ME403 Design of Composite Structures Lamina -Strain Relationships

Asst.Prof.Dr.Turgut AKYÜREK

Çankaya University, Ankara Schematic Analysis of Laminated Composites A lamina is a thin layer of a composite material that is generally of a thickness on the order of 0.005 in. (0.125 mm). A laminate is constructed by stacking a number of such laminae in the direction of the lamina thickness. Mechanical structures made of these laminates, such as a leaf spring suspension system in an automobile, are subjected to various loads, such as and twisting. The design and analysis of such laminated structures demands knowledge of the stresses and strains in the laminate. Also, design tools, such as failure theories, stiffness models, and optimization algorithms, need the values of these laminate stresses and strains.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 2/75 Introduction The basic building block of a composite structure is the lamina, which usually consists of one of the fiber/ configurations shown. For the purposes of mechanics analysis, however, the “unidirectionally reinforced” or “unidirectional” lamina with an arrangement of parallel, continuous fibers is the most convenient starting point. As shown in subsequent chapters, the stress–strain relationships for the unidirectional lamina form the basis for the analysis of not only the continuous fiber , but also of woven fiber, chopped fiber, and hybrid chopped/continuous fiber composites.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 3/75 Introduction A composite material is obviously heterogeneous at the constituent material level, with properties possibly changing from point to point. For example, the stress–strain relationships at a point are different for a point in the fiber material than for a point in the matrix material. If we take the composite lamina as the basic building block, however, the “macro-mechanical” stress–strain relationships of the lamina can be expressed in terms of average stresses and strains and effective properties of an equivalent homogeneous material. This chapter is concerned with the development and manipulation of these macro-mechanical stress– strain relationships.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 4/75 Effective Moduli in Stress-Strain Relationships A general three-dimensional (3D) state of stress at a point in a material can be described by nine stress components σij (where i, j = 1, 2, 3). According to the conventional subscript notation, when i = j, the stress component σij is a normal stress; and when i ≠ j, the stress component is a shear stress. The first subscript refers to the direction of the outward normal to the face on which the stress component acts, and the second subscript refers to the direction in which the stress component T.Akyürekitself acts. ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 5/75 Effective Moduli in Stress-Strain Relationships Corresponding to each of the stress components, there is a strain component εij describing the at the point. Normal strains (i = j) describe the extension or contraction per unit length along the xi direction, and shear strains (i ≠ j) describe the distortional deformations associated with lines that were originally parallel to the xi and xj axes. It is very important to distinguish between the “” strain εij and the “engineering” strain �ij. In the case of normal strain, the engineering strain is the same as the tensor strain, but for shear strain, εij = �ij/2. Thus, the engineering shear strain �ij describes the total distortional change in the angle between lines that were originally parallel to the xi and xj axes, but the tensor shear strain εij describes the amount of rotation of either of the lines.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 6/75 Effective Moduli in Stress-Strain Relationships In the most general stress–strain relationship at a point in an elastic material, each stress component is related to each of the nine strain components by an equation of the form. σ ij = fij (ε11,ε12 ,ε13,ε21,ε22 ,ε23,ε 31,ε 32 ,ε 33 ) where the functions fij may be nonlinear. For linear elastic material, the most general linear stress–strain relationships at a point in the material (excluding effects of environmental conditions) are given by equations of the form. 9 9x9 9

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 7/75 Effective Moduli in Stress-Strain Relationships

9 9x9 9 where [C] is a fully populated 9 × 9 matrix of stiffnesses or elastic constants (or moduli) having 81 components. Note that the first two subscripts on the elastic constants correspond to those of the stress, whereas the last two subscripts correspond to those of the strain. If no further restrictions are placed on the elastic constants, the material is called anisotropic and Equation is referred to as the generalized Hooke’s law for anisotropic materials. In practice, there is no need to deal with this equation and its 81 elastic constants because various symmetry conditions simplify the equations T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 8/75 considerably. Isotropic vs Anisotropic Comparison of Deformation

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 9/75 Effective Moduli in Stress-Strain Relationships As shown in any mechanics of materials book, both stresses and strains are symmetric (i.e., σij = σji and εij = εji), so that there are only six independent stress components and six independent strain components. This means that the elastic constants must be symmetric with respect to the first two subscripts and with respect to the last two subscripts (i.e., Cijkl = Cjikl and Cijkl = Cijlk, where i, j, k, and l each take on the values 1, 2, and 3) and that the number of nonzero elastic constants is now reduced to 36. These simplifications lead to a contracted notation that reduces the number of subscripts based on the following changes in notation: 6 6x6 6 ⎡ ⎤ ⎧σ ⎫ C11 C12 C13 C14 C15 C16 ⎧ε ⎫ 1 ⎢ ⎥ 1 ⎪σ ⎪ C C C C C C ⎪ε ⎪ ⎪ 2 ⎪ ⎢ 21 22 23 24 25 26 ⎥⎪ 2 ⎪ ⎢ ⎥ ⎪σ 3 ⎪ C31 C32 C33 C34 C35 C36 ⎪ε 3 ⎪ ⎨ ⎬ = ⎢ ⎥⎨ ⎬ ⎪τ 23 ⎪ ⎢ C41 C42 C43 C44 C45 C46 ⎥⎪γ 23 ⎪ ⎢ ⎥ ⎪τ 31 ⎪ C C C C C C ⎪γ 31 ⎪ ⎪ ⎪ ⎢ 51 52 53 54 55 56 ⎥⎪ ⎪ τ ⎢ C C C C C C ⎥ γ ⎩ 12 ⎭ ⎣ 61 62 63 64 65 66 ⎦⎩ 12 ⎭

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 10/75 Effective Moduli in Stress-Strain Relationships With this contracted notation, the generalized Hooke’s law can now be written as σ i = Cijε j i, j = 1,2,...,6 {σ } = [C]{ε} where the elastic constant matrix or stiffness matrix [C] is now 6 × 6 with 36 components and the stresses {σ} and strains {ε} are column vectors, each having six elements.

Alternatively, the generalized Hooke’s law relating strains to stresses⎡ ⎤ can be ⎧ε ⎫ S11 S12 S13 S14 S15 S16 ⎧σ ⎫ 1 ⎢ ⎥ 1 written as ⎪ε ⎪ S S S S S S ⎪σ ⎪ εi = Sijσ j ,i, j = 1,2,..6 ⎪ 2 ⎪ ⎢ 21 22 23 24 25 26 ⎥⎪ 2 ⎪ ⎢ ⎥ ⎪ε 3 ⎪ S31 S32 S33 S34 S35 S36 ⎪σ 3 ⎪ ⎨ ⎬ = ⎢ ⎥⎨ ⎬ γ ⎢ S S S S S S ⎥ τ in matrix form as {ε} = [S]{σ } ⎪ 23 ⎪ 41 42 43 44 45 46 ⎪ 23 ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ γ 31 ⎢ S51 S52 S53 S54 S55 S56 ⎥ τ 31 ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎩γ 12 ⎭ S61 S62 S63 S64 S65 S66 ⎩τ12 ⎭ where [S] is the compliance matrix, which is the ⎣ inverse of the stiffness matrix ([S] ⎦ = [C]−1). For a brief review of matrix operations, please refer to Appendix A. As shown later, due to the existence of the strain energy density, the stiffness and compliance matrices are symmetric. Note that nothing has been said thus far about any symmetry that the material itself may have. All real materials have some form of symmetry, however, and no known T.Akyürekmaterial is completely anisotropic.ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 11/75 Effective Moduli in Stress-Strain Relationships Before discussing the various simplifications of the stress–strain relationships, it is appropriate to deal with the problem of heterogeneity in the composite material. Recall that the stress–strain relationships presented up to now are only valid at a point in the material, and that the stresses, strains, and elastic moduli will change as we move from point to point in a composite (i.e., the elastic moduli for the matrix material are different from those of the fiber). In order to analyze the macro- mechanical behavior of the composite, it is more convenient to deal with averaged stresses and strains that are related by “effective moduli” of an equivalent homogeneous material. Figure shows schematically how the stresses in a heterogeneous composite may be nonuniform even though the imposed strain is uniform.

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If the scale of the inhomogeneity in a material can be characterized by some length dimension, d, then the length dimension, L, over which the macro- mechanical averaging is to take place, must be much larger than d if the average stresses and strains are to be related by effective moduli of an equivalent homogeneous material.

We now define the average stresses, σ i

and the average strains, (i = 1, 2, …, εi 6) to be averaged over a volume V, σ i dV σ i dV ε dV ε dV which is characterized by the ∫ ∫ ∫ i ∫ i σ = V = V V V dimension L, so that i εi = = ∫ dV V ∫ dV V where i = 1, 2, …, 6 and σi and εi are the V V position-dependent stresses and strains at a point, respectively. If these averaged stresses and strains are used in place of the stresses and σ i = Cijε j strains at a point, the generalized Hooke’s law becomes

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The elastic moduli Cij then become the “effective moduli” of the equivalent homogeneous material in volume V. Similarly, the “effective compliances” Sij may be defined by εi = Sijσ j

For example, in Figure, the scale of the inhomogeneity is assumed to be the diameter of the fiber, d, and the averaging dimension, L, is assumed to be a characteristic lamina dimension such that L ≫ d. The effective modulus σ dV σ dV ε dV ε dV C22 of the lamina is thus defined as ∫ i ∫ i ∫ i ∫ i σ V V V V 2 σ i = = εi = = C22 = dV V dV V ε2 ∫ ∫ In the remainder of this course, lamina V V properties are assumed to be effective properties as described above.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 14/75 Symmetry in Stress-Strain Relationships In this section, the generalized anisotropic Hooke’s law will be simplified and specialized using various symmetry conditions. The first symmetry condition, which has nothing to do with material symmetry, is strictly a result of the existence of a strain energy density function. 1 1 The strain energy density function, W W = σ ε = C ε ε 2 j j 2 ji i j By taking first derivative ∂W = σ j = C jiεi ∂ε j 2 By taking second derivative ∂ W = C ji ∂ε j ∂εi and by reversing the order of differentiation, we find that 1 1 ∂W 2W = σ = C ε ∂ W = σ iεi = Cijε jεi i ij j = Cij 2 2 ∂εi ∂ε ∂ε Cij = Cji, the stiffness matrix is symmetric. i j

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 15/75 Symmetry in Stress-Strain Relationships 1 1 1 1 W = ε σ = S σ σ W = ε σ = S σ σ 2 j j 2 ij i j 2 i i 2 ji j i Similarly, W can be expressed in terms of compliances and stresses, and by taking two derivatives with respect to stresses, it can be shown that Sij = Sji. Thus, the compliance matrix is also symmetric. Due to these mathematical manipulations, only 21 of the 36 anisotropic elastic moduli or compliances are independent, and we still have not said anything about any inherent symmetry of the material itself. According to the above developments, the stiffness matrix for the linear elastic anisotropic material without any material property symmetry is of the form.

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Further simplifications of the stiffness matrix are possible only if the material properties have some form of symmetry. For example, a monoclinic material has one plane of material property symmetry. It can be shown that since the Cij for such a material must be invariant under a transformation of coordinates corresponding to reflection in the plane of symmetry, the number of independent elastic constants for the monoclinic material is reduced to 13. The direction perpendicular to the plane of symmetry is called the principal direction. Feldspar is an example of a monoclinic material.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 17/75 Symmetry in Stress-Strain Relationships Deformation of a cubic element made of monoclinic material.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 18/75 Symmetry in Stress-Strain Relationships As shown in Figure, a unidirectional composite lamina has three mutually orthogonal planes of material property symmetry (i.e., the 12, 23, and 13 planes) and is called an orthotropic material. Anisotropic The term “orthotropic” alone is not sufficient to describe the form of the stiffness matrix, however. Unlike the anisotropic stiffness matrix (Above Equation), which has the same form (but different terms) for different coordinate systems, the form of the stiffness matrix for the orthotropic material depends on the used.

