Discover what is the Cell Method and how to apply it to physical problems. This presentation will guide you through some interesting computational implementations. Youtube video: https://youtu.be/ZyyRovtibjo

1. The Cell Method
for Computational
Physics
Prof. Enzo Tonti Dr. Giovanni Rinaldin
1
.

2. The quantitative description of physical properties, typical of physical
theories, leads to the introduction of physical quantities.
The measurement of a physical quantity implies always a tolerance:
the term ”exact value” has no place in applied physics. Also the
result of a calculation on physical variables is requested within a
prescribed tolerance.
These statements sound provocative to hears of a mathematician for
which the term “exact” is essential, but we are speaking of “applied”
mathematics not of “pure” mathematics.
The mathematical formulation of physics is made, usually, in terms of
differential calculus, hence the fundamental problem is expressed by a
differential equation.
Since the computer cannot deal with the notion of limit, exclusive feature
of the human mind, in order to solve the fundamental problem of a
physical theory we are obliged to discretize the differential equations.
Premise
2

3. When we read the indication of an amperometer or of a caliber we read
the first two or three significant values. This means that the value of the
physical quantity is approximate.
For this reason also the solution of a physical problem is approximate.
The important fact is that the result satisfies a prescribed tolerance.
Hence the passage from a differential formulation of the fundamental
equation to a discrete formulation in order to solve physical problem, by
computers become a tortuous process that can be avoided starting
directly with a discrete (=algebraic) formulation of the fundamental
equation.
3
Premise
Premise

4. Whatdowewanttodo?
This is the Cell Method. To describe it, let's look at some types of
problems that we may encounter in physics and engineering.
We want to present a method that allows to give an approximate
algebraic formulation to problems which are usually described by exact
differential equations.
In fact, the differential equations, in order to be treated with the computer,
they must be discretized (i.e. transformed into algebraic equations).
This procedure is typical of the traditional methods, like FEM, FVM, BEM,
FDM, LSQ, etc.
The method we present provides a formulation directly discrete
(or algebraic, or finite) without having to pass through the differential
formulation.
4

5. Thefundamentalproblemofafield
The fundamental problem of a physical field, such as:
the thermal field,
the elastic field,
the electromagnetic field
is to determine the “configuration” of the field
once its “sources” are assigned.
Let us look at three physical theories: 5

6. 1)Thermalconduction
The heat Q emitted from heat generators propagates from regions of higer
temperature to those of lower temperature.
The fundamental problem of the thermal conduction is to find
the temperature (=potential) at each point of the region we are studying
once the heat generators, their position and their intensity are assigned.
( , , , ) ?T t x y z temperature
The main physical variable of a thermal field is the temperature: this physical
variable is introduced in order to describe how cold and how worm is any
part of the region or of the body.
The temperature can be considered as the “potential” of the thermal field.
Let us consider a space region or a body: we want to consider its thermal
status.
6

7. 1)Thermalconduction
A law states that heat goes from regions to higher temperatures to those of
lower temperature.
Another law states that the amount of heat that passes through one surface
element depends on the difference in temperature between the two points
who are riding the surface.
This law is espressed by a constitutive equation, and is known as the
Fourier Elementary Law.
What are the laws of thermal conduction?
Another law is the energy balance. This law states that the increase of the
internal energy of a system is the sum of the heat and the work entering
the system.
Increase of internal energy = work entering + heat entering
Combining these laws on obtains the fundamental equation of the
thermal conduction, the Fourier law.
7

8. Consider the deformation of a solid body.
The cause of deformation are the forces acting on the solid.
Since deformation is transmitted from point to point, there must some laws
that govern the way in which it is propagated: this is expressed by a
constitutive equation. If the material is elastic it is Hooke's law.
2) Deformation of a solid body
Once a force is applied, the deformation is transmitted along the body
determining the displacement (effect) of the body's points.
The fundamental problem of deformation theory is to find
the displacement that undergoes every point of the body, starting from an
initial configuration, when are assigned the forces that generate it.
One of these laws imposes the equilibrium, of every piece of body if the
deformation is static; or the balance of momentum if the deformation is
dynamic.
8

9. Consider the electromagnetic field that is generated by electric charges
at rest and in motion. The description of the field is made by the scalar
electric potential and the vector magnetic potential.
3)Electromagnetism
( , , , ) !
( , , , ) !
t x y z
t x y z
 
J
electric charge density
electric current density
The fundamental problem of electromagnetism is the following:
assigned the spatial and temporal distribution of charges and currents,
determine the electric an the magnetic potential at every point of the
field at every moment:
( , , , ) ?
( , , , ) ?
t x y z
t x y z
 
