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MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
Final Exam (160 points total) Wednesday, March 21, noon to 3:00 pm
1. TTT Diagrams
A U.S. steel producer has four “quench baths,” used to quench plates of eutectoid steel to 700˚C,
590˚C, 350˚C, and 22˚C respectively. Using the TTT diagram below, advise the company on
how they can produce steel with the following microstructures. Assume that each bath will
instantaneously allow the steel to reach the bath temperature. (8 pts., 2 pts. each)
a. 50% fine pearlite, 12.5% bainite, 37.5% martensite.
590˚C for 5 seconds, 350˚C for 50 seconds, cool to room temperature.
b. 50% coarse pearlite, 50% martensite.
700˚C for 8,000 seconds, cool to room temperature.
c. 50% bainite, 50% coarse pearlite.
700˚C for 8,000 seconds, 350˚C for at least 200 seconds, cool to room temperature.
d. 85-95% fine pearlite, 5-15% coarse pearlite.
700˚C for 1,000 seconds, 590˚C for at least 10 seconds, cool to room temperature.
e. The company makes a cylinder of eutectoid steel of radius L. Due to heat transfer, the
entire cylinder will not cool at the same rate. The cooling rates are as follows: at the
surface (a), the cooling rate is 200 C°/s, at a distance (b) into the cylinder, the cooling rate
is 140 C°/s, and at a distance (c) into the cylinder the cooling rate is 30 C°/s. Please draw
the microstructure as a function of distance into the cylinder. (12 pts)
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
100% Martensite 100% Pearlite 100% Martensite
Mixture of Martensite and Pearlite
2. Phase diagram: Sketch the following diagram in your bluebook.
a. Determine the melting point of Bi and Pb. Mark it on the diagram. (2 pts)
Tm(Bi)~ 270C
Tm(Pb)~325C
ε + β
a b c
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
b. Write down any invariant points. State what type of invariant point it is, and write down
the reaction and temperature (ex: L−>α+β). (4 pts)
Peritectic: L + (Pb) -> (epsilon Pb) 185.0 oC
Eutectic: L -> (epsilon Pb) + (Bi) 125.9 oC
c. Label all the phase fields. (5 pts)
d. Calculate the number of degrees of freedom for all phase fields. (5 pts)
Single phase field = α, β, ε, and Liquid
• C=2 (Bi + Pd = 2 components), P=1, N=1
• F = 2-1+1 = 2
Two phase field = α + ε, α+ Liquid, β+Liquid , ε+ Liquid, β+ ε
• C=2, P=2, N=1
• F = 2-2+1 = 1
e. Determine the phase composition(s) at 100°C for an overall composition of 50%Pb. (4
pts)
The tie line is illustrated in the above figure.
α phase is 99%Bi-1%Pd
ε phase is 39%Bi-61%Pd
3. Stress and Strain A metal rod with the length 0.4 m and the radius 0.01 m has the following stress-strain curve:
a. Determine the Young’s Modulus. (4 pts)
The slope of the linear portion of the plot gives the elastic modulus. From the inset figure,
E = Δσ/Δε ~ 60MPa/.001 = 60GPa
b. Determine the yield strength for a strain offset of 0.002. (4 pts)
The yield strength is given by the intersection of the curve with the parallel offset line
featured in the figure above. From the figure, σy = 90MPa
c. Determine the maximum load that can be sustained by the cylinder. (6 pts)
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
The tensile strength is given by the maximum of the curve. The figure gives a value of σ ~
1100MPa. The maximum load is then given by F= σA0= (1100MPa)*π(0.01m)2 =
3.45•105 N
d. The rod experiences a tension perpendicular to the circular end with a stress of 160MPa.
Using the modulus of elasticity you found in part a, and assuming entirely elastic
deformation, calculate the elongation. (6 pts)
σ = εE = (Δl/l0)*E
Δl = σl0/E = (160MPa)(0.4m)/60GPa = 0.0016m
4. Failure a. Describe qualitatively the difference between brittle failure and ductile failure at the
fracture interface of a metal. (2 pts)
Brittle failure usually results in a relatively flat interface, with a cross sectional area
comparable to that of the original sample; while ductile failure usually involves necking
and a very small fracture interfacial area.
b. For the two cracks below under equal nominal applied tensile stress along the y axis,
compute the ratio of the maximum stress at the tip of crack A to crack B, and state which
one is larger. (8 pts)
( )
(
)
( )
Crack B has a larger maximum stress.
y-axis
x-axis
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
c. Fracture toughness in a brittle material is given by the expression:
√
√
where E is the modulus of elasticity, the specific surface energy, the critical stress
required for crack propagation, and is one half the length of an internal crack. Suppose
now that crack A in part b is in material A with E = 100 MPa, and crack B from part b is
in material B with E = 50 MPa. If the two cracks retain the same geometry as in part b
(length, radius of curvature), and all other parameters not stated here being equal, which
material will have a higher critical stress, and which one will have higher fracture
toughness? Explain your reasoning. (10 pts)
The fracture toughness is independent of the length of the internal crack which is obvious
once the expression for critical stress is substituted into the expression for fracture
toughness, thus material A will have a higher fracture toughness due to its higher elastic
modulus.
√
√
√
(
)
Material A also has a higher critical stress.
5. Bonding Answer the following questions with arguments related to bonding.
a. Why is H20 more dense as a liquid than a solid? (8 pts)
There is extensive intermolecular hydrogen bonding between water molecules in the solid
phase. The hydrogen bonds give an ordered and rigid, but open final structure of ice. The
open structure is why ice is less dense than liquid water.
Below is the structure of Kevlar, a high strength, low density material commonly used for
applications in body armor. In bold represent is a monomer unit, dashed lines indicate
hydrogen bonds.
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
a. What is the dominant form of bonding between carbon atoms in the backbone? (2 pts)
Covalent, sp2 hybridized HC=CH
b. How does this bonding along the backbone contribute to the high tensile strength of
Kevlar? (2 pts)
Hybridized bonding, strong bonds, highly directional leading to rigid backbone
c. In the C-N bond, will the electron density be higher on the nitrogen or the carbon? (2 pts)
High electron density on N because more electronegative
d. Do you expect Kevlar to have a more crystalline or amorphous intermolecular structure?
(2 pts)
Crystalline, ordered
e. What kind of intermolecular bonding is responsible for the structure you chose in part
(e)? (2 pts)
Large number of hydrogen bonds between the N- and O- groups provide ordering
between chains
f. Do you expect that Kevlar is a thermoplastic or a thermoset? (2 pts)
Thermoset
6. Planes
a. Draw the following planes ( ̅ ), ( ̅) and (210). For reach plane draw the
corresponding perpendicular direction on the same coordinate axis. (3 pts for each plane
and direction)
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
b. For a BCC crystal system, list the primary and secondary families of slip planes. (4 pts)
Primary={110}, Secondary={211} ( {311} also acceptable)
c. Draw the BCC crystal with the secondary family of slip planes. Draw the shortest atom
translation in this slip plane. What is the family of slip directions in this slip plane? (7
pts)
< ̅11> drawn above in red
- -
- -
-
-
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
7. Deformation – Cold Working A non-cold-worked 1040 steel cylindrical bar having an initial diameter of 25.1 mm is to be cold
worked by drawing in order to reduce the cross-sectional area such that the final diameter is
22.5mm. Figures 7.19a (left) and 7.19c (right) from the text are provided.
a. Determine the ductility (%EL) and yield strength of the steel rod after a single drawing
process reducing the diameter from the initial 25.1mm to the final 22.5mm. (8 pts)
First determine the percent cold work ,%CW:
2 2
2 2
2 2
2 2
2
2 2% 100% 100% 100%
2
25.1 22.5% 100% 19.64%
25.1
o d
o d o d
o oo
d d
A A d dCW
A dd
mm mmCW
mm
Reading the figures provided with the %CW found gives us the ductility and yield
strength.
For %CW = 19.64% (~20%) we find: %EL = 12.5% and σy = 740MPa
b. If a recrystallization heat treatment were performed on the bar after the processing in part
a, to what would the mechanical properties (ductility and yield strength) change? (2 pts)
The recrystallization process returns the mechanical properties to the non-cold-worked
values (those corresponding to 0%CW)
%EL = 26%
σy = 450MPa
c. The final 1040 steel rod with a 22.5mm diameter must have a yield strength of at least
600 MPa and a ductility of at least 20%. The yield strength found in part a from a single
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
drawing process is above the required value. However, the ductility is below the required
20%. In order to produce the 1040 steel rod with a 22.5mm diameter with the required
properties an alternative process must be used. In this case we will draw the steel to
some intermediate diameter, perform a recrystallization heat treatment and then draw the
steel rod to the desired final diameter. Determine the intermediate diameter to which the
steel must be drawn before recrystallization in order for it to have the required
mechanical properties after the second cold-work draw. (10 pts)
The effect of the first draw process is not of concern due to the recrystallization heat
treatment which follows the draw returning the mechanical properties to the non-cold-
worked values. We must examine the figures to determine a %CW which results in both
desired values for the mechanical properties for the second drawing process.
The yield strength is not at or above 600MPa until at %CW > 4%.
