Micromechanical Analysis of a Lamina 213 Wc Pa-Pa (3.15) Poel Pa Substituting Equation (3.13)and Equation (3.15)in Equation (3.11),the volume fraction of the voids is (3.16) =Pa-Pac Pa Example 3.2 A graphite/epoxy cuboid specimen with voids has dimensions of a x b x c and its mass is M After it is put it into a mixture of sulfuric acid and hydrogen peroxide,the remaining graphite fibers have a mass M.From independent tests,the densities of graphite and epoxy are pand P respectively.Find the volume fraction of the voids in terms of a,b,c,MoM Po and P Solution The total volume of the composite v is the sum total of the volume of fiber Ur,matrix v,and voids v: 0e=Ur+0m+0。 (3.17) From the definition of density, L 0- (3.18a) P M.-Mr 0m= (3.18b) Pm The specimen is a cuboid,so the volume of the composite is Ve=abc. (3.19) Substituting Equation(3.18)and Equation(3.19)in Equation(3.17)gives 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 213 (3.15) Substituting Equation (3.13) and Equation (3.15) in Equation (3.11), the volume fraction of the voids is (3.16) Example 3.2 A graphite/epoxy cuboid specimen with voids has dimensions of a × b × c and its mass is Mc. After it is put it into a mixture of sulfuric acid and hydrogen peroxide, the remaining graphite fibers have a mass Mf . From independent tests, the densities of graphite and epoxy are ρf and ρm, respectively. Find the volume fraction of the voids in terms of a, b, c, Mf , Mc, ρf , and ρm. Solution The total volume of the composite vc is the sum total of the volume of fiber vf , matrix vm, and voids vv: (3.17) From the definition of density, (3.18a) (3.18b) The specimen is a cuboid, so the volume of the composite is (3.19) Substituting Equation (3.18) and Equation (3.19) in Equation (3.17) gives v w v c ce ct ce ct = ⎛ − ⎝ ⎜ ⎞ ⎠ ⎟ ρ ρ ρ ρ . V v v v v c ct ce ct = = ρ ρ − ρ . vvv v c f mv =++ . v M f f f = ρ , v M M m c f m = − ρ . v abc c = . 1343_book.fm Page 213 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
214 Mechanics of Composite Materials,Second Edition _MLM.-ML+ abc P Pm and the volume fraction of voids then is ==1-1M+M-M (3.20) abc abc Ps Alternative Solution The preceding problem can also be solved by using Equation(3.16).The theoretical density of the composite is Pa=P/V+Pm(1-V), (3.21) where V/is the theoretical fiber volume fraction given as volume of fibers volume of fibers+volume of matrix M Pr Vi-ML,M-M (3.22) Pr Pim The experimental density of the composite is Pee= M (3.23) abc Substituting Equation(3.21)through Equation(3.23)in the definition of void volume fractions given by Equation (3.16), (3.24) Experimental determination:the fiber volume fractions of the constituents of a composite are found generally by the burn or the acid digestion tests.These tests involve taking a sample of composite and weighing it.Then the density 2006 by Taylor Francis Group,LLC
214 Mechanics of Composite Materials, Second Edition , and the volume fraction of voids then is (3.20) Alternative Solution The preceding problem can also be solved by using Equation (3.16). The theoretical density of the composite is , (3.21) where Vf ′ is the theoretical fiber volume fraction given as (3.22) The experimental density of the composite is (3.23) Substituting Equation (3.21) through Equation (3.23) in the definition of void volume fractions given by Equation (3.16), (3.24) Experimental determination: the fiber volume fractions of the constituents of a composite are found generally by the burn or the acid digestion tests. These tests involve taking a sample of composite and weighing it. Then the density abc M MM v f f c f m = + v − + ρ ρ V v abc abc M MM v v f f c f m = =− + ⎡ − ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 1 ρ ρ ρρ ρ ct f f m f = V V ′ + − ( ) 1 ′ ′ = + V volume of fibers volume of fibers volume f of matrix ′ = + − V M f M MM f f f f c f m ρ ρ ρ . ρce Mc abc = . V abc M MM v f f c f m =− + ⎡ − ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 1 ρ ρ . 1343_book.fm Page 214 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 215 of the specimen is found by the liquid displacement method in which the sample is weighed in air and then in water.The density of the composite is given by P:= 0一Pw' (3.25) 0e-0: where w.=weight of composite w;=weight of composite when immersed in water P=density of water (1000 kg/m3 or 62.4 1b/ft3) For specimens that float in water,a sinker is attached.The density of the composite is then found by Pe=- Pw (3.26) 0e+0,-0m where w.=weight of composite w;=weight of sinker when immersed in water w=weight of sinker and specimen when immersed in water The sample is then dissolved in an acid solution or burned.2 Glass-based composites are burned,and carbon and aramid-based composites are digested in solutions.Carbon and aramid-based composites cannot be burned because carbon oxidizes in air above 300C(572F)and the aramid fiber can decompose at high temperatures.Epoxy-based composites can be digested by nitric acid or a hot mixture of ethylene glycol and potassium hydroxide;polyamide-and phenolic resin-based composites use mixtures of sulfuric acid and hydrogen peroxide.When digestion or burning is complete, the remaining fibers are washed and dried several times and then weighed. The fiber and matrix weight fractions can be found using Equation(3.2).The densities of the fiber and the matrix are known;thus,one can use Equation (3.4)to determine the volume fraction of the constituents of the composite and Equation(3.8)to calculate the theoretical density of the composite. 