# Calculation of Energy Release Rate in Mode I Delamination of Angle Ply Laminated Composites

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2 62 Copyright c 2007 ICCES ICCES, vol.1, no.2, pp.61-67, 2007 The double cantilever beam (DCB) test [7] is the most widely used method for determining the Mode I toughness of composite materials. The test specimen, the applied load, P, to the two arms and the corresponding load line displacement, δ, are schematically illustrated in Fig. 1. A model based on an Euler-Bernoulli beam on an elastic Winkler foundation for analysing the DCB specimen was originally developed by Kanninen [8,9] for isotropic materials and by Williams [10] for transversely isotropic materials. This model is extended to angle-ply laminates by Ozdil and Carlsson [11]. The beam displacements derived from this model are used to calculate the compliance and strain energy release rate of the DCB specimen. Recently, Hamed et al. [12], introduced an improved analytical model for delamination in composite beams under general edge loading which also takes into account the shear-thickness deformation. In the present study, Hamed model together with the compliance equation has been used to develop a method of determining the Mode I interlaminar fracture toughness in thick angle ply laminated composites. Figure 1: Angle ply DCB specimen divided into three regions Theoretical Analysis Consider the DCB specimen in Fig. 1 made of an angle ply laminate and divided into three regions I, II and III. Assuming a second order shear thickness deformation beam theory (SSTDBT), the displacement field is u 1 (x,y,z)=u(x)+zψ x (x)+z 2 η x (x), u 2 (x,y,z)=0, (1) u 3 (x,y,z)=w(x)+zψ z (x), in all three regions, where u 1, u 2 and u 3 are the displacement components of a point in the x, y and z directions, respectively. Note that the superscripts I, II and III will be given to all of the displacement components to distinct them for the regions I, II and III. SubstitutingEq. 1 into the strain-displacement relations of elasticity we obtain ε x = ε1 0 +zκ0 1 +z 2 κ1, 1 ε y = 0, ε z = ε3, 0 γ yz = 0, γ xz = ε5 0 +zκ5 0, γ xy = 0 (2)

3 Calculation of Energy Release Rate 63 where ε 0 1 = u, κ 1 0 = ψ x, κ 1 1 = η x, ε 0 3 = ψ z, ε 0 5 = ψ x +w, κ 0 5 = 2η x +ψ z (3) and a prime denotes differentiation with respect to x. Since we are developing a beam theory, a state of plane stress is presented and therefore σ y = σ yz = 0. (4) Next by imposing Eq. 4 on three dimensional Hooke s law of elasticity, the plane stress constitutive law for the kth layer of each region is σ x σ z σ xy (k) C 11 C 13 C 16 = C 13 C 33 C 36 C 16 C 36 C 66 (k) ε x ε z γ xy (k), σ (k) xz = C (k) 55 γ (k) xz, (5) where C ij = C ij C i2c j2 i, j = 1,3,6 and C 55 = C 55 C (6) C 22 C 44 In Eq. 6, C ij denote off-axis stiffness coefficients of the kth layer [13]. Using the principle of minimum total potential energy, and considering the displacement field in Eq. 1, the equilibrium equations together with the boundary conditions (B.C. s) at boundaries and continuity conditions (C.C. s) and the equilibrium conditions (E.C. s) at the intersection of three regions will be obtained. Further by substituting the laminate constitutive relations into the equilibrium equations, the governingequilibriumequationsintermsof displacement componentsfor thethree regions will be found as A 11 u A 13 ψ z B 11 ψ x D 11 η x = 0, B 11 u B 13 ψ z D 11ψ x E 11η x +A 55(ψ x +w )+B 55 (2η x +ψ z )=0, D 11 u D 13 ψ z E 11 ψ x F 11 η x +2B 55 (ψ x +w )+2D 55 (2η x +ψ z)=0, A 55 (ψ x +w ) B 55 (2η x +ψ z )=0, A 13 u +A 33 ψ z +B 13 ψ x +D 13 η x B 55 (ψ x +w ) D 55 (2η x +ψ z )=0, (7) where A ij, B ij, D ij, E ij and F ij are laminate stiffness coefficients [12]. Giving the superscripts I, II and III to all of the displacement components and also the stiffness coefficients in Eq. 7, a system of ordinary differential equations (ODE) of order 30 will be resulted. This system of ODEs together with the B.C. s, C.C. s and E.C. s can be solved and all displacement components for thethree regions can be obtained.

