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Composites Design and Analysis Stress-Strain Relationship Prof Zaffar M. Khan Institute of Space Technology Islamabad Next Generation Aerospace Vehicle Requirements 2 Composite design and analysis Laminate Theory Manufacturing Methods Materials Composite Materials Design / Analysis Engineer Design Guidelines Design Data Quality Assurance MATERIAL SELECTION 4 The design/analysis engineer should be able to determine for the following question: How do composite laminates behave under load? Which composite material should be used? What are the guidelines for design and the analysis methods? What design data is required and how is it obtained? Can the composite product be fabricated easily? How can the quality of the composite product be assured? The purpose of the material presented here is to introduce the reader to the principles of the mechanical behavior of thin laminates (how laminates behave under loads). This includes both symmetric and unsymmetric laminates relative to the mid-plane, with both in-plane and flexural loads. Also included are the effects of temperature and moisture, and laminate strength. A liberal use is made of flow charts for aids in analysis Composite Analysis 2D vs. 3D State of Stress 6 Required elastic material properties for composites • Metals (isotropic materials) – E, G, ν – 2 independent properties – G = __E____ 2(1 + ν) • Composite lamina (unidirectional layer, ply) – In plane: E1, E2, G12, ν12 – Out of plane : E3, G13, G23, ν13, ν23 • But for transverse isotropy (2 = 3): E2 = E3 G12 = G13 ν12 = ν13 G23 = __E23___ 2(1 + ν 23) Therefore 5 independent properties: E1 E2 G12 ν 12 G23 Terminology used in micromechanics • • • • • Ef, Em - Young’s modulus of fiber and matrix Gf, Gm - Shear modulus of fiber and matrix νf, νm - Poisson’s ratio of fiber and matrix Vf, Vm - Volume fraction of fiber and matrix Wf, Wm – Weight fraction of fiber and matrix Elastic properties: Rule of mixtures approach • Parallel model - E1 and ν12 (Constant Strains) E1 = Ef1Vf + EmVm ν12 = νf Vf + νmVm Predictions agree well with experimental data • Series model – E2 and G12 (Constant Stress) E2 = G12 = Ef2 Em________ Ef2 Vm + Em Vf __ __ Ef2 Em Ef2 Vm + Em Vf __ __ ________ Experimental results predicted less accurately Micromechanics example: Volume fraction changes Knowns : Carbon: E = 34.0 x 106 psi Epoxy : E = 0.60 x 106 psi • How much does the longitudinal modulus change when the fiber volume fraction is changed from 58% to 65%? E = E V +E V 1 - For Vf = 0.58: - For Vf = 0.65: E1 = E1 = f1 f m m (34.0 x 106 psi)(.58) + (0.60 x 106 psi)(0.42) = 20.0 x 106 psi (34.0 x 106 psi)(.65) + (0.60 x 106 psi)(0.35) = 22.3 x 106 psi Thus, raising the fiber volume fraction from 58% to 65% increases E1 by 12% Design and analysis of composite laminates: Laminated Plate Theory (LPT) • Used to determine the response of a composite laminate based on properties of a layer (or ply) Laminate Ply Orientation Code • • • • • • Designate each ply by it’s fiber orientation angle List plies in sequence starting from top of laminate Adjacent plies are separated by “/” if their