Macromechanical Analysis of Laminate Hygrothermal Loads

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Explore the macromechanical analysis of laminates under hygrothermal loads presented by Dr. Autar Kaw from the University of South Florida. This analysis delves into the mechanics of composite materials, focusing on key properties and behaviors. Gain insights into the interdisciplinary field of mechanical engineering through this comprehensive study.

  • Macromechanical
  • Laminate
  • Hygrothermal Loads
  • Composite Materials
  • Mechanical Engineering
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  1. Chapter 4 MacromechanicalAnalysis of a Laminate Hygrothermal Loads Dr. Autar Kaw Department of Mechanical Engineering University of South Florida, Tampa, FL 33620 Courtesy of the Textbook Mechanics of Composite Materials by Kaw

  2. M x T x T x x x = M y T y = T y y y T M xy T xy xy xy xy k k k k k

  3. Q Q Q T x M x 11 12 16 T y M y = Q Q Q 2 2 1 2 26 T xy M xy Q Q Q 16 26 66 k k k

  4. T x / 2 h T y 0 dz = / 2 h T xy T x n h = k 0 dz k - T y 1 k = h 1 T xy k

  5. Q Q Q M x 16 n 11 12 h k M y dz 0 = Q Q Q 12 22 26 h 1 = k 1 k M xy Q Q Q 16 26 66 k k

  6. Q Q Q T x 16 n 11 12 x h k T y 0 dz = Q Q Q y 12 22 26 h 1 k = 1 k T xy Q Q Q xy k 16 26 66 k k Q Q Q 0 x T x 16 n 11 12 x h k 0 y T y 0 + z dz = Q Q Q y 12 22 26 h 1 k = 1 k 0 xy T xy Q Q Q xy 16 26 66 k k

  7. Q Q Q 0 x 16 n 11 12 x h k 0 y + z dz = Q Q Q y 12 22 26 h 1 = k 1 k 0 xy Q Q Q xy 16 26 66 k Q Q Q T x 16 n 11 12 h k T y dz Q Q Q 12 22 26 h 1 = k 1 k T xy Q Q Q 16 26 66 k

  8. Q Q Q 0 x 16 n 11 12 x h k 0 y + z dz = Q Q Q y 12 22 26 h 1 = k 1 k 0 xy Q Q Q xy 16 26 66 k Q Q Q 16 n 11 12 x h k T dz Q Q Q y 12 22 26 h 1 = k 1 k xy Q Q Q k 16 26 66 k

  9. o x T x N A A A B B B 11 12 16 11 12 16 x o y T y + = N A A A B B B 12 22 26 12 22 26 y A A A B B B T xy o xy N 16 26 66 16 26 66 xy Q Q Q T x N n 11 12 16 x T T y ( ) [ ] = = T Q Q Q N N h h 1 y k k- 12 22 26 1 k = T xy N Q Q Q xy k 16 26 66 k

  10. o x T x M B B B D D D 11 12 16 11 12 16 x o y T y + = B B B D D D M 12 22 26 12 22 26 y B B B D D D T xy o xy M 16 26 66 16 26 66 xy Q Q Q T x M n 11 12 16 x 1 2 k 2 k - T T y ( ) [ ] = = T Q Q Q h h M M 1 y 12 22 26 2 1 k = T xy M Q Q Q xy k 16 26 66 k

  11. T 0 A | B N = T | B D M Q Q Q T x N n 11 12 16 x T T y ( ) [ ] = = T Q Q Q N N h h 1 y k k- 12 22 26 1 k = T xy N Q Q Q xy k 16 26 66 k Q Q Q T x M n 11 12 16 x 1 2 k 2 k - ( ) T T y [ ] = = T Q Q Q h h M M 1 y 12 22 26 2 1 k = T xy M Q Q Q xy k 16 26 66 k

  12. M x T x x M y T y = y M xy T xy xy k k k T x x = T y y T xy xy k k

  13. Calculate the residual stresses at the bottom surface of the 90o ply in a two ply [0/90] Graphite/Epoxy laminate subjected to a temperature change of -75oC. Use the unidirectional properties of Graphite/Epoxy lamina from Table 2.1. Each lamina is 5 mm thick.

