ES 240 Solid Mechanics
22. 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:






 xy x y






1
  
E
1

2
  



1


0
 
1

0

0
0
0 .
5
 










 x y xy




.
23. 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:
 
 x ij  b i

0 , in V j
 ij n
 j t i
, on S
(a1)
(a2)
Instead of requiring that these equations hold for every point in volume and on surface, it is equivalent that we require that

 
 x j ij for all virtual displacement field
 b i
 u i dV
  t
 i
  ij n j

 u i dA

0 (b)
 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

 
 x j ij  u i dV
 


 ij
 x
 u j i

   ij n j
 u i dA



 ij

  u
 x ij j i
 x
 dV
  u i dV j
Recall that
 ij

1
2
 
 x u i j

  u j
 x i and that the stress is a symmetric tensor. We obtain that
  ij
 
 x u j i dV
   ij
 ij dV . апрель 12, 2020 1

ES 240 Solid Mechanics
Combining the above steps, we obtain that
  ij
 ij dV =
 b i
 u i dV
  t i
 u i dA
This is PVW. The statements (a), (b) and (c) are all equivalent.
. (c)
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:
J
  k

T
J x y
  k
 x

T
 y
J z
  k

T
 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 c


T
 t


J
 x x


J
 y y 

J
 z z

0 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 n x
J x
 n y
J y
 n z
J z
 h

T

T a
 on the surface of the body. Here
 n x
, n y
, n z

is the unit vector normal to the surface, and T is a 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


 c


T
 t
 
J x
 
 x

J y
 
 y

J z
 
 z

 dV +
 h

T

T a
  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.
24. 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 . апрель 12, 2020 2

ES 240 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, S . On the other part of the t u surface, S , the displacement is prescribed. The work u potential is
WP
   b i u i dV
  t i u i dA .
The first integral extends over the volume of the body.
The second integral extends over the part of the surface
P
S t
, 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:
 ij

1
2
 u i
 x j

 u j
 x i
.
The elastic energy in the body is
U

1
2

C ijpq
 ij
 pq dV .
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 
C ijpq
 ij
 pq dV
  b i 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 S t
. 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 S . t
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 . For the virtual u displacement u + h , the strain is calculated by inserting u + h in the usual displacement relation.
The linearity requires that
  u
 h
    
. A direct calculation shows that апрель 12, 2020 3

ES 240 Solid Mechanics

 u
 h

 
 

1
2
 c ijpq

 u i
 x
 j h i
 
 u
 h
 p p  c ijpq  x q
 u i
 x j
 u p
 x q dV
  b i h i dV
  t i h i dA
1
2
 c ijpq
 h i
 x j
  c ijpq
 u p
 x q
  t i h i dA

 h p
 x q dV
 h i
 x j dV
  b i h i dV
Because u is the equilibrium displacement,
 ij
 c ijpq
 u p
/
 x q
. According to the PVW, the last three terms cancel each other. Consequently,
  u
 h
     
1
2
 c ijpq
 h i
 x j
 h p
 x q dV .
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 h
1
, h
2
,…
h n
be a set of known virtual displacement fields. A linear combination is also a virtual displacement field: u
 a
1 h
1
 a
2 h
2

...
 a n h n
, where a
1
, a
2
,..., a n
are arbitrary numbers. This gives us a family of functions. In the jargon of linear algebra, we say that h
1
, h
2
,…,
h n
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 a , a ,...,
1 2 a n to minimize the potential energy.
Linearity requires that the strain be given as the sum
  
 a
1
  
1
 a
2
  
2

...
 a n
   n
.
A direct calculation gives that
 
1
2
 j i ,
K ij a i a j
  i
F i a i
, where
K ij
F i

 
 
   i

T h T i b dV
Minimizing the potential energy, we set
D

 


  j
 dV
/
 h i
T
 a m t dA

0
.
for m = 1, 2, …, n . This gives a set of algebraic equation:
 j
K ij a j

F i
.
Solve for a , a ,...,
1 2 a n
, and we obtain an approximate equilibrium displacement field: u
 a
1 h
1
 a
2 h
2

...
 a n h n
. апрель 12, 2020 4

ES 240 Solid Mechanics
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
 
. (b) Functions like r

1
, r

2
,...
are all virtual displacement fields. Assume a virtual displacement field u

A
. r
2
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.
25. 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

 x
1
, y
1
 node 2 (0, 1)
 x
2 node 2
, y
2
 y node 3 node1
 x
3
, y
3
 (0,0)
x
(1,0)
 x node 3
(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
 x
1
, y
1
  x
2
, y
2
  x
3
, y
3

.
Map a triangle on a different plane,
 
, to the triangle in the
 
plane. The following functions map a point in the
 
plane to a point in the
 
plane: x y


N
1 x
1

N
1 y
1

N
2 x
2
N
2 y
2


N
3 x
3
N
3 y
3
Show that the shape functions are
N
1
 
, N
2
 
, N
3

1
   
.
(b) The Jacobian matrix.
Calculate the Jacobian matrix of the coordinate transformation, namely,
J






 x


 x
 
 y


 y
 





.
Also calculate J J

1
. апрель 12, 2020 5

ES 240 Solid Mechanics

(c) Displacement field and its gradient.
Let the displacements at the three nodes be u v
2 2

,
 u v
3 3

. Interpolate the displacement ( u,v ) of a point inside the triangle as
 u
1
, v
1

, u

N
1 u
1

N
2 u
2

N
3 u
3 v

Calculate the displacement gradients in the
N
1 v
1

 
N
2 v
2

N
3 v
3
, plane:
 u /
 x ,
 u /
 y ,
 v /
 x ,
 v /
 y .
(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. апрель 12, 2020 6