Assumption 1: Uniform distribution by volume

In this case, the fraction of the total aggregate volume represented by particles with volumes in the range (V,V+dV), contained in the i'th sieve, is given by

$$p_i(V) dV = \frac{c_i dV}{(V_{i+1} - V_i)}$$ (11)

so that the integral over the interval (V_i+1, V_i ) will be equal to c_i. If N is the total number of aggregate particles used per the total concrete volume V_TOT, so that $\rho \equiv N / V_{TOT}$ , V_agg is the total aggregate volume, c_agg = V _agg / V_TOT, the fraction of the total number of aggregate particles with volumes in the range (V,V+dV), contained in the i'th sieve, is given by

$$n_i(V) dV = \frac{c_{agg} c_i dV}{\rho V (V_{i+1} - V_i)}$$ (12)

where V is the volume of a particle in this range. If we now convert to radius, using V = 4 r ³/ 3 and dV = 4 r ²dr, the equivalent expression in terms of the particle radius is

$$n_i(r) dr = \frac{9 c_{agg} c_i r^{-1} dr}{4 \pi \rho (r_{i+1}^3 - r_i^3)}$$ (13)

Integrating over each sieve's endpoints and summing over each sieve must give 1 for this expression:

$$1 = \sum_{i=1}^M \int_{r_i}^{r_{i+1}} n_i(r) dr$$ (14)

This normalization determines the value of $\rho$ :

$$\rho = \sum_{i=1}^M \frac{9 c_{agg} c_i }{4 \pi (r_{i+1}^3 - r_i^3)} \ln \left (\frac{r_{i+1}}{r_i} \right )$$ (15)

Therefore, the average of Rⁿ over the particle number density is then

$$< R^n> = \sum_{i=1}^M \frac{9 c_{agg} c_i }{4 \pi \rho (r_{i+1}^3 - r_i^3)} \int_{r_i}^{r_{i+1}} r^{n-1} dr$$ (16)

Note that the quantity c_agg drops out of eq. (16), as it appears in the numerator and in the denominator, in $\rho$. This is as it should be, since < Rⁿ > is the same for any representative amount of aggregate, and is independent of the total amount added to a concrete. The value of $\rho$ does depend on c_agg, however, since it is the number fraction of particles in the total concrete volume.