restart;

interface(imaginaryunit=II):

with(LinearAlgebra):

The Euler-Bernoulli beam equation
.

We wish to determine the natural modes of vibration of

a possibly multi-span Euler-Bernoulli beam.

Separate the variables by setting .   We get
-

whence
.

Let .  Then

 and

The idea behind the Krylov-Duncan technique is to express  

in terms an alternative (and equivalent) set of basis
functions  through ,, as
,

where the functions  through  are defined in the next section.

In some literature the symbols , are used for these

functions but I find it more sensible to use the indexed function

notation.

The Krylov-Duncan approach is particularly effective in formulating
and finding a multi-span beam's natural modes of vibration.

The Krylov-Duncan functions

The K[i](x) defined by this proc evaluates to the th

Krylov-Duncan function.

Normally the index  will be in the set, however the proc is

set up to accept any integer index (positive or negative).  The proc evaluates

the index modulo 4 to bring the index into the set .   For

instance, K[5](x) and K[-3](x)i are equivalent to K[1](x) .

Here are the Krylov-Duncan basis functions:

and here is what they look like.  All grow exponentially for large
but are significantly different near the origin.

The cyclic property of the derivatives: 
We have .  Let's verify that:

The fourth derivative of each   function equals itself. This is a consequence of the cyclic property:

The essential property of the Krylov-Duncan basis function is that their

zeroth through third derivatives at  form a basis for :

As noted earlier, in the case of a single-span beam, the modal  shapes

are expressed as
.

Then, due to the cyclic property of the derivatives of the Krylov-Duncan

functions, we see that:
.
.
.
Let us note, in particular, that
,
,
,
.

A general approach for solving multi-span beams

In a multi-span beam, we write  for the deflection of the th span, where

 and where  is the span's length.  The  coordinate indicates the

location within the span, with  corresponding to the span's left endpoint.

Thus, each span has its own  coordinate system.

 
We assume that the interface of the two adjoining spans is supported on springs

which (a) resist transverse displacement proportional to the displacement (constant of

proportionality of   (d for displacement), and (b) resist rotation proportional to the
slope (constant of proportionality of   (t for torsion or twist). The spans are numbered

from left to right. The interface conditions between spans  and +1 are

The special case of a pinned support corresponds to  and .
In that case, condition 3 above implies that ,
and condition 4 implies that

Let us write the displacements  and  in terms of the Krylov-Duncan

functions as:

Then applying the cyclic properties of the Krylov-Duncan functions described

earlier, the four interface conditions translate to the following system of four
equations involving the eight coefficients .

,

which we write as a matrix equation
.

That  coefficient matrix plays a central role in solving

for modal shapes of multi-span beams.  Let's call it .

Note that the value of  enters that matrix only in combinations with
 and .  Therefore we introduce the new symbols

,    .

The following proc generates the matrix .  The parameters  and  

are optional and are assigned the default values of infinity and zero, which

corresponds to a pinned support.

The % sign in front of each Krylov function makes the function inert, that is, it
prevents it from expanding into trig functions.  This is so that we can

see, visually, what our expressions look like in terms of the  functions.  To

force the evaluation of those inert function, we will apply Maple's value function,

as seen in the subsequent demos.

Here is the interface matrix for a pinned support:

And here is the interface matrix for a general springy support:

Note:  In Maple's Java interface, inert quantities are shown in gray.

Note:  The  in this matrix is the length of the span to the left of the interface.
Recall that it is  , not , in the derivation that leads to that matrix.

In a beam consisting of  spans, we write the th span's deflection  as

Solving the beam amounts to determining the  unknowns

which we order as

At each of the  interface supports we have a set of four equations as derived
above, for a total of  equations.  Additionally, we have four user-supplied

boundary conditions -- two at the extreme left and two at the extreme right of the

overall beam.  Thus, altogether we have  equations which then we solve for the

 unknown coefficients .   

The user-supplied boundary conditions at the left end are two equations, each in the
form of a linear combination of the coefficients .  We write  for the
 coefficient matrix of that set of equations.  Similarly, the user-supplied boundary
conditions at the right end are two equations, each in the form of a linear combination
of the coefficients .  We write  for the  coefficient matrix of
that set of equations.   Putting these equations together with those obtained at the interfaces,

we get a linear set of equations represented by a  matrix which can be assembled

easily from the matrices , and .  In the case of a 4-span beam the

assembled matrix looks like this:

The pattern generalizes to any number of spans in the obvious way.

For future use, here we record a few frequently occurring  and  matrices.

The following proc builds the overall matrix in the general case.  It takes
two or three arguments.  The first two arguments are the  matrices
which are called  and  in the discussion above.  If the beam
consists of a single span, that's all the information that need be supplied.
There is no need for the third argument.

