23. Chapter 23 - Transient Analyses _a4

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<ul><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 1/10</p><p>Chapter 23 The Transient Dynamic Analysis_____________________________________</p><p>230</p><p>CHAPTER 3</p><p>THE TRANSIENT DYNAMIC ANALYSIS</p><p>The transient dynamic analysis is used in civil engineering to determine the</p><p>time-varying displacements, strains, stresses and forces of a structure during</p><p>time-dependent loads, as earthquake or wind. Furthermore, when nonlinear</p><p>material properties are associated to the model, the process highlights the</p><p>regions where the linear-elastic behavior is exceeded (for frame structures,</p><p>the birth of so called plastic hinges) and also the order in which these</p><p>nonlinear regions appear over the model. When exceeding the linear-elastic</p><p>behavior, the amount of residual strains reveals the quantity of energydissipated during the plastic deformation.</p><p>The transient dynamic analysis requires generally more computer resources</p><p>because, at least, it is equivalent to a multi-step static analysis. Three</p><p>solution methods are usually available: thefull method, thereduced method</p><p>and the mode superposition method. Each one has its own advantage,</p><p>according to the analysis aims, computing time and the required</p><p>preprocessing work.</p><p>The transient dynamic equilibrium equation is the equation of motion</p><p>)(ta</p><p>FKCM =++ &amp;&amp;&amp; (23.1)</p><p>where</p><p> M, Cand Kare the mass, damping and stiffness matrices;</p><p> &amp;&amp; is the nodal acceleration vector, &amp; the nodal velocity vector and</p><p>the displacement vector;</p><p> )(taF is the applied load vector.</p><p>The procedure used for solving the linear equation is the time integration</p><p>method (the Newmark method). It uses finite differences expansions over</p><p>the time interval t, in which the following relationships are assumed:</p><p>( ) taa nnnn ++= ++ 11 1 &amp;&amp;&amp;&amp;&amp;&amp; (23.2)</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 2/10</p><p>______________________Basics of the Finite Element Method Applied in Civil Engineering</p><p>231</p><p>2</p><p>1121 tbbt nnnnn </p><p> +</p><p> ++= ++ &amp;&amp;&amp;&amp;&amp; (23.3)</p><p>where</p><p> a, bare the Newmark integration parameters</p><p> nn ttt = +1 </p><p> n&amp;&amp; , n</p><p>&amp; , n are the nodal acceleration, velocity and displacement</p><p>vectors at time nt </p><p> 1+n&amp;&amp; , 1+n</p><p>&amp; , 1+n are the nodal acceleration, velocity and</p><p>displacement vectors at time 1+nt </p><p>Since the first intend is to calculate the displacement 1+n and the seismic</p><p>load is usually applied as ground acceleration d&amp;&amp; , the governing equation at</p><p>time 1+nt is:</p><p>1111 ++++ =++ nnnn dMKCM &amp;&amp;&amp;&amp;&amp;</p><p> (23.4)</p><p>Rearranging the equations (23.4), using some notations:</p><p>( ) nnnnn aaa &amp;&amp;&amp;&amp;&amp; 32101 = ++ </p><p>1761 ++ ++= nnnn aa &amp;&amp;&amp;&amp;&amp;&amp; (23.5)</p><p>20</p><p>1</p><p>tba</p><p>= ;</p><p>tb</p><p>aa</p><p>=1 ;</p><p>tba</p><p>=</p><p>12 ; 1</p><p>2</p><p>13 =</p><p>ba ; 14 =</p><p>b</p><p>aa ;</p><p>= 2</p><p>25</p><p>b</p><p>ata ; ( )ata = 16 ; taa =7 </p><p>and substituting 1+n&amp;&amp; , the general form becomes:</p><p>( ) =++ +110 naa KCM </p><p>) )nnnnnnn aaaaaa CdM &amp;&amp;&amp;&amp;&amp;&amp;&amp;&amp; 5413201 ++++++= + (23.6)</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 3/10</p><p>Chapter 23 The Transient Dynamic Analysis_____________________________________</p><p>232</p><p>Once the solution is obtained for 1+n , the accelerations and the velocitiesare updated with the equations (23.5). The solution is unconditionally stable</p><p>for</p><p>2</p><p>2</p><p>1</p><p>4</p><p>1</p><p>+ ab ;</p><p>2</p><p>1a and 0</p><p>2</p><p>1&gt;++ ba *. The Newmark method</p><p>becomes the constant average acceleration method for2</p><p>1=a and</p><p>4</p><p>1=b .