ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 12 ppsx

372 Advanced mechanics of composite materials laminate element with the aid of Eqs. (5.3) and (5.14), i.e., ε 0 xT = ∂ u ∂ x ,ε 0 yT = ∂ v ∂ y ,γ 0 xyT = ∂ u ∂ y + ∂ v ∂ x (7.28) κ xT = ∂ θ x ∂ x ,κ yT = ∂ θ y ∂ y ,κ xyT = ∂ θ x ∂ y + ∂ θ y ∂ x (7.29) γ xT = θ x + ∂ w ∂ x ,γ yT = θ y + ∂ w ∂ y (7.30) It follows from Eqs. (7.23) that in the general case, uniform heating of laminates induces, in contrast to homogeneous materials, not only in-plane strains but also changes to the laminate curvatures and twist. Indeed, assume that the laminate is free from edge and surface loads so that forces and moments in the left-hand sides of Eqs. (7.23) are equal to zero. Since the CTE of the layers, in the general case, are different, the thermal terms N T and M T in the right-hand sides of Eqs. (7.23) are not equal to zero even for a uniform temperature field, and these equations enable us to find ε T ,γ T , and κ T specifying the laminate in-plane and out-of-plane deformation. Moreover, using the approach described in Section 5.11, we can conclude that uniform heating of the laminate is accompanied, in the general case, by stresses acting in the layers and between the layers. As an example, consider the four-layered structure of the space telescope described in Section 7.1.1. First, we calculate the stiffness coefficients of the layers, i.e., • for the internal layer of aluminum foil, A (1) 11 = A (1) 22 = E f = 76.92 GPa,A (1) 12 = ν f E f = 23.08 GPa • for the inner skin, A (2) 11 = A (2) 22 = E i x = 34.87 GPa,A (2) 12 = ν i xy E i x = 5.23 GPa • for the lattice layer, A (3) 11 = 2E r δ r a r cos 4 φ r = 14.4 GPa A (3) 22 = 2E r δ r a r sin 4 φ r = 0.25 GPa A (3) 12 = 2E r δ r a r sin 2 φ cos 2 φ = 1.91 GPa • for the external skin, A (4) 11 = E e 1 cos 4 φ e +E e 2 sin 4 φ e +2  E e 1 ν e 12 +2G e 12  sin 2 φ e cos 2 φ e = 99.05 GPa Chapter 7. Environmental, special loading, and manufacturing effects 373 A (4) 22 = E e 1 sin 4 φ e +E e 2 cos 4 φ e +2  E e 1 ν e 12 +2G e 12  sin 2 φ e cos 2 φ e = 13.39 GPa A (4) 12 = E e 1 ν e 12 +  E e 1 +E e 2 −2  E e 1 ν e 12 +2G e 12  sin 2 φ e cos 2 φ e = 13.96 GPa Using Eqs. (7.18), we find the thermal coefficients of the layers (the temperature is uniformly distributed over the laminate thickness)  A T 11  1 =  A T 22  1 = E f α f T = 1715 ·10 −6 T GPa/ ◦ C  A T 11  2 =  A T 22  2 = E i x  1 +ν i xy  α i x T = 32.08 ·10 −6 T GPa/ ◦ C  A T 11  3 = 2E r δ r a r α r cos 2 φ r T = 4.46 ·10 −6 T GPa/ ◦ C  A T 22  3 = 2E r δ r a r α r sin 2 φ r T = 1.06 ·10 −6 T GPa/ ◦ C  A T 11  4 =  E e 1  α e 1 +ν e 12 α e 2  cos 2 φ +E e 2  α e 2 +ν e 21 α e 1  sin 2 φ  T = 132.43 · 10 −6 T GPa/ ◦ C  A T 22  4 =  E e 1  α e 1 +ν e 12 α e 2  sin 2 φ +E e 2  α e 2 +ν e 21 α e 1  cos 2 φ  T = 317.61 · 10 −6 T GPa/ ◦ C Since the layers are orthotropic, A T 12 = 0 for all of them. Specifying the coordinates of the layers (see Fig. 5.10) i.e., t 0 = 0mm,t 1 = 0.02 mm,t 2 = 1.02 mm,t 3 = 10.02 mm,t 4 = 13.52 mm and applying Eq. (7.27), we calculate the parameters J (r) mn for the laminate J (0) 11 =  A T 11  1 ( t 1 −t 0 ) +  A T 11  2 ( t 2 −t 1 ) +  A T 11  3 ( t 3 −t 2 ) +  A T 11  4 ( t 4 −t 3 ) = 570 · 10 −6 T GPa mm/ ◦ C J (0) 22 = 1190 · 10 −6 T GPa mm/ ◦ C J (1) 11 = 1 2  A T 11  1  t 2 1 −t 2 0  +  A T 11  2  t 2 2 −t 2 1  +  A T 11  3  t 2 3 −t 2 2  +  A T 11  4  t 2 4 −t 2 3  = 5690 · 10 −6 T GPa mm/ ◦ C J (1) 22 = 13150 · 10 −6 T GPa mm/ ◦ C 374 Advanced mechanics of composite materials To determine M T mn , we need to specify the reference surface of the laminate. Assume that this surface coincides with the middle surface, i.e., that e = h/2 = 6.76 mm. Then, Eqs. (7.25) yield N T 11 = J (0) 11 = 570 · 10 −6 T GPa mm/ ◦ C N T 22 = J (0) 22 = 1190 · 10 −6 T GPa mm/ ◦ C M T 11 = J (1) 11 −eJ (0) 11 = 1840 · 10 −6 T GPa mm/ ◦ C M T 22 = 5100 · 10 −6 T GPa mm/ ◦ C Thus, the thermal terms entering the constitutive equations of thermoplasticity, Eqs. (7.23), are specified. Using