ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 14 pps

442 Advanced mechanics of composite materials Fig. 7.51. For this type of loading, N x = 1 2 pR , N y = pR , N xy = 0 where N x and N y are the circumferential and axial stress resultants, respectively, p the internal pressure and R is the cylinder radius. Thus, we have n y = 2 and λ = 1/3. Since N xy = 0, the structure of the laminate is symmetric with respect to the cylinder meridian, and Eqs. (8.12)–(8.14) can be reduced to h = 3pR 2σ 1 (8.16) k  i=1 h i (3 cos 2 φ i −1) = 0 (8.17) Comparing Eq. (8.16) with the corresponding expression for the thickness of a metal pressure vessel, which is h m = pR/σ , we can see that the thickness of an optimal composite vessel is 1.5 times more than h m . Nevertheless, because of their higher strength and lower density, composite pressure vessels are significantly lighter than metal ones. To show this, consider pressure vessels with radius R = 100 mm made of different materials and designed for a burst pressure p = 20 MPa. The results are listed in Table 8.1. As can be seen, the thickness of a glass–epoxy vessel is the same as that for the thickness of a steel vessel, because the factor 1.5 in Eq. (8.16) is compensated by the composite’s strength which is 1.5 times greater than the strength of steel. However, the density of a glass–epoxy composite is much lower than the density of steel, and as a result, the mass of unit surface area of the composite vessel is only 27% of the corresponding value for a steel vessel. The most promising materials for pressure vessels are aramid and carbon composites, which have the highest specific tensile strength (see Table 8.1). Consider Eq. (8.17) which shows that there can exist an infinite number of optimal laminates with one and the same thickness specified by Eq. (8.16). Table 8.1 Parameters of metal and composite pressure vessels. Parameter Material Steel Aluminum Titanium Glass– epoxy Carbon– epoxy Aramid– epoxy Strength, σ, σ 1 (MPa) 1200 500 900 1800 2000 2500 Density, ρ (g/cm 3 ) 7.85 2.7 4.5 2.1 1.55 1.32 Thickness of the vessel, h m ,h(mm) 1.67 4.0 2.22 1.67 1.5 1.2 Mass of the unit surface area, ρh (kg/m 2 ) 13.11 10.8 10.0 3.51 2.32 1.58 Chapter 8. Optimal composite structures 443 The simplest is a cross-ply laminate having k = 2, φ 1 = 0 ◦ , h 1 = h 0 , and φ 2 = 90 ◦ , h 2 = h 90 . For this structure, Eq. (8.17) yields h 90 = 2h 0 . This result seems obvious because N y /N x = 2. For symmetric ±φ angle-ply laminate, we should take k = 2, h 1 = h 2 = h φ /2, φ 1 =+φ, φ 2 =−φ. Then, cos 2 φ = 1 3 ,φ= φ 0 = 54.44 ◦ As a rule, helical plies are combined with circumferential plies as in Fig. 7.51. For this case, k = 3, h 1 = h 2 = h φ /2, φ 1 =−φ 2 = φ, h 3 = h 90 , φ 3 = 90 ◦ , and Eq. (8.17) gives h 90 h φ = 3 cos 2 φ −1 (8.18) Since the thickness cannot be negative, this equation is valid for 0 ≤ φ ≤ φ 0 . For φ 0 ≤ φ ≤ 90 ◦ , the helical layers should be combined with an axial one, i.e., we should put k = 3, h 1 = h 2 = h φ /2, φ 1 =−φ 2 = φ and h 3 = h 0 ,φ 3 = 0 ◦ . Then, h 0 h φ = 1 2 (1 −3cos 2 φ) (8.19) The dependencies corresponding to Eqs. (8.18) and (8.19) are presented in Fig. 8.3. As an example, consider a filament wound pressure vessel whose parameters are listed in Table 6.1. The cylindrical part of the vessel shown in Figs. 4.14 and 6.22 consists of a ±36 ◦ angle-ply helical layer and a circumferential layer