  14. 7 - 7 0 200 - 0 200 . . 10 10 x 1 4 - o 0 225 = . m/m C 4 - o 10 0 225 / = . m/m C 10 / y 2 0 0 xy 12 o 0 4 - 0 225 . 10 x 7 - o 0 200 = . m/m C 10 / y 0 xy o 90

  15. 181 8 2 897 0 . . Q [ ] 2 897 10 35 0 = . . GPa 0 0 0 7 . 17 10 35 2 897 0 . . Q [ ] 2 897 181 8 0 = . . GPa 90 0 0 7 . 17

  16. 7 - 0 200 . 10 ( ) T x N 181 8 . 2 897 . 0 9 4 T y - ( 75 ) 2 897 . 10 35 . 0 10 ( ) 0 225 . 10 ( ) 0 [ 000 . ( 0 005 . )] = + - - N ) 0 0 7 17 . 0 T xy N 4 0 225 . 10 ( 10 35 . 2 897 . 0 9 7 ( 75 ) 2 897 . 181 8 . 0 10 ( ) 0 200 . 10 ( ) 0 [ 005 . 0 000 . ] + 0 0 7 17 . 0 5 . 131 1 10 5 . 131 1 = Pa - m. 10 0

  17. T x 7 - 0 200 . M 10 181 8 . 2 897 . 0 1 2 2 9 4 - T y ( 75 ) 2 897 . 10 35 . 0 10 ( ) 0 225 . [( 0 000 . ) ( 0 005 . ) ] = 10 M 2 0 0 7 17 . 0 T xy M 4 - 0 225 . 10 10 35 . 2 897 . 0 1 . 0 ( 2 2 9 7 - ( 75 ) 2 897 . 10 35 . 0 10 ( ) 0 200 . [( 0 005 . ) 000 ) ] + 10 2 0 0 7 17 . 0 2 1 538 . 10 2 1 538 = . Pa m 10 0

  18. 0 8 7 9 608 2 897 . . 10 10 0 7 8 [ = ] 2 897 9 608 A . . Pa -m 10 10 0 0 7 7 . 170 10 0 0 6 . 143 2 - 10 2 [ = ] 0 0 B Pa - 6 . 143 2 m 10 0 0 0 0 3 2 8 007 2 414 . . 10 10 0 2 3 3 [ = ] 2 414 8 007 D . . Pa - 10 10 m 0 0 2 5 975 . 10

  19. 0 0 0 0 8 7 6 9 608 2 897 . 143 2 . . - 10 10 10 x 5 . 131 1 - 10 0 0 0 0 7 8 6 2 897 9 608 . 143 2 . . 10 10 10 5 . 131 1 - y 10 0 0 0 0 0 7 0 7 . 170 0 10 xy = 0 0 0 2 6 3 2 1 538 . 143 2 8 007 2 414 - . - . . 10 10 10 10 x 0 0 0 2 6 2 3 1 538 . 143 2 2 414 8 007 . . . 10 10 10 10 y 0 0 0 0 0 0 2 5 975 . 10 xy 0 4 - 3 907 - . x 10 0 4 - 3 907 - . y 10 = T 0 A B m/m N 0 0 T B D xy M = 1 - 1 276 - . 10 x 1 /m 1 - 1 276 . 10 y 0 xy

  20. 4 1 - 3 907 . 1 276 . 10 10 x 4 1 - - 3 907 . 0 ( 005 . ) 1 276 . = + 10 10 y 0 0 xy o 90 3 - 1 029 . 10 4 - 2 475 = . m/m 10 0

  21. T x 4 - 0 225 . 10 7 T y - 0 200 . ( 75 ) = 10 0 T xy 2 - . 16875 0 10 5 - . 15000 0 = m/m 10 0

  22. 3 - 0 6585 . M x 10 3 2 - - 1 029 . 0 16875 . 10 10 3 - 0 2490 . = 4 5 M y - - 2 475 . 0 1500 . = 10 10 10 0 0 0 M xy 6 7 535 . 10 3 - 0 6585 . 10 10 35 . 2 897 . 0 x 9 3 7 - 2 897 . 181 8 . 0 10 ( ) 0 2490 . 4 718 = = . Pa. 10 10 y 0 0 7 17 . 0 0 xy 900

  23. Global Strains for Example 4.3 xy y x Ply # Position 1 (00) 2.475 10-4 -1.029 10-3 Top 0.0 -7.160 10-5 -7.098 10-4 Middle 0.0 -3.907 10-4 -3.907 10-4 Bottom 0.0 2 (900) -3.907 10-4 -3.907 10-4 Top 0.0 -7.098 10-4 -7.160 10-5 Middle 0.0 -1.029 10-3 2.475 10-4 Bottom 0.0

  24. Global Stresses for Example 4.3 y y xy Ply # Position 1(00) 4.718 107 7.535 106 Top 0.0 -9.912 106 9.912 106 Middle 0.0 -6.701 107 1.229 107 Bottom 0.0 2(900) 1.229 107 -6.701 107 Top 0.0 9.912 106 -9.912 106 Middle 0.0 7.535 106 4.718 107 Bottom 0.0

