Chemical and Materials Engineering

MAT E 202 Fall 2019

Midterm Examination
25% of final grade
October 23rd , 2019, 6:30-8:00 PM
GRADE:

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Lecture Section in which Instructor Exam Room

you are REGISTERED

A1
Dr. Wiskel ETLC 1- 013
10am NREF 2-003

A2
Dr. Hodder ETLC 1- 017
10am NREF 2-001

A3
Dr. Rajani ETLC 2-001 and 2-002
1pm NREF 2-001

Photo Identification (ONE card) is required.

There are 13 questions on this exam - all are mandatory.
MARKS
Points for each question are shown in the question

TIPS
Read through the entire test first before starting the exam.
Do not leave blanks - attempt all problems, you cannot get marks for blanks.
Give concise answers. Point form is OK.

***You are only allowed a university approved non-programmable calculator, a ruler, and
writing utensils. You cannot share any items with others. Nothing else is allowed on your
desk – no phones, no pencil cases, etc.
This exam covers Units 1-4 and Lab 1 inclusive.

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NO WORK ON THIS PAGE WILL BE GRADED.

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 Material A and Material B are tested in tension.
Material A is loaded to the proportional limit which occurs at an elongation of 0.1 mm.
Determine the load at which the resilience of B is equal to the resilience of A.
Assume that the elastic modulus of A = 200 GPa and the elastic modulus of B is 70 GPa.
The diameter of both tensile samples is 10 mm and the gauge length is 25 mm.
Show all work. (6)

We first determine the resilience of A at 0.1 mm elongation:

1
𝑈𝑟 = 2 𝜎𝜀 (1 point for this formula). (Eq.1)
Using the Hooke’s law for stress (𝜎 = 𝐸𝜀), we can rewrite Eq.1 as:
1
𝑈𝑟 = 𝐸𝜀 2 (Eq.2)
2
The strain (𝜀) for A at 0.1 mm elongation is given by:
∆𝑙
𝜀𝐴 = (Eq.3)
𝑙𝑜
0.1 𝑚𝑚
Where ∆𝑙 = 0.1 𝑚𝑚, 𝑙𝑜 = 25 𝑚𝑚, and thus: 𝜀𝐴 = 25 = 0.004 (𝑚𝑚) (1 point for correct strain value)
Plugging 𝜀𝐴 = 0.004 into Eq.2 for A with an elastic modulus of 200 GPa yields the resilience of A at
0.1 mm elongation:
1 𝐺𝐽
𝑈𝑟𝐴 = 2 (200)(0.004)2 = 0.0016 (𝑚3 ) (1 point for correct resilience value)
We now consider the resilience of material B (Eq.1) to determine the stress at which the resilience of A
equals the resilience of B:
𝜎
Using the Hooke’s law for strain (𝜀 = 𝐸), we can rewrite Eq.1 as:
1 𝜎2
𝑈𝑟 = (Eq.4)
2𝐸
Since 𝑈𝑟𝐴 = 𝑈𝑟𝐵 , and the elastic modulus for B equals 70 GPa we will have:
1 𝜎2
0.0016 = 2 70 or 𝜎 = √0.224 = 0.47 𝐺𝑃𝑎 or 470 MPa (1 point for correct stress value)
We finally turn stress to load:
𝐹
𝜎 = 𝐴 or 𝐹 = 𝐴𝑜 𝜎 (Eq.5)
𝑜
The cross sectional surface area for material B is:
𝐷2 252
𝐴𝑜 = 𝜋 4 where 𝐷 = 25 𝑚𝑚 and thus: 𝐴𝑜 = 𝜋 4 = 490.87 𝑚𝑚2
Plugging the stress value of 470 MPa and the surface area of 490.87 mm2 into Eq.5:
𝐹 = 470 × 490.87 = 240526.3 𝑁 or 240.5263 KN (1 point for the work towards converting stress to
load and 1 point for correct final value)

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 The measured value of %El in a tensile test (Test 1) for a metal = 25%. The gauge length of
the tensile sample was 50 mm. The same metal was tested in tension (Test 2) using a gauge
length of 25 mm. The measured %El = 30%.

Determine the plastic strain at the UTS for Test 1. Show all work (6).

