D.3Determination of the original cross-sectional area (S₀)

The original cross-sectional area of the test piece shall be determined to the nearest ± 1 %.

The original cross-sectional area of the length of tube or longitudinal or transverse strip may be determined from the mass of the test piece, the length of which has been measured, and from its density.

The original cross-sectional area (S₀) of a test piece consisting of a longitudinal or transverse strip shall be calculated according to the following equation

$S_{\mathrm{o}}=\frac{b}{4}\left(D^{2}-b^{2}\right)^{1 / 2}+\frac{D^{2}}{4} \arcsin \frac{b}{D}-\frac{b}{4}\left[\left(D-2 a\right)^{2}-b^{2}\right]^{1 / 2}-\left(\frac{D-2 a}{2}\right)^{2} \arcsin \frac{h}{D-2 a}$

where

• a is the thickness of the tube wall;
• $b$ is the average width of the strips;
• $D$ is the external diameter.

The following simplified equations can be used for longitudinal or transverse test pieces:

$S_{\mathrm{o}}=a b\left[1+\frac{b^{2}}{6 D(D-2 a)}\right]$ when $\frac{b}{D}<0,25$

$S_{\mathrm{o}}=a b$ when $\frac{b}{D}<0,17$.

In the case of a length of tube, the original cross-sectional area (S₀) shall be calculated as follows:

$S_{o}=\pi a(D-a)$.