The philosophy behind the exterior matrix method (EMM) is to represent each segment of a complicated structure with a super-sized matrix,which embeds all eigenfrequency information on the segment under all possible initial conditions. These exterior matrices can then be combined together to produce an equation for which the eigenfrequencies of the entire structure will satisfy. For example,if the segment is a single Euler-Bernoulli beam,the exterior matrix can be solved exactly,allowing any two-dimensional structure composed of Euler-Bernoulli beams to be analyzed^[1],even three-dimensional structures^[2]. Unlike finite element methods,which are simulation models,the EMM is a symbolic model,giving a mathematical representation of the vibrations,which can be used to prove the results about the locations of the eigenfrequencies^[3]. Although it may seem like an incredible amount of work computing the exterior matrix for a certain type of segment,one must realize that once this matrix is computed,any structure involving any number of such types of segments can be analyzed by multiplying the exterior matrices together. Paulsen and Manning^[4] showed that it was actually possible to compute the exterior matrix for a segment without first solving the equations governing the vibrations.

The differential equations determining the in-plane dynamics of an elastic catenary do not admit an exact solution^[5]. Therefore,analyzing the dynamics of a structure consisting of severalinclined cable spans becomes a formidable problem. In Ref. [6],the vibrations of a single span of semi-taut inclined cable were analyzed,in which the leading order approximations had to be expressed in terms of the Airy functions and hyper-geometric functions. In Ref. [7],these results were used to analyze a cable span with clamped endpoints,and the equation for the eigenfrequencies to the leading order was determined. A similar analysis was done in Ref. [8], in which the Bessel functions instead of the Airy functions were used. Ironically,all of the Airy functions and the Bessel functions were combined by use of the Wronskian property. Therefore,the equation for the eigenfrequencies simplified to a form which did not involve higher transcendental functions. This suggested that there ought to be a way of finding the eigenfrequencies to the leading order by which the wave motion did not have to be determined first.

One of the earliest papers which analyzed a structure with more than one inclined cable was Ref. [9],which used the EMM,but only worked out the leading order. The 4 × 4 transfer matrix determined from Ref. [6] was converted to a 6×6 exterior matrix for the inclined cable. Once the exterior matrix for a cable segment was computed,a structure involving several cables could be analyzed by multiplying the exterior matrices together. Once again,the Airy functions were combined by the Wronskian property. Therefore,the exterior matrix did not involve the Airy functions,and in fact only involved the elementary functions. This gave the evidence to show that,though the exterior matrices are larger than the transfer matrices,they are not as complicated.

In this paper,we use the results of Ref. [4] to find the exterior matrix of an inclined cable up to the third-order corrections. We then use this exterior method to find the eigenfrequencies of a particular example of a cable structure involving two cable spans and a point mass. Therefore, we obtain the eigenfrequencies to a high level of precision. Although this paper only goes up to the third-order,it is clear how the perturbation series can be extended to any order of precision.

The purpose of this paper is twofold. Not only do we determine how to find the eigenfrequencies of any cable structure,but also do we demonstrate how many complex vibration problems can be converted to a regular perturbation problem. Hence,this paper is of interest not only to those dealing with cable structures but also to those dealing with vibrations of any compound structure.

The dynamics of an inclined cable are highly nonlinear. Therefore,we focus on the small vibrations which would not significantly change the static performance. Let us consider a section of cable that is at an angle Φ,in which the unstretched cable length is C. If the curve of the static solution is extended,it will eventually have a local minimum. We let s denote the unstretched arc length,measured from this local minimum. Thus,the actual cable begins at a point with s = s₁,and go to the value s₂ = s₁ + C. Note,however,that the static solution is the same if we extend the cable to s = 0. We also let s = σ be the point where the vertical sag is at a maximum. This will be the point where the slope is Φ (see Fig. 1).

Fig. 1 Definition diagram for inclined cable

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The horizontal component of the static tension H is constant throughout the span. If we let W be the weight minus the buoyancy per unit length,then the vertical component of the tension can be given by Ws,since this is the weight of the cable below the point s. Thus,the static tension is

Also,we can find the static angle at any point in the span by φ₀(s) = tan⁻¹(Ws/H). Then,

Note that since the cable stretches a little,φ′0(s) is not exactly the same as the curvature.

The exact static solution computed in Ref. [10] is

It can be used to determine the static solution,but we will not need this in determining the governing equations.

