Plastics Engineering 3E Episode 4 doc

Mechanical Behaviour of Plastics 89 Fig. 2.37 Response of KelvinNoigt model (ii) Relaxation If the strain is held constant then equation (2.38) becomes U=C*E That is, the stress is constant and supported by the spring element so that the predicted response is that of an elastic material, Le. no relaxation (see Fig. 2.37) (iii) Recovery If the stress is removed, then equation (2.38) becomes 0 = C.E+ r)€ Solving this differential equation with the initial condition E = E’ at the time of stress removal, then i! ~(t) = de- (2.4) This represents an exponential recovery of strain which is a reversal of the predicted creep. (c) More Complex Models It may be seen that the simple Kelvin model gives an acceptable first approx- imation to creep and recovery behaviour but does not account for relaxation. The Maxwell model can account for relaxation but was poor in relation to creep 90 Mechanical Behaviour of Plastics and recovery. It is clear therefore that some compromise may be achieved by combining the two models. Such a set-up is shown in Fig. 2.38. In this case the stress-strain relations are again given by equations (2.27) and (2.28). The geometry of deformation yields. Total strain, E = e1 + €2 + &k (2.41) 52 Fig. 2.38 Maxwell and Kelvin models in series where &k is the strain response of the Kelvin Model. From equations (2.27), (2.28) and (2.41). From this the strain rate may be obtained as (2.42) (2.43) The response of this model to creep, relaxation and recovery situations is the sum of the effects described for the previous two models and is illustrated in Fig. 2.39. It can be seen that although the exponential responses predicted in these models are not a true representation of the complex viscoelastic response of polymeric materials, the overall picture is, for many purposes, an acceptable approximation to the actual behaviour. As more and more elements are added to the model then the simulation becomes better but the mathematics become complex. Example 2.12 An acrylic moulding material is to have its creep behaviour simulated by a four element model of the type shown in Fig. 2.38. If the creep curve for the acrylic at 14 MN/m2 is as shown in Fig. 2.40, determine the values of the four constants in the model. Mechanical Behaviour of Plastics 91 0 tl 12 Time Fig. 2.39 Response of combined Maxwell and Kelvin models 0 0 100 200 300 400 500 Time (hours) Fig. 2.40 Creep curve for acrylic at 20°C Solution The spring element constant, 61, for the Maxwell model may be obtained from the instantaneous strain, ~1. Thus oo 14 , = - 2800 MN/m2 ' - e1 0.005 The dashpot constant, VI, for the Maxwell element is obtained from the slope of the creep curve in the steady state region (see equation (2.32)). = 1.2 x io7 MN.iu/m2 a0 14 E 1.167 x ?)I=-= = 4.32 x 10" MN.s/m2 92 Mechanical Behaviour of Plastics The spring constant, 62, for the Kelvin-Voigt element is obtained from the maximum retarded strain, ~2, in Fig. 2.40. The dashpot constant, q2, for the Kelvin-Voigt element may be determined by selecting a time and corresponding strain from the creep curve in a region where the retarded elasticity dominates (i.e. the knee of the curve in Fig. 2.40) and substituting into equation (2.42). If this is done then q2 = 3.7 x lo8 MN.s/m2. Having thus determined the constants for the model the strain may be predicted for any selected time or stress level assuming of course these are within the region where the model is applicable. (d) Standard Linear Solid Another model consisting of elements in series and parallel is that attributed to Zener. It is known as the Standard Linear Solid and is illustrated in Fig. 2.41. The governing equation may be derived as follows. I Stress. u Fig. 2.41 The standard linear solid Stress-Strain Relations As shown earlier the stress-strain relations are 01 = 41&1 02 = E2&2 (73 = 7l3E3 (2.44) (2.45) (2.46) Equilibrium Equation In a