The main reinforcement of an RC slab consists of 10 mm bars at 10 cm spacing. If it is desired to replace 10 mm bars by 12 mm bars, then the spacing of 12 mm bars should be

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BPSC AE Paper 5 (Civil Engineering) 2019 Official Paper
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  1. 10 cm
  2. 12 cm
  3. 14.40 cm
  4. 16 cm

Answer (Detailed Solution Below)

Option 3 : 14.40 cm
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Detailed Solution

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Concept:

Spacing between the bars for 1 m(1000 mm) width:

\({s} = \frac{a}{{{A_{st}}}} \times 1000\)

Where, a = Area of one bar = \(\frac{\pi \ \times \ ϕ^2}{4}\)

ϕ = Diameter of bar

s = spacing of bars

Ast = Area of total main reinforcement

Calculations:

Given, 

Case 1: when ϕ = 10 mm then spacing(s1) = 10 cm = 100 mm

Case 2: when ϕ = 12 mm then spacing(s2) = ?

Case 1:

When the main reinforcement of an RC slab consists of 10 mm bars at 10 cm spacing, Ast will be

\({s_1} = \frac{a}{{{A_{st}}}} \times 1000\)

\(\Rightarrow {A_{st}} = \frac{{\frac{{\pi \times 10^2 }}{4}}}{{100}} \times 1000 = 785.4\;m{m^2}\)

Case 2:

⇒ If 10 mm bars is to be replaced by 12 mm bars, then the spacing of 12 mm bars

As the Area of the main reinforcement will be the same so Ast = 785.4 mm2

\({s_2} = \frac{a}{{{A_{st}}}} \times 1000\)

\(\Rightarrow \;{s_2} = \frac{{\frac{{\pi \times 12^2}}{4}}}{{785.4}} \times 1000 = 14.40\;cm\)

Shortcut Trick

\(S_2 \ ={\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} \times {S_1} = {\left( {\frac{{12}}{{10}}} \right)^2} \times 10 = 14.4\;cm\)

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