In the case of a multi-span beam, in the third argument we supply the
list of the interface matrices  , as in , listed in order
of the supports,  from left to right.   An empty list is also
acceptable and is interpreted as having no internal supports,
i.e., a single-span beam.

For instance, for a single-span cantilever beam of length  we get the following  matrix:

For a two-span beam with with span lengths of  and , and all three
supports pinned,  we get the following  matrix:

The matrix  represents a homogeneous linear system (i.e., the right-hand side vector

is zero.)  To obtain a nonzero solution, we set the determinant of  equal to zero.

That gives us a generally transcendental equation in the single unknown .  Normally

the equation has infinitely many solutions.  We call these  

Remark: In the special case of pinned supports at the interfaces, that is, when
, the matrix  depends only on the span lengths .
It is independent of the parameters  that enter the Euler-Bernoulli
equations.  The frequencies , however, depend on those parameters.

This proc plots the calculated modal shape corresponding to the eigenvalue .
The params argument is a set of equations which define the  numerical values

of all the parameters that enter the problem's description, such as the span

lengths.

It is assumed that in a multi-span beam, the span lengths are named  etc.,
and in a single-span beam, the length is named .

A single-span pinned-pinned beam

Here we calculate the natural modes of vibration of a single span

beam, pinned at both ends.  The modes are of the form

The matrix  is:

The characteristic equation:

We conclude that the eigenvalues are .

A non-trivial solution of the system  is in the null-space of :

Here are the weights that go with the Krylov functions:

and here is the deflection:

We see that the shape functions are simple sinusoids.

A single-span free-free beam

Here we calculate the natural modes of vibration of a single span

beam, free at both ends.  The modes are of the form
.

The reasoning behind the calculations is very similar to that in the

previous section, therefore we don't comment on many details.

The characteristic equation:

Let .  Then the characteristic equation takes the form

Here are the graphs of the two sides of the characteristic equation:

The first three roots are:

A single-span clamped-free cantilever

We have a cantilever beam of length .  It is clamped at the

left end, and free at the right end.

Let .  Then the characteristic equation takes the form

Here are the graphs of the two sides of the characteristic equation:

A dual-span pinned-pinned-free cantilever beam

We have a two-span beam of span lengths  and , with the left end of the
first span pinned, the right end of the second span free, and the interface

between the spans on a pinned support.  .

The characteristic equation:

That equation does not seem to be amenable to simplification.  The special case of , however,

is much nicer:

Let :

That expression grows like , so we divide through by that to obtain

a better-behaved equation

Here are the first three roots:

A dual-span clamped-pinned-free cantilever beam

We have a two-span beam of span lengths  and , with the left end of the
first span clamped, the right end of the second span free, and the interface

between the spans on a pinned support.  This is different from the previous

case only in the nature of the left boundary condition.

The characteristic equation:

That equation does not seem to be amenable to simplification.  The special case of , however,

is much nicer:

Let :

That expression grows like , so we divide through by that to obtain

a better-behaved equation

Here are the first three roots:

A triple-span free-pinned-pinned-free beam

We have a triple-span beam with span lengths of .  The beam is supported

on two internal pinned supports.  The extreme ends of the beam are free.

The graphs of the first three modes agree with those

in Figure 3.22 on page 70 of the 2007 article of
Henrik Åkesson, Tatiana Smirnova, Thomas Lagö, and Lars Håkansson.

In the caption of Figure 2.12 on page 28 the span lengths are given
as 3.5, 5.0, 21.5.

The characteristic equation

That graphs grows much too fast to be useful.  We moderate it by dividing through
the fastest growing cosh term:

Here are the first three roots:

A triple-span free-spring-spring-free beam

We have a triple-span beam with span lengths of .  The beam is supported

on two internal springy supports.  The extreme ends of the beam are free.
The numerical data is from the worksheet posted on July 29, 2020 at
https://www.mapleprimes.com/questions/230085-Elasticfoundation-Multispan-EulerBernoulli-Beamthreespan#comment271586

The problem is pretty much the same as the one in the previous section, but the

pinned supports have been replaced by spring supports.

This section's calculations require a little more precision than

Maple's default of 10 digits:

Calculate the determinant of .  The result is quite large, so we terminate the command
with a colon so that not to have to look at the result.  If we bothered to peek,  however, we
will see that the determinant has a factor of .  But that quite obvious by looking at the
entries of the matrix shown above. Two of its rows have  in them and another two have
. When multiplied, they produce the overall factor of .

Here are the parameters that the determinant depends on:

So we provide values for those parameters:

Here is the characteristic equation.  We multiply it by  to remove the singularity at

We can't see anything useful in that graph.  Let's limit the vertical range: