EXAMPLE 3.4 The transverse modulus is roughly estimated by equation(3.36): modulus of matrix. inverse rule of mixtures: 0482 H0.82X10mP61(6.65GP E:-82X10000.506-+05X10000.482-16.43x100P81118GPg Solution.The longitudinal modulus is given by equation(3.23) maximum possible values of Er and E2? moduli of the composite.Given these fiber and matrix materials,what are the GPa),and Em=0.5x 106 psi(3.45 GPa).Estimate the longitudinal and transverse 3.2 have the properties En=32.0 x 106 psi(220 GPa),Ep=2.0 x 106 psi(13.79 The constituent materials in the composite described in example 3.1 and example tions will be discussed in the following sections. better equations for estimating the in-plane shear modulus.Such equa- are in fact not equal.As with the transverse modulus,we need to find that used in section 3.2.2 can be used here to show that the shear stresses stresses are not equal as assumed.A strain energy approach similar to As we might expect,this equation is not very accurate because the shear where Ga2 is the shear modulus of fiber in the 12 plane and Gm=shear assumption of equal shear stresses in fibers and matrix,leads to another is,geometric compatibility of the shear deformations,along with the approach similar to that which was used for the transverse modulus.That An equation for the in-plane shear modulus can be derived using an Y2=2E the average engineering shear strain in the 12 plane. where oc2 is the average composite shear stress.in the 12 plane and The effective in-plane shear modulus is defined as(fig.3.5[d]) metric compatibility relationships leading to the solution are valid. design purposes.As in the case for the longitudinal modulus,the geo- Equation(3.41)is generally accepted as being sufficiently accurate for Principles of Composite Material Mechanics 943 3.3) EXAMPLE 3.5 based on invalid assumptions.and agreement with experimental resuilts is models for E2 and Gi2 are of questionable value,however,because they are models for E and vi2 are good enough for design use.The corresponding As shown in the previous section,the elementary mechanics of materials Improved Mechanics of Materials Models to the ratio of fiber volume fraction to the matrix volume fraction. to matrix stress is ratio would increase dramatically to 624,however,since it is proportional same since it is independent of the fiber volume fraction.The strain energy possible fiber volume fraction of 0.907,the stress ratio would remain the that is almost the same as the stress ratio.If the composite had the maximum EmEm! 3200500-00 0.5(0.482) and equation(3.22),the ratio of fiber strain energy to matrix strain energy is higher than the matrix modulus.From equation 3.25(b),equation 3.25(c) Thus,the fiber carries most of the stress since the fiber modulus is always 820-640 Solution.From equation(3.20)and equation(3.22),the ratio of fiber stress with that stored in the matrix. in the fiber and matrix materials.Compare the strain energy stored in the fibers For longitudinal loading of the composites in example 3.4,compare the stresses stiffness-to-weight ratios than conventional metallic structural materials. about 20%of the density of steel.Composites typically have much greater the same as the modulus of steel,but the density of the composite is only the longitudinal modulus of the graphite/epoxy composite is now about reinforcement is usually necessary in practical applications.Note also that content,the transverse modulus is still very low.Thus,some transverse E2=1.56 x 106 psi (10.75 GPa).Note that even with this maximum fiber properties are still highly anisotropic,with E1=29x 106 psi(200 GPa)and a close-packed triangular array (eq.[3.13]),the corresponding composite composite has the theoretical maximum fiber volume fraction of 0.907 for As expected,the composite is highly anisotropic,with E1>>E2.If the Effective Moduli of a Continuous Fiber-Reinforced Lamina 三