Beam

SIMPACK

Beam SIMPACK

Release 8.6

24th September 2003/SIMDOC v8.607

COPYRIGHT COPYRIG HT 2003 c



BEAM:0.0 -2

BEAM A PRE-PROCESSOR FOR MODE SHAPE ANALYSIS OF STRAIGHT BEAM STRUCTURES AND GENERATION OF THE SID FILE FOR SIMPACK AND OTHER MBS CODES User Manual written by Dr. Oskar Wallrapp D-82234 Wessling, FRG Version 3.3 - August 1996

BEAM:0.0 -2

Hint:

Beam is not part of a basic distribution of SIMPACK.

Contents 1 Ne New w in Ver Versio sion n 3.3 3.3

1.0 -5

2 Wh What at is Be Beam am? ?

2.1 -7

2.11 2.

Introd In troduct uctio ion n . . . . . . . . . . . . . . . . . . . . . . . . . 2. 1 -7

2.2 Theo Theoreti retical cal Bac Backgr kground ound . . . . . . . . . . . . . . . . . . 2.2 -8 Longitudinal Vibration . . . . . . . . . . . . . . . . . . . 2.2 -9 Torsional Vibrations Vibrations . . . . . . . . . . . . . . . . . . . . 2.2 -10 Bending Vibrati Vibration onss . . . . . . . . . . . . . . . . . . . . . 2.2 -11 2.3 Lim Limitat itations ions of Model Modellin lingg . . . . . . . . . . . . . . . . . . 2.0 -13 2.4 Sys System tem Req Require uiremen ments ts . . . . . . . . . . . . . . . . . . . . 2.0 -13

3 St Star arti ting ng of BE BEAM AM

3.2 -15

3.1 Gen General eral Inf Informa ormatio tion n . . . . . . . . . . . . . . . . . . . . 3.2 -15 3.2 Exam Example ple of of a Simpl Simplee Cantil Cantilev ever er Beam 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

4

3.3

Exampl Exa mplee of a Beam Beam with with Secti Sections ons Usin Usingg Differen Differentt Geometr Geometry y3.3 -31

3.4

Exampl Exa mplee of a Beam Beam with with Conce Concent ntrate rated d Mass Mass and and Spring Spring 3.4 -35

3.5

Beam Bea m Structu Structure re with with Measur Measured ed Torsi Torsional onal Stiff Stiffness ness Data 3.5 -40

Refe Re fere renc nces es

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 usage and modelling capabilities capabilities with respect to version 3.0. The major topics are as follows:

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

• The markers for the MBS code may be attached outside of the

beam’s centerline but within the sections of the beam structure. This means that a markers is now defined by data of the x, y and z position, see section BEAM:3.5 BEAM:3.5.. For this markers, markers, BEAM evaluates the mode shape values within the assumption of ”rigid arms” 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 implemented ment ed . From these values values BEAM computes the modal damping matrix De. The input is illustrated in section BEAM:3.5.

• In the past, beam structures with one homogenous section and

the boundary conditions free-free results in an error due to the solution of the eigenvalue eigenvalue equation. equation. In version 3.3 the exact solution soluti on is implem implemented ented..

• The SI SID D

file is optimized . No zer file zero o ele elemen ments ts are written on the SID file now. now. A key allow allowss to generat generatee the SID file for SIMPACK v5.xx versions, too.

Hint:

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

BEAM:1.0 -6

BEAM:1 BEA M:1.. NEW IN IN VERSIO VERSION N 3.3

BEAM:2. BEAM: BE AM:2. 2.1 1

What is Beam?

Intr In troduc oducti tion on BEAM is a computer program for the evaluation of the mode shapes of straight flexible beam structures as shown in Fig. BEAM:2.1.1. From this mode shapes, BEAM generates the SID (Standard Input Data) file contai con tainin ningg inp input ut data for MBS (MultiBody (MultiBody System) System) codes codes.. BEA BEAM M is a so-called pre-processor for SIMPACK and other MBS codes. BEAM z x y

Figure BEAM:2 BEAM:2.1.1: .1.1: A straight beam structure The structure has two sections, boundaries at the left end, concentrated springs at left and right, as well as a concentrated mass between section 1 and 2. solves the eigenvalue problem of beam structures with various sections of mass and geometrical properties, fixed and flexible boundary conditions tio ns at nodes nodes,, and concentrat concentrated ed nodal mas masses ses.. BEA BEAM M uses an eas easy y input description like FEM codes. A beam is modelled by the Euler-Bernoulli hypothesis, for which the mode shapes of

• longitu longitudinal dinal vibrati vibrations ons (in x-dire x-direction), ction), • 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 section BEAM:2.2. Using the mode shapes of a beam, the submatrices of the system equations of a flexible body as a part of the MBS are evaluated based on a modal approximation. The submatrices represent the input data of a flexible body of a MBS, which are written on the Standard Input Data (SID) file – especially – for a Taylor expansion up to the firstt ord firs order. er. MB MBS S cod codes es with flexibl flexiblee mem members bers,, wh whic ich h are based based on a formulation of the deformations assuming small displacements and a modal approximation, approximation, they can take those data as input to generate the system equations. The SID file contains all submatrices for a complete linearization of the equations of deformations, see (Wallrapp 1993). BEAM allows the user to select a various number of modes shapes and markers markers for the MBS dat dataa gen genera erati tion. on. Th Thee com compu putat tatio ion n of the geometric stiffening terms can be activated or not.

BEAM:2.2 -8

Theoretical Background

BEAM can be used either in an interactive modus, where the beam data comes from the standard input or where the beam input data are taken from a file. The output may be test prin prints ts for detailed discussions discussions of the results, the BEAM input data stored on a ASCII file for future computations, and the SID on a formatted ASCII file for the transfer to MBS codes running on various devises. Fig. BEAM:2.1.2 shows the data flow flow of BEA BEAM. M. BEA BEAM M solves solves the eig eigen env valu aluee prob problem lem of beam

INPUT

Input Data .....

BEAM Input Data

OR

BEAM

OUTPUT

Test Prints .....

BEAM Input Data

SID

Figure BEAM:2.1.2: Data flow of BEAM structures with various sections of mass and

BEAM:2 BEA M:2.2 .2

Theo The oret retica icall Bac Backgro kground und The following section gives an introduction into the vibration analysis of homogenous beams represented by one section and simple boundary condi con diti tions ons,, e. g. th thee free - free situa situati tion. on. Th There erefo fore, re, an analyt analytic ical al solution of the equations of motions is possible which one finds e. g. in (Meirovitch, 1967). For problems with different sections and boundary conditions as shown in Fig. BEAM:2.1.1, the differential equations are more difficult which one have to solve numerically using the Kolousek method. Both methods are implemented in BEAM.

BEAM:2.2 -9


Longitudinal Vibration Let us consider a thin homogenous rod with a longitudinal flexibility measured by u(x, t) with the unit [m [m]. The mass density density per unit length is ? [kg/m [kg/m], ], the longitudinal stiffness is EA with the unit [N [N ], ], where E is the Young’s modulus and A is the cross cross secti section on of ar area ea.. Th Thee time derivative may be denoted by a dot and the space derivative with respect to x by ’, the equation of longitudinal motion is represented by the differential equation (BEAM:2.1 (BEAM:2.1)) 

EAu (x, t) = µu ¨(x, t),

(BEAM:2.1)

≤ x ≤ L.

which must be satisfied over the domain 0

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

(BEAM:2.2) (BEAM:2.3) (BEAM:2.4)

= =

0 0

Applying the separation of variables method (Meirovitch, 1967) one writes 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 rad/sec]] and U ((x) is the unknown function U function of the mode shapes. Substitu Substituting ting (4) into 2 (1) to (3) and denoting u¨(x, t) = ω U (x)q(t) yields the differential equation of the eigenvalue problem

−

,,

= µω 2U U ((x)

−EAU (x)

(BEAM:2. M:2.6) 6) U ,, (x) + λ2 U U ((x) = 0 (BEA

⇒

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



ω = λ

EA µ

(BEAM:2.7)

and the boundary conditions

(BEAM:2.8) (BEAM:2.9) (BEAM:2.10)

EAU , (0) EAU , (l)

= =

0 0

The general solution of (BEAM:2.6) (BEAM:2.6) is U ((x) = C 1cosλx + C 2 sinλx U

(BEAM:2.11)

BEAM:2.2 -10


where C 1 and C 2 are amplitudes and which have to specialized by the boundary conditions (BEAM:2.8 (BEAM:2.8)) and (BEAM:2.9 (BEAM:2.9). ). From (BEAM:2.8 (BEAM:2.8)) one gets C 2 = 0 and from (BEAM:2.9 (BEAM:2.9)) the frequency equation sinλL = 0

(BEAM:2.12)

which is satisfied for eigenvalues π λi = i , L

i = 0, 1, 2, . . .

(BEAM:2.13)

The first value λ0 = 0 represent representss the lon longitu gitudin dinal al rigid body mot motion. ion. The other values describes the orthogonal, natural mode shapes for longitudinal longitu dinal deformat deformations ions as U i (x) = cosλi x,

i = 1, 2 , . . .

(BEAM:2.14)

where C 1 is the amplitudes, which has the same value for all modes and which may be defined by the normalization of the modal mass conditions L

M qij =



µU i U j dx =

o

1 f or or i = j 0 f oorr i = j



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

For (BEAM:2.14) (BEAM:2.14) one finds with (BEAM:2.15 (BEAM:2.15)) C 1 = sqrt

2 µL

i = 1, 2, . . .

