energy
x
K
Spatial position
L(Rm , Rn )
The set of m × n tensors
Z
The set of integers
`
The spatial velocity gradient tensor
λ
Stretch ratio
{λi }
Principal stretches
M
Moment
b
All content © 2016-2018, Brandon Runnels
D
5
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes
https://canvas.uccs.edu/courses/22031
Examples
Example 1.1: Gateaux derivative . . . . . . . . . . . . . . . . . .
Example 1.2: Gradient of norm squared . . . . . . . . . . . . . .
Example 1.3: Relationship of Gateaux derivative to gradient . .
Example 1.4: Derivative of function with respect to tensor . . .
Example 1.5: Cylindrical polar coordinates . . . . . . . . . . . .
Example 1.6: Gradient in cylindrical polar coordinates . . . . . .
Example 1.7: Divergence in cylindrical polar coordinates . . . .
Example 1.8: Curl in cylindrical polar coordinates . . . . . . . .
Example 2.1: Time-dependent deformation of unit cube . . . . .
Example 2.2: Material derivative . . . . . . . . . . . . . . . . . .
Example 2.3: Relationship of stretch, angle to components of C
Example 2.4: Metric changes in a planar deformation . . . . . .
Example 2.5: Shear twinning . . . . . . . . . . . . . . . . . . . .
Example 2.6: Pure shear metric changes . . . . . . . . . . . . .
Example 4.1: Minimum distance . . . . . . . . . . . . . . . . . .
Example 4.2: Brachistochrone problem . . . . . . . . . . . . . .
Example 4.3: Thermodynamic potentials mnemonic . . . . . .
All content © 2016-2018, Brandon Runnels
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3.3
.
3.4
.
4.1
.
4.1
.
4.3
.
4.4
.
5.1
.
5.2
.
6.2
.
6.3
.
7.1
.
7.4
. 10.3
. 11.3
. 19.2
. 19.3
. 25.5
5
Lecture 1
0
Introduction and index notation
Introduction
Welcome to continuum mechanics. In this course we will develop the mathematical framework for describing
precisely the deformation of solids, fluids, and gasses, and describing the physical laws that govern their motion.
We will develop the mathematical formulation of the equations of motion, elasticity, viscoelasticity, plasticity, etc.
The course will be organized in the following way:
(1) Tensor analysis: index notation, tensor algebra and calculus, curvilinear coordinates and transformation
rules.
(2) Kinematics of deformation: deformation mappings, local deformation, metric changes, decompositions,
compatibility, linearized kinematics.
(3) Balance laws: conservation of mass, linear momentum, angular momentum, energy.
(4) Constitutive modeling and the thermodynamics of solids: constitutive models, second law of thermodynamics and dissipative systems.
(5) Computational mechanics: finite elements, Galerkin method, Ritz method.
0.1
Motivation
Consider a bar subjected to a tensile load as shown in the following figure. How do we describe this process?
`0
Governing equations
Undeformed
A
Deformed
f
Kinematics:
=
`−`0
`0
Balance law:
σ=
f
A
Constitutive model: σ = E ε
`
The above equations are fairly straightforward for this simple system, and we are familiar with them from statics
and mechanics of materials. But we are in the business of mechanics of bodies with arbitrary shape, loading,
constraints, etc:
Undeformed
Deformed
What is ε, σ, E for this complex case? How do we formulate our equations of kinematics, balance laws, and consitituve models here? Can we use what we know about the system to determine what the deformed configuration is
under applied loads and displacements?
All content © 2016-2018, Brandon Runnels
1.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
0.2
Course Notes - Lecture 1
https://canvas.uccs.edu/courses/22031
Notation
Continuum mechanics is built on the mathematical framework of differential geometry. As a result, the convention
is to use standard math notation when formulating continuum mechanics. Additionally, as we will see, it is generally
necessary to maintain a level of mathematical precision beyond that typically found in engineering disciplines.
0.2.1
Sets
A set is a collection of objects. Examples:
• The integers Z = {... , −1, 0, 1, 2, 3, ...}
• The real numbers R
• The complex numbers C
• n-dimensional vectors Rn
To indicate that an item is in a set, we use the ∈ symbol. For instance,
x ∈ R3
(0.1)
indicates that x is a 3D vector. To indicate that a set is a subset, we use the ⊂ symbol. For instance
Z⊂R
(0.2)
indicates that the integers are a subset of the real numbers. Another common use is to denote a 3D body:
Ω ⊂ R3
(0.3)
is an arbitrary region in 3D space.
0.2.2
Proof notation
We will not be doing any serious proofs in this course, but we frequently use some of the proof notation to simplify
definitions and theorems.
• ∃ is read “there exists.” For instance, ∃x ∈ R3 states that there is at least one item in the R3 set, or that the R3
set is not empty.
• ∀ is read “for all.” For instance,
x − x = 0 ∀x ∈ Rn
(0.4)
tells us that the statement x − x = 0 is true for every possible vector.
Example:
∀x ∈ R3 ∃a ∈ R3 s.t. x − a = 0
(0.5)
can be read “for all 3D vectors there exists another 3D vector such that their difference is equal to zero.”
1
Tensor Analysis
Let us consider the space of three dimensional vectors, R3 . A vector r ∈ R3 can be represented in two different
ways: in terms of its components, or in terms of basis vectors {g1 , g2 , g3 }
All content © 2016-2018, Brandon Runnels
1.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
z
c
a
Course Notes - Lecture 1
https://canvas.uccs.edu/courses/22031
ĝ3
" #
a
r= b
c
r = r1 ĝ1 + r2 ĝ2 + r3 ĝ3
b
ĝ2
y
ĝ1
x
For example, we may have
" #
1
g1 = 0
0
" #
0
g2 = 1
0
" #
0
g3 = 0
1
(1.1)
where we see that the basis vectors correspond to the familiar î, ĵ, k̂ notation. For maximum generality, however,
we do not define the basis vectors explicitly. In subsequent sections will talk about changes of basis.
Note also that we have dropped the familiar x, y , z notation in favor of 1, 2, 3. In general, we will stick with this
convention exclusively; the reason for this will become apparent in the next section.
1.1
Index notation and the Einstein summation convention
Let us consider r defined in the previous section as
r = r1 g1 + r2 g2 + r3 g3
(1.2)
We can write this more simply using summation notation:
r=
3
X
ri gi
(1.3)
i=1
It turns out that we write sums like this a lot, and it becomes cumbersome to write the summation symbol every
time. Thus, we introduce the Einstein summation convention, and we drop the explicit sum. This allows us to simply
write
r = ri gi
(1.4)
This leads us to define the rules of the summation convention. For a vector equation in Rn , expressed using index
notation:
Rule 1:
An index appearing once in a term must appear in every term in the equation, and is not summed. It
is referred to as a free index.
Rule 2:
An index appearing twice must be summed from 1 to n. It is referred to as a dummy index.
(Dummy indices can be changed arbitrarily, that is, e.g. ri gi = rj gj )
Rule 3:
No index may appear more than twice in any term.
(If an index does appear more than twice, we go back to using a summation symbol. Alternatively, if
an index appears twice but is not a dummy index, we use parentheses to denote this, e.g. ui = λ(i) v(i) .
Usually, when a rule gets broken, it means that some algebra got messed up.)
These rules may seem a bit strange, and they usually take a little bit of time to get used to. To help solidify them,
let us look at a couple of algebraic examples.
(Fun fact: the Einstein summation convention was introduced by Albert Einstein to simplify the equations of general
relativity. In fact, the formulation of continuum mechanics has a number of similarities to general relativity.)
All content © 2016-2018, Brandon Runnels
1.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.1.1
Course Notes - Lecture 1
https://canvas.uccs.edu/courses/22031
Vector equality
Consider two vectors u, v ∈ Rn with components u1 , u2 , ..., v1 , v2 , .... In invariant/symbolic notation, we say the two
vectors are equal if
or
u=v
ui gi = vi gi
(1.5)
This tells us that each component of the vector is equal; in other words,
ui = vi
(1.6)
Does this obey the summation convention? Yes it does: i is a free index that appears exactly once in every term of
the equation.
1.1.2
Inner product
Let us again consider u, v ∈ Rn . The inner product (or “dot product”) is defined as
u · v = uT v = |u| |v| cos θ
(1.7)
where | · | is the magnitude of · and θ is the angle between the two vectors. In matrix notation, we evaluate this as
 
v1
n
X
v
uT v = [u1 u2 ... un ]  ...2  = u1 v1 + u2 v2 + ... + un vn =
ui vi = ui vi
(1.8)
i=1
vn
Note that there are no free indices, only dummy indices.
1.1.3
Kronecker delta
The Kronecker Delta is defined in the following way:
(
δij =
1
0
i =j
i 6= j
(1.9)
Consider the basis vectors that we described above. We know that they are orthonormal, so gi · gj is 1 if i = j and
0 otherwise; that is,
gi · gj = δij
(1.10)
Let us use this technology in the context of the dot product. Let u = ui gi , v = vi gi . Then we might write the dot
product as
u · v = (ui gi ) · (vi gi )
(1.11)
But wait: this breaks one of our rules, that an index cannot repeat more than twice. To fix this, we will replace the
is in the second term with js:
u · v = (ui gi ) · (vj gj )
(1.12)
u · v = ui vj (gi · gj ) = ui vj δij
(1.13)
Now, let us distribute these terms:
This term has two summed indices, so if we expand it out, we would have n2 terms. However, we know that only
the terms where i = j survive. Thus, the effect of the Kronecker Delta is to turn one of the dummy indices into the
other: in this case, if we “sum over j”
ui vj δij = ui vi
All content © 2016-2018, Brandon Runnels
(1.14)
1.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.1.4
Course Notes - Lecture 1
https://canvas.uccs.edu/courses/22031
Components of a vector
To extract a specific component of a vector, we can dot with the corresponding unit vector. That is:
u · gi = (uj gj ) · gi = uj (gi · gj ) = uj δij = ui
1.1.5
(1.15)
Norm of a vector
As you may recall, the norm of a vector is given by
|u| =
√
(1.16)
u·u
In index notation, this becomes
|u| =
1.1.6
√
(1.17)
ui ui
Permutation tensor
The next order of business is to introduce the cross product in tensor notation. To work with cross products, we
need to introduce a new bit of machinery, called the permutation tensor, also referred to as the Levi-Civita tensor.
(Note: it’s not actually a tensor. However, it is frequently referred to as one, so we will stick with convention here.)
Here it is:


ijk = 123, 231, 312 = "even permutation"
1
(1.18)
ijk = −1 ijk = 321, 132, 213 = "odd permutation"


0
otherwise
Let’s make a couple of notes here:
• ijk is zero if any of the two indices take the same value.
• Flipping two indices changes the sign of ε; that is, e.g.
ijk = −jik = −ikj = −kji
(1.19)
These identities will come in handy in the future.
1.1.7
Cross product
Let us consider the cross product of unit vectors. We know that
g1 × g2 = g3
g2 × g1 = −g3
(1.20)
g2 × g3 = g1
g3 × g2 = −g1
(1.21)
g3 × g1 = g2
g1 × g3 = −g2
(1.22)
Let us attempt to express this using the permutation tensor. Try:
gi × gj = ijk gk
(1.23)
Does this work? Let’s plug in i = 1, j = 2. Then we have
:g10 + :g20 + 123 g3 = g3
121
122
g1 × g2 = 12k gk = (1.24)
as expected. Plugging in other values for i, j shows that we can recover all of the identities expressed above. Now,
let us see what happens when we take the cross product between u, v:
u × v = (ui gi ) × (vj gj ) = ui vj (gi × gj ) = ui vi ijk gk
(1.25)
(u × v)k = ijk ui vj
(1.26)
Alternatively,
All content © 2016-2018, Brandon Runnels
1.5
Lecture 2
1.1.8
Mappings and tensors
Notation
Let us clarify some of the notation that we have been using:
Equality
Dot product
Cross Product
Example
Invariant/Symbolic Notation
u=v
u·v
u×v
(u · v) w
Full Component Notation
ui gi = vi gi
ui vi
ijk ui vj gk
uk vk wi gi
Termwise Index Notation
u i = vi
ui vi
ijk uj vk
u k vk w i
In general:
• Invariant/symbolic notation is independent of coordinate system, which means that invariant expressions
are more general. However, there are some operations that are too complex to be represented in invariant
notation, and it can more easily get confusing.
• Full component notation is slightly less general than invariant notation, but is the best for working in almost
any coordinate system, especially ones with non-constant unit vectors.
• Termwise index notation is very nimble and convenient when working in a constant, orthonormal coordinate
system. This is frequently what we use, so we will use it a lot. However, it is dangerous to use when unit
vectors are non-constant.
1.2
Mappings and tensors
A mapping is a machine that takes a thing of one type and turns it into a thing of another type. For instance,
f (x) = x 2 takes a real number and turns it into a positive real number. We use the notation
f :U →V
(1.27)
to denote a mapping; in this case, if x ∈ U, then f (x) ∈ V.
A linear mapping f : Rn → Rm is a mapping that satisfies
• f (α x) = α f (x) ∀ x ∈ Rn , α ∈ R
• f (x + y) = f (x) + f (y) ∀ x, y ∈ Rn
1.2.1
Second order tensors
A second order tensor (S.O.T.) is a linear mapping from vector spaces to vector spaces. The set of second order
tensors mapping n-dimensional vectors to m-dimensional vectors is referred to as L(Rn , Rm ). We are familiar with
thinking of them as n × m matrices: For example, if A ∈ L(Rn , Rm ) and u ∈ Rn , v ∈ Rm then we could write

v  A
A12 ... A1n u1 
11
1
A21 A22 ... A2n  u2 
 v2  
v = Au
(1.28)
..
.. 
 ..  = 
.
..
 ...
.
.
.
.  ..
vm
un
Am1 Am2 ... Amn
We can write all possible linear mappings from Rn to Rm in matrix form. Therefore, in general, we can think of
second order tensors as being similar to matrices. Let us make a few notes:
• If m = n the matrix is said to be square. For the most part, we will work with square matrices.
All content © 2016-2018, Brandon Runnels
2.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 2
https://canvas.uccs.edu/courses/22031
• If u = A u then A = I is said to be the identity mapping.
• The difference between tensors and matrices is subtle. A matrix is just a collection of numbers, but a tensor is something that must be transformed with a change of coordinates. Thus we say that tensors have
transformation properties but matrices do not.
1.2.2
Index notation
Index notation makes it very convenient to write second order tensors and tensor-vector multiplication. In the above
example, we have
 v   A u + A u + ... + A u 
11 1
12 2
1n n
1
A21 u1 + A22 u2 + ... + A2n un 
 v2  

(1.29)
..
 ..  = 


.
.
vm
Am1 u1 + Am2 u2 + ... + Amn un
and so we can write
(1.30)
vi = Aij uj
1.2.3
Dyadic product
We have expressed tensor-vector multiplication using invariant notation and termwise index notation. How can we
express a tensor using full component notation? To do this, we introduct the dyadic product:
u 
u v u v
... un v1 
1
1 1
2 1
u
u
v
u
v
...
un v2 
2 2
 2
 1 2
u ⊗ v = u vT =  ..  [v1 v2 ... vn ] =  ..
(1.31)
..
.. 
..
.
.
.
.
.
un
u1 vn u2 vn ... un vn
In tensor notation, we simply write
(1.32)
(u ⊗ v)ij = ui vj
We can use the dyadic product with unit vectors to extract a specific component of a tensor. That is,
(1.33)
gi ⊗ gj
is the zero matrix except for a 1 in the ij column. For example:
" #
"
0
0
g2 ⊗ g3 = 1 [0 0 1] = 0
0
0
0
0
0
#
0
1
0
(1.34)
Therefore, we can express a tensor A as:
A = Aij gi ⊗ gj
(1.35)
How do we write a tensor operating on a vector? Suppose A acts on v:
Av = (Aij gi ⊗ gj )(vk gk ) = Aij vk (gi ⊗ gj )gk
(1.36)
What do we do with this? Recall that we can write u ⊗ v as uvT . Then we have
(gi ⊗ gj )gk = gi gjT gk = gi (gj · gk ) = gi δjk
(1.37)
A v = Aij vk gi δjk = Aij vj gi
(1.38)
Substituting, we get
All content © 2016-2018, Brandon Runnels
2.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 2
https://canvas.uccs.edu/courses/22031
as expected.
We will take this opportunity to reiterate the identity we described earlier: namely, for u, v, w ∈ Rn ,
(u ⊗ v) w = u (v · w)
(1.39)
(ui vj )wj = ui (vj wj )
(1.40)
This is easily seen using index notation:
1.2.4
Tensor components
We can extract components of a tensor A in the following way:
gi · A gj = gi · (Apq gp ⊗ gq ) gj = Apq gi · (gp ⊗ gq ) gj = Apq gi · gp (gq · gj ) = Apq δip δjq = Aij
1.2.5
(1.41)
Higher order tensors
We can express higher order tensors in the following way:
(1.42)
A = Aij...k gi ⊗ gj ⊗ ... ⊗ gk
When we start working with contitutive theory, we will frequently see fourth-order tensors (the elasticity tensor). I
don’t know of any cases where we work with anything higher than fourth order.
1.2.6
Transpose
The transpose of a tensor is identitcal to the transpose of a matrix: the ij term is swapped with the ji term. How
can we express this in index notation?
(AT )ij = Aji
(1.43)
Aij = Aji
(1.44)
Aij = −Aji
(1.45)
A tensor A is called symmetric iff
A tensor is called antisymmetric iff
(Note that the diagonal terms in an antisymmetric tensor must be zero.) Any tensor can be decomposed into its
symmetric and antisymmetric parts in the following way:
A=
1
1
1
1
(A + A) + (AT − AT ) = (A + AT ) + (A − AT )
2
2
2
|
{z
} |2 {z
}
symmetric
1.2.7
(1.46)
antisymmetric
Trace (first invariant)
The trace of a tensor is defined in the folowing way. For u, v ∈ Rn ,
tr(u ⊗ v) = u · v
(1.47)
tr(A) = tr(Aij gi ⊗ gj ) = Aij tr(gi ⊗ gj ) = Aij (gi · gj ) = Aij δij = Aii
(1.48)
Let A ∈ L(Rn , Rn ). Then the trace is given by
One can think of this as the sum of the diagonal terms in the tensor. Note: the trace of a tensor is called the first
invariant of the tensor. This means that the trace of the tensor does not change under rotation. The significance
of this will become apparent later on.
All content © 2016-2018, Brandon Runnels
2.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.2.8
Course Notes - Lecture 2
https://canvas.uccs.edu/courses/22031
Determinant (third invariant)
A quantity that we use frequently is the determinant of a tensor. For A ∈ L(R3 , R3 ),
A11
det(A) = A21
A31
A12
A22
A32
A13
A23 = AA11 AA22 AA33 +−AA12 AA23 AA31 +−AA13 AA21 AA32 −
11 23 32
12 21 33
13 22 31
A33
(1.49)
How can we represent this using index notation? In 3D, we can write it as
det(A) = ijk A1i A2j A3k
(1.50)
Alternatively, we can write it in a slightly more satisfying way as
det(A) =
1
ijk pqr Aip Ajq Akr
6
(1.51)
Some things to note: for A, B ∈ L(Rn , Rn )
det(A) = det(AT )
(1.52)
det(AB) = det(A) det(B)
(1.53)
In three dimensions, the determinant is the third invariant of the tensor, which means it is the third of three important
quantities that do not change under rotation.
All content © 2016-2018, Brandon Runnels
2.4
Lecture 3
1.2.9
Tensor calculus
Inverse
Let A ∈ L(Rn , Rn ). The inverse of A, A−1 satisfies, ∀u ∈ Rn ,
(A−1 )A u = I u = u
(1.54)
If the inverse of a tensor exists it is said to be invertible. One can prove that a tensor A is invertible iff det(A) 6= 0.
In index notation,
A−1
ij Ajk = Iik = δik
(1.55)
(AB)−1 = B−1 A−1
(1.56)
For composite mappings,
What is the determinant of the inverse of a tensor? For A defined above, we know that AA−1 = I. So we can say
that
det(I) = 1 = det(AA−1 ) = det(A) det(A−1 )
(1.57)
so
det(A−1 ) =
1
det(A)
(1.58)
What if det(A) = 0? Then A−1 cannot exist, so its determinant is naturally ill-defined.
1.2.10
The special orthogonal group
Consider the set of tensors A ∈ L(R3 , R3 ) such that AT A = I. We call this group of tensors the orthogonal group,
and we denote it as S(n) where n is the dimensions. So for A ∈ S(n) we see that
• A−1 = AT
• 1/ det(A) = det(A−1 ) = det(AT ) = det(A). This implies that det(A) = ±1
Now, let us consider only those tensors in the orthogonal group that satisfy det(A) = 1. We will call this the special
orthogonal group, and denote it by SO(d). It turns out that SO(d) is exactly the same as the group of all rotation
matrices. In 3D, we will refer to tensors that are in SO(3) very frequently.
1.3
Tensor calculus
Tensor calculus is the language of continuum mechanics. So far we have talked a lot about vectors and tensors.
Now, we are going to talk about vector and tensor fields. There are three kinds of fields that we will use a lot:
• Scalar fields f : Rn → R; e.g. temperature, pressure
• Vector fields v : Rn → Rn ; e.g. displacement, velocity
• Tensor fields T : Rn → L(Rn , Rn ); e.g. stress, strain
We will be looking at a wide variety of differentiation operations on these types of fields. Note: for this section, we
are assuming that we are working in a constant Cartesian basis. That is, we assume that the basis vectors gi are
constants. This is not always true! When we work with curvilinear coordinates, we will have to be very careful about
taking derivatives of basis vectors.
All content © 2016-2018, Brandon Runnels
3.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.3.1
Course Notes - Lecture 3
https://canvas.uccs.edu/courses/22031
Gradient
Suppose we have a scalar field φ : Rn → R, φ(x). We define the gradient of φ to be
grad(φ) =
∂φ
∂φ
∂φ
∂φ
g1 +
g2 + ... +
gn =
gi
∂x1
∂x2
∂xn
∂xi
(1.59)
Note that gradient operator turns a scalar field into a vector field.
Now consider a vector field u : Rn → Rn . The gradient of u is defined to be
grad(u) =
∂u
∂ui
⊗ gi =
gi ⊗ gj
∂xi
∂xj
(1.60)
Note that the gradient operator here turns a vector field into a tensor field.
We can generalize this in the following way:
grad(·) =
∂(·)
⊗ gi
∂xi
(1.61)
where we drop the dyadic product if · is a scalar field. In general, we only generally care about gradients on scalar
and vector fields. However, there are some models that depend on tensor field gradients, such as strain gradient
plasticity and ductile fracture.
1.3.2
Divergence
For a vector field u : Rn → Rn , the divergence is defined to be
div(u) =
∂uj
∂uj
∂ui
∂u
· gi =
gj · gi =
δij =
∂xi
∂xi
∂xi
∂xi
(1.62)
Note that the divergence turns a vector field into a scalar field. Now, consider a tensor field T : Rn → L(Rn , Rn ).
The divergence of the tensor field is
div(T ) =
dT
dTjk
dTjk
dTjk
dTji
gi =
(gj ⊗ gk )gi =
gj (gk · gi ) =
gj δik =
gj
dxi
dxi
dxi
dxi
dxi
(1.63)
Note that the divergence turns a tensor field into a vector field. We can generalize this in the same way we generalized the gradient:
div(·) =
∂(·)
· gi
∂xi
(1.64)
where we drop the dot for tensor fields. Note also that we cannot take the divergence of a scalar field.
1.3.3
Laplacian
The Laplacian is the composition of the gradient and the divergence operators on a scalar field: for φ : R → R:
∆φ = div(grad(φ))
(1.65)
In Cartesian coordinates, this comes out to be
∆φ =
∂2φ
∂xi ∂xi
(1.66)
Note that we do not write ∂xi2 in order to be in keeping with the summation convention.
All content © 2016-2018, Brandon Runnels
3.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.3.4
Course Notes - Lecture 3
https://canvas.uccs.edu/courses/22031
Curl
The curl operator is defined on a vector field u : R3 → R3 in the following way:
curl(u) = −
∂u
∂uj
∂uj
∂uj
× gi = −
gj × gi = −
jik gk =
ijk gk
∂xi
∂xi
∂xi
∂xi
(1.67)
Note that the curl acts on a vector field and produces a vector field, so we cannot take the curl of a scalar field. In
generalized form we have
curl(·) = −
1.3.5
∂(·)
× gi
∂xi
(1.68)
Gateaux derivatives
A more general type of derivative is the “Gateaux derivative,” which will prove very useful later on. Consider some
field (scalar, vector, or tensor, etc.) φ : Rn → V (where V is the set of scalars, vectors, tensors, etc.) and a vector
v ∈ Rn . The Gateaux derivative is defined as
Dφ(x)v =
d
φ(x + εv)
dε
(1.69)
ε→0
that is, the derivative is taken with respect to ε which is then set to 0.
Example 1.1: Gateaux derivative
Let φ(x) = xi xi = |x|2 , and compute Dφ(x)v . We evaluate this simply by substituting φ into (1.69):
Dφ(x)v = lim
ε→0
1.3.6
d
d
((xi + εvi )(xi + εvi )) = lim
(xi xi + 2εxi vi + ε2 vi vi ) = lim (2xi vi + 2εvi vi ) = 2xi vi (1.70)
ε→0 dε
ε→0
dε
Notation
Here it is important to make a couple of remarks about notation.
(1) We avoid the use of the ∇ operator (e.g. ∇· for divergence, ∇× for curl) because it is difficult to impossible
to express certain vector operations.
(2) We work with a lot of derivatives in continuum mechanics and it frequently becomes cumbersome to write
∂
∂xi . Therefore, we adopt comma notation:
∂
(·) = (·),i
(1.71)
∂xi
For example, we can write grad/div/curl compactly for scalar/vector/tensor fields φ, v, T
grad(φ)i = φ,i
div(v) = vi,i
curl(v)i = ijk vk,j
grad(v)ij = vi,j
div(T)i = Tij,j
∆φ = φ,ii
(1.72)
(1.73)
(3) We will occasionally use the ∂ symbol to denote differentiation. Examples of usage include:
∂x ≡
∂
∂x
All content © 2016-2018, Brandon Runnels
∂θ ≡
∂
∂θ
∂i ≡
∂
∂xi
(1.74)
3.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.3.7
Course Notes - Lecture 3
https://canvas.uccs.edu/courses/22031
Evaluating derivatives
How do we evaluate a derivative with respect to x in terms of x? For instance, how would we compute the gradient
of φ(x) = x · x?
In index notation, we have
∂xi
= δij
∂xj
or, in symbolic notation,
∂x
=I
∂x
(1.75)
Let us look at a couple of examples:
Example 1.2: Gradient of norm squared
Compute the gradient of φ(x) = x · x. The first step is to write φ in index notation: φ(x) = xk xk . Then, we use
the formula:
∂
grad(φ) =
(xk xk ) gi
(1.76)
∂xi
We can use the product rule exactly like we would normally:
= (δik xk + xk δik ) gi = 2δik xk gi
(1.77)
= 2xi gi = 2x
(1.78)
Summing over i we get
All content © 2016-2018, Brandon Runnels
3.4
Lecture 4
Curvilinear coordinates and tensor transformations
Example 1.3: Relationship of Gateaux derivative to gradient
Let φ : R3 → R. Show that Dφ(x)v = grad(φ) · v. To do this we again evaluate the Gateaux derivative:
d
φ(x + ε v)
ε→0 dε
(1.79)
Dφ(x)v = lim
We use the chain rule exactly as we would normally:
= lim
ε→0
∂φ(x + ε v) d(xi + ε vi )
∂φ(x + ε v)
∂φ
= lim
vi =
vi = grad(φ) · v
ε→0 ∂(xi + ε xi )
∂(xi + ε xi )
dε
∂xi
(1.80)
In addition to taking derivatives with respect to vectors (such as x) we may take derivatives with respect to tensors.
For example, given φ : L(Rn , Rn ) → R, we may wish to calculate
dφ(T )
dTij
(1.81)
Similarly to the case of vectors, we have
dTij
= δip δjq
dTpq
or, in symbolic notation,
dT
=I⊗I
dT
(1.82)
Example 1.4: Derivative of function with respect to tensor
Let A : L(Rn , Rn ) → L(Rn , Rn ) with A(T) = TT T. Find the derivative of A with respect to T:
∂Aij
∂
∂Tki
∂Tkj
=
(Tki Tkj ) =
Tkj + Tki
= δkp δiq Tkj + Tki δkp δjq = Tpj δiq + Tpi δjq
∂Tpq
∂Tpq
∂Tpq
∂Tpq
(1.83)
How do we represent this in symbolic notation? It’s actually rather tricky, and is much easier to leave things
in index notation.
Note: when taking vector or tensor derivatives, it is always important to use a fresh new free index. Do not reuse
existing ones.
1.4
The divergence theorem
We have developed enough machinery to introduce the single most important theorem in all of continuum mechanics: the divergence theorem.
V(x)
Ω
n
All content © 2016-2018, Brandon Runnels
4.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 4
https://canvas.uccs.edu/courses/22031
Theorem 1.1 (The divergence theorem). Let Ω ⊂ Rn , and V : R3 → R3 be a differentiable vector field: Then:
Z
Z
Z
Z
div(V) dv =
V · n da
Vi,i dv =
Vi ni da
(1.84)
Ω
∂Ω
Ω
∂Ω
where ∂Ω is the boundary of the body.
We have a similar theorem for tensor fields: Let T : R3 → L(R3 , R3 ) be a tensor field. Then
Z
Z
div(T) dv =
Tn da
Ω
Using index notation in an Cartesian coordinate system, we can write
Z
Z
Z
Z
Vi,i dv =
Vi ni da
Tij,j dv =
Ω
(1.85)
∂Ω
∂Ω
Ω
(1.86)
Tij nj da
∂Ω
What does this mean? We are relating a volume integral of the divergence to a surface integral – or, in this case, a
flux integral. For the case of a vector field, the integral of the divergence over the body can be intuitively thought
of as the total amount of compression/expansion in the vector field. The flux integral can be thought of as the
total amount of vector field entering or leaving the body. Thus, you can think of the divergence theorem as the
mathematical formulation of the statement “the total compression of the vector field is equal to the rate of flux
through the boundary.” We will make extensive use of the divergence theorem in this course.
1.5
Curvilinear coordinates
Up until now we have worked within a simple Cartesian basis, let us call it ei . However, it is frequently convenient
to switch to a more natural coordinate system:
ĝθ
ĝ3
ĝr
r = r1 ĝ1 + r2 ĝ2 + r3 ĝ3
ĝ2
ĝ1
Consider a new set of coordinates {θ1 , θ2 , θ3 , ... , θn }. Position as a function of these coordinates is expressed as
(1.87)
x(θ1 , θ2 , ... , θn )
Let us define a new basis: {a1 , a2 , ... , a3 } defined as
a1 =
∂x
∂θ1
a2 =
∂x
∂θ1
...
ai =
∂x
∂xj
=
ej
∂θi
∂θi
(1.88)
We will refer to {ai } as the covariant basis vectors.
1.5.1
The metric tensor
The metric tensor G is defined as
Gij = ai · aj
(1.89)
Notes:
• The metric tensor is symmetric
• If the metric tensor is diagonal then the new coordinate system {ai } is said to be orthogonal
• If G = I then the coordinate system is said to be orthonormal
All content © 2016-2018, Brandon Runnels
4.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.5.2
Course Notes - Lecture 4
https://canvas.uccs.edu/courses/22031
Orthonormalized basis
There is no guarantee that our new basis {ai } will be normalized, that is, we don’t know that |ai | = 1. But we can
make sure that they are by defining scale factors hi = |ai |. Then we define a new basis
a(i)
gi =
no summation over i
(1.90)
h(i)
Example 1.5: Cylindrical polar coordinates
Cylindrical Polar Coordinates: we can specify any point using x1 , x2 , x3 , but we can also specify it using the
coordinates r , θ, z.
z
r
θ
Let us compute x(r , θ, z):
x1 = r cos θ
x2 = r sin θ
Now we can compute our basis vectors:
"
#
"
#
cos θ
−r sin θ
∂xi
∂xi
ar =
ei = sin θ
aθ =
ei = r cos θ
∂r
∂θ
0
0
x3 = z
" #
0
∂xi
az =
ei = 0
∂z
1
(1.91)
(1.92)
Our scale factors are
h1 = |ar | = 1
h2 = |aθ | = r
(1.93)
h3 = |az | = 1
so we have
"
#
cos θ
gr = sin θ
0
"
− sin θ
gθ = cos θ
0
#
" #
0
gz = 0
1
(1.94)
Important note: while {e1 , e2 , e3 } are independent of x1 , x2 , x3 , {gi } is not necessarily independent of {θi }. In our
above example, we see that
"
#
"
#
− sin θ
− cos θ
∂
∂
gr = cos θ = gθ
gθ = − sin θ = −gr
(1.95)
∂θ
∂θ
0
0
1.5.3
Change of basis
Let {gi } be an orthonormal basis for Rn , and let v ∈ Rn . Suppose we wish to find {vi } such that v = vi gi . To do
this, we use the orthogonality property of the basis:
v · gj = vi gi · gj = vi δij = vj =⇒ v = (v · gj ) gj
(1.96)
Suppose we have another basis {ei }. Then we can relate the two bases by writing
ei = (ei · gj ) gj
gi = (gi · ej ) ej
(1.97)
These relationships will be useful as we start discussing curvilinear coordinates.
All content © 2016-2018, Brandon Runnels
4.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.6
Course Notes - Lecture 4
https://canvas.uccs.edu/courses/22031
Calculus in curvilinear coordinates
Now that we have defined a framework for working in other coordinate systems, we need to know what our calculus
operations look like in those systems. Before we do that, however, we need to introduce a couple of important
identities.
(1) Earlier we learned that
gi =
1 ∂xj
1
a(i) =
ej
h(i)
h(i) ∂θ(i)
no sum on i
(1.98)
How can we express ei in terms of {gj }? To do this, we’ll pull a trick. Remember that {gi } forms an orthonormal
basis for Rn : that means that we can express any vector in terms of that basis. For instance, a vector v can
be written as v = vi gi . How do we find vi ? It’s nothing other than v · gi . So we can write
(1.99)
v = (v · gi ) gi
We can do exactly the same thing for our original basis vectors
ei = (ei · gj ) gj =
X 1 ∂xi
X 1
X 1 ∂xk
X 1
∂xk
(ei · aj ) gj =
(ei ·
ek ) gj =
(ei · ek ) gj =
gj
hj
hj
∂θj
hj ∂θj | {z }
hj ∂θj
j
j
j
(1.100)
j
δik
(Note that we broke one of our summation convention rules. To compensate for this we drop the summation
notation and use an explicit sum.)
(2) There is an important theorem called the inverse function theorem that states:
h ∂θ i
∂x
=
h ∂x i−1
∂θ
=⇒
∂θ ∂x
=I
∂x ∂θ
∂θi ∂xk
= δij
∂xk ∂θj
or, in index notation,
(1.101)
We will use both of these rules to derive expressions for the familiar divergence, gradient, and curl in curvilinear
coordinates.
1.6.1
Gradient
We want to express the gradient as computed in the above
grad(f (θ)) =
X 1 ∂f
∂
∂f ∂θj
∂f ∂θj X 1 ∂xi X 1 ∂f ∂θj ∂xi
gk =
(f (θ)) ei =
ei =
gk =
gk (1.102)
∂xi
∂θj ∂xi
∂θj ∂xi
hk ∂θk
hk ∂θj ∂xi ∂θk
hk ∂θk
k
k
k
| {z }
δjk
Notice how this is almost identical to our original expression for the gradient, except that {xi }, {ei } have been
replaced with {θi }, {gi }. The only difference is the presence of the scale factors. Let’s solidify this with an example:
Example 1.6: Gradient in cylindrical polar coordinates
Let us continue with our example of cylindrical polar coordinates. Let f = f (r , θ, z). Then we have:
grad(f ) =
1 ∂f
1 ∂f
1 ∂f
∂f
1 ∂f
∂f
gr +
gθ +
gz =
gr +
gθ +
gz
hr ∂r
hθ ∂θ
hz ∂z
∂r
r ∂θ
∂z
All content © 2016-2018, Brandon Runnels
(1.103)
4.4
Lecture 5
1.6.2
Kinematics of deformation, frames
Divergence
Let v = vi (θ) gi be a vector field that is defined exclusively using the {θi }, {gi } coordinate system. What is the
divergence of this vector field? We can follow the exact same procedure as when computing the gradient:
div(v) =
X 1 ∂v ∂θj ∂xi
X 1 ∂v
∂
∂v ∂θj
∂v ∂θj X 1 ∂xi
(v) · ei =
· ei =
·
gk =
·gk =
δjk · gk
∂xi
∂θj ∂xi
∂θj ∂xi
hk ∂θk
hk ∂θj ∂xi ∂θk
hk ∂θj
k
k
k
|{z} |{z}
(1.104)
X 1 ∂v
· gk
hk ∂θk
(1.105)
Jji
=
Jik−1
k
Notice again how we arrive at an almost identical formula except that we include scale factors. We must also make
one very important note: how do we evaluate the following?
∂
∂
∂vi
∂gi
(v) =
(vi gi ) =
g i + vi
∂θk
∂θk
∂θk
∂θk
(1.106)
In Cartesian coordinates the basis vectors are constant so their derivatives vanish. However, this is not true in
most other curvilinear coordinates! To illustrate this, let’s do an example:
Example 1.7: Divergence in cylindrical polar coordinates
Compute the divergence of a vector field in cylindrical polar coordinates: v = vr gr + vθ gθ + vz gz .
1 ∂v
1 ∂v
1 ∂v
· gr +
· gθ +
· gz
hr ∂r
hθ ∂θ
hz ∂z
∂
1 ∂
∂
= (vr gr + vθ gθ + vz gz ) · gr +
(vr gr + vθ gθ + vz gz ) · gθ +
(vr gr + vθ gθ + vz gz ) · gz
∂r
r ∂θ
∂z
∂v
∂vθ
∂vz
r
gr +
gθ +
gz · gr
=
∂r
∂r
∂r
∂vθ
∂vz
∂
∂
∂ 1 ∂vr
+
gr +
gθ +
gz + vr gr + vθ gθ + vz gz · gθ
r ∂θ
∂θ
∂θ
∂θ
∂θ
∂θ
∂v
∂vθ
∂vz r
gr +
gθ +
gz · gz
+
∂z
∂z
∂z
∂vr
1 ∂vθ
∂vz
=
+
gθ + vr gθ − vθ gr · gθ +
∂r
r ∂θ
∂z
∂v
∂vr
1 ∂vθ
z
=
+
+ vr +
(1.107)
∂r
r ∂θ
∂z
div(v) =
Notice how we picked up a couple of extra terms: this is a result of our choice of coordinate system. This
is a tedious process, but fortunately we only have to do it a couple of times.
All content © 2016-2018, Brandon Runnels
5.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
1.6.3
Course Notes - Lecture 5
https://canvas.uccs.edu/courses/22031
Curl
Hopefully this is starting to seem familiar. Starting with our original expression for curl and converting to curvilinear
coordinates, we have
X 1 ∂v ∂θj ∂xi
X 1 ∂v
∂
∂v ∂θj X 1 ∂xi
(v) × ei = −
×
gk = −
× gk = −
δjk × gk
∂xi
∂θj ∂xi
hk ∂θk
hk ∂θj ∂xi ∂θk
hk ∂θj
k
k
k
X 1 ∂v
× gk
=−
hk ∂θk
curl(v) = −
(1.108)
(1.109)
k
Once again, we see that we recover a very similar expression except for the presence of scale factors.
Example 1.8: Curl in cylindrical polar coordinates
Find the expression for the curl in cylindrical polar coordinates. We can reuse quite a bit of what we computed earlier; we just need to be careful about which vectors we cancel out.
h 1 ∂v
i
1 ∂v
1 ∂v
× gr +
× gθ +
× gz
hr ∂r
hθ ∂θ
hz ∂z
∂v
∂vθ
∂vz r
=−
gr +
gθ +
gz × gr
∂r
∂r
∂r
1 ∂vr
∂vθ
∂vz
−
gr +
gθ +
gz + vr gθ − vθ gr × gθ
r ∂θ
∂θ
∂θ
∂v
∂vθ
∂vz r
−
gr +
gθ +
gz × gz
∂z
∂z
∂z
∂v
∂v
∂vz
∂vr
∂vz
∂vθ 1
r
θ
gz −
gθ +
−
gz +
gr + vθ gz +
gθ −
gr
=
∂r
∂r
r
∂θ
∂θ
∂z
∂z
1 ∂v
∂vθ
∂vz
∂vr
1 ∂(r vθ ) ∂vr
z
=
−
−
−
gr +
gθ +
gz
r ∂θ
∂z
∂z
∂r
r
∂r
∂θ
curl(v) = −
1.7
(1.110)
Tensor transformation rules
Suppose we have two orthonormal {ei }, {gi }. We recall that orthonormality allows us to write each basis in terms
of the other:
ei = (ei · gp ) gp
gq = (gq · ei ) ei
(1.111)
Now, let us suppose we have a tensor A ∈ L(Rn , Rn ) with components Aij in the {ei } basis, that is, A = Aij ei ⊗ ej .
How can we express A in terms of the other basis? To do that, we simply substitute
Aij ei ⊗ ej = Aij [(ei · gp ) gp ] ⊗ [(ej · gq ) gq ] = (gp · ei ) Aij (ej · gq ) gp ⊗ gq = QpiT Aij Qjq gp ⊗ gq = Âpq gp ⊗ gq
| {z }
| {z }
QpiT
Qjq
(1.112)
where Âpq are the components of A in the other basis. Or, put more simply, we have that
Âpq = QpiT Aij Qjq
T
Aij = Qip Âpq Qqj
(1.113)
Note that we are not transforming the tensor itself: we are merely changing the components of the tensor to fit
with the assigned basis. This is called a tensor transformation property.
All content © 2016-2018, Brandon Runnels
5.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 5
https://canvas.uccs.edu/courses/22031
Let’s take another look at the Q matrices. Specifically, let’s look at QT Q:
[QT Q]pq = QpiT Qiq = (gp · ei )(ei · gq ) = [(gp · ei ) ei ] ·gq = gp · gq = δpq = [I]pq
|
{z
}
(1.114)
gp
Because QT Q = I, we conclude that QT = Q−1 . We can also write Q = [g1 ... gn ]. If the new basis is right-handed,
then we have det(Q) = 1. Thus, we have shown that Q ∈ SO(n), or that Q is a rotation matrix.
2
Kinematics of Deformation
We are now ready to introduce the machinery that we need to describe the deformation of solid bodies. Let us
introduce a few definitions:
Definition 2.1. A body is a set of material particles occupying a region in Euclidean space; generally denoted Ω ⊂ R3 .
Definition 2.2. A configuration is a specific correspondance between particles of the body and points in space.
Definition 2.3. A deformation mapping is an injective1 mapping that describes a configuration of the body.
We will also refer to the deformed and undeformed configurations. The following figure illustrates the general setup
for describing the deformation of a solid body:
φ
G3
X
g3
x = φ(X)
Ω
G1
φ(Ω)
g2
G2
g1
• {GI } are the basis vectors in the undeformed configuration
• {gi } are the basis vectors in the deformed configuration.
**Note: this is completely general, but we often (usually) keep the same in both configurations.
• φ : R3 → R3 is the deformation mapping.
• X = XI GI is the location of a point in the undeformed configuration.
• x = xi gi = φi (X) gi is the location of point X in the deformed configuration.
Note that we adopt the convention that uppercase symbols correspond to quantities the undeformed configuration;
whereas lowercase symbols correspond to quantities in the deformed configuration. We will even use uppercase
and lowercase indices to indicate components in the undeformed and deformed frames. We will stick to this
convention consistently, and it will be useful in helping us to keep track of which coordinate system we are in.
To illustrate, let us consider the following examples:
(i) Stretching of a unit cube:
What is the deformation mapping for the following stretched cube?
1 “injective” – no two points can be mapped to the same location, that is, if f
All content © 2016-2018, Brandon Runnels
: U → V is injective, then for x, y ∈ U , f (x) = f (y ) =⇒ x = y .
5.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 5
https://canvas.uccs.edu/courses/22031
g3
G3
φ
λ3
1
1
1
G2
g1
G1
λ2
λ1
g2
(Notice that g1 = G1 , etc.) We identify the deformation mapping simply as
x1 = λ1 X1
x2 = λ2 X2
x3 = λ3 X3
(2.1)
Or, we can describe this as
"
λ1
x = φ(X) = 0
0
0
λ2
0
#
0
0 X = FX
λ3
(2.2)
(ii) Shearing of a unit cube:
What is the deformation mapping for the following cube subjected to pure shear?
G3
g3
φ
1
γ
1
1
1
G2
1
1
G1
g2
g1
We identify the deformation mapping to be
x1 = X1
x2 = X2 + γ X3
Or, we can describe it as
"
1
x = φ(X) = 0
0
0
1
0
x3 = X3
#
0
γ X = FX
1
(2.3)
(2.4)
(iii) An affine deformation is one in which “straight lines remain straight” – as in the following figure:
φ
φ
affine
undeformed
non-affine
What form must this mapping take? Consider the line between two vectors X1 , X2 , and the resulting line
between the mapped vectors x1 , x2 .
All content © 2016-2018, Brandon Runnels
5.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 5
https://canvas.uccs.edu/courses/22031
φ
α
1−α
X1 + (
X1
αx 1
)X 2
+
α
(1 −
)x 2
x2
x1
X2
x0
The set of vectors αX1 + (1 − α)X2 are all in the line in the undeformed configuration. By the definition of
affine mappings, αx1 + (1 − α)x2 must be must form the points in the deformed line. In other words:
φ(αX1 + (1 − α)X2 ) = α φ(X1 ) + (1 − α) φ(X2 ) + x0 ∀ X1 , X2 ∈ Ω
(2.5)
Since this must hold for all X, we conclude that the part of the mapping depending on X must be linear. Since
all linear maps can be represented in tensor form, we conclude that affine maps can be represented in the
form
φ(X) = F X + x0 F ∈ L(Rn , Rn ), φ affine
(2.6)
Because this is a real deformation, we know that it must be invertible. That is, we can construct a φ−1 such
that X = φ−1 (x). But if x = φ(X) = F X, then the inverse mapping must be
X = F−1 x =⇒ φ−1 (x) = F−1 x
(2.7)
This means that F must be invertible, which means that det(F) 6= 0. The set of all invertible matrices is called
the general linear group
GL(n) = {F ∈ L(Rn , Rn ) : det(F) 6= 0} ⊂ L(Rn , Rn )
(2.8)
so we say that a mapping is affine if and only if ∃F ∈ GL(n) and x0 ∈ Rn such that
φ = FX + x0
All content © 2016-2018, Brandon Runnels
(2.9)
5.5
Lecture 6
Material derivative, metric changes
(iv) A rigid body mapping in Rn is, formally, an “orientation-preserving isometry of Rn .” What does that mean?
First, let’s define the term “isometry:”
Definition 2.4. A mapping φ : Rn → Rn is an isometry if
(2.10)
|φ(X)| = |X| ∀x ∈ Rn
In other words, an isometry is a mapping that does not change the length of any vector. Let us make the
ansatz (i.e. starting assumption) that φ is an affine isometry, that is, φ = F X = x. What are the conditions
for φ to be an isometry?
q
√
√
√
!
(2.11)
|φ| = |x| = xT x = (F X)T (F X) = XT F T F X = XT X ∀X ∈ Rn
What does this imply about F T F ? It must equal the identity. So F T F = I =⇒ F T = F −1 , or F ∈ O(n), the
orthogonal group.
What about the other part? Without going into extensive detail, “orientation-preserving” simply means that
the body cannot be reflected or turned inside-out. This is equivalent to stating that det(F ) > 0. Thus, for φ to
be orientation-preserving, F ∈ SO(3); that is, F must be a rotation.
There is one more aspect of affine and rigid-body mapping that we have not yet discussed: rigid body translation. A rigid body translation mapping can simply be expressed as
(2.12)
φ(x) = x0 + X
where x0 is the translation vector.
Thus, the general expression for a rigid body mapping is
(2.13)
φ(X) = x0 + F X x0 ∈ Rn , F ∈ SO(n)
2.1
Eulerian and Lagrangian frames
Let’s go back to our generalized form of the deformation mapping.
φ
G3
X
g3
x = φ(X)
Ω
G1
φ(Ω)
g2
G2
g1
As stated before, we have adopted the convention of using uppercase variables (and indices) to describe the material in the undeformed configuration, and lowercase for the deformed configuration. The reason for doing this,
as we’ll see, is that we will be able to formulate almost everything analagously in terms of either set of variables.
Definition 2.5. The Lagrangian / material frame refers to the quantities defined in the undeformed configuration.
Definition 2.6. The Eulerian / spatial frame refers to the quantities defined in the deformed configuration.
All content © 2016-2018, Brandon Runnels
6.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 6
https://canvas.uccs.edu/courses/22031
As we go along and derive various equations, we will frequently formulate those equations in both the Lagrangian
and Eulerian frames. (You can think of this as finding the “uppercase” and “lowercase” versions of the equations.)
The following are a couple of examples of the convention that we will use.
• Variables and unit vectors: X are the locations of the material points in the Lagrangian frame, x = φ(X) are
the locations of the points in the Eulerian frame.
X = XI GI = φ−1
I (x) GI
x = xi gi = φi (X) gi
(2.14)
∂f
∂xi
∂vi
div(v(x))i =
∂xi
(2.15)
• Calculus:
∂F
∂XI
∂VI
Div(V(X))I =
∂XI
grad(f (x))i =
Grad(F(X))I =
(2.16)
and similarly for curl, the Laplacian, etc. Note that Div, Grad are not necessarily the same as div, grad!
2.2
Time-dependent deformation
Let us now consider a body whose deformation varies with time: that is, x(t) = φ(X, t). What is the velocity of the
material? Let us define the Lagrangian velocity field as
V(X, t) =
∂
φ(X, t)
∂t
Vi (X, t) =
∂
φi (X, t)
∂t
(2.17)
Ai (X, t) =
∂
Vi (X, t)
∂t
(2.18)
Similarly, the Lagrangian acceleration as
A(X, t) =
∂
V(X, t)
∂t
Suppose we want to get the velocity and acceleration as a function of the deformed location? To do this we define
the Eulerian velocity field as
v(x, t) = V(φ−1 (x), t)
vi (x, t) = Vi (φ−1 (x), t)
(2.19)
ai (x, t) = Ai (φ−1 (x), t)
(2.20)
and the Eulerian acceleration field as
a(x, t) = A(φ−1 (x), t)
Example 2.1: Time-dependent deformation of unit cube
A unit cube is undergoing a time-dependent uniaxial stretching deformation as shown below.
1 + λt
All content © 2016-2018, Brandon Runnels
6.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 6
https://canvas.uccs.edu/courses/22031
We wish to find the material and spatial velocity and acceleration fields. First, we must find the deformation
mapping:
x1 = (1 + λt) X1
x2 = X2
x3 = X3
The material velocity and acceleration fields are given by straight-up differentiation:
"
# "
#
λX1
∂ (1 + λt) X1
∂
X2
V=
V=0
= 0
A=
∂t
∂t
0
X3
(2.21)
(2.22)
Now, we need an inverse relationship to convert to the spatial velocity and acceleration. It is pretty easy to
find: we see that
X1 =
1
X1
1 + λt
X2 = X2
X3 = X3
(2.23)
Now, all we do is substitute:
"
#
λx1 /(1 + λt)
0
v = V(φ−1 (x)) =
0
2.2.1
a=0
(2.24)
The material derivative
We have expressed Eulerian and Lagrangian time derivatives in terms of Lagrangian derivatives; that is, we always
get the Eulerian version by back-substituting in the deformation mapping into our Lagrangian version. How can we
express the Eulerian time derivatives exclusively in terms of Eulerian coordinates? To do this, we use the material
derivative, which is nothing more than an application of the chain rule. Suppose we have two scalar fields
F :Ω→R
f : φ(Ω, t) → R
(2.25)
Computing the time derivative of F is easy:
dF
d
= F (X)
dt
dt
(2.26)
However, computing the time derivative of f is more complicated, and we have to use the chain rule:
d
d
∂f ∂xi ∂f
∂f
∂f
∂f
f (x, t) = f (x(X, t), t) =
+
=
vi +
= grad(f ) · v +
dt
dt
∂xi |{z}
∂t
∂t
∂xi
∂t
∂t
(2.27)
=vi
Notice that we no longer have any dependence on our deformation mapping φ, only its derivative. This is very
convenient for when φ is not available, such as in fluid flow. So though this derivative is nothing other than an
application of the chain rule, it is used so frequently that it gets a special name:
Df
∂f
∂f
=
vj +
≡ The Material Derivative
Dt
∂xj
∂t
(2.28)
where f can be a scalar, vector, or tensor field operating in the Eularian configuration. As an example: Eulerian
acceleration is given by the material derivative of the Eulerian velocity:
a=
Dv
∂
= grad(v) · v + v
Dt
∂t
ai =
∂vi
∂vi
vj +
= vi,j vj + vi,t
∂xj
∂t
(2.29)
We will show how this works with an example:
All content © 2016-2018, Brandon Runnels
6.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 6
https://canvas.uccs.edu/courses/22031
Example 2.2: Material derivative
Continuing with our previous example, let us prove that we recover a using the material time derivative.
0
0
λ λx 7
7
λx1
∂v1
∂v1
∂v1
∂v1
1
+ −
(λ)
a1 =
v1 + v2 + v3 +
=
∂x1
∂x2
∂x3
∂t
1 + λt
1 + λt
(1 + λt)2
λ 2 x1
λ 2 x1
−
=0 X
=
2
(1 + λt)
(1 + λt)2
(2.30)
(2.31)
The other components are all zero, so their derivatives will be zero as well.
2.3
Kinematics of local deformation
Let us consider the case of all deformations, affine and otherwise. Affine deformations are fairly easy to quantify:
the material deforms the same way at every material point, and we can easily represent the mapping using a tensor.
The general case is more complicated, but it is what we are interested in.
Before discussing local deformation, let us briefly discuss Taylor series. You may recall from Caclulus that any
sufficiently smooth function, f (x), can be represented as a Taylor expansion about a point a as follows:
∞
X 1
1
f (x) = f (a) + f 0 (a)(x − a) + f 00 (a)(x − a)2 + ... =
f (n) (a)(x − a)n
2
n!
n=0
(2.32)
We can do a similar thing with multivariate functions: let f : Rn → R. Then the expansion of f about a point X0 ∈ Rn
is
f (X) = f (X0 ) +
∂f
1 ∂2f
(XI − XI0 ) +
(XI − XI0 )(XJ − XJ0 ) + higher order terms
∂XI
2 ∂XI ∂XJ
(2.33)
or, using invariant notation,
1
= f (X0 ) + Grad(f )(X − X0 ) + (X − X0 )T Grad(Grad(f ))(X − X0 ) + higher order terms
2
(2.34)
(As you can see, the problem with invariant notation is that we don’t really have a way to express higher order terms;
on the other hand, index notation handles it easily.)
Now, let us apply this to deformation mappings. We will restrict ourselves (for now) to smooth mappings, that is,
mappings with continuous derivatives.
φ
undeformed
non-affine
locally affine
For mappings of this type, at every point, we can always find a “neighborhood” of the point at which the mapping
is locally affine. Here, let us consider the neighborhood around a point X0 that is mapped to x0 , and observe what
happens to a small vector ∆X that is in the locally affine neighborhood.
∆X
X0
All content © 2016-2018, Brandon Runnels
φ
∆x
x0
6.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 6
https://canvas.uccs.edu/courses/22031
We expand the mapping out as a Taylor series:
xi0 + ∆xi = φi (X0 + ∆X) = φi (X0 ) +
∂φi
∂φi
((XJ0 + ∆XJ ) − XJ0 ) + h.o.t. = xi0 +
∆XJ + h.o.t.
∂XJ
∂XJ
(2.35)
(2.36)
If we ignore the higher-order terms, we can write
∂φi
∆XJ = FiJ ∆XJ
∂XJ
(2.37)
∂φi
≡ Deformation Gradient Tensor
∂XJ
(2.38)
∆xi =
where
FiJ =
is the deformation gradient tensor. In full component notation, we would write
F = Grad(x) =
∂xi
∂
(xi gi ) ⊗ GJ =
gi ⊗ GJ = FiJ gi ⊗ GJ
∂XJ
∂XJ
(2.39)
Because F has both g and G components, it is referred to as a two-point tensor. We say that it is partially in the
deformed configuration, and partially in the undeformed configuration, and that it turns undeformed vectors into
deformed vectors.
All content © 2016-2018, Brandon Runnels
6.5
Metric changes, eigenvalues, eigenvectors
Lecture 7
2.4
Metric changes
We can use this new technology to describe how various quantities in the body (length, area, angle, etc.) change
as they are acted on by the deformation mapping.
2.4.1
Change of length
Let us begin by considering a small vector ∆X in the neighborhood of X as we had before. What is the change in
length as ∆X is acted on by φ? Let F = F(X) be the local deformation gradient. Then
q
√
√
p
p
|∆x| = ∆xk ∆xk = (FkI ∆XI )(FkJ ∆XJ ) = ∆XI FIkT FkJ ∆XJ = ∆XT F T F ∆X = ∆XT C ∆X
(2.40)
where
C = FT F CIJ = FkI FkJ ≡ Right Cauchy-Green Deformation Tensor
(2.41)
Notice that C has strictly uppercase indices: this implies that it lives entirely in the Lagrangian frame. If we divide
both sides by ∆x, we have
|∆x|2
= NI CIJ NJ = NT C N = λ2 (N) ≡ Stretch Ratio
|∆X|2
(2.42)
where N = ∆X/|∆X| is the unit vector in the direction of ∆X, and the stretch ratio is simply the ratio of the deformed
length to the undeformed length.
2.4.2
Change of angle
Consider two vectors ∆X, ∆Y.
∆Y
Θ
∆X
φ
∆y
θ
∆x
We can find the angle between them in the undeformed configuration by using the inner product:
cos Θ =
∆X · ∆Y
|∆X||∆Y|
(2.43)
In the deformed configuration, the angle between them is given by
cos θ =
∆xT ∆y
∆XT FT F∆Y
NT
X C NY
=
=
|∆x||∆y|
(λ(NX )|∆X|)(λ(NY )|∆Y|)
λ(NX ) λ(NY )
(2.44)
where NX = ∆X/|∆X|, NY = ∆Y/|∆Y| are the unit vectors in the directions of ∆X, ∆Y.
All content © 2016-2018, Brandon Runnels
7.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 7
https://canvas.uccs.edu/courses/22031
Example 2.3: Relationship of stretch, angle to components of C
We can express C in full component notation as
(2.45)
C = CIJ GI ⊗ GJ = CIJ GI GT
J
What is the change of length of basis vector G1 ?
q
q
p
p
T )G =
T
(C
G
G
CIJ (GT
CIJ δ1I δ1J = C11
λ(G1 ) = GT
IJ
I
1
1
1 GI )(GJ G1 ) =
J
(2.46)
and similarly for G2 , G3 , and so on. What about the change in angle between unit vectors? Consider vectors
GP , GQ . The change in angle is
GT
P C GQ
no sum on P, Q
λ(GP )λ(GQ )
CPQ
=p
no sum on P, Q
CPP CQQ
(2.47)
cos θPQ =
(2.48)
We notice a general trend here, as with the deformation gradient: diagonal terms relate to elongation,
whereas off-diagonal terms relate to shear.
2.4.3
Determinant identities
Recall the formulae for the determinant of a tensor F ∈ L(R3 , R3 ):
det(F) = ijk Fi1 Fj2 Fk3
det(F) =
1
ijk IJK FiI FjJ FkK
6
(2.49)
= det(F) IJK
(2.50)
Identity:
ijk FiI FjJ FkK
2.4.4