If we let u(s,t) and w(s,t) be the dynamic displacements along the static tangent direction and the static normal directions,respectively,e(s,t) be the dynamic stretching,φ(s,t) be the dynamic angle,and τ(s,t) be the dynamic effective tension,then the linearized governing equations for the vibrations are^[7]:

where m is the mass per unit length,M is the cable mass augmented by the added mass per unit length,A is the unstretched cross sectional area of the cable,and E is Young’s modulus. In order to make the equations unitless,we introduce ℓ as a fixed length whose magnitude is of the order of the lengths of each cable span,e.g.,the average cable span. Thus,x = s/ℓ will be a unitless independent variable.

By assuming that the general solution is a linear combination of the functions as follows:

and then converting λ to the unitless variable

we get the system of ordinary differential equations as follows:

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We also introduced the unitless variables as follows:

The perturbation parameter is ε,and a simple analysis shows that x = O(1/ε). The parameter δ is small,satisfying δ = O(ε^k) for some k > 0 (see Ref. [4] for the details).

Unfortunately,there is no exact solution to this equation. The approximate solutions given in Refs. [6] and [7] involve the hyper-geometric functions and the Airy functions. However,when the eigenfrequencies are computed for a single cable span,these complicated functions will cancel out so that the eigenfrequencies can be expressed in terms of the elementary functions. Since the eigenfrequencies are determined by the exterior matrices,if we can determine the set of equations for which the exterior matrix satisfies,we will have a new way of computing the eigenfrequencies.

The EMM was formally introduced as a way to asymptotically find the eigenfrequencies for a system of serially connected elements in 2007^[11]. However,the idea behind the EMM went back much further. In Ref. [1],exterior matrices were used to find the approximate in-plane eigenfrequencies for a sequence of Euler-Bernoulli beams,even if the connections had angles in them. In Ref. [12],the approximate eigenfrequencies of a curved Euler-Bernoulli beam were analyzed by considering it as the limit of the small segments with small angles between each segment. In Ref. [13],the EMM was used to study the structures with damping components. In Ref. [3],exterior matrices were used to prove the spectral gap property for a boundary controlled system of the Euler-Bernoulli beams.

We begin by first defining the exterior matrix. If

then

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where ext(M) is called the exterior matrix of M. The important properties of the exterior matrices are

where

This last property is called the pseudo-orthogonal property,since if for a non-singular matrix M,then A^T · J · A = J. The final property of the exterior matrices is that if we have a 2 × 4 matrix F and a 4 × 2 matrix B,then

where f is the row vector defined by

and b is the column vector defined by

The name “exterior matrix” comes from the fact that the linear transformation determined by the matrix M induces a linear transformation on the alternating covariant tensors or the exterior forms of the order n/2. The exterior matrix simply converts this induced linear transformation back into a matrix,using an appropriate basis.

What do exterior matrices have to do with the finding eigenfrequencies? Each cable in the structure involves the system of 4 first-order equations given in (1) for x₁ ≤ x ≤ x₂. If we let v = (p,q,T,ψ),we can express this system of equations as v′ = G · v,where G is the matrix defined by

If we let v_i (1 ≤ i ≤ 4) be the solutions satisfying the initial conditions as follows:

we can form the matrix

Then,M(x2) becomes the transfer matrix for the system v′ = G · v from x₁ ≤ x ≤ x₂. That is,if v represents the values of all variables at x₁,then M · v can represent the values of the variables at x₂.

The power of the exterior matrices is revealed when we consider the adding boundary conditions to the problem. Usually,the boundary conditions amount to setting 2 of the 4 variables to 0 at each endpoint. For example,if the two initial conditions at x₁ are

then we can say that

for some w₃ and w₄. Therefore,

If we have two initial conditions at x₂,i.e.,

then we have

It is easy to see how this can be generalized for different sets of initial conditions so that the eigenfrequencies can be found by setting det(F ·M · B) = 0 for the appropriate 2 × 4 matrix F and the 4 × 2 matrix B. However,this can be expressed as f · ext(M) · b,where f is a row vector with only one non-zero element,and b is a column vector with only one non-zero element. Thus,the eigenfrequencies can be solved by setting one of the elements of the exterior matrix to zero. In fact,since there are 6 ways of setting 2 of the 4 variables to zero at each endpoint,the exterior matrix can be thought of as the compilation of all possible boundary value problems for this set of equations.

However,there is another advantage of using the exterior matrices. If there are several components to the structure that are sequentially coupled,then we can find the exterior matrix for each component and use the homomorphism property ext(M·N) = ext(M) ·ext(N). Therefore, we can multiply the component exterior matrices to find the exterior matrix for the entire structure and set the appropriate element to zero to solve for the eigenfrequencies. Furthermore, since we no longer have to worry about the cancelation errors from the determinants,we can make the first-order approximations to the component exterior matrices before multiplying them together. We will still be able to find the first-order asymptotic approximations to the eigenfrequencies.

