CITE THIS NOTEBOOK: Analytical solution of the eigenvalue problem for the elastic cable loaded with a mass point by Markus Wenin, Wolfram Community, SEP 28 2022
ORIGINAL ARTICLE: Wenin, M. Analytical solution of the eigenvalue problem for the elastic cable loaded with a mass point. Arch Appl Mech (2022). https://doi.org/10.1007/s00419-022-02254-7
Original Article Abstract: In this paper, we investigate the oscillations of an elastic cable loaded with a mass point analytically. The eigenvalue problem for the linearized system with small sag/span ratio, yielding the eigenfrequencies and eigenmodes, is formulated and solved. For this purpose, we use a computer algebra system to handle the long expressions arising throughout the derivation and in the final results. We compare the analytical solution (valid for arbitrary masses and positions within the span) with numerical simulations. A main new result of this work is the derivation of a function f(ω), for which the zeros correspond to the eigenfreqeuncies for the inplane oscillations, valid as long as the vertical displacement is much greater than the horizontal displacement. Furthermore, we compare our outcomes with those of Cheng and Perkins (J Acoust Sos Am 91(5):2654–2662, 1992; J Appl Mech 59:923–928, 1992) and recover the classical results of the linear theory for the unloaded cable.

Introduction

The eigenfrequencies and eigenmodes of an elastic cable loaded with a point mass are computed within the linear cable theory of Irvine [1]. A detailed description of the theory is given in our publication [2]. The cable is spanned horizontally with horizontal component tension H and loaded with a mass point at position
X
m
. The mass point divides the span in a left and right region. We solve for both regions the equations of motion. The boundary conditions at the left and right anchors and
theposition
X
m
are used to find the eigenfrequencies. The displacement from equilibrium is {u,v,w}. The horizontal displacement u is calculated as in [1] using the compability equation, which relates elastic strength with geometrical cable length. Import assumptions are:
- the span is flat and the parabola approximation is applicable,
- the additional (to static) cable tension (related to elastic lengthening) is piecewise constant within the left and right region,
- the vertical displacement v ≫ u (must be check a posteriori).
We use the Fourier transform u(x,t) =
iωt
e

u
(x), v(x,t) =
iωt
e

v
(x). The out of plane oscillations w are not considered here (see [2] for this).
In[]:=

Outline of analytical calculations. Definitions, basic equations

Definition of cable equilibrium configuration, vertical displacement and auxilliary quantities. Note that the second derivative is equal for both regions
(
ρ
0
= mass/length, g = earth acceleration, H = horizontal component of cable tension, m = mass of attached load
In[]:=
ClearAll["Global`*"]​​
y
l
[x_]:=-
ρ
0
g
2H
(
2
x
-
X
1
x)-
mg
H
X
m
X
1
-1x​​
y
r
[x_]:=-
ρ
0
g
2H
(
2
x
-
X
1
x)-
mg
H
x
X
1
-1
X
m
​​dyldx[x_]=
y
l
'[x]//Simplify;​​dyrdx[x_]=
y
r
'[x]//Simplify;​​d2yldx2=dyldx'[x];
Solution of the equation of motion for the vertical displacement (left and right).
A
l
,

h
l
,
A
r
,

h
r
are integration constants
In[]:=

v
l
[x_]:=
A
l
Sin[kx]+

h
l
ρ
0
g
2
H
2
k
(1-Cos[kx])​​

v
r
[x_]:=
A
r
Sin[kx]-
A
r
Tan[k
X
1
]+

h
r
ρ
0
g
2
H
2
k
Cos[k
X
1
]
Cos[kx]+

h
r
ρ
0
g
2
H
2
k
Check boundary conditions and continuity at
X
m
In[]:=
y
l
[0]​​
y
r
[
X
1
]​​
y
l
[
X
m
]-
y
r
[
X
m
]
Out[]=
0
Out[]=
0
Out[]=
0
Check differential equations and boundary conditions
In[]:=
H

v
l
''[x]+

h
l
d2yldx2+
ρ
0
H
2
k
ρ
0

v
l
[x]//Simplify​​H

v
r
''[x]+

h
r
d2yldx2+
ρ
0
H
2
k
ρ
0

v
r
[x]//Simplify​​

v
l
[0]​​

v
r
[
X
1
]//Simplify
Out[]=
0
Out[]=
0
Out[]=
0
Out[]=
0

Setup of the four linear equations and extraction of coefficients

Setup of eq. 1, continuity of

v
[
X
m
]
In[]:=
eq1=Numerator[Together[(

v
l
[
X
m
]-

v
r
[
X
m
])]]//Simplify
Out[]=
2
H
2
k
Sin[k
X
m
]
A
l
+
2
H
2
k
Sec[k
X
1
]Sin[k(
X
1
-
X
m
)]
A
r
+g
ρ
0
-(-1+Cos[k
X
m
])

