Ebook A first course in the finite element method (4th edition) Part 2

CHAPTER

8

Development
of the Linear-Strain
Triangle Equations

Introduction
In this chapter, we consider the development of the stiffness matrix and equations for
a higher-order triangular element, called the linear-strain triangle (LST). This element
is available in many commercial computer programs and has some advantages over
the constant-strain triangle described in Chapter 6.
The LST element has six nodes and twelve unknown displacement degrees of
freedom. The displacement functions for the element are quadratic instead of linear
(as in the CST).
The procedures for development of the equations for the LST element follow the
same steps as those used in Chapter 6 for the CST element. However, the number of
equations now becomes twelve instead of six, making a longhand solution extremely
cumbersome. Hence, we will use a computer to perform many of the mathematical
operations.
After deriving the element equations, we will compare results from problems
solved using the LST element with those solved using the CST element. The introduction of the higher-order LST element will illustrate the possible advantages of higherorder elements and should enhance your general understanding of the concepts
involved with finite element procedures.

d

8.1 Derivation of the Linear-Strain
Triangular Element Stiffness Matrix
and Equations

d

We will now derive the LST stiffness matrix and element equations. The steps used
here are identical to those used for the CST element, and much of the notation is the
same.
398


8.1 Derivation of the Linear-Strain Triangular Element Stiffness Matrix and Equations

d

399

Step 1 Select Element Type
Consider the triangular element shown in Figure 8–1 with the usual end nodes and
three additional nodes conveniently located at the midpoints of the sides. Thus, a
computer program can automatically compute the midpoint coordinates once the
coordinates of the corner nodes are given as input.

Figure 8–1 Basic six-node triangular element showing degrees of freedom

The unknown nodal displacements are now given by
8 9
u1 >
>
>
>
>
>

>
>
>
v1 >
>
>
>
>
>
>
>
>
>
u2 >
>
>
>
8 9 >
>
> >
>
d
v
>
>
1
2
>
>
>

>
>
>
>
>
>
>
>
>
>
>
>
>
d
u
>
>
>
>
2
3
>
>
>
>
>
>
>
=

3
3
¼
fdg ¼
>
d4 >
>
> >
> u4 >
>
>
>
> >
> >
> >
>
>
>
>
d
v
>
>
>
>
5
4
>
>
; >

> >
: >
>
>
>
d6
u5 >
>
>
>
>
>
>
>
>
>
>
v
>
>
5
>
>
>
>
>u >
>
>
6
>

>
>
;
: >
v6

ð8:1:1Þ

Step 2 Select a Displacement Function
We now select a quadratic displacement function in each element as
uðx; yÞ ¼ a1 þ a2 x þ a3 y þ a4 x 2 þ a5 xy þ a6 y 2
vðx; yÞ ¼ a7 þ a8 x þ a9 y þ a10 x 2 þ a11 xy þ a12 y 2

ð8:1:2Þ

Again, the number of coefficients ai ð12Þ equals the total number of degrees of freedom
for the element. The displacement compatibility among adjoining elements is satisfied
because three nodes are located along each side and a parabola is defined by three
points on its path. Since adjacent elements are connected at common nodes, their displacement compatibility across the boundaries will be maintained.
In general, when considering triangular elements, we can use a complete polynomial in Cartesian coordinates to describe the displacement field within an element.


400

d

8 Development of the Linear-Strain Triangle Equations

Figure 8–2 Relation between type of plane triangular element and polynomial
coefficients based on a Pascal triangle

Using internal nodes as necessary for the higher-order cubic and quartic elements, we
use all terms of a truncated Pascal triangle in the displacement field or, equivalently,
the shape functions, as shown by Figure 8–2; that is, a complete linear function is
used for the CST element considered previously in Chapter 6. The complete quadratic
function is used for the LST of this chapter. The complete cubic function is used for
the quadratic-strain triangle (QST), with an internal node necessary as the tenth node.
The general displacement functions, Eqs. (8.1.2), expressed in matrix form are
now
9
8
a1 >
>
>
>
>
& '
!>
u
0 0 < a2 =
1 x y x 2 xy y 2 0 0 0 0
ð8:1:3Þ
fcg ¼
¼
.
v
0 0 0 0
0 0 1 x y x 2 xy y 2 >
>
> .. >

>
>
;
:
a12
Alternatively, we can express Eq. (8.1.3) as
fcg ¼ ½M à Šfag

ð8:1:4Þ

where ½M à Š is defined to be the first matrix on the right side of Eq. (8.1.3). The coefficients
a1 through a12 can be obtained by substituting the coordinates into u and v as follows:
38
8 9 2
9
1 x1 y1 x12 x1 y1 y12 0 0 0 0
0
0 >a >
>
>
u
1
1>
>
>
>
7>
> >
> 6
>

>
>
6 1 x2 y2 x22 x2 y2 y22 0 0 0 0
>
>
>
>
0
0 7>
a
u
>
>
>
>
2
2
6
7>
>
>
>
>
>
>
>
.
.
.
.

.
.
.
.
.
.
.
.
.
.
6
7
>
>
>
>
>
>
>
>
.
.
.
.
.
.
.
.
.
.

.
.
.
.
>
>
>
>
6
7
.
.
.
.
.
.
.
.
.
.
.
.
.
.
>
>
>
>
>
>

>
6
7
<a >
=
2
2
7
y
x
x
y
y
0
0
0
0
0
0
6
6
6
6
6
6
6
6
7
¼6

ð8:1:5Þ
2
2
6
7
>
a >
v > 60 0 0 0
0
0 1 x1 y1 x1 x1 y1 y1 7>
>
>
> .7 >
> .1 >
> 6
>
>
>
>
>
7
.. 7>
..
..
..
.. .. ..
..
..
..
>

.. >
.. >
>
>
>
6 ... ...
>
>
>
>
. 7>
.
.
.
. . .
.
.
.
>
>
>
>
6
>
>
>
>
>
>
>

>
6
7
2
2
v
a
>
>
>
5>
11 >
0
0
0
0
0
0
1
x
y
x
x
y
y
>
>
>
4
5

5
5
5
5
5
5 >
>
>
:a >
:v ;
;
2
2
6
12
0 0 0 0
0
0 1 x6 y6 x6 x6 y6 y6


8.1 Derivation of the Linear-Strain Triangular Element Stiffness Matrix and Equations

Solving for the ai ’s, we have
2
9
8
1 x1 y1 x12
a
1
> 6. .

