beam

SIMPACK

Beam

SIMPACK Release 8.6

24th September 2003/SIMDOC v8.607

COPYRIGHT 2003 c©

BEAM:0.0 -2

BEAMA PRE-PROCESSOR

FOR MODE SHAPE ANALYSISOF STRAIGHT BEAM

STRUCTURESAND GENERATION OF THE SID

FILEFOR SIMPACK

AND OTHER MBS CODESUser Manual

written by

Dr. Oskar Wallrapp

D-82234 Wessling, FRG

Version 3.3 - August 1996

BEAM:0.0 -2

Beam is not part of a basic distribution of SIM-Hint:PACK.

Contents

1 New in Version 3.3 1.0 -5

2 What is Beam? 2.1 -7

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 -7

2.2 Theoretical Background . . . . . . . . . . . . . . . . . . 2.2 -8

Longitudinal Vibration . . . . . . . . . . . . . . . . . . . 2.2 -9

Torsional Vibrations . . . . . . . . . . . . . . . . . . . . 2.2 -10

Bending Vibrations . . . . . . . . . . . . . . . . . . . . . 2.2 -11

2.3 Limitations of Modelling . . . . . . . . . . . . . . . . . . 2.0 -13

2.4 System Requirements . . . . . . . . . . . . . . . . . . . . 2.0 -13

3 Starting of BEAM 3.2 -15

3.1 General Information . . . . . . . . . . . . . . . . . . . . 3.2 -15

3.2 Example of a Simple Cantilever Beam . . . . . . . . . . . 3.2 -16

Model Description . . . . . . . . . . . . . . . . . . . . . 3.2 -16

Input Data and Run of BEAM . . . . . . . . . . . . . . 3.2 -16

Input file of BEAM . . . . . . . . . . . . . . . . . . . . . 3.2 -23

SID File of the Simple Beam . . . . . . . . . . . . . . . . 3.2 -25

3.3 Example of a Beam with Sections Using Different Geometry3.3 -31

3.4 Example of a Beam with Concentrated Mass and Spring 3.4 -35

3.5 Beam Structure with Measured Torsional Stiffness Data 3.5 -40

4 References 4.0 -49

BEAM:0.0 -4 CONTENTS

BEAM:1. New in Version 3.3

The version 3.3 of BEAM contains significant extensions for the usageand modelling capabilities with respect to version 3.0. The major topicsare as follows:

• The stiffness of tension, bending and torsion motions may bedefined direct by input data, e. g. given by measured values,see section BEAM:3.5.

• The markers for the MBS code may be attached outside of thebeam’s centerline but within the sections of the beam structure.This means that a markers is now defined by data of the x, yand z position, see section BEAM:3.5. For this markers, BEAMevaluates the mode shape values within the assumption of ”rigidarms” perpendicular to the beam’s centerline.

• In addition to the damping ratio proportional to the modal stiff-ness values, the important natural damping (Lehr ratio) is imple-mented . From these values BEAM computes the modal dampingmatrix De. The input is illustrated in section BEAM:3.5.

• In the past, beam structures with one homogenous section andthe boundary conditions free-free results in an error due tothe solution of the eigenvalue equation. In version 3.3 the exactsolution is implemented.

• The SID file is optimized . No zero elements are writtenon the SID file now. A key allows to generate the SID file forSIMPACK v5.xx versions, too.

Note that the new SID file requires the new readHint:routine for SID in the MBS code.

BEAM:1.0 -6 BEAM:1. NEW IN VERSION 3.3

BEAM:2. What is Beam?

BEAM:2.1 Introduction

BEAM is a computer program for the evaluation of the mode shapes ofstraight flexible beam structures as shown in Fig. BEAM:2.1.1. Fromthis mode shapes, BEAM generates the SID (Standard Input Data) filecontaining input data for MBS (MultiBody System) codes. BEAM isa so-called pre-processor for SIMPACK and other MBS codes. BEAM

x

y

z

Figure BEAM:2.1.1: A straight beam structure The structure hastwo sections, boundaries at the left end, concentrated springs at leftand right, as well as a concentrated mass between section 1 and 2.

solves the eigenvalue problem of beam structures with various sectionsof mass and geometrical properties, fixed and flexible boundary condi-tions at nodes, and concentrated nodal masses. BEAM uses an easyinput description like FEM codes.

A beam is modelled by the Euler-Bernoulli hypothesis, for which themode shapes of

• longitudinal vibrations (in x-direction),• bending vibrations in two directions (in y and z -direction), and• torsion vibrations (about x-axes)

can be computed exactly from the continuum equations, see sectionBEAM:2.2. Using the mode shapes of a beam, the submatrices of thesystem equations of a flexible body as a part of the MBS are evaluatedbased on a modal approximation. The submatrices represent the inputdata of a flexible body of a MBS, which are written on the StandardInput Data (SID) file – especially – for a Taylor expansion up to thefirst order. MBS codes with flexible members, which are based ona formulation of the deformations assuming small displacements and amodal approximation, they can take those data as input to generate thesystem equations. The SID file contains all submatrices for a completelinearization of the equations of deformations, see (Wallrapp 1993).

BEAM allows the user to select a various number of modes shapesand markers for the MBS data generation. The computation of thegeometric stiffening terms can be activated or not.

BEAM:2.2 -8 Theoretical Background

BEAM can be used either in an interactive modus, where the beamdata comes from the standard input or where the beam input data aretaken from a file. The output may be test prints for detailed discussionsof the results, the BEAM input data stored on a ASCII file for futurecomputations, and the SID on a formatted ASCII file for the transferto MBS codes running on various devises. Fig. BEAM:2.1.2 shows thedata flow of BEAM. BEAM solves the eigenvalue problem of beam

Input Data..... BEAM

Input Data

BEAMInput Data S I D

Test Prints.....

O R

INPUT

OUTPUT

B E A M

Figure BEAM:2.1.2: Data flow of BEAM

structures with various sections of mass and

BEAM:2.2 Theoretical Background

The following section gives an introduction into the vibration analysisof homogenous beams represented by one section and simple boundaryconditions, e. g. the free - free situation. Therefore, an analyticalsolution of the equations of motions is possible which one finds e. g. in(Meirovitch, 1967). For problems with different sections and boundaryconditions as shown in Fig. BEAM:2.1.1, the differential equations aremore difficult which one have to solve numerically using the Kolousekmethod. Both methods are implemented in BEAM.

Theoretical Background BEAM:2.2 -9

Longitudinal Vibration

Let us consider a thin homogenous rod with a longitudinal flexibilitymeasured by u(x, t) with the unit [m]. The mass density per unit lengthis ? [kg/m], the longitudinal stiffness is EA with the unit [N ], whereE is the Young’s modulus and A is the cross section of area. Thetime derivative may be denoted by a dot and the space derivative withrespect to x by ’, the equation of longitudinal motion is represented bythe differential equation (BEAM:2.1)

EAu′′(x, t) = µu(x, t), (BEAM:2.1)

which must be satisfied over the domain 0 ≤ x ≤ L.

