Micro Mechanical Analysis of a Lamina: Introduction, Evaluation of the four elastic moduli by Rule of mixture, Numerical problems. Macro Mechanics of a Lamina: Hooke's law for different types of materials, Number of elastic constants, Two - dimensional relationship of compliance and stiffness matrix.

BEHAVIOUR OF UNIDIRECTIONAL COMPOSITESElastic properties of unidirectional continuous fiber laminaLongitudinal strength and stiffness:Considering the mechanics of materials approach to analyze the composite, the following assumptions are used: Fibres are uniformly distributed throughtout the matrix Fibres have uniform properties and diameter Perfect bonding exist between the fibres and matrix The matrix is free of voids, the fibres and matrix are linearly elastic The load applied is either parallel to or normal to the fibre directionConsder a unidirectional laminate as shown in figure 3.1 subjected to a tensile load Pc. Let Af = Area of c/s of fibre, f = Strain, Ef= Youngs Modulus, f= density, Vf= volume fraction, Wf= weight fraction, Pf= load sheared by the fibre respectively.Similarly Am, m , Em ,m ,Vm ,Wm ,Pm be the respective propertives of matrix and Ac, c ,Ec ,c ,Vc ,Wc, Pc be the composite materials.

Figure 3.1: Unidirectional laminae

For this model, according to strength of materials approach the strain experienced by the fibre, matrix and composite materials is same along its length, i.e. constant strain model. i.e. c= f = m (1)and total load of the composites is sheared by both the matrix and fibre materials. i.e. Pc= Pm + Pf (2)W.K.TPc= c. AcPf= f. AfPm= m. AmEq 2 impliesc. Ac = f. Af + m. Amc = f. (Af / Ac )+ m. (Am / Ac)(3)But for composites with parallel fibres the volume fractions are equal to area fraction such that Vf= (Af / Ac ) and Vm= (Am / Ac )Therefore, c = f. Vf + m. VmDifferentiating the above equation wrt to strain (d c/d)= (d f/ d) Vf + (d m/ d) Vm(4)Where (d /d) represents the slope of corresponding stress- strain curve at the given strain.Since the material assumed are elasticity linear, the slope of (d /d) are constants and are called elastic modules. (d /d)=ETherefore, Ec= Ef Vf + Em Vm(5)And also Ec = ElThe equation 3 and 5 indicate that the contribute of the fibres and the matrix to the average properties are proportional to their volume fractions. Such a relationship is called rule of mixtures. The equation 3 and 5 can be generalized by c = Ec = Load shearing/ fraction of load carried by fibres in axial directions. c = f. Vf + m. VmBut Vf +Vm = 1Then Vm = 1 - Vfc = f. Vf + m (1 - Vf)Similarly, Ec = Ef Vf + Em VmEc = Ef Vf + Em (1 - Vf)Ec = Em + Vf (Ef - Em )(Pf/Pc)= ((f. Vf) / (f. Vf + m. Vm))(Pf/Pc)= ((f. Vf) / (f. Vf + m. (1 - Vf)))(Pf/Pc)= ((Ef. Vf) / (Ef. Vf + Em(1-Vf))

Fig 3.2 : Longitudinal stress strain diagram for longitudinal composite with linear elastic materialTRANSVERSE YOUNGS MODULUSTransverse Strength and stiffness A simple mathematical model may be constructed for studying the transverse properties which is as shown in figure. The composite is stressed in transverse direction i.e. the direction perpendicular to axis of the fibres. Each layer of matrix material and fibre is perpendicular to the direction of loading and has the same area on which the load acts. It is clear that each layer will carry the same load and experiences equal stresses. i.e. f = c = m(1)Pc= Pm = PfEach layer is also assumed uniform in thickness; this volume fraction will be proportional to respective thickness fraction.Further the total elongation of the composite structure is the sum of elongation of the layer of fibre and matrix materials.Shear Modulus of an unidirectional laminaeThe in plane shear modulus of an unidirectional composites may be derived by the same model used for transverse module in the previous section. The model with the shear loading is as shown in figure a and b. The shearing stress on the fiber and matrix are equal. Let f, m, c be the shear stress in fibre, matrix and composites respectively. Let c , f, m be the shear deformation. Therefore f =m =c (1)The total shear deformation of the composite (c) is the sum of shear deformation of the fibres f and the matrix (m)c = f + m (2)Shear deformation in each of the material can be written as the product and corresponding shear strain () and the cumulative thickness of the material.

