Problem Statement
Determine the voids ratio and porosity of a uniformly graded sand of perfectly spherical grains arranged in a rhombohedral array with an orientation angle \( \alpha = 75^\circ \).
Solution
1. Volume of Solids (\( V_s \))
Volume of solids remains constant (spherical grains):
\( V_s = \frac{\pi}{6} \approx 0.5236 \)
2. Revised Total Volume (\( V \))
For rhombohedral packing with \( \alpha = 75^\circ \):
\( V = 1 + 2\cos\alpha(1 – \cos\alpha) \)
Substitute \( \alpha = 75^\circ \):
\( \cos75^\circ \approx 0.2588 \)
\( V = 1 + 2(0.2588)(1 – 0.2588) = 0.9131 \)
3. Voids Ratio (\( e \)) and Porosity (\( n \))
Volume of voids:
\( V_v = V – V_s = 0.9131 – 0.5236 = 0.3895 \)
Voids ratio:
\( e = \frac{V_v}{V_s} = \frac{0.3895}{0.5236} \approx 0.7439 \)
Porosity:
\( n = \frac{e}{1 + e} = \frac{0.7439}{1.7439} \approx 42.66\% \)
Results:
- Voids ratio: \( e \approx 0.7439 \)
- Porosity: \( n \approx 42.66\% \)
Explanation
For a rhombohedral array with \( \alpha = 75^\circ \):
- The total volume decreases compared to loose cubical packing but remains higher than the densest packing (\( \alpha = 60^\circ \)).
- The voids ratio (\( e \)) and porosity (\( n \)) are intermediate values between loose and dense states.
Physical Meaning
This calculation shows how particle arrangement affects soil properties:
- Voids ratio indicates the balance between particle packing and void space.
- Porosity influences permeability and compressibility, critical for drainage and stability in construction.
