Straus r7-Software Dynamics Analysis

Dynamic Analysis with
Straus7
Presented by
G+D Computing Pty Limited

Presented by G+D Computing Pty Limited

Table of Contents
Discussion: Categories of Dynamic Problems and the Corresponding Straus7 Solvers .........1
Discussion: Modelling Considerations for Dynamic Analysis ..................................................5
Discussion: Natural Frequency Analysis .................................................................................9
Torsional Vibration of a Shaft with Disc Attached ...................................................................15
Normal Modes Analysis of a Simply Supported Beam ...........................................................17
Discussion: Mass Participation Factor ...................................................................................19
Discussion: The Use of Symmetry in Natural Frequency Analysis ........................................23
Stress Stiffening Effects on Frequency ...................................................................................27
Discussion: Damping in finite element analysis .....................................................................29
Discussion: Harmonic Response Analysis ............................................................................35
Discussion: The Mode Superposition Method .......................................................................41
Rotating Out-of-Balance Mass ................................................................................................45
Frame on a Shaker Table .......................................................................................................49
Discussion: Mass Matrix Formulation - Consistent vs Lumped .............................................53
Discussion: Transient Dynamics ............................................................................................55
Modelling Moving Loads .........................................................................................................57
Single Degree of Freedom System .........................................................................................61
Viscous Damping Coefficient of a Cantilever ..........................................................................67
Masses Falling on Two Cantilever Beams ..............................................................................69
Discussion: Modelling Shock Problems in Straus7 ................................................................73
Shock Qualification of an Instrumentation Frame ...................................................................77
Drop test on an instrumentation frame ...................................................................................81
Discussion: Modelling Rotating or Pretensioned Structures ..................................................85
Discussion: Spectral Response .............................................................................................87
Discussion: Earthquake Analysis using Straus7 .....................................................................93
A Simple Example of Seismic Analysis ................................................................................107
PSD Spectral Response .......................................................................................................111
PSD - Base Excitation ..........................................................................................................113
References ...........................................................................................................................115

Presented by G+D Computing Pty Limited 1
Discussion: Categories of Dynamic Problems and the
Corresponding Straus7 Solvers
Overview
Generally dynamic problems can be categorised into the following four groups:
1. Eigenvalue problems
The dynamic behaviour of a structure is closely related to its natural frequencies and
corresponding mode shapes. A well known phenomenon is that when a structure is subjected to
a sinusoidal force and the forcing frequency approaches one of the natural frequencies of the
structure, the response of the structure will become dynamically amplified i.e. resonance occurs.
Natural frequencies and their corresponding mode shapes are related directly to the structure’s
mass and stiffness distribution (for an undamped system).
An eigenvalue problem allows the calculation of the (undamped) natural frequencies and mode
shapes of a structure. A concern in the design of structures subject to dynamic loading is to avoid
or cope with the effects of resonance.
Another important aspect of an eigenvalue solution is in its mathematical significance - that is, it
forms the basis of the technique of mode superposition (an effective solution strategy to decouple
a coupled dynamic matrix equation system). The mode shape matrix is used as a transformation
matrix to convert the problem from a physical coordinate system to a generalized coordinate
system (mode space).
In general for an FE model, there can be any number of natural frequencies and corresponding
mode shapes. In practice, only a few of the lowest frequencies and mode shapes may be
required.
Natural Frequency and Period spectrum for a number of common structures.

2 Presented by G+D Computing Pty Limited
2. Forced Vibration Problems (under Sinusoidally Varying Load)
Sinusoidally varying load is common in engineering analysis. For example rotating machinery
subject to a mass imbalance. Another example is that of a vibration test table driven by a
sinusoidal base excitation.
When a sinusoidal excitation is applied to a structure, the structure will initially vibrate in an
irregular manner often referred to as the transient stage. The irregular part decays to zero over
time due to damping. After the transient stage, the structure will vibrate in a sinusoidal fashion at a
frequency identical to the frequency of the applied excitation, but the phase of the response may
be different from the applied load. This stage of the response is called the steady state response.
For a given excitation intensity, the amplitude of the steady state response changes with the
different frequencies of the applied excitation.
The forced variation analysis (or harmonic response analysis) is used to calculate the peak values
(amplitudes) of the steady state response of a structure at different frequency points within a
frequency range.
3. Transient analysis problems
Transient dynamic analysis is used to calculate the entire time history, from the starting point of
loading, of the dynamic response of a structure subjected to external dynamic loading of an
arbitrary time function and initial conditions.
This kind of analysis is often used to analyse a structure under a shock loading which has a short
action duration but perhaps wide frequency range.
The calculation of a transient analysis uses numerical integration methods, such as the Newmark
and Wilson methods which are used in Straus7.
4. Spectral analysis problems
Basically, spectral analysis is a fast method to get the dynamic response information of a structure
which is subjected to an non-deterministic (i.e. random) load.
There are two types of spectral analysis problems:
(1) Response spectrum analysis which estimates the maximum possible response of a structure
based on given spectral curves. This method is widely used in earthquake analysis.
(2) Power spectral density analysis where the loading is a stationary random process and a
statistical estimation of the response is sought.
General Equation of Motion
The governing equation for all four types of problems can in most of situations, be expressed as:
For an applied load,
(1)
For a base excitation,
MU
··
t( ) CU
·
t( ) KU t( )+ + P t( )=

(2)
where
M - Global mass matrix
C - Global damping matrix
K - Global stiffness matrix
P(t) - Applied external excitation vector
U(t) - Unknown nodal displacement vector
The dots on top of the U(t) represent first and second order time derivatives respectively.
is the base movement.
Depending on the types of external loads and analysis requirements, one or more of the four
analyses are conducted.
(1) When there is no external load, i.e. P(t)=0 and also the damping matrix C=0, equation (1) can
be turned into an eigenvalue problem.
(2) When the external load is of sinusoidal form, a harmonic analysis problem is formed.
(3) When the external load is a general form of time function and the whole time history of the
response of the structure is of interest, transient analysis is needed.
(4) If the external load is non-deterministic, spectral analysis is carried out.
Comparison with Static Analysis
Dynamic analysis may be required in addition to, or replacing altogether, static analysis by
considering the following points:
(1) Loading and response is time dependent (P=p(t), U=u(t)).
(2) Inertial forces become significant and cannot be neglected.
(3) A dynamic problem is often considered in the frequency domain.
Summary of Dynamic Solvers Available in Straus7
Corresponding to the four categories introduced above, Straus7 has 4 dedicated dynamic solvers:
Natural Frequency Solver
Calculates the undamped natural (or resonant) frequencies of a structure. Although the solver
can calculate any number of natural frequencies for a given model (depending on the number of
MU
··
t( ) C U
·
t( ) Y
·
t( )–( ) K U t( ) Y t( )–( )+ + 0=
Y t( ) Y0 ωtcos=

degrees of freedom contained in the model), in practice only a relatively small number of modes is
calculated.
Harmonic Response Solver
Calculates the steady state response of a structure subjected to a sinusoidally varying load. The
solver gives deflections, stresses, etc., in the frequency domain.
Transient Dynamic Solver - full system/mode superposition
Calculates the response of a structure subject to an arbitrary time varying load. The solver gives
displacements, stresses, etc., as a function of time. Both linear and nonlinear analysis can be
performed.
Spectral Response Solver
Calculates the response of a structure subjected to a random dynamic loading (e.g. an earthquake
represented by its response spectrum or a mechanical vibration represented by its Power Spectral
Density). The solver gives estimated maximum deflections, stresses, etc. or statistical estimations
(PSD).

Discussion: Modelling Considerations for Dynamic
Analysis
Introduction
Most Straus7 users have a good understanding of the basic requirements for the design and
construction of models used in static structural analysis. In general, the techniques used to build
models for use in dynamic analysis are similar but there are a number of issues that need special
consideration.
Number of Elements and Mesh Density
The overall mesh density required for dynamic analysis is in general higher than that for a static
structural analysis, although localised refinement near fillets etc., is usually not required.
The higher modes of many structures are very complicated and many elements are required to
provide a good representation of these. All the elements in the Straus7 element library have shape
functions of a fixed order. These shape functions define the deformed shape of the element. For
example the beam element has a cubic shape function.
The modes of vibration for a beam with simple supports at both ends are of sinusoidal shapes. If a
single beam element is used then the higher order modes cannot be calculated and possibly even
the frequencies and mode shapes of some of the lower modes may be incorrect. A single beam
cannot represent the sinusoidal mode shapes because of the inability of a single cubic equation to
approximate more than half of a sine curve. In this case many cubic beams are required to provide
a piecewise cubic approximation to the sinusoidal mode shape. If only the lower modes are of
interest, the mesh can be relatively coarse. However for harmonic, transient dynamic and spectral
analysis the higher modes are frequently important as they may be excited by high frequency
excitation of the structure. Decisions on the mesh density clearly require a sound understanding of
the likely behaviour of the structure and the requirements of the analysis.
The other point to consider when designing meshes for dynamic analysis is that in general there is
less of a need to refine the mesh locally around areas of stress concentration, particularly if the
mesh is not being used to calculate stresses in a separate linear static analysis. In dynamic
analysis the global inertial and stiffness characteristics of the model are usually more important
than local behaviour. There are however some special cases where local modes are important
and the mesh may require some local refinement in order the capture these.
Representation of Mass in Dynamic Analysis
In dynamics the stiffness and mass of a structure both play an equally important role in the
determination of the frequencies and mode shapes. This is evident in the simple equation for the
natural frequency of a mass on a spring:
m
k
=ω

This means that when we build a finite element model for use in dynamic analysis, it is important
to ensure that the model provides a correct representation of both the stiffness and mass of the
structure.
Modelling of Non-Structural Mass
Often a structure being modelled for a dynamic analysis will be a support frame for some sort of
equipment. There will be many parts of the structure that can be referred to as non structural mass
- that is, items of equipment and other dead weights that contribute mass but no stiffness to the
structure. In a typical linear static analysis these masses might be represented with equivalent
forces and pressure loads but in a dynamic analysis the actual mass and its distribution must be
represented accurately. In many cases non-structural masses can be represented using point
masses. A portion of the non structural mass is lumped at each of its attachment points on the
structure. This approximation assumes that the item has mass but no stiffness.
Often items of non-structural mass provide some additional stiffness between the attachment
points. The way in which the stiffness of these items is modelled depends on the relative stiffness
of the non structural mass and the structure.
If the items of mass have very large stiffness in comparison to the structure, a point mass can be
lumped at the centre of gravity of the item. This is connected to the attachment points on the
structure with rigid links. An example of a mass that would be modelled in this manner is an
engine mounted in a frame.
In other cases the stiffness of the non-structural mass, between the attachment points, is similar to
the stiffness of the structure. In this case there is no option but to include a coarse finite element
representation of the item producing the mass. This mesh can be crude because it is only being
used to provide an approximate representation of stiffness and inertia, not to calculate stresses
and deflections. Furthermore, the use of a crude mesh helps to keep the model to a reasonable
size. A coarse mesh also helps to suppress any local modes of the non structural mass since
these are generally of little interest in the analysis.
There are two methods commonly used to include the mass of the item in the unrefined finite
element model:
• A point mass is often placed at the centre of gravity of the item and connected to the
attachment points on the structure with the finite element representation of the non-structural
mass. In this case the elements used to model the non-structural mass are not assigned
density.
• In other cases a density is assigned to the properties for the elements used to model the non-
structural mass. This density is factored until the total mass of the item is correct.

