The first equation and the boundary condition N(0, t) = 0 implies that N(x, t) = 0.
Then differentiate the third equation with respect to x, insert the first and the second
equation, insert N = 0 and M = EIu′′, and obtain:
EIu
0000
þ qA€ u ¼ 0:
ð2:32Þ
Assuming a time harmonic solution:
uðx; tÞ ¼ uðxÞ sinðxt þ wÞ;
ð2:33Þ
an ordinary differential equation results:
u
0000
¼ k
4 u; k
4
qAx
2
=EI:
ð2:34Þ
The boundary conditions are u(0, t) = u′′(0, t) = u(l, t) = u′(l, t) = 0, or,
expressed in terms of u(x):
uð0Þ ¼ u
00
ð0Þ ¼ uðlÞ ¼ u
0
ðlÞ ¼ 0
ð2:35Þ
Equations (2.34)–(2.35) constitute an EVP. The general solution of (2.34):
uðxÞ ¼ c 1 sinðkxÞ þ c 2 cosðkxÞ þ c 3 sinhðkxÞ þ c 4 coshðkxÞ;
ð2:36Þ
must satisfy the boundary conditions (2.35). This gives c 2 = c 4 = 0, and:
c 1 sinðklÞ þ c 3 sinhðklÞ ¼ 0
c 1 cosðklÞ þ c 3 coshðklÞ ¼ 0:
ð2:37Þ
For nontrivial solutions (c 1 , c 3 ) to exist the determinant of coefficients must
vanish; This gives the frequency equation:
tan a ¼ tanh a; a kl:
ð2:38Þ
Solutions to this transcendent equation are to be found numerically. The lowest
three are a 1 = 3.9266023ÁÁÁ, a 2 = 7.0685716ÁÁÁ and a 3 = 10.210164. The eigenvalues k j are given by k j = a j /l, and the natural frequencies x j of the beam are then
given by the definition of k in (2.34):
x j ¼ a j
l
À
Á 2 ffiffiffiffiffiffiffiffiffiffiffiffiffi
EI=qA
p
; j ¼ 1; 1:
ð2:39Þ
The corresponding eigenfunctions – with vibration problems better known as
mode shapes or normal modes – are obtained by inserting c 2 = c 4 = 0, and one of
the equations in (2.37) into (2.36), giving:
2.5 Vibration-Related EVPs
61