Cm Lecture1

7/28/2019 Cm Lecture1

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Romain TeyssierContinuum Mechanics 2011

Continuum Mechanics

Lecture 1

Solid mechanics

Prof. Romain Teyssier

http://www.itp.uzh.ch/~teyssier


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- Point mechanics

- Fundamentals laws of mechanics

- Continuum mechanics

- Kinetic energy theorem

- Rigid body motions

- Internal forces

- Stress field and stress tensor

- Displacement field and strain tensor

Outline


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A point particle is defined by its position in space with respect to anarbitrary reference point, O, accompanied by a coordinate system: theframe of reference.

Space geometry is Euclidian. Time is considered as absolute andhomogeneous.

Fundamental quantities for point kinematics are:

Position:

Velocity:

Acceleration:

Derivatives are taken along the particle trajectory.

Trajectory: the path of a particle through space in the frame of reference.


Point mechanics

OM

v =d

dt

OM

a =d

dt

v


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The frame of reference is usually defined by 4 objects

the origin and 3 unit vectors forming the frame of a vector space.

The 3 vectors form an orthonormal basis.

for i different from j

Both A and the vector basis can be fixed or can vary with time.


The frame of reference

(A,n1,n2,n3)

(n1,n2,n3)

ni ni = 1

ni nj = 0


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Coordinates of the point particle are function of time.

Since the frame is fixed,

The velocity writes

In vector notations:

OM= x(t)i+ y(t)j + z(t)k

i = 0 j = 0k = 0

v =d

dt

OM = xi + yj + zk

v = ( x, y, z) a = (x, y, z)


Cartesian frame

(O,i,j,k)

A fixed frame of reference


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OM = r(t)r(t) + z(t)k

= ( sin , cos , 0)

r =dr

d

= =d

d

= r


Cylindrical coordinates

The frame is now moving with the particle

We introduce the radial and azimuthal vectors that nowdepend on time.

(O,r, ,k)

r = (cos , sin , 0)


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OM = (r, 0, z)

v = ( r, r, z)

a = (r r2, 2r2 + r, z)


Kinematics in cylindrical coordinates


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cos sin

sin

0 Romain TeyssierContinuum Mechanics 2011

Spherical coordinates

(O,r, , )

r = sin cossin sin

cos

= cos cos

= sincos

The moving frame is now defined by

We introduce the radial, azimuthal and polar vectors.


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OM = (r, 0, 0)

v = r, r, r sin a = for homework


Kinematics in spherical coordinates


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The trajectory is known: we define thecurvilinear coordinate by the length of the

curve from A to M:

The tangent vector is defined as . We have

The normal vector is defined as . We have

Finally, the binormal vector is just

The velocity is obtained as

and the acceleration by

M

s(t) = AM(t)

b = t n


Curvilinear coordinates and the Frenet frame

A

Ris the curvatureradius of the trajectory

t =d

ds

OM

t

= 1

n =1

dt

ds

dt

dsn t = 0

v = s(t)t

a = s(t)t +v2

Rn


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A particle trajectory is only defined relative to a frame.

A frame is defined by a reference point and 3 units orthogonal vectors.

We have seen one example of fixed frame:

the Cartesian frame with Cartesian coordinates

and 3 moving frames:

with cylindrical coordinates

with spherical coordinates

and with curvilinear coordinates

(x,y,z)

(s, 0, 0)


Conclusion on point kinematics

(r, , z)

(r, ,)


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The principle of inertia (Galileo 1630) or Newtons first law of motion:

Every body remains in a state of rest or uniform motion unless it is actedupon by an external unbalanced force.

For an isolated particle,

This property is called inertia.

It is true only in a certain set of frames of reference, the so-called Galileanor Newtonian or inertial frames defined as fixed or non-accelerated frames.

v =

cst


The fundamental laws of mechanics


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The fundamental principle of mechanics (Newton 1680) or Newtonssecond law of motion:

A body of mass m subject to a net force Fundergoes and acceleration asuch as F=ma. If we define the momentum of the particle asp=mv, thenNewtons second law states that the total force applied to the particle is

equal to the time derivative of the momentum.

Warning: the second law is also valid only in an inertial frame.

