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Macromechanics of a Laminate Textbook: Mechanics of Composite Materials Author: Autar Kaw Figure 4.1 Fiber Direction x y z CHAPTER OBJECTIVES Understand the code for laminate stacking sequence Develop relationships of mechanical and hygrothermal loads applied to a laminate to strains and stresses in each lamina Find the elastic stiffnesses of laminate based on the elastic moduli of individual laminas and the stacking sequence Find the coefficients of thermal and moisture expansion of a laminate based on elastic moduli, coefficients of thermal and moisture expansion of individual laminas, and stacking sequence Laminate Behavior • elastic moduli • the stacking position • thickness • angles of orientation • coefficients of thermal expansion • coefficients of moisture expansion x P P z (a) M M x z (b) M M x P x P z (c) Figure 4.2 Figure 4.3 x x z z Mxy Myx Nx y Nyx Nxy My y Ny (a) (b) Mx Classical Lamination Theory Each lamina is orthotropic. Each lamina is homogeneous. A line straight and perpendicular to the middle surface remains straight and perpendicular to the middle surface during deformation. ( γxz = γyz = 0 ) . The laminate is thin and is loaded only in its plane (plane stress) (σz = τxz = τyz = 0 ) . Displacements are continuous and small throughout the laminate (| u |, | v |, | w | | h |) , where h is the laminate thickness. Each lamina is elastic. No slip occurs between the lamina interfaces. Figure 4.4 U0 h/2 Mid-Plane h/2 x z A z A z Cross-Section Before Loading Cross-Section after Loading wo Global Strains in a Laminate εx εy γ xy 2 w0 u0 2 x x 2 w0 v0 = . + z 2 y y u0 2 v0 w 0 + 2 x y xy 0 κx ε x 0 ε y + z κ y . κ xy γ0xy Figure 4.5 Mid-Plane z Laminate Strain Variation Stress Variation Figure 4.6 1 2 h0 3 h1 h/2 h2 h3 Mid-Plane hk-1 hk tk k-1 k k+1 hn-1 hn n h/2 z Stresses in a Lamina in a Laminate σ x σ y = τ xy k Q11 Q12 Q16 ε x Q Q ε y Q 22 26 12 γ Q16 Q 26 Q 66 k xy k Q11 Q12 Q16 = Q12 Q 22 Q 26 Q Q Q 26 66 k 16 ε0 Q11 Q12 Q16 x 0 + z Q Q Q ε y 22 26 12 0 Q Q Q γ xy 26 66 k 16 κ x κ y . κ xy Forces and Stresses N x N y = N xy N x N y = N xy h/ 2 -h/ 2 σ x σ y dz, τ xy n hk k=1 h k- 1 σ x σ y dz, τ xy k Forces and Strains N x N y = N xy Q11 Q12 Q16 ε0x 0 dz Q Q Q 26 ε y 12 22 0 Q16 Q 26 Q66k γ xy hk n k = 1 h k-1 n + hk k = 1 h k 1 Q11 Q12 Q16 Q Q Q 22 26 12 Q16 Q 26 Q66k κ x κ y z dz κ xy Forces and Strains 0 Q11 Q12 Q16 N x x hk n 0 N y = Q12 Q 22 Q 26 dz y h k =1 k -1 0 N xy xy Q16 Q 26 Q66 k Q11 Q12 Q16 n x hk + Q12 Q 22 Q 26 z dz y h k =1 k 1 xy Q Q Q 26 66 k 16 Integrating terms hk dz = ( hk hk - 1) , hk - 1 hk hk - 1 1 2 zdz = ( hk h2k - 1) , 2 Forces and Strains N x A11 N y = A12 N xy A16 A12 A22 A26 0 A16 x B11 B12 B16 x 0 A26 y + B12 B22 B26 y A66 0 B16 B26 B66 xy xy n Aij = [( Qij ) ] k ( hk - hk - 1 ), i = 1,2,6; j = 1,2,6, k =1 1 Bij = 2 n k =1 [( Qij ) ] k ( h2k - h2k - 1 ), i = 1,2,6; j = 1,2,6 Moments and Strains M x B11 B12 B16 ε0x D11 D12 