Wxample - Solution Final Exam Question SKMM3623 1 - Updated
Wxample - Solution Final Exam Question SKMM3623 1 - Updated
Wxample - Solution Final Exam Question SKMM3623 1 - Updated
The presence of stress risers on the component has been identified as one of the major factors that
affect fatigue strength of a metal. As an engineer, how could you design engineering components
that can reduce failures due to fatigue loading. With the aid of suitable sketches wherever
possible, discuss any two (2) theses factors
(4 marks)
(i)
(ii)
Surface effect
Surface finish, surface properties, residual stress : fatigue cracks frequently start at or
near the materials surface
(iii)
Microstructural factors
Small grain size improves fatigue life. Small grain size contains large number of grain
boundaries impede/delay crack initiation
(iv)
Mean stress
Increasing mean stress will improves the fatigue life
(b)
(i)
With the aid of sketches describe the S-N curve for stainless steel component. If the
component is surface hardened, how would be the profile of the S-N curve? Discuss
briefly.
(4 marks)
The fatigue strength decreases as the number of stress cycles increase until reaching a
certain value of cycles where no decrease in fatigue strength occurs. This is called
fatigue limit.
If the component is surface hardened, the fatigue strength will be increased, this will
increase the leveling off of the S-N curve or it will increase the fatigue limit of the
stainless steel
Increasedfatiguelimit
Fatiguelimit
-3SKMM3623
(ii)
Figure 1
Fmax =
Fmin =
max d o2
4
min d o2
4
(210 x10
=
( 110x10
=
d
F
, and A o = o
A
2
o
) (
) (
N / m 2 () 12.5 x10 3 m
= 25771 N
4
N / m 2 () 12.5 x10 3 m
= 13499 N
4
then
-4SKMM3623
(c)
A large plate is subjected to constant amplitude uniaxial cyclic tensile and compressive stresses.
If the initial and critical crack lengths in the plate are 1.25 and 12 mm respectively, determine the
maximum tensile stress that will produce the fatigue life of 2.0 x 106 cycles. Assume: m = 3.0,
A = 6.0 x 10-11 and Y = 1.2.
)[
( )]
dN =
1
da
10
3
5.774 10
max a 3 2
( )
da dN = A.K m
(7 marks)
)(
)( )
3
= 5.774 10 10 max
a3 2
Nf
dN =
1
3
5.774 10 10 max
)(
af
ao
a 3 2 da
0.012
a 3 2+1
1
Nf =
3
3 2 + 1 0.00125
5.774 10 10 max
)(
3
max
= 33169.7
max = 32.13MPa
(d)
Most of the fatigue failure is normally initiated at the surface of the engineering components.
However there are circumstances that the failure is initiated from the inside the component. Give
your opinion why this is happening.
(4 marks)
The presence of second phase particles in an engineering material, and although they increase
the overall strength, decrease the fatigue life because they act as stress concentrators and hence
accelerate the crack initiation
-5SKMM3623
(i)
Creep phenomenon cannot be avoided totally in materials when they are subjected to
static loading in high temperature environment. Discuss briefly ways to minimize this
phenomenon with appropriate examples. Limit your discussion to two (2) methods only.
(4 marks)
1. Choose alloy an alloy with a very low creep rate, with high dimensional stability
at high temperature such as refractory metal, super alloy etc.
2. Modify alloy properties via:
o Precipitation hardened,
o Directional solidified alloy
(ii)
There are at least three (3) important parameters or variables that significantly
influenced the creep rate. Briefly describe two (2) of these parameters and elaborate how
each of them affects the creep rate.
(4 marks)
Load/stress
Temperature
Time
-6SKMM3623
(b)
A specimen 760 mm long of a S590 austenitic steel alloy is to be exposed to a tensile stress of 80
MPa at 815OC. With the aid of Figure 2, determine its elongation after 5000 hours.
(5 marks)
Figure 2
From the 815OC line in Figure 2, the steady state creep rate, d/dt, is about 3.0 x 10-5 h-1 at 80 MPa.
The steady state creep strain, s, therefore, is just the product of d/dt and time as
s = d dt (time)
l s = l o s = (760 mm)(0.15)=114 mm
(c)
An alloy is evaluated for potential creep deformation in a short term laboratory experiment. The
creep rate is found to be 1% per hour at 800OC and 5.5 x 10-2 % per hour at 700OC when exposed
to a tensile stress of 50 MPa. (R = 8.314 J/mol.K)
(i)
800 C Ce Q R (1073 K )
=
700 C Ce Q R (973 K )
o
1%
= e [8.314 J / mol .K ] 1073 973
2
5.5 x10 %
-7SKMM3623
(ii)
500
( 8.314 )(1073)
= (1.80 x1012 %.hr 1 )e ( 252 x10 ) (8.314 )( 773) = 1.75 x10 5%.hr 1
(8 marks)
(d)
A nickel alloy turbine blade for a jet engine which operates at high temperature will
eventually initiate crack in service. With the aid of suitable sketches explain the mechanism of
failure if the crack grows and leads to fracture.
(4 marks)
Due to grain boundary sliding. At high temperature the grains are able to slide relative
to each other which produce wedge crack as well as small holes at the grain boundaries
due to movement of vacancies.
(2)
(3)
-8SKMM3623
(ii)
(iii) zinc
Rank the corrosion potential of the pairs from the highest to the lowest.
Zinc and titanium will generate the highest corrosion potential whilst Inconel
(active) and nickel (active) will have the lowest corrosion potential.
(5 marks)
(b)
A cylindrical steel (Fe) tank is coated with thick layer of zinc on the inside. The tank is 50 cm in
diameter, 70 cm high and filled to the 45 cm level with aerated water. If the corrosion current
density is 5.8 x 10-5 A/cm2, how much zinc in grams per minute is being corroded? (Atomic
weight of Fe = 65.38 g/mol, Faradays constant = 96500 A.s/mol)
(6 marks)
(c)
(i)
How does the Pilling Bedworth (PB) ratio is used to determine the characteristic of oxide
layer for corrosion protection?
(3 marks)
PB ratio is determined by the ratio of the oxide produced to metal used for oxidation
PB < 1 : oxide layer is porous and non-protective
PB > 1 : oxide is protective
PB > 2 : oxide is in compressive stress, easy to crack, non protective
-9SKMM3623
(ii)
For each of the metals listed in Table 2, compute the Pilling-Bedworth ratio and specify
whether the oxide that forms on the metal surface to be protective or not. Justify your
answer. (Atomic weight : Zr = 91.22 g/mol, Bi = 208.98 g/mol, O = 16 g/mol)
(5 marks)
PB ratio = W . d
n . D. w
For Zr:
WZrO2 = (91.22g/mol) + 2(16 g/mol) = 123.22 g/mol
PB ratio = (123.22 g/mol)(6.51 g/cm3) = 1.49 (Protective)
(1) (5.89 g/cm3)(91.22 g/mol)
For Sn:
WBi2O3 = 2(208.98 g/mol) + 3(16 g/mol) = 465.96 g/mol
PB ratio = (465.96 g/mol)(9.80 g/cm3) = 1.23 (Protective)
2 (8.90 g/cm3)(208.98 g/mol)
Cast
Iron and steel
Aluminum alloys
Cadmium
Commercially pure aluminum
Zinc
Magnesium and magnesium alloys
Table 2
- 10 SKMM3623
Zr
Metal Density
(g/cm3)
6.51
Bi
9.80
Metal
(d)
ZrO2
Oxide Density
(g/cm3)
5.89
Bi2O3
8.90
Metal Oxide
Figure 3 shows the corroded stainless steel sink. Suggest the type and discuss the mechanism of
corrosion that has occurred. Also, describe how it can be controlled.
(6 marks)
Type of corrosion : Pitting
Mechanism : 1. Local breakdown of passive film (initiation) act as anode
2. The unbroken film (protective film) acts as cathode
3. Pits develop at the anodic region.
4. Presence of Cl- reduces the pH inside the electrolyte of the growing pit to about
1 (acidic) increase corrosion (autocatalytic process)
Control : 1. Decrease the aggressiveness of the environment - By decreasing the Clcontent, acidity and temperature
2. Increase the resistance of materials
3. The best protection against pitting corrosion is to select a material with
adequate pitting resistance
Figure 3
- 11 SKMM3623
(i)
List the advantages and disadvantages of thermoset over the thermoplastic polymers
and give two (2) examples for each type of polymer.
(6 marks)
Answer:
Thermosetting
Advantages
-
1mark
disadvantages
-
Cannot be re-melted and reformed into another shape but degrade after heated to a high temp.