+ det(F) IJK = 123, 231, 312
= − det(F) IJK = 321, 132, 213


0
else
Change of volume
Consider the parallelpiped defined by three vectors ∆X, ∆Y, ∆Z. What is the volume of this parallelpiped? We
know that we can get it by taking the triple scalar product, that is, ∆X · (∆Y × ∆Z). In index notation, this comes
out to be
∆V = ∆X · (∆Y × ∆Z) = (∆XI GI ) · (∆YJ GJ ) × (∆ZK GK ) = ∆XI ∆YJ ∆ZK GI · (JKL GL )
(2.51)
(2.52)
= JKL ∆XI ∆YJ ∆ZK δLI = IJK ∆XI ∆YJ ∆ZK
Now, suppose the volume is mapped to a deformed configuration by a mapping φ:
∆Z
∆z
φ
∆v
∆x
∆V
∆Y
∆y
∆X
All content © 2016-2018, Brandon Runnels
7.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 7
https://canvas.uccs.edu/courses/22031
What is the volume of the deformed element in terms of the undeformed element?
(2.53)
∆v = ∆x · (∆y · ∆z) = ijk ∆xi ∆yj ∆zk
= ijk (FiI ∆XI )(FjJ ∆YJ )(FkK ∆ZK ) = ijk FiI FjJ FkK ∆XI ∆YJ ∆ZK
(2.54)
= det(F ) IJK ∆XI ∆YJ ∆ZK = J ∆V
|
{z
}
(2.55)
∆V
where
J = det(F) =
∆v
≡ Jacobian
∆V
(2.56)
is the ratio of the deformed volume to the undeformed volume. Notes:
(1) Nonzero determinant ≡ no vanishing mass ≡ invertibility of the deformation mapping
(2) The constraint J > 0 makes sense in this sense; negative volume only exists if the volume is turned inside
out.
This allows us to make the connection between the invertibility of the deformation gradient with the fact that the
volume of a section of the body cannot go to zero. It should also be noted that in (2.56) ∆v and ∆V denote
differential (i.e. small) final and initial volumes, respectively, and not changes in volume, which could be mistaken
due to the use of deltas.
2.4.5
Change of area
Consider two vectors ∆X, ∆Y. What is the area of the parallelogram that they span? We get that by the cross
product, so the area vector is
(2.57)
∆A = IJK ∆XI ∆YJ Gk
We note that ∆A = |∆A| where ∆A is the magnitude of the area. Then
(2.58)
∆A = N ∆A
where N is the unit vector normal to the surface. Now, consider the action of a deformation mapping on the vectors:
φ
∆A
N
∆a
n
∆x
∆Y
∆X
∆a
∆A
∆y
What is the area of the new vectors? Let us follow a procedure similar to that for the change of volume:
∆ak = (∆x × ∆y)k = ijk ∆xi ∆yj = ijk (FiI ∆XI )(FjJ ∆YJ ) = ijk FiI FjJ ∆XI ∆YJ
(2.59)
Now,we’re going to pull a trick: multiply both sides by FkK . This gives us
∆ak FkK = ijk FiI FjJ FkK ∆XI ∆YJ = J IJK ∆XI ∆YJ = J ∆AK
(2.60)
Now, we can rewrite the above in invariant notation
FT ∆a = J∆A
All content © 2016-2018, Brandon Runnels
(2.61)
7.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 7
https://canvas.uccs.edu/courses/22031
(Why the F T ? Out of necessity: in order to make everything line up like it’s supposed to in invariant notation, we
had to shuffle things around a little.) Isolating the expression for ∆a gives us
∆a = J F−T ∆A
≡ Piola Transform
∆ai = J F−T
iJ ∆AJ
(2.62)
The Piola transform, or “Nanson’s formula” will be useful later on when we start talking about forces per unit deformed or per unit undeformed area.
Note: Covariance and contravariance of vectors
We are now at an ideal point to introduce an important (yet frequently neglected) subject: the covariance
and contravariance of vectors in Rn . We will motivate the need for this distinction by means of an example.
Let us consider two of the metric changes that we have introduced: change of length and change of area.
In particular, let us consider two vectors T and N, where both have the same components, but T is a length
vector, while A = AN is an area vector. In the figure below, both transform under φ, where φ is a simple shear
deformation:
X2
φ, F
x2
t
a
T
A
X1
x1
Notice that the distance vector t has a different direction and magnitude, whereas the area vector a is unchanged. We recall that this is because they transform differently:
t = FT
aT = J AT F−1
(2.63)
This gives us a strong indication that there are two kinds of vectors. The first kind (T, t) transform by the
deformation gradient and are called contravariant vectors. The second kind (A, a) transform by the inverse
of the deformation gradient and are called covariant vectors. A couple things are worth noting:
(1) Contravariant vectors are usually just called vectors. Covariant vectors are frequently referred to as
“covectors,” “dual vectors,” or “1-forms.”
(2) Contravariant vectors have a natural representation as column vectors, while covectors have a natural
representation as row vectors. This is why the above transformation is written in terms of AT , aT .
(3) Contravariant vectors frequently correspond to line elements, whereas covectors frequently correspond to surface elements. As a result, it is very important to distinguish between directional vectors
and normal vectors.
It is quite possible to go all the way through continuum mechanics and vector calculus without any notion
of the distinction between vectors and covectors, and the finer details are outside of the scope of this class.
However, you will see that there is a subtle yet deep and fundamental difference between them. They begin
to introduce the theme of duality that is intrinsic to constitutive theory, and makes some strong ties between
constitutive modeling and the geometry of physical quantities.
All content © 2016-2018, Brandon Runnels
7.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 7
https://canvas.uccs.edu/courses/22031
Example 2.4: Metric changes in a planar deformation
Consider the following deformation
X2
γ
λ
1
θ
X1
1
The deformation mapping is given by
x1 = X1 + γX2
x2 = λX2
X3 = X3
And from that we can quickly find the deformation gradient and the Cauchy-Green stretch tensor
"
#
"
#"
# "
#
1
γ
0
1 γ 0
1 0 0 1 γ 0
2
2
F= 0 λ 0
C= γ λ 0 0 λ 0 = γ γ +λ 0
0 0 1
0 0 1 0 0 1
0
0
1
All content © 2016-2018, Brandon Runnels
(2.64)
(2.65)
7.5
Lecture 8
Tensor properties and the spectral theorem
Example 2.4 Continued
From this we can compute the following metric changes:
(i) Angle change θ: we can obtain this by using G1 , G2 as our unit vectors. First, compute the stretches:
q
q
p
p
CG
=
C
=
1
λ(G
)
=
GT
γ 2 + λ2
(2.66)
λ(G1 ) = GT
1
1
2
1
2 CG2 =
Now, use the angle formula:
cos θ12 =
λ
GT
1 CG2
=p
2
λ(G1 )λ(G2 )
γ + λ2
(2.67)
(ii) Length change: let us compute
the deformed length of the diagonal line. The unit vector corresponding
√
to this is N = (G1 + G2 )/ 2, so the stretch ratio is
r √
1
T
λ(N) = N CN =
(1 + γ)2 + λ2
(2.68)
2
p
√
So, if the original length of the diagonal was 2, the deformed length is just (1 + γ)2 + λ2 . (A quick
geometric calculation will verify this.)
(iii) Volume change: very easy, all we have to do is compute the determinant:
v
= v = det(F ) = λ
V
(iv) Area change: to do this, we must use the Piola transform. Undeformed area vector is
" #
0
A = (1)(N) = 0
1
(2.69)
(2.70)
To use the Piola transform, we must invert the deformation gradient. There are a number of ways to
do this, but the easiest way here is to use the cofactor method:
"
# "
#
λ −γ
0
1 −γ/λ 0
1
−1
0 1
0
= 0 1/λ 0
(2.71)
F =
det F 0 0 det(F )
0
0
1
"
#
1
0
0
−T
F
= −γ/λ 1/λ 0
(2.72)
0
0
1
Now we can compute our new area vector:
"
a = JF
−T
#" # " #
1
0
0 0
0
A = (λ) −γ/λ 1/λ 0 0 = 0
λ
0
0
1 1
(2.73)
Thus, the magnitude of the area is a = λ. Looking back, the answer is obvious: with the thickness
constant, the area change would be the same as the volume change. We also know that the normal
vector would not change since everything was in 2D. So, we see that we recover the right answer using
the Piola transform.
All content © 2016-2018, Brandon Runnels
8.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
2.5
Course Notes - Lecture 8
https://canvas.uccs.edu/courses/22031
Tensor decomposition
We have introduced the tensors F (deformation gradient) and C = FT F (Cauchy-Green deformation tensor). To
analyze these tensors, we need to review and introduce a couple of linear algebra concepts.
2.5.1
Eigenvalues and Eigenvectors
Recall that we can think of a tensor as a machine that turns a vector into another vector. In general, the tensor
can change both the vector’s magnitude and its direction. But what happens if the tensor changes the vector’s
magnitude only? For instance if we have T ∈ GL(3), u ∈ R3 , then we might write
Tu = λu
(2.74)
with λ ∈ C, where C is the set of all complex numbers, and of course we recall that R ⊂ C, so λ could be real.
Vectors that satisfy this relationship are called eigenvalues, and the value λ are called eigenvalues. We care a lot
about eigenvalues and eigenvectors, because it’s much easier to work with scalars acting on vectors than tensors.
How do we go about finding these eigenvalues and eigenvectors? We know that they will satisfy the equation
Tu − λu = (T − λI)u = 0
(2.75)
Of course, one solution is that u is just zero, but this isn’t very interesting at all because the result would be trivial.
Instead, we notice that the matrix T − λI is able to take a non-zero vector and spit out zero. We say that u is in the
“nullspace” or “kernel” of T − λI, and there’s a nice theorem (called the “Rank-Nullity theorem” that unfortunately
we don’t have time to prove here) that tells us that the tensor T − λI can map a vector u to zero if and only if
det(T − λI) = 0. This means that we can now solve the algebraic equation
det(T − λI) = 0
(2.76)
which is a nth order polynomial. We know that we will find at least one, and no more than three eigenvalues {λi } by
solving this equation. We can then find our eigenvectors {ui } by solving
Tui = λ(i) u(i)
(2.77)
An interesting side note is the following: we can always write
det(T − λI) = c3 λ3 + c2 λ2 + c1 λ + c0 = 0
(2.78)
where (2.78) is the characteristic equation and
• c3 = 1
• c2 = I1 = tr(T) ≡ the first invariant of T
• c1 = I2 = 21 (tr(T)2 − tr(T2 )) ≡ the second invariant of T
• c0 = −I3 = − det(T) ≡ the (negative) third invariant of T
We will continue with our introduction to tensor decomposition. The following theorem is an interesting side note
to the discussion of a tensor’s characteristic equation:
Theorem 2.1 (Cayley-Hamilton). Every square tensor T ∈ L(Rn , Rn ) satisfies its own characteristic equation.
We will also introduce another theorem without proof:
Theorem 2.2. Every square tensor T ∈ GL(Rn , Rn ) has n linearly independent eigenvectors.
This will come in handy in following sections.
All content © 2016-2018, Brandon Runnels
8.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
2.5.2
Course Notes - Lecture 8
https://canvas.uccs.edu/courses/22031
Symmetric and positive definite tensors
We recall the definition for a symmetric matrix:
Definition 2.7. A square tensor T is symmetric if T = TT . If T is symmetric and has components {Tij } in an
orthonormal basis, then Tij = Tji
We note the following important identity: if T is symmetric, then for u, v ∈ Rn we can always write
uT Tv = uT TT v = (Tu)T v
(2.79)
Let us make another very important definition:
Definition 2.8. A square tensor T ∈ GL(Rn , Rn ) is positive definite if
uT Tu > 0 ∀u ∈ Rn , |u| > 0
(2.80)
uT Tu = 0 =⇒ |u| = 0
(2.81)
and
(An important equivalence to note here is that a matrix is positive definite if and only if all of its eigenvalues are
real and strictly positive.) Finally, we will make one more similar definition:
Definition 2.9. A square tensor T ∈ L(Rn , Rn ) is positive semidefinite if
uT Tu ≥ 0 ∀u ∈ Rn
(2.82)
Note that the only difference is that this allows for nonzero vectors to be mapped to zero. Matrices are positive
semidefinite if and only iff all of its eigenvalues are real and nonnegative, i.e. it can have eigenvalues with value
zero.
2.5.3
Spectral theorem (symmetric tensors)
The spectral theorem is one of the most powerful theorems in applied mathematics. It has widespread applications
beyond matrices, such as the Fourier and Laplace transforms, or Sturm-Liouville theory. If you haven’t taken a
(graduate level) linear algebra course, you should definitely consider taking one.
The first theorem we will introduce without proof:
Theorem 2.3 (Real eigenvalues). If a tensor T is symmetric and real-valued, all of its eigenvalues are also real.
The proof of this theorem is quite easy, but it involves the use of complex analysis which is a bit beyond the scope
of the course right now. The next theorem is very important, and so we will prove it.
Theorem 2.4 (Orthogonal eigenvectors). If a tensor T ∈ GL(Rn , Rn ) is symmetric, it has n orthonormal eigenvectors.
Proof. We will look at two cases.
(1) Distinct eigenvalues: let {ui } be the eigenvectors of T corresponding to the distinct eigenvalues {λi }, that is,
no two eigenvalues are the same. Now, let us consider two eigenvectors ui , uj with eigenvalues λi , λj , λi 6= λj .
Because the matrix is symmetric, we can write
T
T
T
0 = uT
i Tuj − (Tui ) Tuj = ui (λ(j) uj ) − (λ(i) u(i) ) uj = (λ(i) − λ(j) )(u(i) u(j) ) = 0
(2.83)
Since λi 6= λj , the only way for this to be true is for ui to be orthogonal to uj .
All content © 2016-2018, Brandon Runnels
8.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 8
https://canvas.uccs.edu/courses/22031
(2) Degenerate eigenvalues: let ui , uj be two eigenvectors of T both with eigenvalue λ. Then we can write, ∀α, β ∈
R,
T(αui + βuj ) = αTui + βTuj = λ(αui + βuj )
(2.84)
In other words, every vector αui + βuj is an eigenvector of T. More generally, we can write that
Tu = λu ∀u ∈ span(ui , uj )
(2.85)
that is, that there is a 2D space of eigenvectors of T with eigenvalue λ. We know from linear algebra that
there exists an orthonormal basis for this space; this orthonormal basis gives the orthogonal eigenvectors
for T . The case of 3+ degenerate eigenvalues proceeds by induction.
Now that we know that every symmetric matrix T has a set of orthonormal eigenvectors, we can use these eigenvectors to construct an orthonormal basis. We call this the eigenbasis of T.
All content © 2016-2018, Brandon Runnels
8.4
Spectral and polar decomposition
Lecture 9
Theorem 2.5 (Spectral decomposition). A symmetric matrix T can be expressed strictly in terms of rotation matrices
constructed from its eigenvectors and a diagonal matrix with each diagonal entry containing an eigenvalue of T.
Proof. Let us consider the action of T on any vector x ∈ Rn . By casting x into the eigenbasis of T, then by finding
the action of T on its eigenvectors, and finally casting back, we obtain
X
X
X
T
Tx = T(ei xi ) = Tvj (vjT ei )xi =
λj vj (vjT ei )xi =
vj λj (vjT ei )xi =
ek (eT
(2.86)
k vj )λj (vj ei )xi
j
j
j
Introducing a couple of Kronecker deltas and rearranging indices gives
= ei (eT
vp )
| i{z }
Qip
X
r
|
δpr λr δrq (vqT ej ) xj
| {z }
{z
} QqjT
(2.87)
Λpq
concluding the proof.
in other words, the components of T in the original basis can be expressed as
(2.88)
T
Tij = Qip Λpq Qqj
where Q are orthogonal transformation matrices whose components are given in terms of the eigenvectors v by
"
#
e1 · v1 e1 · v2 e1 · v3
Qij = ei · vj
Q = e2 · v1 e2 · v2 e2 · v3 = [v1 v2 v3 ]
(2.89)
e3 · v1 e3 · v2 e3 · v3
and Λ is a diagonal matrix with the eigenvalues λi along the diagonal. This is called diagonalization; it can also be
referred to as spectral decomposition or eigendecomposition.
So what are we actually doing when we do an eigendecomposition? The important thing to remember is that we
are not changing the tensor at all. Rather, we are finding a different way to express the tensor by decomposing it
into seperate stages. Consider the following illustration:
e2
u2
e1
u1
QT
e2
u2
u1
Λ
e1
Q
A tensor T applies shear and normal deformation to a square. We can express this deformation as a rotation
into the eigenspace of T, applying pure stretching, and rotating back. Note that the diagonals of the square are
orthogonal, and do not change direction under the deformation. These are the eigenvectors of the deformation,
and their stretches are the eigenvalues.
Theorem 2.6. All of the eigenvalues of a symmetric positive definite tensor A are positive.
The proof of this theorem is straightforward.
All content © 2016-2018, Brandon Runnels
9.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
2.5.4
Course Notes - Lecture 9
https://canvas.uccs.edu/courses/22031
Spectral theorem (general case)
We will not go into this in as much detail, but let us consider a nonsymmetric tensor A. Then A has n linearly
independent eigenvectors, but they are not necessarily orthogonal. We say that A is diagonalizable if, given that A
has components Aij in the reference frame,
(2.90)
A = UΛU−1
where Λ is the diagonal matrix of eigenvalues, U is the matrix of eigenvectors, and U−1 is the inverse of U. Note
that if A is orthogonal, U−1 becomes UT , and we recover the earlier result.
Not all tensors are diagonalizable. If a tensor is not diagonalizable it is calle defective. In general, for purposes of
this course, however, we don’t need to worry about those cases.
2.5.5
Functions of tensors
The spectral decomposition has a wide range of applications, and one of the most useful is to describe functions
on matrices. Just like with scalar numbers, we can raise a matrix to a power. Let us consider a diagonalizable
tensor A. What is An ?

 n
λ1
n
λ2
−1
−1
−1
−1
U
(2.91)
... Λ} U = U−1 
An = |AAAA
{z ... A} = (U ΛU)(U ΛU) ... (U ΛU) = U ΛΛ
| {z
..
.
n times
n times
so we see that we can raise the matrix to the n-th power simply by raising the eigenvalues to the n-th power. This
means that we can conceivably have polynomials of tensors.
Now, let us suppose we have f : R → R, a smooth function that admits the Taylor Series expansion
f (x) =
∞
X
1 (n)
f (0)x n
n!
n=0
(2.92)
For instance, we know that exponential function has the Taylor series expansion
exp(x) = e x =
∞
X
xn
n=0
n!
(2.93)
It turns out that we can use the Taylor series expansion to define functions of matrices. Suppose we have a function
f . Then we can describe the action of f on a diagonalizable tensor A by expanding in Taylor series form:
P∞ 1 (n)

(0)λn1 P
∞
n=0 n! f
X
∞ 1 (n)
1 (n)


(0)λn2
f (A) =
f (0)An = U−1 
(2.94)
n=0 n! f
U
n!
..
n=0
.


f (λ1 )
f (λ2 )
 U = U−1 f (Λ) U
= U−1 
(2.95)
..
.
That is, we can describe the effect of a scalar function on a tensor, simply by taking the spectral decomposition
and evaluating the function on each of the eigenvalues. For instance, if A ∈ L(R3 , R3 ) is diagonalizable, we have