The ductility is not at or above 20% until at a %CW < 4%
This shows us that the second draw process must be conducted from an intermediate
diameter to the final 22.5mm diameter such that %CW = 4% 2 2
2 2
2 2
2 2% 100% 100% 100%
2
22.5 22.522.96
% 4% 1 0.041 1
100% 100%
o d
o d o d
o oo
do
d d
A A d dCW
A dd
d mm mmd mm
CW
The entire process would be a cold-work draw from 25.1mm to 22.96mm followed by a
recrystallization heat treatment which is finally followed by a cold- work draw from
22.96mm to 22.5mm.
8. Dislocations and Plastic Deformation a. If the total length of the dislocations contained in 10 cm
3 sample of a material is equal to
the equatorial circumference of the Earth, what is the dislocation density of that sample?
The circumference of the Earth around the equator is about 4 x 104 km. (3 pts)
Dislocation density is defined as the number of dislocation intersecting a randomly
chosen area in a sample (number per area) which is the same as the sum of the length of
all dislocations contained in a given volume (total length per volume). We find:
b. You have an ideal material that has no dislocations. Can this material elastically deform?
(1 pt) Can it plastically deform? (1 pts)
Elastic deformation does not depend on the presence of dislocations, so the material can
elastically deform. However, plastic deformation can only occur if dislocations are
present and able to move, so this material cannot plastically deform unless new
dislocations are formed.
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
c. You have a material that has many dislocations. The material has been processed such
that fracture occurs before there is enough stress to cause the dislocations to move. Can
this material elastically deform? (1 pt) If the nucleation of new dislocations is suppressed
can the material plastically deform? (1 pts)
Elastic deformation occurs without the motion of dislocations, so it may deform
elastically. However, due to fracture before dislocation movement the material cannot
plastically deform.
d. You have a ceramic material with some non-zero concentration of dislocations. You also
have a metal with the same non-zero concentration of dislocations. Which material do
you expect to plastically deform easier (with a lower applied load)? (1 pt)
Briefly explain why? (1 pts)
I would expect the metal to deform easier. Since there is an equal amount of
dislocations, the material which allows for easier dislocation motion will plastically
deform more easily. Due to covalent or ionic bonding in ceramics which hinder
dislocation motion greater than metallic bonding, dislocations are generally harder to
move than in metals.
Use the following figure showing a plane of a crystal with a dislocation for parts e – h.
The dislocation line runs perpendicular to the plane shown.
e. What type of dislocation is depicted in the figure? (1 pt)
This is an edge dislocation.
f. Does the Burgers vector for this dislocation run parallel or perpendicular to the
dislocation line? (2 pts)
The Burgers vector for edge dislocations run perpendicular to the dislocation line. In
this case the Burgers vector would point to the right in the plane (using a clockwise
Burgers loop) while the dislocation line is perpendicular to the plane.
g. Imagine there is an applied load such that the dislocation feels a net force to the right.
Draw the figure depicting the same crystal and dislocation such that the dislocation has
moved 2 ‘steps’ to the right. (2 pts)
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
h. Draw the figure depicting the same crystal after the dislocation has escaped to the right
side assuming no more atoms exist past the edge of the figure. (2 pts)
i. A Force, F, is applied to the crystal from parts e-h at an angle inclined 30˚ from the
normal to the slip plane (see diagram below). Slip can occur as described above along
the slip plane with a slip direction to the right (90˚ off the normal to the slip plane ). F is
the only force which has an effect on the dislocation. If the load is removed the
dislocation feels no shear stress. The cross-section over which the force is applied has
an area of 100 mm2. The yield strength of the material is 200MPa. Find the critical
resolved shear stress for the slip system? (2pts)
If the applied force is 10,000N will the sample deform plastically? (1 pt)
For plastic deformation to occur, the resolved shear stress on the slip plane must meet or
exceed the critical resolved shear stress. We are given the inclination of the plane,
ɸ=30˚. We can find the inclination of the slip direction for this system: λ = 90˚- 30˚=
60˚.
max
cos cos
crssy
200 cos 30 cos 60 86.6crss MPa MPa
MATERIALS 101 INTRODUCTION TO STRUCTURE AND PROPERTIES WINTER 2012
2
10000100
100
F NMPa
A mm
The applied load is half the yield strength so the sample will not plastically deform due
under the 10000N load. We can test this comparing resolved shear stress as well:
100 cos 30 cos 60 43.3R MPa MPa
From R crss we can again see that plastic deformation will not occur for the applied
force of 10000N.
j. A simple cubic crystal with lattice parameter a = 0.3 nm is subject to a shear stress which
causes 150 edge dislocations moving along the same slip plane to escape the crystal to
the right along a lattice direction. How far does the new ‘step’ which has formed
protrude beyond the original boundary of the crystal? (1 pts)
Every dislocation that escapes increases the protrusion size by the lattice parameter in
the direction of dislocation movement. For a simple cubic crystal all the lattice
parameter are the same so the new ‘step’ protrudes 150 x a = 1