similar manner to the previous models, equilibrium of forces yields. (TI = a3 (J = a1 + a2 (2.47) Mechanical Behaviour of Plastics 93 Geometry of Deformation Equation In this case the total deformation, E, is given by From equation (2.48) but from equation (2.47) ~II~U~=CT-U~ E = E2 = El + E3 &=&1+&3 a1 =a - a2 Rearranging gives (2.48) (2.49) This is the governing equation for this model. The behaviour of this model can be examined as before (9 creep If a constant stress, a,, is applied then the governing equation becomes &{q3(6l + t2)1 + 6162e - 6100 = 0 The solution of this differential equation may be obtained using the boundary condition E = a,/(ij1 + 62) at t = 0. So (2.50) It may be seen in Fig. 2.42 that this predicts the initial strain when the stress is first applied as well as an exponential increase in strain subsequently. (ii) Relaxation If the strain is held constant at E', then the governing equation becomes v3a + 610 - 6162.s' = 0 This differential equation may be solved with the boundary condition that a = a, = ~'(61 + 62) when the strain is first kept constant. Stress, a(t) = ~ 61 +h (2.51) This predicts an exponential decay of stress as shown in Fig. 2.42. 94 Stress =0 d -I. &I Mechanical Behaviour of Plastics w Time Time 1 Fig. 2.42 Response of standard linear solid (iii) Recovery equation becomes The solution of this differential equation may be obtained using the boundary condition that when the stress is removed, the strain is given by If the stress is at a value of 0’ and then completely removed, the governing 1)3(e1 + e2)& + e1e2& = 0 E’ = a’/(e1 + (2) (2.52) This predicts an instantaneous recovery of strain followed by an exponential decay. It may be observed that the governing equation of the standard linear solid has the form + U,O = bli + bo& Mechanical Behaviour of Plastics 95 where al, a,, bl and bo are all material constants. In the more modem theories of viscoelasticity this type of equation or the more general form given in equation (2.53) is favoured. The models described earlier are special cases of this equation. 2.12 Intermittent Loading The creep behaviour of plastics considered to date has assumed that the level of the applied stress is constant. However, in service the material may be subjected to a complex pattern of loading and unloading cycles. This can cause design problems in that clearly it would not be feasible to obtain experimental data to cover all possible loading situations and yet to design on the basis of constant loading at the maximum stress would not make efficient use of material or be economical. In these cases it is useful to have methods of predicting the extent of the recovered strain which occurs during the rest periods of conversely the accumulated strain after N cycles of load changes. There are several approaches that can be used to tackle this problem and two of these will be considered now. 2.12.1 Superposition Principle The simplest theoretical model proposed to predict the strain response to a complex stress history is the Boltzmann Superposition Principle. Basically this principle proposes that for a linear viscoelastic material, the strain response to a complex loading history is simply the algebraic sum of the strains due to each step in load. Implied in this principle is the idea that the behaviour of a plastic is a function of its entire loading history. There are two situations to consider. (a) Step Changes of Stress tl, then the creep strain, &(t), at any subsequent time, t, may be expressed as When a linear viscoelastic material is subjected to a constant stress, u1, at time, (2.54) where E(t - tl) is the time-dependent modulus for the elapsed time (t - tl). Then suppose that instead of this stress q, another stress, a2 is applied at some arbitraxy time, t2, then at any subsequent time, t, the stress will have been applied for a time (t - t2) so that the strain will be given by 1 - t2) &(t) = .fJ1 96 Mechanical Behaviour of Plastics Now consider the Situation in which the stress, 01, was applied at time, tl, and an additional stress, 02, applied at time, r2, then Boltzmanns’ Superposition Principle states that the total strain at time, t, is the algebraic sum of the two independent responses. 