whose thicknesses h 1 = h φ and h 2 = h 90 are presented in Table 6.1. The ratio h 90 /h φ for two experimental vessels is 0.97 and 1.01, whereas Eq. (8.18) gives for this case h 90 /h φ = 0.96 which shows that both vessels are close to optimal structures. Laminates reinforced with uniformly stressed fibers can exist under some restrictions imposed on the acting forces N x ,N y , and N xy . Such restrictions follow from Eqs. (8.13) and (8.14) under the conditions that h i ≥ 0, 0 ≤ sin 2 φ i , cos 2 φ i ≤ 1 and have the form 0 ≤ λ ≤ 1, − 1 2 ≤ λn xy ≤ 1 2 In particular, Eqs. (8.13) and (8.14) do not describe the case of pure shear for which only the shear stress resultant, N xy , is not zero. This is quite natural because the strength condition σ (i) 1 = σ 1 under which Eqs. (8.12)–(8.14) are derived is not valid for shear inducing tension and compression in angle-ply layers. To study in-plane shear of the laminate, we should use both solutions of Eq. (8.7) and assume that for some layers, e.g., with i = 1, 2, 3, ,n−1,σ (i) 1 = σ 1 whereas for the other layers (i = n, n +1, n +2, , k), σ (i) 1 =−σ 1 . Then, Eqs. (8.1) can be reduced to 444 Advanced mechanics of composite materials 0 0.4 0.8 1.2 1.6 2 0 153045607590 f 0 h 0 h f f° h 0 h f h 90 h f h 90 h f , Fig. 8.3. Optimal thickness ratios for a cylindrical pressure vessel consisting of ±φ helical plies combined with circumferential (90 ◦ ) or axial (0 ◦ ) plies. the following form N x +N y = σ 1 (h + −h − ) (8.20) N x −N y = σ 1  n−1  i=1 h + i cos 2φ i − k  i=n h − i cos 2φ i  (8.21) N xy = 1 2 σ 1  n−1  i=1 h + i sin 2φ i − k  i=n h − i sin 2φ i  (8.22) where h + = n−1  i=1 h + i ,h − = k  i=n h − i are the total thicknesses of the plies with tensile and compressive stresses in the fibers, respectively. For the case of pure shear (N x = N y = 0), Eqs. (8.20) and (8.21) yield h + = h − and φ i =±45 ◦ . Then, assuming that φ i =+45 ◦ for the layers with h i = h + i , whereas φ i =−45 ◦ for the layers with h i = h − i , we get from Eq. (8.22) h = h + +h − = 2N xy σ 1 Chapter 8. Optimal composite structures 445 The optimal laminate, as follows from the foregoing derivation, corresponds to a ±45 ◦ angle-ply structure as shown in Fig. 8.2b. 8.2. Composite laminates of uniform strength Consider again the panel in Fig. 8.1 and suppose that unidirectional plies or fabric layers, that form the panel are orthotropic, i.e., in contrast to the previous section, we do not now neglect stresses σ 2 and τ 12 in comparison with σ 1 (see Fig. 3.29). Then, the constitutive equations for the panel in a plane stress state are specified by the first three equations in Eqs. (5.35), i.e., N x = B 11 ε x +B 12 ε y +B 14 γ xy N y = B 21 ε x +B 22 ε y +B 24 γ xy N xy = B 41 ε x +B 42 ε y +B 44 γ xy (8.23) where, in accordance with Eqs. (4.72), (5.28), and (5.42) B 11 = k  i=1 h i  E (i) 1 cos 4 φ i +E (i) 2 sin 4 φ i +2E (i) 12 sin 2 φ i cos 2 φ i  B 12 = B 21 = k  i=1 h i  E (i) 1 ν (i) 12 +  E (i) 1 +E (i) 2 −2E (i) 12  sin 2 φ i cos 2 φ i  B 22 = k  i=1 h i  E (i) 1 sin 4 φ i +E (i) 2 cos 4 φ i +2E (i) 12 sin 2 φ i cos 2 φ i  B 14 = B 41 = k  i=1 h i  E (i) 1 cos 2 φ i −E (i) 2 sin 2 φ i −E (i) 12 cos 2φ i  sin φ i cos φ i B 24 = B 42 = k  i=1 h i  E (i) 1 sin 2 φ i −E (i) 2 cos 2 φ i +E (i) 12 cos 2φ i  sin φ i cos φ i B 44 = k  i=1 h i  E (i) 1 +E (i) 2 −2E (i) 1 ν (i) 12  sin 2 φ i cos 2 φ i +G (i) 12 cos 2 2φ i  (8.24) and E (i) 