ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 7 potx

Chapter 4. Mechanics of a composite layer 197 t xy t xy s y s x Fig. 4.44. A cross-ply layer in a plane stress state. ∗ ∗ s 1 , s 2 , t 12 e 1 , e 2 , g 12 s 1 s 2 s 2 s 2 s 2 s 1 s 1 s 2 s 1 t 12 t 12 Fig. 4.45. Stress–strain diagrams of a unidirectional ply simulating its behavior in the laminate and allowing for cracks in the matrix. (1) For the first stage of loading (before the cracks appear), the strains are calculated with the aid of Eqs. (4.114) and (4.115) providing ε (1) x (σ ), ε (1) y (σ ), and γ (1) xy (σ ), where σ = (σ x ,σ y ,τ xy ) is the given combination of stresses. Using Eqs. (4.112), we find stresses σ 1 , σ 2 and τ 12 in principal material coordinates for all the plies. (2) We determine the combination of stresses σ ∗ 1k , σ ∗ 2k , and τ ∗ 12k which induce the first failure of the matrix in some ply and indicate the number of this ply, say k, applying the appropriate strength criterion (see Section 6.2). Then, the corresponding stresses σ ∗ = (σ ∗ x ,σ ∗ y ,τ ∗ xy ) and strains ε (1) x (σ ∗ ), ε (1) y (σ ∗ ), and γ (1) xy (σ ∗ ) are calculated. (3) To proceed, i.e., to study the material behavior for σ>σ ∗ , we need to consider two possible cases for the layer stiffnesses. For this purpose, we should write Eqs. (4.114) for stiffness coefficients in a more general form, i.e., 198 Advanced mechanics of composite materials A 11 = m  i=1 E (i) 1 h (i) 0 + n  j=1 E (j) 2 h (j) 90 ,A 22 = m  i=1 E (i) 2 h (i) 0 + n  j=1 E (j) 1 h (j) 90 A 12 = m  i=1 ν (i) 12 E (i) 1 h (i) 0 + n  j=1 ν (j) 12 E (j) 1 h (j) 90 ,A 44 = m  i=1 G (i) 12 h (i) 0 + n  j=1 G (j) 12 h (j) 90 (4.135) where h (i) 0 = h (i) 0 /h and h (j) 90 = h (j) 90 /h. (a) If σ 2k > 0inthekth ply, it can work only along the fibers, and we should calculate the stiffnesses of the degraded layer taking E k 2 = 0, G k 12 = 0, and ν k 12 = 0in Eqs. (4.135). (b) If σ 2k < 0inthekth ply, it cannot work only in shear, so we should take G k 12 = 0in Eqs. (4.135). Thus, we find coefficients A (2) st (st = 11, 12, 22, 44) corresponding to the second stage of loading (with one degraded ply). Using Eqs. (4.116) and (4.115) we can determine E (2) x , E (2) y , G (2) xy , ν (2) xy , ν (2) yx and express the strains in terms of stresses, i.e., ε (2) x (σ ), ε (2) y (σ ), γ (2) xy (σ ). The final strains corresponding to the second stage of loading are calculated as ε f x = ε (1) x (σ ∗ ) +ε (2) x (σ − σ ∗ ), ε f y = ε (1) y (σ ∗ ) +ε (2) y (σ − σ ∗ ) γ f xy = γ (1) xy (σ ∗ ) +γ (2) xy (σ − σ ∗ ) To study the third stage, we should find σ 1 , σ 2 , and τ 12 in all the plies, except the kth one, identify the next degraded ply and repeat step 3 of the procedure which is continued up to failure of the fibers. The resulting stress–strain curves are multi-segmented broken lines with straight segments and kinks corresponding to degradation of particular plies. The foregoing procedure was described for a cross-ply layer consisting of plies with different properties. For the layer made of one and the same material, there are only three stages of loading – first, before the plies degradation, second, after the degradation of the longitudinal or the transverse ply only, and third, after the degradation of all the plies. As a numerical example, consider a carbon–epoxy cylindrical pressure vessel consisting of axial plies with total thickness h 0 and circumferential plies with total thickness h 90 . The vessel has the following parameters: radius R = 500 mm, total thickness of the wall h = 7.5mm,h 0 = 2.5mm,h 90 = 5mm. The mechanical characteristics of a carbon–epoxy unidirectional ply are E 1 = 140 GPa, E 2 = 11 GPa, ν 12 = 0.0212, ν 21 = 0.27, σ + 1 = 2000 MPa, σ + 2 = 50 MPa. Axial, σ x , and circumferential, σ y , stresses are expressed as (see Fig. 4.46) σ x = pR 2h ,σ y = pR h (4.136) where p is the internal pressure. Chapter 4. Mechanics of a composite layer 199 x y s y s x Fig. 4.46. Element of a composite pressure vessel. Using Eqs. (4.114) and (4.116), we calculate first the stiffness coefficients. The result is as follows A 11 = 54.1 GPa, A 12 = 3 GPa, A 22 = 97.1 GPa E x = 54 GPa, E y = 97 GPa, ν xy = 0.055,ν yx = 0.031 (4.137) Substituting stresses, Eqs. (4.136) into the constitutive equations, Eqs. (4.115), we obtain ε (1) x (p) = pR h  1 2E x − ν xy E y  = 0.58 × 10 −3 p ε (1) y (p) = pR h  1 E y − ν yx 2E x  = 0.66 × 10 −3 p where p is measured in mega pascals. For axial plies, ε x = ε 1, 0 and ε y = ε 2, 0 . The corresponding stresses are σ (1) 1, 0 (p) = E 1 (ε 1, 0 +ν 12 ε 2, 0 ) = 83.2p, σ (1) 2, 0 (p) = E 2 (ε 2, 0 +ν 21 ε 1, 0 ) = 9.04p For circumferential plies, ε x = ε 2, 90 ,ε y = ε 1, 90 and σ (1) 1,90 (p)=E 1 (ε 1,90 +ν 12 ε 2,90 ) =94.15p, σ (1) 2,90 (p)=E 2 (ε 2,90 +ν 21 ε 1,90 ) =8.4p As can be seen, σ (1) 2, 0 >σ (1) 2, 90 . This means that the cracks appear first in the axial plies under the pressure p ∗ that can be found from the equation σ (1) 2, 0 (p ∗ ) = σ + 2 . The result is p ∗ = 5.53 MPa. To study the second stage of loading for p>p ∗ , we should put E 2 = 0, and ν 12 = 0in Eqs. (4.135) for the axial plies. Then, the stiffness coefficients and elastic constants become A 11 = 54.06 GPa, A 12 = 2 GPa, A 22 = 93.4 GPa E x = 54 GPa, E y = 93.3 GPa, ν xy = 0.037,ν yx = 0.021 200 Advanced mechanics of composite materials The strains and stresses in the plies are ε (2) x (p) = 0.59 × 10 −3 p, ε (2) y p) = 0.7 × 10 −3 p, σ (2) 1, 0 (p) = 82.6p σ (2) 1, 90 p) = 99.8p, σ (2) 2, 90 (p) = 8.62p The total transverse stress in the circumferential plies can be calculated as σ 2, 90 = σ (1) 2, 90 (p ∗ ) +8.62(p −p ∗ ) Using the condition σ 2, 90 (p ∗∗ ) = σ + 2 , we find the pressure p ∗∗ = 5.95 MPa at which cracks appear in the matrix of the circumferential plies. For p ≥ p ∗∗ , we should take E 2 = 0 and ν 12 = 0 for all the plies. Then A 11 = 46.2 GPa, A 12 = 0,A 22 = 93.4 GPa E x = 46.2 GPa, E y = 93.4 GPa, ν xy = ν yx = 0 (4.138) ε (3) x (p) = 0.72 × 10 −3 p, ε (3) y (p) = 0.71 × 10 −3 p σ (3) 1,0 (p) = 100.8p, σ (3) 1,90 (p) = 99.4p The total stresses acting along the fibers are σ 1, 0 (p) = σ (1) 1, 0 (p ∗ ) +σ (2) 1, 0 (p ∗∗ −p ∗ ) +σ (3) 1, 0 (p − p ∗∗ ) = 100.8p −105 σ 1, 90 (p) = σ (1) 1, 90 (p ∗ ) +σ (2) 1, 90 (p ∗∗ −p ∗ ) +σ (3) 1, 90 (p − p ∗∗ ) = 99.4p −28.9 To determine the ultimate pressure, we can use two possible strength conditions – for axial fibers and for circumferential fibers. The criterion σ 1,0 (p) = σ + 1 yields p = 20.9 MPa, whereas the criterion σ 1,90 (p) = σ + 1 gives p = 20.4 MPa. Thus, the burst pressure governed by failure of the fibers in the circumferential plies, is p = 20.4 MPa. The strains can be calculated for all three stages of loading using the following equations • for p ≤ p ∗ ε x,y (p) = ε (1) x,y (p) • for p ∗ <p≤ p ∗∗ ε x,y (p) = ε (1) x,y (p ∗ ) +ε (2) x,y (p − p ∗ ) • for p ∗∗ <p≤ p ε x,y (p) = ε (1) x,y (p ∗ ) +ε (2) x,y (p ∗∗ −p ∗ ) +ε (3) x,y (p − p ∗∗ ) Chapter 4. Mechanics of a composite layer 201 0 5 10 15 20 25 0 0.5 1 1.5 p, MPa e y , % Fig. 4.47. Dependence of the axial and the circumferential strains of the carbon–epoxy pressure vessel on pressure: model allowing for cracks in the matrix; model ignoring cracks in the matrix; model ignoring the matrix. For the pressure vessel under study, the dependency of the circumferential strain on pressure is shown in Fig. 4.47 (solid line). The circles correspond to failure of the matrix and fibers. For comparison, consider two limiting cases. First, assume that no cracks occur in the matrix, and the material stiffness is specified by Eqs. (4.137). The corresponding diagram is shown in Fig. 4.47 with a dashed line. Second, suppose that the load is taken by the fibers only, i.e., use the monotropic model of a ply introduced in Section 3.3. Then, the material stiffnesses are given by Eqs. (4.138). The corresponding result is also presented in Fig. 4.47. It follows from this figure that all three models give close results for the burst pressure (which is expected since σ + 2  σ + 1 ), but different strains. 4.4.3. Two-matrix composites The problem of the analysis of a cracked cross-ply composite laminate has been studied by Tsai and Azzi (1966), Vasiliev and Elpatievckii (1967), Vasiliev et al. (1970), Hahn and Tsai (1974), Reifsnaider (1977), Hashin (1987), and many other authors. In spite of this, the topic is still receiving repeated attention in the literature (Lungren and Gudmundson, 1999). Taking into account that matrix degradation leads to reduction of material stiffness and fatigue strength, absorption of moisture and many other consequences that are difficult to predict but are definitely undesirable, it is surprising how many efforts have been undertaken to study this phenomenon rather than try to avoid it. At first glance, the problem looks simple – all we need is to synthesize unidirectional composite whose ultimate elongations along and across the fibers, i.e., ε 1 and ε 2 are the same. Actually, 202 Advanced mechanics of composite materials the problem is even simpler, because ε 2 can be less than ε 1 by a factor that is equal to the safety factor of the structure. This means that matrix degradation can occur but at the load that exceeds the operational level (the safety factor is the ratio of the failure load to the operational load and can vary from 1.25 up to 3 or more depending on the application of a particular composite structure). Returning to Table 4.2, in which ε 1 and ε 2 are given for typical advanced composites, we can see that ε 1 > ε 2 for all the materials and that for polymeric matrices the problem could be, in principle, solved if we could increase ε 2 up to about 1%. Two main circumstances hinder the direct solution of this problem. The first is that being locked between the fibers, the matrix does not show the high elongation that it has under uniaxial tension and behaves as a brittle material (see Section 3.4.2). To study this effect, epoxy resins were modified to have different ultimate elongations. The corre- sponding curves are presented in Fig. 4.48 (only the initial part of curve 4 is shown in this figure, the ultimate elongation of this resin is 60%). Fiberglass composites that have been fabricated with these resins were tested under transverse tension. As can be seen in Fig. 4.49, the desired value of ε 2 (that is about 1%) is reached if the matrix elonga- tion is about 60%. However, the stiffness of this matrix is relatively low, and the second circumstance arises – matrix material with low stiffness cannot provide sufficient stress diffusion in the vicinity of damaged or broken fibers (see Section 3.2.3). As a result, the main material characteristic – its longitudinal tensile