Question

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)

 

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.

Answer 1

Correct Answer: 0.41

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\(^3\).