>
>
>
..
..
>
.. >
6 .. ..
>
>
>
>
.
.
>
>
6
.
>
>
>
= 6
<a >
2
6
1
x
6
6 y6 x6
¼6

6
>
> 60 0 0 0
> a7 >
>
>
>
.. >
> 6
>
..
..
>
6 .. ..
>
>
. >
>
>
.
.
; 4. .
:
a12
0 0 0 0

x 1 y1
..
.

y12
..
.

0
..
.

x 6 y6

y62

0

0

0

0

0

0
..
.

0
..
.

1
..
.

x1
..
.

y1
..
.

x12
..
.

x1 y1
..
.

0

0

1

x6

y6

x62

x6 y6

0
..
.

0
..
.

0
..
.

d

401

3À1
8 9
0
u1 >
7
>
>
.. 7 >
>
>

>
>
>
. 7 > ... >
>
>
>
>
7 >
=
<
7
u
07
6
7
y12 7 >
>
> v1 >
>
>
> . >
.. 7
.. >
>
>
7 >
>
>
. 5 >

;
: >
v
6
y62

0
..
.

ð8:1:6Þ
or, alternatively, we can express Eq. (8.1.6) as
fag ¼ ½X ŠÀ1 fdg

ð8:1:7Þ

where ½X Š is the 12  12 matrix on the right side of Eq. (8.1.6). It is best to invert the
½X Š matrix by using a digital computer. Then the ai ’s, in terms of nodal displacements,
are substituted into Eq. (8.1.4). Note that only the 6  6 part of ½X Š in Eq. (8.1.6)
really must be inverted. Finally, using Eq. (8.1.7) in Eq. (8.1.4), we can obtain the general displacement expressions in terms of the shape functions and the nodal degrees of
freedom as

where

fcg ¼ ½NŠfdg

ð8:1:8Þ

½NŠ ¼ ½M à Š½X ŠÀ1

ð8:1:9Þ

Step 3 Define the Strain=Displacement and Stress=Strain
Relationships
The element strains are again given by
8
>
>
>
>
>
>
>
<

qu
qx
qv
qy

9
>
>
>
>
>
>
>
=

9
8
>
=
< ex >
ey
¼
feg ¼
>
>
>
; >
>
:g >
>
>
>
>
xy
>
>
>
>
qv
qu
>
;
: þ >
qx qy
or, using Eq. (8.1.3) for u and v in Eq. (8.1.10), we obtain

2

0
feg ¼ 4 0
0

1
0
0

0 2x
0 0
1 0

y
0
x

0
0
2y

0 0
0 0
0 1

0
1
0

0
0
2x

0
x
y

ð8:1:10Þ

8
3>
>
0 >
<
2y 5
>
0 >
>
:

a1
a2
..
.
a12

9
>
>

>
=
>
>
>
;

ð8:1:11Þ

We observe that Eq. (8.1.11) yields a linear strain variation in the element. Therefore,
the element is called a linear-strain triangle (LST). Rewriting Eq. (8.1.11), we have
feg ¼ ½M 0 Šfag

ð8:1:12Þ


402

d

8 Development of the Linear-Strain Triangle Equations

where ½M 0 Š is the first matrix on the right side of Eq. (8.1.11). Substituting Eq. (8.1.6)
for the ai ’s into Eq. (8.1.12), we have feg in terms of the nodal displacements as
feg ¼ ½BŠfdg

ð8:1:13Þ

where ½BŠ is a function of the variables x and y and the coordinates ðx1 ; y1 Þ through
ðx6 ; y6 Þ given by

½BŠ ¼ ½M 0 Š½X ŠÀ1

ð8:1:14Þ

where Eq. (8.1.7) has been used in expressing Eq. (8.1.14). Note that ½BŠ is now a
matrix of order 3 Â 12.
The stresses are again given by
(
)
(
)
ex
sx
sy ¼ ½DŠ ey
¼ ½DŠ½BŠfdg
ð8:1:15Þ
gxy
txy
where ½DŠ is given by Eq. (6.1.8) for plane stress or by Eq. (6.1.10) for plane strain.
These stresses are now linear functions of x and y coordinates.

Step 4 Derive the Element Stiffness Matrix and Equations
We determine the stiffness matrix in a manner similar to that used in Section 6.2 by
using Eq. (6.2.50) repeated here as
ððð
½kŠ ¼
½BŠ T ½DŠ½BŠ dV
ð8:1:16Þ
V

However, the ½BŠ matrix is now a function of x and y as given by Eq. (8.1.14). Therefore, we must perform the integration in Eq. (8.1.16). Finally, the ½BŠ matrix is of the
form
2
3
b1 0 b2 0 b3 0 b4 0 b5 0 b6 0
1 6
7
ð8:1:17Þ
½BŠ ¼
4 0 g1 0 g2 0 g3 0 g4 0 g5 0 g6 5
2A
g1 b 1 g2 b 2 g3 b 3 g4 b 4 g5 b 5 g6 b 6
where the b’s and g’s are now functions of x and y as well as of the nodal coordinates,
as is illustrated for a specific linear-strain triangle in Section 8.2 by Eq. (8.2.8).
The stiffness matrix is then seen to be a 12 Â 12 matrix on multiplying the matrices
in Eq. (8.1.16). The stiffness matrix, Eq. (8.1.16), is very cumbersome to obtain in
explicit form, so it will not be given here. However, if the origin of the coordinates
is considered to be at the centroid of the element, the integrations become amenable
[9]. Alternatively, area coordinates [3, 8, 9] can be used to obtain an explicit form
of the stiffness matrix. However, even the use of area coordinates usually involves
tedious calculations. Therefore, the integration is best carried out numerically.
(Numerical integration is described in Section 10.4.)


