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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 1 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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