Mechanics Analysis Composite Materials Episode 5 pot

126 Mechanics and analysis of composite materials The name “complementary” becomes clear if we consider a bar in Fig. 1.1 and the corresponding stress-strain curve in Fig. 4.8. The area OBC below the curve represents U in accordance with the first equation in Eqs. (4.8), while the area OAB above the curve is equal to U,. As was shown in Section 2.9, dU in Eqs. (4.8) is an exact differential. To prove the same for dU,, consider the following sum: which is obviously an exact differential. Since dU in this sum is also an exact differential, dU, should have the same property and can be expressed as Comparing this result with Eq. (4.9), we arrive at Castigliano’s formulas (4.10) which are valid for any elastic solid (for a linear elastic solid, U, = U). Complementary potential, U,, in general, depends on stresses, but for an isotropic material, Eq. (4.10) should yield invariant constitutive equations that do not depend on the direction of coordinate axes. This means that U, should depend on stress invariants 11~12, I3 in Eqs. (2.13). Assuming different approximations for function Uc(Zl, Z2, I3)we can construct different classes of nonlinear elastic models. Existing experimental verification of such models shows that dependence U, on 13 can be neglected. Thus, we can present complementary potential in a simplified form U, (ZI, Z2)and expand this function into the Taylor series as E C E de Fig. 4.8. Geometric interpretation of elastic potential. U. and complementary potential, U,. Chapter 4. Mechanics of a composite layer 127 1 1 1 2 3! 4! u, = c0 + c~~I~ + -clzif + -cI31; +-cI4if + + C2li2 + -c2& + ??I; + -c,4i; + 1 1 I 2 3! 4! + -CI~ZIZII~ 1 +-c1221iji2 1 3 1 +-~112211122 2 3! 3! 1 1 1 4! 4! 4! +-cI3,1.l:i2 +-c12,21:i; +-c1123i1i;3 + , (4.1 1) where Constitutive equations follow from Eq. (4.10) and can be written in the form au, ai, au, aiz E ‘I - ail aoij ai, aoii . (4.12) Assuming that for zero stresses U, = 0 and cij = 0 we should take co = 0 and CI I = 0 in Eq. (4.1 1). Consider a plane stressed state with stresses o.~, o,,, z.~! shown in Fig. 4.5. Stress invariants in Eqs. (2.13) entering Eq. (4.12) are Linear elastic material model is described with Eq. (4.1 1) if we take u, = fC12i; + C2II2 . (4.14) Using Eqs. (4.12)-(4.14) and engineering notations for stresses and strains, we arrive at 8.r = c12(o.v +ox) - C,Iql., 4; = c12(o., + oy) - c2lo.r’ y.vj. = 2C21Z,,. These equations coincide with the corresponding equations in Eqs. (4.6) if we take 1 I +V E’ c,1 = - E ’ c12 = - To describe nonlinear stress-strain diagram of the type shown in Fig. 4.6, wc can generalize Eq. (4.14) as u. - -c12i; 1 + C2lZ2 + -C14Z1 1 4 1 + -c22z, 2 . ‘-2 4! 2 128 Mechanics and analysis of composite materials Then, Eqs. (4.12) yield the following cubic constitutive law: The corresponding approximation is shown in Fig. 4.6 with a solid line. Retaining more higher-order terms in Eq. (4.1l), we can describe nonlinear behavior of any isotropic polymeric material. To describe nonlinear elastic-plastic behavior of metal layers, we should use constitutive equations of the theory of plasticity. As known, there exist two basic versions of this theory - the deformation theory and the flow theory that are briefly described below. According to the deformation theory of plasticity, the strains are decomposed into two components - elastic strains (with superscript 'e') and plastic strains (superscript 'p'), i.e., E,I = E& + &; . (4.15) We again use the tensor notations of strains and stresses (Le., cij and ou) introduced in Section 2.9. Elastic strains are linked with stresses by Hooke's law, Eqs. (4.1), which can be written with the aid of Eq. (4.10) in the form (4.16) where U, is the elastic potential that for the linear elastic solid coincides with complementary potential U, in Eq. (4.10). Explicit expression for U, can be obtained from Eq. (2.5 1) if we change strains for stresses with the aid of Hooke's law, i.e., Now present plastic strains in Eqs. (4.15) in the form similar to Eq. (4.16): (4.17) (4.18) where Up is the plastic potential. To approximate dependence of Up on stresses, a special generalized stress characteristic, i.e., the so-called stress intensity 0, is introduced in classical theory of plasticity as Chapter 4. Mechanics of a composize layer I29 Transforming Eq. (4.19) with the aid of Eqs. (2.13) we can reduce it to the following form: This means that 0 is an invariant characteristic of a stress state, i.e., that it does not depend on position of a coordinate frame. For a unidirectional tension as in Fig. 1.1, we have only one nonzero stress, e.g., 011. Then Eq. (4.19) yields rs = 01 I. In a similar way, strain intensity E can be introduced as (4.20) Strain intensity is also an invariant characteristic. For a uniaxial tension (Fig. 1.1) with stress CTI I and strain