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

Presented by G+D Computing Pty Limited 1

Discussion: Categories of Dynamic Problems and theCorresponding 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 andcorresponding mode shapes. A well known phenomenon is that when a structure is subjected toa sinusoidal force and the forcing frequency approaches one of the natural frequencies of thestructure, 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’smass and stiffness distribution (for an undamped system).

An eigenvalue problem allows the calculation of the (undamped) natural frequencies and modeshapes of a structure. A concern in the design of structures subject to dynamic loading is to avoidor cope with the effects of resonance.

Another important aspect of an eigenvalue solution is in its mathematical significance - that is, itforms the basis of the technique of mode superposition (an effective solution strategy to decouplea coupled dynamic matrix equation system). The mode shape matrix is used as a transformationmatrix to convert the problem from a physical coordinate system to a generalized coordinatesystem (mode space).

In general for an FE model, there can be any number of natural frequencies and correspondingmode shapes. In practice, only a few of the lowest frequencies and mode shapes may berequired.

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 machinerysubject to a mass imbalance. Another example is that of a vibration test table driven by asinusoidal base excitation.

When a sinusoidal excitation is applied to a structure, the structure will initially vibrate in anirregular manner often referred to as the transient stage. The irregular part decays to zero overtime due to damping. After the transient stage, the structure will vibrate in a sinusoidal fashion at afrequency identical to the frequency of the applied excitation, but the phase of the response maybe 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 thedifferent 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 afrequency range.

3. Transient analysis problems

Transient dynamic analysis is used to calculate the entire time history, from the starting point ofloading, of the dynamic response of a structure subjected to external dynamic loading of anarbitrary time function and initial conditions.

This kind of analysis is often used to analyse a structure under a shock loading which has a shortaction duration but perhaps wide frequency range.

The calculation of a transient analysis uses numerical integration methods, such as the Newmarkand 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 structurewhich 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 structurebased 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 astatistical 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)

whereM - Global mass matrixC - Global damping matrixK - Global stiffness matrixP(t) - Applied external excitation vectorU(t) - Unknown nodal displacement vectorThe 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 fouranalyses are conducted.

(1) When there is no external load, i.e. P(t)=0 and also the damping matrix C=0, equation (1) canbe 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 theresponse 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 byconsidering 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 solvercan 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 iscalculated.

Harmonic Response Solver

Calculates the steady state response of a structure subjected to a sinusoidally varying load. Thesolver 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 givesdisplacements, stresses, etc., as a function of time. Both linear and nonlinear analysis can beperformed.

Spectral Response Solver

Calculates the response of a structure subjected to a random dynamic loading (e.g. an earthquakerepresented by its response spectrum or a mechanical vibration represented by its Power SpectralDensity). The solver gives estimated maximum deflections, stresses, etc. or statistical estimations(PSD).


Discussion: Modelling Considerations for DynamicAnalysis

Introduction

Most Straus7 users have a good understanding of the basic requirements for the design andconstruction of models used in static structural analysis. In general, the techniques used to buildmodels for use in dynamic analysis are similar but there are a number of issues that need specialconsideration.

Number of Elements and Mesh Density

The overall mesh density required for dynamic analysis is in general higher than that for a staticstructural 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 toprovide a good representation of these. All the elements in the Straus7 element library have shapefunctions of a fixed order. These shape functions define the deformed shape of the element. Forexample 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 asingle beam element is used then the higher order modes cannot be calculated and possibly eventhe frequencies and mode shapes of some of the lower modes may be incorrect. A single beamcannot represent the sinusoidal mode shapes because of the inability of a single cubic equation toapproximate more than half of a sine curve. In this case many cubic beams are required to providea piecewise cubic approximation to the sinusoidal mode shape. If only the lower modes are ofinterest, the mesh can be relatively coarse. However for harmonic, transient dynamic and spectralanalysis the higher modes are frequently important as they may be excited by high frequencyexcitation of the structure. Decisions on the mesh density clearly require a sound understanding ofthe 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 isless of a need to refine the mesh locally around areas of stress concentration, particularly if themesh is not being used to calculate stresses in a separate linear static analysis. In dynamicanalysis the global inertial and stiffness characteristics of the model are usually more importantthan local behaviour. There are however some special cases where local modes are importantand 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 thedetermination of the frequencies and mode shapes. This is evident in the simple equation for thenatural 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 importantto ensure that the model provides a correct representation of both the stiffness and mass of thestructure.

Modelling of Non-Structural Mass

Often a structure being modelled for a dynamic analysis will be a support frame for some sort ofequipment. 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 thestructure. In a typical linear static analysis these masses might be represented with equivalentforces and pressure loads but in a dynamic analysis the actual mass and its distribution must berepresented accurately. In many cases non-structural masses can be represented using pointmasses. A portion of the non structural mass is lumped at each of its attachment points on thestructure. This approximation assumes that the item has mass but no stiffness.

Often items of non-structural mass provide some additional stiffness between the attachmentpoints. The way in which the stiffness of these items is modelled depends on the relative stiffnessof 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 belumped at the centre of gravity of the item. This is connected to the attachment points on thestructure with rigid links. An example of a mass that would be modelled in this manner is anengine mounted in a frame.

In other cases the stiffness of the non-structural mass, between the attachment points, is similar tothe stiffness of the structure. In this case there is no option but to include a coarse finite elementrepresentation of the item producing the mass. This mesh can be crude because it is only beingused to provide an approximate representation of stiffness and inertia, not to calculate stressesand deflections. Furthermore, the use of a crude mesh helps to keep the model to a reasonablesize. A coarse mesh also helps to suppress any local modes of the non structural mass sincethese are generally of little interest in the analysis.

There are two methods commonly used to include the mass of the item in the unrefined finiteelement model:

• A point mass is often placed at the centre of gravity of the item and connected to theattachment points on the structure with the finite element representation of the non-structuralmass. In this case the elements used to model the non-structural mass are not assigneddensity.

• 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 andstiffness of a piece of electronic equipment is shown inthe adjacent figure. This model is an idealization of arack of integrated circuit boards. The density of thevarious components (i.e. rack, boards, etc) wasfactored to get the correct overall mass. Note thecrudeness of the model. This mesh would clearly betoo coarse for use in a linear static structural analysisor a dynamics analysis of the component itself. It ishowever, sufficient for including the mass and stiffnesseffect of the circuit board rack on the overall behaviourof the structure to which it is attached.

Often the centre of gravity of equipment or other itemsof non-structural mass are offset significantly from theattachment points on the structure. It is very important that the centre of gravity of all items ofmass be correctly located.

If the centre of gravity offsets are to be included in the model, it is common to offset the mass fromthe attachment point by a rigid link of an appropriate length. Alternatively if the non-structuralmass is modelled using the coarse finite element model approach, the centre of gravity offset willbe included automatically. Any such finite element approximation should be checked to verify thatthe centre of gravity is in the correct location. This can be done by using the Summary/Modeloption 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 ofan element is automatically calculated provided a density has been assigned. This mass isassumed to be distributed uniformly over the element.

In the finite element method all mass is eventually assigned to the nodes. This means that thecontinuously distributed mass of the elements must be converted to an equivalent set of massesat 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 thelumped mass approximations.

In the lumped mass approximation, mass is lumped to the nodes of the elements in a simpledistribution such that the sum of these nodal masses equals the total mass of the structure. For a2-node beam, it is intuitive to lump half the mass at each node. In this typical finite elementapproach, usually only translational inertias are represented directly at the nodes, omitting termsrelated to the rotational inertia. Overall rotational inertia is accommodated by the fact that thenodal translational masses are distributed over a large geometric region - a bit like a governorwhere the overall rotational inertia is a function of the translational masses and the distancebetween them. In Straus7, the lumped mass approach generates a very small (diagonal) matrixwhich means that compared with a linear static analysis, only a small amount of extra space isneeded.

The consistent mass approach is more accurate and the distribution is based on determining amass lumping scheme that gives both translational and rotational inertias. The distribution isbased on the same integrations that are used to calculate the element stiffness matrix and thisgenerally results in a distribution that is not very intuitive. Furthermore, because the mass matrixis 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 withconsistent mass, the solver is slower.

The choice of the lumped or consistent mass approximation can, in some special cases, have asignificant 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 massis always treated as a nondiagonal mass. This is because the general case of a rotational inertiaabout 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 matrixassemblies. However, if you choose Lumped, but the analysis requires consistent (e.g. becauseyou have a rotational mass or a beam or plate offset, etc.) then for those elements/nodes, thematrix 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 equationreduces to:

This has a solution in the form of a simple harmonic motion, where the displacements are givenby:

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 sin2odd −=��

ω 2πf=

[ ]{ } [ ]{ }oo dMdK 2ω=

ω2d0{ }

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 canbe 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 (naturalfrequencies).

For each eigenvalue there is a corresponding eigenvector which is a set of displacementsdefining the mode shape. It is important to realize that the displacements in the eigenvector arenot absolute values of displacement. It is the relative magnitude of the displacements that isimportant in defining the mode shape. The actual amplitude of a mode depends on the magnitudeof an excitation force. Since the natural frequency solver is solving for unforced naturalfrequencies only, information on the magnitude of the displacements associated with a particularmode shape is not available.

The values of displacement (i.e. the eigenvector) in the Straus7 output are normalized such thatthe 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 themodal 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 tosolve this in a reasonable time some approximations must be made to reduce the size of theproblem. One solution method (and the one used by Straus7) is Sub-Space iteration.

ω 2π fi

[ ]{ } [ ]{ }iii dMdK λ=

λ i ωi2

= ith

di{ } ith

K[ ] λ i M[ ]–( ) di{ } 0{ }=

λ i

wi λ i=

fiwi

2π------=

d{ }

{ } [ ]{ } 1dMd iTi =


The basic concept of this method is that it takes the stiffness and mass matrices for the fullstructure and reduces these to a manageable size so that it can be solved by direct eigenvalueextraction methods.

Convergence of the Natural Frequency Solution

The default convergence tolerance in the natural frequency solver panel is 1.0E-5. This meansthat in order for the solution to terminate, the least accurate eigenvalue must be changing by lessthan this amount between successive iterations. This is a tight tolerance and well below normalengineering accuracy. Users are often tempted to increase the convergence tolerance to valuessuch as 1.0E-3 to decrease the solution time. In many cases this practice will work but it shouldbe used with caution. The reason the default value is set to such a high tolerance is based on theresults of extensive experience with the solver. Often additional modes are found between thecurrent eigenvalues as the solution proceeds. This usually occurs when there are many closelyspaced modes. These slot in between the modes that had been calculated to this point in thesolution 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 solutionprocess may not have adequately represented all the modes. As the solution proceeds, randomadjustment introduces new degrees of freedom into the sub space. This can allow previouslyunknown modes to be identified and captured. Reducing the convergence tolerance can cut thesolution process short and not allow sufficient time for the detection of all modes including theadditional modes not identified by the initial starting vectors. The convergence tolerance shouldonly 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 defaultconvergence criterion at least once to assist with this verification. In addition to this the Sturmcheck can be used to verify that all of the modes are being located by the solution with relaxedconvergence 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 notbe fully converged. If unconverged eigenvectors are used in mode superposition analysis(harmonic, spectral, etc.), the results can be erroneous. This is sometimes manifested by spectralruns producing mass participation factors in excess of 100%.

Another way of controlling convergence is to reduce the maximum number of iterations (whichdefaults to 20).

Further Notes on Eigenvalues

There are a number of questions concerning eigenvalues that inevitably arise when doing naturalfrequency analysis.

Multiple eigenvalues

In many models some of the calculated eigenvalues will appear as identical pairs. The reason forthis is that many structures are symmetric and thus have orthogonal pairs of modes. That is, thestructure has an equal tendency to vibrate in two perpendicular planes. These planes need not bethe 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 eithertranslates or rotates as a completely rigid body with no relative displacement between thenodes on the structure. This is a common result whenever you model the natural frequenciesof an unrestrained structure - e.g. an aeroplane in level flight.