these results, we can determine the apparent coefficients of thermal expansion for the space telescope section under study (see Fig. 7.3). We can assume that, under uniform heating, the curvatures do not change in the middle part of the cylinder so that κ xT = 0 and κ yT = 0. Since there are no external loads, the free body diagram enables us to conclude that N x = 0 and N y = 0. As a result, the first two equations of Eqs. (7.23) for the structure under study become B 11 ε 0 xT +B 12 ε 0 yT = N T 11 B 21 ε 0 xT +B 22 ε 0 yT = N T 22 Solving these equations for thermal strains and taking into account Eqs. (7.20), we get ε 0 xT = 1 B  B 22 N T 11 −B 12 N T 22  = α x T ε 0 yT = 1 B  B 11 N T 22 −B 12 N T 11  = α y T where B = B 11 B 22 −B 2 12 . For the laminate under study, calculation yields α x =−0.94 ·10 −6 1/ ◦ C,α y = 14.7 · 10 −6 1/ ◦ C Return to Eqs. (7.13) and (7.20) based on the assumption that the coefficients of thermal expansion do not depend on temperature. For moderate temperatures, this is a reasonable approximation. This conclusion follows from Fig. 7.6, in which the experimental results of Sukhanov et al. (1990) (shown with solid lines) are compared with Eqs. (7.20), in which T = T − 20 ◦ C (dashed lines) represent carbon–epoxy angle-ply laminates. However, for relatively high temperatures, some deviation from linear behavior can be observed. In this case, Eqs. (7.13) and (7.20) for thermal strains can be generalized as ε T =  T T 0 α(T )dT Chapter 7. Environmental, special loading, and manufacturing effects 375 −30 −20 −10 10 20 30 −100 −50 50 100 ±10° ±10° 0° 0° 90° 90° ±45° ±45° T,°C 10 5 ε T x Fig. 7.6. Experimental dependencies of thermal strains on temperature (solid lines) for ±φ angle-ply carbon– epoxy composite and the corresponding linear approximations (dashed lines). Temperature variations can also result in a change in material mechanical properties. As follows from Fig. 7.7, in which the circles correspond to the experimental data of Ha and Springer (1987), elevated temperatures result in either higher or lower reduction of material strength and stiffness characteristics, depending on whether the corresponding material characteristic is controlled mainly by the fibers or by the matrix. The curves presented in Fig. 7.7 correspond to a carbon–epoxy composite, but they are typical for polymeric unidirectional composites. The longitudinal modulus and tensile strength, being controlled by the fibers, are less sensitive to temperature than longitudinal com- pressive strength, and transverse and shear characteristics. Analogous results for a more temperature-sensitive thermoplastic composite studied by Soutis and Turkmen (1993) are presented in Fig. 7.8. Metal matrix composites demonstrate much higher thermal resis- tance, whereas ceramic and carbon–carbon composites have been specially developed to withstand high temperatures. For example, carbon–carbon fabric composite under heat- ing up to 2500 ◦ C demonstrates only a 7% reduction in tensile strength and about 30% reduction in compressive strength without significant change of stiffness. Analysis of thermoelastic deformation for materials whose stiffness characteristics depend on temperature presents substantial difficulties because thermal strains are caused not only by material thermal expansion, but also by external forces. Consider, for example, a structural element under temperature T 0 loaded with some external force P 0 , and assume that the temperature is increased to a value T 1 . Then, the temperature change will cause a thermal strain associated with material expansion, and the force P 0 , being constant, also induces additional strain because the material stiffness at temperature T 1 is less than its stiffness at temperature T 0 . To determine the final stress and strain state of the structure, 376 Advanced mechanics of composite materials + s 1 − + E 2 E 1 G 12 0 0.2 0.4 0.6 0.8 1 0 50 100 150 200 T,°C t 12 s 2 s 1 Fig. 7.7. Experimental dependencies of normalized stiffness (solid lines) and strength (dashed lines) character- istics of unidirectional carbon–epoxy composite on temperature. 