F = mdv

dt=

dp

dt




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JO =

OM

P

d

dt

JO =

OM

F =

TO(F)



Angular momentum of a particle relative to the frame center O:

We can show that the angular momentum variation is equal to the torque of

the forces acting on the particle


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The law of action-reaction (Newton 1680) or Newtons third law:

The mutual forces of action and reaction between two bodies are equal,opposite and collinear. This means that whenever a first body exerts a forceFon a second body, the second body exerts a force -Fon the first body.

r ij

F ij = 0



F ij =

F ji


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r ij

F ij = 0



F ij =

F ji

F=

m

dv

dt=

dp

dt

F = 0 v =

cst1.

2.

3.


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We now consider a system of point particles moving together.

Continuum mechanics is the macroscopic description of a system ofmicroscopic particles (defined by their masses and coordinates).

We assume that we can separate the macroscopic scales (the size of the

solid, the depth of the liquid) from the microscopic scales (the size of theparticles, the average distance between them).

A particular point: the center of mass

A particular frame: the center of mass frame moves at the center of massvelocity


The fundamental laws of continuum mechanics

mi

OMi = mi

OG

mi

OMi

vG =

d

dt

OG


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JO =

J 0,i

OMi =

OG+

GMi

JO =

JG +

OG MvG



Similarly, we can define the total, macroscopic angular momentum as:

vi = vG + v

i


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F =

F i,ext +

F i,int



Forces can be divided into 2 categories: external and internal

Newtons 3rd law in weak form: internal forces add up to zero.

Newtons 2nd law can be generalized to the center of mass:

F=

F i,ext=

M

dvG

dt


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d

dt

JO =

TO (Fext)

TO (Fext) =

OMi

F i,ext



Fi =

Fi,ext +

Fi,int

Forces can be divided into 2 categories: external and internal

Using Newtons 3rd law in the strong form, we can show that

where we define the total external torque w. r. t. point O.

Recall that Newtons 3rd law in strong form states that:

F ij =

F ji r ij


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When a point particle moves, the force acting on the particle moves too.

We define the work done by the force during an infinitesimal displacement

During this infinitesimal path, the force can be considered as constant.

Defining the kinetic energy of a point particle as

We can prove that when the particle moves from point A to B:

Warning: in general, the work done depends on each particular path, notonly on the starting and final points.

Ki =1

2v2


The kinetic energy theorem

Wi =

Fi

OMi

Ki =1

2

v2B v

2A

=WAB


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Macroscopic approach using the center of mass.

We use the center of mass frame of reference.

We can prove that

The total kinetic energy is the sum of the kinetic energy of translation andthe internal kinetic energy (or internal energy)

vi = vG + v

i

K=1

2Mv

2

G +Kint

Kint = 1

2mi (v

i)2

K=

Ki =

1

2

miv2

i


Kinetic energy for systems


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Again, we decompose the total force on each particle into

Fi =

Fi,ext +

Fi,int

K= KB KA

= WAB(Fext) + WAB(Fint)

Kint =WAB(Fint)


The kinetic energy theorem for systems

We can show that the difference in kinetic energy between point A and B is

Warning: in general, the work of internal forces is not zero.

For an isolated system, we have:


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F = gradU = U F =

Ux

, Uy

, Uz

Wi

=

F

i

OM

i

= U x = dU

WAB = (UB UA)

K+U = 0


A particular case: force deriving from a potential

In electromagnetism or for gravity, the force is given by

The infinitesimal work done by the force for a small displacement is

The work is now a differential form (like the kinetic energy).

Integrating along the full path, we have

The total (mechanical) energy is now conserved), independently on the exact path

f


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F ij =

F ji

Wi =

Fji

OMi

Wj =

F ij

OMj

|MiMj | = cst

F ij

MiMj

Wi =

F ij

MiMj = 0


A particular case: solid with zero deformation

Newtons 3rd law in weak form:

Compute work for both particle

Using Newtons law in strong form

and the no deformation assumption

We can show that

For a non deforming solid, the work of internal forces is zero.

Ri id b d ti


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d

dt

MiMj = vj vi = 0

X =

X

v i =

v 0 +

OMi


Rigid body motions

Can we define a particular type of motion that preserve the rigidity of the body ?

Trivial case: uniform velocity. vi= v

G

We require any orthonormal basis in the body to be preserved (proof).