angle is different Designate groups of plies with same angle using subscripts Enclose complete laminate in brackets Use subscript “S” to denote mid plane symmetry, or “T” to denote total laminate • Bar on the top of the ply indicates mid-plane Special types of laminates • Symmetric laminate – for every ply above the laminate mid plane, there is an identical ply(material and orientation) an equal distance below the mid plane • Balanced laminate – for every ply at a +θ orientation, there is another ply at the – θ orientation somewhere in the laminate • Cross-ply laminate – composed of plies of either 0o or 90o (no other ply orientations) • Quasi-isotropic laminate – produced using at least three different ply orientations, all with equal angles between them. Exhibits isotropic extensional stiffness properties (have the same E in all in-plane directions) The response of special laminates • • • Balanced, unsymmetric, laminate – Tensile loading produces twisting curvature – Ex: [+θ/0/- θ]τ Symmetric, unbalanced laminate – Tensile loading produces in-plane shearing – Ex: [+θ/0/- θ]τ Unsymmetric cross-ply laminate – Tensile loading produces bending curvature – Ex: [0/90]τ • Balanced and symmetric laminate – Tensile loading produces extension – Ex: [+θ/- θ]s • Quasi isotropic laminate: [+60/0/- 60]s and [+45/0/+45/90]s – Tensile loading produces extension loading, independent of angle Stress-Strains Relationships in Lamina (1) (2) The lamina is homogeneous, orthotropic and linear-elastic The lamina is in a state of plane stress (very thin), therefore, the stresses associated with the z-direction are negligible: σz = τxz = τyz = 0 The normal strain in the x-direction for the lamina in Figure 2a is the sum of the normal strains in the x-direction for the laminae in Figures 2b, 2c and 2d: From Figure 2b, εx = 0 From Figure 2c, Ex= x or εx = x x x Ex From Figure 2d, νy = − or εx = − νy εy = − νy y y Ey Consequently, the total normal strain in the x-direction is εx = 0 + or εx = x Ex x Ex − νy − νy y Ey y Ey The normal strain in the y-direction for the lamina is determined from superposition as follows: From Figure 2b, εy = 0 From Figure 2c, νx = − From Figure 2d, Ey= y x y y or εy = − νx εx = − νx or εy = x Ex y Ey Consequently, the total normal strain in the y-direction is εy = 0 − νx or εy = − νx x Ex x Ex + + y Ey y (2) Ey The shear strain, which is independent of the normal strains, is determined as follows: From Figure 2b, Es = s s or εs = s Es From Figure 2c, εs = 0 From Figure 2d, εs = 0 Therefore, the shear strain associated with the x-y coordinate system is; εs = s Es (3) If we express Equations (1), (2) and (3) in matrix form, we have x y = s 1 Ex x Ex 0 y Ey 1 Ey 0 0 0 1 Es x S xx y = S yx s 0 S xy S yy 0 0 0 S ss x y s (4) Equation (4) may also be expressed as follows: x y s Ex 1 x y xE y = 1 x y 0 [ε] = [C] [σ] [σ] = [M] [ε] yE x 1 x y Ey 1 x y 0 0 0 Es x y = s Q xx Q yx 0 Q xy Q yy 0 0 0 Q ss x y s (5) Elastic Constants-x P σx = P/A εx εy lbs psi in/in in/in 0 0 0 0 105 1872 0.000520 − 0.000073 0.140 200 3565 0.001005 − 0.000140 0.139 300 5348 0.001495 − 0.000210 0.140 400 7398 0.002165 − 0.000245 0.136 