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Structural Dynamics
Introduction
Newton’s first law, or more particularly its corollary, is the cornerstone of static
analysis: if a body is in equilibrium then the some of the forces acting on the body
must be zero.
Similarly, Newton’s second law is the cornerstone of dynamic analysis. Newton’s
second law states that the rate of change of momentum of a body, in a particular
direction, is equal to the applied force in that direction.
d (mv )
= Force
dt
Equation (1)
If the mass of the body is constant then this equation simplifies to
m
dv
= Force
dt
or simply
mass × acceleration = Force
D’Alembert introduced the concept of inertial force,
inertial _ force = − mass × acceleration
so that the dynamic equation given above could be formulated as an equilibrium
equation i.e.
Force + inertial _ force = 0
Force − mass × acceleration = 0
which is
Clearly these formulations are identical.
Simple Harmonic Motion
Consider the behaviour of the simple dynamic system shown in Figure 1.
m
k
Figure 1.
If the mass is given an initial displacement from its static position and then released.
Then the mass will oscillate about its equilibrium position.
x
m
k
Figure 2.
If the spring is linear then the force applied to the mass by the spring is;
− kx = Force
Equation (2)
and hence, from Newton’s second law, the acceleration of the mass is &x& = −
k
x,
m
which has the form;
&x& = −ω 2 x
ω=
where
Equation (3)
k
m
Equation (4)
Thus the acceleration is a constant times the displacement. If the vertical
displacement of the mass is plotted against time then the result is as shown in Figure
3.
2 pi
A
wt
Figure 3.
The displacement of the mass can is described by the expression
x = A sin ωt
where ω is the circular frequency (units of radians/sec).
Equation (5)
Note:
A is the amplitude (maximum displacement)
The natural period of the oscillation is
τ=
2π
= 2π
m
k
Equation (6)
1
2π
k
m
Equation (7)
&x& + ω n 2 x = 0
Equation (8)
ω
and the natural frequency of the oscillation
fn =
1
τ
=
The general solution to the differential equation
is given by
x=
x& (0)
ωn
sin ω n t + x(0 ) cos ω n t
Equation (9)
Energy Formulation
The fundamental differential for the dynamic system in the previous example was
obtained directly. Other methods of formulating the differential equations of dynamic
equations exist. One potential approach is the energy approach. A conservative
system is one in which energy is not lost. Un-damped elastic systems can often be
considered as conservative systems. In such a system the total energy does not
change. Thus, if T is the total kinetic energy of the system and U is the potential
energy of the system then
d
(T + U ) = 0
Equation (10)
dt
In particular, if the reference position for the potential energy is chosen such that U=0
at the static position then the equation can be formulated as
T + U = Constant
and hence,
Tmax = U max
Equation (11)
Multi Degree of Freedom Systems
The previous section explored a single degree of freedom system. However, most
dynamic systems cannot be described in terms of a single degree of freedom.
However, by using D’Alembert’s principle we can formulate the dynamic equation by
developing equivalent static equilibrium equations that incorporate inertial forces.
Thus, we obtain the general equilibrium equation for a multi-degree-of-freedom
system, which is similar in form to the standard static equation;
[K ]{x} = {F }
Equation (12)
Where the forces acting on the body include the inertial forces, thus
[K ]{x} = { f }− [M ]{&x&}
Equation (13)
A general system will include damping, which is typically taken to be dependent on
{x&}. Thus, the general dynamic equation for a linear multi-degree-of-freedom system
is described by the equation;