X=A

XWe then look for a general linear transform such as

We obtain that the matrix A is a rotation, defined by the rotation vector

Rigid body motions are then defined by

W k d b i id b d di l t


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OMi =

v it

OMi

t=

F i v i

W

t =

F i

v 0 +

F i

OMi

W

t=

F v 0

TO


Work done by a rigid body displacement

We consider an infinitesimal displacement from a rigid body velocity field.

The work per unit time for each particle is

The total work integrated over the body is

After some calculation, we get the following simple relation (proof)

We immediately see that if the net force and the net torque are both zero, thetotal work of the rigid body is zero.

For internal forces, Newton thirds law gives us zero net force and torque(proof).

F di t t ti h i


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So far, we can treat our system as a set of point particles, or a an ensembleof volume elements.

We can identify with

Replace everywhere

In Cartesian coordinates:

In cylindrical coordinates:

In spherical coordinates:

mi

Xi =

XdV

dV = dr rd r sin d

dV = dr rd dz


From discrete to continuum mechanics

mi dV

V= x y z

Wh t t l d i t l f ?


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What are external and internal forces ?

The most important step: define properly the system.

An internal force can become external to the system, depending on how thesystem boundaries are chosen.

2 types of forces:

- body, volume, long distance forces such as gravity or electromagnetic forces

- contact, short-range forces such as atomic or molecular bounds in crystals.Usually, internal forces are of the second kinds.

Ch i l b d i lid d li id


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U = UR + UA = 4U0

r0

r

n

r0

r

m


Chemical bounds in solids and liquids

Strong atomic bounds in crystal structure such as

glasses, concrete, metals. Resistant to compressionand shear.

Weak chemical bounds in liquids or rubbers(hydrogen, Van der Waals interaction). Resistant tocompression only.

In the general case, the binding energy isdescribed by a Lennard-Jones potential (usuallyone considers n=12 and m=6).

Equilibrium state for interatomic separation .requ 1.12r0

R t i f i L d J t ti l


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Close to equilibrium, the bounds act as a spring with a fixedspring constant

The value of the spring constant is related to the atomicmicroscopic properties of the material by

F(r) =

U

r

U

r2 (r

requ)

k U0

r20


Restoring force in Lennard-Jones potentials

Atomic forces are derived as gradient of the potential energy

F = k0x

From microscopic to macroscopic forces


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We want to move away from the atomic bound, microscopic view.

Lets consider a material of section S.

The number of atomic bounds on the surface is

The total macroscopic force acting on the surface is

If we define as in the following the force surface density

and the relative elongation ,

we obtain the Young modulus, defined by .

The Young modulus is homogeneous to a pressure (energy/volume).

EU0

r30


From microscopic to macroscopic forces

S

N S

r20

F =

Fi S

r20

k0(r r0)

= F/S = (r r0)/r0

E= /

Typical values for the Young modulus


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Typical values for the Young modulus

Bound type k0 (N/m) E (GPa)

C-C bound (diamond,carbon nanotube)

180 1000

Ionic bound (Na-Cl) 10-20 30-70

Metallic bound (Cu) 15-40 30-150

Hydrogen bound (H20) 2 8

Van der Waals 1 2

Rubber 1,00E-03 0.01-0.1

Stress field


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V

gdV+

S

TdS=

Fext

T is the stress field of stress vector: it is a force surface density on the bodysurface and/or on any surface cut through the body.

In full generality, T is a function of the position on the surface, but it alsodepends on the unit vector normal to the surface. The stress vector depends onthe orientation of the surface with respect to some frame of reference.

T(M, n)


Stress field

A body in mechanical equilibrium must have no net forceand no net torque. External forces are of 2 kinds: bodyforce (gravity) and surface forces.

Surface forces are in fact internal (contact) forces acting on the surface.

Stress tensor and Cauchys theorem


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n1 =MBC

SABCn2 =

MAC

SABCn3 =

MAB

SABC

S

TdS = 0

T(M, n)SABC +

T1(M,x1)SMBC +

T2(M,x2)SMAC +

T3(M,x3)SMAB = 0

T(M,x) =

T(M, x)

T(M, n) = T1(M, x1)n1 +

T2(M, x2)n2 +

T3(M, x3)n3

T i = (1i,2i,3i)

T(M, n) = n


Stress tensor and Cauchys theorem

We consider a small volume element in mechanicalequilibrium. For sake of simplicity, we consider thefollowing tetrahedron, sharing three faces with aCartesian frame.