√
"
#
λ1 √
e λ1
ln(λ1 )
√
 U, ln(A) = U−1
ln(λ2 )
A = U−1 
U,
e λ2
λ2 √  U, e A = U−1 
λ3
ln(λ3 )
e
λ3
(2.96)
and so on. Many of the same properties of f hold when extended to the matrix case, although it’s not always safe
to make this generalization. For instance, if two tensors A, B have the same eigenbasis, you can even show that
some properties of f such as e A+B = e A e B , ln(AB) = ln(A) + ln(B) hold; however, this is trickier if they do not.
All content © 2016-2018, Brandon Runnels
9.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
2.5.6
Course Notes - Lecture 9
https://canvas.uccs.edu/courses/22031
Polar decomposition
We work with a lot of very nice symmetric positive definite tensors, but what about when it is not symmetric or
positive definite? Specifically, the deformation gradient F does not satisfy either of those properties. For this case,
we will use the polar decomposition.
Theorem 2.7 (Right polar decomposition). For every tensor T ∈ GL(n) with det(T) > 0 there exists a symmetric
positive-definite tensor U ∈ GL(n) and a rotation R ∈ SO(n) such that T = RU
Proof. Proceed formally:
(1) TT T is symmetric positive definite, because vT TT Tv = |Tv| = 0 only if Tv = 0. But if there exists a nonzero
v ∈ Rn such that Tv = 0, then det(T) = 0, a contradiction with the fact that T ∈ GL(n). Therefore TT T is
positive definite. It is symmetric by inspection.
√
(2) Because TT T is symmetric positive definite, there exists a symmetric positive definite tensor U = TT T,
where UU = TT T.
(3) Suppose T = RU: then R ∈ SO(n). R = TU−1 , and RT R = U−T TT TU−1 = U−1 UUU−1 = II = I, proving
that RT = R−1 and therefore R ∈ O(n). Now, show that det(R) = 1: det(R) = det(T) det(U−1 ) > 0, implying
that det(R) = 1. Therefore R ∈ SO(n).
In other words, even if a tensor lacks the nice properties of being symmetric (or even positive definite), we can
always decompose it into a “nice” tensor and a “rotation” tensor. Intuitively, we can interpret this as meaning that
all of the nastiness of an asymmetric deformation gradient (i.e. a def. grad. that has complex eigenvalues) is due
to rotation.
Now, we can find a similar decomposition such that the rotation matrix is on the left:
Theorem 2.8 (Left polar decomposition). For every tensor T ∈ GL(n) with det(T) > 0 there exists a symmetric
positive-definite tensor V ∈ GL(n) and a rotation R ∈ SO(n) such that T = VR
Proof. By the previous theorem, we know T = RU, where R is the rotation for the left polar decomposition. Let
V = TRT = RURT . Then VT = RTT UT RT = RURT = V, verifying symmetry, and uT Vu = uT RURT u =
(RT u)T U(R T u) > 0, verifying positive definiteness.
√
Remark: VV = RURT RURT = RUURT = RU(RU)T = TTT so V = TTT
Intuitively, this means that we can get the same result whether we rotate first then deform, or deform first then
rotate.
2.6
Principal deformations
Now that we’ve developed a complete set of tools for decompositing tensors, let us apply them to our deformation
tensors.
Consider a mapping with a local deformation gradient F. We know by example that the deformation gradient is not
symmetric or positive definite. This is because the deformation gradient includes a rigid-body rotational component
– the part that actually goes towards deforming the material should be symmetric positive definite. We can use
the polar decomposition to decouple these two components.
We know that F ∈ GL(3) and satisfies det(F) > 0, therefore there exists R ∈ SO(3) such that
√
√
√
√
F = UR = RV
U = FT F = C
V = FFT = B
(2.97)
where we introduce
B = FFT ≡ Left Cauchy-Green deformation tensor
(2.98)
What does this mean intuitively? This decomposition is best illustrated pictorally. Consider a cube undergoing
simple shear.
All content © 2016-2018, Brandon Runnels
9.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 9
https://canvas.uccs.edu/courses/22031
R
U
n2
N2
N1
n1
F = RU = VR
R
V
Note that the two vectors N1 , N2 transform into λ1 n1 , λ2 n2 , where λ1 = λ(N1 ), λ2 = λ(N2 ) are the stretches, and
n1 , n2 remain orthogonal. Not all vectors transform this way, but (as we’ll see) we will always be able to find 2 (or
3 in 3D) vectors that remain orthogonal to each other under transformation. These special vectors are called the
principal directions, and are referred to as the material principal directions (N1 , N2 , N3 ) in the undeformed configuration, and as the spatial principal directions n1 , n2 , n3 in the deformed configuration. Moreover, these directions
are related to each other via the rotation tensor, that is
ni = RNi
(2.99)
where we obtain R from the polar decomposition. The length change λi are referred to as the principal stretches,
and we know that
FNi = λ(i) n(i)
(2.100)
How do we go about finding our principal directions and stretches? There are two ways: beginning with the above
expression:
FN = λn =⇒ RUN = λn =⇒ UN = λRT n =⇒ UN = λN
(2.101)
FN = λn =⇒ VRN = λn =⇒ Vn = λn
(2.102)
√
√
So, we see that Ni are eigenvectors
√ of C√and ni are eigenvectors of B. Furthermore, we see that the principal
stretches are eigenvalues of both C and B. We recall from the spectral theorem that
√
√
C = UT ΛU
(2.103)
√
where
Λ is the diagonal matrix of the square root of eigenvalues of C. Hence, we conclude that the eigenvalues
√
of C are the square root of the eigenvalues of C, and that the eigenvectors are the same. To recap, here is a
summary of the principal stretches/directions and how to find them:
{λi }
{Ni }
{ni }
√
= eigenvalues of C
= eigenvectors of C, U
= eigenvectors of B, V
All content © 2016-2018, Brandon Runnels
≡ Principal Stretches
≡ Material Principal Directions
≡ Spatial Principal Directions
(2.104)
9.4
Lecture 10
2.7
Continuous and Hadamard compatibility
Compatibility
We have seen that the deformation gradient F is an extremely handy way of working with deformations. In fact,
we will often think of deformations as being defined by their deformation gradient at each point. However, just
because we have a deformation gradient F(X) does not mean that it is “realistic” – that is, it may not correspond
to an actual deformation. So, we introduce the notion of compatibility:
Definition 2.10. A deformation gradient field over a domain B is said to be compatible is there exists a mapping φ
such that F = Grad(φ).
One way to ensure that a deformation gradient is compatible is by finding the deformation mapping φ. However, we
may not always be able to do this; it is generally easier to find local conditions on F that do not require integration.
2.7.1
Continuous case
We will begin with the case of smooth mappings, where φ is assumed to be twice differentiable. The following
theorem provides a powerful way to determine compatiblity by differentiation.
Theorem 2.9. A deformation gradient field is compatible if and only if Curl(F) = 0.
Proof. This theorem states that zero curl is a necessary and sufficient condition; therefore we must prove it both
ways.
Necessary: Use the symmetry of differentiation to show that
FiJ,K =
∂ 2 φi
∂ 2 φi
=
= FiK ,J
∂XJ ∂XK
∂XK ∂XJ
(2.105)
Now we pull a trick: using the fact that these two terms are equal, we multiply by permutation tensor and
swap indices to show that
FiJ,K − FiK ,J = 0
=⇒ (FiJ,K − FiK ,J )IJK = (IJK FiJ,K − IJK FiK ,J ) = (IJK FiJ,K + IJK FiJ,K ) = Curl(F )iI = 0
(2.106)
(2.107)
So if a deformation mapping exists, the curl must be zero.
Sufficient: Stoke’s theorem is similar to the divergence theorem (indeed, both are special cases of a single generalized theorem) that states the following: for a surface A with boundary ∂S, for some vector or tensor field
denoted generally as F
Z
Z
Curl(F )iJ NJ dA =
FiJ dXJ
(2.108)
S
∂S
Now we apply this to the deformation gradient. We are given that Curl(F) = 0 everywhere, so we have
Z
Z
0=
IJK FiJ,K dAK =
FiJ dXJ
(2.109)
S
∂S
with the implication that integrating F over every closed contour returns zero.
All content © 2016-2018, Brandon Runnels
10.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 10
https://canvas.uccs.edu/courses/22031
B
Γ0
S
Γ00
A
Ω
Consider two points a, b in the undeformed configuration, and two contours Γ0 , Γ00 going from a to b, forming
a closed contour so that
Z
Z
Z
Z
FiJ dXJ −
FiJ dXJ = 0 =⇒
FiJ dXJ =
FiJ dXJ
(2.110)
Γ0
Γ00
Γ0
Γ00
Now, let us define
Z
φ(X) =
(2.111)
FiJ d X̂J .
γ[A,X]
where γ can be any differentiable contour, and X̂J distinguishes the variable of integration from the argument
of φ. If we can show that Grad(φ) = F, we will be done. To do this, we evaluate the Gateaux derivative of φ
for some arbitrary vector V:
Z
Z
Z
i
d
1h
FiJ d X̂J
FiJ d X̂J −
FiJ d X̂J
= lim
(2.112)
Dφ(X)V = lim
ε→0 ε
ε→0 dε γ[A,X+εV]
ε=0
ε=0
γ[A,X+εV]
γ[A,X]
Path independence allows us to break up the integral:
1h
ε→0 ε
Z
= lim
Z
γ[A,X]
Z
FiJ d X̂J −
FiJ d X̂J +
γ[X,εV]
FiJ d X̂J
γ[A,X]
i
=
h
ε=0
1
ε→0 ε
Z
FiJ d X̂J
lim
γ[X,εV]
i
ε=0
(2.113)
Let γ[X, X + εV] be linear, and introduce the parameterization X̂I = XI + sVI , d X̂I = VI ds:
Z
i
i
hd Z ε
h
i
1 ε
FiJ (X + sV)VJ ds
FiJ (X + sV)VJ ds
=
= FiJ (X + εV)VJ
ε→0 ε 0
dε 0
ε=0
ε=0
ε=0
= FiJ (X)VJ
=
h
lim
(2.114)
(2.115)
whence we identify FiJ = Grad(φ), resulting from the fact that the above holds for all V.
Thus the condition Curl(F) = 0 is both necessary and sufficient for compatibility.
This theorem is quite useful because we don’t always know if a deformation mapping exists. By this, we see that
we only have to check that the curl is zero to ensure that it the deformation gradient for a real mapping.
A notable exception/special case of this is dislocations. Dislocations introduce a discontinuity into the deformation
mapping, and as such, they create a singularity. This means that the curl of the deformation gradient is nonzero
whenever there is a dislocation. In fact, we can actually use Curl F to determine the dislocation content of a sample.
All content © 2016-2018, Brandon Runnels
10.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
2.7.2
Course Notes - Lecture 10
https://canvas.uccs.edu/courses/22031
Discontinuous case (Hadamard)
In the above section we have considered only deformations that are twice differentiable, and consequently fairly
smooth. There are a number of cases where the deformation is not smooth, however. One example is the formation of laminate microstructure. Sometimes in order to conform to an imposed deformation gradient, a material
“finds it” more energetically favorable to create alternating variants of microstructure that approximate the imposed
deformation, as in the following figure:
The resulting deformation map is only C0 continuous, because the deformation gradient jumps across the boundary. In the following figure, a material undergoes a discontinous deformation:
X2
x2
N
F+
t
T
F−
X1
x1
Suppose we only know F + , F − . What constraints must exist on F+ and F− to ensure that they correspond to a
deformation mapping? The answer is quite simple: consider a vector T in the boundary. We know that
or alternatively that
F+ T = F− T
!
(2.116)
(F+ − F− )T = [[F]]T = 0 ∀T ∈ S
(2.117)
where S is the boundary plane and the double bracket [[·]] denotes “jump in.” We know that S is two-dimensional;
therefore because [[F ]] maps all vectors in a 2D space to zero, it has a kernel of dimension 2. By the Rank-Nullity
theorem, this implies that it is rank 1.
Definition 2.11 (Hadamard compatibility). For a discontinuous deformation gradient across a boundary with normal
N in the undeformed configuration, the deformation gradients are Hadamard or Rank-1 Compatible if there exists
an a ∈ R3 such that
[[F]] = a ⊗ N
(2.118)
Let’s check this: what happens when [[F]] acts on T ∈ S?
[[F]]T = (a ⊗ N)T = a (NT T) = a(N · T) = 0
(2.119)
because N is normal to the plane containing T.
All content © 2016-2018, Brandon Runnels
10.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 10
https://canvas.uccs.edu/courses/22031
Example 2.5: Shear twinning
Consider a material that approximates a shear deformation of γ by creating two layers that shear with known
values γ1 , γ2 , respectively, as shown.
x2
γ2
γ
λ2
γ1
1
λ1
x1
1
Check that the interface is compatible and find λ1 , λ2
First, let us write down the deformations that we know:
h
i
h
i
1 γ
1 γ
F= 0 1
F1 = 0 11
Check compatibility:
h
0
[[F]] = F2 − F1 = 0
h
1
F2 = 0
γ2
1
i
(2.120)
i h
i
γ2 − γ1 0 1
γ2 − γ1
=
[
] X
0
0
| {z } | {z }
(2.121)
γ1 λ1 + γ2 λ2 = γ
(2.122)
N
a
We also know the following
λ1 + λ2 = 1
which allow us to solve for λ1 , λ2 :
λ1 =
γ2 − γ
γ2 − γ1
λ2 =
γ − γ1
γ2 − γ1
(2.123)
Given that 0 ≤ λ1 , λ2 ≤ 1, we also conclude that γ2 ≤ γ ≤ γ1
2.8
Other deformation measures
We are now familiar with using F and C as measures of local deformation. However, you may notice that these are
somewhat different from the traditional measure of strain. In particular, the C and F tensors corresponding to zero
deformation are the identity, rather than the zero tensor. While this is not really problematic, some prefer to use
tensors for which zero deformation corresponds to the zero tensor. The first example is the Green-Lagrange strain
tensor:
E=
1
(C − I) ≡ Green-Lagrange Strain Tensor
2
(2.124)
Notice that F = I =⇒ C = I =⇒ E = 0. Also, note that for R ∈ SO(3), if F = R then
E=
1 T
1
(R R − I) = (I − I) = 0
2
2
(2.125)
Furthermore, we see that E inherits invariance properties because it depends on C only: if F = RU, the R0 will
always disappear leaving U only.
All content © 2016-2018, Brandon Runnels
10.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 10
https://canvas.uccs.edu/courses/22031
A slightly less ad hoc deformation measure is the log strain tensor:
ln(C) ≡ Logarithmic Strain Tensor
(2.126)
Note that if F = I, ln(C) = 0. Another satisfying property of this is the following: if we write out ln(C ), we see that
"
#
2 ln(λ1 )
T
2 ln(λ2 )
ln(C) = Q
Q
(2.127)
2 ln(λ3 )
If λ1 , λ2 , λ3 go to zero, this causes ln(C) to go to infinity. This causes the non-zero-deformation property of F to be
“built-in” to the deformation tensor, which is very convenient. However, the log strain tensor comes with a couple
of problems. First, it is somewhat costly to compute computationally as it requires a full eigendecomposition of
C . Second, and more importantly, we know that ln(x)p for 1 ≤ p < ∞ will always scale sublinearly. This means
that we cannot construct an energy function W (ln(C)) that is convex – the log term will always restrict the growth
of the energy with deformation.
All content © 2016-2018, Brandon Runnels
10.5
Lecture 11
2.9
Linearized kinematics
Linearized kinematics
Up until now we have left everything completely general: the measures of deformation that we have formulated will
hold for any continuous deformation. This is great, but the problem is that these deformation measures are often
a bit unwieldy to use. On the other hand, there are many cases where the amount of deformation that the material
undergoes is very small. As a result, it is often advantageous to linearize our deformation measures.
Suppose G is some arbitrary function of a deformation measure, such as the deformation gradient or the Jacobian.
We express the linearization of G about a deformation φ in terms of the Gateaux derivative:
G [φ(X) + u] = G (φ) + DG (φ)u + higher order terms
(2.128)
where u is a displacement from φ that is small enough so that higher order terms are negligible.
For example, the linearized deformation gradient would be
F(φ + u) ≈ F(φ) + DF(φ)u = F0 +
d
Grad(F + εu)
dε
ε→0
= F0 + Grad u
(2.129)
where F0 is the large deformation and u is the small displacement. It is possible to do the entire analysis with an
arbitrary “large” deformation combined with a small deformation; however, we will assume here that F = I, so that
(2.130)
F ≈ I + Grad u
This has a number of implications, so when we say we are working with “small strains:”
(1) We assume that the material and spatial coordinates coincide
(2) We describe deformation in terms of the displacement u and its gradient rather than the deformation mapping
φ and its gradient.
(3) We assume that u is small enough so that higher powers of u is neglected; every result should be linear in u
or its gradient.
We now proceed to linearize the other measures of deformation that we have introduced. One possibility for doing
this is to apply (2.128) to every deformation measure that we have used; we would indeed obtain the correct result.
An easier (though less general) procedure is to simply substitute (2.130) into our rexpressions, eliminate higherorder terms, and Taylor expand when necessary.
Beginning with the right Cauchy Green tensor:
T
C = FT F = (I + Grad u)T (I + Grad u) = I + Grad uT + Grad u + Grad
u{z
Grad u} = I + Grad u + Grad uT
|
higher order term
(2.131)
Note that we drop out the Grad u Grad uT term because it is higher order in Grad u.
Let us apply this to the Green-Lagrange strain tensor:
E=
1
1
1
(C − I) = (I + Grad uT + Grad u − I) = (Grad uT + Grad u) = sym(Grad u)
2
2
2
(2.132)
i.e., E is the symmetric part of Grad u. It turns out that the linearized Green-Lagrange strain tensor is a very convenient choice for measuring deformation in small strains, and so it is given a special symbol:
ε=
1
(Grad u + Grad uT ) ≡ Small Strain Tensor
2
All content © 2016-2018, Brandon Runnels
(2.133)
11.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 11
https://canvas.uccs.edu/courses/22031
ε is symmetric, as we would expect. What about the antisymmetric part of Grad u? We define
r=
1
(Grad u − Grad uT ) ≡ Infinitesimal Rotation Tensor
2
(2.134)
The infinitesimal rotation tensor gives an indication of the rotation at a point, although it breaks down under large
deformation as does the small strain tensor. Note that ε + r = Grad u. This is actually the small-strain analog to
the polar decomposition. For future decompositions of F (such as the elastic-plastic decomposition) we will see
that the tensor-tensor multiplication in large strain boils down to a much friendlier additive decomposition in small
strain.
2.9.1
Linearized metric changes
What happens to our metric change measures when we apply linearization? Let’s begin by considering the stretch.
Recalling our formula for λ(n) where n is the direction of stretch, and substituting our linearized measures of
deformation, we get
q
p
p
√
(2.135)
λ(n) = nT Cn = nT (I + Grad u + Grad uT )m = nT n + 2nT ε n = 1 + 2nT ε n
And now we’re stuck, because we have a nonlinear function of ε. But we know that ε is small, so we can apply a
linearization of the square root function about 1:
1
= 1 + (1 + 2nT ε n − 1) = 1 + nT ε n
2
(2.136)
Rearranging this expression, we can write
∆`
(n) = nT ε n ≡ Engineering Strain
`0
(2.137)
Note that this implies that ε11 is the strain in the x1 direction, and so on; in other words, the diagonals of the small
strain tensor describe the elongation in the x1 , x2 , x3 directions.
Let us follow a similar procedure for angle change. Consider two orthogonal vectors m, n. Applying the formula
developed earlier, we obtain:
cos θ(m, n) =
mT Cn
≈ mT Cn ≈ mT n + 2mT εn = 2mT εn
λ(m)λ(n)
(2.138)
Consider the angle subtended by the deformation of n – we’ll call it γ.
n
γ
θ
m
To first order, we can express γ as
cos θ = sin γ ≈ γ(m, n)
(2.139)
Note that γ is symmetric regarding m, n; this would change for large deformation. Now we can express
γ(m, n) = 2 mT ε n ≡ Shear Angle
All content © 2016-2018, Brandon Runnels
(2.140)
11.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 11
https://canvas.uccs.edu/courses/22031
We note that plugging in the basis vectors shows that the off-diagonals of ε determine the shear, just as they did
for C .
Finally, consider the volume change. Substuting our formula, we find
V
1
= det(F ) = ijk IJK FiI FjJ FkK
V0
6
1
= ijk IJK (δiI + ui,I )(δjJ + uj,J )(δkK + uk,K )
6
1
1
1
1
= ijk IJK (δiI )(δjJ )(δkK ) + ijk IJK (δiI )(δjJ )(uk,K ) + ijk IJK (δiI )(uj,J )(δkK ) + ijk IJK (δiI )(uj,J )(uk,K )
6
6
6
6
1
1
1
1
+ ijk IJK (uiI )(δjJ )(δkK ) + ijk IJK (uiI )(δjJ )(uk,K ) + ijk IJK (uiI )(uj,J )(δkK ) + ijk IJK (uiI )(uj,J )(uk,K )
6
6
6
6
1
1
1
= 1 + ijk IJK (δiI )(δjJ )(uk,K ) + ijk IJK (δiI )(uj,J )(δkK ) + ijk IJK (uiI )(δjJ )(δkK )
6
6
6
1
= 1 + ijk IJK (uiI )(δjJ )(δkK )
2
= 1 + i23 I 23 (uiI ) + 1i3 1I 3 (uiI ) + 12i 12I (uiI )
= 1 + u1,1 + u2,2 + u3,3 = tr(Grad(u))
(2.141)
= 1 + tr(ε)
which can be expressed as
∆V
= tr(ε) ≡ Percent Volume Change
V0
(2.142)
All of our measures of deformation can be expressed in terms of ε, so we see that in small strain, ε plays the same
role as C.
Example 2.6: Pure shear metric changes
Consider a square sample undergoing plane strain shear deformation in 2D.
X2
κ
1
γ
θ
1
X1
Compute length, volume, and angle change in small and large deformation.
"
#
X1 + κX2
X2
φ=
X3
"
#
1 κ 0
F= 0 1 0
0 0 1
"
#
1
κ
0
C = κ 1 + κ2 0
0
0
1
All content © 2016-2018, Brandon Runnels
"
#
κX2
u(X ) = φ − X = 0
0
"
#
0 κ 0
grad u = 0 0 0
0 0 0
"
#
0
κ/2 0
ε = κ/2
0
0
0
0
0
(2.143)
(2.144)
(2.145)
11.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 11
https://canvas.uccs.edu/courses/22031
(1) Change of length: in large and small strain we have
q
p
C11 = 1
λ(G1 ) = GT
1 CG1 =
p
p
λ(G2 ) = C22 = 1 + κ2
∆`
(G1 ) = GT
1 εG1 = 0
`
∆`
(G2 ) = GT
2 εG2 = 0
`
(2.146)
(2.147)
For G1 we see that both the large and small strain measures show no stretch, as expected. For G2 , the
large deformation theory shows an elongation. However, we note that κ must be small for the small
strain to hold, in which case λ(G1 ) = 1, recovering the small strain prediction.
(2) Now we consider the change of angle. Note that sin γ(G1 , G2 ) = cos θ(G1 , G2 ). We have, in large and
small strain,
sin γ(G1 , G2 ) =
κ
GT
1 CG2
=√
λ(G1 )λ(G2 )
1 + κ2
γ(G1 , G2 ) = 2GT
1 εG2 = κ
(2.148)
If κ is small then the large strain recovers sin γ ≈ γ = κ, recovering the small strain approximation.
(3) Change of area: in large and small strain we have
V
= det(F) = 1
V0
∆V
= tr(ε) = 0
V0
(2.149)
The large and small strain predictions are consistent.
2.9.2
Small strain compatibility
Consider the case of purely 2D strain. We know that
ε11 = u1,1
ε12 = ε21 =
1
(u1,2 + u2,1 )
2
ε22 = u2,2
(2.150)
Let us consider the following equation:
ε11,22 − 2ε12,12 + ε22,11 = u1,122 − u1,212 − u2,112 + u2,211
(2.151)
Because of the symmetry of differentiation, we know that this equals
= u1,122 − u1,122 − u2,211 + u2,211 = 0
(2.152)
As with large deformation, a strain field ε is said to be compatible if there exists a displacement field u such that
εij =
1
(ui,j + uj,i )
2
(2.153)
As before, we may only have a strain field without having a displacement field; verifying that (2.151) is equal to zero
is a necessary and sufficient condition for 2D compatibility. In 3D, we have six such equations
All content © 2016-2018, Brandon Runnels
ε11,22 + ε22,11 − 2ε12,12 = 0
(2.154a)
ε22,33 + ε33,22 − 2ε23,23 = 0
(2.154b)
ε33,11 + ε11,33 − 2ε13,13 = 0
(2.154c)
ε11,23 + ε23,11 − ε31,21 − ε12,31 = 0
(2.154d)
ε22,31 + ε23,12 − ε31,22 − ε12,32 = 0
(2.154e)
ε33,12 + ε23,13 − ε31,23 − ε12,33 = 0
(2.154f)
11.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 11
https://canvas.uccs.edu/courses/22031
We can alternatively express the above equations compactly as
curl curl ε = 0
(2.155)
which is necessary and sufficient for compatibility.
All content © 2016-2018, Brandon Runnels
11.5
Conservation laws, Reynold’s transport theorem
Lecture 12
2.10
The spatial/Eulerian picture
We can use the machinery that we’ve developed for linearization to do something analagous. Consider some timedependent deformation mapping φ(X, t), and some measure on φ, we’ll call it G [φ]. We can express the rate of
change of G as the following derivative
Ġ =
∂
∂
G [φ(X, t)] = (grad G ) φ = (grad G )V = DG V .
∂t
∂t
(2.156)
In other words, the time rate of change of G is nothing other than the Gateaux derivative of G in the direction of the
