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Solid Mechanics
Z. Suo
Elements of Linear Elasticity
A classic textbook is Theory of Elasticity, by S.P. Timoshenko and J.N. Goodier,
McGraw-Hill, New York.
Material particle. A solid is made of atoms, each atom is made of electrons, protons and
neutrons, and each proton or neutron is made of... This kind of description of matter is too
detailed, and is not used in the theory of elasticity. Instead, we develop a continuum theory, in
which a material particle contains many atoms, and represents their average behavior.
Displacement field. Any configuration of the material particles can be used as a
reference configuration. The displacement is the vector by which a material particle moves
relative to its position in the reference configuration. If all the material particles in the body
move by the same displacement vector, the body as a whole moves by a rigid-body translation. If
the material particles in the body move relative to one another, the body deforms. For example,
in a bending beam, material particles on one face of the beam move apart from one another
(tension), and material particles on the other face of the beam move toward one another
(compression). In a vibrating rod, the displacement of each material particle is a function of
time.
We label each material particle by its coordinates x, y, z  in the reference configuration.
At time t, the material particle x, y, z  has the displacement ux, y, z, t  in the x-direction,
vx, y, z, t  in the y-direction, and wx, y, z, t  in the z-direction. A function of spatial coordinates
is known as a field. The displacement field is a time-dependent vector field. It is sometimes
convenient to write the coordinates of a material particle in the reference configuration as
x1, x2 , x3  , and the displacement vector as u1 x1, x2 , x3 , t  , u2 x1, x2 , x3 , t  , u3 x1, x2 , x3 , t  . If we
place markers on a body, the motion of the markers visualizes the displacement field and its
variation with time.
Strain field. Given a displacement field, we can calculate the strain field. Consider two
material particles in the reference configuration: particle A at x, y, z  and particle B at
x  dx, y, z  . In the reference configuration, the two particles are distance dx apart. At a given
time t, the two particles move to new locations. The x-component of the displacement of particle
A is ux, y, z, t  , and that of particle B is ux  dx, y, z, t  . Consequently, the distance between
the two particles elongates by ux  dx, y, z, t   ux, y, z, t  . The axial strain in the x-direction is
x 
u  x  dx, y, z , t   u  x, y, z , t  u

.
dx
x
This is a strain of material particles in the vicinity of x, y, z  at
time t. The function  x x, y, z, t  is a component of the strain field
of the body.
ux, y  dy, z, t 
C
dy
ux, y, z, t 
The shear strain is defined as follows. Consider two lines
B
A
of material particles. In the reference configuration, the two lines
dx
are perpendicular to each other. The deformation changes the
included angle by some amount. This change in the angle defines the shear strain,  . We now
translate this definition into a strain-displacement relation. Consider three material particles A,
B, and C. In the reference configuration, their coordinates are A x, y, z  , B x  dx, y, z  , and
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C x, y  dy, z  . In the deformed configuration, in the x-direction, particle A moves by ux, y, z, t 
and particle C by ux, y  dy, z, t  . Consequently, the deformation rotates line AC about axis z by
an angle
u x, y  dy, z, t   u x, y, z, t  u
.

dy
y
Similarly, the deformation rotates line AB about axis z by an angle
v x  dx, y, z , t   v x, y, z , t  v

.
dx
x
Consequently, the shear strain in the xy plane is the net change in the included angle:
 xy 
u v
 .
y x
For a body in the three-dimensional space, the strain state of a material particle is
described by a total of six components. The strains relate to the displacements as
u
v
w
x  , y  , z 
x
y
z
u v
v w
w u
 xy 
 ,  yz 

,  zx 

y x
z y
x z
Another definition of the shear strain relates the definition here by  xy   xy / 2 . With this
new definition, we can write the six strain-displacement relations neatly as
1  u u 
 ij   i  j  .
2  x j xi 
The state of stress at a material particle. Imagine a three-dimensional body. The body
may not be in equilibrium (e.g., the body may be vibrating). The material property is unspecified
(e.g., the material can be solid or fluid). Imagine a material particle inside the body. What state
of stress does the material particle suffer from?
To talk about internal forces, we must expose them by drawing a free-body diagram.
Represent the material particle by a small cube, with its edges parallel to the coordinate axes.
Cut the cube out from the body to expose all the forces on its 6 faces. Define stress as force per
unit area. On each face of the cube, there are three stress components, one normal to the face
(normal stress), and the other two tangential to the face (shear stresses). Now the cube has six
faces, so there are a total of 18 stress components. A few points below get us organized.
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 zz
 yx
 yy
 zx
 zy
 xx
 xy
 yz
x
y



 xz
 xz
 zy
z

 yz
 xx
 zx
 yy
 yx
 xy
 zz
Notation:  ij . The first subscript signifies the direction of the stress component. The second
subscript signifies the direction of the vector normal to the face. We label the coordinates as
x, y, z  . It is sometimes convenient to write the coordinates as x1, x2 , x3  .
Sign convention. On a face whose normal is in the positive direction of a coordinate axis, the
stress component is positive when it points to the positive direction of the axis. On a face
whose normal is in the negative direction of a coordinate axis, the stress component is
positive when it points to the negative direction of the axis.
Equilibrium of the cube. As the size of the cube shrinks, the forces that scale with the
volume (gravity, inertia effect) are negligible. Consequently, the forces acting on the cube
faces must be in static equilibrium. Normal stress components form pairs. Shear stress
components form quadruples. Consequently, only 6 independent stress components are
needed to describe the state of stress of a material particle.
Write the six stress components in a 3 by 3 symmetric matrix:
 11  12  13 


 12  22  23  .
 13  23  33 
Hooke's law. For an isotropic, homogeneous solid, only two independent constants are
needed to describe its elastic property: Young’s modulus E and Poisson’s ratio . In addition, a
thermal expansion coefficient  characterizes strains due to temperature change. When
temperature changes by T , thermal expansion causes a strain T in all three directions. The
combination of multi-axial stresses and a temperature change causes strains
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