[M ]{&x&}+ [C ]{x&}+ [K ]{x} = { f }
Equation (14)
Where both the forcing function { f (t )} and the displacements {x(t )} are functions of
time.
Before considering how this dynamic equation can be solved let us consider the
various terms, or at least some of the components of the terms. The matrix [K ] is the
stiffness matrix of the structure, hence, it is completely defined by the structure and
should be relatively easy to calculate, either by hand or by extracting the stiffness
matrix from a structural analysis package having constructed an appropriate structural
model. The mass matrix [M ] is also defined by the mass of the components of the
structure and again can be calculated either by hand, which is relatively easy if a
lumped mass model is used, or taken from a structural analysis package. The
damping matrix [C ] is more difficult to quantify and for this reason, and because
damping will generally tend to limit the response of the structure, damping is often
omitted and the un-damped response sought. Finally, it is assumed that the forcing
function { f (t )} is known. However, many useful analyses can be carried out for the
particular case where there is no external forcing function.
[M ]{&x&}+ [C ]{x&}+ [K ]{x} = 0
Equation (15)
In this case the free vibration of the structure is considered.
A number of approaches to tackling the fundamental equation for a multi-degree-offreedom linear system will now be considered. These approaches are not necessarily
independent.
Time Stepping
The fundamental equation, Equation (14),
[M ]{&x&}+ [C ]{x&}+ [K ]{x} = { f }
gives a snapshot in time. At a particular instant in time the forces, real and inertial
must balance. However, what is generally sought are expressions for {x(t )}. There a
many approaches to getting such results from the fundamental equation but perhaps
the most versatile, robust and conceptually simple approach is that of time stepping.
Put simply, if we know the displacements, velocities and accelerations of the system
at a particular instant then we can calculate the displacements, velocities and
accelerations at a short time interval later.
This is most easily seen by referring to a single degree of freedom system. If the
velocity, acceleration and displacement of the mass a simple system such as that
shown below are known at time t. At a time t + ∆t later, the new displacement of the
system can be described by a Taylor series
x(t + ∆t ) = x(t ) + x& (t ) ⋅ ∆t + &x&(t ) ⋅
∆t 2
∆t 3
+ &x&&(t )
+K
2!
3!
Equation (16)
Expressed simply, if we know our position, velocity and acceleration at an instant
then we can make a very good guess at where we will be an instant later.
In the general multi-degree-of-freedom case the fundamental equation can be
rearranged into the following form.
{&x&} = [M ]−1 [{ f }− [C ]{x&}− [K ]{x}]
Equation (17)
Thus, if we know the velocity and displacements at an instant we can calculate the
accelerations. Knowing the accelerations at time t we can then predict the velocity
and displacement of the system at an instant ∆t later. The crudest approach would be
to assume that the force doesn’t change during the time-step, a better approximation
assumes the force to vary linearly during the time step, with more sophisticated
algorithms modelling the variation of the acceleration more accurately.
Runge-Kutta
One popular algorithm for undertaking this time stepping is the Runge-Kutta method.
By introducing a dummy variable y = x& , Equation (17) can be expressed at two
equations, namely
x& = y
y& = f ( x, y, t )
Equation (18a)
Equation (18b)
In the neighbourhood of xi and yi, x and y can be expressed using Taylor expansions.
Thus, letting h = ∆t
 d 2 x  h2
 dx 
x = xi +   h +  2 
+K
dt
2
 dt i
i

