162 Advanced mechanics of composite materials As an example, consider a boron–aluminum unidirectional composite whose experi- mental stress–strain diagrams (Herakovich, 1998) are shown in Fig. 4.17 (circles) along with the corresponding approximations (solid lines) plotted with the aid of Eqs. (4.65). 4.3. Unidirectional anisotropic layer Consider now a unidirectional layer studied in the previous section and assume that its principal material axis 1 makes some angle φ with the x-axis of the global coordinate frame (see Fig. 4.18). An example of such a layer is shown in Fig. 4.19. 4.3.1. Linear elastic model Constitutive equations of the layer under study referred to the principal material coor- dinates are given by Eqs. (4.55) and (4.56). We need now to derive such equations for the global coordinate frame x, y, and z (see Fig. 4.18). To do this, we should transfer stresses σ 1 , σ 2 , τ 12 , τ 13 , τ 23 acting in the layer and the corresponding strains ε 1 , ε 2 , γ 12 , γ 13 , γ 23 into stress and strain components σ x , σ y , τ xy , τ xz , τ yz and ε x , ε y , γ xy , γ xz , γ yz using Eqs. (2.8), (2.9) and (2.21), (2.27) for coordinate transformation of stresses and strains. According to Fig. 4.18, the directional cosines, Eqs. (2.1), for this transformation are (we take x  = 1, y  = 2, z  = 3) l x 1 x = c, l x 1 y = s, l x 1 z = 0 l y 1 x =−s, l y 1 y = c, l y 1 z = 0 l z 1 x = 0,l z 1 y = 0,l z 1 z = 1 (4.66) where c = cos φ and s = sin φ. Using Eqs. (2.8) and (2.9), we get σ 1 = σ x c 2 +σ y s 2 +2τ xy cs σ 2 = σ x s 2 +σ y c 2 −2τ xy cs τ 12 = (σ y −σ x )cs +τ xy  c 2 −s 2  τ 13 = τ xz c +τ yz s τ 12 =−τ xz s +τ yz c (4.67) The inverse form of these equations is σ x = σ 1 c 2 +σ 2 s 2 −2τ 12 cs σ y = σ 1 s 2 +σ 2 c 2 +2τ 12 cs Chapter 4. Mechanics of a composite layer 163 −300 −250 −200 −150 −100 −50 50 100 150 −2 −1.8 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 (a) (b) 0 012345678 20 40 60 80 100 120 140 t 12 , MPa g 12 , % e 2 , % s 2 , MPa Fig. 4.17. Calculated (solid lines) and experimental (circles) stress–strain diagrams for a boron–aluminum composite under transverse loading (a) and in-plane shear (b). 164 Advanced mechanics of composite materials t 23 t yz t xz t xy t 13 s 1 s y s x s 2 t 12 f 2 3 1 z y x Fig. 4.18. A composite layer consisting of a system of unidirectional plies with the same orientation. Fig. 4.19. An anisotropic outer layer of a composite pressure vessel. Courtesy of CRISM. τ xy = (σ 1 −σ 2 )cs +τ 12  c 2 −s 2  (4.68) τ xz = τ 13 c −τ 23 s τ yz = τ 13 s +τ 23 c The corresponding transformation for strains follows from Eqs. (2.21) and (2.27), i.e., ε 1 = ε x c 2 +ε y s 2 +γ xy cs ε 2 = ε x s 2 +ε y c 2 −γ xy cs Chapter 4. Mechanics of a composite layer 165 γ 12 = 2(ε y −ε x )cs +γ xy  c 2 −s 2  (4.69) γ 13 = γ xz c +γ yz s γ 23 =−γ xz s +γ yz c or ε x = ε 1 c 2 +ε 2 s 2 −γ 12 cs ε y = ε 1 s 2 +ε 2 c 2 +γ 12 cs γ xy = 2(ε 1 −ε 2 )cs +γ 12  c 2 −s 2  γ xz = γ 13 c −γ 23 s γ yz = γ 13 s +γ 23 c (4.70) To derive constitutive equations for an anisotropic unidirectional layer, we substitute strains, Eqs. (4.69), into Hooke’s