At the borders x = 0 and x = L boundary conditions are given, e.g.for the free-free situation

EAu,(x = 0, t) = 0(BEAM:2.2)

EAu,(x = l, t) = 0(BEAM:2.3)

(BEAM:2.4)

Applying the separation of variables method (Meirovitch, 1967) onewrites the displacements u(x, t) in the form

u(x, t) = U(x)q(t) (BEAM:2.5)

where q(t) is a harmonic function with the frequency ω [rad/sec] andU(x) is the unknown function of the mode shapes. Substituting (4) into(1) to (3) and denoting u(x, t) = −ω2U(x)q(t) yields the differentialequation of the eigenvalue problem

−EAU ,,(x) = µω2U(x) ⇒ U ,,(x) + λ2U(x) = 0 (BEAM:2.6)

with the eigenvalues λ, which are correlated to the frequencies ω as

ω = λ

√EA

µ(BEAM:2.7)

and the boundary conditions

EAU ,(0) = 0(BEAM:2.8)

EAU ,(l) = 0(BEAM:2.9)

(BEAM:2.10)

The general solution of (BEAM:2.6) is

U(x) = C1cosλx + C2sinλx (BEAM:2.11)


where C1 and C2 are amplitudes and which have to specialized by theboundary conditions (BEAM:2.8) and (BEAM:2.9). From (BEAM:2.8)one gets C2 = 0 and from (BEAM:2.9) the frequency equation

sinλL = 0 (BEAM:2.12)

which is satisfied for eigenvalues

λi = iπ

L, i = 0, 1, 2, . . . (BEAM:2.13)

The first value λ0 = 0 represents the longitudinal rigid body motion.The other values describes the orthogonal, natural mode shapes forlongitudinal deformations as

Ui(x) = cosλix, i = 1, 2, . . . (BEAM:2.14)

where C1 is the amplitudes, which has the same value for all modesand which may be defined by the normalization of the modal massconditions

Mqij=

∫ L

o

µUiUjdx =1 for i = j0 for i = j

i = 1, 2, . . . (BEAM:2.15)

For (BEAM:2.14) one finds with (BEAM:2.15)

C1 = sqrt2

µLi = 1, 2, . . . (BEAM:2.16)

The first three mode shapes of (BEAM:2.14) are plotted in Fig.BEAM:2.2.3.

0 0.2 0.4 0.6 0.8 1-1

-0.5

0

0.5

1

Mode 2

Mode 3

Mode 1

Figure BEAM:2.2.3: Three longitudinal mode shapes of a rod withfree-free borders

Torsional Vibrations

Let us focus at torsional vibrations about the x-axis of a homogenousbar with the torsional stiffness GJT [Nm2/rad] and the mass moment

Theoretical Background BEAM:2.2 -11

of inertia per unit length ix [kgm/rad]. If θ(x, t) is the torsional angleof the bar’s center line, the differential equation of motion is given as

GJT θ,,(x, t) = ixθ(x, t) (BEAM:2.17)

which must be satisfied over the domain 0 ≤ x ≤ L. At the bordersx = 0 and x = L boundary conditions are given, e.g. for the free-freesituation

GJT θ,,(x = 0, t) = 0 (BEAM:2.18)

and

GJT θ,,(x = L, t) = 0 (BEAM:2.19)

The solution of (BEAM:2.17) to (BEAM:2.19) is in analogy to thelongitudinal vibration: one has to substitute u(x, t) by θ(x, t), EA byGJT , and µ by ix . The results are the natural frequencies

ωi = λiGJT

ix, i = 0, 1, 2, . . . (BEAM:2.20)

and the torsional modes shapes

Ti(x) = C1cosλix, i = 1, 2, . . . (BEAM:2.21)

where λi is given in (BEAM:2.13).

Bending Vibrations

Bending motions may be considered in y- and z-direction. If v(x, t) isthe displacement in y-axis of the beam’s center line, EJz [Nm2] thebending stiffness and µ [kg/m] the mass density per unit length, theequation of motion is given as

EJxv,,,, = µv(x, t), (BEAM:2.22)

which must be satisfied over the domain 0 ≤ x ≤ L. The bendingstiffness is the product of the Young’s modulus E and the area momentof inertia Jz with respect to the z-axis. To obtain separation of variablesmethod and write the displacements as

v(x, t) = V (x)q(t) (BEAM:2.23)

where q(t) is a harmonic function with the frequency ω [rad/sec]and V (x) is the unknown function of the mode shapes. Substituting(BEAM:2.23) into (BEAM:2.22) and denoting v(x, t) = −ω2V (x)q(t)yields the differential equation of the eigenvalue problem

−EJzv,,,,(x) = µω2V (x) → V ,,,,(x) + λ4V (x) = 0 (BEAM:2.24)


where the eigenvalues ωi, are given by

λ4 = ω2 µ

EJz

→ ω = λ2

√EJz

µ(BEAM:2.25)

The mode shapes V (x) must be satisfy four boundary conditions. Inthe case of a free-free supported beam, one finds the conditions at theborders x = 0 and x = L

EJzV,,(x = 0, t) = 0 , EJzV

,,,(x = 0, t) = 0,

(BEAM:2.26)

EJzV,,(x = L, t) = 0 , EJzV

,,,(x = L, t) = 0.

Consequently, one obtain the frequency equation

cosλLcosλL = 1 (BEAM:2.27)

with the solutions λ0 = λ1 = 0 as well as the eigenvalues of thebending motion, i = 2, 3, . . . which have to find numerically from(BEAM:2.27). The natural frequencies follows with (BEAM:2.25).Modes i = 0 and i = 1 represent the rigid body motion as translationin the y-axis and rotation about the z-axis. The solution of the eigen-value equation (BEAM:2.24) with respect to the boundary conditions(BEAM:2.26) and the infinity sequence of eigenvalues λi, i = 2, 3, . . .yields the bending modes shapes

Vi(x) = C1icoshλix+ C2isinhλix+ C3icosλix+ C4isinλix, i = 2, 3, . . .(BEAM:2.28)

where the coefficients C1 to C4 are obtained from the normalizationcondition of the modal masses

Mqij=

∫ L

o

µViVjdx =1 for i = j0 for i = j

i = 1, 2, . . . (BEAM:2.29)

One finds

C1i = C3i = Ai(−sinλiL+ sinhλiL) =1√µL

,

(BEAM:2.30)

C2i = C4i = Ai(+cosλiL− coshλiL) =Bi√µL

.

where values Bi are near the value 1 but very sensitive. The first fourbending mode shapes of (BEAM:2.28) are plotted in Fig. BEAM:2.2.4.

Limitations of Modelling BEAM:2.0 -13

0 0.2 0.4 0.6 0.8 1-2

-1

0

1

2

Mode 1

Mode 3

Mode 2

Mode 4

Figure BEAM:2.2.4: Four bending mode shapes of a beam with free-free borders

BEAM:2.3 Limitations of Modelling

The solution of beam structures applied by BEAM V3.3 are limited asfollow:

1 Beam sections are oriented in a straight line. The x-axis is alwaysthe beam’s centerline.

2 Boundary conditions and concentrated masses and stiffness areonly allowed at nodes. A section has two nodes. If more nodesrequired, more sections have to be introduced.