1, 2 = E (i) 1, 2 1 −ν (i) 12 ν (i) 21 , E (i) 12 = E (i) 1 ν (i) 12 +2G (i) 12 . In the general case, the panel can consist of layers made of different composite materials. Using the optimality criterion developed in the previous section for fibrous structures, 446 Advanced mechanics of composite materials we suppose that the fibers in each layer are directed along the lines of principal strains, or principal stresses because τ (i) 12 = G 12 γ (i) 12 for an orthotropic layer and the condition γ (i) 12 = 0 is equivalent to the condition τ (i) 12 = 0 (see Section 2.4). Using the third equation in Eqs. (4.69), we can write these conditions as 2(ε y −ε x ) sin φ i cos φ i +γ xy cos 2φ i = 0 (8.25) This equation can be satisfied for all the layers if we take ε x = ε y = ε, γ xy = 0 (8.26) Then, Eqs. (8.23) yield N x = (B 11 +B 12 )ε, N y = (B 21 +B 22 )ε, N xy = (B 41 +B 42 )ε These equations allow us to find the strain, i.e., ε = N x +N y B 11 +2B 12 +B 22 (8.27) and to write two relationships specifying the optimal structural parameters of the laminate (B 11 +B 12 )N y −(B 21 +B 22 )N x = 0 (B 41 +B 42 )(N x +N y ) −(B 11 +2B 12 +B 22 )N xy = 0 Substitution of B mn from Eqs. (8.24) results in the following explicit form of these conditions k  i=1 h i  E (i) 1  1 +ν (i) 12  N x sin 2 φ i −N y cos 2 φ i  + E (i) 2  1 +ν (i) 21  N x cos 2 φ i −N y sin 2 φ i  = 0 k  i=1 h i   N x +N y   E (i) 1 −E (i) 2  sin φ i cos φ i −N xy  E (i) 1  1 +ν (i) 12  + E (i) 2  1 +ν (i) 21  = 0 (8.28) To determine the stresses that act in the optimal laminate, we use Eqs. (4.69) and (8.26) that specify the strains in the principal material coordinates of the layers as ε 1 = ε 2 = ε, Chapter 8. Optimal composite structures 447 γ 12 = 0. Applying constitutive equations, Eqs. (4.56), substituting ε from Eq. (8.27) and writing the result in explicit form with the aid of Eqs. (8.24), we arrive at σ (i) 1 = E (i) 1 S i  1 +ν (i) 12   N x +N y  σ (i) 2 = E (i) 2 S i  1 +ν (i) 21   N x +N y  τ (i) 12 = 0 (8.29) where S i = k  i=1 h i  E (i) 1  1 +ν (i) 12  + E (i) 2  1 +ν (i) 21  is the laminate stiffness coefficient. If all the layers are made from the same material, Eqs. (8.28) and (8.29) are simplified as k  i=1 h i  N x sin 2 φ i −N y cos 2 φ i +n  N x cos 2 φ i −N y sin 2 φ i  = 0 k  i=1 h i  m  N x +N y  sin φ i cos φ i − ( 1 +n ) N xy  = 0 (8.30) σ (i) 1 = σ 1 = N x +N y h(1 +n) ,σ (i) 2 = σ 2 = n(N x +N y ) h(1 +n) ,τ (i) 12 = 0 (8.31) in which n = E 2 (1 +ν 21 ) E 1 (1 +ν 12 ) ,m= E 1 −E 2 E 1 (1 +ν 12 ) ,h= k  i=1 h i Laminates of uniform strength exist under the following restrictions n 1 +n ≤ N x N x +N y ≤ 1 1 +n ,     N xy N x +N y     ≤ 1 −n 2(1 +n) For the monotropic model of a unidirectional ply considered in the previous section, n = 0, m = 1, and Eqs. (8.30) reduce to Eqs. (8.9) and (8.10). To determine the thickness of the optimal laminate, we should use Eqs. (8.31) in con- junction with one of the strength criteria discussed in Chapter 6. For the simplest case, 448 Advanced mechanics of composite materials using the maximum stress criterion in Eqs. (6.2), the thickness of the laminate can be found from the following conditions σ 1 = σ 1 or σ 2 = σ 2 , so that h 1 = N x +N y (1 +n)σ 1 ,h 2 = n(N x +N y ) (1 +n)σ 2 (8.32) Obviously, for the optimal structure, we would like to have h 1 = h 2 . However, this can happen only if material characteristics meet the following condition σ 2 σ 1 = n = E 2 (1 +ν 21 ) E 1 (1 +ν 12 ) (8.33) The results of calculations for typical materials whose properties are listed in Tables 3.5 and 4.4 are presented in Table 8.2. As can be seen, Eq. (8.33) is approximately valid for fabric composites whose stiffness and strength in the warp and fill directions (see Section 4.6) are controlled by fibers of the same type. However, for unidirectional polymeric and metal matrix composites, whose longitudinal stiffness and strength are governed by the fibers and transverse characteristics are determined by the matrix properties, σ 2 /σ 1  n. In accordance with Eqs. (8.32), this means that h 1  h 2 , and the ratio h 2 /h 1 varies from 12.7 for glass–epoxy to 2.04 for boron–epoxy composites. Now, return to the discussion presented in Section 4.4.2 from which it follows that in laminated composites, transverse stresses σ 2 reaching their ultimate value, σ 2 , cause cracks in the matrix, which do not result in failure of the laminate whose strength is controlled by the fibers. To describe the laminate with cracks in the matrix (naturally, if cracks are allowable for the structure under design), we can use the monotropic model of the ply and, hence, the results of optimization are presented in Section 8.1. Consider again the optimality condition Eq. (8.25). As can be seen, this equation can be satisfied not only by strains in Eqs. (8.26), but also if we take tan 2φ i = γ xy ε x −ε y (8.34) Since the left-hand side of this equation is a periodic function with period π, Eq. (8.34) determines two angles, i.e., φ 1 = φ = 1 2 tan −1 γ xy ε x −ε y ,φ 2 = π 2 +φ (8.35) Table 8.2 Parameters of typical advanced composites. Parameter Fabric–epoxy composites Unidirectional-epoxy composites Boron–A1 Glass Carbon Aramid Glass Carbon Aramid Boron σ 2 /σ 1 0.99 0.99 0.83 0.022 0.025 0.012 0.054 0.108 n 0.85 1.0 1.0 0.28 0.1 0.072 0.11 0.7 Chapter 8. Optimal composite structures 449 Thus, the optimal laminate consists of two layers, and the fibers in both layers are directed along the lines of principal stresses. Suppose that the layers are made of the same com- posite material and have the same thickness, i.e., h 1 = h 2 = h/2, where h is the thickness of the laminate. Then, using Eqs. (8.24) and (8.35), we can show that B 11 = B 22 and B 24 =−B 14 for this laminate. After some transformation involving elimination of γ 0 xy from the first two equations of Eqs. (8.23) with the aid of Eq. (8.34) and similar transfor- mation of the third equation from which ε 0 x and ε 0 y are eliminated using again Eq. (8.34), we get N x = (B 11 +B 14 tan 2φ)ε 0 x +(B 12 −B 14 tan 2φ)ε 0 y N y = (B 12 −B 14 tan 2φ)ε 0 x +(B 11 +B 14 tan 2φ)ε 0 y N xy = (B 44 +B 14 cot 2φ)γ 0 xy Upon substitution of coefficients B mn from Eqs. (8.24) we arrive at N x = h 2   E 1 +E 2  ε 0 x +  E 1 ν 12 +E 2 ν 21  ε 0 y  N y = h 2   E 1 ν 12 +E 2 ν 21  ε 0 x +  E 1 +E 2  ε 0 y  N xy = h 4  E 1 ( 1 −ν 12 ) + E 2 ( 1 −ν 21 )  γ 0 xy Introducing average stresses σ x = N x /h, σ y = N y /h, and τ xy = N xy /h and solving these equations for strains, we have ε 0 x = 1 E (σ x −νσ y ), ε 0 y = 1 E (σ y −νσ x ), γ 0 xy = τ xy G (8.36) where E = 1 2(E 1 +E 2 )  2E 1 E 2 + E 2 1  1 −ν 2 12  +E 2 2  1 −ν 2 21  1 −ν 12 ν 21  ν = E 1 ν 12 +E 2 ν 21 E 1 +E 2 ,G= E 2(1 +ν) (8.37) Changing strains for stresses in Eqs. (8.35), we can write the expression for the optimal orientation angle as φ = 1 