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File: Lecture2
phy411 lecture notes part 2 alice quillen october 2 2018 contents 1 canonical transformations 2 1 1 poisson brackets 2 1 2 canonical transformations 3 1 3 canonical transformations are ...

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                                                                                                            1/11/2017
                                       Lecture 2: One‐dimensional 
                                                      Problems
                                          APL705 Finite Element Method
                                           Steps in FEM Problems
                              1.   System idealization –the given actual problem is broken 
                                   down into idealized elements
                              2.   Element equilibrium – the equilibrium requirements are 
                                   established in terms of displacements or the state 
                                   variables (or primary unknowns)
                              3.   Assembly of elements –the element interconnections 
                                   are used to develop a set of simultaneous equations in 
                                   the unknown state variables
                              44.  CalculaCalculattionion ofof rreesponsesponse –– byby solvingsolving thethe sysysstteemm ofof 
                                   equations, the response of each element and thereby 
                                   that of the entire system under consideration is 
                                   determined
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                                                                                                                  1/11/2017
                                            Direct Stiffness Method
                                • The direct aprroach
                                ThThe process off obtbtaiiniing ththe tttotall or systtem stiftifffness matitrix bby 
                                summation of individual element stiffness matrices is called 
                                direct stiffness method.
                                                             n
                                                       K=∑k(j)
                                                            i=1
                                       (j) 
                                Here k are element stiffness matrices. Through the examples 
                                discussed later, we will see that this approach is general and can 
                                be applied to other non‐structural problems also. This method is 
                                simple and gives a basic idea about obtaining the behaviour of a 
                                finite element of a continuum.
                                                    Some Examples 
                                • A single element model
                                HerHeree wewe cacann coconnssiiddeerr aa rorodd, springspring, trusstruss membermember, beambeam, pipepipe oror 
                                any such simple structural element for analysis. Consider the 
                                following single linear spring model as an example.
                                              i               k              j
                                              U                              U
                                               i                             f j
                                              fi                             j
                                • HerHeree ii andand jj araree thethe twtwoo nodesnodes wherwheree thethe 
                                   displacements and forces are present and k is 
                                   the stiffness (spring constant) of the linear 
                                   spring element under consideration
                                                                                                                             2
                                                                                                                                                                                                               1/11/2017
                                                                                            Simple Examples
                                                          •    Force – displacement relation
                                                          Let us consider the equilibrium of forces at nodes iand j:
                                                                                      f =k(u −u )=ku −ku
                                                                                        i            i       j            i          j
                                                                                      f =k(u −u)=ku −ku
                                                                                        j             j       i            j         i
                                                          Expressing these in a matrix form:
                                                                        ⎡                       ⎤⎧ u ⎫ ⎧ f ⎫
                                                                               k       −k ⎪ i ⎪=⎪ i ⎪⇒ku= ff
                                                                        ⎢⎢   −kk         kk     ⎥⎥⎨ u         ⎬ ⎨ ff ⎬
                                                                        ⎣⎣                      ⎦⎦⎪⎪      j   ⎪⎪     ⎪⎪       j   ⎪⎪
                                                                                                  ⎩           ⎭ ⎩                 ⎭
                                                          • Here [k] –element stiffness matrix, {u}‐ nodal 
                                                               displacement vector and {f}‐ element force vector
                                                                                             Direct Approach
                                                          • Two‐element model
                                                          NowNow lelett usus coconnssiiddeerr thethe ffoollowingllowing linearlinear elaselastictic springssprings ccoonnectnnecteded endend 
                                                          to end as follows:
                                                               1              x              K
                                                                                               1                                          K                            3
                                                               U                                                                            2                          U
                                                               f 1                           (1)                        2                 (2)                             3
                                                                1                                                       U                                              f3
                                                                                                                        f 2
                                                                                                                         2
                                                          • HerHeree agagainain thethe nomenclanomenclaturturee isis similarsimilar atat nodesnodes 
                                                               and elements. Now we have 3 nodes and two 
                                                               elements with their stiffness as k1 and k2
                                                               respectively. 
                                                                                                                                                                                                                                   3
                                                                                                                                                                                                                1/11/2017
                                                                                             Direct Approach
                                                          •     Simple Two‐element Example
                                                          Let us extend the idea of to a two element system now:
                                                          For Element 1                    ⎡                        ⎤⎧             ⎫ ⎧ (1) ⎫
                                                                                                 k         −k         ⎪ u ⎪ ⎪ f                           ⎪
                                                                                           ⎢       1            1 ⎥⎨          1    ⎬=⎨ 1 ⎬
                                                                                           ⎢ −k             k       ⎥⎪ u ⎪ ⎪ f(1) ⎪
                                                          For Element 2                    ⎣        1         1     ⎦⎩        2    ⎭ ⎩ 2 ⎭
                                                                                          ⎡     k          −k ⎤⎧ u ⎫ ⎧ f(2) ⎫
                                                                                          ⎢        2            2    ⎥⎪        2    ⎪=⎪ 1                  ⎪
                                                                                          ⎢⎢   −kk          kk       ⎥⎥⎨ u          ⎬ ⎨ ((22)) ⎬
                                                                                          ⎣         2          2     ⎦⎪⎪       3    ⎪⎪     ⎪⎪    f2        ⎪⎪
                                                                                                                       ⎩            ⎭ ⎩                    ⎭
                                                          • For assembling the system or total stiffness matrix, 
                                                                there are two approaches: (i) Equilibrium of forces 
                                                                (ii) Superposition of element matrices
                                                                          Assembling Stiffness Matrix
                                                          •     (i) Considering Equilibrium of forces
                                                                           NNodde11 ff ((11)) = RR ⇒kk u −kk u
                                                                                             1            1         1 1        1   2
                                                                           Node2 f(1)+ f(2) = R ⇒ku +(k +k )u −k u
                                                                                              2          1             2         1 1           1       2      2       2 3
                                                                           Node3 f(2) = R ⇒k u +k u
                                                                                              2            3         2 2          2 3
                                                                             ⎡      k            −k               0 ⎤⎧ u ⎫ ⎧ R ⎫
                                                                             ⎢        1               1                  ⎥⎪        1    ⎪ ⎪ 1 ⎪
                                                                             ⎢⎢   −k          k +k             −k ⎥⎥⎨⎨ u                ⎬⎬ = ⎨⎨      R ⎬⎬
                                                                             ⎢         1        1       2           2    ⎥⎪        2    ⎪ ⎪ 2 ⎪
                                                                                    0           −k               k               u                   R
                                                                             ⎢                        2            2     ⎥⎪        3    ⎪ ⎪ 3 ⎪
                                                                             ⎣                                           ⎦⎩             ⎭ ⎩                  ⎭
                                                          • (ii) Superposition of element matrices
                                                                                                                                                                                                                                    4
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...Phy lecture notes part alice quillen october contents canonical transformations poisson brackets are symplectic generating functions for example transformation action angle coordinates the harmonic oscillator some geometry tangent bundle dierential forms and wedge product form geometrically vectors generate flows trajectories hamiltonian ow two extended phase space ows preserve bracket on connection to lagrangian discretized systems integrators surfaces of section following a lie derivative examples orbits in plane galaxy or around massive body epicyclic motion jacobi integral shearing sheet symmetries conserved quantities that commute with noether s theorem integrability it is straightforward transfer coordinate using formulation as minimization can be done any system however hamil tonian only hamilton equations dened here those equivalently also search con served equivalent nice preserves consider function f q p t h where mo menta time dependence df fq fp dt we write this wecan short...

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