1150
ME 382 HOMEWORK 6 SOLUTION
4
890
700
0.25
2.75
1. (a) As shown in the diagram, 0.25%2.75%.
(b) As shown in the diagram, the alloy should be held between 700 Cand 890 C.
2
ME 382 HOMEWORK 6 SOLUTION
500
2. After 100 hours
ME 382 HOMEWORK 2 SOLUTION
1.
xx = 3.0 103
yy = 1.5 103
xy = 2.0 103
The strain gauges are on the free surface of a steel structure,
E = 196 GPa = 1.96 105 MPa
= 0.3
E
196
G=
=
75.38GPa
2(1 + )
2 (1 + 0.3)
Since this is free surface zz = 0, xz = yz = 0
3.
4. (Ashby & Jones 1: 13.2) A large thick plate of steel is examined by X-ray methods, and found
to contain no detectable cracks. The equipment can detect a single edge-crack of depth a
= 5 mm or greater. The steel has a fracture toughness Kc of 40 MN m
ME 382 Fall 2012
Homework 1
(Due on Friday, September 14th, 2012)
Question numbers are from third editions of the textbooks.
(available on reserve in the Engineering Library and online)
Answers given have generally been rounded, so you may get slightly di
ME 382 HOMEWORK 10 SOLUTION
1.
(i)
ar =
680
C1
=
61 MPa
.15
Nf
(107 ). 15
max = 61 MPa < Y
Therefore, no yielding.
(ii)
max = 100 MPa < Y
Therefore, no yielding.
Nf = (
C1 1/.15
680 1/.15
)
=(
)
3.5 105 cycles
ar
100
(iii)
max = 50 + 100 = 150 MPa < Y
T
ME 382 HOMEWORK 2 SOLUTION
1.
xx = 3.0 103
yy = 1.5 103
xy = 2.0 103
The strain gauges are on the free surface of a steel structure,
E = 196 GPa = 1.96 105 MPa
= 0.3
E
196
G=
=
75.38GPa
2(1 + )
2 (1 + 0.3)
Since this is free surface zz = 0, xz = yz = 0
ME 382 HOMEWORK 2 SOLUTION
1.
xx = 3.0 103
yy = 1.5 103
xy = 2.0 103
The strain gauges are on the free surface of a steel structure,
E = 196 GPa = 1.96 105 MPa
= 0.3
E
196
G=
=
75.38GPa
2(1 + )
2 (1 + 0.3)
Since this is free surface zz = 0, xz = yz = 0
ME 382 Fall 2010 HW 9 Solutions
2.
Dowling 11.28.
For 7075-T6 aluminum, Y = 523 MPa, KIC = 29 MPam, and the Paris Law for
fatigue crack growth here is given by da/dN = 1.38 x 10-11K3.70 m/cycle, where
K is in MPam. You may assume that yielding does not oc
ME 382 Fall 2012
Homework 3
(Due on Friday, September 28th, 2012)
Question numbers are from third editions of the textbooks
(available online and on reserve in the Engineering Library)
Answers given have generally been rounded, so you may get slightly dif
ME 382 - Winter 2016
Homework 2
Due on January 22, 2016
The$questions$identified$with$*$are$intended$to$be$straight4forward$applications$of$what$has$
been$covered$in$class$or$required$background$from$ME211.$
The$answers$may$have$been$rounded.$
You will ne
ME 382 Fall 2012
Homework 6
(Due Friday, October 26, 2012)
The questions identified with * are intended to be straightforward applications or
review of what has been covered in class. Explain your answers, with ske
Problem 1:
(a) Derive an expression for the critical radius needed for the homogeneous nucleation of a
cylindrical precipitate (with circular basis of radius r and height h=Cr, where C is a
constant) of in a matrix of , in terms of the surface energy !" ,
given
ME 382 HW1 Solutions
>
ME 382 HW1 Solutions
ME 382 HW1 Solutions
ME 382 HW1 Solutions
ME 382 HW1 Solutions
is shown as follows:
ME 382 HW1 Solutions
Note: Another solution provided from a separated pdf file
ME 382 HW1 Solutions
ME 382 HW1 Solutions
HW02 Solution
A: Strength of the attractive potential
B: Strength of the repulsive potential
m: Steepness of the attractive slope
n: Steepness of the repulsive slope
ME 382 - Fall 2015
Homework 2
(Due on Friday, January 20, 2017)
The questions identified with * are intended to be more challenging / interesting
The answers may have been rounded.
1.
A set of strain gauges on the free surface of an elastic steel stru
Lecture 03 - ME211 Review
ME 211 Review
Stresses
Stress is a second-order tensor:
#Ti
#A i $ 0 # A j
"ij = Lim
!
where dTi is traction in i direction on j face
!
#" " " & #" ) ) &
% xx xy xz ( % xx xy xz (
"ij = %"yx "yy "yz ( = %) yx "yy ) yz (
%
( %
(
$
ME 382
Lecture 05
January 13, 2017
Forces between atoms and molecules
Concept of Gibbs free energy
Physical origin of modulus
Primary and secondary bonding in solids
Structure of polymers and glass-transition temperature
Force and stiffness
Force can be d
HW04 Solution
The complete solubility indicates that Cu and Pt have the same crystal structure.
( It's sufficient to say they are of the same structure. In this case, Cu has FCC
structure, so Pt must also have FCC structure.
The statement is false.
The
ME 382 Winter 2017
Homework 5
(Due Friday, February 17, 2017)
1.
Explain the following observations:
(i) Nickel can be strengthened by adding small hard particles of thoria.
(ii) At room temperature, the yield strength of pure silver (~50 MPa) is much
low
Lecture 08 - Introduction to equilibrium phase diagrams
STRUCTURES OF SOLUTIONS AND COMPOUNDS
Few metals used in pure state (usually alloyed with another element)
E.g.
Steel (Fe & C),
2xxx Al (Al +4%Cu+Mg, Si, Mn)
Titanium (Ti-6%Al-4%V)
Brass (Cu + 20% Zn
ME 382
Lecture 02
September 11, 2015
Stiffness / strength / toughness
Tensile test
Yield strength, tensile strength, modulus
True stress & strain
Work hardening
Uniaxial tensile test to measure mechanical properties
1. Measure length and area
2
Lecture 05 - Bonding & physical origin of modulus
PROPERTIES OF ENGINEERING MATERIALS
Material properties depend on:
(a) Type of atomic bonding
(b) Three dimensional arrangements of atoms and molecules (crystal structure)
(c) Atomic-scale defects in the c
ME 382 Winter 2017
Homework 3
(Due on Friday, January 27th, 2017)
1. It is possible to measure the density of a material to a high degree of accuracy (4
significant figures). A sample of aluminum cooled rapidly from a temperature just below
the melting po
ME 382 Fall 2015
Homework 4
(Due Friday, February 3, 2017)
The questions identified with * are intended to be more challenging / interesting
The answers may have been rounded.
1.
Use the phase diagram for copper and platinum shown below to answer
this que
Phase Diagrams
a Review
Topic 2
Review of
Phase Transformation
Diagrams
Solution and Solubility
Example: Solubility of salt in water
There exists a maximum amount of salt that can be
completely dissolved in water; excess of salt stays as solid.
This maxim