18Standard Mass

Matrices For PlaneBeam Elements

18–1

Chapter 18: STANDARD MASS MATRICES FOR PLANE BEAM ELEMENTS

TABLE OF CONTENTS

Page§18.1 The Plane BE Beam . . . . . . . . . . . . . . . . . . 18–3

§18.2 The Plane Timoshenko Beam . . . . . . . . . . . . . . 18–4

18–2

§18.1 THE PLANE BE BEAM

§18.1. The Plane BE Beam

The two-node plane BE-beam element has length �, cross section area A and uniform mass densityρ. Only the translational inertia due to the lateral motion of the beam is considered in the kineticenergy T = 1

2

∫ �

0 ρv̇(x̄)2 dx̄ of the element, whereas its rotational inertia is ignored. The freedomsare arranged as ue = [ v1 θ1 v2 θ2 ]T . The natural coordinate ξ varies from ξ = −1 at node 1(x = 0) to ξ = +1 at node 2 (x = �), whence dx/dξ = 1

2� and dξ/dx = 2/�. The well knowncubic shape functions in terms of ξ are collected in the shape function matrix

Ne = [ 14 (1 − ξ)2(2 + ξ) 1

8�(1 − ξ)2(1 + ξ) 14 (1 + ξ)2(2 − ξ) − 1

8�(1 + ξ)2(1 − ξ) ](18.1)

The CMM obtained by analytical integration is

MeC M M = ρ A

∫ 1

−1J (Ne)T Ne dξ = me

420

156 22� 54 −13�

22� 4�2 13� −3�2

54 13� 156 −22�

−13� −3�2 −22� 4�2

. (18.2)

in which the Jacobian J = dx/dξ = �/2 and me = ρ A �. The mass matrices obtained with Gaussintegration rules of 1, 2 and 3 points are

C1

16 4� 16 −4�

4� �2 4� −�2

16 4� 16 −4�

−4� −�2 −4� �2

, C2

86 13� 22 −5�

13� 2�2 5� −�2

22 5� 86 −13�

−5� −�2 −13� 2�2

, C3

444 62� 156 −38�

62� 11�2 38� −9�2

156 38� 444 −62�

−38� −9�2 −62� 11�2

,

(18.3)

in which C1 = me/64, C2 = me/216 and C3 = me/1200. Their eigenvalue analysis shows that allthree are singular, with rank 1, 2 and 3, respectively. The result for 4 and more points agrees with(18.2), which has full rank. The main purpose of this example is to illustrate the rank property statedin §16.6: each Gauss point adds one to the rank up to 4, since the problem is one-dimensional.

The matrix (18.2) conserves linear and angular momentum. So do the reduced-integration massmatrices (18.3) if the number of Gauss points is 2 or greater.

To get a diagonally lumped mass matrix is trickier. Obviously the translational nodal masses mustbe the same as that of a bar: 1

2ρ A�. See Figure 18.1. But there is no easy road on rotational masses.To accommodate these variations, it is convenient to leave the latter parametrized as follows

M̄eL = me

12 0 0 00 α�2 0 00 0 1

2 00 0 0 α�2

, α ≥ 0. (18.4)

Here α is a nonnegative parameter, typically between 0 and 1/100. The choice of α has been arguedin the FEM literature over several decades, but the whole discussion is largely futile. Matchingthe angular momentum of the beam element gyrating about its midpoint gives α = −1/24. Thisviolates the positivity condition. It follows that the best possible α — as opposed to possible best— is zero. This choice gives, however, a singular mass matrix. This is undesirable in scenarioswhere a mass-inverse appears.

18–3

Chapter 18: STANDARD MASS MATRICES FOR PLANE BEAM ELEMENTS

Total mass ρA

ρA ρA

x

αρA 3αρA 3

-

1 2

Figure 18.1. Direct mass lumping for two-node plane BE beam element.

This result can be readily understood physically. The me/2 translational end node masses grosslyoverestimate (in fact, by a factor of 3) the angular momentum of the element. Hence adding anyrotational lumped mass only makes things worse.