3 A fixed support of the beam is only allowed at the end nodes ofthe beam structure, not at nodes within the structure. In thiscase, divide the structure into two parts.

4 A structure with a revolute joint between two sections is not solv-able. The development group is endeavour to satisfy the customerwishes and to extend the BEAM code.

BEAM:2.4 System Requirements

BEAM V 3.3 has a line oriented user interface and is coded in FOR-TRAN 77. At present BEAM runs on different UNIX-platforms, PCand Macintosh. The SID file is portable over various platforms.The runtime memory is about 2 MB.

Please note that both the input file of BEAMHint:and the SID file have to be located in the sameworking directory.

BEAM:2.0 -14 System Requirements

BEAM:3. Starting of BEAM

The following chapter describes how to use BEAM. The variants aredivided into four classes: a simple cantilever beam, a beam with twosections, a beam with elastic borders and concentrated masses at nodes,and a beam with measured stiffness.

BEAM:3.1 General Information

Note at first, that the x-axis of the reference frame is identical to thebeam’s centerline and second, that the y- and z-axis are perpendicularto the x-axis, like a Cartesian frame as shown in Fig. BEAM:2.1.1. Theorigin is defined at the coordinate x = 0, therefore, nodes with positiveor negative values are allowed.

The units of the beam data should be taken from the ISO norm: massby [kg], length by [m], and time by [sec]. Therefore, the unit of theforce is [N ] = [kgm/sec2]. The SID depend also on these units. If otherunits are used, the whole data set of the MBS should be changed tothe new units.

The code is limited to a maximum number of sections of the beamstructure, number of modes, and number of desired markers for theMBS simulation. The actual limitation is printed out in the headerof the BEAM code as shown below and may be different from yourinstallation.

Example:B E A M

a preprocessor

for mode shape analysis of straight beam

structures

and generation of the SID file for MBS codes

Version 3.3 - March 1996

Copyright by Oskar Wallrapp

D-82234 Wessling

=======================================================

Note: Model limited to

Max. number of sections = 10

Max. number of modes = 30

Max. number of markers = 30

If a larger model is desired, consult your BEAM distributor.

BEAM:3.2 -16 Example of a Simple Cantilever Beam

BEAM:3.2 Example of a Simple Cantilever Beam

Model Description

Let first consider a simple beam as shown in Fig.BEAM:3.2.1. Thebeam is clamped to its left end, the right end is free. The beam’s dataare given in Fig.BEAM:3.2.1. Due to the homogeneous structure of thebeam, only one section with two major nodes at the ends are requiredto describe the beam’s properties. The nodes are measured by valuesat the x-axis only.

L = 10 m

*

Marker Major node

x

y

z

* *

Figure BEAM:3.2.1: A simple cantilever beam

Datalength L = 10 m height = 0.0774 mcrosssectionarea A = 0.0004 m2 massdensity = 3000 kg/m3

areamomentofinertia Jx = 4 ∗ 10−7 m4

Jy = 2 ∗ 10−7 m4

Jz = 2 ∗ 10−7 m4

Y oung′smodulus E = 7 ∗ 1010 N/m2 Shearmodulus G = 2.5 ∗ 1010 N/m2

Input Data and Run of BEAM

Before starting BEAM, a folder of your project should be created. ThenBEAM can be started by execution of Beam. The prompt is as follows:

Example:B E A M

a preprocessor

for mode shape analysis of straight beam

structures




D-82234 Wessling

=======================================================

Note: Model limited to

Example of a Simple Cantilever Beam BEAM:3.2 -17




=======================================================

Test output on screen(1), no test output(0)

0

Data of general beam structure

input interactive (0)

or

from file with mode selections etc. (1)

or

all data from file and all modes for SID (2)

or

as option 2 but SID for SIMPACK v5.xx (3)

or

Data of measured beam structure


or


or


or


or

Stop the program (99)

?

0

The code asks you at first about test prints. Test prints are importantfor detailed information on the computation only. Usually zero is used.Always, the natural frequencies and mode shape values are stored inthe input file.

The next prompt defines the input media. If there is no data input filefor the beam properties, use the interactive modus, i.e. set 0, otherwiseuse 1 or 2. Is the input 9, the code stops. For input 1, the code willask you about mode selection, consideration of geometric stiffening byincorporating geometric stiffening terms.

The next block of input data describes the geometrical properties of thebeam. This data can already exist on the beam input file or have to betold the program. Here, we have chosen the interactive modus. See thenext program prompts and the inputs, where the data are taken fromthe chapter of Fig. BEAM:3.2.1.

Provide model description, not to exceed 80Example:characters

Simple Beam: Date 19.03.94/OW

Number of different beam sections (ns) = ?


1

=== Write down x-position of each node

(node=ns+1)

x-Position in [m] of 1-th node = ?

0

x-Position in [m] of 2-th node = ?

10

=== Write down mass, geometric & material

properties

For 1-th section

Mass density [kg/m**3] = ?

3000

Cross sectional area(A) [m**2] = ?

4e-4

Area moments of inertia(Jt,Jy,Jz) [mm**4] = ?

4e-7 2e-7 2e-7

Young’s modulus(E) [N/m**2] = ?

7e10

Shear modulus(G) [N/m**2] = ?

2.5e10

The beam data starts with a description of the model. In our example,the beam consists only of one section with major nodes at both endsat x = 0.0 and x = 10m. For this section the mass density, the crosssection area, the area moments of inertia about the x-, y-, and z-axis arerequired. Additionally, Young’s modulus and Shear modulus are neededmeasured in the corresponding units. For this section, the values areconstant.

The next block describes the boundary conditions, additional node stiff-ness and nodal masses. This block can also be written on the beaminput file. The program prompts and the inputs are:

Sum of nodes with boundary conditions(nb) = ?Example:1

Sum of nodes with additional masses(nam) =

?

0

Sum of nodes with additional springs(nas) =

?

0

=== Write down boundary conditions


For 1-th boundary condition

Node no., tx,ty,tz,rx,ry,rz(free = 0, fixed =

1)

1 1 1 1 1 1 1

As shown in Fig. BEAM:3.2.1, the beam structure has one boundary,no additional nodal masses and no additional spring suspensions. For anode up to 6 constraints can be defined. Here, at node one all directionsare locked - three translational motions and three rotational motionsabout the reference frame. Additional masses and springs at nodes willbe discussed in detail in section BEAM:3.4. Here, the inputs are zeros.

The next beam data tells the code the number of modes to be computedand the corresponding modal damping rates. The program prompts andthe inputs are:

=== No. of modes in longitudinal(x)Example:vibration,

in y-bending vibration,

in z-bending vibration, and

in torsional vibration = ?

