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NUC 323 Material Science
M4.4: Physical Properties and Material Testing (Version C)
1. Describe how the Knoop Hardness test is performed. (10 points)
The Knoop Hardness test consists of indenting a material with a diamond indenter. With a force ranging from grams to kilograms. The load is applied for a specific duration to ensure consistency. After the specified duration, the load is removed, and the indention length is measured using a microscope.
2. A piece of steel has a grain diameter of 5.0 x 10
-3
mm and is heated to 350 ⁰C for 12 minutes. After 12 minutes, the grain size increases to 9.2 x 10
-3
mm. Calculate the grain diameter (in mm) when a specimen of the original material is heated to 350 ⁰C for 45 minutes. Assume the grain diameter exponent “n” is 2.0. (20 points)
#2: wrong math (-5pt)
First, we have to solve for K:
d
2
−
d
0
2
=
kt
d
0
=
5
x
10
−
3
mm
¿
d
=
9.2
x
10
−
3
mm
d
2
−
d
0
2
=
(
9.2
x
10
−
3
)
2
−
(
5
x
10
−
3
)
2
=
K
∗
12
k
=
8.464
x
10
−
5
−
2.5
x
10
−
5
12
=
4.97
x
10
−
6
mm
We can then use the above answer to solve for 45 minutes:
4.97
x
10
−
6
(
45
)
+
5
x
10
−
3
d
2
=
5.224
x
10
−
3
d
=
7.23
x
10
−
2
3. What is the angle between the [2,1,2] and [1,0.5,0] lattice directions in a cubic system? (20 points)
cosδ
=
uu
1
+
v v
1
+
w w
1
(
√
u
2
+
v
2
+
w
2
)
¿¿
Plug in and solve:
(
2
) (
1
)
+
(
1
) (
0.5
)
+
(
2
) (
0
)
(
√
(
2
2
)
+
(
1
2
)
+
(
2
2
)
)
(
√
(
1
)
2
+
(
0.5
)
2
+
(
0
)
2
)
=
¿
cos
= 2.5
√
9
√
1.25
=> cos
= 0.745
cos
-1
(0.745) = 0.730 radians
or 41.84
o
NUC 323 Material Science
M4.4: Physical Properties and Material Testing (Version C)
4. A Knoop hardness (HK) test is performed per ASTM E384 on a gold foil. A load of 0.1 kg produces a pyramidal indentation whose outer legs measure l = 0.143 mm. Calculate the HK of the gold foil in Kg/mm
2
. (P in kg, D and l in mm) (20 points)
Using the equation in module 4 we can plug in the knowns and solve:
HK = 14.2
P/l
2
Given:
P =
0.1 kg
l = 0.143 mm
HK = 14.2(0.1)/(0.143)
2
HK = 69.44Kg/mm
2
5. Calculate the resilience (ability of a material to absorb energy when it is deformed elastically and release that energy upon unloading in Pa) of tungsten that has a yield stress of 750 MPa (@Strain 0.200 %). (10 points)
#5: read module notes "Resilience" and Table 3-
1 in the notes, the modulus of elasticity is
denominator. (-5pt)
U
=
1
2
*
y
*
y
Given:
= 750 MPa
= 0.002
Plug in numbers
U = ½*750*10
6
*0.002
U = 0.75*10
6
Pa
6. A bar of copper has 25 percent elongation and final cross-sectional area of 9 mm
2
following cold working. Using Figure 4-1 in the Module Notes in your solution and convert percent elongation (ductility) to percent cold work, calculate the initial (original or A
o
) area (in mm
2
)
of the cross-section. (20
points)
%CW = (A
o
-A
d
/A
o
)*100
From Figure 4-1, we know that a 25% elongation corresponds to approximately 12% cold work for
copper. We can substitute these values into the formula and solve for A
o
12 = (A
o
-9/A
o
)*100
A
o
= 9/(1-12/100)
A
o
= 10.227 mm
2
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Related Questions
Figure 1 shows the tensile testing results for different materials. All specimens have an initial
diameter of 12 mm and an initial gauge length of 50 mm.
