9 ANISOTROPIC ELASTIC MEDIA 9.1 REVIEW OF NOTATIONS 9.1.1 Continuum Mechanics We consider the following classical notions and notations of the mechanics of continuous media: Ⅲ State of stress at a point: This is defined by a second order tensor with the symbol  . The 3 by 3 matrix associated with this tensor is symmetric. In this matrix, there are six distinct terms, which are denoted as s ij : s 11 ; s 22 ; s 33 ; s 23 ; s 13 ; s 12 Ⅲ State of strain at a point: This is defined as a second order tensor ee ee . The 3 by 3 matrix for this tensor is symmetric. It consists of six distinct terms denoted as e ij : e 11 ; e 22 ; e 33 ; e 23 ; e 13 ; e 12 Ⅲ Linear elastic medium: The strains are linear and homogeneous functions of the stresses. The corresponding relations are: 1 Ⅲ Homogeneous medium: In this case, the matrix terms j ijkl characterizing the elastic behavior of the medium are not point functions. They are the same at all points in the considered medium. 1 For example: e 11 = j 1111 s 11 + j 1112 s 12 + j 1113 s 13 + j 1121 s 21 + j 1122 s 22 + j 1123 s 23 + j 1131 s 31 + j 1132 s 32 + j 1133 s 33 . e ij j ijkᐉ s kᐉ ¥= TX846_Frame_C09 Page 207 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC 9.1.2 Number of Distinct jj jj ijk ഞ Terms The above stress–strain relation can be written in matrix form as: Ⅲ Due to the symmetry of the stresses ( ), the corresponding coefficients are the same, i.e., . Ⅲ Due to the symmetry of the strains ( e ij = e ji ), the corresponding coefficients are the same, i.e., . In other words, the knowledge of only the coefficients of the 6 ¥ 6 matrix written above is required. Ⅲ In addition, application of the theorem of virtual work on the stresses shows that the coefficients j ijk ᐉ are symmetric, meaning: . 2 Therefore, the 6 ¥ 6 matrix mentioned previously is symmetric. The number of distinct coefficients is: 2 Consider two simple stress states: • State No. 1: One single stress, ( s k ᐉ ) 1 , which causes the strain: ( e ij ) 1 = j ijk ᐉ ( s k ᐉ ) 1 • State No. 2: One single stress, ( s pq ) 2 , which causes the strain: ( e mn ) 2 = j mnpq ( s pq ) 2 One can write that the work of the stress in state No. 1 on the strain in state No. 2 is equal to the work of the stress in state No. 2 on the strain in state No. 1, as: which means: from which one has: j k ᐉ pq = j pqk ᐉ s kᐉ s ᐉk = j ijkᐉ j ijᐉk = j ijkᐉ j jikᐉ = j ijkᐉ j kᐉij = s kᐉ () 1 e kᐉ () 2 ¥ s pq () 2 e pq () 1 ¥= s kᐉ () 1 j kᐉpq s pq () 2 ¥¥ s pq () 2 j pqkᐉ s kᐉ () 1 ¥¥= 66 1+() 2 21 coefficients= TX846_Frame_C09 Page 208 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC Ⅲ In summary: The previous stress–strain relation can then be written as: The matrix above does not have the general symmetry as in the general form (9 ¥ 9) presented previously. (Note the coefficients 2 in this matrix). One can get around this inconvenience by doubling the terms e 23 , e 13 , e 12 , introducing the shear strains: from which the stress–strain behavior can then be written in a symmetric form as: (9.2) 9.2 ORTHOTROPIC MATERIALS Definition: An orthotropic material is a homogeneous linear elastic material having two planes of symmetry at every point in terms of mechanical properties, these two planes being perpendicular to each other. stress reciprocity: strain definition: symmetry: There remain 21 