A
scalar potential
vector potential
9

10. Fieldfunctions
We said that in each field the goal is to determine the configuration of the field.
The configuration is described by place and time functions of scalar
or vector type.
( , , , ) ?t x y z v
( , , , ) ?t x y z u
( , , , ) ?T t x y z 
A(t,x, y,z)=?
( , , , ) ?t x y z 
temperature
displacement
velocity
electric scalar-potential
magnetic vector potential
For historical reasons,
these laws are expressed by
partial differential equations.
We raise the question:
is it easy to solve
the partial differential equations
that we encounter
in Physics and Engineering?
10

11. ...solvethefundamentalequations
it is not easy at all !
Approximations need to be made
and to use numerical techniques
which force the use of the computer.
This requires transforming the differential equations
into approximated algebraic equations
through one of the many methods of discretization.
The answer is:
11

12. Some Mathematical
Considerations
12

13. use the four
fundamental
operations:
sums
differences
products
divisions
From that moment they
introduce:
the derivative (limit of a ratio)
and the integral (limit of a sum)
and therefore the whole
differential calculus;
the ordinary differential equations
and those with partial derivatives
algebra infinitesimal analysis
introduces the notion of
limit
mathematics
13
The two great brances of mathematics

14. Why can not a computer treat infinity?
1/2
1/4
1/8
1/16
1/32
1 1 1 1 1
1 ...
2 4 8 16 32
     
the notion of limit is an ideal notion, that belongs to men,
not to the computers.
the computer would provide: 1.99999999999909
1
Let's propose to evaluate the sum
2
14

15. Thedifferentialformulationofphysicallaws
It is thought to discretize the differential equations to obtain
an appproximate algebraci formulation.
At the beginning using the finite differences, later other using
other discretization methods.
What happened then?
-occurred about three centuries ago (1687)
-physical laws have been mathematically formulated
in terms of differential equations.
The advent of computers required an algebraic
description of physical laws because the computer
ignores the notion of limit
It was not thought to go back to experimental facts
to obtain a direct algebraic formulation of the physical laws.
Philosophiae
Naturalis
Principia
Mathematica
I.Newton
5+7=12
15

16. from discrete to the differential ...
and then to go back to the discrete!
...
This lecture
BEM
FDM
physical
field
approximated
solution
algebraic
equations
FVM
FEM
differential
equations
16

17. It's possible
a DIRECT algebraic formulation
of physical fields?
The variety of discretization methods of differential equations
it raises the following question:
Finiteformulation(=discrete=algebraicformulation)
17
is easy !
is possible !
is intuitive !
is ready for numerical solution !
We want to prove that a DIRECTALGEBRAIC formulation of field laws ...

18. Apologyofthe”balance"
The fundamental equations of physical phenomena comes from a balance:
• in the static of deformable solids, the balance of forces is essential;
• in the thermal conduction the energy balance is essential;
• in the fluid dynamics, the balance of momentum is essential;
as well as the balance of mass;
• chemical reactions expresses the mass balance;
• in the theory of electric circuits the balance of the currents at the nodes
is essential;
• in electromagnetism the balance of charge is essential;
• etc.
18

19. Think for example to
a bottle factory.
The freshly formed bottles
come out of the furnace:
some of these are stored
on the shelves of the plant
and a part is sent out.
Whatisabalance?
A balance is a relatiom about some of extensive physical variables for
which we can talk about storage, production and outgoing.
Bproduced
= DtBstored
+ Bsent out
Bproduced
19

20. For the equilibrium of a body, the sum of the of volume and surface forces
must vanish.
A balance is valid for any size any shape of the body portion to which
it is applied, there is no need to shrink to an infinitesimal volume.
Hence, why we apply it to an infinitesimal region
to get a differential equation?
A balance is valid in the finite setting
as well as in the infinitesimal
20
surface forces
volume forces

21. 21
A preliminary Example:

22. Stationary thermal conduction 1/7
22
Let us consider three rooms in each the temperature is uniform.
They are separated by a wall of thickness d and area A. Each room contain
an heater and let us denote with Pk the heat emitted for unit time.
Suppose that the side walls, the ceiling and the floor are insulating.
The heat for unit time passing throug each wall is given by the Fourier
first law, the constitutive equation
T
1P 2P 3P
20° 16T2
dT1
T1 T2
T3 T3
G1 = T1 - 20o
G2 = T2 -T1 G3 = T3 -T2 G4 =16o
-T3
d d d d

23. fundamental equation
23
1 2 3, ,P P P
3 heat generation rates
assigned
1 2 3, ,T T T
3 temperatures
unknowns
Stationary thermal conduction 2/7
We propose to find the temperature Tk in each room.
F
T = temperature;
G = temperature difference
P = heat generation rate
= heat current
T
1P 2P 3P
20° 16T2
dT1
T1 T2
T3 T3
G1 = T1 - 20o
G2 = T2 -T1 G3 = T3 -T2 G4 =16o
-T3
d d d d
As is common in thermodynamics, we consider positive the heat entering
into the system (here each room).