Equation (19a)
 d 2 y  h2
 dy 
y = yi +   h +  2 
+K
 dt  i
 dt  i 2
Equation (19b)
Truncating the series after the first derivative but using an average value for the first
derivative gives,
 dx 
x = xi + h 
 dt  Average _ over (i −i +h )
Equation (20a)
 dy 
y = yi + h 
 dt  Average _ over ( i −i + h )
Equation (20b)
Clearly truncating the series this early is not a great idea in general unless the values
for the first derivatives are good averages for the average value during the time step,
i.e. not just the value at the start of the time step. Remember too that y is in fact x& and
dy
thus the average value of
is in essence an average value for &x& .
dt
 dx 
 dy 
Good values for the averages of   and   can be obtained by using Simpson’s
 dt 
 dt 
rule. Thus,
1  dy 
 dy 
 dy 
 dy  

=
+
+
4
 
 
 
  
 dt  Average _ over ( i −i + h ) 6  dt ti
 dt  ti + h 2  dt  ti + h 
Equation (21)
 dy 
is evaluated
The Runge-Kutta procedure is very close to this except that  
 dt  ti + h 2
twice, the second time using an updated estimate for the force. The procedure is best
explained using the following table taken from Theory of Vibration with applications
by W.T. Thomson, published by Unwin Hyman.
Table 1. Runge-Kutta – From Theory of Vibration with application by W.T. Thomson
Using the Runge-Kutta scheme x and y are evaluated four times during the time-step.
T1 at the start of the time-step, T2 and T3 half way through the time-step, and T4 at the
end of the time-step. At T1 the forces f (T1 , X 1 , Y1 ) , which govern the accelerations F1,
 dy 
i.e. the   values, are evaluated on the basis of the initial values of xi and yi. At
 dy 
time T2 the xi + h and yi + h values half-way through the time-step are calculated based
2
2
on the accelerations at the start of the time-step. At time T3 the values of xi + h and
2
yi + h
2
are re-evaluated using the accelerations based on the displacements and
velocities calculated at T2. Finally, at T4 the value of xi + h and yi + h are calculated
using accelerations based on the T3 accelerations.
Finally, the values of xi + h and yi + h are taken to be
h
[Y1 + 2Y2 + 2Y3 + Y4 ]
6
h
= yi + [F1 + 2 F2 + 2 F3 + F4 ]
6
xi + h = xi +
Equation (22a)
yi + h
Equation (22b)
Note:
The Runge-Kutta is not unconditionally stable and the output from this routine has to
be vetted. Nevertheless, it tends to give good results and can be used for non-linear
analysis. Furthermore, this method is self-starting. In general, the duration of the timestep should be limited to less than one tenth of the period of oscillation.
Natural Response – Free Vibration
The homogenous equation, Equation (15)
[M ]{&x&}+ [C ]{x&}+ [K ]{x} = 0
describes the response of a linear system in the absence of external forces. One trivial
solution to this system of equations is {x}, {x&} and {&x&} equal to zero. However, there
are other solutions. The natural modes of the system are the patterns of oscillation that
the structure will tend to oscillate in if disturbed.
The natural frequencies of a system are very important. If an external loading function
has a periodic component matching one or more of the natural frequencies then
resonance may occur with the effect that the amplitude of the oscillation will increase
until it is either restricted by damping, or the structure fails. See the video clip on the
Tacoma Narrows collapse in the Advanced Structures directory)
The natural modes of vibration and their corresponding frequencies can be found by
solving the equation
[[K ] − ω [M ]]{x} = 0
2
(or
equivalently
[A] − λ [I ] = 0 ).
[M ]−1 [K ] − ω 2 [I ] = 0 ,
which
Equation (23)
is
sometimes
written
as
For the non-trivial case where {x} ≠ 0 this equates to finding the
values for ω 2 for which the determinate of
[K ] − ω 2 [M ] = 0
Equation (24)
Once the values of the natural frequencies, ω s, have been found the corresponding
mode shapes (eigenvectors) can be obtained by substituting the known value of ω
back into Equation (23).
For example, for a three degree of freedom system and the fundamental natural
frequency (smallest value of ω 2 ) Equation (23) becomes,
 k11 − ω12 m11

2
k 21 − ω1 m21
 k31 − ω12 m31

2
k13 − ω1 m13   x1  0
   
2
k 23 − ω1 m23   x2  = 0
2
k33 − ω1 m33   x3  0
k12 − ω1 m12
2
k 22 − ω1 m22
2
k32 − ω1 m32
2
This equation has no unique solution because the right hand side equals zero. The
mode shape gives the relative magnitudes of the displacement at the different degrees
of freedom. Therefore it is necessary to seed the problem by setting the magnitude of
one of the displacements to 1. Thereafter it is possible to solve for the other
magnitudes uniquely. Thus for example,
1

