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Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition

Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition helps you reinforce learning with in-depth, accurate solutions.

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    Introduction to Finite Elements in Engineering Tirupathi R. Chandrupatla Rowan University Glassboro, New Jersey Ashok D. Belegundu The Pennsylvania State University University Park, Pennsylvania Solutions Manual Prentice Hall, Upper Saddle River, New Jersey 07458

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    CONTENTS Preface Chapter 1 Fundamental Concepts 1 Chapter 2 Matrix Alge bra and Gaussian Elimination 18 Chapter 3 One - Dimensional Problems 26 Chapter 4 Trusses 61 Chapter 5 Beams and Frames 86 Chapter 6 Two - Dimensi onal Problems Using Constant 103 Strain Triangles Chapter 7 Axisym metric Solids Subjected to 151 Axisymmetric Loading Chapter 8 Two - Dimens ional Isoparametric Elements 181 and Numerical Integration Chapter 9 Three - Dim ensional Problems in Stress 207 Analysis Chapter 10 Scalar Field Problems 218 Chapter 11 Dynamic Considerations 264 Chapter 12 Prepr ocessing and Postprocessing 282

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    PREFACE This solutions manual serves as an aid to professors in teaching from the book Introduction to Finite Elements in Engineering , 4 th Edition . The problems in the book fall into the following categories: 1. Simple problems to understand the concepts 2. Derivations and direct solutions 3. Solutions requiring computer runs 4. Solutions requiring program modifications Our basic philosophy in the development of this manual is to provide a complete guidance to the teacher in formulating, modeling, and solving the problems. Complete solutions are given for problems in all categories stated. For some larger problems such as those in three dimensional stress analysis, complete formulation and modeling aspects are discussed. The students should be able to proceed from the guidelines provided. For problems involving distributed and other types of loading, the nodal loads are to be calculated for the input data. The programs do not generate the loads. This calculation and the boundary condition decisions enable the student to develop a physical sense for the problems. The students may be encouraged to modify the programs to calculate the loads automatically. The students should be introduced to the programs in Chapter 12 right from the point of solving problems in Ch apter 6 . This will enable the students to solve larger problems with ease. The input data file for each program has been provided. Data for a problem should follow this format. The best strategy is to copy the example file and edit it for the problem under consideration. The data from program MESHGEN will need some editing to complete the information on boundary conditions, loads, and material properties. We thank you for your enthus iastic response to our first three editions of the book. We look forward to receive your feedback of your experiences, comments, and suggestions for making improvements to the book and this manual. Tirupathi R. Chandrupatla P.E., CMfgE Department of Mechanical Engineering Rowan University, Glassboro, NJ 08028 e - mail: chandrupatla@rowan.edu Ashok D. Belegundu Department of Mechanical and Nuclear Engineering The Pennsylvania State University University Park, PA 16802 e - mail: adb3@psu.edu

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    1 CHAPTER 1 FUNDAMENTAL CONCEPTS 1.1 We use the first three steps of Eq. 1.11 E E E E E E E E E z y x z z y x y z y x x                              Adding the above, we get   z y x z y x E              2 1 Adding and subtracting E x   from the first equation,   z y x x x E E             1 Similar expressions can be obtained for  y , and  z . From the relationship for  yz and Eq. 1.12,   yz yz E      1 2 etc. Above relations can be written in the form  = D  where D is the material property matrix defined in Eq. 1.15.  1.2 Note that u 2 ( x ) satisfies the zero slope boundary condition at the support. 

