MIT2_094S11_lec9

MIT2_094S11_lec9 - 2.094 — Finite Element Analysis of...

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Unformatted text preview: 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 9 - u/p formulation Prof. K.J. Bathe MIT OpenCourseWare We want to solve Reading: Sec. 4.4.3 I. Equilibrium � τij,j + fiB = 0 in Volume S τij nj = fi f on Sf (9.1) II. Compatibility III. Stress-strain law Use the principle of virtual displacements � �T C � dV = R (9.2) V We recognize that if ν → 0.5 �V → 0 (�V = �xx + �yy + �zz ) E →∞ κ= 3(1 − 2ν ) p = −κ�V must be accurately computed (9.3) (9.4) (9.5) Solution τij = κ�V δij + 2G�� ij (9.6) where � 1 δij = Kronecker delta = 0 i=j i=j � (9.7) Deviatoric strains: �V �� = �ij − δij ij 3 τij = −pδij + 2G�� ij (9.8) � p=− τkk � 3 (9.9) (9.2) becomes � � �T � � � C � dV + �V κ�V dV = R V V � � T �� C � �� dV − �T p dV = R V V (9.10) (9.11) V 37 MIT 2.094 9. u/p formulation We need another equation because we now have another unknown p. p + κ�V = 0 (9.12) p (p + κ�V ) dV = 0 �V � p� − p �V + dV = 0 κ V (9.13) � (9.14) For an element, ˆ u = Hu � ˆ � = BD u ˆ �V = BV u (9.16) (9.17) ˆ p = Hp p (9.18) (9.15) Plane strain (�zz = 0) 4/1 element Example �V = �xx + �yy ⎡ 1 �xx − 3 (�xx + �yy ) 1 ⎢ � − (� + �yy ) �� = ⎢ yy 3 xx ⎣ γxy − 1 (�xx + �yy ) 3 Reading: Ex. 4.32 in the text (9.19) ⎤ ⎥ ⎥ ⎦ (9.20) Note: �zz = 0 but �� z = 0! z� ˆ p = Hp p = [1]{p0 } (9.21) p(x, y ) = p0 (9.22) We obtain from (9.11) and (9.14) � �� �� � ˆ Kuu Kup u R = ˆ Kpu Kpp p 0 (9.23) � T BD C � BD dV V � T =− BV Hp dV V � T =− Hp BV dV V � T1 =− Hp Hp dV κ V Kuu = (9.24a) Kup (9.24b) Kpu Kpp (9.24c) (9.24d) 38 MIT 2.094 9. u/p formulation In practice, we use elements that use pressure interpolations per element, not continuous between elements. For example: Then, unless ν = 0.5 (where Kpp = 0), we can use static condensation on the pressure dof’s. ˆ ˆ ˆ Use p equations to eliminate p from the u equations. � � − ˆ Kuu − Kup Kpp1 Kpu u = R (9.25) (In practice, ν can be 0.499999. . . ) The “best element” is the 9/3 element. (9 nodes for displacement and 3 pressure dof’s). p(x, y ) = p0 + p1 x + p2 y (9.26) The inf-sup condition Reading: Sec. 4.5 ⎡ ⎤ =�V � �� � � ⎢ qh � · vh dVol ⎥ ⎢ ⎥ inf sup ⎢ Vol ⎥≥β>0 ���� ���� ⎣ �qh � �vh � ⎦ � �� � qh ∈Qh vh ∈Vh (9.27) for normalization Qh : pressure space. If “this” holds, the element is optimal for the displacement assumption used (ellipticity must also be satisfied). Note: infimum = largest lower bound supremum = least upper bound For example, inf {1, 2, 4} = 1 sup {1, 2, 4} = 4 inf {x ∈ R; 0 < x < 2} = 0 sup {x ∈ R; 0 < x < 2} = 2 (9.23) rewritten (κ = ∞, full incompressibility). Diagonalize using eigenvalues/eigenvectors. For a mesh of element size h we want βh > 0 as we refine the mesh, h → 0 39 MIT 2.094 9. u/p formulation For (entry [3,1] in matrix) assume the circled entry is the minimum (inf) of . Also, all entries in the matrix not shown are zero. Case 1 βh = 0 ⎧ 0 · uh |i = 0 (from the bottom equation) ⎨ ⇒ α · uh |i + 0 · ph |j = Rh |i (from the top equation) ⎩ ���� �=0 ⇒ no equation for ph |j ⇒ spurious pressure! (any pressure satisfies equation) Case 2 βh = small = � � · uh |i = 0 ∴�· ⇒ uh |i = 0 ph |j + uh |i · α = Rh |i Rh |i ⇒ ph |j = � � ⇒ displ. = pressure → 0 large � as � is small The behavior of given mesh when bulk modulus increases: locking, large pressures. See Example 4.39 textbook. 40 MIT OpenCourseWare http://ocw.mit.edu 2.094 Finite Element Analysis of Solids and Fluids II Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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This note was uploaded on 12/29/2011 for the course ENGINEERIN 2.094 taught by Professor Prof.klaus-jürgenbathe during the Spring '11 term at MIT.

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