300
250
Low carbon steel
Network polymer
200
Crystalline polymer
150
Amorphous polymer
100
50
5
10
15
20
25
30
Strain (%)
Figure 1: Stress-strain curve
b. Determine the following parameters for each material:
• the tensile strength
the 0.2% offset yield strength
the modulus of elasticity
• the ductility
Stress (MPa)
LO
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(d) Following experimental data (TableQ1b) has obtained from the
torsion test for the specimen dimension given in Table Q1a.
Determine the material of a circular bar. (Take the material
Poisson's Ratio as 0.32)
Table Q1a Specimen Dimensions
Dimensions
Material
Diameter (mm)
6
Length (mm)
77.15
TableQ1b Experimental Results from Torsion test
Angular Deflection
Degree
Radian
Torque/Nm
0.5
0.008727
0.06
1
0.017453
0.47
1.5
0.02618
0.76
2
0.034907
1.01
2.5
0.043633
1.32
3
0.05236
1.65
3.5
0.061087
1.99
4
0.069813
2.32
4.5
0.07854
2.66
5
0.087267
3.00
5.5
0.095993
3.33
6
0.10472
3.64
10
0.174533
5.92
20
0.349066
7.74
30
0.523599
8.07
40
0.698132
8.17
50
0.872665
7.93
60
1.047198
8.10
70
1.221731
8.2
80
1.396264
8.4
90
1.570797
8.24
100
1.74533
8.7
110
1.919863
8.7
120
2.094396
8.7
180
3.141594
9.02
240
4.188792
9.11
300
5.23599
9.48
360
6.283188
9.51
420
7.330386
9.71
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1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves
were generated, both curves generated from the same set of data. Use the graphs to fill in the
mechanical properties of the material tested in the box below. Don't forget units!
Stress vs Strain
Stress, psi
Stress, psi
80000
70000
60000
50000
40000
30000
20000
10000
0
0.00
80000
70000
60000
50000
40000
30000
20000
10000
0.02
0
0.000 0.002
0.04
0.004
0.06
0.006
0.08
0.10
Strain
Stress vs Strain
0.008
0.12
Elastic Modulus, E:
0.2% Offset Yield Strength, oo:
Tensile Strength, ou:
Breaking Strength, of:
% Elongation:
0.14
0.010 0.012 0.014
Strain
0.16
0.18
0.016 0.018
0.20
0.020
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hings
The following results were obtained during a tensile test to destruction on a mild steel test piece
of diameter 15.96mm and 80mm gauge length
Extension for a load of 40KN =
0.08mm
Maximum load applied during test = 93kN
Diameter at fracture = 12.85mm
%3D
Final length between gauge points 106mm
From these results, determine:
i) The modulus of elasticity for mild steel
ii) The tensile strength
iii) Percentage of elongation
iv) Percentage of reduction in area
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6. State your answers to the following questions.Strain Gauge represents the deformation of a material through a change in resistance. If so, explain how temperature will affect the strain gauge in the experimental environment.①:In this experiment, the Strain Gauge measures the strain in micro units. Explain one possible error factor when applying a load by hanging a weight on the material with the strain gauge attached. (Hint: It is easy to shake by hanging the weight using a thread)①:
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I need the answer as soon as possible
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I need the answer as soon as possible
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Define the given terms related to mechanical properties of materials.
1. Nanoindentation
2. Neckling
3. Newton’s material
4. Notch sensitivity test
5. Offset strain value
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1. Plot the engineering stress & strain diagram of
an alloy having a tensile test result found in
Table 1. The tensile test specimen has a
diameter of 12.5mm and a gage length of
50.0mm.
The given alloy is used to make a 30.0mm diameter
cylinder, which is placed inside a hardened circular
steel casement with a 30.01mm inner diameter.
Table 1: Tensile test results of an alloy
Change In Length
(mm)
Change In Diameter
(mm)
Load (kN)
0.000
0.0000
0.0000
4.364
0.0254
0.0019
-0.0057
13.092
0.0762
21.819
0.1270
0.0095
30.547
32.729
0.1778
0.7620
34.911
3.0480
30.01 mm O
F
Rigid Plate
Cylindrical Alloy
-Steel casement
Figure k Section view of the steel casement
encapsulating the cylindrical alloy
2. Determine the required minimum value of F
such that the cylindrical alloy would touch the
walls of the steel casement.
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1. Plot the engineering stress & strain diagram of
an alloy having a tensile test result found in
Table 1. The tensile test specimen has a
diameter of 12.5mm and a gage length of
50.0mm.