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    2 1. 3 Plane strain condition implies that 0 y x z z E E E         Which gives ( ) z x y       On substitution of the given values z  = 0.3 (3000 – 15000) = 4500 psi 1. 4 u = ( x 2 + 4 y 2 – 16 xy ) × 10 - 4 2 4 ( 5 8 ) 10 v y x y      4 10 u x     (2 x – 16 y ) 4 (8 16 ) 10 u y x y       4 5 10 u y       u y   = (2 y + 8) × 10 - 4 u x u y u u y x                              At x = 1, y = 0 4 2 8 10 21 x y xy                             

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    3 1 . 5 From the figure P 1.5, it is revealed that the displacements u and v are given by u ( x ) = 0.2 y + 6 v ( y ) = 0  u 1 = 0.2 × 0 + 6 = 6 v 1 = 0 u 2 = u 1 + [0.2 × 0 + 6] = 12 v 2 = 0 u 4 = 0.2 × 15 + 6 = 9 u 3 = u 4 + [0.2 × 15 + 6] = 15 The strain values are computed as : ε x = u x   = 0 ε y = v y   = 0 xy u v x y        = 0.2  1. 6 u = 2 + 2 x + 4 x 2 + 3 xy 2 v = xy = 8 x 2 a) Expressions for ε x , ε y and γ xy ε x = u x   = 2 + 8 x + 3 y 2 ε y = v y   = x xy u v x y        = 6 xy + y – 16 x

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    4 (b) In order to draw the contours of the strain field using MATLAB, we need to create a script file, which may be edited as a text file and save with “.m” extension . The file for plotting ε x ε y and γ xy is given below b) MATLAB code for ε x , ε y , γ xy : [X,Y] = meshgrid( - 1:.1:1, - 1:.1:1); Z = 2.+8.*X + 3.*Y.^2; k = X.; L = 6.*X.*Y+Y. - 16.*X; [C,h]= contour(X,Y,Z); clabel(C,h); On running the program, the contour map is shown as follows:

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    5 c) The maximum value of ε x is 12 in the location (0.9, 0.9) of the square region.  1.7 a) 0.2 0.2 0 1 u y u y v     b) 0 0 0.2 x y xy u v u v x y y x                   ( x , y ) ( u , v )

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    6 1.8 x  = 120 MPa, y  =55MPa, z  = – 85MPa yz  = 33 MPa, xz  = 75 MPa and xy  = – 55MPa. n = 1 1 1 , , 2 2 2 T       From Eq. 1.8 we get T x = x  x n + xy y xz z n n    = 85.50 MPa T y = xy x y y yz z n n n      = 23.33 MPa T Z = xy x yz y z z n n n       = – 6.10 MPa The normal stress n x x y y z z T n T n T n     n  = 50 MPa  1. 9 From the derivation made in P1.1, we have                 yz yz v x x z y x x E E E                              1 2 and 2 1 2 1 1 form in the written be can which 1 2 1 1 Lame’s constants  and  are defined in the expressions

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    7                         1 2 2 1 1 , inspection On 2 E E yz yz x v x  is same as the shear modulus G .  1. 10 5 2 10     T  = 45°C E = 123GPa 6 16.2 10 /     °C From Eq. 1.20 and 1.21, 4 7.29 10 o T         = E ( ε – ε o )  = – 87.2MPa 1.11 The strain at any point ε x = 5 + 4 x 2 = du dx δ= 3 4 5 3 L L O O du dx x x dx          2 4 5 3 L L          1.12 With reference to Eq.1.29, the global stiffness matrix can be formed as;

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    8 1 2 30 25 50 50 45 50 50 30 q q                        Solving 105 q 1 + 50 q 2 = 45 – (i) – 50 q 1 + 50 q 2 = 30 – (ii) Adding (i) and (ii) 55 q 1 = 75  q 1 = 1.363 mm and on substitution of q 1 , q 2 = 1.963 mm 1.13 When the wall is smooth, 0 x   . T  is the temperature rise. a) When the block is thin in the z direction , it corresponds to plane stress condition. The rigid walls in the y direction require 0 y   . The generalized Hooke’s law yields the equations y x y y T E T E               From the second equation, setting 0 y   , we get y E T      . x  is then calculated using the first equation as   1 T     . b) When the block is very thick in the z direction, plain strain condition prevails. Now we have 0 z   , in addition to 0 y   . z  is not zero.