1
 x    y   z   T
E
1
 y   y    z   x   T
E
1
 z   z    x   y   T
E
x 
The relations for shear are
21  
21  
21  
 xy 
 xy ,  yz 
 yz ,  zx 
 zx .
E
E
E
Recall the notation  xy   xy / 2 , and we have
1 
1 
1 
 xy ,  yz 
 yz ,  zx 
 zx
E
E
E
Index notation and summation convention. The six stress-strain relation may be
written as
1 
 ij 
 ij   kk ij .
E
The symbol  ij stands for 0 when i  j and for 1 when i  j . We adopt the convention that a
 xy 
repeated index implies a summation over 1, 2 and 3. Thus,  kk  11   22   33 .
Homogeneity. When talking about homogeneity, you should think about at least two
length scales: a large (macro) length scale, and a small (micro) length scale. A material is said
to be homogeneous if the macro-scale of interest is much larger than the scale of microstructures.
A fiber-reinforced material is regarded as homogeneous when used as a component of an
airplane, but should be thought of as heterogeneous when its fracture mechanism is of interest.
Steel is usually thought of as a homogeneous material, but really contains numerous voids,
particles and grains.
Isotropy. A material is isotropic when response in one direction is the same as in any
other direction. Metals and ceramics in polycrystalline form are isotropic at macro-scale, even
though their constituents—grains of single crystals—are anisotropic. Woods, single crystals,
uniaxially fiber reinforced composites are anisotropic materials.
Example: a rubber layer pressed between two steel plates. A very thin elastic layer,
of Young's modulus E and Poisson's ratio , is well bonded between two perfectly rigid plates. A
thin rubber layer between two thick steel plates is a good approximation of the situation. The
thin layer is compressed between the plates by a known normal stress  z . Calculate all the stress
and strain components in the thin layer.
Solution. The stress at the edges of the elastic layer is complicated. We will neglect this
edge effect, and focus on the field away from the edges, where the field is uniform. This
emphasis makes sense if we are interested in, for example, the displacement of one plate relative
to the other. Of course, this emphasis is misplaced if we are concerned of debonding of the layer
from the plates, as debonding may initiate from the edges, where stresses are high.
By symmetry, the field has only the normal components and has no shear components.
Also by symmetry, we note that
x y
Because the elastic layer is bonded to the rigid plate, the two strain components vanish:
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x  y  0 .
Using Hooke’s law, we obtain that
0  x 
1
 x   y   z  ,
E
or
x 

1 
z .
Using Hooke’s law again, we obtain that
1   1  2  .
1
 z   z   y   x  
z
1  E
E
Consequently, the elastic layer is in a state of uniaxial strain, but all three stress components are
nonzero. When the elastic is incompressible,   0.5 , it cannot be strained in just one direction,
and the stress state will be hydrostatic.
Traction vector. Imagine a plane inside a material. The plane has the unit normal vector
n, with three components n1 , n2 , n3 . The force per area on the plane is called the traction. The
traction is a vector, with three components:
 t1 
t  t2  .
t3 
Question: There are infinite many planes through a point. How do we determine the
traction vectors on all these planes?
Answer: You can calculate the traction vector from
 t1   11  12  13   n1 
t   
 
 2   12  22  23  n2 
t3   13  23  33   n3 
We now see the merit of writing the components of a state of stress as a matrix. Also, the six
components are indeed sufficient to characterize a state of stress, because the six components
allow us to calculate the traction vector on any plane.
z
 xx
 yx
y
tx
 zx
x
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Proof of the stress-traction relation. This traction-stress relation is the consequence of
the equilibrium of a tetrahedron formed by the particular plane and the three coordinate planes.
Denote the areas of the four triangles by A, Ax , Ay , Az . Recall a relation from geometry:
Ax  Anx , Ay  Any , Az  Anz
Balance of the forces in the x-direction requires that
t x A   xx Ax   xy Ay   xz Az .
This gives the desired relation
t x   xx nx   xy n y   xz nz .
This relation can be rewritten using the index notation:
t1   11n1   12n2   13n3 .
It can be further rewritten using the summation convention:
t1   1 j n j .
Similar relations can be obtained for the other components of the traction vector:
t2   21n1   22n2   23n3
t3   31n1   32n2   33n3
The three equations for the three components of the traction vector can be written
collectively in the matrix form, as give in the beginning of this section. Alternatively, they can
be written as
ti   ij n j .
Here the summation is implied for the index j. The above expression represents three equations.
Example: stress and traction. A material particle is in a state of stress with the
following components:
1 2 5 
2 3 6 .


5 6 4
(a) Compute the traction vector on a plane intersecting the axes x, y and z at 1, 2 and 3,
respectively.
(b) Compute the magnitude of the normal stress on the plane.
(c) Compute the magnitude of the shear stress on the plane.
(d) Compute the direction of the shear stress on the plane.
Solution. We need to find the unit vector normal to the plane. This is a problem in
analytical geometry. The equation of a plane intersecting the axes x, y and z at 1, 2 and 3 is
x y z
  1
1 2 3
Alternatively, a plane can be defined by a given point on the plane, x0, and a unit vector normal
to the plane, n. For any point x on the plane, x – x0 is a vector lying in the plane, so that
n  x  x0   0 , or
nx x  x0   n y  y  y0   nz z  z0   0
A comparison of the two equations of the plane shows that the normal vector is in the direction
1 1 1 
1 , 2 , 3  . Normalizing this vector, we obtain the unit vector normal to the plane:
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6 / 7 
n  3 / 7  .
2 / 7
(a) The traction vector on the plane is
t x  1 2 5 6 / 7 22 / 7
t   2 3 6 3 / 7   33 / 7 
 y 

 

t z  5 6 4 2 / 7 56 / 7 
(b) Normal stress on the plane is the traction vector projected on to the normal direction
of the plane
22 6 33 3 56 2
n  t n        7
7 7 7 7 7 7
(c) and (d) The shear stress in the plane is a vector:
 20
1
  t   nn   12  .
7
 42 
The direction of this vector is the direction of the shear stress on the plane. The magnitude of the
shear stress is 6.86.
 zx x, y, z  dz, t 
z
dz
 xx x, y, z, t 
 yx x, y, z , t 
x, y, z 
x
 yx x, y  dy, z, t 
y
 zx x, y, z, t 
 xx x  dx, y, z, t 
dy
dx
Stress field. Imagine the three-dimensional body again. At time t, the material particle
x, y, z  is under a state of stress  ij x, y, z, t  . Denote the distributed external force per unit
volume by bx, y, z, t  . An example is the gravitational force, bz  g . The stress and the
displacement are time-dependent fields. Each material particle has the acceleration vector
 2ui / t 2 . Cut a small differential element, of edges dx, dy and dz. Let  be the density. The
mass of the differential element is dxdydz . Apply Newton’s second law in the x-direction, and
we obtain that
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dydz xx x  dx, y, z, t    xx x, y, z, t 

Z. Suo

 dxdz  yx x, y  dy, z, t    yx x, y, z, t 
 dxdy zx x, y, z  dz, t    zx x, y, z, t   bx dxdydz  dxdydz
 2u x
t 2
Divide both sides of the above equation by dxdydz, and we obtain that
 xx  yx  zx
 2u


 bx   2x .
x
y
z
t
This is the momentum balance equation in the x-direction. Similarly, the momentum balance
equations in the y- and z-direction are
 yx
x

 yy
y

 zy
z
 by  
 2u y
t 2
 xz  yz  zz
 2u


 bz   2z
x
y
z
t
When the body is in equilibrium, we drop the acceleration terms from the above equations.
Using the summation convention, we write the three equations of momentum balance as
 ij
 2u
 b j   2i .
x j
t
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Summary of Elasticity Concepts
Players: Fields
Stress tensor: stress state must be represented by 6 components (directional property).
Stress field: stress sate varies from particle to particle (positional property).
Stress field is represented by 6 functions,  xx x, y, z;t  ,  xy x, y,z;t …
Displacement field is represented by 3 functions, ux, y, z;t ,vx, y,z;t , wx, y, z;t  .
Strain field is represented by 6 functions.
Rules: 3 elements of solid mechanics
Momentum balance
Deformation geometry
Material law
Complete equations of elasticity. Partial differential equations.
Boundary conditions
Prescribe displacement.
Prescribe traction.
Initial conditions
For dynamic problems (e.g., vibration and wave propagation), one also need prescribe initial
displacement and velocity fields.
Solving boundary value problems.
ODE and PDE.
Idealization, analytical solutions.
S.P. Timoshenko and J.N. Goodier, Theory of Elasticity, McGraw-Hill, New York.
Handbook solutions.
R.E. Peterson, Stress Concentration Factors, John Wiley, New York, 1974. 2nd edition by
W.D. Pilkey, 1997.
Brute force, numerical methods.
Finite Element Methods.
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3D Elasticity: Collected Equations
Stress-traction relation
t1   11  12  13 n1 
t2   21  22  23 n2 

t 3 
 
 31  32  33 

n3 

Momentum balance
 x  xy  xz
 2u


 bx   2
x
y
z
t
 yx  y  yz
 2v


 by   2
x
y
z
t
 zx  zy  z
 2w


 bz   2
x
y
z
t
Strain-displacement relation
Hooke's Law
x 
u
,
x
1 v w 
 yz   
2 z y 
y 
v
,
y
 zx 
z 
w
,
z
1 u v 
 xy   
2 y x 
1 w u 

2 x z 
1
    y   z  T,
E x
1
 y    y    z   x   T,
E
1
 z    z    x   y   T,
E
x 
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1 
 zx
E
1
 xy 
 xy
E
1 
 yz 
 yz
E
z x 
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3D Elasticity: Equations in other coordinates
Cylindrical Coordinates (r, , z)
u, v, w are the displacement components in the radial, circumferential and axial directions,
respectively. Inertia terms are neglected.
 r 1  r  rz  r   



0
r r 
z
r
 r 1    z 2 r



0
r
r 
z
r
 rz 1  z  z  rz



0
r
r 
z
r
r 
u
,
r
 
1 v
 u,


r 
z 
w
,
z
z 
 zr 
1 v 1 w 

2 z r  
1 w u 

2  r z 
 r 
1 1 u v v 


2 r  r r 
Spherical Coordinates (r, , )
 is measured from the positive z-axis to a radius;  is measured round the z-axis in a righthanded sense. u, v, w are the displacements components in the r, ,  directions, respectively.
Inertia terms are neglected.
 r 1  r
1  r 1


 2 r         r cot    0
r r 
rsin  
r
 r 1  
1   1


      cot   3 r   0
r
r  rsin  
r
 r 1  
1   1


 3 r  2  cot    0
r
r 
rsin   r
 r  u ,
  1  w  w cot   1 v 
r
2r  
sin   
1
u 
  r1  v  u  ,
r  1  w  wr 

2  r
r sin   
1  w
 
 u sin   v cos   ,
 r  1  r1 u  v  vr 
r sin   

2

r
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Lamé Problem. As an example of boundary value problems, consider a spherical cavity
in a large body, remote hydrostatic tension. The symmetry of the problem makes spherical
coordinate system convenient.
List nonzero quantities.
 the radial displacement u ,
 the radial stress  r , two equal hoop stresses     ,
 the radial strain  r , two equal hoop strains     .
 They are all functions of r.
List equations. Use the basic equation sheet. Simplify to the special symmetry.
d r
  