(BEAM:2.16)

The fir The first st th thre reee mo mode de sh shap apes es of (BEAM:2.14 BEAM:2.14)) ar aree pl plot otte ted d in Fi Fig. g. BEAM:2.2.3. 1 Mode 2

0.5

Mode 3

0

Mode 1

-0.5 -1 0

0.2

0.4

0.6

0.8

1

Figuree BEA Figur BEAM: M:2. 2.2. 2.3: 3: Thr Three ee lo longi ngitud tudin inal al mod modee sh shapes apes of a rod wi with th free-free borders

Torsional Vibrations Let us focus at torsional vibrations about the x-axis of a homogenous bar with the torsional stiffness GJ T T [Nm2/rad] and the mass moment

BEAM:2.2 -11


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

(BEAM:2.17)

which must be satisfied over the domain 0 x borderss L. At the border x = 0 and x = L bound boundary ary condition conditionss are given, given, e.g. for the free free-fr -free ee situation

≤ ≤

,, GJ T T θ (x = 0, t) = 0

(BEAM:2.18)

,, GJ T T θ (x = L, t) = 0

(BEAM:2.19)

and

The solution of (BEAM:2.17 (BEAM:2.17)) to (BEAM:2.19 (BEAM:2.19)) is in analogy to the longitudinal longitu dinal vibration: vibration: one has to substit substitute ute u(x, t) by θ(x, t), EA by GJ T T , and µ by ix . The results are the natural frequencies ωi = λi

GJ T T , ix

i = 0, 1, 2, . . .

(BEAM:2.20)

and the torsional modes shapes T i (x) = C 1 cosλi x,

i = 1, 2, . . .

(BEAM:2.21)

where λi is given in (BEAM:2.13 (BEAM:2.13). ).

Bending Vibrations Bending motions Bending motions may be cons conside idered red in y- and z-d z-direc irection tion.. If If vv (x, t) is the displacement in y-axis of the beam’s center line, EJ z [Nm2] the bending stiffness and µ [kg/m kg/m]] the mass density per unit length, the equation of motion is given as EJ x v ,,,, = µv¨(x, t),

(BEAM:2.22)

which must be satisfied over the domain 0 x L. Th Thee bend bendin ingg stiffness is the product of the Young’s modulus E and the area moment of inertia J z with respect to the z-axis. To obtain separation of variables method and write the displacements as

≤ ≤

v (x, t) = V V ((x)q (t)

(BEAM:2.23)

where q (t) is a ha harm rmon onic ic fu func ncti tion on wi with th th thee fr freq eque uenc ncy y ω [rad/sec] and V V ((x) is the unknown unknown function function of the mode shapes. Subs Substitu tituting ting 2 (BEAM:2.23) into (BEAM:2.22 (BEAM:2.22)) and denoting v¨(x, t) = ω V V ((x)q (t) yields the differential equation of the eigenvalue problem

−

−EJ v z

,,,,

(x) = µω2 V V ((x)

→

V ,,,, (x) + λ4 V V ((x) = 0 (BE (BEAM: AM:2.2 2.24) 4)

BEAM:2.3 -12


where the eigenvalues ωi , are given by λ4 = ω 2

µ EJ z

ω = λ2

→



EJ z µ

(BEAM:2.25)

The mode shapes V V ((x) must be sati satisfy sfy four boundary boundary conditio conditions. ns. In the case of a free-free supported beam, one finds the conditions at the borders x = 0 and x = L EJ z V ,, (x = 0, t) = 0 , EJ z V ,,, (x = 0, t) = 0, (BEAM:2.26) EJ z V ,, (x = L, t) = 0 , EJ z V ,,, (x = L, t) = 0. Consequently, one obtain the frequency equation cosλLcosλL = 1

(BEAM:2.27)

with the solu with solutio tions ns λ0 = λ1 = 0 as wel elll as th thee ei eige gen nval alue uess of th thee bending motion, i = 2, 3, . . . which have to find numerically from (BEAM:2.27) BEAM:2.27).. Th Thee na natu tura rall fr freq eque uenc ncie iess fo foll lloows wi with th (BEAM:2.25) BEAM:2.25).. Modes i = 0 and i = 1 represent the rigid body motion as translation in the y-axis and rotation about the z-axis. The solution of the eigenvalue equation (BEAM:2.24 (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

V i (x) = C 1i coshλi x + C 2i sinhλi x + C 3i cosλi x + C 4i sinλi x, i = 2, 3, . . . (BEAM:2.28) where the coefficients C 1 to C 4 are obtained from the normalization condition of the modal masses L

M qij =

 o

µV i V j dx =

1 f or or i = j 0 f oorr i = j



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

One finds

C 1i = C 3i = Ai ( sinλi L + sinhλi L) =

−

√1µL ,

(BEAM:2.30) C 2i = C 4i = Ai (+ (+cosλ cosλi L

− coshλ L) i

=

√BµL . i

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

BEAM:2.0 -13

Limitations Limita tions of Modelli Modelling ng

2 Mode 1 1 Mode 3 0 Mode 2 -1

Mode 4

-2 0

0.2

0.4

0.6

0.8

1

Figure BEAM:2.2 Figure BEAM:2.2.4: .4: Four bending mode sha shapes pes of a beam with freefree borders

BEAM: BE AM:2. 2.3 3

Limit Li mitat ation ionss of Mode Modelli lling ng The solution of beam structures applied by BEAM V3.3 are limited as follow: 1 Beam sections sections are oriented oriented in a straight line. line. The x-axis x-axis is always always the beam’s centerline. 2 Boundary conditions and concentrated masses and stiffness are only onl y allowe allowed d at nodes nodes.. A section section has two two nodes. If more nodes required, more sections have to be introduced. 3 A fixed support of the beam is only allowed at the end nodes of thee beam structu th structure, re, not at nod nodes es within within the str struct ucture ure.. In this case, divide the structure into two parts. 4 A structure with a revolute revolute joint between between two sections sections is not solvable. The development development group is endeavour endeavour to satisf satisfy y the customer wishes and to extend the BEAM code.


Syst Sy stem em Re Requ quire ireme ment ntss BEAM V 3.3 has a line oriented user interface and is coded in FORTRAN TRA N 77. At present present BEAM runs on diff differen erentt UNI UNIX-p X-platf latforms orms,, PC and Macintosh. The SID file is portable over various platforms. The runtime memory is about 2 MB.

Hint:

Please note that both the input file of BEAM and the SID file have to be located in the same working directory.

BEAM:2.0 -14

System Requirements

BEAM:3.

Starting of BEAM

The following chapter describes how to use BEAM. The variants are divide div ided d in into to fou fourr cla classes sses:: a sim simple ple cantile cantileve verr beam, a beam with tw twoo sections,, a beam with elastic borders and concen sections concentrated trated masses at nodes, and a beam with measured stiffness.


Gene Ge nera rall Inf Info orma rmatio tion n Note at first, that the x-axis of the reference frame is identical to the beam’s centerline and second, that the y- and z-axis are perpendicular to the x-axis, like a Cartesian frame as shown in Fig. BEAM:2.1.1. The origin is defined at the coordinate x = 0, therefore, nodes with positive or negative values are allowed. The units of the beam data should be taken from the ISO norm: mass by [kg kg], ], length by [m [m], and time by [sec [sec]. ]. Th There erefor fore, e, the unit of the force is [N [N ] = [kgm/sec2 ]. The SID depend also on these units. If other units are used, the whole data set of the MBS should be changed to the new units. The code is limited to a maximum number of sections of the beam structure, number of modes, and number of desired markers for the MBS simula simulatio tion. n. The actual limita limitation tion is printed printed out in the head header er of the BEAM code as shown below and may be different from your installation.

Example: B E A M a prepr preproces ocessor sor for mod mode e sha shape pe ana analys lysis is of str straig aight ht bea beam m structures and an d ge gene nera rati tion on of th the e SI SID D fi file le fo for r MB MBS S co code des s Versio Vers ion n 3. 3.3 3 - Ma Marc rch h 19 1996 96 Copyri Cop yright ght by Osk Oskar ar Wal Wallra lrapp pp D-82234 D-822 34 Wessl Wessling ing ======================================================= ===================================== ================== Note: Not e: Mod Model el lim limite ited d to Max ax. . numb number er of of sect sectio ions ns = 10 Max. number of modes = 30 Max. nu number of of ma markers = 30

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

BEAM:3.2 -16


Example of a Simple Cantilever Beam

Exam Ex ample ple of of a Simp Simple le Can Canti tile leve verr Beam Beam

Model Description Let first consider a simple beam as shown in Fig.BEAM:3.2.1. Fig.BEAM:3.2.1. The beam is clamped to its left end, the right end is free. The beam’s data are given in Fig.BEAM:3.2.1. Fig.BEAM:3.2.1. Due to the homogeneous structure of the beam, only one section with two major nodes at the ends are required to describe describe the beam’s properties properties.. The nodes are meas measured ured by values values at the x-axis only. z

Marker

*

*

* x

Major node

L = 10 m

y

Figure BEAM:3.2.1: A simple cantilever beam

Data l e ng t h L = 10 m cr osssectionarea A = 0.0004 m2 aream are amom omen ento toff in inert ertia ia J x = 4 10 7 m4 J y = 2 10 7 m4 J z = 2 10 7 m4 Y oung smodulus E = 7 1010 N/m2 

