Splitting tensile strength of cylinder and cube (500 kN)

Formulae: cylinder $f_{sp} = \dfrac{2P}{\pi L d}$; cube (load along the centre line of a face) $f_{sp} = \dfrac{2P}{\pi a^2}$, where $a$ is the side.

$P = 500$ kN $= 500\,000$ N.

Cylinder 300 mm × 150 mm dia

$$f_{sp} = \frac{2\times500000}{\pi\times300\times150} = \frac{1000000}{141371.7} = 7.07\ \text{N/mm}^2$$

Cube 150 mm

$$f_{sp} = \frac{2\times500000}{\pi\times150\times150} = \frac{1000000}{70685.8} = 14.15\ \text{N/mm}^2$$

Answer: cylinder 7.07 N/mm$^2$; cube 14.15 N/mm$^2$. The cube gives the higher value because the same load acts over a smaller splitting plane (150 × 150 mm against 300 × 150 mm). The cylinder result is the standard one.