</p><p>23.1 The stiffness matrix</p><p>The procedures to calculate the global stiffness matrix of the structure were</p><p>already described in the case of static analyses. The same procedures are</p><p>available for transient dynamic analyses. As before, the stiffness matrix is amatrix of constants in which the geometrical characteristics and the material</p><p>propertiesEdand (the dynamic Youngs modulus and the Poissons ratio)</p><p>are involved. It is common to assign different values for the Youngs</p><p>modulus in dynamic analyses as for static analyses.</p><p>23.2 The mass matrix</p><p>The mass matrix assessment is explained in chapter 22. The procedure is</p><p>also available for the transient dynamic analysis.</p><p>23.3 The damping matrix</p><p>The damping matrix of a finite element is expressed in terms of a damping</p><p>force which is proportional with its mass and also in terms of the</p><p>deformation velocity. The first dependence is expressed by the parameter </p><p>as a force vector in the integral expression of the functional, while the</p><p>second one corresponds to a stress vector &amp;dd E= which is due to</p><p>elements stiffness. In a general form, the element damping matrix is:</p><p>( )dVVe</p><p>d</p><p>TT</p><p>e += BEBNNc (23.7)</p><p>The total damping matrix of the structure Cis obtained following the sameassembling procedures as for the global stiffness matrix.</p><p>*Zienkiewicz, O. C.- The Finite Elements Method, Mc Graw Hill, 1979</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 4/10</p><p>______________________Basics of the Finite Element Method Applied in Civil Engineering</p><p>233</p><p>Using the notations = and EE d = the general form of the Rayleigh</p><p>damping model outcomes:</p><p>KMC += (23.8)</p><p>where and are scalar multipliers.</p><p>The values of and are generally not known directly, but are calculated</p><p>from the modal damping ratios i , which are the ratios of actual damping to</p><p>critical damping for each vibration mode i. If i is the natural circular</p><p>frequency of mode i, and satisfy the relation:</p><p>22</p><p>i</p><p>i</p><p>i</p><p> += (23.9)</p><p>In practical structural problems the mass damping may be ignored (= 0).</p><p>Because only one value of can be prescribed during a load step, the</p><p>dominant frequency should be chosen. When using both and multipliers,</p><p>it is commonly assumed that the sum of their terms are nearly constant over</p><p>a frequency range. Therefore, knowing the damping ratio and a circular</p><p>frequency range i to j , two simultaneous equations can be solved for </p><p>and .</p><p>A more complete expression of the damping matrix takes into account the</p><p>damping properties assigned to materials and the element damping matrices</p><p>available for some elements in the element library:</p><p>==</p><p>+++=nelem</p><p>k</p><p>kj</p><p>nmat</p><p>j</p><p>j11</p><p>CKKMC (23.10)</p><p>where j is the stiffness matrix multiplayer for materialj, Kjthe structures</p><p>region with assigned materialj, nmatthe number of materials with dampingproperties, Ck the element damping matrices and nelem the number of</p><p>elements with specified damping.</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 5/10</p><p>Chapter 23 The Transient Dynamic Analysis_____________________________________</p><p>234</p><p>23.4 Transient dynamic solution methods</p><p>The most applied solution methods for solving equation (23.6) are the full</p><p>solution method, the reduced solutionmethod and the mode superposition</p><p>method.</p><p>Thefull solution methodsolves the equation directly without any additional</p><p>assumptions. The initial (or start) values of 0 , 0&amp; , 0</p><p>&amp;&amp; must be known. If</p><p>non-zero initial conditions are required, they are assigned by performing a</p><p>static analysis load step.