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

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Presentation on theme: "Composites Design and Analysis Stress-Strain Relationship Prof Zaffar M. Khan Institute of Space Technology Islamabad."— Presentation transcript:

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

2 2 Next Generation Aerospace Vehicle Requirements

3 Composite design and analysis

4 4 MATERIAL SELECTION

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6 Composite Analysis 2D vs. 3D State of Stress 6

7 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: E 1, E 2, G 12, ν 12 – Out of plane : E 3, G 13, G 23, ν 13, ν 23 But for transverse isotropy (2 = 3): E 2 = E 3 G 12 = G 13 ν 12 = ν 13 G 23 = __E 23 ___ 2(1 + ν 23 ) Therefore 5 independent properties: E 1 E 2 G 12 ν 12 G 23

8 Terminology used in micromechanics E f, E m - Young’s modulus of fiber and matrix G f, G m - Shear modulus of fiber and matrix ν f, ν m - Poisson’s ratio of fiber and matrix V f, V m - Volume fraction of fiber and matrix W f, W m – Weight fraction of fiber and matrix

9 Elastic properties: Rule of mixtures approach Parallel model - E 1 and ν 12 (Constant Strains) E 1 = E f1 V f + E m V m ν 12 = ν f V f + ν m V m Predictions agree well with experimental data Series model – E 2 and G 12 (Constant Stress) E 2 = __ __ E f2 E m ________ E f2 V m + E m V f G 12 = __ __ E f2 E m ________ E f2 V m + E m V f Experimental results predicted less accurately

10 Micromechanics example: Volume fraction changes Knowns : Carbon: E = 34.0 x 10 6 psi Epoxy : E = 0.60 x 10 6 psi How much does the longitudinal modulus change when the fiber volume fraction is changed from 58% to 65%? E 1 = E f1 V f + E m V m -For V f = 0.58: E 1 = (34.0 x 10 6 psi)(.58) + (0.60 x 10 6 psi)(0.42) = 20.0 x 10 6 psi -For V f = 0.65: E 1 = (34.0 x 10 6 psi)(.65) + (0.60 x 10 6 psi)(0.35) = 22.3 x 10 6 psi Thus, raising the fiber volume fraction from 58% to 65% increases E 1 by 12%

11 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)

12 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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14 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 0 o or 90 o (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)

15 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

16 Stress-Strains Relationships in Lamina

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20 Elastic Constants-x Pσ x = P/Aεxεx εyεy ν x = − ε y / ε x lbspsiin/in 0000 10518720.000520− 0.0000730.140 20035650.001005− 0.0001400.139 30053480.001495− 0.0002100.140 40073980.002165− 0.0002450.136 50089130.002515− 0.0003400.135 600106950.003022− 0.0004050.134 700124780.003545− 0.0004600.130 800142600.004050− 0.0005200.128

21 Elastic Constants-y Pσ y = P/Aεxεx εyεy ν y = − ε x / ε y lbspsiin/in 0000 1001751− 0.0000750.0005400.139 2003503− 0.0001450.0010900.133 3005254− 0.0002100.0016280.129 4007005− 0.0002850.0022180.128 5008757− 0.0003500.0027930.125 60010508− 0.0004150.0033510.124 70012259− 0.0004700.0039430.119 80014011− 0.0005400.0046200.117

22 Elastic Constants-s Pσ s = P/2Aεμεμ ενεν ε s = ε μ − ε ν lbspsiin/in 00000 1008750.001075− 0.0005610.001632 20017480.002212− 0.0012290.003441 30026220.003527− 0.0020650.005592 40034970.005175− 0.0031640.008339 50043710.007219− 0.0046150.011834 60052450.010547− 0.0072500.017797 70061190.013412− 0.0096300.023042 80069930.019082− 0.0147100.033792

23 Determination of Elastic Constants TypeMaterialExEx EyEy νxνx EsEs νfνf Specific gravity Typical thickness GPa meters T300/5208Graphite/Epoxy1811.30.287.170.701.60.000125 B(4)/5505Boron/Epoxy20418.50.235.590.502.00.000125 AS/3501Graphite/Epoxy1388.960.307.10.661.60.000125 Scotchply 1002Glass/Epoxy38.68.270.264.140.451.80.000125 Kevlar 49/EpoxyAramid/Epoxy765.50.342.30.601.460.000125 TypeMaterialExEx EyEy νxνx EsEs t Msi inches T300/5208Graphite/Epox y 26.251.490.281.040.005 B(4)/5505Boron/Epoxy29.592.680.230.810.005 AS/3501Graphite/Epox y 20.011.300.301.030.005 Scotchply 1002Glass/Epoxy5.601.200.260.600.005 Kevlar 49/EpoxyAramid/Epoxy11.020.800.340.330.005