strength – decreases. Experimental results corresponding to composites with resins 1, 2, 3, and 4 are presented in Fig. 4.50. Thus, a significant increase in transverse elongation is accompanied with an unacceptable drop in longitudinal strength (see also Chiao, 1979). One of the possible ways for synthesizing composite materials with high transverse elongation and high longitudinal strength is to combine two matrix materials – one with 0 20 40 60 80 100 120 0 4 8 12 16 20 4 2 3 1 60% s m , MPa e m , % Fig. 4.48. Stress–strain curves for epoxy matrices modified for various ultimate elongations. Chapter 4. Mechanics of a composite layer 203 0 10 20 30 40 50 0 0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 s 2 , MPa e 2 , % Fig. 4.49. Stress–strain curves for transverse tension of unidirectional fiberglass composites with various epoxy matrices (numbers on the curves correspond to Fig. 4.48). 1000 1100 1200 1300 1400 1500 0 204060 1 2 3 4 s 1 , MPa e m , % Fig. 4.50. Dependence of the longitudinal strength on the matrix ultimate elongation (numbers on the curve correspond to Figs. 4.48 and 4.49). 204 Advanced mechanics of composite materials high stiffness to bind the fibers and the other with high elongation to provide the appro- priate transverse deformability (Vasiliev and Salov, 1984). The manufacturing process involves two-stage impregnation. At the first stage, a fine tow is impregnated with a high- stiffness epoxy resin (of the type 2 in Fig. 4.48) and cured. The properties of the composite fiber fabricated in this way are as follows • number of elementary glass fibers in the cross section – 500; • mean cross-sectional area – 0.15 mm 2 ; • fiber volume fraction – 0.75; • density – 2.2 g/cm 3 ; • longitudinal modulus – 53.5 GPa; • longitudinal strength – 2100 MPa; • longitudinal elongation – 4.5%; • transverse modulus – 13.5 GPa; • transverse strength – 400 MPa; • transverse elongation – 0.32%. At the second stage, a tape formed of composite fibers is impregnated with a highly deformable epoxy matrix whose stress–strain diagram is presented in Fig. 4.51. The microstructure of the resulting two-matrix unidirectional composite is shown in Fig. 4.52 (the dark areas are cross sections of composite fibers, the magnification is not sufficient to see the elementary glass fibers). Stress–strain diagrams corresponding to transverse tension, compression, and in-plane shear of this material are presented in Fig. 4.16. The main mechanical characteristics of the two-matrix fiberglass composite are listed in Table 4.3 (material No. 1). As can be seen, two-stage impregnation results in relatively low fiber volume content (about 50%). Material No. 2 that is composed of composite fibers and a conventional epoxy matrix has also low fiber fraction, but its transverse elongation is 10 times lower than that of material No. 1. Material No. 3 is a conventional 0 4 8 12 16 20 0 20406080100120 e m , % s m , MPa Fig. 4.51. Stress–strain diagram of a deformable epoxy matrix. Chapter 4. Mechanics of a composite layer 205 Fig. 4.52. Microstructure of a unidirectional two-matrix composite. Table 4.3 Properties of glass–epoxy unidirectional composites. No. Material components Fiber volume fraction Longitudinal strength σ + 1 (MPa) Ultimate transverse strain ε + 2 (%) Density ρ (g/cm 3 ) Specific strength σ + 1 /ρ ×10 3 (m) 1 Composite fibers and deformable matrix 0.51 1420 3.0 1.83 77.6 2 Composite fibers and high-stiffness matrix 0.52 1430 0.3 1.88 76.1 3 Glass fibers and high- stiffness matrix 0.67 1470 0.2 2.07 71.0 4 Glass fibers and