The normal to the fourth face has coordinates

Applying Newtons first law

Applying Newtons third law

we get the general result:

We define the stress tensor as

In matrix form, we have

Stress tensor components


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Consider a face perpendicular to the x-axis

The stress vector is

The first normal component corresponds to tension

or compression along the normal to the face.

The other 2 components are tangential to the normal.

They correspond to shear forces.


Stress tensor components

=

11 12 13

21 22 23

31

32

33

n = (1, 0, 0)

T = (11,21,31)

11 > 0

11 < 0

=

xx xy xz

yx yy yz

zx

zy

zz

or

Symmetry of the stress tensor


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Similarly, from the other component of the torque, we obtain the generalresult that the stress tensor must be symmetric.

or

We get for the z-component of the total torque

=

11 12 13

21 22 23

31

32

33

=

xx xy xz

yx yy yz

zx

zy

zz

S

r

TdS= 0


Symmetry of the stress tensor

Now consider a cube in mechanical equilibrium,

small enough so that the stress tensor is constant

The stress field formula ensures zero net force.

We now require zero net torque around the center.

or

21L3 12L

3= 0

ij = ji =t

General case


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T(M, n)

T1(M, x1)n1

T2(M, x2)n2

T3(M, x3)n3 =VMABC

SABCdv

dt g

=1

3H

dv

dt g

H 0

T(M,n)SABC +

T1(M,x1)SMBC +

T2(M,x2)SMAC+

T3(M,x3)SMAB

= VMABC

dv

dt g


General case

For the tetrahedron with external body forces and acceleration:

Dividing by the surface of the base, we get:

We now go to the continuum limit and shrink the tetrahedron to a point

and we recover the previous result.The same apply for the cube and the zero torque constraint (exercise).

This property holds because internal forces are surface/contact forces,

for which relevant quantity is the stress, or the force per unit surface.

Invariant properties of the stress tensor


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The stress tensor thus depends on 6 independent quantities, the 3 eigenvalues and

the 3 Euler angles of the local rotation matrix to the eigenvectors.

The 3 eigenvalues are solutions of

The 3 coefficients of this polynomial form are invariants: they are the same in anyframe of reference.

Another set of invariants is more commonly used:

I

3= Tr(

3)I

2= Tr(

2)I

1= Tr()

I1 =

1 +

2 +

3 = Tr(

)I2=

12+

23+

13

=t

P D P D = diag(1,2,3)

I3 = 123 = Det()


Invariant properties of the stress tensor

The stress tensor is an intrinsic quantity, independent on the frame of reference.

The values of the components depend of the chosen frame.

Since the stress tensor is symmetric, we know it has real eigenvalues and there isalways an orthonormal basis where the stress tensor is diagonal.

3 I1

2+ I2 I3 = 0

Isotropic and deviatoric stress


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Note that the isotropic stress is equal to the invariant .

In fluid mechanics, the isotropic stress is usually referred to as the hydrostatic orthermal pressure.

ij =1

3Tr()ij + ij

p = 1

3Tr()


Isotropic and deviatoric stress

The stress tensor is decomposed into the sum of an isotropic tensor calledhydrostatic stress and a symmetric, with zero trace tensor called anisotropic ordeviatoric stress.

A positive pressure corresponds to a compression.

The deviator has the same eigenvectors than the stress tensor.

It depends now only on 2 invariants, plus the 3 Euler angles.

I1

The divergence theorem


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S

(v n) dS =

V

v

dV = 0


The divergence theorem

The volume total of all sink and sources is equal to the net flow across the boundary.

It comes from Stokes theorem (not trivial).

We can work out simple examples in 1D and in a cubical region in 3D.

Using conformal mapping techniques, we can extend this to arbitrary shaped regions.