2 tan −1 2τ xy σ x −σ y (8.38) It follows from Eqs. (8.36) that a laminate consisting of two layers reinforced along the directions of principal stresses behaves like an isotropic layer, and Eqs. (8.37) specify the 450 Advanced mechanics of composite materials Table 8.3 Effective elastic constants of an optimal laminate. Property Glass– epoxy Carbon– epoxy Aramid– epoxy Boron– epoxy Boron– Al Carbon– carbon Al 2 O 3 – Al Elastic modulus, E (GPa) 36.975.950.3114.8 201.195.2 205.4 Poisson’s ratio, ν 0.053 0.039 0.035 0.035 0.21 0.06 0.176 elastic constants of the corresponding isotropic material. For typical advanced composites, these constants are listed in Table 8.3 (the properties of unidirectional plies are taken from Table 3.5). Comparing the elastic moduli of the optimal laminates with those for quasi-isotropic materials (see Table 5.1), we can see that for polymeric composites the characteristics of the first group of materials are about 40% higher than those for the second group. However, it should be emphasized that whereas the properties of quasi- isotropic laminates are universal material constants, the optimal laminates demonstrate characteristics shown in Table 8.3 only if the orientation angles of the fibers are found from Eqs. (8.35) or (8.38) and correspond to a particular distribution of stresses σ x ,σ y , and τ xy . As follows from Table 8.3, the modulus of a carbon–epoxy laminate is close to the modulus of aluminum, whereas the density of the composite material is lower by a factor of 1.7. This is the theoretical weight saving factor that can be expected if we change from aluminum to carbon–epoxy composite in a thin-walled structure. Since the stiffness of both materials is approximately the same, to find the optimal orientation angles of the structure elements, we can substitute in Eq. (8.38) the stresses acting in the aluminum prototype structure. A composite structure designed in this way will have approximately the same stiffness as the prototype structure and, as a rule, higher strength because carbon composites are stronger than aluminum alloys. To evaluate the strength of the optimal laminate, we should substitute strains from Eqs. (8.36) into Eqs. (4.69) and thence these strains in the principal material coordinates of the layers – into constitutive equations, Eqs. (4.56), that specify the stresses σ 1 and σ 2 (τ 12 = 0) acting in the layers. Applying the appropriate failure criterion (see Chapter 6), we can evaluate the laminate strength. Comparing Tables 1.1 and 8.3, we can see that boron–epoxy optimal laminates have approximately the same stiffness as titanium (but is lighter by a factor of about 2). Boron– aluminum can be used to replace steel with a weight saving factor of about 3. For preliminary evaluation, we can use a monotropic model of unidirectional plies neglecting the stiffness and load-carrying capacity of the matrix. Then, Eqs. (8.37) take the following simple form E = E 1 2 ,ν= 0,G= E 1 4 (8.39) As an example, consider an aluminum shear web with thickness h = 2 mm, elastic constants E a = 72 GPa, ν a = 0.3 and density ρ a = 2.7g/cm 3 . This panel is loaded with Chapter 8. Optimal composite structures 451 shear stress τ. Its shear stiffness is B a 44 = 57.6 GPa ·mm and the mass of a unit surface is m a = 5.4kg/m 2 . For the composite panel, taking σ x = σ y = 0 in Eq. (8.38) we have φ = 45 ◦ . Thus, the composite panel