24. 1 2 3T ,T ,T
1 2 3 4G ,G ,G ,G
G1
:=T1
- T0
G2
:=T2
-T1
G3
:=T3
-T2
G4
:=T4
-T3
ì
í
ï
ï
î
ï
ï
1 2 3P ,P ,P
1 2 3 4, , ,   
2 1 1
3 2 2
4 3 3
P
P
P
   

   
   
F1
= -l
A
d
G1
F2
= -l
A
d
G2
F3
= -l
A
d
G3
F4
= -l
A
d
G4
ì
í
ïï
î
ï
ï
24
Stationary thermal conduction 3/7
The fundamental equation is obtained
combining three kinds of equations:
the equations that define the temperature
differences; the constitutive equations and
the equations of balance.
constitutive equations
equationsofDefining
equationsofbalance
T
1P 2P 3P
20° 16T2
dT1
T1 T2
T3 T3
G1 := T1 - 20o
G2 := T2 -T1 G3 := T3 -T2 G4 :=16o
-T3
d d d d

25. 1
1
2
2
3
3
4
1 1 0 0
0 1 1 0
0 0 1 1
P
P
P
 
                         
1 1 0
2 2 1
3 3 2
4 4 3
=T - T
G =T -T
G =T -T
G =T -T
G





2 1 1
3 2 2
4 3 3
P
P
P
   

   
   
1 1 2 2
3 3 4 4
A A
G G
d d
A A
G G
d d
 
 

     

     

G1
G2
G3
G4
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
=
1 0 0
-1 1 0
0 -1 1
0 0 -1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
T1
T2
T3
ì
í
ï
î
ï
ü
ý
ï
þ
ï
+
-20
0
0
16
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
1 1
2 2
3 3
4 4
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
G
GA
Gd
G

    
            
    
         
defining matrix
balance matrix
constitutive
matrix
25
Stationary thermal conduction 4/7T
1P 2P 3P
20° 16T2
dT1
T1 T2
T3 T3
G1 := T1 - 20o
G2 := T2 -T1 G3 := T3 -T2 G4 :=16o
-T3
d d d d

26. F1
F2
F3
F4
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
= -l
A
d
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
G1
G2
G3
G4
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
1
1
2
2
3
3
4
1 1 0 0
0 1 1 0
0 0 1 1
P
P
P
 
                         
defining matrix balance matrix
Here some general properties are revealed:
constitutive matrix
minus signG1
G2
G3
G4
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
:=
1 0 0
-1 1 0
0 -1 1
0 0 -1
é
ë
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
T1
T2
T3
ì
í
ï
î
ï
ü
ý
ï
þ
ï
+
-20
0
0
16
ì
í
ï
ï
î
ï
ï
ü
ý
ï
ï
þ
ï
ï
2) the constitutive matrix is symmetric.
26
transpose with the minus sign
symmetric
1) the balance matrix is the transpose of the defining matrix with the minus sign.
Stationary thermal conduction 5/7

27. defining matrix
minus sign
0G = T +GD
G C
T
= P D
constitutive matrix
balance matrix
The fact that the matrix on the right side is the transpose of the matrix on the
left side, with the minus sign, is a property common to all physical theories.
The symmetry of the constitutive matrix is another property common to many
physical theories: C = CT.
From these two properties it follows that the matrix of the fundamental
equation is symmetric. This symmetry is the root of the variational principles
of physics.
27
source
“vector”
temperature
“vector”
T
0( T +G ) =P D C D
fundamental equation
Stationary thermal conduction 6/7
algebraic formulation (useful for computational purposes)

28. By connecting the source with the configuration variable by means of the
balance-constitutive-defining equations, we obtain the fundamental equation.

constitutive equation
(Fourier)
balance
equation
defining
equation
temperature
thermal
gradient
heat
source
heat current
density
P
L
T
g:= ÑT
-q g
Ñi q=s
qg
div qdiv -lg( )=sdiv -lgrad T( )=s-ldiv grad T( )=s
28
Stationary thermal conduction 7/7
differential formulation
2
- T  
Fundamental equation
Fourier

29. 29

30. Energy balance (first principle of thermodynamics):
the increase in internal energy over a period of time
it is equal to the sum of the heat and of the work supplied to the system
in the same time interval.
F = - k A
T2
-T1
d
The two laws of thermal conduction
given to the system produced
 ingoing
Q Q Q
30
2) Constitutive equation: (Fourier elementary law)
the heat that crosses a “wall” for unit time
(heat current) is proportional to the area of the
“wall”, to temperature jump across the “wall” and
is inversely proportional to the thikness of the
“wall” and proceeds from heat to cold:
F
DtU= Qgiven to the system
+ Wentering
F
A
d