∗ ∗ ∗ ∗

h e ig h t massdensity

= 0.0774 m = 30 3000 00 kg/m3

Shear She armo modu dulu luss

G = 2.5 1010 N/m2

−

−

−

∗

Input Data and Run of BEAM Before starting starting BEAM, a folder of your project should be created. Then BEAM can be started by execution of Beam. The prompt is as follows: Example: B E A M a prepr preproces ocessor sor for mod mode e sha shape pe ana analys lysis is of str straig aight ht bea beam m structures and an d ge gene nera rati tion on of th the e SI SID D fi file le fo for r MB MBS S co code des s Versio Vers ion n 3. 3.3 3 - Ma Marc rch h 19 1996 96 Copyri Cop yright ght by Osk Oskar ar Wal Wallra lrapp pp D-82234 D-822 34 Wessl Wessling ing ======================================================= ===================================== ================== Note No te: : Mo Mode del l limit limited ed to to


BEAM:3.2 -17

Max ax. . numb number er of of sect sectio ions ns = 10 Max. number of modes = 30 Max. nu number of of ma markers = 30 ===================================== ================== ===================================== ================== Test Tes t out output put on scr screen een(1) (1), , no tes test t out output put(0) (0) 0 Data of gen Data genera eral l bea beam m str struct ucture ure input interactive or from file with mode selections etc. or all al l da data ta from from file file and all all mo mode des s fo for r SI SID D or as op option 2 but SI SID fo for SI SIMPACK v5 v5.xx or Data Dat a of mea measur sured ed bea beam m str struct ucture ure input interactive or from file with mode selections etc. or all al l da data ta fr from om fi file le an and d al all l mo mode des s fo for r SI SID D or as opt ptio ion n 12 but but SI SID D for SIM SIMPA PAC CK v5 v5.x .xx x or Stop the program ? 0

(0) (1) (2) (2 ) (3)

(10) (11) (12) (1 2) (13 13) ) (99)

The code asks you at first about test prints. Test prints are important for detailed information on the computation only. Usually zero is used. Always, the natural frequencies and mode shape values are stored in the input file. The next prompt defines the input media. If there is no data input file for the beam properties, use the interactive modus, i.e. set 0, otherwise usee 1 or 2. Is the input us input 9, the code stops stops.. For input input 1, the code will will ask you about mode selection, consideration of geometric stiffening by incorporating geometric stiffening terms. The next block of input data describes the geometrical properties of the beam. This data can already exist on the beam input file or have to be told the program. Here, we have chosen the interactive modus. See the next program prompts and the inputs, where the data are taken from the chapter of Fig. BEAM:3.2.1. Example:

Provide Provid e mod model el des descri cripti ption, on, not to exc exceed eed 80 characters Simple Sim ple Beam: Beam: Dat Date e 19.03.9 19.03.94/O 4/OW W Numb Nu mber er of di diff ffer eren ent t be beam am se sect ctio ions ns (n (ns) s) = ?

BEAM:3.2 -18


1 === Wr === Writ ite e do down wn xx-po posi siti tion on of ea each ch no node de (node=ns+1) x-Posi x-Po siti tion on in [m [m] ] of 11-th th no node de = ? 0 x-Posi x-Po siti tion on in [m [m] ] of 22-th th no node de = ? 10 === Wri Write te dow down n mas mass, s, geo geomet metric ric & mat materi erial al properties For 1-t 1-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 3000 Cross Cro ss sec sectio tional nal are area(A a(A) ) [m* [m**2] *2] = ? 4e-4 Area Are a mom moment ents s of ine inerti rtia(J a(Jt,J t,Jy,J y,Jz) z) [mm [mm**4 **4] ] = ? 4e-7 4e7 2e2e-7 7 2e2e-7 7 Young’ You ng’s s mod modulu ulus(E s(E) ) [N/ [N/m** m**2] 2] = ? 7e10 Shear She ar mod modulu ulus(G s(G) ) [N/ [N/m** m**2] 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 ends at x = 0.0 and x = 10 10m m. For this sec section tion the mass dens densit ity y, the cross section area, the area moments of inertia about the x-, y-, and z-axis are required. Additionally, Young’s modulus and Shear modulus are needed measured meas ured in the correspondi corresponding ng uni units. ts. For thi thiss sect section, ion, the value aluess are constant. The next block describes the boundary b oundary conditions, conditions, additional node stiffnesss an nes and d nod nodal al masses. masses. Thi Thiss bl block ock can also be wr writ itten ten on the beam input inp ut file. The program program prom prompts pts and the inputs are: Example:

Sum of no Sum node des s wi with th bo boun unda dary ry co cond ndit itio ions ns(n (nb) b) = ? 1 Sum of nod nodes es wit with h add additi itiona onal l mas masses ses(na (nam) m) = ? 0 Sum of nod nodes es wit with h add additi itiona onal l spr spring ings(n s(nas) as) = ? 0 === Wri Write te dow down n bou bounda ndary ry con condit dition ions s

BEAM:3.2 -19


For 1-t 1-th h bou bounda ndary ry con condit dition ion Node Nod e no. no., , tx, tx,ty, ty,tz, tz,rx, rx,ry, ry,rz( rz(fre free e = 0, fix fixed ed = 1) 1 1 1 1 1 1 1

As shown in Fig. BEAM:3.2.1 BEAM:3.2.1,, the beam structure has one boundary, no additional nodal masses and no addition additional al spring suspensions. For a node up to 6 constrai constraints nts can be defined. Here, at node one all directions are locked - three translational motions and three rotational motions about the reference frame. Additional masses and springs at nodes will be 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 computed and the corresponding modal damping damping rates. The program prompts and the inputs are: Example:

=== == = No No. .

of modes modes in longi longitu tudi dina nal( l(x) x)

vibration, in y-ben y-bending ding vibration, vibration, in z-b z-bend ending ing vib vibrat ration ion, , and in to tors rsio iona nal l vi vibr brat atio ion n = ? 2 3 3 2 === Dam Dampin ping g rat ratio io ( pos. pos. va valu lues es mea mean n prop propor ort. t. st stif iffn fnes ess, s, neg. ne g. va valu lues es mean mean na natu tura ral l da damp mpin ing g ) for longi longitudin tudinal al vibra vibration tion, , y-be ybend ndin ing g vibr vi brat atio ion, n, z-be zbend ndin ing g vibr vi brat atio ion, n, torsional vibration = ? 0.01 0.02 0.02 0.001

Here two modes in longitudinal direction, three bending modes in y and z direction, as well as two torsional mode should be computed. The modal damping coefficients will be multiplied with the stiffness term to get a damping value in the equation of deformations of the MBS. All modes of one vibration direction are multiplied by the same coefficient. For the evaluation of the SID, which are used in the MBS code, additional nodes - here denoted by mark markers ers - (often also called attachment attachment points) are required in the MBS simulation, e.g. for evaluations of the kinematic, attachment points of joints, action points of forces, etc. So, within the range of the major nodes, markers for the MBS simulation have to be defined now. They may have also y and z values. Example:

=== Writ === Write e do down wn th the e to tota tal l nu numb mber er > 0 of ma mark rker ers s for fo r MB MBS S code code 3 For Fo r al all l ma mark rker ers: s: na name me (max (max 8 ch char ar in st stri ring ngs) s)

BEAM:3.2 -20


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 - position of 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 can be incorporated in the bending motion, when the key is on. The default is zero. For the computation of the modes shapes, a nonlinear equation in eigenvalues eigenvalues has to be solv solved. ed. For this iterati iteration on solution written by KOLLOUSEK, starting values have to be computed by SOTIROPULUS. For both methods, iteration boundaries epskol and epssot have to be defined. Additionally, elements in the mass integrals of the equations of deformation can be set to zero when the value is smaller than an borde b orderr epsm epsmass ass.. Usu Usuall ally y, the defaults defaults can be used used,, then the input is zero, see below. Example:

=== Four key keys s to con consid sider: er: mass mass mom mom.o. .o.ine inerti rtia a for ben bendin ding g (0/ (0/1), 1), boundaries bound aries epskol for mode comp computati utations, ons, boundaries bound aries epssot for mode comp computati utations, ons, epsm ep smas ass s fo for r ze zero ro ma mass ss el elem emen ents ts ? Zero Ze ro in inpu put t me mean ans s de defa faul ults ts = 0 1. 1.00 000E 0E-6 -6 1.000E-3 1.000 E-3 1.00 1.000E-6 0E-6 0 0 0 0

Because the interactive modus is used, BEAM ask the user to save the data in a inpu inputt file for corr correcti ections ons and addi additio tional nal computati computations. ons. The program prompts and inputs are: Example:

Save of inp Save input ut dat data a and pro progra gram m con contin tinuat uation ion 0 = co comp mput ute e wi with thou out t sa savi ving ng da data ta 1 = co comp mput ute e wi with th sa savi ving ng da data ta 3 = save data and exit 4 = do not save data and exit 1 Name of th Name the e in inpu put t da data ta fi file le wi with thin in 60 characters in ca case se of s> fi file le na name me = be beam am.d .dat at is assumed > Data Da ta ar are e sa save ved d on fi file le = be beam am.d .dat at


BEAM:3.2 -21

With the current input data, BEAM computes at first the desired eigenvalues for the defined vibration directions. The code will print the following statements and results given by the mode shape description and natural frequenci frequencies: es: Example:

