Solution Manual for A First Course in the Finite Element Method 6th Edition Logan / All Chapters 1 - 16 / Full Complete 2023

Contents Chapter 1: ................................ ................................ ................................ ............................ 1 Chapter 2: ................................ ................................ ................................ ............................ 3 Chapter 3: ................................ ................................ ................................ .......................... 25 Chapter 4: ................................ ................................ ................................ ........................ 137 Chapter 5: ................................ ................................ ................................ ........................ 203 Chapter 6: ................................ ................................ ................................ ........................ 315 Chapter 7: ................................ ................................ ................................ ........................ 363 Chapter 8 : ................................ ................................ ................................ ........................ 383 Chapter 9: ................................ ................................ ................................ ........................ 397 Chapter 10 :................................ ................................ ................................ ...................... 423 Chapter 11 :................................ ................................ ................................ ...................... 449 Chapter 12 :................................ ................................ ................................ ...................... 477 Chapter 13 :................................ ................................ ................................ ...................... 499 Chapter 14 :................................ ................................ ................................ ...................... 539 Chapter 15 :................................ ................................ ................................ ...................... 561 Chapter 16 :................................ ................................ ................................ ...................... 591 Appendix A : ................................ ................................ ................................ ..................... 629 Appendix B : ................................ ................................ ................................ ..................... 635 Appendix D : ................................ ................................ ................................ ..................... 641 629 © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or post ed to a publicly accessible website, in whole or in part. Appendix A A.1 (a) [A] + [B] = 4 0 2 0 6 0
1 8 2 4 3 12                (b) [A] + [C], Nonsense, [ A] and [ C] not same order (c) [A] [C]T, Nonsense, columns [ A]  rows [ C]T (d) [D] [E] = 3 5 2 1
2 10 0 2
1 0 5 1    = 5(3) 2(2) 1(1)
2(3) 10(2) 0(1)
1(3) 0(2) 5(1)
 = 20
26
8
 (e) [D] [C], Nonsense, columns [ D]  rows [ C] (f) [C] [D] = 5 2 13 2 02 10 01 0 21 0 5    = 3(5) (2)(2) 0 6 20 0 3 0 0
5 0 2 2 0 0 1 0 10     
         = 19 26 3
3 2 9
 A.2 [A] = 10
14 [A]–1 = []
|[ ] |TC
A C11 = (–1)1 + 1 (8) = 8 , C12 = (–1)1 + 2 (1) = -1 C21 = (–1)2 + 1 (0) = 0, C22 = (–1)2 + 2 (4) = 4 [C] = 81
04
 | [A] | = A11 C11 + A12 C12 = (4) (8 ) + (0) ( -1) = 32 [C]T = 80
14
  [A]–1 = 80
14
32
 = 1
4
11
32 80

 630 © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or post ed to a publicly accessible website, in whole or in part. Verify by multiplying [ A] [A]–1 = [I] A.3 [D]–1 = []
|[ ]|TC
D [D] = 5 2 1
2 10 0
1 0 5


 Cofactor matrix, [C] = 50 10 10
10 24 2
10 2 46

 | [D] | = 5(50) + (2 )(-10) + (1)( -10) = 220 [D]–1 = 5 1 1
22 22 22
1 6 1
22 55 110
1 1 23
22 110 11050 10 10
110 24 222010 2 46

      A.4 Nonsense A.5 [B] = 20
24
 (1) 2 0 1 0
2 4 0 1
 divide 1st row by 2 (2) 1
21 0 0
2 4 0 1
 st row by -2 and add to row 2 (3) 1
21 0 0
0 4 1 1
 divide 2nd row by 4 (4) 1
2
11
441 0 0
01
  [B]–1 = 1
2
11
440
 A.6 [D]–1 by row reduction 5 2 1 5 2 1 1 0 0
[D] 2 10 0 2 10 0 0 1 0
1 0 5 1 0 5 0 0 1   
      
       a. Divide R 1 by 5 2 1 1
5 5 51 0 0
2 10 0 0 1 0
1 0 5 0 0 1



 b. Multiply R1 by -2 + Add to R 2 631 © 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or post ed to a publicly accessible website, in whole or in part. 2 1 1
5 5 5
1 22
5 551 0 0
0 9 1 0
1 0 5 0 0 1



 c. Multiply R1 by -1 + Add to R 3 2 1 1
5 5 5
1 2 2
5 5 5
2 4 1
5 5 51 0 0
0 9 1 0
0 4 0 1

 d. Multiply R 2 by 1/23 + Add to R 3 2 1 1
5 5 5
1 2 2
5 5 5
110 5 1
23 23 231 0 0
0 9 1 0
0 0 1

  e. Multiply R 3 by 23/275 + Add to R2 2 1 1
5 5 5
23 276 23 1
5 55 275 275
110 5 1
23 23 231 0 0
0 9 0
0 0 1

  f. Multiply R 3 by -23/550 + Add to R 1 23 23 21
5 110 550 550
23 276 23 1
5 55 275 275
110 5 1
23 23 2310
0 9 0
0 0 1 

  g. Multiply R 2 by -1/23 + Add to R 1 5 11
22 22 22
46 23 276 23
5 55 275 275
110 5 1
23 23 231 0 0
00
0 0 1 

  h. Divide R 2 by 46/5 + R 3 by 110/23 5 11
22 22 22
6 11
22 55 110
23 11
22 110 1101 0 0
0 1 0
0 0 1 

  [D]–1 = 5 11
22 22 22
6 11
22 55 110
23 11
22 110 110

 A.7 Show that ([ A] [B])T = [B]T [A]T by using [A] = 11 12
21 22aa
aa [B] = 11 12 13
21 22 23b b b
b b b