Equilibrium equation


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S

TxdS=

V

xdV

x =

xx

xy

xz

y =

yx

yy

yz

z =

zx

zy

zz

S

TxdS=

V

xdV

+F = 0

+

F =

dv

dt


Equilibrium equation

We start with the total net force being zero. We define the following vectors

We treat each component of the force separately:

Using the divergence theorem, we get:

Since this is valid for arbitrary volumes, we obtain the equilibrium equation

S

TdS+

V

FdV = 0

The stress vector component is given by Tx = x n

From Newtons 2nd law, we get similarly the dynamical equilibrium equation

Deformations


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A small region Q0 around the initial position is deformedin a new region Q around the final position.

To compute the magnitude of the deformation, we usethe gradient operator on each component:

We define the initial (unperturbed) positions of the body particles as

and the final positions after deformation as .

We have a mapping between the final and initial positionx =

x1(

X)x2(

X)

x3(

X)

dxi =xi d

X =

xi

X

dX

gij =xi

Xjdx = g d

X

u = x X


Deformations

X =

OP0

x =

OP

Gathering all components, we obtain de gradient tensor

Usually, instead of the position, we use the displacement field defined by

Similarly, we have the gradient operator where .du = G d

X G = g I

Small deformations


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The length of the infinitesimal vector is modified onlyby pure deformation, and should not change during rigid solid motions.

We have . .

Size changes are due to a symmetric tensor characterizing true deformations.

Lets now consider small perturbation such as .

We can neglect the quadratic term and we obtain

103

=

tG+G

2


Small deformations

L2

= d X d X =t

d Xd X

l2 =t dxdx =t d Xtg gd X= L2 +t d XtG + G +t G G

d X

l = L (1 + )

td X

L

tG+G

2

d X

L

For most material, typical applications deal with small deformations

We are in the so-called elastic regime.The symmetric part of the gradient tensor is called the strain tensor.

The strain tensor


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The displacement field infinitesimally close to point Po is given by

In the general case, we can decompose the gradient tensor into a symmetric and anantisymmetric part .

ij =1

2

ui

Xj+

uj

Xi

ij =

1

2

ui

Xj

uj

Xi


The strain tensor

G = +

= G

t

G2

= G+t

G

2

and

We have and .

The antisymmetric part has only 3 independent components.

It is a rotation matrix that can be rearranged as

The symmetric part has 6 independent components (like the stress tensor).

This is the strain tensor and it corresponds to pure deformations.

rigid body motion deformation or strain

du = d

X

u = u0 + d

X+ d

X

Strain tensor components


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Strain tensor components

=

11 12 13

21 22 23

31 32 33

=

xx xy xz

yx yy yz

zx zy zz

or

Consider a unit vector parallel to the x-axis

The displacement vector is

The displaced vector is therefore

Its length is given byThe normal component affects the length of the line element.

The corresponding strain is called extension or contraction.

The angle between the 2 vectors is given byIt is equal to the tangential component.

The corresponding strain is called distortion.

u = (xx, yx, zx)

N = (1, 0, 0)

n = N+ u = (1 + xx, yx, zx)

|n| =

(1 + xx)2

+ 2yx +

2zx

1 + xx

xx

tan =

2yx

+ 2zx

(1 + xx)2 + 2yx + 2zx

tx =

2yx +

2zx

Properties of the strain tensor


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Since the strain tensor is symmetric, invariance properties of the stress tensorare also valid here.

Similarly, the strain tensor can be decomposed into an isotropic tensor and adeviator tensor.

The isotropic part is an frame invariant quantity, equal to the divergence of thedisplacement field. It measures the relative variation of the volume element atconstant shape

The anisotropic tensor has zero trace. It measures the change of shape atconstant volume.

=

11 12 13

21 22 23

31 32 33

=

xx xy xz

yx yy yz

zx zy zz

ij = 13Tr()ij + ij

Tr() =u1

X1

+u2

X2

+u3

X3

= u =

dV

V


Properties of the strain tensor

or

Summary for stress and strain tensors


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du = d

X

Tr() = u =

dV

V

Tr() = 3p

Summary for stress and strain tensors

=G+t G

2

ij =1

2

ui

Xj

+uj

Xi

T(M, n) = n

+

F =

dv

dt

ij =1

3Tr()ij + ij

ij =1

3Tr()ij + ij

Stress field Displacement field

Gradient tensorDynamical equilibrium

Isotropic/deviatoric stress

Hydrostatic pressure

Isotropic/deviatoric strain

Compressibility

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