Let c , f and m be the shear strains of composite, fiber and matrix materials. i.e. c = c tcf = f tfm = m tmSubstituting these in equation 2c tc = f tf + m tmDivide by tcc = f (tf/ tc) + m (tm/ tc)Then c = f Vf + m Vm (3)As per Hookes law for a linear elastic material shear stress is proportional to shear strain. Then,G = (/) or = (/G)Substituting this in 3, (c / Gc) = (f / Gf)*Vf + (m / Gm)*VmAnd from equation 1, f =m =c (1/ Gc) = (Vf / Gf) + (Vm / Gm)Or Gc = Gf Gm / (Gm Vf + GfVm)

Derivation for Poissons ratioFor inplane loading of a unidirectional composite two Poissons ratio are defined 1) Major Poisson ratio: which relates the longitudinal stress and transverse strain and is denoted by Lt2) Minor Poisson ratio: which relates the transverse stress and the longitudinal strain (tl)The major Poisson ratio can be derived from the model as shown in figure. The load is applied parallel to the fibers.Transverse strains in the fiber, matrix and composite can be written in terms of longitudinal strains & the Poisson ratio as follows:FIGURE

Let (T)f,( T)m,(T)c,(c)f,(L)m and (L)c be the transverse and longitudinal strains of fiber, matrix and composite materials respectively.

i.e. (T)f = - f (L)f(T)m = - m (L)m(T)c = - Lt (L)cWhere f, m, Lt are Poissons ratio of fiber, matrix and composite respectively.Let c , f and m be the transverse deformation of composites, fibre and matrix.The transverse deformation can be written as the product of strain and cumulative thickness.f = tf (L)f = - tf (L)f fm = tm (L)m= - tm (m)f mc= tc(L)c = - tc (L)c cThe total deformation of the composite is the sum of deformation of the fibers and matrix. Therefore, c = f + m (1)Substituting above equation in 1i.e. - tc (L)c c = - tm (m)f m = - tf (L)f f (2)Since the longitudinal strains in the fiber, matrix and composite due to the longitudinal stress are equal.i.e. (L)c = (m)f = (L)f The equation (2) becomes tc Lt = tf f + tmmDivide throughout by tcLt = (tf/ tc) f + (tm/ tc )mas the fibers are parallel, thickness ratio are equal to volume ratiosi.e. Lt = Vf f + Vm mThis is the rule of mixture for the Major Poisson ratio of a unidirectional composites. The Minor Poisson ratio can be obtained from the knowledge of values for EL, ET, and Lt. Derivation of the following relations between four elastic constants will be (Lt/ EL) = (Tl/ ET)

Analysis of an orthographic laminaA single layer of a laminated composite material is generally referred to has a ply or lamina. It usually contains a single layer of reinforcement, unidirectional or ulti directional sereral laminates are bonded together to form a structure term be a laminate.The properties of a laminate may be predicted by knowing the properties of its constituent laminae. Thus analysis or design of a laminate requires a complete knowledge of the behavior and the laminae, microscopicallty fiber composites are heterogeneous materials. Their properties and behavior are controlled by their microstructure and the properties of the constituent.From the mechanics stand point fiber composites are among the class of materials called orthographic materials whose behavior lies between that of isotropic and that of anisotropic materials. The difference these materials explain through the response to tensile and shear loading.Consider rectangular specimens made up of isotropic, anisotropic and orthotropic materials.

Fig: (a) Isotropic (b) Special orthotropic (c) OrthotropicIsotropic materialsA uniaxial T-load will produce on elongated in the load direction and shortening in the perpendicular direction. And the adjacent sides remains parallel. A pure shear load will produce distorsion of the specimen through equal changes in the angle between adjacent sides but will cause no change in length when the direction of load is changed the material response does not change.i.e. Equal loads applied in different directions produce equal changes in length and angles. Thus the deformation behavior of isotropic material is direction independent and is characterized by normal stresses produce normal strain only but no shear strain.Anisotropic materials Uniaxial tension will produce changes in length as well as in angles. Similarly pure shear will also produce changes in length as well as angles. Further when the direction of applied load is changed the material response does not change.i.e equal loads applied in different direction produce unequal changes in length and angles The deformation behavior of anistropic material is direction dependent.Orthotropic materials:Deformations response of on orthotropic material, in general is similar to that of the anisotropic materials. i.e. it is direction dependent normal stresses and shear stresses are alike. However in special cases when the loads are applied in some specific directions the material response is similar to that of isotropic materials in that the normal stress produce normal strain only and shear stress produce shear strain only. These directions with special behavior are the axis of symmetry of the material. In a unidirectional composite, longitudinal and transverse directions are the axis of symmetry. The geral 3D orthographic material has 3 mutually perpendicular axes of symmetry. A unidirectional composites is an orthographic material but has more than 3 axes of symmetry. The longitudinal direction and all directions perpendicular to it. Because of these additional axes of symmetry the unidirectional composites are transversely isotropic.

Volume and weight fractionsOne of the most important factors determining the properties of composites is the relative proportions of the matrix and reinforcing materials. The relative proportions can be given as the weight fractions or volume fractions. All the experimental method uses weight fraction. However the volume fraction are exclusively used in the theoretical analysis of composite material.Consider a composite consisting of fiber and matrix. Let vc,,f,,m = volume of composite, fiber, and matrix, respectivelyc,f,m = density of composite, fiber, and matrix, respectively.Fiber volume fractionvf and the matrix volume fraction vm asVf =( vf/vc) and Vm =( vm/vc)The sum of volume fractions is Vf + Vm =1