Hutton - Fundamentals of Finite Element Analysis (523155), страница 74

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In these simple cases, we neglect the mass of the springelements in comparison to the concentrated masses. In the general case of solidstructures, the mass is distributed geometrically throughout the structure and theinertia properties of the structure depend directly on the mass distribution. Toillustrate the effects of distributed mass, we first consider longitudinal (axial)vibration of the bar element of Chapter 2.The bar element shown in Figure 10.7a is the same as the bar element introduced in Chapter 2 with the very important difference that displacements and applied forces are now assumed to be time dependent, as indicated.

The free-bodydiagram of a differential element of length dx is shown in Figure 10.7b, wherecross-sectional area A is assumed constant. Applying Newton’s second law to thedifferential element gives∂␴∂ 2u␴+dx A − ␴ A = (␳ A dx ) 2(10.51)∂x∂tHutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.4 Bar Elements: Consistent Mass Matrixu(x, t)1␴2u1(x1, t)u2(x2, t)⫹x(a)␴⫹dx(b)Figure 10.7(a) Bar element exhibiting time-dependent displacement. (b) Free-body diagram of adifferential element.where ␳ is density of the bar material. Note the use of partial derivative operators,since displacement is now considered to depend on both position and time. Substituting the stress-strain relation ␴ = E ε = E (∂ u/∂ x ) , Equation 10.51 becomesE∂ 2u∂ 2u=␳∂x2∂t2(10.52)Equation 10.52 is the one-dimensional wave equation, the governing equationfor propagation of elastic displacement waves in the axial bar.In the dynamic case, the axial displacement is discretized asu(x , t ) = N 1 (x )u 1 (t ) + N 2 (x )u 2 (t )(10.53)where the nodal displacements are now expressed explicitly as time dependent,but the interpolation functions remain dependent only on the spatial variable.Consequently, the interpolation functions are identical to those used previouslyfor equilibrium situations involving the bar element: N 1 (x ) = 1 − (x /L ) andN 2 (x ) = x /L .

Application of Galerkin’s method to Equation 10.52 in analogy toEquation 5.29 yields the residual equations as"L∂ 2u∂ 2uN i (x ) E 2 − ␳ 2 A dx = 0i = 1, 2(10.54)∂x∂t0Assuming constant material properties, Equation 10.54 can be written as"L␳A0∂ 2uN i (x ) 2 dx = A E∂t"LN i (x )0∂ 2udx∂x2i = 1, 2403(10.55)Mathematical treatment of the right-hand side of Equation 10.55 is identical tothat presented in Chapter 5 and is not repeated here, other than to recall that theresult of the integration and combination of the two residual equations in matrixform is A E 1 −1u1f1=⇒ [k]{u} = { f }(10.56)u2f2L −1 1⭸␴dx⭸xHutton: Fundamentals ofFinite Element Analysis40410.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsSubstituting the discretized approximation for u(x , t ), the integral on the leftbecomes"L␳A0∂ 2uN i (x ) 2 dx = ␳ A∂t"LN i ( N 1 ü 1 + N 2 ü 2 ) dxi = 1, 2(10.57)0where the double-dot notation indicates differentiation with respect to time. Thetwo equations represented by Equation 10.57 are written in matrix form as "L ␳ AL 2 1N 12N1 N2ü 1ü 1␳Adx== [m]{ü}(10.58)ü 21 2ü 2N1 N2N 2260and the reader is urged to confirm the result by performing the indicated integrations.

Also note that the mass matrix is symmetric but not singular. Equation 10.58 defines the consistent mass matrix for the bar element. The term consistent is used because the interpolation functions used in formulating the massmatrix are the same as (consistent with) those used to describe the spatial variation of displacement. Combining Equations 10.56 and 10.58 per Equation 10.55,we obtain the dynamic finite element equations for a bar element as A E 1 −1␳ AL 2 1ü 1u1f1+=(10.59)ü 2u2f21 26L −1 1or[m]{ü} + [k]{u} = { f }(10.60)and we note that ␳ A L = m is the total mass of the element. (Why is the sign ofthe second term positive?)Given the governing equations, let us now determine the natural frequencies of a bar element in axial vibration.

