Assignment 5 solutions

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 5 Section 2.7, Problem 1: How many independent solutions to Au = 0. Draw them and find solutions u = (uH , uV , . . . , uH , uV ). What shapes are A and AT A? First rows? 1 1 4 4 6 bars � n = 2(6) − 4 = 8 unknown disp. So Au = 0 has 8 − 6 = 2 independent solutions (mechanisms). 1 1� 2 2� � � � � � � � � u=� � � � � � � � � ⎪ uH 1 ⎡� uV ⎡ � 1⎡ � ⎡� uH ⎡ � 2⎡ � ⎡� uV ⎡ � 2 ⎡=� ⎡� uH ⎡ � 3 ⎡� ⎡� uV ⎡ � 3 ⎡� uH ⎡ � � 4⎢ uV 4 ⎜ ⎪ 1 1 ⎡ 0⎡ ⎡ ⎡ 1⎡ ⎡ ⎡ 0⎡ ⎡ ⎡3 0⎡ ⎡ 0⎡ ⎡ ⎡ ⎡ 0⎢ 0 ⎜ 2 � � � � � � � � � u=� � � � � � � � ⎪ uH 1 ⎡� ⎡ uV ⎡ � 1 ⎡ 1⎡ � ⎡ ⎡� ⎡ uH ⎡ � 1 ⎡ 2⎡ � ⎡ ⎡ ⎡� u V ⎡ � −1 ⎡ 2 ⎡. ⎡=� ⎡ ⎡� uH ⎡ � 1 ⎡ 3 ⎡� ⎡ ⎡ ⎡� uV ⎡ � 1 ⎡ 3 ⎡� ⎡ ⎡ uH ⎡ � 1 ⎢ � 4⎢ −1 uV 4 ⎜ ⎪ 1 ⎜ 3 4 4 A = 6 × 8 matrix First row of A: AT = 8 × 6 matrix AT A = 8 × 8 matrix [ −1 0 1 0 0 0 0 0 ]. First row of AT A: [ 1 0 −1 0 0 0 0 0 ]. Section 2.7, Problem 2: 7 bars, N = 5 so n = 2N − r = 2(5) − 2 = 8 unknown displacements. a) What motion solves Au = 0? b) By adding one bar, can A become square/invertible? c) Write out row 2 of A (for bar 2 at 45� angle). H d) Third equation in AT w = f with right side f2 ? Solution: a) 8 − 7 = 1 rigid motion, rotation about node 5. b) No � rigid motion only, so adding a bar won’t get rid of motion. Needs another support. � � � � � �� �� �� �� ⎫� 2 2 c) 0 0 − cos � sin � cos � − sin � 0 0 = 0 0 − 22 − 22 2 2 2 2 2 2 H d) w1 + f2 + w4 cos � + w2 cos � = 0. H −f2 = w1 + w4 cos � − w2 cos �, 00 � . → cos � = 1/ 2. 1 Section 2.7, Problem 3: 1 2 3 a) Find 8 − 4 independent solutions to Au = 0. b) Find 4 sets of f ’s so AT w = f has a solution. c) Check that uT f = 0 for those four u’s and f ’s. a) Solutions: horizontal: vertical: 1 2 3 4 4 u1 u2 = = = � � 1 0 1 0 1 0 0 ⎟ 1 0 0 1 1 0 0 0 ⎟ 0 1 0 0 1 0 0 1 rotation (about node 3): u3 � ⎫ ⎫ 1 −1 1 ⎭ 0 0 −1 0 0 ⎫ . � 1 mechanism: u4 = ⎭ ⎟ ⎭ ⎟ ⎭ 1 0 0 ⎝ 1⎣ ⎝0 ⎝ 0⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎝ −1 ⎣ ⎝ 0⎣ ⎝ ⎝ ⎣ ⎝ ⎣ ⎝0 ⎝ 0⎣ ⎝ 0⎣ ⎝ ⎝ ⎣ ⎝1 ⎣ b) f1 = ⎝ ⎝ 0 ⎣ f2 = ⎝ 0 ⎣ f3 = ⎝ 0 ⎝ ⎣ ⎝ ⎣ ⎝ ⎝ ⎣ ⎝ ⎝ 0⎣ ⎝ ⎣ ⎝ −1 ⎣ ⎝0 ⎞ 0⎠ ⎞0 ⎞ 0⎠ 0 0 −1 c) Each uT fj = 0 i 0 ⎣ ⎝ 0⎣ ⎣ ⎝ ⎣ ⎣ ⎝ ⎣ ⎣ ⎝ 0⎣ ⎣ ⎝ 0⎣ ⎣ f4 = ⎝ ⎣ ⎣ ⎝ ⎣ ⎣ ⎝ −1 ⎣ ⎣ ⎝ 0⎣ ⎣ ⎝ ⎣ ⎠ ⎞ 1⎠ 0 ⎫ Section 2.7, Problem 5: Is AT A positive definite? Semidefinite? Draw complete set of mechanisms. 