Monoclinic Orthotropic

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 19/75 Symmetry in Stress-Strain Relationships Deformation of a cubic element made of orthotropic material.

D

A

ε1 = S13σ 3

ε2 = S23σ 3

ε 3 = S33σ 3

γ 23 = 0

γ 31 = 0

γ 12 = 0

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The 1, 2, 3 coordinate axes in Figure are referred to as the principal material coordinates since they are associated with the reinforcement directions. Invariance of the Cij under transformations of coordinates corresponding to reflections in two orthogonal planes may be used to show that the stiffness matrix for a so-called specially orthotropic material associated with the principal material coordinates is of the form

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21 A stiffness matrix of this form in terms of engineering constants will be obtained in the next section using observations from simple experiments. Note that there are only 12 nonzero elastic constants and 9 independent elastic constants for the specially orthotropic material. Table summarizes similar results for the different combinations of materials and coordinate systems.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 22/75 Symmetry in Stress-Strain Relationships If the stress–strain relationships for the specially orthotropic material are transformed to a non- principal coordinate system such as x, y, z as shown, the material is called generally orthotropic. Specially Orthotropic Note that the number of independent coefficients in the stress–strain relationships for a material is the same, 9 coefficients regardless of whether the material is described in principal or non-principal coordinates.

21 coefficients

Generally Orthotropic 21 coefficients, 9 independent

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 23/75 Symmetry in Stress-Strain Relationships The general forms of the stiffness matrices are given here for completeness. In most composites, the fiber-packing arrangement is statistically random in nature, so that the properties are nearly the same in any direction perpendicular to the fibers (i.e., properties along the 2 direction are the same as those along the 3 direction, and properties are invariant to rotations about the 1 axis), and the material is transversely isotropic. If the fibers were packed in a regular array, such as a square array or a triangular array, the material would not be transversely isotropic. As shown later in Chapter 3, real composites typically have random fiber packing, ε ε ε γ γ γ and can be considered transversely isotropic. 1 2 3 23 31 12 For such a material, we would expect that C22 = C33, C12 = C13, C55 = C66, and that C44 would not be independent from the other stiffnesses. It can be shown that the complete stiffness matrix for a specially orthotropic, transversely isotropic material is of τ τ τ σ 1 σ 2 σ 3 23 31 12 the form.

T.AkyürekSimilarly, compliance matrix ME403 Design of Composite Structures - for a specially orthotropic, Lamina Stress-Strain Relaonships 24/75 transversely isotropic material: Symmetry in Stress-Strain Relationships

where the 23 plane and all parallel planes are assumed to 21 be planes of . Later, a stiffness matrix of the same form will be derived, except that the so-called engineering constants will be used instead of the Cij. Note that now there are still 12 nonzero elastic moduli but that only 5 are independent. T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 25/75 Symmetry in Stress-Strain Relationships The simplest form of the ε ε ε γ γ γ stress–strain relationship 1 2 3 23 31 12 occurs when the material is isotropic and every coordinate axis is an axis of symmetry. Now we would expect that C11 = C22 = C33, C12 = C13 = C23, that C44

E(1−ν ) νE C11 − C12 E C = C12 = = = C55 = C66, and that C44 again 11 (1− 2ν )(1+ν ) (1− 2ν )(1+ν ) 2 2(1+ν ) σ σ σ τ τ τ would not be independent from 1 2 3 23 31 12 the other Cij. The isotropic stiffness matrix is of the form Now, there are still 12 nonzero elastic constants, but only two are independent. T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 26/75 Example 2.1

Problem: A representative volume element (RVE) from a fiber-reinforced composite lamina is shown in Figure, along with the longitudinal stress and strain distributions across the fiber and matrix materials in the section. The fiber has a uniform longitudinal stress of 4000 psi (along the 1 direction) and a diameter of 0.0003 in., whereas the matrix has a uniform longitudinal stress of 1000 psi. A uniform longitudinal strain of 0.000423 in./in. acts over the entire section. The RVE has edge dimensions 0.0004 in. × 0.0004 in. in the 23 plane. Assume that the cross-sectional dimensions of the section do not change along the longitudinal direction, and use the concept of the effective modulus of an equivalent homogeneous material to find the numerical value of the effective longitudinal modulus, E1 (or E11) of the composite. T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 27/75 Example 2.1

Solution: σ 4000 E = f = = 9.456(10)6 psi f ε 0.000423 The effective longitudinal modulus is given by σ 1 f E1 = σ 1000 E = m = = 2.364(10)6 psi where as the average stress is ε m 1 ε m 0.000423 σ dV σ l dA ∫ 1 ∫ 1 1 σ A +σ A F + F σ A +σ A F σ = V = A = f f m m = f m = f f m m = 1 = σ υ +σ υ = σ 1 dV l dA A + A A + A A A f f m m 1 ∫ ∫ 1 f m f m 1 1 V A σ = σ υ +σ υ = E υ + E υ ε = E ε 1 f f m m ( f f m m ) 1 1 1 = 2325 psi E1 = E fυ f + Emυm and the stresses and cross-sectional areas of the fiber and matrix are

denoted by the subscripts f and m, respectively. The average strain is ε1 = 0.000423 so the effective modulus is σ 1 2325 6 E1 = = = 5.496(10 )psi ε1 0.00423

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 28/75 Example 2.2 Problem: If a specially orthotropic, transversely isotropic material is subjected to a pure shear strain γ23 = ε4, use the contracted notation to determine all of the resulting stresses associated with 1, 2, 3 axes. Solution: Substituting the stiffness matrix for a specially orthotropic, transversely isotropic material in the stress–strain relationships of the form {σ} = [C]{ε} and solving for the stresses resulting from the shear strain γ23 = ε4, it is seen that �23 = σ4 = C44ε4 = ((C22 − C23)/2)ε4 and all other stresses are equal to zero. ⎡ ⎤ C11 C12 C12 0 0 0 ⎧σ 1 ⎫ ⎧σ 1 ⎫ ⎢ ⎥⎧ε1 ⎫ ⎢ C12 C22 C23 0 0 0 ⎥ ⎪σ ⎪ ⎪σ ⎪ ⎪ε ⎪ ⎪ 2 ⎪ ⎪ 2 ⎪ ⎢ C C C 0 0 0 ⎥⎪ 2 ⎪ ⎢ 12 23 22 ⎥ C ⎪σ 3 ⎪ ⎪σ 3 ⎪ ⎪ε 3 ⎪ {σ } = [ ]{ε} = = ⎢ ⎥ ⎨ ⎬ ⎨ ⎬ (C22 − C23 ) ⎨ ⎬ ⎪σ 4 ⎪ ⎪τ 23 ⎪ ⎢ 0 0 0 0 0 ⎥⎪ε 4 ⎪ ⎢ 2 ⎥ ⎪σ 5 ⎪ ⎪τ 31 ⎪ ⎪ε5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ 0 0 0 0 C55 0 ⎥⎪ ⎪ σ 6 τ12 ⎢ ⎥ ε6 ⎩ ⎭ ⎩ ⎭ 0 0 0 0 0 C ⎩ ⎭ ⎣⎢ 55 ⎦⎥

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 29/75 Orthotropic Engineering Constants In the previous section, symmetry conditions were shown to reduce the number of elastic constants (the Cij or Sij) in the stress–strain relationships for several important classes of materials and the general forms of the relationships were presented. When a material is characterized experimentally, however, the so-called “engineering constants” such as Young’s modulus (or modulus of ), , and Poisson’s ratio are usually measured instead of the Cij or the Sij. The engineering constants are also widely used in analysis and design because they are easily defined and interpreted in terms of simple states of stress and strain. In this section, several simple tests and their resulting states of stress and strain will be used to develop the 3D and 2D stress–strain relationships for orthotropic and isotropic materials.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 30/75 Orthotropic Engineering Constants Consider a simple uniaxial tensile test consisting of an applied longitudinal normal stress, σ1, along the reinforcement direction (i.e., the 1 direction) of a specimen from an orthotropic material, as shown on the Simple states of stress used to define lamina engineering constants. (a) Applied longitudinal right. normal stress, (b) applied transverse normal stress, It is assumed that all other stresses and (c) applied in-plane shear stress. are equal to zero. Typical experimental stress–strain data from such a test on a carbon/ epoxy composite are shown, where both longitudinal strains ε1 and transverse Poisson strains ε2 are shown. Typical stress–strain data from (a) longitudinal tensile test and (b) transverse tensile test of All the values are effective values. carbon/epoxy composite.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 31/75 Orthotropic Engineering Constants Within the linear range the experimental observation is that the resulting strains associated with the 1, 2, 3 axes can be expressed empirically in terms of σ1 σ “engineering constants” as ε = 1 σ 1 E ε = 1 1 1 E σ 1 ε = −ν 1 σ 1 2 12 Red- Measured ε2 = −ν12ε1 = −ν12 γ 12 = γ 23 = γ 13 = 0 E1 E1 Blue- Found σ 1 σ 1 ε3 = −ν13 ε 3 = −ν13ε1 = −ν13 E1 E1 where E1 = longitudinal modulus of elasticity associated with the 1 direction, and νij = −εj/εi is the Poisson’s ratio, the ratio of the strain in the j direction to the strain in the perpendicular i direction when the applied stress is in the i direction.

All the values are effective values.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 32/75 Orthotropic and Isotropic Engineering Constants For isotropic materials, no subscripts are needed on properties such as the modulus of elasticity and Poisson’s ratio because the properties are the same in all directions. This is not the case with orthotropic Simple states of stress used to define lamina engineering constants. (a) Applied longitudinal materials, however, and subscripts are normal stress, (b) applied transverse normal needed on these properties because of stress, and (c) applied in-plane shear stress. their directional nature. For example, E1 ≠ E2 and ν12 ≠ ν21. Note that, as with isotropic materials, a negative sign must be used in the definition of the Poisson’s ratio.

Typical stress–strain data from (a) longitudinal tensile test and (b) transverse tensile test of carbon/epoxy composite.

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A property like ν12 is usually called a major Poisson’s ratio, whereas a property like ν21 is called a Poisson’s ratio. As with isotropic materials, a normal stress induces only normal strains, and all shear strains are equal to zero. This lack of shear/normal interaction is Simple states of stress used to define lamina observed only for the principal material engineering constants. (a) Applied longitudinal coordinate system, however. normal stress, (b) applied transverse normal stress, and (c) applied in-plane shear stress. For any other set of coordinates, the so- called “shear-coupling” effect is present. This effect will be discussed in detail later.

Typical stress–strain data from (a) longitudinal tensile test and (b) transverse tensile test of carbon/epoxy composite.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 34/75 Orthotropic Engineering Constants Now consider a similar experiment where a transverse normal stress, σ2, is applied to the same material as shown, with all other stresses being equal to zero. Typical stress–strain data from such a test on a carbon/epoxy composite are shown on the right. Now the experimental observation is that the resulting strains can be expressed as σ2 σ σ ε = −ν ε = −ν 2 2 1 21 2 21 ε1 = −ν21 E2 γ 12 = γ 23 = γ 13 = 0 E σ 2 Red- Measured ε = 2 σ Blue- Found 2 ε = + 2 E2 2 σ E2 ε = −ν ε = −ν 2 σ 3 23 2 23 E ε = −ν 2 2 3 23 E where E2 is the transverse modulus of elasticity associated with the 2 2 direction. A similar result for an applied transverse normal stress, σ3, can be obtained by changing the appropriate subscripts in the above equation.