2 3 3 2

=== Damping ratio

( pos. values mean proport. stiffness,

neg. values mean natural damping )

for longitudinal vibration,

y-bending vibration,

z-bending vibration,

torsional vibration = ?

0.01 0.02 0.02 0.001

Here two modes in longitudinal direction, three bending modes in yand z direction, as well as two torsional mode should be computed.The modal damping coefficients will be multiplied with the stiffnessterm to get a damping value in the equation of deformations of theMBS. All modes of one vibration direction are multiplied by the samecoefficient.

For the evaluation of the SID, which are used in the MBS code, addi-tional nodes - here denoted by markers - (often also called attachmentpoints) are required in the MBS simulation, e.g. for evaluations of thekinematic, attachment points of joints, action points of forces, etc. So,within the range of the major nodes, markers for the MBS simulationhave to be defined now. They may have also y and z values.

=== Write down the total number > 0 of markersExample:for MBS code

3

For all markers: name (max 8 char in strings)


and

x, y, z - position in [m]

marker : 1

’m1’ 0 0 0

marker : 2

’m2 ’ 5 0 0

marker : 3

’m3’ 10 0 0

Here, markers are called m1, m2 and m3 and are located at x - positionof zero, at 5 m, and 10 m. The coordinates y and z are zero.

Next, in the mass integral evaluation, the mass moment of inertia of canbe incorporated in the bending motion, when the key is on. The defaultis zero. For the computation of the modes shapes, a nonlinear equationin eigenvalues has to be solved. For this iteration solution written byKOLLOUSEK, starting values have to be computed by SOTIROPU-LUS. For both methods, iteration boundaries epskol and epssot haveto be defined. Additionally, elements in the mass integrals of the equa-tions of deformation can be set to zero when the value is smaller thanan border epsmass. Usually, the defaults can be used, then the inputis zero, see below.

=== Four keys to consider:Example:mass mom.o.inertia for bending (0/1),

boundaries epskol for mode computations,

boundaries epssot for mode computations,

epsmass for zero mass elements ?

Zero input means defaults = 0 1.000E-6

1.000E-3 1.000E-6

0 0 0 0

Because the interactive modus is used, BEAM ask the user to save thedata in a input file for corrections and additional computations. Theprogram prompts and inputs are:

Save of input data and program continuationExample:0 = compute without saving data

1 = compute with saving data

3 = save data and exit

4 = do not save data and exit

1

Name of the input data file within 60

characters

in case of <blanks> file name = beam.dat is

assumed

<one blank>

Data are saved on file = beam.dat


With the current input data, BEAM computes at first the desired eigen-values for the defined vibration directions. The code will print the fol-lowing statements and results given by the mode shape description andnatural frequencies:

********* Mode Shape Calculation of BeamExample:Structure *********

Longitudinal analysis in x direction

====================================

Bending analysis in y direction

================================

Bending analysis in z direction

================================

-------------------------------------------------

| |

| Mode number Type of mode Natural frequency[Hz]|

|-------------------------------------------------|

| |

| 1 Longitudinal 1 120.76 |


| 3 Bending y 1 .60443 |

| 4 Bending y 2 3.7879 |

| 5 Bending y 3 10.606 |

| 6 Bending z 1 .60443 |

| 7 Bending z 2 3.7879 |

| 8 Bending z 3 10.606 |

| 9 Torsion 1 72.169 |

| 10 Torsion 2 216.51 |

-------------------------------------------------

Results are (also) listed in file: beam.dat

1

The above list of mode shapes will be also written on the input filefor reason of saving the data. For the following evaluation of the MBSinput data, specific modes have to be selected by the user, which areimportant for the MBS simulation model. Only the desired modes willbe incorporated for SID. The number must be greater zero, but it maybe chosen in a serious related to frequencies. The BEAM prompts andthe inputs are listed next:

=== write down the total number of modes to beExample:chosen

4

=== write down 3 different mode numbers

selected

3 6 9 1


Here, the first bending modes in y - direction and z - direction as wellas the first torsional and longitudinal modes are desired given by thenumbers 3, 6, 9, and 1.

Note that the serious of the selected number willHint:be taken for the serious of the SID file.

After these inputs, BEAM calculates the modal mass integrals, modalmass matrix, modal stiffness matrix, and modal damping matrix. Thestiffness calculations can be done with or without consideration of ge-ometric stiffening terms, see the prompts and program input below.Geometric stiffening has to be added if the structure is very flexibleand the loads are high, therefore simulations will appear in the rangeof buckling or stiffening due to these loads.

******** Mass Terms Calculation ********Example:

******** Stiffness Calculations ********

Choice of geometric stiffening calculations:

0=no, 1=yes

1

*** stiffness calculation due to force at

marker 1 ***


marker 2 ***


marker 3 ***

*** stiffness calculation due to long. accel.

ax ***

*** stiffness calculation due to centrifugal

accel. ***

The geometric stiffness matrices are calculated for all possible unitloads: for longitudinal forces at all markers, a longitudinal accelera-tion ax of the reference frame, and for a centrifugal acceleration dueto rotations about y- and z-axis of the reference frame. Here, in thismodel, geometric stiffening is included.

The last step of BEAM is to save the MBS data on a SID file, if the userwill choose it. The name of the file is an input data, see the dialoguebelow.

Generation of SID file?Example:(0 = no, 1 = for SIMPACK 6.xx, 2 = for SIMPACK

5.xx)

1

Provide the name of the SID file.

It will be added by .SID Beam

blanks means default <SID Beam>

<one blank>


SID file is denoted: SID Beam

The SID file is successfully written

FORTRAN STOP

If the writing of the SID file was successful, BEAM gives a prompt andstops.

Input file of BEAM

Please check the data of the BEAM input file, if a saving of the datawas chosen. All data of the beam, the natural frequencies and the modeshapes results are stored on the BEAM input file - see the printed beaminput file ¡beam.dat¿ next.

Note that phi(1,2,3) are the displacements in x, y, z -axis, psi(1,2,3) arerotation angle about x, y, z -axis.

Simple Beam: Date 19.03.94/OW !!Example:Model Description

1 !!

Number of sections (ns)

.000000 !!

x-Position of node 1

10.0000 !!

x-Position of node 2

3000.00 !!

Mass density [kg/m**3] of sect. 1

4.000000E-04 !!

Cross sectional area(A) of sect. 1

4.0E-07 2.0E-07 2.0E-07 !!

Area moment of inertia (Jt,Jy,Jz) of sect.1

7.000000E+10 !!

Young’s Modulus (E) of sect. 1

2.500000E+10 !!

Shear Modulus (G) of sect. 1

1 !!

Number of nodes with boundary cond.

0 !!

Number of nodes with additional masses

0 !!

Number of nodes with additional springs

1 1 1 1 1 1 1 !!

Node no., tx,ty,tz,rx,ry,rz(free=0, fixed=1)

of bound.cond. 1

2 3 3 2 !!

No. of modes for long., bend. , bend.z, &

torsion

0.10000E-01 0.20000E-01 0.20000E-01 0.10000E-02 !!