</p><p>The initial displacements are:</p><p>if there is no previous load step</p><p>displacement vector from the static load step</p><p>The initial velocities are:</p><p>in the case of an initial static load case; t is the timeincrement</p><p>otherwise</p><p>The initial acceleration is 0&amp;&amp; = 0.</p><p>The total force applied on each node is computed as a sum of inertia,</p><p>damping and static loads over all elements connected at that location. The</p><p>element inertia load is computed by eem</p><p>e &amp;&amp;MF = , where eM is the element</p><p>mass matrix and e&amp;&amp; the element acceleration vector. While the acceleration</p><p>of each nodal DOF is given by equation (23.3) for time 1+nt , e&amp;&amp; is an</p><p>average acceleration between 1+nt and nt , since it is assumed that the</p><p>average acceleration is the true one.</p><p>The damping load part of the element is computed by eec</p><p>e &amp;CF = , where</p><p>eC is the element damping matrix and e&amp;</p><p>the element vector velocity, givenby equation (23.5).</p><p>=s</p><p>00</p><p>=0t</p><p>s</p><p> 0&amp;</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 6/10</p><p>______________________Basics of the Finite Element Method Applied in Civil Engineering</p><p>235</p><p>The nodal reaction loads are computed as the negative of the sum of inertia,damping and static effects over all elements which are connected to a given</p><p>fixed displacement node.</p><p>A faster solution can be obtained applying the reduced method. In the</p><p>general motion equation (23.4) the characteristic matrices M, Cand Kare</p><p>computed considering two main assumptions:</p><p>- only a selected set of masterDOF are taking into account (the DOF</p><p>considered essential to characterize the response of the structure);</p><p>- the matrices are constant during the loading time interval, thus</p><p>nonlinear effects are suppressed (plasticity, large deflections, etc).</p><p>Supplementary assumptions as a constant time step size and the use of only</p><p>concentrated loads applied on master DOF are also considered.</p><p>The reduced motion equation system yields:</p><p>( ) =++ +110 naa KCM))))</p><p>nnnnnnn aaaaaa CdM)&amp;&amp;</p><p>)&amp;</p><p>)))&amp;&amp;</p><p>)&amp;</p><p>))&amp;&amp;</p><p>)</p><p>5413201 ++++++= + (23.11)</p><p>where the ^ symbol denotes the reduced matrices and vectors.</p><p>The reduced solution is obtained by inverting the left-hand side of the</p><p>equation and performing a matrix multiplication at each time step. The full</p><p>solution is computed by expandingthe reduced solution to the other DOF of</p><p>the model, which are calledslaveDOF. Once the expansion pass is done by</p><p>computing all nodal displacements, the elements stresses are evaluated.</p><p>The mode superposition methodsums factored mode shapes obtained from a</p><p>modal analysis to calculate the dynamic response. The mass and stiffness</p><p>matrices are constant during the analysis, as well as the time step size. The</p><p>element damping matrices are neglected.</p><p>The equations of motion may be expressed as:</p><p>snd sFFKCM +=++ &amp;&amp;&amp; (23.12)</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 7/10</p><p>Chapter 23 The Transient Dynamic Analysis_____________________________________</p><p>236</p><p>wherend</p><p>F are the time varying nodal forces (which can be also expressedas</p><p>nddM &amp;&amp; when inertia forces), sF the load vector from the modal analysis</p><p>andsa load scale factor.