ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 10 ppsx

302 Advanced mechanics of composite materials the facings are made of one and the same material (only the thicknesses are different), Eqs. (5.113) and (5.114) yield e = h 2 1 +h 3 (h 3 +2h 1 +2h 2 ) 2(h 1 +h 3 ) Returning to the general case, we should emphasize that the reference plane providing C mn = 0 for all the mn values does not exist in this case only if the laminate structure is given. If the stacking-sequence of the layers is not pre-assigned and there are sufficient number of layers, they can be arranged in such a way that C mn = 0. Indeed, consider a laminate in Fig. 5.32 and suppose that its structure is, in general, not symmetric, i.e., z  i = z i and k  = k. Using plane z = 0 as the reference plane, we can write the membrane– bending coupling coefficients as C mn = 1 2 k/2  i=1 A (i) mn h i (z i +z i−1 ) − 1 2 k  /2  i  =1 A (i  ) mn h  i  z  i +z  i−1  where, z i ≥ 0 and z  i ≥ 0. Introduce a new layer coordinate z i = (z i + z i−1 )/2, which is the distance between the reference plane of the laminate and the middle plane of the ith layer. Then, the condition C mn = 0 yields k/2  i=1 A (i) mn h i z i = k  /2  i  =1 A (i  ) mn h  i z  i k ′ i′ z′ i−1 z′ i z i−1 z i z i k e x Fig. 5.32. Layer coordinates with respect to the reference plane. Chapter 5. Mechanics of laminates 303 Now assume that we have a group of identical layers or plies with the same stiffness coefficients A mn and thicknesses. For example, the laminate could include a 1.5 mm thick 0 ◦ unidirectional layer which consists of 10 plies (the thickness of an elementary ply is 0.15 mm). Arranging these plies above ( z i ) and below (z  i ) the reference plane in such a way that 10  j=1  z j −z  j  = 0 (5.115) we have no coupling for this group of plies. Doing the same with the other layers, we arrive at a laminate with no coupling. Naturally, some additional conditions following from the fact that the laminate is a continuous structure should be satisfied. However even with these conditions, Eq. (5.115) can be met with several systems of ply coordinates, and symmetric arrangement of the plies ( z j = z  j ) is only one of these systems. The general analysis of the problem under discussion has been presented by Verchery (1999). Return to laminates with pre-assigned stacking-sequences for the layers. It follows from Eq. (5.114), we can always make one of the coupling stiffness coefficients equal to zero, e.g., taking e = e st where e st = I (1) st I (0) st (5.116) we get C st = 0 (the rest of coupling coefficients are not zero). Another way to simplify the equations for stiffnesses is to take e = 0, i.e., to take the surface of the laminate as the reference plane. In this case, Eqs. (5.28) take the form B mn = I (0) mn ,C mn = I (1) mn ,D mn = I (2) mn In practical analysis, the constitutive equations for laminates with arbitrary structure are often approximately simplified using the method of reduced or minimum bending stiffnesses described, e.g., by Ashton (1969), Karmishin (1974), and Whitney (1987). To introduce this method, consider the corresponding equation of Eqs. (5.28) for bending stiffnesses, i.e., D mn = I (2) mn −2eI (1) mn +e 2 I (0) mn (5.117) and find the coordinate e delivering the minimum value of D mn . Using the minimum conditions d de D mn = 0, d 2 de 2 D mn > 0 304 Advanced mechanics of composite materials we have e = e mn = I (1) mn I (0) mn (5.118) This result coincides with Eq. (5.116) and yields C mn = 0. Thus, calculating I (1) mn and I (0) mn , we use for each mn = 11, 12, 22, 14, 24, 44 the corresponding