Damping ratio for long., bend.y, bend.z,&

torsion


3 !!

number > 0 of markers for MBS code

’m1 ’ .00000 .00000 .00000 !!

marker description and x-position

’m2 ’ 5.000 .00000 .00000 !!


’m3 ’ 10.000 .00000 .00000 !!


0 1.00000E-06 1.00000E-03 1.00000E-06 !!

Inertia key, epskol, epssot, epsmass

-------------------------------------------------

| |

| Mode shapes computed by BEAM - V 3.3 |

| |

-------------------------------------------------

| |

| Mode number Type of mode Natural

frequency[Hz]|

-------------------------------------------------



| 3 Bending y 1 .60444 |

| 4 Bending y 2 3.7879 |

| 5 Bending y 3 10.606 |

| 6 Bending z 1 .60443 |

| 7 Bending z 2 3.7879 |

| 8 Bending z 3 10.606 |

| 9 Torsion 1 72.169 |

| 10 Torsion 2 216.51 |

-------------------------------------------------

4 modes are selecet for MBS data generation:

They are:

Bending y 1

Bending z 1

Torsion 1

Longitudinal 1

Mode shapes for mode i and marker k

i k x y z | phi(1) phi(2) phi(3) psi(1)

-----------------------------------------------------------------

1 1 .00 .00 .00 .0000 .0000 .0000 .0000

1 2 5.00 .00 .00 .0000 .1960 .0000 .0000

1 3 10.00 .00 .00 .0000 .5774 .0000 .0000

2 1 .00 .00 .00 -.0000 .0000 .0000 .0000

2 2 5.00 .00 .00 -.0000 .0000 .1960 .0000

2 3 10.00 .00 .00 -.0000 .0000 .5774 .0000


3 1 .00 .00 .00 .0000 .0000 -.0000 .0000

3 2 5.00 .00 .00 .0000 .0000 -.0000 9.1287

3 3 10.00 .00 .00 .0000 .0000 -.0000 12.9099

4 1 .00 .00 .00 .0000 .0000 .0000 .0000

4 2 5.00 .00 .00 .2887 .0000 .0000 .0000

4 3 10.00 .00 .00 .4082 .0000 .0000 .0000

SID File of the Simple Beam

An outline of the SID-file written by BEAM is listed below. The gener-ated SID file is an ASCII file and is readable using an standard editor.

It should be noted that changes on the file canHint:disturb the readability of SID by SIMPACK.

3 4 = No. nodes & No. modes of model =SimpleExample:Beam: Date 19.03.94/OW

|SID generated by BEAM-V.3.3 including Geo

Stiff=yes from Beam input file=beam.dat

15-Aug-96; part

new modal = m

refmod

mass = 1.20000000000D+01

nelastq = 4

ielastq ( 1) = Bending y 1

ielastq ( 2) = Bending z 1

ielastq ( 3) = Torsion 1

ielastq ( 4) = Longitudinal 1

end refmod

frame

new node = m1 >>> marker

m1

rframe = body ref

origin

order = 1

nrow = 3

ncol = 1

nq = 4

nqn = 0

structur = 0

end origin

phi

order = 1

nrow = 3

ncol = 4

nq = 4

nqn = 0

structur = 0

end phi

psi


order = 0

nrow = 3

ncol = 4

nq = 4

nqn = 0

structur = 0

end psi

AP

order = 0

nrow = 3

ncol = 3

nq = 4

nqn = 0

structur = 4

end AP

end node

...

...