2 r
0.
Equilibrium equation:
dr
r
du
u
 r  ,   .
Deformation geometry:
dr
r
Material law (elasticity, Hooke’s law):
1
1
 r   r  2     1       r  .
E
E
Reduce to a single ODE. The above are a set of 5 equations for 5 functions of r. You can
follow a number of approaches to solve them. I’ll follow the approach below. I want to obtain a
single equation for the radial stress,  r . From the equilibrium equation, I express  in terms of
r :
r d r
   r 
.
2 dr
Then I use the material law to express both strains in terms of  r :
1
d 
 r  1  2  r  r r 
E
dr 
1
r d r 
   1  2  r  1   
E
2 dr 
I can eliminate u from the two equations of deformation geometry, and resulting equation is in
terms of the two strains,
 r  d r  / dr .
Express this equation in terms of the radial stress, and I have
d 2 r 4 d r

 0.
dr 2
r dr
Solve the ODE. This is an equidimensional equation. The solution is of form r  rm .
Substitute r  rm into the ODE, and we find two roots: m = 0 and m = -3. Consequently, the
full solution is
B
r  A  3 ,
r
where A and B are constants to be determined by the boundary conditions. The hoop stress is
given by
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B
.
2r 3
The boundary conditions are
 Prescribed remote stress:
r  S as r   .
 Traction-free at the cavity surface: r  0 as r  a
Upon determining the two constants A and B, we obtain the stress distribution
  a 3 
 1  a 3 
 r  S 1    ,    S 1     .
  r  
 2  r  
Plot the stresses as a function of r.
Stress concentration factor. Note that the hoop stress is nonzero near the cavity surface,
where the hoop stress reaches the maximum. The stress concentration factor is the ratio of the
maximum stress over the applied stress. In this case, the stress concentration factor is 3/2.
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Principal Stress
Imagine a material particle in a state of stress. The state of stress is fixed, but we can
represent the material particle in many ways by cutting cubes in different orientations. For any
given state of stress, it is always possible to cut a cube in a suitable orientation, such that the
stress components on all the cube faces are normal to the faces, and there are no shear stresses on
the cube faces. These cube faces are called the principal planes, the normal vectors to these
faces the principal directions, and the stresses on them the principal stresses.
Examples
 Uniaxial tension
 Equal biaxial tension
 Hydrostatic pressure
 Pure shear is the same state of stress as the combination of pulling and pressing in 45°.
Given a stress state, how to find the principal stresses? When a plane is the principal
plane, the traction on the plane is normal to the plane, namely, the traction vector t must be in the
same direction as the unit normal vector n. Let the magnitude of the traction be  . On the
principal plane, the traction vector is in the direction of the normal vector, t  n . Write in the
matrix notion, and we have
 t1 
 n1 
t    n  .
 2
 2
t3 
 n3 
Here both t and n are vectors, but  is a scalar representing the magnitude of the principal stress.
Recall that the traction vector is the stress matrix times the normal vector, so that
 11  12  13   n1 
 n1 




 
 21  22  23  n2    n2  .
 31  32  33   n3 
 n3 
This is an eigenvalue problem. When the stress state is known, i.e., the stress matrix is given,
solve the above eigenvalue problem to determine the eigenvalue  and the eigenvector n. The
eigenvalue  is the principal stress, and the eigenvector n is the principal direction.
Linear algebra of eigenvalues. Because the stress tensor is a 3 by 3 symmetric matrix,
you can always find three real eigenvalues, i.e., principal stresses,  a , b , c . We distinguish
three cases:
(1) If the three principal stresses are unequal, the three principal directions are orthogonal (e.g.,
pure shear state).
(2) If two principal stresses are equal, but the third is different, the two equal principal stresses
can be in any directions in a plane, and the third principal direction is normal to the plane
(e.g., pure tensile state).
(3) If all the three principal stresses are equal, any direction is a principal direction. This stress
state is called a hydrostatic state.
Maximum normal stress. Why do we care about the principal stress? Chalks are made
of a brittle material: they break by tensile stress, not by shear stress. When a chalk is under
bending, the tensile stress is along the axial direction of the chalk, so that the chalk breaks on a
plane normal to the axial direction. When a chalk is under torsion, the maximum tensile stress is
45° from the axial direction, so that the chalk breaks in a direction 45° from the axial direction.
(The fracture surface of the chalk under torsion is not a plane, because of some 3D effects.) We
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care about the principal stress because brittle materials fail by tensile stress, and we want to find
the maximum tensile stress.
Let’s order the three principal stresses as  a   b   c . This ordering takes into
consideration the signs: a compressive stress (negative) is smaller than a tensile stress (positive).
On an arbitrary plane, the traction vector may be decomposed into two components: one
component normal the plane (the normal stress), and the other component parallel to the plane
(the shear stress). Obviously, when you look at a plane with a different normal vector, you find
different normal and shear stresses. You will be delighted by the following theorem:
Of all planes, the principal plane corresponding to  c has the maximum normal stress.
Maximum shear stress. Why do we care about the maximum shear stress? Most metals
are ductile materials: they fail by plastic yielding. When a material is under a complex stress
state, it is known empirically that yielding first occurs on a plane with maximum shear stress. To
find the maximum shear stress and the particular plane, you are helped by the following theorem:
The maximum shear stress is  max   c   a  / 2 .
 max acts on a plane with the normal vector 45° from the principal directions n a and n c .
A proof of the above theorems is outlined below. Consider a system of coordinates that
coincide with three orthogonal directions of the principal stresses,  a , b , c . Then consider an
arbitrary plane whose unit normal vector has components n1 , n2 , n3 in this coordinate system.
The components of the stress tensor in this coordinate system is
 a 0 0 
0 
0  .
b

