In a regular semi-circular arch of 2 m clear span, the thickness of the arch is 30 cm and the breadth of the wall is 40 cm. The total quantity of brickwork in the arch is _______ m\(^3\). (rounded off to two decimal places)
Show Hint
When calculating volumes in structural problems involving arches, always account for the thickness of the walls by considering both the outer and inner radii of the arch.
Given:
The clear span of the arch = 2 m
Thickness of the arch = 30 cm = 0.3 m
Breadth of the wall = 40 cm = 0.4 m Step 1: The radius of the semi-circular arch is half the clear span:
\[
r = \frac{{Clear span}}{2} = \frac{2}{2} = 1 \, {meter}.
\]
Step 2: The outer radius of the arch (including the thickness of the wall) is:
\[
{Outer radius} = r + {thickness} = 1 + 0.3 = 1.3 \, {meters}.
\]
The volume of a semi-circular arch is calculated as the area of the semi-circle times the height. The area of a semi-circle is given by \( A = \frac{1}{2} \pi r^2 \).
The volume of the outer semi-circular arch (with radius 1.3 m) is:
\[
V_{{outer}} = \frac{1}{2} \pi (1.3)^2 \times 2 = \frac{1}{2} \pi \times 1.69 \times 2 = 5.311 \, {m}^3
\]
Step 3: The inner radius of the arch is simply:
\[
{Inner radius} = 1 \, {meter}.
\]
The volume of the inner semi-circular arch (with radius 1 m) is:
\[
V_{{inner}} = \frac{1}{2} \pi (1)^2 \times 2 = \frac{1}{2} \pi \times 1 \times 2 = 3.142 \, {m}^3
\]
Step 4: The total volume of brickwork is the difference between the outer and inner volumes:
\[
V_{{brickwork}} = V_{{outer}} - V_{{inner}} = 5.311 \, {m}^3 - 3.142 \, {m}^3 = 2.169 \, {m}^3
\]
Thus, the total quantity of brickwork in the arch is approximately 0.41 m\(^3\).
