9.7 REVIEW QUESTIONS
1.
Can 3D solid elements be used for solving 2D plane stress and plane strain
problems? Give justification to your answer.
Suggested answer (Hint: Section 9.1): Yes, the 3D solid elements can be used for
solving 2D plane stress and plain strain problems since the 2D problems are
basically special cases of the 3D problem. It is the special characteristics of a
particular 3D problem that allows us to simplify the geometry to a 2D (or 1D) one.
For example, if the structure has a small thickness as compared to the other lateral
dimensions, then it can be simplified into a 2D, plane stress problem. Of course,
3D elements can also be used to model such a problem taking into account the
whole 3D structure (including the thickness direction). Nevertheless, the efforts
in geometry creation and meshing would no doubt be increased.
2.
What is the difference between using tetrahedron elements and hexahedron
elements derived using assemblage of tetrahedron elements? Can they give the
same results for the same problem? Give justification to your answer.
Suggested answer: Basically there is no difference between using tetrahedron
elements and hexahedron elements derived using the assembly of tetrahedron
elements. This is so because, the shape functions used would be the same as that
of the tetrahedron elements and the total number of nodes will also be the same.
The assembly is done in the same way when we assemble elements sharing the
same node(s), therefore they basically give the same results. If the hexahedron is
derived using the tri-linear shape function instead, then the results might be
different because the tetrahedron elements actually have a constant strain matrix,
which means that it is less accurate generally than the hexahedron element
derived using the tri-linear shape functions.
3.
Can one develop pentahedron elements? How?
Suggested answer: A pentahedron element has 5 faces and takes a form similar to
a pyramid as shown.
To formulate equations for such an element, one can always begin from first
principles by constructing the shape functions from a polynomial function (See
Chapter 3 and example in Chapter 4). Nevertheless, another simpler way of
formulating such an element is by using the assembly of two tetrahedron elements.
5
5
4
4
3
1
3
5
4
2
2
1
2
It should be noted that there is more than one way of splitting a pentahedron into
two tetrahedron. The assembly is done in a way similar to the assembly between
elements.
4.
How many Gauss points should be used for evaluating mass and stiffness matrices
for 4-node tetrahedron elements? Give justification to your answer.
Suggested answer: The stiffness matrix of the linear tetrahedron element contains
the integration of a constant matrix (since the strain matrix is constant) as shown
in Eq. (9.19). Therefore, there is no need to use the gauss quadrature to perform
the integration. But if the gauss quadrature is to be used, one can use one gauss
point (point a in the diagram below) in the center of the tetrahedron to perform the
integration.
The mass matrix consists of the integrand NTN, which is a product of two linear
functions making the integrand a quadratic polynomial (assuming the density is
constant). The integration can actually be easily evaluated using the mathematical
formula of Eisenberg and Malvern (Eq. (9.22)). If one is to use the Gauss
quadrature, then 4 gauss points (b1, b2, b3, b4) as shown can be used to obtain the
solution of the quadratic integrand exactly. The values of the weight coefficients
and the position of the gauss points (in terms of volume coordinates – see Section
9.2.l) are shown in the table below. Note that a mapping of the physical
coordinates (x, y, z) to the volume coordinates is normally carried out so that the
gauss integration can be applied easily.
4
b4 
b1
1

b
a
b2 
2
3
3
m
1
4
5.
Table 4-1 Gauss points and weight coefficients for tetrahedron element
Position of gauss points in terms of
Gauss
Accuracy,
volume coordinates, Li
wj
point(s)
n
L1
L2
L3
L4
a
0.25
0.25
0.25
0.25
1
1
b1
0.5854
0.1382
0.1382
0.1382
0.25
b2
0.1382
0.5854
0.1382
0.1382
0.25
3
b3
0.1382
0.1382
0.5854
0.1382
0.25
b4
0.1382
0.1382
0.1382
0.5854
0.25
How many Gauss points should be used for evaluating mass and stiffness matrices
for 8-node hexahedron elements? Give justification to your answer.
Suggested answer: The stiffness matrix and the mass matrix consist of integrands
containing the determinate of the Jacobian. And the strain matrix, B, in the
stiffness matrix is also made up of the Jacobian. Since the components in the
Jacobian matrix are fraction functions, this makes the integrands fraction
functions too. The Gauss quadrature is meant for polynomial functions and hence
is not able to provide an exact solution for fraction functions. Nevertheless, 2 x 2
x 2 gauss points can be used to give an approximate solution.
6.
If higher order shape function is used, do Eqs. (9.30) and (9.63) still hold? Give
justification to your answer.
Suggested answer: If higher order shape function is used, Eq. (9.30) will still
hold but bear in mind that the shape function defined in terms of the natural
coordinate system in Eq. (9.27) is not valid.
Eq. (9.63) will not be valid in the sense that when a higher order element is used,
there will firstly be more nodes and hence the dimension of the force vector will
be different. For example, for a 20-node hexahedron element, the full force
vector will comprise of a 60 (20 nodes x 3 dofs/node) by 1 column vector. And
for higher order elements, the distribution of the uniform distributed load on the
edge may not be such that it’s equally distributed to the number of nodes on that
edge. Note that in Eq. (9.63), the load is distributed equally to the two nodes of
the edge.