31. Thefundamentalproblemofstationarythermalconduction
thermal
source
assigned a body;
specified the physical nature of the materials that make up;
specify the thermal sources in extension and intensity;
specify the conditions at the boundary of the region;
determine the temperature at each point of the region.
temperature
assigned
temperature
assigned
unknown
temperature
temperature
unlnown
unknown heat
unknown heat
heat assigned
heat
assigned
heath
assigned
31

32. We build in the region a complex of triangular-shaped cells.
Let's triangulate, so that the triangles lean against the surface of
separation between the two materials. Let us assign a number to each
vertex and a number to each cell. This is automatically done by the mesh
generator used.
1
2
3 5 7
4
6
8
9
10
11
12
13
14
15
16
17
Complex of triangular cells
32
temperature
assigned
temperature
assigned
heat assigned
heat
assigned
1
2
7
11
3
9
4
5
6
8
1012
13
14
15
16
17
18
19
20

33. A cell complex in two dimensions imply to consider it as a transversal
section of a cilindrical solid. Along each straigth fiber the temperature
is uniform so that heat propagates only in the transversal direction with
respect to the fibers.
What really means a two dimensional problem?
33
Hence every triangle is really a triangular prism, as shown below.
h
i
j
s
We will denote with s the thickness of the layer.

34. A first type of nodal influence areas
is made up of polygons whose sides are the axes of the sides of the triangles.
The resulting polygons (in red) are called polygons of Voronoi.
To write the energy balance, for each node we choose a polygon that
contains the node. These polygons give rise to a second subdivision of the
region that is called dual of the first one.
Node influence areas: Voronoi polygons
34
1
2
3 5 7
4 6
8
9
10
11
12
13
14
15
16
17
1 2
7
11
3
9
4
5
6
81213
14
15
16
17
18
19
20
10

35. 35
Association of physical variables
with space elements

36. 36
Physical variables are associated
with “space elements”
All geometrical objects are composed of four “space elements”:
points, lines, surfaces, and volumes.
Physical quantities can be divided into two classes (with few ambiguities):
physical constants and physical variables. To the first class belong the
universal constants, such as the Plank constant, the speed of light in vacuum,
the Avogadro number, etc. To this class belong also all material constants
such as the elastic modulus of a material, the thermal conductivity of a material,
the permittivity of a material, etc.
Physical variables are those quantities that may assume whatever value inside
a prescribed interval. Such is the temperature, the speed, the displacement, the
mass, the momentum, the force, etc. Of course the physical variables may be
scalars, vectors, tensors, etc.
The index of refraction of a transparent material is usually a constant, but for a gas
depends on its pressure, hence in this case it behaves as a physical variable.

37. 37
Mass, enegy, entropy are associated with volumes; fluxes of energy, of electric
charge, heat are associated with surfaces; elongation is associated with a line;
temperature, electric potential, gravitational potential are associated with points
etc.
Every space element can be endowed with orientation. There are two kinds of
orientation, the inner orientation and the outer orientation. So a line segment
connecting two points, A and B may be oriented from A to B or from B to A. These
are the two opposite inner orientations. Line segments that belong to a same
plane may also endowed with a transversal (or outer) orientation: this is the
inner orientations of a line segment that crosses the original line segment.
A A
B B
inner orientations
A A
B B
transversal (=outer) orientation
Inner and outer orientation

38. 38
Inner and outer orientation in space 1/2
The notions of inner and outer orientation of space elements in the
threedimensional space are illustrated in the next two slides
It is important to remark that a physical variable associated with
an oriented space element changes its sign when we reverse the orientation
of the space element.
Then in fluid dynamics the vorticity is linked to a small surface element and
is associated with the inner orientation of the surface element.
The same happens for magnetic flux. On the contrary, a flow of matter and
an energy flow are associated with a small surface element with an external
orientation and change sign when we invert the external orientation of the
surface element.
The voltage along a line from A to B has the opposite sign of the
voltage from B to A. In a fluid flow the velocity circulation changes sign when
let's go from B to A.

39. 39
Inner and outer orientation in space 2/2
If we consider a point P and a point Q, in this order, the temperature
difference is T(Q)-T(P). If we consider as first the point Q and as second
the point P, the temperature difference is T(P)-T(Q). The same happens
with the potential difference between two points.
The voltage along a line from A to B has the opposite sign of the
voltage from B to A. In a fluid flow the velocity circulation changes sign
when we go from B to A instead that from A to B.
The notion of inner and outer orientation is valid also for points.
This means that a line connecting P with Q the inner orientation is
from P to Q then the arrow that denotes the orientation direction is
”outgoing” from P and ”ingoing" in point Q, then Q is considered as
a well and P as a source.