0 0.2 0.4 0.6 0.8 1 0 40 80 120 T,°C E 1 E 2 s − 1 s + 2 s + 1 Fig. 7.8. Experimental dependencies of normalized stiffness (solid lines) and strength (dashed lines) character- istics of unidirectional glass–polypropylene composite on temperature. Chapter 7. Environmental, special loading, and manufacturing effects 377 we should describe the process of loading and heating using, e.g., the method of successive loading (and heating) presented in Section 4.1.2. 7.2. Hygrothermal effects and aging Effects that are similar to temperature variations, i.e., expansion and degradation of properties, can also be caused by moisture. Moisture absorption is governed by Fick’s law, which is analogous to Fourier’s law, Eq. (7.1), for thermal conductivity, i.e., q W =−D ∂ W ∂ n (7.31) in which q W is the diffusion flow through a unit area of surface with normal n, D is the diffusivity of the material whose moisture absorption is being considered, and W is the relative mass moisture concentration in the material, i.e., W = m m (7.32) where m is the increase in the mass of a unit volume material element due to mois- ture absorption and m is the mass of the dry material element. Moisture distribution in the material is governed by the following equation, similar to Eq. (7.2) for thermal conductivity ∂ ∂ n  D ∂ W ∂ n  = ∂ W ∂ t (7.33) Consider a laminated composite material shown in Fig. 7.9 for which n coincides with the z axis. Despite the formal correspondence between Eq. (7.2) for thermal conductivity and Eq. (7.32) for moisture diffusion, there is a difference in principle between these problems. This difference is associated with the diffusivity coefficient D, which is much lower than z W m W m W m h h x x z (b)(a) Fig. 7.9. Composite material exposed to moisture on both surfaces z = 0 and z = h (a), and on the surface z = 0 only (b). 378 Advanced mechanics of composite materials the thermal conductivity λ of the same material. As is known, there are materials, e.g., metals, with relatively high λ and practically zero D coefficients. Low D-value means that moisture diffusion is a rather slow process. As shown by Shen and Springer (1976), the temperature increase in time inside a surface-heated composite material reaches a steady (equilibrium) state temperature about 10 6 times faster than the moisture content approaching the corresponding stable state. This means that, in contrast to Section 7.1.1 in which the steady (time-independent) temperature distribution is studied, we must consider the time-dependent process of moisture diffusion. To simplify the problem, we can neglect the possible variation of the mass diffusion coefficient D over the laminate thickness, taking D = constant for polymeric composites. Then, Eq. (7.33) reduces to D ∂ 2 W ∂ z 2 = ∂ W ∂ t (7.34) Consider the laminate in Fig. 7.9a. Introduce the maximum moisture content W m that can exist in the material under the preassigned environmental conditions. Naturally, W m depends on the material nature and structure, temperature, relative humidity (RH)ofthe gas (e.g., humid air), or on the nature of the liquid (distilled water, salted water, fuel, lubricating oil, etc.) to the action of which the material is exposed. Introduce also the normalized moisture concentration as w(z, t) = W(z,t) W m (7.35) Obviously, for t →∞, we have w → 1. Then, the function w(z, t) can be presented in the form w(z, t) = 1 − ∞  n=1 w n (z)e −k n t (7.36) Substitution into Eq. (7.34), with due regard to Eq. (7.35), yields the following ordinary differential equation w  n +r 2 n w n = 0 in which r 2 n = k n /D and ()  = d()/dz. The general solution is w n = C 1n sin r n z +C 2n cos r n z The integration constants can be found from the boundary conditions on the surfaces z = 0 and z = h (see Fig. 7.9a). Assume that on these surfaces W = W m or w = 1. Then, in accordance with Eq. (7.36), we get w n (0,t)= 