• The structure is a mechanism. This means that insufficient stiffness is provided to preventsome 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 orexpected. If this occurs the most likely cause is incorrect zipping of the model. It may be necessaryto increase the value of zip tolerance that has been used to ensure that all parts of the model arecorrectly 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 thefollowing 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 frequencyresults will include many local modes in which there is no interest. In general these modes willoccur 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 tothe first n modes required by the user. Occasionally some low order modes may be missed andhigher modes found in their place. An eigenvalue may be missed by the solver if the initial startingvector in the sub-space does not include degrees of freedom that provide an adequaterepresentation of the mode. The degrees of freedom contained in the initial starting vector must becapable 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 orthogonalmodes, both in the plane and out of the plane of the page. If the degrees of freedom excited by thestarting vector are only in the vertical direction then these cannot represent the out of plane modesand some of the modes will be missed.

The likelihood of Straus7 missing eigenvalues is low since special precautions are taken duringthe solution process to continually introduce new degrees of freedom into the trial vectors thatspan the subspace.

It should be noted that the Sturm check can only determine the number of eigenvalues in aspecific range, it does not calculate the value of the eigenvalues. It is however a useful check onthe output data. The Sturm check is a very stable and reliable method for determining the numberof eigenvalues in a given range. This stability results from the fact that the method only relies onthe signs of numbers and not the actual values. Thus rounding error and other errors will have lesseffect 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 thisincluded 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 normalstiffness 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 tosome vibrating machinery). In these cases we can use the shift value (in Hertz) to ask thesolver to calculate only modes near the shift value. The eigenvalue search is centred on thegiven shift and the solver will find the eigenvalues closest to the shift, both above and below theshift.

2. The shift can also be used for finding the natural frequencies of a structure which is notrestrained, (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[ ] λ i M[ ]–( ) di{ } 0{ }=

λo

K[ ] λ i λo+� �� M[ ]–( ) di{ } 0{ }=

[ ] [ ]( ) [ ]( ){ } { }0dMMK iio =−− λλ

λ i

λ i λo λ i


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 stainlesssteel. 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.

Hzl

JJ

GJf

sd

p 7091.271

321 =

��

��

� +=

π

E2 1 v+� �--------------------- 7.4806202 10

4×

Jsπds

4

32----------- 9.817477 10

2× mm4

Jsπds

4psls

32-------------------- 3.9269908 10

3–×= = tonnes·mm2

Jdπdd

4pdld

32---------------------- 4.9087385 10

2–×= = tonnes·mm2

πd2pdld4

-------------------= 1.5707963 104–×= tonnes


• Construct three models ofthe rod side by side in theone model window. Use10 beam elements tomodel the shaft.

• The first model has thedisc modelled using ashort beam. The secondmodel represents the discas a point mass withrotational inertia (RY mass

= 0.049 Tmm2) and thethird model has the disc asa point mass withtranslational 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 andcalculate the first three modes using both thelumped and consistent mass option. To swapbetween the lumped and consistent massmatrix option, go to the Defaults tab page inthe Natural Frequency solver dialogue. Clickthe Elements button on the left.

Results

Summarize the results in the following table:

Modelling TechniqueSolution

Number ofElements Short Beam Rotational Mass

TranslationalMass

Strand7 (Mass Matrix Lumped) 10 271.67 271.67 1527.38Strand7 (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 310UB46.2universal beam are examined. In particular, acomparison is made between the analyticalsolution, lumped mass matrix and theconsistent mass matrix methods. The meshdensity and shear area are also evaluated. Thebeam is simply supported with a length of5000 mm and the global freedom condition isset 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 thesecond into ten elements.

• Set the global freedoms to 2D Beam.

• Set the beam property. Set the material of the beam to StructuralSteel and the section to 310UB46.2 in the BHP - Universal Beamssection database. In the Sections tab page, set the section areasto zero.

• Run the natural frequency solver and solve for 10 modes.

• Rerun the solution using the Consistent Mass Matrix option.

wn 2πf n2π2 EI

ρAl4

-----------

ωn 2πf π E

ρl2-------


Results

Note that at least two elements are required to calculate the natural frequency of the flexural andlongitudinal modes using the lumped mass approximation. This is because there are insufficientmass degrees of freedom.

The consistent mass approximation however uses the element displacement shape functions, andoften can better represent the real mass distribution over the structural element. For beamelements, 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 asnon-zero (note that the analytical results assume thin-beam theory).

Mode of VibrationSolution Number of

Elements 1st Bending 2nd Bending 3rd Bending 1st Axial 4th Bending

Analytical n.a. 41.1844 164.7376 370.65956 504.7545 658.9503Strand7 (Lumped) 2 40.8854 n.a. n.a. 454.4387 n.a.Strand7 (Consistent) 2 41.2099 180.4273 445.6487 556.5714 794.1139Strand7 (Lumped) 10 41.1841 164.7180 370.4102 502.6813 657.3207Strand7 (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 astructure, which can be useful in any mode superposition analysis. The aim of this section is tohelp 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 modeshas 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; andR - 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 asufficient number of relevant modes included in the analysis for the given global movement vectorR. As more and more relevant modes are included, the value of the sum should approach 1. Asa general rule for each excitation direction, the sum of the participation factors of the modesshould be greater than 90%.

The relevant modes refer to those that have nonzero values of mass participation factor and havesome contribution to the global movement. For example, consider a vertical rod. If the baseexcitation is in a horizontal direction, only the mode shape vectors which have nonzerocomponents in the horizontal direction will have some contribution, while mode shape vectorswhich have only components in the vertical direction will make no contribution. Mathematically, arelevant mode shape vector is closer or more correlated to the global movement vector R than anirrelevant mode.

For the relevant modes, their contribution to the total structural response depends on themagnitude of the response of the individual mode to the modal force. This magnitude is also afunction of the mode frequency under the given modal force. Hence, although it is often used as agood indicator, the mass participation factor cannot be used on its own for determining thenumber of modes to include for mode superposition.

The global movement vector R represents the corresponding rigid movements of all thetranslational degrees of freedom in the FE model under a given base excitation movement. Thevector is formed based on specifying the direction factor vector.

PFi

φiT

MR( )2

RT

MR------------------------=

φi


Example

The following example is used to illustrate the above discussion. A vertical rod is modelled by 10beam elements, restrained at the base. Material is structural steel from the Straus7 materiallibrary.

• To better illustrate the problem, 2D Beam freedom conditions are set such that all modeshapes 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 globalmovement 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 MassParticipation summary reports that the total mass participation factor is 99.499% indicating almostall 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 modeshape of this mode can be examined by looking at the Straus7 natural frequency result andplotting 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 modeor vertical movement which is irrelevant to X direction movement, hence the mass participationfactor 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 totalmass 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 areoften sufficient to obtain an accurate result.


Discussion: The Use of Symmetry in NaturalFrequency Analysis

Introduction

When modelling symmetric structures it is common practice to reduce the size of the model byusing the principle of symmetry. Appropriate freedom conditions are applied on the plane ofsymmetry so that half of the structure modelled behaves as though it is still attached to the otherhalf of the structure.

The nature of a symmetry boundary condition means that a structure must deform symmetricallyabout the plane of symmetry. This normally means that in addition to the structural geometry beingsymmetric, the loading must also be symmetric. Whilst most analysts are comfortable with theconcept of symmetry in linear static problems, experience shows that this is not necessarily thecase with regard to natural frequency and buckling analysis (note that the equations solved forlinear 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 asstraightforward as it is for linear static analysis.

A symmetry model with symmetric boundary conditions will yield the symmetric buckling andvibration modes only. To obtain the anti-symmetric modes it is necessary to run the model asecond time with anti-symmetric boundary conditions applied to the geometric symmetry plane ofthe structure. For very large models it may be better to use the symmetry approach, since runningthe half model twice will usually be faster than running the full model once.

Anti-symmetric boundary conditions are simply the opposite of symmetric conditions - anydegrees 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 withdimensions (in metres) as shown. The model uses Structural Steelas the material property and BHP - Universal Beam 530UB92.4 asthe section. Three models are constructed: the full model, asymmetric 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 Model1 9.567 9.5672 45.78 45.783 81.534 81.5344 90.973 90.9735 146.334 146.3346 150.701 150.7017 154.025 154.0258 169.834 169.8349 188.927 188.92710 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 ofradius a = 381 mm,thickness t = 0.254 mm issimply supported along itsedge and is subjected to anin-plane radial pre-stress of68.971125 MPa. Determinethe first 3 axisymmetricnatural frequencies oflateral vibration of themembrane.

Model the membrane usingthe axisymmetric 8 nodeplate elements. Use 20elements as shown in the figure.

Modelling Procedure


• Create a beam 0.254mm long in the Y direction. Extrude this beam to atotal length of 381mm in 20 steps. Ensure that the beams are convertedto 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 themembrane.

• Run the Linear Static solver.


• Run the NaturalFrequency solver tocalculate for 4 modes ofthe structure withoutpreload. Results shouldappear as shown.

• Record the frequencies inthe table below.

• Run the Natural Frequency solveragain, this time include the effect of thepreload. Do this by using the linear staticresults 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 notaxisymmetric. In this model, those modes have been ignored.

Mode TheoryStraus7withoutPreload

Straus7withPreload

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 manymechanisms responsible for damping, e.g. material damping, friction at contact surfaces, etc.

In Straus7, damping is represented by different linear damping models. Three models areavailable: 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 atthe element level. This last option is useful for determining the overall damping ratio of a systemcomposed of parts with different damping characteristics.

Viscous Damping

The viscous damping model uses the following expression to calculate the element dampingmatrix:

where

- Material viscous damping coefficient with the units of force/velocity per unitvolume;

N - Element shape function matrix; andVe - Element volume domain.

The global damping matrix is obtained by assembling all element damping matrices. The elementdamping matrix can come from the viscous damping coefficient assigned to the elementproperties, as shown in the dialog box below, or in the case of a spring element, it can come fromthe damper coefficient.

This type of damping is only relevant to the Full System Transient solvers (both Linear andNonlinear). To activate the viscous damping contribution for all but the spring elements, you needto select the option Viscous on the Added Damping settings on the solver dialog (as shown).

Ce µNT

N Vd

Ve�=

µ


For spring-damper elements, the damping part (the so-called discrete damper) is alwaysassembled, irrespective of the setting of the Added Damping option.

Rayleigh Damping

Rayleigh damping, also known as proportional damping, assumes that the damping matrix is alinear combination of the stiffness and mass matrices:

where and are proportional constants. One of the important advantages of Rayleighdamping 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 dynamicequilibrium 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 tothis 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 ζ1ζ2 ω1 ω2

ζ 12---

αω---- βω+� �� =

α2ω1ω2 ζ2ω1 ζ1ω2–( )

ω12 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 dampingvalues can be assigned to each mode. Consider the case where the analysed structure iscomposed of different materials that when combined together deliver different damping propertiesper vibration mode (natural frequency). For example, an FE model of a concrete structure sittingon a soil foundation may contain two groups of modes: modes where the movement is dominatedby the structural deformation of the concrete and modes where the concrete structure moves as arigid body on the soil foundation. In this case, the concrete deformation modes would have a lowdamping 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.

β2 ζ1ω1 ζ2ω2–( )

ω12 ω2

2–

----------------------------------------=

ω1 ω2

ω1 ω2

α βζ1 ζ2

ζ ici 2ζ iωni=


Modal damping can be used with anyStraus7 solver that uses ModalSuperposition. This includes the harmonicand spectral solvers and the linear transientdynamic solver with the Superpositionoption.

Use of modal damping involves the specification of thefrequency file along with mode damping coefficients.