********* ****** *** Mod Mode e Sha Shape pe Cal Calcul culati ation on of Bea Beam m Structure Struc ture ***** ********* **** Longitudi Longit udinal nal ana analys lysis is in x dir direct ection ion ==================================== Bending Bendin g ana analys lysis is in y dir direct ection ion ================================ Bending Bendin g ana analys lysis is in z dir direct ection ion ================================ ------------------------------------------------| | | Mode Mode numbe number r Ty Type pe of mode mode Na Natu tura ral l frequ frequen ency cy[H [Hz] z]| | |-------------------------------------------------| | | | 1 Longitudinal 1 120.76 | | 2 Longitudinal 2 362.28 | | 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 | ------------------------------------------------Resu Re sult lts s ar are e (a (als lso) o) li list sted ed in fi file le: : be beam am.d .dat at

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

=== wr === writ ite e do down wn th the e to tota tal l nu numb mber er of mo mode des s to be chosen 4 === write dow down n 3 dif differ ferent ent mod mode e num number bers s selected 3 6 9 1

BEAM:3.2 -22


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

Note that the serious of the selected number will be taken for the serious of the SID file. After these inputs, BEAM calculates the modal mass integrals, modal mass matrix, modal stiffness matrix, and modal damping matrix. The stiffness calculations can be done with or without consideration of geometric stiffening terms, see the prompts and program input below. Geometric stiffening has to be added if the structure is very flexible and the loads are high, therefore simulations will appear in the range of buckling or stiffening due to these loads.

Example:

*** *** *** *** **

Mas Ma ss Ter erm ms Cal alc cul ula ati tion on

*** ** *** **** *** *

*** **** **** *** *

Sti tif ffn fne ess Ca Cal lcu cul lat ati ion ons s

*** **** **** *** *

Choice of geome Choice geometric tric stiff stiffening ening calcu calculati lations: ons: 0=no, 1=yes 1 *** sti stiffn ffness ess mar marke ker r 1 ** *** * *** sti stiffn ffness ess mar marke ker r 2 ** *** * *** sti stiffn ffness ess mar marke ker r 3 ** *** * *** ** * st stif iffn fnes ess s ax ** *** * *** sti stiffn ffness ess acce ac cel. l. ** *** *

calcul cal culati ation on due to for force ce at calcul cal culati ation on due to for force ce at calcul cal culati ation on due to for force ce at calc ca lcul ulat atio ion n du due e to lo long ng. .

acce ac cel. l.

calcul cal culati ation on due to cen centri trifug fugal al

The geom geometri etricc sti stiffne ffness ss matr matrice icess are cal calcula culated ted for all possi p ossible ble unit loads: loa ds: for lon longit gitudin udinal al forc forces es at all markers, markers, a lon longitu gitudin dinal al acce acceler leraation ax of the reference frame, and for a centrifugal acceleration due to rotations rotations about y- and z-ax z-axis is of the reference reference frame. frame. Her Here, e, in thi thiss model, geometric stiffening is included. The last step of BEAM is to save the MBS data on a SID file, if the user willl choose it. The name of the file is an inp wil input ut data, see the dia dialogu loguee below. Example:

Generatio Genera tion n of SID fil file? e? (0 = no, 1 = for SIMPACK 6.xx, 2 = for SIMPACK 5.xx) 1 Prov Pr ovid ide e th the e na name me of th the e SI SID D fi file le. . It wi will ll be ad adde ded d by .S .SID ID Be Beam am blanks bla nks mea means ns def defaul ault t m> >

BEAM:3.2 -23


SID SI D fi file le is denot denoted ed: :

SID SI D Be Beam am

The SID fil file e is suc succes cessfu sfully lly wri writte tten n FORTRAN FORTR AN STOP

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

Input file of BEAM Please check the data of the BEAM input file, if a saving of the data was chosen. chosen. All data of the beam, the natural frequencies and the mode shapes results are stored on the BEAM input file - see the printed beam input file ¡beam.dat¿ next. Note that phi(1,2,3) are the displacements in x, y, z -axis, psi(1,2,3) are rotation angle about x, y, z -axis. Example:

Simple Beam: Date 19.03.94/OW Model Descr Descriptio iption n 1 Number Num ber of sec sectio tions ns (ns (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 Cro ross ss sec ecti tion ona al are area( a(A) A) of sec sect t. 1 4.0E-07 2.0E-07 2.0E-07 Area Are a mom moment ent of ine inerti rtia a (Jt (Jt,Jy ,Jy,Jz ,Jz) ) of sec sect.1 t.1 7.000000E+10 Young’s Modulus (E) of sect. 1 2.500000E+10 Shear Modulus (G) of sect. 1 1 Number Num ber of nod nodes es wit with h bou bounda ndary ry con cond. d. 0 Number Num ber of nod nodes es wit with h add additi itiona onal l mas masses ses 0 Number Num ber of nod nodes es wit with h add additi itiona onal l spr spring ings s 1 1 1 1 1 1 1 Node no., tx,ty tx,ty,tz,r ,tz,rx,ry, x,ry,rz(fr rz(free=0, ee=0, fixe fixed=1) d=1) of bou bound nd.c .con ond. d. 1 2 3 3 2 No. of mode modes s for for lon ong g., ben bend. , bend bend.z .z, , & torsion 0.10000E-0 0.100 00E-01 1 0.200 0.20000E-0 00E-01 1 0.20 0.20000E000E-01 01 0.10 0.10000E000E-02 02 Dampin Dam ping g rat ratio io for lon long., g., ben bend.y d.y, , ben bend.z d.z,& ,& torsion

!! !! !! !! !! !! !! !! !! !! !! !! !!

!!

!!

BEAM:3.2 -24


3 numb nu mber er > 0 of ma mark rker ers s fo for r MB MBS S co code de ’m1 ’ .00000 .00000 .00000 marker marker descr descriptio iption n and x-pos x-position ition ’m2 ’ 5.000 .00000 .00000 marker marker descr descriptio iption n and x-pos x-position ition ’m3 ’ 10.000 .00000 .00000 marker marker descr descriptio iption n and x-pos x-position ition 0 1.00000E-06 1.00000E-03 1.00000E-06 Inerti Ine rtia a key key, , eps epskol kol, , eps epssot sot, , eps epsmas mass s

!! !! !! !! !!

------------------------------------------------| | Mode shapes computed by BEAM - V 3.3 | ------------------------------------------------| | Mo Mode de nu numb mber er Ty Type pe of mo mode de Na Natu tura ral l frequency[Hz]| ------------------------------------------------| 1 Longitudinal 1 120.76 | 2 Longitudinal 2 362.28 | 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 mo mode des s ar are e se sele lece cet t fo for r MB MBS S da data ta ge gene nera rati tion on: : They The y are are: : Bending y 1 Bending z 1 Torsion 1 Lon ongi gitu tud din ina al 1

Mode Mo de sh shap apes es fo for r mo mode de i an and d ma mark rker er 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

BEAM:3.2 -25


3 3 3 4 4 4

1 .00 2 5.00 3 10.00 1 .00 2 5.00 3 10.00

.00 .00 .00 .00 .00 .00

.00 .00 .00 .00 .00 .00

.0000 .0000 .0000 .0000 .2887 .4082

.0000 .0000 .0000 .0000 .0000 .0000

-.0000 -.0000 -.0000 .0000 .0000 .0000

SID File of the Simple Beam An outline of the SID-file written by BEAM is listed below. The generated SID file is an ASCII file and is readable using an standard editor. Hint:

Example:

It should be noted that changes on the file can disturb the readability of SID by SIMPACK. 3 4 = No. nodes & No. modes of mod model =Si =Simple Beam: Bea m: Dat Date e 19.0 19.03.9 3.94/O 4/OW W |SID |SI D gen genera erated ted by BEA BEAM-V M-V.3. .3.3 3 inc includ luding ing Geo Stiff=yes Stiff =yes from Beam input file=beam.dat file=beam.dat 15-Aug-96; 15-Au g-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) 4) = Longitudinal 1 end ref refmod mod frame new node = m1 >>> marker m1 rframe = body ref origin order = 1 nrow = 3 ncol = 1 nq = 4 nqn = 0 structur = 0 end ori origin gin phi order = 1 nrow = 3 ncol = 4 nq = 4 nqn = 0 structur = 0 end en d ph phi i psi

.0000 9.1287 12.9099 .0000 .0000 .0000

BEAM:3.2 -26


order nrow ncol nq nqn structur end en d ps psi i AP order nrow ncol nq nqn structur end en d AP end nod node e ... ... new node rframe origin order nrow ncol nq nqn structur m0( 1, 1) in x for m3 m1( 1, 4, 1) m1( 2, 1, 1) m1( 3, 2, 1) end ori origin gin phi order nrow ncol nq nqn structur m0( 1, 4) m0( 2, 1) m0( 3, 2) m1( 1, 1, 1) m1( 1, 2, 2) end en d ph phi i psi order nrow ncol

= = = = = =

0 3 4 4 0 0

= = = = = =

0 3 3 4 0 4

= m3 = body ref = 1 = 3 = 1 = 4 = 0 = 3 = 1.00000000000D+01 position = 4.08248290463D-01 = 5.77350269190D-01 = 5.77350269190D-01