40. Inner orientation
inner orientation of a line
is an order of its bounding points.
A
B
A
B
inner orientation of a volume = a compatible (*)
inner orientation of its bounding faces.
inner orientation of a surface = a compatible (*)
inner orientation of its bounding edges;
or a direction of rotation.
40
P
L
S
V
(*) – “compatible” means that two adjacent elements induce opposite
orientation on the common boundary.
inner orientation of a point; the
lines pointing inwards being positive, or viceversa
(the points are “sinks” or “sources”).
“sinks” “sources”
The space elements endowed with inner orientation will be denoted with
a bar over the letter.

41. Outer orientation
outer orientation of a surface:
an order of its faces (rear / front)
outer orientation of a volume:
from the inside to the outside or viceversa
outer orientation of a line:
a direction of rotation around the line
41
outer orientation of a point:
a consistent orientation of the lines
with one extreme to the point;
i.e. lines endowed with outer orientation.
part 3/8
The space elements endowed with outer orientation will be denoted with
a tilde over the letter.

42. 42
Triangulation = “simplicial complex”
We oganize our presentation of the Cell Method by a division of the plane region
considered into triangles: the complex arising is called “simplicial complex”.
Since we wont to consider the intersection of the axes of the sides of each
triangle, i.e. its circumcenter, we require that the circumcircles of every triangle
do not contain other vertices of the mesh. Such a mesh is called Delaunay mesh.
There are mesh generators that authomatically generate a Delaunay complex for
a given two-dimensional region and also for a three dimensional region.

43. 43
Numeration of the nodes
The mesh generetor assigned a number to the vertices of a cell complex (= nodes)
and a number to every triangle.
The side of a (two-dimensional) mesh will be endowed with inner orientation:
the orientation of each side goes from the node of lower number to the one
of greater number.
1
2 3
5
6
7 8
9
10
42
1 3
6
4
8
5
10
9
7
11

44. 44
Now, let us consider a triangle in a cartesian system. Let us denote with
h,i,j its vertices, with Lh, Li, Lj its sides, with Ah, Ai, Aj three vectors normal
to the sides and of equal length of the corresponding side.
h
i
j
y
Lh
Li
Lj
Ah
Aj
Ai
We approximate the temperature inside each triangle by a function with
linear behaviour, i.e. with an affine function:
a + gx xh + gy yh = Th
a + gx xi + gy yi = Ti
a + gx xj + gy yj = Tj
ì
í
ï
î
ï
Imposing that in the three nodes the
funtion T(x,y) assumes the nodal
temperatures, we can find the coefficients
a, gx, gy:
T(x,y) = a+ gx x + gy y
We have used the letters gx an gy because
they are the cartesian components of the
gradient of the function: gx =
¶T
¶x
gy =
¶T
¶y
x
Defining equation 1/4

45. ...
45
a + gx xh + gy yh = Th
a + gx xi + gy yi = Ti
a + gx xj + gy yj = Tj
ì
í
ï
î
ï
subtracting the second equation from the third
one; the first equation from the second one and
putting
Lhx = xj - xi Lhy = yj - yi
Ljx = xi - xh Ljy = yi - yh
we obtain
Ljx Ljy
Lhx Lhy
é
ë
ê
ù
û
ú
gx
gy
é
ë
ê
ù
û
ú
Ti -Th
Tj -Ti
é
ë
ê
ù
û
ú=
At this point we must invert the matrix
gx
gy
é
ë
ê
ù
û
ú
Ljx Ljy
Lhx Lhy
é
ë
ê
ù
û
ú=
Ti -Th
Tj -Ti
é
ë
ê
ù
û
ú
-1
Defining equation 2/4

46. 46
Let us remember how to invert a 2 x 2 matrix.
a b
c d
é
ë
ê
ù
û
úM = the transpose is
a c
b d
é
ë
ê
ù
û
ú the minor are
d - b
-c a
é
ë
ê
ù
û
ú
putting D = det(M) = ad -bc the inverse matrix is M-1
=
1
D
d - b
-c a
é
ë
ê
ù
û
ú
We can test the result:
M M-1
=
1
D
a b
c d
é
ë
ê
ù
û
ú
d - b
-c a
é
ë
ê
ù
û
ú =
1
ad - bc
ad - bc 0
0 ad - bc
é
ë
ê
ù
û
ú =
1 0
0 1
é
ë
ê
ù
û
ú
that is the unit matrix.
Defining equation 3/4