value e mn specified by Eq. (5.118). Substitution yields D r mn = I (2) mn −  I (1) mn  2 I (0) mn ,C r mn = 0 (5.119) and the constitutive equations, Eqs. (5.5) become uncoupled. Naturally, this approach is only approximate because the reference plane coordinate should be the same for all stiffnesses, but it is not in the method under discussion. It follows from the foregoing derivation that the coefficients D r mn specified by Eqs. (5.119) do not exceed the actual values of bending stiffnesses, i.e., D r mn ≤ D mn . So, the method of reduced bending stiffnesses leads to underestimation of the laminate bending stiffness. In conclusion, it should be noted that this method is not formally grounded and can yield both good and poor approximation of the laminate behavior, depending on the laminate structure. 5.11. Stresses in laminates The constitutive equations derived in the previous sections of this chapter relate forces and moments acting on the laminate to the corresponding generalized strains. For compos- ite structures, forces and moments should satisfy equilibrium equations, whereas strains are expressed in terms of displacements. As a result, a complete set of equations is formed allowing us to find forces, moments, strains, and displacements corresponding to a given system of loads acting on the structure. Since the subject of structural mechanics is beyond the scope of this book and is discussed elsewhere (Vasiliev, 1993), we assume that this problem has already been solved, i.e., we know either generalized strains ε, γ , and κ entering Eqs. (5.5) or forces and moments N and M. If this is the case, we can use Eqs. (5.5) to find ε, γ , and κ. Now, to complete the analysis, we need to determine the stress acting in each layer of the laminate. To do this, we should first find strains in any ith layer using Eqs. (5.3) which yield ε (i) x = ε 0 x +z i κ x ,ε (i) y = ε 0 y +z i κ y ,γ (i) xy = γ 0 xy +z i κ xy (5.120) where z i is the layer normal coordinate changing over the thickness of the ith layer. If the ith layer is orthotropic with principal material axes coinciding with axes x and y, Chapter 5. Mechanics of laminates 305 (e.g., made of fabric), Hooke’s law provides the stresses we need, i.e., σ (i) x = E (i) x  ε (i) x +ν (i) xy ε (i) y  ,σ (i) y = E (i) y  ε (i) y +ν (i) yx ε (i) x  ,τ (i) xy = G (i) xy γ (i) xy (5.121) where E (i) x,y = E (i) x,y /  1 −ν (i) xy ν (i) yx  and E (i) x ,E (i) y ,G (i) xy ,ν (i) xy ,ν (i) yx are the elastic constants of the layer referred to the principal material axes. For an isotropic layer (e.g., metal or polymeric), we should take in Eqs. (5.121), E (i) x = E (i) y = E i ,ν (i) xy = ν (i) yx = ν i ,G i xy = G i = E i /2(1 +ν i ). Consider a layer composed of unidirectional plies with orientation angle φ i . Using Eqs. (4.69), we can express strains in the principal material coordinates as ε (i) 1 = ε (i) x cos 2 φ i +ε (i) y sin 2 φ i +γ (i) xy sin φ i cos φ i ε (i) 2 = ε (i) x sin 2 φ i +ε (i) y cos 2 φ i −γ (i) xy sin φ i cos φ i γ (i) 12 = 2  ε (i) y −ε (i) x  sin φ i cos φ i +γ (i) xy cos 2φ i (5.122) and find the corresponding stresses, i.e., σ (i) 1 = E (i) 1  ε (i) 1 +ν (i) 12 ε (i) 2  ,σ (i) 2 = E (i) 2  ε (i) 2 +ν (i) 21 ε (i) 1  ,τ (i) 12 = G (i) 12 γ (i) 12 (5.123) where E (i) 1,2 = E (i) 1,2 /  1 −ν (i) 12 ν (i) 21  and E (i) 1 ,E (i) 2 ,G (i) 12 ,ν (i) 12 ,ν (i) 21 are the elastic constants of a unidirectional ply. Thus, Eqs. (5.120)–(5.123) allow us to find in-plane stresses acting in each layer or in an elementary composite ply. Compatible deformation of the layers is provided by interlaminar stresses τ xz , τ yz , and σ z . To find these stresses, we need to use the three-dimensional equilibrium equations, Eqs. (2.5), which yield ∂ τ xz ∂ z =−  ∂ σ x ∂ x + ∂ τ xy ∂ y  , ∂ τ yz ∂ z =−  ∂ σ y ∂ y + ∂ τ xy ∂ x  , ∂ σ z ∂ z =−  ∂ τ xz ∂ x + ∂ τ xz ∂ y  (5.124) Substituting stresses σ x , σ y , and τ xy from Eqs. (5.4) and integrating Eqs. (5.124) with due regard to the forces that can act on the laminate surfaces, we can calculate the transverse shear and normal stresses τ xz , τ yz , and σ z . 