velocity V. This means that the procedure for deriving rates of G is exactly the same as deriving linearizations of
G . As such, we can dispense with the full derivation and jump directly to definition by analogy:
Linearized kinematics
Rates
Displacement
Displacement gradient
Strain tensor
Infinitesimal rotation tensor
Rotation vector
Spatial velocity
Spatial velocity gradient
Rate of strain tensor
Spin tensor
Vorticity vector
ui
βij = ui,j
εij = 21 (βij + βji )
rij = 21 (βij − βji )
αk = 12 rji ijk
vi
`ij = vi,j
dij = 21 (`ij + `ji )
wij = 12 (`ij − `ji )
ωk = 12 wji ijk
Note that even though all of the above quantities are analagous to small strain deformation measures, we are
no longer making the small strain approximation. We are merely showing that there are similarities between the
large strain Eulerian rate quantities and the small strain linearized quantities. It is using these analogies that many
connections are drawn between solid and fluid mechanics.
As a side note, how do we relate these new large deformation measures to our familiar deformation measures
such as F ? The following identity is invaluable in expressing v , `, d, w , etc., in terms of F :
ḞiJ FJj−1 =
∂ ∂xi ∂XJ
∂ ∂xi ∂XJ
∂vi ∂XJ
∂vi
=
=
=
= vi,j = `ij
∂t ∂XJ ∂xj
∂XJ ∂t ∂xj
∂XJ ∂xj
∂xj
(2.157)
Written symbolically, we have the relationship
(2.158)
` = ḞF−1
Let’s use the fact that we can draw analogies between small strain and rate deformation to derive the rate version
of metric changes. Beginning with volume change, we expect that
∆V
= tr(ε)
V
=⇒
V̇
= tr(d)
V
(2.159)
Let’s check to be sure. To do this, let us compute the derivative explicitly.
J̇ =
d
d(det F)
−1
−1
det(F) =
ḞiJ = (JF−T
iJ )ḞiJ = J (ḞiJ FJi ) = J tr(ḞF ) = J tr(`) = J tr(d)
dt
dFiJ
(2.160)
Or, rearranging, we obtain
J̇
V̇
=
= tr(d)
J
V
All content © 2016-2018, Brandon Runnels
(2.161)
12.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 12
https://canvas.uccs.edu/courses/22031
as expected. As a side note, we use the fact that tr(d) = dii = `ii = vi,i = div(v) to obtain the useful identity
(2.162)
J̇ = J div(v)
This will come in handy when deriving time-dependent balance laws.
Having verified that the analogy holds, we can use our small-strain derivations to construct rates of metric changes:
• Rate of stretching: for a vector n the rate of change of its length is given by analogy to be
∆`
(n) = nT εn
`
`˙
= nT d n
`
=⇒
(2.163)
• Rate of angle change: for two vectors m, n that are instantaneously orthogonal to each other, the rate of
change of shear angle is given by analogy to be
γ(m, n) = 2mT εn
γ̇(m, n) = 2mT d n
=⇒
(2.164)
Other rates of metric changes can be derived in a similar fashion.
3
Conservation Laws
We have spent a great deal of time describing and quantifying metrics for the deformation of materials under
mappings φ(X) or time-dependent mappings φ(X, t). Everything that we have done so far is strictly geometric;
that is, we have not imparted any physical laws or restrictions on φ. Now, we introduce the physical restrictions
of the conservation of mass, conservation of linear/angular momentum, and conservation of energy. We will also
introduce a rigorous formulation of the second law of thermodynamics in a continuum mechanics setting.
Before we begin, let us introduce a couple of lemmas and theorems that will prove invaluable as we derive our
conservation laws.
Lemma 3.1 (Fundamental lemma of the calculus of variations). Let Ω be a (Lebesgue measurable) subset of Rn ,
and f : Ω → Rm as vector (or scalar)-valued function over Ω. Then
Z
f (X) · g (X)dX = 0 ∀g : Ω → Rm
(3.1)
Ω
if and only if
(3.2)
f (X) = 0
We call (3.1) the weak form and (3.2) the strong form of the equation f (X) = 0. A similar lemma is easily derived
from the above:
Lemma 3.2. Let Ω be a (Lebesgue measurable) subset of Rn , and f : Ω → Rm as vector (or scalar)-valued function
over Ω. Then
Z
f (X) dX = 0 ∀B ⊂ Ω (Lebesgue measurable)
(3.3)
B
if and only if f (X) = 0
We now introduce a very important theorem. Consider a body that is undergoing a time-dependent deformation.
φ
Ω
φ(Ω, t)
X
All content © 2016-2018, Brandon Runnels
x(t)
12.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 12
https://canvas.uccs.edu/courses/22031
Suppose there is a field defined over the body in the Eulerian configuration, we want to evaluate the time rate of
change of an integral quantity evaluated over the deformed body. How do we do it? We will show how this is done
for the general case by introducing the Reynold’s transport theorem:
Theorem 3.1 (Reynold’s Transport Theorem). Let Ω be a (Lebesgue measurable) subset of Rn , let φ : Ω × R → R3
be a time-dependent mapping, and let f : φ(Ω, t) × [t1 , t2 ] → Rn be a spatial/Eulerian function. Then
Z
Z
df
d
f dv =
+ f div(v) dv
(3.4)
dt φ(Ω,t)
φ(Ω,t) dt
Proof. Let us begin by considering the left hand side:
Z
d
f dv
dt φ(Ω,t)
(3.5)
We want to take the derivative inside the integral, but we can’t because the bound on the integral itself is timedependent. To remedy this, we will perform a change of variables to evaluate the integral in the undeformed configuration. Then we have
φ(Ω, t) → Ω
dv = J dV
(3.6)
Our integral thus becomes
d
dt
Z
Z
f J dV =
Ω
Ω
d
(f J) dV
dt
We now evaluate the derivative using the product rule, and recalling (), we obtain
Z Z Z df
df
df
=
J + f J̇ dV =
J + f J div(v) dV =
+ f div(v) J dV
Ω dt
Ω dt
Ω dt
(3.7)
(3.8)
Changing variables back, we obtain
Z
=
φ(Ω,t)
df
dt
+ f div(v) dv
(3.9)
concluding the proof.
With this machinery in hand, we will now proceed to derive balance laws.
All content © 2016-2018, Brandon Runnels
12.3
Lecture 13
3.1
Conservation of mass
Conservation of Mass
To develop conservation of mass, we must introduce two new quantities:
R(X, t) ≡ Lagrangian Mass Density
ρ(x, t) ≡ Eulerian Mass Density
(3.10)
The Lagrangian mass density is the mass per unit undeformed volume and is defined in the undeformed configuration, whereas the Eulerian mass density is the mass per unit deformed volume and is defined in the deformed
configuration. We can relate the two in the following way:
R=
dm
dm
dm
=
=J
=Jρ
dV
dv /J
dv
Furthermore, for a region B ⊂ Ω contained in the body, we can express the total mass in that body as
Z
Z
m(B) =
R dV =
ρ dv
B
(3.11)
(3.12)
φ(B,t)
Now, we introduce mass conservation by stating that for every B ⊂ Ω (measurable) and for all time,
d
m(B) = 0
dt
(3.13)
Pictorally, consider a body that is undergoing a deformation as follows:
φ
φ(Ω, t)
Ω
B
φ(B, t)
Conservation of mass simply states that the mass in some sub-region cannot change. Let us take this relationship
and use it to derive some identities. In the Lagrangian frame, we have
Z
Z
dR
d
RdV =
dV = 0
(3.14)
0=
dt B
B dt
Because this holds for all B ⊂ Ω, we have a strong-form relationship:
dR
= 0 ≡ Conservation of Mass (Lagrangian Frame)
dt
(3.15)
In the Eulerian frame, things are a little more complicated. Fortunately, we can use the Reynolds transport theorm
to skip several steps:
Z
Z
dρ
d
0=
ρdv =
+ ρ div(v) dv = 0
(3.16)
dt φ(B,t)
φ(B,t) dt
All content © 2016-2018, Brandon Runnels
13.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 13
https://canvas.uccs.edu/courses/22031
Because this holds for all φ(B, t) ⊂ φ(Ω, t), we can again state conservation of mass in strong form:
dρ
+ ρ div v = 0
dt
(3.17)
Note that we are taking the total time derivative (i.e. the material derivative) of ρ: this is no problem in the Lagrangian
frame but it is a little tricker in the Eulerian frame. Let’s write it out explicitly, using index notation and recalling the
definition of the material derivative:
0 = ρ̇ + ρ vi,i = ρ,t + ρ,i vi + ρ vi,i = ρ,t + (ρ vi ),i = 0
(3.18)
In symbolic notation, we recover
∂ρ
+ div(ρ v) = 0 ≡ Conservation of Mass (Eulerian Frame)
∂t
3.1.1
(3.19)
Control volume
In fluid flow it is frequently convenient to use control volume analysis. Consider the motion of a material (e.g. a
fluid) that is described by its velocity in the Eulerian frame, v(x, t). Also, consider a fixed volume V , such that the
material can flow through it.
x2
v(x, t)
x1
V
Let us consider the time rate of change of mass contained in V , noting that V does not depend on time and therefore
time derivatives can be taken inside:
Z
Z
Z
Z
∂ρ
d
∂
∂
m(V , t) =
m(V , t) =
ρ dv =
dv
=
−
ρ div(v)dv = −
ρ v · n da
(3.20)
dt
∂t
∂t V
cons. of mass
div. thm.
V ∂t
V
∂V
So we have the following expression for conservation of mass:
d
m(V , t) = −
dt
Z
∂V
Conservation of mass
ρ v · n da ≡ for a control volume
(3.21)
where n is the outward-facing normal vector to the control volume. The intuition is that the total velocity flux through
the boundary determines the rate of mass change in the volume, or that “change of mass equals mass flow in minus
mass flow out.”
3.2
Conservation of linear momentum
Next we will introduce the conservation of linear momentum for a continuous body. However, before we can do
this, it is necessary to introduce the notion of forces.
3.2.1
Forces, tractions, and stress tensors
Consider a body that is undergoing deformation as shown in the following figure.
All content © 2016-2018, Brandon Runnels
13.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 13
https://canvas.uccs.edu/courses/22031
x3
X3
∆A
∆f
N
∆a
∆f
n
x2
X2
x1
X1
Intuitively we know that the deformation is effected by forces acting on it. In this case, we see that a force ∆f
is acting on the face with normal vector n in the undeformed configuration. From undergraduate mechanics of
materials, we know that the magnitude of the force is not generally very interesting; rather, it is the force per unit
area that determines material response.
But now we have a question: which area? We must allow for the possibility (actually, the probability) that the area
will change with deformation. So, as with all of the other quantities we have defined so far, we will define a material
and a spatial version of this quantity.
T(N) =
dfi
df
Material
Ti (N) =
≡ Traction Vector
dA
dA
t(n) =
dfi
df
Spatial
ti (n) =
≡ Traction Vector
da
da
(3.22)
We pause here to make a couple of important notes about these quantities:
(1) These quantities are “forces per areas,” so why don’t we call them stresses? They are indeed very similar,
and we will use them to define stress in a moment, but stress is naturally a tensorial quantity, not a vector
quantity.
(2) What is going on with the indices – specifically, why do we use lowercase instead of uppercase? The reason
is that (in general) we will assume that forces live in the deformed configuration. As a result, forces will be
indexed with lowercase indices.
This is not always the case, and there are many instances where we map forces from the deformed configuration back to the undeformed. In these cases, forces would be indexed with uppercase variables. However,
for current purposes, we will always think of them as being deformed.
Note that the traction vectors are written as somewhat mysterious functions of a unit normal vector; that is, we feed
them a normal vector to a surface and they give us the force acting on that surface. What other kind of construct
do we know of that turns a vector into another vector: tensors. Therefore it stands to reason that we should be
able to find a tensor representation of the traction vectors. To do this, consider the following figure:
G3
N
∆A1
∆A
G2
∆A2
∆A3
G1
We begin by noting that we can describe the areas in the following way in terms of ∆A
∆AI = (GI · N)∆A = NI ∆A
(3.23)
Neglecting volumetric forces (reasonable to do as long as the volume is small, since volume forces scale with the
volume and therefore decrease much more quickly), we can write
T(N)∆A = T(GI )∆AI = T(GI )(GI · N)∆A = T(GI )NI ∆A
All content © 2016-2018, Brandon Runnels
(3.24)
13.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 13
https://canvas.uccs.edu/courses/22031
Or, noting that this holds for any ∆A, we have
T(N) = T(GI )NI = P N
(3.25)
P = [T(G1 ) T(G2 ) T(G3 )] ≡ The Piola-Kirchoff Stress Tensor
(3.26)
where we define
The Piola-Kirchoff tensor allows us to express the tractions (that are a function of the unit normal vector in the
undeformed configuration) in matrix form:
T(N) = P N
Ti = PiJ NJ
(3.27)
Note that, like the deformation gradient, the Piola-Kirchoff tensor is a “two point” tensor. Unlike with F , the implication is fairly straight forward in that it is the forces in the deformed configuration per unit area in the undeformed
configuration. Additionally we note that it turns a normal vector in the undeformed configuration to a force vector
in the deformed configuration. We can think of P as the tensor version of engineering stress.
Now, let us consider forces in the deformed configuration per unit area in the deformed configuration. Recall that
we defined the Piola transform: n∆a = JF−T N∆A Multiplying (3.27) by ∆A we have
T(N)∆A = P N∆A =
1 T
PF n ∆a = σ n ∆a = t(n)∆a
J
(3.28)
where we define
σ=
1 T
PF ≡ Cauchy Stress Tensor
J
(3.29)
and note that
t(n) = σ n
ti = σij nj
(3.30)
Note that σ lives entirely in the deformed configuration. σ is the tensor analog to true stress.
P and σ are tensors, so what is the physical significance of their components? The following picture is illustrative
in demonstrating the concept of a stress tensor:
X3
x3
P33
σ33
P23
P13
P32
P31
P21
P12
P11
σ13
P22
σ23
σ32
σ22
σ31
σ21
X2
X1
σ11
σ12
x2
x1
We can think of colum n of the stress tensor as the force vector acting on the n-facing face. Alternatively we can
think of the ij-th component of the tensor as the ith component of force acting in the j direction.
Let us introduce one additional type of force: body forces.
B=
df
dfi
Material
Bi =
≡ Body Force
dM
dM
b=
df
dfi
Spatial
bi =
≡ Body Force
dm
dm
(3.31)
The material body force is the force per unit undeformed mass, and the spatial is the force per unit deformed mass.
Common examples of a body force include gravity and electromagnetic forces.
All content © 2016-2018, Brandon Runnels
13.4
Lecture 14
3.2.2
Linear momentum balance
Balance laws
Recall that momentum is mass times velocity. What is momentum for a body with variable density R(X, t), ρ(x, t)
and variable velocity V(X, t), v(x, t)? We can write it in integral form in either the material or spatial frames as
Z
Z
L(B, t) =
R V dV =
ρ v dv
(3.32)
B
φ(B,t)
We can express Newton’s second law, the conservation of linear momentum, for all B ⊂ Ω as
d
L(B, t) = Ftot (B, t)
dt
where Ftot is the resultant of all forces acting on or in B ⊂ Ω,
Z
Z
Z
Ftot (B, t) =
T (N)dA +
R B dV =
∂B
B
(3.33)
Z
t(n)da +
ρ b dv
φ(∂B,t)
(3.34)
B
Let us begin by considering the conservation of linear momentum in the Lagrangian frame only: we have
Z
Z
Z
d
R V dV =
T (N)dA +
R B dV
dt B
∂B
B
(3.35)
What do we do with the partial derivative? We can take it right inside the integral.
d
dt
Z
Z
R V dV =
B
B
d
(R V) dV =
dt
Z
B
0
Z
(Ṙ
V
+
R
V̇)
dV
=
R A dV
(3.36)
B
Now, let’s consider the right hand side. In particular, let’s consider the surface traction term. Because we can write
the surface traction term as a matrix-vector multiplication with the Piola-Kirchoff stress tensor, we can apply the
divergence theorem to get
Z
Z
Z
T(N)dA =
P NdA =
Div(P)dV
(3.37)
∂B
∂B
B
Note that we now have three volume integrals. We can use this to collect terms and write our conservation equation
as
Z
[R A − Div(P) − R B] dV = 0
(3.38)
B
Because this must hold true for all B ⊂ Ω, applying the fundamental lemma of the Calculus of Variations, we can
write the above in strong form:
Div(P) + RB = RA ≡ Conservation of Linear Momentum (Lagrangian Frame)
PiJ,J + R Bi = R Ai
Now, let us consider the Eulerian frame:
Z
Z
Z
d
ρ v dv =
t(n)da +
ρ b dv
dt φ(B,t)
φ(∂B,t)
φ(B,t)
All content © 2016-2018, Brandon Runnels
(3.39)
(3.40)
14.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 14
https://canvas.uccs.edu/courses/22031
As before, let us consider the left hand quantity first. In a similar procedure to that which we followed with conservation of mass, we apply the Reynold’s Transport Theorem to take the derivative inside the integral:
Z
Z
Z
i
hd
d
ρ v dv =
(ρ v) + ρ v div(v) dv =
[ρv̇ + ρ̇ v + ρ v div(v)]dv
(3.41)
dt φ(B,t)
φ(B,t) dt
φ(B,t)
Z
Z
:0
ρ a dv
(3.42)
=
[ρ v̇ + v(
ρ̇ + ρ div(v)) ]dv =
}
| {z
φ(B,t)
φ(B,t)
conservation of mass
Now, let us consider the surface tractions term on the right hand side. Again, we can apply the divergence theorem
to obtain
Z
Z
Z
t(n)da =
σ n da =
div(σ) dv
(3.43)
φ(∂B,t)
φ(∂B,t)
φ(B,t)
Combining all of the terms, which are now in integral form, we have
Z
[ρa − div(σ) − ρb] dv = 0
(3.44)
φ(B,t)
By the fundamental lemma, we can write the above in strong form as
div(σ) + ρb = ρa
σij,j + ρ bi = ρai ≡ Conservation of Linear Momentum (Eulerian Frame)
3.2.3
(3.45)
Navier-Stokes momentum equations
The above form for linear momentum balance is convenient for solid mechanics, but it can easily be rewritten in a
number of ways. A particularly popluar method (especially when doing fluid mechanics) is to start by writing the
time derivative in a different way. Let us return to Equation (3.41): using index notation,
Z
Z
Z
hd
i
h∂
i
∂
d
ρvi dv =
(ρvi ) + ρvi vj,j dv =
(ρ vi ) +
(ρvi )vj + ρ vi vj,j dv
(3.46)
dt φ(B,t)
∂xj
φ(B,t) dt
φ(B,t) ∂t
Z
i
h∂
∂
(ρ vi ) +
(ρvi vj ) dv
(3.47)
=
∂xj
φ(B,t) ∂t
In symbolic form, this can be expressed as
Z
i
h∂
(ρ v) + div(ρv ⊗ v) dv
φ(B,t) ∂t
(3.48)
The Cauchy stress tensor is frequently decomposed into two components: a hydrostatic pressure component and
a deviatoric stress component,
σ = −pI + τ
(3.49)
where p is the scalar hydrostatic pressure, and is negative because pressure is compressive. (Note that the sign is
frequently reversed in fluid mechanics formulations.) Then the stress divergence term reduces to
Z
h
i
− div(pI) + div(τ ) dV
(3.50)
φ(B,t)
Finally, make the assumption that the only body force is gravity, so the body force term reduces to
Z
ρ gdV
(3.51)
φ(B,t)
All content © 2016-2018, Brandon Runnels
14.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 14
https://canvas.uccs.edu/courses/22031
Making all of these substitutions and rearranging a bit, the balance of linear momentum can be expressed in the
following form:
∂
(ρv) + div(ρv ⊗ v + pI) = div(τ ) + ρ g ≡ Navier-Stokes Momentum Equation
∂t
(3.52)
Even though the equation looks quite different, the mechanics are still the same. The biggest simplification made
when writing the Navier-Stokes equations is in decoupling the stress tensor. Solid bodies are capable of sustaining
stress states that are much more complex than that sustained by fluids, so we will prefer to leave σ in its nondecoupled form.
All content © 2016-2018, Brandon Runnels
14.3
Conservation of angular momentum, energy
Lecture 15
3.3
Conservation of angular momentum
Angular momentum for a body B ⊂ Ω is given y
Z
Z
G=
x × RV dV =
B
(3.53)
x × ρv dv
φ(B,t)
and the moments generated from the applied body forces and surface tractions are
Z
Z
Z
Z
M=
x × T(N) dA +
x × R B dv =
x × t(n) da +
∂B
B
φ(∂B,t)
x × ρ b dv
(3.54)
φ(B,t)
We note the important fact that we use the spatial position x for both the Lagrangian and Eulerian formulation. This
is a result of the fact that our forces live in the deformed configuration, so the moments generated by them must
also live in the deformed configuration.
Having defined angular momentum and moments for a continuous body, we can now express the conservation of
angular momentum as
d
G(B, t) = M(B, t)
dt
(3.55)
Let us begin with the Lagrangian frame. We start by simplifying the time derivative, which is easy here as it is taken
right inside the integral. Using the product rule, noting that V × V = 0, and applying conservation of mass, we arrive
at the expression
d
dt
Z
Z
x × RVdV =
B
B
0
Z
[( ẋ × RV) + (x × ṘV)
+ (x × R V̇)]dV =
x × RAdV
=V
(3.56)
B
As with linear momentum balance, we want to attempt to write the surface integral in volume form. The presence
of the cross product makes the application of the divergence theorem tricky, so we will switch to index notation
here. Applying the divergence theorem and the product rule, and applying the previously obtained conservation of
momentum equation, we have
Z
Z
Z
Z
[x × T(N)]k dA =
[x × P N]k dA =
ijk xi PjP NP dA = (ijk xi PjP ),P dV
(3.57)
∂B
∂B
B
Z
Z∂B
= (ijk xi,P PjP + ijk xi PjP,P ) dV = (ijk FiP PjP + [x × (RA − RB)]k ) dV
(3.58)
|{z}
| {z }
B
B
FiP
[x×Div P]K
Combining the above we have
Z
Z h
Z
[x × RA]k dV =
ijk FiP PjP + [x × (RA − RB)]k dv + [x × R B]k dV
|B
{z
} |B
{z
} |B
{z
}
momentum derivative
surface tractions
(3.59)
body force
Almost everything cancels except for one term. Because this holds ∀B ⊂ Ω, we can write the remaining term in
strong form:
T
ijk FiP PPj
=0
All content © 2016-2018, Brandon Runnels
(3.60)
15.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 15
https://canvas.uccs.edu/courses/22031
We can now pull an index notation trick to simplify the above even further:
1
1
T
T
T
T
(ijk FiP PPj
+ ijk FiP PPj
)
=
(ijk FiP PPj
− jik FiP PPj
)
2
swap indices 2
1
1
T
T
T
T
(ijk FiP PPj
− ijk FjP PPi
) = ijk (FiP PPj
− PiP FPj
)=0
2
2
T
0 = ijk FiP PPj
=
=
rename
(3.61)
(3.62)
We see that this implies
FPT = PFT ≡ Conservation of Angular Momentum (Lagrangian Frame)
(3.63)
Now let us consider the spatial frame. As before, we begin by using the transport theorem to simplify the time
derivative term. We also apply the product rule and conservation of mass, arriving at:
Z
Z
i
hd
d
x × ρv dv =
(x × ρv) + (x × ρv) div(v) dv
(3.64)
dt φ(B,t)
φ(B,t) dt
Z
h
i
=
(ẋ × ρv) + (x × ρ̇v) + (x × ρv̇) + (x × ρv) div(v) dv
(3.65)
φ(B,t)
Z
: 0i
:+0 (x × ρv̇) + (x × v) [ρ̇ + ρdiv(v)]
dv
( ẋ×ρv)
=v
{z
}
|
h
=
φ(B,t)
Mass conservation
Z
(3.67)
x × ρa dv
=
(3.66)
φ(B,t)
Now let’s look at the applied moment terms. In particular, we want to try to apply the divergence theorem to convert
the surface term to a volume term: Again, we’ll find that it is much easier to use index notation here.
Z
Z
hZ
i
hZ
i
∂
x × t(n) da =
x × σn da =
ijk xi σjp np da =
(ijk xi σjp ) dv
(3.68)
∂x
k
k
p
φ(∂B,t)
φ(∂B,t)
φ(∂B,t)
φ(B,t)
Z
Z
=
(ijk xi,p σjp + ijk xi σjp,p ) dv =
(ijk σji + [x × div σ]k ) dv
(3.69)
|{z}
φ(B,t)
φ(B,t)
δip
Z
(3.70)
(ijk σji + [x × (ρa − ρb)]k ) dv
=
φ(B,t)
Combining all of the terms in the conservation equation, we obtain
Z
Z
Z
[x × ρa]k dv =
(ijk σji + [x × (ρa − ρb)]k )dv +
φ(B,t)
|
φ(B,t)
{z
momentum rate
}
|
[x × ρb]k dv
(3.71)
φ(B,t)
{z
surface tractions
}
|
{z
body forces
}
Now we see that almost everything cancels except for one term:
Z
ijk σji dv = 0
(3.72)
φ(B,t)
Because this holds for all B, we can write it locally as
ijk σji = 0
(3.73)
Using a procedure identical to the above, we arrive at the following expression for conservation of angular momentum:
σ = σ T ≡ Conservation of Angular Momentum (Eulerian Frame)
(3.74)
Recalling the relationship between P and σ, we note that this is consistent with the Eulerian frame.
All content © 2016-2018, Brandon Runnels
15.2
Lecture 16
3.4
Conservation of energy, second law
Conservation of energy
We have established continuum mechanical formulations of the conservation of mass, linear momentum, and
angular momentum. The final conservation law is the conservation of energy. Energy conservation is interesting
because it is where we begin to make connections with other mechanisms such as mass or thermal transport. In
fact, it is the great strength of energy conservation that almost all mechanisms have an energetic formulation that
allows the coupling of these phenomena together.
Before we begin, let us remind ourselves of a couple of important thermodynamical definitions:
Definition 3.1. An extensive quantity for a system is a quantity that is equal to the sum of the quantities of all
subsystems. (Defined in integral form – a “nonlocal” quantity.)
Definition 3.2. An intensive quantity for a system is a quantity whose value is independent of the subsystem containing it. (Defined pointwise – a “local” quantity.)
3.4.1
Energetic quantities
To describe our energetic conservation laws, we need to make some definitions. Some are familiar, while some
require the definition of new quantities.
• Kinetic energy: kinetic energy for a body Ω is an extensive quantity, given by an integral over the entire volume.
We recall that the kinetic energy for a particle is 12 mv 2 . For a body with variable density and with variable
internal speed, the total kinetic energy is
Z
Z
1
1
2
K (Ω) =
R|V| dV =
ρ|v|2 dv ≡ Kinetic Energy
(3.75)
2 Ω
2 φ(Ω)
(We note that V, v are mean velocities, not velocities due to thermal fluctuations. We can do this because
we are working at the continuum level and we will account for thermal fluctuations with temperature. Note,
this is a little trickier with turbulent fluids, because fluctuations on the continuum level transfer down to the
microscopic level; this is called the energy cascade.)
• Heat: we have not dealt with heat yet, so in order to quantify it in a continuum setting we need to introduce a
couple of new variables. Q(Ω) denotes the total heat of the body, with Q̇ representing the rate of change of
heat. We define Sn (X), sn (x) to be the heat generated in the material per unit undeformed mass and per unit
deformed mass, respectively. Finally, we let H, h be the heat flux vectors per unit deformed area and per unit
undeformed area, respectively. With these definitions, we can write a simple heat balance law for our body in
both the Lagrangian and Eulerian configurations:
Z
Z
Z
Z
Q̇(Ω) =
R Sn dV −
H · N dA =
ρ sn dv −
h · n da
(3.76)
φ(Ω)
φ(∂Ω)
| Ω {z }
| ∂Ω {z
}
internal generation
outward flux
We note that the divergence theorem allows us also to write
Z
Z
Q̇(Ω) = [R Sn − Div(H)] dV =
[ρ sn − div(h)]dv
Ω
(3.77)
φ(Ω)
This is not of much use to us yet, because we have no relationship between these variables and the variables
that we are using for everything else (P, F , etc.). However it will be useful when we construct our energy
balance law.
All content © 2016-2018, Brandon Runnels
16.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 16
https://canvas.uccs.edu/courses/22031
• External power: we define the external power to be the rate of work done by all external forces. Remember
that work done by a force is f · ∆x, so the power exerted by a force is f · v. Therefore, to capture the power
of the applied body forces and surface tractions, we dot them with the velocity of the material at the point of
application, then integrate over the body:
Z
Z
Z
Z
E
T(N) · VdA =
ρb · vdv +
t(n) · vda
(3.78)
P (Ω) =
RB · VdV +
φ(Ω)
φ(∂Ω)
| ∂Ω {z
| Ω {z
}
}
power of body forces
power of surface tractions
• Deformation power: the deformation power is the portion of the external power that goes towards deforming
the material, rather than going towards changing the kinetic energy. For instance, if you throw a football, your
applied surface tractions cause the football to start moving; on the other hand, if you squeeze the football,
the kinetic energy doesn’t change (much) so all of your energy goes into deforming the ball. This quantity is
actually defined rather easily as the difference between external power and change in kinetic energy:
P D (Ω) = P E (Ω) − K̇ (Ω)
Z
Z
=
RBi Vi dV +
Z
1 d
RVi Vi dV
(3.79)
2 dt Ω
Ω
∂Ω
0 conservation of mass
Z
Z
Z
1
=
RBi Vi dV + (Vi PiJ ),J dV −
[ṘV
i Vi + 2RVi Ai ]dV
2 Ω
Ω
Ω
Z h
Z h
:i0
i
=
RBi Vi + Vi,J PiJ + Vi PiJ,J − RVi Ai dV =
Vi,J PiJ + Vi (RBi
+
P
−
RA
iJ,J
i ) dV
{z
}
|
Ω
Ω |{z}
Vi PiJ NJ dA −
=ḞiJ
Z
=
conservation of momentum
Z
(3.80)
Ḟ · P dV
ḞiJ PiJ dV =
Ω
Ω
(Note that the · symbol between two tensors is used to indicate a termwise product, so that the result is a
scalar.) Now let us, as usual, do the same thing in the Eulerian configuration:
Z
Z
Z
1 d
P D (Ω) =
ρbi vi dv +
vi σij nj da −
ρvi vi dv
2 dt φ(Ω)
φ(Ω)
φ(∂Ω)
Z
Z
Z
:0
1
=
ρbi vi dv +
(vi σij ),j dv −
(2ρvi ai + ρ̇v
+
ρv
i vi
i vi vj,j )dv
2 φ(Ω)
{z
}
|
φ(Ω)
φ(Ω)
conservation of mass
:0
=
[ρbi vi
−
ρv
i ai + vi σij,j + vi,j σij ]dv
|
{z
} |{z}
φ(Ω) =dij
conservation of momentum
Z
Z
=
dij σij dv =
d · σ dv
Z
φ(Ω)
(3.81)
φ(Ω)
where we recall that d was the rate of strain tensor. So, we can now write
Z
Z
D
P (Ω) =
Ḟ · P dV =
d · σ dv ≡ Deformation Power
Ω
(3.82)
φ(Ω)
• Internal Energy: let us define the intensive variables U(X) and u(x), the internal energies per unit undeformed
mass and deformed mass, respectively. Then the internal energy of a body Ω is given by
Z
Z
E (Ω) =
R U dV =
ρ u dv ≡ Internal Energy
(3.83)
Ω
All content © 2016-2018, Brandon Runnels
φ(Ω)
16.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
3.4.2
Course Notes - Lecture 16
https://canvas.uccs.edu/courses/22031
Balance laws
We are now in a position to write the first law of thermodynamics in a continuum mechanics setting. We will state
it as:
[change in internal energy] + [change in kinetic energy] = [external power] + [heat flow & generation]
(3.84)
Using what we have developed above, we write
Ė (Ω) + K̇ (Ω) = P E (Ω) + Q̇(Ω) ≡ Conservation of Energy
(3.85)
for all Ω. Alternatively, using our definition of deformation power, we can write
Ė (Ω) = P D (Ω) + Q̇(Ω)
All content © 2016-2018, Brandon Runnels
(3.86)
16.3
Second law, Clausius-Duhem inequality
Lecture 17
Since we have a nice and compact form for P D , the above expression is a little more useful. We already have
expressions for P D and Q̇ in Lagrangian and Eulerian forms, now we just need to evaluate Ė . Hopefully this process
is starting to seem fairly familiar: in the Lagrangian form,
d
Ė (Ω) =
dt
Z
Z
R U dV =
Ω
Ω
0
Z
[Ṙ
U
+
R
U̇]dV
=
R U̇ dV
(3.87)
Ω
In the Eulerian form
Z
:0
[ρu̇ + ρ̇ u+ρ
ρ u̇ dv
u
div(v) ] dv =
|
{z
}
φ(Ω)
φ(Ω)
(3.88)
Now, putting it all together, we have in the material/Lagrangian and spatial/Eulerian frames:
Z
Z
Z
R U̇ dV =
Ḟ · P dV + [R Sn − Div(H)] dV
Ω
Z Ω
ZΩ
Z
ρ u̇ dv =
d · σ dv +
[ρ sn − div(h)]dv
(3.89)
Ė (Ω) =
d
dt
Z
Z
ρ u dv =
φ(Ω)
conservation of mass
φ(Ω)
φ(Ω)
(3.90)
φ(Ω)
By the fundamental lemma, since the above holds for all subbodies Ω, we can finally write the above locally as
3.4.3
R U̇ = Ḟ · P + R Sn − Div(H)
≡ Conservation of Energy (Lagrangian Frame)
R U̇ = ḞiJ PiJ + R Sn − HK ,K
(3.91)
ρ u̇ = d · σ + ρ sn − div(h)
ρ u̇ = dij σij + ρ sn − hk,k ≡ Conservation of Energy (Eulerian Frame)
(3.92)
Power-conjugate pairs
As a side note (and sanity check), let’s see if we can relate these two sets of power conjugate pairs. We begin by
recalling the definitions of the rate of strain and Cauchy stress tensors:
dij =
1
1
−1
−1
(`ij + `ji ) = (ḞiK FKj
+ ḞjK FKi
)
2
2
σij =
1
T
PiK FKj
J
(3.93)
Now, let us substitute these definitions into the integrand:
dij σij =
1
−1
−1
PiL FLjT + ḞjK FKi
PiL FLjT )
(ḞiK FKj
2J
(3.94)
T
We recall from the conservation of angular momentum that PiL FLjT = FiL PLj
; substituting we get
=
1
1
1
−1
−1
T
(ḞiK PiL FKj
FjL +ḞjK FKi
FiL PLj
)=
(ḞiK PiK + ḞjK PjK ) = ḞiJ PiJ
2J
2J
J
|
{z
}
| {z }
δKL
(3.95)
δKL
So we see that we recover what we have already shown, namely:
ḞiJ PiJ dV = dij σij J dV = dij σij dv
All content © 2016-2018, Brandon Runnels
(3.96)
17.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 17
https://canvas.uccs.edu/courses/22031
We have seen that the pairs (Ḟ , P) and (d, σ) have a power-conjugate relationship; that is, if we “dot” them together
and integrate over the volume, we obtain a measure of the deformation power in the material. Let us define
S = F−1 P = J F−1 σ F−T
−T ≡ The Second Piola-Kirchoff Stress Tensor
SIJ = FIk−1 PkJ = J FIk−1 σkp FpJ
(3.97)
Notice that S has only uppercase indices. We have worked a lot with σ (deformed forces per deformed area) and P
(deformed forces per undeformed area); now, S completes the picture as the undeformed forces per undeformed
area.
Why did we introduce S? We are looking for something that power-conjugate to C, and now we are going to suppose
that S is conjugate to 21 Ċ, and therfore also to E = 21 (C − I). To show this, we begin by computing the derivative of
C. By the product rule,
(3.98)
Ċ = ḞT F + FT Ḟ
Now, to show that S and Ė are power-conjugate, we will evaluate their product directly
ĖIJ SIJ =
1
1
1
T
T
ĊIJ SIJ = (FIp−1 PpJ )(ḞIkT FkJ + FIkT ḞkJ ) = (FpI−T ḞIkT FkJ PJp
)
+ FpI−T FIkT ḞkJ PJp
2
2
2
| {z } | {z }
T
PkJ FJp
=
(3.99)
δpk
1 T
1
1
T −T
T
T
(Ḟ PkJ FJp
FpI +ḞkJ PJk
) = (ḞkI PkI + ḞkJ PkJ ) = ḞkI PkI X
) = (ḞIkT PkI + ḞkJ PJk
2 Ik
2
2
| {z }
(3.100)
δJI
To summarize, here are the pairs that we have found so far:
Rate of Deformation Measure
Strain rate tensor
Deformation gradient
Cauchy-Green / Green Lagrange
3.5
Stress Measure
d
Ḟ
Ė, 21 Ċ
Cauchy stress tensor (def. force / def. area)
1st PK stress tensor (undef. force / def. area)
2nd PK stress tensor (undef. force / undef. area)
σ
P
S
Second law of thermodynamics
The second law of thermodynamics is probably the most confusing in mechanics, because (i) it is an inequality
rather than an equality, and (ii) it is coupled with the notion of entropy, one of the most unintuitive quantities in
physics. As with the other balance laws, we will develop a continuum mechanics version of the second law. But,
before we do this, let’s talk a little bit about entropy. To do that, we will introduce a little bit of statistical mechanics.
3.5.1
Introduction to statistical thermodynamics and entropy
Let us consider a box full of, say, 10 particles each of mass m as shown in the following figure. Consider the box of
particles in three different scenarios: first, where only one particle is moving at a speed v0 , then with four particles
moving at speed v0 /2, and finally with nine of the particles moving with speed v0 /3.
|v| =
|v| =
v0
2
v0
3
|v| = v0
All content © 2016-2018, Brandon Runnels
17.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 17
https://canvas.uccs.edu/courses/22031
What is the energy of the set of particles? Let’s compute:
E1 = 1 × m(v0 )2 = mv02
E2 = 4 × m
v 2
0
2
= mv02 = E1
E3 = 9 × m
v 2
0
3
= mv02 = E1
(3.101)
Notice that even though the configuration of the particles in each case was completely different, the total amount
of energy remained the same. This illustrates a fundamental concept in statistical thermodynamics: macrostates
and microstates.
Definition 3.3. A particular configuration of each particle in a system is called a microstate
Definition 3.4. A macroscopic property of the system is called a macrostate
What are some other possible macrostates of our above system? Possibilities are the total number of particles in
the box, or the total mass of the system. A key idea in statistical thermodynamics is determining the number of
microstates associated with a given macrostate.
All content © 2016-2018, Brandon Runnels
17.3
Lecture 18
Calculus of variations
Let us begin by considering two systems, (1) and (2).
System 1: N1 , V1 , E1
System 2: N2 , V2 , E2
We will impose the following constraints: system 1 has N1 atoms, volume V1 and total energy E1 ; system 2 has N2
atoms, volume V2 and a total energy E2 . (Sidenote: that is, we are considering the microcanonical or NVE ensemble.)
Let us define the following:
Ω(E ) ≡ # of microstates with energy E ≡ Microcanonical Partition Function
(3.102)
That is, Ω is a number that tells us how many different ways our system can have total energy E . (Note: this is a
slight abuse of notation; Ω has been used previously to denote the body. Ω will be used this way in this section
only, unless otherwise explicitly indicated.)
So, for our two systems we have Ω(E1 ), Ω(E2 ). Now, let us suppose that we are considering the total number of
combined configurations. Given that system (1) has Ω1 (E1 ) possible configurations and system (2) has Ω2 (E2 )
possible configurations, then the total number of possible configurations of the combined system must be
Ω(E1 , E2 ) = Ω1 (E1 )Ω2 (E2 )
(3.103)
(For instance, if system (1) has two possible configurations and system (2) has three, then the total number of
combined configurations would be six.)
Now, let us assume that our system is in equilibrium, so that dΩ = 0. Then, from the form above, we have
dΩ = Ω2
∂Ω1
∂Ω2
dE1 + Ω1
dE2 = 0
∂E1
∂E2
(3.104)
Now, we have constrained our system so that no energy is allowed to leave. As a result, we know that
dE1 + dE2 = 0
(3.105)
Substituting this into our relationship above, we get
Ω2
dΩ2
dΩ1
dE1 − Ω1
dE1 = 0
dE1
dE2
(3.106)
Or, rearranging a bit, we get
1 dΩ1
1 dΩ2
=
Ω1 dE1
Ω2 dE2
(3.107)
We’ll pull a calculus trick here, and rewrite the equation above as
d
d
log Ω1 =
log Ω2
dE1
dE2
All content © 2016-2018, Brandon Runnels
(3.108)
18.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 18
https://canvas.uccs.edu/courses/22031
Let us define β to be
β1 =
d
log Ω1
dE1
β2 =
d
log Ω2
dE2
(3.109)
(Again, we restrict our overloading of the term β to this section only.) This allows us to write our equilibrium equation
as β1 = β2 . Now, we notice from the definition that
βdE = d log Ω or dE =
1
d log Ω
β
(3.110)
Recall that both of our systems have fixed volume and a fixed number of atoms. As a result, we know that the first
law of thermodynamics is given by dE = dQ; in other words, all changes in internal energy must be the result of
heat transfer. We also know that dQ = TdS. This means we can write
1
d log Ω = TdS
β
(3.111)
Here, we identify that β1 ∼ T and d log Ω ∼ dS. In fact, for classical reasons, a constant kB (Boltzmann’s constant)
is introduced, giving us
β=
1
≡ Reciprocal Temperature
kB T
S = kB log Ω ≡ Entropy
(3.112)
where β is the statistical thermodynamical temperature and S is the total Gibbs entropy of the system. There are
some important notes to be made here:
(1) If S1 = kB log Ω1 and S2 = kB log Ω2 , then the combined entropy is S = kB log(Ω1 Ω2 ) = kB log Ω1 + kB log Ω2 =
S1 + S2 , verifying that entropy is an extensive property of the system.
(2) Entropy is interpreted as the number of possible microstates for our system to maintain its current macrostate.
This is consistent with the interpretation of entropy as being a “measure of disorder.” For instance, if a coffee
cup has fewer possible “microstates” if its macrostate is “being all in one piece” than if its macrostate is
“shattered in pieces on the floor.”
If you don’t feel comfortable with the things discussed in this section, that’s ok. The purpose of this section is just
to hint at where entropy comes from, in the hope that it will help provide a little bit of intuition. At this point, we will
switch back to continuum mode, where we will work with entropy in a continuum setting.
3.5.2
Internal entropy generation
In the previous section we concluded that
dS =
dQ
T
or Ṡ =
Q̇
T
(3.113)
In other words, this shows that the rate of increase of entropy is correlated to the rate of heat flowing into the
system. This is true when the process happens quasistatically and all entropy change in the system is due to
external contributions. In this case, we say that the process is reversible, because we can return to the original
entropy state by simply removing the previously added heat.
For irreversible processes, we allow for the possibility of the system to generate entropy on its own. Thus, we say
that we have
Ṡ
|{z}
total entropy change
=
Ṡ int
|{z}
internal entropy generation
+
Q̇
T
|{z}
(3.114)
entropy generated externally
As an example, let us consider an adiabatic system (no heat transfer in or out of the system) divided into two
regions with two seperate temperatures as follows:
All content © 2016-2018, Brandon Runnels
18.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 18
https://canvas.uccs.edu/courses/22031
T1
T2
Q̇12
Let us suppose that there is heat flowing from one region to the other at a rate Q̇12 . Now, let us look at the rate of
change of entropy in the total system, keeping in mind that S = S1 + S2 :
Ṡ = Ṡ int
0
1
Q̇
Q̇12
Q̇12
1 + = S int = Ṡ1 + Ṡ2 = −
= Ṡ int
+
= Q̇12
−
T1
T2
T2
T1
T
(3.115)
adiabatic
Notice that Ṡ int is definitely nonzero. We know from observation that heat does not flow from cold to hot, so
because T2 < T1 we conclude that Ṡ int > 0. This leads us to the formal statement of the second law:
Ṡ int = Ṡ −
3.5.3
Q̇
≥ 0 ≡ Second Law of Thermodynamics
T
(3.116)
Continuum formulation
As with the other balance laws, we need to introduce a few more terms. Let T (X), Θ(x) = T (φ−1 (x)) be the
temperatures in the undeformed and deformed configurations, let N(X), η(x) = N(φ−1 (x)) be the entropy per unit
undeformed mass and per unit deformed mass. Then we write the entropy of the system as
Z
Z
S(Ω) =
R N dV =
ρ η dv
(3.117)
Ω
φ(Ω)
Then the rate of change of the entropy of the system is:
Z
Z
Z
d
d
Ṡ(Ω) =
R N dV =
(R N) dV =
R Ṅ dV
dt Ω
dt
Ω
Z
ZΩ
Z
d
=
ρ η dv =
[ρ η̇ + ρ̇ η + ρ η div v ] dv =
ρ η̇ dv
|
{z
}
dt φ(Ω)
φ(Ω)
φ(Ω)
(3.118)
(3.119)
=0 (cons. of mass)
Recalling the definition of Q̇, we let the contribution of entropy from external sources be
Z
Z
Z
Z
R Sn
H·N
ρ sn
h·n
ext
Ṡ (Ω) =
dV −
dA =
dv −
da
T
T
Θ
Ω
∂Ω
φ(Ω)
φ(∂Ω) Θ
(3.120)
Combining all of the above terms with the statement of the second law, we have
Ṡ −
Z
R Sn
H·N
dV +
dA ≥ 0
T
T
Ω
Ω
∂Ω
Z
Z
Z
ρ sn
h·n
=
ρ η dv −
dv +
da ≥ 0
φ(Ω)
φ(Ω) Θ
φ(∂Ω) Θ
Q̇
=
T
All content © 2016-2018, Brandon Runnels
Z
Z
R N dV −
(3.121)
(3.122)
18.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 18
https://canvas.uccs.edu/courses/22031
Applying the divergence theorem and using the fundamental lemma, we obtain the local expression of the second
law in the Lagrangian and Eulerian frames:
H
R Sn
≥ 0 ≡ Clausius-Duhem Inequality (Lagrangian Frame)
+ Div
T
T
h
ρsn
ρη̇ −
≥ 0 ≡ Clausius-Duhem Inequality (Eulerian Frame)
+ div
Θ
Θ
R Ṅ −
3.6
(3.123)
(3.124)
Review and summary
In this section, we used the kinematic framework that had been developed in the previous section to describe the
familiar laws of the conservation of mass, momentum, and energy in a continuum setting. We began by describing
these laws in bulk (integral) form, and then showed by means of the Reynolds transport and divergence theorems
that these balance laws can be expressed locally as differential equations. We found local versions for both the
Lagrangian and Eulerian frames. The following table summarizes the results:
Balance Law
Lagrangian Form
Conservation of mass
Div(P) + R B = R A
Eulerian Form
∂ρ
+ div(ρ v) = 0
∂t
div(σ) + ρ b = ρ a
P FT = F PT
σ = σT
R U̇ = Ḟ · P + R Sn − Div(H)
H
R Sn
+ Div
R Ṅ −
≥0
T
T
ρ u̇ = d · σ + ρ sn − div(h)
h
ρ sn
+ div
ρ η̇ −
≥0
Θ
Θ
Ṙ = 0
Conservation of linear momentum
Conservation of angular momentum
Conservation of energy
Second law of thermodynamics
If we want to solve the system, how many unknowns and how many equations do we have?
Givens:
Body forces B, b
Internal heat generation Sn , sn
Unknowns: Deformation mapping + spatial derivatives (φ, F) – 12 unknowns
Density (R, ρ) – 1 unknown
Internal energy (U, u) – 1 unknown
Heat flux (H, h) – 3 unknowns
Equations: Conservation of mass – 1 equation
Conservation of linear momentum – 3 equations
Conservation of angular momentum – 3 equations (nine total, but six are redundant)
Conservation of energy – 1 equation
We have a total of 17 unknowns but only 8 equations, so we need an additional 9 equations to close the system.
These additional equations are the constitutive relationship of the material, and may take the form
P = P(F)
σ = σ(ε)
(3.125)
H = H(T )
h = h(T )
(3.126)
This is a total of 9+3=12 equations, which is more than we need. However, it turns out that natural restrictions on
P(F ) reduce the number of actual equations to six, so that the system is not overconstrained. The thermal relationship is usually H(T ) = −k Grad T , Fourier’s law of heat conduction, where k is the thermal conductivity coefficient.
The stress-deformation relationship is, in 1D small strain, just σ = E ε; however, for the large deformation 3D case,
it will prove to be significantly more complex.
All content © 2016-2018, Brandon Runnels
18.4
Lecture 19
4
Calculus of variations
Constitutive Theory
We began our discussion of continuum mechanics by introducing kinematics of deformation, the geometric structure of bodies under continuous deformation. Everything that was developed there is true by geometry – no physics
was introduced at this point.
We then introduced the familiar concepts of mass, momentum, and energy conservation within a continuum setting. For each of these laws, we introduced their formulation in the familiar bulk setting (the “global” formulation),
and then derived differential formulations of these laws (the “local” formulation). The results are true by the laws
of physics.
We concluded the last section with the observation that the number of equations we have comes up short when we
try to solve for all of the variables in our problem. Constitutive theory introduce equations that describe materialspecific behavior, and are said to “close” the mechanical system. We will develop a framework for formulating
constitutive models, as well as restrictions on their physical admissibility; however, the equations derived from
constitutive models are inherently models – they are derived either from experimental observation or highly simplified theory.
4.1
Introduction to the calculus of variations
We’re going to take a bit of a break here to introduce another bit of mathematical machinery that will be invaluable
in formulating constitutive theory. The calculus of variations (or “variational calculus”) is the mathematical theory
of minimizing a functional. So first, what is a functional?
Definition 4.1. A functional on V is a mapping from V to R.
In other words, a functional is a machine that takes a thing (that thing could be a scalar, vector, tensor, or even a
function) and turns it into a real number. What is special about functions that return real numbers? It is the fact
that they can be optimized – that is, for a functional f [x], we can try to find the x that gives us the smallest value of
f . For instance,
(4.1)
f (x) = |x|
is a functional on Rn , minimized by x = 0. Now, let us consider a more complex example: find the function y (x)
that minimizes the distance between the two points (a, ya ), (b, yb ). In other words, we want to find the shortest path
connecting those two points:
yb
ya
b
a
We want to minimize a functional that takes a function y (x) and returns a distance – a scalar value. What form
might this functional take?
Z bp
L[y ] =
1 + y 02 (x)dx
(4.2)
a
All content © 2016-2018, Brandon Runnels
19.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 19
https://canvas.uccs.edu/courses/22031
So, we can state our mathematical problem in the following way:
Z
inf
y
b
p
1 + y 02 (x)dx
such that y (a) = ya , y (b) = yb
(4.3)
a
How do we go about solving this problem? Rather than solve this problem explicitly, let us consider a more general
problem:
Z
inf L[y ] = inf
y
y
b
f (x, y , y 0 ) dx
such that y (a) = ya , y (b) = yb
(4.4)
a
(Notice that we have merely replaced our specific integrand with a more general one. ) In first-year calculus, we
learn that we can optimize a function by finding the point at which its derivative is equal to zero. We can do a similar
thing here–only this time, we will use the Gateaux derivative.
yb
y (x) + εη(x)
y (x)
ya
η(x)
b
a
Let us introduce a function η(x) such that η(a) = η(b) = 0. Suppose that we’ve already found a minimizing function
y (x). If that is the case then we know that
L[y ] ≤ L[y + εη] ∀ε ∈ R, ∀ admissible η
(4.5)
In fact, we could say that the above is minimized when ε = 0. This amounts to saying that
d
L[y + εη]
dε
=0
ε→0
(4.6)
Does that look familiar? It’s nothing other than the Gateaux derivative in functional form. Let’s see if we can actually
evaluate this derivative:
Z b
Z b
d
d
d
0
0
L[y ]
=
f (x, y + εη, y + εη ) dx
=
f (x, y + εη, y 0 + εη 0 ) dx
