MODELLING OF MECHANICAL SYSTEM VOLUME 2 Episode 6 ppsx

178 Structural elements
Figure 3.16. Local and global coordinate systems
structure since the local reference frames differ necessarily along the horizontal
beam and along the legs. It is precisely this transformation which produces the
coupling terms between bending and axial terms which arise in the [K] matrix. Let
us consider the beam element of Figure 3.16, which is assumed to lie in the Oxz
plane. The local frame can be transformed into the global frame by the product of
the translation
−→
IO and the plane rotation α. Neither the forces nor the displacements
are modified by a translation, which thus can be disregarded. Obviously, the same
is not true as far as the rotation is concerned. The rotation is described by the matrix
[R] which relates the displacements by [q]=[R][q
′
].
The transformation rule of the node labelled J is:









X
J
Y
J
Z
J

ψ
xJ
ψ
yJ
ψ
zJ









=







cos α 0 sin α 000
010 000
−sin α 0 cos α 000
000cosα 0 sin α
000 010
000−sin α 0 cos α


















X
′
J
Y
′
J
Z
′
J
ψ
′
xJ
ψ
′

yJ
ψ
′
zJ










[3.110]
More generally the rotation matrix takes the form:
[R]=

[R][0]
[0][R]

; with [R]=


ℓ
x
′
m
x
′

n
x
′
ℓ
y
′
m
y
′
n
y
′
ℓ
z
′
m
z
′
n
z
′


[3.111]
Straight beam models: Hamilton’s principle 179
where ℓ
i
′
, m
i

′
, n
i
′
are the director cosines of the local axes as expressed in the global
system.
Since energy is invariant through a transformation of axes, the element matrices
and force vectors are found to comply with the following conditions:
[
q
I
q
J
][K
(n)
]

q
I
q
J

=[
q
′
I
q
′
J
][K

′(n)
]

q
′
I
q
′
J

⇒[K
(n)
]=[R]
T
[K
′(n)
][R]
[
˙q
I
˙q
J
][M
(n)
]

˙q
I
˙q
J


=[
˙q
′
I
˙q
′
J
][M
′(n)
]

˙q
′
I
˙q
′
J

⇒[M
(n)
]=[R]
T
[M
′(n)
][R]
[
q
I
q

J
]

Q
I
Q
J

=[
q
′
I
q
′
J
]

Q
′
I
Q
′
J

⇒

Q
I
Q
J


=[R]
T

Q
′
I
Q
′
J

[3.112]
Once the element matrices and vectors are suitably transformed into the global
system of coordinates, the finite element model can be assembled according to the
same rules as described in subsection 3.4.2.5.
Colour Plate 3 shows the elastic response of a portal frame to a concentrated
force as computed by using the finite element software CASTEM 2000 [CAS 92].
The finite element model uses two nodes beam elements. The material is steel and
the beams have uniform rectangular cross-sections 20 cm × 10 cm. The height of
the vertical legs is 8 m and the length of the horizontal beam (lintel) is 12 m. The
legs are assumed to be clamped at the lower ends. A concentrated force is applied at
some point of the lintel, load magnitude is 2.5 kN. which is assumed successively to
be vertical (case a), along the horizontal beam (case b), and finally in the transverse
Oy direction (case c). For each loading case, two figures are shown. The figures
on the left-hand side show the global reference frame in green, the finite element
mesh in black, the external force represented as a red arrow and finally the support
reactions as blue arrows for the forces and magenta arrows for the moments. The
length of the arrows is determined by a magnifying factor 1000. The figures on
the right-hand side show the undeformed and the deformed frame (the black and
red lines respectively). Real displacements are magnified by a factor 100 to make

deformations clearly noticeable.
In case (a), thelintel and the legs are deflected. As expected, maximumdeflection
occurs along the lintel. The reaction forces comprise a vertical component equal
to the shear forces at the ends of the lintel, transmitted to the legs as compressive
forces, and horizontal components in the plane of the frame, which are equal to
the shear forces at the clamped ends of the legs. The moments are perpendicular to
the plane of the frame. They are clearly related to the bending of the legs. As the
loaded point is nearer to one leg than the other, the two legs are not flexed by the
same amount, but in the ratio of the lengths of the lever arms defined by the loaded
point and the extremities of the lintel.
180 Structural elements
In case (b), maximum deflection occurs in the legs, which are deflected by
the same amount. The bending moments at the junction between the legs and the
lintel induce in-plane bending of the latter. Deflection is found to be antisymmetric
because the moments exerted at the ends of the lintel are the same. As a consequence,
one leg is compressed and the other is stretched.
In case (c), the legs and the lintel are deflected out of the frame plane. However,
due to the asymmetrical position of the loaded point with respect to the two legs,
the portal frame is also twisted about the vertical direction. Accordingly, though
the major component of the moment reactions is horizontal and, in the plane of the
frame, equal to half the load times the length of the legs, they also have a vertical
component which differs from one leg to the other and a horizontal component in
the out-of-plane direction.
3.4.3.4 Transverse load resisted by string and bending stresses in
a roof truss
The example of the portal frame was used in the last subsection to bring out that in
an assembly of connected straight beams transverse shear stresses can be transmitted
as longitudinal stresses to another part of the assembly due to geometrical effects.
This leads also to a coupling of the basic modes of deformation of the individual
beams. However, because of the large difference which usually exist between the