8.2 Example LST Stiffness Determination

d

403

The element body forces and surface forces should not be automatically lumped
at the nodes, but for a consistent formulation (one that is formulated from the same
shape functions used to formulate the stiffness matrix), Eqs. (6.3.1) and (6.3.7), respectively, should be used. (Problems 8.3 and 8.4 illustrate this concept.) These forces can
be added to any concentrated nodal forces to obtain the element force matrix. Here
the element force matrix is of order 12 Â 1 because, in general, there could be an x
and a y component of force at each of the six nodes associated with the element. The
element equations are then given by
9
8
2
3
f1x >
k11 . . . k1; 12
>
>
>
>
>
>
6 k
=
< f1y >
k2; 12 7
6 21
7
6
.. 7
..
6 ..
7

>
>
¼
. 5
. >
>
4 .
>
>
>
>
;
:
k12; 1 . . . k12; 12
f6y
ð12 Â 12Þ
ð12 Â 1Þ

8 9
u1 >
>
>
>
>
=
<v >
1
.
.. >
>

>
>
>
;
: >
v6
ð12 Â 1Þ

ð8:1:18Þ

Steps 5–7
Steps 5–7, which involve assembling the global stiffness matrix and equations, determining the unknown global nodal displacements, and calculating the stresses, are
identical to those in Section 6.2 for the CST. However, instead of constant stresses in
each element, we now have a linear variation of the stresses in each element. Common
practice was to use the centroidal element stresses. Current practice is to use the
average of the nodal element stresses.

d

8.2 Example LST Stiffness Determination

d

To illustrate some of the procedures outlined in Section 8.1 for deriving an LST
stiffness matrix, consider the following example. Figure 8–3 shows a specific LST
and its coordinates. The triangle is of base dimension b and height h, with midside
nodes.

Figure 8–3 LST triangle for evaluation of
a stiffness matrix

404

d

8 Development of the Linear-Strain Triangle Equations

Using the first six equations of Eq. (8.1.5), we calculate the coefficients a1
through a6 by evaluating the displacement u at each of the six known coordinates of
each node as follows:
u1 ¼ uð0; 0Þ ¼ a1
u2 ¼ uðb; 0Þ ¼ a1 þ a2 b þ a4 b 2
u3 ¼ uð0; hÞ ¼ a1 þ a3 h þ a6 h 2
u4 ¼ u



 2
 2
b h
b
h
b
bh
h
;
þ a5 þ a6
¼ a1 þ a2 þ a3 þ a4
2 2

2
2
2
4
2

ð8:2:1Þ



 2
h
h
h
u5 ¼ u 0;
¼ a1 þ a3 þ a6
2
2
2
u6 ¼ u



 2
b
b
b
; 0 ¼ a1 þ a2 þ a4
2
2

2

Solving Eqs. (8.2.1) simultaneously for the ai ’s, we obtain
a1 ¼ u1

a2 ¼

4u6 À 3u1 À u2
b

a4 ¼

2ðu2 À 2u6 þ u1 Þ
b2

a6 ¼

2ðu3 À 2u5 þ u1 Þ
h2

a5 ¼

a3 ¼

4u5 À 3u1 À u3
h

4ðu1 þ u4 À u5 À u6 Þ
bh

ð8:2:2Þ

Substituting Eqs. (8.2.2) into the displacement expression for u from Eqs. (8.1.2), we
have
!
!
!
4u6 À 3u1 À u2
4u5 À 3u1 À u3
2ðu2 À 2u6 þ u1 Þ 2
u ¼ u1 þ
x
xþ
yþ
b2
b
h
!
!
4ðu1 þ u4 À u5 À u6 Þ
2ðu3 À 2u5 þ u1 Þ 2
þ
ð8:2:3Þ
y
xy þ
bh
h2
Similarly, solving for a7 through a12 by evaluating the displacement v at each of the
six nodes and then substituting the results into the expression for v from Eqs. (8.1.2),
we obtain

!
!
!
4v6 À 3v1 À v2
4v5 À 3v1 À v3
2ðv2 À 2v6 þ v1 Þ 2
v ¼ v1 þ
x
xþ
yþ
b2
b
h
!
!
4ðv1 þ v4 À v5 À v6 Þ
2ðv3 À 2v5 þ v1 Þ 2
þ
ð8:2:4Þ
y
xy þ
bh
h2


8.2 Example LST Stiffness Determination

d

405

Using Eqs. (8.2.3) and (8.2.4), we can express the general displacement expressions in
terms of the shape functions as
8 9
> u1 >
>
> >
!>
& '
u
N1 0 N2 0 N3 0 N4 0 N5 0 N6 0 < v1 =
¼
..
0 N1 0 N2 0 N3 0 N4 0 N5 0 N6 >
v
>
> . >
>
;
: >
v6
ð8:2:5Þ
where the shape functions are obtained by collecting coefficients that multiply each ui
term in Eq. (8.2.3). For instance, collecting all terms that multiply by u1 in Eq. (8.2.3),
we obtain N1 . These shape functions are then given by
N1 ¼ 1 À

3x 3y 2x 2 4xy 2y 2
À þ 2 þ
þ 2

b
h
bh
b
h

N3 ¼

Ày 2y 2
þ 2
h
h

N6 ¼

4x 4x 2 4xy
À 2 À
b
bh
b

N4 ¼

4xy
bh

N5 ¼

N2 ¼

Àx 2x 2
þ 2
b
b

4y 4xy 4y 2
À
À 2
h
bh
h

ð8:2:6Þ

Using Eq. (8.2.5) in Eq. (8.1.10), and performing the differentiations indicated on u
and v, we obtain
e ¼ Bd

ð8:2:7Þ

where B is of the form of Eq. (8.1.17), with the resulting b’s and g’s in Eq. (8.1.17)
given by
4hx
4hx
þ 4y
b2 ¼ Àh þ
b3 ¼ 0
b1 ¼ À3h þ
b
b

8hx
À 4y
b5 ¼ À4y
b6 ¼ 4h À
b4 ¼ 4y
b
ð8:2:8Þ
4by
4by
g2 ¼ 0
g1 ¼ À3b þ 4x þ
g3 ¼ Àb þ
h
h
8by
g6 ¼ À4x
g4 ¼ 4x
g5 ¼ 4b À 4x À
h
These b’s and g’s are specific to the element in Figure 8–3. Specifically, using Eqs.
(8.1.1) and (8.1.17) in Eq. (8.2.7), we obtain
1
½b u1 þ b2 u2 þ b3 u3 þ b4 u4 þ b5 u5 þ b6 u6 Š
2A 1
1
½g v1 þ g2 v2 þ g3 v3 þ g4 v4 þ g5 v5 þ g6 v6 Š
ey ¼
2A 1
1
½g u1 þ b 1 v1 þ Á Á Á þ b6 v6 Š

gxy ¼
2A 1
ex ¼

The stiffness matrix for a constant-thickness element can now be obtained on
substituting Eqs. (8.2.8) into Eq. (8.1.17) to obtain B, then substituting B into