  25. To find coefficients of thermal and moisture expansion of laminates Symmetric laminates [B] = 0 No bending occurs under thermal hygrothermal loads Assuming T = 1 and C = 0 B A A A 16 xy o x T x C x N N A A A B B B 11 12 16 11 12 16 x + + = o y T y C y N N A A A B B B 12 22 26 12 22 26 y B B T xy C xy o N N 16 26 66 26 66 xy 0 x T x * 11 * 12 * 16 N A A A x 0 y T y * 12 * 22 * 26 = N A A A y * 16 * 26 * 66 T xy A A A 0 xy N xy 0 = C 1 = T

  26. To find coefficients of thermal and moisture expansion of laminates Symmetric laminates [B] = 0 No bending occurs under thermal hygrothermal loads Assuming T = 0 and C = 1 B A A A 16 xy o x T x C x N N A A A B B B 11 12 16 11 12 16 x + + = o y T y C y N N A A A B B B 12 22 26 12 22 26 y B B T xy C xy o N N 16 26 66 26 66 xy 0 x C x * 11 * 12 * 16 N A A A x 0 y C y * 12 * 22 * 26 = N A A A y * 16 * 26 * 66 C xy A A A 0 N 0 = T xy xy 1 = C

  27. coefficien the Find ts of thermal moisture and expansion of a [0/ 90] properties the Use laminate. poxy Graphite/E of S unidirecti Graphite/E onal poxy lamina from table 2.1.

  28. 0 10 11 - - 5 353 2 297 . - . 10 10 1 0 11 10 - - * [ = ] 2 297 9 886 - . . 10 10 A Pa-m 0 0 9 - 9 298 . 10

  29. T = 1C Q Q Q T x N 11 12 16 x 3 T y ( ) = T - Q Q Q N h h 1 y k k - 12 22 26 1 k = 200 . T xy N Q Q Q 897 . xy 0 k 16 26 66 k 7 - 10 181 8 . 2 0 9 4 - ) 1 ( 2 897 . 10 35 . 0 10 ( ) 0 225 . [ 0 0025 . 0 0075 . )] = - - (- 10 0 0 0 7 17 . - 0 4 0 225 . 10 10 35 . 2 897 . 9 7 - ) 1 ( 2 897 . 181 35 . 0 10 ( ) 0 200 . 0 [ 0025 . ( 0 0025 . )] + - - 10 0 0 7 17 . 0 0 3 1 852 . 10 7 - 200 . 10 181 8 . 2 897 . 0 3 2 673 = . Pa - m 10 9 4 - ) 1 ( 2 897 . 10 35 . 0 10 ( ) 0 225 . 0 [ 0075 . 0 0025 . ] + - 10 0 0 0 7 17 . 0

  30. 0 0 x 10 - 11 - 3 5 353 . 2 297 . 1 852 . - 10 10 10 0 11 - 10 - 3 0 y 2 297 . 9 886 . 2 673 . = - 10 10 10 0 0 0 9 - 9 298 . 10 0 xy 7 - 9 303 . 10 6 - 2 600 = . m/m. 10 0 0 x 7 - 9 303 . 10 x = 6 0 y - 2 600 / = . m/m C 10 y 0 0 xy xy 0 = C 1 = T

  31. Q Q Q C x N 11 12 16 x 3 C = 1 kg/kg C y ( ) = C h - k Q Q Q N h 1 - k y 12 22 26 1 = k C xy N Q Q Q xy 16 26 66 k 181 8 . 2 897 . 0 0 9 ) 1 ( 2 897 . 10 35 . 0 10 ( ) 0 6 . [ 0 0025 . 0 0075 . )] = - - (- 0 0 7 17 . 0 10 35 . 2 897 . 0 0 6 . 9 ) 1 ( 2 897 . 181 35 . 0 10 ( ) 0 0 [ 0025 . ( 0 0025 . )] + - - 0 7 4 842 . 0 0 7 17 . 0 10 181 8 . 2 897 . 0 7 7 077 = . Pa - m 10 9 ) 1 ( 2 897 . 10 35 . 0 10 ( ) 0 6 . 0 [ 0075 . 0 0025 . ] + - 0 0 0 7 17 . 0

  32. 0 0 x 10 - 11 - 7 5 353 . 2 297 . 4 842 . - 10 10 10 0 11 - 10 - 7 0 y 2 297 . 9 886 . 7 077 . = - 10 10 10 0 0 0 9 - 9 298 . 10 0 xy 2 - 2 430 . 10 2 - 6 885 = . m/m 10 0 0 x 2 - 2 430 . 10 x = 2 0 y - 6 885 / / . m/m kg kg 10 y 0 0 xy xy = C 1 = T 0

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