consists of +45 ◦ and −45 ◦ unidirectional layers of the same thickness. The total thickness of the laminate is h = 2 mm, i.e., the same as for an aluminum panel. Substituting E 1 = 140 GPa and taking into account that ρ = 1.55 g/cm 3 for a carbon–epoxy composite which is chosen to substitute for aluminum we get B c 44 = 70 GPa · mm and m c = 3.1kg/m 3 . The stresses acting in the fiber directions of the composite plies are σ c 1 =±2τ . Thus, the composite panel has a 21.5% higher stiffness and its mass is only 57.4% of the mass of a metal panel. The composite panel also has higher strength because the longitudinal strength of unidirectional carbon–epoxy composite under tension and compression is more than twice the shear strength of aluminum. The potential performance of the composite structure under discussion can be enhanced if we use different materials in the layers with angles φ 1 and φ 2 specified by Eqs. (8.35). According to the derivation of Obraztsov and Vasiliev (1989), the ratio of the layers’ thicknesses is h 2 h 1 = E (1) 1 −E (1) 2 E (2) 1 −E (2) 2 and the elastic constants in Eqs. (8.37) are generalized as E = E 1 −ν 2 = E (1) 1 E (2) 1 −E (1) 2 E (2) 2 E (1) 1 +E (2) 1 −E (1) 2 −E (2) 2 ν = E (1) 1 E (2) 1  ν (1) 12 +ν (2) 12  − E (1) 2 E (2) 2  ν (1) 21 +ν (2) 21  E (1) 1 E (2) 1 −E (1) 2 E (2) 2 Superscripts 1 and 2 correspond to layers with orientation angles φ 1 and φ 2 , respectively. 8.3. Application to optimal composite structures As stated in the introduction to this chapter, there exists special composite structures for which the combination of the specific properties of modern composites with the appropriate design concepts and potential of composite technology provide a major improvement of these structures in comparison with the corresponding metal prototypes. Three such special structures, i.e., geodesic filament-wound pressure vessels, composite flywheels, and an anisogrid lattice structure are described in this section. 8.3.1. Composite pressure vessels As the first example of the application of the foregoing results, consider filament-wound membrane shells of revolution, that are widely used as pressure vessels, solid propellant [...]... Optimal composite structures z/R n = 1.9 n=2 n = 1.8 n = 1.6 n = 1.4 n = 1.2 n=1 n = 0.8 n = 0.4 n=0 r/R Fig 8.5 Meridians of optimal composite shells 457 458 Advanced mechanics of composite materials Substituting h(r) from Eq (8.58) in Eq (8.51), we arrive at the following equation for the tape orientation angle r dφ sin φ[n − (1 − n) cos2 φ] =1 · dr cos φ[1 − (1 − n) cos2 φ] The solution of this... transformed into a disk, the resin in the composite material is cured The maximum kinetic energy that can be stored and the mass of the disk are R E = πω2 ρ 0 hr 3 dr, R M = 2πρ hrdr 0 Chapter 8 Optimal composite structures 469 y x Fig 8.15 Fiber patterns in the spinning optimal composite disk Fig 8.16 Carbon–epoxy flywheel 470 Advanced mechanics of composite materials Substituting h in accordance with... thickness (the height of the rib cross section), h, • the angle of helical ribs with respect to the shell meridian, φ, 472 Advanced mechanics of composite materials Helical ribs dh dc 2 dc 2 ah 2f Hoop ribs ac lh Fig 8.19 Lattice structure • the widths of the helical and the circumferential (hoop) ribs cross sections, δh and δc (for the structure in Fig 8.19, δc is the total width of the adjacent hoop...452 