law, Eqs. (4.56), and the derived stresses – into Eqs. (4.68). The final result is as follows σ x = A 11 ε x +A 12 ε y +A 14 γ xy σ y = A 21 ε x +A 22 ε y +A 24 γ xy τ xy = A 41 ε x +A 42 ε y +A 44 γ xy τ xz = A 55 γ xz +A 56 γ yz τ yz = A 65 γ xz +A 66 γ yz (4.71) The stiffness coefficients in these equations are A 11 = E 1 c 4 +E 2 s 4 +2E 12 c 2 s 2 A 12 = A 21 = E 1 ν 12 +(E 1 +E 2 −2E 12 )c 2 s 2 A 14 = A 41 =  E 1 c 2 −E 2 s 2 −E 12  c 2 −s 2   cs A 22 = E 1 s 4 +E 2 c 4 +2E 12 c 2 s 2 A 24 = A 42 =  E 1 s 2 −E 2 c 2 +E 12  c 2 −s 2   cs (4.72) A 44 = (E 1 +E 2 −2E 1 ν 12 )c 2 s 2 +G 12  c 2 −s 2  2 A 55 = G 13 c 2 +G 23 s 2 A 56 = A 65 = (G 13 −G 23 )cs A 66 = G 13 s 2 +G 23 c 2 166 Advanced mechanics of composite materials where E 1, 2 = E 1, 2 1 −ν 12 ν 21 ,E 12 = E 1 ν 12 +2G 12 ,c= cos φ, s = sin φ The dependence of stiffness coefficients A mn in Eqs. (4.72) on φ has been studied by Tsai and Pagano (see, e.g., Tsai, 1987; Verchery, 1999). Changing the powers of sin φ and cos φ in Eqs. (4.72) for multiple-angle trigonometric functions, we can reduce these equations to the following form (Verchery, 1999) A 11 = S 1 +S 2 +2S 3 cos 2φ +S 4 cos 4φ A 12 =−S 1 +S 2 −S 4 cos 4φ A 14 = S 3 sin 2φ +S 4 sin 4φ A 22 = S 1 +S 2 − 2S 3 cos 2φ +S 4 cos 4φ A 24 = S 3 sin 2φ −S 4 sin 4φ A 44 = S 1 −S 4 cos 4φ A 55 = S 5 +S 6 cos 2φ A 56 = 4S 6 sin 2φ A 66 = S 5 −S 6 cos 2φ (4.73) In these equations, S 1 = 1 8  A 0 11 +A 0 22 −2A 0 12 +4A 0 44  S 2 = 1 4  A 0 11 +A 0 22 +2A 0 12  S 3 = 1 4  A 0 11 −A 0 22  S 4 = 1 8  A 0 11 +A 0 22 −2A 0 12 −4A 0 44  S 5 = 1 2  A 0 55 +A 0 66  S 6 = 1 2  A 0 55 −A 0 66  Chapter 4. Mechanics of a composite layer 167 where A 0 n are stiffness coefficients corresponding to φ = 0. It follows from Eqs. (4.72) that, A 0 11 = E 1 ,A 0 12 = E 1 ν 12 ,A 0 14 = A 0 24 = A 0 56 = 0 A 0 22 = E 2 ,A 0 44 = G 12 ,A 0 55 = G 13 ,A 0 66 = G 23 As can be seen in Eqs. (4.73), there exist the following differential relationships between tensile and coupling stiffnesses (Verchery and Gong, 1999) dA 11 dφ =−4A 14 , dA 22 dφ = 4A 24 It can be directly checked that Eqs. (4.73) provide three invariant stiffness characteristics whose forms do not depend on φ, i.e., A 11 (φ) +A 22 (φ) +2A 12 (φ) = A 0 11 +A 0 22 +2A 0 12 A 44 (φ) −A 12 (φ) = A 0 44 −A 0 12 A 55 (φ) +A 66 (φ) = A 0 55 +A 0 66 (4.74) Any linear combination of these equations is also an invariant combination of stiffness coefficients. The inverse form of Eqs. (4.71) can be obtained if we substitute stresses, Eqs. (4.67), into Hooke’s law, Eqs. (4.55), and the derived strains in Eqs. (4.70). As a result, we arrive at the following particular form of Eqs. (2.48) and (2.49) ε x = σ x E x −ν xy σ y E y +η x,xy τ xy G xy ,ε y = σ y E y −ν yx σ x E x +η y,xy τ xy G xy γ xy = τ xy G xy +η xy, x σ x E x +η xy, y σ y E y ,γ xz = τ xz G xz +λ xz,yz τ yz G yz γ yz = τ yz G yz +λ yz,xz τ xz G xz (4.75) in which the compliance coefficients are 1 E x = c 4 E 1 + s 4 E 2 +  1 G 12 − 2ν 21 E 1  c 2 s 2 ν xy E y = ν yx E x = ν 21 E 1 −  1 E 1 + 1 E 2 + ν 21 E 1 − 1 G 12  c 2 s 2 η x,xy G xy = η xy,x E x =2  c 2 E 1 − s 