Hutton - Fundamentals of Finite Element Analysis (523155), страница 80

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Here, we describe Newmark’s method [7] alsoreferred to as the constant acceleration method. In the Newmark method, it is assumed that the acceleration during an integration time step t is constant and anaverage value. For constant acceleration, we can write the kinematic relations˙ )t + ␦¨av␦(t + t ) = ␦(t ) + ␦(t˙ + t ) = ␦(t˙ ) + ␦¨av t␦(tt 22(10.156)(10.157)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.10Transient Dynamic ResponseThe constant, average acceleration is¨ + t ) + ␦(t¨ )␦(t␦¨av =2(10.158)Combining Equations 10.156 and 10.158 yields˙ )t + [␦(t¨ + t ) + ␦(t¨ )]␦(t + t ) = ␦(t ) + ␦(tt 24(10.159)which is solved for the acceleration at t + t to obtain44¨ + t ) =˙ ) − ␦(t¨ )␦(t[␦(t + t ) − ␦(t )] −␦(t(10.160)t 2tIf we also substitute Equations 10.158 and 10.160 into Equation 10.157, we findthe velocity at time t + t to be given by2˙ + t ) =˙ )␦(t[␦(t + t ) − ␦(t )] − ␦(t(10.161)tEquations 10.160 and 10.161 express acceleration and velocity at t + t interms of known conditions at the previous time step and the displacement att + t.

If these relations are substituted into Equation 10.155, we obtain, after abit of algebraic manipulation,24[M ]{␦(t + t )} +[C ]{␦(t + t )} + [K ]{␦(t + t )}2tt 44¨ )} +˙ )} += {F (t + t )} + [M ] {␦(t{␦(t)}{␦(ttt 22˙ )} +(10.162)+ [C ] {␦(t{␦(t )}tEquation 10.162 is the recurrence relation for the Newmark method.

While therelation may look complicated, it must be realized that the mass, damping, andstiffness matrices are known, so the equations are just an algebraic system in theunknown displacements at time t + t. The right-hand side of the system isknown in terms of the solution at the previous time step and the applied forces.Equation 10.162 is often written symbolically as[ K̄ ]{␦(t + t )} = {Feff (t + t )}(10.163)42[M ] +[C ] + [K ]t 2t(10.164)with[ K̄ ] ={Feff (t + t )} = {F (t + t )}44¨ )} +˙ )} ++ [M ] {␦(t{␦(t{␦(t)}tt 22˙ )} +{␦(t )}+ [C ] {␦(tt(10.165)433Hutton: Fundamentals ofFinite Element Analysis43410. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsThe system of algebraic equations represented by Equation 10.163 can be solvedat each time step for the unknown displacements.

For a constant time step t,matrix [ K̄ ] is constant and need be computed only once. The right-hand side{Feff (t + t )} must, of course, be updated at each time step. At each time step, thesystem of algebraic equations must be solved to obtain displacements. For this reason, the procedure is known as an implicit method. By back substitution throughthe appropriate relations, velocities and accelerations can also be obtained.The Newmark method is known to be unconditionally stable [8]. While thedetails are beyond the scope of this text, stability (more to the point, instability) ofa finite difference technique means that, under certain conditions, the computeddisplacements may grow without bound as the solution procedure “marches” intime.

Several finite difference methods are known to be conditionally stable,meaning that accurate results are obtained only if the time step t is less than aprescribed critical value. This is not the case with the Newmark method. Thisdoes not mean, however, that the results are independent of the selected time step.Accuracy of any finite difference technique improves as the time step is reduced,and this phenomenon is a convergence concern similar to mesh refinement in afinite element model.

For dynamic response of a finite element model, we mustbe concerned with not only the convergence related to the finite element meshbut also the time step convergence of the finite difference method selected. Asdiscussed in a following section, finite element software for the transient dynamicresponse requires the user to specify “load steps,” which represent the changein loading as a function of time. The software then solves the finite element equations as if the problem is one of static equilibrium at the specified loading condition. It is very important to note that the system equations represented by Equation 10.163 are based on the finite element model, even though the solutionprocedure is that of the finite difference technique in time.10.11 BAR ELEMENT MASS MATRIXIN TWO-DIMENSIONAL TRUSSSTRUCTURESThe bar-element-consistent mass matrix defined in Equation 10.58 is valid onlyfor axial vibrations.

When bar elements are used in modeling two- and threedimensional truss structures, additional considerations are required, and the massmatrix modified accordingly. When a truss undergoes deflection, either staticallyor dynamically, individual elements experience both axial and transverse displacement resulting from overall structural displacement and element interconnections at nodes. In Chapter 3, transverse displacement of elements was ignoredin development of the element stiffness matrix as there is no transverse stiffnessowing to the assumption of pin connections, hence free rotation. However, in thedynamic case, transverse motion introduces additional kinetic energy, whichmust be taken into account.Hutton: Fundamentals ofFinite Element Analysis10.

