Page 125 - Plastics Engineering

108 Mechanical Behaviour of Plastics

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) - Er(t) =I-- Er (t)
EC(W 0.747 *
Also, the Reduced Time is given by
t-T
200-100
tR=-- - =1
T 100
Then from equation (2.63)
Fr = 1 +ti - (tR + 1)"

E,(?) = 0.044%

Alternatively, since recovery may be regarded as the reverse of creep this
part of the problem may be solved as follows.
The projected creep strain after 200 hours at 10.5 MN/m2 is

=
~~(200) 0.51(200)0.083 = 0.792%
(see Fig. 2.51 to confirm this value of strain).
However, since the stress was removed after 100 hours the recovered strain
after a further 100 hours will be the same as E,(T), i.e. 0.747%. Thus the