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).
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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.
Step 1: Use the cross-sectional area method.
Instead of finding the waist slab and the steps as two separate solids, we can find the total cross section area of the RCC (waist slab plus the 10 step-wedges) in the vertical plane of the stair, then multiply by the 1.5 m width to get the volume.
Step 2: Area of the waist slab strip.
The waist slab strip in cross section is a long, thin strip of thickness 0.075 m running along the sloping length. Slope length $L = \sqrt{3^2+1.5^2} = \sqrt{11.25} = 3.3541$ m.
$$A_{waist} = 3.3541 \times 0.075 = 0.25156 \text{ m}^2$$
Step 3: Area of the 10 step wedges.
Each step wedge in cross section is a right triangle with legs 0.3 m (tread) and 0.15 m (riser).
$$A_{one\ step} = \frac{1}{2}(0.3)(0.15) = 0.0225 \text{ m}^2$$
For 10 steps:
$$A_{steps} = 10 \times 0.0225 = 0.225 \text{ m}^2$$
Step 4: Total cross-section area and volume.
$$A_{total} = 0.25156 + 0.225 = 0.47656 \text{ m}^2$$
Multiply by the 1.5 m width of the stair:
$$V_{total} = 0.47656 \times 1.5 = 0.71484 \text{ m}^3$$
This agrees with the volume found by treating the waist slab and steps separately in three dimensions.