4 64 Copyright c 2007 ICCES ICCES, vol.1, no.2, pp.61-67, 2007 Hamed [12] calculated the energy release rate using the J-integral. Here the classical Irwin-Kies expressions [14] will be used to obtain the strain energy release rate (SERR) from the beam theory solution, i.e., G = P2 δc 2b δa. (8) where in Eq. 8, P is the load, b is the specimen width, a is the crack length and C the compliance is defined as C = δ P, (9) To obtain the G, we must first express C in terms of the crack length a. FromFig. 1, it is obvious that δ = u I 3 u II x = a 3 (10) x = a z = 0 z = 0 Because of complicated computations, however, the problem cannot be solved explicitly in terms of a. Therefore the DCB problem has been solved for a range of crack length and then a least-square regression was fitted to obtain C as a function of crack length, a C(a)=C 3 a 3 +C 2 a 2 +C 1 a +C 0 (11) The above equation can be differentiated with respect to crack length a and G is calculated from Eq. 8. Numerical Examples and Discussions Glass/polyester DCB specimens consisting of anti-symmetric angle ply laminates of the form [±30 ] 5 and [±45 ] 5 was chosen as those in [11] where h=7.3 mm, b=20 mm, a=35 mm and l= mm. AlsoE 1 = 34.7GPa, E 2 = 8.5GPa, ν 12 = ν 13 = 0.27, ν 23 = 0.5, G 12 = G 13 = 4.34GPa, G 23 = 2.83GPa. Furtherfor θ = 45, E 3 = 9.85GPa and for θ = 30, E 3 = 9.37GPa. Considering these properties, the corresponding C i are presented in Table 1. Table 1: Polynomial coefficients of compliance in Eq. 11. Lay C 0 C 1 C 2 C 3 up [±30 ] e [±45 ] e Fig. 2a compares SERR obtained from the compliance equation with those obtained from J-Integral [12] for two anti-symmetric angle ply laminates of θ = 30 and θ = 45,and for various values of crack length at constant applied load of P=40 N. The results are in good agreement with each other. In similar crack lengths,

5 Calculation of Energy Release Rate G [J/m 2 ] G [J/m 2 ] θ = 30 ο -Compliance Equation θ = 45 ο -Compliance Equation θ = 30 ο -J Integral θ = 45 ο -J Integral 50 0 q = 30 o - Compliance Equation q = 45 o - Compliance Equation q = 30 o - J Integral q = 45 o - J Integral (a) Figure 2: Comparison of pure mode I strain energy release rate, G, obtained from J-Integral and compliance equation for [±θ] 5 specimens versus (a) Crack length at constant applied load of P=40 N, (b) Applied load P at constant crack length of a=40 mm. (b) P [N] = = 45 Compliance [ m/n] SSTDBT Experiment Compliance [ m/n] SSTDBT Experiment Figure 3: Comparison of Calculated and measured Compliance = 30 = 45 G c [J/m 2 ] SSTDBT EFM G c [J/m 2 ] 200 SSTDBT EFM (a) Figure 4: Comparison of pure mode I fracture energy, G c, obtained from elastic foundation model, EFM [11] and SSTDBT using compliance equation for [±θ] 5 specimens (a) θ = 30 (b) θ = 45. SERR for θ = 45 are greater than when θ = 30. In Fig. 2b, SERR are plotted versus applied load for twoangle plylay-ups at constantcrack length ofa=40 mm. (b)

6 66 Copyright c 2007 ICCES ICCES, vol.1, no.2, pp.61-67, 2007 Similar to Fig. 2a, the results obtained from compliance equation are in accordance with the results from J-Integral. In Figs. 3 the calculated compliances from SSTDBT theory are compared with the experimental results obtained by Ozdil and Carlsson [11]. It is observed that when θ = 30, SSTDBT underestimates the compliance value but for θ = 45 the results are very close. In Figs. 4 pure mode I fracture energy, G c, obtained from elastic foundation model, EFM [11] and SSTDBT using compliance equation for [±θ] 5 specimens at two angle-plies of θ = 30 and θ = 45 are compared. It is shown and verified that the second order shear thickness deformation beam theory (SSTDBT) together with compliance equation accurately estimates the mode I fracture toughness and strain energy release rate for any angle-ply laminates. The method is robust and general and other mode of fracture can be modelled using this method. References 1. Nicholls DJ, Gallagher JP. Determination of GI C in angle-ply composites using a cantilever beam test method. J. Reinf. Plast. Compos. 1983;2: Chai H. The characterization of mode I delamination failure in nonwoven, multidirectional laminates. Composites 1984;15: Laksimi A, Benzeggagh ML, Jing G, Hecini M, Roelandt JM. Mode I interlaminar fracture of symmetrical cross-ply composites. Compos. Sci. Technol. 1991;41: Robinson P, Song DQ. A modified DCB specimen for mode-i testing of multidirectional laminates. J. Compos. Mater. 1992;26: Johnson WS, editor. Delamination and debonding of materials, ASTM STP 876, Philadelphia, PA, Prel VJ, Davies P, Benzeggagh ML, de Charentenay FX. Mode I and Mode II delamination of thermosetting and thermoplastic composites. ASTM STP 1012, Philadelphia, PA, 1989, pp ASTM Test Method D a Mode I Interlaminar Fracture Toughness of Unidirectional Continuous Fiber Reinforced Composite Materials, Annual Book of ASTM Standards, Vol 15.03, American Society for Testing and Materials, West Conshohocken, PA. 8. Kanninen MF. An augmented double cantilever beam model for studying crack propagation and arrest. Int J Fracture 1973; 9(1): Kanninen MF. A dynamic analysis of unstable crack propagation and arrest in the DCB test specimen. Int J Fracture 1974;10(3):415.

7 Calculation of Energy Release Rate Williams JG. End corrections for orthotropic DCB specimens. Compos Sci Tech 1989;35: Ozdil F, Carlsson LA. Beam analysis of angle-ply laminate DCB specimens. Compos Sci Technol 1999; 59: Hamed MA, Nosier A, Farrahi GH. Separation of delamination modes in composite beams with symmetric delaminations. J. Materials & Design, 2006;27(10): Herakovich, Carl T. Mechanics of fibrous composites. NewYork: Wiley; Irwin GR, Kies JA. Critical energy release rate analysis of fracture strength. Weld J Res Suppl 1954; 33:193 8.

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