new node = m3

rframe = body ref

origin

order = 1

nrow = 3

ncol = 1

nq = 4

nqn = 0

structur = 3

m0( 1, 1) = 1.00000000000D+01 position

in x for m3

m1( 1, 4, 1) = 4.08248290463D-01

m1( 2, 1, 1) = 5.77350269190D-01

m1( 3, 2, 1) = 5.77350269190D-01

end origin

phi

order = 1

nrow = 3

ncol = 4

nq = 4

nqn = 0

structur = 3

m0( 1, 4) = 4.08248290463D-01

m0( 2, 1) = 5.77350269190D-01

m0( 3, 2) = 5.77350269190D-01

m1( 1, 1, 1) =-3.87314860533D-02

m1( 1, 2, 2) =-3.87314860533D-02

end phi

psi

order = 0

nrow = 3

ncol = 4


nq = 4

nqn = 0

structur = 3

m0( 1, 3) = 1.29099444873D+01

m0( 2, 2) =-7.94725812117D-02

m0( 3, 1) = 7.94725812117D-02

end psi

AP

order = 1

nrow = 3

ncol = 3

nq = 4

nqn = 0

structur = 3

m0( 1, 1) = 1.00000000000D+00

m0( 2, 2) = 1.00000000000D+00

m0( 3, 3) = 1.00000000000D+00

m1( 2, 1, 1) = 7.94725812117D-02

m1( 3, 2, 1) = 7.94725812117D-02

m1( 1, 1, 2) =-7.94725812117D-02

m1( 3, 3, 2) = 1.29099444873D+01

m1( 1, 2, 3) =-7.94725812117D-02

m1( 2, 3, 3) -1.29099444873D+01

end AP

end node

end frame

mdCM

order = 1

nrow = 3

ncol = 1

nq = 4

nqn = 0

structur = 3

m0( 1, 1) = 6.00000000000D+01 center

of mass

m1( 1, 4, 1) = 3.11878720494D+00

m1( 2, 1, 1) = 2.71236300674D+00

m1( 3, 2, 1) = 2.71236300674D+00

end mdCM

J

order = 1

nrow = 6

ncol = 1

nq = 4

nqn = 0

structur = 3

m0( 1, 1) = 1.20000000000D-02 mass

mom. o. inertia

m0( 2, 1) = 4.00000000000D+02

m0( 3, 1) = 4.00000000000D+02


m1( 2, 4, 1) = 3.97096320094D+01

m1( 3, 4, 1) = 3.97096320094D+01

m1( 4, 1, 1) =-1.97040089549D+01

m1( 5, 2, 1) =-1.97040089549D+01

end J

Ct

order = 1

nrow = 4

ncol = 3

nq = 4

nqn = 0

structur = 3

m0( 1, 2) = 2.71236300674D+00

m0( 2, 3) = 2.71236300674D+00

m0( 4, 1) = 3.11878720494D+00

m1( 1, 1, 1) =-1.57087820334D-01

m1( 2, 2, 1) =-1.57087820334D-01

end Ct

Cr

order = 1

nrow = 4

ncol = 3

nq = 4

nqn = 0

structur = 3

m0( 1, 3) = 1.97047017752D+01

m0( 2, 2) =-1.97047017752D+01

m0( 3, 1) = 9.86247110499D-02

m1( 1, 2, 1) =-1.00000000000D+00

m1( 2, 1, 1) = 1.00000000000D+00

m1( 2, 4, 2) =-9.58641445425D-01

m1( 4, 2, 2) = 9.58641445425D-01

m1( 1, 4, 3) = 9.58641445425D-01

m1( 4, 1, 3) =-9.58641445425D-01

end Cr

Me

order = 0

nrow = 4

ncol = 4

nq = 4

nqn = 0

structur = 1

m0( 1, 1) = 1.00000000000D+00 modal

mass

m0( 2, 2) = 1.00000000000D+00

m0( 3, 3) = 1.00000000000D+00

m0( 4, 4) = 1.00000000000D+00

end Me

Gr

order = 0


nrow = 3

ncol =12

nq = 4

nqn = 0

structur = 3

m0( 2, 1) =-3.94094035503D+01

m0( 2, 8) = 3.97096320094D+01

m0( 3, 2) =-3.94094035503D+01

m0( 3,12) = 3.97096320094D+01

end Gr

Ge

order = 0

nrow = 4

ncol =12

nq = 4

nqn = 0

structur = 3

m0( 1, 2) =-2.00000000000D+00

m0( 1,12) = 1.91728289085D+00

m0( 2, 1) = 2.00000000000D+00

m0( 2, 8) =-1.91728289085D+00

m0( 4, 6) = 1.91728289085D+00

m0( 4, 9) =-1.91728289085D+00

end Ge

Oe

order = 1

nrow = 4

ncol = 6

nq = 4

nqn = 0

structur = 3

m0( 1, 4) = 1.97040089549D+01

m0( 2, 6) = 1.97040089549D+01

m0( 4, 2) =-1.98548160047D+01

m0( 4, 3) =-1.98548160047D+01

m1( 1, 1, 1) =-1.00000000000D+00

m1( 2, 2, 1) =-1.00000000000D+00

m1( 1, 1, 2) = 1.19333638409D+00

m1( 2, 2, 2) = 1.93336384089D-01

m1( 4, 4, 2) =-1.00000000000D+00

m1( 1, 1, 3) = 1.93336384089D-01

m1( 2, 2, 3) = 1.19333638409D+00

m1( 4, 4, 3) =-1.00000000000D+00

m1( 1, 4, 4) = 9.58641445425D-01

m1( 4, 1, 4) = 9.58641445425D-01

m1( 1, 2, 5) = 1.00000000000D+00

m1( 2, 1, 5) = 1.00000000000D+00

m1( 2, 4, 6) = 9.58641445425D-01

m1( 4, 2, 6) = 9.58641445425D-01

end Oe


ksigma

order = 0

nrow = 4

ncol = 1

nq = 4

nqn = 0

structur = 0

end ksigma

Ke

order = 0

nrow = 4

ncol = 4

nq = 4

nqn = 0

structur = 1

m0( 1, 1) = 1.44227572630D+01 modal

stiffness

m0( 2, 2) = 1.44227572630D+01

m0( 3, 3) = 2.05616758357D+05

m0( 4, 4) = 5.75726923399D+05

end Ke

De

order = 0

nrow = 4

ncol = 4

nq = 4

nqn = 0

structur = 1

m0( 1, 1) = 2.88455145261D-01 modal

damping

m0( 2, 2) = 2.88455145261D-01

m0( 3, 3) = 2.05616758357D+02

m0( 4, 4) = 5.75726923399D+03

end De

end modal

end part

The outline is supplemented by comments to give an interpretation ofthe entities and the structure of the file. At the beginning of the file thefour natural frequencies of the selected modes can be recognised. Thecoordinates for the markers (in SID are called nodes) are relevant forthe simulation with SIMPACK. See marker m3 for an example. Alsothe mass and the mass moments of inertia can be found in the SID-file.At the end the modal stiffness and damping matrix are contained.

Usually the user should not be bothered with the SID-file. The onlyentities which are relevant to the user in connection with the set-up ofthe MBS-model are the coordinates of the nodes of the FEM-structure.

Congratulations on finishing your first BEAM example!

Example of a Beam with Sections Using Different Geometry BEAM:3.3 -31

The purpose of this first example has to demonstrate the handling ofBEAM. However, the next examples are more complicated with respectto the beam’s geometry and the boundary conditions.

BEAM:3.3 Example of a Beam with Sections UsingDifferent Geometry

As mentioned before, a beam section has constant beam parameters.Therefore, a structure with variable mass and geometrical propertieshas to be subdivided (and approximated) into a finite number of sec-tions. The example as shown in BEAM:3.3.2 is a simplified model ofa helicopter rotor blade. The blade is modelled by four sections: Thefirst section represents the arm of the blade, the next two sections thetransition and the last section the blade area. The data of the exampleare given in BEAM:3.3.2. Here, the bending motions in y-direction areof interest only. The aim of this simulation is to show the influence ofthe rotation Ω about the y-axis on the bending motion.

x

z

y

x

Section 1 2 3 Section 4

** **

Figure BEAM:3.3.2: Beam model of an aliminium helicopter rotorblade

Data of the helicopter rotor bladeSection Length Height Area Area m.o. Inert. Mass Density ρ Y oung′s Modulus

n [m] [cm] [cm2] Jz [cm4] [Kg/m3] [N/m2]

1 0.60 6.0 42 126.0 2787 7.13 ∗ e10

2 0.12 5.0 50 104.2 2787 7.13 ∗ e10

3 0.08 3.4 51 49.0 2787 7.13 ∗ e10

4 4.00 2.5 50 26.0 2787 7.13 ∗ e10

Additional data: Shear modulus G = 2.5 E10, Area m.o.Inert. Jx and Jy are given by 0,

because the data are unknown and not required in the computation for bending in y-direction

The input data for BEAM are listed below, where four sections withfive major nodes are introduced. Moreover, four markers at 0.0, 0.8,2.8, and 4.8 m are set for the SID file.

Provide model description, not to exceed 80Example:

BEAM:3.3 -32 Example of a Beam with Sections Using Different Geometry

characters

Helicopter rotor Blade, y-bending, with

geo.Stiff


4


(#node=ns+1)

x-Position of 1-th node = ?

0


0.6


0.72


0.8


4.8


properties

For 1-th section


2787

Cross sectional area(A) = ?

0.0042

Area moments of Inertia(Jx,Jy,Jz) = ?

0 0 126e-8

Young’s modulus(E) = ?

7.31e10

Shear modulus(G) = ?2.5e10

For 2-th section


2787


0.0050


0 0 104.2e-8


7.31e10

Shear modulus(G) = ?

2.5e10

For 3-th section


2787


Example of a Beam with Sections Using Different Geometry BEAM:3.3 -33

0.0051


0 0 49e-8


7.31e10


2.5e10

For 4-th section


2787


0.0050


0 0 26e-8


7.31e10


2.5e10

==========================================

Sum of nodes with boundary conditions(nb) = ?

1

Sum of nodes with additional masses(nam) = ?

0

Sum of nodes with additional springs(nas) = ?

