EM111 Fundamentals of Engineering Tutorials COMPLETE

8/2/2019 EM111 Fundamentals of Engineering Tutorials COMPLETE

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EM111 Materials & Energy:-

Engineering Materials Tutorial Questions

Dr. Joseph Stokes

Solutions will be only given out in Tutorials

Question 1 [25 Marks]

a) Sketch the tensile stress-strain curve for a ductile material and for a brittle material and

define the following:

1. Yield Point and Fracture Point

2. Elastic Range and Plastic Range

3. Permanent elongation and Elastic Recovery

4. Youngs Modulus of Elasticity (E) [8 Marks]

b) Define hardness, and describe briefly how it is measured [5 Marks]

c) Explain how the toughness of a material is measured experimentally [5 Marks]

d)

The following data has been measured from two experimental tests, explain the relevanceof the results

Experiment Aluminium Steel Nylon

1 75 J 60 J 2 J

2 550 Hv 910 Hv 30 Hv

Table 1 [7 Marks]


ANSWER


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a) Sketch the tensile stress-strain curve for a ductile material and for a brittle material and

define the following:

1. Yield Point (#) and Fracture Point(#)

2. Elastic Range (1) and Plastic Range (2)

3. Permanent elongation (2) and Elastic Recovery(1)

4. Youngs Modulus of Elasticity (E) [8 Marks]

Modulus of Elasticity (E) =Strain

StressN/m

2

b) Define hardness, and describe briefly how it is measured [5 Marks]

The hardness of a material is a measure of its resistance to abrasion or indentation. The

hardness of a material may be specified in terms of some standard test involving indentation

or scratching of the surface of the material.

c) Explain how the toughness of a material is measured experimentally [5 Marks]

An alternative way of considering toughness is the ability of a material to withstand shock loads.A measure of the ability to withstand suddenly applied forces is obtained by impact tests, such as

the Charpy and Izod tests. Another measure of toughness that can be used is fracture toughness.

Stress

Strain0

(a)

Stress

Strain0

(b)

Breaks

Breaks

(#)

(#)

1

1

2


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Fracture toughness can be defined as a measure of the ability of a material to resist the

propagation of a crack.

d) The following data has been measured from two experimental tests, explain the relevance

of the results

Experiment Aluminium Steel Nylon

1 75 J 60 J 2 J

2 550 Hv 910 Hv 30 Hv

Table 1 [7 Marks]

Aluminium is a tough material, followed by steel, however nylon is very poor. A tough material

can be considered to be one that resists breaking. This we can take as meaning that a tough

material requires more energy to break it than a less tough one. Aluminium must be very ductile.

Steel is very hard, with Aluminium following behind. Nylon should not be tested using the

Vickers Hardness test, but it is poor as a hard material.


(a) A pipe has an outside diameter of 25 mm, an inside diameter of 15 mm and length 0.40 m

and it supports a compressive load of 40 kN. The pipe shortens by 0.5 mm when the load is

applied. Determine (1) the compressive stress, (2) the compressive strain in the pipe whensupporting this load.

[12 Marks]

(b) Sketch typical load/extension curves for (1) an elastic non-metallic material, (2) a brittle

material and (3) a ductile material. Give an example of each type of material.

[13 Marks]


ANSWER

(a) A pipe has an outside diameter of 25 mm, an inside diameter of 15 mm and length 0.40 m

and it supports a compressive load of 40 kN. The pipe shortens by 0.5 mm when the load is


4/23

applied. Determine (1) the compressive stress, (2) the compressive strain in the pipe when

supporting this load. [12 Marks]

%)125.0(00125.04.0

0005.0

3.127

10142.3

40000

24

l

xstraineCompressiv

MPaAreaForceeCompressiv

mArea

NforceeCompressiv

(b) Sketch typical load/extension curves for (1) an elastic non-metallic material, (2) a brittle

material and (3) a ductile material. Give an example of each type of material. [13

Marks]

Question 3 [25 Marks]:

a) You as an engineer will test the material described in Table Q1.

Table Q1. Properties of a certain type of material.

Yield

Strength

@ 0.2%

Strain

Tensile

Strengt

h

Origin

al

Length

Resistan

ce

Cross-

section

al Area

Breakdow

n Voltage

Density Specific

Heat

Capacity

200MPa 220MP

a

100mm 5x1013 10mm

x

10mm

0.6x106 V 1200

Kg/m3

1000

J/KgK

For each test result determine:

Extension ExtensionExtension

Load

Load

Load

Polyethylene Cast iron Mild steel

Extension ExtensionExtension

Load

Load

Load

Polyethylene Cast iron Mild steel


5/23

(i) Modulus of Elasticity and briefly comment on the meaning of its result

[2 Marks]

(ii)Compressive Strength if a force of 30kN broke the sample and briefly comment on themeaning of its result [2 Marks]

(iii)Percentage Elongation if the sample broke at an extension of 1mm and briefly comment

on the meaning of its result [2 Marks]

(iv)Electrical Resistivity and briefly comment on the meaning of its result[2 Marks]

(v)Electrical Conductivity [1 Mark]

(vi)Dielectric Strength and briefly comment on the meaning of its result[2 Marks]

(vii) Linear expansion if the sample extends by 2mm when the temperature is varied

by 10oC [1 Mark]

(viii)

Mass of the sample [1 Mark](ix)Amount of heat required to raise the sample by 10

oC [1 Mark]

(x)Specific Strength and briefly comment on the meaning of its result [2 Marks]

(xi)Vickers Hardness if the diagonal length of the indent was found to be 2x10-4 m under a

force of 100N. [2 Marks]

(xii) Do the above results describe any particular type of material? [7 Marks]


ANSWER

b) You as an engineer will test the material described in Table Q1.

Table Q1. Properties of a certain type of material.

Yield

Strength

@ 0.2%

Strain

Tensile

Strength

Origin

al

Length

Resistan

ce

Cross-

section

al Area

Breakdow

n Voltage

Density Specifi

c Heat

Capaci

ty

200MPa 220MPa 100mm 5x1013

10mm

x10mm

0.6x106

V 1200

Kg/m3

1000

J/KgK

For each test result determine:

ANSWER


6/23

(i) Modulus of Elasticity and comment on the meaning of its result [2 Marks]

E = Yield/Strain = [200x106

N/m2

]/ [0.2%/100% strain)] = 100 x109

N/m2

or100GPa

Proof stress possible used to get result, low stiffness value

(ii)Compressive Strength if a force of 30kN broke the sample and comment on the meaning

of its result[2 Marks]

CS = F/A = [30 x103

N]/[10x10/(1000)2

m2] = 300 x10

6N/m

2or 300MPa

The Comp Stress is higher than Tensile (220MPa) so material may be a ceramic

(iii)Percentage Elongation if the sample broke at an extension of 1mm and comment on the

meaning of its result[2 Marks]

%El. = Delta L / L0 x100% = [1mm/100mm] x100% = 1%

So brittle material eg. ceramic

(iv)Electrical Resistivity and comment on the meaning of its result [2 Marks]

= RA/L0 = {5x1013 x[10x10/(1000)2 m2]}/[100/(1000) m] =5x1010m

This is an insulator, eg. ceramic has value of 1x1010m

(v)Electrical Conductivity [1 Mark]

Conductivity =[1/] =1/5x1010m = 0.2x10-10-1m-1

(vi)Dielectric Strength and comment on the meaning of its result

[2 Marks]

DS = [Breakdown Volt]/[thick] = [0.6x106

V]/[10/1000 m] = 6 x107

V/m

Requires a DS similar to plastics to breakdown the material

(vii) Linear expansion if the sample extends by 2mm when the temperature is variedby 10oC [1 Mark]

= [DeltaL/L0]/[Delta T] = [2mm/100mm]/[10K] = 2 x 10-3 K-1

(viii) Mass of the sample [1 Mark]

= M/Vol = > 1200 Kg/m3

= M/[100x10x10x10-9

m3], M = 12 x10

-3Kg or 12g

(ix)Amount of heat required to raise the sample by 10oC [1 Mark]

c = [Q]/[M x deltaT] => 1000J/KgK = Q/[12 x10-3

Kg x 10K], Q = 120J

(x)Specific Strength and comment on the meaning of its result [2 Marks]

S.S. = [T.S./] =[220MPa]/ 1200 Kg/m3 = 0.183 MPa/ Kgm-3 or 183.3 MPa/ Mgm-3

This SS is higher than steel so good material

(xi)Vickers Hardness if the diagonal length of the indent was found to be 2x10-4 m under a

force of 100N. [2 Marks]


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Area = d2/1.854 = [2x10

-4m]

2/1.854 = 2.157x10

-8m

2, Stress (HV) =F/A

HV=100N/( 2.157x10-8

m2) = 4.64 x10

9Hv (N/m

2)

(xii) Do the above results describe any particular type of material?[7 Marks]

The above material is a ceramic, with high elastic modulus, very low % elongation,compressive strength higher than tensile, insulator, high dielectric strength and needs Vickers

hardness test to measure its high hardness


(a)Describe briefly the effects of adding the following elements to alloy steels:

(i) 1% Aluminium

(ii) Chromium

(iii) Nickel(iv) Silicon.

(v) Tungsten[10 Marks]

(b)Figure Q2(a) shows a schematic taken from your Material Science notes, label the type of

Solid Solution produced in A and in B.

A B

[2 Marks]

(c)Figure Q2(b) shows the Copper-nickel phase equilibrium diagram. Describe what is

happening and how many phases exist; (i) Above the Liquidus Line, (ii) Between theLiquidus and the Solidus Line and (iii) Below the Solidus Line.

[9 Marks]

(d)Sketch what you would expect to see at room temperature if you looked through a

microscope at a sample with weight composition; (i) 80%Cu 20%Ni, (ii) 50%Cu 50%Niand (iii) 20%Cu 80%Ni

[3 Marks]

Figure Q2(a)


8/23

Figure Q2(b)


ANSWER

(a)Describe briefly the effects of adding the following elements to alloy steels:

(i) 1% AluminiumEnables steels to have a hard, wear-resistant skin by the process of nitriding

(ii) Chromium

The presence of small amounts of chromium stabilises the formation of hard carbides,and improves the susceptibility of steels to heat treatment. Unfortunately the presence of

chromium also promotes grain growth. The presence of large amounts of chromium

improves corrosion resistance and heat resistance of steels (stainless steel).

(iii) NickelThe presence of nickel in alloy steels results in increased strength by grain refinement. It

also improves corrosion resistance of steel. Unfortunately, nickel is a powerful

graphitiser and reduces the stability of any carbide. Nickel and chromium are often used

to complement each other's properties.(iv) Silicon

The presence of up to 0.3 % silicon improves the fluidity of casting steels without the

reduction in mechanical properties associated with phosphorus. Up to 1 % siliconimproves the heat resistance of steels.

(v) Tungsten

The presence of tungsten in alloy steels promotes the formation of very hard carbides,and induces sluggishness into the heat treatment of transformations. This enables the

1500-

1400-

1300-

1200-

1100-

1000-

900-

800-

-

-1500

-1400

-1300

-1200

-1100

-1000

-900

-800

-

0 20 40 60 80 100% Ni

100 80 60 40 20 0% Cu

Temperature (oC)

Composition

Molten solution of Copper (Cu)

plus Nickel (Ni)

Solid solution of Copper

plus Nickel

-1455

1084-

Liquidus (L)

Solidus (S)

L +S

1135oC

1190

o

C

1380oC

1410oC


9/23

steels to retain their hardness at high temperature. Tungsten is mainly found in high-

speed steels which are used for cutting tools and in high-duty die steels which have tooperate at high temperatures.

(b)Figure Q2(a) shows a schematic taken from your Material Science notes, label the type of

Solid Solution produced in A and in B.

A B

[2Marks]A = Substitutional and B = Interstitial

(c)Figure Q2(b) shows the Copper-nickel phase equilibrium diagram. Describe what is

happening and how many phases exist; (i) Above the Liquidus Line, (ii) Between the

Liquidus and the Solidus Line and (iii) Below the Solidus Line.

[9 Marks](i) Above the Liquidus line both Copper and Nickel are soluble in each other.

(ii) Between the solidus and the liquidus is a solution of molten copper and nickel

together with crystals of a solid solution of copper and nickel.

(iii) Below the solidus the alloy consists entirely of crystals of copper and nickel in solidsolution.

[9 Marks]

Figure Q2(a)


10/23

(d)Sketch what you would expect to see at room temperature if you looked through a

microscope at a sample with weight composition; (i) 80%Cu 20%Ni, (ii) 50%Cu

50%Ni and (iii) 20%Cu 80%Ni

[3 Marks](d)

[3 Marks]

1500-

1400-

1300-

1200-

1100-

1000-

900-

800-

-

-1500

-1400

-1300

-1200

-1100

-1000

-900

-800

-

0 20 40 60 80 100% Ni

100 80 60 40 20 0% Cu

Temperature (oC)

Composition

Molten solution of Copper (Cu)

plus Nickel (Ni)

Solid solution of Copper

plus Nickel

-1455

1084-

Liquidus (L)

Solidus (S)

L +S

1135oC

1190o

C

1380o

C

1410oC

Same amount of eachPrimary or mostly Cu Primary or

mostly Ni

(i) (ii) (iii)


11/23


(a) Define Phase [3 Marks]

(b) Below in Table Q3 are the solidus and liquidus for the germanium-silicon system.

Construct the phase diagram on graph paper for this system and label each region.

Table Q3

Composition Solidus Temperature Liquidus Temperature

(wt% Si) (0C) (

0C)

0 938 938

10 1005 1147

20 1065 122630 1123 1278

40 1178 1315

50 1232 134660 1282 1367

70 1326 1385

80 1359 139790 1390 1408

100 1414 1414

[16 Marks]

(c) Indicate the following compositions as lines from liquid temperature to their solidtemperature; (i) 80%Gr 20%Si, (ii) 50%Gr 50%Si and (iii) 20%Gr 80%Si

[3 Marks]

(d) Sketch what you would expect to see at room temperature if you looked through a

microscope at a sample with weight composition; (i) 80%Gr 20%Si, (ii) 50%Gr 50%Si and(iii) 20%Gr 80%Si

[3 Marks]

Question 5

ANSWER

The term "phase" may be defined as:

A portion of a system which is of uniform composition and texture throughout, and which isseparated from the other phases by clearly defined surfaces. [3 Marks]


12/23

Graph plus 3 lines = 19 Marks

(c)

[3 Marks]

0

200

400

600

800

1000

1200

1400

1600

0 10 20 30 40 50 60 70 80 90 100

Composition (wt % Si)

Temperature(oC)

Liquidus

Solidus

Liquid + Solid

Same amount of eachPrimary or mostly Gr Primary or

mostly Si

(i) (ii) (iii)


13/23


(a) What are the characteristics of a phase? [5 Marks]

(b) When a liquid solution of two metals (binary alloy) solidifies one of three things occur,

describe them? [5 Marks]

(c) Figure 1 (see overleaf) shows the iron carbide phase equilibrium diagram describe thefollowing:

1. The 6 phases which are shown on the diagram [5 Marks]

2. How the microstructure of an alloy of composition 0.5 wt% C, develops as the alloycools slowly from 1000 oC [9 Marks]


ANSWER

(a) What are the characteristics of a phase? [5 Marks]

The term "phase" may be defined as:A portion of a system which is of uniform composition and texture throughout, and which is

separated from the other phases by clearly defined surfaces.

(b) When a liquid solution of two metals (binary alloy) solidifies one of three things occur,describe them?

[5 Marks]

Temperature

(oC)

723oC

1400-

1200-

1000-

800-

600-

400-

0 1 2 3

Carbon content (%)

+ Fe3C

+ Fe3C

+ liquid

+

Figure 1


14/23

With regard to metal alloys, when a liquid solution of two metals (binary alloy) solidifies one of

the following will occur:

Metal, which are soluble in the liquid state, may become totally insoluble in the solidstate and separate out as grains in the pure state. Thus there will be two phases present,

with each phase consisting of many grains of the same composition.

Metals, which are soluble in the liquid state, may remain totally soluble in the solid stateresulting in a "solid solution". Thus a single-phase solid solution will be presentconsisting of many grains of the same composition.

The two metals may react together chemically to form an "intermetallic compound".Again a single phase consisting of many grains of the same composition.

(c) Figure 1 (see overleaf) shows the iron carbide phase equilibrium diagram describe the

following:

The 6 phases which are shown on the diagram [5 Marks]

(i) Ferrite (), (ii) Austenite (), (iii) Ferrite () + Austenite (), (iv) Austenite () & liquid

matrix, (v) Austenite () & Cementite (Fe3C), (vi) Ferrite () + Cementite (Fe3C) Ferrite (-phase) this is the weakest solution of carbon in BCC crystals of iron. There is a

maximum of 0.03% carbon in solid solution at 723 C falling to 0.006% carbon in solid

solution at room temperature. Ferrite is very soft, ductile and of relatively low strength.

Austenite (-phase) this is a much more concentrated solid solution of carbon in iron than

ferrite. Austenite is formed when carbon dissolves in FCC crystals of iron in the solidstate. The maximum amount of carbon, which can be held in solution with iron in the

solid state, is 1.7% at 1150 oC. Although this is the upper limit of carbon which can be

present in plain carbon steels, for all practical purposes there is no advantage to

increasing the carbon content beyond about 1.2 -1.4%.

Cementite (iron-carbide phase) an excess of carbon combines with iron to form iron

carbide (Fe3C). Each molecule of iron carbide contains three atoms of iron chemicallycombined with one atom of carbon. This is true up to the limit of 1.7% at room

temperature, beyond which the excess carbon is precipitated out as "free" or uncombinedflakes of graphite.

How the microstructure of an alloy of composition 0.5 wt% C, develops as the alloy

cools slowly from 1000oC [9 Marks]

In this example the steel contains 0.5% carbon. Again, the steel will commence to solidify at

temperature T1 = 1000 oC. The steel now consists entirely of crystal of the solid solution -phaseaustenite. No further changes occur until the steel reaches temperature T2 (800). When the steel

reaches T2 crystals of ferrite will start to grow in the austenite so that both -phase and -phase

crystals will be present. Since the -phase crystals contain ferrite rather less than 0.03% carbonin solid solution, the carbon content of the remaining phase austenite, increases progressively as

more and more ferrite is formed until 723 C, the structure will contain ferrite and austenitewhich is eutectoid in composition. Thus at T3 (723 C) the austenite will change suddenly into

the eutectoid composition of pearlite, and the final composition below T3 will consist of crystals


15/23

of ferrite and crystals of pearlite. Figure shows a typical microstructure for annealed 0.5% carbon

steel.


(a) Briefly describe the following Heat Treatment Processes; Annealing, Normalising, and

Stress Relieving

[15 Marks]

(b) Discuss the effect of carbon on the properties of plain carbon steels and outline the uses of

the different classifications of plain carbon steel. (Use Figure Q2)

Figure Q2

[10 Marks]

900-

750-

600-

450-

0,100-

50-

0-

-300

-200

-100

-0

0 0.2 0.4 0.6 0.8 1.0 1.2

Ultimate

strength (MPa)Brinell

hardness

Carbon content (%)

Pearlite

Ferrite

Ductility

(High)

(Low)

Hardness

Strength

CementitePearlite content (%)

FerritePearlite


16/23

Question 7 [25 Marks]:SOLUTION

(a) Briefly describe the following Heat Treatment Processes;

ANSWER:

Annealing: In this process the component is heated above the upper critical temperature(temperature at which the steel is fully austenite) and "soaked" to ensure uniform heating.

The furnace is then switched off and allowed to cool or the cooling rate may be

controlled by gradually reducing the furnace temperature. Annealing makes a componentas soft as possible.

Normalising: When a material is formed by cold rolling, hot rolling, forging, stresses areset up in the material. Normalising is the process that will remove these internal stresses.

The metal is heated to a prescribed temperature and allowed to "soak" until it is heatedthroughout. It is then allowed to cool in air. The cooling rate is slightly faster than in

annealing. This gives a fine grain structure, which is free from internal stresses.

Stress Relieving: This is the process in which the component is reheated and held at a lowtemperature for a period of time and then cooled slowly, to remove all the thermal

stresses in the steel.

[15 marks]

(b) Discuss the effect of carbon on the properties of plain carbon steels and outline the uses of

the different classifications of plain carbon steel (Use Figure Q2).

ANSWER

900-

750-

600-

450-

0,100-

50-

0-

-300

-200

-100

-0

0 0.2 0.4 0.6 0.8 1.0 1.2

Ultimate

strength (MPa)Brinell

hardness

Carbon content (%)

Pearlite

Ferrite

Ductility

(High)

(Low)

Hardness

Strength

CementitePearlite content (%)


17/23

Plain carbon steel can be classified as to their uses as follows:

Low carbon steels, up to 0.15% carbon. Their main property is ductility and they are used where

cold forming is necessary. They are easily welded.

Structural steels, 0.15-0.3% carbon. These are less easily cold-formed, but they are stronger and

can still be easily welded. They are used for girders, ship plate and containers for gases and

liquids.

Medium carbon steels, 0.3-0.6% carbon. These steels are usually hot forged and are likely to

crack if welded. They have good strength and ductility and are used as axles, crankshafts and

connecting rods.

High carbon steels, 0.6-0.8% carbon. The main property is hardness and wear resistance. They

are shaped by hot forging and are used as springs, hammers, railway wheels and rails and wire

ropes.

Tools steels, 0.8-1.2% carbon. In cast or normalised state, these steels have very low toughness

due to a grain boundary network of iron carbide. Thermo-mechanical treatments allow these

steels to be used as chisels, cutting tools, saws, punches, files and knives.

[10 marks]

Question 8 [25 Marks](a)Define a Cermet? [5 marks]

(b)Each Polymeric/Elastomeric material has a specific application, unique to itself. Table

Q3 shows a list of Polymeric/Elastomeric materials, with the wrong application (uses)

across from it. List the corresponding Material letters with its correct correspondingapplication number. (e.g. list as K-12, etc.)

Table Q3

POLYMERIC & ELASTOMERIC

MATERIALS

APPLICATIONS

A. Polyethylene (PET) 1. Motor Tyres

B. Polypropylene (PP) 2. Gears and Bearings

C. Polystyrene (PS) 3. Lenses and Dentures

D. PolyVinyl Chloride (PVC) 4. Electrical Plugs and Insulators


18/23

E. PolyTetraFluoroEthylene (PTFE) 5. High Temperature Seals

F. PolyMethyl MethAcrylate (PMMA) 6. Squeezy Bottles

G. Polyamide (Nylon) 7. Plant Pots

H. Phenol-Formaldehyde (PF) 8. Non-Stick Coating for CookingUtensils

I. Silicon Rubber 9. Domestic Hardware

J. Polyisoprene 10.Foam Cups

[20 Marks]


Answer

(a)Define a Cermet? These are materials, which combine that ductility and toughness of a

matrix with the hardness and compressive strength of ceramic particles to provide particlereinforcement. Certain metal oxides and carbides can be bonded together and sintered into a

metal powder matrix to form important composites called cermets. This is short for CEramic

Reinforced METalS. [5 Marks]

(b) Place the corresponding Polymeric/Elastomeric material letter with its correspondingapplication number (e.g. K-12)

POLYMERIC & ELASTOMERIC

MATERIALS

APPLICATIONS

A. Polyethylene (PET) 6. Squeezy Bottles

B. Polypropylene (PP) 9. Domestic Hardware

C. Polystyrene (PS) 10. Foam Cups

D. PolyVinyl Chloride (PVC) 7. Plant Pots

E. PolyTetraFluoroEthylene (PTFE) 8. Non-Stick Coating for Cooking Utensils

F. PolyMethyl MethAcrylate (PMMA) 3. Lenses and Dentures

G. Polyamide (Nylon) 2. Gears and Bearings

H. Phenol-Formaldehyde (PF) 4. Electrical Plugs and Insulators


19/23

I. Silicon Rubber 5. High Temperature Seals

J. Polyisoprene 1. Motor Tyres

Or: A-6, B-9, C-10, D-7, E-8, F-3, G-2, H-4, I-5, J- 1 [20 Marks]


(a) Name the two molecular structures found in Thermoplastic materials (support with

diagrams), and briefly describe the effects of increasing the temperature on these structures andthe material itself?

[10 Marks]

(b) Name three properties of the Polyethylene (PET) Thermoplastic material and list four

applications of the material. [10 Marks]

(c) What happens to Polyethylene when it is exposed to strong sunlight, unless it contains thepigment carbon black. [5 Marks]


ANSWER

(a) Name the two molecular structures found in Thermoplastic materials (support withdiagrams), and briefly describe the effects of increasing the temperature on these structures and

the material itself? =

1. Linear Chain Molecules

2. Branched Chain Molecules

Structure of thermoplastic polymers.

