Page 318 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 10.1 Element Stiffness and Mass 305

The first and third rows of the stiffness matrix, as it stands, result in kj^u^ - Uj) =

kfj{u2 —W3) = 0, which indicates that no relative motion between coordinates takes
place, a situation corresponding to rigid-body translation.
If we fix point 1 so that W = 0, then the first column of the matrices can be
j
deleted. The second and third rows then result in the following equation for the
longitudinal vibration of the two-section bar:
1
6 2M,

Special case. IÎ A ^ = A_ K ^ ^ the
previous problem becomes that of a uniform bar of total length L and total mass
M, solved as a 2-DOF system with coordinates at the midpoint and the free end.
The equation of the previous problem then becomes
M ■ 4 r lEA 2 -1 0
12 1 2 . r n + L -1 1 _ ^ ^0

“ 3
If we let A = o)^ML/2AEA, the characteristic equation for the natural

frequencies becomes
(2 - 4A) -(1 -f A)
= 0
-(1 +A) (1 -2A)
or
= 0
7
Its solution results in
EA
1.6115
0.1082 ME
)
A = O =
1.3204 T Â
5.6293
ML
The natural frequencies of the uniform bar in longitudinal vibration are
known and are given by the equation = (2n + W it/ 2)yJEA /ML . Results
computed from this equation for the first two modes are
EA
1.5708
ML
4.7124
ML
Comparison between the two indicates that the agreement between the results of
the 2-DOF finite element model and that of the continuous model is 2.6 percent
high for the first mode and 19.5 percent high for the second mode. A three-ele­
ment model will of course be expected to give closer agreement.