EI I in the loading direction, we have ~22 = ~33 = -Y,,EI I, where vp is the elastic-plastic Poisson's ratio which, in general, depends on 01 I. For this case, Eq. (4.20) yields &=;(I + \'p)Ell . (4.21) For an incompressible material (see Section 4.1.1), vp = 1/2 and E = EII. Thus, numerical coefficients in Eqs. (4.19) and (4.20) provide 0 = 011 and E = I:II for uniaxial tension of an incompressible material. Stress and strain intensities in Eqs. (4.19) and (4.20) have an important physical meaning. As known from experiments, metals do not demonstrate plastic properties under loading with stresses 0, = or = 0: = 00 resulting only in the change of material volume. Under such loading, materials exhibit only elastic volume deformation specified by Eq. (4.2). Plastic strains occur in metals if we change material shape. For a linear elastic material, elastic potential U in Eq. (2.51) can be reduced after rather cumbersome transformation with the aid of Eqs. (4.3), (4.4) and (4.19), (4.20) to the following form: U=:a"&o+~aE - . (4.22) The first term in the right-hand side part of this equation is the strain energy associated with the volume change, while the second term corresponds to the change of material shape. Thus, CT and E in Eqs. (4.19) and (4.20) are stress and strain 130 Mechanics and analysis of composite materials characteristics associated with the change of material shape under which it demonstrates the plastic behavior. In the theory of plasticity, plastic potential Up is assumed to be a function of stress intensity a, and according to Eqs. (4.18), plastic strains are (4.23) Consider further a plane stress state with stresses a,, a),, and zxv in Fig. 4.5. For this case, Eq. (4.19) acquires the form (4.24) Using Eqs, (4.15H4.17) and (4.23), (4.24) we finally arrive at the following constitutive equations: where 1 dU, a da o(a) = . (4.25) (4.26) To find ~(a), we need to specify dependence of U, on a. The most simple and suitable for practical applications is the power approximation u,=ca”, (4.27) where C and n are some experimental constants. As a result, Eq. (4.26) yields To determine coefficients C and n we introduce the basic assumption of the plasticity theory concerning the existence of the universal stress-strain diagram (master curve). According to this assumption, for any particular material there exists the dependence between stress and strain intensities, i.e., a = (P(E) (or E = f(a)), that is one and the same for all the loading cases. This fact enables us to find coefficients C and n from the test under uniaxial tension and extend thus obtained results to an arbitrary state of stress. Chapter 4. Mechanics of a composite layer 131 Indeed, consider a uniaxial tension as in Fig. 1.1 with stress 61 1. For this case, a = and Eqs. (4.25) yield 01- CY = - E + o(o,)a., , (4.29) V 1 E. 2 cy = av - -o(a,)a., , y.Yv = 0 . Solving Eq. (4.29) for co(ax), we get 1 1 Es(0.r) E o(a.,) = , (4.30) (4.3 1) where E, = is the secant modulus introduced in Section 1.1 (see Fig. 1.4). Using now the existence of the universal diagram for stress intensity r~ and taking into account that cr = a., for a uniaxial tension, we can generalize Eq. (4.31) and write it for an arbitrary state of stress as (4.32) To determine E,(o) = a/E, we need to plot the universal stress-strain curve. For this purpose, we can use an experimental diagram o,(c,) for the case of uniaxial tension, e.g., the one shown in Fig. 4.9 for an aluminum alloy with a solid line. To plot the universal curve o(E), we should put 6 = a, and change the scale on the strain axis in 0, ,6, MPU 250 200 150 100 50 0 0 1 2 3 4 Fig. 4.9. Experimental stress-strain diagram for an aluminum alloy under uniaxial tension (solid line), the universal stress-strain curve (broken line) and its power approximation (circles). 132 Mechanics and analysis of composite materials accordance with Eq. (4.21). To do this, we need to know the plastic Poisson's ratio vp that can be found as vp = -E,,/&,. Using Eqs. (4.29) and (4.30) we arrive at As follows from this equation vp = v if E, =E and vp + 112 for E, + 0. Dependencies of Es and vp on E for the aluminum alloy under consideration are presented in Fig. 4.10. With the aid of this figure and Eq. (4.21) in which we should take 81 I = E,, we can calculate E and plot the universal curve shown in Fig. 4.9 with a broken line. As can be seen, this curve is slightly different from the diagram corresponding to a uniaxial tension. For the power approximation in Eq. (4.27), from Eqs. (4.26) and (4.32) we get E l 0 E' a(.) = - - - Matching these results we find E = - + Cncrn-' . (4.33) E This is a traditional approximation for a material with a power hardening law. Now, we can find C and n using Eq. (4.33) to approximate the broken line in Fig. 4.9. The results of approximation are shown in this figure with circles that correspond to