coefficients of the longitudinal and flexure stiffness matrices [3.98] and [3.106],
suitably scaled by the coefficients K
ℓ
= ES/ℓ and K
b
= EI/ℓ
3
respectively,
coupling is barely detectable when looking at the deformed portal frame. Indeed,
the axial displacements of the vertical legs are found to be quite negligible in
comparison with the transverse deflection due to bending. To analyse this point
further, it is instructive to study the assembly depicted in Figure 3.17 which stands
Figure 3.17. Beam assembly used as roof truss
Straight beam models: Hamilton’s principle 181
for a roof truss made of two identical straight beams (A
1
A
2
) and (A
2
A
3
) tilted
by the angle α with respect to the horizontal Ox direction. The beams are rigidly
connected to each other at (A
2
) and the rooting points (A
1
) and (A
3

) are assumed
to be either pinned or clamped. A vertical force

F
0
is applied to (A
2
).Ifα = 0,
the assembly responds to the load as a pinned-pinned, or clamped-clamped flexed
beam and if α = 90
◦
, it responds as a compressed beam. It is thus of interest to
analyse the response of the roof truss when α is varied from 0 to 90
◦
. An analytical
solution of this problem can be obtained conveniently by using a finite element
model reduced to two beam elements and to the nodal displacements of interest.
Let us consider first the pinned-pinned configuration. In the local coordinate
system of the first element (A
1
A
2
), the displacement variables are denoted U
1
, U
2
(axial displacements), W
1
, W
2

(transverse displacements in the Oxz plane), ϕ
1
, ϕ
2
(small rotations about the Oy axis). These variables define the vector displacement
[U
1
W
1
ϕ
1
U
2
W
2
ϕ
2
]
T
. By using the matrix [3.109], the stiffness matrix
of the first element is written as:
K
ℓ








10 0−10 0
012γ −6γℓ 0 −12γ −6γℓ
0 −6γℓ 4γℓ
2
06γℓ 2γℓ
2
−10 0 1 0 0
0 −12γ 6γℓ 012γ 6γℓ
0 −6γℓ 2γℓ
2
06γℓ 4γℓ
2







where γ =
K
b
K
ℓ
=
I
Sℓ
2
≃
1

12η
2
≪ 1
η is the slenderness ratio of the element, assumed to be much larger than one.
In the same way, the vector displacement of the second element (A
2
A
3
) is
written as:

U
2
W
2
ϕ
2
U
3
W
3
ϕ
3

T
The stiffness matrix is the same as that of the first element.
However, due to the boundary conditions, U
1
= U
3

= W
1
= W
3
= 0 and the
displacement vectors and stiffness matrices can be reduced as follows:

U
1
W
1
ϕ
1
U
2
W
2
ϕ
2

T
⇒

ϕ
1
U
2
W
2
ϕ

2

T
182 Structural elements
K
ℓ







10 0−10 0
012γ −6γℓ 0 −12γ −6γℓ
0 −6γℓ 4γℓ
2
06γℓ 2γℓ
2
−10 0 1 0 0
0 −12γ 6γℓ 012γ 6γℓ
0 −6γℓ 2γℓ
2
06γℓ 4γℓ
2








⇒ K
(1)
= K
ℓ




4γℓ
2
06γℓ 2γℓ
2
010 0
6γℓ 012γ 6γℓ
2γℓ
2
06γℓ 4γℓ
2





U
2
W
2
ϕ

2
U
3
W
3
ϕ
3

T
⇒

U
2
W
2
ϕ
2
ϕ
3

T
K
ℓ








10 0−10 0
012γ −6γℓ 0 −12γ −6γℓ
0 −6γℓ 4γℓ
2
06γℓ 2γℓ
2
−10 0 1 0 0
0 −12γ 6γℓ 012γ 6γℓ
0 −6γℓ 2γℓ
2
06γℓ 4γℓ
2







⇒ K
(2)
= K
ℓ




1000
012γ −6γℓ −6γℓ
0 −6γℓ 12γ 2γℓ

2
0 −6γℓ 2γℓ
2
4γℓ
2




Then, K
(1)
is rotated by the angle α by using the coordinate transformation rule:



ψ
1
X
2
Z
2
ψ
2



=




10 0 0
0 C −S 0
0 SC0
00 0 1






ϕ
1
U
2
W
2
ϕ
2



; where C = cos α and S = sin α
The rotated stiffness matrix is found to be:
K
(1)
α
= K
ℓ