D16 κ x 0 M y = B12 B22 B26 ε y + D12 D22 D26 κ y M xy B16 B26 B66 γ0 xy D16 D26 D66 κ xy 1 Dij = 3 n k =1 [( Qij ) ] k ( h3k - h3k - 1 ) i = 1, 2, 6; j = 1, 2, 6. Forces, Moments, Strains, Curvatures N x A11 A12 A16 B11 B12 B16 N y A12 A22 A26 B12 B22 B26 N xy A16 A26 A66 B16 B26 B66 = M x B11 B12 B16 D11 D12 D16 M y B12 B22 B26 D12 D22 D26 M xy B16 B26 B66 D16 D26 D66 ε0x ε0y γ0 xy κ x κ y κ xy Steps 1. Find the value of the reduced stiffness matrix [Q] for each ply using its four elastic moduli, E1, E2, v12, G12 in Equation (2.93). 2. Find the value of the transformed reduced stiffness matrix [ Q ] for each ply using the [Q] matrix calculated in Step 1 and the angle of the ply in Equation (2.104) or Equations (2.137) and (2.138). 3. Knowing the thickness, tk of each ply, find the coordinate of the top and bottom surface, hi, i = 1, . . . . . . . , n of each ply using Equation (4.20). 4. Use the [ Q ] matrices from Step 2 and the location of each ply from Step 3 to find the three stiffness matrices [A], [B] and [D] from Equation (4.28). 5. Substitute the stiffness matrix values found in Step 4 and the applied forces and moments in Equation (4.29). Steps 6. Solve the six simultaneous Equations (4.29) to find the midplane strains and curvatures. 7. Knowing the location of each ply, find the global strains in each ply using Equation (4.16). 8. For finding the global stresses, use the stress-strain Equation (2.103). 9. For finding the local strains, use the transformation Equation (2.99). 10. For finding the local stresses, use the transformation Equation (2.94). Figure 4.7 z = -7.5mm 0o 5mm 30o 5mm z = -2.5mm z = 2.5mm -45o z = 7.5mm z 5mm Problem A [0/30/-45] Graphite/Epoxy laminate is subjected to a load of Nx = Ny = 1000 N/m. Use the unidirectional properties from Table 2.1 of Graphite/Epoxy. Assume each lamina has a thickness of 5 mm. Find a)the three stiffness matrices [A], [B] and [D] for a three ply [0/30/-45] Graphite/Epoxy laminate. b)mid-plane strains and curvatures. c) global and local stresses on top surface of 300 ply. d)percentage of load Nx taken by each ply. Solution • A) From Example 2.4, the reduced stiffness matrix for the 00 Graphite/Epoxy ply is 0 181.8 2.897 9 [Q] = 2.897 10.35 0 (10 ) Pa 0 0 7.17 0 • From Equation (2.99), the transformed reduced stiffness matrix for each of the three plies are 0 181.8 2.897 [Q ]0 = 2.897 10.35 0 (109) Pa 0 0 7.17 109.4 32.46 54.19 [Q ]30 = 32.46 23.65 20.05 (109) Pa 54.19 20.05 36.74 56.66 42.32 - 42.87 [Q]-45 = 42.32 56.66 - 42.87 (109) Pa - 42.87 - 42.87 46.59 The total thickness of the laminate is h = (0.005)(3) = 0.015 m. The mid plane is 0.0075 m from the top and bottom of the laminate. Hence using Equation (4.20), the location of the ply surfaces are h0 = -0.0075 m h1 = -0.0025 m h2 = 0.0025 m h3 = 0.0075 m 3 Aij = [Qij]k ( hk - hk - 1) k =1 0 181.8 2.897 [A] = 2.897 10.35 0 (109) [(-0.0025)- (-0.0075)] 0 0 7.17 From Equation (4.28a), the extensional stiffness matrix [A] is 109.4 32.46 54.19 + 32.46 23.65 20.05 (109) [0.0025- (-0.0025)] 54.19 20.05 36.74 56.66 42.32 - 42.87 + 42.32 56.66 - 42.87 (109) [0.0075- 0.0025] - 42.87 - 42.87 46.59 The [A] matrix 1.739(109 ) 3.884(108 ) 5.663(107 ) 8 8 8 [A] = 3.884(10) 4.533(10) 1.141(10) Pa - m 5.663(107 ) 1.141(108 ) 4.525(108 ) From Equation (4.28b), the coupling stiffness matrix [B] is 1 3 2 2 = [ ] ( Qij k hk hk - 1) Bij 2 k =1 181.2 2.897 0 1 [B] = 2.897 10.35 0 (109 ) [(-0.0025)2 - (-0.0075)2 )] 2 0 7.17 0 109.432.4654.19 1 + 32.4623.6520.05 (109 ) (0.0025)2 - (-0.0025)2 2 54.1920.0536.74 42.32 42.87 56.66 1 + 42.32 56.66 42.87 (109 ) [(0.0075)2 - (0.0025)2 ] 2 42.87 42.87 46.59 The B Matrix 3.129106 9.855105 1.072106 5 6 6 [B] = 9.85510 1.15810 1.07210 Pa-m2 1.072106 1.072106 9.855105 3 From Equation (4.28c), the bending stiffness matrix [D] is 1 3 3 = [ ( Qij ]k hk hk - 1) Dij 3 k =1 181.8 1 [D] = 2.897 3 0 109.4 1 + 32.46 3 54.19 0 10.35 0 ( 109 ) (-0.0025 )3 - (-0.0075 )3 0 7.17 2.897 32.46 54.19 23.65 20.05 ( 109 ) ( 0.0025 )3 - (-0.0025 )3 20.05 36.74 42.32 42.87 56.66 1 + 42.32 56.66 42.87 ( 109 ) ( 0.0075 )3 - ( 0.0025 )3 3 42.87 42.87 46.59 The [D] matrix 3.343104 6.461103 5.240103 3 3 3 [D] = 6.46110 9.32010 5.59610 Pa - m 3 5.240103 5.596103 7.663103 B) Since the applied load is Nx = Ny = 1000N/m, the midplane strains and curvatures can be found by solving the following set of simultaneous linear equations (Equation 4.29). 1.739(109 ) 3.884(108) 5.663(107 ) - 3.129(106) 9.855(105) 1000 3.884(108) 4.533(108) - 1.141(108) 9.855(105) 1.158(106) 1000 0 5.663(107 ) - 1.141(108) 4.525(108) - 1.072(106) - 1.072(106) = 0 - 3.129(106) 9.855(105) - 1.072(106) 3.343(104) 6.461(103) 0 9.855(105) 1.158(106) - 1.072(106) 6.461(103) 9.320(103) 0 6 6 5 3 3 - 1.072(10 ) - 1.072(10 ) 9.855(10 ) - 5.240(10 ) - 5.596(10 ) - 1.072(106) ε0x 0 6 - 1.072(10 ) εy 5 0 9.855(10 ) γxy - 5.240(103) κ x - 5.596(103) κ y 7.663(103) κ xy Mid-plane strains and curvatures ε0x 3.123(10-7) 0 εy 3.492(10-6) m/m γ0 - 7.598( -7) 10 xy = 2.971(10-5) κx - 3.285( -4) 1/m 10 κ y -4 4.101( ) 10 κ xy C) The strains and stresses at the top surface of the 300 ply are found as follows. First, the top surface of the 300 ply is located at z = h1 = -0.0025 m. From Equation (4.16), εx 3.123(10-7) 2.971(10-5) -6 -4 εy = 3.492(10 ) + (-0.0025)- 3.285(10 ) γxy 0 - 7.598(10-7) 4.101(10-4) 30 , top 2.380(10-7) -6 = 4.313(10 ) m/m - 1.785(10-6) Table 4.1 Global strains (m/m) in Example 4.3 εx εy Ply # Position 1 (00) Top Middle Bottom 8.944 (10-8) 1.637 (10-7) 2.380 (10-7) 5.955 (10-6) 5.134 (10-6) 4.313 (10-6) -3.836 (10-6) -2.811 (10-6) -1.785 (10-6) 2 (300) Top Middle Bottom 2.380 (10-7) 3.123 (10-7) 3.866 (10-7) 4.313 (10-6) 3.492 (10-6) 2.670 (10-6) -1.785 (10-6) -7.598 (10-7) 2.655 (10-7) 3(-450) Top Middle Bottom 3.866 (10-7) 4.609 (10-7) 5.352 (10-7) 2.670 (10-6) 1.849 (10-6) 1.028 (10-6) 2.655 (10-7) 1.291 (10-6) 2.316 (10-6) xy Using the stress-strain Equations (2.98) for an