MFEMD.Ch04

.

4PMP Dynamical

Systems:Matrix EOM Formulation

41

Chapter 4: PMP DYNAMICAL SYSTEMS: MATRIX EOM FORMULATION 42

TABLE OF CONTENTS

Page4.1. Matrix Equations of Motion 43

4.1.1. Approaches to EOM Formulation . . . . . . . . . . . 43

4.2. EOM Derivation by Force Equilibrium 434.2.1. PMP Dynamic System Example . . . . . . . . . . . 434.2.2. Equilibrium Equations . . . . . . . . . . . . . . 454.2.3. Matrix Form . . . . . . . . . . . . . . . . . 464.2.4. Matrix Symmetry Check . . . . . . . . . . . . . . 464.2.5. Center of Mass Motions Check . . . . . . . . . . . 474.2.6. PMP Equations of Motion for Numerical Data . . . . . . 47

4.3. EOM Derivation by the Finite Element Method 484.3.1. Element-Level Calculations . . . . . . . . . . . . 484.3.2. Assembly of Master Mass Matrix . . . . . . . . . . . 494.3.3. The FEM Equations of Motion . . . . . . . . . . . 49

4.4. Comparison Between Formulation Approaches 4104. Notes and Bibliography. . . . . . . . . . . . . . . . . . . . . . 411

42

43 4.2 EOM DERIVATION BY FORCE EQUILIBRIUM

4.1. Matrix Equations of Motion

This Chapter describe how to construct the dynamic equations of motion (EOM) in matrix form.Several simplifying assumptions are made for clarity of exposition:

The dynamical model is a point-mass particle (PMP) system, with point masses linked bycollinear interaction forces.

Motions from the reference configuration are small in the sense that the initial geometry can beused throughout

Interaction forces depend linearly on displacements and may be idealized as extenssionalsprings. Damping effects are ignored although they are briefly covered in other Chapters.

The last two assumptions ensure that the EOM are linear in displacements and accelerations.The first assumption (dynamic model is PMP) is partially lifted when considering the equivalencebetween two ways of constructing the matrix EOM: by equilibrium or via FEM, as outlined next.

4.1.1. Approaches to EOM Formulation

Sections 4.2 and 4.3 cover two approaches to the formulation of matrix EOM:

Equilibrium Method. Writing down force balance expressions for each point-mass particle via FreeBody Diagrams (FBD) that account for dynamic effects modeled as forces.

Finite Element Method. Assembling element-level EOM provided by a FEM discretization.

The two methods are quite different in philosophy as well as procedure. The equilibrium methodreflects the direct application of Newtons laws to free bodies isolated from the rest of the systemby appropriate external and internal forces. Technically those are known as Free Body Diagramsor FBD. By contrast, FEM equations are generally derived in a variational framework, such as thePrinciple of Virtual Work, Lagranges equations, or Hamiltons principle.

Nonetheless, for the PMP systems considered in this Chapter, the two methods produce exactlythe same matrix EOM. Why then bother to go over both? The answer is that each has distinctadvantages as well as shortcomings. Those are contrasted in 4.4.

4.2. EOM Derivation by Force Equilibrium

The force equilibrium approach to deriving matrix EOM will be illustrated in the two-dimensionalPMP system pictured in Figure 4.1. A reader familiar with IFEM may note its similarity with theexample truss used in [106, Chapters 23]. That is not accidental: the stiffness matrix equationsderived there will be reused for part of the EOM derivation via FEM carried out in 4.3.

4.2.1. PMP Dynamic System Example

The two-dimensional PMP system of Figure 4.1(a) is referred to a fixed RCC frame {x, y}, withorigin as shown in the Figure. The geometry is defined by the reference position of the three pointmasses, labeled 1,2 and 3.

43


(a)

45o

45o 2(10,0)

3(10,10)

1(0,0)x

y

x3x3f , u

y2y2f , uy1y1f , u

x2x2f , ux1x1f , u

y3 y3f , u

(1)

(2)(3)

(b)

45o

45o

1 2

3

(1)

(2)(3)

m = 3 2 m = 5 1

m = 4 3

k = 5 (2)

k = 10 (1)

k = 20 (3) x3 y3External forces:f =2H(t), f =H(t),others zero. Here H(t) is the Heaviside unit-step function

ext ext

Figure 4.1. Example PMP system to illustrate derivation of matrix EOM by force equilibrium:(a) reference geometry and physical properties; (b) kinematic DOF and associate forces.