4γℓ
2
6γℓS 6γℓC 2γℓ
2
6γℓS C
2
+ 12γS
2
CS(12γ −1) 6γℓS
6γℓC CS(12γ −1)S
2
+ 12γC
2
6γℓC
2γℓ
2
6γℓS 6γℓC 4γℓ
2




Straight beam models: Hamilton’s principle 183
K
(2)
is rotated by the angle −α by using the coordinate transformation rule:




X
2
Z
2
ψ
2
ψ
3



=



C +S 00
−SC00
0010
0001






U
2
W
2

ϕ
2
ϕ
3



The rotated stiffness matrix is found to be:
K
(2)
α
= K
ℓ




C
2
+ 12γS
2
−CS(12γ −1) 6γℓS 6γℓS
−CS(12γ −1)S
2
+ 12γC
2
−6γℓC −6γℓC
6γℓS −6γℓC 4γℓ
2
2γℓ

2
6γℓS −6γℓC 2γℓ
2
4γℓ
2




The finite element model of the problem is assembled as:






4γℓ
2
6γℓS 6γℓC 2γℓ
2
0
6γℓS 2(C
2
+ 12γS
2
) 012γℓS 6γℓS
6γℓC 02(S
2
+ 12γC
2

) 0 −6γℓC
2γℓ
2
12γℓS 08γℓ
2
2γℓ
2
06γℓS −6γℓC 2γℓ
2
4γℓ
2











ψ
1
X
2
Z
2
ψ
2

ψ
3





=





0
0
−F
0
0





Finally, the solution can be conveniently obtained by noticing that because
of the symmetry of the problem with respect to the Oz axis, it is necessary that
X
2
= 0 and ψ
3
=−ψ

1
. As a consequence:
Z
2
(α) =
−F
2(S
2
+ 3γC
2
)K
ℓ
=
−F
2(K
ℓ
(sin α)
2
+ 3K
b
(cos α)
2
)
[3.113]
and
ψ
3
=−ψ
1
=

3Z
2
cos α
2ℓ
The solution [3.113] brings out that the structure resists the vertical force essen-
tially through its longitudinal stiffness as soon as the tilt angle is larger than
1/η. Such a result is of paramount importance for designing structures resistant
184 Structural elements
Figure 3.18. Deflection versus the tilt angle of the roof truss
Figure 3.19. Roof truss in pinned-pinned configuration, from the above to the bottom
plots: unloaded and loaded structure, plots of the vertical displacement in meters and
flexure angle in radians
Straight beam models: Hamilton’s principle 185
Figure 3.20. Roof truss in the clamped-clamped configuration: unloaded and loaded
structure
Figure 3.21. Roof truss α =45
◦
, in pinned-pinned configuration: unloaded and loaded
structure, plots of the vertical displacement in meters and flexure angle in radians
to transverse loads, as will be further detailed in Chapters 7 and 8 devoted
to curved structures. Figure 3.18 refers to a roof truss made of wood beams
(E = 10
9
Pa, ℓ = 10 m, square cross-sections a = 20 cm) loaded by a vertical
force F = 10 kN applied to (A
2
). It plots the vertical displacement of (A
2
) versus
the tilt angle in degree. The dashed lines stand for the limit cases α = 0 and

α = 90
◦
. As soon as α>5
◦
magnitude of the displacement is less than 1% than
its value at α = 0.
Figure 3.19 refers to the pinned-pinned configuration. It shows the deflection
of the whole structure for α = 5
◦
, as computed by using a finite element model
comprising 40 two node beam elements. The load vector and the non-deformed
structure are also represented. Vertical displacement and flexure angle are plotted
along the beams.
Figure 3.20 refers to the clamped-clamped configuration. The magnitude of
the maximum deflection is less than in the pinned-pinned configuration (10.9 cm
instead of 14.7 cm) as the result of a stiffening effect in bending. A simple analytical
186 Structural elements
calculation carried out in the same manner as above gives for the clamped-clamped
configuration:
Z
2
(α) =
−F
2(K
ℓ
(sin α)
2
+ 12K
b
(cos α)

2
)
and
ψ
1
= ψ
2
= ψ
3
= 0
Finally, Figure 3.21 refers to a roof truss α = 45
◦
, in pinned-pinned con-
figuration. The deflection is drastically reduced in comparison with the case
α = 5
◦
.
3.4.4 Saving DOF when modelling deformable solids
The methodology followed to model deformable solids as equivalent beams is
worth brief mention to emphasize the practical importance of minimizing judi-
ciously the number of degrees of freedom when modelling continuous systems.
This can be sketched graphically, as shown in Figure 3.22. Figure 3.22a sketches
the ‘real’ structure as a 3D solid, supporting a load distributed as a 2D pressure
field. In Figure 3.22b we take advantage of the slenderness of the body to reduce
the size of the problem by shifting from a 3D medium to an equivalent 1D medium,
via a beam model. Finally, in Figure 3.22c the beam is discretized in finite ele-
ments. In this way, the final model to be solved has a finite number of DOF
1
Z
1

