Plastics Engineering 3E Episode 5 ppt

1 24 Mechanical Behaviour of Plastics Example 2.19 During tensile tests on 4 mm thick acrylic sheets of the type shown in Fig. 2.63(a), the force-displacement characteristics shown in Fig. 2.64(a) were recorded when the crack lengths were as indicated. If the sheet containing a 12 mm long crack fractured at a force of 330 N, determine the fracture toughness of the acrylic and calculate the applied force necessary to fracture the sheets containing the other crack sizes. Solution The compliance (S/F) of each sheet may be determined from the slope of the graph in Fig. 2.64(a). A plot of compliance, C, against crack dimension, a, is shown in Fig. 2.64(b) and from this the parameter dC/da may be obtained. This is also shown plotted on Fig. 2.64(b). Using this, for a = 6 mm, it may be seen that dC/da = 115 x N-I. Thus, from equation (2.91) As this is a material constant, it may be used to calculate F, for the other crack sizes. For example, for a = 2 mm, dC/da = 7 x N-' so Fc = \/2(0.004)1.56 7 x 10-6 x lo3 = 1.34 kN Similarly for a = 3,4,5 and 5.5 mm the fracture loads would be 1.15 kN, 0.98 kN, 0.71 kN and 0.5 kN respectively. 1.6 1.4 1.2 A 1.0 B 8 o*8 0.6 0.4 0.2 1 0 10 20 30 40 50 60 Displacement (pm) Fig. 2.64(a) Force-displacement characteristics for cracked plate Mechanical Behaviour of Plastics 125 1 2 3 4 5 6 a (mm) Fig. 2.w) Compliance and rate. of change of compliance for various crack lengths An alternative energy approach to the fracture of polymers has also been developed on the basis of non-linear elasticity. This assumes that a material without any cracks will have a uniform strain energy density (strain energy per unit volume). Let this be UO. When there is a crack in the material this strain energy density will reduce to zero over an area as shown shaded in Fig. 2.65. This area will be given by &a2 where k is a proportionality constant. Thus the loss of elastic energy due to the presence of the crack is given by -U = &a2BUo and (2.92) (2.93) Comparing this with equation (2.84) and assuming that the external work is G, = 2hU, (2.94) Now, for the special case of a linear elastic material this is readily expressed zero then it is apparent that where U, is the value of strain energy density at which fracture occurs. in terms of the stress, a,, on the material and its modulus, E. (2.95) 126 F Mechanical Behaviour of Plastics Fig. 2.65 Loading of cracked plate So, combining equations (2.94) and (2.95) and in practice it is found that k 2: R, so nu;a E G, = - (2.96) This is an alternative form of equation (2.91) and expresses the fundamental material parameter G, in terms the applied stress and crack size. From a knowl- edge of G, it is therefore possible to specify the maximum permissible applied stress for a given crack size, or vice versa. It should be noted that, strictly speaking, equation (2.96) only applies for the situation of plane stress. For plane strain it may be shown that material toughness is related to the stress system by the following equation. nu:a 2 GI, = -(1 - v ) E (2.97) where v is the lateral contraction ratio (Poissons ratio). Note that the symbol GI, is used for the plane strain condition and since this represents the least value of toughness in the material, it is this value which is usually quoted. Table 2.2 gives values for GI, for a range of plastics. Mechanical Behaviour of Plastics 127 Table 2.2 spital fracture toughness parameters for a range of materials (at 20°C) Material Ductility Factor (in mm) GI, KI~ (5) ($)'I2 (kJ/mz) (m/m3I2) ABS Acetal Acrylic EPOXY Glass reinforced polyester LDPE MDPEMDPE Nylon 66 Polycarbonate Polypropylene copolymer Polystyrene UPVC Glass Mild Steel 5 1.2-2 0.35-1.6 0.1 -0.3 5-7 6.5 3.5-6.5 0.25-4 0.4-5 0.3-0.8 1.3-1.4 0.01 -0.02 100 8 2-4 4 0.9-1.6 0.3-0.5 5-7 1 0.5-5 3 1-2.6 3-4.5 0.7-1.1 1-4 0.75 140 0.13 0.08 0.014-0.023 0.005-0.008 0.12 0.125 0.025-0.25 0.06 0.02-0.5 0.15-0.2 0.02 0.03-0.13 0.01 0.5 17 6 0.2-0.5 0.02-0.06 14 16 5-100 3.6 0.4-2.7 22-40 0.4 1.1-18 0.1 250 2.18 Stress Intensity Factor Approach to Fracture Although Griffith put forward the original concept of linear elastic fracture mechanics (LEFM), it was Irwin who developed the technique for engineering materials. He examined the equations that had been developed for the stresses in the vicinity of an elliptical crack in a large plate as illustrated in Fig. 2.66. The equations for the elastic stress distribution at the crack tip are as follows. cos (4) { 1 + sin (:) sin ( ) } B cos (f) { 1 -sin (f) sin (y)} - (2nr)l/2 K By = - (2nr) 112 (2nr) 1 12 txy =- sin (!) cos (i) cos ( y) and for plane strain fH\ u cos (2) - (2Irr)1/2 or for plane stress, a, = 0. Irwin observed that the stresses are proportional to (nu)'/2 where 'u' is the half length of the crack. On this basis, a Stress Intensity Factor, K, was 128 Mechanical Behaviour of Plastics Normal stnsro ftt hriation of uy along x axis (O=O) Fig. 2.66 Stress distribution in vicinity of a crack defined as K = o(Ru)'/~ The stress intensity factor is a means of charactensing the elastic stress distribution near the crack tip but in itself has no physical reality. It has units of MN m-3/2 and should not be confused with the elastic stress concentration factor (K,) referred to earlier. In order to extend the applicability of LEFM beyond the case of a central crack in an infinite plate, K is usually expressed in the more general form K = Yo(Ru)'/~ (2. loo) where Y is a geometry factor and 'u' is the half length of a central crack or the full length of an edge crack. Fig. 2.67 shows some crack configurations of practical interest and expres- sions for K are as follows. (a) Central crack of length 2u in a sheet of finite width (b) Edge cracks in a plate of finite width (2.101) (2.102) Mechanical Behaviour of Plastics 129 (b) Double edge crack la (a) Finite width plate t" to (c) Single edge crack 1. (d) Internal penny crack 02 IL 01 10 15 20 4 (e) Elliptical surface crack (1) Three point bending Fa. 2.67 Mal crack configurations (c) Single edge cracks in a plate of finite width K = ~(ZU)''~ { 1.12 - 0.23 (5> + 10.6 (;)* - 21.7 ( ;)3 + 30.4 = ( G)4} (2.103) Note: in most cases (i) is very smd so Y = 1.12. 130 Mechanical Behaviour of Plastics (d) Penny shaped internal crack assuming a << D (e) Semi-elliptical surface flaw 1.12 K = ~(na)'/~ ( ) (0 Three point bending 3FL K = ~ 2~~312 { 1.93 (G) 'I2 - 3.07 ( $)312 + 14.53 ( $)*I2 (2.105) -25.11 (G)'l2+25.8 ($)9/2} or (2.104) (2.106) (2.107) Thus the basis of the LEFM design approach is that (a) all materials contain cracks or flaws (b) The stress intensity value, K, may be calculated for the particular loading (c) failure is predicted if K exceeds the critical value for the material. The critical stress intensity factor is sometimes referred to as the fracture toughness and will be designated K,. By comparing equations (2.96) and (2.99) it may be seen that K, is related to G, by the following equation and crack configuration (EG,)'/~ = K, (2.108) This is for plane stress and so for the plane strain situation (2.109) Table 2.2 gives typical values of K1, for a range of plastics Example 2.20 A cylindrical vessel with an outside radius of 20 mm and an inside radius of 12 mm has a radial crack 3.5 mm deep on the outside surface. If the vessel is made from polystyrene which has a critical stress intensity factor of 1.0 MN m-3/2 calculate the maximum permissible pressure in this vessel. Mechanical Behaviour of Plastics 131 Solution The stress intensity factor for this configuration is The information given in the question may be substituted directly into this equation to give the bursting pressure, PB, as 1.0(8 x 10-3)(32 x PB = = 5.6 MN/m2 3.0qi2 x 10-3)2(Ir x 3.5 x 10-3)1/2 2.19 General Fracture Behaviour of Plastics If the defect or crack in the plastic is very blunt then the stress intensification effect will be small and although failure will originate from the crack, the failure stress based on the net section will correspond to the failure stress in the uncracked material. If the stress on the material is based on the gross area then what will be observed is a reduction in the failure stress which is directly proportional to the size of the crack. This is shown as line A in Fig. 2.68. " oo Crack size ( 20 I Fig. 2.68 Brittle and ductile failure characteristics for plastics If, however, the defect or crack is sharp then the picture can change significantly. Although ABS and MDPE are special cases, where the materials 