REINFORCEMENT IN CONCRETE, FOUNDATIONS, CONCRETE FLOORS AND QUANTITIES (SPECIFIC)
6.1 Various options are given as possible answers to the following questions - NSC Civil Technology Construction - Question 6 - 2020 - Paper 1
Question 6
REINFORCEMENT IN CONCRETE, FOUNDATIONS, CONCRETE FLOORS AND QUANTITIES (SPECIFIC)
6.1 Various options are given as possible answers to the following questions. Choo... show full transcript
Worked Solution & Example Answer:REINFORCEMENT IN CONCRETE, FOUNDATIONS, CONCRETE FLOORS AND QUANTITIES (SPECIFIC)
6.1 Various options are given as possible answers to the following questions - NSC Civil Technology Construction - Question 6 - 2020 - Paper 1
Step 1
6.1.1 Steel that is used as reinforcement in concrete is available in the following thicknesses:
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Answer
The correct answer is: A 8 mm, 25 mm and 40 mm.
This standard thickness is widely used in various concrete reinforcement applications.
Step 2
6.1.2 Reinforcement rods and bars are available in lengths of up to …
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Answer
The correct answer is: C 12 000 mm.
This length is sufficient for most construction purposes.
Step 3
6.1.3 Bars are used to prevent the reinforcement from …
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Answer
The correct answer is: D All the above-mentioned.
Reinforcement bars are crucial for maintaining dimension stability under various forces.
Step 4
6.1.5 The purpose of minimum concrete cover is to ensure …
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Answer
The correct answer is: D All the above-mentioned.
Minimum concrete cover is essential for ensuring durability and safety.
Step 5
6.2 Give TWO reasons for the installation of pile foundations.
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Answer
Poor soil conditions: Pile foundations help stabilize structures where the surface soil is weak or unstable, such as soft or loose soil.
Load distribution: They distribute loads more evenly to deeper, more stable soil layers, reducing the risk of settling.
Step 6
6.3 Draw a neat freehand sketch in your ANSWER BOOK showing the first step of the installation of a driven in-situ pile.
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The sketch should include:
Steel cable: Depict it correctly at the top of the pile.
Steel pipe-casing: Shown as surrounding the pile.
Drop hammer: Illustrate it correctly as the tool used for driving the pile.
Any ONE label: Clearly label the components to show understanding.
Step 7
6.4.1 Predict TWO consequences of installing the rib and block floor as shown.
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Structural instability: The block may not support proper weight distribution, leading to potential collapse or fall-through.
Risk of excessive weight: Without sufficient openings, the block can accumulate undue stress from lack of ventilation and moisture.
Step 8
6.4.2 Draw a neat freehand drawing in your ANSWER BOOK and rectify the faults in FIGURE 6.4.
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Answer
The drawing should fix highlighted errors:
Ensure proper reinforcement in the ribs.
Illustrate correct rebates or supports to strengthen the structure.
Step 9
6.4.3 What is the minimum recommended width of the load-bearing walls that support this type of floor construction?
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Answer
The minimum recommended width for load-bearing walls supporting rib and block floors is 220 mm.
Step 10
6.5 Use ANSWER SHEET 6.5 and draw a neat sectional view of a round reinforced concrete column with eight main bars in good proportion.
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Answer
The sectional view should show:
Column: Depicted as a circular shape.
Main bars: Eight evenly spaced bars inside the column.
Stirrups: Included as necessary reinforcement.
Cover depth: Clearly indicated with appropriate measurements.
Step 11
6.6.1 The area of the floor
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Answer
To calculate the area:
Using the dimensions:
Length: 6 500 mm
Width: 4 500 mm
The area is calculated as:
extArea=extLengthimesextWidth=6500imes4500=29,250,000extmm2=29.25extm2
Step 12
6.6.2 The volume of concrete needed for the floor. Round off your answer to TWO decimal places.
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Answer
To calculate the volume of concrete:
Using the area calculated and the thickness:
Area: 29.25extm2
Thickness: 0.075extm
The volume is:
extVolume=extAreaimesextThickness=29.25imes0.075=2.194extm3
Rounding off to two decimal places: 2.19 m³.
Step 13
6.6.3 The number of tiles needed.
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Answer
To find the number of tiles:
Size of one tile: 350extmmimes350extmm=0.1225extm2
Total area to be covered: 29.25extm2
The number of tiles is calculated as:
ext{Number of tiles} = rac{ ext{Total area}}{ ext{Area of one tile}} = rac{29.25}{0.1225} = 239.22 ext{ tiles}
Rounding up gives 240 tiles needed.
