Plastics Engineering 3E Episode 7 pot

194 Mechanical Behaviour of Composites when the stresses a,, ay and txy are applied. Calculate also the strains in the global X - Y directions. E1 = 125O0O1 MN/m2, E2 = 7800 MN/m2, u12 = 0.34 a, = 10 MN/m2, ay = -14 MN/m2, txy = -5 MN/m2 Solution The Stress Transformation Matrix is G12 = 4400 MN/m2, c2 s2 2sc 0.671 0.329 0.94 ) T, = [ s2 c2 -2sc ] or T, = ( 0.329 0.671 -0.94 -sc sc (2 - 2) -0.47 0.47 0.342 The stresses parallel and perpendicular to the fibres are then given by [E:] =To. [ 21 t12 =xY so, a1 = -2.59 MN/m2 02 = -1.4 MN/m2 t12 = -12.98 MN/m2 In order to get the strains in the global directions it is necessary to determine the overall compliance matrix [SI. This is obtained as indicated above, ie [SI = [a]-' where [a] = [To]-' [e]. [T,] The local compliance matrix is - 0 0 1 - and Q=S-' The Strain Transformation matrix is c2 s2 SC -2sc 2sc (2 -2) T, = [ s2 c2 -sc ] - so, Q=T;'.Q.T, and s=a-' Then 6.64 x -2.16 x -7.02 x ["]=3.["] - s = [ -2.16 x 10-~ 1.07 x 10-~ -4.27 x 10-~ YXY TXY -7.02 x 10-~ -4.27 x 10-~ 1.51 x 10-~ Mechanical Behaviour of Composites 195 Directly by matrix manipulation E, = 1.318 x lod3 cy = -1.509 x yxy = 8.626 x or by multiplying out the terms E, = [(s11). (a,) + (512) (ay)] + (516). (txy) E, = 1.318 x and similarly for the other two strains. 3.8 General Deformation Behaviour of a Single Ply The previous section has considered the in-plane deformations of a single ply. In practice, real engineering components are likely to be subjected to this type of loading plus (or as an alternative) bending deformations. It is convenient at this stage to consider the flexural loading of a single ply because this will develop the method of solution for multi-ply laminates. (i) Loading on Fibre Axis Consider a unidirectional sheet of material with the fibres aligned in the x- direction and subjected to a stress, u,. If the sheet has thickness, h, as shown in Fig. 3.15 and we consider unit width, then the normal axial force N, is given by !! 2 N, = /ox& h 2 _- Fig. 3.15 General forces and moments acting on a single ply 196 Mechanical Behaviour of Composites Or more generally for all the force components N,, N, and N,, h 2 [NI = JI.1. h 2 _- Similarly the bending moment, MI, per unit width is given by h -2 Once again, all the moments M,, My and M,,, can be expressed as h 2 [MI = /[alzdz -2 h (3.27) (3.28) Now in order to determine [a] as a function of z, consider the strain E(Z) at any depth across the section. It will be made up of an in-plane component (E) plus a bending component (z/R) which is normally written as Z.K, where K is the curvature of bending. Hence, The stresses will then be given by E(Z) = E -k Z.K dz) = [Ql * [&I + [Ql * Z.