= = = = = =

1 3 4 4 0 3 = 4.08248290463D-01 = 5.77350269190D-01 = 5.77350269190D-01 =-3.87314860533D-02 =-3.87314860533D-02

= 0 = 3 = 4


BEAM:3.2 -27

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 en d ps psi i 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 en d AP end nod node e end fra frame me mdCM order = 1 nrow = 3 ncol = 1 nq = 4 nqn = 0 structur = 3 m0( 1, 1) = 6.00000000000D+01 center of ma mass ss m1( 1, 4, 1) = 3.11878720494D+00 m1( 2, 1, 1) = 2.71236300674D+00 m1( 3, 2, 1) = 2.71236300674D+00 end mdC mdCM M 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

BEAM:3.2 -28


m1( 2, 4, 1) m1( 3, 4, 1) m1( 4, 1, 1) m1( 5, 2, 1) end J Ct order nrow ncol nq nqn structur m0( 1, 2) m0( 2, 3) m0( 4, 1) m1( 1, 1, 1) m1( 2, 2, 1) end en d Ct Cr order nrow ncol nq nqn structur m0( 1, 3) m0( 2, 2) m0( 3, 1) m1( 1, 2, 1) m1( 2, 1, 1) m1( 2, 4, 2) m1( 4, 2, 2) m1( 1, 4, 3) m1( 4, 1, 3) end en d Cr Me order nrow ncol nq nqn structur m0( 1, 1) mass m0( 2, 2) m0( 3, 3) m0( 4, 4) end en d Me Gr order

= 3.97096320094D+01 = 3.97096320094D+01 =-1.97040089549D+01 =-1.97040089549D+01

= = = = = =

1 4 3 4 0 3 = 2.71236300674D+00 = 2.71236300674D+00 = 3.11878720494D+00 =-1.57087820334D-01 =-1.57087820334D-01

= = = = = =

1 4 3 4 0 3 = 1.97047017752D+01 =-1.97047017752D+01 = 9.86247110499D-02 =-1.00000000000D+00 = 1.00000000000D+00 =-9.58641445425D-01 = 9.58641445425D-01 = 9.58641445425D-01 =-9.58641445425D-01

= 0 = 4 = 4 = 4 = 0 = 1 = 1.00000000000D+00 modal = 1.00000000000D+00 = 1.00000000000D+00 = 1.00000000000D+00

= 0

BEAM:3.2 -29


nrow ncol nq nqn structur m0( 2, 1) m0( 2, 8) m0( 3, 2) m0( 3,12) end en d Gr Ge order nrow ncol nq nqn structur m0( 1, 2) m0( 1,12) m0( 2, 1) m0( 2, 8) m0( 4, 6) m0( 4, 9) end en d Ge Oe order nrow ncol nq nqn structur m0( 1, 4) m0( 2, 6) m0( 4, 2) m0( 4, 3) m1( 1, 1, 1) m1( 2, 2, 1) m1( 1, 1, 2) m1( 2, 2, 2) m1( 4, 4, 2) m1( 1, 1, 3) m1( 2, 2, 3) m1( 4, 4, 3) m1( 1, 4, 4) m1( 4, 1, 4) m1( 1, 2, 5) m1( 2, 1, 5) m1( 2, 4, 6) m1( 4, 2, 6) end en d Oe

= 3 =12 = 4 = 0 = 3 =-3.94094035503D+01 = 3.97096320094D+01 =-3.94094035503D+01 = 3.97096320094D+01

= 0 = 4 =12 = 4 = 0 = 3 =-2.00000000000D+00 = 1.91728289085D+00 = 2.00000000000D+00 =-1.91728289085D+00 = 1.91728289085D+00 =-1.91728289085D+00

= = = = = =

1 4 6 4 0 3 = 1.97040089549D+01 = 1.97040089549D+01 =-1.98548160047D+01 =-1.98548160047D+01 =-1.00000000000D+00 =-1.00000000000D+00 = 1.19333638409D+00 = 1.93336384089D-01 =-1.00000000000D+00 = 1.93336384089D-01 = 1.19333638409D+00 =-1.00000000000D+00 = 9.58641445425D-01 = 9.58641445425D-01 = 1.00000000000D+00 = 1.00000000000D+00 = 9.58641445425D-01 = 9.58641445425D-01

BEAM:3.2 -30


ksigma order nrow ncol nq nqn structur end ksi ksigma gma Ke order nrow ncol nq nqn structur m0( 1, 1) stiffness m0( 2, 2) m0( 3, 3) m0( 4, 4) end en d Ke De order nrow ncol nq nqn structur m0( 1, 1) damping m0( 2, 2) m0( 3, 3) m0( 4, 4) end en d De end mod modal al end par part t

= = = = = =

0 4 1 4 0 0

= 0 = 4 = 4 = 4 = 0 = 1 = 1.44227572630D+01 modal = 1.44227572630D+01 = 2.05616758357D+05 = 5.75726923399D+05

= 0 = 4 = 4 = 4 = 0 = 1 = 2.88455145261D-01 modal = 2.88455145261D-01 = 2.05616758357D+02 = 5.75726923399D+03

The outline is supplemented by comments to give an interpretation of the entities and the structure of the file. At the beginning of the file the four natural natural frequenci frequencies es of the selected selected modes can be reco recogni gnised. sed. The coordinates for the markers (in SID are called nodes) are relevant for the simulation with SIMPACK. See marker m3 for an exampl example. e. Als Alsoo the 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 Usuall y the use userr should not be bother bothered ed with the SID-file. SID-file. The only entities which are relevant to the user in connection with the set-up of the MBS-model are the coordinates of the nodes of the FEM-structure. Congratulations on finishing your first BEAM example!

BEAM:3.3 -31

Example of a Beam with Sections Using Different Geometry

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

BEAM BE AM:3 :3.3 .3

Exampl Exam ple e of of a Bea Beam m wit with h Sec Secti tion onss Usi Using ng Different Geometry As mentioned before, a beam section has constant beam parameters. Therefore, a structure with variable mass and geometrical properties has to be subdivided (and approximated) into a finite number of sections. tio ns. The example example as shown shown in BEAM:3.3.2 is a simplified model of a hel helicop icopter ter rotor blade. blade. The blade is modelled modelled by four sections sections:: The first section represents the arm of the blade, the next two sections the transition and the last section the blade area. The data of the example are given in BEAM:3.3.2. Here, the bending motions in y-direction are of interest only. The aim of this simulation is to show the influence of the rotation Ω about the y-axis on the bending motion. Section 1

*

2 3

x

*

Section 4

*

*

y z x

Figuree BE Figur BEAM AM:3 :3.3 .3.2 .2:: Be Beam am mod model el of an al alim imin iniu ium m hel helic icop opter ter rot rotor or blade

Sectioon Secti n 1 2 3 4

Len engt gth h [m] 0.60 0.12 0.08 4.00

Data of the helicopter rotor blade Heigh Hei ghtt Are rea a Are rea a m. m.o. o. Ine Inert rt.. Mas Masss De Dens nsit ityy ρ Y ou oun ng s Modulus 2 4 3 [cm cm]] [cm ] J z [cm ] [Kg/m ] [N/m2 ] 6.0 42 126.0 2 787 7.13 e10 5.0 50 104.2 2 787 7.13 e10 3.4 51 49.0 2 787 7.13 e10 2.5 50 26.0 2 787 7.13 e10 

Additi Add itiona onal l data:

Shear She ar modulus modulus G = 2.5 E10, E10, Area m.o.Iner m.o.Inert. t.

Jx and Jy are given by 0,

∗ ∗ ∗ ∗

becaus bec ause e the data are unknown unknown and not required required in the computati computation on for ben bendin ding g in y-di y-direc rectio tion n

The input data for BEAM are listed below, where four sections with fivee ma jor nodes are in fiv introduc troduced. ed. More Moreov over, er, four markers markers at 0.0 0.0,, 0.8, 2.8, and 4.8 m are set for the SID file. Example:

Provid Pro vide e mod model el des descri cripti ption, on, not to exc exceed eed 80

BEAM:3.3 -32


characters Helicopter Helic opter rotor Blade Blade, , y-be y-bending nding, , with geo.Stiff Number Numb er of di diff ffer eren ent t be beam am se sect ctio ions ns (n (ns) s) = ? 4 === Wr === Writ ite e do down wn xx-po posi siti tion on (#node=ns+1) x-Po xPosi siti tion on of 11-th th no node de = 0 x-Po xPosi siti tion on of 22-th th no node de = 0.6 x-Po xPosi siti tion on of 33-th th no node de = 0.72 x-Po xPosi siti tion on of 44-th th no node de = 0.8 x-Po xPosi siti tion on of 55-th th no node de = 4.8

of ea each ch no node de ? ? ? ? ?

=== Wri Write te dow down n mas mass, s, geo geomet metric ric & mat materi erial al properties For 1-t 1-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 2787 Cross Cro ss sec sectio tional nal are area(A a(A) ) = ? 0.0042 Area Are a mom moment ents s of Ine Inerti rtia(J a(Jx,J x,Jy,J y,Jz) z) = ? 0 0 12 126e 6e-8 -8 Young’ You ng’s s mod modulu ulus(E s(E) ) = ? 7.31e10

Shear modulus(G) = ? 2.5e10 For 2-t 2-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 2787 Cross Cro ss sec sectio tional nal are area(A a(A) ) = ? 0.0050 Area Are a mom moment ents s of Ine Inerti rtia(J a(Jx,J x,Jy,J y,Jz) z) = ? 0 0 10 104. 4.2e 2e-8 -8 Young’ You ng’s s mod modulu ulus(E s(E) ) = ? 7.31e10 Shear She ar mod modulu ulus(G s(G) ) = ? 2.5e10 For 3-t 3-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 2787 Cross Cro ss sec sectio tional nal are area(A a(A) ) = ?


