Page 126 - Plastics Engineering

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 on-stress off. If this sequence is
repeated many times then this superposition approach becomes rather complex.
In these cases the analytical solution shown below is recommended but it
should be remembered that the equations used were derived on the basis of the
superposition approach illustrated above.
(b) If the stress is applied for 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
=
~~(800) ~~(100)
x=l [ ( y)fl - (7 - I)']
x=4

x=l
= 0.747(0.059 + 0.0265 + 0.0174 + 0.0131)
= 0.0868%
Thus the total creep strain after the 5th load application for 100 hours would
be 0.0868 + 0.747 = 0.834%.
(c) The residual strain after the lOOOth cycle may be calculated as shown
in (b) with the limits x = 1 to x = 1OOO. Clearly this repetitive calculation is
particularly suitable for a computer and indeed there are many inexpensive
desk-top programmable calculators which could give the solution quickly and
easily. Thus for a time, t, of lo00 x 200 hours.

= 0.342%.

However, in the absence of a programmable calculator or computer the
problem may be solved as follows. If the residual strain is calculated for,
say 10 cycles then the value obtained is

&(io3) = 0.121%
The total creep strain after the stress of 10.5 MN/m2 has been applied for the
1 lth time would be 0.121 + 0.747 = 0.868%. Now tests have shown that a plot
of total creep strain plotted against the logarithm of the total creep time (i.e.
ignoring the recovery times) is a straight line which includes the point E,(T).