= 0
′ ′
Z
1
= 0F
n
2N
Z
N +1
= 0
′ ′
Z
N +1
= 0
N+1
(c)
f(x)
(b)
P
(a)
Figure 3.22. From the deformable solid to the finite element model
Straight beam models: Hamilton’s principle 187
and can be solved by using a computer. The whole process justifies the interest
in discretizing a continuum; one other specificity of this modelling procedure is
its ability to identify the most pertinent DOF in order to minimize their number.
This step is partially fulfilled by a good appreciation of the structure behaviour.
In the following chapter another efficient method for suppressing superfluous
degrees of freedom is described, based on the natural modes of vibration of the
structures.
Chapter 4

Vibration modes of straight beams and
modal analysis methods
As already outlined in Chapter 1, modes of vibration arise as a natural concept
in the study of the free vibrations of mechanical systems, discrete or continuous. In
the continuous case, they are defined mathematically as many solutions of an eigen-
value problem involving the stiffness and mass operators introduced in Chapter 3.
Provided the system is self-adjoint and statically stable, the discrete and infinite
sequence of eigenvalues are positive and the related eigenvectors are real. From
a physical viewpoint, they describe standing waves in which the material of the
structure vibrates harmonically about a static and stable position of equilibrium,
at specific frequencies and according to specific space shapes. Amongst several
interesting properties, the most important one certainly is that the mode shapes can
be used as an orthonormal vector basis to transform the partial derivative equations
of motion into a set of time differential equations described by modal stiffness
and mass matrices which operate on the so-called natural, or modal, coordinates
of the material system. This new discretization procedure gives rise to a so called
modal model, in which the response properties of the structure are characterized
by a set of modal oscillators, instead of a set of finite elements. Many linear and
even nonlinear problems of continuous mechanics are much more efficiently solved
starting from a modal model, instead of a finite element model.
Modal analysis methods 189
4.1. Introduction
From the mathematical viewpoint, the natural modes of vibration of elastic
structures arise as the solutions of an eigenvalue problem written in the canonical
form as:

K(r) − ω
2
M(r)


ϕ(r) =

0
+ self-adjoint homogeneous boundary conditions [4.1]
where K(r) and M(r) are the linear stiffness and mass operators of the mater-
ial system.
The differential system [4.1] is the continuous counterpart of the algebraic modal
equation encountered in the analysis of discrete systems. It can be demonstrated
that, when provided with a suitable set of self-adjoint boundary conditions, it has
an infinite sequence of nontrivial solutions called natural, or normal, modes of
vibration, which can be used to determine an orthonormal basis with respect to
the stiffness and mass operators of the associated Hilbert space. These infinitely
numerous solutions are enumerated by using one, two or three integer indices,
in agreement with the dimension of the Euclidean space of the structural model.
Considering here the 1D case of beams, the eigenvalues are denoted ω
2
n
, n =
1, 2, and the related eigenvectors are denoted ϕ
n
(r). Furthermore, the ω
n
are
positive or eventually nil in the absence of supports, and the ϕ
n
(r) are real vectors.
The relations of orthogonality of mode shapes are formulated as follows:

ϕ
i

(r), M[ϕ
j
(r)]

(V)
=

M
n
> 0; if i = j = n
0; if i = j

ϕ
i
(r), K[ϕ
j
(r)]

(V)
=

K
n
≥ 0; if i = j = n
0; if i = j
[4.2]
These mathematical properties, of paramount importance from both the theor-
etical and the practical standpoints, extend to the continuous case those already
established in the modal analysis of discrete systems (see [AXI 04], Chapters 5
and 6). The only new feature is that n extends to infinity. A formal proof of such

results requires however rather advanced mathematical developments in the spec-
tral theory of partial derivative operators which are beyond the scope of this book.
The reader interested in the subject is referred to mathematically oriented textbooks
for instance [COL 63], [STA 70].
The vibration equations of straight beams are simple enough to provide us with
convenient structural examples in order to study the methods of modal analysis
in the case of continuous systems. In particular, it will be shown that many prob-
lems can be efficiently solved by projecting first the local equations of motion on
190 Structural elements
a suitable modal basis. Using this procedure, the problem is restated in terms of dis-
crete, but still infinitely many unknowns, termed modal displacements or normal
coordinates, which can be understood as a particularly convenient set of general-
ized displacements. At this step, the remaining problem is to truncate judiciously
the modal series to a finite number of terms. It turns out that, in many applications,
the dimension of the functional space can be efficiently reduced to a few DOF only,
by truncating the modal series in accordance with the physical peculiarities of the
specific problem to be solved.
Section 4.2 is concerned with the natural modes of vibration of straight beams.
In section 4.3, the basic principles of mode shape expansion methods are intro-
duced and criteria to truncate the modal series suitably are established. As could
be anticipated, such criteria are based on the spatial and spectral features of the
excitation signals in relation to those of the natural modes of vibration. Finally,
the rate of convergence of modal series related to displacements and stresses are
briefly discussed. All these notions are illustrated by means of a few examples.
In section 4.4 the so called substructuring method is introduced, according to
which a complex structure may be partitioned into simpler substructures, each
substructure being finally discretized by using a truncated modal basis. Finally, in
section 4.5 a few impacting beam problems are worked out to demonstrate that
the modal analysis method can be also recommended for dealing with nonlinear
problems, especially when nonlinearities are concentrated in a few specific places

