Question

The section of a R.C.C. staircase flight is shown in the figure below. Each riser and tread of the staircase is 150 mm and 300 mm, respectively. The steps are supported by a 75 mm thick waist slab.

The width of the staircase is 1500 mm. The flight consists of 10 steps. Consider reinforcement to be 0.75% by volume of the total R.C.C. The density of steel is 7800 kg/m\(^3\). The amount of reinforcement required to construct the staircase flight (in kg) is (rounded off to two decimal places).

Hint:

Find the waist slab volume (sloping length x width x 0.075 m thickness) plus the volume of the 10 triangular step wedges, then take 0.75% of the total for steel and multiply by 7800 kg/m3.

Answer 1

Correct Answer: 41.82

Step 1: Understanding the Question.
We need the weight of steel reinforcement in a staircase flight. The flight has 10 steps, each with a 150 mm riser and 300 mm tread, resting on a 75 mm thick sloping waist slab. The staircase is 1500 mm wide. Reinforcement is 0.75% of the total concrete (R.C.C.) volume by volume, and steel density is 7800 kg/m\(^3\).

Step 2: Find the length of the waist slab.
The waist slab runs along the slope of the stair. Its horizontal projection (going) equals the tread multiplied by the number of steps, and its vertical projection (rise) equals the riser multiplied by the number of steps.
Going = \(10 \times 0.3 = 3.0\) m
Rise = \(10 \times 0.15 = 1.5\) m
The sloping length of the waist slab is found using Pythagoras' theorem:
\[ L = \sqrt{3.0^2 + 1.5^2} = \sqrt{9 + 2.25} = \sqrt{11.25} = 3.3541 \text{ m} \]

Step 3: Find the volume of the waist slab.
The waist slab is 75 mm (0.075 m) thick and 1500 mm (1.5 m) wide, running the full sloping length.
\[ V_{waist} = L \times \text{width} \times \text{thickness} = 3.3541 \times 1.5 \times 0.075 = 0.3773 \text{ m}^3 \]

Step 4: Find the volume of the steps.
Each step is a triangular wedge sitting on the waist slab, with a right-triangle cross section of base equal to the tread (0.3 m) and height equal to the riser (0.15 m), running the full 1.5 m width of the stair.
\[ V_{one\ step} = \frac{1}{2} \times 0.3 \times 0.15 \times 1.5 = 0.03375 \text{ m}^3 \]
For 10 steps:
\[ V_{steps} = 10 \times 0.03375 = 0.3375 \text{ m}^3 \]

Step 5: Find the total R.C.C. volume and the steel volume.
\[ V_{total} = V_{waist} + V_{steps} = 0.3773 + 0.3375 = 0.7148 \text{ m}^3 \]
Reinforcement is 0.75% of this volume:
\[ V_{steel} = 0.0075 \times 0.7148 = 0.005361 \text{ m}^3 \]

Step 6: Find the mass of steel.
Multiply the steel volume by the density of steel (7800 kg/m\(^3\)).
\[ M_{steel} = 0.005361 \times 7800 = 41.82 \text{ kg} \]

Final Answer:
The reinforcement required to construct the staircase flight is about 41.82 kg. \[ \boxed{41.82 \text{ kg}} \]