Macromechanics of a Laminate Tetboo: Mechanics of Composite Materials Author: Autar Kaw
Figure 4.1 Fiber Direction θ z
CHAPTER OJECTIVES Understand the code for laminate stacing sequence Develop relationships of mechanical and hgrothermal 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 stacing sequence Find the coefficients of thermal and moisture epansion of a laminate based on elastic moduli, coefficients of thermal and moisture epansion of individual laminas, and stacing sequence
Laminate ehavior elastic moduli the stacing position thicness angles of orientation coefficients of thermal epansion coefficients of moisture epansion
P P z (a M M z (b M M P P (c z Figure 4.
Figure 4. z z M M N M N N M N (a (b
Classical Lamination Theor 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. ( γz = γz =. The laminate is thin and is loaded onl in its plane (plane stress ( σz = τz = τz =. Displacements are continuous and small throughout the laminate ( u, v, w << h, where h is the laminate thicness. Each lamina is elastic. No slip occurs between the lamina interfaces.
Figure 4.4 U h/ Mid-Plane z h/ z A A α α w o zα Cross-Section efore Loading Cross-Section after Loading
Global Strains in a Laminate. w w w + z v + u v u = γ ε ε. κ κ κ + z γ ε ε =
Figure 4.5 Mid-Plane z Laminate Strain Variation Stress Variation
Figure 4.6 h 1 h 1 h/ h h Mid-Plane h h -1 t -1 +1 h/ z h n h n-1 n
Stresses in a Lamina in a Laminate γ ε ε = τ σ σ 66 11. κ κ κ + z γ ε ε = 66 11 66 11
Forces and Stresses dz, τ σ σ = N N N h/ -h/ dz, τ σ σ = N N N h h n = - 1 1
Forces and Strains dz γ ε ε = N N N h h n = - 66 11 1 1 z dz κ κ κ + h h n = 66 11 1 1
Forces and Strains γ ε ε h h 66 11 n = 1 dz = N N N - 1 κ κ κ h h 66 11 n = 1 z dz + 1
Integrating terms h h h h - - 1 1 dz = ( h h - 1, 1 1 zdz = ( h h -,
Forces and Strains N N N A = A A 11 A A A A A A 66 ε ε + γ 11 66 κ κ κ n A ij - = 1 = [( ] ( h - h, i = 1,, 6; j = 1,, 6, ij 1 1 n = ij [( ] ( h - h ij - 1 = 1, i = 1,, 6; j = 1,, 6
Moments and Strains κ κ κ D D D D D D D D D + γ ε ε = M M M 66 11 66 11.,, ; j =,, i = h - h ( ] [( = D - ij n = ij 6 1 6 1 1 1 1
Forces, Moments, Strains, Curvatures κ κ κ γ ε ε D D D D D D D D D A A A A A A A A A = M M M N N N 66 66 11 11 66 66 11 11
Steps 1. Find the value of the reduced stiffness matri [] for each pl using its four elastic moduli, E 1, E, v, G in Equation (.9.. Find the value of the transformed reduced stiffness matri [ ] for each pl using the [] matri calculated in Step 1 and the angle of the pl in Equation (.14 or Equations (.17 and (.18.. Knowing the thicness, t of each pl, find the coordinate of the top and bottom surface, h i, i = 1,......., n of each pl using Equation (4.. 4. Use the [ ] matrices from Step and the location of each pl from Step to find the three stiffness matrices [A], [] and [D] from Equation (4.8. 5. Substitute the stiffness matri values found in Step 4 and the applied forces and moments in Equation (4.9.
Steps 6. Solve the si simultaneous Equations (4.9 to find the midplane strains and curvatures. 7. Knowing the location of each pl, find the global strains in each pl using Equation (4.. 8. For finding the global stresses, use the stress-strain Equation (.1. 9. For finding the local strains, use the transformation Equation (.99. 1. For finding the local stresses, use the transformation Equation (.94.
Figure 4.7 z = -7.5mm z = -.5mm z =.5mm z = 7.5mm o o -45 o z 5mm 5mm 5mm
Problem A [//-45] Graphite/Epo laminate is subjected to a load of N = N = 1 N/m. Use the unidirectional properties from Table.1 of Graphite/Epo. Assume each lamina has a thicness of 5 mm. Find athe three stiffness matrices [A], [] and [D] for a three pl [//-45] Graphite/Epo laminate. bmid-plane strains and curvatures. cglobal and local stresses on top surface of pl. dpercentage of load N taen b each pl.