47. 47
gx
gy
é
ë
ê
ù
û
ú
1
D
Lhy - Ljy
-Lhx Ljx
é
ë
ê
ù
û
ú=
Ti - Th
Tj -Ti
é
ë
ê
ù
û
úif we apply the previous formula we get
gx
gy
é
ë
ê
ù
û
ú
1
D
+Lhy Ti -Th( )- Ljy Tj -Ti( )
-Lhx Ti -Th( )+ Ljx Tj -Ti( )
é
ë
ê
ê
ù
û
ú
ú
=
1
D
Lhy + Ljy( )Ti - LhyTh - LjyTj
Lhx + Ljx( )Ti + LhxTh + LjxTj
é
ë
ê
ê
ù
û
ú
ú
=
gx
gy
é
ë
ê
ù
û
ú =
1
D
-LhyTh - LiyTi - LjyTj
+LhxTh + Lix Ti +LjxTj
é
ë
ê
ù
û
ú =
1
D
-Lhyh - Liy - Ljy
+Lhx + Lix +Ljxj
é
ë
ê
ù
û
ú
Th
Ti
Tj
é
ë
ê
ê
ê
ê
ê
ê
ù
û
ú
ú
ú
ú
ú
ú
since Lh +Li +LJ = 0 it follows that
We remark that the determinant is double of the area A of the triangle,
hence we can write
D
D = 2A.
Defining equation 4/4

48. 48
The heat current
Let us start with an analogy: let us consider a river and a region in which
the water flow is uniform. If we introduce in the region a plate that is free
to rotatate around a vertical axis, a force is exerted by the water on it.
Of course the force depends on the inclination of the plate and si maximum
when the plate is orthogonal to the lines of heat current.
lines of current
In analogous way, if we consider a uniform heat flow and a plane surface
(not a plate) the amount of heat that crosses the plate depends on the
inclination of the plate.
Let us consider the direction in which the heat crossing the surface is
maximum. Let us consider the heat current (in Watt) crossing the surface
and divide it by the area of the surface. Let us denote by q such a ratio.
There will be a direction for which q is maximum. We can consider a vector
q whose direction is orthogonal to the plane surface for which q is
maximum:
q = qmaxn

49. A first type of nodal influence areas
is made up of polygons whose sides are the axes of the sides of the triangles.
The resulting polygons (in red) are called polygons of Voronoi.
To write the energy balance, for each node we choose a polygon that
contains the node. These polygons give rise to a second subdivision of the
region that is called dual of the first one.
Node influence areas: Voronoi polygons
49
1
2
3 5 7
4 6
8
9
10
11
12
13
14
15
16
17
1 2
7
11
3
9
4
5
6
81213
14
15
16
17
18
19
20
10

50. We will write the heat balance for each dual polygon.
Voronoi polygons have the advantage of having the sides of the polygons
orthogonal to the sides of the triangles: this makes it very simple
the writing of constitutive equations.
Let us first consider internal polygons (whole)
Whole interior polygons
50
1
2
3 5 7
4 6
8
9
10
11
12
13
14
15
16
17
1 2
7
11
3
9
4
5
81213
14
15
16
17
18
19
20
10

51. We will limit ourselves to the stationary case: the energy accumulation is
null and therefore the outgoing heat is equal to that produced.
Qoutgoing
=Qproduced
Energy balance on dual polygons
51
We will denote with P the heat generated for unit time.
1
2
3 5 7
4 6
8
9
10
11
12
13
14
15
16
17
1 2
7
1
1
3
9
4
5
81213
14
15
16
17
18
19
20
1
02
Q
2,1
0
Q2,11
Q2,3
Q2,4
Q1.2
Q2,15
Q2,14
+ Q1.2 - Q2,3 - Q2,4- Q2,15 - Q2,14- Q2,10 - Q2,11 = P2poligono 2:

52. Estimatetheheatcomingout
The heat crossing one side of the Voronoi polygon, considered
positive when it enters, depends on the temperature difference
measured on the two sides of the faces.
L2,1
A2,1
The Fourier Heat Elementary Law states that in a uniform flow region, the
heat passing through a surface in the unit of time is proportional to the
temperature difference per unit of length according to the formula
We will make the approximation to consider
the surrounding region on each side of
the polygon as a region of uniformity.
This will be the only approximation we will make!
1
2
4
10
11
14
15
3
52

53. Basicequationforinteriorpolygons
constitutive equations
(approximate)
F2,1 » -k
A2,1
L2,1
(T2-T1)
...........
F2,15 » -k
A2,15
L2,15
(T2-T15)
ì
í
ï
ï
ï
î
ï
ï
ï
you get the approximate
algebraic equation
   2,1 2,3 2,4 2,15 2+ + +...+ = P
f2,1T1+f2,2T2+f2,3T3+...+f2,15T15 » P2
balance equation
(exact)
53
L2,1
A2,1
1
2
4
10
11
14
15
3