Strain before necking is the same for both tests. The difference in strain is due to the differences in
the necking strain after UTS. (1)
∆𝐿2 ∆𝐿2
− = 0.3 − 0.25 (2)
25 𝑚𝑚 50 𝑚𝑚
Where ΔL2 is the elongation due to necking, which is the same for both samples. ΔL 2 = 2.5 mm
εp1 = plastic strain prior to necking (at UTS) (1)

2.5 𝑚𝑚
𝜀𝑝1 = 25% − ∗ 100% = 𝟐𝟎% (2)
50 𝑚𝑚

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 Using the image below :
a) Sketch, the SN curve for chromium (Cr) and copper (Cu) (2)
b) Label both axes on the figure. Include the appropriate units for each axes (2)

a) and b)
Since Chromium has BCC crystal structure, it shows fatigue limit. 1 point for an S-N curve with an obvious
fatigue limit (horizontal line at high cycles) for chromium.
Since Copper has FCC crystal structure, it does not have fatigue limit. 1 point for an S-N curve without
fatigue limit for copper.
The horizontal axis is “Cycles to failure N” or “number of cycles N”. 1 point. Mentioning the logarithmic
scale is not necessary.
The vertical axis is “Stress amplitude S”. MPa is the appropriate unit. 1 point

c) The fracture surface from a failed rotating shaft is shown in the schematic below.
Sketch and label on this graph ALL the fracture features that would be observed (2)

c)
Label (1 point): At least three different zones must be labeled. Below are the acceptable labels for these
regions:
A: Crack origin, origin, crack initiation, stress riser
B: Crack propagation, slow crack growth, beachmarks. Note that striations is not acceptable.
C: Final rupture (failure), rapid rupture (failure), static rupture (failure).
Sketch (1 point): Beachmarks must be shown in region B and around the crack origin. If students show any
sign of ductile (for instance river marks) or brittle fracture in region C, it is Ok, but not necessary.

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 Determine the Miller (Planar) Indices for the plane in a).

Label the axis origin (0,0,0) . Show all work.

Determine the directional Indices for the direction in b). Show all work.

a)
The proper origin is shown in the figure (1 point). The corners that pass through the plane are not acceptable.
A possible orientation for the coordinate system at this origin is shown above. For this coordinate system, we
find the intercepts and take the reciprocal (1 point):
For a: fractional intercept=1 and the reciprocal of 1=1
For b: fractional intercept=-1 and the reciprocal of -1=-1
For c: fractional intercept=1 and the reciprocal of 1=1
Put the digits in parentheses without any commas: (11̅1) (1 point)
Note that for the same origin, different orientations of the coordinate system could lead to other indices
including (1̅11), (111), (1̅1̅1), (1̅11̅), (11̅1̅), (1̅1̅1̅). Therefore, these indices are acceptable too if the proper
coordinate orientation and calculations are shown.

b)
We first find the coordinates of the head and tail points (1 point):
Head point: 0,1,0
Tail point: 0,0,1
Subtract tail from head and that yields: 0, 1, -1 (1 point)
Use the bracket notation with no commas: [011̅] (1 point)

a) Miller (Planar) Indices = (11̅1) (3) b) Directional Indices = [011̅] (3)

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 The number of vacancies in lead at 25 °C is 1.66x10 19/m3 and at 227°C the number of
vacancies is 9.4x1022/m3. Determine the activation energy (in J/mol) for vacancy formation.
Show all work (6)

−𝑄𝑣
T1= 298K 𝑁𝑣 = 𝑁 ∗ 𝑒𝑥𝑝( ) (1)
(1) 𝑅𝑇
T2 = 500K
𝑙𝑛(𝑁𝑣 ) − 𝑙𝑛(𝑁) = −𝑄𝑣 /𝑅𝑇 (1)
NV1=1.66E19/m3
𝑄𝑣 1 1 (1)
𝑙𝑛(𝑁𝑣2 ) − 𝑙𝑛(𝑁𝑣1 ) = ( − )
NV2 = 9.4E22/m3 𝑅 𝑇1 𝑇2

Solving for Qv = 53 kJ/mol (2)

a) Sketch the net force vs. distance curve, for a covalent bond. (1)
b) Directly beneath, sketch the corresponding net energy vs. distance curve. (1)
c) Label on the force curve, the equilibrium separation distance, ro (1) and So (1)

a)
A proper net force vs distance curve is shown above (1 point). The vertical axis must be labeled as force.
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b)
A proper energy vs distance curve is shown below the force curve. The vertical axis must be labeled as
energy and the minimum point on the energy curve must be aligned with the intersection point on the force
curve (1 point).

c)
ro which is the equilibrium separation distance, is the intersection point on the force-distance curve (1 point).
This must be clearly shown on the curve.
So which is the atomic bond stiffness, is the slope of the tangent line at the intersection point. Students must
show the slope of the tangent line at ro on the force-distance curve as So. (1 point)

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