Advanced mechanics of composite materials z T r0 −f +f r a p r R b Fig 8.4 Axisymmetrically loaded membrane shell of revolution rocket motor cases, tanks for gases and liquids, etc (see Figs 4 .14 and 7.51) The shell is loaded with uniform internal pressure p and axial forces T uniformly distributed along the contour of the shell cross section r = r0 as in Fig... of the shells whose meridians are presented in Fig 8.9 are determined by Eq (8.50) in which we should take n = 0 Substituting h from Eq (8.58), z/R 0.6 r0=0 r0=0.1 r0=0.2 r0=0.3 r0=0.4 r0=0.5 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 r/R Fig 8.9 Meridians of isotensoids corresponding to various normalized radii of the polar openings r 0 = r0 /R 464 Advanced mechanics of composite materials. .. To find the fiber trajectory, consider Fig 8 .14 showing the tape element in Cartesian x, y and polar r, β coordinate frames As follows from the figure, x = r sin β, y = r cos β, tan φ = rdβ dr (8.82) 468 Advanced mechanics of composite materials y x f rdb dr r b db y x Fig 8 .14 A tape element in Cartesian and polar coordinate frames Applying the last equation of Eqs (8.82) and using Eq (8.81) for φ, we... the previous section the optimal shell is reinforced along the lines of principal stresses, i.e., in such a way that 454 Advanced mechanics of composite materials τ12 = 0 In accordance with the last equation of Eqs (8.43), for such a shell γ12 = 0 and, as follows from Eqs (8.44), εα = εβ = ε1 = ε2 Putting τ12 = 0 in the last equation of Eqs (8.45), we can conclude that for the optimal shell Nβ 1 − (1... Maximum values of the circumferential velocities for fibrous composite disks of uniform strength Composite material Glass–epoxy Carbon–epoxy Aramid–epoxy v R (m/s) 1309 1606 1946 Chapter 8 Optimal composite structures 471 Fig 8.17 Winding of a composite lattice structure Fig 8.18 Removal of elastic coating After curing, elastic coating is pulled out of the structure as shown in Fig 8.18 Cylindrical anisogrid... n) 1 , 1−λ 2 456 Advanced mechanics of composite materials Here, Bx is the β-function (or the Euler integral of the first type) The constant of integration is found from the condition z(r = R) = 0 Meridians corresponding to various n-numbers are presented in Fig 8.5 For n = 1 the optimal shell is a sphere, whereas for n = 2 it is a cylinder As follows from Eq (8.52), the thickness of the spherical (n... get σh = P 2πDhδ h c2 (8.91) 474 Advanced mechanics of composite materials lh lh (b) (a) Fig 8.20 Local buckling of helical ribs Thus, the strength constraint σh ≤ σ , in which σ is the strength of helical rib under compression, can be written as P ≤ 2πDhδ c c2 σ (8.92) Helical ribs can experience local buckling under compression which shows itself as a local bending of the rib segments between the . can consist of layers made of different composite materials. Using the optimality criterion developed in the previous section for fibrous structures, 446 Advanced mechanics of composite materials we. strength of steel. However, the density of a glass–epoxy composite is much lower than the density of steel, and as a result, the mass of unit surface area of the composite vessel is only 27% of the. thickness of the optimal laminate, we should use Eqs. (8.31) in con- junction with one of the strength criteria discussed in Chapter 6. For the simplest case, 448 Advanced mechanics of composite materials using