Quantity of a semi-circular arch

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.

        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).