--- title: "Quantity of a semi-circular arch" subject: "Estimating and Costing" chapter: 2 key: "semester_7/estimating_and_costing:semicircular-arch-quantity" source: https://www.bctengineer.com/q/semester_7/estimating_and_costing/semicircular-arch-quantity site: BCT Engineering --- # Quantity of a semi-circular arch Estimating and Costing · Chapter 2: Method of Estimating - Asked in: 2071 Chaitra, 4 marks - Video: [Semi Circular Arch || Estimation & Costing of arch || Quantity of arch || Estimates of Arch](https://www.youtube.com/watch?v=4ogd4G6eeY8) — Exam Guru g - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/estimating_and_costing/semicircular-arch-quantity - Whole chapter: https://www.bctengineer.com/md/semester_7/estimating_and_costing/chapter-2 ## Question Explain with neat sketches how to work out the quantity of a semi-circular arch (span, thickness and rise of arch given). ## Short answer For a semicircular arch, rise equals half span; intrados radius r = span/2 and extrados radius R = r + t. Area is pi/2(R^2 - r^2), volume equals area times wall thickness. With span 2 m, t = 0.30 m and b = 0.23 m, arch masonry is 0.249 m3. ## Answer A semicircular arch has a **rise equal to half the span**. The brickwork quantity of the arch is the volume of the arch ring (the area between the intrados and the extrados) multiplied by the wall thickness. ```text extrados R = r + t .-----------. .' .---------. '. / / intrados \ \ | | r | | |__|____________|__| <---- span ---> S = 2r ``` #### Method 1. Span $S = 2r$, so the radius of the intrados is $r = S/2$ and the rise equals $r$. 2. Radius of the extrados: $R = r + t$, where $t$ is the thickness (depth) of the arch ring. 3. Area of the arch ring (half annulus): $$ A = \frac{\pi}{2}\left(R^2 - r^2\right) $$ which is the same as $\pi \times$ mean radius $\times t$, i.e. $A = \pi\left(r + \frac{t}{2}\right)t$. 4. Quantity of arch masonry: $V = A \times b$, where $b$ is the thickness of the wall (the width of the arch along the wall). 5. Deductions from the wall: when the wall masonry is measured gross, deduct from it the opening (the rectangle $S \times$ height up to the springing line, plus the semicircle $\frac{\pi r^2}{2} \times b$ above it) and also the arch ring volume $V$, because the arch masonry is measured separately at its own rate. The centering for the arch is measured as a separate item. #### Worked example (assumed values) Span 2.0 m ($r = 1.0$ m), arch thickness $t = 0.30$ m ($R = 1.30$ m), wall thickness $b = 0.23$ m. $$ A = \frac{\pi}{2}(1.30^2 - 1.00^2) = 1.0838\ \text{m}^2, \qquad V = 1.0838 \times 0.23 = 0.249\ \text{m}^3 $$ Opening to be deducted from the wall (semicircle above the springing line): $\frac{\pi}{2}(1.0)^2 \times 0.23 = 0.361$ m³. **Answer: Arch ring masonry = 0.249 m³** (for the assumed dimensions). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a semicircular arch of span S, what is the rise? A. Equal to the span B. Half the span (S/2 = r) C. One third of the span D. Equal to the arch thickness Answer: B. A semicircular arch has radius r = S/2, so the rise equals r. (Key point: Rise) 2. How is the extrados radius R found? A. R = r - t B. R = 2r C. R = r + t, intrados radius plus arch thickness D. R = S + t Answer: C. The ring thickness t adds to the inner radius. (Key point: Extrados radius) 3. What is the area of the arch ring? A. pi R^2 B. (pi/2)(R - r) C. pi (R^2 + r^2) D. (pi/2)(R^2 - r^2) Answer: D. The ring is a half annulus. (Key point: Area formula) 4. For span 2.0 m, t = 0.30 m and wall thickness 0.23 m, the arch ring masonry is: A. 0.249 m3 B. 1.084 m3 C. 0.361 m3 D. 0.498 m3 Answer: A. A = (pi/2)(1.69 - 1) = 1.0838 m2; V = 1.0838 x 0.23 = 0.249 m3. 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