 0 0  c 
Thus, on the plane with unit vector ( n1 , n2 , n3 ), the traction vector is (  a n1, b n2 , c n3 ). The
normal stress on the plane is
 n   a n12   b n22   c n32 .
We need to maximize  n under the constraint that n12  n22  n32  1 .
The shear stress on the plane  is given by
 2   a n1 2   b n2 2   c n3 2   a n12   b n22   c n32  .
2
We need to maximize  under the constraint that n12  n22  n32  1 .
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Change of Coordinates
The direction-cosine matrix relating two bases. In the 3D space, let e1, e2 and e3 be an
orthonormal basis, namely,
ei  e j   ij .
The base vectors are ordered to follow the right-hand rule.
Let e'1 , e'2 , e'3 be a new orthonormal basis, namely,
e  e    .
Let the angle between the two vectors e i and e' be  i . Denote the direction cosine of
the two vectors by
li  ei  e'  cosi .
We follow the convention that the first index of li refers to a coordinate in the old basis, and the
second to a coordinate in the new basis. For the two bases, there are a total of 9 direction
cosines. We can list li as a 3 by 3 matrix. By our convention, the rows refer to the old basis,
and the columns to the new basis.
Note that li is the component of the vector e' projected on the vector e i . We can
express each new base vector as a linear combination of the three old base vectors:
e'  l1 e1  l2 e2  l3 e3 .
If you are tired of writing sums like this, you abbreviate it as
e'  li ei ,
with the convention that a repeated index implies summation over 1,2,3. Because the sum is the
same whatever the repeated index is named, such an index is called a dummy index.
Similarly, we can express the old basis as a linear combination of the new basis:
ei  li1e'1 li 2e'2 li 3e'3 .
Using the summation convention, we write more concisely as
ei  li e' .
Transformation of components of a vector due to change of basis. Let f be a vector.
It is a linear combination of the base vectors:
f  f i ei ,
where f1 , f 2 , f 3 are the components of the vector, and are commonly written as a column.
Consider the vector pointing from Cambridge to Boston. When the basis is changed, the vector
between Cambridge and Boston remains unchanged, but the components of the vector do change.
Let f '1 , f '2 , f '3 be the components of the vector f in the new basis, namely,
f  f ' e'
Recall the transformation between the two sets of basis, ei  li e' , we write that
f  fiei  fili e'
A comparison between the two expressions gives that
f '  fili .
Thus, the component column in the new basis is the transpose of the direction-cosine matrix
times the component column in the old basis.
Similarly, we can show that
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fi  li f '
The component column in the old basis is the direction-cosine matrix times the component
column in the old basis.
Transformation of stress components due to change of basis. The stress tensor, ,
describes the state of stress suffered by a material particle. Represent the material particle by a
cube. The stress components are the force per unit area on 6 faces of the cube. The stress tensor
is represented by a 3 by 3 symmetric matrix. The state of stress of a material particle is a
physical object, and is independent of your choice of the basis (i.e., how you cut a cube to
represent the particle). However, the components of the stress tensor do depend on your choice
of the basis. How do we transform the stress components when the basis is changed?
Consider the stress state of a material particle, and the traction vector on a given plane. In
the old basis e1, e2 and e3, denote the components of the stress state by  ij , the components of the
unit vector normal to the plane by n j , and the components of the traction vector on the plane by
ti . Using the summation convention, we write the traction-stress equations as
ti   ij n j .
Recall that we obtained this relation by the balance of forces on a tetrahedron. In the language of
linear algebra, we call the stress as a linear operator that maps the unit normal vector of a plane
to the traction vector acting on the plane.
Similarly, in the new basis e'1 , e'2 , e'3 , denote the components of the stress state by  ' ,
the components of the unit vector normal to the plane by n ' , and the components of the traction
vector on the plane by t ' . Force balance requires that
t '   ' n' .
(a)
We now examine the relations between the components in the old basis and those in the
new basis. The traction is a vector, so that its components transform as t '  li ti . Insert
ti   ij n j into the above, and we obtain that t '  li  ij n j . The unit normal vector transforms as
n j  l j n' . Consequently, we obtain that
t '  li  ijl j n' .
(b)
Equations (a) and (b) are valid for any choice of the plane. Consequently, we must require that
 '  li ijl j .
Thus, the stress-component matrix in the new basis is the product of three matrices: the
transpose of the direction-cosine matrix, the stress-component matrix in the old basis, and the
direction-cosine matrix.
Scalars, vectors, and tensors. When the basis is changed, a scalar (e.g., temperature,
energy, and mass) does not change, the components of a vector transform as
f '  fili ,
and the components of a tensor transform as
 '   ijli l j .
This transformation defines the second-rank tensor. By analogy, a vector is a first-rank tensor,
and a scalar is a zeroth-rank tensor. We can also similarly define tensors of higher ranks.
Multilinear algebra. The rule of the above transformation is based on only one fact: the
stress is a linear map from one vector to another vector. You can generate a new tensor from a
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linear map from one tensor to another tensor. You can also generate a tensor by a bilinear form,
e.g., a bilinear map from two vectors to a scalar. Of course, a multilinear map of several tensors
to a tensor is yet another tensor. Consequently, all tensors follow a similar rule under a change of
basis. We write this rule again for a third-rank tensor:
g '  gijkli l j lk .
Invariants of a tensor. A scalar is invariant under any change of basis. When the basis
changes, the components of a vector change, but the length of the vector is invariant. Let f be a
vector, and f i be the components of the vector for a given basis. The length of the vector is the
square root of
fi fi .
The index i is dummy. Thus, this combination of the components of a vector is a scalar, which is
invariant under any change of basis. For a vector, there is only one independent invariant. Any
other invariant of the vector is a function of the length of the vector.
This observation can be extended to high-order tensors. By definition, an invariant of a
tensor is a scalar formed by a combination of the components of the tensor. For example,
for a symmetric second-rank tensor  ij , we can form three independent invariants:
 ii ,  ij ij ,  ij jk ki .
In each case, all indices are dummy, resulting in a scalar. Any other invariant of the tensor is a
function of the above three invariants.
Exercise. For a nonsymmetric second-rank tensor, give all the independent invariants.
Write each invariant using the summation convention, and then explicitly in all its terms.
Exercise. Give all the independent invariants of a third-rank tensor.
A special case: the new basis and the old basis differ by an angle  around the axis
e3. The sign convention for  follows the right-hand rule. The direction cosines are
e1  e'1  cos , e1  e'2   sin  , e1  e'3  0,
e2
e
'
2
e  e'  sin  , e  e'  cos , e  e'  0,
2
1
2
2
2
3
e3  e'1  0, e3  e'2  0, e3  e'3  1.
Consequently, the matrix of the direction cosines is
cos  sin 
li    sin  cos
 0
0
The components of a vector transform as
 f '1   cos
 f '    sin 
 2 
 f '3   0
sin 
cos
0
0
0
1
0  f1 
0  f 2 
1  f3 
Thus,
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f '1  f1 cos  f 2 sin 
f '2   f1 sin   f 2 cos
f '3  f 3
The components of a stress state transform as
 '11  '12  '13   cos sin  0  11  12  13  cos
 '  '
 '23    sin  cos 0  21  22  23   sin 
22
 21
 '31  '32  '33   0
0
1  31  32  33   0
Thus,
 '11 
 11   22
 sin 
cos
0
0
0
1
 11   22
cos 2   12 sin 2
2
2
   22  11   22
 '22  11

