ES 240
Solid Mechanics
Z. Suo
Homework
Due Friday, 3 November
26. Stress-strain relations under the plane strain conditions
Starting from a familiar form of Hooke’s law in three dimensions, namely,
 x   x   y   z / E ,  xy  21    xy / E , etc., show that under the plane strain conditions
the stress-strain relation takes the following form:
 x 

0   x 
1  
E
 
 

1 
0    y  .
  y   1   1  2   
 xy 
 0
0
0.5     xy 
 
27. Getting weak: derive weak statements from differential equations
If we call the principle virtual work (PVW) a weak statement of equilibrium, we may as
well call the equilibrium equations a strong statement. In class I stated the PVW without telling
you how it is discovered. I don’t really know the history, but a more useful thing to tell you is
how to construct a weak statement if you already know the differential equation. This way, you
can extend what you have learned in solid mechanics to other fields of engineering.
Stress field. As an example, start with the equilibrium equations:
 ij
 bi  0 , in V
(a1)
x j
 ij n j  ti , on S
(a2)
Instead of requiring that these equations hold for every point in volume and on surface, it is
equivalent that we require that
  ij

(b)
  x j  bi ui dV   t i   ij n j ui dA  0


for all virtual displacement field u i (i.e., any field what so ever). The first integral extends over
the volume of the body, and the second integral extends over the surface of the body. The left
hand side of (b) is called the weighted residual.
Recall the divergence theorem, and we obtain that
   iju i 
 ij
u i 
 x j ui dV    x j   ij x j dV


u i
   ij n j u i dA    ij
dV
x j
Recall that
1  ui u j 
ij  

2  x j
xi 
and that the stress is a symmetric tensor. We obtain that
u i
  ij x j dV    ijij dV .
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Solid Mechanics
Combining the above steps, we obtain that
  ijij dV =  biui dV   tiui dA .
Z. Suo
(c)
This is PVW. The statements (a), (b) and (c) are all equivalent.
Temperature field. The above method can be applied to other fields. To have some
appreciation of these applications, consider heat conduction in a three dimensional body.
First recall the governing equations. Let T x, y, z, t  be the temperature field, and
Jx, y, z, t  be the vector field of heat flux (i.e., energy across unit area per unit time).
Material law. Fourier’s law relates the heat flux to the temperature gradient:
T
J x  k
x
T
J y  k
y
T
J z  k
z
where k is the heat conductivity.
Energy balance. Let  be the mass density, and c be the heat capacity (i.e., the energy
needed to increase the temperature per unit mass per unit degree). For a unit volume of the body
to change temperature by dT, the energy needed is cdT . Energy balance requires that
T J x J y J z
c



0
t
x
y
z
in the volume of the body.
The body exchanges heat with the ambient by convection. Let h be the convection
coefficient (energy per area per time per degree). Energy balance requires that
nx J x  n y J y  nz J z  hT  Ta 
on the surface of the body. Here nx , n y , nz  is the unit vector normal to the surface, and Ta is the
ambient temperature.
Action items. (a) Start with a weighted residual and derive the equivalent statement.
Let  be an arbitrary scalar field. Show that the energy balance equations are equivalent to
requiring that the equation
 T


 
  c t   J x x  J y y  J z z dV +  hT  Ta dA =0,
hold true for every field  . The first integral is over the volume of the body. The second
integral is over the surface of the body. (b) Outline a finite element method for heat conduction.
28. Potential energy and the Rayleigh-Ritz Method
Here is yet another way to do pretty much the same thing. Applying a constant external
force to a body is equivalent to hanging a weight. For example, consider the structure in the
figure. Assume the rope that hangs the weight is inextensible. When the node moves by
displacement u, the weight P drops by the same distance, so that the potential energy of the
weight changes by  Pu . Now regard the body and the weights as a single system. The potential
energy of this system,  , is the sum of the elastic energy in the body and the potential energy of
the weights. We will call the potential energy due to the fixed load the work potential.
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Solid Mechanics
Consider a three-dimension elastic body, subject
to the body force b in the volume, and the traction t on
one part of the surface of the body, St . On the other part
of the surface, S u , the displacement is prescribed. The
work potential is
WP    bi u i dV   t i u i dA .
Z. Suo
u
P
The first integral extends over the volume of the body.
The second integral extends over the part of the surface St , where the traction is prescribed.
Note that b and t are known external forces. The displacement field, u, is unknown, except for
its prescribed values on the part of the surface of the body, S u . For any displacement field u, one
can calculate the work potential. A relation that maps a field to a number is known as a
functional. The work potential is a functional of the displacement field.
The strain field relates to the displacement field as usual:
u j 
1  u
.
 ij   i 
2  x j xi 
The elastic energy in the body is
1
U   C ijpq ij  pq dV .
2
The integral extends over the volume of the body. The elastic energy is also a functional of the
displacement field.
Potential energy is a functional of displacement. By definition, the potential energy is
the sum of the elastic energy and the work potential:
1
 u    Cijpq ij  pq dV   bi u i dV   t i u i dA .
2
The potential energy is a functional of the displacement field. The body force is prescribed over
the volume of the body, and the traction is prescribed on the surface St . The displacement need
not be the actual displacement field occurring in the body, but can be any field that satisfies the
prescribed displacement on the surface S u . The first two integral extends over the volume of the
body. The third integral extends over the area St .
Principle of minimum potential energy. The principle of minimum potential energy
states that, of all displacement fields that satisfy the prescribed values on the surface of a body,
the displacement field corresponding to equilibrium minimizes the potential energy.
Proof. Let u be the displacement field corresponding to equilibrium, and h be a variation
in the displacement field. The variation is arbitrary, except that h = 0 on S u . For the virtual
displacement u + h, the strain is calculated by inserting u + h in the usual displacement relation.
The linearity requires that  u  h   u   h . A direct calculation shows that
u i  hi   u p  h p 
u i u p 
1 
c