54. We will now write the equation of balance for each dual edge polygon
(broken).
1
...
2
4
10
11
14
15
3
Basicequationfortheonboardpolygons
The term B indicates the heat coming from the edge of the broken polygon
for unit time (B stands for boundary)
54

55. For each edge polygon we write that the
sum of the output streams is equal to the
incoming flow rate B from the edge.
Also using comstitequations here is the approximate
algebraic equation
Balance on polygon 1
1B 1
2
4
10
11
15
3
55
Energy balance on split dual polygons

56. Thefundamentalalgebraicsystem
In this way you get to an algebraic system
of n equations in n unknowns containing
the nodal temperatures:
Since we have a dual polygon
for each node, writing a balance equation
for each dual polygon, we will have so
many equations how many knots are and
how many are the unknowns.
56
1
2
4
10
11
15
3

57. Thefundamentalalgebraicsystem
This algebraic system
constitutes the finite equivalent of Fourier's heat conduction equation
S
T
k assigned
n "

 

S
T assigned'

57
plus boundary conditions:

58. Insteadthedifferentialformulation...
But then there is a need to make the limit, consider infinitesimal volumes
and so come to a differential equation that we can not solve?
gives us two beautiful differential equations, one for each material
(assumed homogeneous), forces us to write the equations of junction on
separation surfaces between two different materials.
C1
C1
C2
C2
A
B
58

59. From where is the approximation born?
It is evident that the smaller the size of the cells,
the more the field can be considered uniform in each of them.
All it was worth to avoid passing to the limit and keeping us
in algebraic formulation from the beginning!
At this point comes the temptation to make the limit, or to reduce volumes
so that the law of constitution becomes ”exact".
Doing so we fall back into the differential formulation!
And since the calculator needs an algebraic formulation,
we are forced to discretize the differential equation...
Here it is the only approximation of numeric solution
59
Because a constitutive equation is valid in regions of uniform field,
when applied in non-uniform field regions, becomes approximate.

60. Simplicialcomplex
20020 40 60 80 100 120 140 160 180 200
concentrated source
distribuite source
Region with material 2
Region with Material 1
60
Cell method can work with:
 Different materials
 Different kind of sources
 Complex meshes

61. ItsVoronoidual
20 40 60 80 100 120 140 160 180 20061
It’s always possible to
find the Voronoi dual of
a primal mesh.
Dual mesh always allows
equibrium for any kind
of sources.

62. 62
A detailed presentation of the problem of finding the temperature in
a stationary thermal conduction can be found in the author paper
http://www.discretephysics.org/papers/TONTI/The%20Cell%20Method.pdf
Tonti E., A Direct Discrete Formulation of Field Laws: The Cell Method
CMES, Vol. I, no .1, p.11-26
The paper can be freely downloadesd from the web site

63. 63

64. 64
Using Cell Method for elasto-static problems
The formulations with CM (Cell Method) of a regular plane-stress
quadrilateral element is shown in the following.
From an overall point of view, the entire procedure follows the
diagram depicted below:
For the stress matrix, a Voigt’s notation is used, such to have a 3 components
vector.

65. 65
Using Cell Method for elasto-static problems
Formulation of a regular quadrilateral cell for
mechanical plane stress analysis (i.e.
membrane behaviour):
1. Choose a displacement interpolation
function (bilinear in this case) as to describe all
the possible deformation into the element:
Deformation matrix
( , )
( , )
u x y a bx cy dxy
v x y a bx cy dxy
   
    The internal stresses
are evaluated
in points A,B,C,D
Such matrix connects
deformations and
displacements:
u
U
v
 
  
BU 

66. 66
Sample problem: plane stress analysis of a wall
2. Use constitutive equations
With:
E as elastic (Young’s) modulus
G as shear modulus
n as Poisson’s coefficient
 as deformation component
Deformation matrix is the
equivalent, in FEM, of the matrix
collecting the derivatives of
shape functions

67. 67
Sample problem: plane stress analysis of a wall
3. Write equilibrium in internal points (A,B,C,D) as to obtain
stiffness matrix. Let be:
• t the thickness of the membrane
• p the length of half the edge
We can write the equilibrium on the dual cells (red edges) as
follows:
In which:
N is the number of dual cells
T is the resultant of surface forces acting from the outside
F is the resultant of volume forces (e.g. self-weight)
B is the extenal surface force eventually present if the node in
which we write the equilibrium lies on the boundary.

