Step 1: Understanding the Question: To resist the same design moment with the same grade of steel and effective depth, the total area of steel reinforcement per unit width of the slab must remain constant. We need to find the new spacing for larger diameter bars that provides the same steel area per unit width.
Step 2: Key Formula or Approach: The area of steel per unit width is given by \( \frac{A_s}{s} \), where \(A_s\) is the area of a single bar and \(s\) is the spacing. For the condition to hold, we must have: \[ \frac{A_{s1}}{s_1} = \frac{A_{s2}}{s_2} \] Where:
- \(A_{s1}\) and \(s_1\) are the area and spacing of the initial bars.
- \(A_{s2}\) and \(s_2\) are the area and spacing of the new bars.
The area of a bar is \( A_s = \frac{\pi d^2}{4} \).
Step 3: Detailed Explanation: Given data:
- Initial bar diameter, \(d_1 = 10\) mm
- Initial spacing, \(s_1 = 150\) mm
- New bar diameter, \(d_2 = 12\) mm
Calculate the areas of the bars: \[ A_{s1} = \frac{\pi (10)^2}{4} = 25\pi \text{ mm}^2 \] \[ A_{s2} = \frac{\pi (12)^2}{4} = 36\pi \text{ mm}^2 \] Now, set up the equivalence relation and solve for the new spacing, \(s_2\): \[ \frac{25\pi}{150} = \frac{36\pi}{s_2} \] The \(\pi\) terms cancel out: \[ \frac{25}{150} = \frac{36}{s_2} \] \[ s_2 = \frac{36 \times 150}{25} \] \[ s_2 = 36 \times 6 = 216 \text{ mm} \]
Step 4: Final Answer: The required centre-to-centre spacing for the 12 mm diameter bars is 216 mm.