All the values are effective values.

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Next, consider a shear test where a pure shear stress, σ12 = τ12, is applied to the material in the 12 plane, as shown. Now the experimental observation is that the resulting strains can be written as Red- Measured τ 12 γ 12 = ε1 = ε2 = ε 3 = γ 13 = γ 23 = 0 Blue- Found G12

where G12 is the shear modulus associated with the 12 plane. Similar results can be obtained for pure shear in the 13 and 23 planes by changing the appropriate subscripts in the above Equation. Again, notice that there is no shear/normal interaction (or shear coupling). As before, however, this is only true for the principal material axes.

All the values are effective values.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 36/75 Orthotropic Engineering Constants Finally, consider a general 3D state of stress consisting of all possible normal and shear stresses associated with the 1, 2, 3 axes as shown. Since we are dealing with linear behavior, it is appropriate to use superposition and add all the resulting strains due to the simple uniaxial and shear tests, as described above. The resulting set of equations is given below: Red- Measured σ1 σ2 σ3 Blue- Found

σ τ 23 σ 1 σ 2 3 γ = −ν 31 23 ε1 = −ν 21 G E3 23 E1 E2 σ 3 σ τ 31 1 σ 2 −ν 32 ε = −ν γ 31 = 2 12 + E3 E G31 1 E2 σ σ σ ε = −ν 1 2 3 τ 3 13 −ν23 + 12 E γ 12 = 1 E2 E3 G12 ⎡ 1 −ν21 −ν 31 ⎤ ⎢ 0 0 0 ⎥ E1 E2 E3 ⎢ −ν 1 −ν ⎥ ⎢ 12 32 0 0 0 ⎥ ⎢ E E E ⎥ ⎧ ε1 ⎫ 1 2 3 ⎧ σ 1 ⎫ ⎪ ε ⎪ ⎢ −ν13 −ν23 1 ⎥⎪ σ ⎪ 2 ⎢ 0 0 0 ⎥ 2 ⎪ ε 3 ⎪ E1 E2 E3 ⎪ σ 3 ⎪ ⎨ γ ⎬ = ⎢ 1 ⎥⎨ τ ⎬ ⎪ 23 ⎪ ⎢ 0 0 0 0 0 ⎥⎪ 23 ⎪ γ 31 τ 31 ⎪ ⎪ ⎢ G23 ⎥⎪ ⎪ ⎩ γ 12 ⎭ ⎢ 1 ⎥⎩ τ12 ⎭ 0 0 0 0 0 ⎢ G ⎥ ⎢ 31 ⎥ T.Akyürek ME403 Design of Composite Structures - 1 Lamina Stress-Strain Relaonships 37/75 ⎢ 0 0 0 0 0 ⎥ ⎣⎢ G12 ⎦⎥ All the values are effective values. Orthotropic Engineering Constants Note that the compliance matrix is of the same form as the stiffness matrix for a specially orthotropic material as it should be because [S] = [C]−1. Note also that due to symmetry of the compliance matrix, Sij = Sji, νij/Ei = νji/Ej and only nine of the engineering constants are independent. If we now consider a simple uniaxial tensile test consisting of an applied normal stress, σx, along some arbitrary x axis, we find that the full complement of normal strains and shear strains are developed. The generation of shear strains due to normal stresses and normal strains due to shear stresses is often referred to as the “shear-coupling effect.” As a result of shear coupling, all the zeros disappear in the compliance matrix and it becomes fully populated for the general 3D state of stress associated with the arbitrary x, y, z axes; this is the T.Akyürekgenerally orthotropic materialME403 Design of Composite Structures - . Lamina Stress-Strain Relaonships 38/75 Orthotropic Engineering Constants The stiffness or compliance matrices for the generally orthotropic material are of the same form as those for the generally anisotropic material, although the material still has its orthotropic symmetries with respect to the principal material axes. Obviously then, the experimental characterization of such a material is greatly simplified by testing it as a specially orthotropic material along the principal material directions. As shown later, once we have the stiffnesses or compliances associated with the 1, 2, 3 axes, we can obtain those for an arbitrary off-axis coordinate system such as x, y, z by transformation equations involving the angles between the axes.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 39/75 Specially Orthotropic and Transversely Isotropic If the material being tested is specially orthotropic and transversely isotropic as shown in Figure, the subscripts 2 and 3 in below Equation are interchangeable, and we have G13 = G12, E2 = E3, �21 = �31, and �23 = �32. In addition, the familiar relationship among the isotropic engineering constants is now valid for the engineering constants associated with the 23 ⎡ 1 −ν −ν ⎤ Blue- Independent ⎡ 1 ν ⎤ ⎢ 21 21 0 0 0 ⎥ ⎢ − 21 0 ⎥ ⎢ E1 E2 E2 ⎥ plane, so that ⎢ ⎥ E E −ν 1 −ν ⎢ 1 2 ⎥ ⎧ ⎫ ⎢ 12 32 0 0 0 ⎥⎧ ⎫ Plane Stress ⎧ε ⎫ ⎧σ ⎫ ε1 ⎢ ⎥ σ1 1 ⎢ ⎥ 1 ⎪ ⎪ E1 E2 E2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ν12 1 ⎥ ε2 ⎢ ⎥ σ 2 ⎪ ⎪ −ν12 −ν23 1 ⎪ ⎪ Assumption ⎨ε2 ⎬ = − 0 ⎨σ 2 ⎬ ⎪ ε ⎪ ⎢ 0 0 0 ⎥⎪ σ ⎪ ⎢ E E ⎥ E ⎪ 3 ⎪ ⎢ E1 E2 E2 ⎥⎪ 3 ⎪ ⎪ ⎪ 1 2 ⎪ ⎪ 2 ⎨ ⎬ = ⎨ ⎬ γ ⎢ ⎥ τ γ ⎢ 1 ⎥ τ ⎩ 12 ⎭ ⎩ 12 ⎭ G = ⎪ 23 ⎪ ⎢ ⎥⎪ 23 ⎪ 1 23 ⎪ ⎪ 0 0 0 0 0 ⎪ ⎪ ⎢ ⎥ γ ⎢ G23 ⎥ τ 0 0 2 1+ν ⎪ 21 ⎪ ⎢ ⎥⎪ 21 ⎪ ⎢ ⎥ ( 32 ) ⎪ ⎪ 1 ⎪ ⎪ G12 γ ⎢ ⎥ τ ⎪ 12 ⎪ 0 0 0 0 0 ⎪ 12 ⎪ ⎣⎢ ⎦⎥ ⎩ ⎭ ⎢ G ⎥⎩ ⎭ ν13 = ν12 ⎢ 12 ⎥ ν13 ν 23 ν12 ν 23 ⎢ 1 ⎥ ε 3 = − σ 1 − σ 2 ε 3 = − σ 1 − σ 2 ⎢ 0 0 0 0 0 ⎥ G E1 E2 E1 E2 ⎣⎢ 12 ⎦⎥ Now, the compliance matrix is of the same form as Equation 2.17 and only five of the engineering constants are independent. Finally, for the isotropic material, there is no need for subscripts and G13 = G23 = G12 = G, E1 = E2 = E3 = E, �12 = �23 = �13 = �, and G = E/2(1 + �). Now the compliance matrix is of the same form as Equation 2.18 and only two of the T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 40/75 engineering constants are independent. Specially Orthotropic Lamina As shown later in the analysis of laminates, the lamina is often assumed to be in a simple 2D state of stress (or plane stress). In this case, the specially orthotropic stress–strain relationships can be simplified by letting σ3 = τ23 = τ31 = 0, so that ⎡ 1 −ν −ν ⎤ ⎢ 21 21 0 0 0 ⎥ Blue- Independent ⎢ E1 E2 E2 ⎥ ⎢ ⎥ −ν 1 −ν ⎧ ⎫ ⎢ 12 32 0 0 0 ⎥⎧ ⎫ ε1 ⎢ ⎥ σ1 ⎪ ⎪ E1 E2 E2 ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ε2 ⎢ ⎥ σ 2 ⎡ ⎤ ⎪ ⎪ −ν12 −ν23 1 ⎪ ⎪ ⎧ε ⎫ S S 0 ⎧σ ⎫ ⎪ ε ⎪ ⎢ 0 0 0 ⎥⎪ σ ⎪ 1 11 12 1 ⎪ 3 ⎪ ⎢ E1 E2 E2 ⎥⎪ 3 ⎪ ⎢ ⎥ ⎨ ⎬ = ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ γ ⎢ ⎥ τ ⎪ 23 ⎪ ⎢ 1 ⎥⎪ 23 ⎪ ⎪ ⎪ 0 0 0 0 0 ⎪ ⎪ ε = ⎢ S S 0 ⎥ σ γ ⎢ G23 ⎥ τ ⎨ 2 ⎬ 12 22 ⎨ 2 ⎬ ⎪ 21 ⎪ ⎢ ⎥⎪ 21 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎢ ⎥ γ 12 ⎢ 0 0 0 0 0 ⎥ τ12 ⎪ ⎪ ⎪ ⎪ ⎩⎪ ⎭⎪ ⎢ G ⎥⎩⎪ ⎭⎪ γ 0 0 S τ ⎢ 12 ⎥ ⎩ 12 ⎭ ⎢ 66 ⎥⎩ 12 ⎭ ⎢ 1 ⎥ ⎣ ⎦ 0 0 0 0 0 ⎢ G ⎥ ⎣⎢ 12 ⎦⎥ where the compliances Sij and the engineering constants are related by the equations 1 1 ν 21 ν12 1 S11 = S22 = S12 = S21 = − = − S66 = E1 E2 E2 E1 G12 Thus, there are five nonzero compliances and only four independent ⎡ 1 ν ⎤ ⎢ − 21 0 ⎥ ⎢ E E ⎥ compliances for the specially orthotropic lamina. ⎧ ⎫ 1 2 ⎧ ⎫ ε1 ⎢ ⎥ σ 1 ⎪ ⎪ ⎪ ⎪ ⎢ ν12 1 ⎥ Plane Stress ⎨ε ⎬ = − 0 ⎨σ ⎬ 2 ⎢ E E ⎥ 2 ⎪ ⎪ 1 2 ⎪ ⎪ γ ⎢ ⎥ τ ⎩ 12 ⎭ ⎢ 1 ⎥⎩ 12 ⎭ 0 0 ⎢ G ⎥ ⎣⎢ 12 ⎦⎥