cos 2   12 sin 2
2
2
   22
 '12   11
sin 2   12 cos 2
2
 '13   13 cos   23 sin 

 '23   13 sin    23 cos
 '33   33
State of plane stress. A even more special case is that the stress components out of the
plane are absent, namely,  13   23   33  0 .
   22  11   22
 '11  11

cos 2   12 sin 2
2
2
   22  11   22
 '22  11

cos 2   12 sin 2
2
2
   22
 '12   11
sin 2   12 cos 2
2
In Beer’s Section 7.4, these equations are represented by a graph (i.e., Mohr’s circle).
In this case, e3 is one principal direction, and the principal stress in this direction is zero.
To find the other two principal directions, we set  '12  0 , so that the two principal directions are
at the angle  p from the e1 and e2 directions. This angle is given by
tan 2 p 
2 12
.
 11   22
The two principal stresses are given by
 11   22
    22 
2
  11
   12 .
2
2


We need to order these two principal stresses and the zero principal stress in the e3 direction.
The maximum shear stress is the half of the difference between the maximum principal stress
and the minimum principal stress.
2
The state of strain of a material particle is a second-rank tensor. In the old basis, the
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coordinates of a material particle are x1, x2 , x3  , and the components of the displacement field
are ui x1 , x2 , x3 , t  . In the new basis, the coordinates of the same material particle are
x'1 , x'2 , x'3  , and the components of the displacement field are u' x'1 , x'2 , x'3 , t  . Recall that
u '  li ui and x j  l j x' , so that
u '
u x j
u
 li i
 li l j i .
x'
x j x'
x j
Thus, the gradients of the displacement field, ui / x j , form the components of a second-rank
tensor. Consequently, the state of strain, being the symmetric part of the displacement gradient,
is a second-rank tensor. When the basis is changed, the components of the strain state transform
as
 '   ijli l j .
Helmholtz free energy of a rod under uniaxial tension. Consider a rod, initial length
L0 and cross-sectional area A0 . When a loading machine applies a force f to the rod, the length
of the rod becomes L. We record the function f L experimentally, which need not be linear.
When the length changes from L to L  dL , the loading machine does work fdL to the rod.
We model an elastic solid with a Helmholtz free energy, F L,T  . When the length
increases by dL, and the temperature increases by dT, the free energy increases by
dF  fdL  SdT .
Here S is the entropy, which can also be measured experimentally by measuring the heat
absorbed by the rod in a slow loading process, dS  dQ / T .
Define the stress and strain as
L  L0
f
 ,

.
L0
A0
Define the free energy density, w, as the free energy per unit volume, namely
F
w
.
A0 L0
We can similarly define the entropy density
S
s
.
A0 L0
Here we have used the initial area and initial length to define the stress, the strain, and the free
energy density.
With these definitions, we can rewrite dF  fdL  SdT as
dw  d  sdT .
The free energy density is a function of the strain and the temperature:
w  w ,T  .
For a given solid, the function w ,T  is determined experimentally. Once we know this
function, we can obtain the stress-strain relation by taking the differentiation:
w , T 

.

Similarly, we can also calculate the entropy density by
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w , T 
.
T
We next restrict ourselves to small strains, so that we can expand the function w ,T 
into a Taylor series in the strain:
1
w , T   w0  w1  w2 2 .
2
The coefficients w0 , w1 , w2 are functions of the temperature. We will only go up to the quadratic
term in strain.
The stress is obtained by taking partial differentiation:
  w1  w2 .
Thus, we identify w2 as Young’s modulus E. We also identify w1 as residual stress. Recalling
the experimental observation of thermal expansion, we may with to write this residual stress as
w1   E T  Tref  ,
s
where  is the coefficient of thermal expansion, and Tref is a reference temperature at which the
strain is set to be zero when the stress vanishes. In most circumstances, E and  only depend
on the temperature weakly, and are regarded as material constants.
The entropy density is given by
w , T  dw0 dw1
1 dw0 2
s



