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ENGR 25 Delta College Stress Strain Curves Questions
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wbuaan1234
Engineering
ENGR 25
Delta College
ENGR
Description
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ENGR 25
Fall 2021
Midterm Exam 2
November 10, 2021
This is an open book and open notes examination, however the work submitted must be your
own. Students are given 24-hours to complete the exam. There are 3 problems, each with
multiple parts, for a total of 30 possible points. There is one extra credit problem on the ninth
page of this exam, worth 2 extra points. The exam is worth 15% of your final grade for the
class. There are a total of 11 pages including this cover page, the list of equations, and the
periodic table.
Show all of your work for each question.
Name:____________________________________________________(please print)
Honor Code: All students agree, individually and collectively, that they will not give or receive
unpermitted aid during examinations. Engaging in such unpermitted actions is
considered cheating.
By my signature, I affirm that I have adhered to the letter and spirit of the Honor Code.
Signature:____________________________________________________
1/11
Problem 1 Stress-Strain Curves (10 points)
Problem 3 (8 points)
a) This titanium specimen was deformed in tension. The specimen failed at the point marked
a) X.
This
was stress-strain
deformed in tension.
failed at the
point marked
with an
Is titanium
this an specimen
engineering
curve The
or aspecimen
true stress-strain
curve?
How do you
with
an
X.
Is
this
an
engineering
stress–strain
curve
or
a
true
stress–strain
curve.
How
do you
know? (1 point)
know? (1 point)
b) Would you say this material is tough? Why or why not? Refer to features of the stress–strain
curve in your answer. (1 point)
b) Approximate the maximum load in Newton’s that this specimen sustained if the initial crosssectional area was 10 mm2. (1 point)
c) Approximate the true stress that corresponds to the ultimate tensile strength. (1 point)
c) Approximate the true stress that corresponds to the ultimate tensile strength. (1 point)
d) Approximate the maximum load in newtons that this specimen sustained if the initial cross–
sectional area was 10 mm2. (1 point)
5/7
2/11
Problem 1 continued
d) At the moment of fracture, what was the final length of the specimen if the initial length was
50 mm? (1 point)
Problem 3 continued
e) At the moment of fracture, what was the final length of the specimen if the initial length was
50 mm? (1 point)
e) Using the magnification of the elastic region shown below, approximate the elastic modulus.
(1 point)
f) Using the magnification of the elastic region shown above, approximate the elastic modulus.
(1 point)
f) Approximate the 0.2% offset yield stress. Show your work on the plot above for full credit.
(1 point)
g) Approximate the 0.2% offset yield stress. Show your work on the plot for full credit.
(1 point)
h) What is the amount of plastic strain at failure sustained by this specimen? (1 point)
3/11
Problem 1 continued
g) Calculate the total elongation ∆l of the specimen when subjected to 200 MPa if the initial
length was 50 mm. You will need to use the elastic modulus computed in part e). (1 point)
h) What is the amount of plastic strain at failure sustained by this specimen? You will need to
use the elastic modulus computed in part e). (1 point)
i) Compute the percent elongation of the specimen if the initial length was 50 mm. You will
need to use the final length of the specimen computed in part d). (1 point)
j) Suppose the specimen is loaded to a stress of 500 MPa, then the specimen is completely
unloaded, and without interrupting the test, it is loaded again to failure. Draw the path that is
followed by the data from start to finish on the graph below. You will need to deviate from
the Problem
path shown.
Draw on top of the given data where appropriate. (1 point)
3 (8 points)
a) This titanium specimen was deformed in tension. The specimen failed at the point marked
with an X. Is this an engineering stress–strain curve or a true stress–strain curve. How do you
4/11
know? (1 point)
Problem 2 Defects (10 points)
a) Pure nickel has a higher hardness than pure copper. The hardnesses of pure nickel and copper
are indicated in the plot below. Draw a schematic curve showing how you might expect the
hardness of a nickel–copper alloy to vary from pure nickel to pure copper as a function of
composition. Explain the shape of your curve with 1–2 sentences. (2 points)
Ultimate Tensile Strength (MPa)
b) The plot below shows the ultimate tensile strength of a metal that was cold-worked and then
annealed at increasingly higher temperatures. Identify the stages of annealing that you see in
the data, give the approximate temperature ranges for these stages, and explain the features of
the plot that led you to these answers. (3 points)
450
400
350
300
250
200
150
100
500
550
600
650
700
Temperature
5/11
750
(oC)
800
850
Problem 2 continued
Consider the data in the table below. Samples 1 and 2 represent the same material that has been
processed slightly differently to produce the different properties. Consider each part below as a
separate question that has no relationship to previous parts of the question. A one or two
sentence answer should be sufficient.