ν13 ν23 ν = ν ε = − σ − σ 13 12 ν12 ν23 T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships3 1 2 ε = − σ − 41σ/75 E1 E2 3 1 2 E1 E2 Specially Orthotropic Lamina The lamina stresses in terms of tensor strains are given by ⎡ ⎤ ⎧σ ⎫ Q11 Q12 0 ⎧ε ⎫ ⎪ 1 ⎪ ⎢ ⎥⎪ 1 ⎪ ⎨σ 2 ⎬ = ⎢ Q12 Q22 0 ⎥⎨ε2 ⎬ ⎪ ⎪ ⎢ ⎥⎪ ⎪ Blue- Independent τ 0 0 Q γ ⎩ 12 ⎭ ⎣⎢ 66 ⎦⎥⎩ 12 ⎭ where the Qij are the components of the lamina stiffness matrix, which are related to the compliances and the engineering constants by S22 E1 S12 ν12E2 ⎡ E E ⎤ Q Q 1 ν12 2 Q11 = 2 = 12 = − 2 = = 21 ⎢ 0 ⎥ S11S22 − S12 1−ν12ν 21 S11S22 − S12 1−ν12ν 21 ⎧ ⎫ 1 1 ⎧ ⎫ σ 1 ⎢ −ν12ν21 −ν12ν21 ⎥ ε1 ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎨σ ⎬ = ν E E ⎨ε ⎬ S E 2 ⎢ 12 2 2 0 ⎥ 2 11 2 1 ⎪ ⎪ ⎢ ⎥⎪ ⎪ Q22 = 2 = τ 1−ν ν 1−ν ν γ Q66 = = G12 ⎩ 12 ⎭ 12 21 12 21 ⎩ 12 ⎭ S11S22 − S12 1−ν12ν 21 S ⎢ ⎥ 66 ⎢ 0 0 G ⎥ ⎣ 12 ⎦ As shown later, the experimental characterization of the orthotropic lamina involves the measurement of four independent engineering constants such as E1, E2, G12, and ν12.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 42/75 Specially Orthotropic Lamina Typical values of these properties for several composites are shown in Table.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 43/75 Example 2.3 Problem: An orthotropic, transversely isotropic material is subjected to a 2D stress condition along the principal material axes 1, 2, 3 as shown and the engineering constants for the material are as follows: 6 6 6 E1 = 20x10 psi E2 = 1.5x10 psi G12 = 1x10 psi ν12 = 0.3 ν 23 = 0.2 Determine all strains associated with the 1,

2, 3 axes. ⎡ 1 ν ⎤ ⎢ − 21 0 ⎥ ⎢ E E ⎥ ⎧ ⎫ 1 2 ⎧ ⎫ ν13 = ν12 ε1 ⎢ ⎥ σ 1 ν ν ⎪ ⎪ ν 1 ⎪ ⎪ 13 23 ν ν ⎢ 12 ⎥ ε3 = − σ 1 − σ 2 12 23 ⎨ε2 ⎬ = − 0 ⎨σ 2 ⎬ ε = − σ − σ Plane Stress ⎢ E E ⎥ E1 E2 3 1 2 ⎪ ⎪ 1 2 ⎪ ⎪ E E γ ⎢ ⎥ τ 1 2 ⎩ 12 ⎭ ⎢ 1 ⎥⎩ 12 ⎭ 0 0 ⎢ G ⎥ Blue- Independent Solution: ⎣⎢ 12 ⎦⎥ ν 21 ν12 ν ν = ε = − 13 σ − 23 σ Due to symmetry of the compliance matrix in above Equation 3 1 2 E2 E1 E1 E2 and due to , ν13 = ν12. Therefore, from above Equation, the strains are

So even though the stress condition is 2D, the resulting strains are 3D due to the T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 44/75 Poisson strain ε3. Balanced Orthotropic Lamina The balanced orthotropic lamina shown schematically in Figure often occurs in practice when the fiber reinforcement is woven or cross-plied at 0° and 90°. In this case, the number of independent elastic constants in Equations below is reduced to 3 because of the double symmetry of properties with respect to the 1 and 2 axes (1 and 2 are interchangeable). Thus, for the balanced orthotropic lamina, we have E1 = E2, Q11 = Q22, and S11 = S22. ⎡ 1 ν ⎤ ⎧ ⎫ ⎡ S S 0 ⎤⎧ ⎫ ⎢ − 12 0 ⎥ ε1 11 12 σ 1 Blue- Independent ⎢ E E ⎥ ⎢ ⎥ ⎧ ⎫ 1 1 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ε1 ⎢ ⎥ σ 1 ⎪ ⎪ ⎪ ⎪ ε = ⎢ S S 0 ⎥ σ ⎢ ν12 1 ⎥ ⎨ 2 ⎬ 12 11 ⎨ 2 ⎬ ⎨ε ⎬ = − 0 ⎨σ ⎬ ⎪ ⎪ ⎢ ⎥⎪ ⎪ 2 ⎢ E E ⎥ 2 ⎪ ⎪ ⎢ 1 1 ⎥⎪ ⎪ γ 12 ⎢ 0 0 S ⎥ τ 12 γ 12 τ 12 ⎩ ⎭ ⎣ 66 ⎦⎩ ⎭ ⎩ ⎭ ⎢ 1 ⎥⎩ ⎭ 0 0 1 1 1 ν 21 ν12 ⎢ G ⎥ S = S = S66 = S = S = − = − ⎢ 12 ⎥ 11 E 22 G 12 21 ⎣ ⎦ 1 E2 12 E2 E1

⎧σ ⎫ ⎡ Q Q 0 ⎤⎧ε ⎫ ⎡ E ν E ⎤ 1 ⎢ 11 12 ⎥ 1 ⎢ 1 12 1 0 ⎥ ⎪ ⎪ ⎪ ⎪ ⎧σ ⎫ ⎢ 1−ν 2 1−ν 2 ⎥⎧ε ⎫ 3 ⎢ Q Q 0 ⎥ 1 12 12 1 ⎨σ 2 ⎬ = 12 11 ⎨ε2 ⎬ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎨σ ⎬ = ν E E ⎨ε ⎬ ⎪ ⎪ ⎪ ⎪ 2 ⎢ 12 1 1 0 ⎥ 2 τ 0 0 Q γ ⎪ ⎪ 2 2 ⎪ ⎪ 12 ⎢ 66 ⎥ 12 τ ⎢ 1−ν 1−ν ⎥ γ ⎩ ⎭ ⎣ ⎦⎩ ⎭ ⎩ 12 ⎭ ⎢ 12 12 ⎥⎩ 12 ⎭ ⎢ 0 0 G ⎥ ⎣ 12 ⎦

S22 E1 S12 ν12E2 Q11 = 2 = Q12 = − 2 = = Q21 S11S22 − S12 1−ν12ν 21 S11S22 − S12 1−ν12ν 21

S11 E2 1 1 T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 45/75 Q22 = 2 = Q66 = = G12 2 S11S22 − S12 1−ν12ν 21 S66 Generally Orthotropic Lamina In the analysis of laminates having multiple laminae, it is often necessary to know the stress–strain relationships for the generally orthotropic lamina in non principal coordinates (or “off-axis” coordinates) such as x and y in Figure. Fortunately, the elastic constants in these so-called “off-axis” stress–strain relationships are related to the four independent elastic constants in the principal material coordinates and the lamina orientation angle. The sign convention for the lamina orientation angle, θ, is given in Figure. The relationships are found by combining the equations for transformation of stress and strain components from the 1, 2 axes to the x, y axes.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 46/75 Generally Orthotropic Lamina Relationships for transformation of stress components between coordinate axes may be obtained by writing the equations of static equilibrium for the wedge-shaped differential element in Figure. For example, the force equilibrium along the x direction is given by 2 2 ∑Fx =σ xdA −σ 1dAcos θ −σ 2dAsin θ + 2τ12dAsinθ cosθ = 0 which, after dividing through by dA, gives an equation relating σx to the stresses in the 1, 2 system: 2 2 σ x = σ 1 cos θ +σ 2 sin θ −2τ 12 sinθ cosθ Using a similar approach, the complete set of transformation equations for the stresses in the x, y-coordinate system can be developed and written in matrix form as ⎧ ⎫ ⎡ 2 2 ⎤⎧ ⎫ ⎧ ⎫ σ x c s −2cs σ 1 σ 1 ⎪ ⎪ 1 ⎪ ⎪ ⎢ 2 2 ⎥⎪ ⎪ − ⎪ ⎪ σ = s c 2cs σ = ⎡T ⎤ σ ⎨ y ⎬ ⎢ ⎥⎨ 2 ⎬ ⎣ ⎦ ⎨ 2 ⎬ ⎪ ⎪ ⎢ 2 2 ⎥⎪ ⎪ ⎪ ⎪ τ cs −cs c − s τ τ ⎩⎪ xy ⎭⎪ ⎣ ⎦⎩ 12 ⎭ ⎩ 12 ⎭ and the stresses in the 12 system can be ⎧ ⎫ written as ⎧σ 1 ⎫ σ x ⎪ ⎪ ⎪ ⎪ ⎨σ 2 ⎬ = [T ]⎨σ y ⎬ −1 T.Akyürek ⎪τ ⎪ ME403 Design of Composite Structures - ⎪ ⎪ ⎡T ⎤ s=-sLamina Stress-Strain Relaonships⎡T ⎤ 47/75 ⎩ 12 ⎭ ⎩τ xy ⎭ ⎣ ⎦ ⎣ ⎦ Generally Orthotropic Lamina where c = cos θ , s = sin θ, and the transformation matrix, [T], is defined as ⎡ c2 s2 2cs ⎤ ⎡ c2 s2 −2cs ⎤ ⎢ ⎥ −1 1 2 2 s=-s − ⎢ 2 2 ⎥ [T ] = ⎢ s c −2cs ⎥ ⎡T ⎤ ⎡T ⎤ ⎡T ⎤ = ⎢ s c 2cs ⎥ ⎢ 2 2 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ −cs cs c − s ⎢ cs −cs c2 − s2 ⎥ ⎣ ⎦ ⎣ ⎦ It can also be shown that the tensor strains transform the same way as the stresses, ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ε1 ε1 ε x ε x and that ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ −1 ⎪ ⎪ ε = ⎡R⎤ ε ε = ⎡R⎤ ε ⎨ 2 ⎬ ⎣ ⎦⎨ 2 ⎬ ⎨ y ⎬ ⎣ ⎦ ⎨ y ⎬ ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ⎫ ⎧σ 1 ⎫ σ x ⎧ε1 ⎫ ε x γ γ /2 /2 ⎧ε ⎫ ε ⎩ 12 ⎭ ⎩ 12 ⎭ ⎪γ xy ⎪ ⎪γ xy ⎪ 1 x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 1 ⎪ ⎪ σ = T σ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ ⎨ 2 ⎬ [ ]⎨ y ⎬ ⎨ε2 ⎬ = T ⎨ε y ⎬ ε = ⎡R⎤⎡T ⎤⎡R⎤ ε [ ] ⎨ 2 ⎬ ⎣ ⎦⎣ ⎦⎣ ⎦ ⎨ y ⎬ ⎪τ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 12 ⎭ τ xy γ / 2 ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ 12 ⎭ γ xy / 2 γ γ ⎩ ⎭ ⎩ 12 ⎭ ⎩⎪ xy ⎭⎪

⎧ ⎫ ⎡ Q Q 0 ⎤⎧ ⎫ ⎧ ⎫ ⎧ ⎫ σ 1 11 12 ε1 ⎧σ ⎫ ⎧ε ⎫ σ ε ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ x ⎪ ⎪ x ⎪ ⎪ x ⎪ ⎪ x ⎪ ⎢ Q Q 0 ⎥ ⎪ ⎪ −1 ⎪ ⎪ ⎪ ⎪ −1 −1 ⎪ ⎪ ⎨σ 2 ⎬ = 12 22 ⎨ε2 ⎬ ⎡T ⎤ σ = ⎡Q⎤⎡R⎤⎡T ⎤⎡R⎤ ε σ = ⎡T ⎤ ⎡Q⎤⎡R⎤⎡T ⎤⎡R⎤ ε ⎢ ⎥ ⎣ ⎦⎨ y ⎬ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦ ⎨ y ⎬ ⎨ y ⎬ ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦ ⎨ y ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ τ 12 ⎢ 0 0 Q ⎥ γ 12 ⎩ ⎭ 66 ⎩ ⎭ ⎪τ xy ⎪ ⎪γ xy ⎪ τ γ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎩⎪ xy ⎭⎪ ⎩⎪ xy ⎭⎪ where [R] is the Reuter matrix. ⎧σ ⎫ ⎧ε ⎫ ⎡ 1 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎪ x ⎪ ⎪ x ⎪ −1 −1 −1 ⎪ ⎪ ⎪ ⎪ ⎡Q⎤ ⎡T ⎤ ⎡Q⎤⎡R⎤⎡T ⎤⎡R⎤ ⎢ ⎥ ⎢ ⎥ ⎨σ ⎬ = ⎡Q⎤⎨ε ⎬ ⎣ ⎦ = ⎣ ⎦ ⎣ ⎦⎣ ⎦⎣ ⎦⎣ ⎦ ⎣⎡R⎦⎤ = ⎢ 0 1 0 ⎥ ⎣⎡R⎦⎤ = ⎢ 0 1 0 ⎥ y ⎣ ⎦ y ⎢ 0 0 2 ⎥ ⎢ 0 0 0.5 ⎥ ⎪ ⎪ ⎪ ⎪ ⎣ ⎦ ⎣ ⎦ ⎪τ xy ⎪ ⎪γ xy ⎪ ⎩ ⎭ ⎩ ⎭ ⎣⎡Q⎦⎤⎣⎡S ⎦⎤ = ⎣⎡I ⎦⎤

