Page 317 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 317
304 Introduction to the Finite Element Method Chap. 10
we find
d dT A 1 •• 1 ..
= ml\
dt dii| +
d dT ,l 1 •• 1 ..
= ml\^-^u^ + 3 U2
dt dû 2
which establishes the mass matrix for the axial element as
r ^ \ 2 1
(10.1-7)
“6” [ l 2
The terms of the mass matrix are also available from
m, = / <P<P dm
j
i
Example 10.1-1
Determine the equation of motion for the longitudinal vibration of the two-section
bar shown in Fig. 10.1-6.
Solution: Numbering the joints as 1, 2, and 3, we have two axial elements, 1-2, and 2-3,
with displacements Wj, U2, and Uy Although is zero, we at first allow it to be
unrestrained and later impose its zero value.
The element mass and stiffness terms from Eqs. (10.1-7) and (10.1-3) are as
follows:
2 1 k 1 -1
6 1 2 -1 1
K 2 1 Ir 1 -T
Element b:
6 1 2 -1 1
where = £/!„//„, ki, = EAf,/l^, M„ = mj^, and A/j, = mjf,.
The element matrices have a common coordinate U2, and by superimposing
them, they can be assembled into a 3 X 3 matrix as follows:
2M„ ! 1 0
Mass matrix 2M, + 2M,
0 2M,
k. \ 0
Stiffness matrix
-
0
We note here that the stiffness matrix is singular and does not have an inverse.
This is to be expected because no limitations have been placed on the displacements.
t -
1“
3 Figure 10.1-6.