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Unformatted text preview: Homework 1 Anirban Chaudhuri 1. Using the stiffness representation models discussed in Section 1.2.2, show that the transverse stiffness, E 2 , of a fiber-reinforced composite material is given by Eq. (1.2.6). Solution: Eq. 1.2.6 is 2 1 f m f m V V E E E = + where, V f and V m are the volume fractions and E f and E m are Youngs Moduli of fiber and matrix material respectively. Fig 1: Rule of mixtures model of a transverse stiffness of a composite layer (Fig 1.5 in the book) In this case the stress on the fiber and matrix material is the same which gives us f m f m P P P A A A = = (1) where, P is the total force and A is the cross sectional area. The indices f and m represent the respective quantities for fiber and matrix respectively. If each constituent is assumed to be an axial bar then force displacement relation gives PL AE = . In this case the length is proportional to the volume fractions. , f f m m L V L L V L = = (2) It can also be seen from the figure 1 that f m = + . or, 2 f f m m f f m m P L P L PL AE A E A E = + 2 f m f m L L L E E E = + [from (1)] 2 f m f m V L V L L E E E = + [from (2)] or, 2 1 f m f m V V E E E = + which is the same as equation 1.2.6. 2. Assuming the packing of the fibers to be represented by the unit volume shown in Fig. 1.8, plot the longitudinal and transverse specific stiffness, E 1 / and E 2 / , and the weight density of the composite, , as a function of the fiber radius (0.0 r f r f max ) for a unidirectional fiber- reinforced Graphite/Epoxy composite. Solution: Graphite properties: E f = 230GPa, f = 17.2 kN/m 3 Epoxy properties: E m = 3.45GPa, m = 12.0 kN/m 3 Fig. 2: Unit volume r f is the fiber radius. Fiber volume fraction, 2 2 f f f A V r A = = ...(1) Specific moduli in the longitudinal direction and weight density can be found by using equation1.2.2....
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- Spring '08