Point masses are connected by linearly elastic extensional springs, which are labelled as (1), (2)and (3) for convenience.1 The springs possess extensional stiffnesses k(1), k(2) and k(3) with valuesindicated as in the Figure. The system only moves on the {x, y} plane. Consequently it has 6kinematic degrees of freedom (DOF), which are taken to be the point mass displacements in thecoordinate directions. These DOF are collected in a system displacement 6-vector with a mass-by-mass arrangement:

u = [ ux1 uy1 ux2 uy2 ux3 uy3 ]T . (4.1)The velocity and acceleration 6-vectors are configured accordingly as

u = [ ux1 u y1 ux2 u y2 ux3 u y3 ]T ,u = [ ux1 u y1 ux2 u y2 ux3 u y3 ]T .

(4.2)

The 6-vector of generic forces associated to (4.1) is

f = [ fx1 fy1 fx2 fy2 fx3 fy3 ]T (4.3)Unlike displacements, several kinds of forces may be considered: external, internal, interaction,constraint (also called reaction), inertial, effective, residual, and so on. When there is need to dis-tinguish force type, the generic symbols of (4.3) are adorned with appropriate 3-letter superscripts.For example,

f ext = [ f extx1 f exty1 f extx2 f exty2 f extx3 f exty3 ]T ,f int = [ f intx1 f inty1 f intx2 f inty2 f intx3 f inty3 ]T ,

(4.4)

denote vectors of external and internal forces, respectively, for the example system. To distinguishinternal from interaction, the latter is identified with superscript iac, as in f iac. For consistencywith NFEM [107], residual forces may be denoted by r instead of f res to reduce clutter.

1 This follows the element-identification convention of IFEM [106, Chapter 2], since in 4.3 the springs of Figure 4.1 areshown to be equivalent to bar elements in a truss FEM model. Enclosing parentheses distinguish these from point-masslabels, while avoiding confusion with exponents when placed as superscripts of spring properties.

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(a)

21

3 x3f

x3

y2f

y2

y1f

y1

y3f

y3

(b)

1 2

3

m2 m u2

m u2

m1 m u1 m u1

m 3 m u 3

m u 3

x2 x1f = k (u u ) (1) (1)s

f (1)s f (1)s

f (2)s

f (1)s

f (2)s

f (2)s

f (3)s

f (1)s

f (3)s

f (3)s f (2)sf (3)s

y3 y2f = k (u u ) (2) (2)s

y3

y2

x3

x2

oo

o

o

(3) (3)sf = k (u cos 45 + u sin 45

u cos 45 u sin 45 )

x2f

x2

x1f

x1 ..

..

..

..

..

..

ext

ext

ext

ext

ext

ext

Figure 4.2. Auxiliary diagrams in equilibrium EOM derivation for PMP system of Figure 4.1: (a) spring-interaction forces in terms of displacements; (b) point-mass FBD diagrams. External, spring-interaction

and inertial forces are pictured in black, blue and red, respectively, for visualization clarity.

Derivations are carried out first for a floating PMP system, which has no motion constraints. InChapter 5 the EOM will be modified to account for kinematic constraints of various types.

4.2.2. Equilibrium Equations

Begin by disconnecting the point-masses. Replace the springs by interaction forces f (1)s , f(2)s

and f (3)s . These are depicted in Figure 4.2(a), in which their expressions in terms of point-massdisplacements is listed. The + sense of the interaction forces is governed by an easy-to-rememberconvention: assume that each spring is in tension. To complete the Free Body Diagrams (FBD)shown in Figure 4.2(b), two more force sets are added:

The external forces f extx1 through f exty3 . The inertial forces2 f inex1 = m1 ux1 through f iney3 = m3 u y3. Their displayed sense is dictated

by a simple rule: the pertinent acceleration is assumed positive. Since masses are positive,these forces point in the opposite sense of a + acceleration. See Figure 4.2(b).

The three force sets: external, interaction and inertial, are pictured in different colors in Figure 4.2(b)for easier visualization. Next, force equilibrium conditions for each degree of freedom (DOF)are written down. Interaction and inertia forces are expressed in terms of displacements andaccelerations. For example, force balance of the motion of mass 1 along x gives

fx1 = f extx1 + f (1)s + f (3)s cos 45 + f inex1

= f extx1 + k(1)(ux2ux1) + k(3)(ux3 cos 45

+ uy3 sin 45 ux1 cos 45 uy1 sin 45

)cos 45 m1 ux1

= f extx1 + k(1)(ux2ux1) + 12 k(3)(ux3+uy3ux1uy1

) m1 ux1 = 0.(4.5)

2 Also known as DAlembert forces, pseudo forces, or effective forces in the literature.

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At this point it is convenient to revert the sign of the LHS of (4.5) so as to make the inertia termm1 ux1 positive. While doing this, common displacements are collected, and the external forceterm passed to the RHS. The rearranged expression is

m1 ux1 + (k(1)+ 12 k(3)) ux1 k(1) ux2 + 12 k(3) uy1 12 k(3) ux3 12 k(3) uy3 = f extx1 . (4.6)

Following exactly the same pattern, five more balance equations are obtained. The end result is sixequilibrium equations, one for each DOF.

