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This phenomenonshould come as no surprise. Mass is an absolute scalar property and therefore independent of coordinate system. A similar development leads to the same conclusion when a bar element is used in modeling three-dimensional truss structures.The complication described for including the additional transverse inertiaeffects of the bar element are also applicable to the one-dimensional beam (flexure) element. The mass matrix for the beam element given by Equation 10.78is applicable only in a one-dimensional model.

If the flexure element is used inmodeling two- or three-dimensional frame structures, additional considerationmust be given to formulation of the element mass matrix owing to axial inertia437Hutton: Fundamentals ofFinite Element Analysis43810. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamicseffects. For beam elements, most finite software packages include axial effects(i.e., the beam element is a combination of the bar element and the twodimensional flexure element) and all appropriate inertia effects are included informulation of the consistent mass matrix.EXAMPLE 10.10As a complete example of modal analysis, we return to the truss structure of Section 3.7,repeated here as Figure 10.19.

Note that, for the current example, the static loads appliedin the earlier example have been removed. As we are interested here in the free-vibrationresponse of the structure, the static loads are of no consequence in the dynamic analysis.With the additional specification that material density is ␳ = 2.6(10) −4 lb-s2/in.4, wesolve the eigenvalue problem to determine the natural circular frequencies and modalamplitude vectors for free vibration of the structure.As the global stiffness matrix has already been assembled, the procedure is notrepeated here. We must, however, assemble the global mass matrix using the elementnumbers and global node numbers as shown. The element and global mass matrices forthe bar element in two dimensions are given by Equation 10.178 as2 (e) ␳ AL  0m=6 10020110200102As elements 1, 3, 4, 5, 7, and 8 have the same length, area, and density, we have M (1) = M (3) = M (4) = M (5) = M (7) = M (8)2 0 1 0(2.6)(10) −4 (1.5)(40) 0 2 0 1=61 0 2 00 1 0 25.2 0 2.6 0 0 5.2 0 2.6 −32= 2.6 0 5.2 0  (10) lb-s /in.0 2.6 0 5.2while for elements 2 and 6 2.6(10)M (2) = M (6) =−42 0 1 0√(1.5)(40 2) 0 2 0 11 0 2 060 1 0 27.3603.680 07.3603.68 (10) −3 lb-s2 /in.= 3.6807.360 03.6807.36Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.11 Bar Element Mass Matrix in Two-Dimensional Truss Structures3746226440 in.1853140 in.540 in.Figure 10.19 Eight-element truss ofExample 10.10.The element-to-global displacement relations are as given in Chapter 3. Using the directassembly procedure, the global mass matrix is12.56000 2.6 03.6800000 012.56 000 2.603.680000  005.2 0002.600000  000 5.2 0002.60000  2.6000 7.8 02.602.6000  02.6000 7.802.602.600  (10) −3[M] = 02.6 0 2.6 0 22.5203.6802.60  3.68 03.680 2.6 0 2.6022.5203.6802.6  0000 2.6 03.68017.7602.60  00000 2.603.68017.7602.6  0000002.602.6010.40 00000002.602.6010.4Applying the constraint conditions U 1 = U 2 = U 3 = U 4 = 0, the mass matrix for theactive degrees of freedom becomes7.8 02.602.6000 0 7.802.602.600  2.6 0 22.5203.6802.60  0 2.6022.5203.6802.6 [Ma ] =  (10) −3 lb-s2 /in. 2.6 03.68017.7602.60  0 2.603.68017.7602.6  002.602.6010.40 0002.602.6010.4439Hutton: Fundamentals ofFinite Element Analysis44010.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsExtracting the data from Section 3.7, the stiffness matrix for the active degrees of freedom is7.5000−3.75000 03.750−3.750000 010.150−1.325 1.325 −3.750  0−3.7506.41.325 −1.32500  105[K a ] =  0lb/in. −3.750−1.325 1.3255.075 −1.32500  001.325 −1.325 −1.325 5.0750−3.75  00−3.750003.750 00000−3.7503.75The finite element model for the truss exhibits 8 degrees of freedom; hence, the characteristic determinant|−␻ 2 [M ] + [K ]| = 0yields, theoretically, eight natural frequencies of oscillation and eight correspondingmodal shapes (modal amplitude vectors).

