Question

The required centre-to-centre spacing of 10 mm diameter bars is 150 mm to resist the design moment in a concrete slab. Instead of 10 mm diameter bars, if 12 mm diameter bars of the same grade are used, the required centre-to-centre spacing (in mm) to resist the same design moment becomes (rounded off to the nearest integer).

Hint:

Keep the steel area per metre width of slab constant; spacing scales as the square of the bar diameter.

Answer 1

Correct Answer: 216

Step 1: Set up the design condition.
A reinforced concrete slab is designed to resist a fixed design moment. The moment capacity of a lightly reinforced slab section depends directly on the area of steel provided per metre width of the slab, \(A_{st}\). To resist the same moment with a different bar diameter, this steel area per metre width must stay the same in both cases.

Step 2: Write the area of steel per metre width for the 10 mm bars.
For bars of diameter \(d\) placed at centre-to-centre spacing \(s\), the steel area supplied per metre width of slab is
\[ A_{st} = \frac{\pi}{4} d^2 \times \frac{1000}{s} \]
For the 10 mm bars, \(d_1 = 10\) mm and \(s_1 = 150\) mm, so
\[ A_{st,1} = \frac{\pi}{4}(10)^2 \times \frac{1000}{150} = 78.54 \times 6.667 = 523.6 \ \text{mm}^2/\text{m} \]

Step 3: Equate steel areas for the 12 mm bars.
For the 12 mm bars, \(d_2 = 12\) mm and the spacing \(s_2\) is unknown. Since \(A_{st,2}\) must equal \(A_{st,1}\),
\[ \frac{\pi}{4}(12)^2 \times \frac{1000}{s_2} = 523.6 \]
\[ \frac{\pi}{4}(12)^2 = 113.1 \ \text{mm}^2 \]
\[ s_2 = \frac{113.1 \times 1000}{523.6} = 216.0 \ \text{mm} \]

Step 4: Confirm using a direct proportion.
Since \(A_{st} \propto d^2/s\), holding \(A_{st}\) constant means \(s \propto d^2\). So
\[ s_2 = s_1 \times \left(\frac{d_2}{d_1}\right)^2 = 150 \times \left(\frac{12}{10}\right)^2 = 150 \times 1.44 = 216 \ \text{mm} \]
This matches Step 3.

Final Answer:
The required centre-to-centre spacing of the 12 mm bars is 216 mm.
\[ \boxed{s_2 = 216 \ \text{mm}} \]