132 Mechanical Behaviour of Plastics are insensitive to notch condition, other thermoplastics will exhibit brittle failure if they contain sharp cracks of significant dimensions. Polycarbonate is perhaps the most notoriously notch-sensitive of all thermo- plastics, although nylons are also susceptible to ductilehrittle transitions in failure behaviour caused by notch sharpening. Other plastics such as acrylic, polystyrene and thermosets are always brittle - whatever the crack condition. For brittle failures we may use the fracture mechanics analysis introduced in the previous sections. From equations (2.96) and (2.99) we may write K% m2a G =- =constant E E c- (2.110) From this therefore it is evident that the failure stress, af, is proportional to u-'j2. This relationship is plotted as line B on Fig. 2.68. This diagram is now very useful because it illustrates the type of ductilehrittle transitions which may be observed in plastics. According to line B, as the flaw size decreases the failure stress tends towards infinity. Clearly this is not the case and in practice what happens is that at some defect size (w) the material fails by yielding (line A) rather than brittle fracture. This diagram also helps to illustrate why the inherent fracture toughness of a material is not the whole story in relation to brittle fracture. For example, Table 2.2 shows that polystyrene, which is known to be a brittle material, has a K value of about 1 MN m-3/2. However, LDPE which has a very high resistance to crack growth also has a K value of about 1 MN m-3/2. The explanation is that polyethylene resists crack growth not because it is tough but because it has a low yield strength. If a material has a low yield stress then its yield locus (line A in Fig. 2.68) will be pulled down, possibly below the brittle locus as happens for polyethylene. Fig. 2.69 illustrates some of the variations which are possible in order to alter the ductilehrittle characteristics of plastics. The brittle failure line can be shifted by changes in chemical structure, use of alloying techniques, changes in processing conditions, etc. The yield locus line can be shifted by the use of additives or changes in the ambient temperature or straining rate. It is apparent therefore that a materials resistance to crack growth is defined not just by its inherent toughness but by its ratio of toughness to yield stress. Some typical values of Kl,/a, are given in Table 2.2. Another approach to the question of resistance to crack growth is to consider the extent to which yielding occurs prior to fracture. In a ductile material it has been found that yielding occurs at the crack tip and this has the effect of blunting the crack. The extent of the plastic zone (see Fig. 2.70) is given by 2 1 rp = - (E) 27t ar for plane stress. The plane strain value is about one third of this. (2.111) Mechanical Behaviour of Plastics 133 *Yl - $ 3 1% (3 ii t 0 1 cb a1 02 Crack depth Fig. 2.69 Effect of varying stress field on flaw size for ductilehittle transition (K = constant) ,Plastic zone Fig. 2.70 Extent of plastic zone at crack tip The size of the plastic zone can be a useful parameter in assessing toughness and so the ratio (K~,/O,.)~ has been defined as a ducrilityfactor. Table 2.2 gives typical values of this for a range of plastics. Note that although the ratio used in the ductility factor is conceptually related to plastic zone size, it utilises KI,. This is to simplify the definition and to remove any ambiguity in relation to the stress field conditions when related to the plastic zone size. It is important that consistent strain rates are used to determine K1, and aY, particularly when materials are being compared. For this reason the values in Table 2.2. should not be regarded as definitive. They are given simply to illustrate typical orders of magnitude. [...]... 0.468 0. 354 0.287 0.233 0.187 1.7 15 1.112 0.631 0. 