Hutton - Fundamentals of Finite Element Analysis (523155), страница 75

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In the lumped matrix approach, half the total mass of theelement is assumed to be concentrated at each node and the connecting materialis treated as a massless spring with axial stiffness. The lumped mass matrix for abar element is then␳ AL 1 0[m] =(10.65)0 12Use of lumped mass matrices offers computational advantages. Since the element mass matrix is diagonal, assembled global mass matrices also are diagonal.On the other hand, although more computationally difficult in use, consistentmass matrices can be proven to provide upper bounds for the natural circular frequencies [3].

No such proof exists for lumped matrices. Nevertheless, lumpedmass matrices are often used, particularly with bar and beam elements, to obtainreasonably accurate predictions of dynamic response.10.5 BEAM ELEMENTSWe now develop the mass matrix for a beam element in flexural vibration. First,the consistent mass matrix is obtained using an approach analogous to that for thebar element in the previous section. Figure 10.9 depicts a differential element ofa beam in flexure under the assumption that the applied loads are time dependent.As the situation is otherwise the same as that of Figure 5.3 except for the use of407Hutton: Fundamentals ofFinite Element Analysis40810. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsq(x, t)yM⫹MxVdxV⫹⭸Mdx⭸x⭸Vdx⭸xFigure 10.9 Differential element of a beamsubjected to time-dependent loading.partial derivatives, we apply Newton’s second law of motion to the differentialelement in the y direction to obtain∂V∂ 2vFy = ma y ⇒ V +dx − V − q (x , t ) dx = (␳ A dx ) 2(10.66)∂x∂twhere ␳ is the material density and A is the cross-sectional area of the element.The quantity ␳ A represents mass per unit length in the x direction.

Equation 10.66simplifies to∂V∂ 2v− q (x , t ) = ␳ A 2(10.67)∂x∂tAs we are dealing with the small deflection theory of beam flexure, beam slopes,therefore rotations, are small. Therefore, we neglect the rotational inertia of thedifferential beam element and apply the moment equilibrium equation. The resultis identical to that of Equation 5.37, repeated here as∂M= −V(10.68)∂xSubstituting the moment-shear relation into Equation 10.67 gives∂2M∂ 2v−−q(x,t)=␳A(10.69)∂x2∂t2Finally, the flexure formula∂ 2vM = E Iz 2(10.70)∂xis substituted into Equation 10.69 to obtain the governing equation for dynamicbeam deflection as∂2∂ 2v∂ 2v− 2 E I z 2 − q (x , t ) = ␳ A 2(10.71)∂x∂x∂tUnder the assumptions of constant elastic modulus E and moment of inertia Iz,the governing equation becomes␳A∂ 2v∂ 4v+EI= −q (x , t )z∂t2∂x4(10.72)Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.5 Beam ElementsAs in the case of the bar element, transverse beam deflection is discretizedusing the same interpolation functions previously developed for the beam function. Now, however, the nodal displacements are assumed to be time dependent.Hence,v(x , t ) = N 1 (x )v1 (t ) + N 2 (x )␪1 (t ) + N 3 (x )v2 (t ) + N 4 (x )␪2 (t )(10.73)and the interpolation functions are as given in Equation 4.26 or 4.29. Applicationof Galerkin’s method to Equation 10.72 for a finite element of length L results inthe residual equations"L∂ 2v∂ 4vN i (x ) ␳ A 2 + E I z 4 + q = 0i = 1, 4(10.74)∂t∂x0As the last two terms of the integrand are the same as treated in Equation 5.42,development of the stiffness matrix and nodal force vector are not repeated here.Instead, we focus on the first term of the integrand, which represents the terms ofthe mass matrix.For each of the four equations represented by Equation 10.74, the first integralterm becomes v̈1 LL"" ␪¨ 1␳ A Ni ( N1 v̈1 + N2 ␪¨1 + N3 v̈2 + N4 ␪¨2 ) dx = ␳ A Ni [N ] dxi = 1, 4v̈2 00␪¨2(10.75)and, when all four equations are expressed in matrix form, the inertia termsbecome  v̈1 v̈1 "L ␪¨   ␪¨1 1T(e)␳ A [N ] [N ] dx= m(10.76)v̈ v̈  2 20␪¨2␪¨2The consistent mass matrix for a two-dimensional beam element is given by"L (e) m= ␳ A [N ] T [N ] dx(10.77)0Substitution for the interpolation functions and performing the required integrations gives the mass matrix as15622L54−13L (e) ␳ AL  22L4L 213L−3L 2 m=(10.78)4205413L156 −22L −13L −3L 2 −22L4L 2and it is to be noted that we have assumed constant cross-sectional area in thisdevelopment.409Hutton: Fundamentals ofFinite Element Analysis41010.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsCombining the mass matrix with previously obtained results for the stiffnessmatrix and force vector, the finite element equations of motion for a beam element are v v̈1 −V1 (t) 1"L¨ (e) ␪1 ␪1−M1 (t)= − [N ]T q(x, t) dx +m+ k (e)(10.79)v̈  v2  V2 (t)  20M2 (t)␪2␪¨2and all quantities are as previously defined. In the dynamic case, the nodal shearforces and bending moments may be time dependent, as indicated.Assembly procedures for the beam element including the mass matrix areidentical to those for the static equilibrium case. The global mass matrix is directlyassembled, using the individual element mass matrices in conjunction with theelement-to-global displacement relations.