2 E 2 −  1 2G 12 − ν 21 E 1   c 2 −s 2   cs 168 Advanced mechanics of composite materials 1 E y = s 4 E 1 + c 4 E 2 +  1 G 12 − 2ν 21 E 1  c 2 s 2 (4.76) η y,xy G xy = η xy,y E y =2  s 2 E 1 − c 2 E 2 +  1 2G 12 − ν 21 E 1   c 2 −s 2   cs 1 G xy =4  1 E 1 + 1 E 2 + 2ν 21 E 1  c 2 s 2 + 1 G 12  c 2 −s 2  2 1 G xz = c 2 G 13 + s 2 G 23 , λ xz,yz G yz = λ yz,xz G xz =  1 G 13 − 1 G 23  cs, 1 G yz = s 2 G 13 + c 2 G 23 There exist the following dependencies between the coefficients of Eqs. (4.71) and (4.75) 1 E x = 1 D 1  A 22 A 44 −A 2 24  , ν xy E y = ν yx E x = 1 D 1 (A 12 A 44 −A 14 A 24 ) η x,xy G xy = η xy,x E x = 1 D 1 ( A 12 A 24 −A 22 A 14 ) , 1 E y = 1 D 1  A 11 A 44 −A 2 14  η y,xy G xy = η xy,y E y = 1 D 1 (A 12 A 14 −A 11 A 24 ), 1 G xy = 1 D 1  A 11 A 22 −A 2 12  1 G xz = A 66 D 2 , 1 G yz = A 55 D 2 , λ xz,yz G yz = λ yz,xz G xz =− A 56 D 2 Here, D 1 = A 11 A 22 A 44 −A 11 A 2 24 −A 22 A 2 14 −A 44 A 2 12 +2A 12 A 14 A 24 D 2 = A 55 A 66 −A 2 56 and A 11 = 1 −η y,xy η xy, y D 3 E y G xy ,A 12 = ν xy −η x,xy η xy, y D 3 E y G xy A 14 =− η x,xy +ν xy η y,xy D 3 E y G xy ,A 22 = 1 −η x,xy η xy, y D 3 E x G xy A 24 =− η y,xy +ν yx η x,xy D 3 E x G xy ,A 44 = 1 −ν xy ν yx D 3 E x E y A 55 = 1 D 4 G yz ,A 56 =− λ xz,yz D 4 G yz ,A 66 = 1 D 4 G xz Chapter 4. Mechanics of a composite layer 169 where D 3 = 1 E x E y G xy (1 −ν xy ν yx −η x,xy η xy, x −η y,xy η xy, y −ν xy η y,xy η xy, x −ν yx η x,xy η xy, y ) D 4 = 1 G xz G yz (1 −λ xz,yz λ yz,xz ) As can be seen in Eqs. (4.71) and (4.75), the layer under study is anisotropic in plane xy because the constitutive equations include shear–extension and shear–shear coupling coefficients η and λ. For φ = 0, the foregoing equations degenerate into Eqs. (4.55) and (4.56) for an orthotropic layer. The dependencies of stiffness coefficients on the orientation angle for a carbon–epoxy composite with properties listed in Table 3.5 are presented in Figs. 4.20 and 4.21. Uniaxial tension of the anisotropic layer (the so-called off-axis test of a unidirectional composite) is often used to determine material characteristics that cannot be found in tests with orthotropic specimens or to evaluate constitutive and failure theories. Such a 0 0 153045607590 20 40 60 80 100 120 140 A mn , GPa A 11 A 22 A 44 f° Fig. 4.20. Dependencies of tensile (A 11 , A 22 ) and shear (A 44 ) stiffnesses of a unidirectional carbon–epoxy layer on the orientation angle. 170 Advanced mechanics of composite materials 0 0 153045607590 10 20 30 40 A 12 A 14 A mn , GPa A 24 f° Fig. 4.21. Dependencies of coupling stiffnesses of a unidirectional carbon–epoxy layer on the orientation angle. test is shown in Fig. 4.22. To study this loading case, we should take σ y = τ xy = 0in Eqs. (4.75). Then, ε x = σ x E x ,ε y =−ν xy σ x E x ,γ xy = η xy, x σ x E x (4.77) As can be seen in these equations, tension in the x-direction is accompanied not only with transverse contraction, as in orthotropic materials, but also with shear. This results in the deformed shape of the sample shown in Fig. 4.23. This shape is natural because the material stiffness in the fiber direction