S2020-HW7-Solutions-v1
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MATS 322 Homework #7 Due May 26
th
(Tue) at 7 pm
(Submit PDF through Canvas-gradescope
) Name: ___________________________________ Student ID #: _____________________________________ From lecture 1.
Based on the figure to the right (Callister Fig. 8.16), what is the maximum carbon content possible for a plain carbon steel that must have an impact energy of at least 100 J at 0
°
C? Solution From the curves in the figure, the impact energies (and therefore, ductile-to-brittle temperatures) are greater than 200 J for only the 0.22 wt% C, 0.11 wt% C and 0.01 wt% C steels. Therefore, 0.22 wt%
is the maximum carbon concentration. 2.
A polystyrene component must not fail when a tensile stress of 1.25 MPa (180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/m
2
. Assume a modulus of elasticity of 3.0 GPa. Solution Polystyrene is a brittle plastic. The maximum allowable surface crack length for polystyrene may be determined by re-arranging the expression for the critical stress for crack propagation in a brittle material (Equation 8.3 of Callister). Solving for a and inserting the given tensile strength, modulus, and surface energy leads to: = 6.1 ×
10
-
4
m = 0.61 mm
3
. Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.5 mm and a tip radius of curvature of 5 ×
10
–3
mm, when a stress of 1035 MPa is applied.
Solution In order to estimate the theoretical fracture strength of this material it is necessary to calculate σ
m
using Equation 8.1 given that σ
0
= 1035 MPa, a
= 0.5 mm, and ρ
t
= 5 ×
10
–3
mm. Thus, 4. If the specific surface energy for aluminum oxide is 0.90 J/m
2
, then using data in Table below (Callister Table 12.5), compute the critical stress required for the propagation of an internal crack of length 0.40 mm. Solution We may determine the critical stress required for the propagation of an internal crack in aluminum oxide using Equation 8.3. Taking the value of 393 GPa (Table 12.5) as the modulus of elasticity, and realizing that values for γ
s
(0.90 J/m
2
) and 2
a
(0.40 mm) are given in the problem statement, leads to
5. Calculate the maximum internal crack length allowable for a 2024-T3 aluminum alloy (see Table below) component that is loaded to a stress one half of its yield strength. Assume the value of Y is 1.35. Solution For the alloy of interest, . The applied stress is σ
= σ
y
/2 = (345 MPa)/2 = 172.5 MPa. Solving for 2a
c
, the length of a critical internal crack, using the fracture mechanics expression for critical crack length (Equation 8.7, Callister) yields:
2
࠵࠵?
࠵࠵?
=
2
࠵་࠵?
�
࠵ా࠵?
࠵఼࠵?࠵࠵?
࠵ౌ࠵?σ
�
2
=
2
࠵་࠵?
�
24 MPa
√
m
1.35 × 172.5 MPa
�
2
= 0.00676 ࠵ౚ࠵?
= 6.8 ࠵ౚ࠵?࠵ౚ࠵?
(0.27 ࠵ౖ࠵?࠵ౖ࠵?
. )
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From lab 1.
Describe the procedures for the Charpy V-notch impact tests on furnace cooled (FC) 1018 steel. Include both a description of the materials and the step-by-step process (week 4 and week 7 lab demos). Include the standard sample dimensions and mass and scale for the Charpy test (ASTM ----
-). A description of the standard is posted to Canvas. Solution The description should include: •
Homogenization and furnace cooling •
ASTM standard: -- •
specimen dimensions: -- •
mass and scale of Charpy test •
The four temperatures of the test: lN2 cooled, dry ice cooled, room temperature, boiling water. 2.
The Charpy data from the week 7 lab demo for the FC 1018 steel and for water quenched (WQ), oil quenched (OQ) and air quenched (AQ) is posted on Canvas. Use this data to create a single 2D scatter plot of impact energies vs. the testing temperature for each heat treatment. Present all the data on one plot. For full credit, include axis labels with units, a figure number, and descriptive caption. Note
: If you would like to connect the data points from one set, straight line connections are preferred over a fitted function that is not associated with a physical model. Solution . 3.
Give a plausible range for the ductile-to-brittle transition temperature (DBTT) for the FC, AQ, and OQ specimens. Solution .
4.
In general, metals become more brittle as temperature decreases. Why does this occur? (explain in 1-3 sentences) Solution .
5.
In general, metals become more brittle as strain rate increases. Why does this occur? (explain in 1-3 sentences) Solution .
6.
Briefly describe why some DBTT is seen bcc metals, like 1018 steel, and not fcc metals. Solution
. 7.
Explain why the FC, AQ, and OQ specimen had a ductile-to-brittle transition and why the WQ specimen did not. For full credit, explain how the mechanical behavior relates to the expected microstructure of each specimen. You may find it convenient to refer to your plot for question 2 to describe the mechanical behavior of the specimens. Solution . 8.
What are possible sources of these errors and uncertainties in the assessment of the DBTT found through the Charpy test? Solution . 9.
What should be done to obtain a better DBTT plot? Solution . 10.
Briefly describe the materials used in the SEM fractography demo. Describe the preparation of the materials for SEM and the describe the imaging mode used. (week 8 lab demo) Solution .
11.
Below is an SEM image from the furnace-cooled 1018 steel Charpy bar that was tested after cooling in liquid nitrogen. Does the fracture surface suggest that brittle or ductile fracture occurred? To support your answer, describe identifying characteristics in the fracture surface. Solution .
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12.
Below is an SEM image from the furnace-cooled 1018 steel Charpy bar that was tested at room temperature. Does the fracture surface suggest that the brittle or ductile fracture occurred? To support your answer, describe identifying characteristics in the fracture surface. (maximum 5 sentences) Solution .
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