Per the foregoing discussion of freevibration, we set the nodal force vector to zero and write the frequency equation as|[k] − ␻ 2 [m]| = 0to obtain k − ␻2 m3 m − k + ␻26=0mk − ␻23(10.61)m− k+␻62Expanding Equation 10.62 results in a quadratic equation in ␻22 22m2m− k+␻=0k−␻36(10.62)(10.63)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.4 Bar Elements: Consistent Mass Matrixor␻2k␻ − 12m405=02(10.64)Equation 10.64 has roots ␻ 2 = 0 and ␻ 2 = 12k/m . The zero root arises becausewe specify no constraint on the element; hence, rigid body motion is possibleand represented by the zero-valued natural circular frequency. The nonzero natural circular frequency corresponds to axial displacement waves in the bar,which could occur, for example, if the free bar were subjected to an axial impulseat one end.

In such a case, rigid body motion would occur but axialvibra√␻=12k/m=tion wouldsimultaneouslyoccurwithcircularfrequency1√(3.46/L ) E /␳ . The following example illustrates determination of natural circular frequencies for a constrained bar.EXAMPLE 10.4Using two equal-length finite elements, determine the natural circular frequencies of thesolid circular shaft fixed at one end shown in Figure 10.8a.■ SolutionThe elements and node numbers are shown in Figure 10.8b. The characteristic stiffness ofeach element isk=AE2AE=L /2Lso that the element stiffness matrices are 2AEk (1) = k (2) =L1−1−11The mass of each element ism=␳ AL2and the element consistent mass matrices are ␳ AL 2m (1) = m (2) =12 112A, E1x1LL兾2(a)Figure 10.8(a) Circular shaft of Example 10.4.

(b) Model using two bar elements.22(b)L兾23Hutton: Fundamentals ofFinite Element Analysis40610. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsFollowing the direct assembly procedure, the global stiffness matrix is1 −1 02AE −1 2 −1 [K ] =L0 −1 1and the global consistent mass matrix is2 1 0␳ AL 1 4 1[M] =120 1 2The global equations of motion are then    2 1 0  Ü1 1 −1 0  U1   0 ␳ AL 2AE −1 2 −1  U2 = 01 4 1  Ü2 +    12L0 1 20 −1 1U30Ü3Applying the constraint condition U 1 = 0, we have␳ AL 412 112Ü 2Ü 32AE+L−112−1U2U3 0=0as the homogeneous equations governing free vibration. For convenience, the last equation is rewritten as4112Ü 2Ü 324 E+␳ L22−1−11U2U3 0=0Assuming sinusoidal responsesU 2 = A 2 sin(␻t + ␾)U 3 = A 3 sin(␻t + ␾)differentiating twice and substituting results in−␻ 24112A2A3sin(␻t + ␾) +24 E␳ L22−1−11A2A3sin(␻t + ␾) = 00Again, we obtain a set of homogeneous algebraic equations that have nontrivial solutionsonly if the determinant of the coefficient matrix is zero.

Letting ␭ = 24 E /␳ L 2 , thefrequency equation is given by the determinant 2␭ − 4␻ 2 −␭ − ␻ 2−␭ − ␻ 2 =0␭ − 2␻ 2 which, when expanded and simplified, is7␻ 4 − 10␭␻ 2 + ␭2 = 0Treating the frequency equation as a quadratic in ␻ 2 , the roots are obtained as␻12 = 0.1082 ␭␻22 = 1.3204 ␭Substituting for ␭, the natural circular frequencies are1.611␻1 =LE␳5.629␻2 =LErad/sec␳Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.5 Beam ElementsFor comparison purposes, we note that the exact solution [2] for the natural circularfrequencies of a bar in axial vibration yields the fundamental natural circular frequency√√as 1.571/L E/␳ and the second frequency as 4.712/L E/␳ . Therefore, the error for thefirst computed frequency is about 2.5 percent, while the error in the second frequency isabout 19 percent.It is also informative to note (see Problem 10.12) that, if the lumped mass matrixapproach is used for this example, we obtain1.531␻1 =LE␳3.696␻2 =LErad/sec␳The solution for Example 10.4 yielded two natural circular frequencies forfree axial vibration of a bar fixed at one end.

Such a bar has an infinite number ofnatural frequencies, like any element or structure having continuously distributedmass. In finite element modeling, the partial differential equations governingmotion of continuous systems are discretized into a finite number of algebraicequations for approximate solutions. Hence, the number of frequencies obtainable via a finite element approach is limited by the discretization inherent to thefinite element model.The inertia characteristics of a bar element can also be represented by alumped mass matrix, similar to the approach used in the spring-mass examplesearlier in this chapter.

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