deformable matrix 0.65 1100 1.2 2.02 54.4 glass–epoxy composite that has the highest longitudinal strength and the lowest transverse strain. Comparing materials No. 1 and No. 3, we can see that although the fiber volume fraction of the two-matrix composite is lower by 24%, its longitudinal strength is less than that of a traditional composite by 3.4% only (because the composite fibers are not damaged in the processing of composite materials), whereas its specific strength is a bit higher (due to its lower density). Material No. 4 demonstrates that direct application of a highly deformable matrix allows us to increase transverse strains but results in a 23% reduction in longitudinal specific strength. Thus, two-matrix glass–epoxy composites have practically the same longitudinal strength as conventional materials but their transverse elongation is greater by an order of magnitude. Comparison of a conventional cross-ply glass–epoxy layer and a two-matrix one is presented in Fig. 4.53. Line 1 corresponds to a traditional material and has, typical for this material, a kink corresponding to matrix failure in the transverse plies (see also Fig. 4.37). A theoretical diagram was plotted using the procedure described above. Line 2 corresponds to a two-matrix composite and was plotted using Eqs. (4.60). As can be seen, there is no kink on the stress–strain diagram. To prove that no cracks appear in the matrix 206 Advanced mechanics of composite materials 0 100 200 300 400 500 0 0.4 0.8 1.2 1.6 2 s, MPa e x ,% 1 2 Fig. 4.53. Stress–strain diagrams of a conventional (1) and two-matrix (2) cross-ply glass–epoxy layers under tension: theoretical prediction;  experiment. Fig. 4.54. Intensity of acoustic emission for a cross-ply two-matrix composite (above) and a conventional fiberglass composite (below). of this material under loading, the intensity of acoustic emission was recorded during loading. The results are shown in Fig. 4.54. Composite fibers of two-matrix materials can also be made from fine carbon or aramid tows, and the deformable thermosetting resin can be replaced with a thermoplastic matrix (Vasiliev et al., 1997). The resulting hybrid thermoset–thermoplastic unidirectional com- posite is characterized by high longitudinal strength and transverse strain exceeding 1%. Having high strength, composite fibers are not damaged in the process of laying-up [...]... equation of Eqs (4. 173 ) yields in conjunction with the first equation of Eqs (4. 174 ) γxy = ∂ux du = dy ∂y Using this expression and substituting ε from Eq (4. 170 ) in Eq (4. 172 ), we arrive at τxy = G+ xy 1−η du − ηxy, x ε dy (4. 177 ) + + where η = ηx, xy ηxy, x (txy + ∂txy ∂y (txy + sxd txz dy dx txyd dy)d ∂txy ∂x sxd txyd Fig 4 .77 Forces acting on the infinitesimal element of a ply dx)d Chapter 4 Mechanics of. .. 30 45 60 75 90 −0.2 Fig 4 .70 Dependencies of the normalized stresses in the plies on the ply orientation angle 226 Advanced mechanics of composite materials layer cannot take the load Indeed, putting E2 = G12 = ν12 = 0 in Eqs (4 .72 ), we obtain the following stiffness coefficients A11 = E1 cos4 φ, A22 = E1 sin4 φ, A12 = E1 sin2 φ cos2 φ With these coefficients, the first equation of Eqs (4.1 47) yields... (4.152) 216 Advanced mechanics of composite materials sx , MPa 80 60 40 20 0 0 0.4 0.8 1.2 ex, % (a) sx , MPa 16 12 8 4 0 0 1 2 3 4 5 ex, % (b) sx , MPa 12 8 4 0 0 0.4 0.8 1.2 1.6 ex, % (c) Fig 4.62 Theoretical (solid lines) and experimental (dashed lines) stress–strain diagrams for ±30◦ (a), ±45◦ (b), and 75 ◦ (c) angle-ply two-matrix composites under uniaxial tension Chapter 4 Mechanics of a composite. .. total thickness of the layer Stresses with superscripts ‘+’ and ‘−’ are related to strains εx , εy , and γxy (which are presumed to be the same for all the plies) 210 Advanced mechanics of composite materials F 1 0.995 0.99 0.95 0.9 0.94 0.92 0.8 0 .79 0.81 t = 0.68 0 .7 0 .74 0.68 0.68 0.68 0.59 0.6 0.58 0.5 0.49 0.4 a = 0.9 0.3 0.26 0.3 0.2 a = 0.5 0.19 3.15 1.8 0 1 0.03 3 3.6 4 2 0.16 0. 