E = 71.4 GPa, n = 6, and C = 6.23 x Thus, constitutive equations of the deformation theory of plasticity are specified by Eqs. (4.25) and (4.32). These equations are valid only for active loading that can (MPa)-5. E 100 80 60 40 20 0 "D 0.5 0.4 0.3 09 0.1 0 E, 0 1 2 3 4 Fig. 4.10. Dependencies of the secant modulus (Es), tangent modulus (Et), and the plastic Poisson's ratio (v,) on strain for an aluminum alloy. Chapter 4. Mechanics of a composite layer 133 be identified by the condition do > 0. Being applied for unloading (i.e., for do < 0), Eqs. (4.25) correspond to nonlinear elastic material with stress-strain diagram shown in Fig. 1.2. For elastic-plastic material (see Fig. 1.5), unloading diagram is linear. So, if we reduce the stresses by some increments AO.~,Abv, AT^?, the corresponding increments of strains will be Direct application of nonlinear equations (4.25) substantially hinders the problem of stress-strain analysis because these equations include function o(0) in Eq. (4.32) which, in turn, contains secant modulus E,(a). For the power approximation corresponding to Eq. (4.33), E, can be expressed analytically, i.e., 1 1 _- +cnon-= . Es E However, in many cases E, is given graphically as in Fig. 4.10 or numerically in the form of a table. Thus, Eqs. (4.25) sometimes cannot be even written in the explicit analytical form. This implies application of numerical methods in conjunction with iterative linearization of Eqs. (4.25). There exist several methods of such linearization that will be demonstrated using the first equation in Eqs. (4.25), i.e., (4.34) In the method of elastic solutions (Ilyushin, 1948), Eq. (4.34) is used in the following form: (4.35) where s is the number of the iteration step and For the first step (s = l), we take qo = 0 and solve the problem of linear elasticity with Eq. (4.35) in the form Finding the stresses, we calculate yl and write Eq. (4.35) as (4.36) 134 Mechanics and analysis of composite materials where the first term is linear, while the second term is a known function of coordinates. Thus, we have another linear problem resolving which we find stresses, calculate q2 and switch to the third step. This process is continued until the strains corresponding to some step become close within the given accuracy to the results found at the previous step. Thus, the method of elastic solutions reduces the initial nonlinear problem to a sequence of linear problems of the theory of elasticityfor the same material but with some initial strains that can be transformed into initial stresses or additional loads. This method readily provides a nonlinear solution for any problem that has a linear solution, analytical or numerical. The main shortcoming of the method is its poor convergence. Graphical interpretation of this process for the case of uniaxial tension with stress (r is presented in Fig. 4.1 la. This figure shows a simple way to improve the convergence of the process. If we need to find strain at the point of the curve that is close to point A, it is not necessary to start the process with initial modulus E. Taking E' < E in Eq. (4.36) we can reach the result with much less number of steps. According to the method of elastic variables (Birger, 1951), we should present Eq. (4.34) as (4.37) Fig. 4.1 1. Geometric interpretation of (a) the method of elastic solutions, (b) the method of variable elasticity parameters, (c) Newton's method, and (d) method of successive loading. Chapter 4. Mechanics of a composite layer 135 In contrast to Eq. (4.39, stresses d, and v;. in the second term correspond to the current step rather than to the previous one. This enables us to write Eq. (4.37) in the form analogous to Hooke's law, i.e., where (4.38) (4.39) are elastic variables corresponding to the step with number s - 1. The iteration procedure is similar to that described above. For the first step we take EO = E and vo = v in Eq. (4.38). Find e:, et. and 61, determine El, VI, switch to the second step and so on. Graphical interpretation of the process is presented in Fig. 4.1 Ib. Convergence of this method is by an order higher than that of the method of elastic solutions. However, elastic variables in the linear constitutive equation of the method, Eq. (4.38), depend on stresses and hence, on coordinates whence the method has got its name. This method can be efficiently applied in conjunction with the finite element method according to which the structure is simulated with the system of elements with constant stiffness coefficients. Being calculated for each step with the aid of Eqs. (4.39), these stiffnesses will change only with transition from one element to another, and it practically does not hinder the finite element method calculation procedure. The iteration process having the best convergence is provided by the classical Newton's method requiring the following form of Eq. (4.34): &;. = c-1 + c;;'(o-; - a;:') + qg(a;, - CT:') +c?;;yT& - ?;;I) ] (4.40) where Because coefficients c are known from the previous step (s - I), Eq. (4.40) is linear with respect to stresses and strains corresponding to step number s. Graphical interpretation of this method is presented in Fig. 4.1 IC. In contrast to the methods discussed above, Newton's method has no physical interpretation and being characterized with very high convergence, is rather cumbersome for practical applications. [...]