BEAM:3.3 -33

0.0051 Area Are a mom moment ents s of Ine Inerti rtia(J a(Jx,J x,Jy,J y,Jz) z) = ? 0 0 49e-8 Young’ You ng’s s mod modulu ulus(E s(E) ) = ? 7.31e10 Shear She ar mod modulu ulus(G s(G) ) = ? 2.5e10 For 4-t 4-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 2787 Cross Cro ss sec sectio tional nal are area(A a(A) ) = ? 0.0050 Area Are a mom moment ents s of Ine Inerti rtia(J a(Jx,J x,Jy,J y,Jz) z) = ? 0 0 26e-8 Young’ You ng’s s mod modulu ulus(E s(E) ) = ? 7.31e10 Shear She ar mod modulu ulus(G s(G) ) = ? 2.5e10 ========================================== Sum Su m of no node des s wi with th bo boun unda dary ry co cond ndit itio ions ns(n (nb) b) = ? 1 Sum Su m of no node des s wi with th ad addi diti tion onal al ma mass sses es(n (nam am) ) = ? 0 Sum Su m of no node des s wi with th ad addi diti tion onal al sp spri ring ngs( s(na nas) s) = ? 0

=== Write down boundary conditions For 1-t 1-th h bou bounda ndary ry con condit dition ion Node Nod e no. no., , tx, tx,ty, ty,tz, tz,rx, rx,ry, ry,rz( rz(fre free e = 0, fix fixed ed = 1) 1 1 1 1 1 1 1 No. of modes modes in lon longit gitudi udinal nal(x) (x) vibrat vibration ion, , in y-ben y-bending ding vibra vibration, tion, in z-b z-bend ending ing vib vibrat ration ion, , and in to tors rsio iona nal l vi vibr brat atio ion n = ? 0 7 0 0 === Dam Dampin ping g rat ratio io ( pos. pos. va valu lues es mea mean n prop propor ort. t. st stif iffn fnes ess, s, neg. ne g. va valu lues es mean mean na natu tura ral l da damp mpin ing g ) for longi longitudin tudinal al vibra vibration tion, , y-be ybend ndin ing g vibr vi brat atio ion, n, z-be zbend ndin ing g vibr vi brat atio ion, n, torsional vibration = ? 0 0.03 0 0

BEAM:3.3 -34


=== Wr === Writ ite e do down wn th the e to tota tal l nu numb mber er > 0 of ma mark rker ers s for fo r MB MBS S co code de 4 For Fo r al all l ma mark rker ers: s: na name me (max (max 8 ch char ar in st stri ring ngs) s) 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 === Fo === Four ur ke keys ys to co cons nsid ider er: : mass mass mom mom.o. .o.ine inerti rtia a for ben bendin ding g (0/ (0/1), 1), boundaries bound aries epskol for mode comp computati utations, ons, boundaries bound aries epssot for mode comp computati utations, ons, epsm ep smas ass s fo for r ze zero ro ma mass ss el elem emen ents ts ? Zero Ze ro in inpu put t me mean ans s de defa faul ults ts = 0 1. 1.00 000E 0E-6 -6 1.000E-3 1.000 E-3 1.00 1.000E-6 0E-6 0 0 0 0 Example:

======================================================== ====================================== ================== Save Sav e of inp input ut dat data a and pro progra gram m con contin tinuat uation ion 0 = co comp mput ute e wi with thou out t sa savi ving ng da data ta 1 = co comp mput ute e wi with th sa savi ving ng da data ta 3 = save data and exit 4 = do not save data and exit 1 Name of th Name the e in inpu put t da data ta fi file le wi with thin in 60 characters in ca case se of s> fi file le na name me = be beam am.d .dat at is assumed hrotor.dat

Now for the rotor blade, BEAM calculates the following seven natural frequencies, where only the first four modes are taken into account in the SID generation. Example:

------------------------------------------------Mode nu Mode numb mber er Ty Type pe of mo mode de Natu Na tura ral l fr freq eque uenc ncy[ y[Hz Hz] ] ------------------------------------------------1 Bending y 1 1.1553 2 Bending y 2 6.9340 3 Bending y 3 18.589

Example of a Beam with Concentrated Mass and Spring

BEAM:3.4 -35

4 Bending y 4 34.995 5 Bending y 5 56.565 6 Bending y 6 84.193 7 Bending y 7 118.32 ------------------------------------------------Resu Re sult lts s ar are e (a (als lso) o) li list sted ed in fi file le: :

hrot hr otor or.d .dat at

=== wr === writ ite e do down wn th the e to tota tal l nu numb mber er of mo mode des s to be cho chosen sen 4 === write dow down n 4 dif differ ferent ent mod mode e num number bers s selected 1 2 3 4

After the last data input, BEAM calculates the mass and stiffness matrices and if desired, the results are saved on the SID file. This part of computations and the program dialogue has been described in section BEAM:3.2.

BEAM BE AM:3 :3.4 .4

Exampl Exam ple e of of a Bea Beam m wit with h Con Conce cent ntra rate ted d Mass and Spring From thi thiss exa examp mple le the us user er sh shoul ould d le learn arn,, ho how w to use con concen centra trated ted mass ma sses es or sp spri ring ngss at maj major or nod nodes es.. In the ca case se that there there is a co conncentrated mass on the beam, a major node must be introduced at the masss poin mas point. t. Add Additi itional onally ly,, conc concent entrate rated d sti stiffne ffness ss sus suspensi pensions ons suc such h as translational or rotational springs have to be considered in a similar way. The example as illustrated in Fig. BEAM:3.4.3 shows a flexible rod with a 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. Thee beam has one se Th secti ction on with two two majo majorr nod nodes. es. For the mod model el all vibra vi brati tion on di direc recti tions ons are of in inter teres est: t: He Here, re, in thi thiss exa examp mple le we wa want nt to calculate 2 modes in x-direction, 2 bending modes in y-direction, 4 ben bendi ding ng mod modes es in z-direc z-directi tion on and 2 tor torsi sion onal al mod modes. es. Mar Mark kers are defined 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 are used: [kg [kg]] for mass, [cm [cm]] for length, and [kgcm/s [kgcm/s2 ] = N/ N/100] 100] for forces. 2 Thus, the Young’s modulus of 2. 2.1e11 11N/m N/m represents 2. 2.1e9kg/cms2 .

BEAM:3.4 -36


*

x

h = 0.05 cm

b = 1.32 cm

*

z

Tip Mass

*

y L = 1 4 4 c m m

Figure BEAM:3.4.3: Flexible rod with tip mass

Data l e ng t h cr osssectionarea area ar eamo mome ment ntofin ofinert ertia ia Y oungsmodulus massdensity tipma tip mass ssM M om om.o .o.Iner .Inertia tia tipma tip mass ssM M om om.o .o.Iner .Inertia tia

L = 14 cm A = 0.066 cm2 I y = 1.375 375ee 6 cm4 E = = 2.1e + 9 N/cms2  = 7.85 85ee 3 kg/cm3 I x = 2.2e 2 kgcm2 I z = 1.25 25ee 2 kgcm2

−

− − −

height h = 0.05 cm area ar eamo mome ment ntofin ofinert ertia ia I x = 9.597 597ee 3 cm4 area ar eamo mome ment ntofin ofinert ertia ia I z = 9.583 583ee 3 cm4 S hearmodulus G = 0.8e + 9kg/cms 9kg/cms2 tipmass m = 0.026 026kg kg tipm ti pmas assMom. sMom.o. o.Inerti Inertia a I y = 1.25 25ee 2kgcm2

The input data for an interactive session is given below. Example:

Provide Provid e mod model el des descri cripti ption, on, not to exc exceed eed 80 characters Flex Fl exib ible le rod wi with th tip ma mass ss incl. incl. Ge Geom omet etri ric c Stiffening Number Numb er of di diff ffer eren ent t be beam am se sect ctio ions ns (n (ns) s) = ? 1 === Wr === Writ ite e do down wn xx-po posi siti tion on of ea each ch no node de (#node=ns+1) x-Po xPosi siti tion on of 11-th th no node de = ? 0 x-Po xPosi siti tion on of 22-th th no node de = ? 14 === Wri Write te dow down n mas mass, s, geo geomet metric ric & mat materi erial al properties For 1-t 1-th h sec sectio tion n Mass Ma ss de dens nsit ity y [k [kg/ g/m* m**3 *3] ] = ? 7.85e-3 Cross Cro ss sec sectio tional nal are area(A a(A) ) = ? 0.066 Area Are a mom moment ents s of Ine Inerti rtia(J a(Jx,J x,Jy,J y,Jz) z) = ? 9.597e-3 9.597 e-3 1.37 1.375e-6 5e-6 9.58 9.583e-3 3e-3 Young’ You ng’s s mod modulu ulus(E s(E) ) = ?