Sample
Ultimate Tensile Strength (MPa)
1
2
620
400
c) Samples 1 and 2 are identical except that they have different grain sizes. Which sample has
the smaller grain size? How do you know? (1 point)
d) Samples 1 and 2 are identical except that they have different dislocation densities. Which
sample has a lower dislocation density? How do you know? (1 point)
e) Samples 1 and 2 are identical except that one has a small alloying addition. Which sample
has been alloyed? How do you know? (1 point)
f) Which sample is weaker? How do you know? (1 point)
g) Samples 1 and 2 are identical except that one has been work hardened. Which sample has
been work-hardened? (1 point)
6/11
Problem 1 (11 points)
Problem 3 Phase Diagrams (10 points)
The binary phase diagram for the gallium and indium system is shown below. The composition
The binary phase diagram for the gallium and indium system is shown below. The composition
of 100% indium is given by the right–hand axis.
of 100% indium is given by the right-hand axis.
a) Label the phases present in each region of the phase diagram. Denote α as the Ga–rich solid
and β as
the In–rich
solid
phase.
to denote
the liquid
phase.
(2 the
points)
a) Labelphase
the phases
present
in each
region
ofUse
the Lphase
diagram.
Denote
α as
Ga-rich solid
phase and β as the In-rich solid phase. Use L to denote the liquid phase. (2 points)
b) What is the what
composition
the solid
with a melting
temperature
350
K? indium?
(1 point) (1 point)
b) Approximately
are the of
melting
temperatures
of pure
galliumof
and
pure
c) Draw a schematic diagram of the microstructure of the alloy with an overall composition of
80 atomic percent In when cooled to 288 K. Label your diagram with the phases. (2 points)
c) What is the composition of the first solid to form at a temperature of 350 K? (1 point)
2/7
7/11
Problem 3 continued
d) Draw a schematic diagram of the microstructure of the alloy with an overall composition of
80 atomic percent In when cooled to 288 K. Label your diagram with the phases. (2 points)
e) What are the compositions and amounts of the phases present for an overall composition of
80 atomic percent In at 288 K? Make any necessary approximations from the phase diagram.
(2 points)
f) What is the overall alloy composition required to have a material consisting of 75% liquid
and 25% β at a temperature of 300 K? (2 points)
8/11
Extra Credit (+2 points; optional)
Consider the interstitial diffusion of nitrogen in FCC iron. If D0 = 0.0034 cm2/s and the activation
energy is 34,600 cal/mol, calculate the diffusion coefficient at 950°C.
9/11
ENGR 25
Equations and Constants
N A = 6.022 × 10 23
1 eV = 1.602 × 10–-. J
1 N = 1 Pa × m2
Chapter 5
Arrhenius-type equations:
𝑄
Rate = 𝑐4 exp 7− <
𝑅𝑇
𝑄
𝑛> = 𝑛 exp 7− <
𝑅𝑇
𝑄
𝐷 = 𝐷4 exp 7− <
𝑅𝑇
Fick’s first law:
𝑑𝑐
𝐽 = −𝐷
𝑑𝑥
Fick’s second law:
𝑐C − 𝑐D
𝑥
= erf 7
<
𝑐C − 𝑐4
2√𝐷𝑡
Chapter 6
Engineering Stress (S):
𝐹
𝑆=
𝐴4
Engineering Strain (e):
𝛥𝑙
𝑒=
𝑙4
True Stress (𝜎):
𝜎 = 𝑆(1 + 𝑒)
True Strain (𝜀):
𝜀 = ln(1 + 𝑒)
Hooke’s Law:
𝑆 = 𝐸𝑒
𝜎 = 𝐸𝜀
10/11
11/11
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