0




1)

1 1 1 1 1 1 1

No. of modes in longitudinal(x) vibration,




0 7 0 0

=== Damping ratio







0 0.03 0 0

BEAM:3.3 -34 Example of a Beam with Sections Using Different Geometry

=== Write down the total number > 0 of markers

for MBS code

4


and

x, y, z - position

marker : 1

’m1’ 0 0 0

marker : 2

’m2’ 0.8 0 0

marker : 3

’m3’ 2.8 0 0

marker : 4

’m4’ 4.8 0 0

=== Four keys to consider:

mass mom.o.inertia for bending (0/1),





1.000E-3 1.000E-6

0 0 0 0

========================================================Example:Save of input data and program continuation

0 = compute without saving data




1


characters


assumed

hrotor.dat

Now for the rotor blade, BEAM calculates the following seven naturalfrequencies, where only the first four modes are taken into account inthe SID generation.

-------------------------------------------------Example:

Mode number Type of mode Natural frequency[Hz]

-------------------------------------------------

1 Bending y 1 1.1553



Example of a Beam with Concentrated Mass and Spring BEAM:3.4 -35





-------------------------------------------------

Results are (also) listed in file: hrotor.dat

=== write down the total number of modes to

be chosen

4

=== write down 4 different mode numbers

selected

1 2 3 4

After the last data input, BEAM calculates the mass and stiffness ma-trices and if desired, the results are saved on the SID file. This part ofcomputations and the program dialogue has been described in sectionBEAM:3.2.

BEAM:3.4 Example of a Beam with ConcentratedMass and Spring

From this example the user should learn, how to use concentratedmasses or springs at major nodes. In the case that there is a con-centrated mass on the beam, a major node must be introduced at themass point. Additionally, concentrated stiffness suspensions such astranslational or rotational springs have to be considered in a similarway.

The example as illustrated in Fig. BEAM:3.4.3 shows a flexible rod witha tip mass. The tip mass is much heavier than the rod itself, therefore,the tip mass has to be taken into account into the eigenvalue analysis.The beam has one section with two major nodes. For the model allvibration directions are of interest: Here, in this example we wantto calculate 2 modes in x-direction, 2 bending modes in y-direction,4 bending modes in z-direction and 2 torsional modes. Markers aredefined for positions at 0, 7, and 14 cm in x-direction.

For this example as shown in Fig. BEAM:3.4.3, the following units areused: [kg] for mass, [cm] for length, and [kgcm/s2] = N/100] for forces.Thus, the Young’s modulus of 2.1e11N/m2 represents 2.1e9kg/cms2.

BEAM:3.4 -36 Example of a Beam with Concentrated Mass and Spring

L = 14 cm

b = 1.32 cmh = 0.05 cmx

y

z Tip Mass

*

*

*

Figure BEAM:3.4.3: Flexible rod with tip mass

Datalength L = 14 cm height h = 0.05 cmcrosssectionarea A = 0.066 cm2 areamomentofinertia Ix = 9.597e− 3 cm4

areamomentofinertia Iy = 1.375e− 6 cm4 areamomentofinertia Iz = 9.583e− 3 cm4

Y oungsmodulus E = 2.1e+ 9 N/cms2 Shearmodulus G = 0.8e+ 9kg/cms2

massdensity 4 = 7.85e− 3 kg/cm3 tipmass m = 0.026kgtipmassMom.o.Inertia Ix = 2.2e− 2 kgcm2 tipmassMom.o.Inertia Iy = 1.25e− 2kgcm2

tipmassMom.o.Inertia Iz = 1.25e− 2 kgcm2

The input data for an interactive session is given below.

Provide model description, not to exceed 80Example:characters

Flexible rod with tip mass incl. Geometric

Stiffening


1


(#node=ns+1)


0


14


properties

For 1-th section


7.85e-3


0.066


9.597e-3 1.375e-6 9.583e-3



2.1e9


0.8e9

==========================================

Sum of nodes with boundary conditions(nb) = ?

1


1


0




1)

1 1 1 1 1 1 1

=== Write down additional masses

For 1-th additional mass

Node no., mass, mass moment of

inertia(Ix,Iy,Iz)

2 0.026 2.2e-2 1.25e-2 1.25e-2

No. of modes in longitudinal(x) vibration,




2 2 4 2

=== Damping ratio

( + values = proport. Stiffness, - values =

Lehr )





0 0.02 0 0


for MBS code

3


and

x, y, z - position

marker : 1

’m1’ 0 0

marker : 2

’m2’ 7 0

BEAM:3.4 -38 Example of a Beam with Concentrated Mass and Spring

marker : 3

’m3’ 14 0







1.000E-3 1.000E-6

0 0 0 0

========================================================

Save of input data and program continuation





1


characters


assumed

flexrul.dat

Data are saved on file = flexrul.dat

With the defined data, BEAM calculates the natural frequencies asshown below, where the first four bending modes in z-direction and thefirst torsion mode are considered in the generation of the SID. All othermodes are not of interest due to the much higher frequencies.

-------------------------------------------------Example:

Mode number Type of mode Natural frequency[Hz]

-------------------------------------------------

1 Longitudinal 1 2968.4

2 Longitudinal 2 18979.



5 Bending z 1 1.6943




9 Torsion 1 788.33

10 Torsion 2 11464.

---------------------------------------------------

Results are (also) listed in file:

flexrul.dat


From these results, the user can chose various mode shapes for evalua-tion of SID.

Beam structure with spring suspensions

As mentioned before, beam structures attached by spring suspensions indifferent directions and various major nodes can be modelled in BEAM.At the attachment point, a major node must be introduced. Then, thestiffness coefficient are given by the BEAM input data. Let us considera translational spring in z-direction with a value of 100[N/cm] at node2. Then, the BEAM dialogue due to this spring may be changed asfollows:

==========================================Example:Sum of nodes with boundary conditions(nb) = ?

1


1


1




1)

1 1 1 1 1 1 1

=== Write down additional masses

For 1-th additional mass

Node no., mass, mass moment of

inertia(Ix,Iy,Iz)

2 0.026 2.2e-2 1.25e-2 1.25e-2

=== Write down additional springs

For 1-th additional spring

Node no., tkx,tky,tkz,rkx,rky,rkz

2 0 0 100 0 0 0

Referring to this additional spring, the natural frequencies of the bend-ing motion in z-direction will be changed. The results of BEAM noware shown below.

-------------------------------------------------------Example:

Mode number Type of mode Natural

frequency[Hz]

-------------------------------------------------------

1 Longitudinal 1 2968.4

2 Longitudinal 2 18979.


BEAM:3.5 -40 Beam Structure with Measured Torsional Stiffness Data






9 Torsion 1 788.33

10 Torsion 2 11464.

-----------------------------------------------------

A comparison with the results without spring shows that only the firstbending mode in z- direction are significant changed.

BEAM:3.5 Beam Structure with Measured TorsionalStiffness Data

The torsional flexibility of a frame may be considered in the MBSsimulation. Therefore, the torsional stiffness GJT is required as theinput data for BEAM. Here, this stiffness may be taken from an ex-periment, where the frame is twisted about an angle θ [rad] due to atorque T [Nm], see Fig. BEAM:3.5.4. For the frame as given in Fig.

T

L

Figure BEAM:3.5.4: Experiment to find out the torsional stiffnessGJT = L T

thetaNm2/rad

BEAM:3.5.5, a SID file should be now generated with respect to thefollowing conditions:

The body reference frame is in the middle of the frame. Therefore, node1 is located at -5 m and node 2 at +5 m. The boundary conditionsare taken from the free-free borders. Markers are here located at someof the corners of the prismatic body in addition to markers at thecenterline, see Fig. BEAM:3.5.5. The damping ratio via Lehr is 10%.

Beam Structure with Measured Torsional Stiffness Data BEAM:3.5 -41

z

y

xL

beam's centerline

node 1

node 2

a b

marker 1

marker 2

marker 3

marker 5

marker 4

marker 6

Figure BEAM:3.5.5: Frame structure with torsional flexibility

Datalength L = 10 m height b = 0.2 mtorsionalstiffness GJT = 0.066 Nm2/rad width a = 0.5 mY oungsmodulus E = 21e+ 10 N/m2 ShearModulus G = 7.8e+ 10 N/m2

bodymass m = 200 Kg Shearmodulusmom.o.Inertia Ix = 5.0 kgm2 Mom.o.Inertia Iy = 400.0 kgm2

mom.o.Inertia Iz = 600.0 kgm2

In the case of the input of stiffness, the body mass data with respectto the reference frame are required. The input data for an interactivesession is listed below.

Example:=======================================================

B E A M

a p r e p r o c e s s o r

for mode shapes analysis of straight beam

structures




D-82234 Wessling

=======================================================

Note: Model Limitations are





=======================================================

Test output on screen(1), no test output(0)

0

Data of general beam structure


or


or


or


or

Data of measured beam structure


or


or


or


or

Stop the program (99)

?

10

Provide model description, not to exceed 80

characters

Frame with torsional flexibility, 1.6.95

====================================================

General Beam Structure Data

Note the frame definitions: z y

| /

|/

+------ x = longitudinal

Origin is at the coordinates 0/0/0

Nodes may be at the positive and negative

x-axis

====================================================

=== Write down


1


(#node=ns+1)

x-Position of 1-th node =?


-5


5


properties

Body mass [kg]

200

Body mass moments of inertia (Ix,Iy,Iz)

[kg*m**2]

5 400 600

Stiffness: longitudinal in x, bending in y,

bending in z, torsion about x

0 0 0 1e5


21e10


7.8e10

==========================================

Sum of nodes with boundary conditions(nb) =

?

0

Sum of nodes with additional masses(nam) =

?

0

Sum of nodes with additional springs(nas) =

?

0

=== No. of modes in longitudinal(x)

vibration,




0 0 0 3

=== Damping ratio







0 0 0 -0.1



for MBS code

6


and

x, y, z - position

marker : 1

’m1’ 0 0 0

marker : 2

’m2’ -5 -0.5 0.2

marker : 3

’m3’ -5 0.5 0.2

marker : 4

’m4’ 5 0 0

marker : 5

’m5’ 5 -0.5 -0.2

marker : 6

’m6’ 5 0.5 -0.2







1.000E-3 1.000E-6

0 0 0 0

========================================================

Save of input data and program continuation





1


characters


assumed

TorsionFrame.dat

Data are saved on file = TorsionFrame.dat

The results of the first three natural frequencies and the mode shapesevaluated at the markers 1 to 6 are printed out in the input file of thisexample and they are given next.

-------------------------------------------------Example:


| |

| Mode shapes computed by

BEAM - V 3.3 |

-------------------------------------------------

| |

| Mode number Type of mode Natural frequency[Hz]|

-------------------------------------------------

| |

| 1 Torsion 1 22.361 |

| 2 Torsion 2 44.721 |

| 3 Torsion 3 67.082 |

-------------------------------------------------

3 modes are selected for MBS data generation:

They are:

Torsion 1

Torsion 2

Torsion 3

Modes shapes for modes i and markers k

i k x y z | phi(1) phi(2) phi(3) psi(1)

-----------------------------------------------------------------

1 1 .00 .00 .00 .0000 .0000 .0000 .0000

1 2 -5.00 -.50 .20 .0000 .1265 .3162 .6325

1 3 -5.00 .50 .20 .0000 .1265 -.3162 .6325

1 4 5.00 .00 .00 .0000 .0000 .0000 -.6325

1 5 5.00 -.50 -.20 .0000 .1265 -.3162 -.6325

1 6 5.00 .50 -.20 .0000 .1265 .3162 -.6325

2 1 .00 .00 .00 .0000 .0000 .0000 -.6325

2 2 -5.00 -.50 .20 .0000 .1265 .3162 .6325

2 3 5.00 .50 .20 .0000 .1265 -.3162 .6325

2 4 5.00 .00 .00 .0000 .0000 .0000 .6325

2 5 5.00 -.50 -.20 .0000 -.1265 .3162 .6325

2 6 5.00 .50 -.20 .0000 -.1265 -.3162 .6325

3 1 .00 .00 .00 .0000 .0000 .0000 .0000

3 2 -5.00 -.50 .20 .0000 .1265 .3162 .6325

3 3 -5.00 .50 .20 .0000 .1265 -.3162 .6325

3 4 5.00 .00 .00 .0000 .0000 .0000 -.6325

3 5 5.00 -.50 .20 .0000 .1265 -.3162 -.6325

3 6 5.00 .50 .20 .0000 .1265 .3162 -.6325

Please note that the table shows the modeHint:shape values of markers, which are outside ofthe beam’s center line. phi(1,2,3) are the dis-placements in x,y,z -axis, psi(1,2,3) are rotationangle about x,y,z -axis.

The presented examples demonstrates the power of the program BEAMto calculate the eigenvalues and mode shapes of straight beam struc-


tures and to evaluate the data of flexible bodies modelled in a multibodysystem using the standard input data (SID) description.

Error Messages

Referring to the model data, BEAM may not find the eigenvalues ofthe vibration problem. A common message may occur:

Stiffness matrix not positive definiteExample:problem not solvable, change input: ierg=1

or

Newton’s method does not convert, II,w= 1 NaNExample:problem not solvable, change input: ierg=1

In this case, first - increase the boundaries for epskol and epssot. If theerror is still present, please stop the program and change the boundaryconditions of the beam structure. Moreover, springs with a smallstiffness may be help to get a convergence of the iterative solution, e.g. for tx = 1.0e−6N/m. Do this for corresponding vibration directions.A variation of this so-called numerical stabilisation factors brings theexact solution in the range of digits of the printed natural frequencies.

Error messages due to non-consistent geometry data are self-explanatory. Save the data, change the values and start BEAM again.

BEAM:4. References

Wallrapp, O. (1993). Standard Input Data of Flexible Bod-ies for Multibody Codes. Internal Report IB 515-93-4, DeutscheForschungsanstalt fr Luft- und Raumfahrt (DLR), Inst. Robotik undSystem Dynamik, Oberpfaffenhofen.

Meirovitch, L. (1967). Analytical Methods in Vibrations. NewYork, Macmillan Company.

beam

Documents

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mode shape

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mode shape