500 8913 0.002515 − 0.000340 0.135 600 10695 0.003022 − 0.000405 0.134 700 12478 0.003545 − 0.000460 0.130 800 14260 0.004050 − 0.000520 0.128 νx = 0.135 (average) νx = − εy / εx Elastic Constants-y σy vs εy 14000 12000 P σy = P/A εx εy lbs psi in/in in/in νy = − εx / εy 10000 8000 0 100 0 1751 0 − 0.000075 0 0.000540 0.139 200 300 400 500 600 700 800 3503 5254 7005 8757 10508 12259 14011 − 0.000145 − 0.000210 − 0.000285 − 0.000350 − 0.000415 − 0.000470 − 0.000540 0.001090 0.001628 0.002218 0.002793 0.003351 0.003943 0.004620 0.133 0.129 0.128 0.125 0.124 0.119 0.117 νy = 0.127 (average) σy (psi) Ey = σy / εy = 3.05 x 106 psi 6000 4000 2000 0 0 0.001 0.002 0.003 εy (in/in) 0.004 0.005 Elastic Constants-s P σs = P/2A εμ εν εs = ε μ − ε ν lbs psi in/in in/in in/in 0 0 0 0 0 100 875 0.001075 − 0.000561 0.001632 200 1748 0.002212 − 0.001229 0.003441 300 2622 0.003527 − 0.002065 0.005592 400 3497 0.005175 − 0.003164 0.008339 500 4371 0.007219 − 0.004615 0.011834 600 5245 0.010547 − 0.007250 0.017797 700 6119 0.013412 − 0.009630 0.023042 800 6993 0.019082 − 0.014710 0.033792 Determination of Elastic Constants Type Type Material Ex Ey GPa GPa νx Es νf Specific gravity GPa Material Ex Ey Msi Msi νx Es t Msi inches Typical thickness meters T300/5208 Graphite/Epoxy 181 1.3 0.28 7.17 0.70 1.6 0.000125 B(4)/5505 Boron/Epoxy 204 18.5 0.23 5.59 0.50 2.0 0.000125 AS/3501 Graphite/Epoxy 138 8.96 0.30 7.1 0.66 1.6 0.000125 Scotchply 1002 Glass/Epoxy 38.6 8.27 0.26 4.14 0.45 1.8 0.000125 Kevlar 49/Epoxy Aramid/Epoxy 76 5.5 0.34 2.3 0.60 1.46 0.000125 T300/5208 Graphite/Epox y 26.25 1.49 0.28 1.04 0.005 B(4)/5505 Boron/Epoxy 29.59 2.68 0.23 0.81 0.005 AS/3501 Graphite/Epox y 20.01 1.30 0.30 1.03 0.005 Scotchply 1002 Glass/Epoxy 5.60 1.20 0.26 0.60 0.005 Kevlar 49/Epoxy Aramid/Epoxy 11.02 0.80 0.34 0.33 0.005 Transformation of Stress and Strain Area Stresses Stresses Forces These three equilibrium equations may be combined in matrix form as follows: The equations for the transformation of strain are the same as those for the transformation of stress: Stress-Strain Relationships in Global Coordinates Develop the relationship between the stresses and strains in global coordinates. [σxys] = [Qxys] [εxys] [σxys] = [T] [ σ126] [εxys] = [Ť] [ε126] (1) (2) (3) Substitution of Equations (2) and (3) into (1) yields [T] [ σ126] = [Qxys] [Ť] [ε126] (4) Pre multiplying both sides of Equation (4) by [T] −1: [T] −1 [T] [σ126] = [T] −1 [Qxys] [Ť] [ε126] or [ σ126] = [T] −1 [Qxys] [Ť] [ε126] or = [T] −1 [Qxys] [Ť] (5) (6) (7) or or = [Q126] [ σ126] = [Q126] [ε126] where [Q126] = [T] −1 [Qxys] [Ť] Note that [T(+θ)] −1 = [T(−θ)] The elements of the matrix [Q126] are as follows: Q11 = Qxx cos4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy sin4θ Q22 = Qxx sin4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy cos4θ Q12 = Q21 = (Qxx + Qyy − 4Qss) sin2θ cos2θ + Qxy (sin4θ +cos4θ) Q66 = (Qxx + Qyy − 2 Qxy − 2Qss) sin2θ cos2θ + Qss (sin4θ +cos4θ) Q16 = Q61 = (Qxx – Qxy − 2Qss) sinθ cos3θ + (Qxy – Qyy + 2Qss) sin3θ co Q26 = Q62 = (Qxx – Qxy − 2Qss) sin3θ cosθ + (Qxy – Qyy + 2Qss) sinθcos3θ [Q126] −1 [ σ126] = [Q126] −1 [Q126] [ε126] or [Q126] −1 [ σ126] = [ε126] or [ε126] = [Q126] −1 [ σ126] or [ε126] = [S126] [ σ126] or The elements of the matrix [S126] are as follows: S11 = (Q22 Q66 – Q262) / ∆ S12 = S21 = (Q16 Q26 – Q12 Q66) / ∆ S16 = S61 = (Q12 Q26 – Q22 Q16) / ∆ S22 = (Q11 Q66 – Q162) / ∆ S26 = S62 = (Q12 Q16 – Q11 Q26) / ∆ S66 = (Q11 Q22 – Q122) / ∆ where ∆ = Q11Q22 Q66 + 2 Q12Q26 Q61 − Q22Q162 – Q66Q122 – Q11Q622 Summary: Lamina Analysis The Symmetric Laminate with In-Plane Loads A. Stress Resultants and Strains: The “A” Relationship between strains and the stress resultants. Related by constants Aij Deriving expressions for these constants: Expression for is: Expression for N1: N1 = (σ1)1 (h1-h2) + (σ1)2 (h2-h3) + (σ1)3 (h3-h4) + … Substitute the expressions for stress: Above equation can be rewritten as: Similarly for N2 and N6: Above can be combined into matrix form as: Or Where A11 = A12 = A16 = A22 = A26 = A66 = (Q11)k tk (Q12)k tk (Q16)k tk (Q22)k tk (Q26)k tk (Q66)k tk Example Determine the elements of [A] for a symmetric laminate of 6 layers: [0/45/90]s . The material of each lamina is Gr/Ep T300/5208 and t=0.005” = 0.000125 m 0° Lamina Q11 = Qxx = 181.8 Q22 = Qyy = 10.34 Q12 = Qxy = 2.897 Q66 = Qss = 7.17 Q16 = 0 Q26 = 0 90° Lamina Q11 = Qyy = 10.34 Q22 = Qxx = 181.8 Q12 = Qxy = 2.897 Q66 = Qss = 7.17 Q16 = 0 Q26 = 0 Q11 = Qxx cos4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy sin4θ Q22 = Qxx sin4θ + 2 (Qxy + 2 Qss) sin2θ cos2θ + Qyy cos4θ Q12 = Q21 = (Qxx + Qyy – 4 Qss) sin2θ cos2θ + Qxy (sin4θ + cos4θ) Q66 = (Qxx + Qyy – 2Qxy – 2 Qss) sin2θ cos2θ + Qss (sin4θ + cos4θ) Q16 = Q61 = (Qxx – Qxy – 2 Qss) sinθ cos3θ + (Qxy – Qyy + 2 Qss) sin3θ cosθ Q26 = Q62 = (Qxx – Qxy – 2 Qss) sin3θ cosθ + (Qxy – Qyy + 2 Qss) sinθ cos3θ Q11 = 0.25 Qxx + 2 (Qxy + 2 Qss) (0.25) + 0.25 Qyy = 56.65 Q22 = 0.25 Qxx + 2 (Qxy + 2 Qss) (0.25) + 0.25 Qyy = 56.65 Q12 = (Qxx + Qyy – 4 Qss) (0.25) + Qxy (0.5) = 42.31 Q66 = (Qxx + Qyy – 2 Qxy –2Qss) (0.25) + Qss (0.25) = 46.59 Q16 = (Qxx – Qxy – 2 Qss) (0.25) + (Qxy – Qyy +2 Qss) (0.25) = 42.87 Q26 = (Qxx – Qxy – 2 Qss) (0.25) + (Qxy – Qyy +2 Qss) (0.25) = 42.87 The Laminate A11 = 0.000125 (181.8 + 10.34 + 56.25) 2 = 0.0622 A22 = 0.000125 (10.34 + 181.8 + 56.65) 2 = 0.0622 A12 = 0.000125 (2.897 + 2.897 + 42.31) 2 = 0.0120 A66 = 0.000125 (7.17 + 7.17 + 46.59) 2 = 0.0152 A16 = 0.000125 (0 + 0 + 42.87) 2 = 0.0107 A26 = 0.000125 (0 + 0 + 42.87) 2 = 0.0107 GPa-m Equivalent Engineering Constants for the Laminate Equation for the composite laminate is Or [N] = [A] [ε] Matrix equation for a single orthotropic lamina or layer is Or [σ] = [M] [ε] The properties of the single “equivalent” orthotropic layer can be determined from the following equation: or 1 [A] = [M] A11 = E 1 1 h 1 A22 = h 1 1 2 1 A21 = h 2 1 1 1 h 2 E2 1 1 where h = laminate thickness 1 A =E A12 = E 66 6 h 2 1E 2 1 1 2 Solving last equation, we have the following elastic constants for the single “equivalent” orthotropic layer: A 12 A 12 A 11 E1 = (1 – ) E2 = A (1 – ) 2 2 22 A 11 A 22 h E6 = 1 h A66 A 11 A 22 h ν1 = A 12 A 22 ν2 = A 12 A 11 The equation for the composite laminate is or [ε] = [a] [N] where [a] = [A] -1 The equation for a