175 0.1 0 0.4... the matrix However, Fig 4 .71 A failure mode of ±30◦ angle-ply specimen Chapter 4 Mechanics of a composite layer 2 27 sx , MPa 100 80 ± 60° 60 ± 75 ° 40 20 0 0 0.2 0.4 0.6 0.8 ex, % Fig 4 .72 Experimental (solid lines) and calculated (dashed lines) stress–strain diagrams for ±60 and 75 ◦ angle-ply carbon–epoxy layers this is not the case To explain why, consider the last equation of Eqs (4.168), i.e., tan... Fig 4 .76 A model simulating the plies interaction 230 Advanced mechanics of composite materials Integration of the first equation yields for the +φ and −φ plies u+φ = ε · x + u(y), x u−φ = ε · x − u(y) x (4. 174 ) where u(y) is the displacement shown in Fig 4 .76 This displacement results in the following transverse shear deformation and transverse shear stress γxz = 2 u(y), δ τxz = Gxz γxz (4. 175 ) where... transverse shear modulus of the ply specified by Eqs (4 .76 ) Consider the equilibrium state of +φ ply element shown in Fig 4 .77 Equilibrium equations can be written as δ ∂τxy = 0, ∂x δ ∂τxy − 2τxz = 0 ∂y (4. 176 ) The first of these equations shows that τxy does not depend on x Since the axial stress, σx , in the middle part of a long specimen also does not depend on x, Eqs (4. 171 ) and (4. 173 ) allow us to conclude... 0.16 0. 175 0.1 0 0.4 a = 0.1 0.1 0.1 0 5 6 7 0.06 0.05 0.03 8 9 0.034 0.014 lc 10 Fig 4.56 Dependencies of function F on the normalized width of the strip for α = 0.1, 0.5, and 0.9 lc 4 3.6 3 3.15 2.6 2 1.8 1 0 0 0.2 0.4 0.6 0.8 1 a Fig 4. 57 Normalized width of the strip l c as a function of the relative thickness of transverse ply α Chapter 4 Mechanics of a composite layer +f 211 s+ x t+ xy t xz sx... infinitesimal element of a ply dx)d Chapter 4 Mechanics of a composite layer 231 Substitution of Eqs (4. 175 ) and (4. 177 ) into the second equation of Eqs (4. 176 ) provides the following governing equation for the problem under study d2 u − k2u = 0 dy 2 (4. 178 ) in which 4Gxz (1 − η) G+ δ 2 xy k2 = Using the symmetry conditions, we can present the solution of Eq (4. 178 ) as u = C sinh ky The constant C can be determined... by uniaxial tension 214 Advanced mechanics of composite materials where Ex = + Ex + + 1 − ηx, xy ηxy, x = + Ex 1− (4.149) G+ xy + 2 + (ηxy, x ) Ex is the modulus of the ±φ angle-ply layer Under pure shear of an angle-ply layer, its plies are loaded with the additional normal stresses These stresses can be found if we take εx = 0 and εy = 0 in the first two equations of Eqs (4 .75 ) The result is σx = −τxy . and deformable matrix 0.51 1420 3.0 1.83 77 .6 2 Composite fibers and high-stiffness matrix 0.52 1430 0.3 1.88 76 .1 3 Glass fibers and high- stiffness matrix 0. 67 1 470 0.2 2. 07 71.0 4 Glass fibers and deformable. Stress–strain diagram of a deformable epoxy matrix. Chapter 4. Mechanics of a composite layer 205 Fig. 4.52. Microstructure of a unidirectional two-matrix composite. Table 4.3 Properties of glass–epoxy. 93.3 GPa, ν xy = 0.0 37, ν yx = 0.021 200 Advanced mechanics of composite materials The strains and stresses in the plies are ε (2) x (p) = 0.59 × 10 −3 p, ε (2) y p) = 0 .7 × 10 −3 p, σ (2) 1,