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    9 0 0 y z x y z y y z z T E E T E E T E E                                 From the last two equations, we get 1 2 1 1 y z E T E T                 x  is now obtai ned from the first equation.  1.14 For thin block, it is plane stress condition. Treating the nominal size as 1, we may set the initial strain 0 0.1 1 T      in part (a) of problem 1.13. Thus 0.1 y E    .  1.15 The potential energy π is written as 2 1 . . . 2 L L o o du EA dx u g A dx dx            - 1 Let us consider a polynomial eqn. for the displacement u = a ( – 2 x + x 2 ) du dx = 2 a ( – 1 + x ) Substituting the expression in the first term of E qn 1 and integrating: 2 2 0 1 [2 ( 1)] 2 EA a x dx     11 4 10 3 a         The second term of the eqn ( 1 ) becomes; 2 2 2 0 0 . . 9.81 ( 2 ) u g Adx a x x dx      

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    10 2 3 2 0 9.81 3 x a x           4 (9.81) 3 a         Eqn ( 1 ) becomes: 11 2 4 4 10 9.81 3 3 a a                  Then 11 4 10 9.81 3 a a              Minimizing the potential energy yields: n a   = 0 = 4 3 [ 2 a × 10 11 + 9.81] = 0  a = – 4.90 × 10 - 11 The displacement of the mid node is; u x = 1 = – 4.90 × 10 - 11 ( – 2 + 1) u x = 1 = 4.90 × 10 - 11 m We use the displacement field by u = a 0 + a 1 x + a 2 x 2 u = 0 at x = 0  a 0 = 0 u = 0 at x =1  a 1 + a 2 = 0  a 2 = – a , We have u = a 1 x (1 – x ) & du dx = a 1 (1 – x ) 2 1 1 0 0 1 2 du E dx f u dx dx                1 1 11 2 1 1 0 0 2 10 1 2 (1 ) 2 a x x x a x x dx        2 11 1 1 2 10 11 2 3 60 a a           0 d da  

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    11 11 1 2 11 10 0 3 60 a     a 1 = 11 11 2 10 60 3    a 1 = 0.275 × 10 - 11 Displacement = u = 0.275 x (1 – x ) × 10 - 11 Stress Edu dx   = 2×10 11 ×0.275 ×10 - 11 (1 – x ) = 0.55 (1 – x ) 1.16 We use the displacement field defined by u = a 0 + a 1 x + a 2 x 2. u = 0 at x = 0  a 0 = 0 u = 0 at x = 1  a 1 + a 2 = 0  a 2 =  a 1 We then have u = a 1 x (1  x ), and du / dx = a 1 (1  x ). The potential energy is now written as         30 6 6 1 5 1 3 4 2 4 1 2 1 4 4 1 2 1 1 2 1 2 1 2 1 1 2 1 1 2 1 1 0 1 0 5 4 1 2 2 1 1 0 1 0 1 3 2 2 1 1 0 1 0 2 a a a a dx x x a dx x x a dx x x a x dx x a fudx dx dx du                                            f = x 3 E = 1 A = 1 x =0 x =1

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    12 0 30 1 3 0 1 1        a a This yields, a 1 = 0.1 Displacemen u = 0.1 x (1  x ) Stress  = E du / dx = 0.1( 1  x )  1.1 7 Let u 1 be the displacement at x = 300 mm piece wise linear displacement that is continuous in the interval 0 ≤ x ≤ 800 is represented as shown in the figure below: For element ( 1 ) u = 0 at x = 0; a 1 = 0 u = u 1 at x = 300; a 2 = 1 300 u  The linear displacement field 1 300 u u x        1 300 u du dx              For element ( 2 ) u = 0 at x =800  a 3 + 800 a 4 = 0 - (i) u = u 1 at x =800  a 3 + 300 a 4 = u 1 - (ii) From (i) and (ii) 1 4 500 u a         3 1 8 5 a u         Substitution of a 3 and a 4 in the linear displace at field