Thus,the goal becomes to determine the exterior matrix for a component of a structure. If we can find the approximate 6×6 exterior matrix for each component of the structure,we can multiply these matrices together,along with the vectors f and b,and set the result to be zero so as to obtain the approximate eigenfrequencies. We will in fact compute the first three terms for the matrix,which shows how further terms can be computed.

The problem with computing the exterior matrix from the transfer matrices is that the transfer matrices have to be computed fairly accurately in order to ensure that there is at least one first-order accuracy in the exterior matrices. For many applications,we will only be able to computeM approximately,and this can result in problems if we try to compute the approximate exterior matrix from the approximate transfer matrix Hence,the subdominant terms of the transfer matrix affect the leading term of the exterior matrix.

An alternative will be to bypass the transfer matrices altogether,and instead compute the exterior matrix directly from the governing equations. This is possible through Proposition 1 in Ref. [4].

Let M be the transfer matrix for the system dv/(dx) = G· v for a 4×4 matrix G,for which all of the entries are differentiable. Let

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Then,we can create a system of 6 first-order equations in 6 variables by letting

and form the equation dw/(dx) = Z · w. If we let wi (1 ≤ i ≤ 6) be the solutions satisfying the conditions as follows:

then the matrix P(x) = (w₁,w₂,w₃,w₄,w₅,w₆) can be formed,satisfying

Using this proposition on the cable equations,we can find that

Thus,we must solve,at least approximately,the system of equations as follows:

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It might seem silly to convert a system of 4 first-order equations to a system of 6 first-order equations. However,we will be able to solve these equations approximately without introducing the Airy functions,unlike the original system of 4 equations. Furthermore,we can use perturbation techniques to gain further degrees of accuracy. Therefore,we are confident that the precision order in the exterior matrix is the precision order for the computed eigenfrequencies.

In this section,we will find six approximate solutions to (4). There is in fact an exact solution to this set of equations,i.e.,

For the other solutions,we can assume,by proper rescaling,that the orders of w₁,w₂,w₃,w₄, w₅,and w₆ are at most 1.

To eliminate the square roots from the system of equations,we can expand

and all similar terms in a Taylor series. We will let x₀ be a fixed point somewhere between x1 and x₂,so that εx₀ will be of the order 1. We will later show that the final answer does not depend on x0. (We cannot just set x0 to be the midpoint

since the derivative of the midpoint with respect to x₁ and x₂ is non-zero). If we can expand the six equations in (4) in a Taylor series centered at x₀ up to the first three terms,we can get

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To save space,we let ,and T₀ = εx₀,which are all of the order 1. It is clear that each term of the Taylor series is of the order ε less than the preceding one.

Since w^′₂ and w^′₅ must be of the order ε or less,w₁,w^′₁ ,w₃,w^′₃ ,w₄,w^′₄ ,w₆,and w^′₆ are of the order 1,whereas w₂,w^′₂ ,w₅,and w^′₅ are of the order ε. If we make these assumptions,the leading order of (5) becomes

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Using the first and third parts of (6),we can see that S²₀w₁^′′ ~ −η²w₁,which immediately gives us w₁ ~ Ce^±iηt/S₀ . Since we only want one solution,we can assume that w₁ ~ e^±iηt/S₀ ,and then the complex conjugate will give us a second solution. With this assumption w³ ~ w^′₁ ~ iηe^iηt/S₀/S₀, the fourth and sixth parts of (6) can produce w^′′ ₆ + η²w⁶/S₂ 0 ~ 2ihη³e^iηt/S₀/S₀. Then,we can easily solve this via the variation of the parameters,and use the convention of choosing the arbitrary constants so as to minimize the number of the terms in the solution. Therefore, we have w₆ ~ hη²te^iηtS₀/ and w₄ ~ −ihη³te^iηt/S₀/S₀. Finally,the second and fifth parts of (6) produce w^′₂ ~ ε(iη/S³₀ + hη²t/S⁴₀) e^iηt/S₀ and w^′₅ ~ ε ( iη/S⁵ ₀ + hη²t/S⁶ ₀ ) e^iηt/S₀ ,which can be integrated to produce w₂ ~ ε ((h + 1)/S² ₀ − ihηt/S³ ₀ ) e^iηt/S₀ and w₅ ~ ε ( (h + 1)/S⁴ ₀ − ihηt/S⁵ ₀ ) e^iηt/S0 . We can now use the perturbation theory to find w₁ through w6 up to the order ε². This gives us one vector solution as follows:

Then,a second solution y² can be obtained by taking the complex conjugate.