§18.2. The Plane Timoshenko Beam

The Timoshenko beam (Ti-beam) incorporates two refinements over the Bernoulli-Euler (BE)model:

1. For both statics and dynamics: plane sections remain plane but not necessarily normal to thedeflected midsurface. See Figure 24.4 for the kinematics. This assumption allows the averagedshear distortion to be included in both strain and kinetic energies.

2. In dynamics: the rotary inertia is included in the kinetic energy.

This model is more important for dynamics and vibration than BE, and indispensable for rapidtransient and wave propagation analysis. More specifically, the BE beam has infinite phase velocity,because the EOM is parabolic, and thus becomes useless for high-fidelity wave propagation.

According to the second assumption, the kinetic energy of the Ti-beam element is given by

T = 12

∫ �

0

(ρ A v̇(x)2 + ρ IR θ̇ (x)2

)dx . (18.5)

Here IR is the second moment of inertia to be used in the computation of the rotary inertia andθ = v′ + γ is the cross-section rotation angle shown in Figure 24.4; γ = V/(G As) being thesection-averaged shear distortion. The element DOF are ordered ue = [ v1 θ1 v1 θ2 ]T . Thelateral displacement interpolation is

v(ξ) = v1 N ev1(ξ) + v′

1 N ev′1(ξ) + v2 N e

v2(ξ) + v′2 N e

v′2(ξ), ξ = 2x

�− 1, (18.6)

in which cubic interpolation functions are used. A complication over BE is that the rotationalfreedoms are θ1 and θ2 but the interpolation (18.6) is in terms of the neutral surface end slopes:v′

1 = (dv/dx)1 = θ1 − γ and v′2 = (dv/dx)2 = θ2 − γ . From a kinmatic analysis we can derive

the relation [v′

1v′

2

]= 1

1 + �

[ −��

1 + �2

��

−�2

−��

−�2

��

1 + �2

]

v1

θ1

v2

θ2

, (18.7)

18–4

§18.2 THE PLANE TIMOSHENKO BEAM

in which the dimensionless parameter � = 12E I/(G As �2) characterizes the ratio of bending andshear rigidities. The end slopes of (18.7) are replaced into (18.6), the interpolation for θ obtained,and v and θ inserted into the kinetic energy (18.5).

After lengthy algebra the CMM emerges as the sum of two contributions:

MeC M M = Me

CT + MeC R =

CT

1335+ 7

10�+ 13�2 ( 11

210+ 11120�+ 1

24�2)� 970+ 3

10�+ 16�2 −( 13

420+ 340�+ 1

24�2)�

( 1105+ 1

60�+ 1120�2)�2 ( 13

420+ 340�+ 1

24�2)� −( 1140+ 1

60�+ 1120�2)�2

1335+ 7

10�+ 13�2 ( 11

210+ 11120�+ 1

24�2)�

symmetric ( 1105+ 1

60�+ 1120�2)�2

+ CR

65 ( 1

10 − 12�)� − 6

5 ( 110 − 1

2�)�

( 215 + 1

6� + 13�2)�2 (− 1

10 + 12�)� −( 1

30 + 16� − 1

6�2)�2

65 (− 1

10 + 12�)�

symmetric ( 215 + 1

6� + 13�2)�2

.

(18.8)

in which CT = ρ A �/(1 + �)2 = me/(1 + �)2 and CR = ρ IR/((1 + �)2 �). Matrices MCT andMC R account for translational and rotary inertia, respectively. Caveat: the I in � = 12E I/(G As �2)

is the second moment of inertia that enters in the elastic flexural elastic rigidity. If the beam ishomogeneous IR = I , but that is not necessarily the case if, as sometimes happens, the beam hasnonstructural attachments that contribute rotary inertia.

The scale factor of MeC R can be further transformed to facilitate parametric studies by introducing

r2R = IR/A as cross-section gyration radius and � = rR/� as element slenderness ratio. Then

CR = ρ IR/((1 + �)2 �) = ρ A � �2/(1 + �)2 = me �2/(1 + �)2. If � = 0 and � = 0, MeC R

vanishes and MeCT in (18.8) reduces to (18.2).

A DLMM can be obtained through the HRZ scheme explained in §D.1. The optimal lumped massis derived in §24.2.5 via templates.

18–5