sample Finite Element Analysis of Composite Materials Using ANSYS 2nd edition solution manual pdf_1 (2)

1. cl i ck h ere t o d ow nl oad solutions MAnuAl FoR Finite eleMent AnAlysis oF CoMposite MAteRiAls using Ansys® 2nd edition by ever J. Barbero @solutionmanual1 K15077_SM_Cover.indd 1 14/02/15 4:22 pm

2. cl i ck h ere t o d ow nl oad @solutionmanual1 K15077_SM_Cover.indd 2 14/02/15 4:22 pm

3. cl i ck h ere t o d ow nl oad solutions MAnuAl FoR Finite eleMent AnAlysis oF CoMposite MAteRiAls using Ansys® 2nd edition by ever J. Barbero Boca Raton London New York CRC Press is an imprint of the Taylor & Francis Group, an informa business @solutionmanual1 K15077_SM_Cover.indd 3 14/02/15 4:22 pm

4. cl i ck h ere t o d ow nl oad CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150211 International Standard Book Number-13: 978-1-4665-1691-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com @solutionmanual1 K15077_SM_Cover.indd 4 14/02/15 4:22 pm

5. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition Chapter 1 Solution to Problems in Chapter 1 of FEACM Using ANSYS—2nd Edition © Ever J. Barbero (2014) all rights reserved Problem 1.1 The PVW says that: ? ? V Assuming u(x) as: ? ? 1 0 C x C C x u ? ? ? Using the boundary condition ? 0 ? x u F E x L ? ? ? ? ? ? S ? V ? ?? ? ? ? ? ? dV t u dS f u dV 0 ij ij i i i i 2 x 2 ? ? ? ? ? ? 0? THEN 0 C 0 Because of it’s a case of uniaxial load without body forces (fi = 0), the PVW can be rewritten as: ? ? 0 So, we can calculate: ? ? u x ? ? ? ?? L ? ? ? ? ?? ? ? ? ? A dx F u x L 0 x x x ? ? ? ? ? 2 u x x C x C 1 2 ? ? ? ? C 2 C x 1 2 x ? ? ? C ? 2 x C x 1 2 ? ? ? x 2 ? ? ? E E C 2 C x x 1 And ? ? ? L x x ? ? A A A x 1 2 L L ? ?x ? A A ? ? 1 2 A A x 1 L Now, © Ever J. Barbero (2014) all rights reserved. P a g e | 1 @solutionmanual1 K15077_SM_Cover.indd 5 14/02/15 4:22 pm

6. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition ? ?? ? 2 2 1 2 1 2 1 ?? ? ? L ? L ? ? ? ? A A ? ? ? ? ? ? ? ? ? ? ? 2 1 2 ?? E C C x C x C A x dx F L C L C 0 1 2 0 ? ? L ? L ? ? ? ? ? ? A A ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 1 2 ?? ?? E C C C x C C x C C x C A 2 2 4 x dx F L C L C 0 1 1 1 2 2 1 2 2 1 1 2 0 ? ? L ? L ? ? ? ? ? ? A A ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 1 2 ?? ?? E 4 C x C 2 C C C C x C C A x dx F L C L C 0 2 2 1 2 2 1 1 1 1 1 2 0 L ? ? dx ? ? ? ? ? ? ? ? ? ? 2 EA 4 C x C 2 C C C C x C C 1 2 2 1 2 2 1 1 1 0 ? ? L ? L ? ? ? ? A A ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 3 2 2 1 2 E 4 2 0 C x C C C C C x C C x dx F L C L C 2 2 1 2 2 1 1 1 1 2 0 ? ? 4 ? ? ? ? ? ? ? ? ? 3 2 EA ?? L C C L C C C C LC C ?? 1 2 2 1 2 2 1 1 1 3 A ? ? ? L ? ? ? ? A 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 3 2 2 1 2 E ?? L C C L C C C C L C C ?? F L C L C 0 2 2 1 2 2 1 1 1 1 2 3 2 Reordering the equation to express it like a linear combination of the variations of Ci’s ? 2 ? ? ? ? A E C L EA C L EA C ? ? ? 1 2 ? ? ? ? ? ? ? ? ? ? 2 2 ?? ?? C EA L C EA LC E A A LC E A A L C FL 1 1 2 1 1 1 2 1 1 2 2 3 ? 4 2 ? ? ? ? ? ? ? ? ? 3 2 3 2 2 A L C E A A L C FL 0 ?? ?? 2 1 2 1 1 1 2 2 1 2 1 3 3 Since ?Ci’s are independent variations of Ci’s values there are two equations involved in the above one. The solution of those equations (2x2 system) will give the values of coefficients Ci’s. ? ? 1 2 1 1 2 ? ? ? ? 1 2 ? ? ? ? ? ? ? ? 2 2 ?? ?? ?? ?? EA L E A A L C EA L E A A L C FL 1 1 2 2 3 ? ? ? ? 2 4 ? ? ? ? ? ? ? ? ? ? 2 2 3 3 2 ?? ?? ?? ?? EA L E A A L C EA L E A A L C FL 1 1 2 1 1 1 2 2 3 3 Calling © Ever J. Barbero (2014) all rights reserved. P a g e | 2 @solutionmanual1 K15077_SM_Cover.indd 6 14/02/15 4:22 pm

7. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition 1 ? ? ? ? ? ? EA E A A 1 1 2 2 2 ? ? ? ? ? ? EA E A A 1 1 2 3 4 ? ? ? ? ? ? EA E A A 1 1 2 3 Then ? ? ? ? C LC F 1 2 ? ? LC ? ? LC F 1 2 The solution for this system is: ? ? ? ? ? F ? C ? ? 2 ? ? ?? 2 L ? ? ?? ? ? ? ? ? F ? C ? 1 ? 2 Finally, the displacement solution is: ? ? ?? ? ? ? ? ? ? ? ? ? ? F F ? ? x ? ? 2 u x x ? ? ? ? ? ? ?? 2 2 L In terms of the known data ? ? ? ? ? 6 A 3 A 4 A Fx ? ? x ? ? 1 2 2 u Fx ? ? ? ? L ? ? 2 1 2 2 2 1 2 2 E A 4 A A A E A A A A 1 2 1 2 Fx When A1 = A2 = A the well known result ? ? ? u x is gotten. EA Problem 1.2 The PVW says that: x L © Ever J. Barbero (2014) all rights reserved. P a g e | 3 @solutionmanual1 K15077_SM_Cover.indd 7 14/02/15 4:22 pm

8. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition ? ? V Assuming ?(x) as: ? ? 1 0 x C C x ? ? ? ? Using the boundary condition ? ? x ? ? ? ? ? ? ? S ? V ? ?? ? ?? ? ?? ? dV T dS f dV 0 ij ij i i i i 2 C x 2 ? ? ? ? ? ? 0? THEN 0 0 C 0 Because of it’s a case of pure torsion without body forces (fi = 0), the PVW can be rewritten as: ? 0 L ? ? ? ? ? ? ? ? ? ? ?? ? ? A dx T x L 0 1 2 xy xy x So, we can calculate: ? ? x ?? ? ? ? ? 2 x C x ? C 1 ? 2 ? ? u v ? ? ?? ? ?? ? ? ? ? ? ? 2 r C C x xy x 1 2 ? ? y x ? ? u ? ? ? ?? ? y 0 0 u ? y ? ? v ? ? x ? ? ? ? ?? ? ? ? v r r C 2 C ? x x x 1 2 ? x ? ? r 2 ? ? ? ? ? ? ? ? ? x C 2 x C 1 2 xy 1 2 ? ? x 2 ? ? ? ? G Gr C 2 C xy xy x 1 And ? ? ? L x x ? ? A A A x 1 2 L L ? ? ? A A ? ? ? ? 2 A A 1 2 x r x 1 x L ? ?x ? 2 2 r r ? ? 2 2 r r 1 2 x 1 L Now, L ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 Gr C 2 C x r C 2 x C r dx T L C L C 0 1 x 1 2 x 1 2 x 1 2 2 0 Solving the integral and organizing the result © Ever J. Barbero (2014) all rights reserved. P a g e | 4 @solutionmanual1 K15077_SM_Cover.indd 8 14/02/15 4:22 pm

9. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition ? ? ? ? ? ? ? ? 1 1 60 2 5 C LT C T ? ? ? ? ? ? ? ? r ? ? ? ? ? ? ? ? 4 2 2 4 4 2 2 4 2 2 C 3 C ? ? G ? C r r r ? r LC r r r r 1 1 C 1 1 1 2 2 r 2 1 1 2 2 12 ? ? ? ? ? ? ? 4 2 2 4 2 4 2 2 4 2 3 4 3 6 G LC r r r L r r r 1 1 2 2 2 1 1 2 2 1 2 The coefficients of the terms ?Ci’s must be equal on RHS and LHS, therefore a 2x2 system can be written. LC C ? ? 4 5 1 ? ? 2 ? ? ? ? ? 1 2 ? 5 C LC 2 with ? ? ? ? 4 2 2 4 r r r r ? 1 1 2 2 ? ? ? 4 2 2 4 r 2 r r 3 r 1 1 2 2 ? ? ? ? 4 2 2 4 r 12 3 r r 6 r 1 1 2 2 T ? ? ? G Solving the system ? ? ? ? 5 ? 4 ? ? ?? ? ?? ? C 1 ? ? 5 ?? 8 ? 2 ? ? ? 2 2 ? ? 5 ? ?? ? ?? ? C 2 ? 5 ? ?? 8 L 2 Tx ? ? ? ? ? r r r When the well known result is gotten. 1 2 4 G r Problem 1.3 Initial system z y x (a)The reflection of z-axis about the plane x-y can be reached by one of the following four rotations: ? = (+?) or (-?) about the x-axis or y-axis. y x © Ever J. Barbero (2014) all rights reserved. P a g e | 5 z’ @solutionmanual1 K15077_SM_Cover.indd 9 14/02/15 4:22 pm

10. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition (b)The reflection of y-axis about the plane x-z can be reached by one of the following four rotations: ? = (+?) or (-?) about the x-axis or z-axis. (c)x’-axis represents the reflection of x-axis about the plane y-z. It can be reached by one of the following four rotations: ? = (+?) or (-?) about the y-axis or z-axis. Notice that the new system doesn’t accomplish the right hand rule. ? ? ? ? 1 0 0 y x z’ x’ y z’ ? 0 ? 1 0 0 ? 1 0 ij a Rotation matrix: ? ? ? ? ? ? ? Problem 1.4 Rotation about the x-axis ? ? sin 0 ? ? ? 1 0 0 1 0 0 ? ? ? ? ? ? ? ? ?? ? ? ? 0 ? ? 0 cos sin 0 1 0 a a ? ? ? ? ij ij ? ? ? ? ? ? ? ? ? ? ? ? cos 0 1 Rotation about the y-axis ? ? sin ? ? ? 0 ? ? ? ? ? cos 0 sin 1 0 0 ? ? ? ? ? ? ? ? ?? ? ? 0 1 0 1 0 a a ? ? ? ? ij ij ? ? ? ? ? ? ? ? ? ? 0 cos 0 0 1 © Ever J. Barbero (2014) all rights reserved. P a g e | 6 @solutionmanual1 K15077_SM_Cover.indd 10 14/02/15 4:22 pm

11. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition Rotation about the x-axis ? ? ? ? 1 0 0 ? ? ? ? ? ? cos sin 0 1 0 0 ? ? ? ? ? ? ? ? ? ?? ? ? ? sin cos 0 0 1 0 a a ? ? ? ? ij ij ? ? ? ? ? ? ? ? 0 0 1 Problem 1.5 Scilab™ is a free, open source software with a syntax virtually identical to MATLAB® Scilab can be downloaded from http://scilab.org The verification requested in this problem can be done and understood best with a short Scilab program. Note there is a function defined to construct the T matrix (1.34). MATLAB code to produce (1.34) in symbolic form is provided in the textbook immediately below (1.34) but requires you to have the symbolic package, which like everything else cost money, so I have switched to Scilab for simple numerical code, and to Maxima™ (also free) for symbolic manipulation. // Program P1p5.sce http://scilab.org clc; funcprot(0); mode(0); //displays every line unless suppressed by ; (MATLAB default) sigma = [[10,2,1];[2,5,1];[1,1,3]] a = [[sqrt(3)/2,1/2,0];[-1/2,sqrt(3)/2,0];[0,0,1]]//Ex.1.2 // (1.26) sigmap = a * sigma * a' // (1.29) sigmaVoigt = [sigma(1,1);sigma(2,2);sigma(3,3);... sigma(2,3);sigma(1,3);sigma(1,2)] T = Transform(a) sigmapp = T * sigmaVoigt //end program function [T,Tbar] = Transform(a) // T(6,6), Tbar(6,6) : stress and strain transformation matrices // a(3,3) vector transformation matrix // tensor indexes i,j,p,q = 1..3 // Voigt contracted notation indexes alpha,bbeta = 1..6 // transformation matrix [T] T(1:6,1:6) = 0; for i=1:1:3 for j=1:1:3 if i==j; alpha = j; else alpha = 9-i-j; end for p=1:1:3 for q=1:1:3 if p==q bbeta = p; else bbeta = 9-p-q; end T(alpha,bbeta) = 0; if alpha<=3 & bbeta<= 3; T(alpha,bbeta)=a(i,p)*a(i,p); end © Ever J. Barbero (2014) all rights reserved. P a g e | 7 @solutionmanual1 K15077_SM_Cover.indd 11 14/02/15 4:22 pm

12. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition if alpha> 3 & bbeta<= 3; T(alpha,bbeta)=a(i,p)*a(j,p); end if alpha<=3 & bbeta>3; T(alpha,bbeta)=a(i,q)*a(i,p)+a(i,p)*a(i,q);end if alpha>3 & bbeta>3; T(alpha,bbeta)=a(i,p)*a(j,q)+a(i,q)*a(j,p);end end end end end R = eye(6,6); R(4,4)=2; R(5,5)=2; R(6,6)=2;// Reuter matrix Tbar = R*T*R^(-1) endfunction Problem 1.6 The transversely isotropic material can be considered as a simplification of the orthotropic material, when it has the same properties in the transverse-to-the-fibers plane (2-3 plane). For this case the set of elastic properties describing the material reduces from 13 to 5, namely E1, E2, G12, ?12, ?23 (in engineering terms). The simplification can be shown using the compliance matrix for two materials. ? ? ? ? ? 44 0 0 0 0 0 S ? ? ? ? ? ? ? 0 0 0 0 0 0 S S S 1 31 21 E E 2 E 11 12 13 1 3 ? ? ? ? ? ? ? 0 0 0 0 0 0 S S S 1 32 12 ? ? ? ? E E E 12 22 23 1 2 3 ? ? ? ? ? ? ? 0 0 0 0 0 0 S S S 1 13 23 ? ? S E E E 13 0 23 0 33 0 ? ? ? 1 2 3 ? ORTHO 0 0 0 0 0 0 0 S 1 ? ? ? ? G 23 ? ? ? ? 0 0 0 0 0 S 1 G 55 0 ? ? ? ? ? ? 13 0 0 0 0 0 0 0 0 0 ? ? ? ? 1 G 66 12 ? ? S S S 0 0 0 11 12 12 ? ? S S S 0 0 0 ? ? 12 22 23 ? ? S S S 0 0 0 ? ? STRANS 12 0 23 0 22 0 ? ? ? ; But S44=2(S22-S23). S 0 0 ? ? 44 0 ? ? S 0 0 0 0 66 0 ? ? ? ? S 0 0 0 0 66 In terms of the Engineering Constants ? ? ? ? ? ? 0 0 0 1 21 21 E E E 1 2 2 ? ? ? ? ? ? 0 0 0 1 32 12 ? ? E E E 1 2 3 ? ? ? ? ? ? 0 0 0 1 E 23 12 ? ? S E E E ? ? 2 ? ? G ? 1 2 2 ; And ? ? TRANS 23 ? 0 0 0 0 0 2 1 1 ? ? G 23 23 ? ? 0 0 0 0 0 1 G ? ? 12 0 0 0 0 0 ? ? ? ? 1 G 12 A MATLAB® routine, named P1p6.m (code attached), was built to construct both stiffness and compliance matrices for a transversely isotropic single layer laminate, provided that its five properties are given in a file © Ever J. Barbero (2014) all rights reserved. P a g e | 8 @solutionmanual1 K15077_SM_Cover.indd 12 14/02/15 4:22 pm

13. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition named props.dat as a column vector in the following order: E1, E2, G12, ?12, ?23. The routine calculate the compliance matrix from the elastic properties and invert it to get the stiffness matrix. The routine was verified using the following data set: Material: E-Glass / Isophthalic Polyster (took from [4]) E1 = 37.9 GPa; E2 = 11.3 GPa; G12 = 3.3 GPa; ?12 = 0.3; ?23 = 0.38 (from the matrix) The results were: CODE PROBLEM 1.6 (P1p6.m) % Routine that solve the Problem 1.6. It builds up the Stiffness and % Compliance matrix for a transversely isotropic single layer composite % provided its five elastic properties in the file props.dat clear; clc; % Read data load props.dat % Engineering properties E1 = props(1); E2 = props(2); G12 = props(3); v12 = props(4); v23 = props(5); % Compliance matrix [S] S(1:3,1:3) = [1/E1 -v12/E1 -v12/E1; -v12/E1 1/E2 -v23/E2; -v12/E1 -v23/E2 1/E2]; S(4,4) = 2*(1+v23)/E2; S(5,5) = 1/G12; S(6,6) = S(5,5); % Stiffness matrix [C] C = inv(S); % Write data out1 = fopen('matrices.dat','w'); fprintf(out1,'Compliance_matrix_[S]\n'); fprintf(out1,'%8.3f %8.3f %8.3f %8.3f %8.3f %8.3f\n',S); fprintf(out1,'\n'); fprintf(out1,'Stiffness_matrix_[C]\n'); © Ever J. Barbero (2014) all rights reserved. P a g e | 9 @solutionmanual1 K15077_SM_Cover.indd 13 14/02/15 4:22 pm

14. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition fprintf(out1,'%8.3f %8.3f %8.3f %8.3f %8.3f %8.3f\n',C); fclose(out1); Problem 1.7 A rotation of an angle ? around the z-axis is defined by: ? ? ? ? cos sin 0 ? ? ? ? ? ? sin cos 0 ij a ? ? ? ? ? ? 0 0 1 From aij the values mi, li and ni (director cosines) can be obtained: ? ? ? ? ? l m n cos ? sin 0 1 1 1 ? l ? ? ? ? l m n sin ? cos ? 0 2 2 m 2 ? n 0 0 1 3 3 3 Now, the transformations matrices [T] and [T-bar] can be obtained using: ? 1 1 m l ? 2 2 2 1 2 2 2 l m n m n l n l m 1 1 1 1 1 1 ? ? 2 2 2 2 2 2 2 2 2 n m n l n l m ? ? 2 2 2 2 2 2 ? ? 2 3 l 2 3 m 2 3 n n l n l m 2 n 2 2 l m n m ? ? T ? 3 3 3 3 n 3 3 ? ? ? ? ? l m n m n m l n l l m m l ? ? 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 ? ? ? ? ? l l m m n n m n n m l n n l l m m l 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 ? ? ? ? ? ? ? ? ? l l m m n n m n n m l n n l l m m l 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 ? ? 2 1 2 1 2 1 l m n m n l n l m 1 1 1 1 1 1 ? ? 2 2 2 2 2 2 l m n m n l n l m ? ? 2 2 2 2 2 2 ? ? ? ? T 2 3 2 3 m 2 3 l m n m n l n l m ? 3 3 n 3 3 n 3 3 m ? ? ? ? ? l l m n n m n m l n l l m l 2 2 2 ? ? 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 ? ? ? ? ? l l m m n n m n n m l n n l l m m l 2 2 2 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 ? ? ? ? ? ? ? ? ? l l m m n n m n n m l n n l l m m l 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 Due to the material is transversely isotropic both the compliance and stiffness matrix are defined like in the problem 1.6. Their transformations are given by: * ? ? ? ? ? ? ? ? S * ' ? ? ? ? ? ? ? ? S ? T ? C T C ' * T T T * T A MATLAB®’s routine was built (code attached) which reads the information about [C’], [S’] and ? from a file named matrices.dat, written with the following format: © Ever J. Barbero (2014) all rights reserved. P a g e | 10 @solutionmanual1 K15077_SM_Cover.indd 14 14/02/15 4:22 pm

15. cl i ck h ere t o d ow nl oad SOLUTIONS MANUAL for Finite Element Analysis of Composite Materials Using ANSYS® 2nd Edition Text: Compliance_matrix_[S] Data: Formatted array 6x6 Text: <<<Blank>>> Text: Stiffness_matrix_[S] Data: Formatted array 6x6 Text: <<<Blank>>> Text: Rotation_angle_theta Data: Single value of theta in Degrees Using the matrices gotten from the previous problem and making a rotation of 45 degrees over z-axis it’s obtained: CODE PROBLEM 1.7 (P1p7.m) % Routine that solves the problem 1.7. It transforms the Stiffness and % Compliance matrix for a tranversely isotropic single layer composite % from its material coordiantes to another system of coordinates clear; clc; % Read data inp1 = fopen('matrices.dat','r'); dummy = fscanf(inp1,'%s',1); % first text line % Compliance matrix S = fscanf(inp1,'%f %f %f %f %f %f',[6,6]); dummy = fscanf(inp1,'%s',1); % second text line % Stiffness matrix C = fscanf(inp1,'%f %f %f %f %f %f',[6,6]); dummy = fscanf(inp1,'%s',1); % third text line % Rotation angle (theta) the = fscanf(inp1,'%f',1); fclose(inp1); % Director cosines l1 = cos(the*pi/180); m1 = sin(the*pi/180); n1 = 0; l2 = -m1; m2 = l1; n2 = n1; l3 = 0; m3 = 0; n3 = 1; © Ever J. Barbero (2014) all rights reserved. P a g e | 11 @solutionmanual1 K15077_SM_Cover.indd 15 14/02/15 4:22 pm