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Chapter12 hw

# Chapter12 hw - Chapter 12 Problem 20 A unidirectional...

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Unformatted text preview: Chapter 12, Problem 20 A unidirectional carbon-ﬁber—epoxy-resin composite contains 68 percent by volume of carbon ﬁber and 32 percent epoxy resin. The density of the carbon ﬁber is 1.79 g."cl:n3 and that of the epoxy resin is 1.20 gfcrng. (a) What are the weight percentages of carbon ﬁbers and epoxy resin in the composite? (b) What is the average density of the composite? Chapter 12, Solution 20 Applying a basis of 1 cm3 of composite material, we have 0.68 cm3 of carbon ﬁber and 0.32 cm3 of epoxy resin. (a) The weight percentages are based upon the individual and total mass of the components. Mass of carbon ﬁber = mcf = ,0”.ch = (1.79 g/cm3)(0.68 cms) =1.217 g Mass of epoxy resin = mgr = perVgr = (1.20 g/cm3)(0.32 cm3) = 0.384 g Total mass = 1.601 g Since we are dealing with a ratio of weights, the gravity terms cancel. Thus, (1.217 g) (1.601 g) (0.384 g) (1.601 g) Wt % carbon ﬁbers = X 100% = 76.0% Wt % carbon ﬁbers = X100% = 24.09/11 (b) The average density of the composite is: 1.601 g _ 3 3 — 1.60 gz'cm p_E_ c V 1cm Chapter 12, Problem 21 The average density of a carbon-ﬁber-epoxy composite is 1.615 g/cmg. The density of the epoxy resin is 1.21 gl’crn3 and that of the carbon ﬁbers is 1.74 gfcmg. (a) What is the volume percentage of carbon ﬁbers in the composite? (b) What are the weight percentages of epoxy resin and carbon ﬁbers in the composite? Chapter 12, Solution 21 Using a basis ofl cms, the total mass ofthe composite is m = pV = (1.615 g/crn3)(l cm3) =1.615 g. (a) The volume percentages can be derived from the density equation. p =pch'cf-FperVigr lief-Flier 1.74 cus +1.21 cm3V 1_615g/cm3=( g/ M; (3 g/ )5.r 1cm 1.615 g =(1.74g/cm3)I/;f +(1.21 g/cm3)Vgr But ch = 1— P2,. Substituting, 1.615 g =(1.74g/cm3)(1—Vg,)+(1.21 g/cm3)V€r 0.125 =0.5317;r V2, = 0.236 and 1;). = 0.764 (b) The weight percentages are calculated based upon the masses. Mass of carbon ﬁber = mg, = pchCf = (1.74 g/cm3)(0.764 cm3) = 1.329 g Mass ofepoxy resin = m —mcf = 1.615 g —1.329 g = 0.286 g (1.329 g) (1.615 g) (0.286 g) (1.615 g) Wt % carbon ﬁbers = X100% = 82.39/31 Wt % carbon ﬁbers = x100% = 17.7%} Chapter 12, Problem 23 Calculate the tensile modulus of elasticity of a unidirectional carbon-ﬁber—reinforced—plastic composite material that contains 64 percent by volume of carbon ﬁbers and is stressed under isostrain conditions. The carbon ﬁbers have a tensile modulus of elasticity of 54.0 X 10‘5 psi and the epoxy matrix a tensile modulus of elasticity of 0.530 X 10‘5 psi. Chapter 12, Solution 23 E6 = 13fo +5me = (54.0><106 psi)(0.64)+(0.53><106 psi)(0.36) = 34.75 X 10‘5 psi = 239.6 GPa Chapter 12, Problem 24 If the tensile strength of the carbon ﬁbers of the 64 percent carbon-ﬁber—epoxy composite material of Prob. 12.23 is 0.31 X 106 psi and that of the epoxy resin is 9.20 X 103 psi, calculate the strength of the composite material in psi. \Vhat ﬁaction of the load is carried by the carbon ﬁbers? Chapter 12, Solution 24 The strength of the composite material, using 7/, = 1 as a basis, is 0'ch = 0'fo + 0'me a, (1) = (0.31x106 psi)(0.64) + (9.2cux103 psi)(0.36) a, = 2.02 x 105 psi = 1.39 GPa The fraction of the load carried by the carbon ﬁbers is: Pf = 13fo = (54.0mm6 psi)(0.64) P £ch (34.75x1o6 psi)(l.0) C = 0.995 Chapter 12, Problem 25 Calculate the tensile modulus of elasticity of a unidirectional Kevlar 49-ﬁber—epoxy composite material that contains 63 percent by volume of Kevlar 49 ﬁbers and is stressed under isostrain conditions. The Kevlar 49 ﬁbers have a tensile modulus of elasticity of 27. 5 X 10‘5 psi and the epoxy matrix a tensile modulus of elasticity of 0.550 X 10IS psi. Chapter 12, Solution 25 E, = .13fo +3me = (27.5x106 psi)(0.63)+(0.55><106 psi)(o.37) = 17.53 X 106 psi = 120-9 GPa Chapter 12, Problem 26 If the tensile strength of the Kevlar 49 ﬁbers is 0.550 X 106 psi and that of the epoxy resin is 11.0 X 103 psi, calculate the strength of the composite material of Prob. 11.25. What fraction of the load is carried by the Kevlar 49 ﬁbers? Chapter 12, Solution 26 The strength of the composite material, using V, = 1 as a basis, is 0ch = 0'fo + (:me crc (1) = (0.55x106 psi)(0.63)+(11.0><106 psi)(0.37) o, = 4.42 x 10‘ psi = 30.48 GPa The fraction of the load carried by the Kevlar 49 ﬁbers is P, _ :3fo 2 (275x106 psi)(0.63) _ 20.988 P, E617; (17.53X106psi)(1.0) Chapter 12, Problem 83 A11 MMC is made with an A1 2024 alloy with 20 vol percent SiC whiskers. If the density of the composite is 2.90 gj'cm3 and that of the SiC ﬁbers is 3.10 gfems, what must the density of the A1 2024 alloy be? Chapter 12, Solution 83 pmmp = fSiCpSic +fmpm where fSiC = 0-20: fAl =0-80 2.90 g/cm3 = (0.20)( 3.10 g/cnzna3)+0.8pAl pAl = 2.85 gfcms ...
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