dε
dε a
dε
ε→0
ε→0
ε→0
a
Z bh
0
0 i
∂f
d(y + εη)
∂f
d(y + εη )
=
+
dx
0
0
∂(y + εη)
dε
∂(y + εη )
dε
ε→0
a
Z bh
i
∂f
∂f
=
η + 0 η 0 dx
∂y
∂y
a
Z b
Z b
∂f b
∂f
d ∂f
η dx + 0 η −
=
η dx
∂y
∂y
dx
∂y 0
a
a
a
Z bh
d ∂f i
∂f
=
−
η dx
∂y
dx ∂y 0
a
∀η =⇒
∂f
d ∂f
−
= 0 ≡ Euler-Lagrange Equation
∂y
dx ∂y 0
(4.7)
So we see that we began with a global minimization principle, and arrived at a local differential equation.
All content © 2016-2018, Brandon Runnels
19.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 19
https://canvas.uccs.edu/courses/22031
Example 4.1: Minimum distance
Let’s apply this result to our minimum distance problem: if f =
satisfy
p
1 + y 02 , then the minimizing path must
d y0
y 02
d ∂f
y 00
y 02 y 00
1
∂f
00
p
p
p
=
−
−
=
−
+
=
y
−
= 0 (4.8)
∂y
dx ∂y 0
dx
(1 + y 02 )3/2
(1 + y 02 )3/2
1 + y 02
1 + y 02
1 + y 02
There could potentially be multiple solutions, but the most obvious one is that y 00 (x) = 0. This implies that
the minimizing function y is linear – exactly what we expect.
Example 4.2: Brachistochrone problem
Consider a ball of mass m under the action of gravity g traveling from (xa , ya ) to (xb , yb ) along a path defined
by y (x). What path y (x) minimizes the transit time?
xa
xb
ya
y (x)
yb
Our functional T [y ] returns transit time as a function of path; we need a form for the functional. Use the
fact that potential energy is constant, in other words, we can say without loss of generality that
T + U = const =⇒
p
ds
ds
1
m v 2 − m g y = 0 =⇒ v =
= 2gy =⇒ dt = √
2
dt
2gy
(4.9)
But we also know that
p
ds =
dx 2 + dy 2 =
p
1 + y 02 dx
(4.10)
so our functional is given by
Z
T =
tf
Z
b
s
dt =
ti
a
1 + y 02
dx
2gy
(4.11)
The integrand of our functional is given by
s
f (x, y , y 0 ) =
1 + y 02
2gy
(4.12)
and we can simply substitute into the Euler Lagrange equation to solve for y (x). The actual solution is a bit
messy, and it is easier to describe parametrically. The form of the optimal path is a cycloid:
x = r (θ − sin θ) + c
y = r (1 − cos θ)
(4.13)
where r , c are determined based on the boundary conditions.
There is a very wide range of applications of variational calculus. Some applications include dynamics (Lagrangian
All content © 2016-2018, Brandon Runnels
19.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 19
https://canvas.uccs.edu/courses/22031
mechanics), optimal control theory, mathematical finance. We will seek to find a variational (functional minimization) formulation of our balance laws.
All content © 2016-2018, Brandon Runnels
19.4
Lecture 20
4.1.1
Variational balance law formulation
Stationarity condition
Let us consider our usual body, Ω, and this time we will imagine that it is subjected to some external loading, body
forces, and a displacement condition as indicated below.
∂1 Ω
∂2 Ω
Ω
Note that we have split out boundary into two portions, so that ∂Ω = ∂1 Ω ∪ ∂2 Ω. We say that we have prescribed
displacement on ∂1 Ω whereas we prescribed forces on ∂2 Ω. Alternatively we could say that we prescribe Dirichlet
boundary conditions on ∂1 Ω and Neumann boundary conditions on ∂2 Ω. With this picture in mind, let us introduce
a powerful theorem for functional minimization on Ω:
Theorem 4.1. Let Ω ⊂ Rn be Lebesgue measurable. Let φ : Ω → Rn , and let
f : Ω × φ(Ω) × GL(n) → R
g : ∂Ω × φ(∂Ω) × GL(n) → R
(4.14)
be functions defined over the body and the body’s boundary, respectively. Let the boundary of the body be divided
into two regions, ∂1 Ω, ∂2 Ω such that φ(X) = X ∀X ∈ ∂1 Ω. Finally, let
Z
Z
L[φ, Grad φ] =
f (X, φ, Grad φ)dV +
g (X, φ, Grad φ)dA
(4.15)
Ω
∂Ω
If φ solves
(4.16)
inf L[φ] = 0
φ
then the following Euler-Lagrange equations hold for φ:
∂f
d ∂f
−
= 0 ∀X ∈ Ω
∂φi
dXJ ∂φi,J
∂g
= 0 ∀X ∈ ∂1 Ω
∂φi,J
∂g
∂f
+
NJ = 0 ∀X ∈ ∂2 Ω
∂φi
∂φi,J
(4.17)
Proof. As before, we will prove this using a special case of the Gataeux derivative called the variational or functional
derivative. The key aspect of the proof is that if φ minimizes L, then
d
L[φ + εη]
dε
ε=0
= 0 ∀ admissible η : Ω → Rn
(4.18)
What does admissible mean? From the beginning we constrained φ such that there is no deformation on ∂1 φ. (This
is called a Dirichlet boundary condition.) On the other hand, it is free to move over the remainder of the boundary,
so the derivative must be zero. (This is called a Neumann boundary condition.) That means that our test function
η must satisfy
η(X) = 0 ∀X ∈ ∂1 Ω
All content © 2016-2018, Brandon Runnels
Grad η(X) = 0 ∀X ∈ ∂2 Ω
(4.19)
20.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 20
https://canvas.uccs.edu/courses/22031
This will come in very handy in a moment. Now, let us begin by evaluating the derivative:
Z
d
d
L[φ + εη, Grad(φ + εη)]
=
f (X, φ + εη, Grad φ + ε Grad η)dV |ε=0 +
dε
ε=0
Ω dε
Z
d
g (X, φ + εη, Grad φ + ε Grad η)dA
+
dε
ε=0
∂Ω
Z h
∂f
d(φi + εηi )
∂f
d(φi,J + εηi,J ) i
=
dV
+
dε
∂(φi,J + εηi,J )
dε
ε=0
Ω ∂(φi + εηi )
Z h
i
∂g
d(φi + εηi )
∂g
d(φi,J + εηi,J )
+
dA
+
dε
∂(φi,J + εηi,J )
dε
ε=0
∂Ω ∂(φi + εηi )
Z h
Z h
i
i
∂f
∂f
∂g
∂g
=
ηi +
ηi,J dV +
ηi +
ηi,J dA
(4.20)
∂φi,J
ηi,J
Ω ∂φi
∂Ω ∂φi
Consider the third term. Using the product rule and integration by parts:
Z
Z
Z
Z
Z
∂f
∂ ∂f
d ∂f ∂f
d ∂f ηi dV =
ηi dV
ηi,J dV =
ηi dV −
ηi NJ dA −
Ω ∂φi,J
Ω ∂XJ ∂φi,J
Ω dXJ ∂φi,J
∂Ω ∂φi,J
Ω dXJ ∂φi,J
Substituting back into the above expression and gathering terms, we have
Z h
Z h
i
∂f
d ∂f i
∂g
∂g
∂f
=
−
ηi dV +
ηi +
ηi,J +
ηi NJ dA
dXJ ∂φi,J
ηi,J
∂φi,J
Ω ∂φi
∂Ω ∂φi
(4.21)
Here is where the admissibility condition is useful: let us apply the condition that ηi,J = 0 everywhere on the
boundary except for ∂1 Ω. This leaves
Z
Z h
Z h
i
∂g
∂g
d ∂f i
∂f
∂f
−
ηi dV +
ηi,J dA +
+
NJ ηi dA
(4.22)
=
dXJ ∂φi,J
∂φi,J
∂1 Ω ηi,J
∂2 Ω ∂φi
Ω ∂φi
And here is the final payoff: we note that this has to be true for all admissible η. And so, we can pass from the weak
form to the strong form by the fundamental lemma. The result is that
∂f
∂ ∂f
−
= 0 ∀X ∈ Ω
∂φi
∂XJ ∂φi,J
∂g
= 0 ∀X ∈ ∂1 Ω
∂φi,J
∂g
∂f
+
NJ = 0 ∀X ∈ ∂2 Ω
∂φi
∂φi,J
(4.23)
This concludes the proof.
As you can see, the proof of the above theorem is a little bit technical, but it’s really nothing more than a fully 3D
version of the simple 1D Euler-Lagrange equation that we found earlier. Notice that we left the proof as general as
possible, yet the results are still quite simple. We can now take what we found and apply it directly to continuum
mechanics.
4.2
Variational formulation of linear momentum balance
Let us introduce the potential energy functional:
Z
Z
Π[φ] =
W (F)dV −
φi RBi dV
−
Ω
Ω
|
{z
}
|
{z
}
strain energy
work done by body forces
Z
φi Ti dA
∂Ω
|
{z
}
≡
Potential Energy Functional
(Lagrangian Frame)
(4.24)
work done by surface tractions
were we note, of course, that F = Grad φ.
We’ve introduced one new character into this equation: W (F), called the strain energy density. We will define what
this is in just a moment. For now, you can think of it simply as the energy of a body (per unit volume) associated
with deforming a body by a deformation gradient F.
All content © 2016-2018, Brandon Runnels
20.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 20
https://canvas.uccs.edu/courses/22031
The principle of minimum potential energy states that, for a body subjected to surface tractions T and body forces
B, that
φ = arginf φ Π[φ] ≡ Principle of Minimum Potential Energy
(4.25)
In other words, we can figure out what φ is simply by minimizing Π. This is very convenient computationally, because
it is frequently convenient to formulate problems in terms of minimization. This is called a variational principle. Let
us use the theorem we derived above to show that the principle of minimum potential energy returns our familiar
balance laws. If we write Π in terms of f , g , we have
f = W (F) − φi RBi
(4.26)
g = −φi Ti
Since we proved the above theorem for the general case, we know that:
−RBi −
∂ dW
=0
∂XJ dFiJ
0=0
−Ti +
dW
NJ = 0
dFiJ
(4.27)
Let us rearrange the first and last Euler-Lagrange equations. Writing them in invariant/symbolic notation, we get:
Div
dW dF
−RB=0
dW dF
N=T
(4.28)
Does this look familiar? You may notice that it has a very similar form to that of the local balance of linear momentum in the Lagrangian frame–if we suppose P = dW /dF we have
Div(P) + RB = 0 ∀X ∈ Ω
PN = T ∀X ∈ ∂Ω
(4.29)
exactly the formulation for linear momentum that we derived earlier.
In fact, this is exactly how we will introduce our constitutive model. Define the following scalar function of the
deformation gradient:
W : GL(3) → R such that P =
dW
≡ Elastic Free Energy Density
dF
(4.30)
This is how we describe material behavior.
All content © 2016-2018, Brandon Runnels
20.3
Lecture 21
4.3
Constitutive theory and elastic tensor
Material frame indifference
Consider the energy of a body under the action of a deformation gradient F and a rotation R:
F
W (I)
R
W (F)
W (RF)
Because the rotation is rigid, we expect that W (RF) = W (F) for all R ∈ SO(3). Recall that any F can be decomposed
into F = RU by the polar decomposition. We may say that for all W there exists a function Ŵ such that
(4.31)
W (F) = Ŵ (U)
In other
√ the strain energy depends on the pure deformation only, not the subsequent rotation. But because
√ words,
U = C = FT F we could also say that must exist a Ŵ such that
(4.32)
W (F) = Ŵ (C)
In other words, W (F) = γ F11 would violate material frame indifference, whereas W (F) = γ tr(FT F) would not.
Note that while W must depend on C only, P does not.
Because we generally have W (C), it is frequently handy to differentiate with respect to C instead of F. Using the
chain rule:
dW
dW dCkL
dW d
dW
dW
dW
=
=
(FpK FpL ) =
(δip δJK FpL + FpK δip δJL ) = FiL
+ FiK
dFiJ
dCKL dFiJ
dCKL dFiJ
dCKL
dCJL
dCKJ
(4.33)
Recall that C is symmetric, so CIJ = CJI ; the following simplifies to
dW
dW
= 2F
dF
dC
(4.34)
dW
1
dW
1
1
= F−1
= F−1 P = S
dC
2
dF
2
2
(4.35)
We notice that
We recall that (Ḟ , P) and ( 12 Ċ , S) are conjugate pairs, and we see that this relationship is repeated again here:
P=
4.4
dW
dF
S=
dW
dW
=
dE
d( 12 C)
(4.36)
Elastic modulus tensor
Let us consider a linearization of W (F) about F = I, where we know that W (I) = 0, = P(I) = 0. Expanding to
second order, we have
0
0 dW >
1
dW
1
*
W (F) = W
(I) + (I)(FiJ − δiJ ) + (FiJ − δiJ )
(I)(FkL − δkL ) + h.o.t. ≈ (F − I) · C (F − I)
dF
2
dF
dF
2
iJ
kL
iJ
All content © 2016-2018, Brandon Runnels
(4.37)
21.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 21
https://canvas.uccs.edu/courses/22031
The second derivative tensor
CiJkL =
d 2W
dFiJ dFkL
(4.38)
is called the elastic modulus tensor. It is a useful quantity to compute when solving problems numerically.
dW
dW
= 2FiI
(4.39)
dFiJ
dC IJ
dW
d dW
dW
d 2 W dCKP
dW
d 2W
d 2W
= 2δik δIL
+ 2FiI
= 2δik
+ 2FiI
= 2δik
+ 4FiI FkK
(4.40)
dFiJ dFkL
dC IJ
dFkL dCIJ
dCJL
dCIJ dCKP dFkL
dCJL
dCIJ dCKL
dW
d 2W
= δik 1
+ FiI FkK 1
(4.41)
d( 2 CJL )
d( 2 CIJ )d( 12 CKL )
(4.42)
= δik SJL + FiI FkK CIJKL
where CIJKL can also be expressed as
CIJKL = 4
d 2W
d 2W
=
dCIJ dCKL
dEIJ dEKL
(4.43)
We see that CIJKL has both “major symmetry” (CIJKL = CKLIJ ) and “minor symmetry” (CIJKL = CJIKL = CIJLK )
4.5
Elastic material models
We see from the above that the only thing necessary to define a material model is the free energy function W (F )
– given a W , it is always possible to derive P and C. We will illustrate this with a couple of examples.
4.5.1
Useful identities
We will be doing a great deal of matrix differentiation, and we frequently have to evaluate complex expressions.
The following identities will prove useful:
• Derivative of the determinant
d(det F)
= det(F)FiJ−T
dFiJ
(4.44)
• Derivative of the matrix inverse
dF −1
dF −1
d
d
!
−1
−1
−1
δJM =
(FJn
FnM ) = Jn FnM + FJn
δkn δLM = Jn FnM + FJk
δLM = 0
dFkL
dFkL
dFkL
dFkL
−1
dFJn
−1
−1
FnM F −1 = −FJk
δLM FMi
dFkL | {z Mi}
(4.45)
(4.46)
δni
dFJi−1
dFkL
All content © 2016-2018, Brandon Runnels
−1 −1
−T −T
= −FJk
FLi = −FkJ
FiL
(4.47)
21.2
Lecture 22
4.5.2
Constitutive models
Pseudo-Linear
A simplistic model that guarantees material frame indifference is the following, where C is a rank-4 tensor and
obeys major symmetry (CIJKL = CKLIJ ) and minor symmetry (CIJKL = CJIKL = CIJLK = CJILK ).
1
1
1 T
T
W (F) = E · C E = (FT F − I) · C (FT F − I) = (FPm
FmQ − δPQ ) CPQRS (FRn
FnS − δRS )
2
8
8
(4.48)
From this definition, we compute the Piola-Kirchoff stress tensor by differentiation
i dC
d h1
1
d
dW
AB
T
=
(C − I) C (C − I)
= CABRS (FRn
(F T F`B )
FnS − δRS )
dFiJ
dCAB 8
dFiJ
4
dFiJ A`
1
T
FnS − δRS )(δi` δJA F`B + F`A δi` δJB )
= CABRS (FRn
4
1
T
= (FiB CJBRS + FiA CAJRS ) (FRn
FnS − δRS )
4
DW (F)iJ =
(4.49)
We compute the elasticity tensor by differentiating again
h1
i dC dC
i d 2C
d 2W
d2
W h1
AB
CD
AB
=
(C − I) C (C − I)
+
(C − I) C (C − I)
dFiJ dFkL
dCAB dCCD 8
dFiJ dFkL
dCAB 8
dFiJ dFkL
δik
δik
1
(CJLRS + CLJRS )FnR FnS −
(CJLRR + CLJRR )
= (CJALD + CJADL + CAJLD + CAJDL )FiA FkD +
4
4
4
(4.50)
DDW (F )iJkL =
The result is somewhat unwieldy, so let us apply symmetry conditions on C to simplify. The result is
1
DDW (F )iJkL = CAJDL FiA FkD + δik (CLJRS FnR FnS − CJLRR )
2
(4.51)
As a sanity check, we can verify that the stress at F = I must be equal to zero. Substituting, we get
:0
P(I)iJ = (δiB CJBRS + δiA CAJRS ) (δRn
δ
−
δRS ) = 0 X
nS
(4.52)
as expected. What happens if we substitute the identity for the elasticity tensor? We obtain:
1
C(I)iJkL = CAJDL δiA δkD + δik (CLJRS δnR δnS − CJLRR )
2
:0
1
C
= CiJkL + δik (CLJnn
−
) = CiJkL
JLRR
2
(4.53)
whence we see that the elastic constants correspond to the tangent modulus in the undeformed state.
4.5.3
Compressible neo-Hookean
The neo-Hookean model is a simple model that does a fairly good job of modeling rubbery materials. Two material
constants are used, the shear modulus µ and the bulk modulus κ. We define the material by its free energy W (F ):
W (F) =
κ
κ
µ tr(FT F)
µ FpQ FpQ
− 3 + (det(F) − 1)2 =
− 3 + (det(F) − 1)2
2/3
2/3
2 det(F)
2
2 det(F )
2
All content © 2016-2018, Brandon Runnels
(4.54)
22.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 22
https://canvas.uccs.edu/courses/22031
We find the Piola Kirchoff stress tensor which is given by the first derivative
µ (FpQ FpQ ),iJ
2 FpQ FpQ d det(F) d det(F)
dW
=
−
+ κ(det(F) − 1)
2/3
5/3
dFiJ
2 det(F)
3 det(F)
dFiJ
dFiJ
F
1
F
F
iJ
pQ pQ
=µ
−
F −T + κ(det(F) − 1) det(F)FiJ−T = PiJ
3 det(F)2/3 iJ
det(F)2/3
DW (F)iJ =
We find the elastic modulus tensor which is given by the second derivative
i
1 FpQ FpQ −T d h FiJ
d 2W
−
FiJ
=
µ
+ κ(det(F) − 1) det(F)FiJ−T
2/3
2/3
dFiJ dFkL
dFkL
3 det(F)
det(F)
h
i
2
2
2
µ
1
−T
−T
−T −T
−T −T
δ
δ
−
F
F
F
(F
F
)F
(F
F
)F
−
F
+
F
F
=
+
ik
JL
iJ
pQ
pQ
pQ
pQ
kL
kL
iJ
iL
kL
kJ
3
3 iJ
9
3
det(F)2/3
h
−T −T
−T −T i
+ κ det(F) 2 det(F) − 1 FiJ FkL + 1 − det(F) FiL FkJ = CiJkL
(4.55)
DDW (F)iJkL =
Substituting F = I shows quickly that the stress is zero in that state. Substituting into the tangent modulus, we
get
DDW (I) =µ δik δJL + δiL δkJ
2
− δkL δiJ
3
+ κ δiJ δkL
(4.56)
where we see that µ and κ correspond to the shear and bulk moduli in small deformation.
All content © 2016-2018, Brandon Runnels
22.2
Lecture 23
4.6
Internal constraints
Internal constraints
We made a point of noting that the above was compressible neo-Hookean. We’ve also noted that it is generally
“easier” to work with compressible rather than incompressible materials. Incompressibility is an example of an
internal constraint – a material restriction on admissible geometric material deformations. The variational structure of the linear momentum balance law provides a nice structure for enforcing internal constraints by means of
Lagrange multipliers.
4.6.1
Review of Lagrange multipliers
Consider the following minimization problem:
y
f (x, y ) = x + y
x
x2 + y2 = 1
Find the minimum point of f (x, y ) = x +y subject to the constraint that x 2 +y 2 = 1. In the language of optimization,
we write
inf f (x, y ) subject to x 2 + y 2 = 1
(x,y )
(What if we tried to optimize f (x, y ) without the constraint? The stationarity conditions are
d
d
d
d
f (x, y ) = 0
f (x, y ) = 0 =⇒
f (x, y ) = 1 = 0
f (x, y ) = 1 = 0
dx
dy
dx
dy
(4.57)
which is completely unsolvable for x, y . This is to be expected: f (x, y ) has no minimizer – it can grow negatively
to infinity. Minimization only makes sense when there is an imposed constraint. )
How do we write our stationarity conditions when the constraint is active? To do this, we modify our problem to
include an additional variable, called a Lagrange multiplier:
h
i
inf sup f (x, y ) + λ(x 2 + y 2 − 1)
(4.58)
x,y
λ
(We recall that “sup” is basically the same as “max”.) Why do this? Let’s make a few notes:
(1) We’re restricting x 2 + y 2 = 1, so there will be no resulting contribution to the optimized value.
(2) In addition to optimizing with respect to x, y , we are optimizing with respect to λ as well. What is the stationarity condition for λ? We get this by evaluating
i
d h
f (x, y ) + λ(x 2 + y 2 − 1) = x 2 + y 2 − 1 = 0
dλ
(4.59)
In other words, the stationarity condition on λ is identical to our internal constraint.
All content © 2016-2018, Brandon Runnels
23.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 23
https://canvas.uccs.edu/courses/22031
(3) Why are we maximizing with respect to λ? Let’s look at the inside part:
(
h
i
0 x2 + y2 = 1
2
2
sup f (x, y ) + λ(x + y − 1) =
∞ else
λ
(4.60)
In other words, this inside part will blow up to infinity except if x 2 + y 2 = 1. Recall that on the outside, we are
minimizing with respect to x, y – this means that all non-infinite solutions must satisfy the constraint.
With these notes in mind, let’s attempt to solve the above optimization problem. We have three stationarity conditions, which we’ll use to solve for x, y , λ
dΦ
= 1 + 2λx = 0
dx
=⇒ x = −
dΦ
= 1 + 2λy = 0
dy
1
2λ
=⇒ y = −
1
2λ
dΦ
= x2 + y2 − 1 = 0
dλ
1
1
1
1
+ 2 = 2 = 1 =⇒ λ = ± √
=⇒
4λ2
4λ
2λ
2
(4.61)
(4.62)
Substituting back, we get two solutions:
1
x = ±√
2
1
y = ±√
2
(4.63)
corresponding to the two points along the x = y line passing through the circle. Exactly what we expect!
4.6.2
Examples of internal constraints
Let us express internal constraints for materials such that
(4.64)
ψ(F) = 0
Two examples:
• Incompressibility: ψ(F) = det(F) − 1 = 0
• Inextensible fibers: ψ(F) = λ2 (N) − 1 = NT FT F N − 1 = 0
(Notice that we can write the above constraints entirely in terms of C as well.)
4.6.3
Lagrange multipliers in the variational formulation of balance laws
Recall that the equilibrium configuration is given by
Z
Z
hZ
φ = arg inf Π[φ] = arg inf
W (F)dV −
φi RBi dV −
φ
φ
Ω
Ω
i
φi Ti dA
(4.65)
∂Ω
Let us suppose that we are working with a system that has an internal constraint ψ(F ) = 0. How would we include
this in our optimization problem? Let’s try:
Z
Z
Z
h
i
hZ
i
arg inf sup Π[φ] +
λ φ(F) dV = arg inf sup
W (F) + λψ(F) dV −
φi RBi dV −
φi Ti dA
(4.66)
φ
λ
Ω
φ
λ
Ω
Ω
∂Ω
What are our stationarity conditions? For φ, they are nothing other than our usual Euler-Lagrange equations, which
come out to be
dW
dψ
dψ dW
+λ
+ RB = 0
+λ
N=T
(4.67)
dF
dF
dF
dF
All content © 2016-2018, Brandon Runnels
23.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 23
https://canvas.uccs.edu/courses/22031
This prompts us to define the stress tensor to be
P=
dW
dψ
+λ
dF
dF
(4.68)
Let us look at a couple of examples: first, consider an incompressible material with free energy density W0 . Then
ψ(F) = det(F) − 1, and the stress is given by
P=
dW
d
+ λ (det(F) − 1) = P0 + λ J F−T
dF
dF
(4.69)
(where P0 is the stress with the incompressibility constraint removed.) What does this look like in the deformed
configuration? Apply the relationship between σ and P to obtain
1
1
1 T
PF = P0 FT + λ J F−T FT = σ0 + λ I
J
J
J
σ=
(4.70)
In other words, our result is a combination of the regular stress σ0 pluss a “hydrostatic” stress λ I with magnitude
λ. This kind of stress state is identical to hydrostatic pressure, and we generally rename λ to p. Let’s interpret this
result by noting a couple of points:
(1) The material resists all volumetric compression; the way it resists compression is by exerting hydrostatic
pressure. Thus we can interpret Lagrange multipliers as forces exerted by the material to enforce the material
restraint.
(2) What are the units of p? Recall that λψ(F) must have units of energy per unit volume, and that ψ(F) = det(F)−1
is unitless. Pressure has units of [force]/[area] which can be written as [force][length]/[volume].
4.7
Linearized constitutive theory
At the end of the section on kinematics, we introduced a framework for linearized kinematics, in which we replace
the deformation mapping φ with the displacement field u, powers and gradients of order greater than two are
sufficiently small to ignore.
Let us assume that we can express the energy of our material under small strain as W (ε). Then the principal of
minimum potential energy becomes
Z
Z
hZ
i
u = arg inf
W (ε)dV −
u · b dV −
u · tdS
(4.71)
u
Ω
Ω
∂Ω
The Euler-Lagrange for this problem (in the static case) give
div
dW dε
(4.72)
+ ρb = 0
We compare this with the local balance of linear momentum, which is given by
(4.73)
div(σ) + ρb = ρü
(note that ü = a.) This implies that the Cauchy stress is given by
σ=
dW
dε
σij =
dW
dεij
(4.74)
What form must W (ε) take? Let us Taylor expand about ε = 0:
W (ε) = W (0) +
All content © 2016-2018, Brandon Runnels
dW
1 d 2W
(0) εij +
(0) εij εkl + h.o.t.
dεij
2 dεij dεkl
(4.75)
23.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 23
https://canvas.uccs.edu/courses/22031
We can state without loss of generality that W (0); also, we assume that our material is purely elastic, so DW (0) = 0.
Thus we are left with
W (ε) =
1
1 d 2W
εij εkl = Cijkl εij εkl
2 dεij dεkl
2
(4.76)
where Cijkl is a fourth order tensor containing the constant elastic moduli of the material. This is amenable to
differentiation, so that our stress tensor is given by
σij =
dW
= Cijkl εkl
dεij
(4.77)
Finally, we write our momentum balance equation in the following way:
Cijkl εkl,j + ρbi = ρüi
4.7.1
(4.78)
Major & minor symmetry and Voigt notation
We have established that in small strain, the elastic modulus tensor is defined by
Cijkl =
d 2W
dεij dεkl
(4.79)
This is an order 3 × 3 × 3 × 3 tensor, meaning that it can contain up to 34 = 81 elastic constants – far too many to
keep track of easily. Indeed, the above is too general: we can use symmetry arguments to reduce the number of
possible constants:
Theorem 4.2. In small strain, the elastic modulus tensor has both major symmetry
Cijkl = Cklij
(4.80)
Cijkl = Cjikl = Cijlk
(4.81)
and minor symmetry
Proof. It is simple to prove the above: because the energy function is continuous, we have
Cijkl =
d 2W
d 2W
=
= Cklij
dεij dεkl
dεkl dεij
(4.82)
proving major symmetry. In small strain, ε is symmetric, so
d 2W
d 2W
d 2W
=
=
dεij dεkl
dεji dεkl
dεij dεlk
(4.83)
proving minor symmetry.
How does symmetry reduce the number of possible independent constants? Recall that a n × n tensor has n2
constants, but a symmetric n × n tensor has
1
n (n + 1)
2
(4.84)
independent constants. Think of C as a second order 9 × 9 tensor. If it is symmetric, (by major symmetry) then
it only has 12 9 × (10) = 45 independent constants. But minor symmetry means that C is really a 6 × 6 symmetric
tensor. Then the number of possible terms is
1
6 × 7 = 21
2
All content © 2016-2018, Brandon Runnels
(4.85)
23.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 23
https://canvas.uccs.edu/courses/22031
So C can have no more than 21 independent constants. Alternatively, we can think of C as being represented in the
following way:
  C
σ11
1111
σ22  C2211
σ33  C3311
σ23  = C
   2311
σ31
C3111
σ12
C1211
|
C1122
C2222
C3322
C2322
C3122
C1222
C1133
C2233
C3333
C2333
C3133
C1233
C1123
C2223
C3323
C2323
C3123
C1223
C1131
C2231
C3331
C2331
C3131
C1231
C1112   ε11 
C2212   ε22 
C3312   ε33 