In these equations,x₀ is an arbitrary constant,with the only provision that (x−x₀) is of the order 1. If we consider two such solutions and subtract them,we will get a solution for which w₄,w^′₄ ,w₆,and w^′₆ are of the order 1,whereas w₁,w^′₁,w₂,w^′₂ ,w₃,w^′₃ ,w⁵,and w^′₅ are of the order at most ǫ. We can also assume that w3 and w′3 are of the same order. Then,εw₃ ＜＜w^′₃ . If we make these assumptions in (5),we will find that w₆^′′ ~ −η²w₆/S² ₀ . Thus,we can assume that w₆ ~ e^iηt/S₀ . With the perturbation theory,we can derive another solution as follows:

Again,we can take the complex conjugate to obtain a fourth solution y₄.

For both the sets of the solutions,we can integrate to find w₂ and w₅ without adding an arbitrary constant. We can write these arbitrary constants as the additional solutions for which only w2 and w5 are of the order 1. We already have one such solution that is exact,but since the two arbitrary constants are independent,there must be a second such solution. In fact,by incorporating the exact solution,we can assume that only w5 is of the order 1,and w₁,w^′₁ , w₂,w^′₂ ,w₃,w^′₃ ,w₄,w^′₄ ,w^′₅ ,w₆,and w^′₆ are all of the order ǫ or smaller. The fifth part of (5) indicate w^′₅ ~−εT₀w₅/S⁴ ₀. This can be easily solved to produce a solution

Apparently,this solution will not involve the exponential. Therefore,we can assume that the derivatives decrease the order of the function. Hence,w^′₃ is of the order ε² or less. With this information,the third part of (5) to the order ε becomes

Hence,Likewise,by assuming that w^′₆ << ε,the sixth part of (4) to the order ε gives us w₄ ~ hη²w₁ − εw₅/S²₀. Therefore,

If we plug this value of w4 into the fourth part of (4),we can get

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This shows that w₃ and w₆ are of the same order. Therefore,from the first part of (4),we have

Then,we can use (7) to find

Finally,we can use the second part of (4) to compute w^′₂ and integrate

We can use the perturbation theory to find the corrections to the order ε² and produce the vector. Writing this solution as a single vector yields

Finally,we can find the Taylor series of the exact solution expanded about x = x₀ up to ε²,i.e.,

We can assemble these vectors to form a matrix Q,whose columns are six linearly independent solutions,i.e.,

By Proposition 3 of Ref. [4],if we form the matrix S by computing

where F(x) is any anti-derivative of tr(G),then S will not depend on x. Since the trace of G is

an anti-derivative would be

We can expand e^−F(x) in a Taylor series about x₀ to get

i.e.,

Note that this indeed does not depend on x or t. Since this is a non-singular matrix,we can compute C = S⁻¹,i.e.,

We can now use Proposition 4 in Ref. [4] to find P(x₁,x₂),the exterior matrix for the segment going from x₁ to x₂,i.e.,

Although this matrix is a complicated mess,MATHEMATICA can show that it indeed does not depend on x₀. That is,if we replace x₀ with x₀ + Δx,T₀ with T₀ + εΔx,and S₀ with

and re-expand to the order ǫ,then the matrix remains unchanged. Thus,we can replace x0 with the midpoint (x₁ + x₂)/2,which simplifies the matrix considerably.

However,there is a way to simplify it even more with the pseudo-orthogonal property as follows:

In fact,we can use the way that we constructed P to compute

Thus,if we normalize P by multiplying by then

will be pseudo-orthogonal,i.e.,

We can now display the entries of the matrix . Because of the pseudo-orthogonal property, many of the elements will differ only by minus signs. Therefore,we can display two at one time. Whenever two matrix entries are mentioned,the first will use the upper part of the ± or signs,whereas the second will use the lower part of the signs. We will use the abbreviations

To understand the meaning of this matrix,let us consider the boundary conditions. If both ends of the cable are clamped,we have p(s₁) = 0,q(s₁) = 0,p(s₂) = 0,and q(s₂) = 0. Thus, the transfer matrix equation for a single cable can be expressed as follows:

where * represents any value. This will have a non-trivial solution if and only if

Using the exterior matrices,we can find that f_clamped · P_cable · b_clamped = 0,where

Thus,the eigenfrequencies of a single cable span with both ends clamped can be found by setting P_1,4 = 0. Other matrix entries represent the 35 other ways corresponding the boundary conditions to the cable span.