406

d

8 Development of the Linear-Strain Triangle Equations

Eq. (8.1.16) and using calculus to set up the appropriate integration. The explicit expression for the 12 Â 12 stiffness matrix, being extremely cumbersome to obtain, is
not given here. Stiffness matrix expressions for higher-order elements are found in
References [1] and [2].

d

8.3 Comparison of Elements

d

For a given number of nodes, a better representation of true stress and displacement is
generally obtained using the LST element than is obtained with the same number of
nodes using a much finer subdivision into simple CST elements. For example, using
one LST yields better results than using four CST elements with the same number of
nodes (Figure 8–4) and hence the same number of degrees of freedom (except for the
case when constant stress exists).

We now present results to compare the CST of Chapter 6 with the LST of this
chapter. Consider the cantilever beam subjected to a parabolic load variation acting
as shown in Figure 8–5. Let E ¼ 30 Â 10 6 psi, n ¼ 0:25, and t ¼ 1:0 in.
Table 8–1 lists the series of tests run to compare results using the CST and LST
elements. Table 8–2 shows comparisons of free-end (tip) deflection and stress sx
for each element type used to model the cantilever beam. From Table 8–2, we can observe that the larger the number of degrees of freedom for a given type of triangular
element, the closer the solution converges to the exact one (compare run A-1 to run
A-2, and B-1 to B-2). For a given number of nodes, the LST analysis yields some what
better results for displacement than the CST analysis (compare run A-1 to run B-1).

Figure 8–4 Basic triangular element: (a) four-CST and (b) one-LST

Figure 8–5 Cantilever beam used to compare the CST and LST elements
with a 4 Â 16 mesh


8.3 Comparison of Elements

d

407

Table 8–1 Models used to compare CST and LST results for the cantilever beam
of Figure 8–5

Series of Tests Run

Number
of Nodes

Number of Degrees
of Freedom, nd

Number of
Triangular Elements

A-1 4 Â 16 mesh
A-2 8 Â 32
B-1 2 Â 8
B-2 4 Â 16

85
297
85
297

160
576
160
576

128 CST
512 CST
32 LST
128 LST

Table 8–2 Comparison of CST and LST results for the cantilever beam of Figure 8–5

Run

nd

Bandwidth1
nb

Tip Deflection
(in.)

sx (ksi)

Location (in.),
x; y

A-1
A-2
B-1
B-2

160
576
160
576

14
22
18
22

À0.29555
À0.33850

À0.33470
À0.35159

67.236
81.302
58.885
69.956

2.250, 11.250
1.125, 11.630
4.500, 10.500
2.250, 11.250

À0.36133

80.000

0, 12

Exact solution
1 Bandwidth is described in Appendix B.4.

However, one of the reasons that the bending stress sx predicted by the LST
model B-1 compared to CST model A-1 is not as accurate is as follows. Recall that
the stress is calculated at the centroid of the element. We observe from the table that
the location of the bending stress is closer to the wall and closer to the top for the
CST model A-1 compared to the LST model B-1. As the classical bending stress is a
linear function with increasing positive linear stress from the neutral axis for the
downward applied load in this example, we expect the largest stress to be at the very
top of the beam. So the model A-1 with more and smaller elements (with eight elements through the beam depth) has its centroid closer to the top (at 0.75 in. from the

top) than model B-1 with few elements (two elements through the beam depth) with
centroidal stress located at 1.5 in. from the top. Similarly, comparing A-2 to B-2 we
observe the same trend in the results—displacement at the top end being more accurately predicted by the LST model, but stresses being calculated at the centroid making the A-2 model appear more accurate than the LST model due to the location
where the stress is reported.
Although the CST element is rather poor in modeling bending, we observe from
Table 8–2 that the element can be used to model a beam in bending if a sufficient
number of elements are used through the depth of the beam. In general, both LST
and CST analyses yield results good enough for most plane stress/strain problems,
provided a sufficient number of elements are used. In fact, most commercial programs
incorporate the use of CST and/or LST elements for plane stress/strain problems,


408

d

8 Development of the Linear-Strain Triangle Equations

Figure 8–6 Plates subjected to parabolically distributed edge loads; comparison of
results for triangular elements. (Gallagher, R. H. Finite Element Analysis: Fundamentals,
( 1975, pp. 269, 270. Reprinted by permission of Prentice Hall, Inc., Englewood Cliffs, NJ)

although these elements are used primarily as transition elements (usually during mesh
generation). The four-sided isoparametric plane stress/strain element is most frequently used in commercial programs and is described in Chapter 10.
Also, recall that finite element displacements will always be less than (or equal
to) the exact ones, because finite element models are normally predicted to be stiffer
than the actual structures when the displacement formulation of the finite element
method is used. (The reason for the stiffer model was discussed in Sections 3.10 and
7.3. Proof of this assertion can be found in References [4–7].
Finally, Figure 8-6 (from Reference [8]) illustrates a comparison of CST and

LST models of a plate subjected to parabolically distributed edge loads. Figure 8–6
shows that the LST model converges to the exact solution for horizontal displacement
at point A faster than does the CST model. However, the CST model is quite acceptable even for modest numbers of degrees of freedom. For example, a CST model
with 100 nodes (200 degrees of freedom) often yields nearly as accurate a solution as
does an LST model with the same number of degrees of freedom.
In conclusion, the results of Table 8–2 and Figure 8–6 indicate that the LST
model might be preferred over the CST model for plane stress applications when relatively small numbers of nodes are used. However, the use of triangular elements of
higher order, such as the LST, is not visibly advantageous when large numbers of
nodes are used, particularly when the cost of formation of the element stiffnesses,
equation bandwidth, and overall complexities involved in the computer modeling are
considered.


