MSE250 ASU Material Science: Structure and Properties of Materials Analysis We discussed in class about defects in solids and mechanical propersties. You c

MSE250 ASU Material Science: Structure and Properties of Materials Analysis We discussed in class about defects in solids and mechanical propersties. You can use use the resources attached. You can do it by your handwriting, but it must be clear. MSE 250 HW 3 Due 09/24/2019 (Tuesday)
Reading: Callister Chapter 5 and 6
1) a) Explain why metals are ductile and ceramics are brittle, in terms of their dislocation motion.
b) Draw 2 pictures to sequentially show an edge dislocation gliding through a crystal, using dots to represent the atoms.
c) What factors limit dislocation mobility?
d) What positive and negative effects does limiting dislocation mobility have on mechanical properties?
2) a) Draw what would happen to the grain structure if a metal with small grains were heated to a high temperature (before and
after picture).
b) How will this affect its mechanical properties (strength, ductility)? Hint: think about how the motion of dislocations is blocked
by grain boundaries.
3) Determine the rate of diffusion of Cu in Al at 300K, using D = D o exp (-ED/RT), where Do = 0.15 cm2/sec, and ED = 125
kJ/mole. Note that 1 kJ = 1000 J.
4) Determine the temperature at which the diffusivity D of Cu in Al is twice the diffusivity at 500K.
5) Using your result for D of Cu in Al at 500K, estimate the distance that a Cu impurity would diffuse in Al at 500 K after 1 hour.
Use the simple estimate x = (Dt)1/2
6) The following stress-strain points are generated for a titanium alloy for aerospace applications:
? (in MPa)
? (unitless)
???
????????
???
????????
???
????????
Calculate Young’s modulus for this alloy. (Hint: think about whether you should use only some of the points or all of the points).
7) An aluminum wire (1 inch in diameter, 1000 m long) is used to pull a cable car up a mountain. If the weight of the unloaded
cable car is 1 ton, and if there are 6 people in the car, and each weighs an average of 150 pounds:
a) calculate the stress (in pounds per sq. inch),
b) calculate the strain of the wire when it begins to lift the cable car,
c) calculate the change in length of the cable.
d) determine the change in the diameter of the wire (Poisson’s ratio is 0.31 for Al) – be sure to indicate the proper sign (positive for
increase, negative for decrease)
State your assumptions. EAl = 10 x 106 psi (pounds per square inch). [hint: be careful with the units]
?) A steel rod of radius 0.01 m and length 2. M is pulled on with a tensile force of 200,000 N, and the bar stretches to a length of
2.08 m. The radius of the bar is reduced to 0.0099 m.
Determine the stress, strain, Young’ modulus (E), and Poisson’s ratio (?) from this test. Use the Young’s modulus and Poisson’s
ratio to determine the Bulk Modulus and Shear Modulus, using the equations given in class.
MSE 250: Structure and Properties of Materials
8. Mechanical Properties II
Instructor: Prof. Yang Jiao (yang.jiao.2@asu.edu)
Office: Engineering Research Center Rm 387
Phone: 480-965-4560
Expected Outcome
Key concepts:
Stress and strain
Strength of materials
Mechanisms for strengthening a material
Skills:
Draw stress-strain curve for typical metal
Determine Elastic moduli, yield stress, ultimate stress,
ductility
Explain and apply different strengthening mechanisms
Mechanical Properties
Stress: ? =
Force
Area
Strain:
?=
?Length
Length
Applied stress results in strain of material
?
?
Deformation can be temporary (elastic) or permanent (plastic)
Uniaxial Tensile Stress:
?
?l
2
?=
?l
l
l
?
?l
2
?
Young’s Modulus = stiffness of material = E =
?
Contraction in other two directions given by Poisson’s ratio =
?=?
lateralstrain
tensilestrain
Typically, = 0.2 ~ 0.4 for solids.
Typical Stress-Strain Curve for a Metal
U
yield:
stress which results in permanent strain of 0.2%
ultimate: largest stress required to deform material (tensile strength)
fracture: stress at fracture (lower than u due to necking, void formation)
Elastic vs. Plastic Deformation
Ductility
Ductility can be defined as the permanent strain at fracture
Strengthening metals
Pure metals are generally soft and need to be strengthened for
structural applications
4 Primary methods
1) Solid solution strengthening: individual impurities slow
dislocation motion
2) Precipitation strengthening: clusters of impurities stop
dislocation motion
3) Grain boundaries: blocks dislocations from moving
between grains
4) Strain Hardening: deforming metals at low temperatures
increases density of dislocations, which block one another
All 4 methods relate to blocking dislocations
increase yield strength,
decrease ductility, little effect on elastic modulus
Solid solution strengthening
b
Precipitation strengthening
Fe
glide
Fe3C
Grain boundaries
Grain Boundary Strengthening

? ?
?
?

! ”
#
Cold Work Strengthening
b
Activity:
From the figure above, estimate % change in strength,
ductility and Young’s modulus for each metal.
Typical Thermoplastic
0OF
100OF
?
200OF
300OF
?
Question: Explain how increasing the temperature affects
Young’s Modulus
Strength
Ductility
Ductility depends on temperature
Example: The Titanic was built in warm waters in Belfast harbor in
Northern Ireland. Fracture toughness of the steel was high. In the
frigid North Atlantic Ocean, the fracture toughness was reduced, so
the collision with the iceberg tore the hull open.
Expected Outcome
Key concepts:
Stress and strain
Strength of materials
Mechanisms for strengthening a material
Skills:
Draw stress-strain curve for typical metal
Determine Elastic moduli, yield stress, ultimate stress,
ductility
Explain and apply different strengthening mechanisms
MSE 250: Structure and Properties of Materials
7. Mechanical Properties I
Instructor: Prof. Yang Jiao (yang.jiao.2@asu.edu)
Office: Engineering Research Center Rm 387
Phone: 480-965-4560
Expected Outcome
Key concepts:
Stress and strain
Stress-strain curve
Elastic moduli, yield stress, ultimate stress, ductility
Tensile test
Skills:
Draw stress-strain curve for typical metal
Determine Elastic moduli, yield stress, ultimate stress,
ductility
Compute stress/strain in a material under given loading
conditions
Which one develops largest
elongation?
L, 2A
F
L, A
F
2L, A
F
Mechanical Properties
Stress: ? =
Force
Area
Strain:
?=
?Length
Length
Applied stress results in strain of material
?
?
Deformation can be temporary (elastic) or permanent (plastic)
F
F
Uniaxial Tensile Stress:
?l
2
?
?=
l
?l
l
?l
2
?
?
Young’s Modulus = stiffness of material = E =
?
? = E?
? =? /E
Poisson’s ratio
Contraction in other two directions given by Poisson’s ratio =
? =?
lateral strain
??D / D
=?
tensile strain
?l / l
Typically, = 0.2 ~ 0.4 for solids.
Poisson’s ratio
•
Bulk modulus
•
•
•
•
Volume change: dV
Volume strain: = dV/V
Applied pressure (normal stress): P
Bulk modulus: K = – P/
•
K = E/[3(1-2v)], E- Young’s modulus, v – Possion’s ratio
Shear
?
: shear stress
: shear strain
?=
?
aa’ displacement
=
unit length
ad
Shear Modulus:
a a’
?
G=
?
and
d
c
E = 2G (1+ )
Temperature effects
E, G, and K all decrease as temperature increases (material softens)
From 0K to ½ Tmp, metals lose ~ ½ E
• A steel wire with a cross-section area 2.5X10-3 m2 and a length of 10
m is used to pull a truck on the road. The truck is 10 tons and the
friction coefficient between the tires and road is 0.25N/kg. Steel has
a Young’s modulus 250MPa and Poisson’s ratio 0.3.
•
•
•
•
(1) compute the stress of the wire
(2) compute the strain of the wire
(3) compute the elongation of the wire
(4) compute the % of change of the cross section area
Stress-strain curve: Tensile Test
• http://www.youtube.com/watch?v=D8U4G
5kcpcM
Typical Stress-Strain Curve for a Metal
U
yield:
stress which results in permanent strain of 0.2%
ultimate: largest stress required to deform material (tensile strength)
fracture: stress at fracture (lower than u due to necking, void formation)
Ductility
Ductility can be defined as the permanent strain at fracture
A steel has a Young’s modulus 250 MPa, yield stress 1000 Pa, tensile
stress 1500 Pa with strain 0.05, fracture stress 1200 Pa and ductility
0.1.
Draw a stress-strain curve for this steel based on the given properties.
Hardness
Hardness is determined by pressing a hard indentor into a material, and
measuring the size of the indentation
It is a quick, easy, non-destructive test, so it is widely used. Hardness
values can be used to estimate tensile strength.
Nano-indentation techniques
Expected Outcome
Key concepts:
Stress and strain
Stress-strain curve
Elastic moduli, yield stress, ultimate stress, ductility
Tensile test
Skills:
Draw stress-strain curve for typical metal
Determine Elastic moduli, yield stress, ultimate stress,
ductility
Compute stress/strain in a material under given loading
conditions
MSE 250: Structure and Properties of Materials
4. Defects in Solids I
Instructor: Prof. Yang Jiao (yang.jiao.2@asu.edu)
Office: Engineering Research Center Rm 387
Phone: 480-965-4560
Expected Outcome
Key concepts:
Types of defects in solids
Different types of point defects
Different types of line defects (dislocations)
Burgers vector
Skills:
Determine the size of a vacancy in given solid
Predict vacancy concentration
Determine the type of a given dislocation
Draw illustrations for different types of defects
A
c
B
A
a
a = 2R
Defects are not “deficiencies”
How vacancies are formed?
Both atoms have low electronegativity (less than 1.8), so valence
(outermost) electrons are only weakly attracted by specific atoms,
and hence are shared between the atoms, e.g., Fe +Fe, Fe+Ni
Why vacancies are formed?
S = k Ln( )
Why vacancies are formed?
F= U–T S
Equilibrium Concentration of Vacancies =
nv
G fv
k
– Number of vacancies
– “Energy cost” to form vacancy
– Boltzmann’s constant
nv
Gvf
= exp(? )
N
kT
N – number of atom sites
T – Temperature
nv
? 10 ?3
At 99% of Tmp,
N
nv
?10
At 50% of Tmp, N ? 10
ln(nv/N) = -Gvf/(kT)
ln(nv/N) = – (Gvf/k) (1/T)
nv
N
ln
T
nv
N
1/T
Example 1
• Calculate the vacancy concentration in aluminum at 80%
of TMP=923 K. Gvf = 0.66 eV, and k = 8.62 x 10-5 eV/K.
nv
Gvf
= exp(?
)
N
kT
T = 0.8Tmp = 0.8X923 K=738.4 K
kT = 738.4 (K) X8.62X 10-5 eV/K = 6.37X10-2 eV = 0.0637 eV
nv/N = exp(-Gvf/(kT)) = exp(-0.66/0.0637) = 3.16 x 10-5
Example 2
• The vacancy concentration in Al at 90% of TMP=923 K is
9.93 x 10-5 .Compute the activation energy Gvf . k = 8.62
x 10-5 eV/K.
nv
Gvf
= exp(?
)
N
kT
Ln(nv/N) = -Gvf/KT
Gvf = – kT Ln(nv/N)
T = 0.9Tmp = 0.9X923 K = 830.7 K
kT = 0.0716 eV
Ln(nv/N) = -9.217
Gvf = -9.217 X 0.0716 eV = 0.66 eV
Interstitials: Usually small atoms (H, C, N, O, B) in crystal of large atoms
Example: C in FCC Fe
2R+2r = a
4R = sqrt(2)a
•
Determine the largest size of the interstitial hole in BCC Fe.
RFe = 0.124 nm.
The radius of a carbon atom: Rc = 0.071 nm.
Dislocations
Dislocations are lines of defective bonding in crystals. There are 3 types: edge,
screw and mixed dislocations.
Edge Dislocation
b
B
A
An edge dislocation can be
thought of as the bottom of
an extra half-plane inserted
into a crystal. (See line)
Front view of an edge
dislocation with the
dislocation line coming out
of the paper
Burgers vector b
Screw dislocation
Expected Outcome
Key concepts:
Types of defects in solids
Different types of point defects
Different types of line defects (dislocations)
Burgers vector
Skills:
Determine the size of a vacancy in given solid
Predict vacancy concentration
Determine the type of a given dislocation
Draw illustrations for different types of defects
MSE 250: Structure and Properties of Materials
5. Defects in Solids II
Instructor: Prof. Yang Jiao (yang.jiao.2@asu.edu)
Office: Engineering Research Center Rm 387
Phone: 480-965-4560
Expected Outcome
Key concepts:
Edge dislocations and screw dislocations
Motions of dislocations
Planer defects in solids (grain boundaries)
Diffusion of point defects
Skills:
Draw illustrations for different motions of dislocations
Determine and describe defects in a give solid structure
Compute diffusion coefficient (diffusivity)
Compute the average diffusion distance
b
Dislocations
Dislocations are lines of defective bonding in crystals. There are 3 types: edge,
screw and mixed dislocations.
Edge Dislocation
b
B
A
An edge dislocation can be
thought of as the bottom of
an extra half-plane inserted
into a crystal. (See line)
Front view of an edge
dislocation with the
dislocation line coming out
of the paper
Burgers vector b
Screw dislocation
Edge dislocation
Screw dislocation
Dislocation Motion
Dislocations in a crystal can move when stress is applied, thereby
permanently deforming the crystal.