3.3 Evaluation of the Four Elastic Moduli As shown in Section 2.4.3,there are four elastic moduli of a unidirectional lamina: 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 215 of the specimen is found by the liquid displacement method in which the sample is weighed in air and then in water. The density of the composite is given by , (3.25) where wc = weight of composite wi = weight of composite when immersed in water ρw = density of water (1000 kg/m3 or 62.4 lb/ft3) For specimens that float in water, a sinker is attached. The density of the composite is then found by , (3.26) where wc = weight of composite ws = weight of sinker when immersed in water ww = weight of sinker and specimen when immersed in water The sample is then dissolved in an acid solution or burned.2 Glass-based composites are burned, and carbon and aramid-based composites are digested in solutions. Carbon and aramid-based composites cannot be burned because carbon oxidizes in air above 300°C (572°F) and the aramid fiber can decompose at high temperatures. Epoxy-based composites can be digested by nitric acid or a hot mixture of ethylene glycol and potassium hydroxide; polyamide- and phenolic resin-based composites use mixtures of sulfuric acid and hydrogen peroxide. When digestion or burning is complete, the remaining fibers are washed and dried several times and then weighed. The fiber and matrix weight fractions can be found using Equation (3.2). The densities of the fiber and the matrix are known; thus, one can use Equation (3.4) to determine the volume fraction of the constituents of the composite and Equation (3.8) to calculate the theoretical density of the composite. 3.3 Evaluation of the Four Elastic Moduli As shown in Section 2.4.3, there are four elastic moduli of a unidirectional lamina: ρ ρ c c c i w w w w = − ρ ρ c c csw w w www = + − 1343_book.fm Page 215 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
216 Mechanics of Composite Materials,Second Edition Longitudinal Young's modulus,E Transverse Young's modulus,E2 Major Poisson's ratio,V12 In-plane shear modulus,G12 Three approaches for determining the four elastic moduli are discussed next. 3.3.1 Strength of Materials Approach From a unidirectional lamina,take a representative volume element*that consists of the fiber surrounded by the matrix(Figure 3.3).This representa- tive volume element(RVE)can be further represented as rectangular blocks. The fiber,matrix,and the composite are assumed to be of the same width, h,but of thicknesses frfm,and f,respectively.The area of the fiber is given by Ar=tih. (3.27a) The area of the matrix is given by Am =tmh (3.27b) and the area of the composite is given by A=th. (3.27c) The two areas are chosen in the proportion of their volume fractions so that the fiber volume fraction is defined as V= A A (3.28a) and the matrix fiber volume fraction V is A representative volume element(RVE)of a material is the smallest part of the material that represents the material as a whole.It could be otherwise intractable to account for the distribu- tion of the constituents of the material. 2006 by Taylor Francis Group,LLC
216 Mechanics of Composite Materials, Second Edition • Longitudinal Young’s modulus, E1 • Transverse Young’s modulus, E2 • Major Poisson’s ratio, ν12 • In-plane shear modulus, G12 Three approaches for determining the four elastic moduli are discussed next. 3.3.1 Strength of Materials Approach From a unidirectional lamina, take a representative volume element* that consists of the fiber surrounded by the matrix (Figure 3.3). This representative volume element (RVE) can be further represented as rectangular blocks. The fiber, matrix, and the composite are assumed to be of the same width, h, but of thicknesses tf, tm, and tc, respectively. The area of the fiber is given by . (3.27a) The area of the matrix is given by (3.27b) and the area of the composite is given by (3.27c) The two areas are chosen in the proportion of their volume fractions so that the fiber volume fraction is defined as (3.28a) and the matrix fiber volume fraction Vm is * A representative volume element (RVE) of a material is the smallest part of the material that represents the material as a whole. It could be otherwise intractable to account for the distribution of the constituents of the material. A th f f = A th m m = , A th c c = . V A A t t f f c f c = = , 1343_book.fm Page 216 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Micromechanical Analysis of a Lamina 217 3 2 ◆1 0 0 O 0 0 t.2」 t12 FIGURE 3.3 Representative volume element of a unidirectional lamina. Vm= A. (3.28b) te =1-V The following assumptions are made in the strength of materials approach model: The bond between fibers and matrix is perfect. The elastic moduli,diameters,and space between fibers are uniform. The fibers are continuous and parallel. 2006 by Taylor Francis Group,LLC
Micromechanical Analysis of a Lamina 217 (3.28b) The following assumptions are made in the strength of materials approach model: • The bond between fibers and matrix is perfect. • The elastic moduli, diameters, and space between fibers are uniform. • The fibers are continuous and parallel. FIGURE 3.3 Representative volume element of a unidirectional lamina. 3 Lc t m/2 t r t m/2 t c t c h h 2 1 V A A t t V m m c m c f = = = −1 . 1343_book.fm Page 217 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC