MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering IFall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � � � � � � � � � � � � � � � � � � �� � �� � �� � �� � � 3 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 = (u1 , u1 , . . . , u4 , uH V H V 4 ). What shapes are A and ATA? First rows? 6 bars � n = 2(6) − 4 = 8 unknown disp. So Au = 0 has 8 − 6 = 2 independent solutions (mechanisms). H H⎪ ⎜ ⎪ ⎜ ⎪ ⎜ ⎪ ⎜ 1 1u u1 1 11 � 22 � � 1 2 3 4 V V⎡ � ⎡ ⎡ � ⎡ 4 ⎡ � 0 ⎡ � 1 ⎡u u1 1⎡ � ⎡ � ⎡ � ⎡ � ⎡ H H⎡ � ⎡ u ⎡ � ⎡ 1 ⎡ � 1 ⎡⎡ u2 2⎡ � � ⎡ � ⎡ � ⎡ � ⎡ V V u ⎡ � ⎡ ⎡ � ⎡0 −1u2 ⎡ � ⎡ 2 ⎡ � ⎡ �⎡u = = �⎡u = = � ⎡ ⎡ . ⎡H H0 1u ⎡ � ⎡ � ⎡ ⎡ � ⎡ u3 3 ⎡ � ⎡ � ⎡ � ⎡ � ⎡ � ⎡ V V0 ⎡ � 1 ⎡ u ⎡ � ⎡ u ⎡ � ⎡3 3 ⎡ � ⎡ � ⎡ � ⎡ � ⎡ � ⎡ H H0 1⎡ � ⎡ ⎡ � ⎡ u u4 4⎢ � ⎢ ⎢ � ⎢ V V0 −1u u4 4 A = 6 × 8 matrix AT = 8 × 6 matrix ATA = 8 × 8 matrix First row of A: First row of ATA: [ −1 [ 1 0 0 1 −1 0 0 0 0 0 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). d) Third equation in ATw = f with right side fH 2 ? 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 �2 �2 � ⎫ 2c) 0 0 sin − sin 0 0 = 0 0 0 0− cos cos . 2 2 2− − 22 2 2 Hd) w1 + f + w4 cos � + w2 cos � = 0. 2 −fH 2 = w1 + w4 cos � − w2 cos �, cos � = 1/→2. 1� Section 2.7, Problem 3: 11 2 3 4 a) Find 8 − 4 independent solutions to Au = 0. b) Find 4 sets of f’s so ATw = f has a solution. c) Check that uTf = 0 for those four u’s and f’s. 2 3 4 ⎫ a) Solutions: horizontal: vertical: rotation (about node 3): mechanism: u1 = 1 0 1 0 1 0 1 0 � ⎫ u2 = 0 1 0 1 0 1 0 1 � ⎫ u3 = 1 0 1 −1 0 0 0 −1 � ⎫ u4 = 1 0 1 0 0 0 0 0 . ⎟ ⎭ ⎟ ⎭ ⎟⎭ ⎝ ⎝ ⎝ ⎝ ⎝ ⎝b) f1 = ⎝ ⎝ ⎝ ⎝ ⎞ c) Each uT i⎟ ⎭ 1 0 0 0 0 1 0 0⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎣ ⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎣ ⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎣−1 ⎣ ⎝ 0 ⎣ ⎝ 0 ⎣ ⎝ 0 ⎣ ⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎣0 0 1 0 ⎣ f2 = ⎝ ⎣ f3 = ⎝ ⎣ f4 = ⎝ ⎣ ⎣ ⎝ ⎣ ⎝ ⎣ ⎝ ⎣0 ⎣ ⎝ 0 ⎣ ⎝ 0 ⎣ ⎝ −1 ⎣ 0 ⎣ ⎝ ⎣ ⎝ 0 ⎣ ⎝ 0 ⎣ ⎣ ⎝ −1 ⎣ ⎝ ⎣ ⎝ ⎣ 0 ⎠ ⎞ 0 ⎠ ⎞ 0 ⎠ ⎞ 1 ⎠ 0 0 −1 0 fj = 0 Section 2.7, Problem 5: Is ATA positive definite? Semidefinite? Draw complete set of mechanisms. 7 8 1 2 3 4 5 3 1 2 4 5 6 8 bars, 7 nodes, 4 fixed displacements (ua b H 6 , uV 6 , uH 7 , uV 7 ). So n = 2(7) − 4 = 10, and 10 − 8 = 2 solutions. There are 2 solutions, therefore the truss is unstable � not positive definite. ATA 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 ⎛ 2 � xw(x) = − 0 (1 − s)ds + C1 = −x +2 + C1 ⎛ � 1 1 w(1) = 0 � − 1 − 2+ C1 = 0 � C1 =2 x2 1 � w(x) = − x + 2 2 x x ⎛ 2 � ⎛ 3 2 � w(s) 1 s 1 1 x x x u(x) = c(s) ds = c 2 − s +2 ds = c 6 − 2+2 + C2. 0 0 Case 1: u(0) = 0 � 0 + C2 = 0 � C2 = 0. 1 ⎛ x3 x2 x � u(x) = + c 6 − 2 2 Case 2: u(1) = 0 and u(0) = 0 (fixed, fixed). ⎛ � � 1 � x s2 1 s3 s2 �x u(x) = c 2 − s + C1 ds = c 6 − 2+ C1s 00 1 ⎛ x3 x2 � = + C1x + C2 c 6 − 2 u(0) = 0 C2 = 0 � ⎛ � 1 1 1 1 u(1) = 0 � c 6 − 2+ C1 = 0 � C1 =3 ⎛ � 1 x3 x2 x u(x) = + c 6 − 2 3 Section 3.1, Problem 5: f = constant, c jumps from c = 1 for x � 1 to c = 2 for x > 1 . Solve − dw = f2 2 dx duwith w(1) = 0 as before, then solve c dx = w with u(0) = 0. � 1 w(x) = fdx = (1 − x)f x ⎬ � For 0 � x � 1 , �u = (1 − x)f, u(0) = 0 � u(x) = � x(1 − x)fdx = x 2 f. So u � 1 � = 3 f.2 �x 0 x − 2 2 8 For 1 � x � 1, �u = (1 − x)f, u � 1 � = 3 f � u(x) = f …