Question

A slab panel with an effective depth of 250 mm is reinforced with 0.2\% main reinforcement using 8 mm diameter steel bars. The uniform center-to-center spacing (in mm) at which the 8 mm diameter bars are placed in the slab panel is \_\_\_\_\_ (rounded off to the nearest integer).

Hint:

In reinforced concrete design, ensuring correct spacing of reinforcement bars is crucial for structural integrity, adherence to design codes, and to facilitate concrete placement and compaction.

Answer 1

Step 1: Calculate the cross-sectional area of one steel bar. The diameter of each bar is 8 mm, so the radius is 4 mm. The area \( A \) of one bar is given by: \[ A = \pi \times (\text{radius})^2 = \pi \times (4 \text{ mm})^2 = 50.27 \text{ mm}^2 \] Step 2: Determine the total area of steel required per meter width of the slab. The total reinforcement area \( A_s \) per meter width of the slab is given by: \[ A_s = 0.002 \times 250 \text{ mm} \times 1000 \text{ mm} = 500 \text{ mm}^2 \] Step 3: Calculate the spacing of the bars. The number of bars per meter can be calculated as: \[ n = \frac{A_s}{A} = \frac{500 \text{ mm}^2}{50.27 \text{ mm}^2} \approx 9.95 \text{ bars per meter} \] The spacing \( s \) between the bars is given by: \[ s = \frac{1000 \text{ mm}}{n} \approx \frac{1000 \text{ mm}}{9.95} \approx 100.5 \text{ mm} \]