EMA 4223
Mechanical Behavior of Materials
February 19, 2010
Test 2A
Name _
1. The bond energy curves for three engineering materials are shown in the figure. Select the best material for
use in each application described below by circling your choice.
i.

8.2
Explain why FCC metals show a ductile fracture even at low temperatures,
while BCC metals do not.
FCC metals show ductile fracture at low temperatures while BCC metals do not
because of the fact that there are many more slip systems in FCC metals that

9.1 A Charpy machine with a hammer weighing 200N has a 1-m long arm. The initial
height ho is equal to 1.2m. The Charpy specimen, (See Figure 9.2), absorbs 80J of energy
in the fracturing process. Determine:
(a) The velocity of the hammer upon impact with

7.24
Using the Weibull Equation, establish the tensile strength, with a 50% survival
probability, of specimens with a length of 60mm and a diameter of 5mm. Uniaxial
tensile tests carried out on specimens with a length of 20mm and the same
diameter yielded

EMA 4223
Mechanical Behavior of Materials
February 19, 2010
Test 2A
Name _
1. The bond energy curves for three engineering materials are shown in the figure. Select the best material for
use in each application described below by circling your choice.
i.

7.1 In a polyvinyl chloride (PVC) plate, there is an elliptical, tbrough-the-thickness
cavity. The dimensions of the cavity are:
Major axis = Imm,
Minor axis = O.lmm.
Compute the stress concentrationfactor Kt, at the extremities of the cavity.
Solution
K

f ,t t ", Z F ailure b y g eneraly ield o r f ast f racture
p lotted a gainst c rack
Figure 16.7 s hows t he l oci o f g eneral y ielding a nd f ast fracture
a nd i s s imPlY g iven b 1'
size. T he yield l ocus i s o bviousiy i ndependent o f c rack size,

7.1 In a polyvinyl chloride (PVC) plate, there is an elliptical, tbrough-the-thickness
cavity. The dimensions of the cavity are:
Major axis = Imm,
Minor axis = O.lmm.
Compute the stress concentrationfactor Kt, at the extremities of the cavity.
Solution
K

EMA 4223
Mechanical Behavior of Materials
Date
Test 1B
Name _
1. .[15 points] (a) Sketch the specific volume as a function of temperature for an amorphous material and for a
crystalline material for two different cooling rates, i.e., rapid and slow. Indic

EMA 4223 Test 2
Mechanical Behavior of Materials
Example
Name _
Each Question is worth 20 points unless noted.
1. A polycrystalline metal has a plastic stress-strain curve that obeys the Hollomon-Ludwik equation, = K n.
Determine n, knowing that the flow

Homogeneous Nucleation of Dislocations
Grain Boundary as a Source of Dislocations
Emission of
dislocations
from ledges in
grain boundary, as observed in transmission electron microscopy
durin
g
heating by electron beam. (Courtesy of L. E. Murr.)

Frank-Read Mechanism
Formation of dislocation loop by the FrankRead mechanism.
Dislocation Pileups
Pileup of dislocations against a
barrier.
Pileup of dislocations against grain
boundaries (or dislocations being emitted
from grain boundary sources?) in co

Dislocation Interactions
(a) Edge dislocation traversing forest
dislocation.
(b)
S
c
rew dislocation traversing forest dislocations.
*Courtesy of Prof. Mark Weaver UA

Peierls-Nabarro Stress
(a) Movement of dislocation away from its
equilibrium position.
(b) Variation
of PeierlsNabarro stress with distance. (Reprinted with
permissi
o
n from H. Conrad, J. Metals, 16 (1964), 583.)
*Courtesy of Prof. Mark Weaver UA

Problem 1.15 Different polymorphs of a material can have different mechanical properties. Give some
examples.
A.
There are a number of answers possible. One example is zirconia. Zirconia at room
temperature is monoclinic and easily cracks. Zirconia transi

3.1 A polycrystalline metal has a plastic stress-strain curve that obeys Hollomons equation.
= K n
Determine n, knowing that the flow stresses of this material at 2% and 10% plastic deformation (offset) are equal to 175
and 185 MPa, respectively.
At 2% s

6.16
The flow stress is given:
i G b () 1/2
Experimental observation (p.351) = k /D where k is a constant and D is grain size
i G b (k/D) 1/2
This is in the form of
ik (D) -1/2 where k = G b k1/2
Q.E.D.
6.18
= 25 MPa + 200 0.5 MPa
The area under the true

Equations for Test 2 Examination
u = ln [(A0)/Au]
1 + 2 + 3 = 0
K (ddt)m
F/A = H = 30
+ K n (Holloman-Ludwik Eq.)
t = (dt / dt ) [at necking]
u = n
G = E / [2(1+]
= 0 + G b ()1/2
t = [P/A0] exp(t) = e (1 + e)
= [d/dt]= (dv/dy)
t = ln (1 + e) = ln (L/L

Plot the stress-strain curve for alumina in tension, knowing that the density of
microcracks increases linearly with stress (N= k ). The grain size is 20 m. The
failure stress is 1 GPa; given k = 5.45 x 104 m3/Pa, Eo =420 GPa.
2.48
Given: Grain size: 20 m