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    13 1 1 8 5 500 u u u x               1 500 u du dx   2 2 300 800 1 2 1 0 300 1 1 . 8000 2 2 br du du E A dx Ecu A dx u dx dx                   2 2 1 1 1 2 1 1 1 300 500 8000 2 300 2 500 br cu u u E A E A u                  2 1 2 1 1 1 8000 2 300 500 br cu E A E A u u           1 2 1 0 8000 0 300 500 br cu E A E A u u              Substituting the values of E br , E cu , A 1 and A 2 in the above eq atio n., u 1 = 0.015 mm 1.18 In the Galerkin method, we start from the equilibrium equation 0   g dx du EA dx d Following the steps of Example 1.3, we get       2 0 2 0 dx g dx dx d dx du EA Introducing     1 2 1 2 2 and , 2       x x u x x u where u 1 and  1 are the values of u and  at x = 1 respectively,     0 2 2 1 2 0 2 0 2 2 1 1               dx x x dx x u On integrating, we get

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    14 0 3 4 3 8 1 1           u This is to be satisfied for every  1 , which gives the solution u 1 = 0.5  1.1 9 We use 2 at 0 0 at 0 3 4 2 3 2 1         x u x u x a x a x a a u This implies that 4 3 2 1 1 8 4 2 0 0 a a a a      and         4 3 1 2 4 2 2 4 3 3 4 2 3         x a x a dx du x x a x x a u a 3 and a 4 a re considered as independent variables in                  2 0 4 3 2 2 4 3 3 2 4 3 1 2 2 1 a a dx x a x a on expanding and integrating the terms, we get 4 3 4 3 2 4 2 3 6 2 8 8 . 12 333 . 1 a a a a a a       We differentiate with respect to the variables and equate to zero. 0 6 6 . 25 8 0 2 8 667 . 2 4 3 4 4 3 3               a a a a a a On solving, we get a 3 =  0.74856 and a 4 =  0.00045. On substituting in the expression for u , at x = 1, u 1 = 0.749 This approximation is close to the value obtained in the example problem. 

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    15 1.20 (a)   udx x T Adx L L        0 0 T 2 1 dx du E      and On substitution,   udx udx x dx dx du udx T udx T dx dx du EA                            60 30 30 0 60 0 2 6 60 30 30 0 60 0 2 300 10 10 60 2 1 2 1 (b) Since u = 0 at x = 0 and x = 60, and u = a 0 + a 1 x + a 2 x 2 , we have     60 2 60 2 2     x a dx du x x a u On substituting and integrating, 2 2 2 10 8775000 10 216 a a     Setting d  / da 2 = 0 gives   60 2 935 . 60 10 03125 . 2 6 2          x dx du E a Plots of displacement and stress are given below:

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    16 0 10 20 30 40 50 60 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10 -3 Displacement u 0 10 20 30 40 50 60 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 Stress .  1. 21 y = 20 at x = 60 implies that   2 1 0 2 1 0 180 3 1 20 yields which , 3600 60 20 a a a a a a       Substituting for k , h , L , and a 0 in I , we get             7605000 10 702 10 117 10 45 612000 45600 39 18 3 5000 4 4 10 800 180 3 1 20 25 2 1 2 10 2 4 1 4 2 2 5 2 1 2 1 60 0 2 2 1 2 2 2 2 1 2 1 60 0 2 2 1 2 2 1                        a a a a a a I a a dx a x a xa a I a a dx x a a I 0 10 702 10 90 612000 0 10 117 612000 912000 4 2 5 1 2 4 2 1 1            a a da dI a a da dI

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    17 On solving, a 2 = 0.1699 a 1 =  13.969 Substituting into the expression for a 0 , we get a 0 = 246.538 .  1.22 Since u = 0 at x = 0, the displacement satisfying the boundary condition is u = a 1 x . Also the coordinates are x 2 = 1, and x 3 = 3. The potential energy for the problem is 2 3 2 2 3 3 0 1 2 du EA dx P u Pu dx            We have u 2 = a 1 , u 3 = 3 a 1 , E = 1, A = 1, and 1 du a dx  . Thus   3 2 2 1 1 1 1 1 0 1 3 3 4 2 2 a dx a a a a        . For stationary value, setting 1 0 d da   , we get 3 a 1 – 4 = 0, which gives a 1 = 0.75. The approximate solution is u = 0.75 x .  1.23 U se Galerkin approach with a pproximation 2 u a bx cx    to solve   3 0 1 0 1 du u x x dx u     

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    18 The week form is obtained by multiplying by  satisfying   0 0   . 1 0 3 0 du u x dx dx            We now set 2 1 u bx cx    satisfying   0 1 u  and 2 1 2 a x a x    . On introducing these into the above integral,        1 2 2 1 2 0 1 1 2 2 2 3 2 2 3 3 3 4 1 2 0 0 2 3 3 3 0 3 3 2 3 3 3 2 3 0 a x a x b cx bx cx x dx a bx x x bx cx cx dx a bx x x bx cx cx dx                        On integrating, we get 1 2 1 2 3 1 2 3 1 3 3 1 0 2 2 3 3 4 3 4 4 2 5 3 17 7 13 11 3 0 2 12 6 12 10 4 b c c b b c c a b a a b c a b c                                           This must be satisfied for every a 1 and a 2 . Thus the equations to be solved are 3 17 7 0 2 12 6 13 11 3 0 12 10 4 b c b c       The solution is b = – 1.9157, c = 1.2048. Thus 2 1 1.9157 1.2048 u x x    .  1.24 The deflection and slope at a due to P 1 are 3 1 3 Pa EI  and 2 1 2 Pa EI  . Using this the deflection and slope at L due to load P 1 are

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    19   2 3 1 1 1 2 1 1 3 2 2 Pa L a Pa v EI EI Pa v EI        The deflection and slope due to load P 2 are 3 2 2 2 2 2 3 2 P L v EI P L v EI      We then get 1 2 1 2 v v v v v v         1.25 (a) The displacement of B is given by ( – 0.1, 0.1) and A , C , and D remain in their original position. Consider a displacement field of the type 1 2 3 4 1 2 3 4 u a a x a y a xy v b b x b y b xy         The four constants can be eva luated using the known displacements At A (0, 0) 1 1 0 0 a b   (0,0) (1,0) (1,1) (0,1)

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    20 At B (1, 0) 1 2 1 2 0.1 0.1 a a b b      At C (1, 1) 1 2 3 4 1 2 3 4 0 0 a a a a b b b b         At D (0, 1) 1 3 1 3 0 0 a a b b     The solution is a 1 = 0 , a 2 = – 0.1, a 3 = 0, a 4 = 0.1 b 1 = 0, b 2 = 0.1, b 3 = 0, b 4 This gives 0.1 0.1 0.1 0.1 u x xy v x xy      (b) The shear strain at B is     0.1 0.1 0.1 0.1 1 0.1 0.1 0 0.2 B u v x y y x                