[K] where [Q] is the stiffness matrix as defined earlier. Now, from equation (3.27), the forces [N] are given by VI = [AI[&] + [Bl[Kl (3.29) where [A] is the Extensional Stiffness matrix (= [Qlh) and [B] is a Coupling Matrix. It may be observed that in the above analysis [B] is in fact zero for Mechanical Behaviour of Composites 197 this simple single ply situation. However, its identity has been retained as it has relevance in laminate analysis, to be studied later. Returning to equation (3.28), the moments may be written as h h [MI = [Bl[~l+ PI[KI (3.30) where [D] is the Bending Stiffness Matrix (= [Q]h3/12). Equations (3.29) and (3.30) may be grouped into the following form [E] = [: 4 [:I (3.31) and these are known as the Plate Constitutive Equations. the strains by For this special case of a single ply, [B] = 0 and so the forces are related to [NI = [AI[&] (3.32) (which is effectively [a] = [Q][E] as shown previously) or the strains are related to the forces by [E] = [a][N] where [a] = [A]-' (which is effectively [E] = [S][a] as shown previously). by Also, for the single ply situation, the moments are related to the curvature [MI = [DI[Kl (3.33) or the curvature is related to the moments by [K] = [d][M] where [d] = [D]-' It should be noted that it is only possible to utilise [a] = [AI-' and [dl = [DI-' for the special case where [B] = 0. In other cases, the terms in the [a] and [d] matrices have to be determined from A B -' [i :I=[, D] (3.34) 198 Mechanical Behaviour of Composites (ii) Loading off Fibre Axis For the situation where the loading is applied off the fibre axis, then the above approach involving the Plate Constitutive Equations can be used but it is necessary to use the transformed stiffness matrix terms 0. Hence, in the above analysis [AI = @I * h 103 = @I. h3/12 The use of the Plate Constitutive Equations is illustrated in the following Examples. Example 3.9 For the 2 mm thick unidirectional carbon fibre/PEEK composite described in Example 3.6, calculate the values of the moduli, Poisson’s Ratio and strains in the global direction when a stress of a, = 50 MN/m2 is applied. You should use (i) the lamina stiffness and compliance matrix approach and (ii) the Plate Constitutive Equation approach. Solution (i) As shown in Example 3.6, for the loading in Fig. 3.16 I 4.12 10-~ -2.58 x 10-~ -6.24 x 10-~ -2.58 x lo-’ 7.87 x 1.77 x lo-’ m2h4N -6.24 10-5 1.77 10-5 1.43 10-4 yk n I ’/ 0 Fig. 3.16 Single ply composite subjected to plane stress Mechanical Behaviour of Composites 199 and 1 1 s11 522 Ex = =- = 24.26 GN/m2, E, = =- = 12.7 GN/m2 1 2 Gxy = = 6.9 GN/m , vxy = -Ex??21 = 0.627 s66 vYx = -EyS12 = 0.328 Also, Ex = 2.06 x ly = -1.29 = -3.12 10-~ Note that As cry = 0 in this Example vxy = E' = 0.627 as above EX (ii) Alternatively using the Plate Constitutive Equation Approach. N, = oxh = 100 N/m, N, = 0, Nxy = 0. A= D= 4.86 x 104 3.76 x 104 1.65 x 104 N/mm 2.06 x 16 4.86 x lo4 8.37 x 104 8.37 x 104 1.65 x 104 4.84 x lo4 1.62 x lo4 1.25 x 104 5.51 x 16 N mm 6.86 x lo4 1.62 x 104 2.79 x lo4 1 1 2.79 x io4 5.51 x io3 1.61 x io4 a = A-' and d = D-' (since B = 0) -1.29 x 3.93 x 8.89 x mm/N -3.12 x 8.89 x 7.16 x 1 2.06 10-~ -1.29 x 10-~ -3.12 x 10-~ 200 Mechanical Behaviour of Composites -a12 v,, = -, 1 6.18 x 10-~ -3.87 x 104 9.37 x 104 -3.87 x 1.18 x 2.66 x (N mm)-’ -9.37 x 10-5 2.66 x 10-5 2.14 x 10-4 a22 x h’ wxh’ a1 1 1 1 1 E, = - G,, = - E, = - all x h’ -a12 vp = - a22 E, = 24.26 GN/m2, E, = 12.7 GN/m2 Gxy = 6.98 