−1 −1 ⎡S ⎤ = ⎡R⎤⎡T ⎤ ⎡R⎤ ⎡S⎤⎡T ⎤ −1 −1 −1 −1 T.Akyürek ⎣ ⎦ {⎣ ⎦⎣ ME403 Design of Composite Structures - ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦} {⎣⎡TLamina Stress-Strain Relaonships⎦⎤ ⎣⎡Q⎦⎤⎣⎡R⎦⎤⎣⎡T ⎦⎤⎣⎡R⎦⎤ }{⎣⎡R⎦⎤⎣⎡T ⎦⎤ ⎣⎡R⎦⎤ ⎣⎡S⎦⎤⎣⎡T ⎦⎤} = ⎣⎡I ⎦⎤ 48/75 Generally Orthotropic Lamina Carrying out the indicated matrix multiplications and converting back to ⎧ ⎫ ⎡ Q Q Q ⎤⎧ ⎫ engineering strains, we find that σ x 11 12 16 ε x ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎨σ y ⎬ = ⎢ Q12 Q22 Q26 ⎥⎨ε y ⎬ ⎪ ⎪ ⎢ ⎥⎪ ⎪ τ ⎢ Q Q Q ⎥ γ ⎩⎪ xy ⎭⎪ ⎣ 16 26 66 ⎦⎩⎪ xy ⎭⎪ where the are the components of the transformed lamina stiffness matrix, Qij 4 4 2 2 which are defined as follows: Q11 = Q11c + Q22s + 2(Q12 + 2Q66 )s c Q = Q + Q − 4Q s2c2 + Q c4 + s4 4 4 2 2 12 ( 11 22 66 ) 12 ( ) Q22 = Q11s + Q22c + 2(Q12 + 2Q66 )s c 3 3 3 3 Q16 = (Q11 − Q12 − 2Q66 )c s − (Q22 − Q12 − 2Q66 )cs Q26 = (Q11 − Q12 − 2Q66 )cs − (Q22 − Q12 − 2Q66 )c s

2 2 4 4 Q66 = (Q11 + Q22 − 2Q12 − 2Q66 )s c + Q66 (s + c ) Although the transformed lamina stiffness matrix now has the same form as that of an anisotropic material with nine nonzero coefficients, only four of the coefficients are independent because they can all be expressed in terms of the four independent lamina stiffnesses of the specially orthotropic material. That is, the material is still orthotropic, but it is not recognizable as such in the off-axis coordinates. As in the 3D case, it is obviously much easier to characterize the lamina experimentally in the principal material coordinates than in the off-axis coordinates. Recall that the engineering constants, the properties that are normally measured, are related to the ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships T.Akyüreklamina stiffnesses. 49/75 −1 −1 Generally Orthotropic Lamina ⎣⎡S ⎦⎤ = {⎣⎡R⎦⎤⎣⎡T ⎦⎤ ⎣⎡R⎦⎤ ⎣⎡S⎦⎤⎣⎡T ⎦⎤} Alternatively, the strains can be expressed in terms of the stresses as ⎡ 1 ν ⎤ ⎧ ⎫ ⎡ S S S ⎤⎧ ⎫ ⎢ − 21 0 ⎥ ⎧ε ⎫ ⎧σ ⎫ ε x 11 12 16 σ x ⎢ E E ⎥ x x ⎪ ⎪ ⎢ ⎥⎪ ⎪ 1 2 ⎪ ⎪ −1 −1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ν12 1 ⎥ ⎨ε y ⎬ = ⎡R⎤⎡T ⎤ ⎡R⎤ ⎡S⎤⎡T ⎤⎨σ y ⎬ ε = ⎢ S S S ⎥ σ ⎡S⎤ = − 0 ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎨ y ⎬ 12 22 26 ⎨ y ⎬ ⎣ ⎦ ⎢ E E ⎥ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ 1 2 γ τ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ xy ⎪ ⎪ xy ⎪ S S S ⎢ 1 ⎥ ⎩ ⎭ ⎩ ⎭ γ xy ⎢ 16 26 66 ⎥ τ xy 0 0 ⎩⎪ ⎭⎪ ⎣ ⎦⎩⎪ ⎭⎪ ⎢ G ⎥ ⎣⎢ 12 ⎦⎥ where the are the Sij components of the transformed lamina compliance −1 matrix that are defined by , ⎣⎡S ⎦⎤ = ⎣⎡Q⎦⎤ or in expanded form 4 2 2 4 4 2 2 4 S11 = S11c + (2S12 + S66 )s c + S22s S22 = S11s + (2S12 + S66 )s c + S22c 4 4 2 2 3 3 S12 = S12 (s + c ) + (S11 + S22 − S66 )s c S16 = (2S11 − 2S12 − S66 )sc − (2S22 − 2S12 − S66 )s c 2 2 4 4 3 3 S66 = 2(2S11 + 2S22 − 4S12 − S66 )s c + S66 (s + c ) S26 = (2S11 − 2S12 − S66 )s c − (2S22 − 2S12 − S66 )sc The lamina engineering constants can also be transformed from the principal material axes to the off-axis coordinates. For example, the off-axis longitudinal modulus of elasticity associated with uniaxial loading along the x σ x σ x 1 direction as shown in Figure is defined as Ex = = = ε x S11σ x S11 where the strain εx in the denominator has been found by substituting the stress conditions σx ≠ 0, σy = τxy = 0 in the top Equation. ⎡ 1 ν η ⎤ ⎢ − yx xy ,x ⎥ ⎢ Ex E y Gxy ⎥ ⎧ε ⎫ ⎢ ⎥⎧σ ⎫ ⎪ x ⎪ ⎪ x ⎪ ⎪ ⎪ ⎢ ν xy 1 ηxy ,y ⎥⎪ ⎪ ⎨ε ⎬ = ⎢ − ⎥⎨σ ⎬ y E E G y ⎪ ⎪ ⎢ x y xy ⎥⎪ ⎪ ⎪γ xy ⎪ ⎢ ⎥⎪τ xy ⎪ ⎩ ⎭ η η 1 ⎩ ⎭ T.Akyürek⎢ x ,xy y ,xy ⎥ ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 50/75 ⎢ E E G ⎥ ⎣ x y xy ⎦ Off-axis loading for determination of off-axis longitudinal modulus of elasticity E x . Generally Orthotropic Lamina

4 2 2 4 1 1 1 S = S c + 2S + S s c + S s ν 21 ν12 11 11 ( 12 66 ) 22 S11 = S22 = S12 = S21 = − = − S66 = E1 E2 E2 E1 G12 1 1 σ σ 1 E x x x = 4 2 2 4 = Ex = = = ⎛ ⎞ S11c + 2S12 + S66 s c + S22s 1 4 ν12 1 2 2 1 4 ε x S11σ x S11 ( ) c + −2 + s c + s −1 E ⎜ E G ⎟ E ⎡ 1 ⎛ 1 2ν ⎞ 1 ⎤ 1 ⎝ 1 12 ⎠ 2 E = ⎢ c4 + − 12 s2c2 + s4 ⎥ x E ⎜ G E ⎟ E Similarly ⎣⎢ 1 ⎝ 12 1 ⎠ 2 ⎦⎥ 4 2 2 4 ⎧ε ⎫ ⎡ S S S ⎤⎧σ ⎫ S = S s + 2S + S s c + S c x 11 12 16 x 22 11 ( 12 66 ) 22 ⎪ ⎪ ⎢ ⎥⎪ ⎪ −1 ε = ⎢ S S S ⎥ σ ⎡ 1 ⎛ 1 2ν ⎞ 1 ⎤ ⎨ y ⎬ 12 22 26 ⎨ y ⎬ σ y σ y 1 4 12 2 2 4 ⎢ ⎥ E = = = Ey = ⎢ s + − s c + c ⎥ ⎪ ⎪ ⎪ ⎪ y E ⎜ G E ⎟ E S S S ε y S22σ y S22 ⎣ 1 ⎝ 12 1 ⎠ 2 ⎦ ⎪γ xy ⎪ ⎢ 16 26 66 ⎥⎪τ xy ⎪ 2 2 4 4 ⎩ ⎭ ⎣ ⎦⎩ ⎭ S66 = 2(2S11 + 2S22 − 4S12 − S66 )s c + S66 (s + c ) −1 ⎡ ⎤ τ xy τ xy 1 1 4 4 ⎛ 1 1 2ν12 1 ⎞ 2 2 G = = = Gxy = ⎢ (s + c ) + 4 + + − s c ⎥ xy G ⎝⎜ E E E 2G ⎠⎟ γ xy S66τ xy S66 ⎣ 12 1 2 1 12 ⎦ 4 4 2 2 4 2 2 4 S12 = S12 (s + c ) + (S11 + S22 − S66 )s c S11 = S11c + (2S12 + S66 )s c + S22s

ε y S12 ⎡ν12 4 4 ⎛ 1 1 1 ⎞ 2 2 ⎤ ν xy = − = − ν xy = Ex ⎢ (s + c ) − ⎜ + − ⎟ s c ⎥ ε x S11 E1 ⎝ E1 E2 G12 ⎠ 4 ⎣4 2 2 ⎦ 4 2 2 4 S12 = S12 s + c + (S11 + S22 − S66 )s c ( ) S22 = S11s + (2S12 + S66 )s c + S22c ⎡ ⎤ 1 ν yx ηxy ,x ε S ⎡ν 4 4 ⎛ 1 1 1 ⎞ 2 2 ⎤ ⎢ − ⎥ x 12 12 ⎢ E E G ⎥ ν yx = − = − ν yx = Ey s + c − + − s c x y xy ⎢ ( ) ⎜ ⎟ ⎥ ⎧ε ⎫ ⎢ ⎥⎧σ ⎫ ε y S22 E1 ⎝ E1 E2 G12 ⎠ ⎪ x ⎪ ⎪ x ⎪ ⎣ ⎦ ⎪ ⎪ ⎢ ν xy 1 ηxy ,y ⎥⎪ ⎪ ⎨ε ⎬ = ⎢ − ⎥⎨σ ⎬ y E E G y ⎪ ⎪ ⎢ x y xy ⎥⎪ ⎪ where c = cos θ and s = sin θ as before. The rest of the above Equation for γ τ ⎩⎪ xy ⎭⎪ ⎢ η η ⎥⎩⎪ xy ⎭⎪ ⎢ x ,xy y ,xy 1 ⎥ the T.Akyürekoff-axis transverse modulus EME403 Design of Composite Structures - y, the off-axis shear modulus GLamina Stress-Strain Relaonshipsxy, and the ⎢ E E G ⎥ 51/75 ⎣ x y xy ⎦ off-axis major Poisson’s ratio νxy can be obtained from similar derivations. Generally Orthotropic Lamina The variation of these properties with lamina orientation for several composites is shown graphically in Figure. As intuitively expected, Ex varies from a maximum at θ = 0° to a minimum at θ = 90° for this particular material. What may not be intuitively expected is the sharp drop in the off-axis modulus Ex as the angle changes slightly from 0° and the fact that over much of the range of lamina orientations the modulus Ex is very low. This is why transverse reinforcement is needed in unidirectional fiber composites, which are subjected to multi-axial loading. The maximum value of the off-axis shear modulus Gxy at θ = 45° and minimum values of Gxy at θ = 0° and θ = 90° indicate that off-axis reinforcement is also essential for good shear stiffness in unidirectional composites.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 52/75 Variations of off-axis engineering constants with lamina orientation for unidirectional carbon/epoxy, boron/aluminum, and glass/epoxy composites. Generally Orthotropic Lamina A good analogy here is the effect of diagonal bracing in a truss structure—without such bracing, the structure would be much more susceptible to collapse under shear loading. For filament-wound composite power transmission shafts under torsional loading, “off- axis reinforcement is required for acceptable shear stiffness. Later in Chapter 7, it will become clear that off-axis laminae are essential in the design of laminates, which are subjected to shear loading. The shear-coupling effect has been described previously as the generation of shear strains by off-axis normal stresses and the generation of normal strains by off-axis shear stresses. One way to quantify the degree of shear coupling is by defining dimensionless shear- coupling ratios or mutual influence coefficients or shear-coupling coefficients. For example, when the state of stress is defined as σx ≠ 0, σy = τxy = 0, the ratio