single orthotropic lamina is or [ε] = [C] [σ] The properties of the single “equivalent” orthotropic layer is determined by [a] h = [C] or a11h = a21h = 1 E1 1 E1 a12h = 2 E2 a66h = 1 E6 1 a22h = E2 elastic constants for the single “equivalent” orthotropic layer: E1 = 1 a 11 h ν1 = – a 12 a 11 E2 = 1 a 22 h ν2 = – a 12 a 22 E66 = 1 a 66 h Special Cases: • Cross-Ply Laminates: All layers are either 0° or 90°, which results in A16 = A26 = 0. This is sometimes called specially orthotropic w. r. t. in-plane force and strains. Refer to Figure 1 and Table 1. • • h = total laminate thickness Figure 1 In-Plane Modulus and Compliance of T300/5208 CrossPly Laminates. • Balanced Angle-Ply Laminates: There are only two orientations of the laminae; same magnitude but opposite in sign. With an equal number of plies with positive and negative orientations. Refer to Figure 2 and table 2. • h = total laminate thickness. Figure 2 In-Plane Modulus and Compliance of T300/5208 Angle-Ply Laminates. The balanced angle-ply laminate is another case of “special orthotropy” w.r.t. inplane force and strains; that is A16 = A26 = 0. The values of Table 2 apply for un-symmetric laminate as well as symmetric laminates. For example, the values for [+30, −30, +30, −30] or [+30, +30, −30, −30] are the same as those for [+30, -30, -30, +30]. If the laminate is not balanced, such as [+30, −30, +30], then the terms A16 and A26 are not zero. Also remember that A16 = A26 for all cross-plies, whether they are symmetric or unsymmetric. Example: [0, 90] [0, 90, 0] [+30, −30, −30, +30] [+30, −30, +30, −30] [+30, −30, +30] un-symmetric cross-ply, A16 = A26 = 0 symmetric cross-ply, A16 = A26 = 0 symmetric balanced angle-ply, A16 = A26 = 0 un-symmetric balanced angle-ply, A16 = A26 = 0 symmetric un-balanced angle-ply, A16, A26 ≠ 0 3) Quasi-Isotropic Laminates: Has ‘m’ ply groups spaced at ply orientations of 180/m degrees. Note: This does not apply for m = 1 and m = 2. The laminate need not be symmetric to be quasi-isotropic. For example, the laminate [0, 45, −45, 90] is quasi-isotropic. Examples: [0/60/−60]s [0/90/45/−45]s [0/60/−60/60/0/−60]s m=3 m=4 m=3 180/m = 60° 180/m = 45° 180/m = 60° The modulus of the quasi-isotropic laminate has the following properties: A11 = A22 A16 = A26 = 0 1 A66 = (A11 – A22) which is equivalent to G = E for an isotropic material. 2 1 Summary: Laminate Analysis 1. Determine the modulus for the laminate [0/+45/−45/90]s where the 0° and 90° layers are Scotch-ply 1002, and the +45° and −45° layers are T300/5208. The thickness of each layer is 0.00125m. Obtain the modulus by (1) hand calculations, and (2) computer program. 0 . 402 10 8 N 1 Answer: N 2 = 0 . 222 10 8 N 6 0 0 . 222 10 8 0 . 402 10 8 0 0 8 0 . 254 10 0 1 2 6 2. A stress analysis of a structure results in the stresses shown at a particular point in a composite laminate. The laminate is the same as in problem 5. a. Determine the strains in global coordinates. Answers: 0.00358, -0.00198, 0.00158 b. Determine the stresses in the 0° lamina in local coordinates. c. Determine the stresses in the 45° lamina in local coordinates. Answers: 2.887 MPa, 4.68 MPa, -39.84 MPa