− −

−


BEAM:3.4 -37

2.1e9 Shear She ar mod modulu ulus(G s(G) ) = ? 0.8e9 ========================================== Sum Su m of no node des s wi with th bo boun unda dary ry co cond ndit itio ions ns(n (nb) b) = ? 1 Sum Su m of no node des s wi with th ad addi diti tion onal al ma mass sses es(n (nam am) ) = ? 1 Sum Su m of no node des s wi with th ad addi diti tion onal al sp spri ring ngs( s(na nas) s) = ? 0 === Wri Write te dow down n bou bounda ndary ry con condit dition ions s For 1-t 1-th h bou bounda ndary ry con condit dition ion Node Nod e no. no., , tx, tx,ty, ty,tz, tz,rx, rx,ry, ry,rz( rz(fre free e = 0, fix fixed ed = 1) 1 1 1 1 1 1 1 === Wri Write te dow down n add additi itiona onal l mas masses ses For 1-t 1-th h add additi itiona onal l mas mass s Node No de no no., ., ma mass ss, , ma mass ss mo mome ment nt of inertia(Ix,Iy,Iz) 2 0.0 0.026 26 2.2 2.2e-2 e-2 1.2 1.25e5e-2 2 1.2 1.25e5e-2 2 No. of modes modes in lon longit gitudi udinal nal(x) (x) vibrat vibration ion, , in y-ben y-bending ding vibra vibration, tion, in z-b z-bend ending ing vib vibrat ration ion, , and in to tors rsio iona nal l vi vibr brat atio ion n = ? 2 2 4 2 === Dam Dampin ping g rat ratio io ( + va valu lues es = pr prop opor ort. t. St Stif iffn fnes ess, s, - va valu lues es = Lehr ) for longi longitudin tudinal al vibra vibration tion, , y-be ybend ndin ing g vibr vi brat atio ion, n, z-be zbend ndin ing g vibr vi brat atio ion, n, torsional vibration = ? 0 0.02 0 0 === == = Wr Writ ite e do down wn th the e to tota tal l nu numb mber er > 0 of ma mark rker ers s for fo r MB MBS S co code de 3 For Fo r al all l ma mark rker ers: s: na name me (max (max 8 ch char ar in st stri ring ngs) s) and x, y, z - position marker : 1 ’m1’ 0 0 marker : 2 ’m2’ 7 0

BEAM:3.4 -38


marker : 3 ’m3’ 14 0 === == = Fo Four ur ke keys ys to co cons nsid ider er: : mass mass mom mom.o. .o.ine inerti rtia a for ben bendin ding g (0/ (0/1), 1), boundaries bound aries epskol for mode comp computati utations, ons, boundaries bound aries epssot for mode comp computati utations, ons, epsm ep smas ass s fo for r ze zero ro ma mass ss el elem emen ents ts ? Zero Ze ro in inpu put t me mean ans s de defa faul ults ts = 0 1. 1.00 000E 0E-6 -6 1.000E-3 1.000 E-3 1.00 1.000E-6 0E-6 0 0 0 0 ===================================== ================== ====================================== =================== Save Sav e of inp input ut dat data a and pro progra gram m con contin tinuat uation ion 0 = co comp mput ute e wi with thou out t sa savi ving ng da data ta 1 = co comp mput ute e wi with th sa savi ving ng da data ta 3 = save data and exit 4 = do not save data and exit 1 Name Na me of th the e in inpu put t da data ta fi file le wi with thin in 60 characters in ca case se of s> fi file le na name me = be beam am.d .dat at is assumed flexrul.dat Data Da ta ar are e sa save ved d on fi file le = fl flex exru rul. l.da dat t

With the defined data, BEAM calculates the natural frequencies as shown below, where the first four bending modes in z-direction and the first torsion mode are considered in the generation of the SID. All other modes 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. 3 Bending y 1 141.45 4 Bending y 2 2138.3 5 Bending z 1 1.6943 6 Bending z 2 25.614 7 Bending z 3 62.887 8 Bending z 4 127.16 9 Torsion 1 788.33 10 Torsion 2 11464. --------------------------------------------------Result Res ults s are (al (also) so) lis listed ted in fil file: e: flexrul.dat

BEAM:3.4 -39


From these results, the user can chose various mode shapes for evaluation of SID. Beam structure with spring suspensions

As mentioned before, beam structures attached by spring suspensions in differentt directions and various differen various major ma jor nodes can be modelled in BEAM. At the attachment point, a major node must be introduced. Then, the stiffness coefficient are given by the BEAM input data. Let us consider a translational spring in z-direction with a value of 100[N/cm 100[N/cm]] at node 2. Th Then, en, the BEA BEAM M di dial alogu oguee du duee to this spring spring may be ch chan anged ged as follows: Example:

========================================== Sum Su m of no node des s wi with th bo boun unda dary ry co cond ndit itio ions ns(n (nb) b) = ? 1 Sum Su m of no node des s wi with th ad addi diti tion onal al ma mass sses es(n (nam am) ) = ? 1 Sum Su m of no node des s wi with th ad addi diti tion onal al sp spri ring ngs( s(na nas) s) = ? 1 === Wri Write te dow down n bou bounda ndary ry con condit dition ions s For 1-t 1-th h bou bounda ndary ry con condit dition ion Node Nod e no. no., , tx, tx,ty, ty,tz, tz,rx, rx,ry, ry,rz( rz(fre free e = 0, fix fixed ed = 1) 1 1 1 1 1 1 1 === Wri Write te dow down n add additi itiona onal l mas masses ses For 1-t 1-th h add additi itiona onal l mas mass s Node No de no no., ., ma mass ss, , ma mass ss mo mome ment nt of inertia(Ix,Iy,Iz) 2 0.026 2.2e-2 1.25e-2 1.25e-2 === Wri Write te dow down n add additi itiona onal l spr spring ings s For 1-t 1-th h add additi itiona onal l spr spring ing Node no., tkx,t tkx,tky,tk ky,tkz,rkx z,rkx,rky, ,rky,rkz rkz 2 0 0 100 0 0 0

Referring to this additional spring, the natural frequencies of the bending motion motion in z-di z-direct rection ion will be ch change anged. d. The results results of BEA BEAM M no now w are shown below. Example:

----------------------------------------------------------------------------------------------------------Mode number Type of mode Natural frequency[Hz] ----------------------------------------------------------------------------------------------------------1 2 3

Longitudinal Longitudinal Bending y

1 2 1

2968.4 18979. 141.45

BEAM:3.5 -40

Beam Structure with Measured Torsional Stiffness Data

4 Bending y 2 2138.3 5 Bending z 1 9.6736 6 Bending z 2 25.614 7 Bending z 3 62.887 8 Bending z 4 127.16 9 Torsion 1 788.33 10 Torsion 2 11464. -------------------------------------------------------------------------------------------------------

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


Beam St Beam Struc ructu ture re wit with h Mea Measur sured ed Tors rsio iona nall Stiffness Data The torsional flexibility of a frame may be considered in the MBS simula sim ulation tion.. Ther Therefor efore, e, the torsional torsional stiffness stiffness GJT is requ required ired as the input data for BEAM. Here, this stiffness may be taken from an experiment, where the frame is twisted about an angle θ [rad rad]] due to a torque T [N m], see Fig. BEAM:3.5.4. For the frame as given in Fig.

T L

Figuree BEA Figur BEAM: M:3. 3.5. 5.4: 4: Exp Experi erime ment nt to find ou outt the tor torsi sion onal al st stiff iffnes nesss T 2 GJ T T = L theta N m /rad BEAM:3.5.5, a SID file should be now generated with respect to the following conditions: The body reference frame is in the middle middle of the frame. Therefor Therefore, e, node 1 is loc locate ated d at -5 m an and d node 2 at +5 m. The bounda boundary ry condi conditi tions ons are taken from the free-free borders. Markers are here located at some of the corners of the prismatic body in addition to markers at the centerline, see Fig. BEAM:3.5.5. The damping ratio via Lehr is 10%.

BEAM:3.5 -41


node 1

marker 3

z marker 2

y marker 1 b e ea m a ' s m s c e en t n t e r e rl l i i n e n e

a

b marker 6

L

x marker 5 node 2 marker 4

Figure BEAM:3.5.5: Frame structure with torsional flexibility

Data l en g t h torsi to rsion onals alsti tiff f ne ness ss Y oungs gsm modulus body mass mom.o.I ne ner tia mom.o.I ne ner tia

L = 10 m h e ig h t 2 GJ T width T = 0.066 N m /r ad 2 E = E = 21 21ee + 10 N/m Shear She arModu Modulu luss m = 200 K g S hear modulus Ix = 5.0 kg kgm m2 Mom.o. Mom. o.Iner Inerti tia a 2 Iz = 600. 600.0 kg kgm m

b = 0.2 m a = 0.5 m G = 7.8e + 10 N/m2 I y = 400. 400.0 kg kgm m2

In the case of the input of stiffness, the body mass data with respect to the reference reference frame are requ required ired.. The input data for an in intera teractiv ctivee session is listed below. Example: ===================================== ================== ===================================== ================== B E A M a p r e p r o c e s s o r for mod mode e sha shapes pes ana analys lysis is of str straig aight ht bea beam m structures and an d ge gene nera rati tion on of th the e SI SID D fi file le fo for r MB MBS S co code des s Versio Vers ion n 3. 3.3 3 - Ma Marc rch h 19 1996 96 Copyri Cop yright ght by Osk Oskar ar Wal Wallra lrapp pp D-82234 D-822 34 Wessl Wessling ing ======================================================= ===================================== ================== Note No te: : Mo Mode del l Li Limi mita tati tion ons s ar are e Max. number of sections = 10 Max. number of modes = 20 Max. number of markers = 10