1 1 3 3 4 2 2 4 S = S S16 = (2S11 − 2S12 − S66 )sc − (2S22 − 2S12 − S66 )s c S11 = S11c + (2S12 + S66 )s c + S22s 11 22 = E1 E2 γ ⎡⎛ ⎞ ⎛ ⎞ ⎤ xy S16σ x 2 2ν12 1 3 2 2ν12 1 3 ν ν ηx,xy = = = Ex ⎢ + − sc − + − s c⎥ 21 12 ⎜ ⎟ ⎜ ⎟ σx S12 = S21 = − = − ε x S11σ x ⎣⎝ E1 E1 G12 ⎠ ⎝ E2 E1 G12 ⎠ ⎦ E2 E1 ⎡ ν η ⎤ 1 yx xy ,x ⎡ ⎤ 1 ⎢ − ⎥ ⎧ε ⎫ S11 S12 S16 ⎧σ ⎫ is a measure of the amount of shear x ⎢ ⎥ x ⎢ Ex E y Gxy ⎥ ⎪ ⎪ ⎪ ⎪ S66 = ⎧ ⎫ ⎧ ⎫ ε ⎢ ⎥ σ ⎨ε y ⎬ = ⎢ S12 S22 S26 ⎥⎨σ y = 0 ⎬ strain generated in the xy plane per ⎪ x ⎪ ⎪ x ⎪ G12 ⎪ ⎪ ⎢ ν xy 1 ηxy ,y ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎨ε ⎬ = ⎢ − ⎥⎨σ ⎬ γ xy S16 S26 S66 τ xy = 0 y E E G y ⎩ ⎭ ⎣⎢ ⎦⎥⎩ ⎭ unit normal strain along the direction ⎪ ⎪ ⎢ x y xy ⎥⎪ ⎪ γ τ σ2 ⎩⎪ xy ⎭⎪ ⎢ η η ⎥⎩⎪ xy ⎭⎪ of the applied normal stress, T.Akyürek ME403 Design of Composite Structures - σx. ⎢ x ,xy y ,Lamina Stress-Strain Relaonshipsxy 1 ⎥ 53/75 ⎢ E E G ⎥ ⎣ x y xy ⎦ Generally Orthotropic Lamina

γ xy S16σ x ⎡⎛ 2 2ν12 1 ⎞ 3 ⎛ 2 2ν12 1 ⎞ 3 ⎤ ηx,xy = = = Ex ⎢⎜ + − ⎟ sc − ⎜ + − ⎟ s c⎥ ε x S11σ x ⎣⎝ E1 E1 G12 ⎠ ⎝ E2 E1 G12 ⎠ ⎦ Thus, the shear-coupling ratio is analogous to the Poisson’s ratio, which is a measure of the coupling between normal strains. As shown in Figure, ηx,xy strongly depends on orientation and has its maximum value at some intermediate angle, which depends on the material. Since there is no coupling along principal material directions, ηx,xy = 0 for θ = 0° and θ = 90°. As the shear-coupling ratio increases, the amount of shear coupling increases. Other shear-coupling ratios can be defined for different states of stress. For example, when the stresses are τxy ≠ 0, σx = σy = 0, the ratio

ε y S26 ⎡⎛ 2 2ν12 1 ⎞ 3 ⎛ 2 2ν12 1 ⎞ 3 ⎤ ηxy,y = = = Gxy ⎢⎜ + − ⎟ s c − ⎜ + − ⎟ sc ⎥ γ xy S66 ⎣⎝ E1 E1 G12 ⎠ ⎝ E2 E1 G12 ⎠ ⎦ characterizes the normal strain response along the y direction due to a shear stress in the xy plane. �xy ⎡ 1 ν η ⎤ ⎢ − yx xy ,x ⎥ ⎧ ⎫ ⎡ S S S ⎤⎧ ⎫ ε x 11 12 16 σ x = 0 ⎢ Ex E y Gxy ⎥ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎧ε ⎫ ⎢ ⎥⎧σ ⎫ ε = ⎢ S S S ⎥ σ = 0 ⎪ x ⎪ ⎪ x ⎪ ⎨ y ⎬ 12 22 26 ⎨ y ⎬ ⎪ ⎪ ⎢ ν xy 1 ηxy ,y ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎨ε ⎬ = ⎢ − ⎥⎨σ ⎬ γ S S S τ y E E G y ⎩ xy ⎭ ⎢ 16 26 66 ⎥⎩ xy ⎭ ⎪ ⎪ ⎢ x y xy ⎥⎪ ⎪ ⎣ ⎦ γ τ ⎩⎪ xy ⎭⎪ ⎢ η η ⎥⎩⎪ xy ⎭⎪ ⎢ x ,xy y ,xy 1 ⎥ T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships ⎢ E E G ⎥ 54/75 ⎣ x y xy ⎦ Generally Orthotropic Lamina Finally, for a generally orthotropic lamina under plane stress, the stress–strain relationship for the normal strain εx in terms of off-axis engineering constants can be expressed as

1 ν yx ηxy,x ε x = σ x − σ y + τ xy Ex Ey Gxy with similar relationships for εy and γxy. As with the specially orthotropic case and the general anisotropic case, the stiffness and compliance matrices are still symmetric. So, for example,

the off-axis compliances , or in terms of off-axis engineering constants, (S12 = S21 νyx/Ey)=(νxy/Ex). Off-Axes ⎡ ⎤ Principal Axes ⎡ 1 ν ⎤ 1 ν yx ηxy ,x ⎢ 21 0 ⎥ ⎢ − ⎥ − E E G ⎢ E E ⎥ ⎢ x y xy ⎥ ⎧ ⎫ 1 2 ⎧ ⎫ ⎧ ⎫ ⎢ ⎥⎧ ⎫ ε1 ⎢ ⎥ σ 1 ε x σ x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ν η ⎥⎪ ⎪ ⎢ ν12 1 ⎥ ⎪ ⎪ xy 1 xy ,y ⎪ ⎪ ⎨ε2 ⎬ = − 0 ⎨σ 2 ⎬ ⎨ε ⎬ = ⎢ − ⎥⎨σ ⎬ ⎢ E E ⎥ y E E G y ⎪ ⎪ 1 2 ⎪ ⎪ x y xy γ ⎢ ⎥ τ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎩ 12 ⎭ ⎩ 12 ⎭ γ xy τ xy ⎢ 1 ⎥ ⎩⎪ ⎭⎪ ⎢ η η ⎥⎩⎪ ⎭⎪ 0 0 ⎢ x ,xy y ,xy 1 ⎥ ⎢ G ⎥ 3 ⎢ 12 ⎥ ⎢ E E G ⎥ ⎣ ⎦ ⎣ x y xy ⎦

2

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships4 independent constants 55/75 1 Generally Orthotropic Lamina The effects of lamina orientation on stiffness are difficult to assess from a quick inspection of stiffness transformation equations. In addition, the eventual incorporation of lamina stiffnesses into laminate analysis requires integration of lamina stiffnesses over the laminate thickness, and integration of such complicated equations is also difficult. In view of these difficulties, a more convenient “invariant” form of the lamina stiffness transformation equations has been proposed by Tsai and Pagano. By using trigonometric identities to convert from power functions to multiple angle functions and then using additional mathematical manipulations, Tsai and Pagano showed that Equation 2.36 could also be written as 4 4 2 2 Q11 = U1 +U2 cos2θ +U3 cos4θ Q11 = Q11c + Q22s + 2(Q12 + 2Q66 )s c Q = U −U cos2θ +U cos4θ 4 4 2 2 22 1 2 3 Q22 = Q11s + Q22c + 2(Q12 + 2Q66 )s c Q = U −U cos4θ 2 2 4 4 12 4 3 Q12 = (Q11 + Q22 − 4Q66 )s c + Q12 (c + s ) U2 3 3 Q16 = sin2θ +U3 cos4θ Q16 = (Q11 − Q12 − 2Q66 )c s − (Q22 − Q12 − 2Q66 )cs 2 3 3 U2 Q26 = (Q11 − Q12 − 2Q66 )cs − (Q22 − Q12 − 2Q66 )c s Q26 = sin2θ −U3 cos4θ 2 2 2 4 4 1 Q66 = Q11 + Q22 − 2Q12 − 2Q66 s c + Q66 s + c ( ) ( ) Q66 = (U1 −U4 ) −U3 cos4θ 2

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 56/75 Generally Orthotropic Lamina where the set of “invariants” is defined as ⎡ ⎤ ⎧σ ⎫ Q11 Q12 Q16 ⎧ε ⎫ 1 1 x ⎢ ⎥ x U1 = (3Q11 + 3Q22 + 2Q12 + 4Q66 ) U2 = (Q11 − Q22 ) ⎪ ⎪ ⎪ ⎪ 8 2 ⎨σ y ⎬ = ⎢ Q12 Q22 Q26 ⎥⎨ε y ⎬ 1 1 ⎪ ⎪ ⎢ ⎥⎪ ⎪ U3 = Q11 + Q22 − 2Q12 − 4Q66 τ xy Q16 Q26 Q66 γ xy ( ) U4 = (Q11 + Q22 + 6Q12 − 4Q66 ) ⎩ ⎭ ⎢ ⎥⎩ ⎭ 8 8 ⎣ ⎦ As the name implies, the invariants, which are simply linear combinations of the Qij, are invariant to rotations in the plane of the lamina. Note that there are four independent invariants, just as there are four independent elastic constants. Equation 2.44 is obviously easier to manipulate and interpret than Equation 2.36. For example, all the stiffness expressions except those for the coupling stiffnesses consist of one constant term and terms that vary with lamina orientation. Thus, the effects of lamina orientation on stiffness are easier to interpret.