1 1 E(t) = E(t - tl) a0 + E(r - 12) * 61 This equation can then be generalised, for any series of N step changes of stress, to the form i=N e(t) = Eoi i=l (2.55) where ai is the step change of stress which occurs at time, ti. To illustrate the use of this expression, consider the following example. Example 2.13 A plastic which can have its creep behaviour described by a Maxwell model is to be subjected to the stress history shown in Fig. 2.43(a). If the spring and dashpot constants for this model are 20 GN/m2 and 1000 GNs/m2 respectively then predict the strains in the material after 150 seconds, 250 seconds, 350 seconds and 450 seconds. Solution From Section 2.1 1 for the Maxwell model, the strain up to 100s is given by am E(t) = - + - ttl Also the time dependent modulus E(t) is given by O ttl E(t) = - - ~ - tl+ tt (2.56) Then using equation (2.54) the strains may be calculated as follows: (i) at t = 150 seconds; ai = 10 MN/m2 at tl = 0, a2 = -10 MN/m2 at t2 = 100 s = 0.002 - 0.001 = 0.1% (ii) at 250 seconds; 61, a2 as above, a3 = 5 MN/m2 at t3 = 200 s 1 ~(250) = 10 ttl tl + * (250 - 200) = 0.003 - 0.002 + 0.0005 = 0.15% Mechanical Behaviour of Plastics 97 Time (s) 0.5 0.4 3 0.3 v 9 0.2 0.1 0 0 100 200 300 5 Time (s) Fig. 2.43 Strain predictions using superposition theory K) (iii) at 350 seconds; 01, 02, a3 as above, a4 = 10 MN/m2 at r4 = 300 s SO, ~(350) = 0.003 = 0.3% (iv) and in the same way ~(450) = 0.004 = 0.4% The predicted strain variation is shown in Fig. 2.43(b). The constant strain rates predicted in this diagram are a result of the Maxwell model used in this example to illustrate the use of the superposition principle. Of course superposition is not restricted to this simple model. It can be applied to any type of model or directly to the creep curves. The method also lends itself to a graphical solution as follows. If a stress 01 is applied at zero time, then the creep curve will be the time dependent strain response predicted by equation (2.54). When a second stress, a2 is added then the new creep curve will be obtained by adding the creep due to a2 to the anticipated creep if stress a1 had remained 98 Mechanical Behaviour of Plastics alone. This is illustrated in Fig. 2.44(a). Then if all the stress is removed this is equivalent to removing the creep strain due to 61 and 02 independently as shown in Fig. 2.44(b). The procedure is repeated in a similar way for any other stress changes. i 1 I - 0 4 tL tsm Fig. 2.44(a) Stress history Fig. 2.44(b) Predicted strain response using Boltzmann’s superposition principle (b) Continuous Changes of Stress If the change in stress is continuous rather than a step function then equa- tion (2.55) may be generalis4 further to take into account a continuous loading cycle. So t (2.57) where a(t) is the expression for the stress variation that begins at time, tl. The lower limit is taken as minus infinity since it is a consequence of the [...]... the total creep strain after 40 0 cycles (equivalent to a total time of 9600 hours) would only be 1 .4% This would be equivalent to a stress of 13 MN/m2 being applied continuously for 9600 hours and so the effective creep modulus would be 13/0.0 14 = 929 MN/m2 Mechanical Behaviour of Plastics 107 2 .4 2.2 2.0 18 16 E 1 2 L fj 1 0 0.8 0.6 04 0.2 n yo-l 100 101 102 lo3 lo4 io5 Time (hours) Fig 2.51... 100 hours and removed for 100 hours then in equation (2. 64) ,T = 100 hours and t , = 200 hours Therefore after four cycles i.e t = 800 hours x =4 ~ ~ ( 8 0 0~~(100) = ) x=l [( y)fl - (7- I)'] x =4 x=l + = 0. 747 (0.059 0.0265 + 0.01 74+ 0.0131) = 0.0868% Thus the total creep strain after the 5th load application for 100 hours would be 0.0868 0. 