is much higher than that across the fibers. Such an experiment, in cases where it can be performed, allows us to determine the in-plane shear modulus, G 12 in principle material coordinates using a simple tensile test rather than the much more complicated tests described in Section 3.4.3 and shown in Figs. 3.54 and 3.55. Indeed, if we know E x from the tensile test in Fig. 4.23 and find E 1 , E 2 , and ν 21 from tensile tests along and across the fibers (see Sections 3.4.1 and 3.4.2), we can use the first equation of Eqs. (4.76) to determine G 12 = sin 2 φ cos 2 φ ( 1/E x ) −  cos 4 φ/E 1  −  sin 4 φ/E 2  + ( 2ν 21 /E 1 ) sin 2 φ cos 2 φ (4.78) In connection with this, a question arises as to what angle should be substituted into this equation to provide the most accurate result. The answer is given in Fig. 4.24, which displays the strains in principal material coordinates for a carbon–epoxy layer calculated with the aid of Eqs. (4.69) and (4.77). As can be seen in this figure, the most appropriate angle is about 10 ◦ . At this angle, the shear strain γ 12 is much higher than normal strains ε 1 and ε 2 , so that material deformation is associated mainly with shear. An off-axis test Chapter 4. Mechanics of a composite layer 171 Fig. 4.22. An off-axis test. s x s x g g Fig. 4.23. Deformation of a unidirectional layer loaded at an angle to fiber orientation. [...]... 16 12 8 4 e x, % 0 0 2 4 6 Fig 4.30 Calculated (solid line) and experimental (circles) stress–strain diagram for 45◦ off-axis tension of a two-matrix unidirectional composite Chapter 4 Mechanics of a composite layer 183 sx, MPa 16 12 8 4 ex, % 0 0 1 2 3 4 Fig 4.31 Theoretical (solid line) and experimental (dashed line) stress–strain diagrams for 60 ◦ off-axis tension of a two-matrix unidirectional composite. .. =θ +v ∂y ∂x where ( ) = d ( ) /dx and ε is the elongation of the strip axis These strains are related to stresses by Eqs (4.75) which reduce to εx = τxy σx + ηx, xy Ex Gxy γxy = (4. 96) τxy σx + ηxy, x Gxy Ex y s s V a x l M Fig 4. 26 Off-axis tension of a strip fixed at the ends 178 Advanced mechanics of composite materials The inverse form of these equations is σx = B11 εx + B14 γxy , τxy = B41 εx... following parameters: h1 = 0. 365 mm, h2 = 0.735 mm, E1 = 56 GPa, E2 = 17 GPa, 192 Advanced mechanics of composite materials G13 = 5 .6 GPa, G23 = 6. 4 GPa, ν13 = 0.095, ν23 = 0.35, σ + = 25.5 MPa The 2 distributions of stresses normalized to the acting stress σ are presented in Fig 4.38 As can be seen, there is a stress concentration in the longitudinal plies in the vicinity of the crack, whereas the stress... fabric composite with E1 = 26 GPa, E2 = 22 GPa, G12 = 7.2 GPa, ν12 = 0.11, ν21 = 0.13, we have E1 = 11.5 GPa, 2(1 + ν21 ) E2 = 9.9 GPa, and φ0 = 49.13◦ 2(1 + ν12 ) Ex E1 1 0.9 0.8 0.7 0 .6 1 0.5 0.4 3 0.3 2 0.2 0.1 0 f° 0 10 20 30 40 50 f = 54.31° 60 70 80 90 f = 61 .45° Fig 4.25 Dependencies of Ex /E1 on φ for fiberglass (1), aramid (2) and carbon (3) epoxy composites Chapter 4 Mechanics of a composite. .. stresses causing the failure of longitudinal characteristics of typical advanced