in the physical system. In the following chapters other examples concerning arches,
plates and shells will be given.
4.2. Natural modes of vibration of straight beams
4.2.1 Travelling waves of simplified models
The nature and the speed of the elastic waves propagating in a solid body depend
on the kinematic assumptions made to model it. This arises as a consequence of
the necessity to specify, at least implicitly, the boundary conditions which hold
in the directions not considered in the simplified model, for instance at the lateral
surface bounding a beam, at the faces of a plate, etc.
4.2.1.1 Longitudinal waves
For a longitudinal wave, the equilibrium equation of a straight uniform beam in
which the Poisson effect is neglected is:
−E
∂
2
X
∂x
2
+ ρ
¨
X = 0 [4.3]
Modal analysis methods 191
From the calculation made in Chapter 1, subsection 1.4.3, the phase speed of
the longitudinal wave is readily found to be:
c
ϕ
= c
0
=


E/ρ [4.4]
As this velocity is independent of the pulsation, the model predicts non-
dispersive waves which are the one-dimensional counterpart of the plane dilatation
waves propagating in a three-dimensional elastic solid. The speed given by [4.4]
is however less than that given by [1.82], in the ratio:
c
0
c
L
=

(1 + ν)(1 − 2ν)
(1 − ν)
[4.5]
For instance, in the case of steel ν = 0.3 and c
0
∼
=
0.862c
L
. At first sight, such a
result is somewhat surprising as it could be expected that the stiffness arising from
the beam model would be higher than that arising from the corresponding 3D solid
body; this because of the rigidifying assumption made in the beam model that the
cross-sections are not deformed. Actually it turns out that, to the contrary, by neg-
lecting the Poisson effect, the longitudinal stiffness of the beam is decreased. This
is because no transverse stresses arise in the 1D model to oppose the longitudinal
deformation, as is the case in the 3D solid. Indeed, according to the 3D elastic law
[1.19] we get:
(σ

xx
)
3D
=
(1 − ν)E
(1 + ν)(1 − 2ν)
∂X
∂x
; (σ
yy
)
3D
= (σ
yy
)
3D
=
νE
(1 + ν)(1 − 2ν)
∂X
∂x
whereas according to the 1D model:
(σ
xx
)
1D
= E
∂X
∂x
; (σ

yy
)
1D
= (σ
yy
)
1D
≡ 0
So, in a rigorous analysis of the problem, it would be necessary to use a 3D elastic
law and to prescribe vanishing local stresses at the lateral surface of the beam, in
accordance with the kinematic conditions which prevail in an axially guided wave.
As pointed out in Chapter 1 subsection 1.4.6, this difficult problem has been solved
by Pochammer in the particular case of a circular cylindrical rod. The dispersion
equation thus obtained defines multiple distinct branches k
n
(ω) and only approx-
imations to the first lower branches can be recovered by adopting simplified beam
models. Fortunately, such approximations are found to be sufficiently accurate in
most engineering applications.
On the other hand, it has been outlined in Chapter 3 subsection 3.2.1 that the basic
beam model can be improved by taking into account the inertia related to the lateral
deformations of the beam cross-sections due to the Poisson effect. This additional
inertia becomes significant if the wavelength is not very large in comparison with
192 Structural elements
the transverse dimensions of the beam. This leads to the Love–Rayleigh equation
[3.11], written here in the case of a cylindrical rod as:
c
2
0
∂

2
X
∂x
2
−
¨
X + β
2
∂
2
¨
X
∂x
2
= 0; where β
2
=
ν
2
R
2
2
The phase speed of the Love–Rayleigh model, as expressed in terms of the wave
number k, is found to be:
c
c
0
=

1

1 +
1
2
(kνR)
2
; where k =
2π
λ
[4.6]
These waves are thus found to be dispersive and c decreases as kR increases,
tending to c
0
=
√
E/ρ when kR decreases, as shown in Figure 4.1, where λ/R is
the reduced wavelength. As expected, the phase speed is less than that of the basic
model because the Poisson ratio effect is accounted for in the inertia term only.
Furthermore, as expressed in terms of the pulsation, the phase speed is found to be:
c
c
0
=

1 −
ω
2
β
2
c
2

0
[4.7]
According to [4.7], there is a cut-off frequency f
c
= c
0
/2πβ such that only the
frequencies less than f
c
can propagate.
Figure 4.1. Phase speed of the longitudinal waves according to the Love–Rayleigh model
Modal analysis methods 193
Figure 4.2. Propagation of a longitudinal triangular wave according to the Love–Rayleigh
model
Figure 4.2 refers to the propagation of a longitudinal triangular wave of half-
duration τ = 53.32 µs along a cylindrical steel rod (R = 4cm) according to
the Love–Rayleigh model. The cut-off frequency is f
c
= 94 kHz, whereas the
maximum frequency contained in the triangular signal is f
max
= 1/τ = 19 kHz.
The travelling wave is computed by using the fast Fourier transform algorithm
implemented in MATLAB. The wave at 15 m from the source is marked by a
triangular peak followed by high frequency oscillations of much less magnitude.
4.2.1.2 Flexure waves
Starting from the Bernoulli–Euler model for a uniform beam:
EI
∂
4

Z
∂x
4
+ ρS
¨
Z = 0
The phase speed of the flexure waves is found to be:
c
ψ
= k

EI/ρS = (EI/ρS)
1/4
(ω)
1/2
[4.8]
These waves are highly dispersive. Furthermore, c
ψ
is found to increase without
limit with frequency. Such a behaviour is clearly unrealistic and simply indicates
that the Bernoulli–Euler model is unsuitable at short wavelengths λ/R ≤ 1, as
already discussed in Chapter 3, section 3.2.2. When the wavelength is sufficiently
194 Structural elements
small, contribution of shear deformation and rotatory inertia of the beam cross-
sections must be accounted for. This is the object of the Rayleigh–Timoshenko
model [3.35]:
ρS
∂
2
Z

∂t
2
− ρI

1 +
E
κG

∂
4
Z
∂x
2
∂t
2
+
ρ
2
I
κG
∂
4
Z
∂t
4
+ EI
∂
4
Z
∂x

4
= 0
The related dispersion equation produces the two following branches for the
phase velocity:
c
sb
= c
s




1 +
3 + 2ν
4
(kR)
2
+


1 +
3 + 2ν
4
(kR)
2

2
−
(
1 + ν

)
(kR)
4
2
c
bs
= c
s




1 +
3 + 2ν
4
(kR)
2
−


1 +
3 + 2ν
4
(kR)
2

2
−
(1 + ν)(kR)
4

2
;
where c
s
=

G
ρ
[4.9]
In Figure 4.3, they are plotted versus the reduced wavelength λ/R. The lower
branch identifies with the Bernoulli–Euler branch if λ/R is large enough, as shown
in Figure 4.4. On the other hand, if λ/R is small enough the shear effect prevails,
so the phase speed tends to the shear velocity c
s
. As a consequence, the dispersive
nature of the waves vanishes asymptotically in the range of short wavelengths.
Figure 4.3. Branches of the dispersion equation of the Rayleigh–Timoshenko model
Modal analysis methods 195
Figure 4.4. Phase velocity of the bending-shear coupled waves
Figure 4.5. Propagation of a transverse triangular wave according to the
Rayleigh–Timoshenko model
Figure 4.5 refers to the propagation of a transverse triangular wave of half-
duration τ = 10 ms along a cylindrical steel rod (R = 4cm) according to the
Rayleigh–Timoshenko model. The wave at 15 m from the source is shaped as a
train of oscillations, the frequency of which decreases markedly as time elapses, in
196 Structural elements
agreement with the dispersive law [4.9]. Transverse waves are clearly much more
dispersive than the longitudinal waves.
4.2.2 Standing waves, or natural modes of vibration
The natural modes vibration of a structure are determined according to the

general procedure already introduced in Chapter 1 subsection 1.4.7, which is briefly
summarized here as follows:
1. Find the general solution of the modal equation.
2. Comply with the boundary conditions by adjusting the values of the integration
constants.
A linear and conservative boundary condition is expressed as a linear and homo-
geneous relation between a pair of conjugated displacement and stress variables.
The constant values are specified, except a multiplicative arbitrary factor which can
be set according to a given norm. Thus infinitely numerous natural frequencies and
related mode shapes are produced. However, the problem can be solved analytically
in closed form only in a few cases, in particular when a uniform beam is provided
with standard support conditions at both ends. In what follows, the presentation
will be restricted essentially to those cases. As a general result, it will be noted
that high order modes are found to be rather insensitive to boundary conditions;
this indicates that the ratio of the elastic strain energy of the whole beam over that
induced by the elastic supports increases with the wave number of the mode.
4.2.2.1 Longitudinal modes
According to the basic model, the modal equation is:
d
2
X
dx
2
+

ω
c
0

2

X = 0; with c
0
=

E/ρ [4.10]
The natural circular frequencies and related mode shapes are of the general form:
ω
n
=
̟
n
c
0
L
; ϕ
n
(ξ) = a sin(̟
n
ξ) +b cos(̟
n
ξ); where ξ = x/L [4.11]
The modal stiffness and mass coefficients are given by:
M
n
=
ρSL
2
=
M
2

