Phase diagram in soil mechanics with theory,worked problem and calculator
Problem will be illustrated on soil mechanics.
1. 1 cum of wet soil weighs 20kN. Its dry weight is 18kN. specific gravity of soil, Gs=2.67. Determine below properties
- Water content.
- Void ratio.
- Porosity
- Degree of saturation.
Solution:
The volume and weight of the constituents are now calculated based phase diagram:
The weight of water:
\[\mathop W\nolimits_\omega = \mathop W\nolimits_{s + \omega } - \mathop W\nolimits_s = 20 - 18 = 2kN\]
The volume of solids:
\[\mathop V\nolimits_s = \frac{{\mathop W\nolimits_s }}{{\mathop G\nolimits_s \mathop { \times \gamma }\nolimits_\omega }} = \frac{{18}}{{2.67 \times 9.8}} = 0.6879{m^3}\]
The volume of water:
\[\mathop V\nolimits_\omega = \frac{{\mathop W\nolimits_\omega }}{{\mathop \gamma \nolimits_\omega }} = \frac{2}{{9.8}} = 0.2041{m^3}\]
The volume of voids:
\[\mathop V\nolimits_V = V - \mathop V\nolimits_S = 1 - 0.6879 = 0.3121{m^3}\]
1. Water content.
\[\omega = \frac{{\mathop W\nolimits_\omega }}{{\mathop W\nolimits_s }} = \frac{2}{{18}} = 0.1111 = 11.11\% \]
2. Void ratio.
\[e = \frac{{\mathop V\nolimits_\upsilon }}{{\mathop V\nolimits_s }} = \frac{{0.3121}}{{0.6879}} = 0.45\]
3. Porosity.
\[n = \frac{{\mathop V\nolimits_\upsilon }}{V} = \frac{{0.3121}}{1} = 0.3121 = 31.21\% \]
4. Degree of saturation.
\[S = \frac{{\mathop V\nolimits_\omega }}{{\mathop V\nolimits_\upsilon }} = \frac{{0.2041}}{{0.3121}} = 0.6539 = 65.4\% \]
- Water content.
- Void ratio.
- Porosity
- Degree of saturation.
- Assume unit weight of water γw as
1. Water content: %
2. Void ratio: %
3. Porosity:
4. Degree of saturation: %
Refrence
1. Venkataramaiah, C. (2018). Geotechnical Engineering (6th ed.). New Age International Publishers Pvt Ltd.
2. Punmia, B.C( 2017)Soil mechanics and foundations(17th ed.).Laxmi publications Pvt Ltd.
3. Gopal R, Rao, A, S, R( 2016)Basic and applied Soil mechanics(3rd ed.).New Age International Publishers Pvt Ltd.