k − ω 2 m
1
21
 21
2
 k31 − ω1 m31
0
  x1  1
   
k 23 − ω1 m23   x2  = 0
2
k33 − ω1 m33   x3  0
0
k 22 − ω1 m22
2
k32 − ω1 m32
2
2
Equation (25)
Orthogonal Properties of Eigenvectors
The eigenvectors of a system are orthogonal with respect to both the mass and
stiffness matrices. If the eigenvectors of a system are φ1 , φ 2 , φ3 , K , φ n then,
{φi }T [M ]{φ j } = 0
for all i ≠ j
Equation (26a)
{φi }T [K ]{φ j } = 0
for all i ≠ j
Equation (26a)
and
Modal Analysis
The displaced shape of a structural system can be described in term of its natural
mode shapes (eigenvectors). Thus, if the mode shapes are {φ1 }, {φ 2 }, {φ3 }, K , {φ n } then
the displaced shape {x} can be written in the form
{x} = Y1{φ1} + Y2 {φ2 } + Y3 {φ3 } + K + Yn {φn }
Equation (27a)
where Y1 , Y2 , Y3 , K , Yn are scalar values, or written in matrix notation
{x} = [φ ]{Y }
Equation (27b)
where
[φ ] = [{φ1} {φ2 } {φ3 }
φ1 ( x1 )


L{φ n }] =  M 
φ1 ( xn )
φ 2 ( x1 )


 M 
φ ( x )
 2 n 
φ n ( x1 )
φ3 ( x1 )




 M  L  M 
φ ( x )
φ ( x )
 n n 
 3 n 
Since, the displacement of the system can be expressed in terms of the mode shapes
the response of the system in time can also be expressed in term of the mode shapes.
Thus,
x(t ) = [φ ]Y (t )
Equation (28)
Of course versions of Equations (26) and (27) could be constructed using any set of
independent vectors. However, using expressing the response of a system in terms of
its eigenvectors has two main advantages.
Advantage 1 – Problem Reduction
The eigenvectors of a structural system define the characteristics of the system.
Furthermore, the eigenvector associated with the 1st natural frequency (i.e. the lowest
frequency) is the most indicative of the systems characteristics, followed by the
eigenvector associated with the 2nd natural frequency (2nd lowest frequency) and so on
down to the last eigenvector. In many cases, truncating Equations (26) or (27) so that
the response of the structure is described in terms of a limited number of eigenvectors
does not have a large effect on the overall accuracy of the representation. The
advantage is that a problem with a large number of degrees of freedom can be reduced
to a problem with a very small number of degrees of freedom. This can be
particularly beneficial where techniques such as time-stepping, which may be
computationally expensive, are involved.
Advantage 2 – Separation of the modal responses
Consider Equation (14) for the particular case where the damping is zero
[M ]{&x&} + [K ]{x} = { f }
Equation (29)
This equation can be rewritten in terms of the eigenvectors.
[M ][φ ]{Y&&}+ [K ][φ ]{Y } = { f }
Equation (30)
If all the terms of Equation (30) are now pre-multiplied by [φ ]
particularly interesting.
−1
[φ ]T [M ][φ ]{Y&&}+ [φ ]T [K ][φ ]{Y } = [φ ]T { f }
the result is
Equation (31)
Taking Equation (31) term by term. The first term contains the product
mG1
 0
[φ ]T [M ][φ ] = [M G ] = 


 0
0
mG 2
0
0 
0 
O M 

L mGn 
Equation (32)
which is referred to as the generalised mass matrix. This matrix is diagonal (i.e. it has
the property that it has non-zero entries on the leading diagonal only). The generalised
stiffness matrix,
 k G1
0
T
[φ ] [K ][φ ] = [K G ] = 


0
0
kG 2
0
0 
0 
O M 

L kGn 
Equation (33)
Has the same property. This property arises from the orthogonal properties of the
eigenvectors.
In a similar manner the pre-multiplying the forcing function by [φ ] yields a modal
force vector.
 f G1 
f 
[φ ]T { f } = {FG } =  G 2 
Equation (34)
M
 
 f Gn 
T
Thus Equation (31), which has the same solution as Equation (30), is in a form where
the response of each mode is independent of all the other modes. For example, the
response of the first mode of the system will be governed by the equation,
mM 1Y&&1 + k M 1Y1 = f M 1
Equation (35)
Equation (35) describes the response of a single degree of freedom system. Therefore
identifying the response of the structure is equivalent to summing the response of n
(or less than n) single-degree-of-freedom systems.
The overall response in the natural coordinates {x} can be obtained from the modal
response {Y } via Equation (28).