composites σ (1) x or transverse σ (2) x plies and deformation σ (MPa); ε (%) Glass– epoxy Carbon– epoxy Carbon– PEEC Aramid– epoxy Boron– epoxy Boron– Al Carbon– Carbon Al2 O3 –Al σ (1) x σ (2) x ε1 ε2 ε 1 /ε 2 2 160 69 0 1.43 0.45 3.2 2250 1125 1.5 0.75 2 263 0 590 2 .63 0.2 13.1 1420 840 0 .62 0.37 1 .68 2000 400 0.50 0.1 5 890...172 Advanced mechanics of composite materials 3 e1 / ex , e2 / ex, g12 / ex 2.5 g12 / ex 2 1.5 1 e1/ ex e2 / ex 0.5 f° 0 0 15 30 45 60 75 90 Fig 4.24 Dependencies of normalized strains in the principle material coordinates on the angle of the off-axis test with φ = 10◦ can also be used to evaluate material strength in shear τ 12 (Chamis, 1979) Stresses acting under off-axis tension in... remains constant up to failure of the longitu2 2 dinal plies This means that under transverse tension, a unidirectional ply is in a state of 1 96 Advanced mechanics of composite materials s, MPa 200 150 100 50 0 ex, % 0 0.2 0.4 0 .6 Fig 4.42 Stress–strain diagram for a glass–epoxy cross-ply layer: model tion; 0.8 experiment; theoretical predic- sx2, MPa 30 s2+ 20 10 0 0 0.2 0.4 0 .6 0.8 ex, % Fig 4.43 Stress–strain... = G12 and h0 = h0 , h h90 = h90 h The inverse form of Eqs (4.113) is εx = σy σx − νxy , Ex Ey εy = σy σx − νyx , Ey Ex γxy = τxy Gxy (4.115) where Ex = A11 − νxy A12 = , A11 A2 12 , A22 νyx Ey = A22 − A12 = A22 A2 12 , A11 Gxy = A44 (4.1 16) 1 86 Advanced mechanics of composite materials txz t13 t23 txz t13 txz t23 txz Fig 4.34 Pure transverse shear of a cross-ply layer To determine the transverse shear... φ0 = 61 .45◦ 2(1 + ν12 ) (3) For carbon–epoxy composite with E1 = 140 GPa, E2 = 11 GPa, G12 = 5.5 GPa, ν12 = 0.021, ν21 = 0.27, we have E1 = 55.12 GPa, 2(1 + ν21 ) E2 = 5.39 GPa 2(1 + ν12 ) 1 76 Advanced mechanics of composite materials As can be seen, the conditions in Eqs (4.91) and (4.94) are not satisfied, and angle φ0 does not exist for this material As can be directly checked with the aid of Eqs... Tension of a cross-ply laminate 188 Advanced mechanics of composite materials To simplify the analysis, neglect Poisson’s effect, i.e., taking ν12 = ν21 = 0 Then 0 σx1 = σ1 = σ E1 2(E1 h1 + E2 h2 ) , 0 σx2 = σ2 = σ E2 (4.122) 2(E1 h1 + E2 h2 ) Consider, for example, the case h1 = h2 = 0.5 and find the ultimate stresses corresponding 0 to the failure of longitudinal plies or to the failure of the transverse . (E 1 +E 2 −2E 1 ν 12 )c 2 s 2 +G 12  c 2 −s 2  2 A 55 = G 13 c 2 +G 23 s 2 A 56 = A 65 = (G 13 −G 23 )cs A 66 = G 13 s 2 +G 23 c 2 166 Advanced mechanics of composite materials where E 1, 2 = E 1, 2 1 −ν 12 ν 21 ,E 12 =. equations, S 1 = 1 8  A 0 11 +A 0 22 −2A 0 12 +4A 0 44  S 2 = 1 4  A 0 11 +A 0 22 +2A 0 12  S 3 = 1 4  A 0 11 −A 0 22  S 4 = 1 8  A 0 11 +A 0 22 −2A 0 12 −4A 0 44  S 5 = 1 2  A 0 55 +A 0 66  S 6 = 1 2  A 0 55 −A 0 66  Chapter 4. Mechanics of a composite layer 167 where A 0 n are stiffness coefficients corresponding to. boron–aluminum composite under transverse loading (a) and in-plane shear (b). 164 Advanced mechanics of composite materials t 23 t yz t xz t xy t 13 s 1 s y s x s 2 t 12 f 2 3 1 z y x Fig. 4.18. A composite