7.9 REVIEW QUESTIONS
1.
Show that the stiffness matrix of an isotropic linear triangular element whose
thickness varies linearly in the element is
k e  hAe BT cB
where B is the strain matrix, c is matrix of material constants, Ae is the area of the
triangle and h the average thickness (h1+h2+ h3)/3, where h1, h2 and h3 are the
nodal thickness at the node.
Suggested answer: In this question, the thickness of the element is not uniform
throughout the element, but rather varies in a linear way in the 2D plane of the
element. As such, we can interpolate the thickness at any point in the element
from the thickness at the nodal positions by using the properties of the linear
shape functions. Recall that one of the properties of a shape function is that it is
unity at its home node and zero at other remote nodes. This is the delta function
property of a shape function. Therefore, the usage of the shape function as an
interpolation function is valid too, like that used for the displacements. Thus, the
thickness in the element can be expressed as
h( x, y )  N1h1  N 2 h2  N3h3   N1
 h1 
 
N3  h2 
h 
 3
N2
(1-1)
where Ni and hi are the shape function and nodal thickness for the node i
respectively.
From Eq. (7.39), the element stiffness matrix is rewritten here as
h
k e   BT cBdV   (  dz )BT cBdA   hBT cBdA
0
Ve
Ae
(1-2)
Ae
Substitute the expression for h in Eq. (1-1) into the above equation and knowing
that BTcB is a constant matrix since both B and c are constant matrices, the
element stiffness matrix becomes
k e  BT cB  N1h1  N 2 h2  N 3h3 dA
Ae
(1-3)
The shape functions can be expressed as the area coordinates expressed in Eqs.
(7.25), (7.27) and (7.28) since the area coordinates have the property of partitions
of unity as well as the delta function property. Using the mathematical formula in
Eq. (7.43) and rewritten here:
m !n ! p !
m n p
(1-4)
A L1 L2 L3 dA  (m  n  p  2)! 2 A
the integral in Eq. (1-3) above can be easily evaluated. For example,

Ae
N1h1 dA 
Ah
1!0!0!
2 Ae h1  e 1
(1  0  0  2)!
3
(1-5)
Therefore,
 Ah Ah Ah 
k e  BT cB  e 1  e 2  e 3 
3
3 
 3
 hAe BT cB
2.
(1-6)
Show that the mass matrix of a linear triangular element whose thickness varies
linearly within the plane of the element is
0
6  3
6  41

6  41
0

6  4 2
 hAe 
[m]e 

60 

sy.


0
6 2
6  3
0
0
6  1
6  4 2
0
6  4 3

6   2 
0 

6  1 
0 

6  4 3 
0
where  is the density, Ae is the area, h is the mean thickness and i=hi/ h with
i=1,2,3 for the three nodes.
Suggested answer: For an element with linearly varying thickness, the thickness
can be interpolated from the thickness at the nodes by using the shape functions
like in question 1. Therefore,
h( x, y )  N1h1  N 2 h2  N3h3   N1
N2
 h1 
 
N3  h2 
h 
 3
(2-1)
where Ni and hi are the shape function and nodal thickness for the node I
respectively.
From Eq. (7.41), the element mass matrix is rewritten here as
m e   NT NdV 
Ve

h
0
Ae
dx NT NdA   hNT NdA
(2-2)
Ae
Since the thickness h is not a constant here, it cannot be left out of the integration
and we can rewrite the above equation as
me   
Ae
 hN1 N1
 0

 hN 2 N1

 0
 hN3 N1

 0
0
hN1 N 2
0
hN1 N3
hN1 N1
0
hN1 N 2
0
0
hN 2 N 2
0
hN 2 N 3
hN 2 N1
0
hN 2 N 2
0
0
hN3 N 2
0
hN3 N 3
hN3 N1
0
hN3 N 2
0

hN1 N3 
0 
 dA
hN 2 N 3 
0 

hN3 N 3 
0
Using the mathematical formula in Eq. (7.43) and rewritten here:
m !n ! p !
m n p
A L1 L2 L3 dA  (m  n  p  2)! 2 A
(2-3)
(2-4)
the integrals above can be evaluated easily. For example for the first term in the
matrix,
  hN1 N1dA     N1h1  N 2 h2  N 3h3 N1 N1dA
Ae
Ae
   N13 h1  N12 N 2 h2  N12 N 3h3dA
Ae
h 
h h
  Ae  1  2  3 
 10 30 30 
 Ae

 6h1  2h2  2h3 
60
 Ae
 4h1  2  h1  h2  h3  

60 
 Ae
 4h1  6h 

60 
 hAe

 41  6
60
Similar integration can be carried out for the other terms in the matrix to obtain
0
6  3
6  41

6  41
0

6  4 2
 hAe 
me 

60 

sy.


3.
0
6  2
6  3
0
0
6  1
6  4 2
0
6  4 3

6   2 
0 

6  1 
0 

6  4 3 
0
The thickness variation of a linear rectangular element is given by
4
h( , )   N i ( , )hi
i 1
(2-5)
where Ni is the bi-linear shape function, and hi is the thickness value at node i.
How many Gauss points are required to evaluate exactly the mass and stiffness
matrices.
Suggested answer [Hint: Section 7.3.4]: For the stiffness matrix as shown in Eq.
(7.57), the integrand consists of hBTcB. The strain matrix B is a linear function of
 and . The thickness, h, is expressed as an interpolation from the nodal
thickness by using the bi-linear shape functions (i.e. they are linear in each
direction). Therefore, sampled in each direction, the integrand is a product of
three linear functions and hence produces a cubic function. As such, from Table
7.1, it can be seen that with 2 gauss points in each direction, it is sufficient to
evaluate the stiffness matrix, which has a polynomial order of 3. Hence, for a
rectangular element, a total of 4 gauss points (2 in each direction) is required.
For the mass matrix as shown in Eq. (7.60), the integrand consists of hNTN and
this implies that for bilinear shape functions, it is a product of three linear
functions in each direction. Thereby forming a cubic polynomial function in each
direction and can thus be evaluated exactly using 2 gauss points in each direction.
Hence, for a rectangular element, a total of 4 gauss points (2 in each direction) is
required to evaluate the mass matrix.
4.
If the thickness variation of a linear quadrilateral is the same as the rectangular
element in Problem 3), how many Gauss points are required to evaluate the mass
matrix exactly? How many Gauss points are required to integrate the volume of
the element exactly? How many Gauss points are required to integrate the
stiffness matrix exactly?
Suggested answer: The mass matrix for a linear quadrilateral element is given by
Eq. (7.81) and rewritten here as
h
m e   NT NdV    dx NT NdA   h NT NdA
0
V
A

1
1

1
1
A
(4-1)
h N Ndet J d d
T
Since the thickness variation is linear as in problem no. 3, it can be written as
h( x, y )  N1h1  N 2 h2  N 3h3  N 4 h4   N1
N2
N3
 h1 
h 
 
N4   2 
 h3 
h4 
(4-2)
The thickness is a linear functions of  and . The terms in NTN are of order 2,
and the determinate of the Jacobian is a bilinear function. Therefore, the total
order of the integrand is 4. We can use 3 by 3 Gauss points to obtain the exact
solution.
The volume of the element can be expressed as
Ve   h dA  
1
1
Ae

1
1
h det J d d
(4-3)
The thickness is a linear functions of  and , and the determinate of the Jacobian
is also a bilinear function. Therefore, the total order of the integrand is 2. We can
use 2 by 2 Gauss points to obtain the exact solution.
The stiffness matrix is given in Eq. (7.80) and rewritten here as
ke  
1
1

1
hBTcBdet J d d
1
(4-4)
Because B involves differentiating and then dividing by the Jacobian matrix (see,
Eq. (7.78)), the stiffness matrix may not be able to be evaluated exactly using the
Gauss integration scheme. We can use 2 by 2 Gauss points to obtain an
approximated solution.
5.
Construct the shape functions for the 12-node rectangular element for one corner
node and two side node.

4
10
9
3
11
8
12
7
1
2
6
5

Suggested answer:
1-3 = 0
1+ = 0

4
1
10
9
1- = 0
3
11
8
12
7
5
6

2
1- = 0
To obtain the shape function for a side node, say node 5, we enforce the shape
function to pass through the following four lines as shown in the figure above.
1- = 0  vanishes at nodes 2, 7, 8, 3
1+ = 0  vanishes at nodes 1, 12, 11, 4
1 = 0  vanishes at nodes 3, 9, 10, 4
13 = 0  vanishes at nodes 6, 9
The shape function for node 5 can then be written as
N5  c(1   )(1   )(1   )(1  3 )
 c(1   2 )(1   )(1  3 )
Since at node 5 (=1/3; =1), N5 = 1, therefore,
1
9
c

2
(1  (1/ 3) )(1  (1))(1  3(1/ 3)) 32
Hence, the shape function for node 5 is
9
N 5  (1   2 )(1   )(1  3 )
32
1- = 0

4
10
9
8
11

7
12
92 + 92 –10 =0
1- = 0
3
R=(10/9)
1
5
6
2
To obtain the shape function for a corner node, say node 1, we enforce the shape
function to pass through the following four lines as shown in the figure above.
1 = 0  vanishes at nodes 2, 7, 8, 3
1 = 0  vanishes at nodes 3, 9, 10, 4
92 + 9 2- 10 = 0  vanishes at nodes 6, 5, 12, 11
The shape function for node 1 can then be written as
N1  c(1   )(1   )(9 2  9 2  10)
Since at node 1 (=1; =1), N1 = 1, therefore,
1
1
c

2
2
(1  (1))(1  (1))(9(1)  9(1)  10) 32
Hence, the shape function for node 5 is
1
N1 
(1   )(1   )(9 2  9 2  10)
32