Transcript Lecture 6
MECH593 Introduction to Finite Element Methods Finite Element Analysis of 2-D Problems Dr. Wenjing Ye 2-D Discretization Common 2-D elements: 2-D Model Problem with Scalar Function - Heat Conduction • Governing Equation T ( x, y) T ( x, y) Q( x, y) 0 x x y y • Boundary Conditions Dirichlet BC: Natural BC: Mixed BC: in W Weak Formulation of 2-D Model Problem • Weighted - Integral of 2-D Problem ---- T ( x, y ) T ( x, y ) W w x x y y Q( x, y )dA 0 • Weak Form from Integration-by-Parts ---- w T w T 0 wQ( x, y ) dxdy x x y y W T T w dxdy dxdy w x x y y W W Weak Formulation of 2-D Model Problem • Green-Gauss Theorem ----Ω Ω 𝜕 𝜕𝑇 𝜅𝑤 𝑑𝑥𝑑𝑦 = 𝜕𝑥 𝜕𝑥 𝜕 𝜕𝑇 𝜅𝑤 𝑑𝑥𝑑𝑦 = 𝜕𝑦 𝜕𝑦 Γ Γ 𝜕𝑇 𝜅𝑤 𝑛 𝑑𝑠 𝜕𝑥 𝑥 𝜕𝑇 𝜅𝑤 𝑛𝑦 𝑑𝑠 𝜕𝑦 where nx and ny are the components of a unit vector, which is normal to the boundary G. n nx i n y j i cos j sin Weak Formulation of 2-D Model Problem • Weak Form of 2-D Model Problem ----𝜕𝑤 𝜕𝑇 𝜕𝑤 𝜕𝑇 𝜅 + 𝜅 − 𝑤𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 Ω − 𝑤 Γ 𝜕𝑇 𝜕𝑇 𝜅 𝑛 + 𝜅 𝑛 𝑑𝑠 = 0 𝜕𝑥 𝑥 𝜕𝑦 𝑦 EBC: Specify T(x,y) on G T T NBC: Specify x nx y n y on G T T where qn ( s) i j nxi n y j is the normal x y outward flux on the boundary G at the segment ds. FEM Implementation of 2-D Heat Conduction – Shape Functions Step 1: Discretization – linear triangular element (T3) T1 T T11 T22 T33 Derivation of linear triangular shape functions: T2 1 c0 c1 x c2 y Let T3 Interpolation properties i 1 at ith node i 0 at other nodes 1 1 x Same 2 1 1 x1 y 1 x2 1 x3 c0 c1 x1 c2 y1 1 c0 c1 x2 c2 y2 0 c0 c1 x3 c2 y3 0 y1 y2 y3 x3 y1 x1 y3 y y3 y1 2 Ae x x 1 3 x c0 1 x1 c 1 x 2 1 c 1 x 3 2 1 1 x2 y3 x3 y2 1 x y 0 y y 2 3 2 A e 0 x3 x2 3 1 x1 y2 x2 y1 y y1 y2 2 Ae x x 2 1 x y1 y2 y3 1 1 0 0 FEM Implementation of 2-D Heat Conduction – Shape Functions linear triangular element – local (area) coordinates T1 𝜙1 = A2 T2 1 𝑥 𝑦 2𝐴𝑒 𝑥2 𝑦3 − 𝑥3 𝑦2 𝐴1 𝑦2 − 𝑦3 = =𝜉 𝐴 𝑒 𝑥3 − 𝑥2 1 𝑥 𝑦 2𝐴𝑒 𝑥3 𝑦1 − 𝑥1 𝑦3 𝐴2 𝑦3 − 𝑦1 = =𝜂 𝐴 𝑒 𝑥1 − 𝑥3 1 𝑥 𝑦 2𝐴𝑒 𝑥1 𝑦2 − 𝑥2 𝑦1 𝐴3 𝑦1 − 𝑦2 = =1−𝜉−𝜂 =𝜁 𝐴 𝑒 𝑥2 − 𝑥1 A3 A1 𝜙2 = T3 T1 𝜉=1 𝜂=1 𝜙3 = T2 T3 𝜉=0 𝜂=0 1 3 2 FEM Implementation of 2-D Heat Conduction – Shape Functions quadratic triangular element (T6) – local (area) coordinates T1 T4 T6 𝜙1 = 𝜉 2𝜉 − 1 T2 𝜙2 = 𝜂 2𝜂 − 1 T3 T5 𝜙3 = 𝜁 2𝜁 − 1 𝜙4 = 4𝜉𝜂 T1 𝜂=1 𝜉=1 T4 T6 𝜉 = 0.5 T2 T5 T3 𝜙5 = 4𝜂𝜁 𝜙6 = 4𝜉𝜁 𝜉=0 𝜂=0 Serendipity Family – nodes are placed on the boundary for triangular elements, incomplete beyond quadratic Interpolation Function - Requirements • Interpolation condition • Take a unit value at node i, and is zero at all other nodes • Local support condition • i is zero at an edge that doesn’t contain node i. • Interelement compatibility condition • Satisfies continuity condition between adjacent elements over any element boundary that includes node i • Completeness condition • The interpolation is able to represent exactly any displacement field which is polynomial in x and y with the order of the interpolation function Formulation of 2-D 4-Node Rectangular Element – Bi-linear Element (Q4) Let u( , ) 1u1 2u2 3u3 4u4 4 3 1 1 4 1 𝜙2 = 4 1 𝜙3 = 4 1 𝜙4 = 4 𝜙1 = 2 1−𝜉 1−𝜂 1+𝜉 1−𝜂 1+𝜉 1+𝜂 1−𝜉 1+𝜂 Note: The local node numbers should be arranged in a counter-clockwise sense. Otherwise, the area Of the element would be negative and the stiffness matrix can not be formed. 