Lab 5 and 6
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University of Illinois, Urbana Champaign *
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300
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Industrial Engineering
Date
Apr 3, 2024
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12
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CEE 300/TAM 324: Lab 5&6 Impact testing and Fracture toughness Lab section: Student name: General note: Use space as needed. The answer length is not suggested by the space. All equation numbers listed are from the respective lab in the lab manual. Q1. (6 pts) Complete Table 1. Impact testing Plate penetration Table 1—Plate-penetration data (Polymers)
Measurement or property
Material
Name
Symbol
Units
PMMA
ABS
Specimen dimensions and impact parameters
Specimen shape
—
—
Plate
Plate
Clamp diameter
2
a
mm
76
76
Specimen thickness
h
mm
2.89
3.08
Crosshead height
H
m
760
930
Crosshead weight
W
N
61.8
61.8
Avail. energy (theor.)
WH
J
46968
57474
Measurements of time to failure, maximum load, and energy absorbed
Environment
Tim
e t
fail (mse
c)
Max
. load
P (kN)
Total energ
y
Γ
(J)
Tim
e t
fail (mse
c)
Max
. load
P (kN)
Total energ
y
Γ
(J)
Freezing water (0 °C)
Impact data
1.6
0.532
1.56
5.78
1.16
14.3
Description of fracture surface
Cracks on fracture surface, completely shattered
Cracks on fracture surface, Room temperature ( °C)
Impact data
2.9
0.640
3.91
5.44
1.12
12.9
Description of fracture surface
Cracks on fracture surface
Cracks on fracture surface
Boiling water (100 °C)
Impact data
8.88
3.03
43.7
7.52
2.15
34.1
Description of fracture surface
Plastic deformation before failure
Plastic deformation before failure
Q2. (9 pts) For each polymer tested, plot the values of maximum load and total absorbed energy as functions of test temperature. Then extrapolate the plot to -50 and 150 ºC and show the probable dependence of maximum load and total absorbed energy on temperature. The extrapolation should reflect the observed changes in energy and the expected glass transition behavior of the material. Keep in mind that the minimum value of energy that can be absorbed is zero, and the maximum value is the initial potential energy WH of the impactor, which can be calculated from the data in the header file. When extrapolating to -50 and 150 ºC, the energy should be within these bounds. PMMA: Figure 1. Maximum Load vs Temperature (Plate Penetration): PMMA Figure 2. Total Absorbed Energy vs Temperature (Plate Penetration): PMMA
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ABS: Figure 3. Maximum Load vs Temperature (Plate Penetration): ABS Figure 4. Total Absorbed Energy vs Temperature (Plate Penetration): ABS
Charpy impact Q3. (6 pts) Complete Table 2. Table 2—Charpy impact data
Measurement or property
Material
Name
Symbol
Units
1045 HR Steel
6061 Aluminum
Specimen dimensions and impact parameters
Anvil Spacing
L
mm
40
40
Depth
D
mm
9.37
9.44
Width
w
mm
7.62
8.39
Notch depth
d
mm
7.06
7.06
Notch root radius
r
mm
0.095
0.095
Crosshead height
H
m
738
758
Crosshead weight
W
N
125
125.2
Avail. energy (theor.)
WH
J
92250
94901.6
Measurements of maximum load and energy absorbed
Temperature
T
Max. load
(kN)
Energy
(J)
Max. load
(kN)
Energy
(J)
Freezing water (0 °C)
Impact data
12.1
5.12
7.10
13.5
Description of fracture surface
Shear lip at fracture surface
Shear lip at fracture surface
Room temperature ( °C)
Impact data
16.7
8.41
6.78
12.6
Description of fracture surface
Shear lip at fracture surface
Boiling water (100 °C)
Impact data
13.1
23.4
6.79
11.2
Description of fracture surface
Shear lip at fracture surface
Shear lip at fracture surface
Q4. (9 pts) Using data points, plot maximum load and energy absorbed (in joules) as a function of temperature (in °C) for the samples tested. If the Charpy specimen dimensions are significantly different from the standard values, make an appropriate notation on the plot. Then extrapolate the plot to -50 and 150 °C and show the probable dependence of maximum load and total absorbed energy on temperature. This extrapolation should reflect the observed changes in energy and expected brittle-to-ductile transition behavior of the material. Keep in mind that the minimum energy that can be absorbed is zero, and the maximum energy that can be absorbed is the initial potential energy of the impactor, which you should calculate from data in the header file. When extrapolating to -50 and 150 °C, the energy should be within these bounds. 1045 HR: Figure 5. Maximum Load vs Temperature (Charpy Impact): 1045 Hot-Rolled Steel Figure 6. Total Absorbed Energy vs Temperature (Charpy Impact): 1045 Hot-Rolled Steel
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6061 Al: Figure 7. Maximum Load vs Temperature (Charpy Impact): 6061 Aluminum Figure 8. Total Absorbed Energy vs Temperature (Charpy Impact): 6061 Aluminum
Fracture toughness Q1. (6 pts) Compute the value ࠵?
= ࠵?
, where a is the initial crack length and w is the effective ࠵?
width of the compact tension specimen. Find the value of the geometry factor from Figure 3 of the lab manual. Figure 1. Plot of the function f (
λ
)
for a compact tension specimen.
7075 Al:
6061 Al: Q2. (9pts) Determine the value of the critical load PQ from the load–deflection plot obtained in the laboratory. Determine the type (I, II or III) of the loading curve. 7075 Al: Figure 9. Load-Displacement Curve for 7075 Aluminum The value for PQ is shown in Figure 9, we can see there is a maximum load preceding P
5 that exceeds P5, this means that this is a Type II loading curve.
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6061 Al: Figure 10. Load-Displacement Curve for 6061 Aluminum The value for PQ is shown in Figure 10, we can see that the PQ and P5 value are identical, this means that this is a Type I loading curve.
Q3. (6 pts) Compute a provisional value of the critical stress-intensity factor, KQ, given by Eqn. (2). 7075 Al: 6061 Al:
Q4. (9 pts) If the conditions in Eqn. (3) are met, then the fracture toughness K
IC is equal to the provisional K
Q
. If the conditions in Eqn. (3) are not met, then a thicker specimen (or one with a longer crack, or both) must be used to determine K
IC
.
Assume that the uniaxial yield strength, σ
y of 2024 aluminum is 345 MPa (50 ksi), a reasonable value according to Marks’ Handbook. For 7075 aluminum use the uniaxial yield strength, σ
y from your lab 1 data. Provide calculations and state your final conclusion(s). 7075 Al: T
he conditions in Eqn. (3) are not met due to the value being greater than the m
inimum B
, a for plain strain. This this means a thicker specimen with a longer crack must be used to determine K1c. 6061 Al: T
he conditions in Eqn. (3) are not met due to the value being greater than the m
inimum B
, a for plain strain. This this means a thicker specimen with a longer crack must be used to determine K1c.
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