distinct coefficients (9.1) j ijkᐉ j ijᐉk = j ijkᐉ j jikᐉ = j ijkᐉ j kᐉij = j ijkᐉ e 11 e 22 e 33 e 23 e 13 e 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ j 1111 j 1122 j 1133 2 j 1123 2 j 1113 2 j 1112 j 2211 j 2222 j 2233 2 j 2223 2 j 2213 2 j 2212 j 3311 j 3322 j 3333 2 j 3323 2 j 3313 2 j 3312 j 2311 j 2322 j 2333 2 j 2323 2 j 2313 2 j 2312 j 1311 j 1322 j 1333 2 j 1323 2 j 1313 2 j 1312 j 1211 j 1222 j 1233 2 j 1223 2 j 1213 2 j 1212 s 11 s 22 s 33 s 23 s 13 s 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ = g 23 2 e 23 ; g 13 2 e 13 ; g 12 2 e 12 === e 11 e 22 e 33 2 e 23 2 e 13 2 e 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ g 23 g 13 g 12 j 1111 j 1122 j 1133 2 j 1123 2 j 1113 2 j 1112 j 2211 j 2222 j 2233 2 j 2223 2 j 2213 2 j 2212 j 3311 j 3322 j 3333 2 j 3323 2 j 3313 2 j 3312 2 j 2311 2 j 2322 2 j 2333 4 j 2323 4 j 2313 4 j 2312 2 j 1311 2 j 1322 2 j 1333 4 j 1323 4 j 1313 4 j 1312 2 j 1211 2 j 1222 2 j 1233 4 j 1223 4 j 1213 4 j 1212 s 11 s 22 s 33 s 23 s 13 s 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ = TX846_Frame_C09 Page 209 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC Then one can show that 3 the number of independent elastic constants is nine. The constitutive relation expressed in the so-called “orthotropic” axes, defined by three axes constructed on the two orthogonal planes and their intersection line, can be written in the following form, called the engineering notation because it utilizes the elastic modulus and Poisson ratios: (9.3) where E 1 , E 2 , E 3 are the longitudinal elastic moduli. G 23 , G 13 , G 12 are the shear moduli. n 12 , n 13 , n 23 , n 21 , n 31 , n 32 are the Poisson ratios. In addition, the symmetry of the stress–strain matrix above leads to the following relations: (9.4) 9.3 TRANSVERSELY ISOTROPIC MATERIALS Definition: A transversely isotropic material is a homogeneous linear elastic material such that any plane including a preferred axis, is a plane of mechanical symmetry. One can show that 4 the constitutive relation has five independent elastic constants. For the fiber/matrix composite shown in Figure 9.1 the preferred axis is ᐉ. The fibers are distributed uniformly in the direction along ᐉ. All directions perpendicular to the fibers characterize the transverse direction t. 3 Proof is shown in Section 13.1. 4 Proof is shown in Section 13.2. e 11 e 22 e 33 g 23 g 13 g 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ 1 E 1 n 21 E 2 – n 31 E 3 – 000 n 12 E 1 – 1 E 2 n 32 E 3 – 000 n 13 E 1 – n 23 E 2 – 1 E 3 000 000 1 G 23 00 0000 1 G 13 0 00000 1 G 12 s 11 s 22 s 33 s 23 s 13 s 12 Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ n 21 E 2 n 12 E 1 ; n 31 E 3 n 13 E 1 ; n 32 E 3 n 23 E 2 === TX846_Frame_C09 Page 210 Monday, November 18, 2002 12:24 PM © 2003 by CRC Press LLC 10 ELASTIC CONSTANTS OF UNIDIRECTIONAL COMPOSITES In this chapter we examine a distinct combination of two materials (matrix and fiber), with simple geometry and loading conditions, in order to estimate the elastic properties of the equivalent material, i.e., of the composite. 