1 1 2 4 4 7 6 5 8 3 2 3 5 a b 8 bars, 7 nodes, 4 fixed displacements (uH , uV , uH , uV ). So n = 2(7) − 4 = 10, and 10 − 8 = 2 solutions. 6 6 7 7 There are 2 solutions, therefore the truss is unstable � not positive definite. AT A must be positive semidefinite because A has dependent columns. 2 Section 3.1, Problem 1: Constant c, decreasing f = 1 − x, find w(x) and u(x) as in equations 9-10. Solve with w(1) = 0, u(1) = 0. ⎛ � �x x2 w(x) = − (1 − s)ds + C1 = −x + + C1 2 0 ⎛ � 1 1 w(1) = 0 � − 1 − + C1 = 0 � C 1 = 2 2 x2 1 � w(x) = −x+ 2 2 � ⎛ � �x �⎛ w(s) 1 x s2 1 1 x3 x2 x u(x) = ds = −s+ ds = − + + C2 . c0 2 2 c6 2 2 0 c(s) Case 1: u(0) = 0 � 0 + C2 = 0 � C2 = 0. 1 u(x) = c Case 2: u(1) = 0 and u(0) = 0 (fixed, fixed). ⎛ x3 x2 x − + 6 2 2 � � � �x s2 1 s3 s2 − s + C1 ds = − + C1 s 2 c6 2 0 0 ⎛3 � 2 1x x − = + C1 x + C2 c6 2 u(0) = 0 � C2 = 0 ⎛ 111 1 u(1) = 0 � − + C1 = 0 � C 1 = c62 3 u(x) = 1 c � x ⎛ u(x) = 1 c ⎛ x3 x2 x − + 6 2 3 � 1 2 Section 3.1, Problem 5: f = constant, c jumps from c = 1 for x � du with w(1) = 0 as before, then solve c dx = w with u(0) = 0. w(x) = �1 x to c = 2 for x > 1 . Solve − dw = f 2 dx f dx = (1 − x)f �u �x �u �x 1 For 0 � x � 2 , = (1 − x)f, u(0) = 0 � u(x) = = (1 − x)f, u �1� 2 For 1 2 � x � 1, = 3 f � u(x) = 8 �x 0 ⎬ (1 − x)f dx = x − f 2 x2 2 � 2 f . So u = 7 16 In summary, �x 1/2 (1 − x)f dx + u �1� − 1 f (1 − x)2 . 4 �1� 2 3 = 8f. u(x) = (1 − x)f u(x) = �⎬ 7 16 f x− x2 f, 2 1 − 4 f (1 − � x)2 , 0�x� 1 2 1 2 � x � 1. 3 Section 3.1, Problem 10: Use three hat functions with h = 1 to solve −u�� = 2 with u(0) = u(1) = 0. 4 Verify that the approximation U matches u = x − x2 at the nodes. 1) �1 0 �v c(x) � u � x dx = �x 2) vi = �i 3) Assume c(x) = 1. 4) Let f (x) = 1. �1 0 f (x)v (x)dx. F1 F2 F3 = = = � 3/4 1/4 � 1/2 0 1 · �1 (x)dx 1 · �2 (x)dx 1 · �3 (x)dx c(x) ���1 · x = = = 1 2 bh 1 2 bh 1 2 bh = 1 2 · 1 2 = 1. 