Mechanics of Composite MaterialsThe mechanics of composite materials deals mainly with the analysis of stresses and strains in the laminate. This is usually performed by analyzing the stresses and strains in each lamina first. The results for all the laminas are then integrated over the length of the laminate to obtain the overall quantities. Chaps. 2–6 deal mainly with the analysis of stress and strain in one single lamina. This is performed in the local lamina coordinate system and also in the global laminate coordinate system. Laminate analysis is then discussed in Chaps. 7–9. Chaps. 11 and 12 provide an introduction to the advanced topics of homogenization and damage mechanics in composite materials, respectively. MATLAB Functions for Mechanics of Composite Materials OrthotropicCompliance(E1, E2, E3, NU12, NU23, NU13, G12, G23, G13) OrthotropicStiffness(E1, E2, E3, NU12, NU23, NU13, G12, G23, G13) TransverselyIsotropicCompliance(E1, E2, NU12, NU23, G12) TransverselyIsotropicStiffness(E1, E2, NU12, NU23, G12) IsotropicCompliance(E, NU) IsotropicStiffness(E, NU) E1(Vf, E1f, Em) NU12(Vf, NU12f, NUm) E2(Vf, E2f, Em, Eta, NU12f, NU21f, NUm, E1f, p) G12(Vf, G12f, Gm, EtaPrime, p) Alpha1 (Vf, E1f, Em, Alpha1f, Alpham) Alpha2 (Vf, Alpha2f, Alpham, E1, E1f, Em, NU1f, NUm, Alpha1f, p) E2Modified(Vf, E2f, Em, Eta, NU12f, NU21f, NUm, E1f, p) ReducedCompliance(E1, E2, NU12, G12) ReducedStiffness(E1, E2, NU12, G12) ReducedIsotropicCompliance(E, NU) ReducedIsotropicStiffness(E, NU) ReducedStiffness2 (E1, E2, NU12, G12) ReducedIsotropicStiffness2 (E, NU) z1. z2) Dmatrix(D. z1. Qbar. theta) Tinv2 (theta) Sbar2 (S. NU21. NU21. theta) NUyx(E1. G12. kap xo. z1. NU12. theta) Gxy(E1. NU12. The 1-axis is aligned with the fiber direction. z) Amatrix(A. E2. gam xyo. E2. H) NUbaryx (A. theta) Etaxyx(Sbar) Etaxyy(Sbar) Etaxxy(Sbar) Etayxy(Sbar) Strains(eps xo. T) Ex(E1. H) NUbarxy(A. G12. T) Qbar2 (Q. NU12. Qbar. The 2-axis is in the plane of the layer and perpendicular to the fibers.T(theta) Tinv(theta) Sbar(S. 2. Qbar. kap xyo. theta) Qbar(Q. G12. 3. H) Gbarxy(A. G12. z2) Ebarx (A. theta) Ey(E1. The 3-axis is perpendicular to the plane of the layer and thus also perpendicular to the fibers. E2. eps yo. . z2) Bmatrix(B. H) Ebary(A. E2. E2. kap yo. theta) NUxy(E1. H) Linear Elastic Stress-Strain Relations Orthogonal coordinate systems for single layer fiber reinforced composite: 1. G12. The 1-direction is also called the fiber direction. We assume that the two-material fiber-matrix system is replaced by a single homogeneous material. Such material with different properties in three mutually perpendicular directions is called an orthotropic material. the layer (lamina) is considered to be orthotropic. Therefore. . Obviously. This 1-2-3 coordinate system is called the principal material coordinate system.directions are called the matrix directions or the transverse directions. this single material does not have the same properties in all directions.and 3. The stresses and strains in the layer (also called a lamina) will be referred to the principal material coordinate system. while the 2. as follows: The elements of [C] are not shown here explicitly but are calculated using the MATLAB function Orthotropic Stiffness which is written specifically for this purpose. respectively. Also.Stress-strain relationship: E1. i. and [S] is called the compliance matrix. G23. while G12. and 3 directions. and G13 are the three shear moduli. and E3 are the extensional moduli of elasticity along the 1. respectively. The inverse of the compliance matrix [S] is called the stiffness matrix [C] given.e. C21 = C12. It is shown (see [1]) that both the compliance matrix and the stiffness matrix are symmetric. S23. in general. and S13. 2. 2. C23 = C32. {ε} = [S] {σ} where {ε} and {σ} represent the 6 × 1 strain and stress vectors. j = 1. C13 = C31. 3) are the different Poisson’s ratios. E2. the following expressions can now be easily obtained: . Therefore. and similarly for S21. νij (i. ν) for an isotropic material. Therefore. In addition. we have the following relation: It is clear that there are only two independent material constants (E. E2 = E3. In addition.It should be noted that the material constants appearing in the compliance matrix in (2. A material is called isotropic if its behavior is the same in all three 1-2-3 directions. . This is clear since the compliance matrix is symmetric. ν12. ν23. ν12 = ν23 = ν13 = ν. and G12 = G23 = G13 = G. Thus it is now clear that there are nine independent material constants for an