Civil Engineering Reference
In-Depth Information
Substituting, we obtain the unknown displacements as
1
4
k
F
4
k
+
3
4
0
−
U
2
U
4
−
F
+
3
k
{
U
}=
=
=
1
2
k
2
F
+
2
k
F
k
+
0
The required force at node 3 is obtained by substitution of the displacement into the upper
partition to obtain
5
4
F
+
3
4
k
F
3
=−
Finally, the reaction force at node 1 is
3
4
k
As a check on the results, we substitute the computed and prescribed displacements into
the individual element equations to insure that equilibrium is satisfied.
Element 1
F
4
−
F
1
=−
kU
2
=
k
−
k
−
kk
0
U
2
−
kU
2
kU
2
f
(1
1
f
(1
2
=
=
which shows that the nodal forces on element 1 are equal and opposite as required for
equilibrium.
Element 2
3
k
U
2
U
3
3
k
F
4
k
+
3
4
−
3
k
−
3
k
−
=
−
3
k
3
k
−
3
k
3
k
3
F
4
k
−
3
4
k
f
(2
2
f
(2
3
−
=
=
3
F
4
k
+
3
4
k
which also verifies equilibrium.
Element 3
2
k
f
(3)
3
f
(3)
4
U
3
U
4
2
k
−
2
F
2
F
−
2
k
−
2
k
=
=
=
F
k
+
−
−
2
k
2
k
2
k
2
k
Therefore element 3 is in equilibrium as well.
2.3 ELASTIC BAR, SPAR/LINK/TRUSS ELEMENT
While the linear elastic spring serves to introduce the concept of the stiffness ma-
trix, the usefulness of such an element in finite element analysis is rather limited.
Certainly, springs are used in machinery in many cases and the availability of a
finite element representation of a linear spring is quite useful in such cases. The