306 Advanced mechanics of composite materials 5.12. Example As an example, consider the two-layered cylinder shown in Fig. 5.33 which consists of a ±36 ◦ angle-ply layer with total thickness h 1 = 0.62 mm and 90 ◦ unidirectional layer with thickness h 2 = 0.60 mm. The 200 mm diameter cylinder is made by filament winding from glass–epoxy composite with the following mechanical properties: E 1 = 44 GPa,E 2 = 9.4 GPa,G 12 = 4 GPa,ν 21 = 0.26. Consider two loading cases – axial compression with force P and torsion with torque T as in Fig. 5.33. The cylinder is orthotropic, and to study the problem, we need to apply Eqs. (5.44) with some simplifications specific for this problem. First, we assume that applied loads do not induce interlaminar shear and we can take γ x = 0 and γ y = 0 in Eqs. (5.83) and (5.84). Hence, V x = 0 and V y = 0. In this case, deformations κ x , κ y , and κ xy in Eqs. (5.3) become the changes of curvatures of the laminate. Since the loads shown in Fig. 5.33 deform the cylinder into another cylinder inducing only its axial shortening, change of radius, and rotation of the cross sections, there is no bending in the axial direction (see Fig. 5.3c) or out-of-plane twisting (see Fig. 5.3d) of the laminate. So, we can take κ x = 0 and κ xy = 0 and write constitutive equations, Eqs. (5.44), in the following form N x = B 11 ε 0 x +B 12 ε 0 y +C 12 κ y N y = B 21 ε 0 x +B 22 ε 0 y +C 22 κ y N xy = B 44 γ 0 xy M x = C 11 ε 0 x +C 12 ε 0 y +D 12 κ y M y = C 21 ε 0 x +C 22 ε 0 y +D 22 κ y M xy = C 44 γ 0 xy (5.125) To determine the change of the circumferential curvature κ y , we should take into account that the length of the cross-sectional contour being equal to 2πR before deformation becomes equal to 2πR  1 +ε 0 y  after deformation. Thus, the curvature x, u z, w y, v R T T P P 36° 36° Fig. 5.33. Experimental cylinder. Chapter 5. Mechanics of laminates 307 M y N y M x N x N y N x M x h 1 h 2 M y y t 1 t 2 x z Fig. 5.34. Forces and moments acting on an element of the cylinder under axial compression. change is κ y = 1 R  1 +ε 0 y  − 1 R ≈− ε 0 y R (5.126) The final result is obtained with the assumption that the strain is small (ε 0 y 1). Consider the case of axial compression. The free body diagram for the laminate element shown in Fig. 5.34 yields (see Fig. 5.33) N x =− P 2πR ,N y = 0 As a result, the constitutive equations of Eqs. (5.125) that we need to use for the analysis of this case become B 11 ε 0 x +B 12 ε 0 y =− P 2πR ,B 21 ε 0 x +B 22 ε 0 y = 0 (5.127) M x = C 11 ε 0 x +C 12 ε 0 y ,M y = C 21 ε 0 x +C 22 ε 0 y (5.128) in which B 12 = B 12 − C 12 R , B 22 = B 22 − C 22 R , C 12 = C 12 − D 12 R , C 22 = C 22 − D 22 R (5.129) 308 Advanced mechanics of composite materials The first two equations, Eqs. (5.127), allow us to find strains, i.e., ε 0 x =− P B 22 2πRB ,ε 0 y = PB 21 2πRB (5.130) where B = B 11 B 22 −B 12 B 21 and B 21 = B 12 . The bending moments can be determined with the aid of Eqs. (5.128). The axial moment, M x , has a reactive nature in this problem. The asymmetric laminate in Fig. 5.34 tends to bend in the xz-plane under axial compression of the