GN/m2, vxy = 0.627, up = 0.328 These values agree with those calculated above. Also, for the applied force N, = 50 x 2 = 100 N/mm. [5] =a. [;?I (=a[;] -”) E, = 2.061 x cy = -1.291 x yxy = -3.124 x It is interesting to observe that as well as the expected axial and transverse strains arising from the applied axial stress, a,, we have also a shear strain. This is because in composites we can often get coupling between the different modes of deformation. This will also be seen later where coupling between axial and flexural deformations can occur in unsymmetric laminates. Fig. 3.17 illustrates why the shear strains arise in uniaxially stressed single ply in this Example. Orlginal unidirectional composite Deformed shape Fig. 3.17 Coupling effects between extension and shear Mechanical Behaviour of Composites 20 1 Example 3.10 If a moment of My = 100 Nm/m is applied to the unidirec- tional composite described in the previous Example, calculate the curvatures which will occur. Determine also the stress and strain distributions in the global (x-y) and local (1-2) directions. Solution Using the D and d matrices from the previous Example, then for the applied moment My = 100 N: This enables the curvatures to be determined as K~ = -3.87 m-', K~ = 12 m-', K~~ = 2.67 m-', vYX = 0.328, vXY = 3.05 The bending strains in the global directions are given generally by E = K.Z Hence (&y)max = f Ky (f) = f 0.012 (Yxy)max = f Kxy (f) = f 2.67 x The stresses are then obtained from so a, = 0, ay = 150 MN/m2. tXy = 0 For the local directions, the strain and stress transformation matrices can be used: [ =To. [ "1 t12 TXY so cq = 26.8 MN/m2, a2 = 123.2 MN/m2, txv = 57.5 MN/m2 202 Mechanical Behaviour of Composites [":I =Tea [ :] Y12 YXY so ~1 = -5.14 x ~2 = 8 x y12 = 0.014 These stresses and strains are illustrated in Fig. 3.18. Global strains Global stresses 0 0.39% -1.2% 0 -0.27%0 0 -150 0 0 -0.39% 0 0 1.2% 0 0.27% 0 0 150 0 v 7 Y (3 EX CY rv 0, Local strains Local stresses 0 -0.8% 0 -1.4% 0 -26.80 -123 0 -57.5 0 0 0.8% 0 1.4% 0 26.8 0 123 0 57.5 0 712 €1 €2 r12 01 (32 Fig. 3.18 Stresses and strains, Example 3.10 Note that if both plane stresses and moments are applied then the total stresses will be the algebraic sum of the individual stresses. 3.9 Deformation Behaviour of Laminates (i) Laminates Made from Unidirectional Plies The previous analysis has shown that the properties of unidirectional fibre composites are highly anisotropic. To alleviate this problem, it is common to build up laminates consisting of stacks of unidirectional lamina arranged at different orientations. Clearly many permutations are possible in terms of the numbers of layers (or plies) and the relative orientation of the fibres in each Mechanical Behaviour of Composites 203 layer. At first glance it might appear that the best means of achieving a more isotropic behaviour would be to have two layers with the unidirectional fibres arranged perpendicular to each other. For example, two layers arranged at 0" and 90" to the global x-direction or at +45" and -45" to the x-direction might appear to offer more balanced