C2312 
 2ε23 
2ε31
C3112
2ε12
C1212
}
{z
36 terms, 21 independent constants
(4.86)
This is the form of a constitutive model for a general small-strain anisotropic material. Note that each shear stress
and strain term need only appear once since both stress and strain are symmetric.
4.7.2
Material symmetry
Some materials are said to have cubic symmetry. In that case, one can show that the elastic modulus tensor (in
Voigt notation) must reduce to
C
11
C12
C
C =  012

0
0
C12
C11
C12
0
0
0
C12
C12
C11
0
0
0
0
0
0
C44
0
0
0
0
0
0
C44
0
0 
0 
0 
0 

0
C44
(4.87)
Note that the number of independent constants has been reduced to three: C11 , C12 , C44 . This type of model is
frequently useful when computing the elastic modulus tensor of a single crystal material that has a cubic (e.g. FCC
or BCC) crystal structure.
Macroscopic materials are generally observed to be completely isotropic. It can be shown that isotropic materials
have only two independent parameters. Cijkl can then be defined in the following way:
Cijkl = µ(δik δjl + δil δjk ) + λδij δkl
(4.88)
where µ, λ are referred to as Lamé parameters. In the above case, note that we can then relate σ to ε as
σij = Cijkl εkl = µ(δik δjl + δil δjk )εkl + λδij δkl εkl = µ(εij + εji ) + λδij εkk = 2µ εij + λ tr(εkk ) δij
(4.89)
We can relate the parameters λ and µ to more familiar constants in the following way. Consider a plane stress
configuration with a known applied stress σ0 , so that
"
#
"
#
σ1 0 0
ε1 0 0
σ= 0 0 0
ε = 0 ε2 0
(4.90)
0 0 0
0 0 ε3
Let us compute the strain response ε1 , ε2 , ε3 in terms of σ0 , and determine the corresponding proportionality constants:
"
#
"
#
"
#
σ1 0 0
ε1 0 0
1
0 0 0 = 2µ 0 ε2 0 + λ(ε1 + ε2 + ε3 )
1
(4.91)
0 0 0
0 0 ε3
1
We can write as a system of equations
" # "
2µ + λ
σ1
0 =
λ
0
λ
All content © 2016-2018, Brandon Runnels
λ
2µ + λ
λ
#" #
λ
ε1
ε2
λ
2µ + λ ε3
(4.92)
23.5
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 23
https://canvas.uccs.edu/courses/22031
Solving this system, we get
ε1 =
σ0 (λ + µ)
µ(3λ + 2µ)
ε2 = ε3 = −
1
λσ0
λ
ε1
=−
2 µ(3λ + 2µ)
2(λ + µ)
(4.93)
From this we can identify the elastic modulus E and the Poisson’s ratio ν:
E=
µ(3λ + 2µ)
λ+µ
All content © 2016-2018, Brandon Runnels
ν=
λ
2(λ + µ)
(4.94)
23.6
Lecture 24
4.7.3
Constitutive modeling
The Cauchy-Navier equation and linear elastodynamics
One of the nice things about linearized elasticity is that we can work entirely in displacements and can use the
formulation to solve exact problems. Recall that
1
1
or
εij = (ui,j + uj,i )
Grad(u) + Grad(u)T
ε=
(4.95)
2
2
We also remember that C has minor symmetry. This means we can write σij as
σij =
1
1
Cijkl (uk,l + ul,k ) = (Cijkl uk,l + Cijkl uk,l ) = Cijkl uk,l
2
2
(4.96)
In other words, we can dispense with strain and write everything in terms of displacement only. If we do this, then
we can write our momentum balance equation as
Cijkl uk,lj + ρbi = ρui,tt
(4.97)
Substituting the form for isotropic elasticity
This is referred to in general as the Cauchy-Navier equation. It is a linear second order PDE in space and time that
can be solved relatively easily. Let’s consider a 1D plane strain version of the problem. Then we can write
"
#
"
#
1
u1,1 (t) 0 0
1
0
0 0 + λ u1,1 (t)
σ = 2µ
(4.98)
1
0
0 0
Then
"
#
(2µ + λ)u1,11
0
div σ =
0
(4.99)
If we ignore body forces, then the Cauchy Navier equation becomes
(2µ + λ)
∂2u
∂2u
=ρ 2,
2
∂t
∂x1
(4.100)
equation that can be analyzed easily using the method of characteristics or seperation of variables.
4.8
Thermodynamics of solids and the Coleman-Noll framework
We now return to finite strain mechanics. We have treated the formulation of pure elasticity, which is completely
reversible. However there are multiple types of mechanical phenomena such as plasticity and viscoelasticity that
are irreversible. Additionally, we wish to keep our treatment as general as possible, so we aim to account for thermal
effects in addition to mechanical.
Towards this end, we begin by assuming that a material can locally be defined by
(1) The deformation gradient F and its derivatives (although we will only consider F here)
(2) The material entropy N (per unit undeformed mass) and its derivatives
(3) A set of internal variables that we will denote in vector form by Q. (Note that Q isn’t a vector in the spatial
sense, rather it’s just a collection of individual internal variables.) A good example of an internal variable is
the accumulated plastic deformation in a material.
All content © 2016-2018, Brandon Runnels
24.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 24
https://canvas.uccs.edu/courses/22031
Now, let us assume that there exists an internal energy functional U = U(F, N, Q) that characterizes the materials
response to deformation, entropy, and change in internal variables.
Let us begin by working with the Clasius-Duhem inequaltiy: start by expanding it out using the product rule.
H RS
RSn
1
1
n
−
R Ṅ + Div
=
R Ṅ + Div H − 2 H · Grad T −
≥0
(4.101)
T
T product rule
T
T
T
We spot the appearance of the Div H term: we’ve seen this before in the first law in the Lagrangian frame. Rearranging the first law, we can solve for it:
R U̇ = P · Ḟ + RSn − Div H =⇒ Div H = P · Ḟ + RSn − R U̇
(4.102)
Now, we can substitute this into the Clasius-Duhem inequality, and because T > 0 we can multiply out by T :
R Ṅ +
P · Ḟ + RSn − R U̇
RSn
1
− 2 H · Grad T −
≥0
T
T
T
1
RT Ṅ − R U̇ + P · Ḟ − H · Grad T ≥ 0
T
(4.103)
(4.104)
Now, let’s consider the internal energy functional. Because it’s a function of F, N, Q, we can expand its time derivative out using the chain rule:
U̇ =
∂U
∂U
∂U
· Ḟ +
Ṅ +
· Q̇
∂F
∂N
∂Q
(4.105)
Substituting this relationship back into the inequality, we get
RT Ṅ − R
∂U
∂U
1
∂U
Ḟ − R
Ṅ − R
· Q̇ − P · Ḟ − H · Grad T ≥ 0
∂F
∂N
∂Q
T
And finally, we gather terms:
∂U 1
∂U ∂U
P−R
·Ḟ + RT − R
Ṅ − R
·Q̇ − H · Grad T ≥ 0
∂F
∂N
∂Q
T
|
|
{z
}
|
{z
}
{z
}
| {z }
(4.106)
(4.107)
The arbitrary nature of Ḟ, Ṅ, Q̇ prompt us to identify the following definitions:
∂U
Pe = R
| {z ∂F}
Elastic stress
∂U
T =
∂N
| {z }
Pv = P − Pe
|
{z
}
Viscous stress
Temperature
∂U
Y = −R
∂Q
{z
}
|
(4.108)
Dissipative driving force
Let’s discuss the new quantities briefly:
(1) Pe is the component of the stress that is purely elastic, for instance, the stress computed by Hooke’s law.
(2) Pv is the part of the stress corresponding to dissipation, for instance, the viscous stress generated in a fluid.
(3) Y is a bit more abstract. Components of Y correspond to components of Q, and you can think of them as the
forces that cause the internal variables to change. For instance, for the internal variable of plastic dissipation,
the driving force is the component of the stress that forces the plastic slip.
With these newly defined quantities, we can write the second law of thermodynamics as
1
Pv · Ḟ + Y · Q̇ − H · Grad T ≥ 0
T
| {z } |
| {z }
{z
}
mechanical
internal
(4.109)
thermal
This implies that irreversible (entropy-generating) processes can take three forms: (i) mechanical dissipation (e.g.
viscous drag in fluid flow) (ii) internal dissipation (e.g. plasticity) and (iii) thermal dissipation.
All content © 2016-2018, Brandon Runnels
24.2
Lecture 25
4.8.1
Thermodynamics of solids
Other thermodynamic potentials
We began by defining the internal energy functional U(C, N, Q) as a function of deformation gradient, entropy, and
some vector of internal variables.
(Note: we are using C rather than F because U is not convex in F whereas it is convex in C).
Can we define the internal energy as a function of other variables? That is, could we define A(F, T , Q) that gives
us the same information about the solid?
If N = N(F, T , Q) then the answer is yes, we would simply have
A(F, T , Q) = U(F, N(F, T , Q), Q)
(4.110)
In fact, such a relationship can be found by applying the state postulate. But now it is possible, how do we find
this function, since we don’t know the relationship between N and the other state variables? To do this, we use the
non-standard Legendre transform
All content © 2016-2018, Brandon Runnels
25.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
The Legendre transform
The Legendre transform is a powerful mathematical tool that is perticularly useful when working with energy
methods and with minimization problems. (Note: this is not to be confused with Legendre polynomials,
which are entirely different.) Let us develop the transform by considering a simple function f (x). We will
require that f be C 0 continuous and convex, but will place no other restrictions on it.
f (x)
f (x)
s
f (x)
x
x
x
m
x
∗
The original (or “primal”) way is to use a functional form to match a point in x to f (x); that is, we have a graph
for f (x). Now, let’s consider an alternative approach. Recall that we require that f is convex; this means that
the region above f (x) (called the epigraph) must be convex as well.
Notice how, at each point on the epigraph, there is (at least) one tangent line. Now, let us draw all of the
tangent lines to the epigraph at all of the points along f (x). Notice how we can recapture the shape of the
epigraph using the tangent lines only. This is called the “dual” representation of f (x).
So, we see that we can represent our function either using f (x) or the collection of all the tangent lines to
the epigraph. In particular, let us suppose that we have the following function:
f ∗ : [slope of tangent line] → [negative intercept of tangent line]
(4.111)
So f ∗ generates all possible tangent lines to the epigraph of f (x); we say that f ∗ is the Legendre Transform of
f . This comes as little help, however, if we don’t have a way of constructing f ∗ . Let us consider the following
formulation:
f ∗ (s) = sup [s x − f (x)]
x
(4.112)
To solve this problem we first solve for x ∗ , the minimizing value, using the stationarity condition:
x ∗ = arg sup[s x − f (x)] =⇒ s − f (x ∗ ) = 0 =⇒ f (x ∗ ) = s
x
(4.113)
So x ∗ is the location where f 0 (x) = s; i.e. the location where the line with slope s is tangent to f (x). Evaluating
f ∗ (s), then, we have
f ∗ (s) = s x ∗ − f (x ∗ )
All content © 2016-2018, Brandon Runnels
(4.114)
25.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
Example
Let’s try using the Legendre transformation on a sample function, f (x) = c x 2 . Is this function convex? Yes,
because its second derivative is always positive (assuming c > 0). Therefore, we can indeed apply the
Legendre transformation to this function. Let the slope of the function be s = f 0 . Then, to change variables
from x to the function’s slope, s, we apply the Legendre transformation,
(4.115)
f ∗ (s) = sup[s x − c x 2 ],
x
∂
[s x − c x 2 ] = s − 2c x = 0,
∂x
s
=⇒ x =
.
2c
(4.116)
=⇒
(4.117)
We’ve found the maximizer, so we plug it back into the original function that we are maximizing to get
f ∗ (s) = s
s s 2
s2
−c
=
.
2c
2c
4c
(4.118)
Thus, we’ve successfully found the function f ∗ (s) that is expressed in terms of the slope, that matches the
orignal function f (x). Verifying, we see that indeed
f (3) = c(3)2 = 9c,
f 0 (3) = s(3) = 2c(3) = 6c,
f ∗ (6c) =
(6c)2
= 9c,
4c
(4.119)
(4.120)
=⇒ f (3) = f ∗ (s(3)) = f ∗ (f 0 (3)) = 9c. X
So what do we mean by “non-standard?” This means nothing other than the fact that we change the sign of the
dual function, and infimize rather than supremize. It is exactly the same except for a sign change.
4.8.2
Internal Energy
Here we simply review the internal energy function U(C, N, Q), and recall the relations from Coleman-Noll:
Se = 2 R
∂U
∂C
T =
∂U
∂N
Y = −R
∂U
∂Q
(4.121)
where Se is the elastic part of the second Piola Kirchoff stress tensor and Y is the vector of driving forces corresponding to the internal variables Q.
4.8.3
Helmholtz Free Energy
The Helmholtz free energy A is like the internal energy, except that it is in terms of temperature rather than entropy.
(This makes it much easier to work with.) We perform the change of variables using a non-standard Legendre
transform:
n
o
A(C, T , Q) = inf U(C, N, Q) − TN
(4.122)
N
Notice that the stationarity condition requires
T =
∂U
,
∂N
(4.123)
the second relation for internal energy. Now, can we find relations for this functional similar to that for internal
energy? Let us begin by considering the total differential of A in terms of its three arguments:
dA =
All content © 2016-2018, Brandon Runnels
∂A
∂A
∂A
· dC +
dT +
· dQ
∂C
∂T
∂Q
(4.124)
25.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
On the other hand, from Equation 4.122,
dA = dU − NdT − TdN =
∂U
∂U
∂U
·dC +
dN +
·dQ − N dT − T dN
∂C
∂N
∂Q
|{z}
|{z}
|{z}
S2 /2R
=T
(4.125)
=−Y/R
1
1 e
S · dC − N dT − Y · dQ
=
2R
R
(4.126)
Equating these two expressions gives the relationships
Se = 2R
∂A
∂C
N=−
∂A
∂T
Y = −R
∂A
∂Q
(4.127)
where we see that A and U are identical with respect to Se , Y but now we obtain entropy by differentiation of A with
respect to temperature.
4.8.4
Enthalpy
Enthalpy H(Se , N, Q) is the thermodynamic potential in terms of Se instead of C. It is also obtained through the
following non-standard Legendre transform:
o
n
1
(4.128)
R H(Se , N, Q) = inf R U(C, N, Q) − Se · C
C
2
whence we again identify the stationarity condition
Se = 2 R
∂U
∂C
(4.129)
that is in agreement with the relation above. To find equilibrium relations, we again expand out; this time, however,
we will consider only the terms that involve the transformed variable.
R dH = R
∂U
1
1
∂H
!
·dC − dSe · C − Se · dC + ... = R e · dSe + ...
∂C
2
2
∂S
|{z}
(4.130)
Se /2R
1
∂H
!
= − C · dSe + ... = R e · dSe + ...
2
∂S
(4.131)
Equating gives the following equilibrium relationships
C = −2R
4.8.5
∂H
∂Se
T =
∂H
∂N
Y = −R
∂H
∂Q
(4.132)
Gibbs Free Energy
The Gibbs Free Energy G (Se , T , Q) is the thermodynamic potential in terms of stress and temperature instead of
deformation and entropy, and can be defined in terms of the enthalpy
n
o
G (Se , T , Q) = inf H(Se , N, Q) − NT
(4.133)
N
or in terms of the Helmholtz free energy
n
o
1
RG (Se , T , Q) = inf R A(C, T , Q) − Se · C
C
2
(4.134)
The same procedure can be followed to find that
C = −2R
∂G
∂S e
N=−
∂G
∂T
Y = −R
∂G
∂Q
(4.135)
The differential relationships between A, G , U, H and the state variables C, Se , T , N can be visualized with a mnemonic:
All content © 2016-2018, Brandon Runnels
25.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
C
A
T
U
N
G
1 e
− 2R
S
H
Example 4.3: Thermodynamic potentials mnemonic
Let us suppose we wish to determine the relationship between Gibbs free energy G , temperature T , and
entropy N. Begin with G . Note that G is not adjacent to N, it is only adjacent to Se and T , reminding us that
it is not a function of entropy or the deformation temperature.
We proceed to T . Note that T is at the arrow end of the line, meaning we go in the reverse direction to get
to N. This indicates that there will be a minus sign. The result is then
∂G
= −N
∂T
(4.136)
This can be used to recover any of the eight relationships derived in this section.
4.9
Inelastic constitutive modeling
4.9.1 Crystal plasticity
The first step is to determine the components of the deformation due to elastic vs plastic deformation. The typical
picture is that of the following kinematic decomposition:
F = Fe Fp
Fe
Fp
so that F = Fe Fp . Note that in small strain, the multiplicative decomposition reduces to the additive decomposition
ε = εe + εp . Now we need a constitutive model. The relationship
P=
∂W
F
(4.137)
no longer holds because W is an elastic potential. However, we know from Coleman-Noll that
Fe = 2R
All content © 2016-2018, Brandon Runnels
∂U
∂U
= 2R
,
∂F
∂C
(4.138)
25.5
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
in other words, we can use a potential to determine the elastic portion of the deformation. Now what about Fp ?
We turn to crystallography to determine how pure plastic deformation takes place. FCC materials have four “slip
planes” along which dislocations move, and three slip directions corresponding to each plane.
Slip plane (m)
Slip directions (s)
(111)
[01̄1], [101̄], [1̄10]
(1̄1̄1)
[011], [1̄01̄], [11̄0]
(1̄11)
[01̄1], [1̄01̄],[110]
(11̄1)
[011], [101̄],[1̄1̄0]
Illustrating the slip planes and directions:
x3
x3
dislocation
motion
[1̄10]
x3
(111)
[101̄]
x1
x2
x2
x1
x1
x2
[01̄1]
There is a total of 4 × 3 = 12 combinations of m and s, and each pair is called a slip system. We can then show that
slip along a particular slip system with vectors m and s produces
Ḟp ∝ (s ⊗ m) Fp
(4.139)
So for all combined slip systems we have the model (taken from Bower [1]).
Ḟp Fp−1 =
N
X
γ̇ α s ⊗ m,
(4.140)
α=1
where {γ α } are scalar parameters that determine how much shear has taken place in each slip system. These are
the internal variables for the system so that Q = {γ 1 , ... , γ N }. Now that we have Q, what are the thermodynamic
driving forces? We have
Y = {τ 1 , ... , τ N }
(4.141)
where τ α is the resolved shear stress. The resolved shear stress for a plane is approximated by
τα = s · P m
(4.142)
Now, we need an evolution equation for γ α , which we expect to be in terms of the resolved shear stress. The
following heuristic equation is an example of one such model
!m
|τ α |
α
α
α
(4.143)
γ̇ = γ̇0 sgn(τ )
gα
where γ0α , and m are material parameters, and g α is a hardening parameter. So we’ve solved for one variable and
introduced another - now we need to model the evolution of g α . One such model is given by the following
ġ α =
N
X
hαβ |γ̇ β |
(4.144)
β=1
All content © 2016-2018, Brandon Runnels
25.6
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 25
https://canvas.uccs.edu/courses/22031
where hα,β represents that interaction between different slip systems. If it is diagonal, then there is no interaction
between dislocations in different systems, which we know to be unphysical. The following relations provide a
model for hαβ :
h(γ) = hs + (hs − h0 ) sech
hαβ = qαβ h(γ̄)
h − h 0
s
γ̄
gs − g0
(4.145)
with hs h0 , gs as material constants and γ̄ is the total accumulated slip given by
γ̄ =
N Z
X
α=1
t
|γ̇ α |dt
(4.146)
mα = mβ
else
(4.147)
0
Finally, the matrix qαβ for an FCC has the form
(
qαβ =
1
q
where q is a material property. This material model is complicated, but does the job in many cases. On the other
hand...
“With four parameters I can fit an elephant, and with five I can make him wiggle his trunk.”
John von Neumann
All content © 2016-2018, Brandon Runnels
25.7
Lecture 26
5
Computational mechanics
Computational mechanics
All of the work we have done so far culminated in developing continuous formulations for the mechanics of continuous bodies; these formulations generally take the form of either (i) differential equations or (ii) variational principles. These forms are convenient because they are exact, so any analytical solutions that we find will be precisely
correct.
Most of the time, it’s not practical (or possible) to solve differential equations by hand, especially if the geometry
in question is highly complex. This is where it will be helpful to use a computer to approximate the solutions to the
equations we derived. There are multiple methods for doing this; here, we will focus on the finite element method.
5.1
The finite element method
The finite element method is a dimensionality-reducing technique: rather than solve over the infinite-dimensional
space of all functions, we reduce to a finite dimensional approximation of function space. There are three ingredients to this technique:
(1) Shape functions
(2) The weak formulation of governing equations
(3) Numerical quadrature
A lengthy discussion of all of these topics is enough to fill an entire class; unfortunately, here, a couple of lectures
will have to suffice.
5.1.1
Shape functions
Consider a function f (x) defined over an interval [a, b] (we’ll stick to 1D for the moment.) One way of storing this
function is to store a symbolic expression for f (x) such as x 2 or sin(x) that we can use to evaluate f (x) at any point.
Unfortunately this does not work well in the general case, since most functions f (x) do not have an analytic form.
The next best thing is to choose an interpolation scheme such that instead of defining our function’s value at every
point, we define it at a selection of points, and interpolate between them. We do this via the following steps:
(1) Discretize the domain [a, b] to as set of discrete points {xi } ⊂ [a, b].
(2) Define a set of shape functions φi (x)
(3) Interpolate the function as a linear combination of shape functions
X
f (x) ≈
fi φi (x)
(5.1)
i
What are these shape functions? In 1D, a typical shape function might look like this:
1
0
All content © 2016-2018, Brandon Runnels
φi (x)
... xi−1 xi xi+1 ...
26.1
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
Mathematically, we tend to define shape functions in such a way as to have the following properties:
(1) Weak Dirac: the shape function φi evaluates to 1 at xi and 0 at all other discretization points; in other words,
(5.2)
φi (xj ) = δij
(2) Differentiability: shape functions are at least once differentiable; in other words for (e.g.) [a, b],
(5.3)
{φi } ⊂ C1 [a, b]
(3) Partition of unity: shape functions sum to 1 at every point; in other words for (e.g.) [a, b],
X
φi (x) = 1 ∀x ∈ [a, b]
(5.4)
i
These restrictions on the shape functions mean that we can express our function f (x) as
X
X
f (x) ≈
fi φi (x) =
f (xi ) φi (x)
i
(5.5)
i
In other words, we multiply each shape function φi by the functions value at that point, f (xi ). Graphically, we are
doing the following:
f (x)
{φi (x)}
P
{f (xi )φi (x)}
x0 x1 x2 x3 x4 x5 x6 x7 x8
i
x0 x1 x2 x3 x4 x5 x6 x7 x8
f (xi )φi (x)
x0 x1 x2 x3 x4 x5 x6 x7 x8
Notice how the formalism of shape functions naturally allows us to interpolate the function to an arbitrary degree
of accuracy, depending on the type of shape functions used. This is extensible beyond 1D interpolation: in higher
dimensions, shape functions are defined over small discretized elements; for instance, the following is one possible
shape function scheme defined over a quadrilateral element:
φ1
x1
x3
φ2
x2
x4
x1
x3
φ3
x2
x4
x1
x3
φ4
x2
x4
x1
x3
x2
x4
The wide varieties of shape functions are beyond the scope of this lecture, although it is well worth reading about
them. When you run ANSYS or SolidWorks and select your mesh elements, you are determining what kind of
shape functions you will use. Different shape functions have different advantages in different situations, and it is
important to know what those are.
For now, however, the most important thing is to remember the following:
(1) Shape functions are always defined, so we always know what they are.
(2) In addition to knowing φi , we always know what their derivatives (Grad φi ) are.
(3) Shape functions allow any function f (x) to be described in terms of a set of discrete values fi
(4) If the function f is known, then fi = f (xi ), the value of the function at the discretization point.
All content © 2016-2018, Brandon Runnels
26.2
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
5.1.2
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
Weak formulation
Consider a domain Ω with a set of discretization points {xi } ⊂ Ω and shape functions φi ⊂ C1 (Ω). (If it is helpful,
you can think of Ω = [a, b], the 1D case.)
Now, let us suppose that we wish to solve a linear partial differential equation (in space only, no time dependence)
that is defined over Ω, with boundary conditions defined on ∂Ω. In 1D, an example might be
d 2f
+ f (x) = g (x)
dx 2
(5.6)
We can write a differential equation of this form more generally:
(5.7)
D[f ] − g (x) = 0
where D[f ] is a generalized way of writing all contains all of the derivatives and scalar multipliers on f , and g
contains “everything else.” Recall that we can write f as
X
fi φi (x),
(5.8)
f (x) =
so we can express the above as
D
hX
i
X
fi φi (x) − g (x) =
fi D[φi (x)] − g (x) = 0
i
(5.9)
i
Now, recall the fundamental lemma of the Calculus of Variations? It works both ways: if the above is true, then we
can say that
Z hX
i
fi D[φi (x)] − g (x) η(x) dx = 0 ∀η
(5.10)
Ω
i
This is called the weak form. Here’s the important step: remember that η can be anything – so why not let it be
X
η(x) =
ηi φi (x).
(5.11)
i
If we make this substitution, we have
Z hX
Ω
iX
fi D[φi (x)] − g (x)
ηj φj (x) dx = 0 ∀{ηj }
i
(5.12)
j
It may seem that we just took an easy problem and made it complicated. Let’s rearrange it to try to simplify:
Z
XX
X Z
fi ηj
D[φi (x)] φj (x) dx =
ηj
g (x) φj (x) dx ∀{ηj }
(5.13)
i
Ω
j
Ω
j
Now, recall that the above must hold true for every possible value of ηj . The only way for this to be possible is if
the following is true as well:
Z
X Z
fi
D[φi (x)] φj (x) dx =
g (x) φj (x) dx
(5.14)
i
Ω
Ω
(This is the equivalent of passing from the weak form to the strong form in the discretized case.) So, what is the
advantage of writing our problem this way? Recall that we are trying to find {fi } – if we can find them, then we
can construct an interpolated solution to the problem. If we look through the problem, we will see that we already
All content © 2016-2018, Brandon Runnels
26.3
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
know everything else: we know {φi } so we can compute the integral containing D[φi ], and we know g (x) so we can
compute the second integral as well. We might write the problem this way:
(5.15)
fi Kij = bj
where
Z
Kij =
Z
D[φi (x)] φj (x) dx
bj =
Ω
g (x) φj (x) dx
(5.16)
Ω
K is frequently called the stiffness matrix and b the vector of forces, because the finite element method was historically developed for solid mechanics. Remember, we can compute K and b directly, so the resulting problem is
a simple linear solve – quite doable by computer!
We glossed over some details here, so let’s make a few notes:
• The integral
Z
Kij =
D[φi (x)] φj (x) dx
(5.17)
Ω
can generally be reduced to a more symmetric form using integration by parts.
• Recall that we let our arbitrary function η be described as
X
η(x) =
ηj φj (x)
(5.18)
j
Actually, this was something of a design decision: the selection of η to that form is a particular choice, and is
part of the so-called Galerkin method. There are other discretization methods, but this is generally the most
popular.
5.1.3
Numerical quadrature
Numerical quadrature is nothing other than a fancy term for integral approximation. Recall that we casually stated
that the integral
Z
Kij =
D[φi (x)] φj (x) dx
(5.19)
Ω
was easy to compute, since we know D and {φi }. It would be more accurate to say that it is possible to compute – in
reality, the integral is not necessarily computable easily, especially if we are working with a geometrically complex
mesh.
Quadrature is just a method for approximating a nasty integral by evaluating it a specific points. For instance a
simple so-called “quadrature rule” for a 1D integral might be written this way:
Z
X
f (x) dx ≈
qα f (xα )
(5.20)
Ω
α
where we would call {xα } quadrature points and {qα } quadrature weights. For an integral over an element, there
are conveniently tabulated quadrature points and weights that enable accurate and efficient integral evaluation.
5.2
Linearized kinematics
Let us consider the equation of conservation of linear momentum for the static case in one dimension, where plane
stress is assumed:
d 2u
(5.21)
E 2 + b(x) = 0
dx
All content © 2016-2018, Brandon Runnels
26.4
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
Suppose we have a discretization {xi } and shape functions φi , and our domain is Ω = [a, b] Let us find the finite
element formulation of this equation. First, let
u(x) = ui φi (x)
so that the equation can be written as
X
ui E
i
d 2 φi
+ b(x) = 0
dx 2
Now, we can express the above in weak form as
Z b hX
i
d 2 φi
+
b(x)
η(x) dx = 0 ∀η
ui E
dx 2
a
(5.22)
(5.23)
(5.24)
i
Applying the Galerkin method, we can write the above as
Z b
X
X Z b
d 2 φi
ui ηJ
E
ηj
b(x) φj (x) dx ∀{ηj }
φj (x) dx = −
dx 2
a
a
i
(5.25)
j
Passing back to the strong form, we have
Z b
X Z b d 2 φi
ui
E
φj (x) dx = −
b(x) φj (x) dx
dx 2
a
i
{z
}
|a
(5.26)
Kij
We can use integration by parts to reduce the stiffness matrix a little more:
Z b Z b
b
dφi
dφi dφj d 2 φi
φ
(x)
dx
=
φ
(x)
dx
−
Kij =
E
E
j
j
2
dx 2
dx
dx
a
a
|dx {z a}
(5.27)
=0
Thus, we can write the equilibrium equation as
"
#
Z b
X Z b dφi dφj E
dx uj =
b(x) φi (x) dx
dx
dx
a
a
(5.28)
j
We note that this modified version of the stiffness matrix has one huge advantage: it only requires that the shape
functions be differentiable once, whereas the previous required twice differentiability. It is definitely possible to
use “higher order” shape functions that are differentiable this way, but they frequently come with their own set of
disadvantages.
5.2.1
3D linearized elasticity
Let us now develop the finite element formulation of a fully 3d linear elastic material, for which the equations are
Cijkl uk,lj + ρ bi = ρüi
(5.29)
Given a domain Ω with discretization {xm } and shape functions {φm } ⊂ C1 (Ω), we begin by discretizing our displacement field:
ui (x) = uim φm (x)
(5.30)
(Note that we (i) adopt the summation convention, and (ii) use superscripts to indicate summation over all shape
functions, as opposed to subscripts summing over all dimensions.) Substituting into our governing equation, we
have
m m
ukm Cijkl φm
,lj + ρ bi = ρüi φ
All content © 2016-2018, Brandon Runnels
(5.31)
26.5
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
Again, we write in weak form
Z
m m
[ukm Cijkl φm
,lj + ρ bi − ρüi φ ] ηi (x) dv = 0 ∀η(x)
(5.32)
Ω
(Notice that we are dotting with η.) Adopting a Galerkin discretization (ηi = ηin φn ), we can now write
Z
Z
Z
n
n
n
m n
ukm ηin
Cijkl φm
φ
dv
+
η
ρ
b
φ
dv
=
ü
η
ρ φm φn dv ∀{ηin }
i
,lj
i
i
i
Ω
Ω
Because this holds for all test functions, we have
Z
Z
Z
n
n
m
ukm
Cijkl φm
φ
dv
+
ρ
b
φ
dv
=
ü
ρ φm φn dv
i
,lj
i
Ω
Ω
(5.34)
Ω
Applying the product rule and rearranging,
Z
Z
Z
m
m n
m
m n
uk
Cijkl φ,j φ,l dv + üi
ρ φ φ dv =
ρ bi φn dv
Ω
(5.33)
Ω
Ω
(5.35)
Ω
For the static case, we have a simple linear system to solve. For the dynamic case (with no body forces) we have a
linear eigenvalue problem, where the eigenvalues of the stiffness matrix correspond to resonant frequencies, and
the eigenvectors to resonant modes.
5.3
Newton’s method
Newton’s method is a powerful technique for solving many kinds of variational problems. Suppose that we have a
function f (x) that we wish to optimize, and suppose we have an initial guess x0 . Let us expand f (x) about x0 :
1
f (x) ≈ f (x0 ) + f 0 (x0 )(x − x0 ) + f 00 (x0 )(x − x0 )2
2
(5.36)
What is the optimal point of the above approximated function?
df
f 0 (x0 )
!
= f 0 (x0 ) + f 00 (x0 )(x − x0 ) = 0 =⇒ x = x0 − 00
dx
f (x0 )
(5.37)
Does this equation give us the exact solution? It depends: if it happens that f (x) was quadratic, then yes. Otherwise,
we don’t really know – although it is a good bet that the new point x got us much closer to the actual solution.
What about if we have F (x), a vector functional? We can do exactly the same thing: if our initial guess is x n then
1
F (x) ≈ F (xn ) + DF (xn )p (xp − xpn ) + DDF (xn )pq (xp − xpn )(xq − xqn )
2
(5.38)
Using the stationarity condition we get
dF
!
= DF (xn )i + DDF (xn )ij (xj − xjn ) = 0
dxi
(5.39)
n
xj = xjn − DDF (xn )−1
ji DF (x )i
(5.40)
xn+1 = xn − DDF (xn )−1 DF (xn )
(5.41)
Solving for x gives
The Newton iteration is then defined to be
All content © 2016-2018, Brandon Runnels
26.6
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
5.4
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
Finite kinematics
We introduce the following discretization:
X
φ(X) ≈
φα η α (X)
α
=
summation convention
(5.42)
φα η α (X)
where {φ} are vector-valued constants and {η α (X)} are shape functions that are at least C 0 continuous (so the
gradient can be computed). We recall the variational statement of the
"Z
#
Z
Z
φ = arg inf Π[φ] = arg inf
W (F )dV −
φi RBi dV −
φi Ti dA
(5.43)
φ
φ
Ω
Ω
∂Ω
Our finite element scheme admits the statement of the following discretized problem:
Z
Z
Z
κ κ
κ
κ
φi η RBi dV −
W (φ Grad η )dV −
{φ} = arg inf Π[{φ}] = arg inf
{φ}
{φ}
φκi η κ Ti dA
(5.44)
∂Ω
Ω
Ω
The variational structure of the problem enables a ready solution by means of Newton’s method. Beginning with
an initial guess {φ0 }, the following iteration converges to the solution:
β
α
−1 αβ
φα
i 7→ φi − (DDΠ )ij DΠj
(5.45)
where we compute
Z
κ
κ
Z
φκi η κ RBi dV
Z
W (φ Grad η )dV −
−
φκi η κ Ti dA
Ω
∂Ω
Z
Z
Z
dΠ
κ
κ
α
α
=
DW
(φ
Grad
η
)
Grad
η
dV
−
η
RB
dV
−
η α Ti dA = 0
DΠ[{φ}]α
=
ip
i
,p
i
dφα
Ω
Ω
∂Ω
i
Z
d 2Π
αβ
α
β
DDΠ[{φ}]ij =
=
DDW (φκ Grad η κ )ijpq Grad η,p
Grad η,q
dV = CiJkL
β
dφα
Ω
i dφj
Π[{φ}] =
(5.46)
Ω
(5.47)
(5.48)
The function W (F ) is the elastic free energy of the function in terms of the deformation gradient, and
DW (F )iJ =
dW
= PiJ
dFiJ
d 2W
dFiJ dFkL
DDW (F )iJkL =
(5.49)
are the first Piola-Kirchoff stress tensor, and the tangent modulus of the material.
5.5
Computational fluid dynamics
Computational fluid dynamics (CFD) involves a completely different method for discretizing the governing equations. We will introduce a discretization referred to as the finite volume method, although it is worth mentioning
that many CFD methods are interrelated, and so there may be a lot of crossover.
Let us begin by introducing an arbitrary mesh (we’ll refer to it as an unstructured mesh).
n1
v(x)
v1
v2
v4
v̄
n2
v3
n4
n3
All content © 2016-2018, Brandon Runnels
26.7
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes - Lecture 26
https://canvas.uccs.edu/courses/22031
We will treat each cell of the mesh as a control volume, for which we already know surface normals {ni }, surface
areas {Ai }, and total volume V . We will now use our conservation equations to evolve the velocity field in time via
the following steps.
(1) Recall our conservation of mass expression for a control volume. Here, we will let our discretized cell be the
control volume with volume V , so we write
Z
dm
d(ρV )
=
=−
ρ v · nda
(5.50)
dt
dt
∂V
We assume that the velocities are constant over the faces of the c.v., so we can update the density by the
discretized formula
∆t X
ρi (vi · ni )Ai
(5.51)
ρ = ρold −
V
i
where Ai are the areas of the cells and ρi , vi , ni are the quantities stored at each face.
(2) Recal the expression for the Navier-Stokes momentum equation that we derived (ignoring gravity)
∂
(ρv) + div(ρv ⊗ v + pI) = div(τ )
∂t
Let us express this in volume form also:
Z
Z
Z
∂
(ρv)dV =
div(τ )dV −
div(ρv ⊗ v + pI)dV
V
V ∂t
ZV
=
[τ n − ρ(v ⊗ v)n + ρ p n]dV
(5.52)
(5.53)
(5.54)
∂V
Now, let us assume that the velocities at the faces are constant over each face, and compute the cell averaged
velocity v̄.
ρv̄ = (ρv̄)old +
∆t X
[τ n − ρ(vi ⊗ vi )ni + pi ni ]Ai
V
(5.55)
i
Since we already know ρ at the center, we can write the following updated expression for the cell-averaged
velocity
v̄ =
i
1h
∆t X
(ρv̄)old +
[τ n − ρ(vi ⊗ vi )ni + pi ni ]Ai
ρ
V
(5.56)
i
The viscous stress tensor τ is computed in terms of velocity gradients, so it is a known quantity.
(3) We assume that we have a known equation of state, so we can use it to compute the updated pressure:
p = p(ρ, T )
(5.57)
(4) Finally, we determine pressures {pi }, velocities {vi }, and densities {ρi } at the cell boundaries by interpolation.
We note that this is a fully explicit scheme, in that it relies only on the current timestep’s values to find new ones.
As a result, it cannot conserve energy completely. It is also a highly simplified scheme; there are much better ways
of doing the above computation.
For incompressible flow, special measures have to be taken to find the pressure field, since the equation of state is
ill-defined. One method of doing it is the fractional step or projection method, where the velocity field is estimated,
then used to find a pressure field that is used to correct the original velocity field.
All content © 2016-2018, Brandon Runnels
26.8
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes
https://canvas.uccs.edu/courses/22031
References
[1] Allan F Bower. Applied mechanics of solids. CRC press, 2009.
All content © 2016-2018, Brandon Runnels
9
MAE 5100 - Continuum Mechanics
University of Colorado Colorado Springs
Course Notes
https://canvas.uccs.edu/courses/22031
Index
basis, 1.2
convexity, 25.2
duality, 25.2
dummy index, 1.3
free index, 1.3
Kronecker delta, δ, 1.4
Einstein summation convention , see index notation
piola, 13.4
index notation, 1.3
sets, 1.2
All content © 2016-2018, Brandon Runnels
10