An example of a mesh used to model the mass and
stiffness of a piece of electronic equipment is shown in
the adjacent figure. This model is an idealization of a
rack of integrated circuit boards. The density of the
various components (i.e. rack, boards, etc) was
factored to get the correct overall mass. Note the
crudeness of the model. This mesh would clearly be
too coarse for use in a linear static structural analysis
or a dynamics analysis of the component itself. It is
however, sufficient for including the mass and stiffness
effect of the circuit board rack on the overall behaviour
of the structure to which it is attached.
Often the centre of gravity of equipment or other items
of non-structural mass are offset significantly from the
attachment points on the structure. It is very important that the centre of gravity of all items of
mass be correctly located.
If the centre of gravity offsets are to be included in the model, it is common to offset the mass from
the attachment point by a rigid link of an appropriate length. Alternatively if the non-structural
mass is modelled using the coarse finite element model approach, the centre of gravity offset will
be included automatically. Any such finite element approximation should be checked to verify that
the centre of gravity is in the correct location. This can be done by using the Summary/Model
option in the Straus7 main menu.
Lumped and Consistent Mass
The mass of a structure is simply the sum of the mass of each element. In Straus7, the mass of
an element is automatically calculated provided a density has been assigned. This mass is
assumed to be distributed uniformly over the element.
In the finite element method all mass is eventually assigned to the nodes. This means that the
continuously distributed mass of the elements must be converted to an equivalent set of masses
at the nodes. The method by which this is done can influence the solution speed and accuracy.
There are two ways that this discretisation of mass can be carried out: the consistent and the
lumped mass approximations.
In the lumped mass approximation, mass is lumped to the nodes of the elements in a simple
distribution such that the sum of these nodal masses equals the total mass of the structure. For a
2-node beam, it is intuitive to lump half the mass at each node. In this typical finite element
approach, usually only translational inertias are represented directly at the nodes, omitting terms
related to the rotational inertia. Overall rotational inertia is accommodated by the fact that the
nodal translational masses are distributed over a large geometric region - a bit like a governor
where the overall rotational inertia is a function of the translational masses and the distance
between them. In Straus7, the lumped mass approach generates a very small (diagonal) matrix
which means that compared with a linear static analysis, only a small amount of extra space is
needed.
The consistent mass approach is more accurate and the distribution is based on determining a
mass lumping scheme that gives both translational and rotational inertias. The distribution is
based on the same integrations that are used to calculate the element stiffness matrix and this
generally results in a distribution that is not very intuitive. Furthermore, because the mass matrix
is as populated as the stiffness matrix, the storage requirements are twice those for a linear static

analysis. The work required to manipulate these extra terms in the matrix also means that with
consistent mass, the solver is slower.
The choice of the lumped or consistent mass approximation can, in some special cases, have a
significant effect on the accuracy of the analysis, although in practice, for a relatively large model,
the differences are small, especially for the lower modes.
In Straus7, a point translational mass is always treated as a diagonal mass. Point rotational mass
is always treated as a nondiagonal mass. This is because the general case of a rotational inertia
about an arbitrary axis requires a full 3x3 local matrix at each node.
In Straus7, you have the option of using either Lumped (diagonal) or Consistent (full) mass matrix
assemblies. However, if you choose Lumped, but the analysis requires consistent (e.g. because
you have a rotational mass or a beam or plate offset, etc.) then for those elements/nodes, the
matrix is automatically expanded to include the off-diagonal terms.

Discussion: Natural Frequency Analysis
Examples
• Windmill Blade.
• Bending and Torsional Frequencies of a Crank Shaft.
The Eigenvalue Problem
The equation of motion for a general system is:
where:
[M] = mass matrix
{d} = displacement vector
{ } = velocity vector
{ } = acceleration vector
[C] = damping matrix
[K] = stiffness matrix
{P} = externally applied load vector
If we consider a structure without damping and without externally applied loads then the equation
reduces to:
This has a solution in the form of a simple harmonic motion, where the displacements are given
by:
and
where:
substituting these terms into the equation of motion gives:
This can be recognized as an eigenvalue problem where is the eigenvalue and is the
eigenvector. is also the angular natural frequency in radians per second so that the eigenvalue
[ ]{ } [ ]{ } [ ]{ } { }PdKdCdM =++
d
·
d
··
[ ]{ } [ ]{ } 0=+ dKdM
{ } { } tsindd o ω=
{ } { } tωω sin2
odd −=
ω 2πf=
[ ]{ } [ ]{ }oo dMdK 2
ω=
ω
2
d0{ }
ω

is the square of the natural frequency. If is divided by it gives the cyclic frequency in cycles
per second (Hz).
If the system has n equations then there are n independent solutions to the equation. These can
be written in the form:
where is the eigenvalue and is the corresponding eigenvector.
This can be rewritten in the form:
which is the equation solved by the Straus7 natural frequency solver.
Properties of Eigenvalues and Eigenvectors
The eigenvalues calculated by the solution of the above equation yield the natural frequencies
of the structure as follows:
Angular frequency: (rad/sec).
Cyclic frequency: cycles/sec or Hz.
For a structure with n degrees of freedom there are no more than n eigenvalues (natural
frequencies).
For each eigenvalue there is a corresponding eigenvector which is a set of displacements
defining the mode shape. It is important to realize that the displacements in the eigenvector are
not absolute values of displacement. It is the relative magnitude of the displacements that is
important in defining the mode shape. The actual amplitude of a mode depends on the magnitude
of an excitation force. Since the natural frequency solver is solving for unforced natural
frequencies only, information on the magnitude of the displacements associated with a particular
mode shape is not available.
The values of displacement (i.e. the eigenvector) in the Straus7 output are normalized such that
the modal mass is equal to 1. The following equation is used to carry out this normalization.
This normalization is a very useful way to present the eigenvector because it means that the
modal stiffness of the structure is equal to the frequency.
The Sub-Space Iteration Solver
The eigenvalue problem is a very expensive and time consuming problem to solve. In order to
solve this in a reasonable time some approximations must be made to reduce the size of the
problem. One solution method (and the one used by Straus7) is Sub-Space iteration.
ω 2π fi
[ ]{ } [ ]{ }iii dMdK λ=
λi ωi
2
= i
th
di{ } i
th
K[ ] λi M[ ]–( ) di{ } 0{ }=
λi
wi λi=
fi
wi
2π
------=
d{ }
{ } [ ]{ } 1dMd i
T
i =

The basic concept of this method is that it takes the stiffness and mass matrices for the full
structure and reduces these to a manageable size so that it can be solved by direct eigenvalue
extraction methods.
Convergence of the Natural Frequency Solution
The default convergence tolerance in the natural frequency solver panel is 1.0E-5. This means
that in order for the solution to terminate, the least accurate eigenvalue must be changing by less
than this amount between successive iterations. This is a tight tolerance and well below normal
engineering accuracy. Users are often tempted to increase the convergence tolerance to values
such as 1.0E-3 to decrease the solution time. In many cases this practice will work but it should
be used with caution. The reason the default value is set to such a high tolerance is based on the
results of extensive experience with the solver. Often additional modes are found between the
current eigenvalues as the solution proceeds. This usually occurs when there are many closely
spaced modes. These slot in between the modes that had been calculated to this point in the
solution and all the higher modes are shuffled up one place.
The reason that this occurs is that the initial degrees of freedom used to excite the solution
process may not have adequately represented all the modes. As the solution proceeds, random
adjustment introduces new degrees of freedom into the sub space. This can allow previously
unknown modes to be identified and captured. Reducing the convergence tolerance can cut the
solution process short and not allow sufficient time for the detection of all modes including the
additional modes not identified by the initial starting vectors. The convergence tolerance should
only be reduced when it has been established for a particular problem that this practice is reliable.
The solution process would normally be allowed to continue to completion with the default
convergence criterion at least once to assist with this verification. In addition to this the Sturm
check can be used to verify that all of the modes are being located by the solution with relaxed
convergence tolerance.
A potentially more serious consequence of reducing the tolerance is that although the eigenvalue
(frequency) may be considered adequately represented, the eigenvector (mode shape) may not
be fully converged. If unconverged eigenvectors are used in mode superposition analysis
(harmonic, spectral, etc.), the results can be erroneous. This is sometimes manifested by spectral
runs producing mass participation factors in excess of 100%.
Another way of controlling convergence is to reduce the maximum number of iterations (which
defaults to 20).
Further Notes on Eigenvalues
There are a number of questions concerning eigenvalues that inevitably arise when doing natural
frequency analysis.
Multiple eigenvalues
In many models some of the calculated eigenvalues will appear as identical pairs. The reason for
this is that many structures are symmetric and thus have orthogonal pairs of modes. That is, the
structure has an equal tendency to vibrate in two perpendicular planes. These planes need not be
the global planes.
Zero eigenvalues
If zero eigenvalues are calculated by the solver then this can mean one of two things:

• The freedom conditions applied to the structure are insufficient to restrain the model in space.
The zero mode and associated eigenvalue define a mode in which the structure either
translates or rotates as a completely rigid body with no relative displacement between the
nodes on the structure. This is a common result whenever you model the natural frequencies
of an unrestrained structure - e.g. an aeroplane in level flight.
• The structure is a mechanism. This means that insufficient stiffness is provided to prevent
some part of the structure moving as a rigid body.
Occasionally some part of the model may behave as a rigid body when this is not intended or
expected. If this occurs the most likely cause is incorrect zipping of the model. It may be necessary
to increase the value of zip tolerance that has been used to ensure that all parts of the model are
correctly joined together. The free edge display can be used for this verification.
Missing eigenvalues
In some special cases it is possible that the solver will miss some of the eigenvalues. See the
following section on the Sturm check for a discussion on how to check for missing modes.
Large models with many local modes
In some large natural frequency analyses, such as that on an entire ship, the natural frequency
results will include many local modes in which there is no interest. In general these modes will
occur in cladding panels and will involve diaphragm motion of panels.
The Sturm Check
The iterative nature of the sub-space solver does not guarantee that the solution will converge to
the first n modes required by the user. Occasionally some low order modes may be missed and
higher modes found in their place. An eigenvalue may be missed by the solver if the initial starting
vector in the sub-space does not include degrees of freedom that provide an adequate
representation of the mode. The degrees of freedom contained in the initial starting vector must be
capable of exciting all vibration modes within the range requested by the user.
For example, consider the simple case of a cantilever beam. This will have multiple orthogonal
modes, both in the plane and out of the plane of the page. If the degrees of freedom excited by the
starting vector are only in the vertical direction then these cannot represent the out of plane modes
and some of the modes will be missed.
The likelihood of Straus7 missing eigenvalues is low since special precautions are taken during
the solution process to continually introduce new degrees of freedom into the trial vectors that
span the subspace.
It should be noted that the Sturm check can only determine the number of eigenvalues in a
specific range, it does not calculate the value of the eigenvalues. It is however a useful check on
the output data. The Sturm check is a very stable and reliable method for determining the number
of eigenvalues in a given range. This stability results from the fact that the method only relies on
the signs of numbers and not the actual values. Thus rounding error and other errors will have less
effect on the results.
Loading and Damping
1. The solution of the above equation does not take into account any damping on the structure.

2. If the effects of pre load are to be included (eg. tightening a guitar string will change its fre-
quency), then a linear static analysis can be performed on the model first and the results of this
included in the natural frequency analysis. In this case we solve a slightly different equation,
namely:
is known as the Geometric or Stress Stiffness matrix and is simply added to the normal
stiffness matrix. For an element with zero stress, will be zero.
Shifting the Matrix
1. Often we need to check only on frequencies and modes near a specific frequency (e.g. due to
some vibrating machinery). In these cases we can use the shift value (in Hertz) to ask the
solver to calculate only modes near the shift value. The eigenvalue search is centred on the
given shift and the solver will find the eigenvalues closest to the shift, both above and below the
shift.
2. The shift can also be used for finding the natural frequencies of a structure which is not
restrained, (e.g. an aircraft in flight). Here we apply a "small" shift to make the system non-sin-
gular.
3. The shift is introduced into the natural frequency solution in the following way.
From the above, the basic eigenvalue equation that is solved for the natural frequencies is
Some value of shift can then be introduced as follows:
Rearranging the equation yields:
This equation can be solved in the normal manner for the frequencies . The actual frequencies
of the structure are then = +
}0{}]){[][]([ =−+ ii dMKgK λ
Kg[ ]
Kg[ ]
K[ ] λi M[ ]–( ) di{ } 0{ }=
λo
K[ ] λi λo+
© ¹
§ · M[ ]–( ) di{ } 0{ }=
[ ] [ ]( ) [ ]( ){ } { }0dMMK iio =−− λλ
λi
λi λo λi

Torsional Vibration of a Shaft with Disc Attached
Outcomes
Upon successful completion of this lesson, you will be able to:
• Use the Natural Frequency solver.
• Use and investigate the difference between rotational and translational node masses.
• Use and investigate the difference between lumped and consistent mass matrices.
Problem Description
A 50 mm diameter disc 10 mm thick is suspended by a rod 10 mm in diameter and 500 mm long.
The shaft is fixed at the upper end and the entire assembly is manufactured using 316 stainless
steel. Find the first torsional natural frequency.
The first torsional natural frequency is given by:
where,
Shear Modulus, G = = MPa
Torsional rigidity of shaft, = =
Rotational mass (inertia) of shaft,
Rotational mass (inertia) of disc,
Translational mass of disc,
Modelling Procedure
• Create a new model and set the units to Nmm.
Hz
l
J
J
GJ
f
s
d
p
7091.271
3
2
1
=
¸
¹
·
¨
©
§
+
=
π
E
2 1 v+¢ ²
--------------------- 7.4806202 10
4
×
Js
πds
4
32
----------- 9.817477 10
2
× mm
4
Js
πds
4
psls
32
-------------------- 3.9269908 10
3–
×= = tonnes
·
mm
2
Jd
πdd
4
pdld
32
---------------------- 4.9087385 10
2–
×= = tonnes
·
mm
2
πd
2
pdld
4
-------------------= 1.5707963 10
4–
×= tonnes

• Construct three models of
the rod side by side in the
one model window. Use
10 beam elements to
model the shaft.
• The first model has the
disc modelled using a
short beam. The second
model represents the disc
as a point mass with
rotational inertia (RY mass
= 0.049 Tmm2) and the
third model has the disc as
a point mass with
translational inertia (mass
= 1.57x10-4 T).
• Fully fix the top end of the shaft and globally fix all 3 translations.
• Run the Natural Frequency solver and
calculate the first three modes using both the
lumped and consistent mass option. To swap
between the lumped and consistent mass
matrix option, go to the Defaults tab page in
the Natural Frequency solver dialogue. Click
the Elements button on the left.
Results
Summarize the results in the following table:
Modelling Technique
Solution
Number of
Elements Short Beam Rotational Mass
Translational
Mass
Strand7 (Mass Matrix Lumped) 10 271.67 271.67 1527.38
Strand7 (Mass Matrix Consistent) 10 271.69 271.69 1530.52

Normal Modes Analysis of a Simply Supported Beam
Outcomes
• Use and investigate the difference between lumped and consistent mass matrices.
• Understand the mesh density requirements for calculation of important frequency modes.
• Investigate the effects of varying shear area on the frequency.
Problem Description
The natural frequencies of a 310UB46.2
universal beam are examined. In particular, a
comparison is made between the analytical
solution, lumped mass matrix and the
consistent mass matrix methods. The mesh
density and shear area are also evaluated. The
beam is simply supported with a length of
5000 mm and the global freedom condition is
set to 2D Beam.
Analytically, the modes of the beam are derived as:
For flexural modes, = =
For axial modes, = =
Modelling Procedure
• Create a beam 5000 mm long and set the beam ends as pinned.
• Create a copy of this beam. Subdivide the first into two, and the
second into ten elements.
• Set the global freedoms to 2D Beam.
• Set the beam property. Set the material of the beam to Structural
Steel and the section to 310UB46.2 in the BHP - Universal Beams
section database. In the Sections tab page, set the section areas
to zero.
• Run the natural frequency solver and solve for 10 modes.
• Rerun the solution using the Consistent Mass Matrix option.
wn 2πf n
2
π
2 EI
ρAl
4
-----------
ωn 2πf π
E
ρl
2
-------

Results
Note that at least two elements are required to calculate the natural frequency of the flexural and
longitudinal modes using the lumped mass approximation. This is because there are insufficient
mass degrees of freedom.
The consistent mass approximation however uses the element displacement shape functions, and
often can better represent the real mass distribution over the structural element. For beam
elements, the consistent mass matrix includes terms for rotational inertia.
As further study, investigate the difference in results when leaving the shear areas of the beams as
non-zero (note that the analytical results assume thin-beam theory).
Mode of Vibration
Solution Number of
Elements 1
st
Bending 2
nd
Bending 3
rd
Bending 1
st
Axial 4
th
Bending
Analytical n.a. 41.1844 164.7376 370.65956 504.7545 658.9503
Strand7 (Lumped) 2 40.8854 n.a. n.a. 454.4387 n.a.
Strand7 (Consistent) 2 41.2099 180.4273 445.6487 556.5714 794.1139
Strand7 (Lumped) 10 41.1841 164.7180 370.4102 502.6813 657.3207
Strand7 (Consistent) 10 41.0483 162.6044 360.2227 506.8327 627.4657

Discussion: Mass Participation Factor
Introduction
The Natural Frequency solver can be used to calculate additional dynamic properties of a
structure, which can be useful in any mode superposition analysis. The aim of this section is to
help the reader understand the concept of mass participation factor.
What is the Mass Participation Factor?
The mass participation factor is an important indicator of whether a sufficient number of modes
has been included in a dynamic analysis based on the mode superposition method.
The mass participation factor for the i-th mode is calculated using the following formula:
where
- Mode shape vector of the i-th mode;
M - Global mass matrix; and
R - Global movement vector determined by the excitation direction factor vector.
The sum of the mass participation factors can be used as a guide to determine that there is a
sufficient number of relevant modes included in the analysis for the given global movement vector
R. As more and more relevant modes are included, the value of the sum should approach 1. As
a general rule for each excitation direction, the sum of the participation factors of the modes
should be greater than 90%.
The relevant modes refer to those that have nonzero values of mass participation factor and have
some contribution to the global movement. For example, consider a vertical rod. If the base
excitation is in a horizontal direction, only the mode shape vectors which have nonzero
components in the horizontal direction will have some contribution, while mode shape vectors
which have only components in the vertical direction will make no contribution. Mathematically, a
relevant mode shape vector is closer or more correlated to the global movement vector R than an
irrelevant mode.
For the relevant modes, their contribution to the total structural response depends on the
magnitude of the response of the individual mode to the modal force. This magnitude is also a
function of the mode frequency under the given modal force. Hence, although it is often used as a
good indicator, the mass participation factor cannot be used on its own for determining the
number of modes to include for mode superposition.
The global movement vector R represents the corresponding rigid movements of all the
translational degrees of freedom in the FE model under a given base excitation movement. The
vector is formed based on specifying the direction factor vector.
PFi
φi
T
MR( )
2
R
T
MR
------------------------=
φi