24 Transformation of Stress and Strain

25 Area Stresses Forces Stresses

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

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

28 Stress-Strain Relationships in Global Coordinates [σ xys ] = [Q xys ] [ε xys ] (1) [σ xys ] = [T] [ σ 126 ](2) [ε xys ] = [Ť] [ε 126 ](3) Substitution of Equations (2) and (3) into (1) yields [T] [ σ 126 ] = [Q xys ] [Ť] [ε 126 ](4) Pre multiplying both sides of Equation (4) by [T] −1 : [T] −1 [T] [σ 126 ] = [T] −1 [Q xys ] [Ť] [ε 126 ](5) or [ σ 126 ] = [T] −1 [Q xys ] [Ť] [ε 126 ](6) or = [T] −1 [Q xys ] [Ť] (7) Develop the relationship between the stresses and strains in global coordinates.

29 or = [Q 126 ] or [ σ 126 ] = [Q 126 ] [ε 126 ]where [Q 126 ] = [T] −1 [Q xys ] [Ť] Note that [T (+θ) ] −1 = [T (−θ) ] The elements of the matrix [Q 126 ] are as follows: Q 11 = Q xx cos 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy sin 4 θ Q 22 = Q xx sin 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy cos 4 θ Q 12 = Q 21 = (Q xx + Q yy − 4Q ss ) sin 2 θ cos 2 θ + Q xy (sin 4 θ +cos 4 θ) Q 66 = (Q xx + Q yy − 2 Q xy − 2Q ss ) sin 2 θ cos 2 θ + Q ss (sin 4 θ +cos 4 θ) Q 16 = Q 61 = (Q xx – Q xy − 2Q ss ) sinθ cos 3 θ + (Q xy – Q yy + 2Q ss ) sin 3 θ co Q 26 = Q 62 = (Q xx – Q xy − 2Q ss ) sin 3 θ cosθ + (Q xy – Q yy + 2Q ss ) sinθcos 3 θ

30 [Q 126 ] −1 [ σ 126 ] = [Q 126 ] −1 [Q 126 ] [ε 126 ] or[Q 126 ] −1 [ σ 126 ] = [ε 126 ] or[ε 126 ] = [Q 126 ] −1 [ σ 126 ] or[ε 126 ] = [S 126 ] [ σ 126 ] or The elements of the matrix [S 126 ] are as follows: S 11 = (Q 22 Q 66 – Q 26 2 ) / ∆ S 12 = S 21 = (Q 16 Q 26 – Q 12 Q 66 ) / ∆ S 16 = S 61 = (Q 12 Q 26 – Q 22 Q 16 ) / ∆ S 22 = (Q 11 Q 66 – Q 16 2 ) / ∆ S 26 = S 62 = (Q 12 Q 16 – Q 11 Q 26 ) / ∆ S 66 = (Q 11 Q 22 – Q 12 2 ) / ∆ where ∆ = Q 11 Q 22 Q 66 + 2 Q 12 Q 26 Q 61 − Q 22 Q 16 2 – Q 66 Q 12 2 – Q 11 Q 62 2

31 Summary: Lamina Analysis

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

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

34 Expression for N 1: N 1 = (σ 1 ) 1 (h 1 -h 2 ) + (σ 1 ) 2 (h 2 -h 3 ) + (σ 1 ) 3 (h 3 -h 4 ) + … Substitute the expressions for stress:

35 Above equation can be rewritten as: Similarly for N 2 and N 6:

36 Above can be combined into matrix form as: Or Where A 11 = (Q 11 ) k t k A 12 = (Q 12 ) k t k A 16 = (Q 16 ) k t k A 22 = (Q 22 ) k t k A 26 = (Q 26 ) k t k A 66 = (Q 66 ) k t k

