Formula $\sigma =\frac{3PL}{2b{d}^2}$
Specimen Dimension	$40 mm\times 5 mm\times 2 mm$ Where Total length = 40 mm Gauge length(L) = 25 mm Width of specimen (b) = 5 mm Thickness of specimen(d) = 2 mm
Sample Calculation	$\sigma =\frac{3PL}{2b{d}^2}=\frac{3 \times 0.09 \times 1000 \times 25}{2 \times 5 \times 2^2}=168.75 MPa$
Flexural Modulus	$E_{flexural}=\frac{P{L}^3}{48I\delta }$