Atomic view of glide of edge dislocation
As the dislocation moves, it shifts atoms above the glide plane by its
Burger’s vector (1 atomic distance) relative to those below the glide
plane.
apply shear
Glide of edge dislocation

Question: Why are metals ductile and ceramics brittle?
Metals: delocalized bonding allows dislocations to glide easily
Ceramics: difficult for dislocations to move, due to ionic/covalent bonding
Covalent bonding:
The difference is that covalent bonds are localized (between pairs
of atoms) and directional (prefer certain angles), so it is
unfavorable for dislocations to move.
Ionic bonding
+ – + – + – +
+ – + Before dislocation
moves
+ – + – + – +
+ – + After dislocation moves through
crystal
Similar charges are
adjacent
unfavorable
Summary:
Applying a stress to metals causes dislocations to form and move.
Their motion shifts atoms one atomic distance.
Dislocations move easily in metals, but not easily in ceramics, making
metals ductile and ceramics brittle.
Climb of Edge Dislocation
Vacancy capture by
dislocation line,
dislocation climbs up
Nearby atom moves into
dislocation, leaving a
vacancy nearby
Dislocations are the major source & sink for vacancies. Vacancies
allow dislocations to climb up/down.
Glide vs. Climb
Glide is relatively easy, but climb is slower since vacancies must
diffuse to/from the dislocation.
Fe
glide
Fe3C
Glide can be blocked by large
precipitates. Blocking dislocations
increases strength but lowers
ductility.
glide
Fe
climb
glide
Fe3C
Climb allows dislocations to move
around obstacles
How grain boundaries form?
Grain Boundaries
Structure:
Macro
Atomic-Scale
Typical grains are 0.1-1 mm in length (106 – 107 atoms between grain
boundaries)
Grain Boundaries
Structure:
Macro
Atomic-Scale
Typical grains are 0.1-1 mm in length (106 – 107 atoms between grain
boundaries)
Importance of Grain Boundaries
1.)
2.)
Grain Boundaries block dislocations ( strength, ductility)
Small grain size strengthens metals, since dislocations move
less before being blocked by a Grain Boundary.
3.)
Impurities often segregate there due to extra space. Some
impurities weaken materials, like sulfur in steel
4.)
Grain Growth: heating causes small grains to combine into
larger ones
Question: How does this alter mechanical properties?
Activity: Identify the crystal defects in the schematic below.
Summary of “microstructural” concepts
• 10-9m
• 10-8m
• 10-7m
b
• 10-6m
• Increasing
length scale
Diffusion of point defects
Diffusion of point defects
Vacancies diffuse by neighboring atoms jumping into vacant site
Motion of vacancies
diffusion of atoms
Interstitials: Usually small atoms (H, C, N, O, B) in crystal of large atoms
Example: C in FCC Fe
Interstitials diffuse rapidly
Diffusion
Diffusion is the random motion of atoms due to thermal vibrations.
Atoms diffuse fastest in gases, slower in liquids, and slowest in solids.
Crystals:
< ½ Tmp: diffusion mostly along grain boundaries and dislocations. > ½ Tmp: diffusion mostly by vacancies (or interstitials for small atoms)
Approximate distance diffused:
x=
Dt
x – Distance
t – Time
D – Diffusion coefficient (depends
on atom, host material,
temperature)
Diffusion Distance
x = Dt
Diffusion of point defects
D = D0exp(-ED/RT)
D0: T independent parameter (m^2/s)
ED: activation energy for diffusion (J/mol)
R: gas constant, 8.31 J/(mol K)
T: absolute temperature (K)
D = D0exp(-ED/RT)
D1/D2 = exp(- ED/RT1)/ exp(- ED/RT2 )=exp[-(ED/R)*(1/T1-1/T2)]
Ln(D1/D2) = -(ED/R)*(1/T1-1/T2)
D1/D2 = 100, ED/R = 125/8.31 = 15.04, T2 = 300K
T1 = 335 K
•
Determine the temperature at which the diffusivity D of Cu
in Al is 100 times
the diffusivity at 300K. Ed=125kJ/mol,
X1 = Sqrt(D1*t), X2 = Sqrt(D2*t)
R=8.31 J/(mol K). X1/X2 = Sqrt(D1/D2)
•
On average, how further Cu atom can diffuse at this T
compared to that at 300K?
Expected Outcome
Key concepts:
Motions of dislocations
Planer defects in solids (grain boundaries)
Diffusion of point defects
Skills:
Draw illustrations for different motions of dislocations
Determine and describe defects in a give solid structure
Compute diffusion coefficient (diffusivity)
Compute the average diffusion distance

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