orthotropic material. ν12 = ν13. we have the following equations relating the material constants: The above equations are called the reciprocity relations for the material constants. E1 = E2 = E3 = E. In this case. E2.1) are not all independent. For this case. It should be noted that the reciprocity relations can be derived irrespective of the symmetry of the compliance matrix – in fact we conclude that the compliance matrix is symmetric from using these relations. G12) for a transversely isotropic material. we have the following relation: It is clear that there are only five independent material constants (E1. and G12 = G13. A material is called transversely isotropic if its behavior in the 2direction is identical to its behavior in the 3-direction. G13) – This function calculates the 6 × 6 stiffness matrix for orthotropic materials. ν23. E3. TransverselyIsotropicStiffness(E1. we have anisotropic materials – these materials have nonzero entries at the upper right and lower left portions of their compliance and stiffness matrices. and G13. NU23. OrthotropicStiffness(E1.At the other end of the spectrum. E3. G23. ν13. ν23. G23. Their inputs are the nine independent material constants E1. and G12. NU13. NU) – This function calculates the 6 × 6 compliance matrix for isotropic materials. E3. NU) – This function calculates the 6×6 stiffness matrix for isotropic materials. IsotropicStiffness(E. and G13. ν12. ν12. G12. G12) – This function calculates the 6×6 compliance matrix for transversely isotropic materials. G23. Their inputs are the two independent material constants E and ν. ν23. E2. . E2. ν23. E2. Their inputs are the two independent material constants E and ν. E2. ν12. NU12. G12. G12. G12) – This function calculates the 6×6 stiffness matrix for transversely isotropic materials. TransverselyIsotropicCompliance(E1. NU12. ν12.ν13. E2. and G12. G23. Their inputs are the nine independent material constants E1. E2. NU13. E3. NU23. G13) –This function calculates the 6×6 compliance matrix for orthotropic materials. NU23. E2. E2. IsotropicCompliance(E. Its input are the five independent material constants E1. NU12. MATLAB Functions Used The six MATLAB functions used in this chapter to calculate compliance and stiffness matrices are: OrthotropicCompliance(E1. G12. Its input are the five independent material constants E1. NU23. NU12. the Hooke’s law stress–strain relationships at a point in an x–y–z orthogonal system (Figure 2.9) in matrix form are And .For a linear isotropic material in a three-dimensional stress state. in one plane of material symmetry* (Figure 2. for example. Monoclinic material: If.11). then the stiffness matrix reduces to . direction 3 is normal to the plane of material symmetry.Anisotropic Material The material that has 21 independent elastic constants at a point is called an anisotropic material. The cube will deform in all directions as determined by the normal strain equations. showing that the element will not change shape in those planes Orthotropic Material (Orthogonally Anisotropic)/Specially Orthotropic If a material has three mutually perpendicular planes of material symmetry.And compliance matrix. then the stiffness matrix is given by . The shear strains in the 2–3 and 3–1 plane are zero. This is a commonly found material symmetry unlike anisotropic and monoclinic materials. This is unlike the monoclinic material. showing that the element will not change shape in those planes. .Three mutually perpendicular planes of material symmetry also imply three mutually perpendicular planes of elastic symmetry. 2–3. The shear strains in all three planes (1–2. and 3–1) are zero. The cube will not deform in shape under any normal load applied in the principal directions. Note that nine independent elastic constants are present. in which two out of the six faces of the cube changed shape. 0.0971 0 0 0 0 0 0.0015 -0.1395 0 0 0 .3333 0 0 0 0 0 0 0.0015 -0.2273 0 0 0 0 0 0 0.0055 -0.0583 0.0971 -0.0583 0 0 -0.0015 0.0015 0 0 -0. NU12=NU23. sigma1=input('enter the value of stress in The 1-axis aligned with the fiber direction. NU23=input('enter the value of Poisson`s ratio'). ').tau23. '). disp(' '). . E1=E3. disp(' '). disp(' '). G23=G12. sigma2=input('enter the value of stress in The 2-axis in the plane of the layer and perpendicular to the fibers. S = [1/E1 -NU12/E1 -NU13/E1 0 0 0 . disp(' ').').0 0 0 0 0 1/G12] C= inv(S) input('Enter the value of stresses: ' ).Extra code: elseif q==2 E1=input('enter the value of modulus of elasticity'). tau12=input('enter the value of shear stress in the 1-2 plane'). E2=E3. NU13=NU23. disp(' '). G12=E1/2*(1+NU12). disp(' '). sigma3=input('enter the value of stress in The 3-axis perpendicular to the plane of the layer and thus also perpendicular to the fibers.'). disp(' ').sigma3. tau13=input('enter the value of shear stress in the 1–3 plane'). E1=E2. disp(' '). stress_matrix=[sigma1. 