Problems

d

d

409

References
[1] Pederson, P., ‘‘Some Properties of Linear Strain Triangles and Optimal Finite Element
Models,’’ International Journal for Numerical Methods in Engineering, Vol. 7, pp. 415–430,
1973.
[2] Tocher, J. L., and Hartz, B. J., ‘‘Higher-Order Finite Element for Plane Stress,’’ Journal of
the Engineering Mechanics Division, Proceedings of the American Society of Civil Engineers, Vol. 93, No. EM4, pp. 149–174, Aug. 1967.
[3] Bowes, W. H., and Russell, L. T., Stress Analysis by the Finite Element Method for Practicing Engineers, Lexington Books, Toronto, 1975.
[4] Fraeijes de Veubeke, B., ‘‘Upper and Lower Bounds in Matrix Structural Analysis,’’ Matrix Methods of Structural Analysis, AGAR-Dograph 72, B. Fraeijes de Veubeke, ed.,
Macmillan, New York, 1964.

[5] McLay, R. W., Completeness and Convergence Properties of Finite Element Displacement
Functions: A General Treatment, American Institute of Aeronautics and Astronautics Paper
No. 67–143, AIAA 5th Aerospace Meeting, New York, 1967.
[6] Tong, P., and Pian, T. H. H., ‘‘The Convergence of Finite Element Method in Solving
Linear Elastic Problems,’’ International Journal of Solids and Structures, Vol. 3, pp. 865–
879, 1967.
[7] Cowper, G. R., ‘‘Variational Procedures and Convergence of Finite-Element Methods,’’
Numerical and Computer Methods in Structural Mechanics, S. J. Fenves, N. Perrone, A. R.
Robinson, and W. C. Schnobrich, eds., Academic Press, New York, 1973.
[8] Gallagher, R., Finite Element Analysis Fundamentals, Prentice Hall, Englewood Cliffs, NJ,
1975.
[9] Zienkiewicz, O. C., The Finite Element Method, 3rd ed., McGraw-Hill, New York, 1977.

d

Problems
8.1 Evaluate the shape functions given by Eq. (8.2.6). Sketch the variation of each function over the surface of the triangular element shown in Figure 8–3.
8.2 Express the strains ex ; ey , and gxy for the element of Figure 8–3 by using the results
given in Section 8.2. Evaluate these strains at the centroid of the element; then evaluate
the stresses at the centroid in terms of E and n. Assume plane stress conditions apply.

Figure P8–3

8.3 For the element of Figure 8–3 (shown again as Figure P8–3) subjected to the uniform
pressure shown acting over the vertical side, determine the nodal force replacement
system using Eq. (6.3.7). Assume an element thickness of t.


410

d

8 Development of the Linear-Strain Triangle Equations

8.4 For the element of Figure 8–3 (shown as Figure P8–4) subjected to the linearly varying line load shown acting over the vertical side, determine the nodal force replacement system using Eq. (6.3.7). Compare this result to that of Problem 6.9. Are these
results expected? Explain.

Figure P8–4

8.5 For the linear-strain elements shown in Figure P8–5, determine the strains ex ; ey , and
gxy . Evaluate the stresses sx ; sy , and txy at the centroids. The coordinates of the nodes
are shown in units of inches. Let E ¼ 30 Â 10 6 psi, n ¼ 0:25, and t ¼ 0:25 in. for both
elements. Assume plane stress conditions apply. The nodal displacements are given as
u1 ¼ 0:0

v1 ¼ 0:0

u2 ¼ 0:001 in:

v2 ¼ 0:002 in:

u3 ¼ 0:0005 in:

v3 ¼ 0:0002 in:

u4 ¼ 0:0002 in:

v4 ¼ 0:0001 in:

u5 ¼ 0:0

v5 ¼ 0:0001 in:

u6 ¼ 0:0005 in:

v6 ¼ 0:001 in:

(Hint: Use the results of Section 8.2.)

Figure P8–5


Problems

d

411

8.6 For the linear-strain element shown in Figure P8–6, determine the strains ex ; ey , and
gxy . Evaluate these strains at the centroid of the element; then evaluate the stresses
sx ; sy , and txy at the centroid. The coordinates of the nodes are shown in units of
millimeters. Let E ¼ 210 GPa, n ¼ 0:25, and t ¼ 10 mm. Assume plane stress conditions apply. Use the nodal displacements given in Problem 8.5 (converted to millimeters). Note that the b’s and g’s from the example in Section 8.2 cannot be used here
as the element in Figure P8–6 is oriented differently than the one in Figure 8–3.

Figure P8–6

8.7 Evaluate the shape functions for the linear-strain triangle shown in Figure P8–7. Then
evaluate the B matrix. Units are millimeters.

Figure P8–7

8.8 Use the LST element to solve Example 7.2. Compare the results.
8.9 Write a computer program to solve plane stress problems using the LST element.


CHAPTER

9

Axisymmetric Elements

Introduction
In previous chapters, we have been concerned with line or one-dimensional elements
(Chapters 2–5) and two-dimensional elements (Chapters 6–8). In this chapter, we consider a special two-dimensional element called the axisymmetric element. This element
is quite useful when symmetry with respect to geometry and loading exists about
an axis of the body being analyzed. Problems that involve soil masses subjected to
circular footing loads or thick-walled pressure vessels can often be analyzed using the
element developed in this chapter.
We begin with the development of the stiffness matrix for the simplest axisymmetric element, the triangular torus, whose vertical cross section is a plane triangle.
We then present the longhand solution of a thick-walled pressure vessel to illustrate
the use of the axisymmetric element equations. This is followed by a description of some
typical large-scale problems that have been modeled using the axisymmetric element.

d

9.1 Derivation of the Stiffness Matrix

d

In this section, we will derive the stiffness matrix and the body and surface force matrices for the axisymmetric element. However, before the development, we will first

present some fundamental concepts prerequisite to the understanding of the derivation. Axisymmetric elements are triangular tori such that each element is symmetric
with respect to geometry and loading about an axis such as the z axis in Figure 9–1.
Hence, the z axis is called the axis of symmetry or the axis of revolution. Each vertical
cross section of the element is a plane triangle. The nodal points of an axisymmetric
triangular element describe circumferential lines, as indicated in Figure 9–1.
In plane stress problems, stresses exist only in the x-y plane. In axisymmetric
problems, the radial displacements develop circumferential strains that induce stresses
sr , sy , sz , and trz , where r, y, and z indicate the radial, circumferential, and longitudinal
412


9.1 Derivation of the Stiffness Matrix

d

413

Figure 9–1 Typical axisymmetric element ijm

directions, respectively. Triangular torus elements are often used to idealize the axisymmetric system because they can be used to simulate complex surfaces and are simple to
work with. For instance, the axisymmetric problem of a semi-infinite half-space loaded
by a circular area (circular footing) shown in Figure 9–2(a), the domed pressure vessel
shown in Figure 9–2(b), and the engine valve stem shown in Figure 9–2(c) can be solved
using the axisymmetric element developed in this chapter.