... carbon-epoxy layer - Chapter 4 Mechanics of a composite layer 155 - - Fig 4.22 An off-axis test Fig 4.23 Deformation of a unidirectional layer loaded at an angle to fiber orientation 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 Mechanics and analysis of composite materials 156 calculated with the aid of Eqs... into Hooke’s law, Eqs (4 .55 ), and thus expressed strains - into Eqs (4.70) As a result, we arrive at the following particular form of Eqs (2.48), (2.49): 152 Mechanics and analysis of composite materials (4. 75) where compliance coefficients are (4.76) There exist the following dependencies between coefficients of Eqs (4.71) and (4. 75) : Chapter 4 Mechanics o a eomposite layer f I53 and where As can be... Section 3.4.3 and shown in Figs 3 .54 and 3 .55 Indeed, if we know E, from the tensile test in Fig 4.23 and find E l , E?, v21 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 Mechanics and analysis of composite materials 154 A,,,, ,GPa t 40 120 100 80 60 40 20 0 ‘ I 0 15 30 75 60 45 90 and shear ( A M )stiffnesses... and (2 .53 ), yield (4 .53 ) z,3 t Fig 4.13 An orthotropic layer Fig 4.14 Filament wound composite pressure vessel Chapter 4 Mechanics o a composite layer f 141 where (4 .54 ) where As for an isotropic layer considered in Section 4.1, the terms including transverse normal stress c3 can be neglected in Eqs (4 .53 ) and (4 .54 ) and they can be written in the following simplified forms: (4 .55 ) and (4 .56 ) where... layer under study referred to the principal material coordinates are given by Eqs (4 .55 ) and (4 .56 ) We need now to derive such Mechanics and analysis of composite materials 148 Z 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 equations for the global coordinate frame... polynomial in Eq (4 .58 ) includes 84 ‘c’-coefficients Apparently, this is too many for practical analysis of composite materials To reduce the number of coefficients, we can first use some general considerations Namely, assume that U,= 0 and eij = 0 if there are no stresses (aii = 0) Then, co = 0 and cii = 0 Second, we should take into account that the 144 Mechanics and analysis of composite materials material... (4.71) and (4. 75) , the layer under study is anisotropic in plane because constitutive equations include shear-extension and shear-shear coupling coefficients y and 1 For 5, = 0, the foregoing equations degenerate into Eqs (4 .55 ) and (4 .56 ) for an orthotropic layer Dependencies of stiffness coefficients on the orientation angle for a carbon-epoxy composite with properties listed in Table 3 .5 are presented... in Fig 4. 25 with a solid line Then, uxsy = 0 at 4 = 4,, But for unidirectional advanced composites whose properties are listed in Table 3 .5, the curve is similar to the broken line in Fig 4. 25, and E, reaches its extremum values at d, = 0 and 4 = 90" only Chapter 4 Mechanics o a composite layer f Fig 4. 25 Angle 157 of an off-axis test without shear-extensioncoupling Thus, to describe a real off-axis... E3 is Mechanics and analysis of composite materials 142 usually found testing the layer under compression in the z-direction Transverse 1 2 shear moduli G 3 and G 3 can be obtained by different methods, e.g., by inducing pure shear in two symmetric specimens shown in Fig 4. 15 and calculating shear modulus as G 3 = P/(2Ay), here A is the in-plane area of the specimen 1 w 1 For unidirectional composites,... rn~ma; 4 ~ + 24znoia2 + 2k3z12a: + 2rn2212o: + 2k4z12of+ 4k5zi20: + 2rn3zlzo': + 4rn47i2a: 81 = +azo: C: (4 .59 ) For unidirectional composites, dependence 81(01) is linear which means that we 2 dl5 = 0, kl = - - k5 = 0 Then, the foregoing equations should put d = reduce to 1 As an example, consider a special two-matrix fiberglass unidirectional composite with high in-plane transverse and shear deformation . linear elasticity with Eq. (4. 35) in the form Finding the stresses, we calculate yl and write Eq. (4. 35) as (4.36) 134 Mechanics and analysis of composite materials where the first term. (2 .53 ), yield z,3 t Fig. 4.13. An orthotropic layer. (4 .53 ) Fig. 4.14. Filament wound composite pressure vessel. Chapter 4. Mechanics of a composite layer where 141 (4 .54 ). proportion to one parameter. law CT = Se”. 140 Mechanics and analysis of composite materials 4.2. Unidirectional orthotropic layer A composite layer with the simplest structure consists