0 0 0 0 1/G13 0 . disp(' '). tau23=input('enter the value of shear stress in the 2–3 plane'). 0 0 0 1/G23 0 0 . G13=G23. -NU12/E1 1/E2 -NU23/E2 0 0 0 . sigma2=input('enter the value of stress in The 2-axis in the plane of the layer and perpendicular to the fibers.tau12] strain_matrix=S*stress_matrix elseif q==3 disp('Enter the value of stresses: ' ).sigma2.'). sigma1=input('enter the value of stress in The 1-axis aligned with the fiber direction.tau13. double(S). NU23=NU13. disp('\').-NU13/E1 NU23/E2 1/E3 0 0 0 . G12=G13.00*10^9 % E1=input('enter the value of modulus of elasticity IN LONGITUDINAL DIRECTION').0 0 0 0 0 1/G12] C= inv(S) stress_matrix=[sigma1. disp(' ').tau12] strain_matrix=S*stress_matrix else disp('selected material type is wrong') end . tau13=input('enter the value of shear stress in the 1–3 plane '). G23=E2/2*(1+NU23).sigma3. 0 0 0 0 1/G13 0 . %disp(' '). %NU13=input('enter the value of Poisson`s ratio').0*10^9 %G13 = 7. double(S) S = [1/E1 -NU12/E1 -NU13/E1 0 0 0 . disp(' ').27 G12 = 7. disp(' '). tau12=input('enter the value of shear stress in the 1-2 plane ').-NU13/E1 NU23/E2 1/E3 0 0 0 .sigma2. -NU12/E1 1/E2 -NU23/E2 0 0 0 . '). 0 0 0 1/G23 0 0 . disp(' '). disp(' ').3*10^9 E3 = 10. E2=E3.tau13. tau23=input('enter the value of shear stress in the 2–3 plane ').17*10^9 %G23 = 3.tau23.60 % NU13 = 0. NU12=NU13.3*10^9 NU12 = 0. E1 = 181*10^9 E2 = 10.sigma3=input('enter the value of stress in The 3-axis perpendicular to the plane of the layer and thus also perpendicular to the fibers.28 % NU23 = 0. [ % S11= (S_mo(1.3))*((sin(theta))^3)*(cos(theta))(2*S_mo(2.3))*(sin(theta)*((cos(theta))^3))(2*S_mo(2. relates the shearing stress in the x–y plane to the normal strain in direction-x. Later.2)+S_mo(3.3))*((sin(theta))^2)*((cos(theta))^2)) + S_mo(2.2)-4*S_mo(1.2)-S_mo(3.3))*((sin(theta))^2)*((cos(theta))^2)+ S_mo(3.2)*(sin(theta))^4) %S12=S_mo(1.2)*((sin(theta))^4+(cos(theta))^4)+(S_mo(1.1)-2*S_mo(1.2)-S_mo(3.2)-2*S_mo(1. S12 S22 S23.2)-S_mo(3.2)-2*S_mo(1.2)S_mo(3. mx.1)*((cos(theta))^4) + ((2*S_mo(1. interaction also occurs between the shear strain and the normal stresses.3))*((sin(theta))*(cos(theta))^3) %S33= 2*(2*S_mo(1.3))*((sin(theta))^3)*(cos(theta)) %S23= (2*S_mo(1.3))*((sin(theta))^2)*((cos(theta))^2) + S_mo(2. This is called shear coupling.2)-S_mo(3. unlike in a unidirectional lamina.3)*((sin(theta))^4 + (cos(theta))^4) % S_bar=[S11 S12 S13.1)-2*S_mo(1. . The shear coupling term that relates the normal stress in the x-direction to the shear strain is denoted by mx and is defined as Note that mx is a nondimensional parameter like the Poisson’s ratio.1)+2*S_mo(2.2)*(cos(theta))^4 %S13= (2*S_mo(1.3))*(sin(theta))^2*(cos(theta))^2 %S22=S_mo(1. S13 S23 S33] ] In an angle lamina.1)*(sin(theta))^4 + (2*S_mo(1. note that the same parameter.2) + S_mo(3.2)S_mo(3.1)+S_mo(2. respectively. %disp(' '). This will result in bending moments and shear forces at the clamped ends. it will not allow shearing strain to occur. % sigma3=input('enter the value of stress in The 3-axis perpendicular to the plane of the layer and thus also perpendicular to the fibers. %disp(' ').2° and θ = 53.78°. % tau23=input('enter the value of shear stress in the 2–3 plane'). % disp(' ').The shear coupling term is particularly important in tensile testing of angle plies. if an angle lamina is clamped at the two ends.'). % sigma2=input('enter the value of stress in The 2-axis in the plane of the layer and perpendicular to the fibers. . % tau12=input('enter the value of shear stress in the 1-2 plane'). For example. %tau13=input('enter the value of shear stress in the 1–3 plane').'). %disp(' '). % disp(' ').29. The values of these coefficients are quite extreme. For input of stress values: %disp('Enter the value of stresses: ' ). but cannot be ignored in angle plies. % disp(' '). the shear coupling coefficients mx and my are maximum at θ = 36.28 and Figure 2. showing that the normalshear coupling terms have a stronger effect than the Poisson’s effect. %sigma1=input('enter the value of stress in The 1-axis aligned with the fiber direction.').5 In Figure 2. This phenomenon of shear coupling terms is missing in isotropic materials and unidirectional plies.