A Finite Volume Method for Solids with a Rotational Degree of Freedom Based on the
6-Node Triangle
W. Pan*¹, M.A. Wheel¹ and Y. Qin2
1. Department of Mechanical Engineering, University of Strathclyde, Glasgow, UK.
2. Department of Design, Manufacture Engineering and Management, University of Strathclyde,
Glasgow, UK.
* corresponding author, email: wenke.pan@strath.ac.uk
Abstract: A finite-volume (FV) cell vertex method is presented for determining the
displacement field for solids exhibiting with incompressibility. The solid is discretized
into six-node finite elements and the standard six-node finite-element shape function is
employed for each element. Only control volumes around vertex node of the triangular
element are considered. For considering the material incompressibility, a constant
hydrostatic pressure as an extra unknown variable within each element is assumed. The
force equilibrium in two perpendicular directions and one in-plane moment equilibrium
equation are derived for each control volume. The volume conservation is satisfied by
setting the integration of volumetric strain as zero within each element. By solving the
system control equations, the displacements and rotations of the vertex nodes and the
hydrostatic pressure for each element can be obtained and then the displacements of the
midside nodes can be calculated. The simulation results show that this FV method
passes the patch tests and converges to theoretical results under mesh refinement for
material behaviour incompressibility.
Keywords: Finite Volume Method, Control volume, Vertex centred method, Rotational
degree, Incompressibility.
1
1. Introduction
Finite element (FE) method has dominated solid and structure analysis for many years,
while FV method has been successfully used for fluid dynamic analysis. Over the last
fifteen years, some progresses have been made for employing FV method for solids and
structures. Apart from the application of FV method to elastic [1-6], elastic-plastic [7],
and plate analysis [8,9] for compressible materials. FV method has also been applied to
large deformation compressible and near-incompressible hyperelastic material
analysis[10,11]. For incompressible material deformation analysis, a three-node and
four-node element has been used for analysis 2D elastic media by using the cell central
FV method [12], a bi-linear displacement field was assumed within the quadrilateral
constructed by the connection of cell central and adjoining edge nodes, and also the
constant hydrostatic pressure within each element. Mixed form FV method was also
presented for 2D and 3D and also fracture problems with incompressible linear elastic
materials [13], the non-physical stresses oscillations near crack tip was eliminated. FV
method that included the Allman degree of rotation for 3-node[14] and 6-node FV [15]
analysis for 2D elastic problems were presented, some advantages have been
demonstrated for some cases. In this paper, a mixed form FV cell vertex method is
presented for determining the displacement field for solids exhibiting incompressibility.
The similar discretization method with linear strain distribution six-node triangle
element as [15] is used. To include the rotation degree of freedom, the midside
displacements are replaced by its neighbouring vertex nodes[16], and the constant
hydro-static pressure within each element is assumed although there is no difficulty for
the assumption of a linear distribution of hydrostatic pressure within each element. The
2
force equilibrium in two perpendicular directions and one in-plane moment equilibrium
equation are derived for each control volume. The volume conservation is satisfied by
setting the integration of volumetric strain as zero within each element for small strain
elastic problem. The total number of the force and moment equilibrium equations plus
the number of volume conservation equations is the same as the number of the total
unknown variables due to the special construction of the CV. By solving these
equations, the displacements and rotations of the vertex nodes and the hydrostatic
pressure for each element can be obtained and then the displacements of the middle-side
nodes can be calculated. In section 2, formulation for displacement, strain and stress
filed will be given, and also the compensation equation related with material
incompressibility. The construction of special CV and the system equations are given in
section 3, and the test problems will be followed in section 4, and the final section is the
conclusions.
2. Formulation for displacement, strain and stress fields
The six-node triangle element is considered for the discretization of the 2D
incompressible material problem. The same formulation for displacement and strain
field within each element as [15] are used in this paper and are summarized as
following:
2.1 Displacement fields
The displacement components, u and v, within a six-node triangular element are:
u ( x, y ) a 

