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Wunderlich, W., and Redanz, W., 1995, Die Methode der Finiten Elemente, in Der Ingenieurbau: Grund-
wissen , W. Ernst & Sohn, Berlin.
Zienkiewicz, O.C., 1977, The Finite Element Method in Engineering Science, 3rd ed., McGraw-Hill, New
York,
Problems
6.1 Suppose the structure of Fig. 6.14a, with E
0, is stiffened
with steel rods placed along the outer edge of the structure. If the rods have circular
cross-sections with 0.2 m diameters, compute the distribution of displacements and
stresses throughout the structure. Use the structural properties, e.g., dimensions, of
Example 6.2. For the steel rods, E
=
30 GN/m 2
and
ν =
207 GN/m 2 ,
=
ν =
0
.
3
.
Question: Should these rods be beams or extension bars? If they are beams, how
should the mismatch in DOF between the slopes at the nodes of the beams and the
translations at the nodes of the planar elements be handled?
Answer:
For rods treated as bars with longitudinal motion only.
Node Number
u x (m)
u y (m)
10 6
1
9
.
163
×
0.0
2
6
.
397
×
10 6
2
.
737
×
10 5
10 5
3
0.0
3
.
522
×
4
3
.
747
×
10 7
8
.
572
×
10 6
10 8
10 5
5
9
.
155
×
2
.
885
×
6
0.0
3
.
844
×
10 5
.
×
10 6
.
×
10 5
7
8
926
1
146
10 6
10 5
8
6
.
455
×
3
.
558
×
9
0.0
4
.
917
×
10 5
6.2 Assume that the thin structure of Fig. 6.14a is stiffened with diagonal rods, as well as
bars placed along the outer edge. All of the bars are made of steel
m 2 ,
(
E
=
207 GN
/
with 0.2 m diameters. Continue to use the structural properties of Example
6.2. Compute the distribution of displacement and stresses throughout the struc-
ture. Note that the solution to this problem will involve local-to-global coordinate
transformations.
6.3 Use the displacement method to compute the deformed profile of the structure shown
in Fig. P6.3.
6.4 The structure shown in Fig. P6.4a is modeled with two plane stress elements. Obtain
the shape functions for the elements and calculate the element stiffness matrices.
ν =
0
.
3
)
Hint:
Use the element (local) coordinate system as shown in Fig. P6.4b and assume
that
u x
=
u 1
+
u 2 x
+
u 3 y,
u y
=
u 4
+
u 5 x
+
u 6 y
Answer:
u x 1
u y 1
u x 2
u y 2
u x 3
u y 3
u x
u y
N 1
0
N 2
0
N 3
0
=
0
N 1
0
N 2
0
N 3
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