 .
T
dT
dT
2 dT
Experimentally, the entropy density may be determined by measuring heat capacity c, namely, the
heat absorbed by a unit volume of the solid per unit change in temperature, when the strain is
held constant. Thus,
cdT
ds 
.
T
Using the experimentally measured value of c, we can determine w0 , w1 , w2 .
The above analysis can be generalized in many directions to account for diverse
experimental observations. For the time being, we will focus on the stress strain relation of a
linearly elastic solid, and make one important generalization: we will consider an anisotropic
solid under a multiaxial stress state. We will drop the temperature dependence, and throw in
thermal expansion strain at the end of the analysis. When we are not particularly interested in the
temperature dependence of the free energy, following a large body of literature, we will call the
Helmholtz free energy the elastic energy.
Elastic energy density of a block under a simple shear. Consider a block of height
H0 and cross-sectional area A0 . Subject to a shear force F, the block shear by an angle  . When
the shear angle changes from  to   d , the loading machine does the work FH0d to the
block. For an elastic solid the work is stored as the elastic energy in the block. Define the shear
stress and the (engineering) shear strain as
F
  ,  
A0
The energy per unit volume is a function of the shear strain, w . From the above, we have
dw  d .
The shear stress is the differential coefficient of the energy density function.
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When the block is made of a linearly elastic solid, under shear load, the stress-strain
relation is   G . Consequently, the energy density function is
1
w   G 2
2
This result holds only for linear elastic solid in pure shear condition.
Under a multiaxial stress state, energy density is a quadratic form of the strain
tensor. The advantage of the energy density function becomes clear under the multiaxial stress
state. The energy density is scalar, and the strain is a tensor. The energy density is a function of
all components of the stain tensor:
w  w11, 12 ,... .
(For people familiar with thermodynamics, we deal energy at a constant temperature, so that the
energy is the Helmholz free energy.)
The components of stress tensor are differential coefficients:
dw   pq d pq .
That is
w11, 12 ,...
 pq 
 pq
In linear elasticity, we assume that stress is linear in strain. Thus, the energy density is a
quadratic form of the stain tensor, written as
1
w  Cijkl ij kl .
2
Here Cijkl are the components of a fourth-rank tensor called the stiffness tensor. Without losing
any generality, we can assume the following symmetries:
Cijkl  C jikl  Cijlk  Cklij .
If we count carefully, we should have 21 independent components for a generally anisotropic
elastic solid.
The components of the stress tensor are linear in the components of the strain tensor:
 pq  C pqij ij .
We can also invert this relation to express the strain in terms of the stress:
 pq  S pqij ij
Here S pqij are the components of a fourth-rank tensor called the compliance tensor. They have
the same symmetry properties.
Stress-strain relation in a matrix form. We can also write the above equations in
another form. The state of strain is specified by the six components:
 x , y , z ,  yz , zx , xy
In this order we will label them as 1 ,  2 ,  3 ,  4 ,  5 ,  6 . The six strain components can vary
independently. The elastic energy per unit volume is a function of all the six strain components,
w 1 ,  2,  3 ,  4 , 5 ,  6  . This is the energy density function. When each strain component
changes by a small amount, d i , the energy density changes by
dw  1d1   2d 2   3d 3   4d 4   5d 5   6d 6 .
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Here we use the engineering strains for the shear, rather than the tensor components. We do so
to avoid the factor 2 in the above expression. Each stress component is the differential
coefficient of the energy density function:
w
i 
.
 i
If the function w 1 ,  2,  3 ,  4 , 5 ,  6  is known, we can determine the six stress-strain relations by
the differentiations. Consequently, by introducing the energy density function, we only need to
specify one function, rather than six functions, to determine the stress-strain relations.
Linear elastic solids. The above considerations apply to solids with linear or nonlinear
stress-strain relations. We now examine linear elastic solids. For the stress components to be
linear in the strain components, the energy density function must be a quadratic form of the
strain components:
1
1
w   cij i j  c111 1  c12 1 2  c 21 2 1 ....
2 i, j
2
Here cij are 36 constants. The cross terms come in pairs, e.g., c12  c21 1 2 . Only the
combination c12  c21 will enter into the stress-strain relation, not c12 and c21 individually. We
can call c12  c21 by a different name. A convenient way to say that there is only one independent
constant is to just let c12  c21 . We can do the same for other pairs, namely,
cij  c ji .
The matrix cij is symmetric, with 21 independent elements. Consequently, 21 constants are
needed to specify the elasticity of a linear anisotropic elastic solid. Because the elastic energy is
positive for any nonzero strain state, the matrix cij is positive-definite.
Recall that each stress component is the differential coefficient of the energy density
function,  i  w /  i . The stress relation becomes
 i   cij j .
j
In the matrix notation, we write
 1   c11 c12 c13 c14 c15 c16   1 
  c
 
 2   21 c22 c23 c24 c25 c26   2 
 3  c31 c32 c33 c34 c35 c36   3 
 
 
 4  c41 c42 c43 c44 c45 c46   4 
 5  c51 c52 c53 c54 c55 c56   5 
  
 
 6  c61 c62 c63 c64 c65 c66   6 
The physical significance of the constants cij is now evident. For example, when the solid is
under a uniaxial strain state, 1  0,  2  3   4   5   6  0 , the six stress components on the
solid are 1  c111,  2 c211,... The matrix cij is known as the stiffness matrix.
Inverting the matrix, we express the strain components in terms of the stress components:
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1   s11 s12 s13 s14 s15 s16   1 
   s
 
 2   21 s22 s23 s24 s25 s26   2 
 3   s31 s32 s33 s34 s35 s36   3 
 
 
 4   s41 s42 s43 s44 s45 s46   4 
 5   s51 s52 s53 s54 s55 s56   5 
  
 
 6   s61 s62 s63 s64 s65 s66   6 
The matrix s ij is known as the compliance matrix. The compliance matrix is also symmetric
and positive definite.
The components of the stiffness tensor relate to the corresponding components of the
stiffness matrix as
c11  C1111, c14  C1123, c44  C2323 .
However, the corresponding relations for compliance are
s11  S1111, s14  2S1123, s44  4S2323 .
Anisotropic elasticity. An isotropic, linear elastic solid is characterized by two constants
(e.g., Young’s modulus and Poisson’s ratio) to fully specify the stress-strain relation. Some
solids are anisotropic, e.g., fiber reinforced composites, single crystals. Each stress component is
a function of all six strain components. Consequently, 21 constants are needed to specify the
elasticity of a linear anisotropic elastic solid.
For a crystal of cubic symmetry, such as silicon and germanium, when the coordinates are
along the cube edges, the stress-strain relations are
0
0  1 
 1  c11 c12 c12 0
  c
0
0   2 
 2   12 c11 c12 0
 3  c12 c12 c11 0
0
0   3 
 