Q11 = U1 +U2 cos2θ +U3 cos4θ 4 4 2 2 Q11 = Q11c + Q22s + 2(Q12 + 2Q66 )s c Q = U −U cos4θ 12 4 3 2 2 4 4 Q12 = (Q11 + Q22 − 4Q66 )s c + Q12 (c + s ) Q22 = U1 −U2 cos2θ +U3 cos4θ 4 4 2 2 U2 Q22 = Q11s + Q22c + 2(Q12 + 2Q66 )s c Q16 = sin2θ +U3 sin 4θ 2 3 3 Q16 = (Q11 − Q12 − 2Q66 )c s − (Q22 − Q12 − 2Q66 )cs U2 Q26 = sin2θ −U3 sin 4θ 3 3 2 Q26 = (Q11 − Q12 − 2Q66 )cs − (Q22 − Q12 − 2Q66 )c s 1 ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 2 2 4 57/475 QT.Akyürek66 = (U1 −U4 ) −U3 cos4θ 2 Q66 = (Q11 + Q22 − 2Q12 − 2Q66 )s c + Q66 (s + c ) Generally Orthotropic Lamina Invariant formulations of lamina compliance transformations are also useful. It can be shown that the off-axis compliance components in Equation 2.37 can be written as

S11 = V1 +V2 cos2θ +V3 cos4θ S22 = V1 −V2 cos2θ +V3 cos4θ S12 = V4 −V3 cos4θ

S16 = V2 sin2θ + 2V3 sin 4θ S26 = V2 sin2θ − 2V3 sin 4θ S66 = 2(V1 −V4 ) − 4V3 cos4θ where the invariants are ⎡ ⎤ 1 1 ⎧ε ⎫ S11 S12 S16 ⎧σ ⎫ V 3S 3S 2S S V = S − S ⎪ x ⎪ ⎢ ⎥⎪ x ⎪ 1 = ( 11 + 22 + 12 + 66 ) 2 ( 11 22 ) S S S 8 2 ⎨ε y ⎬ = ⎢ 12 22 26 ⎥⎨σ y ⎬ ⎢ ⎥ 1 ⎪γ ⎪ S S S ⎪τ ⎪ 1 ⎩ xy ⎭ ⎣⎢ 16 26 66 ⎦⎥⎩ xy ⎭ V4 = (S11 + S22 + 6S12 − S66 ) V3 = (S11 + S22 − 2S12 − S66 ) 8 8 Invariant formulations also lend themselves well to graphical interpretation. Stress transformation equations can be combined and manipulated so as to generate the equation of Mohr’s circle. As shown in Figure, the transformation of a normal stress component σx can be described by

the invariant formulation σ x = I1 + I2 cos2θ p

where 2 σ x +σ y ⎛ σ x −σ y ⎞ 2 I1 = I = +τ 2 2 ⎝⎜ 2 ⎠⎟ xy �p is the angle between x axis and the principal stress axis In this case, the invariants are I1, which defines the position of T.Akyürekthe center of the circle, and IME403 Design of Composite Structures - 2, which is the radius of the circle. Lamina Stress-Strain Relaonships 58/75 Generally Orthotropic Lamina

Note the similarity between σx and EquationsQ11 . Invariants consist of a constant term and a σ +σ 2 x y ⎛ σ x −σ y ⎞ 2 term or terms that vary with orientation. σ = I + I cos2θ I1 = I2 = +τ xy x 1 2 p 2 ⎝⎜ 2 ⎠⎟ Similarly, the invariant forms of the stiffness transformations can also be interpreted graphically using Mohr’s circles. For example, Tsai and Hahn have shown that the stiffness

transformation equation Q11 = U1 +U2 cos2θ +U3 cos4θ whereas the distance between the centers of the two circles is given by U1. The radius and angle associated with one circle are U2 and 2θ, respectively, and the radius and angle associated with the other circle are U3 and 4θ, respectively. Thus, the distance between the centers of the circles is a measure of the isotropic component of stiffness, whereas the radii of the circles indicate the strength of the orthotropic component. If U2 = U3 = 0, the circles collapse to points and the material is isotropic.

⎡ ⎤ ⎧σ ⎫ Q11 Q12 Q16 ⎧ε ⎫ ⎪ x ⎪ ⎢ ⎥⎪ x ⎪ ⎨σ y ⎬ = ⎢ Q12 Q22 Q26 ⎥⎨ε y ⎬ ⎢ ⎥ ⎪τ ⎪ Q Q Q ⎪γ ⎪ ⎩ xy ⎭ ⎣⎢ 16 26 66 ⎦⎥⎩ xy ⎭

1 U1 = (3Q11 + 3Q22 + 2Q12 + 4Q66 ) 8 1 1 1 T.Akyürek U = Q − Q ME403 Design of Composite Structures - U Q QLamina Stress-Strain Relaonships2Q 4Q 59/75 2 ( 11 22 ) 3 = ( 11 + 22 − 12 − 66 ) U4 = (Q11 + Q22 + 6Q12 − 4Q66 ) 2 8 8 Example 2.4 Problem: A 45° off-axis tensile test is conducted on a generally orthotropic test specimen by applying a normal stress σx as shown in Figure. The specimen has strain gages attached so as to measure the normal strains εx and εy along the x and y directions, respectively. How many engineering constants for this material can be determined from the measured parameters σx, εx, and εy?

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 60/75 ⎡ 1 ν η ⎤ ⎢ − yx xy,x ⎥ ⎢ Ex Ey Gxy ⎥ ⎧ ⎫ ⎧ ⎫ ε x ⎢ ⎥ σ x ⎪ ⎪ ⎢ ν xy 1 ηxy,y ⎥⎪ ⎪ Example 2.4 ⎨ε y ⎬ = − ⎨σ y = 0 ⎬ ⎢ Ex Ey Gxy ⎥ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎩γ xy ⎭ ⎩τ xy = 0⎭ ⎢ ηx,xy ηy,xy 1 ⎥ ⎢ E E G ⎥ Solution: ⎣⎢ x y xy ⎦⎥ The off-axis modulus of elasticity along the x direction is determined from ε y σ x The off-axis major Poisson’s ratio is given by ν xy = − Ex = ε x ε x Although it may not seem obvious, the in-plane shear modulus G12 may also be found from these measurements by using the appropriate transformations of stress and strain. Using θ = 45° in Equation 2.33, the shear strain γ12 is found from γ γ −ε + ε 12 = − cos450 sin 450 ε + cos450 sin 450 ε + cos2 450 − sin2 450 xy = x y 2 x y ( ) 2 2 Note that the γxy term drops out of the above equation only when θ = 45°. For any other value of θ, we would need to know γxy as well as εx and εy in order to find γ12 from this transformation equation. ⎧ ⎫ ⎧ ⎫ ⎧σ 1 ⎫ σ x ⎧ε1 ⎫ ε x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨σ 2 ⎬ = [T ]⎨σ y ⎬ ⎨ε2 ⎬ = [T ]⎨ε y ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ τ12 τ γ / 2 ⎩ ⎭ ⎩ xy ⎭ ⎩ 12 ⎭ ⎩γ xy / 2⎭

⎡ c2 s2 2cs ⎤ ⎢ 2 2 ⎥ [T ] = ⎢ s c −2cs ⎥ ⎢ −cs cs c2 − s2 ⎥ T.Akyürek⎣ ⎦ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 61/75 Example 2.4 Solution: Likewise, the shear stress τ12 is found by substituting the tensile test conditions σx ≠ 0, σy = τxy = 0 in stress transformation equation: σ τ = − cos450 sin 450σ + cos450 sin 450σ + cos2 450 − sin2 450 τ = − x 12 x y ( ) xy 2 Finally, the shear modulus G12 may now be calculated from the third line of Equation 2.24 as 1 τ12 G12 = = S66 γ 12

Note that this equation holds even though the stresses σ1 ≠ 0 and σ2 ≠ 0 in the first two lines of Equation 2.24. ⎡ ⎤ ⎧ε ⎫ S11 S12 0 ⎧σ ⎫ 1 ⎢ ⎥ 1 1 σ ⎧σ ⎫ ⎪ ⎪ ⎪ ⎪ ⎧ 1 ⎫ x ⎧ ⎫ ⎨ε2 ⎬ = ⎢ S12 S22 0 ⎥⎨σ 2 ⎬ S = ⎪ ⎪ ⎪ ⎪ ⎧ε1 ⎫ ε x 66 G12 ⎨σ 2 ⎬ = [T ]⎨σ y ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ε = T ε ⎩γ 12 ⎭ 0 0 S66 ⎩τ12 ⎭ ⎪ ⎪ ⎪ ⎪ ⎨ 2 ⎬ [ ]⎨ y ⎬ ⎣ ⎦ τ12 τ ⎩ ⎭ ⎩ xy ⎭ ⎪γ / 2⎪ ⎪ ⎪ ⎩ 12 ⎭ ⎩γ xy / 2⎭ ⎡ 1 ν ⎤ ⎢ − 21 0 ⎥ 2 2 ⎢ E E ⎥ ⎡ c s 2cs ⎤ ⎧ ⎫ 1 2 ⎧ ⎫ ε1 ⎢ ⎥ σ 1 ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ 2 2 ⎢ ν12 1 ⎥ T = s c −2cs ⎨ε2 ⎬ = − 0 ⎨σ 2 ⎬ [ ] ⎢ ⎥ ⎢ E E ⎥ ⎪ ⎪ 1 2 ⎪ ⎪ ⎢ 2 2 ⎥ γ ⎢ ⎥ τ −cs cs c − s ⎩ 12 ⎭ ⎢ 1 ⎥⎩ 12 ⎭ T.Akyürek⎣ ME403 Design of Composite Structures - ⎦ 0 0 Lamina Stress-Strain Relaonships 62/75 ⎢ G ⎥ ⎣⎢ 12 ⎦⎥ Example 2.5 Problem: A filament-wound cylindrical vessel of mean diameter d = 1 m and wall thickness t = 20 mm is subjected to an internal pressure, p. The filament-winding angle θ = 53.1° from the longitudinal axis of the pressure vessel, and the glass/epoxy material has the following properties: E1 = 40 GPa, E2 = 10 GPa, G12 = 3.5 GPa, and ν12 = 0.25. By the use of a strain gage, the normal strain along the fiber direction is determined to be ε1 = 1000 � strain. Determine the internal pressure in the vessel.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 63/75 ⎡ 1 ν ⎤ ⎢ − 21 0 ⎥ E E ε ⎢ 1 2 ⎥ σ ⎧ 1 ⎫ ⎢ ⎥⎧ 1 ⎫ ⎪ ⎪ ν12 1 ⎪ ⎪ − 0 Example 2.5 ⎨ε2 ⎬ = ⎢ ⎥⎨σ 2 ⎬ ⎢ E1 E2 ⎥ ⎪γ ⎪ ⎪τ ⎪ ⎩ 12 ⎭ ⎢ 1 ⎥⎩ 12 ⎭ ⎢ 0 0 ⎥ G Solution: ⎣⎢ 12 ⎦⎥

From mechanics of materials, the stresses in a thin-walled cylindrical d r = = 0.5m pressure vessel are given by 2 pr 0.5 p pr 0.5 p σ = = = 12.5 p σ y = = = 25 p τ xy = 0 x 2t 2(0.02) t 0.02 These equations are based on static equilibrium and geometry only. Thus, they apply to a vessel made of any material. Since the given strain is along the fiber direction, we must transform the above stresses to the 12 axes. Recall that in the “netting analysis” in Problems 1.10 and 1.11 of Chapter 1, only the fiber longitudinal normal stress was considered. This was because the matrix was ignored, and the fibers alone cannot support transverse or shear stresses. In the current problem, however, the transverse normal stress, σ2, and the shear stress, τ12, are also considered because the fiber and matrix are now assumed to act as a composite. From Equation 2.31, the stresses along the 12 axes are 2 2 2 2 ⎡ c2 s2 2cs ⎤ σ = σ cos θ +σ sin θ + 2τ sinθ cosθ σ 1 = (12.5 p)(0.6) + (25 p)(0.8) + 0 = 20.5 p(MPa) 1 x y xy ⎢ 2 2 ⎥ [T ] = ⎢ s c −2cs ⎥ 2 2 2 2 sin cos 2 sin cos ⎢ −cs cs c2 − s2 ⎥ σ = 12.5 p 0.8 + 25 p 0.6 − 0 = 17.0 p MPa σ 2 = σ x θ +σ y θ − τ xy θ θ ⎣ ⎦ 2 ( )( ) ( )( ) ( ) 2 2 τ = − 12.5 p 0.8 0.6 + 25 p 0.6 0.8 + 0 = 6.0 p MPa τ12 = −σ x sinθ cosθ +σ y sinθ cosθ +τ xy (cos θ − sin θ ) 12 ( )( )( ) ( )( )( ) ( )

ν 21 ν12 σ ν σ 20.5 p 0.25 17.0 p T.Akyürek = 1ME403 Design of Composite Structures - 12 2 ( Lamina Stress-Strain Relaonships) 64/75 ε1 = − = − = 0.001 p = 2.46MPa E2 E1 3 3 E1 E1 40(10 ) 40(10 ) Example 2.6 Problem: A tensile test specimen is cut out along the x direction of the pressure vessel described in Example 2.5. What effective modulus of elasticity would you expect to get during a test of this specimen?