4.2.3. Matrix Form

Once the equilibrium equations for all DOF are obtained and appropriate arranged as discussedabove, several steps are carried out to put them in matrix form:

The external forces, which are givens, are kept in the RHS and arranged as a vector fext ,configured as per the first of (4.4).

LHS terms that depend on accelerations are organized as a diagonal master mass matrix Mpostmultiplied by the acceleration vector u configured as in the second of (4.2).

LHS terms that depend on displacements are collected and organized as a master stiffness matrixK postmultiplied by the displacement vector u configured as in (4.1).

The resulting matrix EOM have the compact form

M u + K u = f ext . (4.7)

For the example system the mass and stiffness matrices are

M =

m1 0 0 0 0 00 m1 0 0 0 00 0 m2 0 0 00 0 0 m2 0 00 0 0 0 m3 00 0 0 0 0 m3

= diag [ m1, m1, m2, m2, m3, m3 ]. (4.8)

K =

k(1) + 12 k(3) 12 k(3) k(1) 0 12 k(3) 12 k(3)12 k

(3) 12 k

(3) 0 0 12 k(3) 12 k(3)k(1) 0 k(1) 0 0 00 0 0 k(2) 0 k(2)

12 k(3) 12 k(3) 0 0 12 k(3) 12 k(3) 12 k(3) 12 k(3) 0 k(2) 12 k(3) k(2) + 12 k(3)

(4.9)

If the RHS of (4.7) consists entirely of external (applied, given) forces, we will often drop thesuperscript, and write

M u + K u = f, (4.10)

being tacitly understood that f contains only given forces.

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4.2.4. Matrix Symmetry Check

Observe that both M and K in (4.8) and (4.9) are symmetric. For M that is not a surprise, sincereal diagonal matrices are necessarily symmetric. Can that be expected for K? Yes: it is not anaccident. It previews the equivalence of this approach with the Finite Element Method, as covered in4.3. Since FEM equations are automatically symmetric when derived in a variational framework,equivalence implies symmetry. Should K come out unsymmetric, two fixes can be applied:

Reverse sign of individual equations (which become matrix rows) as necessary so that diagonalentries of K and M are positive. If they are not of the same sign, look for derivation errors.

If diagonal signs check out but unsymmetry persists, try row scaling by appropriate factors.

4.2.5. Center of Mass Motions Check

After verifying symmetry, reduction to center-of-mass motion provides another useful EOM check.Assume that x and y translations are rigid by setting ux1 = ux2 = ux3 = ux and uy1 = uy2 =uy3 = uy for all t . Hence ux1 = ux2 = ux3 = ux and u y1 = u y2 = u y3 = u y . Replace these intovectors u and u of (4.10). Add equations 1, 3, and 5, factoring ux and ux , and add equations 2, 4,and 6, factoring uy and u y . The result is

M ux = Fx , M u y = Fy,

in which M = m1 + m2 + m3 is the total mass of the system, whereas Fx = f extx1 + f extx2 + f extx3and Fy = f exty1 + f exty2 + f exty3 are the resultants of the x and y external forces, respectively. Theseare the well known equations for the translational motion of the center of mass, which is locatedat xC = 10(m2 + m3)/M and yC = 10m3/M, as may be easily verified. A similar check can beperformed for an infinitesimal rotation about the center of mass: the result is IC = TC , whereIC is the mass moment about C and TC the torque of the external forces about C .In these reductions the contribution of the stiffness term K u cancels out. This results from therigid-body property of the stiffness matrix: K uR = 0 if uR is a rigid body motion. A directverification can be done by extracting the eigenvalues of K and checking that its rank is 3 [106,Chapter 20], but symbolic eigenvalue verification becomes unwieldy for larger systems.