h
l
+(-1+Cos[k
X
m
]Sec[k
X
1
])

h
r

Setup of eq. 2 (equation of motion for the mass point in vertical direction)
In[]:=
eq2=NumeratorTogetherH(

v
r
'[
X
m
]-

v
l
'[
X
m
])+

h
r
dyrdx[
X
m
]-

h
l
dyldx[
X
m
]+m
2
ω

v
l
[
X
m
]//Simplify
Out[]=
-2
2
H
2
k
(HkCos[k
X
m
]-m
2
ω
Sin[k
X
m
])
A
l
X
1
+2
3
H
3
k
Cos[k(
X
1
-
X
m
)]Sec[k
X
1
]
A
r
X
1
+g2H
2
k
m
X
m


h
l
-

h
r
+H
2
k
2
X
1
ρ
0
-

h
l
+

h
r
-2
X
1
H
2
k
m+(-m
2
ω
+m
2
ω
Cos[k
X
m
]+HkSin[k
X
m
]-H
2
k
X
m
)
ρ
0


h
l
+Hk(-Sec[k
X
1
]Sin[k
X
m
]+k
X
m
)
ρ
0

h
r

Evaluation of the compatibility equation for the left side (boundary condition for

u
l
[0] = 0 already implemented)
In[]:=
dsdx3l[x_]:=1+
3
2
dyldx[x]^2​​​​exprl=

h
l
AE
Integrate[dsdx3l[x],{x,0,
X
m
}]-Integrate[dyldx[x]

v
l
'[x],{x,0,
X
m
}]//Simplify;(*=utl[xm]*)
Evaluation of the compatibility equation for the right side (boundary condition

u
r
[
X
1
] = 0 already implemented)
In[]:=
dsdx3r[x_]:=1+
3
2
dyrdx[x]^2;​​exprr=-

h
r
AE
Integrate[dsdx3r[x],{x,
X
m
,
X
1
}]-Integrate[dyrdx[x]

v
r
'[x],{x,
X
m
,
X
1
}]//Simplify;(*=utr[xm]*)
Setup of eq. 3
In[]:=
eq3=(exprl-exprr)//Simplify;​​eq3=Numerator[Together[eq3]]//Simplify
Out[]=
8AEg
2
H
2
k
Sec[k
X
1
]Sin
1
2
k(-
X
1
+
X
m
)
A
r
X
1
X
1
2Sin
1
2
k(-
X
1
+
X
m
)+kCos
1
2
k(-
X
1
+
X
m
)
X
1
ρ
0
-2kCos
1
2
k(-
X
1
+
X
m
)
X
m
(m+
X
1
ρ
0
)-8AEg
2
H
2
k
Sin
k
X
m
2

