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Composites Design and Analysis Stress-Strain Relationship Prof Zaffar M. Khan Institute of Space Technology Islamabad

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**Next Generation Aerospace Vehicle Requirements**

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**Composite design and analysis**

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MATERIAL SELECTION

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Composite Analysis 2D vs D State of Stress

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**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 = E G12 = G13 ν12 = ν G23 = __E23___ 2(1 + ν 23) Therefore 5 independent properties: E1 E2 G12 ν 12 G23

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**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

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**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 G (Constant Stress) E2 = __ __ Ef2 Em________ Ef2 Vm + Em Vf G12 = __ __ Ef2 Em________ Experimental results predicted less accurately

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**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%? E1 = Ef1 Vf + Em Vm For Vf = 0.58: E1 = (34.0 x 106 psi)(.58) + (0.60 x 106 psi)(0.42) = 20.0 x 106 psi For Vf = 0.65: E1 = (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%

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**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)

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**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

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**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)

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**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 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

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**Stress-Strains Relationships in Lamina**

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**Elastic Constants-x P σx = P/A εx εy νx = − εy / εx lbs psi in/in 105**

105 1872 − 0.140 200 3565 − 0.139 300 5348 − 400 7398 − 0.136 500 8913 − 0.135 600 10695 − 0.134 700 12478 − 0.130 800 14260 − 0.128

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**Elastic Constants-y P σy = P/A εx εy νy = − εx / εy lbs psi in/in 100**

100 1751 − 0.139 200 3503 − 0.133 300 5254 − 0.129 400 7005 − 0.128 500 8757 − 0.125 600 10508 − 0.124 700 12259 − 0.119 800 14011 − 0.117

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**Elastic Constants-s P σs = P/2A εμ εν εs = εμ − εν lbs psi in/in 100**

100 875 − 200 1748 − 300 2622 − 400 3497 − 500 4371 − 600 5245 − 700 6119 − 800 6993 −

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**Determination of Elastic Constants**

Type Material Ex Ey νx Es t Msi inches T300/5208 Graphite/Epoxy 26.25 1.49 0.28 1.04 0.005 B(4)/5505 Boron/Epoxy 29.59 2.68 0.23 0.81 AS/3501 20.01 1.30 0.30 1.03 Scotchply 1002 Glass/Epoxy 5.60 1.20 0.26 0.60 Kevlar 49/Epoxy Aramid/Epoxy 11.02 0.80 0.34 0.33 Type Material Ex Ey νx Es νf Specific gravity Typical thickness GPa meters T300/5208 Graphite/Epoxy 181 1.3 0.28 7.17 0.70 1.6 B(4)/5505 Boron/Epoxy 204 18.5 0.23 5.59 0.50 2.0 AS/3501 138 8.96 0.30 7.1 0.66 Scotchply 1002 Glass/Epoxy 38.6 8.27 0.26 4.14 0.45 1.8 Kevlar 49/Epoxy Aramid/Epoxy 76 5.5 0.34 2.3 0.60 1.46

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**Transformation of Stress and Strain**

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Area Stresses Stresses Forces

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**These three equilibrium equations may be combined in matrix form as follows:**

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The equations for the transformation of strain are the same as those for the transformation of stress:

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**Stress-Strain Relationships in Global Coordinates**

Develop the relationship between the stresses and strains in global coordinates. [σxys] = [Qxys] [εxys] (1) [σxys] = [T] [ σ126] (2) [εxys] = [Ť] [ε126] (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] (5) or [ σ126] = [T] −1 [Qxys] [Ť] [ε126] (6) or = [T] −1 [Qxys] [Ť] (7)

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or = [Q126] or [ 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

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[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 Q Q12Q26 Q61 − Q22Q162 – Q66Q122 – Q11Q622

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**Summary: Lamina Analysis**

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**The Symmetric Laminate with In-Plane Loads A**

The Symmetric Laminate with In-Plane Loads A. Stress Resultants and Strains: The “A”

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**Relationship between strains and the stress resultants**

Relationship between strains and the stress resultants. Related by constants Aij Deriving expressions for these constants: Expression for is:

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**Substitute the expressions for stress:**

Expression for N1: N1 = (σ1)1 (h1-h2) + (σ1)2 (h2-h3) + (σ1)3 (h3-h4) + … Substitute the expressions for stress:

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**Above equation can be rewritten as: Similarly for N2 and N6: **

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**A11 = (Q11)k tk A12 = (Q12)k tk A16 = (Q16)k tk A22 = (Q22)k tk **

Above can be combined into matrix form as: Or Where A11 = (Q11)k tk A12 = (Q12)k tk A16 = (Q16)k tk A22 = (Q22)k tk A26 = (Q26)k tk A66 = (Q66)k tk

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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” = m 0° Lamina 90° Lamina Q11 = Qxx = Q11 = Qyy = 10.34 Q22 = Qyy = Q22 = Qxx = 181.8 Q12 = Qxy = Q12 = Qxy = 2.897 Q66 = Qss = Q66 = Qss = 7.17 Q16 = 0 Q16 = 0 Q26 = 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) Qyy = 56.65

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**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 = ( ) 2 = GPa-m A22 = ( ) 2 = A12 = ( ) 2 = A66 = ( ) 2 = A16 = ( ) 2 = A26 = ( ) 2 =

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**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]

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**[A] = [M] where h = laminate thickness**

The properties of the single “equivalent” orthotropic layer can be determined from the following equation: [A] = [M] where h = laminate thickness or A11 = A12 = A66 = E6 A22 = A21 = Solving last equation, we have the following elastic constants for the single “equivalent” orthotropic layer: E1 = (1 – ) E2 = (1 – ) E6 = A66 ν1 = ν2 =

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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]

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**[a] h = [C] or a11h = a12h = a66h = a21h = a22h = E1 = E2 = E66 =**

The properties of the single “equivalent” orthotropic layer is determined by [a] h = [C] or a11h = a12h = a66h = a21h = a22h = elastic constants for the single “equivalent” orthotropic layer: E1 = E2 = E66 = ν1 = – ν2 = –

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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

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**Figure 1 In-Plane Modulus and Compliance of T300/5208 Cross-Ply Laminates.**

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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.

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**Figure 2 In-Plane Modulus and Compliance of T300/5208 Angle-Ply Laminates.**

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**[0, 90] un-symmetric cross-ply, A16 = A26 = 0 **

The balanced angle-ply laminate is another case of “special orthotropy” w.r.t. in-plane 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 un-symmetric. Example: [0, 90] un-symmetric cross-ply, A16 = A26 = 0 [0, 90, 0] symmetric cross-ply, A16 = A26 = 0 [+30, −30, −30, +30] symmetric balanced angle-ply, A16 = A26 = 0 [+30, −30, +30, −30] un-symmetric balanced angle-ply, A16 = A26 = 0 [+30, −30, +30] symmetric un-balanced angle-ply, A16, A26 ≠ 0

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**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 m = /m = 60° [0/90/45/−45]s m = /m = 45° [0/60/−60/60/0/−60]s m=3 180/m = 60° The modulus of the quasi-isotropic laminate has the following properties: A11 = A22 A16 = A26 = 0 A66 = (A11 – A22) which is equivalent to G = for an isotropic material.

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**Summary: Laminate Analysis**

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