10.1 LONGITUDINAL MODULUS E ᐉ The two materials are shown schematically in Figure 10.1 where m stands for matrix. f stands for fiber. Ⅲ Hypothesis: The two materials are bonded together. More precisely, one makes the following assumptions: Ⅲ Both the matrix m and the fiber f have the same longitudinal strain e l . Ⅲ The interface between the two materials allows the z normal strains in the two materials to be different. The state of stresses resulting from a force F can therefore be written as: and the corresponding state of strains: e z e z π mf S m s ᐉ 00 000 000 Æ m S f s ᐉ 00 000 000 Æ f e m e ᐉ 00 0 e t 0 00 e z Æ m e f e ᐉ 00 0 e t 0 00 e z Æ f TX846_Frame_C10 Page 213 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC Expressing the stresses in terms of the strains for each material yields then: (10.2) Note: Among the real phenomena that are not taken into account in the expression of E l is the lack of perfect straightness of the fibers in the matrix. Also, the modulus E l depends on the sign of the stress (tension or compression). In rigorous consideration, the material is “ bimodulus .” Example: Unidirectional layers with 60% fiber volume fraction ( V f = 0.60) with epoxy matrix: 10.2 POISSON COEFFICIENT Considering again the loading defined in the previous paragraph, the transverse strain for the matrix m and fiber f can be written as: and for the composite (m + f ): The strain in the transverse direction can also be written as: Because e ᐉ has the same value in m and f: (10.3) Kevlar “HR” Carbon “HM” Carbon E ᐉ tension (MPa) 85,000 134,000 180,000 E ᐉ compression (MPa) 80,300 134,000 160,000 E ᐉ e ᐉ E m e ᐉ V m E f e ᐉ V f += E ᐉ E m V m E f V f += e t n E s ᐉ – ne ᐉ –== e t mf+ n ᐉt E ᐉ s ᐉ mf+ ¥ n ᐉt e ᐉ –== e t mf+ D e m e f +() e m e f + De m e m V m De f e f V f +== e t mf+ e t m V m e t f V f += n ᐉt e ᐉ – n m e ᐉ V m – - n f e ᐉ V f = n ᐉt n m V m n f V f += TX846_Frame_C10 Page 215 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC 10.3 TRANSVERSE MODULUS E t To evaluate the modulus along the transverse direction E ᒑ , the two materials are shown in the Figure 10.2. In addition, one uses the following simplifications: Ⅲ Hypothesis: At the interface between the two materials, assume the following: Ⅲ Freedom of movement in the l direction allows for different strains in the two materials: Ⅲ Freedom of movement in the z direction allows for different strains in the two materials: Then, the state of stress created by a load F (see Figure 10.2), can be reduced for each material to the following: Figure 10.2 Transverse Modulus E t e ᐉ m e ᐉ f π e z m e z f π S 000 0 s t 0 000 Æ TX846_Frame_C10 Page 216 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC The strains can be written as: Then for the composite (m + f ), one has On the other hand, using direct calculation leads to (see Figure 10.2) then: (10.4) Remarks: Ⅲ Due to the above simplifications that allow the possibility for relative sliding along the l and z directions at the interface, the transverse modulus E t above may not be accurate. Ⅲ One finds in the technical literature many more complex formulae giving E t . However, none can provide guaranteed good result. Ⅲ Taking into consideration the load applied (see Figure 10.2),: the modulus E f that appears in Equation 10.4 is the modulus of elasticity of the fiber in a direction that is perpendicular to the fiber axis. This modulus can be very different from the modulus along the axis of the fiber, due to the anisotropy that exists in fibers. 