Structural DynamicsText© The McGraw−HillCompanies, 200410.11 Bar Element Mass Matrix in Two-Dimensional Truss Structuresv2u22v (x, t)v1u (x, t)u1dx1(b)(a)Figure 10.17 A bar element in two-dimensional motion:(a) Nodal displacements. (b) Differential element.Consider the differential volume of a bar element undergoing both axial andtransverse displacement, as shown in Figure 10.17. We assume a dynamic situation such that both displacement components vary with position and time.

Thekinetic energy of the differential volume is& 2 '21∂u∂v1dT = ␳ A dx+= ␳ A dx ( u̇ 2 + v̇2 )(10.166)2∂t∂t2and the total kinetic energy of the bar becomes"L"L112T = ␳ A u̇ dx + ␳ A v̇2 dx220(10.167)0Observing that the transverse displacement can be expressed in terms of thetransverse displacements of the element nodes, using the same interpolationfunctions as for axial displacement, we haveu(x , t ) = N 1 (x )u 1 (t ) + N 2 (x )u 2 (t )v(x , t ) = N 1 (x )v1 (t ) + N 2 (x )v2 (t )Using matrix notation, the velocities are written as u̇u̇(x , t ) = [N 1 N 2 ] 1u̇ 2 v̇v̇(x , t ) = [N 1 N 2 ] 1v̇2and element kinetic energy becomes"L"L11TTTT = ␳ A{u̇}[N ] [N ] dx {u̇} + ␳ A{v̇}[N ] T [N ] dx {v̇}2200(10.168)(10.169)(10.170)435Hutton: Fundamentals ofFinite Element Analysis43610. Structural DynamicsC H A P T E R 10Text© The McGraw−HillCompanies, 2004Structural DynamicsExpressing the nodal velocities as˙ ={␦} u̇ 1  v̇ 1(10.171)u̇  2v̇2the kinetic energy expression can be rewritten in the form1 ˙ T (e) T = {␦}m 2 {␦}2N120N1 N20L" 1 ˙ TN120N1 N2  0= {␦}␳A dx{␦}2 N1 N220N20 00N1 N20N22(10.172)From Equation 10.172, the mass matrix of the bar element in two dimensions isidentified asN120N1 N202 0 1 0"L  0 (e) N120N1 N2 dx = ␳ AL  0 2 0 1 m 2 = ␳A N N260N20 1 0 2 0 1 200 1 0 20N1 N20N22(10.173)The mass matrix defined by Equation 10.173 is described in the element(local) coordinate system, since the axial and transverse directions are defined interms of the axis of the element.

How, then, is this mass matrix transformed tothe global coordinate system of a structure? Recall that, in Chapter 3, the elementaxial displacements are expressed in terms of global displacements via a rotationtransformation of the element x axis. To reiterate, the transverse displacementswere not considered, as no stiffness is associated with the transverse motion.Now, however, the transverse displacements must be included in the transformation to global coordinates because of the associated mass and kinetic energy.Figure 10.18 depicts a single node of a bar element oriented at angle ␪ relative to the X axis of a global coordinate system. Nodal displacements in theelement frame are u 2 , v2 and corresponding global displacements are U 3 , U 4 ,respectively.

As the displacement in the two coordinate systems must be thesame, we haveu 2 = U 3 cos ␪ + U 4 sin ␪v2 = −U 3 sin ␪ + U 4 cos ␪oru2v2=cos ␪−sin ␪sin ␪cos ␪U3U4(10.174)(10.175)Hutton: Fundamentals ofFinite Element Analysis10. Structural DynamicsText© The McGraw−HillCompanies, 200410.11 Bar Element Mass Matrix in Two-Dimensional Truss Structuresv2U4u2U3␪Figure 10.18 The relationof element and global displacements at a single node.As the same relation holds at the other element node, the complete transformation is   u00 U1  1  cos ␪ sin ␪U v1−sin␪cos␪002= [R] {U }=(10.176) 00cos ␪ sin ␪   u2  U3 v200−sin ␪ cos ␪U4Since the nodal velocities are related by the same transformation, substitutioninto the kinetic energy expression shows that the mass matrix in the global coordinate system is (e) (e) M 2 = [R] T m 2 [R](10.177)where we again use the subscript to indicate that the mass matrix is applicable totwo-dimensional structures.If the matrix multiplications indicated in Equation 10.177 are performed foran arbitrary angle, the resulting global mass matrix for a bar element is found to be2 0 1 0 (e) ␳ AL  0 2 0 1 M2 =(10.178)6 1 0 2 00 1 0 2and the result is exactly the same as the mass matrix in the element coordinate system regardless of element orientation in the global system.

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