Macromechanical Analysis of a Lamina Part 4: 2D Stiffness Matrix
This chapter discusses the application of stresses to determine the engineering constants of a unidirectional lamina, including pure shear stress and tensile loads in different directions. The stiffness and compliance matrices for a unidirectional lamina are analyzed under various axial and shear loads, providing valuable insights into material behavior and properties.
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Chapter 2 MacromechanicalAnalysis of a Lamina Part 4 2D Stiffness and Compliance Matrix for Unidirectional Lamina Dr. AutarKaw Department of Mechanical Engineering University of South Florida, Tampa, FL 33620 Courtesy of the Textbook Mechanics of Composite Materials by Kaw
(a) Tensile load in direction 1 (b) Tensile load in direction in 2 FIGURE 2.18 Application of stresses to find engineering constants of a unidirectional lamina (c) Pure shear stress in plane of 1-2
Apply a pure axial load in direction 1 0 0 0 , = , = 1 2 12 0 S S = S 1 11 12 1 1 11 1 0 = = S S S 2 12 1 2 12 2 22 0 = 0 0 S 12 12 66 12 1 1 1 = = S E 11 1 S E 1 11 1 S 12 = 2 12 = . S 12 12 S E 1 11 1
Apply a pure axial load in direction 2 0 0 2 1 , , = 0 = 12 0 S S = S 1 11 12 1 1 12 2 = S 0 = S S 2 22 2 2 12 2 22 0 = 0 0 12 S 12 66 12 1 1 = 2 = S E 22 2 E S 2 22 2 S 21 = S 1 12 = . 12 21 E S 2 2 22
Apply a pure shear load in Plane 12 = 0 0 0 , = , 1 2 12 0 S S 0 = 1 11 12 1 1 0 0 = = S S 2 2 12 2 22 =S 0 0 S 66 12 12 12 66 12 1 1 12 = G = S 66 12 G S 66 12 12
0 S S 1 11 12 1 0 = S S 2 12 2 22 0 0 S 12 66 12 1 12 0 E E 1 1 1 1 1 12 0 = 2 2 E 0 E 0 1 2 1 12 12 G 12
0 Q Q 1 11 12 1 0 = Q Q 2 2 12 22 0 0 Q 12 12 66 E E 1 2 S 12 22 = Q 1 1 0 - 11 2 12 S S S 21 12 21 12 11 22 S 1 1 12 = Q E E 12 2 12 S 2 2 12 S S 0 = 22 11 2 2 1 1 21 12 21 12 S 11 = Q 12 22 12 2 12 S S S 11 1 22 0 0 G 12 = Q 66 S 66
1 0 0 0 12 13 E E E 1 1 1 1 1 0 0 0 12 0 21 23 E E E 2 2 2 E E 1 1 1 1 1 1 1 1 0 0 0 2 2 31 32 12 E E E 0 = 3 3 3 3 3 2 2 = E 0 E 0 0 0 0 1 0 0 1 2 23 23 1 G 23 31 31 12 12 0 0 0 0 1 0 12 G 12 12 G 31 0 0 0 0 0 1 G 12
+ + 1 0 0 0 23 32 21 23 31 31 21 32 E E E E E E 2 3 2 3 2 3 E E 1 2 12 1 1 + + 1 0 0 0 1 1 0 - 21 23 31 13 31 32 12 31 21 12 21 12 2 2 E E E E E E 2 3 1 3 1 3 1 1 3 3 = E E + + 1 0 0 0 2 2 12 0 = 31 21 32 32 12 31 12 21 23 23 2 2 E E E E E E 1 1 1 2 2 3 1 3 21 12 21 12 31 31 0 0 0 0 0 G 12 12 23 12 0 0 G 12 0 0 0 0 0 G 31 12 0 0 0 0 0 G 12