Example
The following example is used to illustrate the above discussion. A vertical rod is modelled by 10
beam elements, restrained at the base. Material is structural steel from the Straus7 material
library.
• To better illustrate the problem, 2D Beam freedom conditions are set such that all mode
shapes are found only in the XY plane.
• A 3 x 3 multiview display is selected so that all 9 modes are shown on screen.
• Select the Natural Frequency solver.
• Solve for 9 modes and activate the Sturm check.
• Set Mass Participation with the participation direction vector as Vx=1, Vy=Vz=0 - that is global
movement is in the X direction only.
• Once solved, you should got the following information in the Results Log file.
The Sturm Check reports that all modes within the frequency range are found. The Mass
Participation summary reports that the total mass participation factor is 99.499% indicating almost
all the of the mass is active by using 9 modes. The individual contribution of every mode is listed.
Scanning through the list, it can be seen that the 8th mode has zero contribution. The mode
shape of this mode can be examined by looking at the Straus7 natural frequency result and
plotting the mode shapes.

Mode shapes of a vertical bar
From the figure of the mode shapes, it can be seen that the 8th mode is an axial extension mode
or vertical movement which is irrelevant to X direction movement, hence the mass participation
factor is zero.
Looking back at the mode participation report, it can be seen that by only using the first 4 modes,
a very good result may be obtained as these modes have the greatest contribution to the total
mass participation factor. By only choosing the first 4 modes over 90% of the mass is included,
hence the solution time for mode superposition can be reduced by only including these modes.
This is one of the major advantages of using mode superposition method, in that a few modes are
often sufficient to obtain an accurate result.

Discussion: The Use of Symmetry in Natural
Frequency Analysis
Introduction
When modelling symmetric structures it is common practice to reduce the size of the model by
using the principle of symmetry. Appropriate freedom conditions are applied on the plane of
symmetry so that half of the structure modelled behaves as though it is still attached to the other
half of the structure.
The nature of a symmetry boundary condition means that a structure must deform symmetrically
about the plane of symmetry. This normally means that in addition to the structural geometry being
symmetric, the loading must also be symmetric. Whilst most analysts are comfortable with the
concept of symmetry in linear static problems, experience shows that this is not necessarily the
case with regard to natural frequency and buckling analysis (note that the equations solved for
linear buckling analysis are basically the same as those solved for natural frequency analysis).
Symmetric half models can be used for buckling and natural frequency analysis but this is not as
straightforward as it is for linear static analysis.
A symmetry model with symmetric boundary conditions will yield the symmetric buckling and
vibration modes only. To obtain the anti-symmetric modes it is necessary to run the model a
second time with anti-symmetric boundary conditions applied to the geometric symmetry plane of
the structure. For very large models it may be better to use the symmetry approach, since running
the half model twice will usually be faster than running the full model once.
Anti-symmetric boundary conditions are simply the opposite of symmetric conditions - any
degrees of freedom that are fixed in the symmetric case become free in the anti-symmetric case.
Those that are free in the symmetry case, become fixed in the anti-symmetry case.
Example
The following example consists of a simple portal frame with
dimensions (in metres) as shown. The model uses Structural Steel
as the material property and BHP - Universal Beam 530UB92.4 as
the section. Three models are constructed: the full model, a
symmetric model and an anti-symmetric model.

The table contains the natural frequencies whilst the figures show the first 10 modes.
Full Model
Symmetry Model
Mode Full Model Symmetric Model Anti-symmetric Model
1 9.567 9.567
2 45.78 45.78
3 81.534 81.534
4 90.973 90.973
5 146.334 146.334
6 150.701 150.701
7 154.025 154.025
8 169.834 169.834
9 188.927 188.927
10 219.643 219.643

Anti-Symmetry Model

Stress Stiffening Effects on Frequency
Outcomes
• Investigate the effects of stiffening on the natural frequency of a structure.
Problem Description
An aluminum (alloy 6063 -
T6) circular membrane of
radius a = 381 mm,
thickness t = 0.254 mm is
simply supported along its
edge and is subjected to an
in-plane radial pre-stress of
68.971125 MPa. Determine
the first 3 axisymmetric
natural frequencies of
lateral vibration of the
membrane.
Model the membrane using
the axisymmetric 8 node
plate elements. Use 20
elements as shown in the figure.
Modelling Procedure
• Create a beam 0.254mm long in the Y direction. Extrude this beam to a
total length of 381mm in 20 steps. Ensure that the beams are converted
to Quad8 elements, via the appropriate setting in the Targets tab.
• Apply the restraints shown in the Problem Description section.
• Set the plate property as Axisymmetric and use Aluminium Alloy 6063-
T6 as the material type.
• Apply the pre-stress as a tensile edge stress on the outer edge of the
membrane.
• Run the Linear Static solver.

• Run the Natural
Frequency solver to
calculate for 4 modes of
the structure without
preload. Results should
appear as shown.
• Record the frequencies in
the table below.
• Run the Natural Frequency solver
again, this time include the effect of the
preload. Do this by using the linear static
results as the initial condition, as shown.
• Record the results in tabular form.
Results
Note that a circular disc such as this will exhibit many other frequency modes that are not
axisymmetric. In this model, those modes have been ignored.
Mode Theory
Straus7
without
Preload
Straus7with
Preload
1 160.5
2 368.4
3 577.7

Discussion: Damping in finite element analysis
Introduction
Damping is a term used for the measure of the energy loss in a dynamic system. There are many
mechanisms responsible for damping, e.g. material damping, friction at contact surfaces, etc.
In Straus7, damping is represented by different linear damping models. Three models are
available: Rayleigh damping, modal damping and the viscous damping matrix models. In addition,
an effective modal damping coefficient can be calculated by defining a material damping ratio at
the element level. This last option is useful for determining the overall damping ratio of a system
composed of parts with different damping characteristics.
Viscous Damping
The viscous damping model uses the following expression to calculate the element damping
matrix:
where
- Material viscous damping coefficient with the units of force/velocity per unit
volume;
N - Element shape function matrix; and
Ve - Element volume domain.
The global damping matrix is obtained by assembling all element damping matrices. The element
damping matrix can come from the viscous damping coefficient assigned to the element
properties, as shown in the dialog box below, or in the case of a spring element, it can come from
the damper coefficient.
This type of damping is only relevant to the Full System Transient solvers (both Linear and
Nonlinear). To activate the viscous damping contribution for all but the spring elements, you need
to select the option Viscous on the Added Damping settings on the solver dialog (as shown).
Ce µN
T
N Vd
Ve
³=
µ

For spring-damper elements, the damping part (the so-called discrete damper) is always
assembled, irrespective of the setting of the Added Damping option.
Rayleigh Damping
Rayleigh damping, also known as proportional damping, assumes that the damping matrix is a
linear combination of the stiffness and mass matrices:
where and are proportional constants. One of the important advantages of Rayleigh
damping model is that, like the stiffness and mass matrices, the damping matrix C can be turned
into a diagonal matrix by the normal mode shape matrix . Therefore the general dynamic
equilibrium equation with Rayleigh damping can be de-coupled into independent equations by the
modal transformation matrix , so that the mode superposition technique is applicable. Due to
this property, Rayleigh damping is most commonly used in finite element analysis.
The two constants and are often determined by using two values of the damping ratio(
and ) at two chosen frequencies ( and ) according to the following formula:
Explicitly α and β are given by:
C αM βK+=
α β
Φ
Φ
α β ζ1
ζ2 ω1 ω2
ζ
1
2
---
α
ω
---- βω+
© ¹
§ ·=
α
2ω1ω2 ζ2ω1 ζ1ω2–( )
ω1
2
ω2
2
–
------------------------------------------------------=

Damping ratio for Rayleigh damping
Usually and are chosen such that they cover the whole frequency range of interest, with
being the lowest and the highest frequency.
To define Rayleigh damping, a user can input values of and , or
alternatively input and at two frequency points. Rayleigh
damping can be applied to harmonic, spectral and transient solvers.
Modal Damping
Modal damping is defined in mode space. It can be assigned independently for each vibration
mode. In Straus7, modal damping is input in the form of a damping ratio . So the damping
value is .
Modal damping can provide a better damping approximation to the structure as different damping
values can be assigned to each mode. Consider the case where the analysed structure is
composed of different materials that when combined together deliver different damping properties
per vibration mode (natural frequency). For example, an FE model of a concrete structure sitting
on a soil foundation may contain two groups of modes: modes where the movement is dominated
by the structural deformation of the concrete and modes where the concrete structure moves as a
rigid body on the soil foundation. In this case, the concrete deformation modes would have a low
damping ratio say 5% and the rigid body modes would have a higher damping ratio of say 10% -
20%.
Another advantage of modal damping is that it is easy to correlate it with experimental results,
allowing the evaluation of damping properties from actual test data.
β
2 ζ1ω1 ζ2ω2–( )
ω1
2
ω2
2
–
----------------------------------------=
ω1 ω2
ω1 ω2
α β
ζ1 ζ2
ζi
ci 2ζiωni=

Modal damping can be used with any
Straus7 solver that uses Modal
Superposition. This includes the harmonic
and spectral solvers and the linear transient
dynamic solver with the Superposition
option.
Use of modal damping involves the specification of the
frequency file along with mode damping coefficients.
Effective Damping Coefficients
If a material damping ratio is defined at the element level, then an effective modal damping
coefficient can be calculated based on the material damping ratio. This effective damping
coefficient can be used in the spectral, harmonic response or transient solvers. The effective
modal damping coefficient is computed according to the following formula (Japan Road
Association, 1990):
where
- Mode shape vector of element j of i-th mode
-Stiffness matrix of element j
K - Global stiffness matrix
DCj - Damping ratio of element j
DCi
φj( )
i
T
DCj Ke( )
j
φj( )
i
j 1=
n
¦
φi
T
Kφi
------------------------------------------------------------=
φj( )
i
Ke( )
j

Notes About Damping
1. Damping exists in almost all real structures although its mechanism is often not clear. Damping
can be due to internal friction in the material, Coulomb friction in connection, resistance from
surrounding media of the structure (e.g air or oil in bearings) and so on.
2. Damping dissipates energy and causes the amplitude of free vibration to decay with time. The
function of damping is critical nearer the natural frequencies of the structure because around a
natural frequency, the stiffness force and inertia force tend to cancel each other, leaving only the
damping force to balance the external force. Without damping, theoretically the response of a
structure will become infinite when the forcing frequency is equal to a natural frequency of the
structure.
3. Often damping is small for most structures. The following typical/reference values of damping
ratio are mentioned: from 0.02 for piping systems to about 0.07 for bolted structures and rein-
forced concrete (Cook, 1995); from 0.1 to 0.3 for foundation structures of bridges (Japan Road
Association, 1990). The following table provides typical damping ratios:
Type of Construction ξ
steel frame
welded connections
flexible walls 0.02
steel frame
welded connections
normal floors
exterior cladding 0.05
steel frame
bolted connections
normal floors
concrete frame
flexible internal walls 0.05
concrete frame
flexible internal walls
concrete frame
concrete or masonry shear walls 0.10
concrete or masonry shear wall 0.10
wood frame and shear wall 0.15

Discussion: Harmonic Response Analysis
Examples
• Rotating machinery on concrete structure.
• Plastic fan housing.
• Reciprocating engine mounting.
Background
The Harmonic Response solver calculates the maximum values of a dynamic response of a
structural model due to harmonic loading. The harmonic loads act with identical frequencies and
arbitrary phase angles. In addition to the nodal displacement and velocity, the solver is able to
recover the maximum values of nodal acceleration, reactions, element stresses and phase
angles.
Introduction
The Harmonic Response solver calculates the maximum values of a linear elastic steady state
dynamic response. The structural system is subjected to a set of harmonic forces F(t) with
identical frequencies and different phase angles .
Harmonic Loading
The response is calculated for a set of forcing frequencies, evenly distributed over a user defined
frequency range. If some of the natural frequencies of the structure are within the forcing
frequency range an additional forcing frequency will be introduced, identical to the natural
frequency. This ensures that resonant response is captured. Furthermore, two additional points
are introduced automatically at the half-power points for each natural frequency.
The external forces may be applied on the model in the load cases, in the same manner as the
Linear Static solver. All load cases are assumed to define a single loading condition. All loads
contained in a single load case (e.g. point forces, moments, pressure, etc.) act with the same
phase angle and vary as functions of time in a sinusoidal fashion. The loads from different load
cases act with the same frequency, but can have different amplitudes and different phase angles.
Upon the initial application of loading, the structure will initially vibrate in a random manner, often
referred to as a transient stage. After the initial period all the points of the structure will vibrate in a
sinusoidal fashion with a frequency identical to the forcing frequency , but with different
ω θ
F(t) = F sin (ωt + φ)
F
time t
2π/ω
Structure
φ
ω

amplitudes and different phase angles. This part of the response is known as the steady state
stage. The Harmonic Response solver calculates the maximum values of the steady state
response, i.e. the amplitudes of the sinusoidal steady state response.
For each forcing frequency step an envelope of the maximum values of the response is given in
the results. It is important to note that these maxima will not occur simultaneously; generally each
will occur at different times, out of phase with the others.
The applied harmonic load can act with different frequencies, given as frequency steps on a
frequency range sweep. Each frequency step is treated as an individual loading condition and it is
solved separately. All the results are available for each frequency step. Nodal displacement,
velocity and acceleration may be displayed graphically for the whole frequency range. In addition,
it is also possible to assign a table of load factor vs frequency which then factors the amplitude of
the harmonic load at each step. This is useful for modelling situations such as a machine vibrating
due to an out-of-balance load whose amplitude increases with frequency.
Natural frequency analysis must be performed prior to any harmonic analysis. The Harmonic
Response solver uses the most recent results from the natural frequency analysis to perform a
mode superposition. If the model is modified in any way a new natural frequency analysis must be
performed. Once the natural frequency analysis is carried out many runs of the Harmonic
Response solver may be performed.
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0 5 10 15 20 25 30 35 40
transient stage steady state stage

The number of modes included in the analysis will influence the accuracy of the results. The
method used for calculation of the maximum response is based on superposition of the modal
responses. A greater number of modes included in the analysis will provide results of greater
accuracy.
Basic Theory Review
The total force acting on one particular node is a summation of the forces from all load cases:
where:
F(t) - total force on one node, time dependent
t - time
Fi - force amplitude from load case i
i- phase angle of the forces from load case i
n - total number of load cases
ω - forcing frequency, identical for all load cases
For calculation of the maximum response the Mode Superposition approach is applied. The
maximum response and the phase angle for each mode is calculated by the following
expressions:
x(t) = A sin ( t + )
where:
x(t)- modal response, time dependent
A- amplitude of the modal response
F- modal force
K- modal stiffness
¦=
+=
n
i
ii tFtF
1
)sin()( θω
θ
ω θ
[ ] 21222
21
/
)()(
K
F
A
−
+−= ξββ
»
¼
º
«
¬
ª
−
= −
β
ξβ
θ
1
2
tan 1

ω- forcing frequency
- phase angle of the modal response
- ratio between forcing frequency and natural frequency
- modal damping ratio
The response of the full structure is calculated by superposition of the modal responses
considering the phase angles, the sign and the magnitude of the modal responses with respect to
time. Superposition of three modal responses is shown in the figure below:
Superposition of Modal Response
The maximum response of the displacements, reactions and element stresses is identified
analytically by a closed form solution. The only approximation is the finite number of modes used
in the analysis. When a sufficient number of modes is used, the Harmonic Response solver can
provide an almost exact solution. If all modes for a given model are included, then the exact
solution for that model is obtained.
Results
Before considering any results the log file (extension: 'HRL') should be reviewed. Most of the
global input parameters and input panel settings are listed. Also all warning messages should be
examined carefully.
For each frequency step the nodal displacements, velocities, and accelerations are calculated as
well as displacement phase angles. Reactions and element stresses are only calculated when
requested by the user. All the results, for any frequency step can be shown separately, in a similar
manner to the results for a linear static solution. Also, the maximum displacements, velocities,
accelerations and displacement phase angles over the entire frequency range can be presented
graphically for any node.
The results of the analysis are absolute maximum envelope values of the steady state dynamic
response. The maximum values do not occur at the same time. Consequently the results do not
present an equilibrium state of the structure and they do not correspond to each other. For
instance the maximum displacement results do not correspond to the maximum stress results.
The stress at one particular point is just the maximum value which occurred at that particular point
during the steady state stage of the dynamic response.
Unlike other results, the fibre stresses in beams and principal stresses in bricks are not the
maxima that occur during the dynamic response. The values for beam fibre stresses and brick
θ
β
ξ
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
0 1 2 3 4 5 6 7 8 9 10

principal stresses are calculated from the maximum components. For instance, the beam fibre
stresses are calculated from the maximum M (bending moment) and maximum N (axial force).
But, the maximum M and N may not have occurred at the same time during the response, and the
signs of M and N are unknown.
Similarly, the brick principal stresses are calculated from the maximum components of stresses,
and these may not have occurred at the same time, and their signs are unknown.
The Straus7 Harmonic Response solver calculates an exact solution for the number of modes
used in the analysis. The solution considers the sign and the phase angle of all included modal
responses. When a sufficient number of modes is used, the results are usually of very high
accuracy. In some cases, when the structure has many local modes, even 100 mode shapes may
not provide a sufficient mass participation (see the section on Mass Participation Factor). In this
case the results may underestimate significantly the real behaviour of the structure. It should be
considered that the overall response is a summation of the modal responses, and that the
structure is represented by a finite number of modes.
The contribution of different modes is related to applied loads. The figure below shows the first two
mode shapes of a simple structure and the load. The shape of the modes indicates that the first
mode will be excited by the load, but not the second one because the load is at the position where
the mode shape component is zero. So, the participation for the first mode will be relatively high
(70%), while the participation of the second mode will be relatively low (2%).
Mode Participation
The phase angle of the nodal displacements indicates the time delay of the nodal vibration with
respect to the forcing frequency. For each frequency step all the forces are acting on the structure
with that frequency. All the nodes vibrate with that frequency, but in different phases.
M ode 1 M ode 2. Load
70% 2%

Discussion: The Mode Superposition Method
Introduction
Three of the Straus7 solvers use or have the option to use the mode superposition solution
technique. These are the Transient Dynamics, Harmonic Response and the Spectral
Response solvers.
Basic Theory
The mode superposition method uses the mode shape vectors calculated by the natural
frequency solver to transform the dynamic equilibrium equation into mode space. The response
of each mode to the forcing function is calculated and the results for all modes are superimposed
to get the complete response of the structure.
The motion equilibrium equation of a structure under dynamic loading is expressed as:
(1)
Normally equation (1) represents a set of coupled equations. The mode superposition method
solves the above equations by first de-coupling them using the mode shape vectors , which are
obtained by natural frequency analysis, then solving the de-coupled equations in the modal space
independently and then finally combining the solutions in the modal space to produce the solution
to equation (1).
The nodal displacement vector U(t) can be expressed in terms of mode shape vectors as:
(2)
where q(t) represents the generalised coordinates (also referred to as principal coordinates or
modal coordinates) in the modal space, N is the total degrees of freedom in the FE model.
Substituting (2) into (1) and pre-multiplying the obtained equation by yields:
(3)
or
(4)
where m, c and k are the modal mass, modal damping and modal stiffness matrices, F(t) is the
modal load vector. Matrices m and k are diagonal. For Rayleigh damping and modal damping, c
is diagonal as well.
The elements, mii, kii and cii (i=1, 2,..., N), of matrices m, k and c are called modal mass, modal
stiffness and modal damping respectively. For a mode shape matrix normalized with respect to
the global mass matrix (often referred to as a mathematical mode shape matrix), the
MU
··
t( ) CU
·
t( ) KU t( )+ + P t( )=
Φ
U t( ) Φq t( ) q1 t( )φ1 q2 t( )φ2… qN t( )φN+ += =
Φ
T
Φ
T
MΦq
··
t( ) Φ
T
CΦq
·
t( ) Φ
T
KΦq t( )+ + Φ
T
P t( )=
mq·· t( ) cq· t( ) kq t( )+ + F t( )=

corresponding modal mass mii is equal to 1. If the mode shape matrix is scaled so that the
maximum element of each mode vector is equal to 1 (referred to as the engineering mode shape
matrix), the corresponding modal mass mii is called engineering modal mass. Straus7 can report
this engineering modal mass if required.
If matrices m, k and c are diagonal, equation (4) can be expressed as N independent equations (N
is the total number of mode shape vectors used in the transformation matrix):
i=1,2,..., N (5)
If the mode shape matrix is normalised with respect to the global mass matrix, that is:
(6)
then
(7)
with is the natural frequency of mode i.
Using equations (6) and (7), (5) can be written as:
(8)
where
(9)
is the modal damping ratio. (Note: In (9), mi=1 is used. For a single degree of freedom spring-
mass-dashpot system with parameters k, m, c, the damping ratio, also referred to as damping
factor or viscous damping factor, is expressed as with ).
The damping factor represents a ratio between the damping value of the dashpot and the critical
damping value of the system designated by . is known as
critical damping, a point separating the overdamped case and the underdamped case of the
system.)
miq
··
i t( ) ciq
·
i t( ) kiqi t( )+ + Fi t( )=
Φ
m Φ
T
MΦ
1 0 … 0
0 1 … 0
… …
0 0 … 1
= =
k Φ
T
KΦ
ω1
2
0 … 0
0 ω2
2
… 0
… … … …
0 0 … ωN
2
= =
ωi
q
··
i t( ) 2ζiωniq
·
i t( ) ωni
2
qi t( )+ + Fi t( )=
ζi ci 2ωni( )⁄=
ζ c 2mωn( )⁄= ωn k m⁄=
ccr 2mωn 2 km= = ζ 1=

Solving the N independent differential equations expressed by (5) or (8) then substituting the
solution into equation (2) yields the solution U(t) for equation (1).
When using a solver based on the mode superposition method, it is very important that the
number of modes used in the solution be sufficient to capture the response caused by a particular
excitation. In the majority of problems the response is characterised by the first couple of modes.
In such cases only a couple of modes are needed in the solution. There are however some
problems where this generalisation does not hold true.
If the loading contains high frequency components then some of the higher modes may contribute
significantly to the response. For example with impacts or blasts, the loading can be of a very high
frequency and it is important that the higher frequency modes be included. This means that the
impact frequency of the loading should be calculated and all the natural frequencies in the range
up to and including the impact frequency also included in the analysis.
For spectral analyses involving a seismic spectrum or other broad band excitation, sufficient
modes should be included to ensure that all the natural frequencies of the structure within the
range of the excitation are included in the analysis. Similarly when using the harmonic and
transient solvers, all the modes up to a frequency above the excitation frequency should be
included.
Deciding on the number of modes to include in a solution by the magnitude of the frequencies
alone is not always a reliable criteria. The actual shape of the modes must also be considered.
Consider for instance an analysis being carried out to determine the effect of vertical vibrations on
an instrumentation rack. The lower modes of these sorts of structures are typically swaying or
bending modes. The vertical modes, involving flexing of the shelves and axial excitation of the
columns, are usually the higher modes. It is also important to ensure that a sufficient number of
modes is included in the analysis to represent the behaviour of the structure in the direction of the
excitation. This may require the inclusion of modes with frequencies significantly above the
excitation frequency.
An indicator is used to assist in assessing whether a sufficient number of modes has been
included. This is the mass participation factor. For further details of this see the section on Mass
Participation Factor.
qi t( )

Rotating Out-of-Balance Mass
Outcomes
• Use the harmonic response solver in combination with the Natural Frequency solver.
• Use Rayleigh damping.
Problem Description
The response of a structure to a rotating out-of-balance load is a common problem in structural
dynamics. In this example we calculate the maximum response and stresses in a mine ventilation
fan support structure.
The centre of mass of the fan blades is offset eccentrically
3 mm from the spindle axis and the fan is spinning at
540 RPM (9 Hz or 56.5487 radians/sec). The fan is in a
horizontal plane with a vertical axis of rotation.
The rotating out-of-balance force is the centrifugal force
arising from the rotation of the centre of mass of the fan
on a radius of 3 mm from the rotation axis. This is
calculated using the equation:
F = mr 2
= 10000 * 0.003 * 56.54872
= 95932.6 N.
This force is applied to the structure as a rotating force
vector acting in the radial direction.
Modelling Procedure
The FEA model consists of the fan housing as shown:
• Create a new file and set the units to Nmm.
• Create a plate support structure out of three plates. Each
plate is 5x5 m and the three plates form an open box.
• Subdivide the plates 10x10.
• Create a node centred 3m above the plate support
structure top.
• Create four beams from this node to the top of the plate
support structure.
• Set the beam and plate properties to use Structural
Steelwork as the material.
• Assign CHS 219.1 x 3.0 (in the CHS (350 grade) OD above
100 mm section folder) as the beam section type.
• Assign the plates a thickness of 30 mm.
ω

• Fully fix the base of the plate support structure.
• Assign a nodal translational mass of 10 000 kg to the top centre node to represent the weight
of the fan.
• Create two load cases: In the first, assign a nodal force of 95 932.6 N to the centre node to
represent the out-of-balance force of the fan; in the second, assign the same magnitude force
90 degrees from the 1st applied load in the horizontal plane.
• Run the natural frequency solver to determine the first 12 modes.
Natural Frequency Results
Natural Frequency Results for 12 modes
Harmonic Response Solver
The Harmonic Response solver applies a sinusoidal variation of each load case. The force will
vary from + to - over the specified frequency range. The direction of the force does not change. In
order to simulate a rotating force vector we need to apply two load cases (Load Case 1 and Load
Case 2), both are point forces equal to the centrifugal force but the directions are 90 degrees to
one another. Both forces are in the plane of the rotation. The harmonic solver will apply a vector
summation of these two load cases. If load case 1 is applied 90 degrees out of phase with load
case 2, load case 1 will be maximum when load case 2 is zero and vice versa. The vector
summation of these two forces will then be a rotating vector.

Setup the Harmonic Response solver as shown
For the Frequency File choose the natural
frequency file from the previous solution
Specify the load cases as shown to apply the load
cases out of phase with each other. Specify the
Rayleigh damping to span the 12 natural frequency
modes to be considered along with a damping ratio of
0.05.
Results
Obtain the peak displacement response at the point where the force is applied. Also determine the
maximum tensile principal stress ( ) in the support structure.σ11

Frame on a Shaker Table
Introduction
• Use the Harmonic Response solver.
• Use a base acceleration to model a vibrating base load.
• Use modal damping.
Problem Description
A common test in experimental dynamics involves
mounting a component on a shaker table and
accelerating it back and forth at a particular frequency or
over a range of frequencies. These sorts of tests are
used to determine the displacement response, resonant
frequencies of the structure (by looking for the peak
responses) or often as a fatigue endurance test.
The Straus7 Harmonic Response solver can be used to
model this sort of test and can calculate the stresses and
deflections due to any base acceleration harmonic
loading. Often finite element modelling is used in
conjunction with test results to refine and develop
components.
In this example we calculate the maximum response and
stresses in an instrumentation rack. The frame of the
rack is constructed using a 25 x 25 x 3 square hollow
section and the shelving is 3 mm plating.

Model Setup
The mass of the instruments is included as
lumped masses applied at the points where
the instruments attach to the shelves. This
approach assumes that the actual instruments
themselves do not contribute any stiffness to
the structure.
The feet of the frame are modelled as a
pinned connection. The loading is applied as
a global base acceleration in the X direction.
The acceleration is 1.25g (enter as 1.25*9810
in the X-Direction factor). The harmonic
solver assumes that this is applied
sinusoidally over the frequency range
specified in the Harmonic Response solver
dialog. The specified acceleration is the
amplitude of the sinusoidal acceleration.
Modelling Procedure
• Create two nodes at (0,0,0) and (500 mm,0,0).
• Copy the two nodes by Y=500 mm.
• Extrude all four nodes by Z=500 mm.
• Connect a Quad4 element at the top of the four beams.
• Subdivide the plate into 9x9 Quad4.
• Select all plates and click Tools/Tesselate/Lines
and tesselate a perimeter of beams around the
plates.
• Subdivide the columns by 3.
• Copy all plates and beams by Z=500 mm.
• Assign the plates the correct material properties and a thickness of 3 mm.
• Assign the beams the correct material properties and a 25 mm x 25 mm x 3 mm thick square
hollow section.

• Assign pinned restraints to the table’s footings.
• Assign 4x 2 kg translational masses to the top shelf and 2x 10 kg masses to the lower shelf.
Use Attributes/Node/Translational Mass.
• Run the natural frequency solver to calculate the first 12 natural frequencies (use lumped
mass).
Harmonic Response Solver Setup
• In the Harmonic Response Analysis
dialog box, specify that the Load Type is
to be a Base Acceleration. Set the
Direction Vector to X=12262.5. Set the
Damping to Modal. Set the frequency
range to cover the majority of modes and
ensure that a suitable number of steps is
specified.
Note: The direction factor specifies both the
magnitude and the direction of the base
acceleration. To apply a base acceleration of
1.25g, enter the direction vector as
X=1.25x9810 (where units are consistent
with the model).

• Click Frequency Files and specify that all modes are to be
included with a Damping Ratio set to 0.05 for each mode.
Results
Plot a graph of the DX response at the node
indicated (top corner). Find the peak response and
identify the range of excitation frequencies where
a resonant response occurs. Note that resonance
does not occur for all modes - only those modes
with dominant motion in the direction of the
excitation (X).
Note that the response given by the Straus7
harmonic solver includes both the amplitude and
the phase. You can plot graphs of phase similarly
to graphs of displacement. When plotting graphs
of displacement, it may easier to interpret the results by using the graph option which plots
the absolute Y values.
Additional Work
Investigate the stresses in the structure for the frequency step where resonance occurs. Are these
stresses high enough to cause fatigue if the endurance limit of the steel is 100 MPa? What is the
effect of resonance on the stresses? That is, how much higher are the stresses at resonance
compared to the stresses at other non-resonant frequencies.
Another useful exercise is to evaluate the effect of varying the damping ratio. You should find that
the results are sensitive to damping ratio near the resonant frequencies. Away from the resonant
frequency, the damping ratio should have minimal effect.
Strand7
Maximum Response Dx
(at node indicated)

Discussion: Mass Matrix Formulation - Consistent vs
Lumped
Introduction
In the Defaults tab of the solver dialogue box, you can specify whether to use Lumped or
Consistent mass formulations.
What is a Mass Matrix ?
Mass matrices are used to provide a discrete mass approximation for a structure where the mass
distribution is continuous. This means that the continuously distributed mass within an element is
approximated by lumping a certain amount of mass at each of the nodal points. The mass matrix
specifies how much mass is lumped onto each node.
Mass Matrix Formulations
Straus7 supports two different formulations for the mass matrix that are commonly used in finite
element analysis systems - the consistent mass and the lumped mass approximations.
Lumped Mass Matrix
The simplest representation of the distributed mass within a structure is the lumped mass
approximation.
In the lumped mass approach the mass within each element is assumed to be lumped onto each
of the nodes such that the sum of the nodal masses associated with the translational degrees of
freedom for each global direction equals the total mass of the element. Usually only the
translational inertia effects are included. There is no rotational inertia and no mass coupling
between the different degrees of freedom. This results in a diagonal matrix like that shown below
for a simple two dimensional beam element.
For a bar element the lumped mass matrix is intuitive and simple. In the above example the total
mass of the element is divided by 2 and the resulting mass is lumped onto the nodes at each end
of the element. Note that each degree of freedom at each node has the same mass assigned.
For continuum type elements such as plate and brick elements, the mass matrix cannot be
obtained by using such simple intuitive methods and more refined methods must be used. In the
simplest case the lumped mass matrix for a regular plate with 90 degree corners or a similarly
regular brick is obvious. For a square 4 node plate, 1/4 of the total mass is lumped onto each
corner node. For most real structures the elements are distorted and therefore the mass matrix is
[ ]
»
»
»
»
¼
º
«
«
«
«
¬
ª
=
1000
0100
0010
0001
2
m
m

no longer intuitive - different amounts of mass will have to be lumped onto each of the corner
nodes. It gets even more complicated when we consider higher order quadratic elements.
The lumped mass matrix is commonly used because of its simplicity. The lumped mass matrix is
quicker to assemble and requires less storage space compared with a consistent matrix. The
lumped mass approximation will generally produce a lower level of accuracy than the consistent
mass. This is particularly true if there are significant nodal rotations in the dynamic response such
as those that occur in flexural problems. Errors associated with the use of the lumped mass
approximation can be greatly reduced with mesh refinement so that the lumped approach is
normally the more viable one.
Consistent Mass Matrix
The most accurate method of discretisation of the mass in a continuum is to use the consistent
mass matrix. With this approach, the mass matrix is derived by using the same integrations as are
used in deriving the stiffness matrix so that the mass matrix is consistent with the element shape
functions.
The consistent mass matrix is the same shape as the stiffness matrix and includes off diagonal
terms. It therefore considers the effect of mass coupling between the different degrees of freedom
and the effect of the rotational inertia. The consistent mass matrix for the simple two dimensional
beam of mass m, is shown below:
The calculation for a consistent mass matrix, based on the element shape functions integrated
over volume of the element is as follows:
where = element density, N= shape function, V = volume
Additional Notes
For most finite element models, the lumped mass matrix approach, which generates a diagonal
mass matrix in Straus7, is a reliable option offering very good results, particularly for the low order
modes. Higher order modes will usually exhibit errors of 10% or more compared with a consistent
mass solution. This is illustrated in a number of exercises in this course.
Although the diagonal (lumped) approach is the default option in Straus7, some models need the
consistent (full) mass matrix to correctly model the mass. This includes offset beams and offset
plates, rotational mass at nodes and translational mass assigned to nodes that only connect links
(ie. rigid links, but not elements).
Whenever the Straus7 solver encounters one of these situations, and you have chosen the
lumped mass option, the entries in the mass matrix for those elements/nodes requiring a
consistent (full) mass matrix are automatically expanded. Therefore, the correct influence of the
offset mass is obtained, even when lumped mass is selected.
[ ]
»
»
»
»
¼
º
«
«
«
«
¬
ª
−−−
−
−
−
=
22
22
4L22L3L13L
22L15613L54
3L13L4L22L
13L5422L156
420
m
M
ρ N[ ]
T
N[ ] Vd³
ρ

Discussion: Transient Dynamics
Examples
• Impact loading and drop tests.
• Time history loading.
• Response of a structure suddenly released from a loaded condition.
Background
The Transient Dynamic solver can be used to solve the following equation in the time domain:
where:
= Mass Matrix
= Damping Matrix
= Stiffness Matrix
= Nodal Acceleration Vector
= Nodal Velocity Vector
= Nodal Displacement Vector
= Time Dependent Nodal Force Vector
To solve this equation, a direct integration method or modal superposition method is used
whereby the conditions at time t are assumed to be known and are required after a discrete time
step, Dt.
Direct Integration - Full System
This method generally gives good accuracy provided the time step is sufficiently small. Ref [1]
recommends a time step 0.1 to 0.2 times the period of the structure (i.e. five to ten steps per
period). The method will fail if the time step is too large. With the full system analysis, it is not
necessary to perform a natural frequency analysis if the appropriate time step is known. If the
period of the structure is not known, a natural frequency analysis should be run first to find the
time step. For loads that are varying at a faster rate, the time step may be dictated by the need to
accurately capture the load history.
Mode Superposition - General
Mode Superposition is an approximate method which analyses a reduced structural model. The
structural system is approximated with several independent single degree of freedom systems.
Each single degree of freedom system corresponds to one natural frequency and related mode
shape of the full system. The dynamic response is calculated in two stages. The first stage
involves frequency analysis. One or more mode shapes and corresponding frequencies must be
calculated. Then the Mode Superposition solver will calculate the dynamic response of the model
by summation of all available modal responses. The Mode Superposition solver uses the most
M[ ] x
··
{ } C[ ] x
·
{ } K[ ] x[ ] F t¢ ²{ }=+ +
M[ ]
C[ ]
K[ ]
x
··
{ }
x
·
{ }
x{ }
F t( ){ }

recent results from the frequency analysis. Once the frequency analysis is completed many runs
of the Mode Superposition solver may be performed. If any modification of the model is made a
new frequency analysis must be performed prior to any Mode Superposition analysis.

Modelling Moving Loads
Outlines
• Use the Transient Dynamic solver.
• Use Factor vs Time tables to apply a specified force history.
• Create an animation of a bridge vibrating due to a moving load.
Problem Description
In this example we consider the
problem of a car driving across a
bridge. The bridge is modelled
as a simple square hollow
section (2 m x 2 m x 0.12 m) with
the material properties shown.
The car is traveling at a constant
100 km/h (27.77 m/s). Knowing
this we can calculate the position
of the car at any point in time or
more importantly we can
calculate the time when the car
reaches each point on the
bridge.
To simulate the moving load a series of point
loads are applied along the length of the
beam. Each of these is equal to the static
weight of the car - 9810 N. Each force must be
turned on only when the car is in the vicinity of
the node to which the load is applied. This
means that when the car is within half the
length of the beam elements either side of a
node, the load is applied to that node.
Only one of the forces is actually applied
to the structure at any time. Each force is
a different load case, which means that it
can be factored independently of all
other forces by a load vs time table.
There are 11 load cases and
corresponding to each one of these there
is a load vs. time table.
The load vs time table for each applied
load is a step input. The table that is
applicable turns the load on at the point
in time when the car is mid way between the present node and the previous node and turns the

load off when the car is midway between the present node and the next node. Zero time is
assumed to be when the car drives onto the right hand side of the bridge. If the load tables are
superimposed over one another then it can be seen that the structure is continuously loaded and
that the load moves from one node to another.
Modelling Procedure
• Create a beam from the origin to X=20000mm.
• Subdivide the beam into 10.
• Fully fix one end and set a roller condition on the other.
• Set the global freedom case as 2D Beam.
• Assign the beam the material properties and a square hollow section.
• Apply 9 point loads in 9 separate load cases. Note that a point load applied to a fixed node will
be ignored. Hence, ignore LC1 and LC11 in the previous figure.
• Create 9 Factor vs Time tables (Tables/Factor vs Time) for a 3 second time history.
Solver Setup
• Run the Linear Transient solver for a 3 s
time history. Use a 0.001 s timestep.
Ensure that beam stresses are
calculated.

• In the Load Tables dialog, assign each load case the
appropriate load vs time table.
• Determine the maximum fibre stress and maximum deflection
that occurs when the car drives over the bridge.
• Animate the results.
• As a further exercise, calculate the natural frequencies of the
beam and then apply Rayleigh damping to the structure and
investigate the effects of damping.

Single Degree of Freedom System
Outcomes
• Model a spring/damper system.
• Use the Linear Transient Dynamic solver.
• Use the results of a linear static analysis as the input to a transient dynamic analysis.
Problem Description
In this lesson, two problems will be analysed, both
involving a single degree of freedom (SDOF) system
as shown.
In the first, the undamped SDOF system is
subjected to a harmonic base displacement of two
frequencies - one at 5 rad/s, the other at the
system’s natural frequency. This harmonic
displacement analysis is repeated for a damped
system.
In the second problem, the damped system is firstly applied a force and is then released.
For both problems, we are interested in finding out the displacements of the single degree of
freedom.
The following data are given:
• m: lumped mass of the single degree of freedom system = 0.5 Tonne.
• k: axial stiffness of the spring-type element = 200 N/mm.
• l : spring length = 500 mm.
• w : frequency of the external harmonic base displacement = 5 rad/s and then 20 rad/s.
• Amplitude of the external harmonic base displacement = 20 mm.
SDOF with Support Excitation - Theoretical Solution
S.d.o.f. system without damping
The theoretical solution is obtained by solving the following ordinary linear, second order
differential equation:
)tsin(Ak)t(xkxkxm eg ⋅⋅⋅=⋅=⋅+⋅ ω

The natural frequency of the system is given by:
If the external and the natural frequency are equal then the resonant response will be obtained. In
this case, the displacement of the point B is harmonic but its amplitude increases linearly.
Consider the case with an external frequency equal to 5 rad/s.
The general solution for this case is:
where:
S.d.o.f. system with damping
In this case the equation describing the physical behaviour of the system is:
with the following steady-state response:
where z is the damping ratio.
Modelling Procedure
• Create a node at the origin and another at Y=500mm.
• Create a beam between the two nodes.
• Select the top node and click Attributes/Node/Translational Mass to assign a mass of 0.5 T.
• Select the bottom node and click Attributes/Node/Restraint and assign a fully fixed restraint.
• Set the global freedom condition such that only translation in the Y direction is possible.
• Click Property/Beam and set the beam as a spring/damper with an axial stiffness of 200 N/
mm.
• Select the top node and click Attributes/Node/Restraint and assign a 1mm translation in the
positive Y direction.
rad/s20==
m
k
nω
( ) ( )[ ])sin(sin
1
)( 2
tt
A
tx ne ⋅−⋅
−
= ωβω
β
β
ωe
ωn
------=
)sin()( tAktxkxkxcxm eg ⋅⋅⋅=⋅=⋅+⋅+⋅ ω
( ) ( )
( ) ( ) ( )[ ]ttAtx ee ⋅−⋅−
»
»
¼
º
«
«
¬
ª
+−
⋅= ωζβωβ
ζββ
cos2sin1
21
1
)( 2
222

• Click Tables/Factor vs Time. Click the
equation editor. Setup a sinusoidal time
history with an amplitude of 20 and a
frequency of 5 rad/s over a 3 s period. Use
approximately 80 sampling points.
• Create a second table with a sinusoidal time
history. This time use a frequency of 20 rad/s
and use approximately 500 samping points.
Solver Setup
• Use the linear transient dynamic solver. Click Load Tables and set the freedom condition to
use the first Factor vs Time table such that the enforced displacement acts sinusoidally. Click
Time Steps and set a 3 s time history with time steps that capture roughly 1/20th the period of
the highest frequency of the system.
• Graph the displacements of both the top and
bottom nodes.
• Rerun the solver using the second Factor vs Time table.
• Graph the displacements of this result.
• Assign a damping coefficient to the spring/damper and rerun the solver a third time using a 20
rad/s excitation frequency. The damping coefficient will be set such that it is 0.1 of critical
damping. Critical damping is:
cc = 2mω = 2.(0.5).(20) = 20 Ns/mm
S.d.o.f. without damping, freq=5 rad/s
S.d.o.f. without damping, freq=20 rad/s

Note that the frequency value is in (rad/s). Hence, the damping value is 0.1cc = 2 Ns/mm.
• Graph the displacements of this result.
SDOF with Non-Zero Initial Conditions
Theoretical Solution
The theoretical solution is obtained solving the following ordinary second order differential
equation:
with the following initial conditions:
and
where is the displacement at the top node because of the point load applied.
If we assume c to be as:
the solution has the following expression:
where is the natural frequency of the damped system =
Modelling Procedure
• Use the model from the previous problem.
• Click Attributes/Node/Restraint and fully fix the bottom node.
• Click Attributes/Node/Force and apply a point force of 10 kN at the top node.
• Click Tables/Factor vs Time and setup a table with a 1 value before the release time and 0
after that for a 4 second period.
SDOF with damping, freq=20 rad/s
0=⋅+⋅+⋅ xkxcxm
x t 0=( ) x= x
·
t 0=( ) 0=
x
kmc ⋅⋅≤2
( ) ( ) t
D
D
D etsintcosx)t(x ⋅⋅−
»
¼
º
«
¬
ª
⋅⋅
⋅
+⋅= ωξ
ω
ω
ωξ
ω
ωD ω 1 ξ
2
–⋅

Solver Setup
• Run the Linear Static solver.
• Run the Linear Transient Dynamic solver. Specify as Initial Condition the previous linear
static result file (*.lsa). Use the Time Steps from the previous problem, except run for 4000
steps such that a 4 s history is recorded. In the Load Factors dialog, specify that only the first
load case refers to the created table.
• Graph the results.

Viscous Damping Coefficient of a Cantilever
Outcomes
• Determine the viscous damping coefficient of a component.
• Use the Linear Transient Dynamic solver.
• Use the Linear Static solver to obtain an initial condition for the transient solution.
• Use the Natural Frequency solver to determine a suitable timestep and time length for the
transient solution.
Problem Description
The viscous damping coefficient of a cantilevered beam is to be calculated. This coefficient can
be calculated by trial and error until the appropriate damping has been found such that the free
oscillation of the beam matches what a physical beam does. Alternatively, the coefficient can be
set to match known appropriate damping ratios.
The Linear Transient Dynamic solver can be used to predict the rate of decay of the resulting
free oscillations of the beam. By changing the viscous damping coefficient for the canitlever, the
oscillation rate of decay can be modified.
For this example, a critical damping ratio ( ) of 2%
(nominal for a steel beam) is to be obtained. The
critical damping ratio is related to the logarithmic
decrement of the rate of decay as per the following
equation:
Based on a of 0.02, the ratio of decrement (An/An+1)
is =e0.1257
=1.134.
Modelling Procedure
• Create a horizontal beam, 3 m in length.
• Subdivide the beam into 10.
• Fully fix one end and apply an enforced displacement of 100 mm in the vertical direction at its
tip.
• Set the default freedoms to 2D Beam.
• Assign a Structural Steel material property and a 150UB14.0 section (under BHP Universal
Beams in the section library).
ζ
( )2
1 1
2
ln
ζ
πζ
δ
−
=¸¸
¹
·
¨¨
©
§
=
+n
n
A
A
ζ
e
2πζ( ) 1 ζ
2
–( )⁄( )

• The structural damping of the beam is simulated using viscous damping. Set the viscous
damping coefficient as zero.
Solver Setup
• The Linear Static solver is run to bring the system to an initial rest condition such that the tip
of the beam is deflected by 100 mm.
• Remove the enforced displacement at the tip.
• Run the Natural Frequency solver to determine the natural frequencies of the beam.
• Set up the time steps such that a time history of 0.2 s is recorded. Ensure that the time step
used is sufficiently small to capture the 1st mode. Based on the 1st natural frequency,
1 x 10-3
s is sufficient.
• Ensure that Added Damping is set to Viscous.
Results
• Plot a graph of the vertical displacement of the
tip. The oscillations should be undamped.
• Close the results and set a viscous damping
close to 5 x 10-8 Ns/mm/mm3. Solve the
model again using the Linear Transient
Dynamic solver and view the graph.
• Determine the ratio of rate of decay in the
oscillation (A1/A2). If the ratio is not close to
1.134, then change the viscous damping and
reiterate until a closer value is found.

Masses Falling on Two Cantilever Beams
Outcomes
• Use the Nonlinear Transient Dynamic solver.
• Use zero gap contact elements to model an impact problem.
• Use rigid links to model the thickness of beams.
• Use viscous damping.
• Use a static solution to represent the initial condition for the transient dynamic solution.
Problem Description
Two 150UB14.0 universal beams
3 000 mm in length are
cantilevered and separated by a
50 mm gap. A 200 kg mass is
dropped from a height of
1 000 mm on to the upper beam.
The response of a 2 kg mass,
initially at rest on the lower beam,
is sought.
The beams are sub-divided into
10 equal lengths. The default
freedom condition is set to 2D beam. The 200 kg mass is represented by a beam element 150
mm in diameter and 200 mm in length (density = 5.65884 x 10-8
tonnes/ ). The 2 kg mass is
represented by a beam element 50 mm in diameter and 100 mm in length (density = 1.01859 x 10-
8 tonnes/ ). Point contact elements will be used to simulate contact between masses and
beams.
Modelling Procedure
• Create the two beams: From (-3000,75,0) to (0,75,0) for the left beam and from (0,-125,0) to
(3000,-125,0) for the right beam. Note that the vertical distance is greater than 50 mm - this is
to account for the depth of the beams.
• Subdivide both beams into 10.
• Create the two masses out of beams: From (0,1225,100) to (0,1225,-100) for the 200kg mass
and from (400,-25,50) to (400,-25,-50) for the 2kg mass. Subdivide both into 2.
• Create three point contact elements out of beams - one is required to model the gap between
the two beams, the others are required to model the gap between the masses and the beams.
Create a beam from (0,1150,0) to (0,150,0) for the contact between the 200kg mass and the
left beam and from (0,0,0) to (0,-50,0) for the contact between the two beams. A contact
element is also required to model an infinitesimally small gap between the 2kg mass and the
mm
3
mm
3

Straus r7-Software Dynamics Analysis

Straus r7-Software Dynamics Analysis

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