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    18 CHAPTER 2 MATRIX ALGEBRA AND GAUSSIAN ELIMINATION 2.1                            3 1 2 , 3 3 0 3 4 2 0 2 8 d A (a)                                           8 3 6 3 0 2 6 2 3 3 1 2 3 1 2 1 0 0 0 1 0 0 0 1 T d d I (b) det A = 8 [ (4)(3) – ( – 3) ( – 3) ] – ( – 2) [( – 2)(3) – (0)( – 3)] = 12 (c) The characteristic equation is det ( A -  I ) = 0, or 0 3 3 0 3 4 2 0 2 8 det                      which yields 0 12 55 15 2 3        Handbooks (e.g., CRC Mathematical Handbook) give explicit solutions to cubic equations. Here equations given in Chapter 9 are used, which give formulas for finding the eigenvalues of the (3x3) symmetric stress tensor. Referring to Section 9.3 in the text, we have I 1 = A 11 + A 22 + A 33 = 15, I 2 = 55, I 3 = 12 Thus, a = 20, b = 13, c = 5.164,  = 37.4  whence  1 = 0.2325,  2 = 5.665,  3 = 9.103 Note: Since all  i > 0, A is positive definite . Now, eigenvector y i corresponding to eigenvalue  i is obtained from ( a -  i I ) y i = 0, i = 1,2,3 Thus, y 1 is obtained as                                    0 0 0 7675 . 2 3 0 3 7675 . 3 2 0 2 7675 . 7 3 2 1 y y y Thus,

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    19 7.7675 y 1 – 2 y 2 = 0 - 2 y 1 + 3.7675 y 2 – 3 y 3 = 0 - 3 y 2 + 2.7675 y 3 = 0 ______________________ Only two of the above three equations are independent. We have y 1 = 0.2575 y 2 y 2 = 0.922 y 3 Letting y 3 = 1, we get y 1 = [0.237, 0.922, 1] T The length of the vector is 381 . 1 T 1   y y y . Normalizing y 1 to be a unit vector yields y 1 = [0.172, 0.668, 0.724] T . Similarly, y 2 = [0.495, 0.577, - 0.650] T , y 3 = [0.850, - 0.470, 0.232] T . (d) Solution to A x = b using Algorithm 1 for general matrix : n = 3 First Step ( k = 1) i = 2 (2 nd row)                3 3 3 4 A 0 2 0 2 8 c = a 21 / a 11 = - 2/8 = - 1/4 a 22 (1) = 4 – ( - 1/4)( - 2) = 7/2, a 23 (1) = - 3, d 2 (1) = - 1 - ( - 1/4)(2) = - 1/2 i = 3 (3 rd row) c = 0 a 32 (1) = - 3, a 33 (1) = 3, d 3 (1) = 3 Thus                           3 2 / 1 2 , 3 3 0 3 2 / 7 0 0 2 8 ) 1 ( ) 1 ( d A Second Step ( k = 2) i = 3 (3 rd row) c = - 6/7, a 33 (2) = 3 – ( - 6/7)( - 3) = 3/7, d 3 (2) = 3 – ( - 6/7)( - 1/2)=18/7 Thus                          7 / 18 2 / 1 2 , 7 / 3 0 0 3 2 / 7 0 0 2 8 ) 1 ( ) 2 ( d A

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    20 Back - Substitution                                    7 / 18 2 / 1 2 7 / 3 0 0 3 2 / 7 0 0 2 8 3 2 1 x x x Third row gives: 3/7 x 3 = 18/7 whence x 3 = 6 2 nd row then gives x 2 = 5, 1 st row gives x 1 = 1.5 Thus, solution is x = [1.5, 5, 6 ] T . Solution to A x = b using Algorithm 2 for symmetric, banded matrix A is stored as             0 3 3 4 2 8 n = 3, nbw = 2 First Step ( k = 1) nbk = min (3,2) = 2 2 nd row ( i = 2): i 1 = 2 c = a 12 / a 11 = - 1/4 j 1 = 1 j 2 = 2 a 21 = 4 – ( - 1/4)( - 2) = 7/2 Thus A (1) =             0 3 3 2 / 7 2 8 Second Step ( k = 2) nbk = 2 3 rd row ( i = 3): c = - 6/7 j 1 = 1 j 2 = 2 a 31 = 3/7 Thus A (2) =             0 7 / 3 3 2 / 7 2 8 reduction of right - hand - side vector d and back - substitution is same as in Algorith m 1 above, resulting in the the same solution x = [1.5, 5, 6 ] T . 