747 = 0.8 34% (c) The residual strain after the lOOOth cycle may... = - K l ( t - T ) d 4 ) -= -K1 dt so E(U2) K I U ~ TKIT + = - 17 ~ ( 1 2 5 ) 0.125% = e KlT2 27 1 Mechanical Behaviour of Plastics 102 0.1 0.08 Sp B 0.06 65 0. 04 0.02 0 0 50 100 150 200 (I S Fig 2 .46 Variation of strain w t time ih Stress (MN/m*) 100 0 - Time (s) Fig 2 .47 Stress history: Example 2.15(a) (b) If the stress was completely removed after 100 seconds as shown in Fig 2 .48 then the effective... + t 2 P + 42 41s 115 Mechanical Behaviour of Plastics substituting from equations (2.73) and (2. 74) once again 3aoiweW+s) + tlaoeW+s)~ = Dividing across by EOeiwrand + t 2 ) i w ~ o +~ ~ ~ e 41 t2&0eiwr letting E* = qei(wr+')/qgid E* = then, ( ( 1 v(t1 + t2)iw + t i t 2 t + tloi 1 Multiplying top and bottom by the conjugate of the denominator, we get E* = (-532 + 0 2 ( h + t2'2)w2) ( t 2 0 4 i + 6: and... factor from T1 to T2 - 17 .4( 20 + 10) - 17 .4( 60 + 10) 51.6 + (20 + 10) = -3.62 UT = 2 .4 x 1 0 - ~ 51.6 + (60 + 10) 119 Mechanical Behaviour of Plastics Therefore, using equation (2.75) t = -0.1 (2 .4 x [ E ( t ) l ~= 0.78t-'.' * The modulus at T z ( = 60°C) can now be calculated at any desired time For example, at 1 year (t = 3.15 x lo6 s) [ E ( t ) ] m= 0.78(3.15 x lo6)-'.' = 0. 14 GN/m2 The two modulus... 0 00 1 0.1 1 10 Log (4 Fig 2.57 Variation of E , , Et and loss tangent for standard linear solid 100 116 Mechanical Behaviour of Plastics 2. 14 Time-Temperature Superposition It has been shown throughout this chapter that the properties of plastics are dependent on time In Chapter 1 the dependence of properties on temperature was also highlighted The latter is more important for plastics than it would... 0. 048 6(10.5)(100)0~083 = 0.51(100)0~083 0. 747 % = Mechanical Behaviour of Plastics 108 Time (hours) Fig 2.52 Variation of modulus for continuous and intermittent loading Now from equation (2.60)the Fractional Recovery is given by Fr = ~c(l00) E r ( t ) =I-E C ( W Er ( t ) 0. 747 * Also, the Reduced Time is given by t-T tR=-T - 200-100 100 =1 Then from equation (2.63) Fr = 1 + t i E,(?) - (tR+ 1)" = 0. 044 %... earlier in this chapter, the governing equation for the Maxwell model is given by 1 1 &=-U+-CT 6 1 7 Then from equations (2.73) and (2. 74) Eoiweiar = looiWei(otC6) 4 iw iw=- 6 + aOei(0t+6) 1 D 1 -aoei(ot+a) &Oeio' + ; a,,ei(ot+a) &OeiWf Mechanical Behaviour of Plastics 1 14 where E* is the complex modulus Rearranging gives Using the fact that (a E* = + iw6 E* = 6 oqi bi)(u - b i ) = u2 b2 + iwqe (6 - wqi)... since the stress was removed after 100 hours the recovered strain after a further 100 hours will be the same as E ( ) i.e 0. 747 % Thus the ,T, Mechanical Behaviour of Plastics 109 residual strain after the recovery period may be determined by superposition Er(t) = 0.792- 0. 747 = 0. 045 % This is simpler than the first solution but this approach is only convenient for the simple loading sequence of stress... using the integral expression would be ~ ( 1 0 0 = 0. 149 5% which again agrees ) with the previous example (b) After the time T, the change in stress is given by Hence, where tl = 0 and t2 = T = 100 s - (K1- K2)U2 [- + 6 1 u2 2a 1'a Mechanical Behaviour of Plastics 101 then for K2 = -0.2 MN/m2s, T = 100 seconds and u2 = 125 seconds ~ ( 1 2 5= 0.0 94% ) (c) After the time T ' , the change in stress is . shown in Fig. 2 .44 (b). The procedure is repeated in a similar way for any other stress changes. i 1 I - 0 4 tL tsm Fig. 2 .44 (a) Stress history Fig. 2 .44 (b) Predicted strain. above, a4 = 10 MN/m2 at r4 = 300 s SO, ~(350) = 0.003 = 0.3% (iv) and in the same way ~ (45 0) = 0.0 04 = 0 .4% The predicted strain variation is shown in Fig. 2 .43 (b). The. Relations As shown earlier the stress-strain relations are 01 = 41 &1 02 = E2&2 (73 = 7l3E3 (2 .44 ) (2 .45 ) (2 .46 ) Equilibrium Equation In a similar manner to the previous