Solution A From Eample.4, the reduced stiffness matri for the Graphite/Epo pl is 181.8 9 [] =.897 1.5 (1 7.17.897 Pa
From Equation (.99, the transformed reduced stiffness matri for each of the three plies are 181.8.897 9 [ ] =.897 1.5 (1 Pa 7.17 19.4.46 54.19 9 [ ] =.46.65.5 (1 Pa 54.19.5 6.74 56.66 4. = 4. 56.66-4.87-4.87-4.87-4.87 (1 46.59 9 [ ] -45 Pa
The total thicness of the laminate is h = (.5( =.15 m. The mid plane is.75 m from the top and bottom of the laminate. Hence using Equation (4., the location of the pl surfaces are h = -.75 m h 1 = -.5 m h =.5 m h =.75 m
From Equation (4.8a, the etensional stiffness matri [A] is Aij = [ ] (h - h -1 =1 ij 181.8.897 9 [A] =.897 1.5 (1 [(-.5 - (-.75] 7.17 19.4.46 54.19 9 +.46.65.5 (1 [.5 - (-.5] 54.19.5 6.74 56.66 4. 4. 56.66-4.87-4.87 + 9-4.87-4.87 (1 [.75 -.5] 46.59
The [A] matri [A] = 1.79(1.884(1 5.66(1 9 8 7.884(1 4.5(1 8 8 1.141(1 8 7 5.66(1 8 1.141(1 8 4.55(1 Pa - m
From Equation (4.8b, the coupling stiffness matri [] is h - (h [ ] 1 = 1 - ij =1 ij ] (-.75 - [(-.5 (1 7.17 1.5.897.897 181. 1 [] = 9 [ ] (-.5 - (.5 (1 6.74.5 54.19.5.65.46 54.19.46 19.4 1 + 9 ] (.5 - [(.75 (1 46.59 4.87 4.87 4.87 56.66 4. 4.87 4. 56.66 1 + 9
The Matri ( 6 ( 5 ( 6 9. 855 1 17. 1 ( 5. ( 6. ( 6 1158 1 17 1 ( 6. ( 6. ( 5 17 1 9 855 1 9. 1 [] = 9. 855 1 Pa- m 17. 1
From Equation (4.8c, the bending stiffness matri [D] is D ij [D] = + 1 = 1 1 =1 1818.. 897 19. 4. 46 5419. [ ij. 897 1. 5. 46 5.. 5 ] (h - h -1 ( 1 9 75 717. 5419.. 5 ( 1 9 (. 5 - (-. 5 6. 74 [(-. 5 - (-. ] [ ] + 1 56. 66 4. 4. 87 4. 56. 66 4. 87 4. 87 4. 87 46. 59 ( [(. 75 - (. ] 1 9 5
The [D] matri [D] =.4 1 6.461 1 5.4 1 ( 4 ( 6.461 1 ( 5.4 1 ( ( 9. 1 ( 5.596 1 ( ( 5.596 1 ( 7.66 1 Pa - m
Since the applied load is N = N = 1N/m, the midplane strains and curvatures can be found b solving the following set of simultaneous linear equations (Equation 4.9. 9 1 1.79(1 8 1.884(1 N 7 5.66(1 = M 6 -.9(1 M 5 9.855(1 M 6-1.7(1 8.884(1 8 4.5(1 8-1.141(1 5 9.855(1 6 1.158(1 6-1.7(1 5.66(1 7 8-1.141(1 8 4.55(1 6-1.7(1 6-1.7(1 5 9.855(1 6 -.9(1 5 9.855(1 6-1.7(1.4(1 4 6.461(1-5.4(1 5 9.855(1 6 1.158(1 6-1.7(1 6.461(1 9.(1-5.596(1 6-1.7(1 6-1.7(1 5 9.855(1-5.4(1-5.596(1 7.66(1 ε ε γ κ κ κ
1/m 4.11(1.85(1 -.971(1 m/m 7.598(1 -.49(1.(1 = κ κ κ γ ε ε -4-4 -5-7 -6-7 Mid-plane strains and curvatures
C The strains and stresses at the top surface of the pl are found as follows. First, the top surface of the pl is located at z = h 1 = -.5 m. From Equation (4., ε ε γ, top.(1 =.49(1-7.598(1-7 -6-7 + (-.5.971(1 -.85(1 4.11(1-5 -4-4.8(1 = 4.1(1-1.785(1-7 -6-6 m/m