Effective Damping Coefficients

If a material damping ratio is defined at the element level, then an effective modal dampingcoefficient can be calculated based on the material damping ratio. This effective dampingcoefficient can be used in the spectral, harmonic response or transient solvers. The effectivemodal damping coefficient is computed according to the following formula (Japan RoadAssociation, 1990):

where

- Mode shape vector of element j of i-th mode

-Stiffness matrix of element j

K - Global stiffness matrixDCj - Damping ratio of element j

DCi

φj( )iTDCj Ke( )

jφj( )

i

j 1=

n

�

φiT

Kφi

------------------------------------------------------------=

φj( )i

Ke( )j


Notes About Damping

1. Damping exists in almost all real structures although its mechanism is often not clear. Dampingcan be due to internal friction in the material, Coulomb friction in connection, resistance fromsurrounding 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. Thefunction of damping is critical nearer the natural frequencies of the structure because around anatural frequency, the stiffness force and inertia force tend to cancel each other, leaving only thedamping force to balance the external force. Without damping, theoretically the response of astructure will become infinite when the forcing frequency is equal to a natural frequency of thestructure.

3. Often damping is small for most structures. The following typical/reference values of dampingratio 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 RoadAssociation, 1990). The following table provides typical damping ratios:

Type of Construction ξ

steel framewelded connectionsflexible walls 0.02

steel framewelded connectionsnormal floorsexterior cladding 0.05

steel framebolted connectionsnormal floorsexterior cladding 0.10

concrete frameflexible internal walls 0.05

concrete frameflexible internal wallsexterior cladding 0.07

concrete frameconcrete 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 astructural model due to harmonic loading. The harmonic loads act with identical frequencies andarbitrary phase angles. In addition to the nodal displacement and velocity, the solver is able torecover the maximum values of nodal acceleration, reactions, element stresses and phaseangles.

Introduction

The Harmonic Response solver calculates the maximum values of a linear elastic steady statedynamic response. The structural system is subjected to a set of harmonic forces F(t) withidentical frequencies and different phase angles .

Harmonic Loading

The response is calculated for a set of forcing frequencies, evenly distributed over a user definedfrequency range. If some of the natural frequencies of the structure are within the forcingfrequency range an additional forcing frequency will be introduced, identical to the naturalfrequency. This ensures that resonant response is captured. Furthermore, two additional pointsare 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 theLinear Static solver. All load cases are assumed to define a single loading condition. All loadscontained in a single load case (e.g. point forces, moments, pressure, etc.) act with the samephase angle and vary as functions of time in a sinusoidal fashion. The loads from different loadcases 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, oftenreferred to as a transient stage. After the initial period all the points of the structure will vibrate in asinusoidal fashion with a frequency identical to the forcing frequency , but with different

ω θ

F(t) = F sin (ωt + φ)F

time t2π/ω

Structure

φ

ω


amplitudes and different phase angles. This part of the response is known as the steady statestage. The Harmonic Response solver calculates the maximum values of the steady stateresponse, 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 inthe results. It is important to note that these maxima will not occur simultaneously; generally eachwill occur at different times, out of phase with the others.

The applied harmonic load can act with different frequencies, given as frequency steps on afrequency range sweep. Each frequency step is treated as an individual loading condition and it issolved 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 ofthe harmonic load at each step. This is useful for modelling situations such as a machine vibratingdue to an out-of-balance load whose amplitude increases with frequency.

Natural frequency analysis must be performed prior to any harmonic analysis. The HarmonicResponse solver uses the most recent results from the natural frequency analysis to perform amode superposition. If the model is modified in any way a new natural frequency analysis must beperformed. Once the natural frequency analysis is carried out many runs of the HarmonicResponse 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. Themethod used for calculation of the maximum response is based on superposition of the modalresponses. A greater number of modes included in the analysis will provide results of greateraccuracy.

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. Themaximum response and the phase angle for each mode is calculated by the followingexpressions:

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

iii tFtF

1

)sin()( θω

θ

ω θ

[ ] 21222 21/

)()(K

FA

−+−= ξββ

��

��

�

−= −

βξβθ

12

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 responsesconsidering the phase angles, the sign and the magnitude of the modal responses with respect totime. 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 identifiedanalytically by a closed form solution. The only approximation is the finite number of modes usedin the analysis. When a sufficient number of modes is used, the Harmonic Response solver canprovide an almost exact solution. If all modes for a given model are included, then the exactsolution for that model is obtained.

Results

Before considering any results the log file (extension: 'HRL') should be reviewed. Most of theglobal input parameters and input panel settings are listed. Also all warning messages should beexamined carefully.

For each frequency step the nodal displacements, velocities, and accelerations are calculated aswell as displacement phase angles. Reactions and element stresses are only calculated whenrequested by the user. All the results, for any frequency step can be shown separately, in a similarmanner 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 presentedgraphically for any node.

The results of the analysis are absolute maximum envelope values of the steady state dynamicresponse. The maximum values do not occur at the same time. Consequently the results do notpresent an equilibrium state of the structure and they do not correspond to each other. Forinstance 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 pointduring the steady state stage of the dynamic response.

Unlike other results, the fibre stresses in beams and principal stresses in bricks are not themaxima 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 fibrestresses 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 thesigns 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 modesused in the analysis. The solution considers the sign and the phase angle of all included modalresponses. When a sufficient number of modes is used, the results are usually of very highaccuracy. In some cases, when the structure has many local modes, even 100 mode shapes maynot provide a sufficient mass participation (see the section on Mass Participation Factor). In thiscase the results may underestimate significantly the real behaviour of the structure. It should beconsidered that the overall response is a summation of the modal responses, and that thestructure is represented by a finite number of modes.

The contribution of different modes is related to applied loads. The figure below shows the first twomode shapes of a simple structure and the load. The shape of the modes indicates that the firstmode will be excited by the load, but not the second one because the load is at the position wherethe 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 withrespect to the forcing frequency. For each frequency step all the forces are acting on the structurewith 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 solutiontechnique. These are the Transient Dynamics, Harmonic Response and the SpectralResponse solvers.

Basic Theory

The mode superposition method uses the mode shape vectors calculated by the naturalfrequency solver to transform the dynamic equilibrium equation into mode space. The responseof each mode to the forcing function is calculated and the results for all modes are superimposedto 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 areobtained by natural frequency analysis, then solving the de-coupled equations in the modal spaceindependently and then finally combining the solutions in the modal space to produce the solutionto 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 ormodal 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 themodal load vector. Matrices m and k are diagonal. For Rayleigh damping and modal damping, cis diagonal as well.

The elements, mii, kii and cii (i=1, 2,..., N), of matrices m, k and c are called modal mass, modalstiffness and modal damping respectively. For a mode shape matrix normalized with respect tothe 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+ += =

ΦT

ΦTMΦq·· t( ) ΦT

CΦq· t( ) ΦTKΦq t( )+ + ΦT

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 themaximum element of each mode vector is equal to 1 (referred to as the engineering mode shapematrix), the corresponding modal mass mii is called engineering modal mass. Straus7 can reportthis engineering modal mass if required.

If matricesm, k and c are diagonal, equation (4) can be expressed as N independent equations (Nis 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 thesystem.)

miq··i t( ) ciq

·i t( ) kiqi t( )+ + Fi t( )=

Φ

m ΦTMΦ

1 0 … 0

0 1 … 0… …0 0 … 1

= =

k ΦTKΦ

ω12

0 … 0

0 ω22 … 0

… … … …

0 0 … ωN2

= =

ωi

q··i t( ) 2ζ iωniq·i t( ) ωni

2qi t( )+ + Fi t( )=

ζ i 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 thenumber of modes used in the solution be sufficient to capture the response caused by a particularexcitation. 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 someproblems where this generalisation does not hold true.

If the loading contains high frequency components then some of the higher modes may contributesignificantly to the response. For example with impacts or blasts, the loading can be of a very highfrequency and it is important that the higher frequency modes be included. This means that theimpact frequency of the loading should be calculated and all the natural frequencies in the rangeup to and including the impact frequency also included in the analysis.

For spectral analyses involving a seismic spectrum or other broad band excitation, sufficientmodes should be included to ensure that all the natural frequencies of the structure within therange of the excitation are included in the analysis. Similarly when using the harmonic andtransient solvers, all the modes up to a frequency above the excitation frequency should beincluded.

Deciding on the number of modes to include in a solution by the magnitude of the frequenciesalone 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 onan instrumentation rack. The lower modes of these sorts of structures are typically swaying orbending modes. The vertical modes, involving flexing of the shelves and axial excitation of thecolumns, are usually the higher modes. It is also important to ensure that a sufficient number ofmodes is included in the analysis to represent the behaviour of the structure in the direction of theexcitation. This may require the inclusion of modes with frequencies significantly above theexcitation frequency.

An indicator is used to assist in assessing whether a sufficient number of modes has beenincluded. This is the mass participation factor. For further details of this see the section on MassParticipation 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 structuraldynamics. In this example we calculate the maximum response and stresses in a mine ventilationfan support structure.

The centre of mass of the fan blades is offset eccentrically3 mm from the spindle axis and the fan is spinning at540 RPM (9 Hz or 56.5487 radians/sec). The fan is in ahorizontal plane with a vertical axis of rotation.

The rotating out-of-balance force is the centrifugal forcearising from the rotation of the centre of mass of the fanon a radius of 3 mm from the rotation axis. This iscalculated using the equation:

F = mr 2 = 10000 * 0.003 * 56.54872 = 95932.6 N.

This force is applied to the structure as a rotating forcevector 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. Eachplate is 5x5 m and the three plates form an open box.

• Subdivide the plates 10x10.

• Create a node centred 3m above the plate supportstructure top.

• Create four beams from this node to the top of the platesupport structure.

• Set the beam and plate properties to use StructuralSteelwork as the material.

• Assign CHS 219.1 x 3.0 (in the CHS (350 grade) OD above100 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 weightof the fan.

• Create two load cases: In the first, assign a nodal force of 95 932.6 N to the centre node torepresent the out-of-balance force of the fan; in the second, assign the same magnitude force90 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 willvary from + to - over the specified frequency range. The direction of the force does not change. Inorder to simulate a rotating force vector we need to apply two load cases (Load Case 1 and LoadCase 2), both are point forces equal to the centrifugal force but the directions are 90 degrees toone another. Both forces are in the plane of the rotation. The harmonic solver will apply a vectorsummation of these two load cases. If load case 1 is applied 90 degrees out of phase with loadcase 2, load case 1 will be maximum when load case 2 is zero and vice versa. The vectorsummation of these two forces will then be a rotating vector.


Setup the Harmonic Response solver as shown

For the Frequency File choose the naturalfrequency file from the previous solution

Specify the load cases as shown to apply the loadcases out of phase with each other. Specify theRayleigh damping to span the 12 natural frequencymodes to be considered along with a damping ratio of0.05.

Results

Obtain the peak displacement response at the point where the force is applied. Also determine themaximum 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 involvesmounting a component on a shaker table andaccelerating it back and forth at a particular frequency orover a range of frequencies. These sorts of tests areused to determine the displacement response, resonantfrequencies of the structure (by looking for the peakresponses) or often as a fatigue endurance test.

The Straus7 Harmonic Response solver can be used tomodel this sort of test and can calculate the stresses anddeflections due to any base acceleration harmonicloading. Often finite element modelling is used inconjunction with test results to refine and developcomponents.

In this example we calculate the maximum response andstresses in an instrumentation rack. The frame of therack is constructed using a 25 x 25 x 3 square hollowsection and the shelving is 3 mm plating.


Model Setup

The mass of the instruments is included aslumped masses applied at the points wherethe instruments attach to the shelves. Thisapproach assumes that the actual instrumentsthemselves do not contribute any stiffness tothe structure.

The feet of the frame are modelled as apinned connection. The loading is applied asa global base acceleration in the X direction.The acceleration is 1.25g (enter as 1.25*9810in the X-Direction factor). The harmonicsolver assumes that this is appliedsinusoidally over the frequency rangespecified in the Harmonic Response solverdialog. The specified acceleration is theamplitude 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/Linesand tesselate a perimeter of beams around theplates.

• 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 squarehollow 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 lumpedmass).

Harmonic Response Solver Setup

• In the Harmonic Response Analysisdialog box, specify that the Load Type isto be a Base Acceleration. Set theDirection Vector to X=12262.5. Set theDamping to Modal. Set the frequencyrange to cover the majority of modes andensure that a suitable number of steps isspecified.

Note: The direction factor specifies both themagnitude and the direction of the baseacceleration. To apply a base acceleration of1.25g, enter the direction vector asX=1.25x9810 (where units are consistentwith the model).


• Click Frequency Files and specify that all modes are to beincluded with a Damping Ratio set to 0.05 for each mode.

Results

Plot a graph of the DX response at the nodeindicated (top corner). Find the peak response andidentify the range of excitation frequencies wherea resonant response occurs. Note that resonancedoes not occur for all modes - only those modeswith dominant motion in the direction of theexcitation (X).

Note that the response given by the Straus7harmonic solver includes both the amplitude andthe phase. You can plot graphs of phase similarlyto 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 thesestresses high enough to cause fatigue if the endurance limit of the steel is 100 MPa? What is theeffect of resonance on the stresses? That is, how much higher are the stresses at resonancecompared to the stresses at other non-resonant frequencies.

Another useful exercise is to evaluate the effect of varying the damping ratio. You should find thatthe results are sensitive to damping ratio near the resonant frequencies. Away from the resonantfrequency, the damping ratio should have minimal effect.

Strand7Maximum Response Dx

(at node indicated)


Discussion: Mass Matrix Formulation - Consistent vsLumped

Introduction

In the Defaults tab of the solver dialogue box, you can specify whether to use Lumped orConsistent mass formulations.

What is a Mass Matrix ?

Mass matrices are used to provide a discrete mass approximation for a structure where the massdistribution is continuous. This means that the continuously distributed mass within an element isapproximated by lumping a certain amount of mass at each of the nodal points. The mass matrixspecifies 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 finiteelement 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 massapproximation.

In the lumped mass approach the mass within each element is assumed to be lumped onto eachof the nodes such that the sum of the nodal masses associated with the translational degrees offreedom for each global direction equals the total mass of the element. Usually only thetranslational inertia effects are included. There is no rotational inertia and no mass couplingbetween the different degrees of freedom. This results in a diagonal matrix like that shown belowfor a simple two dimensional beam element.

For a bar element the lumped mass matrix is intuitive and simple. In the above example the totalmass of the element is divided by 2 and the resulting mass is lumped onto the nodes at each endof 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 beobtained by using such simple intuitive methods and more refined methods must be used. In thesimplest case the lumped mass matrix for a regular plate with 90 degree corners or a similarlyregular brick is obvious. For a square 4 node plate, 1/4 of the total mass is lumped onto eachcorner node. For most real structures the elements are distorted and therefore the mass matrix is

[ ]��

�

�

��

�

�

=

1000

0100

0010

0001

2

mm


no longer intuitive - different amounts of mass will have to be lumped onto each of the cornernodes. 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 isquicker to assemble and requires less storage space compared with a consistent matrix. Thelumped mass approximation will generally produce a lower level of accuracy than the consistentmass. This is particularly true if there are significant nodal rotations in the dynamic response suchas those that occur in flexural problems. Errors associated with the use of the lumped massapproximation can be greatly reduced with mesh refinement so that the lumped approach isnormally the more viable one.

Consistent Mass Matrix

The most accurate method of discretisation of the mass in a continuum is to use the consistentmass matrix. With this approach, the mass matrix is derived by using the same integrations as areused in deriving the stiffness matrix so that the mass matrix is consistent with the element shapefunctions.

The consistent mass matrix is the same shape as the stiffness matrix and includes off diagonalterms. It therefore considers the effect of mass coupling between the different degrees of freedomand the effect of the rotational inertia. The consistent mass matrix for the simple two dimensionalbeam of mass m, is shown below:

The calculation for a consistent mass matrix, based on the element shape functions integratedover 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 diagonalmass matrix in Straus7, is a reliable option offering very good results, particularly for the low ordermodes. Higher order modes will usually exhibit errors of 10% or more compared with a consistentmass 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 theconsistent (full) mass matrix to correctly model the mass. This includes offset beams and offsetplates, 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 thelumped mass option, the entries in the mass matrix for those elements/nodes requiring aconsistent (full) mass matrix are automatically expanded. Therefore, the correct influence of theoffset mass is obtained, even when lumped mass is selected.

[ ]��

�

�

��

�

�

−−−−−−

=

22

22

4L22L3L13L

22L15613L54

3L13L4L22L

13L5422L156

420m

M

ρ N[ ] T N[ ] Vd�

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 usedwhereby the conditions at time t are assumed to be known and are required after a discrete timestep, 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 perperiod). The method will fail if the time step is too large. With the full system analysis, it is notnecessary to perform a natural frequency analysis if the appropriate time step is known. If theperiod of the structure is not known, a natural frequency analysis should be run first to find thetime step. For loads that are varying at a faster rate, the time step may be dictated by the need toaccurately capture the load history.

Mode Superposition - General

Mode Superposition is an approximate method which analyses a reduced structural model. Thestructural system is approximated with several independent single degree of freedom systems.Each single degree of freedom system corresponds to one natural frequency and related modeshape of the full system. The dynamic response is calculated in two stages. The first stageinvolves frequency analysis. One or more mode shapes and corresponding frequencies must becalculated. Then the Mode Superposition solver will calculate the dynamic response of the modelby summation of all available modal responses. The Mode Superposition solver uses the most

M[ ] x··{ } C[ ] x·{ } K[ ] x[ ] F t� �{ }=+ +

M[ ]C[ ]K[ ]

x··{ }

x·{ }x{ }F t( ){ }


recent results from the frequency analysis. Once the frequency analysis is completed many runsof the Mode Superposition solver may be performed. If any modification of the model is made anew 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 theproblem of a car driving across abridge. The bridge is modelledas a simple square hollowsection (2 m x 2 m x 0.12 m) withthe material properties shown.

The car is traveling at a constant100 km/h (27.77 m/s). Knowingthis we can calculate the positionof the car at any point in time ormore importantly we cancalculate the time when the carreaches each point on thebridge.

To simulate the moving load a series of pointloads are applied along the length of thebeam. Each of these is equal to the staticweight of the car - 9810 N. Each force must beturned on only when the car is in the vicinity ofthe node to which the load is applied. Thismeans that when the car is within half thelength of the beam elements either side of anode, the load is applied to that node.

Only one of the forces is actually appliedto the structure at any time. Each force isa different load case, which means that itcan be factored independently of allother forces by a load vs time table.There are 11 load cases andcorresponding to each one of these thereis a load vs. time table.

The load vs time table for each appliedload is a step input. The table that isapplicable turns the load on at the pointin 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 isassumed to be when the car drives onto the right hand side of the bridge. If the load tables aresuperimposed over one another then it can be seen that the structure is continuously loaded andthat 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 willbe 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 stime history. Use a 0.001 s timestep.Ensure that beam stresses arecalculated.


• In the Load Tables dialog, assign each load case theappropriate load vs time table.

• Determine the maximum fibre stress and maximum deflectionthat occurs when the car drives over the bridge.

• Animate the results.

• As a further exercise, calculate the natural frequencies of thebeam and then apply Rayleigh damping to the structure andinvestigate 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, bothinvolving a single degree of freedom (SDOF) systemas shown.

In the first, the undamped SDOF system issubjected to a harmonic base displacement of twofrequencies - one at 5 rad/s, the other at thesystem’s natural frequency. This harmonicdisplacement analysis is repeated for a dampedsystem.

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 offreedom.

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 orderdifferential 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. Inthis 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 thepositive Y direction.

rad/s20==m

knω

( ) ( )[ ])sin(sin1

)( 2 ttA

tx ne ⋅−⋅−

= ωβωβ

βωe

ωn------=

)sin()( tAktxkxkxcxm eg ⋅⋅⋅=⋅=⋅+⋅+⋅ ω��

( ) ( )( ) ( ) ( )[ ]ttAtx ee ⋅−⋅−

��

�

��

�

+−⋅= ωζβωβ

ζββcos2sin1

21

1)( 2

222


• Click Tables/Factor vs Time. Click theequation editor. Setup a sinusoidal timehistory with an amplitude of 20 and afrequency of 5 rad/s over a 3 s period. Useapproximately 80 sampling points.

• Create a second table with a sinusoidal timehistory. This time use a frequency of 20 rad/sand use approximately 500 samping points.

Solver Setup

• Use the linear transient dynamic solver. Click Load Tables and set the freedom condition touse the first Factor vs Time table such that the enforced displacement acts sinusoidally. ClickTime Steps and set a 3 s time history with time steps that capture roughly 1/20th the period ofthe highest frequency of the system.

• Graph the displacements of both the top andbottom 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 20rad/s excitation frequency. The damping coefficient will be set such that it is 0.1 of criticaldamping. 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 differentialequation:

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 0after 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

( ) ( ) tD

DD etsintcosx)t(x ⋅⋅−

��

��

�⋅⋅⋅+⋅= ωξω

ωωξω

ωD ω 1 ξ2–⋅


Solver Setup

• Run the Linear Static solver.

• Run the Linear Transient Dynamic solver. Specify as Initial Condition the previous linearstatic result file (*.lsa). Use the Time Steps from the previous problem, except run for 4000steps such that a 4 s history is recorded. In the Load Factors dialog, specify that only the firstload 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 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 thetransient solution.

Problem Description

The viscous damping coefficient of a cantilevered beam is to be calculated. This coefficient canbe calculated by trial and error until the appropriate damping has been found such that the freeoscillation of the beam matches what a physical beam does. Alternatively, the coefficient can beset to match known appropriate damping ratios.

The Linear Transient Dynamic solver can be used to predict the rate of decay of the resultingfree oscillations of the beam. By changing the viscous damping coefficient for the canitlever, theoscillation rate of decay can be modified.

For this example, a critical damping ratio ( ) of 2%(nominal for a steel beam) is to be obtained. Thecritical damping ratio is related to the logarithmicdecrement of the rate of decay as per the followingequation:

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 itstip.

• Set the default freedoms to 2D Beam.

• Assign a Structural Steel material property and a 150UB14.0 section (under BHP UniversalBeams in the section library).

ζ

( )21 1

2ln

ζπζδ−

=��

��

�=

+n

n

A

A

ζ

e2πζ( ) 1 ζ2–( )⁄( )


• The structural damping of the beam is simulated using viscous damping. Set the viscousdamping coefficient as zero.

Solver Setup

• The Linear Static solver is run to bring the system to an initial rest condition such that the tipof 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 stepused 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 thetip. The oscillations should be undamped.

• Close the results and set a viscous dampingclose to 5 x 10-8 Ns/mm/mm3. Solve themodel again using the Linear TransientDynamic solver and view the graph.

• Determine the ratio of rate of decay in theoscillation (A1/A2). If the ratio is not close to1.134, then change the viscous damping andreiterate 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 beams3 000 mm in length arecantilevered and separated by a50 mm gap. A 200 kg mass isdropped from a height of1 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 into10 equal lengths. The defaultfreedom 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 andbeams.

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 isto 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 massand 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 betweenthe 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 theleft beam and from (0,0,0) to (0,-50,0) for the contact between the two beams. A contactelement is also required to model an infinitesimally small gap between the 2kg mass and the

mm3

mm3


right beam. Create a small contact between (400,-50,0) and (400,-49.999,0).

• To correctly model the thickness of the beams (and masses), use rigid links between thebeams (or masses) and the contact elements.

• Create two freedom cases: one for an initial condition, and one for the falling load. Both casesare set to 2D Beam.

• Fix the ends of both cantilevers for both freedom cases.

• For the initial condition, completely fix the 200kg mass and allow the 2kg mass to translate inthe Y direction. The second load case is identical except for allowing both masses to translatein the Y direction.

• Assign both the two cantilevers a Structural Steel material property and a 150UB14.0 section(under BHP Universal Beams in the section library).

• Assign both masses a Structural Steel material property. Both masses are represented with asolid circular section. The 200kg mass has a 150 mm diameter and the 2 kg mass has a 50mm diameter. The density for each mass should be varied to obtain the correct mass. The200 kg mass has a density of 5.65884 x 10-8 T/mm3, while the 2 kg mass has a density of

1.01859 x 10-8 T/mm3.

• All the point contact elements are set as Zero Gapand have a stiffness of 50,000 N/mm.

• The structural damping of the two beams is simulatedusing viscous damping. Set the viscous dampingcoefficient as 5 x 10-8 Ns/mm/mm3.

Solver Setup

• Because contact elements are being used, the non-linear solvers are required.

• The Nonlinear Static solver is run to bring the system to an initial rest condition such that thetwo beams can sag. Ensure that the freedom condition used does not allow the larger mass totranslate.

• The Nonlinear Transient Dynamic solver is then usedto predict the response of the two masses using the non-linear static solution as the initial condition.


• Ensure that the freedom condition used for the transient solveallows for the masses to translate.

• Set up the time steps such that a time history of 5 s is recorded.Ensure that the time step used is sufficiently small to model the

impact of the falling load. Approximately 5 x 10-4 s is sufficientfor this model.

Results

• Create an animation and a graph of the vertical displacement of the two masses.


Discussion: Modelling Shock Problems in Straus7

Introduction

Depending on the exact nature of the test required, a shock analysis can be performed in Straus7using either the transient or spectral solvers. The transient solver can provide a full time history ofthe response of the structure to the shock. It is suited to the analysis of drop tests, tests where apulse load is applied to the structure and base acceleration problems. The spectral solver isideally suited to the majority of shock problems where a short pulse load or acceleration is appliedto the structure. The spectral solver will only yield the maximum response of the structure, itcannot calculate the time history of the response.

There are basically three types of shock test:

1. A drop test. The structure is dropped from a specified height.2. A base acceleration applied in accordance with a specified acceleration/time history (i.e. asinusoidal acceleration pulse).

3. A force is applied to the structure with a specified time history (i.e. a triangular pulse).

Tests 1 and 2 are the most common. The following is a discussion of the methods used to modeleach of these test types.

Use of the Transient Solver for Modelling Drop Tests

Often design specifications require a component to survive, without damage, a drop from a certainheight. This sort of problem is best handled using the Transient Dynamic solver.

If a structure is dropped from a given height we can readily calculate the velocity at the instantbefore impact using the equation

where:

s = drop height (m)

a = gravitational acceleration = 9.81 m/ .

Contact elements are applied to the model at the points where it will contact the ground whendropped. Impact problems require the use of a small time step due to the rapid rates of loading. Asa rough rule of thumb, use a time step equal to approximately 1/100th of the period of the mode ofthe structure that will be excited during the impact. If the time step is too large, the analysis will notcapture the full response of the structure. In particular the higher frequency components of theresponse will be missed or the solution may not converge at all.

)s/m(asV 2=

s2


Use of the Spectral Solver to Model Base Acceleration Prob-lems

Spectral analysis is a method that allows the calculation of the peak response, to a specifiedloading. There are four different spectra types that can be used with the Straus7 spectral solver.These are the seismic acceleration, seismic velocity, seismic displacement and force spectrummethods. The first three spectra are assumed to excite the structure by movement of the base (i.e.the points where freedoms are applied). The force spectrum applies a more general spectralloading at any point on the structure. All the different spectra are used in a similar manner. Theacceleration response spectra is used for most shock problems and will be considered here.

There are two factors which are combined to give the spectral acceleration applied to the base ofthe structure: these are the spectral value and the direction factors in the solver panel. For theacceleration response spectrum the spectral acceleration applied to the model is:

Spectral acceleration = (direction factor) x (spectral value)

The spectral value is a function of the frequency of the structure. This is defined by a spectralcurve in the Tables module.

The components of the direction vector are simply factors that multiply the applied loads. Theydefine the direction and in some cases the magnitude of the seismic acceleration and may beeither normalised (in which case the magnitude of the acceleration must be factored into thespectral table) or non-normalised (which normally means that the factor includes the magnitude ofthe acceleration).

In shock problems using the method outlined here, the magnitude of the acceleration must beincluded in either the spectral values (i.e. spectral table) or in the direction vector. Any globalacceleration applied to the structure has no effect in a base acceleration problem.

As mentioned, the spectral value is defined by a spectral curve entered as a Frequency/PeriodTable. This is also called the Dynamic Amplification Factor by many texts on dynamics. This factordefines a response ratio between the dynamic response and an equivalent static response when asingle degree of freedom is loaded with the peak acceleration. Basically the response of thestructure to a loading will depend on the ratio of the frequency of the applied load and the naturalfrequency of the structure. For a loading frequency much higher than the dominant frequency ofthe structure, the response will in general be less than the response from an equivalent staticacceleration.

In most impact analyses the structure is loaded with a very short pulse of high acceleration. Inthese cases the response of the structure will be considerably less than that which would result if asteady acceleration was applied equal to the peak acceleration during the impact.


The calculation of the spectral value or amplification factor is covered in texts such as 'Dynamicsof Structures' by Clough and Penzien. The following graph shows the spectral value as a functionof the frequency ratio for a number of common impulse loading waveforms.

Equations for these curves are given in Clough and Penzien. The curves are graphed with the Xaxis as the frequency ratio (b = ). This is the normal way of presenting this sort of

data as it is independent of the frequency of the load application and the frequency (or period) ofthe structure. Straus7 requires the spectral table to be defined with the X axis as frequency of thestructure ( ). Thus the X axis of the graph adjacent needs to be converted towstructure. Thisis readily done once the frequency of the load application is known.

For simple examples where one mode of the structure dominates the response a constant valueof spectral acceleration applicable to the dominant frequency can be used (i.e. a table with twopoints defining a constant spectral value).

For a more general problem with say 20 frequencies contributing to the response, a spectral curveis required that covers the entire frequency range and the spectral analysis should include all ofthese frequencies. A spectral value will then be calculated for each of the frequencies and used toexcite the corresponding mode. The responses for each of the modes are then combined to getthe total response using either the CQC or SRSS methods (see Clough or the Straus7 Online Helpfor details of these).

If the spectral curves shown in the above graph are not applicable to a particular problem you cancalculate the spectral curve using a simple single degree of freedom model in Straus7. Basicallythe procedure is as follows:

1. Establish the loading input i.e. magnitude, period and shape of load vs time curve.2. Establish the frequency range of interest i.e. check which modes of the structure will contributeto the response when the loading is applied.

3. For each of the frequencies of interest construct a simple single degree of freedom model suchthat f = √(k/m). Normally this model would be a mass on a spring (beam element). If the spec-tral curve for an acceleration response is being determined the loading will be applied as anacceleration.

4. For each of the models carry out a full time history transient solution with the loading defined bya load vs time table. The loading in all cases should be the input established in step 1. From thetransient response analysis determine the peak acceleration response. This is the spectralacceleration required by Straus7.

ωstructure ωload⁄

ωstructure

ωload


5. The spectral value used in the above method is the ratio between this calculated spectral accel-eration and the applied peak acceleration.

6. Plot a graph of spectral acceleration or spectral value vs the frequency of the structure and fit acurve through this. This is the spectral curve.

An alternative to steps 3 and 4 above is to solve the equation for a single degree of freedom, forthe load input, directly. The equation to be solved is:

This can be solved directly for some simple load functions f(t), or numerically for more complicatedfunctions.

Modelling Impact Loading Problems

When an impact loading is applied to a structure in some general manner such as the structurebeing hit at some point, both the transient and spectral solvers can be used. In both cases, a timehistory of the impact force must be assumed.

As discussed above, the transient solver can calculate the full time history of the response of thestructure to the impact. In this case, the maximum amplitude of the loading is applied to thestructure. The time history of this load is defined by the input of a load vs time table. This table islinked to the appropriate load case in the transient solver panel. The transient solver is run, onceagain using a very small time step to capture the response.

A faster method for this sort of analysis is to use the Spectral Solver with a load spectra. Theprocedure is very similar to that discussed above for base acceleration problems except that thesolver assumes that the loading is applied at some point other than the base. The spectra isderived as described above and in most cases the standard spectra defined in Penzien & Cloughcan be used. The amplitude of load would be entered as a point force. The spectral curve willdefine the spectral value as a function of frequency. Note that in this case the loading is applied tothe model as point forces and accelerations etc. It is not included in the spectral table or thedirection vector as is the case with the base acceleration problems.

x·· t( ) 2ζωx· t( ) ω2x t( )+ + f t( )=


Shock Qualification of an Instrumentation Frame

Outcomes



• Assign an Acceleration vs Time table to model a shock load.

Problem Description

A shock test will be performed on the instrumentation frame used for the Harmonic Responseexample. The frame will be subjected to a double half sinusoidal impulse with a peak shockloading of 20g. The shock will be applied as a base acceleration in the actual shock test.

Modelling Procedure

• Open the model of the instrumentation frame.

• Define a shock spectrumwhich corresponds to adouble half sinusoidalimpulse with a duration of0.03 seconds. For 0-0.01seconds, use: sin(2π*50x);for 0.01-0.03 seconds, use:0.5sin(2π*25x+π/2). ClickTables/Acceleration vsTime and use the equationeditor as shown:

The Acceleration vs Timegraph should appear as shown:


Solver Setup

• Solve using the Linear TransientDynamic solver. Use Rayleigh dampingand only calculate Node Velocity, NodeAcceleration and Beam Force/Stress tominimise solution time.

• Setup Rayleigh Damping to span a rangeof frequencies most likely to be excited by the shock - in this case,from 10-50 Hz. Set both damping ratios as 0.02.

• Set up the Base Acceleration as shown. Note thatthe direction vector multiplies the data in the‘Shock Spectrum’ Acceleration vs Time table suchthat a 20g maximum loading is applied in the Xdirection.

• Set up 400 steps for a total of 2 seconds, savingevery step.


Results

• Determine the maximum stress in the framedue to the shock loading and the maximum Xdisplacement. The graph of the Xdisplacement over the 2 second range isshown.

• Contour of beam fibre stress:

Straus7

Max Displacement (mm)

Max Tensile Fibre Stress in the Frame (MPa)


Drop test on an instrumentation frame

Outcomes


• Use the Nonlinear Transient Dynamic solver.

• Use contact elements to model contact between a model and another surface.

• Use Master/Slave links to enforce a contact condition.

• Model a dropped object.

Problem Description

The instrumentation frame is to be drop tested from a height of300 mm. The frame is rotated slightly such that it drops with someeccentricity. In addition, the frame is restricted from falling over bymodelling a shaft of square cross section that allows for movementof up to 100 mm in all directions (much like a frame falling through achimney with a 100 mm clearance all round). The response ofacceleration levels at the instrumentation locations is sought.

Modelling Procedure

• Open the frame model used in the previous example.

• Click Global/Coordinate Systems andcreate a cylindrical UCS using the points asshown.

• Click Tools/Move/By Increment and selectall nodes and rotate according to the newlycreated cylindrical system by 2 degrees.

• Click Tools/Extrude/Absolute and extrudethe bottom nodes to an absolute value ofY=-300 mm using Beam Property 3.

• Click Property/Beam and assign BeamProperty 3 as a Zero Gap point contact witha stiffness of 1x105 N/mm.

• ClickAttributes/Node/Restraint and deletethe node restraints at the bottom of theframe’s legs. Apply fully fixed restraints to


the bottom nodes at the ends of the point contact elements.

• Restrict X and Z translation at the end of one leg and only X translationat another as shown.

• At the frame’s top corners, extrude point contacts in both the X and Zdirections by 100 mm. This effectively models a ‘chimney’ throughwhich the frame falls.

• Repeat the above procedure, except this time,create master/slave links to enforce thecondition that the point contacts remain normalto the ‘surface’ of the ‘chimney’. All master/slave links should enforce the linked node tofollow Y (vertical) translation and either the X orZ translation (depending on which side the linkis on).

• Because two identical extrusions were applied,duplicate nodes were created. Click Tools/Clean/Mesh to remove these.

• Click Attributes/Node/Restraint and restrainthe newly created nodes at the ends of the linksand contacts. Fix all rotations and thetranslation normal to the ‘surface’ of the ‘chimney’.

• Click Global/Load and Freedom Cases and set a gravity load in the minus Y direction.

• Assign a viscous damping coefficient of 1.075 x 10-7 Ns/mm/mm3 to the frame’s beams.

Solver Setup

The Nonlinear Transient Dynamic solver is required because of the contact elements used in themodel.

• Only Node Velocity and Node Acceleration are to becalculated. Added Damping is set to Viscous. Set theinitial condition to include gravity in the minus Y direction.


• Set up the Time Steps as shown. Note that a coarse timestep is used for the first 0.2 s to record the falling frameprior to it impacting the surface.

Results

• Graph the acceleration and displacement ofthe instrumentation masses over the course ofthe drop.


Discussion: Modelling Rotating or PretensionedStructures

Introduction

Pretension loads or the normal operating loads of structures can have a significant effect on thenatural frequencies and general dynamic response.

In exactly the same way as the frequency or tone of a guitar or drum increases as the tension ofthe string or membrane is increased, the natural frequencies of a structure change as a function ofthe stress state within the structure. This is a particularly important consideration when carryingout natural frequency analysis on a structure that is operating in a state of pre-stress. Examples ofsuch structures include:

• Cable stayed structures with pre-tensioned cables. Cables may be mechanically pre-tensioned or pre-tensioned by the self weight of the structure.

• Fan blades, turbines and flywheels. Significant centrifugal stresses can develop in rotatingstructures. These have the effect of stiffening the structure and therefore raising the bendingfrequencies of the structure.

• A bridge structure subjected to temperature loading.

Inclusion of [Kg]

The effect of the stress state of a structure can be introduced into a natural frequency analysis bythe inclusion of the stress stiffening matrix [Kg]. The formulation of the natural frequency solutionincluding [Kg] is:

This equation follows the same form as the standard natural frequency problem and is solved inexactly the same manner. The only difference is that the stress stiffening matrix [Kg] is added tothe normal stiffness matrix [K] before the solution is carried out.

The stress stiffening matrix [Kg] can also be included in the transient dynamic solver. An option isprovided in the main solver panel for the inclusion of [Kg]. This option is only available when usingthe full system formulation. The transient dynamic solver includes [Kg] in the general equation ofmotion as follows:

The stress stiffening matrix [Kg] is added to the structural stiffness matrix and the problem issolved by the standard technique.

The harmonic and spectral solvers do not have specific options to allow [Kg] to be included in theanalysis. The effect of [Kg] can however be included. These solvers use a modal superpositionmethod to calculate the response. If [Kg] is included in the natural frequency solver whencalculating the natural frequencies, the effect of [Kg] is automatically included in the harmonic orspectral analysis because the mode shape and frequencies include the effect of [Kg].

[ ]( ) [ ]( ){ } { }0dMKK ig =−+ iλ][

[ ]{ } [ ]{ } [ ] [ ]( ){ } ( ){ }tFxKKxCxM g =+++ ��


The same is true for the Linear Transient Dynamic solver when the Mode Superposition methodis used for the solution. In this case, the Include [Kg] option in the Nonlinear Transient Dynamicsolver is not active.


Discussion: Spectral Response

Examples

• Earthquake analysis on buildings.

• Shock analysis of Components.

• Random vibration analysis of mechanical components.

Response Spectrum

General

The spectral analysis solver calculates the linear elastic structural response of a structuresubjected to a dynamic loading. Usually it is used instead of a transient dynamic analysis as aquicker alternative, especially when the dynamic loading is of a random nature (eg. earthquakeloading).

The dynamic loading can be either an earthquake excitation represented by its responsespectrum or a general dynamic load applied on the structure given by a load spectrum. In bothcases the excitation is defined with a spectral curve: Frequency (Hz) versus Spectral Value.

The earthquake spectrum is applied as a translational excitation of the base, equally at allsupports. The earthquake may act in all three global X-Y-Z directions simultaneously. Threetypes of earthquake spectral excitations are available: acceleration spectrum, velocity spectrumand displacement spectrum. The earthquake spectrum is used in the calculation of the maximumstructural response to a seismic action. It can represent only one particular earthquake or it canbe an averaged design spectrum, usually given in the design codes. Rocking seismic excitation ofthe base and multiple support excitation are currently not supported by the spectral solver.

The load spectrum simulates the situation of some random dynamic load applied on the structureat a number of nodes away from the supports. The applied load has the same frequency contentat each of the nodes but the magnitude may differ. Its typical application is in response analysis ofwind loads, ocean wave loads or machinery vibration.

The result of the spectral analysis is given as an envelope of maximum values for nodaldisplacements, element and nodal stresses, element and nodal strains and recovered reactions atconstrained nodes, including elastic forces at all other nodes. The maximum values arecalculated by combining the maximum response of all modes included in the analysis. Twomethods for the modal combination are available: SRSS (Square Root of the Sum of theSquares) and CQC (Complete Quadratic Combination). Maximum response values of nodaldeformations, stresses and reactions are automatically calculated. The maximum values areusually sufficient for most applications.

A natural frequency analysis must be performed prior to the spectral analysis since thefrequencies and corresponding eigenvectors of the structure are used by the spectral solver. It isimportant to consider the implications of using a non zero shift in the natural frequency analysis. Anon zero shift may result in the removal of some lower frequency modes from the frequencyanalysis result. If these modes are significant to the response of the structure for the givenloading, ignoring them may lead to erroneous results.


It is also important to remember that the spectral solver always uses the results of the lastfrequency analysis. When the structure is modified in any way a new frequency analysis must beexecuted before the spectral solver is used.

After successful execution of the analysis it is highly recommended that the log file (filename.SRL)be examined and special attention given to all warning messages. In particular the massparticipation should be checked to ensure that it is at least 90%. Before using the results it shouldbe realized that Spectral Analysis is an approximate method. The solution depends on thenumber of modes included in the analysis and the maximum response is approximated as acombination response of all included modes. The results should be considered as a most likelymaximum response of the structure to the dynamic action given by the input spectral curve.

Note also that in this type of analysis the sign of all result quantities is not significant. Each resultvalue should be considered positive as well as negative. For the graphical presentation of results,the sign of all maximum results is adopted from the mode with the greatest displacementamplitude. For the graphical presentation of results for individual modes, the sign of therespective mode is adopted.

It is important to note that the maximum values given as results will not occur simultaneously.These are only an envelope of the maxima which occurred during the dynamic action at differenttimes. Also, the maximum displacements do not correspond to the maximum stresses andmaximum reactions.

Power Spectral Density

The Power Spectral Density (PSD) option is available in the Spectral Response solver as analternative input excitation. Both excitations can be given as base acceleration, velocity anddisplacement or as a set of forces applied on the structure. The usage of the solver and display ofthe results remains unchanged.

Introduction

PSD analysis is commonly used to determine the response of a structure subjected to a statisticalrandom excitation. The random excitation can be a force, acceleration, velocity or displacement.Because the excitation is random, we cannot determine the structure's response at any particulartime. However, given a sufficiently large time interval, we can find the statistical (or probability)distribution of the response of the structure over this period.

When the PSD option is selected, the Spectral solver estimates a stochastic response of astructural model subjected to a stationary random dynamic excitation given in a form of a singlePSD curve. The results are given as one standard deviation (1 ) of the response. Here, the PSDfunction is a Fourier Transform of the Autocorrelation Function of the random process considered.The process is assumed to be stationary.

Implementation in Straus7

The PSD solver in Straus7 is almost identical to the Spectral Response solver except for thefollowing:

• The Spectral Table entered in the Tables Input Module is treated as a table of Power Spectral

Density vs. Frequency in the PSD solver ( /Hz or /Hz etc. versusFrequency), whereas it is a table of Spectral Response Value (ie. Dynamic Amplification Factorversus Frequency) for the Response Spectrum Solver.

σ

Acceleration2

Velocity2


• The results given by the PSD option are given as one standard deviation of the responsewhereas the Response Spectrum solver gives actual (maximum) results. In other words, theResponse Spectrum works with known (deterministic) data whilst the PSD works withstatistical data.

Standard deviation

Given a time varying function x(t), we can define the variation of the function from its mean, by

using a property known as the variance, . This is simply the integral over the time period T, ofthe square of the difference between the function value and the mean value, divided by T:

The positive root of the above, is known as the standard deviation. This assumes that theresponse is distributed according to a Normal (or Gaussian) distribution which is the familiar Bell-Shaped Curve. The standard deviation then is a measure of the spread of the function about themean value. Given a Straus7 PSD result, say a peak stress of 10 MPa, the result would have thefollowing meaning:

The structure will experience stresses of 10 MPa or less for 68.3% of the time under considera-tion. Alternatively, we could say that the stresses would be less than 20 MPa for 95.4% of thetime (i.e., 2 standard deviations). The percentages given, 68.3%, 95,.4% etc., simply representthe probability of a value being within the given standard deviations for a bell shaped distribu-tion.

Spectral Curve

The input spectral curve should be entered as a spectral table in the Table input. Only one singlesided positive defined PSD spectral curve can be applied. In the case of force spectrum the curvecan be applied on the model at a number of points with different intensities. In the case of seismicexcitation different magnitudes of the PSD curve may be applied in different directionssimultaneously. In the table for each frequency one spectral value should be entered. The unitsfor the frequencies are Hz. The units for the spectral values depend on the excitation type. If theexcitation is Seismic Acceleration, i.e. acceleration of the base, the units for the PSD spectral

values can be: /Hz or /Hz. If the spectral curve is defined as /Hz, the magnitude of g must

be included in the direction vectors in the spectral solver panel. If the acceleration is in /Hz,

the direction vector should be normalized. When the Acceleration spectrum is selected all otherloads on the model, including the global accelerations are ignored. The base acceleration isdefined by the PSD curve multiplied by the directions vector. When force excitation is used, the

units for the spectral values are: /Hz or /Hz. In the case of the load spectrum, the PSD curvefactors the loads applied to the structure.

The PSD curve

The PSD curve is a way of describing the frequency content of a random function in terms of thespectral density of the mean square value of the function. The Mean Square Value of a function isgiven by:

σ2

� −=T

dtxxT 0

22 )(1σ

σ

m

s2----� �� 2 g

2g2

m

s2----� �� 2

N2

kN2

�=T

dtxT

ValueMeanSquare0

21


It can be shown that the Mean Square Value is made up of discrete contributions in eachfrequency interval, Df. That is, the summation of the contribution from all the frequencies whichmake up the function. We call these contributions the power spectrum. The power spectrum isequivalent to the square of the RMS value. The power spectral density is then the power spectrumdivided by Df. The RMS value is simply the square root of the mean square value.

In practice, when measuring random vibrations, the PSD curve can be determined as follows:

• For a given frequency range, we measure the RMS value, , of the vibrations (eg. the

RMS acceleration, therefore g).

• The Power Spectrum is then the RMS value squared , thus ).

• The Power Spectrum Density is then ( )/ .

The units of the PSD curve are (value squared per Hertz). Common sets of units for the power

spectrum are /Hz for acceleration, /Hz for velocity, /Hz for displacement and N2/Hz fora load spectrum.

How is a PSD is generated?

To create an acceleration PSD curve, for a mechanical component the following approach wouldbe used:

• An acceleration PSD curve defines the acceleration of the base of the component - that is theacceleration of the base to which the component is attached. In this case an accelerometer ismounted to the base adjacent to the component for which the PSD curve is required. Theoutput signal from the accelerometer is analyzed using a data logger/FFT analyser.

• The accelerometer will output the random acceleration signal as function of time.

• The random output contains many frequency components.

• The user must decide on a frequency range of interest, say 0 - 1000 Hz. This is set-up in theFFT analyser and only data within this range is collected.

• The frequency analyser is used to do a Fourier Transform on the acceleration vs time data.This produces a graph of acceleration vs frequency or a frequency spectrum of theacceleration.

• This frequency range is divided into a number of narrow frequency bands of bandwidth f. Say100 bands with a bandwidth of 10 Hz. The number of bands can be set on the FFT analyser.

For each of the frequency bands the following process is performed by most modern FFTanalysers.

• Calculate the RMS value of the acceleration within the frequency band.

• Square this value. (We actually need the Mean Square Value).

• Calculate the PSD value by dividing the Mean square value by the bandwidth f.

• This process is illustrated in the following table.

∆f xRMS

xRMS( )2 g2

xRMS( )2 ∆f

g2

ms1–( )

2m2

∆

∆

Frequency Range RMS Value (RMS)2 PSD Value0-10 5 25 2.510-20 4 16 1.620-30 2 4 0.430-40 2 4 0.4etc.


The PSD curve can now be produced by plotting the PSD values vs frequency. The frequencyused for each point is the average frequency for each frequency band. This is shown in thefollowing table:

If the input is a stationary function, which means that the mean value of the function will be thesame (for a reasonably large sample), irrespective of whether the sample goes from to or

some other range to , a single sample is sufficient to produce the PSD curve. If the input is

non-stationary then most FFT analyzers will have provision for taking many samples. IndividualPSD curves are then calculated for each of these samples. The individual curves are thenaveraged to produce the PSD curve for the component.

In the majority of design and analysis situations the PSD curve is defined by specifications anddesign codes such as MIL Handbooks etc.

Freq. PSD (g2/Hz)5 2.515 1.625 0.435 0.4etc.

t1 t2

t3 t4


Discussion: Earthquake Analysis using Straus7

Introduction

The purpose of this lesson is to illustrate and compare the most common methods allowed bydesign codes around the world, for performing seismic analysis of structural systems. The mostcommon methods are:

• Equivalent Static Analysis.

• Spectral Analysis.

• Transient Dynamic Analysis.

The equivalent static analysis is a straightforward method to evaluate horizontal force distributionsin structures by means of simple hand calculations. The method has been common practice in thepast and still today it is a good quick tool for sizing members prior to design. Spectral and transientdynamic analysis are, these days, much more approachable methods that can save aconsiderable amount of time in the verification phase. Throughout this discussion we implicitlyrefer to clauses of Eurocode 8 ENV 1998 as well as Australian standard AS1170.4 – 1993.

The Different Measures of Earthquake

Before describing in depth the above mentioned methods, it is useful to recall the most popularmeasures of earthquake events used in practice. It is common, for example, to speak of anearthquake in terms of “Richter scale” or “Mercalli scale”. The Richter scale is the base 10logarithm of the maximum measured displacement in micrometers, captured by a Wood-Andersonseismograph and corrected in a 100 km radius. Earthquakes whose Richter index is bigger thanfive can produce damages to structures. Another approach (modified Mercalli scale) classifies theearthquake by the effect it has on the community from I (not felt by anyone) to XII (totaldestruction).

These measures, however, are not good enough to quantify seismic events in a rigorous fashion.Instead, a different set of methods based on the behaviour of a “single degree of freedom” systemis used.

Single Degree of Freedom Systems: from Time History toResponse Spectrum.

Direct measures of earthquakes in terms of acceleration time history are often available as aresult of direct records. We want to convert this information to a form suitable for the evaluation ofmaximum actions on structures. To do so, first we decouple the dynamic behaviour of a givenstructure in single oscillators, each fully described by its frequency. The time history is thenapplied to the base of each of these simple systems and the maximum response (in terms ofacceleration, velocity or displacements) stored. The following ordinary differential equationdescribes the behaviour of a single degree of freedom system subject to a base acceleration:

)()()(2)( 2 tatxtxtx b−=⋅+⋅⋅⋅+ ωωξ ��


where ξ is the damping ratio ( < 0.1) , ω the natural frequency and a(t) the acceleration timehistory. It is easy to solve this equation analytically (convolution integral) or numerically by using afinite difference scheme and carefully selecting the time step, considering issues like numericalstability and data sampling rate.

The maximum value obtained will determine a point in the response spectrum graphcorresponding to the frequency considered, and the amount of damping used. Other factors willhave an effect on the response spectrum shape like the class of the foundation soil. Themodification of the elastic spectrum to take into account these factors of practical importance,leads to the definition of “design spectrum”.

Equivalent Static Analysis

Applicability

This analysis can be applied to buildings with the following attributes:

• Structural regularity (plan and elevation).

• First natural period smaller than 2.0 seconds.

The first item assures that the fundamental mode shapes are described in terms of flexure of asimple cantilever and that they excite nearly all the mass of the structure. Torsional naturalfrequencies or modes with a coupled torsional-flexural behaviour are not considered in this type ofanalysis because they generate a more complex force distribution throughout the floors.

The second item is responsible for a nearly linear distribution of forces and horizontaldisplacements in each floor. Note that AS1170.4 allows for the inclusion of structures with higherfundamental frequencies, considering instead a quadratic distribution.

Calculation of the total base shear force

The total base shear force is calculated by multiplying the total dead load (adding a fraction of thelive load as well) for the value of the design spectrum corresponding to the natural frequency ofthe building.

where

• Fb = Base Shear Force

• Sb = Spectral Value at the period in question

• W = Total Weight (or mass depending on units)

Such a frequency has to be obtained by using rigorous analyses or by applying a simplifiedapproach described within the code.

Floor force calculation

WSF db ⋅=


The total shear force is thence distributed to each floor by using the following coefficient:

where z is the height of the i-th floor and Wi the gravitational force associated with it. It is clearthat, if the gravitational force generated by each floor is the same then the above distribution islinear. In AS1170.4 a quadratic distribution is also possible for structures whose fundamentalperiod is bigger than 2.0 seconds, by using the following coefficients:

Torsional effect

The forces generated in the structural members as a result of torsional vibration modes are notdirectly considered in the Eurocodes whereas the Australian standard allows for the evaluation ofa torque for each floor (only for buildings of small importance) by multiplying the horizontalcomponent by the eccentricity between the shear centre and the physical centre of the building.Factors for dynamic amplification and construction imperfections are also used.

Note that this analysis is linear elastic, eventually with combinations that consider the differentdirections of the earthquake. The final design result case is an equilibrated solution of a set oflinear equations.

Spectral Analysis

Introduction

This analysis is a more general approach and allows a much larger number of structuralconfigurations to be dealt with. However, a few assumptions are still needed:

• Linear material and geometry.

• Small structural damping.

Spectral results are based on a natural frequency analysis. These frequencies are linear andundamped.

( )� ⋅⋅=

j jj

iii Wz

WzR

( )� ⋅⋅=

j jj

iii Wz

WzR 2

2


Numerical Issues

A better understanding about how the spectral analysis works comes from the details of itsnumerical implementation. Once the natural frequencies and mode shapes have been calculated,it is possible to decouple the equations of motion into single oscillators.

where y is the modal coordinate vector, and K’ the diagonal normalized stiffness matrix. At thisstage the damping is added to each of the equations and the external force taken from the designspectrum corresponding to the equation’s frequency. If the design spectrum is defined in terms ofacceleration, the right hand side will have the following expression:

where φi is the i-th eigenvalue, M the mass matrix, r seismic direction vector and Si the ordinate inthe design spectrum. If the spectrum has been normalized the peak value is included in thedirection vector r. If the spectrum is defined in terms of an external load and not a groundacceleration then the following right hand side applies:

where R is the load vector for a particular load case.

The decoupled equations are solved and the solution transformed back to physical coordinates:

Number of natural frequencies to be included

It is easy to understand that the accuracy of a spectral analysis is dependent on the number offrequencies included. Design codes consider the number of included modes to be adequate for adynamic analysis if the summation of the modes’ mass participation factor is greater than 90% ofthe total mass.

An important exception that has to be considered arises when analysing the effects of a verticalearthquake. In general vertical structures will be excited by an axial mode at a relatively highfrequency. An important percent of the total mass will also be associated with these members.Hence many more frequencies need to be included to excite a significant amount of mass.However the final results in terms of displacements and forces do not differ much if a smallernumber of frequencies has been included. This means that particular cases exist such that thefinal response will be accurate even with a total mass participation factor less than 90%. The nextexample illustrates a typical case.

LyKy =⋅+ '��

iTii SrML ⋅⋅⋅=φ

RL Tii ⋅=φ

iii yx ⋅=φ


Example

The structure in the figure is a 2D portal frame of 6 m height and 5 m width, subject to a verticalearthquake. The following natural frequencies were calculated with the spectral mass participationfactors as shown:

EXCITATION FACTORSMode Excitation Amplitude Participation (%)1 1.638919E-15 2.448030E-17 0.0002 1.731137E+01 1.495795E-02 8.5403 7.394676E-14 2.817129E-17 0.0004 2.212920E+01 4.995576E-03 13.9555 2.944856E-11 1.244316E-15 0.0006 4.858872E+00 1.762342E-04 0.6737 3.813599E-11 7.171120E-16 0.0008 5.671975E-01 5.109340E-06 0.0099 1.183167E-10 9.264109E-16 0.00010 9.762941E+00 4.490351E-05 2.71611 2.925012E-10 1.107574E-15 0.00012 6.661620E+00 2.416929E-05 1.26513 1.021053E-11 1.917199E-17 0.00014 4.736429E+01 3.397340E-05 63.92815 3.738907E-12 2.675485E-18 0.00016 8.956269E-02 1.439609E-08 0.00017 1.506789E+01 1.232426E-06 6.47018 8.143820E-08 6.660660E-15 0.00019 1.597571E-06 6.696231E-14 0.00020 7.913091E+00 2.601720E-07 1.784TOTAL MASS PARTICIPATION: 99.339%


The 14th mode is associated with the axial vibration of the columns and it excites 64% of themass. Two different solutions are considered: the first includes modes 2 and 4 with a resultantmass participation factor of 22.5%, and the second includes modes 2, 4 and 14 with 86.4% massparticipation. The following figure shows that the final displacements of the two solutions are verysimilar. The response in this case is adequately represented by the flexural modes (modes 2 and4). Mode 14 is an axial mode of the columns, which does not significantly contribute to the result,despite having a large mass participation factor. This indicates that the mass participation alonemay be a too restrictive condition when deciding on whether the spectral response solution isvalid.

If it becomes necessary to increase the mass participation factor in situations such as this, weneed a way of determining only the relevant mode shapes in the natural frequency solution; that is,focus on the vertical modes (in this case), and extract as many vertical modes as possible, withoutsignificantly increasing the set of requested modes.


In Straus7, you can independently ignore mass in any of the three global axis directions (X, Y, Z),by setting the option in the Defaults/Dynamics tab of the solver panel. In this example, you coulddisable X and Z mass, thereby only determining modes that have mass participation in Y.

Superposition of modal results

Once the results for the single modes are found, the problem is how to combine them to obtain themaximum force or displacement within the elements. Note that this is not in general a simpleproblem since the peak amplitude of different modes doesn’t occur at the same point in time. Thusthe simple summation of modal results will in general highly overestimate the physical forcedistribution.

The first proposed method is the SRSS (or Square Root of the Sum of the Squares) combinationin which the final results are obtained as the euclidean norm of the single modal results:

Note that the outcome consists of just positive result quantities. Furthermore this combinationmethod is considered to be valid only if the frequencies are sufficiently spaced, e.g. if fi and fj aretwo natural frequencies with fi < fj then the requirement is that fi < 0.9fj. That is, this method doesnot consider the interaction between close frequencies.

Another method, the CQC (Complete Quadratic Combination) method is widely used in practice.This method combines each frequency with all the others using coefficients that are dependent onthe damping and the natural frequencies, thereby considering the effect of closely spacedfrequencies.

This combination is a generalized version of the SRSS method and the two methods coincide ifthe frequencies are sufficiently spaced.

Note that the final solution, in general, is not equilibrated as it is the result of a combination ofequilibrated modes.

223

22

21 ... nxxxxx +++=

2/1

��

��

�⋅⋅⋅= ��

i jjiijij xxx ρα


Transient Dynamic Analysis

Applicability

This analysis is the most realistic as it considers the structure subject to a time accelerationhistory. The previous hypothesis regarding linear behaviour and small damping are no longerrequired. Material non linearities (e.g. plastic hinges in beams) as well as geometric non linearities(e.g. structures with cables) can be analyzed using this approach. The only problem relates to thesolution time, due to the solution (implicit formulation) of a linear set of equations at any timestep.If the problem is non-linear the time taken is further increased by the iteration process. On theother hand an explicit formulation will decrease the time taken to solve the single time step, but thestability of the method is only assured for a very small time increment.

Example

In the following example the different methods discussed are applied to a simple building structureand the final results are compared. The following picture illustrates the geometry of the simplebuilding:

Evaluation of the Response Spectrum

First of all it is necessary to calculate the response spectrum to be used in the equivalent staticand spectral response. For this example, we have evaluated it directly by starting from theacceleration time history of the “El centro” earthquake. For simplicity, no correction factors havebeen applied to it. A finite difference equation is used to solve the single degree of freedom systemwith a damping ratio of 5% (that is, the equation presented on the first page of this lesson issolved).

The following picture illustrates the comparison of the obtained spectrum with the one given byEurocode 8 for a class A soil. Typical code spectra are simplified (smoothed) with scaling factorsapplied.


Equivalent Static Analysis

The effect of flexural and torsional vibrations is obtained using static forces and moments.Although the geometry of the structure is not regular (lift core not in a central position) the analysiscan be equally carried out in accordance with the clauses in the Australian standards. Accordingto Eurocode 8 a static analysis cannot be used in this case.

The natural periods are:

• First period (Torsional): 1.55585 s

• Second period (Flexural minor axis): 1.26129 s

Different mesh sizes were considered to make sure that the frequencies represent convergedresults.

The total amount of dead load is 705 T. The acceleration of the calculated spectrumcorresponding to the first period (torsional) is approximately 0.1g and for the second period (firstflexural mode) it is around 0.3g. These correspond to approximately 0.6g and 0.8g respectivelyfor the Eurocode spectrum.


The base shear is then distributed along the height using the following relationship:

Using the other fundamental frequency (generating torsion) the forces in each floor are:

The eccentricity between the shear centre and the physical centre of the generic floor is requiredto evaluate torsional effects.

The shear centre can be obtained through the following relationship:

where Ix,i is the moment of inertia of the single structural element referred to its own centroid (witha distance xi from the origin of the reference coordinate system).

The eccentricity is hence 4.66 m.

Once the static forces have been applied to the model, the solution can be run and the resultsviewed.

Spectral Analysis

The following solver parameters have to be set:

Total Base Shear [kN]

Calculated Spectrum Eurocode 8First Period 692 4152Second Period 2074 5536

hh

VF

i i

basefloor ⋅=

�1

1

hh

VF

i i

basefloor ⋅=

�2

2

�� ⋅

=i ix

i iix

I

xIx

,

,


• Load Type - This option allows the selection of different types of external causes (e.g. appliedforce, base acceleration, etc.) which will be subject to the design spectrum.

• Damping - Damping can be modal or Rayleigh. With the former a certain damping ratio canbe entered directly in the decoupled equations of motion. The latter adds a damping to thestructure that is proportional to a combination of its stiffness and mass. This approach assignsa damping to each natural frequency following a non-linear curve defined through twodamping ratio–frequency couples.

• Direction Vector - The direction of action of the earthquake is entered as a vector. If thedesign spectrum is normalized then the peak value has to be specified here.

• Frequency File - The frequencies included in the analysis, together with their damping ratiosand the relative spectra are chosen in this dialog box.

Transient Dynamic Analysis

The acceleration time history must be entered in the solver panel:

• Base Load - Once the time history has been specified, it is possible to choose between eitherabsolute or relative output for displacement, velocity and acceleration results. Forces andstresses are not affected by this choice.

Results Comparisons

The maximum bending moment in the two principal directions from the transient dynamic analysishas been chosen as the comparison parameter. Two different spectral and equivalent staticanalyses were performed using the calculated spectrum and the EC8 spectrum.


The design spectrum given in EC8 has not been factored by any of the importance, site, dampingor statistical factors.

This example illustrates that using a spectral analysis (code design spectra) conservative resultsare obtained in a fraction of the time taken for the full transient dynamic analysis and saving handcalculations needed to calculate statically equivalent forces.


A Simple Example of Seismic Analysis

Outcomes

Upon successful completion of this lesson you will be able to:

• Have a clear idea of what to expect when comparing different approaches for seismic analysis.

Introduction

The example illustrates how to perform three different types of seismic analysis on a very simplebuilding model.

Problem

The first part of this exercise will focus on the selection of a proper response spectrum and of anearthquake time history, such that the results given by each method are directly comparable. Todo this we have to choose a set of multipliers for the unscaled code response spectrum. They are:

• Site Factor: it takes into account the quality of the foundation soil.

• Acceleration Coefficient: it quantifies the peak of acceleration for the relevant earthquake(selected on the basis of a given return period).

• Importance Factor: related to the type of construction and its use.

• Structural response factor: it considers that the real system will exhibit a certain ductility duringthe earthquake.

The choice of a proper acceleration factor that allows comparisons between transient dynamicand spectral results, means that we have to choose also a reference acceleration time history. Thefollowing picture illustrates the Athens September 99 seismic event chosen for this example.

The acceleration peak for this accelerogram is 287.394 cm/s2. Using a discrete range of periodsfor a single degree of freedom system, the response spectrum has been obtained from the time


history and compared with the one given in the Australian code multiplied by the Importance factorand the structural response factor. The result is illustrated in figure.

Once the reference response spectrum has been determined we can start to perform theequivalent static analysis.

Modelling Procedure

The model used is a simple concrete four storey building. Its dimension are 12 m x 6 m with a floorheight of 3 m. The following quantities need to be calculated:

Natural Frequencies

The natural frequency solver is used to evaluate the natural frequencies of the building. Thefrequencies obtained are:

• First natural frequency, flexural oscillation about the building’s minor axis of inertia: 0.518269Hz.

• Second natural frequency, flexural oscillation about the building’s major axis of inertia:0.565712 Hz.

Note that the first two natural frequencies are free of torsional effects. Hence all the threeapproaches are feasible to be applied to the considered geometry.

Total Base Shear

In the Australian standard AS1170.4 - 1993 part 4 the total base shear is calculated using thefollowing formula:

where I is the importance factor, S the site factor, Rf the structural response factor and C theearthquake design coefficient given by:

V IC S⋅Rf

-----------� �� Gg⋅ ⋅=

C1.25 a⋅

T2 3⁄------------------=


with a the acceleration coefficient. The latter value characterizes the peak acceleration of theseismic event for a given return period. In our example it will be equal to the maximum value of thetime history for the Athens earthquake. The total weight of the building is 148.61 Tonnes. Thefollowing table gives coefficients used and the result obtained for each fundamental period.

The values of the shear forces in the two principal directions of inertia of the building thus result:

• 280 kN for the direction associated with the first period.

• 296,7 kN for the direction associated with the second period.

The forces on the single floor is then 3/30, 6/30, 9/30, 12/30 of the total shear just calculated.

Comparison of results

The following graph illustrates the results following different approaches.

Table 1: Coefficients used for the base shear calculation

Coefficient Value

I 1.25

C first period 0.2355

C second period 0.2496

Rf 1.5

a 0.292

S 1.0


PSD Spectral Response

Outcomes


• Use the PSD option in the Spectral Response solver.

• Apply a load spectrum via a PSD curve to a model.

Problem Description

A single degree of freedom system with mass m, natural frequency f and damping is subjectedto a random dynamic force. The random force F is given with a power spectral density.

Geometry

Mass (m) = 60 kg

Natural frequency (f) = 4 Hz

Relative damping ratio ( ) = 0.05

Length of spring (l) = 1 m

Spring stiffness (K) = m(2 f)2

= 37899.3 N/m

F = 1.0

In this analysis, the curve of PSD value is a LoadSpectrum. This will factor the loads applied to themodel. The magnitude of the force is built into thePSD curve and thus a unit load is applied to themodel.

Default Freedom Case: All fixed except DY.

Solver Setup

Run the Natural Frequency solver for 1 mode. Use the Lumped Mass option. The targetfrequency is 4 Hz.

ξ

ξ

π


Spectral Response

• Create a new Factor versus Frequency table and define the PSD curve. Select the SpectralResponse solver and setup as shown:

• Select Frequency File...:

• Select Load Cases:

Results

Displacement = 0.0661 m

Note that this is the one standard deviation response. This means that the displacement will be nomore than 0.066/m for 68.3% of the time.


PSD - Base Excitation

Outcomes


• Apply a PSD curve to specify a base excitation.

Problem Description

A power spectral density analysis is to be carried out on the frame from the Spectral Example. Theframe is to be mounted in an off-road vehicle. The purpose of this analysis is to investigate thestresses when the frame is subjected to a random base excitation that occurs as a result of thevehicle driving over rough ground. The power spectrum of the attachment points in the vehicle hasbeen measured using an accelerometer and a spectrum analyser. The spectrum is based on theaverage of a large number of measurements during the most extreme operations that will occur inpractice.

A graph of the spectrum is shownin the following figure:

This spectrum is assumed to act inthe horizontal global X direction

• Enter the spectrum as SpectralTable 2 using the Tablesoption from the main Straus7menu. (Table 1 already existsfrom Example 1).

• Run the Natural Frequencysolver using the followingparameters:

Solve for 30 ModesLumped Mass

The first 10 calculated frequencies are

Mode Eigenvalue Frequency Frequency(rad/s) (Hertz)

1 3.85384384E+03 6.20793350E+01 9.88023303E+002 3.88046239E+03 6.22933575E+01 9.91429577E+003 1.30269834E+04 1.14135811E+02 1.81652785E+014 2.81202089E+04 1.67690813E+02 2.66888219E+015 7.13699066E+04 2.67151468E+02 4.25184766E+016 7.44108044E+04 2.72783439E+02 4.34148327E+017 7.66895957E+04 2.76928864E+02 4.40745976E+018 7.81961853E+04 2.79635808E+02 4.45054212E+019 2.51547652E+05 5.01545264E+02 7.98234080E+0110 2.51566387E+05 5.01563941E+02 7.98263805E+01


• Run the Spectral Response solver asshown :

• Load table is as shown

Results

The results calculated by the spectral solver are for one standard deviation ( ) - i.e. will not beexceeded 68.3% of the time. Find the maximum stress in the beam elements. Scale this by 2x for2 and 3x for 3 . This will give an indication of the maximum stress that is unlikely to beexceeded 95.4% and 99.7% of the time respectively.

The maximum stress occurs at the upper end of one of the lower legs.

σ

σ σ

R e su lt M a x F ib reS tre ss(M p a )

σσσσ 5 9 .4 92 σσσσ 1 1 8 .9 83 σσσσ 1 7 8 .4 7


References

1. Clough RW and Penzien J. "Dynamics of Structures". McGraw-Hill, 1975

2. Thomson W.T. "Theory of Vibration with Applications" Fourth Edition Chapman and Hall 1993

3. Bruel & Kjaer "Mechanical Vibration and Shock Measurements" 1980

4. NAFEMS "A Finite Element Dynamics Primer" 1992

straus r7-software dynamics analysis

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