Exam Papers Sample 2

Question 1

Describe the failure modes of fibre composites.

Answer 1

Fibre fracture
When the fibres break in one plane, the composite will fail in the plane of fibre fracture
as the soft matrix is unable to carry the load
Fibre pull-out
When the adhesion between the fibres and matrix is not sufficiently strong, the fibres
may be pulled out of the matrix before the composite fails

Question 2

Plywood consists of multiple layers of veneer that are bonded together. State the
reasons below.

Answer 2

The limitation of wood is its directional properties.
Plywood overcomes this limitation by using plies of veneer laminated at 90° to each
other, so the properties are the same in two directions instead of one.

Question 3

Plywood has an odd number of layers (such as 3,5,7….). State the rationale of having
odd numbers of layers.

Answer 3

Odd number of plies is used so that the top and bottom layers of the plywood
will have the same grain orientation (same finishing).

Question 4

State the distinct differences between the filament winding method and the pultrusion
method.

Answer 4

In filament winding, one or more continuous fibres are wrapped around a form or mandrel
to gradually build up a solid hollow shape. It is suitable for cylindrical parts.
Pultrusion is used to extrude a polymer matrix around fibres to form a simple shaped
product with a constant cross section.

Question 5

Name the other TWO common methods of processing composites other than filament
winding and pultrusion.

Answer 5

Hand lay-up method
Spray-up method

Question 6

State the ‘definition’ in radiography inspection.

Answer 6

“Definition” is the clearness and sharpness of the outline of the radiograph image.

Question 7

Name the TWO types of images quality indicators used to determine percentage
sensitivity

Answer 7

Wire IQI and Step hole IQI

Question 8

Corrosion can occur under direct chemical or electrolylic-chemical attack.
Briefly discuss their differences with regard to:
Electron movement

Answer 8

Direct chemical attack - electrons travel through a metal for a very short distance.
Electro-chemical attack - electrons travel through some measurable distance

Question 9

Normal temperature in which they occur.

Answer 9

Direct chemical attack - usually occurs at higher temperature.
Electro-chemical attack - usually occurs at any temperature.

Question 10

How does the EMF series determine the rate of corrosion between two metals in the
series? Which metal is protected and which one is corroded?

Answer 10

The further the two metals are separated in the series.
The greater their potential difference will be, so the greater the rate of corrosion
The more cathodic metal will be protected while the more anodic metal will corrode

Question 11

Name TWO ways in which stress corrosion can be overcome.

Answer 11

-remove the stress through annealing
- redesign the area of high stress concentration
-remove / neutralise the chemical corroding medium

Question 12

What is cathodic protection? With the aid of a simple sketch, show the cathodic
protection of a steel pipeline using magnesium. Label the cathode and anode in your
diagram and indicate clearly the direction of current flow.

Answer 12

Cathodic protection is the protection of a metal against corrosion by making it the cathode
of a galvanic cell.

Question 13

Explain why the steel pipeline in Question 4.2 can be prevented from corrosion and state
the anodic reaction. The chemical symbols for iron and magnesium are Fe and Mg
respectively.

Answer 13

As the steel pipe becomes the cathode and the magnesium the anode of a galvanic cell, the
magnesium ingot corrodes to protect the steel.
The anodic reaction is: Mg  Mg2+ + 2e

Question 14

How can continuous protection of the steel pipeline be ensured in cathodic protection?

Answer 14

A periodic inspection and replacement of the ingot which has dissolved is all that it is
needed to ensure continuous protection.

Question 15

A steel was heated to 910°C during heat treatment and quenched in a quenching
medium to room temperature. After this treatment, the hardness of the steel was found
to be 52 HRC. The data sheet as provided by the supplier showed the following
information: Alloy steel with 1% carbon, Austenitising Temperature: 880°C; Ms:
280°C, Mf : -5°C; Hardness after heat treatment: 62 HRC; Recommended Applications:
Tools and Dies.

(a) Explain briefly why the steel failed to harden to the specified hardness.

Answer 15

The steel failed to harden to the specified hardness as it is only quenche to room
temperature. As Mf is -5°C for this steel, it should be quenched below -5°C.

Question 16

(b) State the microstructure that caused the heat treated steel to have a lower hardness.

Answer 16

It is the presence of retained austenite that lowers the hardness of the steel.

Question 17

(c) State the THREE undesirable effects of the steel with the microstructure mentioned in
Q5.1(b).

Answer 17

-Austenite is much softer than martensite, so reduces the overall hardness of steel
-Spontaneous cracking may occur later when austenite transforms to martensite
-Results in dimensionally unstable product as austenite transforms to martensite

Question 18

(d) Describe a heat treatment method to eliminate the microstructure mentioned in Q5.1(b).

Answer 18

Double Tempering or Sub-zero treatment (either one)

Double tempering: After quenching, the martensite in workpiece is tempered to relieve
stresses, but the austenite is unaffected. The work piece is then cooled. Cooling from
the second tempering process transform some retained austenite to martensite, giving
an increase in hardness. The process may be repeated.

                                                                              OR 

Sub-zero treatment: This cold treatment involves refrigerating the component, after
quenching, to a suitable temperature (below Mf ). The temperature as low as -98°C are
used. The workpiece is then tempered. Sometimes a second cold treatment may be
necessary to ensure complete transformation.

Question 19

(a) Large quantity of small gears made of low carbon steel are required to have a surface
hardness of 60 HRC. Recommend a heat treatment process to achieve the required
hardness.

Answer 19

Gas carburizing process can achieved the required hardness.

Question 20

Describe the heat treatment mentioned in Q5.2(a) and state briefly the mechanisms
involved.

Answer 20

Large quantities of gears are heated to the carburizing temperature of 925°C for 3 to 4
hours in a controlled atmosphere which contains a mixture of hydrocarbon gases,
commonly methane and propane.
Mechanisms involved:
-Generation of CO from endo-gas : CO2 + CH4 => 2CO + 2H2
-Dissociation of CO at steel surface: 2CO => CFe + CO2

Question 21

Highlight and explain difference between the thermochemical flame and Selective
Surface Heating Processes

Answer 21

Thermochemical Processes, in which the surface chemistry of the steel is modified by
the introduction of carbon and/or nitrogen and occasionally other elements. These
include carburising, nitriding and carbonitriding treatments.
Selective Surface Heating Processes, in which only the surface is hardened and the core
is left in its pre-treated condition. These include flame hardening, induction hardening,
laser hardening and electron-beam heat-treating.

Question 22

Highlight and explain difference between the thermochemical flame and Selective
Surface Heating Processes

Answer 22

Thermochemical Processes, in which the surface chemistry of the steel is modified by
the introduction of carbon and/or nitrogen and occasionally other elements. These
include carburising, nitriding and carbonitriding treatments.
Selective Surface Heating Processes, in which only the surface is hardened and the core
is left in its pre-treated condition. These include flame hardening, induction hardening,
laser hardening and electron-beam heat-treating.