Plastics Engineering 3E Episode 3 potx

54 Mechanical Behaviour of Plastics examples. When using this pseudo-elastic design approach it should be remem- bered that the creep curves used to derive modulus values have normally been obtained on test pieces which are essentially isotropic. In practice the manu- facture of the end-product by injection moulding or extrusion, etc. will have resulted in some degree of anisotropy. This may make the predictions inaccu- rate because the creep data for the material is no longer appropriate for the structural morphology introduced by the moulding method. Similar comments could, of course, also be made about metals in that the test data may have been obtained on specimens of the material which do not accurately reflect the nature of the material in the end-product. Therefore, pseudo-elastic design is a valid analytical procedure but one should always be cautious about the way in which the manufacturing method has affected the behaviour of the material. Example 2.1 A ball-point pen made from polypropylene has the clip design shown in Fig. 2.1 1. When the pen is inserted into a pocket, the clip is subjected to a deflection of 2 mm at point A. If the limiting strain in the material is to be 0.5% calculate (i) a suitable thickness, d, for the clip (ii) the initial stress in the clip when it is first inserted into the pocket and (iii) the stress in the clip when it has been in the pocket for 1 week. The creep curves in Fig. 2.5 may be used and the short-term modulus of polypropylene is 1.6 GN/m2. ? +I , width 6 mm Fig. 2.1 1 Ball-point pen clip design Solution Strain, E, is given by the ratio of stress, u, to modulus, E. In the case of the pen clip, it is effectively a cantilever of length 40 mm. bw I = second moment of area (= bd3/12) (i) Hence MY stress = - I where M = bending moment (WL) y = half beam depth (d/2) WLd strain, E = - 2EI (2.11) Mechanical Behaviour of Plastics 55 Also, the classical elastic equation for the end deflection of a cantilever is: WL3 deflection, 6 = - 3EI Combining (2.11) and (2.12) gives 36d strain, E = - 2L2 so (2.12) (2.13) 2(40)* x 0.005 3x2 d= = 2.7 mm (ii) The short-term stress in the material is obtained from the short-term modulus which is given in this question (or could be obtained from the creep/isometric curves, i.e. at 10 seconds, E = 8 x 106/0.5% = 1.6 GN/m2 or from the appro- priate isometric curve). stress = EE = 1.6 x lo9 x 0.005 = 8 MN/m2 (iii) After 1 week (6.1 x lo5 seconds), the isometric curves (Fig. 2.8) derived from the creep curves show that at a strain of 0.5% the stress would have decayed to about 3.3 MN/m2. Example 2.2 A polypropylene beam is 100 mm long, simply supported at each end and is subjected to a load W at its mid-span. If the maximum permis- sible strain in the material is to be 1.5%, calculate the largest load which may be applied so that the deflection of the beam does not exceed 5 mm in a service life of 1 year. For the beam I = 28 mm4 and the creep curves in Fig. 2.5 should be used. Solution The central deflection in a beam loaded as shown in Fig. 2.12 is given by WL3 a=- 48EI 48EI6 w=- L3 W Fig. 2.12 Simply supported beam with central load 56 Mechanical Behaviour of Plastics 0 0 0.5 1 1.5 2 2.5 3 3.5 Strain ("74 Fig. 2.13 Isochronous curve for polypropylene (1 year) The only unknown on the right hand side is a value for modulus E. For the plastic this is time-dependent but a suitable value may be obtained by reference to the creep curves in Fig. 2.5. A section across these curves at the service life of 1 year gives the isochronous graph shown in Fig. 2.13. The maximum strain is recommended as 1.5% so a secant modulus may be taken at this value and is found to be 347 MN/m2. This is then used in the above equation. so 48 x 347 x 28 x 5 = 2.33 N (W3 W= As before, a similar result could have been achieved by taking a section across the creep curves at 1.5% strain, plotting an isometric graph (or a 1.5% modulus/time graph) and obtaining a value for modulus at 1 year (see Fig. 2.8) In this example it has been assumed that the service temperature is 20 "C. If this is not the case, then curves for the appropriate temperature should be used. If these are not available