4 = 1 4 �0 0 ⎥ 1 ⎤ · 1 = 4. ⎤ ⎤ ⎦ ⎤ ⎤ ⎤ � right side 1/2 K11 K12 K13 ⎭ = = K21 = K31 = = � 1/2 �1 0 � v1 � x dx c(x) � �2 · �x �v2 �x � v3 �x �1 0 c(x) � �1 · �x � 1/4 � 1/2 = 0 1 · 1 · 4dx + 1/4 1 · (−4) · 4 dx = 8. �� � 1/2 dx = 1/4 1 · (−4) · 4dx = −16 1 = −4. 4 dx = 0. ⎟⎭ ⎟⎭ ⎟ 8 −4 0 u1 0.25 8 −4 ⎠ ⎞ u2 ⎠ = ⎞ 0.25 ⎠ = F. So K U = ⎞ −4 0 −4 8 u3 0.25 ⎟ 0.1875 Solving yields u = ⎞ 0.250 ⎠. This approximation matches u = x − x2 at the nodes since when 0.1875 ⎭ ⎟ ⎭ ⎟ 0.25 0.1875 x = ⎞ 0.50 ⎠, u(x) = ⎞ 0.250 ⎠ , as desired. 0.75 0.1875 Section 3.1, Problem 18: Fixed-free hanging bar u(1) = 0 is a natural boundary condition. To the N hat functions �i at interior meshpoints, add the half-hat that goes up to UN +1 = 1 at the endpoint 1 x = 1 = (N + 1)h. This �N +1 = VN +1 has nonzero slope h . a) The N by N stiffness matrix K for −uxx now has an extra row and column. How does the new last row of KN +1 represent u� (1) = 0? . . ⎝. ⎝. . ⎞. 00 ⎭ ⎟ 0 0 ··· 1 −h 1 h ⎭ ⎣ ⎣. ⎠ ui−1 + 2ui − ui+1 = fi . For row n + 1, −un + 2un+1 − un+2 = fn+1 . If un+2 = un+1 , then the slope at (n + 1) is zero, and so we have −un + un+1 = fn+1 . 4 b) For constant load, find the new last component FN +1 = � � U with the true mesh values of f0 x − 1 x2 . 2 K11 = � ��1 � v1 c(x) · dx = �x �x � c(x) � 1/3 0 � f0 VN +1 dx. Solve KN +1 U = F and compare 1 · 3f0 · 3 dx + � 2/3 1/3 � 2/3 1/3 1 · (−3) · (−3) dx = 6. K12 K13 F1 = K21 = ��1 � v2 · dx = �x �x 1 · (−3) · (3) dx = −3 = K23 = K32 . F2 F3 = K31 = 0. K33 = 3. � 1 12 1 = �1 f0 dx = f0 = f0 23 3 0 �1 12 1 = �2 f0 dx = f0 = f0 23 3 0 �1 11 1 = �3 f0 dx = f0 = f0 23 6 0 K U = F ⎟ ⎭ u1 f0 /3 u2 ⎠ = ⎞ f0 /3 u3 f0 /6 ⎭ 5/18 U = ⎞ 4/9 1/2 5 f0 18 ⎟⎭ 6 −3 0 ⎞ −3 6 −3 ⎠ ⎞ 0 −3 3 ⎭ ⎟ ⎠ ⎠ f0 ⎟ � � Indeed, considering u = f0 x − 1 x2 , we find that the values exactly match up: 2 x= 1 3 2 3 � � 4 f0 9 1 f0 . 2 x= x=1 � 5 ...
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This note was uploaded on 04/12/2011 for the course MATH 18.085 taught by Professor Staff during the Fall '10 term at MIT.

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