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    21 2.2   2 1     N (a)          3 4 0 1 1  d N (b)                          15 / 16 0 0 3 / 2 ) 1 ( ) 1 ( ) 1 ( 2 2 2 2 2 1 1 T        d N N  2.3 q = x 1 – 6 x 2 + 3 x 1 2 + 5 x 1 x 2 =                             2 1 2 1 2 1 6 1 0 5 . 2 5 . 2 3 x x x x x x  x T Q x + c T x where                  6 1 0 5 . 2 5 . 2 3 c Q and  2.4 The detailed algorithm is given in the text. This is an excercise in computer programming. The solutions are (a) ( – 2.25, – 11.5, – 10.5) (b) (1.55, 5.1, 6.1)  2.5 The minors are M 11 = det 1 3 1 2       = – 1 M 12 = det 2 3 3 2       = – 5 M 13 = det 2 1 3 1       = – 1 M 21 = 1; M 22 = – 7; M 23 = – 5 M 31 = 3; M 32 = – 3; M 33 = – 3 Co factor

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    22 C ij = ( – 1) i + j M ij Thus, the co - factor matrix is A c = 1 5 1 1 7 5 3 3 3                 The inverse of matrix A A - 1 = 1 det A  Ac T yields det ( A ) = (2 – 3) – 2 (4 – 9) + 3 (2 – 3) = 6  A - 1 = 1 6 1 1 3 5 7 3 1 5 3               2.6 Area = 1 2 det 1 2 2 1 7 8 1 11 12           =   6 = 6 s quare unit s . 2.7 A 1 = 625 . 1 5 5 . 2 1 5 . 1 3 1 2 2 1 det 2 1            A 2 = 25 . 1 1 1 1 5 5 . 2 1 2 2 1 det 2 1            A 3 = 75 . 0 5 . 1 3 1 1 1 1 2 2 1 det 2 1            A = A 1 + A 2 + A 3 = 3.625 A 1 / A = 0.448, A 2 / A = 0.345, A 3 / A = 0.207  2.8 A i , j = B i , j - i + 1 for j  i

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    23 Thus, A 11,14 corresponds to B 11,4 and B 6,1   A 6,6  2.9 Full (10x10) matrix BANDED n = 10, nbw = 10 No. of storage locations = ( n ) ( nbw ) = 100 SKYLINE x x x x ... x x x ... x x ... x ... No. of storage locations = no. of column entries = 1 + 2 + 3 + ... + 10 = (10) (10+1) / 2 = 55  2.10 Based o n Eq. 2.2; K = 1 (1 st row) l 11 = 11 a = 4 = 2 K = 2 (2 nd row ) l 21 = 21 11 a l = 2 2 = 1 l 22 = 2 22 21 a l  = 3 K = 3 (3 rd row) l 31 = 31 11 a l = – 3 l 32 = 32 31 21 32 a l l l  = 4 l 33 = 2 2 33 31 32 1 a l l    Thus

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    24 L = 2 0 0 1 3 0 3 4 1            2.11 By expanding the matrix equations, we obtain a set of simultaneous equations for the u and  coefficients, resulting in the solution L = 1 0 0 2 1 0 3 2 1           and U = 1 2 3 0 1 2 0 0 1           2.12 The quadrilateral ABCD is divided in to two triangles with corners ABC and ACD . Area of triangle ABC A 1 = 1 2 det 1 1 1 1 7 2 1 6 6           = 12.5 Area of triangle ACD A 2 = 1 2 det 1 1 1 1 6 6 1 3 7           = 10  Area of the quadrilateral ABCD A = A 1 + A 2 A = 22.5 square unit. 2.13 Setting cos sin sin cos             T , it is easy to show that T 1 0 0 1        TT .  2.14 a) The minors of the given matrix are: M 11 = 7; M 12 = 4; M 13 = – 1 M 21 = 11; M 22 = 17; M 23 = 7

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