0,w n (h, t) = 0 (7.37) Chapter 7. Environmental, special loading, and manufacturing effects 379 The first of these conditions yields C 2n = 0, whereas from the second condition we have sin r n h = 0, which yields r n h = (2n −1)π (n = 1, 2, 3, ) (7.38) Thus, the solution in Eq. (7.36) takes the form w(z, t) = 1 − ∞  n=1 C 1n sin  2n −1 h πz  exp  −  2n −1 h  2 π 2 Dt  (7.39) To determine C 1n , we must use the initial condition, according to which w(0 <z<h,t= 0) = 0 Using the following Fourier series 1 = 4 π ∞  n=1 sin(2n −1)z 2n −1 we get C 1n = 4/(2n −1)π, and the solution in Eq. (7.39) can be written in its final form w(z, t) = 1 − 4 π ∞  n=1 sin(2n −1)πz 2n −1 exp  −  2n −1 h  2 π 2 Dt  (7.40) where z = z/ h. For the structure in Fig. 7.9b, the surface z = h is not exposed to moisture, and hence q W (z = h) = 0. So, in accordance with Eq. (7.31), the second boundary condition in Eqs. (7.37) must be changed to w  (h, t) = 0. Then, instead of Eq. (7.38), we must use r n h = π 2 (2n −1) Comparing this result with Eq. (7.38), we can conclude that for the laminate in Fig. 7.9b, w(z, t) is specified by the solution in Eq. (7.40) in which we must change h to 2h. The mass increase of the material with thickness h is M = A  h 0 mdz where A is the surface area. Using Eqs. (7.32) and (7.35), we get M = AmW m  h 0 wdz 380 Advanced mechanics of composite materials Switching to a dimensionless variable z = z/ h and taking the total moisture content as C = M Amh (7.41) we arrive at C = W m  1 0 w dz where w is specified by Eq. (7.40). Substitution of this equation and integration yields C = C W m = 1 − 8 π 2 ∞  n=1 1 (2n −1) 2 exp  −  2n −1 h  2 π 2 Dt  (7.42) For numerical analysis, consider a carbon–epoxy laminate for which D = 10 −3 mm 2 / hour (Tsai, 1987) and h = 1 mm. The distributions of the moisture concentration over the laminate thickness are shown in Fig. 7.10 for t = 1, 10, 50, 100, 200, and 500 h. As can be seen, complete impregnation of 1-mm-thick material takes about 500 h. The dependence of Con t found in accordance with Eq. (7.42) is presented in Fig. 7.11. An interesting interpretation of the curve in Fig. 7.11 can be noted if we change the vari- able t to √ t. The resulting dependence is shown in Fig. 7.12. As can be seen, the initial 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 w(z,t) z t = 500 hours 200 100 50 10 1 Fig. 7.10. Distribution of the normalized moisture concentration w over the thickness of 1-mm-thick carbon– epoxy composite for various exposure times t. Chapter 7. Environmental, special loading, and manufacturing effects 381 C(t) t, hours 0 0.2 0.4 0.6 0.8 1 0 100 200 300 400 500 Fig. 7.11. Dependence of the normalized moisture concentration C on time t. 0 0.2 0.4 0.6 0.8 1 0 5 10 15 20 25 t)C( hourt, Fig. 7.12. Dependence of the normalized moisture concentration on √ t. part of the curve is close to a straight line whose slope can be used to determine the diffusion coefficient of the material matching the theoretical dependence C(t) with the experimental one. Note that experimental methods usually result in rather approximate evaluation of the material diffusivity D with possible variations up to 100% (Tsai, 1987). The maximum value of the function C(t) to which it tends to approach determines the maximum moisture content C m = W m . [...]... ε1 (p) = ∗ ν ∗ (p) ∗ σ1 (p) − 12 σ (p) ∗ ∗ E1 (p) E2 (p) 2 ∗ ε2 (p) = ∗ ∗ σ2 (p) ν21 (p) ∗ ∗ (p) − E ∗ (p) σ1 (p) E2 1 ∗ 12 (p) = (7.57) ∗ 12 (p) G∗ (p) 12 where ∗ E1 (p) = ∗ 12 (p) E1 , ∗ 1 + C11 (p) ∗ E2 (p) = ∗ 1 + C12 (p) 12 , = ∗ 1 + C22 (p) ∗ ν21 (p) E2 , ∗ 1 + C22 (p) G∗ (p) = 12 ∗ 1 + C21 (p) ν21 = ∗ 1 + C11 (p) G12 ∗ 1 + K12 (p) (7.58) For the unidirectional composite ply whose typical creep... strength of carbon–epoxy composites loaded along (1) and across (2) the fibers s−1 / s 1 0.8 0.6 0.4 0.2 0 log N 0 1 2 3 4 5 6 7 8 Fig 7.34 Normalized fatigue diagram for fabric carbon–carbon composite material (σ static strength), • experimental part of the diagram for the loading frequency of 6 Hz (•) and 330 Hz ( ), extrapolation 406 Advanced mechanics of composite materials Here, N is the number of cycles... 