Wxample - Solution Final Exam Question SKMM3623 1- Updated

-2SKMM3623   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. notches, holes, threaded region etc.

(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

Increased fatigue limit Fatigue limit 

     

-3SKMM3623  

(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

σ max −σ min = 2σ a = (2)(160 MPa) = 320 MPa Since σm = 50 MPa,

σ max +σ min = 2σ m = (2)(50 MPa) =100 MPa Simultaneous solution of these two expressions for σmax and σmin yields σmax = +210 MPa and σmin = -110 MPa. Now, inasmuch as σ =

Fmax =

Fmin =

     

σ max πd o2 4

σ min πd o2 4

(210 x10 =

(− 110x10 =

⎛d ⎞ F , and A o = π ⎜ o ⎟ A ⎝ 2 ⎠ o

) (

)

) (

)

2

6

N / m 2 (π) 12.5 x10 −3 m = 25771 N 4

6

N / m 2 (π) 12.5 x10 −3 m = − 13499 N 4

2

2

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.

(

)[

( )]

da dN = 6.0 × 10 −11 (2.127 )(σ max ) a 1 2

dN =

3

1 da −10 3 5.774 × 10 σ max a 3 2

(

)

( )

∆K = Y .∆σ π .a = (1.2)(σ max ) π .a = 2.127(σ max ) a 1 2

da dN = A.∆K m

(

(7 marks)

)(

(

∫

)

)( )

3 = 5.774 × 10 −10 σ max a3 2

Nf

0

dN =

1 3 5.774 × 10 −10 σ 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 σ max

(

)(

)

3 σ max = 33169.7

σ 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   Question 2 (25 marks) a)

(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

Need to elaborate effect of time, temp and stress on creep

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, dε/dt, is about 3.0 x 10-5 h-1 at 80 MPa. The steady state creep strain, εs, therefore, is just the product of dε/dt and time as

ε s = dε dt × (time)

= (3.0 x 10 -5 h -1 )(5,000 h) = 0.15 Strain and elongation are related; solving for the steady state elongation, ∆ls, leads to

∆l s = l o ε 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)

Calculate the activation energy for creep in this temperature range.

ε 800 C Ce −Q R (1073 K ) = ε 700 C Ce −Q R (973 K ) o

o

−Q

⎛ 1

1 ⎞

− ⎜ ⎟ 1% = e [8.314 J / mol .K ]⎝ 1073 973 ⎠ −2 5.5 x10 %

Q = 2.52 x105 J / mol = 252kJ / mol      

-7SKMM3623   (ii)

Estimate the creep rate to be expected at a service temperature of 500OC

C = ε + Q RT = (1%)e +2.52 x10

ε 500

O

5

( 8.314 )(1073)

= 1.80 x1012 %.hr −1

5

C

= (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.

The cracks will lead to the formation of voids and the number of these voids increase with strain and time in the third stage. The formation of large number of voids will reduce the cross sectional area and weaken the strength of the material which lead to fracture

Question 3 (25 marks) (a)

     

The following pairs of alloys are coupled in sea water. (1)

Aluminium and magnesium

(2)

Inconel (active) and nickel (active)

(3)

Zinc and titanium

-8SKMM3623   Based on the Galvanic series given in Table 1, (i)

Which metal/alloy will corrode for each pair? (i) magnesium (ii) nickel (active)

(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, Faraday’s 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)

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) Copper-Nickel alloys Bronzes (Cu-Sn alloys) Copper Brasses (Cu-Zn alloys) Inconel (active) Nickel (active) Tin Lead 316 Stainless steel (active) 304 Stainless steel (active)

Increasingly inert (cathodic)

Increasingly active (anodic)

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   PART B : Answer ONE question only Question 4 (25 marks) (a)

(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 -

Higher strength than thermoplastics

-

Do not sensitive to heat

1 mark 

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

1 mark 

Examples: Thermoset : electrical moldings, motor housing, epoxies materials, Polyurethane foam, bakelite materials (mounting, telephone)

1 mark 

Thermoplastics: Advantages: -

Can be reheated and reformed into new shapes without significant changes in the properties

-

Can be recycled

-

Easy to color

1 mark 

Disadvantages’: -

Soft

-

Too sensitive to heat

Thermoplastics: bottles, toys, food containers etc…

     

1 mark 

1 mark 

- 12 SKMM3623   (ii)

Describe the general deformation behavior of a thermoplastic materials above and below its glass transition temperature. (3 marks)

Answer: Non-crystalline plastics Above Tg: viscous (rubbery)

1 ½  marks 

Below Tg : glass brittle behavior Semi-crystalline plastics Above Tg: viscous solid (super-cooled liquid) Below Tg : glassy

(iii)

1 ½  marks 

With the aid of sketches show and discuss the effect of temperatures on stress – strain behavior of thermoplastic material. (3 marks)

Answer:

3 marks 

     

- 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:

1 mark 

Improve strength (stiffness), high temperature stability via cross linking sulphur with rubber

2 marks 

If too much sulphur, reduce its extensibility

(ii)

A butadiene-acrylonitrile rubber is made by polymerizing one acrylonitrile monomer (C4H5) with three butadiene monomers (C4H6). How much sulfur (S) is required to react with 100 kg of this rubber to crosslink 20% of the crosslink sites? (Atomic weight: C = 12 g/mol, H = 1.0 g/mol, S = 32 g/mol) (5 marks)

Answer: Mw polybutadiene (C4H6) = (4x12) + (6x1) = 54 g/mol

1 mark 

Mw polyacrylonitrile (C4H5) = (4x12) + (5x1) = 53 g/mol

1 mark 

Mw average copolymer = ¼ (53 g/mol PAN) + ¾ (54 g/mol PB) = 53.75 g/mol

1 mark 

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

     

2 marks 

- 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   Question 5 (25 marks) (a)

(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

3 marks 

Annealed glass „ The glass is heated to the annealing temperature, then slowly cooled to RT Application: table top glass, cabinet door glass and basement windows 3 marks 

(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 .σ f πa 2

2

⎛ 3 .8 ⎞ -6 a=( ) 2 =⎜ ⎟ =7.22x10 m = 7.22 µm 2 π σf ( π) 450 x π ⎝ ⎠ K IC

1

2 marks 

If KIC = 12.5 MPa.√m

σf =

(b)

K IC Y . πa

=

12.5 ( π )( πx7.22 x10 −6

= 1480.78MPa

2 marks 

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) Answer: i. Modulus of elasticity of composite is Ec=EfVf + EmVm = (7.24x104 MPa)(0.6) + (3.1x103MPa)(0.4) = 44.68 GPa      

2 ½  marks 

- 17 SKMM3623   ii. Tensile strength σc = σfVf+σmVm = (2.4 GPa)(0.6) + (0.06GPa)(0.4) = 1.46 GPa

2 ½  marks 

iii. Fraction load carried by fiber is

⎛ 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)

3  marks 

Discuss a method to produce an aircraft part/component made of a carbon-fiber-reinforced-epoxy composite. (5 marks)

Answer: 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