Page 117 - Plastics Engineering

100 Mechanical Behaviour of Plastics

Substituting into equation (2.57)

E(U1) = -
.

then for u1 = 50 seconds and K1 = 0.1 MN/m2 s

=
E(100) = 0.1(100) + E) 0.1%
(20 lo3 2 x lo6
It is interesting to note that if K1 was large (say K1 = 10 in which case
T = 1 second) then the strain predicted after application of the total stress
(10 MN/m*) would be ~(1) = 0.0505%. This agrees with the result in the
previous Example in which the application of stress was regarded as a step
function. The reader may wish to check that if at time T = 1 second, the stress
was held constant at 10 MN/m2 then after 100 seconds the predicted strain
using the integral expression would be ~(100) = 0.1495% 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.

'1
1 u2
- (K1 - K2)U2 [- + - -
6 2a a