The given alloy is used to make a 30.0mm diameter
cylinder, which is placed inside a hardened circular
steel casement with a 30.01mm inner diameter.
Table 1: Tensile test results of an alloy
Change In Length
(mm)
Change In Diameter
(mm)
Load (kN)
0.000
0.0000
0.0000
4.364
0.0254
-0.0019
13.092
0.0762
-0.0057
21.819
0.1270
-0.0095
30.547
0.1778
32.729
0.7620
34.911
3.0480
30.01 mm Ø
F
Rigid Plate
Cylindrical Alloy
- Steel casement
Figure 1: Section view of the steel casement
encapsulating the cylindrical alloy
2. Determine the required minimum value of F
such that the cylindrical alloy would touch the
walls of the steel casement.
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In the First project: you have been asked to perform tensile testing for four different materialsand analyse the results and work on some NDT process selection:a. For the results shown in Table 1 of the tensile testing that you have performed, find thefollowing, if you know that the original length of specimen is 20.8 mm and the original diameteris 6.4 mm. Fill the calculated results in the summary table below (Table 1):1. Plot the engineering stress versus engineering strain for each material and L-D Diagram.2. Compute the modulus of elasticity, E in GPa.3. Determine the yield strength at a strain offset of 0.002.4. Determine the tensile strength in MPa.5. What is the approximate %El ductility, measured by percent elongation?6. Compute the modulus of resilience.7. Determine the fracture stress in MPa.8. Compute the final area (Af) in mm2.
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A tensile test was performed on a metal specimen with a diameter of
1/2 inch and a gage length (the length over which the elongation is meas-
ured) of 4 inches. The data were plotted on a load-displacement graph,
P vs. AL. A best-fit line was drawn through the points, and the slope of
the straight-line portion was calculated to be P/AL = 1392 kips/in. What
is the modulus of elasticity?
BI
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Please please I'm begging you help me in this problem
Need to include sample calculations, solved.
Tabulate/mention the given data and what is supposed to be obtained and how.
All thank and appreciation ?? ?
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Mild steel 1
Young;s modulus 1219.5
Yield strain and stress (0.4101,500.08)
Failure stress and strain :not able to find because the given data shows the experiment did not reach the failure point.
if the material stress and strain does not reach a failure point ,what dose it means , does it means that the material is more stronger?
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Question 2
In designing prosthetic sockets, the latter will need to be experimentally tested for their
structural integrity. Figure 2 shows one such design of a prosthetic socket which is made of
carbon fibre composite. Strain gauges are installed to record the strains at various locations of
the legs during walking and the readings are recorded using a telemetry system to detemine
the critical stressed area. At a particular strain gauge location indicated in Figure 2, the
readings recorded by one of the 45° strain gauge rosettes are:
Ea = 2500 x 10*, es = 1500 x 10°, & = -950 x 10*
Using Mohr's Cicle or otherwise, detemine:
(a) the principal strains and the direction of the maximum principal strain relative to
the gauge "a".
(b) the corresponding principal stresses and sketch the results on a properly oriented
element.
You may assume that the prosthetic socket is made of polypropylene whose Young's
modulus of 1.0 GPa and Poisson ratio of 0.3.
Figure 2
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Question 8. The following figure represents three hardness-testing techniques of materials.
Describe their specifications, how there are conducted and their potential applications?
8 -
Shape of Indentation
Test
Indenter
Side View
Top View
Load
Brinell
10-mm sphere
P
of steel or
tungsten carbide
Vickers
Diamond
136°
microhardness
ругamid
Rockwell and
60 kg)
100 kg } Rockwell
150 kg)
Diamond
120°
superficial
cone:
s - in.-
diameter
Rockwell
15 kg
30 kg } Superficial Rockwell
45 kg)
steel spheres
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q1:
Describe tensile test in detail considering the steps.
. define stress , strain (equation and units)
. elasticity modulus, elastic and plastic zones
. toughness, resilience of matiral
Demonstrate stress strain curve for your answer
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1
The following data are obtained from a tensile test of a copper specimen.
- The load at the yield point is 159 kN.
- Length of the specimen is 22 mm.
- The yield strength is 80 kN/mm².
- The percentage of elongation is 40 %.