44 Let the shell be made of glass–epoxy composite whose mechanical properties are listed in Table 3.5 and creep diagrams are shown in Fig 7.22 To simplify the analysis, we suppose that for the unidirectional composite under study E2 /E1 = 0.22, G12 /E1 = 0.06, and 12 = ν21 = 0, and introduce the normalized shear strain γ = γxy T R 2 hE1 −1 396 Advanced mechanics of composite materials Consider a ±45◦... glass–epoxy ( T°C 100 75 1 50 25 2 N 0 1 · 105 2 · 105 Fig 7.31 Temperature of an aramid–epoxy composite as a function of the number of cycles under tension (1) and compression (2) 404 Advanced mechanics of composite materials sf / s0 1 1 0.8 1 2 0.6 2 0.4 0.2 0 log N 0 1 2 3 4 5 6 7 Fig 7.32 Typical fatigue diagrams for carbon–epoxy composite (solid lines) and aluminum alloy (dashed lines) specimens without... content on time for 1.2-mm-thick carbon–epoxy composite exposed to humid air with 95% RH under temperatures 25◦ C (1), 50◦ C (2), and 80◦ C (3) 384 Advanced mechanics of composite materials C(t),% 16 75°C 12 60°C 40°C 8 T = 20°C 4 0 t, 0 10 20 30 40 hour 50 Fig 7.16 Moisture content as a function of time and temperature for aramid–epoxy composites The cyclic action of temperature, moisture, or sun radiation... single cycle of vibration, i.e., W = |A| = πσ0 ε0 |sin θ| (7.68) Folowing Zinoviev and Ermakov (1994), we can introduce the dissipation factor as the ratio of energy loss in a loading cycle, W , to the value of the elastic potential energy in s s s0 s0 e e e0 (a) e0 (b) Fig 7.29 Stress–strain diagrams for viscoelastic (a) and elastic (b) materials 402 Advanced mechanics of composite materials a cycle,... unidirectional composite ply, referred to axes x and y and making angle φ with the principal material axes 1 and 2 as in Fig 4.18, the dissipation factors are ψx = Ex ψ1 ψ2 cos2 φ − sin2 φ cos 2φ + ψ45 12 sin2 φ cos2 φ E1 E2 ψy = Ey ψ2 ψ1 cos2 φ − sin2 φ cos 2φ + ψ45 12 sin2 φ cos2 φ E2 E1 ψxy = Gxy 2ψ1 2ψ2 12 + − ψ45 12 sin2 φ cos2 φ + cos2 2φ E1 E2 G12 where 12 = 1 − 12 1 − ν21 1 + + E1 E2 G12 Ex ,... transforms of the creep compliance to the relaxation modulus, i.e., 1 = 1 − R ∗ (p) 1 + C ∗ (p) With due regard to Eq (7.55), we can formulate the elastic–viscoelastic analogy or the correspondence principle, according to which the solution of the linear viscoelasticity 394 Advanced mechanics of composite materials problem can be obtained in terms of the corresponding Laplace transforms from the solution of. .. evaluation of composite materials (see, e.g., Skudra et al., 1989), for practical purposes, the experimental dependencies of the ultimate stresses on the time of their action are usually evaluated In particular, these experiments allow us to conclude that fibers, which are the major load-carrying elements of composite materials, possess some residual strength σ ∞ = σ (t → ∞), which is about 50% to 70% of the... corresponding static strength σ 0 = σ (t = 0), depending on the fiber type Typical dependencies of the long-term strength of composite materials on time are presented in Fig 7.28 As can be seen, the time of loading dramatically affects material strength However, being unloaded at any moment of time t, composite materials demonstrate practically the same static strength that they had before long-term loading . state of the structure, 376 Advanced mechanics of composite materials + s 1 − + E 2 E 1 G 12 0 0.2 0.4 0.6 0.8 1 0 50 100 150 200 T,°C t 12 s 2 s 1 Fig. 7.7. Experimental dependencies of normalized. 7.9. Composite material exposed to moisture on both surfaces z = 0 and z = h (a), and on the surface z = 0 only (b). 378 Advanced mechanics of composite materials the thermal conductivity λ of. mm/ ◦ C J (1) 22 = 13150 · 10 −6 T GPa mm/ ◦ C 374 Advanced mechanics of composite materials To determine M T mn , we need to specify the reference surface of the laminate. Assume that this surface coincides