2 2 This point was discussed in Paragraph 3.3.1. e e ᐉ 00 0 e t 0 00 e z m or f Æ e t 1 E t s t = e t D e m e f +() e m e f + e t m V m e t f V f +== 1 E t s t 1 E m s t V m 1 E f s t V f += 1 E t V m E m V f E f or E t + E m 1 1 V f –() E m E f V f + == TX846_Frame_C10 Page 217 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC 10.4 SHEAR MODULUS G ᐉ t Load application that can be used to evaluate the shear modulus G ᐉ t is shown schematically in the Figure 10.3, both with the angular deformations that are produced. The state of stress, identical for both the matrix and fiber material, can be written as: The corresponding strains can be written as: Using the constitutive equation, one has then: Also, from Figure 10.3, one has Figure 10.3 Shear modulus G ഞt S 0 t ᐉt 0 t ᐉt 00 000 Æ e m or f 0 e ᐉt 0 e ᐉt 00 000 Æ e ᐉt 1 n + E t ᐉt t ᐉt 2G == g ᐉt t ᐉt G = g ᐉt mf+ e m e f +() g ᐉt e m m g ᐉt e f f += TX846_Frame_C10 Page 218 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC which can be rewritten as: (10.5) 3 10.5 THERMOELASTIC PROPERTIES 10.5.1 Isotropic Material: Recall When the influence of temperature is taken into consideration, Hooke’s law for the case of no temperature influence: is replaced by the Hooke–Duhamel law: (10.6) where e = Strain tensor S = Stress tensor I = Unity tensor E, n = Elastic constants for the considered material a = Coefficient of thermal expansion 4 DT = Change in temperature with respect to a reference temperature at which the stresses and strains are nil 10.5.2 Case of Unidirectional Composite The coefficient of thermal expansion of the matrix is usually much larger (more than ten times) than that of the fiber. 4 In Figure 10.4, one can imagine that even in the absence of mechanical loading, a change in temperature DT will produce a 3 A few values of the shear modulus G f are shown in Section 3.3.1. 4 See Section 1.6, “Principal Physical Properties.” g ᐉt m + f g ᐉt V m m g ᐉt V f f += t ᐉt G ᐉt t ᐉt G m V m t ᐉt G f V f += 1 G ᐉt V m G m V f G f += G ᐉt G m 1 1 V f –() G m G f V f + = e 1 n + E S n E trace S()I–= e 1 n + E S n E trace S() I a DT I+–= TX846_Frame_C10 Page 219 Monday, November 18, 2002 12:25 PM © 2003 by CRC Press LLC [...]... balanced This means that apart from the midplane symmetry, there are as many and identical plies that make with the x axis an angle +q as those that make 4 The developments of E ij are given in Equation 11 .8 © 2003 by CRC Press LLC TX846_Frame_C12 Page 2 38 Monday, November 18, 2002 12:27 PM with the x axis an angle -q In effect, the coefficients E 13 and E 23 are 6 antisymmetric in q and, therefore, cancel... 2 1 1 1 + s ) – c s Ê - + – -ˆ Ë Eᐉ Et Gᐉt¯ Recall that the matrix of elastic coefficients is symmetric, meaning in particular: hxy /Gxy = hx /Ex and mxy /Gxy = my /Ey See example described in Section 3.1 © 2003 by CRC Press LLC TX846_Frame_C11 Page 2 28 Monday, November 18, 2002 12:26 PM 11.2 STIFFNESS COEFFICIENTS When one inverts Equation 11.1 written in the coordinate axes l,t of a ply, one... 0 Ï aᐉ ¸ Ï sᐉ ¸ Ô Ô Ô Ô 0 Ì s t ˝ + DT Ì a t ˝ Ô Ô Ô Ô Ó0˛ 1 Ó t ᐉt ˛ (10.9) -G ᐉt in which the coefficients Eᐉ, Et, vᐉt, Gᐉt, aᐉ and at have the values given by the Equations 10.2 to 10 .8, respectively © 2003 by CRC Press LLC TX846_Frame_C11 Page 223 Monday, November 18, 2002 12:26 PM 11 ELASTIC CONSTANTS OF A PLY ALONG AN ARBITRARY DIRECTION To