   1 x
v( x, y )  b

y
x2
xy
y2

T
(1)
3
Where a  a1
a2 ... a6  and b  b1 b2 ... b6  are coefficients. By applying
T
T
equations (1) at every node of the triangle, gives:
a  A 1
 
b  0
0 
u
A 1 
(2)
Where u  u1 ... u6
v1 ... v6  , and ui, vi, are displacement components of
T
vertex nodes (for i=1,2,3) and midside nodes (for i=4,5,6) along x and y direction
respectively.
For the introduction of rotation degree of freedom of each vertex nodes, the midside
node displacement will be replaced by the neighbouring vertex node displacements and
rotations. The relationship can be repressed by the transformation matrix T,
u1
... u6
v1 ... v6   Tu1 u 2
T
u3
v1 v2
 
v3 1  2  3   T u uv
T
(3)
Where i, (i=1, 2, 3) are rotations of the vertex nodes of 6-node triangle, and matrix T
can be found in [16].
With the use of relationship between strain and displacement, the direct strain
components  xx ,  yy and shear strain  xy are
 xx  B x
ε   yy    0
  B
 xy   y
B x A 1
0
 a  
B y     0
b
B x    B y A 1
0 

B y A 1 T u uv
B x A 1 
 
(4)
Where A is a constant matrix, the components of vector B x and B y are constant or
linear function of coordinators x or y, and A, B x and B y can be found in [15].
By using the constitutive relationship between stresses and strain and the hydrostatic
pressure , the stress components are
4
 xx  2G 0 0 
σ   yy    0 2G 0 ε 
   0
0 G 
 xy  
1
 
p 1
0
 
(5)
where G is the shear modulus, p is hydrostatic pressure. Put (4) into (5), gives
2GB x
σ   0
 GB y

0 
1
 1
 
uv
2GB y  A T u
 p 1
0
GB x 
 
   
(6)
3. Control volume and the equilibrium equations
3.1 Control volume for cell vertex FV procedure
One difference between FE and FV method is that equilibrium equations must be built
on each individual CV for FV, under the cell centre FV procedure, the CV is the same
as FE element, while for cell vertex FV procedure, the CV is completely different from
FE element. There are some ways to build CV for cell vertex FV procedure.
One method to construct CV for cell vertex FV procedure is to connect the middle point
between neighbouring nodes and the centre of the element to form a multi-side complex
shape CV. Other method was presented in [17], some position of the points needed to be
connected to construct CV was obtained by trial and error method. In this paper, due to
the introduction of rotational degree of freedom, special CV as used in [15] was
adopted. Figure 1(a) shows the typical CVs (constructed by dotted lines) around a
vertex node A and B inside and on the boundary of the model respectively.
Assuming the total number of vertex node and element are ‘nd’ and ‘ne’ respectively,
thus the total numer of the system unknown variables will be ‘3*nd+ne’. There are two
force equilibrium equations along perpendicular directions and one moment equilibrium
equation for each CV, so the total number of equation related with forces and moment
are ‘3*nd’, and it is less than the total number of unknown variables. Therefore, the
5
compensation equation has to be built. The incompressibility condition for each element
can be used for this purpose.

 D
xx


C
y


B
A

P

xx

S


yx
D
xy

O
(a)
C
x
(b)
Figure 1 Typical CVs and inner stresses. Solid circle, open circle and small triangle
represent vertex nodes, midside nodes and geometrical centre of FE element. (a) CVs
around the inside vertex node A and boundary vertex node B (b) inner stresses on one
of the segments of the profile of CV.
3.2 Force and moment equilibrium equations
To obtain the force and moment equilibrium equations for each CV, the resultant forces
and moment have to be integrated along the profile of each CV. From formula (5), it can
be seen that stress components within each CV is linearly distributed, so the resultant
forces and moment can be integrated analytically along the profile of each CV. For one
typical segment of the CV shown in Figure 1(b), the coordinators of any point along the
segment can be expressed as
6
 x P  xC  S sin 

 y P  yC  S cos 
(7)
where the angle between the normal of line CD and the x-axis is  and the distance
from point C to P is S.
By integrating explicitly the stresses components along the profile of each CV, the
resultant force components Fx and Fy acting in x and y directions are.
   