(a) soil mass

(b) domed vessel

(c) engine valve stem

Figure 9-2 Examples of axisymmetric problems: (a) semi-infinite half-space (soil mass)
modeled by axisymmetric elements, (b) a domed pressure vessel, and (c) an engine
valve stem


414

d

9 Axisymmetric Elements

Figure 9–3 (a) Plane cross section of (b) axisymmetric element

Because of symmetry about the z axis, the stresses are independent of the y
coordinate. Therefore, all derivatives with respect to y vanish, and the displacement
component v (tangent to the y direction), the shear strains gry and gyz , and the shear
stresses try and tyz are all zero.
Figure 9–3 shows an axisymmetric ring element and its cross section to represent
the general state of strain for an axisymmetric problem. It is most convenient
to express the displacements of an element ABCD in the plane of a cross section
in cylindrical coordinates. We then let u and w denote the displacements in the radial
and longitudinal directions, respectively. The side AB of the element is displaced an
amount u, and side CD is then displaced an amount u þ ðqu=qrÞ dr in the radial direction. The normal strain in the radial direction is then given by
qu
er ¼
ð9:1:1aÞ
qr
In general, the strain in the tangential direction depends on the tangential displacement v and on the radial displacement u. However, for axisymmetric deformation behavior, recall that the tangential displacement v is equal to zero. Hence, the tangential
strain is due only to the radial displacement. Having only radial displacement u, the

_
new length of the arc AB is ðr þ uÞ dy, and the tangential strain is then given by
ðr þ uÞ dy À r dy u
¼
ð9:1:1bÞ
r dy
r
Next, we consider the longitudinal element BDEF to obtain the longitudinal strain
and the shear strain. In Figure 9–4, the element is shown to displace by amounts u
and w in the radial and longitudinal directions at point E, and to displace additional
amounts ðqw=qzÞ dz along line BE and ðqu=qrÞ dr along line EF. Furthermore, observing lines EF and BE, we see that point F moves upward an amount ðqw=qrÞ dr with respect to point E and point B moves to the right an amount ðqu=qzÞ dz with respect to
point E. Again, from the basic definitions of normal and shear strain, we have the longitudinal normal strain given by
qw
ez ¼
ð9:1:1cÞ
qz
and the shear strain in the r-z plane given by
ey ¼

grz ¼

qu qw
þ
qz qr

ð9:1:1dÞ


9.1 Derivation of the Stiffness Matrix

d

415

Figure 9–4 Displacement and
rotations of lines of element in the r-z
plane

Summarizing the strain/displacement relationships of Eqs. (9.1.1a–d) in one equation
for easier reference, we have
er ¼

qu
qr

ey ¼

u
r

ez ¼

qw
qz

grz ¼

qu qw
þ
qz qr

ð9:1:1eÞ

The isotropic stress/strain relationship, obtained by simplifying the general
stress/strain relationships given in Appendix C, is
3
2
1Àn
n
n
0
8 9
78 9
6
sr >
>
>
>
> er >
6 n
1Àn
n
0 7
>
=
=
<s >
< >
7>
6

E
z
7 ez
6
ð9:1:2Þ
¼
7
6 n
n
1Àn
0 7 > ey >
>
ð1 þ nÞð1 À 2nÞ 6
sy >
>
>
>
>
>
;
>
:
7
6
: ;
4
trz
1 À 2n 5 grz
0
0

0
2
The theoretical development follows that of the plane stress/strain problem
given in Chapter 6.
Step 1 Select Element Type
An axisymmetric solid is shown discretized in Figure 9–5(a), along with a typical triangular element. The element has three nodes with two degrees of freedom per node
(that is, ui , wi at node i). The stresses in the axisymmetric problem are shown in
Figure 9–5(b).
Step 2 Select Displacement Functions
The element displacement functions are taken to be
uðr; zÞ ¼ a1 þ a2 r þ a3 z
wðr; zÞ ¼ a4 þ a5 r þ a6 z

ð9:1:3Þ

so that we have the same linear displacement functions as used in the plane stress,
constant-strain triangle. Again, the total number of ai ’s (six) introduced in the


416

d

9 Axisymmetric Elements

Figure 9–5 Discretized axisymmetric solid

displacement functions is the same as the total number of degrees of freedom for the
element. The nodal displacements are
9

8
ui >
>
>
>
>
>
8 9 >
>
> wi >
>
>
>
>
>
d
=
< i= < u >
j
ð9:1:4Þ
fdg ¼ d j
¼
>
wj >
>
>
:d >
; >
>
>

>
>
m
>
>
>
>
> um >
;
:
wm
and u evaluated at node i is
ð9:1:5Þ
uðri ; zi Þ ¼ ui ¼ a1 þ a2 ri þ a3 zi
Using Eq. (9.1.3), the general displacement function is then expressed in matrix form as
8 9
a1 >
>
>
>
>
>
>
a2 >
>
>
>
>
>
& ' &

'
!>
<
u
a1 þ a2 r þ a3 z
a3 =
1 r z 0 0 0
ð9:1:6Þ
fcg ¼
¼
¼
w
a4 þ a5 r þ a6 z
0 0 0 1 r z >
>
> a4 >
>
>
>
>
>
>
> a5 >
>
;
: >
a6
Substituting the coordinates of the nodal points shown in Figure 9–5(a) into
Eq. (9.1.6), we can solve for the ai ’s in a manner similar to that in Section 6.2. The
resulting expressions are