</p><p>The displacement vector is defined in terms of modal coordinatesyi, so that:</p><p>=</p><p>=n</p><p>i</p><p>ii</p><p>1</p><p>y (23.13)</p><p>where i is the mode shape of mode iand n the number of mode shapes</p><p>taking into account. By substituting into (23.12):</p><p>sndn</p><p>i</p><p>ii</p><p>n</p><p>i</p><p>ii</p><p>n</p><p>i</p><p>ii sFFyKyCyM +=++ === 111</p><p>&amp;&amp;&amp; (23.14)</p><p>Multiplying with Tj</p><p> and applying the orthogonality conditions of the</p><p>natural mode shapes (22.6), the outcome is:</p><p>( )sndTjjjTjjjTjjjTj sFFyKyCyM +=++ &amp;&amp;&amp; (23.15)</p><p>The coefficients of this equation are substituted as follows:- according to the normality condition 1= j</p><p>T</p><p>j M ;</p><p>- it can be demonstrated that iijT</p><p>j 2= C , where i is the</p><p>fraction of critical damping and i the natural circular frequency of</p><p>modej;</p><p>- the third term is22</p><p>jj</p><p>T</p><p>jjj</p><p>T</p><p>j == MK ;</p><p>- the right-hand side term is conveniently noted as fj.</p><p>The motion equation system in modal coordinates yields:</p><p>jjjjjjj fyyy =++</p><p>2</p><p>2 &amp;&amp;&amp;</p><p> (23.16)</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 8/10</p><p>______________________Basics of the Finite Element Method Applied in Civil Engineering</p><p>237</p><p>Since j is an arbitrary mode, the previous system represents a set of nuncoupled equations with the nunknowns yj. Using relationship (23.13), the</p><p>modal coordinates are back-converted into geometric displacements . If the</p><p>initial modal analysis is performed on a reduced model (using only master</p><p>DOF) the matrices and load vectors would be expressed in terms of these</p><p>DOF. In this case an expansion pass is necessary to compute the</p><p>displacements of theslaveDOF.</p><p>23.5 Practical aspects of performing a transient dynamic analysis</p><p>Since the solution of equation (23.4) depends on the external excitation,</p><p>represented by the ground acceleration</p><p>d (variable in time), the acceleration</p><p>vector has to be expressed as a constant value for each time step. Thus, a</p><p>continuous acceleration recording (or an artificial one) has to be transformed</p><p>into a set of discrete values, according to the total duration of the dynamic</p><p>load and the chosen time step, t. Figure 23.1 represents the discretization</p><p>of a recorded accelerogram lasting 0.5 seconds into 10 equal intervals, the</p><p>constant values corresponding to each time step being listed below.</p><p>Time step 1 2 3 4 5 6 7 8 9 10 11</p><p>a (m/s2) 0 1 3 2 0 -2 -1 0 2 1 3</p><p>Fig. 23.1 Discretization of an accelerogram</p><p>Regarding the response acceleration of the structure, it can be assumed to beconstant or with a linear variation over the time step. If the acceleration is</p><p>constant, the response velocities have a linear variation over the load step</p><p>0</p><p>1</p><p>2</p><p>3</p><p>-1</p><p>-2</p><p>0,05 0,10 0,15 0,20 0,25 0,30 0,40 0,500,35 0,45 s </p><p>)(2sma</p><p>t t</p><p>t t</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 9/10</p><p>Chapter 23 The Transient Dynamic Analysis_____________________________________</p><p>238</p><p>and the response displacements is a quadratic one (see figure 23.2). Theinitial conditions may be chosen as:</p><p>t= 0 0 = 0.</p><p> =..</p><p>0 </p><p>Fig. 23.2 Acceleration, velocity and displacement evolutions over the time step</p><p>The time step size t should be chosen small enough for acquiring an</p><p>acceptable numerical precision and to avoid numerical instabilities. The</p><p>precision of the numerical integration is essentially related to the time step</p><p>size. The main factors that should be taken into account are:</p><p>- the magnitude of the lowest natural vibration period which is</p><p>significant for the structural response; the time step size should notexceed 1/10 of its value, in order to achieve a good representation of</p><p>the corresponding mode shape;</p><p>Constant accelerations</p><p>Linear velocities</p><p>Quadratic displacements</p><p>t </p><p>t&amp; </p><p>t&amp;&amp; </p><p>tt +</p><p>tt +&amp; </p><p>tt +&amp;&amp; </p><p>t</p><p>( )ttt ++= &amp;&amp;&amp;&amp;&amp;&amp;2</p><p>1)( </p><p>tttt +=+ )( &amp;&amp;</p><p>..</p><p>tttt +=+ )( &amp;&amp;.. </p><p>t+t</p></li><li><p>8/11/2019 23. Chapter 23 - Transient Analyses _a4</p><p> 10/10</p></li></ul>