For this example, the natural modes were computed using the student edition of the ANSYS program [9], with the results shown inTable 10.1. The corresponding modal amplitude vectors (normalized to the mass matrixas discussed relative to orthogonality) are shown in Table 10.2.The frequencies are observed to be quite large in magnitude. The fundamental frequency, about 122 cycles/sec is beyond the general comprehension of the human eyebrain interface (30 Hz is the accepted cutoff based on computer graphics research [10]).The high frequencies are not uncommon in such structures. The data used in this examplecorrespond approximately to the material properties of aluminum; a+light material withgood stiffness relative to weight. Recalling the basic relation ␻ = k/m , high naturalfrequencies should be expected.The mode shapes provide an indication of the geometric nature of the natural modes.As such, the numbers in Table 10.2 are not at all indicative of amplitude values; instead,Table 10.1 Natural ModesFrequencyMode12345678Rad/secHz767.12082.32958.74504.86790.97975.98664.58977.4122.1331.4470.9716.91080.81269.41379.01428.8Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.11 Bar Element Mass Matrix in Two-Dimensional Truss Structures441Table 10.2 Modal Amplitude VectorsModeDisplacement12345678U5U6U7U8U9U10U11U120.26052.207⫺0.77542.1280.51564.118⫺0.78944.2132.194⫺3.2820.7169⫺2.6863.8552.5560.97122.9011.2133.1252.8881.9571.706⫺1.4594.183⫺1.888⫺3.594⫺2.14122.370⫺0.4322⫺3.9341.1334.9172.818⫺1.4455.826⫺0.142⫺4.274⫺0.0550.9080.7370.604⫺1.802⫺0.934⫺3.8300.5691.9811.6296.077⫺3.4004.7721.058⫺2.174⫺0.341⫺2.781⫺3.3194.3924.828⫺4.3680.727⫺0.4640.4833.956⫺4.407⫺1.2055.344(a)(b)Figure 10.20(a) Fundamental mode shape of the truss in Example 10.10.(b) Second mode shape of the truss.these are relative values of the motion of each node.

It is more insightful to examine plotsof the mode shapes; that is, plots of the structure depicting the shape of the structure if itdid indeed oscillate in one of its natural modes. To this end, we present the mode shapecorresponding to mode 1 in Figure 10.20a. Note that, in this fundamental mode, the trussvibrates much as a cantilevered beam about the constrained nodes. On the other hand,Figure 10.20b illustrates the mode shape for mode 2 oscillation. In mode 2, the structureexhibits an antisymmetric motion, in which the “halves” of the structure move in opposition to one another. Examination of the other modes reveals additional differences in themode shapes.Noting that Table 10.2 is, in fact, the modal matrix, it is a relatively simple matter tocheck the orthogonality conditions by forming the matrix triple products[ A] T [M ][ A] = [I ][ A] T [K ][ A] = ␻ 2 [I ]Within reasonable numerical accuracy, the relations are indeed true for this example. Weleave the detailed check as an exercise.Hutton: Fundamentals ofFinite Element Analysis44210.

Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural Dynamics10.12 PRACTICAL CONSIDERATIONSThe major problem inherent to dynamic structural analysis is the time-consumingand costly amount of computation required. In a finite difference technique, suchas that represented by Equation 10.163, the system of equations must be solved atevery time step over the time interval of interest. For convergence, the time stepis generally quite small, so the amount of computation required is huge.

In modalanalysis, the burden is in computing natural frequencies and mode shapes. Aspractical finite element models can contain tens of thousands of degrees of freedom, the time and expense of computing all of the frequencies and mode shapesis prohibitive. Fortunately, to obtain reasonable approximations of dynamicresponse, it is seldom necessary to solve the full eigenvalue problem. Two practical arguments underlie the preceding statement. First, the lower-valued frequencies and corresponding mode shapes are more important in describing structuralbehavior. This is because the higher-valued frequencies most often representvibration of individual elements and do not contribute significantly to overallstructural response.

Second, when structures are subjected to time-dependentforcing functions, the range of forcing frequencies to be experienced is reasonably predictable. Therefore, only system natural frequencies around that range areof concern in examining resonance possibilities.Based on these arguments, many techniques have been developed that allowthe computation (approximately) of a subset of natural frequencies and modeshapes of a structural system modeled by finite elements. While a complete discussion of the details is beyond the scope of this text, the following discussionexplains the basic premises. (See Bathe [6] for a very good, rigorous descriptionof the various techniques.) Using our notation, the eigenvalue problem that mustbe solved to obtain natural frequencies and mode shapes is written as[K ]{ A} = ␻ 2 [M ]{ A}(10.179)The problem represented by Equation 10.179 is reduced in complexity by staticcondensation (or, more often, Guyan reduction [11]) using the assumption thatall the structural mass can be lumped (concentrated) at some specific degreesof freedom without significantly affecting the frequencies and mode shapes ofinterest.

Using the subscript a (active) to represent degrees of freedom of interest and subscript c (constrained) to denote all other degrees of freedom Equation 10.179 can be partitioned into{ Aa }{ Aa }[K aa ] [K ac ][0]2 [M aa ]=␻(10.180){ Ac}{ Ac}[K ca ] [K cc ][0][0]In Equation 10.180, [Maa] is a diagonal matrix, so the mass has been lumped atthe degrees of freedom of interest. The “constrained” degrees of freedom areconstrained only in the sense that we assign zero mass to those degrees. Thelower partition of Equation 10.180 is[K ca ]{ A a } + [K cc ]{ A c } = {0}(10.181)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 2004Referencesand this equation can be solved as{ A c } = −[K cc ]−1 [K ca ]{ A a }(10.182)to eliminate { A c }.

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