In both of these instances the bonding between adjacent molecules is secondary. Raising the

temperature of the polymer can weaken secondary bonds. Weakening of the bonding causes the

Chain MoleculesLow density polymer

Branched Molecules

High density polymer


20/23

material to become softer. This is the why these polymers can be moulded into different shapes

when they are heated.

Raising and lowering the temperature has no chemical effect on the bonding between themolecules in a thermoplastic. They can, be re-softened and moulded repeatedly. Thermoplastics

are normally flexible. The flexibility decreases as the length of the chain increases and it also

decreases with increased branching.

[10 Marks]

(b) Name three properties of the Polyethylene (PET) Thermoplastic material and list four

applications of the material.

= (1)Tough and (2)Flexible (over a wide range of temperatures), and (3)good dimensional

stability

= pick 4 out of 5; (i)buckets, (ii)bowls, (iii)food containers, (iv)bags or (v)Squeezy bottles

[10 Marks]

(c) What happens to Polyethylene when it is exposed to strong sunlight, unless it contains the

pigment carbon black. = It Degrades [5 Marks]


(a)Explain what is meant be fibre reinforcement and state how different classification of

fibres increase the strength of composites using engineering examples. [10 Marks]

(b)Given the data listed below, calculate: the reinforcement area fraction for the composite:

Average fibre diameter 0.005 mm

Average number of fibres per strand 200 Number of strands 480

Tensile modulus for polyester resin 3.8 GPa

Tensile modulus of glass fibre 70 GPa

Strength of polyester resin 50 MPa

Strength of glass fibre 1450 MPa

Diameter of component 25 mm[15 Marks]


ANSWER

Explain what is meant be fibre reinforcement and state how different classification

of fibres increase the strength of composites using engineering examples.


21/23

FIBRE REINFORCEMENTThis method of reinforcement can range from glass fibres used in GRP (Glass reinforced plastic,explained in following section) plastic mouldings to the thick steel rods used to reinforce

concrete.

Reinforced ConcreteConcrete itself is a particle-reinforced material. It consists of mortar made from cement and sand,

reinforced with an aggregate of chipped stones. The stones are crushed to a rough texture and

sharp comers that will key into and bond with the mortar matrix. This basic concrete has a highcompressive strength but is very weak in tension. To improve its performance overall metal

reinforcing rods are added.

Glass Reinforced Plastic (GRP)The important composite material is produced when a plastic material, usually a polyester resin,

is reinforced with glass fibre in a strand or mat form. The resin is used to provide shape, colour

and finish, whilst the glass fibre, which are laid in all directions, impart mechanical strength.

The amount of reinforcement, which can be used, depends upon the orientation of thereinforcement. With long strands laid parallel to each other the reinforcement area fraction can

be as high as 0.9. The reinforcement area fraction is the cross-sectional area of reinforcementdivided by the total cross-sectional area. With woven strand fabrics the reinforcement area

fraction can be as high as 0.75; with chopped strand mat the reinforcement area fraction is

substantially reduced but a figure of 0.5 should be considered the minimum for satisfactoryreinforcement. Obviously the lower the reinforcement area fraction, the weaker the composite

produced.

[10 Marks]

a.

Given the data listed below, calculate the reinforcement area fraction for the composite:

Average fibre diameter 0.005 mm

Average number of fibres per strand 200

Number of strands 480

Tensile modulus for polyester resin 3.8 GPa

Tensile modulus of glass fibre 70 GPa

Strength of polyester resin 50 MPa

Load LoadMatrix

Reinforcement (area a)

Reinforcement area

fraction =A

an

n = number of reinforcements

a = cross-sectional area of each reinforcement

A = total cross-sectional area of composite

Total cross-sectional area (A)


22/23

Strength of glass fibre 1450 MPa

Diameter of component 25 mm

Reinforcement area fraction =A

an

With n = number of reinforcements, a = cross-sectional area of each reinforcement, A = total

cross-sectional area of composite.

Diameter of a fibre = (0.005/)2 = 1.96 x 10-5 mm2, Strand Area = (1.96 x 10-5 mm2 x 200) =3.926 x 10 -3 mm2, so a = 3.926 x 10 -3 mm2

A = (12.5)2

= 490.874 mm2, n = 480

Reinforcement area fraction =874.490

10926.34803

x= 3.84 x 10

-3which is very low value

[15 Marks]

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EM111 Fundamentals of Engineering Tutorials COMPLETE