68. 68
Sample problem: plane stress analysis of a wall
We can write the equilibrium as follows: Ac: area of
the primal cell
h,i,j,k are the
vertices
x,y reference
planar axis
Tx,Ty are the
resulting
forces

69. 69
Sample problem: plane stress analysis of a wall
Stiffness matrix
We wrote the equilibrium as:
Hence the local stiffness matrix is:
We must turn to the global stiffness matrix to assemble:
where Lc is a location matrix, which allows to transform a
local stiffness matrix kc in global coordinates.
Hence, we get the algebraic equivalent of Navier’s differential
equation of elasto-static.

70. 70
Sample problem: plane stress analysis of a wall
Wall mesh into
quadrilateral cells
Choosing a certain mesh
will influence the accuracy
of results
=>
thicker is the mesh, more
accurate will be the result
The results obtained will be compared with ones from Finite Element
Method (FEM).
Restrained nodes
Applied nodal loadsLet’s consider a practical case:

71. 71
Sample problem: plane stress analysis of a wall
displacement contour
and deformed shape
The whole procedure has been
implemented in a Matlab code,
leading the depicted result.
Boundary conditions were applied
by reducing the global (assembled)
stiffness matrix. Bounded vertices
are obtained by removing the
associated rows and columns.
Result are compared with FEM
FEM (Abaqus): stress Von Mises stresses and deformed shape

72. 72
Sample problem:
plane stress analysis of a wall

73. 73
Sample problem: plane stress analysis of a wall
Results completely agree with FEM approaches with linear shape functions.
For triangular mesh, the result completely
agrees with FEM:
Method
X displacement Y displacement Diff. in X Diff. in Y
[mm] [mm] % %
CEM tria 0.032202701 -0.015655325 0.000% 0.000%
FEM CPS3 0.0322027 -0.0156553 0.000% 0.000%
Method
X displacement Y displacement Diff. in X Diff. in Y
[mm] [mm] % %
CEM quad 0.025277997 -0.012230291 0.000% 0.000%
FEM CPS4R 0.0270812 -0.0134509 7.133% 9.980%
FEM CPS4 0.0255565 -0.0124236 1.102% 1.581%
FEM CPS8 0.026532 -0.0130891 4.961% 7.022%
For quadrilateral meshes, the result presents a
good agreement with FEM, even if the
presented formulation was made for regualar
quadrilateral cells.
CPS4R are reduced integration 4 nodes plane stress
elements,
CPS4 are 4 nodes plane stress elements,
CPS8 are 8 nodes plane stress elements.

74. 74
Sample problem: 3D analysis of a wall
With the same procedure, a regular hexahedral cell has been implemented.
Displacement interpolation functions:
The same load and
boundary conditions as
before has been
applied.
The Matlab code has
been adjusted to
manage exahedral
elements with the
shape functions above.

75. 75
Sample problem: 3D analysis of a wall
Method
X displacement Y displacement Z displacement Diff. in X Diff. in Y
Diff. in
Z
[mm] [mm] [mm] % % %
FEM C3D8 0.002571950 0.000001296 -0.001254150
CEM esa 0.002525352 0.000001258 -0.001221776 1.85% 2.99% 2.65%
A regular hexahedral cell has been implemented.
Results and comparisons with FEM are reported below.

76. ...
Useful hints
76

77. 77
 Planar and solid mesh can be obtained using free tools like:
NETGEN: https://ngsolve.org/showcases/netgen
TetGen: http://wias-berlin.de/software/index.jsp?id=TetGen&lang=1
 In solid mechanics, using Cell Method you obtain the same stiffness of Finite
Elements approach. Assembly of matrices and solving can be conducted in the
same way as FEM, using optimized algorithm.
 The presented approach is able to reproduce the behaviour of linear elements in
FEM. Cell Method can be enriched by adding Gauss points to improve accuracy
 Cell Method is particularly suitable for teaching and for numerical
implementations, see for instance:
Usefulhints
https://www.researchgate.net/publication/224157318_Multiphysics_Problems_via_the_Cell_Method_
The_Role_of_Tonti_Diagrams

78. ...
Conclusions
78

79. the algebraic formulation applies to:
 regions of any shape, with holes, tips, cracks, grooves, etc.
 regions containing different materials
 anisotropic materials
 nonlinear materials
 sintered materials
 it treats naturally concentrated sources
 does not present infinite
 Convergence orders of order higher than the second
 it is applied with simplicity.
...
Characteristicsofalgebraicformulation
79

80. this lecture
...
Letusthereforeavoidauselesspassage!
differential
equations
physical
field
approximate
solution
algebraic
equations
80

81. tontienzo@gmail.com
giovanni.rinaldin@gmail.com
You can find related publications
to algebraic formulation
and the cell method
in the web site:
www.discretephysics.org
81
Authors’ email addresses: