Applied Numerical Methods: Triple Pendulum Vibration Analysis

2018
SKMM3023 APPLIED NUMERICAL METHODS
PROJECT 1 & 2

PROJECT 1
The equation of motion for free vibration of the triple pendulum is given as,
m l2
[ 3 2 1
2 2 1
1 1 1 ] { ̈θ1
̈θ2
̈θ3
}+ mgl [ 3 0 0
0 2 0
0 0 1 ] {
θ1
θ2
θ3
}= {
0
0
0 }
To obtain the standard eigenvalue problem to find the natural frequencies for the vibration,
θi ( t )=θi cos ωt i=1,2,3 ...
PART I
For the different masses given as m1, m2, & m3:
m ̈θ ( t ) +c ̇θ ( t ) + kθ ( t ) =p ( t ) + P
The system has three masses suspended in air and comprising of rigid bodies. The Lagrange’s
equations for the system considering its dynamic nature,
d
dt ( dT
d qn )− dT
d qn
+ dV
d qn
=Qn n=1,2,3 , ... .
For the geometric parameters of the system, the center of mass for the different bars is given as a
function of
θi∧ y∧x
The relationship is given as,
y1= d1
2 cos θi
y2=d1 cos θi+ d2
2 cos θ2
y3=d1 cos θ1+ d2 cos θ2+ d3
2 cos θ3
x1= d1
2 cos θi
1

x2=d1 cos θi + d2
2 cos θ2
x3=d1 cos θ1 +d2 cos θ2 + d3
2 cos θ3
It is governed by the Newtonian second law of motion which states that,
m d2 x
dt =−kx
Where the values of x ( 0 )=X 0 and dx
dt ( 0 ) =V 0
Taking the differential of the equation,
m d2 x
d t2 +kx =0
The solution of the equation is obtained as,
−ω2 mx ( t ) +kx ( t )=0=x ( t )∗(−ω2 m+ k )
The pendulum motion is referenced as,
mL d2 θ
d t2 = 1
L ( T g +T s )
θ(t)= A eiωt
(−mL ω2 + 1
L ( mgL+ k ) )θ ( t )=0
ω= √ mgL+k
m L2
The matrix notation is given as,
A=
[3 2 1
2 2 1
1 1 1 ]
B= [ 3 0 0
0 2 0
0 0 1 ]
2

C= {
0
0
0 }
The eigenvalues are obtained as,
Ax=λx
( a−λI ) x =0
The eigenvalues are the angular frequencies of the vibration and they are obtained as,
λ=ω2
The eigenvectors are the mode shape of the vibration,
x=
[c1
c2
c3 ]
For the system of equation in the triple pendulum, using a free body diagram when there is a load
on the simple pendulum,
θ2 ≅sin θ2 ≅ x2−x1
L
θ3 ≅ sinθ3 ≅ x3−x2
L
The string tensions are given as,
T 1 ≅ 3 mg
T 2 ≅ 2 mg
T 3 ≅ mg
The vertical forces tend to cancel out and the horizontal direction forces follow the second law
such that,
m ̈x2=−T2 sin θ2 +T3 sin θ3
¿−2 mg
L ( x2−x1 ) + mg
L ( x3−x2 )
¿−2 mω0
2 ( x3−x2 )
3