 
0   4 
 4   0 0 0 c44 0
 5   0 0 0
0 c44 0   5 
  
 
0
0 c44   6 
 6   0 0 0
The three constants c11 , c12 and c44 are independent for a cubic crystal. Isotropic solid is a
special case, in which the three constants are related, c44  c11  c12 / 2 .
For a fiber reinforced composite, with fibers in the x3 direction, the material is isotropic
in the x2 and x3 directions. The material is said to be transversely isotropic. Five independent
elastic constants are needed. The stress-strain relations are
 1  c11 c12 c13 0
  c
 2   12 c11 c13 0
 3  c13 c13 c33 0
 
 4   0 0 0 c44
 5   0 0 0
0
  
0
 6   0 0 0
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0
0
0
c44
0
  1 
  
0
 2 
  3 
0
 
0
  4 
  5 
0
 
c11  c12  / 2  6 
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Invariants and isotropic materials. If a material is isotropic, the elastic energy density
w should be a function of the invariants of the strain tensor. For a linear material, w is quadratic
in strain. The strain tensor can form only two invariants quadratic in its components:  ij ij and
 kk 2 .
Consequently, for an isotropic, linearly elastic solid, the elastic energy density takes the
form
1
2
w   ij ij    kk  ,
2
where  and  are known as the Lame constants. Taking differentiation, we obtain the stressstrain relations:
 ij  2 ij   kk  ij .
A comparison with the stress-strain relations in the uniaxial stress state shows that
E
E
.

, 
1   1  2 
21   
Exercise. It is reasonable to require that the elastic energy density be a positive-definite
quadratic form of the strain tensor. That is, w  0 for any state of strain, except that w  0 when
the strain tensor vanishes. For an isotropic, linearly elastic solid, confirm that this positivedefinite requirement is equivalent to require that E  0 and  1    1/ 2 .
Example: stress in an epitaxial film. Both silicon (Si) and germanium (Ge) are crystals
of cubic unit cell. The edge length of the unit cell of Si is aSi  5.428Å , and that of Ge is
aGe  5.658Å . A Ge film 10 nm thick is grown epitaxially (i.e. with matching atomic positions)
on the [100] surface of a 100 m thick Si substrate. Calculate the stress and strain components
in the Ge film. The respective elastic constants are (in GPa)
Si: c11 = 165.8, c12 = 63.9, c44 = 79.6.
Ge: c11 = 128.5, c12 = 48.2, c44 = 66.7.
Solution. Because the Si substrate is much thicker than the Ge film, the strains in the
substrate are much smaller than those in the film. We will neglect these small strains, and
assume that the substrate is undeformed. Let axis 3 be normal to the film surface, and axes 1 and
2 be in the plane of the film, parallel to the cube edges of the crystal cell. To register one atom
on another, Ge must be compressed in directions 1 and 2 to conform to the undeformed atomic
unit cell size of Si. The two in-plane strains in the Ge film are
a a
11   22  Si Ge  4% .
aGe
There will be an elongation normal to the film,  33  0 . All shear strains vanish.
According to the generalized Hooke’s law, the stress normal to the film surface relates to the
strains as
 33  c11 33  c1211  c12 22 .
Physically it is evident that there is no stress normal to the surface of the film,  33  0 . Inserting
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into the above expression, we obtain that
 33  
2c12
11  3% .
c11
The two in-plane stress components are equal, given by
11   22  c1111  c12 22  c12 33 ,
or

 11   22  c11  c12 
2c122 
11 .
c11 

Inserting the numerical values, we obtain that 11   22  5.6GPa . This is a huge stress, it may
generate dislocations in the film.
Ferroelastic phase transition. This part goes beyond linear elasticity. Suppose we have
the following experimental observations. A crystal has a rectangular symmetry at a high
temperature. When the temperature drops below a critical value, Tc , the crystal undergoes a
phase transition. The crystal at a low temperature acquires a spontaneous strain in shear.
Because of the symmetry, the shear strain can go both directions. (Sketch the geometry.)
We model this crystal with a free energy density
1
1
w , T   AT  Tc  2  B 4 ,
2
4
where A and B are positive constants. Due to symmetry, the crystal is equally likely to shear in
two directions, so that we keep the even powers in the strain  . (Sketch w as a function of  at
two temperatures, one above Tc and the other below Tc .)
When T  Tc , the coefficient of the  2 term is positive, so that the crystal behaves like
usual elastic solid, with the shear modulus AT  Tc  . The  4 term is unnecessary to describe the
behavior of the crystal.
When T  Tc , the coefficient of the  2 term is negative, and the energy is no longer
minimal at   0 . Instead, the energy is minimal at two nonezero strains, known as the
spontaneous strains,   s . In this case, the  4 term will ensure that energy goes up again when
the strain is large enough.
The stress-strain relation is
w , T 

 AT  Tc   B 3 .

(Sketch this function. Mark the spontaneous strains and the hysteresis loop.) Setting   0 , we
find the spontaneous strains:
 s   ATc  T  / B .
(Ske
Because the material is nonlinear, the shear modulus is no longer a constant, and is given
by
  , T 

 AT  Tc   3B 2 .

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At the spontaneous strain, the shear modulus is given by
  2 ATc  T  .
Sketch the shear modulus as a function of the temperature, both below and above the critical
temperature.
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