cylinder. However, the cylinder meridian remains straight at a distance from its ends. As a result, a reactive axial bending moment appears in the laminate. The circumferential bending moment, M y , associated with the change in curvature of the cross-sectional contour in Eq. (5.126) is very small. For numerical analysis, we first use Eqs. (4.72) to calculate stiffness coefficients for the angle-ply layer, i.e., A (1) 11 = 25 GPa,A (1) 12 = 10 GPa,A (1) 22 = 14.1 GPa,A (1) 44 = 11.5 GPa (5.131) and for the hoop layer A (2) 11 = 9.5 GPa,A (2) 12 = 2.5 GPa,A (2) 22 = 44.7 GPa,A (2) 44 = 4 GPa (5.132) Then, we apply Eqs. (5.41) to find the I -coefficients that are necessary for the cases (axial compression and torsion) under study: I (0) 11 = 21.2 GPa mm,I (0) 12 = 7.7 GPa mm,I (0) 22 = 35.6 GPa mm, I (0) 44 = 9.5 GPa mm; I (1) 11 = 10.1 GPa mm 2 ,I (1) 12 = 3.3 GPa mm 2 , I (1) 22 = 27.4 GPa mm 2 ,I (1) 44 = 4.4 GPa mm 2 ; I (2) 11 = 21.7 GPa mm 3 , I (2) 12 = 5.9 GPa mm 3 ,I (2) 22 = 94 GPa mm 3 To determine the stiffness coefficients of the laminate, we should pre-assign the coordinate of the reference surface (a cylindrical surface for the cylinder). Let us put e = 0 for simplicity, i.e., we take the inner surface of the cylinder as the reference surface (see Fig. 5.34). Then, Eqs. (5.28) yield B 11 = I (0) 11 = 21.2 GPa mm,B 12 = I (0) 12 = 7.7 GPa mm, B 22 = I (0) 22 = 35.6 GPa mm; C 11 = I (1) 11 = 10.1 GPa mm 2 , C 12 = I (1) 12 = 3.3 GPa mm 2 ,C 22 = I (1) 22 = 27.4 GPa mm 2 ; D 12 = I (2) 12 = 5.9 GPa mm 3 ,D 22 = I (2) 22 = 94 GPa mm 3 Chapter 5. Mechanics of laminates 309 and in accordance with Eqs. (5.129) for R = 100 mm, B 12 = 7.7 GPa mm, B 22 = 35.3 GPa mm, C 12 = 3.2 GPa mm 2 , C 22 = 26.5 GPa mm 2 Calculation with the aid of Eqs. (5.130) gives ε 0 x =−8.1 ·10 −5 P, ε 0 y = 1.8 · 10 −5 P where P should be substituted in kN. Comparison of the obtained results with experimental data for the cylinder in Fig. 5.35 is presented in Fig. 5.36. To determine the stresses, we first use Eqs. (5.120) which, in conjunction with Eq. (5.126) yield ε (1) x = ε (2) x = ε 0 x ,ε (1) y = ε 0 y  1 − z 1 R  ,ε (2) y = ε 0 y  1 − z 2 R  (5.133) where 0 ≤ z 1 ≤ h 1 and h 1 ≤ z 2 ≤ h 1 +h 2 . Since (h 1 +h 2 )/R = 0.0122 for the cylinder under study, we can neglect z 1 /R and z 2 /R in comparison with unity and write ε (1) y = ε (2) y = ε 0 y (5.134) Fig. 5.35. Experimental composite cylinder in test fixtures. 310 Advanced mechanics of composite materials 10 20 30 40 0.10−0.1−0.2−0.3−0.4 P, kN e y 0 , %e y 0 , % Fig. 5.36. Dependence of axial  ε 0 x  and circumferential  ε 0 y  strains of a composite cylinder on the axial force: analysis; ◦ experiment. Applying Eqs. (5.122) to calculate the strains in the plies’ principal material coordinates and Eqs. (5.123) to find the stresses, we get • in the angle-ply layer, σ (1) 1 =−0.26 P Rh ,σ (1) 2 =−0.028 P Rh ,τ (1) 12 = 0.023 P Rh • in the hoop layer, σ (2) 1 = 0.073 P Rh ,σ (2) 2 =−0.089 P Rh ,τ (2) 12 = 0 where h = h 1 + h 2 is the total thickness of the laminate. To calculate the interlaminar stresses acting between the angle-ply and the hoop layers, we apply Eqs. (5.124). Using Eqs. (5.4) and taking Eqs. (5.133) and (5.134) into account, we first find the stresses in the layers referred to the global coordinate frame x, y, z, i.e., σ (i) x = A (i) 11 ε 0 x +A (i) 12 ε 0 y ,σ (i) y = A (i) 21 ε 0 x +A (i) 22 ε 0 y ,τ (i) xy = 0 (5.135) where i = 1, 2 and A (i) mn are given by Eqs. (5.131) and (5.132). Since these stresses do not depend on x and y, the first two equations in Eqs. (5.124) yield ∂ τ xz ∂ z = 0, ∂ τ yz ∂ z = 0 This means that