3À1 8 9
8 9 2
1 ri z i
< a1 =
< ui =
6
7
ð9:1:7Þ
a2 ¼ 4 1 rj z j 5
u
: ;
: j ;
a3
1 rm z m
um
3À1 8
8 9 2
9
1 ri z i
< a4 =
< wi =
6
7
and
¼ 4 1 rj z j 5
ð9:1:8Þ
a
w
: 5;
: j ;

a6
1 rm z m
wm


9.1 Derivation of the Stiffness Matrix

Performing the inversion operations in Eqs. (9.1.7) and (9.1.8), we have
38 9
2
8 9
ai aj am < ui =
< a1 =
1 6
7
a2 ¼
4 b i b j b m 5 uj
: ; 2A
: ;
gi gj gm
a3
um

and

2
8 9
ai
< a4 =
1 6

a5 ¼
4 bi
: ; 2A
gi
a6

aj
bj
gj

38
9
am < w i =
bm 7
5 wj
;
:
gm
wm

d

417

ð9:1:9Þ

ð9:1:10Þ

where
ai ¼ rj z m À z j rm

a j ¼ rm z i À z m ri

a m ¼ ri z j À z i rj

b i ¼ zj À zm

b j ¼ zm À zi

b m ¼ zi À zj

g i ¼ rm À rj

g j ¼ ri À rm

g m ¼ rj À ri

ð9:1:11Þ

We define the shape functions, similar to Eqs. (6.2.18), as
1
ðai þ bi r þ gi zÞ
2A
1
ðaj þ bj r þ gj zÞ
Nj ¼
2A
1
ðam þ bm r þ gm zÞ
Nm ¼

2A
Ni ¼

ð9:1:12Þ

Substituting Eqs. (9.1.7) and (9.1.8) into Eq. (9.1.6), along with the shape function Eqs. (9.1.12), we find that the general displacement function is
9
8
ui >
>
>
>
>
>
>
wi >
>
>
>
>
>
&
'
!>
<
Ni 0 Nj 0 Nm 0
uðr; zÞ
uj =
ð9:1:13Þ
fcg ¼

¼
0 Nm >
0 Ni 0 Nj
wðr; zÞ
wj >
>
>
>
>
>
>
>
>
>
u >
>
;
: m>
wm
or

fcg ¼ ½NŠfdg

ð9:1:14Þ

Step 3 Define the Strain=Displacement and Stress=Strain
Relationships
When we use Eqs. (9.1.1) and (9.1.3), the strains become
9
8

a2
>
>
>
>
>
>
>
>
a6
=
<
feg ¼ a1
a3 z
>
>
þ a2 þ
>
>
>
r >
>
>
;
: r
a3 þ a5

ð9:1:15Þ

418

d

9 Axisymmetric Elements

Rewriting Eq. (9.1.15) with the ai ’s as a separate column matrix, we have
2
38 9
0
1
0
0
0
0
> a1 >
8 9 6
7>
>
> >
er >
>
> a2 >
>
6 0 0 0 0 0 1 7>
>
>
>
>
>

>
7>
=
= 6
<
<e >
6
7
a
z
3
6
7
¼ 61
z
7 > a4 >
> 6
> ey >
1
0 0 0 7>
>
>
>
>
; 6r
>
: >
7>
r
>

> a5 >
grz
>
4
5>
>
;
: >
0 0 1 0 1 0
a6
Substituting Eqs. (9.1.7) and
we obtain
2
bi
6
0
6
1 6
6
feg ¼
ai
gi z
2A 6
6 þ bi þ
r
r
4
gi

ð9:1:16Þ

(9.1.8) into Eq. (9.1.16) and making use of Eq. (9.1.11),
0

bj

0

bm

gi

0

gj

0

0
bi

gj z
aj
þ bj þ
r
r
gj

0
bj

am
g z
þ bm þ m
r
r
gm

9
38
ui >
>
>
>
>
> wi >
7>
>
gm 7 >
>
>
7 < uj =
7
0 7
>
> wj >
7>
>
>
5>

>
>
um >
>
>
;
:
bm
wm
0

ð9:1:17Þ
or, rewriting Eq. (9.1.17) in simplified matrix form,
9
8
ui >
>
>
>
>
>
>
wi >
>
>
>
>
>
=
< u >

j
feg ¼ ½Bi Bj Bm Š
>
wj >
>
>
>
>
>
>
>
>
>
um >
>
>
;
:
wm
3
2
bi
0
7
6
6
0
gi 7
7
6

1 6
where
7
½Bi Š ¼
ai
gi z
6
2A 6 þ bi þ
07
7
r
5
4r
gi
bi

ð9:1:18Þ

ð9:1:19Þ

Similarly, we obtain submatrices Bj and Bm by replacing the subscript i with j and
then with m in Eq. (9.1.19). Rewriting Eq. (9.1.18) in compact matrix form, we have
feg ¼ ½BŠfdg
where

½BŠ ¼ ½Bi

Bj

Bm Š

ð9:1:20Þ
ð9:1:21Þ

Note that ½BŠ is a function of the r and z coordinates. Therefore, in general, the
strain ey will not be constant.
The stresses are given by
fsg ¼ ½DŠ½BŠfdg

ð9:1:22Þ


9.1 Derivation of the Stiffness Matrix

d

419

where ½DŠ is given by the first matrix on the right side of Eq. (9.1.2). (As mentioned in
Chapter 6, for n ¼ 0:5, a special formula must be used; see Reference [9].)
Step 4 Derive the Element Stiffness Matrix and Equations
The stiffness matrix is
½kŠ ¼

ððð

½BŠ T ½DŠ½BŠ dV

ð9:1:23Þ

½BŠ T ½DŠ½BŠr dr dz

ð9:1:24Þ

V

or

½kŠ ¼ 2p

ðð
A

after integrating along the circumferential boundary. The ½BŠ matrix, Eq. (9.1.21), is a
function of r and z. Therefore, ½kŠ is a function of r and z and is of order 6  6.
We can evaluate Eq. (9.1.24) for ½kŠ by one of three methods:
1. Numerical integration (Gaussian quadrature) as discussed in
Chapter 10.
2. Explicit multiplication and term-by-term integration [1].
3. Evaluate ½BŠ for a centroidal point ðr; zÞ of the element
ri þ rj þ rm
zi þ zj þ zm
z¼z¼
r¼r¼
3
3

ð9:1:25Þ

and define ½Bðr; zފ ¼ ½BŠ. Therefore, as a first approximation,

½kŠ ¼ 2prA½BŠ T ½DŠ½BŠ

ð9:1:26Þ

If the triangular subdivisions are consistent with the final stress distribution (that
is, small elements in regions of high stress gradients), then acceptable results can be
obtained by method 3.
Distributed Body Forces
Loads such as gravity (in the direction of the z axis) or centrifugal forces in rotating
machine parts (in the direction of the r axis) are considered to be body forces (as
shown in Figure 9–6). The body forces can be found by
& '
ðð
Rb
f fb g ¼ 2p
½NŠ T
r dr dz
ð9:1:27Þ
Zb
A

Figure 9–6 Axisymmetric element with body
forces per unit volume


420

d

9 Axisymmetric Elements

where Rb ¼ o 2 rr for a machine part moving with a constant angular velocity o about
the z axis, with material mass density r and radial coordinate r, and where Zb is the
body force per unit volume due to the force of gravity.
Considering the body force at node i, we have
& '
ðð
T Rb
f fbi g ¼ 2p
½Ni Š
r dr dz
ð9:1:28Þ
Zb
A

where

Ni
0

½Ni Š T ¼

0
Ni

!

Multiplying and integrating in Eq. (9.1.28), we obtain
& '
2p Rb

f fbi g ¼
Ar
3 Zb

ð9:1:29Þ

ð9:1:30Þ

where the origin of the coordinates has been taken as the centroid of the element, and
Rb is the radially directed body force per unit volume evaluated at the centroid of the
element. The body forces at nodes j and m are identical to those given by Eq. (9.1.30)
for node i. Hence, for an element, we have
8 9
>
Rb >
>
>
>
>
>
>
>
>
Z
>
b>
>
>
>
>

<
2prA Rb =
ð9:1:31Þ
f fb g ¼
3 >
Zb >
>
>
>
>
>
>
>
>
>
Rb >
>
>
>
: >
;
Zb
where

Rb ¼ o 2 rr

ð9:1:32Þ

Equation (9.1.31) is a first approximation to the radially directed body force
distribution.

Surface Forces
Surface forces can be found by
f fs g ¼

ðð

½Ns Š T fTg dS

ð9:1:33Þ

S

where again ½Ns Š denotes the shape function matrix evaluated along the surface where
the surface traction acts.
For radial and axial pressures pr and pz , respectively, we have
& '
ðð
pr
T
f fs g ¼
½Ns Š
dS
ð9:1:34Þ
pz
S

For example, along the vertical face jm of an element, let uniform loads pr and pz
be applied, as shown in Figure 9–7 along surface r ¼ rj . We can use Eq. (9.1.34)

9.1 Derivation of the Stiffness Matrix

d

421

Figure 9–7 Axisymmetric element with surface forces

written for each node separately. For instance, for node j, substituting Nj from Eqs.
(9.1.12) into Eq. (9.1.34), we have
"
#( )
ð zm
 pr
0
1 aj þ b j r þ g j z

f fsj g ¼
ð9:1:35Þ
2prj dz

2A
0
aj þ b j r þ g j z  pz
zj
evaluated at r ¼ rj ; z ¼ z
Performing the integration of Eq. (9.1.35) explicitly, along with similar evaluations for
fsi and fsm , we obtain the total distribution of surface force to nodes i, j, and m as
8 9
0>

>
>
> >
>
>
>
>
0>
>
>
>
>
>
>
2prj ðzm À zj Þ < pr =
ð9:1:36Þ
f fs g ¼
>
2
pz >
>
>
>
>
> >
>
>
p >
>
>

> r>
>
:
;
pz
Steps 5–7
Steps 5–7, which involve assembling the total stiffness matrix, total force matrix,
and total set of equations; solving for the nodal degrees of freedom; and calculating
the element stresses, are analogous to those of Chapter 6 for the CST element, except
the stresses are not constant in each element. They are usually determined by one
of two methods that we use to determine the LST element stresses. Either we determine the centroidal element stresses, or we determine the nodal stresses for the element and then average them. The latter method has been shown to be more accurate
in some cases [2].
Example 9.1
For the element of an axisymmetric body rotating with a constant angular velocity
o ¼ 100 rev/min as shown in Figure 9–8, evaluate the approximate body force
matrix. Include the weight of the material, where the weight density rw is 0.283 lb/in 3 .
The coordinates of the element (in inches) are shown in the figure.
We need to evaluate Eq. (9.1.31) to obtain the approximate body force matrix.
Therefore, the body forces per unit volume evaluated at the centroid of the element
are
Zb ¼ 0:283 lb=in 3


422

d

9 Axisymmetric Elements

Figure 9–8 Axisymmetric element subjected to

angular velocity

and by Eq. (9.1.32), we have


!

rev 
rad 1 min 2 ð0:283 lb=in 3 Þ
2
Rb ¼ o rr ¼ 100
ð2:333 in:Þ
2p
min
rev
60 s
ð32:2 Â 12Þ in:=s 2
Rb ¼ 0:187 lb=in 3
2prA 2pð2:333Þð0:5Þ
¼
¼ 2:44 in 3
3
3
fb1r ¼ ð2:44Þð0:187Þ ¼ 0:457 lb
fb1z ¼ Àð2:44Þð0:283Þ ¼ À0:691 lb

ðdownwardÞ

Because we are using the first approximation Eq. (9.1.31), all r-directed nodal
body forces are equal, and all z-directed body forces are equal. Therefore,

fb2r ¼ 0:457 lb
fb2z ¼ À0:691 lb
fb3r ¼ 0:457 lb

d

fb3z ¼ À0:691 lb

9.2 Solution of an Axisymmetric Pressure Vessel

9

d

To illustrate the use of the equations developed in Section 9.1, we will now solve an
axisymmetric stress problem.
Example 9.2
For the long, thick-walled cylinder under internal pressure p equal to 1 psi shown in
Figure 9–9, determine the displacements and stresses.

Figure 9–9 Thick-walled cylinder subjected to internal
pressure