While system assembly is procedurallystraightforward, the process is tedious when carried out by hand. Consequently, acomplex example is not attempted. Instead, a relatively simple example of naturalfrequency determination is examined.EXAMPLE 10.5Using a single finite element, determine the natural circular frequencies of vibration of acantilevered beam of length L, assuming constant values of ␳ , E, and A.■ SolutionThe beam is depicted in Figure 10.10, with node 1 at the fixed support such that the boundary (constraint) conditions are v1 = ␪1 = 0. For free vibration, applied force and bendingmoment at the free end (node 2) are V2 = M 2 = 0 and there is no applied distributed load.Under these conditions, the first two equations represented by Equation 10.79 are constraint equations and not of interest.

Using the constraint conditions and the known appliedforces, the last two equations are␳ AL156−22L420−22 L4L 2v̈2␪¨2+E Iz123−6LL−6L4L 2v2␪2= 00For computational convenience, the equations are rewritten as156−22 L−22 L4L 2v̈2␪¨2+420 E I z123−6LmL−6L4L 2v2␪2yE, Iz12LFigure 10.10 The cantilevered beam ofExample 10.5 modeled as one element.x= 00Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.5 Beam Elementswith m = ␳ A L representing the total mass of the beam.

Assuming a sinusoidal displacement response, the frequency equation becomes 12␭ − 156␻ 2 −6␭L + 22␻ 2 L−6␭L + 22␻ 2 L =04L 2 (␭ − ␻ 2 ) with ␭ = 420 E I z /m L 3 . After expanding the determinant and performing considerablealgebraic manipulation, the frequency equation becomes5␻ 4 − 102␭␻ 2 + 3␭2 = 0Solving as a quadratic in ␻ 2 , the roots are␻12 = 0.02945 ␭␻22 = 20.37␭Substituting for ␭ in terms of the beam physical parameters, we obtainE Iz␻1 = 3.517m L3␻2 = 92.50E Izrad/secm L3as the finite element approximations to the first two natural circular frequencies.

For comparison, the exact solution gives␻1exactE Iz= 3.516m L3␻2exact= 22.03E Izrad/secm L3The fundamental frequency computed via a single element is essentially the same as theexact solution, whereas the second computed frequency is considerably larger than the corresponding exact value. As noted previously, a continuous system exhibits an infinitenumber of natural modes; we obtained only two modes in this example. If the number ofelements is increased, the number of frequencies (natural modes) that can be computedincreases as the number of degrees of freedom increases.

In concert, the accuracy of thecomputed frequencies improves.If the current example is refined by using two elements having length L/2 and thesolution procedure repeated, we can compute four natural frequencies, the lowest twogiven by␻1 = 3.516E Izm L3␻2 = 24.5E Izm L3and we observe that the second natural circular frequency has improved (in terms of theexact solution) significantly. The third and fourth frequencies from this solution are foundto be quite high in relation to the known exact values.As indicated by the foregoing example, the number of natural frequenciesand mode shapes that can be computed depend directly on the number of degreesof freedom of the finite element model.

Also, as would be expected for convergence, as the number of degrees of freedom increases, the computed frequenciesbecome closer to the exact values. As a general rule, the lower values (numerically) converge more rapidly to exact solution values. While this is discussed411Hutton: Fundamentals ofFinite Element Analysis41210. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicsin more detail in conjunction with specific examples to follow, a general rule ofthumb for frequency analysis is as follows: If the finite element analyst is interested in the first P modes of vibration of a structure, at least 2 P modes shouldbe calculated.

Note that this implies the capability of calculating a subset offrequencies rather than all frequencies of a model. Indeed, this is possible andextremely important, since a practical finite element model may have thousandsof degrees of freedom, hence thousands of natural frequencies. The computational burden of calculating all the frequencies is overwhelming and unnecessary,as is discussed further in the following section.10.6 MASS MATRIX FOR A GENERAL ELEMENT:EQUATIONS OF MOTIONThe previous examples dealt with relatively simple systems composed of linearsprings and the bar and beam elements. In these cases, direct application ofNewton’s second law and Galerkin’s finite element method led directly to the formulation of the matrix equations of motion; hence, the element mass matrices.

Formore general structural elements, an energy-based approach is preferred, as forstatic analyses. The approach to be taken here is based on Lagrangian mechanicsand uses an energy method based loosely on Lagrange’s equations of motion [4].Prior to examining a general case, we consider the simple harmonic oscillator of Figure 10.1. At an arbitrary position x with the system assumed to be inmotion, kinetic energy of the mass is1m ẋ 22(10.80)1k(␦st + x ) 2 − mg(␦st + x )2(10.81)T =and the total potential energy isUe =therefore, the total mechanical energy isE m = T + Ue =11m ẋ 2 + k(␦st + x ) 2 − mg(␦st + x )22(10.82)As the simple harmonic oscillator model contains no mechanism for energyremoval, the principle of conservation of mechanical energy applies; hence,dE m= 0 = m ẋ ẍ + k(␦st + x ) ẋ − mg ẋdt(10.83)m ẍ + k(␦st + x ) = mg(10.84)orand the result is exactly the same as obtained via Newton’s second law in Equation 10.2.Hutton: Fundamentals ofFinite Element Analysis10.

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