study the behavior of a laminate made up of many plies... different orientations, one has: 1 0∞ 90∞ +45∞ -45∞ ¥ A ij = E ij ¥ p 0∞ + E ij ¥ p 90∞ + E ij ¥ p +45∞ + E ij ¥ p -45∞ h 5 6 See Figure 12.1 and figure in the Equation 11 .8 The expressions developed for E ij are given in Equation 11 .8 © 2003 by CRC Press LLC (12 .8) ... s y ˝ = E 21 E 22 E 23 Ô Ô Ó t xy ˛ E 31 E 32 E 33 © 2003 by CRC Press LLC Ï ex ¸ Ô Ô Ì ey ˝ Ô Ô Ó g xy ˛ TX846_Frame_C11 Page 229 Monday, November 18, 2002 12:26 PM Once the calculations are performed, one obtains the following expressions for the stiffness coefficients E ij , where c = cosq and s = sinq Ï sx ¸ E 11 E 12 E 13 Ô Ô s y ˝ = E 21 E 22 E 23 Ì Ô Ô Ó t xy ˛ E 31 E 32 E 33 Ï ex ¸ Ô Ô Ì ey... 11.2 for a ply characterized by moduli Eᐉ and Et with very different 8 values, for example the case of unidirectional layers of fiber/resin 11.3 CASE OF THERMOMECHANICAL LOADING 11.3.1 Compliance Coefficients 9 When considering the temperature variations, one must substitute the stress–strain Equation 11.1 with Equation 10.9: Ï eᐉ ¸ Ô Ô Ô Ô Ì et ˝ = Ô Ô Ô Ô Ó g ᐉt ˛ 8 9 1 -Eᐉ n n tᐉ – -E t ᐉt – -E 1... 3.3.3 See Section 10.5 © 2003 by CRC Press LLC TX846_Frame_C11 Page 231 Monday, November 18, 2002 12:26 PM Then, upon substituting, Ï ex ¸ Ô Ô Ô Ô Ì ey ˝ = [ T ¢ ] Ô Ô Ô Ô Ó g xy ˛ 1 -Eᐉ n n tᐉ – -E 0 t ᐉt – -E 1 Et 0 0 0 1 G ᐉt ᐉ Ï aᐉ ¸ Ï sx ¸ Ô Ô Ô Ô Ô Ô Ô Ô [ T ] Ì s y ˝ + DT [ T ¢ ] Ì a t ˝ Ô Ô Ô Ô Ô Ô Ô Ô Ó0˛ Ó t xy ˛ One finds again in the first part of the second term a matrix of compliance... Ô Ì s y ˝ = [ T l ] n ᐉt E t Et 0 [ T 1¢ ] Ì e y ˝ – DT [ T l ] Ì n ᐉt E t a ᐉ + E t a t ˝ Ô Ô Ô Ô Ô Ô 0 0 0 G ᐉt Ó ˛ Ó t xy ˛ Ó g xy ˛ One finds again, in the first part of the second term, the matrix detailed in Equation 11 .8 The second part of the second term can be developed as follows: c – DT s 2 2 – cs s 2 c 2 cs 2cs Ï E ᐉ a ᐉ + n tᐉ E ᐉ a t ¸ Ô Ô – 2c s Ì n ᐉt E t a ᐉ + E t a t ˝ = º Ô Ô 2 2 0... given by the relations [11 .8] a E 1 = c E ᐉ ( a ᐉ + n tᐉ a t ) + s E t ( n ᐉt a ᐉ + a t ) 2 2 a E 2 = s E ᐉ ( a ᐉ + n tᐉ a t ) + c E t ( n ᐉt a ᐉ + a t ) 2 2 a E 3 = cs [ E t ( n ᐉt a ᐉ + a t ) – E ᐉ ( a ᐉ + n tᐉ a t ) ] c = cos q ; s = sin q E ᐉ = E ᐉ / ( 1 – n ᐉt n tᐉ ) E t = E t / ( 1 – n ᐉt n tᐉ ) © 2003 by CRC Press LLC (11.10) TX846_Frame_C12 Page 235 Monday, November 18, 2002 12:27 PM 12 MECHANICAL... Section 5.2.3 © 2003 by CRC Press LLC TX846_Frame_C12 Page 237 Monday, November 18, 2002 12:27 PM with: th th n ply A 11 =  th n ply  k 11 k E e ; A 12 = st n ply k 12 k E e ; A 13 = st k=1 ply  k E 13 e k st k=1 ply k=1 ply In an analogous manner, one obtains for the Equation 12.2: N y = A 21 e ox + A 22 e oy + A 23 g oxy with: th n ply  A 2j = k E 2j e k st k=1 ply and for the shear stress resultant . LONGITUDINAL MODULUS E ᐉ The two materials are shown schematically in Figure 10.1 where m stands for matrix. f stands for fiber. Ⅲ Hypothesis: The two materials are bonded together combination of two materials (matrix and fiber), with simple geometry and loading conditions, in order to estimate the elastic properties of the equivalent material, i.e., of the composite. 10.1. matrix m and fiber f can be written as: and for the composite (m + f ): The strain in the transverse direction can also be written as: Because e ᐉ has the same value in m and f: (10.3) Kevlar