ENGR 145 — Chemistry of Materials
Page 1 of 7
Practice Problems for Exam 4, Spring 2008
Question 1: Optical Properties in Transmission
a)
(5 points)
OGC problem 20.19 — The name of the organic dye “indanthrene brilliant orange” indicates
the color of light transmitted through it. In what range of wavelengths is the absorption band of this dye?
b)
(5 points)
A solution of indanthrene brilliant orange is placed in a test tube with a diameter of 1.0 cm.
Assume that absorption due to the test tube, and reflection from both the tube and the solution, are
negligible. A measurement finds that 20% of the incident light at a particular wavelength is transmitted
through the specimen. Compute the optical absorption coefficient of the solution at this wavelength.
Solution:
a) Transmitting orange light when white light is incident arises from absorption of blue (the
complementary color to orange). So the absorption band for this compound would fall in the
range of 450500 nm.
b) Use the LambertBeer law (C&R eq. 19.18):
I
t
I
i
=
exp
!
"
x
[
]
=
20%
x
=
1.0
cm
#
"
=
!
ln(0.20)
1.0
cm
=
1.6
cm
!
1
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ENGR 145 — Chemistry of Materials
Page 2 of 7
Practice Problems for Exam 4, Spring 2008
Question 2: Elastic Behavior of Materials
Show your work in the blank space below and transfer your final answers to the table.
A rod, 25.0 cm long and with circular cross section, is to support a load in uniaxial tension along its length.
For each of the candidate materials listed below:
a)
(6 points)
Compute the minimum diameter of the cylinder to ensure that a load of 44.4 kN will not exceed
the listed yield strength.
b)
(6 points)
At a stress equal to the yield strength of each material, compute the elongation of the cylinder.
c)
(8 points)
Compute the modulus of resilience for each material.
Solution:
material
elastic
modulus, GPa
yield strength,
MPa
diameter,
cm
elongation,
mm
modulus of
resilience, MPa
steel
207
355
1.26
0.429
0.304
aluminum oxide
303
337*
1.30
0.278
0.187
nylon
2.69
55
3.21
5.11
0.562
*) fracture strength
a)
σ
= F/A <
σ
y
.
⇒
A
≥
F/
σ
y
(1 pt)
A =
π
d
2
/4
⇒
d
≥
(4F/
π
σ
y
)
1/2
(1 pt).
(1 pt)
per answer in table above.
b) Although yielding has begun (for the metal and the polymer) if the stress reaches the yield
strength, the yield strength is often taken to approximate the stress at which yielding
just begins, and Hooke’s law (
σ
=E
ε
) is estimated to hold up to the yield strength. Under
these assumptions, elongation = l
o
×ε
(1 pt)
= l
o
×σ
/E where stress here is taken to equal
σ
y
(1 pt).
Then
(1 pt)
per answer in the table above.
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 Spring '09
 Aluminium, Tensile strength, yield strength, Chemistry of Materials

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