BEAM:3.5 -42


===================================== ================== ===================================== ================== Test out Test output put on scr screen een(1) (1), , no tes test t out output put(0) (0) 0 Data of gen Data genera eral l bea beam m str struct ucture ure input interactive or from file with mode selections etc. or all al l da data ta from from file file and all all mo mode des s fo for r SI SID D or as op option 2 but SI SID fo for SI SIMPACK v5 v5.xx or Data Dat a of mea measur sured ed bea beam m str struct ucture ure input interactive or from file with mode selections etc. or all al l da data ta fr from om fi file le an and d al all l mo mode des s fo for r SI SID D or as opt ptio ion n 12 but but SI SID D for SIM SIMPA PAC CK v5 v5.x .xx x or Stop the program ? 10

(0) (1) (2) (2 ) (3)

(10) (11) (12) (1 2) (13 13) ) (99)

Provide Provid e mod model el des descri cripti ption, on, not to exc exceed eed 80 characters Frame with torsi torsional onal flex flexibili ibility, ty, 1.6.9 1.6.95 5 ===================================== ================== ================================== =============== Genera Gen eral l Bea Beam m Str Struct ucture ure Dat Data a Note the Note the fra frame me def defin init itio ions ns: : z y | / |/ +----+----- x = lon longit gitudi udinal nal Origin Ori gin is at the coo coordi rdinat nates es 0/0 0/0/0 /0 Node No des s ma may y be at th the e po posi siti tive ve an and d ne nega gati tive ve x-axis ===================================== ================== ================================== =============== === Wri Write te down down Numb Nu mber er of di diff ffer eren ent t be beam am se sect ctio ions ns (n (ns) s) = ? 1 === Wr === Writ ite e do down wn xx-po posi siti tion on of ea each ch no node de (#node=ns+1) x-Posi x-P ositio tion n of 1-t 1-th h nod node e =?


BEAM:3.5 -43

-5 x-Posi x-Po siti tion on of 22-th th no node de = ? 5 === Wri Write te dow down n mas mass, s, geo geomet metric ric & mat materi erial al properties Body mas Body mass s [kg [kg] ] 200 Body Bod y mas mass s mom moment ents s of ine inerti rtia a (Ix (Ix,Iy ,Iy,Iz ,Iz) ) [kg*m**2] 5 400 600 Stif St iffn fnes ess: s: lo long ngit itud udin inal al in x, be bend ndin ing g in y, bend be ndin ing g in z, to tors rsio ion n ab abou out t x 0 0 0 1e5 Young’ You ng’s s mod modulu ulus(E s(E) ) = ? 21e10 Shear She ar mod modulu ulus(G s(G) ) = ? 7.8e10 ========================================== Sum of nod nodes es wit with h bou bounda ndary ry con condit dition ions(n s(nb) b) = ? 0 Sum of nod nodes es wit with h add additi itiona onal l mas masses ses(na (nam) m) = ? 0 Sum of nod nodes es wit with h add additi itiona onal l spr spring ings(n s(nas) as) = ? 0 === No === No. . of modes modes in lo long ngit itud udin inal al(x (x) ) vibration, in y-ben y-bending ding vibra vibration, tion, in z-b z-bend ending ing vib vibrat ration ion, , and in to tors rsio iona nal l vi vibr brat atio ion n = ? 0 0 0 3 === Dam Dampin ping g rat ratio io ( pos. pos. va valu lues es mea mean n prop propor ort. t. st stif iffn fnes ess, s, neg. ne g. va valu lues es mean mean na natu tura ral l da damp mpin ing g ) for longi longitudin tudinal al vibra vibration tion, , y-be ybend ndin ing g vibr vi brat atio ion, n, z-be zbend ndin ing g vibr vi brat atio ion, n, torsional vibration = ? 0 0 0 -0.1

BEAM:3.5 -44


=== Wr === Writ ite e do down wn th the e to tota tal l nu numb mber er > 0 of ma mark rker ers s for fo r MB MBS S co code de 6 For Fo r al all l ma mark rker ers: s: na name me (max (max 8 ch char ar in st stri ring ngs) s) and x, y, z - position marker : 1 ’m1’ 0 0 0 marker : 2 ’m2’ ’m 2’ -5 -0 -0.5 .5 0. 0.2 2 marker : 3 ’m3’ ’m 3’ -5 0. 0.5 5 0. 0.2 2 marker : 4 ’m4’ 5 0 0 marker : 5 ’m5’ ’m 5’ 5 -0 -0.5 .5 -0 -0.2 .2 marker : 6 ’m6’ ’m 6’ 5 0. 0.5 5 -0 -0.2 .2 === Fo === Four ur ke keys ys to co cons nsid ider er: : mass mass mom mom.o. .o.ine inerti rtia a for ben bendin ding g (0/ (0/1), 1), boundaries bound aries epskol for mode comp computati utations, ons, boundaries bound aries epssot for mode comp computati utations, ons, epsm ep smas ass s fo for r ze zero ro ma mass ss el elem emen ents ts ? Zero Ze ro in inpu put t me mean ans s de defa faul ults ts = 0 1. 1.00 000E 0E-6 -6 1.000E-3 1.000 E-3 1.00 1.000E-6 0E-6 0 0 0 0

======================================================== ===================================== =================== Save Sav e of inp input ut dat data a and pro progra gram m con contin tinuat uation ion 0 = co comp mput ute e wi with thou out t sa savi ving ng da data ta 1 = compute with saving data 3 = save data and exit 4 = do not save data and exit 1 Name of th Name the e in inpu put t da data ta fi file le wi with thin in 60 characters in ca case se of s> fi file le na name me = be beam am.d .dat at is assumed TorsionFrame.dat Data Dat a are sav saved ed on fil file e = Tor Torsio sionFr nFrame ame.da .dat t

The results of the first three natural frequencies and the mode shapes evaluated at the markers 1 to 6 are printed out in the input file of this example and they are given next. Example:

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


BEAM:3.0 -45

| | | Mo Mode de sh shap apes es co comp mput uted ed by BEAM - V 3.3 | ------------------------------------------------| | | Mode Mode numbe number r Ty Type pe of mode mode Na Natu tura ral l frequ frequen ency cy[H [Hz] z]| | ------------------------------------------------| | | 1 Torsion 1 22.361 | | 2 Torsion 2 44.721 | | 3 Torsion 3 67.082 | ------------------------------------------------3 mod modes es are sel select ected ed for MBS dat data a gen genera eratio tion: n: They The y are are: : Torsion 1 Torsion 2 Torsion 3 Mode Mo des s sh shap apes es fo for r mo mode des s i an and d ma mark rker ers s 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 Hint:

Please Plea se no note te th that at th the e ta tabl ble e sh sho ows th the e mod mode e shape sh ape val alue uess of ma mark rker ers, s, wh whic ich h ar are e ou outs tsid ide e of the th e bea beam’ m’ss ce cen nte terr lin line. e. ph phi( i(1,2 1,2,3 ,3)) ar are e th the e di dissplacements in x,y,z -axis, psi(1,2,3) are rotation angle about x,y,z -axis. The presented examples examples demonstrates the pow p ower er of the program BEAM to calculate the eigenvalues and mode shapes of straight beam struc-

BEAM:3.0 -46


tures and to evaluate evaluate the data of flexibl flexiblee bodies bo dies modelled in a multibody system using the standard input data (SID) description.

Error Messages Referring to the model data, BEAM may not find the eigenvalues of the vibration problem. A common message may occur: Example:

Stiffness matrix not posit Stiffness positive ive defin definite ite proble pro blem m not sol solvab vable, le, change change input: input: ier ierg=1 g=1

or Example:

Newton’s Newton ’s met method hod does not con conver vert, t, II, II,w= w= 1 NaN proble pro blem m not sol solvab vable, le, change change input: input: ier ierg=1 g=1

In this case, first - increase the boundaries for epskol and epssot. If the error is still present, please stop the program and change the boundary conditions of the beam st struc ructur ture. e. Mor Moreo eove ver, r, sp sprin rings gs wi with th a sm smal alll stiffness may be help to get a convergence of the iterative solution, e. g. for tx = 1.0e 6N/m N/m.. Do this for corresponding vibration vibration directions. A variation of this so-called numerical stabilisation factors brings the exact solution in the range of digits of the printed natural frequencies.

−

Errorr me Erro mess ssag ages es du duee to non-co non-consiste nsistent nt geome geometry try da data ta ar aree se self lf-explanatory. Save the data, change the values and start BEAM again.

BEAM:3.0 -48


BEAM BE AM:4 :4..

Refe Re fere ren nce cess

Wal allr lrap app, p, O. (1 (199 993) 3). St Stan anda dard rd Input Input Data Data of Flex Flexib iblle Bo Boddiess fo ie forr Mu Mult ltibod ibody y Cod Codes es.. In Inter ternal nal Report Report IB 515-93515-93-4, 4, De Deuts utscche Forsch orschungsanst ungsanstalt alt fr Luft- und Raumfahrt (DLR), Inst. Robotik und System Dynami Dynamik, k, Oberpfaffenh Oberpfaffenhofen. ofen.

Meirovit Meiro vitch ch,, L. (19 (1967) 67). An Anal alyt ytic ical al Method Methodss in Vib Vibrat ratio ions. ns. Ne New w York, Macmillan Company.

Beam

Recommend Documents