c

dV   bi hi dV   t i hi dA
ijpq
ijpq
2  
x j
x q
x j x q 
u p hi
h h p
1
  cijpq i
dV   cijpq
dV   bi hi dV   t i hi dA
2
x j x q
x q x j
 u  h    u  
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Solid Mechanics
Z. Suo
Because u is the equilibrium displacement,  ij  cijpqu p / x q . According to the PVW, the last
three terms cancel each other. Consequently,
h h p
1
 u  h    u    cijpq i
dV .
2
x j x q
This difference is always positive unless h vanishes everywhere in the body.
Rayleigh-Ritz Method. The principle of minimum potential energy calls for a search of
the winner among all displacement fields that satisfy the displacement boundary conditions.
Now if we limit the scope of the search to a subset of the admissible displacement fields, we will
not find the real winner, i.e., the actual displacement field, but an approximate. Of the
displacement fields in the subset, the displacement field that we select minimizes the potential
energy.
Here is the Rayleigh-Ritz method that implements the idea. For simplicity, we assume
that the boundary condition that the displacements vanish on S u . Let h1, h2,… hn be a set of
known virtual displacement fields. A linear combination is also a virtual displacement field:
u  a1h1  a2h2  ...  anh n ,
where a1 , a2 ,..., an are arbitrary numbers. This gives us a family of functions. In the jargon of
linear algebra, we say that h1, h2,…, hn is a set of bases, and all their linear combinations form a
space. This is a subspace in that there are other functions that cannot be represented this way.
To select the winner in this subspace, we will vary the coefficients a1 , a2 ,..., an to minimize the
potential energy.
Linearity requires that the strain be given as the sum
 u   a1 h1   a2 h2   ...  an hn  .
A direct calculation gives that
1
 u    K ij ai a j   Fi ai ,
2 i, j
i
where


K ij    h i  D  h j  dV
T
Fi   hTi bdV   hTi tdA .
Minimizing the potential energy, we set  / am  0 for m = 1, 2, …, n. This gives a set
of algebraic equation:
 Kij a j  Fi .
j
Solve for a1 , a2 ,..., an , and we obtain an approximate equilibrium displacement field:
u  a1h1  a2h2  ...  anh n .
Action items. An infinite body has a spherical cavity, radius R and subject to an internal
pressure p.
(a) Express the potential energy of the body as a functional of the radial displacement field u r  .
(b) Functions like r 1 , r 2 ,... are all virtual displacement fields. Assume a virtual displacement
field
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Solid Mechanics
u
Z. Suo
A
.
r2
Use the Rayleigh-Ritz method to determine A.
(c) Calculate the corresponding stress field. Compare with the exact solution to the problem.
Explain your findings.
29. Constant strain triangle
We want to formulate the finite element method to solve a plane elasticity problem.
Divide the plane into triangular elements. Assign a number to each node, and a number to each
element.
node 1

x1 , y1 
node 2
node 2
x2 , y2 
y
x
x3 , y3 
node 3
(0,0)
node 3
(0, 1)
node1
x
(1,0)

(a) Shape functions. Consider one triangle. Label its three nodes, counterclockwise, as
1, 2, 3. In the global coordinate system, the three nodes have coordinates x1 , y1 , x2 , y2 , x3 , y3  .
Map a triangle on a different plane,  ,  , to the triangle in the  x, y  plane. The following
functions map a point in the  ,  plane to a point in the  x, y  plane:
x  N1 x1  N 2 x2  N3 x3
y  N1 y1  N 2 y2  N3 y3
Show that the shape functions are
N1   , N2   ,
(b) The Jacobian matrix.
Calculate
transformation, namely,
 x
 
J
 x
 
N3  1     .
the Jacobian matrix of the coordinate
y 
 
.
y 
 
Also calculate det J and J 1 .
(c) Displacement field and its gradient. Let the displacements at the three nodes be
u1,v1  ,  u2 , v2  , u3 , v3  . Interpolate the displacement (u,v) of a point inside the triangle as
u  N1u1  N 2u2  N3u3
v  N1v1  N 2v2  N3v3
Calculate the displacement gradients in the  x, y  plane: u / x, u / y, v / x, v / y .
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Solid Mechanics
Z. Suo
(d) Strain field. The strain column of a point in the element is linear in the nodal
displacement column. We write
  Bq .
Work out the entries to the matrix B. You will find the strain field in this element is constant.
This element is known as the constant strain triangle.
(e) Stress field. The stress column is also linear in the displacement column:
  DBq .
Give the stiffness matrix.
(f) Stiffness matrix. Calculate the stiffness matrix of the element.
(g) Force column. Calculate the force column of the element.
30. Gaussian quadrature
(a) Determine the points and weights of the three-point Gaussian quadrature. Compare
your results with the values listed in a reference. Please list your reference.
(b) Use the three-point Gaussian quadrature to evaluate the following integral

 sin xdx .
0
Compare with the exact result.
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