A
l
X
1
-2kmCos
k
X
m
2

X
m
+kCos
k
X
m
2

2
X
1
ρ
0
+2
X
1
kmCos
k
X
m
2
+Sin
k
X
m
2
-kCos
k
X
m
2

X
m
ρ
0
+8
3
H
3
k
2
X
1
X
m

h
l
+12
2
g
H
3
k
2
m
2
X
1
X
m

h
l
-24
2
g
H
3
k
2
m
X
1
2
X
m

h
l
+12
2
g
H
3
k
2
m
3
X
m

h
l
-8AE
2
g
km
2
X
1
ρ
0

h
l
+8AE
2
g
kmCos[k
X
m
]
2
X
1
ρ
0

h
l
+8AE
2
g
km
X
1
X
m
ρ
0

h
l
-8AE
2
g
kmCos[k
X
m
]
X
1
X
m
ρ
0

h
l
+12
2
g
H
3
k
m
3
X
1
X
m
ρ
0

h
l
-24
2
g
H
3
k
m
2
X
1
2
X
m
ρ
0

h
l
+12
2
g
H
3
k
m
X
1
3
X
m
ρ
0

h
l
+8AE
2
g
Sin[k
X
m
]
2
X
1
2
ρ
0

h
l
-8AE
2
g
k
2
Sin
k
X
m
2

3
X
1
2
ρ
0

h
l
-8AE
2
g
kCos[k
X
m
]
2
X
1
X
m
2
ρ
0

h
l
+3
2
g
H
3
k
4
X
1
X
m
2
ρ
0

h
l
-6
2
g
H
3
k
3
X
1
2
X
m
2
ρ
0

h
l
+4
2
g
H
3
k
2
X
1
3
X
m
2
ρ
0

h
l
+8
3
H
3
k
3
X
1

h
r
-8
3
H
3
k
2
X
1
X
m

h
r
+12
2
g
H
3
k
2
m
X
1
2
X
m

h
r
-12
2
g
H
3
k
2
m
3
X
m

h
r
-8AE
2
g
km
X
1
X
m
ρ
0

h
r
+8AE
2
g
kmCos[k
X
m
]Sec[k
X
1
]
X
1
X
m
ρ
0

h
r
+12
2
g
H
3
k
m
2
X
1
2
X
m
ρ
0

h
r
-12
2
g
H
3
k
m
X
1
3
X
m
ρ
0

h
r
-8AE
2
g
Sec[k
X
1
]Sin[k
X
m
]
2
X
1
2
ρ
0

h
r
-4AE
2
g
k
3
X
1
2
ρ
0

h
r
-4AE
2
g
kCos[k
X
m
]Sec[k
X
1
]
3
X
1
2
ρ
0

h
r
+
2
g
H
3
k
5
X
1
2
ρ
0

h
r
+8AE
2
g
kCos[k
X
m
]Sec[k
X
1
]
2
X
1
X
m
2
ρ
0

h
r
-3
2
g
H
3
k
4
X
1
X
m
2
ρ
0

h
r
+6
2
g
H
3
k
3
X
1
2
X
m
2
ρ
0

h
r
-4
2
g
H
3
k
2
X
1
3
X
m
2
ρ
0

h
r
+8AE
2
g
2
X
1
2
ρ
0

h
r
Tan[k
X
1
]
Setup of eq. 4 (equation of motion for the mass point in horizontal direction)
In[]:=
dudxl[x_]:=

h
l
AE
dsdx3l[x]-dyldx[x]

v
l
'[x]​​dudxr[x_]:=

h
r
AE
dsdx3r[x]-dyrdx[x]

v
r
'[x]​​​​eq4=TogetherH(dudxr[
X
m
]-dudxl[
X
m
])+

h
r
-

h
l
+m
2
ω
exprl//Simplify;​​eq4=Numerator[eq4]//Simplify
Out[]=
4AEg
3
H
4
k
Cos[k(
X
1
-
X
m
)]Sec[k
X
1
]
A
r
X
1
-
2
X
1
ρ
0
+2
X
m
(m+
X
1
ρ
0
)+4AEg
2
H
2
k
A
l
X
1
2km(-HkCos[k
X
m
]+m
2
ω
Sin[k
X
m
])
X
m
+k(HkCos[k
X
m
]-m
2
ω
Sin[k
X
m
])
2
X
1
ρ
0
+2
X
1
km(HkCos[k
X
m
]-m
2
ω
Sin[k
X
m
])+(m
2
ω
(-1+Cos[k
X
m
])-k(HkCos[k
X
m
]-m
2
ω
Sin[k
X
m
])
X
m
)
ρ
0
-8AE
3
H
3
k
2
X
1