properties in all directions. In fact the lack of symmetry about the centre plane of the laminate causes very complex behaviour in such cases. In general it is best to aim for symmetry about the centre plane. A lami- nate in which the layers above the centre plane are a mirror image of those below it is described as symmetric. Thus a four stack laminate with fibres oriented at 0", 90", 90" and 0" is symmetric. The convention is to denote this as [oo/900/900/o"]T or [0", 90;, Oo]T or [0"/90"],. In general terms any laminate of the type [e, -8, -8, e]T is symmetric and there may of course be any even number of layers or plies. They do not all have to be the same thickness but symmetry must be maintained. In the case of a symmetric laminate where the central ply is not repeated, this can be denoted by the use of an overbar. Thus the laminate [45/ - 45/0/90/0/ - 45/45]T can be written as [f45,0, %lS. In-plane Behaviour of a Symmetric Laminate The in-plane stiffness behaviour of symmetric laminates may be analysed as follows. The plies in a laminate are all securely bonded together so that when the laminate is subjected to a force in the plane of the laminate, all the plies deform by the same amount. Hence, the strain is the same in every ply but because the modulus of each ply is different, the stresses are not the same. This is illustrated in Fig. 3.19. Fig. 3.19 Stresses and strains in a symmetric laminate When external forces are applied in the global x-y direction, they will equate to the summation of all the forces in the individual plies. Thus, for unit width [...]... 4.20 x lo3 1.51 x lo4 0 7. 28 x 1 4 3.44 x 104 4. 17 x lo4 0 3.44~0 2.80~ 14 104 1.98 x 1 4 0 4. 17 x 1 4 1.98 x lo4 3.42 x 1 4 0 0 7. 28 x 104 3.44 x lo4 4. 17 x 104 3.44~ 104 2 8 0 ~ 0 1 9 8 ~ 14 104 4. 17 x 1 4 1.98 x 104 3.42 x 104 0 [ Qs=[ - Q4= - 7. 28 x 104 3.44 x 104 -4. 17 x 1 4 0 3.44 x 104 2.80 x 104 -1.98 x 104 3.42 x lo4 -4. 17 x 104 -1.98 x lo4 1 1 3.44 x 104 -4. 17 x 104 7. 28 x 1 4 0 3.44 x 1... 0 Global stresses 0 0 2 % 4 07% 0 3 0 02%423% 0 - 47 0 0. 07% 0 ._ €1 €2 0 _ 0 57 47 -19.50 - 57 0 0 19.5 'y12 01 71 2 02 (b) Fig 3.23 Stresses and strains, Example 3.12: (a) global; (b) local Note, to assist the reader the values of the terms in the matrices are 17. 52; 105 2.22 x 105 A = 2.22 x 105 1.93 x 1 6 u= D= [ [ 2.01; -2.31 x 5.25 x 2.24 x -1 .75 x 4.34 x -4. 57 x 1.46 x x IN/- B= 0 0 0 0... when loading is applied The reader may wish to check the matrices in this Example: 14. 37: io4 1.41 io4 A = 1.41 x lo4 1. 27 x 104 1.40 x 104 0 6.65 1 0 - ~ -5 .73 1 0 - ~ -6.81 0 -7 0-5] B= [ -6.81 x -2.05 x 1.02 x 0-4 -1.46 x io3 604.5 -1. 67 x io3 604.5 258.2 -79 2 -16 .7 x lo3 -79 2 604.4 2.30 1 0 - ~ -2.09 10-~ 3 .77 1 0 - ~ -2.24 x 1.49 x 5.34 x 9.55 1 0 - ~ 2.01 1 0 - ~ -2.03 x 1 0 - ~ 1 Mechanical Behaviour... Q(e)-I Q T, Q(0) = T,' 6. 27 io4 2 .71 io4 3.66 io4 2 .71 x IO4 2.21 x IO4 1. 87 x IO4 3.66 x io4 1. 87 io4 2.88 io4 Hence, the Extension Stiffness matrix is given by A = 2.Q(O) + 4.e(35) + 4.Q(-35) [7: 530, 105 2.22 105 A = 2 22 x 105 1.93 x 105 ]Nmm 0 2.39 x IO5 The compliance matrix is obtained by inverting [A] [ I a =Au = ] 2.0 ;IOp6 -2.31 x lop6 -2.31 x lop6 7. 