1 2 3 4 Formulation of 2-D 4-Node Rectangular Element – Bi-linear Element (Q4) Physical domain (physical element) 4 3 Reference domain (master element) 4 3 y 1 x 2 1 𝑥 = 𝜙1 𝑥1 + 𝜙2 𝑥2 + 𝜙3 𝑥3 + 𝜙4 𝑥4 𝑦 = 𝜙1 𝑦1 + 𝜙2 𝑦2 + 𝜙3 𝑦3 + 𝜙4 𝑦4 2 FEM Implementation of 2-D Heat Conduction – Element Equation • Weak Form of 2-D Model Problem ----𝜕𝑤 𝜕𝑇 𝜕𝑤 𝜕𝑇 𝜅 + 𝜅 − 𝑤𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 + 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 Ω𝑒 𝑛 Assume approximation: 𝑇= 𝑤𝑞𝑛 𝑑𝑠 = 0 Γ𝑒 𝑇𝑗 𝜙𝑗 𝑗=1 and let w(x,y)=i(x,y) as before, then Ω𝑒 𝜕𝜙𝑖 𝜕 𝜅 𝜕𝑥 𝜕𝑥 𝑛 𝑇𝑗 𝜙𝑗 𝑗=1 𝜕𝜙𝑖 𝜕 + 𝜅 𝜕𝑦 𝜕𝑦 𝑛 𝑇𝑗 𝜙𝑗 − 𝜙𝑖 𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 𝑗=1 𝑛 + 𝜙𝑖 𝑞𝑛 𝑑𝑠 = 0 𝐾𝑖𝑗 𝑇𝑗 = 𝑗=1 Γ𝑒 where i j i j K ij dxdy x x y y We 𝜙𝑖 𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 − Ω𝑒 𝜙𝑖 𝑞𝑛 𝑑𝑠 Γ𝑒 Linear Triangular Element Equation 𝑛 𝐾𝑖𝑗 𝑇𝑗 = 𝑗=1 1 Ω𝑒 x2 y3 x3 y2 y A1 y2 y3 2 Ae x x Ae 3 2 1 x 2 3 𝜙𝑖 𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 − x3 y1 x1 y3 y A2 y3 y1 2 Ae x x Ae 1 3 1 x 1 x1 y2 x2 y1 y A3 y1 y2 2 Ae x x Ae 2 1 x 𝜙𝑖 𝑞𝑛 𝑑𝑠 Γ𝑒 l23 l23 K l23 l31 4 Ae l23 l12 l23 l31 l23 l12 l31 l31 l31 l12 l31 l12 l12 l12 Q1 q1 F Q2 q2 Q q 3 3 where 𝒍𝒊𝒋 is the length vector from the ith node to the jth node. Assembly of Stiffness Matrices 𝑛𝑒 𝐹𝑖 𝑒 = 𝜙𝑖 𝑄 𝑥, 𝑦 𝑑𝑥𝑑𝑦 − Ω𝑒 𝑈1 = 𝑇1 1 , 𝑈2 = 𝑇2 𝜙𝑖 𝑞𝑛 𝑑𝑠 = + 𝑇1 2 , 𝑈3 = 𝑇3 𝑒 𝑇𝑗 𝑒 𝑗=1 Γ𝑒 1 𝐾𝑖𝑗 1 + 𝑇4 2 , 𝑈4 = 𝑇2 2 , 𝑈5 = 𝑇3 2 , Imposing Boundary Conditions The meaning of qi: q1(1) 3 3 1 1 (1) (1) q n 1 ds G1 1 (1) h12 2 (1) (1) q n 2 ds G1 (1) h12 3 1 1 2 (1) h23 (1) h12 qn(1)2(1) ds qn(1)2(1) ds qn(1)2(1) ds (1) h23 (1) h31 (1) h23 (1) (1) q n 3 ds G1 2 (1) h31 qn(1)2(1) ds qn(1)2(1) ds q3(1) 1 (1) h23 (1) h31 q2(1) 3 (1) h12 qn(1)1(1) ds qn(1)1(1) ds 2 qn(1)1(1) ds qn(1)1(1) ds qn(1)1(1) ds (1) h12 qn(1)3(1) ds qn(1)3(1) ds qn(1)3(1) ds qn(1)3(1) ds qn(1)3(1) ds (1) h31 (1) h23 (1) h31 Imposing Boundary Conditions Consider (1) 2 q q ds q ds q1(2) q ds q ds q4(2) (1) (1) n 2 (1) h12 (1) 3 q2 q q q3 q3(1) q4(2) (1) 2 (2) 1 (1) (1) n 2 q (1) h23 (1) (1) n 3 Equilibrium of flux: (1) h23 qn(2) qn(2)1(2) ds qn(2)4(2) ds (2) h41 (2) h34 (1) h31 qn(1) qn(2)4(2) ds (2) h12 (1) h23 (1) (1) n 3 qn(2)1(2) ds (2) h41 ( 2) h41 FEM implementation: qn(1)2(1) ds (1) h23 qn(2)1(2) ds; ( 2) h41 q2 (1) h12 qn(1)2(1) ds qn(1)3(1) ds (1) h23 ( 2) h12 qn(2)1(2) ds qn(2)4(2) ds ( 2) h41 q3 (1) h31 qn(1)3(1) ds ( 2) h34 qn(2)4(2) ds Calculating the q Vector Example: qn 0 T 293K qn 1 2-D Steady-State Heat Conduction - Example A D AB: qn 0 CD: convection h 50W DA and BC: T 180 C o 0.6 m C B 0.4 m y x m C 2 o T 25o C