CHAPTER 15
CHARACTERISTICS, APPLICATIONS, AND PROCESSING OF POLYMERS
PROBLEM SOLUTIONS
15.1
From Figure 15.3, the elastic modulus is the slope in the elastic linear region of the 20
°
C
curve, which is
E =
∆
(stress)
∆
(strain)
=
30 MPa
−
0 MPa
9 x 10
−
3
−
0
= 3.3 GPa
(483,000 psi)
The value range cited in Table 15.1 is 2.24 to 3.24 GPa (325,000 to 470,000 psi).
Thus, the
plotted value is a little on the high side.
The tensile strength corresponds to the stress at which the curve ends, which is 52
MPa (7500 psi).
This value lies within the range cited in the table--48.3 to 72.4 MPa (7,000
to 10,500 psi).
15.2
The reason that it is not necessary to specify specimen gauge length when citing percent
elongation for semicrystalline polymers is because, for semicrystalline polymers that
experience necking, the neck normally propagates along the entire gauge length prior to
fracture;
thus, there is no localized necking as with metals and the magnitude of the percent
elongation is independent of gauge length.
15.3
The explanation of viscoelasticity is given in Section 15.4.
15.4
This problem asks for a determination of the relaxation modulus of a viscoelastic material,
which behavior is according to Equation (15.10)--i.e.,
σ
(t) =
σ
(0) exp
−
t
τ
⎛
⎝
⎜
⎞
⎠
⎟
We want to determine
σ
(10)
, but it is first necessary to compute
τ
from the data provided in
the problem.
Thus,
1

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