84 x lop6 0 0 4. 17 x lop6 (Nmm-I) The stiffness... Behaviour of Composites 226 2.30 x 1 0 - ~ -2.24 x 1 0 - ~ 9.55 1 0 - ~ 2.01 1 0 - ~ x 10-~ 1.49 x 1 0 - ~ 3 .77 x 10-4 5.34 x 10-5 -2.03 x 10-4 1.05 x lo3 235.2 -1 67. 1 239.2 -79 .2 D= 235.2 -1 67. 1 -79 .2 233.0 2.15 x 10-3 -1.60 10-3 1.64 10-3 -1.601 x 6.65 x -4.04 x 1.64 x 1 0 - ~ -4.04 x 1 0 - ~ 8 .74 1 0 - ~ [ [ 1 p = -2.09 1 3.14 Analysis of Short Fibre Composites In order to understand the effect of... since [B] = 0 K~ = 0.435 m-', K~ = -0.4 57 m-', uXy = 1.052 K~~ = -0.1 47 m-l uYx = 0.95 When the bending moment is applied the global stresses and strains in each ply may be obtained as follows: E, = Kx z, Ey = Ky z, yxy = Kxy z At the top surface, Z = -5 mm = -2. 17 x = 2.28 yxy = -7. 34 x 1 0 - ~ and the stresses are given by So that a = - 47 MN/m2, , uy = 5 .7 MN/m2, rxy = 15.4 MN/m2 The local stresses... -2.31 x 5.25 x 2.24 x -1 .75 x 4.34 x -4. 57 x 1.46 x x IN/- B= 0 0 0 0 0 0 [o 0 01 0 2.39 x 105 lo6 -2.31 x 7. 82 x 0 (N/mm)-' 0 4. 17 x lo6 2.24 x lo6 -1 .75 x lo6 lo6 1.84 x lo6 -9.042 x lo5 Nmm, lo6 -9.04 x 1 6 2.38 x lo6 1 0 - ~ -4. 57 x 1 0 - ~ 1.46 x 1 0 - ~ 1.14 x 9 .75 x (NIIMI)-' 1 0 - ~ 9 .75 x io-8 5.63 x 1 0 - ~ ] 1 1 The solution method using the Plate Constitutive Equation is therefore straightforward... 6. 67 MN/m2 2h 2 x 1.2 (or N R = a ~ = 8 N/mm) h - [ 3 , m O x io3 1.244 103 0 = 1.244 x io3 3.444 x io3 0 N/mm2, 0 1.103 x lo3 Q2 956.5 376 .7 = 376 .7 956.5 0 0 285.8 [ ;] 1 N/I-M12 Also, and Then the Extensional Stiffness Matrix is given by 4 ef(hf A= u = A-' - hf-,), (since [BI = 0 ) f=1 The strains are related to the forces (or stresses) by Nx [;!yl Therefore the axial and hoop strains are Ex = 7. 932... and strains in the cross-section Solution As the skins and core are isotropic, we only need to get the values for each: - [ e Foamed Core Skin 4 1 6 : ~lo3 1.6 67 x lo3 Q = 1.6 67 x lo3 4.1 67 x lo3 l 0 Q2= 1.25 x io3I - [ 73 .63 31.63 0 31.63 73 .63 0 o 0 20.98 1 Assuming as before that there are four layers with the centre line at the mid-plane of the cross-section, then ho = - 1 1 , hl = -10, h2 = 0,... ply [ I: =T, t12 [E:] []; [ 21 T*Y =TEf Y12 YXY So that for f = 1, 2, 9 and 10 a = -0.44 MNlm2, 1 a 2 = -6.4 MNlm2, c1 = 1.38 x g2 = -8.15 x txy 7. 3 = y12= MNlm2 1. 67 10-~ For f = 3 , 4 , 7 and 8 01 = -0.96 MN/m2, a = -5.8 MN/m2, 2 = 1.82 E2 = -6.2 x q 2 = -7. 5 MNlm2 y12= -1.8 10-~ For f = 5 a n d 6 a = 4.5 MNIm’, 1 a 2 = -11.3 MNlm2, g1 = 5.25 x g2 = -1.33 x t12 = -0.28 MN/m2 y12= -2.09 10-~ When the . Fig. 3.16 I 4.12 10-~ -2.58 x 10-~ -6.24 x 10-~ -2.58 x lo-’ 7. 87 x 1 .77 x lo-’ m2h4N -6.24 10-5 1 .77 10-5 1.43 10-4 yk n I ’/ 0 Fig. 3.16 Single ply composite. s=a-' Then 6.64 x -2.16 x -7. 02 x ["]=3.["] - s = [ -2.16 x 10-~ 1. 07 x 10-~ -4. 27 x 10-~ YXY TXY -7. 02 x 10-~ -4. 27 x 10-~ 1.51 x 10-~ Mechanical. Composites 2 07 Overall Stiffness Matrix Overall Compliance Matrix - Q(0) = T,' . Q. T, S(e) = Q(e)-I 6. 27 io4 2 .71 io4 3.66 io4 2 .71 x IO4 2.21 x IO4 1. 87 x IO4 3.66