then a linear extrapolation between temperatures which are available is usually sufficiently accurate for most purposes. If the beam in the above example had been built-in at both ends at 20% and subjected to service conditions at some other temperature, then allowance would need to be made for the thermal strains set up in the beam. These could be obtained from a knowledge of the coefficient of thermal expansion of the beam material. This type of situation is illustrated later. Mechanical Behaviour of Plastics 57 For some plastics, particularly nylon, the moisture content can have a signifi- cant effect on the creep behaviour. For such plastics, creep curves are normally available in the wet and dry states and a knowledge of the service conditions enables the appropriate data to be used. For convenience so far we have referred generally to creep curves in the above examples. It has been assumed that one will be using the correct curves for the particular loading configuration. In practice, creep curves obtained under tensile and flexural loading conditions are quite widely available. Obviously it is important to use the creep curves which are appropriate to the particular loading situation. Occasionally it is possible to obtain creep curves for compressive or shear loading but these are less common. If only one type of data is available (e.g. tensile creep curves) then it is possible to make conversions to the other test modes. It should always be remembered, however, that these may not always be absolutely accurate for plastics under all situations. Generally there is a stiffening effect in compression compared to tension. As a first approximation one could assume that tension and compression behaviour are the same. Thomas has shown that typically for PVC, the compression modulus is about 10% greater than the tensile modulus. However, one needs to be careful when comparing the experimental data because normally no account is taken of the changes in cross-sectional area during testing. In tension, the area will decrease so that the true stress will increase whereas in compression the opposite effect will occur. The classical relationship between moduli in tension, compression and flexure is (2.14) where MR = Ec/ET It may be seen that if E, = ET then Eflex = ET. However, if E, = 1.1E~ The classical relationship between the shear modulus G, and the tensile (for example), then Eflex = 1.05E~. modulus, E, for an isotropic material is (2.15) where v = Poissons ratio. is given by Finally, although it is less commonly used for plastics, the bulk modulus, K, I7 c K= 3(1 - 2~) (2.16) 58 Mechanical Behaviour of Plastics The bulk modulus is appropriate for situations where the material is subjected to hydrostatic stresses. The proof of equations (2.15) and (2.16) is given by Benham et al. Example 23 A cylindrical polypropylene tank with a mean diameter of 1 m is to be subjected to an internal pressure of 0.2 MN/m2. If the maximum strain in the tank is not to exceed 2% in a period of 1 year, estimate a suitable value for its wall thickness. What is the ratio of the hoop strain to the axial strain in the tank. The creep curves in Fig. 2.5 may be used. Solution The maximum strain in a cylinder which is subjected to an internal pressure, p, is the hoop strain and the classical elastic equation for this is PR €0 = -(2 - U) 2hE where E is the modulus, R is the cylinder radius and h is the wall thickness (See Appendix C). The modulus term in this equation can be obtained in the same way as in the previous example. However, the difference in this case is the term u. For elastic materials this is called Poissons Ratio and is the ratio of the transverse strain to the axial strain (See Appendix C). For any particular metal this is a constant, generally in the range 0.28 to 0.35. For plastics u is not a constant. It is dependent on time, temperature, stress, etc and so it is often given the alternative names of Creep Contraction Ratio or Lateral Strain Ratio. There is very little published information on the creep contraction ratio for plastics but generally it varies from about 0.33 for hard plastics (such as acrylic) to almost 0.5 for elastomers. Some typical values are given in Table 