Hutton - Fundamentals of Finite Element Analysis (523155), страница 76
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Structural DynamicsText© The McGraw−HillCompanies, 200410.6 Mass Matrix for a General Element: Equations of MotionydydxdzxzFigure 10.11 Differential element of a generalthree-dimensional body.For the general case, we consider the three-dimensional body depicted inFigure 10.11 and examine a differential mass dm = dx dy dz located at arbitrary position (x , y, z) . Displacement of the differential mass in the coordinatedirections are (u, v, w) and the velocity components are ( u̇, v̇, ẇ) , respectively.As we previously examined the potential energy, we now focus on kinetic energyof the differential mass given by11dT = ( u̇ 2 + v̇2 + ẇ 2 ) dm = ( u̇ 2 + v̇2 + ẇ 2 ) dx dy dz(10.85)22Total kinetic energy of the body is then""""""11222T =( u̇ + v̇ + ẇ ) dm =( u̇ 2 + v̇2 + ẇ 2 ) dx dy dz(10.86)22and the integration is performed over the entire mass (volume) of the body.Considering the body to be a finite element with the displacement fielddiscretized asMu(x , y, z, t ) =N i (x , y, z)u i (t ) = [N ]{u}i=1v(x , y, z, t ) =MN i (x , y, z)vi (t ) = [N ]{v}(10.87)i=1w(x , y, z, t ) =MN i (x , y, z)wi (t ) = [N ]{w}i=1(where M is the number of element nodes), the velocity components can beexpressed as∂uu̇ == [N ]{u̇}∂t∂v(10.88)v̇ == [N ]{v̇}∂t∂wẇ == [N ]{ẇ}∂t413Hutton: Fundamentals ofFinite Element Analysis41410.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsThe element kinetic energy expressed in terms of nodal velocities and interpolation functions is then written as"""1(e)T =({u̇} T [N ] T [N ]{u̇} + {v̇} T [N ] T [N ]{v̇}2V (e)+ {ẇ} T [N ] T [N ]{ẇ}) dV (e)Denoting the nodal velocities as(10.89) {u̇} ˙ = {v̇}{␦} {ẇ}(10.90)a 3M × 1 column matrix, the kinetic energy is expressed asT"""[N][N]001 ˙ T(e) ˙T (e) = {␦} dV {␦}0[N ]T [N ]02V (e)00[N ]T [N ]=1 ˙ T (e) ˙{␦} m {␦}2(10.91)and the element mass matrix is thus identified as"""[N ]T [N ]00 (e) (e)m= dV0[N ]T [N ]0V (e)00[N ]T [N ](10.92)Note that, in Equation 10.92, the zero terms actually represent M × M nullmatrices.
Therefore, the mass matrix as derived is a 3M × 3M matrix, which isalso readily shown to be symmetric. Also note that the mass matrix of Equation 10.92 is a consistent mass matrix. The following example illustrates thecomputations for a two-dimensional element.EXAMPLE 10.6Formulate the mass matrix for the two-dimensional rectangular element depicted in Figure 10.12. The element has uniform thickness 5 mm and density = 7.83 × 10 −6 kg/mm3.4(10, 30)3(40, 30)sr1(10, 10)y2(40, 10)xFigure 10.12 The rectangular elementof Example 10.6.Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.6 Mass Matrix for a General Element: Equations of Motion■ SolutionPer Equation 6.56, the interpolation functions in terms of serendipity or natural coordinates areN 1 (r, s) =1(1 − r )(1 − s)4N 2 (r, s) =1(1 + r )(1 − s)4N 3 (r, s) =1(1 + r )(1 + s)4N 4 (r, s) =1(1 − r )(1 + s)4with r = (x − 25)/15 and s = ( y − 20)/10 .
For integration in the natural coordinates,dx = 15 dr and dy = 10 ds. The mass matrix is 8 × 8 and the nonzero terms are defined by""""1 "1T[N ] [N ] dV(e)= t[N ] T [N ](15 dr )(10 ds)−1 −1V (e)"1 "1= 150(5)[N ] T [N ] dr ds−1 −1In this solution, we compute a few terms for illustration, then present the overall results.For example,"1 "1m 11 = 150(5)N 12−1 −1150(5)dr ds =16150(5)(1 − r ) 3 (1 − s)=16333 1−1"1 "1(1 − r ) 2 (1 − s) 2 dr ds−1 −1 4(750)75064==(7.830)(10) −61699= 2.6(10) −3 kgSimilarly,"1 "1m 12 = 150(5)−1 −1==150(5)16150(5)N 1 N 2 dr ds =163r−r3150(5)(7.83)(10) −616(1 − s)3329"1 "1(1 − r 2 )(1 − s) 2 dr ds−1 −131(−1)−1= 1.3(10) −3 kg415Hutton: Fundamentals ofFinite Element Analysis41610.
Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsIf we carry out all the integrations indicated to form the mass matrix, the final result forthe rectangular element is2.6 1.3 0.7 1.3 1.3 2.6 1.3 0.7 0.7 1.3 2.6 1.3 (e) m= 1.3 0.7 1.3 2.60 0 0 00 0 0 00 0 0 00 0 0 000002.61.30.71.300001.32.61.30.70 00 00 00 00.7 1.31.3 0.72.6 1.31.3 2.6 (10) −3 kgWe observe that the element mass matrix is symmetric, as expected.
Also note that storing the entire matrix as shown would be quite inefficient, since only the 4 × 4 submatrixof nonzero terms is needed.Having developed a general formulation for the mass matrix of a finiteelement, we return to the determination of the equations of motion of a structuremodeled via the finite element method and subjected to dynamic (that is, timedependent) loading.
If we have in hand, as we do, the mass and stiffness matrices of a finite element, we can assemble the global equations for a finite elementmodel of a structure and obtain an expression for the total energy in the form11{q̇ } T [M ]{q̇ } + {q } T [K ]{q } − {q } T { f } = E22(10.93)where {q } is the column matrix of displacements described in the global coordinate system and all other terms are as previously defined. (At this point, wereemphasize that Equation 10.93 models the response of an ideal elastic system,which contains no mechanism for energy dissipation.) For a system as described,total mechanical energy is constant, so that dE /dt = 0 .
As the mechanicalenergy is expressed as a function of both velocity and displacement, the minimization procedure requires thatdE∂ E ∂ q̇i∂ E ∂qi=+=0dt∂ q̇i ∂ t∂qi ∂ ti = 1, P(10.94)where we now represent the total number of degrees of freedom of the model asP to avoid confusion with the mass matrix notation [M ]. Application of Equation 10.94 to the energy represented by Equation 10.93 yields a system of ordinary differential equations[M ]{q̈ } + [K ]{q } = {F }(10.95)Equation 10.94 is not necessarily mathematically rigorous in every case. However, for the systems under consideration, in which there is no energy removalmechanism and the total potential energy includes the effect of external forces, theHutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.6 Mass Matrix for a General Element: Equations of Motionresulting equations of motion are the same as those given by both the Lagrangianapproach and variational principles [5].Examination of Equation 10.95 in light of known facts about the stiffness andmass matrices reveals that the differential equations are coupled, at least throughthe stiffness matrix, which is known to be symmetric but not diagonal.
The phenomena embodied here is referred to as elastic coupling, as the coupling termsarise from the elastic stiffness matrix. In consistent mass matrices, the equationsare also coupled by the nondiagonal nature of the mass matrix; therefore, the terminertia coupling is applied when the mass matrix is not diagonal. Obtaining solutions for coupled differential equations is not generally a straightforward prodecure. We show, however, that the modal characteristics embodied in the equationsof motion can be used to advantage in examining system response to harmonic(sinusoidal) forcing functions. The so-called harmonic response is a capability ofessentially any finite element software package, and the general techniques arediscussed in the following section, after a brief discussion of natural modes.In the absence of externally applied nodal forces, Equation 10.95 is a systemof P homogeneous, linear second-order differential equations in the independentvariable time.
Hence, we have an eigenvalue problem in which the eigenvaluesare the natural circular frequencies of oscillation of the structural system, andthe eigenvectors are the amplitude vectors (mode shapes) corresponding to thenatural frequencies. The frequency equation is represented by the determinant|− 2 [M ] + [K ]| = 0(10.96)If formally expanded, this determinant yields a polynomial of order P in the variable 2 . Solution of the frequency polynomial results in computation of P naturalcircular frequencies and P modal amplitude vectors. The free-vibration responseof such a system is then described by the sum (superposition) of the naturalvibration modes as␦i (t ) =P( j)A i sin( j t + j )i = 1, P(10.97)j=1Note that the superposition indicated by Equation 10.97 is valid only for lineardifferential equations.( j)In Equation 10.97, the A i and j are to be determined to satisfy given initialconditions.
In accord with previous examples for simpler systems, we know thatthe amplitude vectors for a given modal frequency can be determined within asingle unknown constant, so we can write the modal amplitude vectors as1 (i) 2 !(i)(i)(i)A= A 1 3i = 1, P(10.98).. . (i) P417Hutton: Fundamentals ofFinite Element Analysis41810. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicswhere the  terms are known constants resulting from substitution of the naturalcircular frequencies into the governing equations for the amplitudes. For asystem having P degrees of freedom, we have 2 P unknown constants A (i)1 andi , i = 1, P in the motion solution.
The constants are determined by applicationof 2 P initial conditions, which are generally specified as the displacements andvelocities of the nodes at time t = 0. While the natural modes of free vibrationare important in and of themselves, application of modal analysis to the harmonically forced response of structural systems is a very important concept. Prior toexamination of the forced response, we derive a very important property of theprincipal vibration modes.10.7 ORTHOGONALITY OFTHE PRINCIPAL MODESThe principal modes of vibration of systems with multiple degrees of freedomshare a fundamental mathematical property known as orthogonality.
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