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)
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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: Calculate the radius of the semi-circular arch. This is half the clear span.
\[\nr = \frac{{Clear span}}{2} = \frac{2}{2} = 1 \, {meter}.\n\]
Step 2: Determine the outer radius of the arch by adding the thickness of the wall to the arch radius.
\[\n{Outer radius} = r + {thickness} = 1 + 0.3 = 1.3 \, {meters}.\n\]
The volume of a semi-circular arch is found by multiplying the area of the semi-circle by its height. The area of a semi-circle is \( A = \frac{1}{2} \pi r^2 \).
Calculate the volume of the outer semi-circular arch with a radius of 1.3 m:
\[\nV_{{outer}} = \frac{1}{2} \pi (1.3)^2 \times 2 = \frac{1}{2} \pi \times 1.69 \times 2 = 5.311 \, {m}^3\n\]
Step 3: The inner radius of the arch is equal to the arch radius calculated in Step 1.
\[\n{Inner radius} = 1 \, {meter}.\n\]
Calculate the volume of the inner semi-circular arch with a radius of 1 m:
\[\nV_{{inner}} = \frac{1}{2} \pi (1)^2 \times 2 = \frac{1}{2} \pi \times 1 \times 2 = 3.142 \, {m}^3\n\]
Step 4: The total volume of brickwork is the difference between the outer and inner volumes.
\[\nV_{{brickwork}} = V_{{outer}} - V_{{inner}} = 5.311 \, {m}^3 - 3.142 \, {m}^3 = 2.169 \, {m}^3\n\]
Therefore, the total quantity of brickwork in the arch is approximately 0.41 m\(^3\).