; K
n
=
1
2
ES
L
̟
2
n
[4.12]
The modal coefficients ̟
n
, a, b, pertinent to standard boundary conditions, are
reported in Table 4.1. It is worth emphasizing the remarkable result that the natural
Modal analysis methods 197
Table 4.1. Longitudinal modes of vibration of a straight beam
B.C. ̟
n
ab
free-free nπ n = 0, 1, 2, 01
fixed-fixed nπ n = 1, 2, 10
free-fixed (1 +2n)
π
2
n = 0, 1, 2, 01
fixed-free (1 +2n)
π
2
n = 0, 1, 2, 10

frequencies form an harmonic sequence and that the mode shapes identify with the
functional vectors used as an orthonormal basis in the Fourier series. As detailed
further in section 4.3, mode shape expansion methods can be rightly understood as
a natural extension of the concept of Fourier series.
example. – fixed-free beam.
The detailed calculation leading to the results gathered in Table 4.1 and formulae
[4.11] is illustrated by taking the case of a uniform beam provided with fixed-free
end conditions. Starting from the equation of motion:
−ES
∂
2
X
∂x
2
+ ρS
∂
2
X
∂t
2
= 0; X(0, t) = ES
∂X
∂x




L
= 0
Its general solution of the harmonic type is written as:

X(x, t) = X
+
e
iω(t−x/c
0
)
+ X
−
e
iω(t+x/c
0
)
At x = 0, the incident wave X
−
is reflected and the sign is changed to com-
ply with the fixed end condition. The free end condition at x = L, gives the
transcendental equation:
cos(ωL/c
0
) = 0 ⇒ ω
n
=
̟
n
c
0
L
; n = 0, 1, 2,
where ̟
n

= (1 + 2n)π/2.
The mode shapes are obtained by superposing the incident and reflected waves,
which gives the standing wave:
ϕ
n
(x) =
1
2i

e
i̟
n
x/L
− e
−i̟
n
x/L

= sin

̟
n
x
L

where the condition adopted to normalize the mode shapes is max
(
|
ϕ
n

(x)
|
)
= 1.
198 Structural elements
Figure 4.6. Longitudinal mode shapes: fixed-free ends
The modal wavelengths are equal to λ
n
= 4L/(1 + 2n).
Defining now the vibration nodes as the locations where the magnitude of dis-
placement is zero and the vibration antinodes as the locations where it is maximum,
it is readily verified that the first longitudinal mode of the fixed-free beam has a
single node at x = 0 and a single antinode at x = L. The n-th mode has n nodes
and n antinodes, as shown in Figure 4.6. More generally, the number of nodes and
antinodes of the mode shapes is found to be proportional to n and the wavelength is
inversely proportional to n. It is also easy to check that the modes are not affected
in any way by providing the beam with additional supports located at the nodes.
It is also interesting to note that by substituting ω
n
into the boundary condition at
x = L, it can be seen that the incident X
+
wave is reflected at the free end without
change of sign:
∂X
∂x





0
= 0 ⇒
iω
c
0
(X
+
− X
−
) = 0
note. – Modal calculation using directly standing waves
A slightly different manner of presenting the modal calculation follows even
more closely the procedure described in [AXI 04]. Again, non-trivial harmonic
solutions ϕ(x)e
iωt
of the equation [4.1] are looked for, which are governed by the
Modal analysis methods 199
modal equation:
d
2
ϕ
dx
2
+

ω
c
0

2

ϕ = 0; ϕ(0) =
dϕ
∂x




L
= 0
However, instead of writing down the general solution as a linear superposition
of travelling waves, it is directly expressed as a superposition of sine and cosine
standing waves:
ϕ(x) = a sin

ωx
c
0

+ b cos

ωx
c
0

Finally, it is easy to check the orthogonality rule [4.2] of the mode shapes. Here
it reduces to the ordinary orthogonality condition between functional vectors, since
the stiffness and mass operators of a uniform beam are uniform:
ρSL

1

0
sin(̟
j
ξ)sin(̟
k
ξ)dξ =

ρSL/2 = M/2; if j = k
0; otherwise
ESL̟
j
̟
k

1
0
sin(̟
j
ξ)sin(̟
k
ξ)sin(̟
k
ξ)dξ =

ρSL̟
2
j
/2; if j = k
0; otherwise
Turning now to the Rayleigh–Love model, the mode shapes are found to be

the same as those related to the basic model, but the modal frequencies are lower,
as could be expected. The modal system for a beam in the free-fixed support
configuration is written in terms of dimensionless variables as follows:
d
2
u
dξ
2
+ ̟
2

u − γ
2
d
2
u
dξ
2

= 0;
du
dξ




0
= u(1) = 0
where ̟ =
ωL

c
0
; γ
2
=
1
2

νR
L

2
; u =
X
L
; ξ =
x
L
u = Ae
kξ
+ Be
−kξ
⇒ k
2
+ ̟
2
− k
2
γ
2

̟
2
= 0 ⇒ k =±
i̟

1 − γ
2
̟
2

1/2
where k = Lk is a dimensionless wave number.
200 Structural elements
The condition ϕ(1) = 0 leads to:
k
n
= i(2n + 1)π/2 ⇒ ϕ
n
(ξ) = cos
(
|k
n
|ξ
)
; ̟
2
n
=
|k
n

|
2
1 +|k
n
|
2
γ
2
In accordance with the variation of the phase speed displayed in Figure 4.1,
the modal frequencies increase nearly as an harmonic sequence only if the dimen-
sionless number
|
k
n
γ
|
is small enough, which means if the modal wavelengths are
much larger than the transverse dimensions of the beam.
4.2.2.2 Torsion modes
Torsion modes are discussed starting from equation [2.41], where the area polar
moment of inertia J is replaced by the torsion constant J
T
to account for the warping
of the cross-sections. The corrective term for warping inertia is usually discarded.
This equation has the same form as the longitudinal equation [4.10]. Then the results
are similar, provided the stiffness and mass coefficients are suitably modified. The
modal equation is:
d
2
ψ

x
dx
2
+

ω
c
T

2
ψ
x
= 0; where c
T
=


G/ρ


J
T
/J

= c
s

J
T
/J

[4.13]
The modal frequencies and mode shapes are immediately found to be:
ω
n
=
̟
n
c
T
L
; ϕ
n
(ξ) = a sin(̟
n
ξ) +b cos(̟
n
ξ) [4.14]
The modal coefficients a, b, ̟
n
are again those found in Table 4.1. The
generalized stiffness and mass coefficients are:
M
n
=
ρJL
2
; K
n
=
GJ

T
̟
2
n
2
[4.15]
4.2.2.3 Flexure (or bending) modes
In so far as the wavelengths are sufficiently larger than the cross-sectional dimen-
sions of the beam (roughly λ
n
/D ≥ 10), the Bernoulli–Euler model is satisfactory,
as shown in Figure 4.4. The vibration equation for bending in the plane Oxz is :
EI
d
4
Z
dx
4
− ω
2
ρSZ = 0 [4.16]
Modal analysis methods 201
Table 4.2. Flexure modes of a straight beam
cos ̟
n
cosh ̟
n
− 1 = 0 a
n
=

cosh ̟
n
− cos ̟
n
sinh ̟
n
− sin ̟
n
̟
n
free-free : ε
n
=+1 4.73; 7.85; 11.0
clamped-clamped ε
n
=−1 14.1; 17.3; (2n +1)π/2
tan ̟
n
+ tanh ̟
n
= 0 ̟
n
free-sliding ε
n
=+1 a
n
=
sinh ̟
n
− sin ̟

n
cosh ̟
n
+ cos ̟
n
2.37; 5.50; 8.64
clamped-sliding ε
n
=−1 11.8; 14.9; (4n − 1)π/4
cos ̟
n
cosh ̟
n
+ 1 = 0 a
n
=
sinh ̟
n
− sin ̟
n
cosh ̟
n
+ cos ̟
n
̟
n
cantilever ε
n
=−1 1.88; 4.69; 7.85
11.0; 14.1; (2n − 1)π/2

tan ̟
n
− tanh ̟
n
= 0 ̟
n
pinned-free: ε
n
=+1 a
n
=
cosh ̟
n
− cos ̟
n
sinh ̟
n
− sin ̟
n
(0); 3.93; 7.07; 10.2
pinned-clamped: ε
n
=−1 13.4; 16.5; (4n +1)π/4
The corresponding natural frequencies and related mode shapes are:
ω
n
=
̟
2
n

c
b
L
2
; where c
b
=

EI
ρS
and ξ = x/L
ϕ
n
(ξ) = cos(̟
n
ξ) +ε
n
cosh(̟
nξ
) − a
n
{
sin(̟
n
ξ) +sinh(̟
n
ξ)
}
[4.17]
The modal stiffness and mass coefficients are of the type:

M
n
= µ
n
M; K
n
= µ
n
EI̟
4
n
/L
3
[4.18]
where µ
n
is a dimensionless coefficients which depends on the mode shapes and
their norm.
The formulas giving the modal coefficients for various standard boundary condi-
tions are gathered in Table 4.2. Whatever the boundary conditions are, if n becomes
sufficiently large, a
n
tends to 1 and ̟
n
to nπ. It is worth noticing that care should
be taken in numerical applications as the results are increasingly sensitive to small
errors in λ
n
as n increases. In Figures 4.7 to 4.12, the shapes of the first three modes
are plotted for different boundary conditions. Most of them are normalized with

respect to the mass, in such a way that the modal mass is constant and equal to the
physical mass of the beam when dimensioned quantities are used. However, the
numerical values reported in the figures are dimensionless.
202 Structural elements
Figure 4.7. Bending mode shapes: free-free ends
Figure 4.8. Bending mode shapes: cantilevered beam