Modulus of rupture of beam (numerical)

Assumption: the beam of length 750 mm rests on supports 75 mm in from each end, so the effective span is $l = 750 - 2\times75 = 600$ mm (IS 516 standard span for a 150 mm beam). $b = d = 150$ mm.

Single-point (centre) loading

$P = 100$ kN $= 100000$ N. Maximum moment $M = Pl/4$, section modulus $bd^2/6$:

$$f_{cr} = \frac{3Pl}{2bd^2} = \frac{3\times100000\times600}{2\times150\times150^2} = 26.67\ \text{N/mm}^2$$

Two-point (third-point) loading

Total load $P = 50 + 50 = 100$ kN, with loads at $l/3$ from each support. The moment is constant, $M = (P/2)(l/3) = Pl/6$, in the middle third (the crack is assumed to occur within it):

$$f_{cr} = \frac{Pl}{bd^2} = \frac{100000\times600}{150\times150^2} = 17.78\ \text{N/mm}^2$$

Answer: modulus of rupture = 26.67 N/mm$^2$ (single point) and 17.78 N/mm$^2$ (two point). If the full 750 mm is taken as the span, the values would be 33.3 and 22.2 N/mm$^2$. The centre-point test gives the higher value because the maximum stress acts at only one section. These values are far above the usual $0.7\sqrt{f_{ck}}$ (about 3 to 4 N/mm$^2$ for ordinary concrete), so the data are unrealistic, but they are used as given.