Table 4.1 Global strains (m/m in Eample 4. Pl # Position ε ε γ 1 ( Top Middle ottom 8.944 (1-8 1.67 (1-7.8 (1-7 5.955 (1-6 5.14 (1-6 4.1 (1-6 -.86 (1-6 -.811 (1-6 -1.785 (1-6 ( Top Middle ottom.8 (1-7. (1-7.866 (1-7 4.1 (1-6.49 (1-6.67 (1-6 -1.785 (1-6 -7.598 (1-7.655 (1-7 (-45 Top Middle ottom.866 (1-7 4.69 (1-7 5.5 (1-7.67 (1-6 1.849 (1-6 1.8 (1-6.655 (1-7 1.91 (1-6. (1-6
Using the stress-strain Equations (.98 for an angle pl, σ σ τ, top = 19.4.46 54.19.46.65.5 54.19.5 6.74 (1 9.8(1 4.1(1-1.785(1-7 -6-6 6.9(1 = 7.91(1.81(1 4 4 4 Pa
Table 4. Global stresses (Pa in Eample 4. Pl # Position σ σ τ 1 ( Top Middle ottom.51 (1 4 4.464 (1 4 5.577 (1 4 6.188 (1 4 5.59 (1 4 4.51 (1 4 -.75 (1 4 -.15 (1 4-1.8 (1 4 ( Top Middle ottom 6.9 (1 4 1.6 (1 5 1.44 (1 5 7.91 (1 4 7.747 (1 4 8. (1 4.81 (1 4 5.9 (1 4 8.4 (1 4 (-45 Top Middle ottom 1.5 (1 5 4.9 (1 4 -.547 (1 4 1.56 (1 5 6.894 (1 4-1.84 (1 4-1.187 (1 5 -.888 (1 4 4.91 (1 4
The local strains and local stress as in the pl at the top surface are found using transformation Equation (.94 as γ ε1 ε / = -.75.5.4.5.75.4 - -7.866.8(1-6.866 4.1(1.5-6 -1.785(1 / ε1 ε γ = 4. 87(1-4. 67(1. 66(1-7 6-6 m/m
Table 4. Local strains (m/m in Eample 4. Pl # Position ε 1 ε γ 1 ( Top Middle ottom 8.944 (1-8 1.67 (1-7.8 (1-7 5.955(1-6 5.14(1-6 4.1(1-6 -.86(1-6 -.811(1-6 -1.785(1-6 ( Top Middle ottom 4.87(1-7 7.781(1-7 1.7(1-6 4.67(1-6.(1-6 1.985(1-6.66(1-6.74(1-6.111(1-6 (-45 Top Middle ottom 1.96(1-6 5.96(1-7 -.766(1-7 1.661(1-6 1.8(1-6 1.94(1-6 -.84(1-6 -1.88(1-6 -4.98(1-7
.81(1 7.91(1 6.9(1.5.4.4 - -.866.75.5.866.5.75 = τ σ σ 4 4 4 1 Pa 1.89(1 4.48(1 9.97(1 = 4 4 4
Table 4.4 Local stresses (Pa in Eample 4. Pl # Position σ 1 σ τ 1 ( Top Middle ottom.51 (1 4 4.464 (1 4 5.577 (1 4 6.188 (1 4 5.59(1 4 4.51 (1 4 -.75 (1 4 -.15 (1 4-1.8 (1 4 ( Top Middle ottom 9.97 (1 4 1.5 (1 5.7 (1 5 4.48 (1 4.56 (1 4.64 (1 4 1.89 (1 4 1.7 (1 4 1.51 (1 4 (-45 Top Middle ottom.586 (1 5 9.786 (1 4-6.85 (1 4. (1 4.1 (1 4 1.898 (1 4-1.68 (1 4-9.954 (1 -.5 (1
D The portion of the load N taen b each pl can be calculated b integrating the stress through the thicness of each pl. However, since the stress varies linearl through each pl, the portion of the load N taen is simpl the product of the stress σ at the middle of each pl (See Table 4. and the thicness of the pl. Portion of load N taen b pl = 4.464(1 4 (5(1 - =. N/m Portion of load N taen b pl = 1.6(1 5 (5(1 - = 51.5 N/m Portion of load N taen b -45 pl = 4.9(1 4 (5(1 - = 45. N/m The sum total of the loads shared b each pl is 1 N/m, (. + 51.5 + 45. which is the applied load in the -direction, N. σ
Percentage of load N taen b pl. = 1 1 =. % Percentage of load N taen b pl 51.5 = 1 1 = 5.15 % Percentage of load N taen b -45 pl 45. = 1 1 = 4.5 %
Figure 4.8 Strip 1, E 1, ν 1 Strip, E, ν