48352 Construction Materials - Assessment Answer at UTS

QUESTION 1

a) List the main recognised constituents of Type GP cement. List the hydrates resulting from the hydration of each of these constituents.

C2S

C3S

C3A

C4AF

3 to 5% Gypsum

Maximum 7.5% of mineral additions (fly ash and/or GGBFS)

Maximum 5% of other inorganic minor constituents (kiln dust)

Hydrates:

C2S and C3S produce CSH and Calcium Hydroxides

C3A and Gypsum produce AFm phases: monosulfoalumnates

C4AF and Gypsum produce AFm phases: Tetracalcium Hydrates

(b) Excess of water in the concrete mix, what consequences on concrete (fresh and hardened)? Fresh properties: risk of bleeding and segregation. Hardened: reduction in strength and other mechanical properties of concrete, increase in porosity and transport properties leading to a reduction in durability. Increase in shrinkage and creep.

(c) Question about aggregate: explain what a grading curve is. The grading curve shows the particle size distribution of aggregates used in concrete

(d) Explain why using SCMs (such as fly ash or slag), to replace a part of the cement, is of a great interest regarding sustainability.

Answer 1.1

a – H2O

b – CO2

c – CaCO3

d – Low quartz

e – Free Lime

f - Belite

g – Alite

Answer 1.2

Generally, when the clinker is removed from the kiln, its temperature is around 1200 degrees Celcius. Therefore, it needs to be immediately cooled for the following reasons,

1. Such that heat energy is not wasted
2. To bring the temperature of the clinker near the suitable temperature for grinding. The temperature needs to be less than 100 degrees Celcius
3. If cooled immediately, the reactive metastable high-temperature minerals of the material are preserved in the clinker. This keeps the strength potential of the material at its maximum.

Answer 1.3

The mandatory component added during the grinding process of the clinker is calcium sulphate. The material is added in the range of 3-5% by weight. This is mandatory because it helps in controlling the chemical reactions happening in the cement and controlling the hardening process of the materials when water is mixed with it and the entire batch is transferred to silos.

Answer 1.4

As seen in the table, the Tricalcium silicate percentage of cement A is 56 whereas that of cement B is 40. The Tricalcium aluminate percentage of cement A is 8% whereas that of cement B is 7%.

According to theory, C3S is the material that contributes the most to achieving the early strength of the cement mix. And C3A is a chemical that generates the most heat when mixed with water and C2S is something that generates the least heat. It is also known that less percentage of C3S leads to high ultimate strength. 

Therefore, It can be stated that cement A with the composition stated in the table below, would provide the highest standard mortar strength one day after batching.

Answer 1.5

Compressive strength of concrete is given as 32 MPa with 2.5% defective rate. We know that if defective rate is greater than 5% we can’t take Fck and as per code 5% is allowed. Therefore, here it means that 2.5 are failing out of 100.

QUESTION 2

The project aims to design a massive foundation of a nuclear power plant. The risk of thermal stress
induced concrete cracking is very high. The concrete will be cast in summer. What type of Standard
Australian cement and what type of chemical admixture would you used in the concrete? Justify
your answer.

The concrete mix design will be carried out according to the British Building Research Establishment
method. Calculate the proportions of cement, water and aggregates for 1m3 of the concrete (in kg).
Also provide the dosage of the chemical admixture based on supplier recommendation: 1% of
cement mass.
The following requirements are specified:
a) Target mean compressive strength = 60 MPa at 28 days.
b) Portland cement class is 42.5.
c) Slump required is 0–10 mm.
d) Maximum aggregate size is 20 mm.
e) Maximum cement content is 450 kg/m3.
Additional information:
- The fine and coarse aggregates to be used are crushed; the bulk density is unknown,
- The fine aggregate has 40% passing a 600 μm sieve,
- Two types of coarse aggregate are used: 10 and 20 mm.

Answer : Concrete Mix Design 

The following procedure of concrete mix design is as per the British Building Research Establishment Method also known as DoE.

These calculations aim to calculate the proportions of cement, water and aggregate for 1m^3 of concrete.