h
l
-8
4
H
3
k
2
X
1

h
l
-12
2
g
2
H
3
k
2
m
2
X
1

h
l
+24
2
g
2
H
3
k
2
m
X
1
X
m

h
l
+8
3
H
3
k
m
2
ω
2
X
1
X
m

h
l
+12
2
g
H
3
k
3
m
2
ω
2
X
1
X
m

h
l
-12
2
g
2
H
3
k
2
m
2
X
m

h
l
-24
2
g
H
3
k
3
m
2
ω
X
1
2
X
m

h
l
+12
2
g
H
3
k
3
m
2
ω
3
X
m

h
l
-8AE
2
g
k
2
m
2
ω
2
X
1
ρ
0

h
l
+8AE
2
g
k
2
m
2
ω
Cos[k
X
m
]
2
X
1
ρ
0

h
l
+8AE
2
g
H
2
k
mSin[k
X
m
]
2
X
1
ρ
0

h
l
-12
2
g
2
H
3
k
m
3
X
1
ρ
0

h
l
+8AE
2
g
k
2
m
2
ω
X
1
X
m
ρ
0

h
l
-8AE
2
g
k
2
m
2
ω
Cos[k
X
m
]
X
1
X
m
ρ
0

h
l
-8AE
2
g
H
2
k
mSin[k
X
m
]
X
1
X
m
ρ
0

h
l
+36
2
g
2
H
3
k
m
2
X
1
X
m
ρ
0

h
l
+12
2
g
H
3
k
2
m
2
ω
3
X
1
X
m
ρ
0

h
l
-24
2
g
2
H
3
k
m
X
1
2
X
m
ρ
0

h
l
-24
2
g
H
3
k
2
m
2
ω
2
X
1
2
X
m
ρ
0

h
l
+12
2
g
H
3
k
2
m
2
ω
X
1
3
X
m
ρ
0

h
l
+8AE
2
g
m
2
ω
Sin[k
X
m
]
2
X
1
2
ρ
0

h
l
-8AE
2
g
km
2
ω
2
Sin
k
X
m
2

3
X
1
2
ρ
0

h
l
+4AE
2
g
H
2
k
Sin[k
X
m
]
3
X
1
2
ρ
0

h
l
-3
2
g
2
H
3
k
4
X
1
2
ρ
0

h
l
-8AE
2
g
km
2
ω
Cos[k
X
m
]
2
X
1
X
m
2
ρ
0

h
l
-8AE
2
g
H
2
k
Sin[k
X
m
]
2
X
1
X
m
2
ρ
0

h
l
+12
2
g
2
H
3
k
3
X
1
X
m
2
ρ
0

h
l
+3
2
g
H
3
k
m
2
ω
4
X
1
X
m
2
ρ
0

h
l
-12
2
g
2
H
3
k
2
X
1
2
X
m
2
ρ
0

h
l
-6
2
g
H
3
k
m
2
ω
3
X
1
2
X
m
2
ρ
0

h
l
+4
2
g
H
3
k
m
2
ω
2
X
1
3
X
m
2
ρ
0

h
l
+8AE
3
H
3
k
2
X
1

h
r
+8
4
H
3
k
2
X
1

h
r
+12
2
g
2
H
3
k
2
m
2
X
m

h
r
+8AE
2
g
H
2
k
mSec[k
X
1
]Sin[k
X
m
]
X
1
X
m
ρ
0

h
r
-12
2
g
2
H
3
k
m
2
X
1
X
m
ρ
0

h
r
+24
2
g
2
H
3
k
m
X
1
2
X
m
ρ
0

h
r
-4AE
2
g
H
2
k
Sec[k
X
1
]Sin[k
X
m
]
3
X
1
2
ρ
0

h
r
+3
2
g
2
H
3
k
4
X
1
2
ρ
0

h
r
+8AE
2
g
H
2
k
Sec[k
X
1
]Sin[k
X
m
]
2
X
1
X
m
2
ρ
0

h
r
-12
2
g
2
H
3
k
3
X
1
X
m
2
ρ
0

h
r
+12
2
g
2
H
3
k
2
X
1
2
X
m
2
ρ
0

h
r
We would like examine all matrix elements independently. Extraction of individual coeffients, eq. 1
Check extraction
Extraction of individual coeffients, eq. 2
Check extraction
Extraction of individual coefficients, eq. 3
Check extraction
Extraction of coefficients, eq. 4
Check extraction

Special case: m = 0

Here we check our result with the cable theory of Irvine [1]. She System matrix now is simpler
Determinant of the system matrix contains both the symmetric and antisymmetric modes respectively
Comparison with Irvine result: antisymmetric modes
Comparison with Irvine result: symmetric modes (some definitions from [1] are introduced)
Difference our result from Irvine
Here we check our result with the special case when the mass point is in the middle of the span [3]. System matrix
1. factor (antisymmetric modes)
2. factor (antisymmetric modes, since it gives the remaining solutions for m = 0)
3. factor (symmetric modes, since it gives Irvines result for m = 0)

A numerical example

1. Equilibrium cable configuration (note that the horizontal component of the cable tension is constant, and not the length of the tensionless cable). Definition of some variables.
2. Definition of the 4 x 4 matrix and the function f(ω)
3. Computation of the zeros using FindRoot command with running initial guess
Computation of poles (used only in Plot below)
4. Plot of the function f(ω) and the zeros. The poles are excluded. Useful (and necessary) to confirm the results of the zero-search process
5. Determination of eigenfunctions
6. Plot of eigenfunctions

References

[1] H. M. Irvine and T. K. Caughey, The linear theory of free vibrations of a suspendet cable, Proc. R. Soc. Lond. A. 341, 299-315 (1972)
[2] M. Wenin, Analytical solution of the eigenvalue problem for the elastic cable loaded with a mass point. Arch Appl Mech (2022). https://doi.org/10.1007/s00419-022-02254-7​
[3] E. Pagani and M. Wenin, Abstract Book of the 25th International Congress of Theoretical and Applied Mechanics, pp. 2389-2390 (2020)