2.1 but do remember that these may change in specific loading situations. Using the value of 0.4 for polypropylene, h= -(2-~) PR ~EE 6.5 0.02 from Fig. 2.7, E = - = 325 MNlm2 0.2 x 0.5 x 103 x 1.6 h= = 12.3 mm 2 x 0.02 x 325 For a cylindrical tank the axial strain is given by PR Ex = -(1 - 2u) Ex 1 - 2v 0.2 2hE 2-u 1.6 2 = (-) = - = 8 so Mechanical Behaviour of Plastics 59 Table 2.1 Qpical tensile and shear moduli for a range of polymers Material Tensile Shear modulus modulus Poisson’s Density (E)? (G) ratio (u) (kg/m3) (GN/m2) (GN/mZ) Polystyrene (PS) Polymethyl Methacrylate (PMMA) Polyvinyl Chloride (PVC) (Unplasticised) Nylon 66 (at 65% RH) Acetal Homopolymer (POM) Acetal Copolymer (POM) Polyethylene - High Density (HDPE) Polyethylene - Low Density (LDPE) Polypropylene Homopolymer (PP) Polypropylene Copolymer (PP) Pol yethersulphone 1050 1180 1480 1140 1410 1410 955 920 910 902 1390 2.65 3.10 3.15 0.99 3.24 2.52 1.05 0.32 1.51 1.13 2.76 0.99 1.16 1.13 0.34 1.15 0.93 0.39 0.1 1 0.55 0.40 0.98 0.33 0.33 0.39 0.44 0.41 0.39 0.34 0.45 0.36 0.40 0.41 - tl00 second modulus at 20°C for small strains (t0.2%) Note that the ratio of the ratio of the hoop stress (pR/h) to the axial stress (pR/2h) is only 2. From the data in this question the hoop stress will be 8.12 MN/m2. A plastic cylinder or pipe is an interesting situation in that it is an example of creep under biaxial stresses. The material is being stretched in the hoop direction by a stress of 8.12 MN/mz but the strain in this direction is restricted by the perpendicular axial stress of OS(8.12) MN/m2. Reference to any solid mechanics text will show that this situation is normally dealt with by calculating an equivalent stress, a,. For a cylinder under pressure a, is given by OSa& where a0 is the hoop stress. This would permit the above question to be solved using the method outlined earlier. Example 2.4 A glass bottle of sparkling water has an acetal cap as shown in Fig. 2.14. If the carbonation pressure is 375 kN/m2, estimate the deflection at the centre of the cap after 1 month. The value of Poissons ratio for acetal may be taken as 0.33. Solution The top of the bottle cap is effectively a plate with clamped edges. The central deflection in such a situation is given by Benham et al. as PP Eh3 where D = 6 = - 640 12(1 - v2) To calculate 6 after 1 month it is necessary to know the 1 month creep modulus. The stresses at the centre of the cap are biaxial (radial and circumferential) both 60 Mechanical Behaviour of Plastics Thickness = 1.2 mm Fig. 2.14 Acetal bottle cap given by 3pR2(1 + u) - 3 x 0.375 x (13)2(1.33) - 8h2 8( 1 .2)2 o = 22 m/m2 o= After 1 month (2.6 x lo6 seconds) at this stress, the strain is obtained from Fig. 2.15 as 2.2%. Hence E = 22 x 106/0.022 = 1 GN/m2. pp 0.375(13)412(1 - 0.332) 6= 640 - 64 x io00 x (1.213 S=lmm Example 2.5 In a small polypropylene pump the flange on the cover plate is 2 mm thick. When the rigid clamping screws are tightened, the flange is reduced in thickness by 0.03 mm. Estimate the initial stress in the plastic and the stress after 1 week Solution The strain in the material is given by x 100 = 1.5% 0.03 2 E=- This is a stress relaxation problem and strictly speaking stress relaxation data should be used. However, for most purposes isometric curves obtained from the creep curves are sufficiently accurate. By considering the 1.5% isometric curve shown in Fig. 2.8 it may be seen that the initial stress is 16 MN/m2 and the stress after 1 week is 7 MN/m2. Mechanical Behaviour of Plastics 61 22 20 MWm2 18 16 14 12 10 8 6 4 Fig. 2.15 Creep curves for acetal (20°C) Accurately performed relaxation tests in which the strain in the material was maintained constant and the decaying stress monitored, would give slightly lower values than those values obtained from the isometric data. It should also be noted that in this case the material was loaded in compre- sion whereas the tensile creep curves were used. The vast majority of creep data which is available is for tensile loading mainly because this is the simplest and most convenient test method. However, it should not be forgotten that the material will behave differently under other modes of deformation. In compres- sion the material deforms less than in tension although the effect is small for strains up to 0.5%. If no compression data is available then the use of tensile data is permissible because the lower modulus in the latter case will provide a conservative design. 2.6 Thermal Stresses and Strains It is quite common in modem engineering designs, for plastics to be used in conjunction with other materials, particularly metals. In such cases it is wise to consider the possibility of thermal stresses being set up due to the differences in the thermal expansion (or contraction) in each material. The change in shape of a material when it is subjected to a change in temperature is determined by the coefficient of thermal expansion, a~. Normally for isotropic materials the value of CYT will be the same in all directions. For convenience this is often taken to be the case in plastics but one always needs 62 Mechanical Behaviour of Plastics to bear in mind that the manufacturing method may have introduced anisotropy which will result in different thermal responses in different directions in the material. The coefficient of thermal expansion, UT, is given by (2.17) where SL is the change in length in the material L is the original length AT is the change in temperature. There are standard procedures for determining UT (e.g. ASTM 696) and typical values for plastics are given in Table 1.2. It may be observed that the coefficients of thermal expansion for plastics are higher than those for metals. Thus if 50 mm lengths of polypropylene and stainless steel are each heated up by 60°C the changes in length would be (a) polypropylene, SZ = 100 x (b) stainless steel, SL = 10 x x 50 x 60 = 0.3 mm x 50 x 60 = 0.03 mm If these changes in length take place freely then we will have a thermally induced strain in the material (= 0.3 x 100/50 = 0.6% in the polypropylene) but no stress. However, if the polypropylene was constrained in some way so that the 0.3 mm expansion could not happen when it is heated by 6O"C, then there would be a thermally induced stress in the material, i.e. stress = modulus x strain If the modulus of the material is 1.2 GN/m2 at the final temperature, then the stress in the material would be given by stress = 1.2 x io9 (;E) - = -7.2 MN/m2 Note that the stress is compressive because the material is effectively compressed by 0.3 mm. Example 2.6 The bobbin shown in Fig. 2.16 has been manufactured by sliding the acetal ring on to the steel inner and then placing the end-plate in position. At 20°C there are no stresses in the acetal and the distance between the metal end-plates is equal to the length of the acetal ring. If the whole assembly is heated to lOO"C, calculate the axial stress in the acetal. It may be assumed that there is no friction between the acetal and the steel. The coeffi- cients of thermal expansion for the acetal and the steel are 80 x 10-6"C-' and 11 x 10-60C-1 respectively. The modulus of the acetal at 100°C is 1.5 GN/m*. Mechanical Behaviour of Plastics 63 Steel end-plate rigidly fixed to steel bobbin Acetal / LG-4 Fig. 2.16 Metal bobbin with plastic sleeve Solution Under free conditions, the acetal would expand more than the steel but in the configuration shown they will both expand to the same extent. Hence, the acetal will effectively be put in compression by an amount given by eqn (2.17) S = (ao - CY,)L. AT = (80 - 1 1)10-6(40)(80) = 0.22 mm x 100 = -0.55% 0.22 strain = 40 0.55 x - 100 stress = EE = -1.5 x 10 = -8.3 MN/m2 Thus there will be a compressive stress of 8.3 MN/m2 in the acetal. It should be noted that the above analysis ignores the effect of the constraining effect which the acetal has on the thermal expansion of the steel. However, as the modulus of the steel is over 100 times greater than the acetal, this constraining [...]... deflection 6=- WL3 8EZ So, strain, Using the information for the aluminium 6= 200 x 8 03 x 12 = 5.5 mm 8 x 70 x lo3 x 50 x 23 Mechanical Behaviour of Plastics 78 Therefore, E= 2 x 5.5 x 2 802 = 0 .34 % From the creep curves for acetal, at 1 month (2.6 x lo6 s) and 0 .34 % strain, ) the modulus ( 0 , ' ~is approximately 1 .3 GN/m2 Therefore, using (2. 23) and ignoring (W/12) on each side 70 x 23 = 1 .3 x D3 D = 7.6... is 0 .3 The safety factor on stress is to be 2 Solution Resolving vertical forces gives V=Ncosa-pNsina = N(cos a - p sin a) Resolving horizontal forces gives 1 F=Nsina+pNcosa = N(sina + pcosa) b Eliminating N from the above equations gives V = F { 1-ptana p+tana } The snap fit may be regarded as a cantilever and for this situation, the vertical deflection, ti, is given by VL3 3EI t i = - {FL3 3EI ti=-... = 22 mm, d = 2.5 mm and b = 12 mm gives an insertion force of 4 N and 4 a maximum stress of 1 MN/m2 which is again acceptable 74 Mechanical Behaviour of Plastics L=10 mm 15 t"Un 40 35 20 mm 30 25 5 20 E 25 mm 5 15 30 mm ' 35 mm 10 40 mm 5 0 0 1 2 3 Beam thickness, d (mm) 4 5 Fig 2.24 Variation of stress with beam dimensions 2.9 Design of Ribbed Sections It will be shown later (Chapters 4 and 5 ) that... which would have the same stiffness when loaded in the same way would need to have the same 2nd moment of area So if its depth is d then 12(d )3 - 1 439 , d = 11 .3 ~ U T I 12 The weight per unit length of the solid beam would be W,= 12 x 11 .3 x x 909 x lo3 = 1 23 g The weight per unit length of the foamed beam is W f = (909 x 2 x 12 x 2 x + (600 x 12 x 8 x = 101.2 g Hence the weight saving is 17.7% Once... transverse stiffness ratio dominates 14 12 3 lo t E= E 6 4 2 0 1 2 3 4 6 5 7 Dh Fig 2 .32 Optimisation of Corrugation depth a 9 10 Mechanical Behaviour of Plastics 84 This would suggest that D / h = 4 offers the best balance of transverse and axial properties In fact the ‘optimum’ ratio is essentially independent of the thickness, h, but it depends on B and b Fig 2 .33 shows this relationship and reflects... two elements &=&1+&2 (2 .30 ) 86 Mechanical Behaviour of Plastics From equations (2.27), (2.28) and (2 .30 ) (2 .31 ) This is the governing equation of the Maxwell Model It is interesting to consider the response that this model predicts under three common timedependent modes of deformation (i) creep If a constant stress, a, is applied then equation (2 .31 ) becomes , 1 €=-*a, (2 .32 ) r) which indicates a... From Fig 2 .35 it may be seen that for the Maxwell model, the strain at any time, t, after the application of a constant stress, a, is given by , 00 € ( t )= - 5 w i a + -t0 r ) t 'E ; i 0 b tl t2 t Fig 2 .35 Response of Maxwell model Time - 87 Mechanical Behaviour of Plastics Hence, the creep modulus, E ( t ) , is given by (2 .33 ) (ii) Relaxation If the strain is held constant then equation (2 .31 ) becomes... in each of the elements, i.e &=E1 =E2 (2 .37 ) From equations (2.27), (2.28) and (2 .36 ) a = 6 * El + qb2 or using equation (2 .37 ) a=f*&+q*& (2 .38 ) This is the governing equation for the Kelvin (or Voigt) Model and it is interesting to consider its predictions for the common time dependent deformations (3 c=p If a constant stress, a, is applied then equation (2 .38 ) becomes , a =6 * E , +qi and this differential... given D = 200/ 935 = 0.214, p, = 935 x g/mm3, W = 0.14 g m / so d = 1.61 mm Using the expression for h from above, the thickness of the foam is given by h = 31 .6 IWII The cost of the solid beam is given by (Cost), = 935 x 10 x 15 x x C , = 0.14 C , The cost of the sandwich beam is given by (C0st)F = pcbhCc + pS2bdCs = 0.159 C , Hence the additional cost = 0.159 - 0.14 = 13. 5% 0.14 It is interesting to... oOe-f/TR (2 .34 ) (2 .35 ) where T R = q / e is referred to as the relaxation r m ie This indicates that the stress decays exponentially with a time constant of q/t (see Fig 2 .35 ) (iii) Recovery When the stress is removed there is an instantaneous recovery of the elastic strain, E ' , and then, as shown by equation (2 .31 ), the strain rate is zero so that there is no further recovery (see Fig 2 .35 ) It can . 139 0 2.65 3. 10 3. 15 0.99 3. 24 2.52 1.05 0 .32 1.51 1. 13 2.76 0.99 1.16 1. 13 0 .34 1.15 0. 93 0 .39 0.1 1 0.55 0.40 0.98 0 .33 0 .33 0 .39 0.44 0.41 0 .39 0 .34 0.45 0 .36 . by bd3 where I = - VL3 ti=- 3EI 12 ti=-{ FL3 1-ptana } 3EI p + tana Insertion force F= (Lfd)l { :-:E:} (2.21) For polypropylene in the situation given, Fig. 2. 23 shows. circumferential) both 60 Mechanical Behaviour of Plastics Thickness = 1.2 mm Fig. 2.14 Acetal bottle cap given by 3pR2(1 + u) - 3 x 0 .37 5 x ( 13) 2(1 .33 ) - 8h2 8( 1 .2)2 o = 22 m/m2