FAxCD  G(2 cos   B x ds  sin   B y ds) G sin   B x ds  A 1 T u uv  S CD cos  * p


SCD
SCD
SCD
(8.1)
   
FAyCD  G cos   B y ds G(cos   B x ds  2 sin   B y ds) A 1 T u uv  S CD sin  * p


SCD
SCD
SCD
(8.2)
The resultant moment about the origin of the coordinate system is:
   
CD
M AO
  G (2 cos   B x yds  sin   B y yds ) G  B x yds  A 1 T u uv


S CD
S CD
S CD
 G (cos   B y xdsA 1  2 sin   B y xds) G sin   B x xds A 1 T u uv


S CD
S CD
S CD
   
 ( cos  
S CD
(8.3)
yds  sin   xds) * p
S CD
By summarizing all the resultants force and moment along the profile of each CV, the
equilibrium equation for each CV can be written as following:
K CV
xu
 CV
K yu
K CV
 Mu
K
K
K
CV
xv
CV
yv
CV
Mv
K
K
K
CV
x
CV
y
CV
M
K
K
K
CV
xp
CV
yp
CV
Mp
u CV 
  CV    FxCV 
 v  
CV 
 θ CV     Fy 
CV 
 

 p CV   M O 


(9)
Where FxCV , FyCV and M CV
O are external forces and moment acting on the CV.
3.3 Compensation equation for incompressibility
For linear elastic material, the incompressibility of each element can be expressed as
7
 (
V
xx
  yy )dV  0
(10)
For simplicity, the strain components at the centre of the triangular element were used
for above formula (10), therefore,
B
x
   
( x  xc , y  yc ) B y ( x  xc , y  yc ) A 1 T u uv  0 (11)
Each element has one compensation equation and one unknown hydrostatic pressure
variable.
3.4 system equations
Combining formulation (9) and (11), the system equation can be assembled as
following:
 K xu
K
 yu
K Mu

 K pu
K xv
K yv
K Mv
K x
K y
K M
K pv
K p
K xp  u    FxE 
K yp   v    FyE 
 

K Mp  θ   M OE 

0  p   0 
(12)
where
u
v θ p  u1 ... u nd
T
F
E
X
FYE
M OE
  F
0
T
E
x1
v1 ... vnd
E
... Fxnd
FyE1
1 ...  nd
E
... Fynp
p1 ... pne 
E
M OE1 ... M Onp
T

0 ... 0
T
3.5. Boundary conditions
For concentrated point force or moment acting on each CV, they can be directly put into
the right-hand side of equations (12). For distributed stresses on boundaries, resultant
forces alone x and y direction and moment to the origin of coordinator can be easily
integrated along the profile of CV, then the same procedure as concentrated forces and
moments can be used. The stiffness matrix of the whole structure shown in equations
8
(12) is a singular matrix, displacement and rotation boundary conditions have to be
applied to the equations, the standard procedure as the treatment of displacement
boundary conditions for FE can be used. That is, let the diagonal element of the stiffness
matrix corresponding to the ith constraint as one, while the others in the same row as
zero, also let the ith element of the right-hand side vector as the exact value of the
constraint. By solving the modified Eqn. (12), the vertex displacements, rotations and
hydrostatic pressure can be obtained, after which, using Eqn. (3), the midside node
displacements can be calculated. From Eqns. (4) and (6), the strain and stress field can
be determined subsequently.
4. Numerical test results
Four 2D plane strain benchmark problems with incompressible material behaviour were
considered. The first and the second examples are isotropic plate under simple tension
and simple shear, the third one is case of a square plate subject to pure bending. The
fourth case is an infinite long tube subjecte to internal pressure. The value of material’s
shear modulus, G, is chosen as 1.0GPa. For each case, the calculated results were
compared with that of theoretical results.
4.1 Test 1 and 2: simple tension and shear of a square plate
An infinitely long square plate with the length and width of 2m*2m subjected to simple
tension stress 1.0MPa is shown by the solid lines in Figure 2. A quarter model is
analysed and the model is discretized into four different size 6-node triangle elements.
The dotted lines in Figure 2 show the expected deformed mesh. The calculated results
show that the constant strain field is exactly predicted. Figure 3 shows the initial and the
deformed mesh of an infinitely long square plate with length and width of 1m*1m
9
subject to simple shear. As the simple tension case, the solid and dotted lines represent
initial and deformed mesh. From this figure, the exact deformation shape is obtained.
xigure 2 deformed and undeformed
mesh of a plate subjected to simple
tension, magnify factor 1e5
Figure 3 deformed and undeformed
mesh of a plate subjected to simple
shear, magnify factor 2e4
4.2 Test problem 3: Square plate subjected to pure bending
The third test example, shown in Figure 4 (a half model), is an isotropic square plate
with length 2.0m subject to pure bending loading and the linearly varying edge normal
stress is given by
 x ( x  L / 2) 
2y
0
L
(13)
Theoretical results of the displacement components are as following