Given,

• Target mean compressive strength = 25 MPa at 7 days.
• Portland cement class is 42.5.
• Slump required is 30–60 mm.
• Maximum aggregate size is 20 mm.
• Maximum cement content is 350 kg/m3.
• The fine and coarse aggregates to be used are crushed
• bulk density = unknown
• The fine aggregate has 40% passing a 600μm sieve
• Two types of coarse aggregates are used: 10 mm and 20 mm

Step 1: Determination of water cement ratio

The only class of cement that is available in Australia = 42.5

Quality of fine and coarse aggregate = crushed

Cement strength class	Type of coarse aggregate	Compressive strength at 3 days	Compressive strength at 7 days	Compressive strength at 28 days
42.5	Uncrushed	22	30	42
42.5	crushed	27	36	49
52.5	Uncrushed	29	37	48
52.5	crushed	34	43	55

Therefore, from the table above,

The compressive strength of concrete at 7 days = 36 MPa

Since, Target mean compressive strength = 25 MPa at 7 days, Now, from the graph below, water cement ratio = 0.6

Step 2: Determine water content (w)

For this, the given requirements are,

Slump = 30–60 mm

Maximum aggregate size = 20 mm

Therefore, from the table below, water content (w) = 210 kg/m^3

Step 3: Determine Cement Content (c)

C = W/0.6

= 210/0.6

= 350 kg/m3 < 450 kg/m3

Therefore, ok.

C = 350 kg/m3

Step 4: Determine total aggregate content

D =2400 kg/m3

Total aggregate content = D – W – C

= 2400 – 210 - 350

Total aggregate content = 1840 kg/m3

Minimum aggregate size: 20mm

 

As per the requirements, fine aggregate has 40% passing a 600 μm sieve.

Therefore,

Fine aggregate content = 0.52 x 1840 = 956.8 kg/m3

Coarse aggregate content = 1840 - 956.8 = 883.2 kg/m3

10 mm Coarse aggregate content (1/3) = 294.4kg/m3

20 mm Coarse aggregate content (2/3) = 288.8 kg/m3

Final values of the concrete mix design

Cement content = 350 kg/m3

Water content = 210 kg/m3

Fine aggregate = 956.8 kg/m3

10mm Course aggregate = 294.4kg/m3

20mm Course aggregate = 288.8 kg/m3

QUESTION 3 Concrete shrinkage

Reinforced concrete slabs will be cast on site using the same concrete as the one specified in
Question 2. Calculate the total final shrinkage strain of the concrete using AS3600-2018 model. The
concrete slab is exposed to an arid environment, the slab cross-section considered is 2000x200mm
(rectangular shape).

Answer 3.1

Step 1: Target mean strength

Given,

Target mean compressive strength, Fm = 25MPa

Margin, M = 5MPa

Now, as per British Building Research Method,

Fm = Fc + M

Fc = 20MPa

As per Australian Standard AS1379

Fcm = mean compressive strength = 25MPa

M = Margin = 5MPa

F’c = characteristic compressive strength = Fcm – M = 20MPa

Answer 3.2

For this calculations, AS3600-2018 model is used.

Given, c/s area of the concrete block = 1000mm x 200mm

Since the concrete is subjected to interior environment, K4 = 0.65

Basic drying shrinkage:

ecsd.b = (0.9 – 0.005f’c) x 800x10-6

= (0.9-0.005*20)*800*10^(-6)

= 0.00064

tn = 2Ac/P

= 2(1000*200)/(2*1000+2*200)

= 166.67

a1 = 1.2835

k1 = 1.2633

Therefore,

Total drying shrinkage

ecsd = k1 x k4 x ecsd.b

=1.2633*0.65*0.00064

=0.00052

eCSE = (0.07*F’c-0.5)50*10^(6)

=((0.07*20-0.5)50*10^(6)

= 0.00045

Total shrinkage strain: ecs = eCSE + ecsd

0.00045+0.00052

=0.00097

Therefore, ecs = 970 me

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