both interlaminar shear stresses do not depend on z. However, on the inner and on the outer surfaces of the cylinder the shear stresses are equal to zero, so τ xz = 0 and τ yz = 0. The fact that τ yz = 0 is natural. Both layers are orthotropic and do not tend to twist under axial compression of the cylinder. Concerning τ xz = 0, a question arises as Chapter 5. Mechanics of laminates 311 to how compatibility of the axial deformations of the layers with different stiffnesses can be provided without interlaminar shear stresses. The answer follows from the model used above to describe the stress state of the cylinder. According to this model, the transverse shear deformation γ x is zero. Actually, this condition can be met if part of the axial force applied to the layer is proportional to the layer stiffness, i.e., as P 1 =−2πσ (1) x h 1 = 2πh 1  A (1) 11 ε 0 x +A (1) 12 ε 0 y  P 2 =−2πσ (2) x h 2 = 2πh 2  A (2) 11 ε 0 x +A (2) 12 ε 0 y  (5.136) Substituting strains from Eqs. (5.130), we can conclude that within the accuracy of a small parameter h/R (which was neglected in comparison with unity when we calculated stresses) P 1 +P 2 =−P, and that the axial strains are the same even if the layers are not bonded together. In the middle part of a long cylinder, the axial forces are automatically distributed between the layers in accordance with Eqs. (5.136). However, in the vicinity of the cylinder ends, this distribution depends on the loading conditions. The corresponding boundary problem will be discussed further in this section. The third equation in Eqs. (5.124) formally yields σ z = 0. However, this result is not correct because the equation corresponds to a plane laminate and is not valid for the cylinder. In cylindrical coordinates, the corresponding equation has the following form (see e.g., Vasiliev, 1993) ∂ ∂ z  1 + z R  σ z  =−   1 + z R  ∂ τ xz ∂ x + ∂ τ yz ∂ y − σ y R  Taking τ xz = 0 and τ yz = 0, substituting σ y from Eqs. (5.135), and integrating, we obtain σ z = R R + z  1 R  z 0  A 21 ε 0 x +A 22 ε 0 y  dz +C  (5.137) where, A mn (mn = 21, 22) are the step-wise functions of z, i.e., A mn = A (1) mn for 0 ≤ z ≤ h 1 A mn = A (2) mn for h 1 ≤ z ≤ h = h 1 +h 2 and C is the constant of integration. Since no pressure is applied to the inner surface of the cylinder, σ z (z = 0) = 0 and C = 0. Substitution of the stiffness coefficients, Eqs. (5.131), (5.132), and strains, Eqs. (5.130), into Eq. (5.137) yields σ (1) z =−0.068 P Rh · z R + z σ (2) z = σ (1) z (z = h 1 ) +0.07 P Rh · z −h 1 R + z (5.138) [...]...312 Advanced mechanics of composite materials h2 h1 0 2 4 6 sz 104 Fig 5.37 Distribution of the normalized radial stress σ z = σz Rh/P over the laminate thickness (2) On the outer surface of the cylinder, z = h and σz = 0 which is natural because this surface is free of any loading The distribution of σz over the laminate thickness is shown in Fig 5.37 As can be seen, interaction of the layers... composites within the accuracy determined by the scatter of experimental results As has been already noted, this criterion ignores the interaction of stresses However, this interaction takes place 326 Advanced mechanics of composite materials s2, MPa 300 200 100 s1, MPa 0 100 200 300 400 500 600 100 −200 −300 Fig 6.6 Failure envelope for glass–epoxy fabric composite in plane (σ1 , σ2 ) ( criterion, Eqs (6.2);... i.e., σ1 ≤ σ + 1 |σ1 | ≤ σ − 1 if σ1 > 0 (6.3) if σ1 < 0 328 Advanced mechanics of composite materials s2, MPa 10 0 5 10 15 20 25 t12, MPa 10 −20 −30 −40 Fig 6 .10 Failure envelope for carbon–carbon unidirectional composite in plane (σ2 , τ12 ) ( stress criterion, Eqs (6.2); ( ) experimental data ) maximum Actually, there exist unidirectional composites with a very brittle matrix (carbon or ceramic) for... Journal of Composite Materials, 30(8) Karmishin, A.V (1974) Equations for nonhomogeneous thin-walled elements based on minimum stiffnesses Applied Mechanics, (Prikladnaya Mekhanika), 10( 6), 34–42 (in Russian) Morozov, E.V (2006) The effect of filament-winding mosaic patterns on the strength of thin-walled composite shells Composite Structures, 76, 123–129 Vasiliev, V.V (1993) Mechanics of Composite Structures... which can only be studied using the maximum strain criterion This is longitudinal compression of a unidirectional ply as discussed earlier 330 Advanced mechanics of composite materials Fig 6.12 Failure modes of a unidirectional glass–epoxy composite under longitudinal compression in Section 3.4.4 Under this type of loading, only longitudinal stress σ1 is induced, whereas σ2 = 0 and τ12 = 0 Nevertheless,... interaction of stresses increases 1 2 material strength under compression, the combination of compressive failure stresses − − |σ1 | = σ1 and |σ2 | = σ2 results in material failure Then σ1 σ− 1 2 σ 1 σ2 − − −+ σ1 σ2 σ2 σ− 2 2 + τ12 τ 12 2 =1 Comparison of this criterion with experimental data is presented in Fig 6.8 (6.16) 334 Advanced mechanics of composite materials Now consider unidirectional composites... criteria and strength of laminates s2, −120 327 MPa 100 −80 −60 −40 −20 0 0 −20 −40 −60 −80 100 −120 s1, MPa Fig 6.8 Failure envelope for glass–phenolic fabric composite loaded with compression in plane (σ1 , σ2 ) ( ) maximum stress criterion, Eqs (6.2); (- - -) polynomial criterion, Eqs (6.16); ( ) experimental data s2, MPa 300 200 100 −300 −200 100 100 200 300 500 s1, MPa 100 −300 Fig 6.9 Failure... thickness of the laminate 5.13 References Aleksandrov, A.Ya., Brukker, L.E., Kurshin, L.M and Prusakov, A.P (1960) Analysis of Sandwich Plates Mashinostroenie, Moscow (in Russian) Ashton, J.E (1969) Approximate solutions for unsymmetrically laminated plates Journal of Composite Materials, 3, 189–191 Chen, H.-J and Tsai, S.W (1996) Three-dimensional effective moduli of symmetric laminates Journal of Composite. .. of the three values provided by Eqs (6.5) for tension or Eqs (6.6) for compression The experimental data of S.W Tsai taken from (Jones, 1999) and corresponding to a glass–epoxy unidirectional composite are presented in Fig 6.11 As can be seen, the maximum stress criterion (solid lines) Chapter 6 Failure criteria and strength of laminates sx, MPa 100 0 329 sx, MPa 100 0 1 500 1 500 3 3 200 100 2 200 100 ... element of the cylinder under torsion Chapter 5 Mechanics of laminates 319 T, kN m 2.5 2 1.5 1 − e45, % −0.3 + e45, % −0.2 −0.1 0 0.1 0.2 ± Fig 5.42 Dependence of ε45 on the torque T for a composite cylinder: 0.3 analysis; ◦ experiment (0) For the cylinder under study with B44 = I44 = 9.5 GPa mm and R = 100 mm, we get ± ε45 = ± T = ±0.84 · 10 6 T 4πR 2 B44 where T is measured in Nm A comparison of the . 5.35. Experimental composite cylinder in test fixtures. 310 Advanced mechanics of composite materials 10 20 30 40 0 .10 0.1−0.2−0.3−0.4 P, kN e y 0 , %e y 0 , % Fig. 5.36. Dependence of axial  ε 0 x  and. z (5.138) 312 Advanced mechanics of composite materials 2046 h 1 h 2 s z . 10 4 Fig. 5.37. Distribution of the normalized radial stress σ z = σ z Rh/P over the laminate thickness. On the outer surface of. laminated plates. Journal of Composite Materials, 3, 189–191. Chen, H J. and Tsai, S.W. (1996). Three-dimensional effective moduli of symmetric laminates. Journal of Composite Materials, 30(8). Karmishin,