ux  2 0

E

2

u y   0 x

E
xy
L
 y 2
L
(14)
where Young’s modulus E=2G(1+) is and =0.5.
y
 0  1.0MPa
A
C
.
B
10
Figure 4 A thick square plate subject to pure bending
4
8
12
16
20
24
28
32
Nx
Vertical disp., m
-2.0E-06
-2.2E-06
Calculated V
Theoretical results
-2.4E-06
-2.6E-06
-2.8E-06
Top right corner V displacement varies with Nx
Figure 5 Top right corner vertical displacement changes with the number of division
along x direction
Horizontal disp.,m
2.6E-06
Nx
2.2E-06
Calculated U
Theoretical results
1.8E-06
1.4E-06
4
8
12
16
20
24
28
32
Top right corner U displacement varies with Nx
Figure 6 Top right corner horizontal displacement changes with the number of division
along x direction
The half plate model is tested using a consistent refinement of a regular mesh with the
same number of division along x and y directions. The boundary conditions are: The
horizontal displacements and in-plane rotations along left edge are zeros; the vertical
displacement at the middle point of the left edge is zero. Figure 5 shows the comparison
between the calculated and theoretical results of vertical displacement of top right
corner under mesh refinement. From this figure, it can be seen that the calculated results
11
converge to the theoretical results. Figure 6 shows the results of the horizontal
displacement at the top right corner, similar trend as Figure 5 can be observed although
the horizontal displacement converges slower than that of vertical displacement.
4.3 Test problem 4: Inflation of an infinitely long tube under internal pressure
Figure 7(a) is a schematic drawing of an infinitely long tube(a quarter model) subjected
to internal pressure, the tube material is linear incomprensible material with shear
modulus 100.0GPa. The analytical solution of radial displacement is
ur 
1 pi ri 2 ro2
2G ro  ri r
(15)
D
C
B
A
(a)
(b)
Figure 7 Infinitely long tube subjected to internal pressure. (a) a quarter model with
loading and boundary shown, (b) the undeformed (solid line) and deformed
mesh(dashed line), solid and empty circle represent nodes attached to undeformed and
deformed mesh respectively, the number of division along the radial and hoop
directions are all 4, displacement amplification factor is6e4.
12
Radial displacement, m
Relative error
20
15
relative error
10
5
0
0
10
20
30
40
Number of divisions along radial and
hoop directions
8.E-07
theoretical results
calculated results
6.E-07
4.E-07
2.E-07
0.08
0.12
0.16
0.20
0.24
Radial distance of the points to the
centre of the tube
Figure 8 The relative error varies with the number of division along radial direction
Figure 9 The comparison between numerical and analytical results
The tube inner and outside radius are 0.1m and 0.4m respectively, the internal pressure
is 1.0MPa. The symmetry boundary conditions were applied for the line AB and CD,
and the rotations along these two lines are all constrained. Figure 7(b) shows a the
undeformed and deformed mesh of a special case with the number of division along
radial and hoop direction are all 4. From this figure, the axi-symmetrical deformation is
observed. Figure 8 shows the relative error of the inner born radial displacement
against the number of divisions along the radial direction. With the mesh refinement,
the relative error becomes smaller and smaller. In Figure 9, with the division along
radial and hoop direction of 32, the comparison between the numerical and analytical
results of radial displacement profile along boundary AB is presented. It can be clearly
seen that the presented method can give good results.
5. Conclusions
13
Six-node triangle element with high order displacement field was used for cell vertex
FV procedure to tackle the 2D problem with material behaviour incompressibility.
Constant hydrostatic pressure within each element was assumed. With special
construction of CV and the replacement of midside node displacement by neighbouring
edge node displacement and rotations, the total number of unknown variables equals to
the total number of force and moment equilibrium equations. The benchmark test
examples shows that the presented FV method can predict the constant strain field
exactly, and can also predict the linear and more complex strain field with the mesh
refinement
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Figure captions
Table captions
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