MAE 3324 Homework 9 - Solution
9.1
Equation 9.1 is employed to solve this problem, as
a
m = 20
t
1/2
2.5 102 mm 1/2
2
= 2404 MPa (354,000 psi)
= (2)(170 MPa)
2.5 104 mm
9.3 Using Equation 9.3; taking the value of 69 GPa (Table 7.1) as the modulus
Chapter 11
Phase Transformations
Controlled Precipitation
Fe-C Time Temperature Transformations
Continuous Cooling Transformations
Slides contain Copyrighted material from Callister text, 2nd ed.
Chapter 11 Phase Transformations
Phase Diagram give us info
TYPES OF CRYSTAL IMPERFECTIONS
Vacancy
Interstitial atoms
(impurity or self)
Substitutional atoms
Point defects: 0-dimensional
Dislocations
Line defects: 1 dimensional
Grain Boundaries
Stacking faults
Area defects: 2 dimensional
Inclusions
Second
Phase Diagrams
1
Definitions: Components, phases and microconstituents
Components: The elements or compounds which are
mixed initially (e.g., Al and Cu, or sugar and water)
Phases: The physically and/or chemically distinct
material regions (e.g., and ,
Binary PD in-class exercise
A-B alloy
1.
2.
3.
4.
5.
Label the axis.
Mark the melting points of pure A and pure B.
Mark all the phase regions
For Co = 0.5 wt. fr. B, at what temperature will the first solid form? What is its conc.?
For Co = 0.5 wt. fr. B,
MAE 3324 - Exam 3
Seat: _
Name _
- Label all drawings so they are unambiguous. For microstructures label phases and microconstituents
when they exist. Show your work, including construction lines on graphs.
- For numerical problems, set up equations, subs
MAE 3324
Homework 8
1. An aluminum bar 125 mm long and having a square cross section 16.5 mm
on an edge is pulled in tension with a load of 66,700 N, and experiences an
elongation of 0.43 mm. Assuming that the deformation is entirely elastic,
calculate th
MAE 3324-001
Spring 2014
Assignment #3
4.2 (a) Compute the repeat unit molecular weight of polypropylene.
(b) Compute the number-average molecular weight for a polypropylene for which the degree of
polymerization is 15,000.
4.6 Sketch portions of a linear
Solutions to suggested homework review problems.
2.8
+
The Na ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the
same as neon (Figure 2.6).
The Cl ion is a chlorine atom that has acquired one extra electr
CHAPTER 3
STRUCTURES OF METALS AND CERAMICS
LEARNING OBJECTIVES
1.
Give a definition of crystalline solid.
2.
Describe the difference in atomic/molecular structure between crystalline and
noncrystalline materials.
3.
Give a brief definition of a unit cell
CHAPTER 4
POLYMER STRUCTURES
LEARNING OBJECTIVES
1.
Define isomerism.
2.
Describe a typical polymer molecule in terms of its chain structure, and, in
addition, how the molecule may be generated from repeat units.
3.
Draw repeat units structures for polyet
CHAPTER 5
IMPERFECTIONS IN SOLIDS
LEARNING OBJECTIVES
1.
Describe both vacancy and self-interstitial crystalline defects.
2.
Given the density and atomic weight for some material, as well as
Avogadro's number, compute the number of atomic sites per cubic
Structure and Mechanical Behavior of Materials
Problem Set No. 1 (Ch.2)
1. Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460
amu, 83.79% of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an at
Phase transformation notes
Types of solid state phase transformations
Diffusion based
Diffusionless
Diffusion based phase transformation consists of nucleation
and growth
Phase nucleation and growth generally will occur when free
energy conditions ar
Face centered cubic structure (FCC)
Close packed directions are face diagonals.
-Note: All atoms are identical; the face-centered
atoms are shaded differently only for ease of viewing.
Adapted from Fig. 3.1(a),
Callister 6e.
Lattice parameters: a
ABCABC.
Ch 3, problems 1, 3, 7, 12, 23, 27, 29, 36, 43, 44
3.1 If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters.
Solution
Lead has an FCC crystal structure (Table 3.1). The FCC unit cell volume may be computed from
E
MAE 3324 - Homework 7
1. For a square cross-section, A0 = b02 , where b0 is the edge length. Combining Equations 7.1, 7.2, and
7.5 and solving for E, leads to = 71.2 109 N/m2 = 71.2 GPa
2. This problem asks us to compute the diameter of a cylindrical spec
MAE 3324 - Solution for Homework 8
8.5
(a) A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation
motion (or slip) occurs.
(b) All metals do not have the same slip system. The reason for this is that for mos
Structures of Metals and
Ceramics: Crystal Structures
Important Terms
Crystalline Material: One in which atoms are situated in a repeating
array over large atomic distances.
Atomic Hard-Sphere Model: Atoms treated as hard spheres; nearest
atoms touching
Phase Diagrams
Important terms
Solutions versus mixtures
Components, constituents and phases
Solubility and solubility limit
Examples: Salt solution (Brine), Sugar
solution (Syrup)
Examples
Find the solubility limit of sugar in water at 20 C and 80 C
Homework 9
Chapter 9, Problems 1, 3, 4, 5, 8, 17, 18, 20
Also recommended: use the extended objectives for chapter 9 to aid in studying concepts
not conducive to homework problems.
MAE 3324-001
Homework 7
1. An aluminum bar 125 mm long and having a square cross section 16.5 mm on an edge is pulled in
tension with a load of 66,700 N, and experiences an elongation of 0.43 mm. Assuming that the
deformation is entirely elastic, calculat
Homework 6
Part A
1. Rewrite the expression for the total free energy change (equation 11.1) for nucleation for the
case of a cubic nucleus of edge length a. Differentiate this expression with respect to a and
solve for both the critical cube edge length
MAE 3324
Homework 5B Solution
= 17.5 % , 82.5 %
Teutectic = 251.7C
b. = 6.2% , 93.8%
= 97.1 % , 2.9 %
= 17.5 % , 82.5 %
1.
a.
2.
a. T = 320C
b.
c.
d.
e.
C = 2 at% Sb, 98 at% Pb
T = 280C
= 13%, 87%
is the only phase present, the composition is 4a
MAE 3324
Homework 5A - Solution
1a) 1310 C
b) 86% Si, 14%Ge
c) In 100 moles total, there are 86 moles Si, 14 moles Ge, so
86 x 28g Si/mole
= 70.3 wt% Si
86 x 28g Si/mole + 14 x 72.6 g Ge/mole
1 70.3= 29.7% wt% Ge
C solid =70.3% Si, 29.7 wt% Ge
d) 1170 C
e
MAE 3324
Homework 5C solution
1. Answer the following
a. In eutectic, the parent phase is liquid, in the eutectoid, the parent phase is solid.
b. Hypo eutectoid is below the eutectoid composition, and hyper is above the eutectoid
composition
c. Proeutecto
MAE 3324
HW 6 Solution
Part A
1. Rewrite the expression for the total free energy change (equation 11.1) for nucleation for the case of a
cubic nucleus of edge length a. Differentiate this expression with respect to a and solve for both the
critical cube
MAE 3324-001
Homework 2 Solutions
Chapter 3:
1a. From the data given in the problem, and realizing that 75.99 = 2, the interplanar spacing for the
(211) set of planes for Nb may be computed using Equation 3.14: d 211 = 0.1347 nm
1b. Equation 3.15 yields t