Each part of full curve 1 mark total 3 marks each

Each part of full curve = 1 mark (total = 3 marks)
• Each point and region = 1 mark (total = 4 marks)
Part 1C: Calculation Questions (8 marks)
Provide the answers and show complete workings for the following question in your
answer booklet.
i) A 10.00 mm round mild steel bar is tested in uniaxial tension and is found to have
the following mechanical properties:
• Engineering elastic limit stress = 315.00 MPa
• Engineering elastic limit strain = 1.50 x 10
-3
• Engineering upper yield strength = 345.00 MPa
• Engineering lower yield strength = 305.00 MPa
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a) Determine the engineering strain when the load applied is 20 kN
Answer (3 marks):
Solving for engineering modulus of elasticity (E
E
) using engineering elastic
limit stress ( !
Eel
) and engineering elastic limit strain ( "
Eel
),
• E
E
= !
Eel
/ "
Eel
(0.5 mark)
• E
E
= 315.00 MPa / 1.50 x 10
-3
• E
E
= 210,000 MPa = 210 GPa (0.5 mark)
Solving for engineering stress ( !
E
) using engineering force (f
E
),
• !
E
= f
E
/ area = f
E
/ ( # x r
2
) (0.5 mark)
• !
E
= 20,000 N / ( # x 5.00 mm x 5.00 mm)
• !
E
= 254.65 MPa $ 255 MPa (0.5 mark)
Solving for engineering strain ( "
E
) using !
E
and E
E
,
• E
E
= !
E
/ "
E
• "
E
= !
E
/ E
E
(0.5 mark)
• "
E
= 254.65 MPa / 210,000 MPa
• "
E
= 0.00121 = 1.21 x 10
-3
(0.5 mark)
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b) Determine the true stress and true strain at 50% of the engineering yield stress
Answer (3 marks):
Solving for !
E
at 50% of engineering yield stress using engineering lower
yield strength ( !
Ely
),
• !
E
= (50 / 100) x !
Ely
• !
E
= (50 / 100) x 305.00 MPa
• !
E
= 152.50 MPa (0.5 mark)
Solving for "
E
using !
E
and E
E
,
• E
E
= !
E
/ "
E
• "
E
= !
E
/ E
E
= 152.50 MPa / 210,000 MPa
• "
E
= 0.000726 = 7.26 x 10
-4
(0.5 mark)
Solving for true stress ( !
T
) using !
E
and "
E
,
• !
T
= !
E
x (1 + "
E
) (0.5 mark)
• !
T
= 152.50 MPa x (1 + 7.26 x 10
-4
)
• !
T
= 152.61 MPa (0.5 mark)
Solving for true strain ( "
T
) using "
E
,
• "
T
= log
e
(1 + "
E
) (0.5 mark)
• "
T
= log
e
(1 + 7.26 x 10
-4
)
• "
T
= 7.26 x 10
-4
(0.5 mark)
c) Determine the ductility if the gauge length after fracture is 70.00 mm
Answer (2 marks):
Solving for original gauge length (l
o
),
• l
o
= 5.65 x (area)
0.5
(0.5 mark)
• l
o
= 5.65 x ( # x 5.00 mm x 5.00 mm)
0.5
• l
o
= 50.07 mm (0.5 mark)
Solving for ductility by percent elongation (% EL) using and maximum gauge
length (l
m
) and l
o
,
• % EL = ((l
m
- l
o
) / l
o
) x 100 (0.5 mark)
• % EL = ((70.00 mm - 50.07 mm) / 50.07 mm) x 100
• % EL = 39.80% (0.5 mark)
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Question Two (20 marks)
Part 2A: Multiple Choice Questions (5 marks)
For each question below, select one correct answer and write down the letter in your
answer booklet.
i) Which compound of Portland cement after hydration contributes the most to the
development of strength of concrete after 7 days ageing?
a) Tricalcium silicate (C
3
S) (1 mark)
b) Dicalcium silicate (C
2
S)
c) Tricalcium aluminate (C
3
A)
ii) Partial replacement of Portland cement with silica fume in concrete will alter the
fresh and hardened properties of concrete by:
a) Decreasing extent of bleeding and retarding strength development
b) Decreasing extent of bleeding and accelerating strength development (1 mark)
c)