Cannot be recycled
Difficult to color
More brittle
1mark
Examples:
Thermoset : electrical moldings, motor housing, epoxies materials, Polyurethane foam, bakelite materials
(mounting, telephone)
1mark
Thermoplastics:
Advantages:
-
Can be reheated and reformed into new shapes without significant changes in the properties
Can be recycled
Easy to color
1mark
Disadvantages:
-
Soft
1mark
1mark
- 12 SKMM3623
(ii)
Answer:
Non-crystalline plastics
Above Tg: viscous (rubbery)
1marks
(iii)
1marks
With the aid of sketches show and discuss the effect of temperatures on stress strain
behavior of thermoplastic material.
(3 marks)
Answer:
3marks
- 13 SKMM3623
(b)
(i)
Natural rubber or latex is the main material in tyres production. But the mechanical
properties of this material at elevated temperature are questionable. Name a process that
commonly use to solve this problem. What are the effects of this process to the
mechanical properties of tyres as the whole?
(3 marks)
Answer
Process: vulcanisation
Effect:
1mark
Improve strength (stiffness), high temperature stability via cross linking sulphur with rubber
2marks
(ii)
Answer:
Mw polybutadiene (C4H6) = (4x12) + (6x1) = 54 g/mol
1mark
1mark
Mw average copolymer = (53 g/mol PAN) + (54 g/mol PB) = 53.75 g/mol
1mark
Therefore if 100 kg rubber used for cross link 20%, the amount of Sulfur, S required is
(100 kg / 53.75 g/mol) x (32 g/mol) x 0.2 = 11.9 kg
2marks
- 14 SKMM3623
(c)
You are given an amount of phenolic which is a thermosetting material to produce engineering
components. Suggest on possible component and explain the processing method of producing
them.
(5 marks)
Answer:
Name component (1 mark)
Motor housings, circuit boards, electrical fixtures
Discuss method (4 marks)
Compression molding
This process was the first to be used to form plastics. It involves four steps:
1. Pre-formed blanks, powders or pellets are placed in the bottom section of a heated mould or die.
2. The other half of the mould is lowered and is pressure applied.
3. The material softens under heat and pressure, flowing to fill the mould. Excess is squeezed from
the mould. If a thermoset, cross-linking occurs in the mould.
4. The mould is opened and the part is removed while they are hot and after curing is complete
- 15 SKMM3623
(i)
Describe how to produce tempered and annealed glass and give an example of application
for each.
(6 marks)
Answer:
Tempered glass
Tempering is achieved by heating the glass to a temperature > Tg, then rapidly cooled to
room temperature.
The surface of the glass cools first and contracts; later the centre cools and attempts to
contract but is prevented from doing so by the rigid and strong surface.
This produces high tensile stresses in the centre but compressive stresses at the surface.
Application: car windows, safety glass doors
3marks
Annealed glass
The glass is heated to the annealing temperature, then slowly cooled to RT
Application: tabletopglass,cabinetdoorglassandbasementwindows
3marks
(ii)
What makes partially stabilized zirconia (PSZ) has higher fracture toughness as
compared to other engineering ceramics?
(2 marks)
Answer:
A fine metastable precipitate with tetragonal structure known as PSZ formed.
As the crack propagates, it creates a local stress field that induces transformation of the tetragonal
structure to the monolithic (or monoclinic) structure in that region.
This transformation is accompanied by a volume expansion, causing a compressive stress locally
and in turn a squeezing effect on the crack and enhancing the fracture toughness
- 16 SKMM3623
(ii)
A partially stabilized zirconia (PSZ) sample has a fracture toughness of KIC = 3.8
MPa.m1/2 when tested on a four point bend test. If the sample fails at a stress of 450 MPa,
what is the size of the largest surface flaw? If this PSZ sampel with the fracture toughness
of KIC = 12.5 MPa.m1/2 has the same surface flaw, calculate the critical stress to cause
failure. (Geometrical factor,Y =
)
(4 marks)
Answer:
K IC = Y . f a
2
3 .8
-6
a=( ) 2
=
=7.22x10 m = 7.22 m
2
f ( )
450
x
K IC
2marks
f =
(b)
K IC
Y . a
12.5
( )( x7.22 x10 6
= 1480.78MPa
2marks
(ii)
(iii)
2marks
- 17 SKMM3623
2marks
Pf
EfVf
=
P
+
E
V
E
V
c
f
f
m
m
(72.4GPa)(0.6)
=
(72.4GPa)(0.6) + (3.1GPa)(0.4)
= 0.97
(c)
3marks
Answer:
Any suitable PMC processing (5 marks)
Hand Lay Up Method