Compare this matrix entry with the results of Ref. [7]. Triantafyllou and Grinfogal7] found the eigenfrequencies by setting

Converting this to our current variables,i.e.,

and scaling by 4ε cos³ Φ/η⁴ yield,after using some trig identities,

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This matches the entry for P1,4 except for two terms,but the discrepancies can be explained. The last term in (8) is of the order ǫ4. Therefore,it will not appear in our analysis until we calculate two more perturbation orders. This term was included in Ref. [7] to show that was not the exact eigenfrequency for the integral k. They calculated higher order terms where η was near these values,a region they called the “cross-over” region. Hence, they kept the leading order terms and the largest term that did not go to zero in the cross-over region. Also,The P1,4 entry contains a δ2 term,which was not included in Ref. [7] because it was a higher order term and went to zero in the cross-over region.

We now want to find the exterior matrix for a point mass M0. We will use the convention that a − superscript refers to the cable span preceding the point mass,and a + superscript refers to the cable span following the point mass (see Fig. 2). We will let θ = φ⁺₀ − φ⁻₀ ,the difference between the static angles at the point mass. From Fig. 2,we can see that φ⁻ is negative.

Fig. 2 Effect of point mass between two cables

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We can assume that the motion is periodic so that we can use e^λt to model this motion. Then,the equations governing the connection are as follows:

(i) The horizontal displacement is

(ii) The vertical displacement is

(iii) The horizontal mass times acceleration is

(iv) The vertical mass times acceleration is

From these equations,we can solve to form the transfer matrix as follows:

From (2),we can form the exterior matrix for a point mass,and convert it to the unitless variables,i.e.,

where R = M0/(Mℓ) is the ratio of the added mass to the average cable mass,and

We now want to find the vibrations of a particular example. A steel spiral strand cable 13mm in the diameter has the Young modulus of E = 150±10 kN/mm². Therefore,EA ≈ 2× 10⁷ N. The mass of the cable is 0.63 kg/m,and the weight of this cable is 6.18N/m. If we have a 50m cable spanning 45m,and add a 100 kg weight somewhere on the span,we can ask how the vibrations of the cable depend on the position of the mass.

We will consider various positions of the mass,either 10m,15m,20m,or 25m from one end (see Fig. 3). We will let ℓ = 25m,since this is the average of the lengths. Then,R = M₀/(Mℓ) will be 6.349. Since the cable is not underwater,we can use h = 1.

Fig. 3 50m cable with 100 kg mass at various positions

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We can now find the eigenfrequencies of an entire structure by combining the exterior matrices of each segment. For the two segments separated by a point mass,with both ends clamped, we can solve the equation as follows:

For other structures,involving more cable spans and/or other components,such as pulleys or dampers,we simply add more matrices to the chain.

Table 1 shows the steady state data needed for each position. We can plug in the parameters from Table 1 into the above matrices,multiply them out,and determine the values of η which are roots of the function produced. The first 18 roots for the four mass positions are given in Table 2.

Table 1 Parameters for various positions of mass in Fig. 3 )

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Table 2 Eigenfrequencies for various mass positions )

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In Table 2,the eigenfrequencies that are close to an eigenfrequency for the single left-hand cable are marked with a ⁺,and the eigenfrequencies that are close to an eigenfrequency for the right hand cable alone are marked with a ⁻. Thus,the vibrations for the eigenfrequencies marked + will mainly occur on the left-side,whereas most of the vibrations corresponding to the eigenfrequencies marked ⁻ occur on the right-side. Since the first eigenfrequencies at all the four positions are neither ⁺ or ⁻,both sides must vibrate significantly. This will indicate that the mass moves significantly in the first node.

Notice that most of the eigenfrequencies of the final position are double roots. This comes from the symmetry of the final position. Even though the first two eigenfrequencies are close,they are not double roots,and both involve the vibration of the mass. In one vibration,the mass will be bobbing up and down,whereas in the other,the mass will be mainly swaying back and forth.

We have analyzed the vibrations of a cable structure with inclined cables. The Airy functions or elliptic functions do not need to be utilized as in Refs. [7] and [6]. These functions are needed in approximating the solution to the governing equations of the inclined cable. Instead, we consider finding the equations governing the exterior matrix of the inclined cable. Even though this produces a system of 6 first-order equations,we are able to approximate it with a perturbation series involving only elementary functions. Since we have found the exterior matrix for a single cable segment,we can analyze many other arrangements involving multiple masses and/or pulleys simply by multiplying the corresponding matrices together. Hence,the EMM is a powerful tool for analyzing vibrations.