37 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° Lamina90° Lamina Q 11 = Q xx = 181.8Q 11 = Q yy = 10.34 Q 22 = Q yy = 10.34Q 22 = Q xx = 181.8 Q 12 = Q xy = 2.897Q 12 = Q xy = 2.897 Q 66 = Q ss = 7.17Q 66 = Q ss = 7.17 Q 16 = 0Q 16 = 0 Q 26 = 0Q 26 = 0 Q 11 = Q xx cos 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy sin 4 θ Q 22 = Q xx sin 4 θ + 2 (Q xy + 2 Q ss ) sin 2 θ cos 2 θ + Q yy cos 4 θ Q 12 = Q 21 = (Q xx + Q yy – 4 Q ss ) sin 2 θ cos 2 θ + Q xy (sin 4 θ + cos 4 θ) Q 66 = (Q xx + Q yy – 2Q xy – 2 Q ss ) sin 2 θ cos 2 θ + Q ss (sin 4 θ + cos 4 θ) Q 16 = Q 61 = (Q xx – Q xy – 2 Q ss ) sinθ cos 3 θ + (Q xy – Q yy + 2 Q ss ) sin 3 θ cosθ Q 26 = Q 62 = (Q xx – Q xy – 2 Q ss ) sin 3 θ cosθ + (Q xy – Q yy + 2 Q ss ) sinθ cos 3 θ Q 11 = 0.25 Q xx + 2 (Q xy + 2 Q ss ) (0.25) + 0.25 Q yy = 56.65

38 Q 22 = 0.25 Q xx + 2 (Q xy + 2 Q ss ) (0.25) + 0.25 Q yy = 56.65 Q 12 = (Q xx + Q yy – 4 Q ss ) (0.25) + Q xy (0.5) = 42.31 Q 66 = (Q xx + Q yy – 2 Q xy –2Q ss ) (0.25) + Q ss (0.25) = 46.59 Q 16 = (Q xx – Q xy – 2 Q ss ) (0.25) + (Q xy – Q yy +2 Q ss ) (0.25) = 42.87 Q 26 = (Q xx – Q xy – 2 Q ss ) (0.25) + (Q xy – Q yy +2 Q ss ) (0.25) = 42.87 The Laminate A 11 = 0.000125 (181.8 + 10.34 + 56.25) 2 = 0.0622 GPa-m A 22 = 0.000125 (10.34 + 181.8 + 56.65) 2 = 0.0622 A 12 = 0.000125 (2.897 + 2.897 + 42.31) 2 = 0.0120 A 66 = 0.000125 (7.17 + 7.17 + 46.59) 2 = 0.0152 A 16 = 0.000125 (0 + 0 + 42.87) 2 = 0.0107 A 26 = 0.000125 (0 + 0 + 42.87) 2 = 0.0107

39 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]

40 The properties of the single “equivalent” orthotropic layer can be determined from the following equation: [A] = [M]where h = laminate thickness or A 11 = A 12 = A 66 = E 6 A 22 = A 21 = Solving last equation, we have the following elastic constants for the single “equivalent” orthotropic layer: E 1 = (1 – )E 2 = (1 – ) E 6 = A 66 ν 1 = ν 2 =

41 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]

42 The properties of the single “equivalent” orthotropic layer is determined by [a] h = [C] or a 11 h = a 12 h = a 66 h = a 21 h = a 22 h = elastic constants for the single “equivalent” orthotropic layer: E 1 = E 2 = E 66 = ν 1 = – ν 2 = –

43 Special Cases: Cross-Ply Laminates: All layers are either 0° or 90°, which results in A 16 = A 26 = 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

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

45 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.

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

47 The balanced angle-ply laminate is another case of “special orthotropy” w.r.t. in- plane force and strains; that is A 16 = A 26 = 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 A 16 and A 26 are not zero. Also remember that A 16 = A 26 for all cross-plies, whether they are symmetric or un- symmetric. Example: [0, 90]un-symmetric cross-ply, A 16 = A 26 = 0 [0, 90, 0] symmetric cross-ply, A 16 = A 26 = 0 [+30, −30, −30, +30] symmetric balanced angle-ply, A 16 = A 26 = 0 [+30, −30, +30, −30]un-symmetric balanced angle-ply, A 16 = A 26 = 0 [+30, −30, +30]symmetric un-balanced angle-ply, A 16, A 26 ≠ 0

48 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]sm = 3 180/m = 60° [0/90/45/−45]sm = 4 180/m = 45° [0/60/−60/60/0/−60]sm=3180/m = 60° The modulus of the quasi-isotropic laminate has the following properties: A 11 = A 22 A 16 = A 26 = 0 A 66 = (A 11 – A 22 ) which is equivalent to G = for an isotropic material.

49 Summary: Laminate Analysis

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