Title Wxample - Solution Final Exam Question SKMM3623 1- Updated Corrosion Fatigue (Material) Thermoplastic Creep (Deformation) 703.7 KB 16
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SKMM3623

PART A : Answer ALL questions

Question 1 (25 marks)

(a) 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) Stress concentration factor

Fatigue strength strongly reduced by introduction of stress concentrations, eg.

(ii) Surface effect

Surface finish, surface properties, residual stress : fatigue cracks frequently start at or

near the material’s 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

Fatigue limit

Increased fatigue limit

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SKMM3623

(ii) A 12.5 mm diameter cylindrical rod fabricated from 2014-T6 alloy is subjected to a

repeated tension-compression load cycling along its axis. By referring to Figure 1,

compute the maximum and minimum loads that should be applied to yield a fatigue life

of 1.0 x 107 cycles. Assume that the data was taken for a mean stress of 50 MPa.

(6 marks)

Figure 1

MPa 320=MPa) (2)(160=2= aminmax σσσ −

Since σm = 50 MPa,

MPa 100=MPa) 50(2)(=2=+ mminmax σσσ

Simultaneous solution of these two expressions for σmax and σmin yields

σmax = +210 MPa and σmin = -110 MPa. Now, inasmuch as σ =
F

Ao
, and Ao = π

do
2

2

then

( ) ( )
N 25771=

4
105.12)(/10210

=
4

=F
23262

max
max

mxmNxd o
−ππσ

( ) ( )
N 34991=

4
105.12)(/10110

=
4

=F
23262

min
min −

π−π − mxmNxd oσ

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SKMM3623

(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)

Table 1: Galvanic series
Platinum
Gold
Graphite
Titanium
Silver
316 Stainless steel (passive)
304 Stainless steel (passive)
Inconel (80Ni-13Cr-7Fe) (passive)
Nickel (passive)
Monel (70Ni-30Cu)

Increasingly inert (cathodic) Copper-Nickel alloys
Bronzes (Cu-Sn alloys)
Copper
Brasses (Cu-Zn alloys)
Inconel (active)
Nickel (active)
Tin

Increasingly active (anodic) 316 Stainless steel (active)
304 Stainless steel (active)
Cast
Iron and steel
Aluminum alloys
Commercially pure aluminum
Zinc
Magnesium and magnesium alloys

Table 2

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SKMM3623

(d) 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 Cl-

content, acidity and temperature

2. Increase the resistance of materials

3. The best protection against pitting corrosion is to select a material with

Figure 3

Metal Metal Density (g/cm3) Metal Oxide
Oxide Density

(g/cm3)
Zr

Bi

6.51

9.80

ZrO2

Bi2O3

5.89

8.90

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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)

aYK fIC πσ.=

2

22

2

450
8.3

)(
)

1
( ⎟

==
ππσπ x

K
a

f

IC =7.22x10-6m = 7.22 µm

If KIC = 12.5 MPa.√m

MPa
xxaY

K IC
f 78.1480

1022.7)((
5.12

. 6
===

−πππ
σ

(b) A composite consists of a continuous glass-fiber-reinforced-epoxy resin produced by using 60

percent by volume of E-glass fibers having a modulus of elasticity of Ef = 7.24 x 104 MPa and a

tensile strength of 2.4 GPa and a hardened epoxy resin with a modulus of Em = 3.1 x 103 MPa and a

tensile strength of 60 MPa.

Calculate:

(i) the modulus of elasticity

(ii) the tensile strength

(iii) the fraction of the load carried by the fiber

for this composite material subjected to stress under isostrain condition.

(8 marks)

i. Modulus of elasticity of composite is

Ec=EfVf + EmVm

= (7.24x104 MPa)(0.6) + (3.1x103MPa)(0.4)

= 44.68 GPa

2 marks

2 marks

2 ½  marks

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SKMM3623

ii. Tensile strength

σc = σfVf+σmVm
= (2.4 GPa)(0.6) + (0.06GPa)(0.4)

= 1.46 GPa

iii. Fraction load carried by fiber is

97.0
)4.0)(1.3()6.0)(4.72(

)6.0)(4.72(

=
+

=

+
=

GPaGPa
GPa

VEVE
VE

P
P

mmff

ff

c

f

(c) Discuss a method to produce an aircraft part/component made of a carbon-fiber-reinforced-epoxy

composite.

(5 marks)

Any suitable PMC processing (5 marks)

Hand Lay – Up Method

1. Mould is treated with mould release agent

2. Thin gel coat (resin, colour) is applied to the outside surface

3. Layers of resin (matrix) and fibres (in the form of mat or cloth). Each layer is rolled to

impregnate the fibre with resin and remove air

4. Part is cured (set)

5. Fully hardened part is removed

2 ½  marks

3  marks