Determine the following
(i) Diameter of the specimen,
ii) Final length of the specimen,
iii) Stress under an elastic load of 15 kN,
iv) Young's Modulus if the elongation is 2.8 mm at 15 kN and
(v) Final diameter if the percentage of reduction in area is 20 %.
Solution
Initial Cross-sectional Area (in mm²)
The Diameter of the Specimen (in mm) -
น
Final Length of the Specimen (in mm).
30
Stress at the elastic load (in N/mm²)
4⁰
Young's Modulus of the Specimen (in N/mm²)
5
Final Area of the Specimen at Fracture (in mm²)
Final Diameter of the Specimen after Fracture (in mm)-
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please solve part b, this is past paper and just iam studying, mechanical engineering
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General Requirements:
External Dimensions:
All models must have the same outer dimensions: height = 15 cm, diameter = 5 cm.
Material Volume:
Ensure each design uses approximately the same material volume (e.g., relative density = 30% of the total volume). If you have any other ideas on how to ensure the test is fair between all 3 samples please note it!
File Format:
Provide the designs as STL files compatible with 3D printing or a step-by-step process.
Design Specifications:
1. Control Cylinder:
A simple hollow cylinder with:
Wall Thickness: 2 mm.
Inner Diameter: 4.6 cm (to maintain the outer diameter and wall thickness).
No additional…
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Draw a typical stress vs strain tensile test curve for the following materials (two seperate graphs) and label the axis.
A ductile metallic test specimen that is stretched to failure displaying a characteristic yield point and show the following parts on the curve.
1- Yield point
2- Ultimate Tensile Strength
3- Breaking point
4- Elastic Region
5- Plastic Region
6- Necking region
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a)Define the following mechanical properties which can be derived and, in each case, state the appropriate unit.
i)Engineering stressii)Yield stress leveliii)Young’s modulusIv) Ultimate tensile stressv)Ductilityvi) Toughness
B) Briefly describe a standard test method which can be used to establish properties (ii), (iii) and (iv) mentioned in part (a) of this question. Also provide a sketch of the apparatus.
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- Figure 1 shows the tensile testing results for different materials. All specimens have an initial diameter of 12 mm and an initial gauge length of 50 mm. 300 250 Low carbon steel Network polymer 200 Crystalline polymer 150 Amorphous polymer 100 50 5 10 15 20 25 30 Strain (%) Figure 1: Stress-strain curve b. Determine the following parameters for each material: • the tensile strength the 0.2% offset yield strength the modulus of elasticity • the ductility Stress (MPa) LOarrow_forward(d) Following experimental data (TableQ1b) has obtained from the torsion test for the specimen dimension given in Table Q1a. Determine the material of a circular bar. (Take the material Poisson's Ratio as 0.32) Table Q1a Specimen Dimensions Dimensions Material Diameter (mm) 6 Length (mm) 77.15 TableQ1b Experimental Results from Torsion test Angular Deflection Degree Radian Torque/Nm 0.5 0.008727 0.06 1 0.017453 0.47 1.5 0.02618 0.76 2 0.034907 1.01 2.5 0.043633 1.32 3 0.05236 1.65 3.5 0.061087 1.99 4 0.069813 2.32 4.5 0.07854 2.66 5 0.087267 3.00 5.5 0.095993 3.33 6 0.10472 3.64 10 0.174533 5.92 20 0.349066 7.74 30 0.523599 8.07 40 0.698132 8.17 50 0.872665 7.93 60 1.047198 8.10 70 1.221731 8.2 80 1.396264 8.4 90 1.570797 8.24 100 1.74533 8.7 110 1.919863 8.7 120 2.094396 8.7 180 3.141594 9.02 240 4.188792 9.11 300 5.23599 9.48 360 6.283188 9.51 420 7.330386 9.71arrow_forward1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020arrow_forward
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- Please show the complete solution. Thank youarrow_forwardDefine the given terms related to mechanical properties of materials. 1. Nanoindentation 2. Neckling 3. Newton’s material 4. Notch sensitivity test 5. Offset strain valuearrow_forward1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 0.0019 -0.0057 13.092 0.0762 21.819 0.1270 0.0095 30.547 32.729 0.1778 0.7620 34.911 3.0480 30.01 mm O F Rigid Plate Cylindrical Alloy -Steel casement Figure k Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.arrow_forward
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