Laminate Ply-by-Ply Failure Analysis in Composite Materials
Explore the detailed design and analysis of a laminate ply-by-ply failure process for a Graphite/Epoxy composite, considering the fiber direction, load application, and material properties. Learn about strain calculations, Tsai-Wu criteria, and ply-specific maximum strain values. Gain insights into determining failure loads for undamaged and damaged plies in this comprehensive study by Dr. Autar Kaw.
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Chapter 5 Design and Analysis of a Laminate Ply by Ply Failure Dr. Autar Kaw Department of Mechanical Engineering University of South Florida, Tampa, FL 33620 Courtesy of the Textbook Mechanics of Composite Materials by Kaw
Fiber Direction x y FIGURE 4.1 Schematic of a lamina z If a ply fails, has the whole laminate failed? Maybe. When the first ply fails, you may get catastrophic failure or you may be able to still apply for more load.
The only load applied is a tensile normal load in the x-direction, that is, the direction parallel to the fibers in the 0o ply. Find the ply-by-ply failure load for a [0/90/0] Graphite/Epoxy laminate. Assume the thickness of each ply is 5 mm and use properties of unidirectional Graphite/Epoxy lamina from Table 2.1. 0o z = -2.5mm z = -7.5mm 5mm 90o 5mm z = 2.5mm 0o 5mm z z = 7.5mm
Assume we apply Nx=1 N/m Mid-plane strains (midplane curvature are zero) 0 x 10 5 353 . 10 11 0 y 2 297 . 10 = 0 0 xy
Ply# Position 1 2 12 1 (00) 9.726 101 9.726 101 9.726 101 1.313 100 1.313 100 1.313 100 Top Middle Bottom 0.0 0.0 0.0 2 (900) -2.626 100 -2.626 100 -2.626 100 5.472 100 5.472 100 5.472 100 Top Middle Bottom 0.0 0.0 0.0 3 (00) 9.726 101 9.726 101 9.726 101 1.313 100 1.313 100 1.313 100 Top Middle Bottom 0.0 0.0 0.0
Ply# Position Maximum Strain Tsai-Wu 1.548 107 (1T) 1.548 107 (1T) 1.548 107 (1T) 7.254 106 (2T) 7.254 106 (2T) 7.254 106 (2T) 1.548 107 (1T) 1.548 107 (1T) 1.548 107 (1T) 1.339 107 1.339 107 1.339 107 7.277 106 7.277 106 7.277 106 1.339 107 1.339 107 1.339 107 1 (00) Top Middle Bottom 2 (900) Top Middle Bottom 3 (00) Top Middle Bottom
Nx=(1 N/m)x(7.277x106) =7.277x106 N/m Nx/h = 7.277x106/0.015 = 485.1 MPa ox = (5.353x10-10)x(7.277x106) = 0.003895
Q for the undamaged plies 181 8 2 897 0 . . Q [ = ] 2 897 10 35 0 . . GPa 0 0 7 . 17 z = -7.5mm Q for the damaged plies 0 0 0 0o 5mm z = -2.5mm 90o = 5mm [ ] 0 0 0 GPa Q z = 2.5mm 0o 5mm 0 0 0 z z = 7.5mm
Assume we apply Nx=1 N/m Mid-plane strains (midplane curvatures are zero) 1 y - = 0 x 10 - 5 525 . 10 0 10 - 547 . 10 0 0 xy
Ply# Position 1 2 12 1 (00) 1.0000 102 1.0000 102 1.0000 102 Top Middle Bottom 0.0 0.0 0.0 0.0 0.0 0.0 2 (900) Top Middle Bottom --- --- --- --- --- --- --- --- --- 3 (00) 1.0000 102 1.0000 102 1.0000 102 Top Middle Bottom 0.0 0.0 0.0 0.0 0.0 0.0
Ply# Position Max Strain Tsai-Wu 1 (00) 1.5000 107 (1T) 1.5000 107(1T) 1.5000 107(1T) 1.5000 107 1.5000 107 1.5000 107 Top Middle Bottom 2 (900) Top Middle Bottom --- --- --- --- --- --- 3 (00) 1.5000 107(1T) 1.5000 107(1T) 1.5000 107(1T) 1.5000 107 1.5000 107 1.5000 107 Top Middle Bottom
Nx=(1 N/m)x(1.5x107) =1.5x107 N/m Nx/h = 1.5x107/0.015 = 1000 MPa ox = (5.525x10-10)x(1.5x107) =0.008288
1500 1200 Normal Stress, Nx/h (MPa) Last Ply Failure 900 117 GPa 600 First Ply Failure 124 GPa 300 0 0 0.005 0.01 FIGURE 5.1 Stress-strain curve showing ply-by-ply failure of a laminated composite Normal Strain, ex
