14.4 Free Vibrations of One-Span Beams with Uniformly Distributed Mass 545
AtxD 0 the Krylov–Duncan functions and their second derivatives equal zero.
According to properties (14.24), onlyT.kx/andV.kx/functions satisfy these con-
ditions. So, the expression for the mode shape is
X.x/DC 2 T.kox/CC 4 V.kx/:ConstantsC 2 andC 4 are calculated from boundary conditions atxDl W
X.l/D 0 IX^00 .l /D 0
X.l/DC 2 T.kl/CC 4 V.kl/D0;
X^00 .l /Dk^2 ŒC 2 V.kl/CC 4 T.kl/D0:(a)A nontrivial solution of the above system is the frequency equation
ˇ
ˇ
ˇ
ˇT.kl/ V.kl/
V.kl/ T.kl/ˇ
ˇ
ˇ
ˇD^0 !T(^2) .kl /V (^2) .kl /D0:
According (14.22), the last formula may be presented as sinklD 0. The roots of
the equation are
DklD ; 2 ;:::
So the frequencies of the free vibration are
!iD
^2 i
l^2
r
EI
m
;! 1 D
2
l^2
r
EI
m
;! 2 D
4 2
l^2
r
EI
m
;:::
The mode shape of vibration is
X.x/DC 2 T.kx/CC 4 V.kx/DC 2
T.kix/C
C 4
C 2
V.kix/
Since the ratioC 4 =C 2 from first and second equations (a) are
C 4
C 2
D
T.kil/
V.kil/
D
V.kil/
T.kil/
;
theni-th mode shape (eigenfunction) corresponding toi-th frequency of vibration
(eigenvalue) is
X.x/DC 2 T.kx/CC 4 V.kx/DC
T.kix/
T.kil/
V.kil/
V.kix/
DC
T.kix/
V.kil/
T.kil/
V.kix/
; (b)
Since the Krylov–Duncan functionsT. /DV. /; T.2 /DV.2 /;:::so the
mode shapes are
Xi.x/DCŒT.kix/V.kix/DCsinkixDCsin
i
l
x; iD1; 2; :::(c)
The first and second modes are shown in Fig.14.15.