Quiz 2 Solutions, Fall 2004

18.085 Quiz 2 November 3, 2006 Your PRINTED name is: 1) (25 pts.) Professor Strang SOLUTIONS Grading 1 2 3 4 This network (square grid) has 12 edges and 9 nodes. 1 2 5 4 7 8 3 3/4 6 1/2 9 1/4 u2 = 1 1/2 u8 = 0 3/4 1/2 1/4 (a) Do not write the incidence matrix ! Do not give me a MATLAB code ! Just tell me: (1) How many independent columns in A ? 8 (2) How many independent solutions to AT y = 0 ? 12 − 8 = 4 (3) What is row 5 (coming from node 5) of AT A ? � ⎦ I do want the whole of row 5. 0 −1 0 −1 4 −1 0 −1 0 (b) Suppose the node 2 has voltage u2 = 1, and node 8 has voltage u8 = 0 (ground). All edges have the same conductance c. On the second picture write all of the other voltages u1 to u9 . Check equation 5 of AT Au = 0 (at the middle node). � � 1 1 1 − − +4 −1−0=0 2 2 2 1 2) (25 pts.) Suppose that square grid becomes a plane truss (usual pin joints at the 9 V H V nodes). Nodes 2 and 8 now have supports so uH 2 = u2 = u8 = u8 = 0. edge 3 1 4 5 2 Any numbering of edges (a) Think about the strain-displacement matrix A. Are there any mecha­ nisms that solve Au = 0 ? If there are, tell me carefully how many and draw a complete set. I believe 3. (b) Suppose now that all 8 of the outside nodes are fixed. Only node 5 is free to move. There are forces f5H and f5V on that node. The bars connected to it (North East South West) have constants cN cE cS cW . What is the (reduced) matrix A for this truss on the right ? What is V the reduced matrix AT CA? What are the displacements uH 5 and u5 ? � 1 0 ⎡ � ⎢ � ⎢ � 0 1⎢ ⎢ A=� � ⎢ � −1 0⎢ � ⎣ 0 −1 (any row order is OK) � AT CA = � cE + c W 0 0 cN + cS ⎡ ⎣ H uH 5 = f5 /(cE + cW ) V uV 5 = f5 /(cN + cS ) For 1 point, is that truss (fixed at 8 nodes) statically determinate or indeterminate ? indeterminate 2 3) (25 pts.) This question is about the velocity field v(x, y) = (0, x) = w(x, y). (a) Check that div w = 0 and find a stream function s(x, y). Draw the streamlines in the xy plane and show some velocity vectors. (b) Is this shear flow a gradient field (v = grad u) or is there rotation ? If you believe u exists, find it. If you believe there is rotation, explain how this is possible with the streamlines you drew in part (a). Solution. (a) div(0, x) = 0 + 0 = 0 0= �s �y x=− �s �x 1 s = − x2 (+C) 2 x = constant are vertical streamlines. � � � � � � � � � � �v1 �v2 =0 = 1. �y �x �v2 �v1 Possible explanation for vorticity (rotation) − = 1. �x �y (b) Not a gradient field because Right side going faster than left side produces rotation. 3 � � 4) (25 pts.) Suppose I use linear finite elements (hat functions �(x) = trial functions V (x)). The equation has c(x) = 1 + x and a point load: Fixed-free with u(0) = 0 � � � � d du 1 − (1 + x) =� x− dx dx 2 and u � (1) = 0 . Take h = 1/3 with two hats and a half-hat as in the notes. (a) On the middle interval from 1/3 to 2/3, U (x) goes linearly from U1 to U2 . Compute � 2/3 ⎥ ⎤2 c(x) U � (x) dx and 1/3 � 2/3 1/3 Write those answers as � ⎡� ⎡ � � 4.5 −4.5 U ⎣� 1⎣ U1 U2 � −4.5 4.5 U2 � � 1 � x− U(x) dx . 2 and � U1 U2 � � ⎡ 1 � 2 ⎣. 1 2 You have found the 2 by 2 “element stiffness matrix” and the 2 by 1 “element load vector.” (b) On the first and third intervals, similar integrations give � �� �� � � �� � and U1 0 ; U1 3.5 U1 � ⎡� ⎡ � ⎡ � � � � 0 5.5 −5.5 U2 ⎣ � ⎣ and U2 U3 � U2 U3 � ⎣ . 0 −5.5 5.5 U3 Assuming your calculations and mine are correct, what would be the overall finite element equation KU = F ? (Not to solve) 4 Solution. (a) � 2/3 1/3 � U2 − U 1 (1 + x) 1/3 �2 �2/3 (1 + x)2 �� 2 dx = � 9(U2 − U1 ) 2 1/3 ( 53 )2 − ( 34 )2 9(U2 − U1 )2 2 � ⎡� ⎡ � � 4.5 −4.5 U ⎣� 1⎣ = U1 U2 � −4.5 4.5 U2 = � 2/3 1/3 � 1 � x− 2 � � � � 1 u(x) dx = U (NOT U(x) dx !!) 2 1 = (U1 + U2 ) (halfway up) 2 � ⎡ � � 1 = U1 U2 � 2 ⎣ 1 2 (b) � 1/3 0 � U1 (1 + x) 1/3 � �2 �1/3 � (1 + x)2 = 9 U12 �� = 3.5 U12 2 0 8 −4.5 0 ⎡ � ⎢ � ⎢ K = � −4.5 10 −5.5 ⎢ � ⎣ 0 −5.5 5.5 5 � ⎡ 1 �2⎢ � ⎢ F = � 12 ⎢ � ⎣ 0