E1 = 40 GPa, E2 = 10 GPa, G12 = 3.5 GPa, and ν12 = 0.25 θ = 53.1°

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 65/75 Example 2.6 Solution: The off-axis modulus of elasticity, Ex, associated with the x direction is given by the first of Equation 2.40 with θ = 53.1°.

−1 ⎡ 1 4 ⎛ 1 2ν12 ⎞ 2 2 1 4 ⎤ Ex = ⎢ c + ⎜ − ⎟ s c + s ⎥ ⎣ E1 ⎝ G12 E1 ⎠ E2 ⎦

⎡ ⎤ or ⎡ 1 ν ⎤ 1 ν yx ηxy ,x ⎢ − 21 0 ⎥ ⎢ − ⎥ ⎢ E E G ⎥ ⎢ E1 E2 ⎥ x y xy E1 = 40 GPa ⎢ ⎥ ⎢ ⎥ ⎢ ν η ⎥ ⎢ ν12 1 ⎥ xy 1 xy ,y E2 = 10 GPa ⎡S⎤ = − 0 ⎡S ⎤ = ⎢ − ⎥ ⎣ ⎦ ⎢ E E ⎥ ⎣ ⎦ E E G ⎢ 1 2 ⎥ ⎢ x y xy ⎥ G12 = 3.5 GPa ⎢ 1 ⎥ ⎢ η η ⎥ 0 0 ⎢ x ,xy y ,xy 1 ⎥ ⎢ G ⎥ ν12 = 0.25 ⎢ 12 ⎥ ⎢ E E G ⎥ ⎣ ⎦ ⎣ x y xy ⎦ ⎡ c2 s2 −2cs ⎤ ⎡ c2 s2 2cs ⎤ −1 ⎢ 2 2 ⎥ ⎢ 2 2 ⎥ [T ] = ⎢ s c 2cs ⎥ [T ] = ⎢ s c −2cs ⎥ ⎢ cs −cs c2 − s2 ⎥ ⎢ −cs cs c2 − s2 ⎥ ⎣ ⎦ ⎣ ⎦ −1 −1 ⎣⎡S ⎦⎤ = ⎣⎡R⎦⎤⎣⎡T ⎦⎤ ⎣⎡R⎦⎤ ⎣⎡S⎦⎤⎣⎡T ⎦⎤ ⎡ 1 0 0 ⎤ ⎡ 1 0 0 ⎤ { } −1 ⎢ ⎥ ⎢ ⎥ ⎣⎡R⎦⎤ = ⎢ 0 1 0 ⎥ ⎣⎡R⎦⎤ = ⎢ 0 1 0 ⎥ ⎢ 0 0 0.5 ⎥ ⎢ 0 0 2 ⎥ ⎣ ⎦ ⎣ ⎦

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 66/75 Example 2.7 Problem: A lamina consisting of continuous fibers randomly oriented in the plane of the lamina is said to be “planar isotropic,” and the elastic properties in the plane are isotropic in nature. Find expressions for the lamina stiffnesses for a planar isotropic lamina.

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 67/75 Example 2.7 Solution: Since the fibers are assumed to be randomly oriented in the plane, the

“planar isotropic stiffnesses” can be found by averaging the transformed π Q dθ lamina stiffnesses as follows: ∫ ij ! 0 Qij = π ∫ dθ where the superscript (~) denotes an averaged property. 0 It is convenient to use the invariant forms of the transformed lamina stiffnesses because they are easily integrated. For example, if we substitute the first of Equation 2.44 in the above equation, we get π π Q dθ U +U cos2θ +U cos4θ dθ ∫ 11 ∫[ 1 2 3 ] Q! = 0 = 0 = U 11 π π 1 ∫ dθ Note that the averaged stiffness equals the isotropic part of the transformed 0 lamina stiffness, and that the orthotropic parts drop out in the averaging process. Similarly, the other averaged stiffnesses can be found in terms of the invariants. The derivations of the remaining expressions are left as an T.Akyürekexercise. ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 68/75 Example 2.8 Problem: Find the following for a 60° angle lamina of graphite/epoxy. Use the properties of unidirectional graphite/epoxy lamina from Table. 1. Transformed compliance matrix 2. Transformed reduced stiffness matrix If the applied stress is σx = 2 MPa, σy = – 3 MPa, and τxy = 4 MPa, also find 3. Global strains 4. Local strains 5. Local stresses 6. Principal stresses 7. Maximum shear stress 8. Principal strains 9. Maximum shear strain

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 69/75 Example 2.8 Solution: 1. c = Cos(60°) = 0.500 s = Sin(60°) = 0.866 1 1 S = 0.5525 10−11 S11 = 11 ( ) E1 Pa 1 1 S = S = 0.9709 10−10 22 E 22 ( ) Pa 2 1 ν 21 ν12 −11 S = S = − = − S12 = −0.1547 10 12 21 E E ( ) Pa 2 1 1 1 S 0.1395 10−9 S66 = 66 = ( ) G12 Pa 4 2 2 4 1 S S c 2S S s c S s −10 11 = 11 + ( 12 + 66 ) + 22 S11 = 0.8053(10 ) Pa −1 −1 1 ⎡S ⎤ = ⎡R⎤⎡T ⎤ ⎡R⎤ ⎡S⎤⎡T ⎤ S = S s4 + c4 + S + S − S s2c2 −11 ⎣ ⎦ {⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦} 12 12 ( 11 22 66 ) S12 = −0.7878(10 ) ( ) Pa 3 3 −10 1 S16 = (2S11 − 2S12 − S66 )sc − (2S22 − 2S12 − Q66 )s c S = −0.3234(10 ) 16 Pa 4 2 2 4 −10 1 S22 = S11s + (2S12 + S66 )s c + S22c S22 = 0.3475(10 ) Pa 3 3 −10 1 ⎡ 1 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎢ ⎥ 1 S = 2S − 2S − S s c − 2S − 2S − S sc S26 = −0.4696(10 ) ⎡R⎤ = 0 1 0 − ⎢ ⎥ 26 ( 11 12 66 ) ( 22 12 66 ) ⎣ ⎦ ⎢ ⎥ ⎣⎡R⎦⎤ = ⎢ 0 1 0 ⎥ Pa ⎢ 0 0 2 ⎥ ⎢ 0 0 0.5 ⎥ ⎣ ⎦ ⎣ ⎦ 2 2 4 4 −9 1 S66 = 2(2S11 + 2S22 − 4S12 − S66 )s c + S66 s + c S66 = 0.1141(10 ) ( ) Pa ⎡ 1 ν ⎤ ⎢ − 21 0 ⎥ ⎢ E E ⎥ 1 2 2 2 2 2 ⎧ ⎫ ⎡ S S S ⎤⎧ ⎫ ⎢ ⎥ ⎡ c s −2cs ⎤ ⎡ c s 2cs ⎤ ε x 11 12 16 σ x ν 1 ⎢ ⎥ ⎡S⎤ = ⎢ − 12 0 ⎥ −1 ⎢ 2 2 ⎥ ⎢ 2 2 ⎥ ⎪ ⎪ ⎪ ⎪ ⎣ ⎦ ⎢ E E ⎥ T = s c 2cs T = s c −2cs 1 2 [ ] ⎢ ⎥ [ ] ⎢ ⎥ ⎨ε y ⎬ = ⎢ S12 S22 S26 ⎥⎨σ y ⎬ ⎢ ⎥ 2 2 2 2 ⎢ 1 ⎥ ⎢ cs −cs c − s ⎥ ⎢ −cs cs c − s ⎥ ⎪ ⎪ ⎢ ⎥⎪ ⎪ T.Akyürek0 0 ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships γ S S S τ70/75 ⎢ G ⎥ ⎣ ⎦ ⎣ ⎦ ⎩ xy ⎭ ⎢ 16 26 66 ⎥⎩ xy ⎭ ⎣⎢ 12 ⎦⎥ ⎣ ⎦ Example 2.8

2. Invert the transformed compliance matrix [ ] to obtain the transformed S reduced stiffness matrix [ Q ]:

3. The global strains in the x–y plane are given as

4. Using transformation, the local strains in the lamina are

⎧ ⎫ ⎧ε1 ⎫ ε x ⎪ ⎪ ⎪ ⎪ ⎨ε2 ⎬ = [T ]⎨ε y ⎬ ⎪γ / 2⎪ ⎪ ⎪ ⎩ 12 ⎭ ⎩γ xy / 2⎭

⎡ c2 s2 2cs ⎤ ⎢ 2 2 ⎥ [T ] = ⎢ s c −2cs ⎥ ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 71/75 T.Akyürek⎢ −cs cs c2 − s2 ⎥ ⎣ ⎦ Example 2.8

5. Using transformation Equation, the local stresses in the lamina are ⎧ ⎫ ⎧σ 1 ⎫ σ x ⎪ ⎪ ⎪ ⎪ ⎨σ 2 ⎬ = [T ]⎨σ y ⎬ ⎪τ ⎪ ⎪ ⎪ ⎩ 12 ⎭ ⎩τ xy ⎭ 6. The principal normal stresses are given by

7. The maximum shear stress is given by

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 72/75 Example 2.8 8. The principal strains are given by

9. The maximum shearing strain is given by

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 73/75 Example 2.9 Problem: As shown in Figure, a 60° angle graphite/epoxy lamina is subjected only to a shear stress τxy = 2 MPa in the global axes. What would be the value of the strains measured by the strain gage rosette — that is, what would be the normal strains measured by strain gages A, B, and C? Use the properties of unidirectional graphite/epoxy lamina from Table. Solution: Example 2.8, the reduced compliance matrix ⎣⎡S ⎦⎤ is

The global strains in the x–y plane

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 74/75 Example 2.9 For a strain gage placed at an angle, φ, to the x-axis, the normal strain recorded by the strain gage is given by 2 2 εφ = ε x cos φ + ε y sin φ +γ xy sinφ cosφ For strain gage A, φ = 0°: εA=-6.468(10-5) For strain gage B, φ = 240°: εB=1.724(10-4) For strain gage C, φ = 120°: εC=1.083(10-5)

T.Akyürek ME403 Design of Composite Structures - Lamina Stress-Strain Relaonships 75/75