angle ply, -7 2.380( ) σx 10 109.4 32.46 54.19 9 σy = 32.46 23.65 20.05 (10 ) 4.313(10-6) 54.19 20.05 36.74 - 1.785(10-6) τxy 0, top 30 6.930(104) = 7.391(104) Pa 3.381(104) Table 4.2 Global stresses (Pa) in Example 4.3 Ply # Position σx σy τxy 1 (00) Top Middle Bottom 3.351 (104) 4.464 (104) 5.577 (104) 6.188 (104) 5.359 (104) 4.531 (104) -2.750 (104) -2.015 (104) -1.280 (104) 2 (300) Top Middle Bottom 6.930 (104) 1.063 (105) 1.434 (105) 7.391 (104) 7.747 (104) 8.102 (104) 3.381 (104) 5.903 (104) 8.426 (104) 3 (-450) Top Middle Bottom 1.235 (105) 4.903 (104) -2.547 (104) 1.563 (105) 6.894 (104) -1.840 (104) -1.187 (105) -3.888 (104) 4.091 (104) The local strains and local stress as in the 300 ply at the top surface are found using transformation Equation (2.94) as -7 ε1 0.7500 0.2500 0.8660 2.380(10 ) -6 ε2 = 0.2500 0.7500 - 0.8660 4.313(10 ) γ12 /2 - 0.4330 0.4330 0.5000 - 1.785(10-6)/ 2 ε1 ε2 γ12 4.837(10- 7 ) = 4.067(10- 6 ) m /m 2.636(10- 6) Table 4.3 Local strains (m/m) in Example 4.3 ε1 ε2 γ12 Ply # Position 1 (00) Top Middle Bottom 8.944 (10-8) 5.955(10-6) -3.836(10-6) 1.637 (10-7) 5.134(10-6) -2.811(10-6) 2.380 (10-7) 4.313(10-6) -1.785(10-6) 2 (300) Top Middle Bottom 4.837(10-7) 4.067(10-6) 2.636(10-6) 7.781(10-7) 3.026(10-6) 2.374(10-6) 1.073(10-6) 1.985(10-6) 2.111(10-6) 3 (-450) Top Middle Bottom 1.396(10-6) 1.661(10-6) -2.284(10-6) 5.096(10-7) 1.800(10-6) -1.388(10-6) -3.766(10-7) 1.940(10-6) -4.928(10-7) 4 6.930( 10 ) σ1 0.7500 0.2500 .8660 4 = 0.2500 0.7500 .8660 7.391( 10 ) σ2 τ12 - 0.4330 0.4330 0.5000 3.381(104) 9.973(104) = 4.348(104) Pa 1.890(104) Table 4.4 Local stresses (Pa) in Example 4.3 Ply # Position σ1 σ2 τ12 1 (00) Top Middle Bottom 3.351 (104) 4.464 (104) 5.577 (104) 6.188 (104) 5.359(104) 4.531 (104) -2.750 (104) -2.015 (104) -1.280 (104) 2 (300) Top Middle Bottom 9.973 (104) 1.502 (105) 2.007 (105) 4.348 (104) 3.356 (104) 2.364 (104) 1.890 (104) 1.702 (104) 1.513 (104) 3 (-450) Top Middle Bottom 2.586 (105) 9.786 (104) -6.285 (104) 2.123 (104) 2.010 (104) 1.898 (104) -1.638 (104) -9.954 (103) -3.533 (103) D) The portion of the load Nx taken by each ply can be calculated by integrating the stress xx through the thickness of each ply. However, since the stress varies linearly through each ply, the portion of the load Nx taken is simply the product of the stress xx at the middle of each ply (See Table 4.2) and the thickness of the ply. Portion of load Nx taken by 00 ply = 4.464(104)(5)(10-3) = 223.2 N/m Portion of load Nx taken by 300 ply = 1.063(105)(5)(10-3) = 531.5 N/m Portion of load Nx taken by -450 ply = 4.903(104)(5)(10-3) = 245.2 N/m The sum total of the loads shared by each ply is 1000 N/m, (223.2 + 531.5 + 245.2) which is the applied load in the x-direction, Nx. Percentage of load Nx taken by 00 223.2 100 ply 1000 = 22.32% Percentage of load Nx taken by 300 ply Percentage of load Nx taken by -450 ply 531.5 100 1000 = 53.15% 245.2 100 1000 = 24.52% Figure 4.8 Strip 1, E1, 1 Strip 2, E2, 2