4.2.6. PMP Equations of Motion for Numerical Data

Numerical values for point masses and spring constants given in Figure 4.1(a) are: m1 = 5,m2 = 3, m3 = 4, k(1) = 10, k(2) = 5, and k(3) = 20. For the external forces, Figure 4.1(b) statesthat f extx1 = f exty1 = f extx2 = f exty2 = 0, f extx3 = 2H(t), and f exty3 = H(t), in which H(t) is theHeaviside unit-step function. Replacing into the foregoing matrix expression yields

5 0 0 0 0 00 5 0 0 0 00 0 3 0 0 00 0 0 3 0 00 0 0 0 4 00 0 0 0 0 4

ux1u y1ux2u y2ux3u y3

+

20 10 10 0 10 1010 10 0 0 10 10

10 0 10 0 0 00 0 0 5 0 5

10 10 0 0 10 1010 10 0 5 10 15

ux1uy1ux2uy2ux3uy3

=

0000

2H(t)H(t)

(4.11)

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1 2

3

x

y

x3x3f , u

y2y2f , uy1y1f , u

x2x2f , ux1x1f , u

y3 y3f , u

(1)

(2)(3)

(3) (3)(3) (3)

E = 100, A = 2 2, L = 10 2, = 3/20

(1)

(1)

(1)

(1)E = 50, A = 2, L = 10, = 1/5

(2)

(2)

(2)

(2)E = 50, A = 1, L = 10, = 1/5

45o

45o

x3 y3

External forces:f =2H(t), f =H(t),others zero. Here H(t) is the Heaviside unit-step function

ext ext

Figure 4.3. The three-member example truss of IFEM [106, Chapters 2-3] reproduced forconvenience. The dynamic model is (for now) floating: it has no supports and thus may

move freely on the {x, y} plane. It is subject to the external forces indicated in the box.

This is a linear system of six second-order ODEs in time. The problem specification is closed byproviding 12 initial conditions: six displacements and six velocities at a start time, which is oftentaken to be t = 0. For example, the rest condition at t = 0 is specified by

ux1(0) = uy1(0) = . . . , uy3(0) = 0, ux1(0) = u y1(0) = . . . , u y3(0) = 0. (4.12)A rest initial condition would be appropriate for a structure hit by a transient event at t = 0, e.g.,earthquake, explosion or impact.

4.3. EOM Derivation by the Finite Element Method

The FEM model for the example truss of IFEM [106, Chapters 23] is reproduced in Figure 4.3.One additional property is now activated: the mass density of the three elements (truss members).This is denoted by e.

4.3.1. Element-Level Calculations

As typical of FEM, computations proceed element by element. In this case two element-levelmatrices are generated: stiffness and mass. The element stiffness matrices are exactly those derivedin IFEM [106, Chapters 23]. The reader is referred there for details.

The mass matrices are produced by a nodal lumping scheme that proceeds as follows. The totalmass of element e is me = e Ae Le, in which Le and Ae are the length and cross section area,respectively. Assign one half of this to each end node, in both the x and y directions. As a resultthe so-called lumped mass matrix of the truss (bar) element is

Me =

12 m

e 0 0 00 12 m

e 0 00 0 12 m

e 00 0 0 12 m

e

= 12 e Ae Le I4, (4.13)

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49 4.3 EOM DERIVATION BY THE FINITE ELEMENT METHOD

1 2

3

m =1 (2)2

m =2 (1)2 m =2+1=3 2 m =2 (1)1

m =3+2=5 1

m =3 (3)1

3m =3 (3)

m =1 (2)3 m =1+3=4 3

(1)

(2)(3)

(a) (b)

Figure 4.4. Mass lumping procedure for the example truss of Figure ?: (a) element-levellumping; (b) addition at nodes to form master mass matrix.

in which I4 denotes the identity matrix of order 4. What happens to Me if the axes are rotated byan angle to be {x, y}? It becomes

Me = (Te)T Me Te = e Ae Le (Te)T I4 Te = e Ae Le (Te)T Te,

where Te is a 44 transformation matrix very similar to that used in [106, 2.8]. But (Te)T Te = I4because Te is orthogonal, and M

e = Me for any . It follows that Me is invariant with respect torotation of axes. Consequence: the machinery for deriving FEM equations in a local system andconverting to global coordinates, as done for the truss element stiffness matrix, can be bypassed forthe mass matrix.

4.3.2. Assembly of Master Mass Matrix

The assembly of the master mass matrix follows exactly the same set of rules used for the masterstiffness matrix. Since Me is diagonal, so will be the master mass matrix. A simple hand calculationproceed by inspecting which elements contribute to a given node. For the example truss:

Node 1 gets half of the mass of elements (1) and (3): m1 = 12 m(1) + 12 m(3).Node 2 gets half of the mass of elements (1) and (2): m2 = 12 m(1) + 12 m(2).Node 3 gets half of the mass of elements (2) and (3): m3 = 12 m(2) + 12 m(3).

