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HW 10
Problem 15.1
a. To Find:
Elastic modulus and tensile strength of poly(methyl methacrylate) at room temperature [20°C
(68°F)]. Compare these with the corresponding values in Table 15.1.
b. Given:
Figure 15.3 is accurate; the elastic modulus and tensile strength have to be evaluated at room
temperature [20°C (68°F)]
c. Assumptions:
Fig.15.3 is accurate; the material poly(methyl methacrylate)
is pure.
d. Solution:
The elastic modulus is the slope of the
linear
elastic region of the 20
C curve in Fig. 15.3.
E
=
(stress)
(strain)
=
30 MPa
0 MPa
9
10
3
0
= 3.3 GPa
(483,000 psi)
The value for E as per Table 15.1 lies in the range 2.24 to 3.24 GPa (325,000 to 470,000 psi).
Thus, the plotted value of E is slightly higher but still close enough to the given value(s).
The tensile strength corresponds to the maximum value of stress on the engineering stress vs.
engineering strain curve. Here, this is the stress at which the curve ends, i.e., 52 MPa (7500 psi).
The range given in Table 15.1 is 48.3 to 72.4 MPa (7000 to 10,500 psi). The value estimated
from the plot falls within this range.
E = 3.3 GPa
Tensile Strength = 52 MPa
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View Full Document Problem 15.6
a. To Find:
E
r
(10) for the given viscoelastic polymer
b. Given:
1.
Stress decays with time according to the following relation:
(
t
) =
(0) exp
t
2.
0
= 0.6
3.
(0) = 2.76 MPa (400 psi)
4.
At t = 60s ,
(t) = 1.72 MPa (250 psi) [ i.e.,
(60) = 1.72 MPa ]
c. Assumptions:
1.
The stress decays according to the given relation
2.
The values of
(0) and
(t) and
0
are measured accurately
3.
The strain is maintained at a constant value after the polymer is suddenly pulled in tension
d. Solution:
Step 1: To determine the value of
in the expression
(
t
) =
(0) exp
t
1.72 = 2.76 exp (60/
)
exp (60/
) = 1.72/2.76
Taking natural log on both sides:
(60/
) = ln(1.72/2.76)
=>
= 126.88 s
Step 2:
E
r
(t) =
(t) /
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This note was uploaded on 12/08/2011 for the course MSE 2090 taught by Professor Leoindzhiglei during the Fall '10 term at UVA.
 Fall '10
 leoindzhiglei

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