450 0.3 45 0.267 0.2 05 2.220 1.423 0.781 0 .53 8 0.398 0.298 0.222 25 50 BD0 ( x lo4 m2) Fig 2.84 Plot of U, against BD0 75 100 Mechanical Behaviour of Plastics 157 Example 2 2 A series of Charpy impact tests on uPVC specimens with a 3 range of crack depths gave the following results Crack length (mm) Fracture Energy (mJ) 1 100 2 62 3 46 .5 4 37 5 31 If... and Manson, J.A., Fatigue of Engineering Plastics, Academic Press, New York (1980) Ogorkiewicz, R.M Engineering Properfies of Plastics, Wiley Interscience, London (1970) Powell, P.C Engineering with Polymers, Chapman and Hall, London (1983) Levy, S and Dubois, J.H Plastics Product Desinn Engineering Handbook, Van Nostrand, New - York (1977) Crawford R.J Plastics and Rubber - Engineering Design and Applications,... (19 85) Moore, D.R., Couzens, K.H and Iremonger, M.J., J Cellular Plastics, May/June (1974) pp 1 35- 139 Benham, P.P., Crawford, R.J., and Armstrong, C.G., Mechanics of Engineering Materials, 2nd Edition, Longmans (1996) Sterrett, T and Miller, E., J Elastomers and Plastics, 20 (1988) p 346 Young, W.C Roark’s Formulas for Stress and Strain 5th Edition, McGraw-Hill (19 75)  158 Mechanical Behaviour of Plastics. .. R.F and Ferry, J.D., J Amer Chem Soc., 77 (1 955 ) p 3701 ilas Thomas, D.A Uniaxial compressive creep studies, Plasrics and Polymers, Oct(1969) p 4 85 Questions Where appropriate the creep curves in Fig 2 .5 should be used 2.1 A plastic beam is to be subjected to load for a period of 150 0 hours Use the 150 0 hour modulus values given below and the data in Table 1 .5 to decide which of the materials listed would... factor of 1 .5 with a safety factor of 2 .5 should be used Solution The alternating stress, a is given by , Alternating load 4 x 100 a = , -(MN/m2) nd2 area Also the mean stress, a , , given by , ,is Steady load 4 x 50 a, = ,, -(MN/m2) area nd2 Mechanical Behaviour of Plastics 1 45 Then using equation (2.116) a = u f (1 , 2) So applying the fatigue strength reduction factor and the factor of safety 2 .5 x 4... both crack initiation and crack propagation 40 I c polyethylene 30 N - E 2 c P 20 g! e u) 'G mp E - 10 - 0 ( 5 1 1. 75 Notch tip radius (mm) Fig 2.79 Variation of impact strength with notch radius for several thermoplastics 150 Mechanical Behaviour of Plastics energy When the very sharp notch (0. 25 mm radius) is used it may be assumed that the energy necessary to initiate the crack is small and the main... sizes are 6.7 6.7 55 L IF Fig 2.83 Charpy test piece It is apparent from equation (2.123) that a graph of BDO against fracture energy U, (using different crack depths to vary 0) will be a straight line, the slope of which is the material toughness, G,  156 Mechanical Behaviour of Plastics Table 2.3 Charpy calibration factor (0) 0 Values alD S/D=6 S/D=8 0.06 0.10 0.20 0.30 0.40 0 .50 0.60 0 S/D=4 1.183... 2.2 25 Fracture M c a i s Approach to Impact ehnc In recent years impact testing of plastics has been rationalised to a certain extent by the use of fracture mechanics The most successful results have been achieved by assuming that LEFM assumptions (bulk linear elastic behaviour and presence of sharp notch) apply during the Izod and Charpy testing of a plastic Mechanical Behaviour of Plastics 155 During... the uPVC is 2 GN/m2 Solution Since B = D = 10 mm and using the values of 0 from Table 2.3 we may obtain the following information a(mm) 1 2 3 4 5 a/D 0.1 0.2 0.3 0.4 0 .5 0 0.781 0.468 0. 354 0.287 0.233 BD@ 78.1 x lop6 46.8 x 35. 4 x 28.7 x 23.3 x U W ) 100 62 46 .5 37 31 A graph of U against BD0 is given in Fig 2.84 The slope of this gives G, = 1.33 M/m2 Then from equation (2.108) the fracture toughness... Behaviour of Plastics 1 45 Then using equation (2.116) a = u f (1 , 2) So applying the fatigue strength reduction factor and the factor of safety 2 .5 x 4 x 100 - 13 nd2 1 .5 - 4 x 50 x 2 .5) nd2 x 40 This may be solved to give d = 6.4 mm 2.21 .5 Effect of S r s S s e tes y t m In the previous sections the stress system has been assumed to be cyclic uniaxial loading since this is the simplest to analyse If, . Steel 5 1.2-2 0. 35- 1.6 0.1 -0.3 5- 7 6 .5 3 .5- 6 .5 0. 25- 4 0.4 -5 0.3-0.8 1.3-1.4 0.01 -0.02 100 8 2-4 4 0.9-1.6 0.3-0 .5 5- 7 1 0 .5- 5 3 1-2.6 3-4 .5 0.7-1.1 1-4 0. 75 140. 0.014-0.023 0.0 05- 0.008 0.12 0.1 25 0.0 25- 0. 25 0.06 0.02-0 .5 0. 15- 0.2 0.02 0.03-0.13 0.01 0 .5 17 6 0.2-0 .5 0.02-0.06 14 16 5- 100 3.6 0.4-2.7 22-40 0.4 1.1-18 0.1 250 2.18 Stress. so Fc = /2(0.004)1 .56 7 x 10-6 x lo3 = 1.34 kN Similarly for a = 3,4 ,5 and 5. 5 mm the fracture loads would be 1. 15 kN, 0.98 kN, 0.71 kN and 0 .5 kN respectively. 1.6