These values are assigned to both x and y directions. Consequently the master mass matrix is

M = 12 diag [ m(1)+m(3), m(1)+m(3), m(1)+m(2), m(1)+m(2), m(2)+m(3), m(2)+m(3) ]= diag [ m1, m1, m2, m2, m3, m3 ] = diag [ 5, 5, 3, 3, 4, 4 ].

(4.14)

This two-step construction of the master mass matrix is schematized in Figure 4.4. The node massesare m1 = 5, m2 = 3 and m3 = 4.Remark 4.1. In later Chapters we will see that the lumped mass matrix is not the only choice; in fact there is aninfinite number of possible element mass matrices. Those other than the lumped one are no longer diagonal.Consequently the assembly process becomes more involved, but can be carried out by the same rules followedfor the master stiffness matrix.

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4.3.3. The FEM Equations of Motion

Combining the master mass matrix derived above with the master stiffness derived in IFEM [106,Chapter 3] and the external forces specified in Figure 4.3, we obtain

5 0 0 0 0 00 5 0 0 0 00 0 3 0 0 00 0 0 3 0 00 0 0 0 4 00 0 0 0 0 4

ux1u y1ux2u y2ux3u y3

+

20 10 10 0 10 1010 10 0 0 10 10

10 0 10 0 0 00 0 0 5 0 5

10 10 0 0 10 1010 10 0 5 10 15

ux1uy1ux2uy2ux3uy3

=

0000

2H(t)H(t)

(4.15)Comparing(4.15) with (4.11) shows these matrix EOM to be identical. Is this a fluke? No. It is aconsequence of two modeling choices.

The mass matrices are the same because of the lumping choice in FEM: conservation conditionsforces diagonal matrix entries to agree. If another choice had been made, the FEM mass matrixwould be nondiagonal and clearly different from that of the PMP model.

Stiffness matrices coalesce because we have set the spring constants of the PMP model of Figure 4.1to be

ke = Ee Ae/Le, e = (1), (2), (3), (4.16)where Ee, Ae and Le are member properties of the FEM truss model of Figure 4.3. As shown inMechanics of Materials textbooks, as well as IFEM [106, Chapter 2], (4.16) provides the equivalentaxial stiffness of a truss (bar) element.

Given that the same EOM are obtained, downstream tasks such as

Application of constraint conditions Modal analysis Direct time integration Internal force recoveryneed not distinguish between the two derivation methods. Those topics are covered in the followingChapters.

4.4. Comparison Between Formulation Approaches

Having gone through the two EOM formulation methods, it is appropriate to summarize theirrelative advantages and shortcomings.

The key advantages of the force equilibrium method are simplicity and physical transparency. Theseemanate from the use of just one modeling tool: Newtons laws, as well as basic linear algebra. Thismakes it the method of choice in undergraduate instruction. Students in those courses have beenexposed to those laws, as well as trained in static FBD and basic linear algebra through introductorylower-division courses, but have not yet encountered the more advanced mathematical machineryused in FEM.

410

411 4. Notes and Bibliography

For modeling dynamic systems in practical projects in science or engineering, the equilibriummethod is recommended in the following scenarios:

Preliminary design. The nature of the method encourages the use of highly simplified dynamicmodels. For example, an experienced engineer designing a car suspension would likely lumpthe car and the wheels as link-connected point masses, thus bypassing the overkill ingrained inusing FEM models too early. Likewise, seismic design of a multistory building often starts witha point-mass stick model. Keeping-it-simple can significantly reduce design flowthrough,while avoiding costly modifications.

Particle model fits problem best. There are some scientific applications where PMP models areproper and natural. The classical example is Astronomy, which is precisely where Newtonianmechanics emerged as system of the world. The key advantage is the ability to discard ab initiointernal particle structural details, which may not be known anyway. This enforces retention ofkey effects, such as inertia and gravitational interaction in that application. A FEM model ofthe Solar System at astronomical scales (or the trajectory of a space vehicle) would be not onlyabsurd but a total waste of time.

The key advantages of FEM are generality and automation. Once system details need to be included,the equilibrium force method rapidly loses its attractive simplicity. FEM is naturally adapted todistributed systems modeled by field (continuum) theories within a variational framework. Itscomputer implementation favors abstraction and modularity, reducing formulation errors whileconcealing internal processing details.

Summarizing: in system design that involves dynamics, the two approaches form a natural hierarchy.The PMP approach may capture gross effects during preliminary design, while FEM can zero indetails in subsequent stages. At the other extreme: verification and validation of existing (in situ)systems, FEM is typically the preferred choice.

Notes and Bibliography

(To be completed)

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