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**Stress calculation in Soil**

The following example shows the stress calculation at different level within the soil.

**Problem 1: Calculate total, effective stress and pore water pressure at different level of the soil as shown in Figure 5.2.**

**Fig. 5.2. (a) Soil profile (b) Stress diagrams**

** Solution:**

The bulk unit weight of the partially saturated sand

\[{\gamma _{t(sand)}}={{{G_s} + Se} \over {1 + e}}{\gamma _w}={{2.6 + 0.5 \times 0.65} \over {1 + 0.65}} \times 10=17.72\;kN/{m^3}\]

The unit weight of the water is taken as 10 kN/m3. The saturated unit weight of sand is

\[{\gamma _{sat(sand)}}={{{G_s} + e} \over {1 + e}}{\gamma _w}={{2.6 + 0.65} \over {1 + 0.65}} \times 10=19.7\;kN/{m^3}\]

The saturated unit weight of the clay-I is.

\[{\gamma _{sat(clay - I)}}={{{G_s} + e} \over {1 + e}}{\gamma _w}={{{G_s} + {G_s}w} \over {1 + {G_s}w}}{\gamma _w}={{2.65 + 2.65 \times 0.55} \over {1 + 2.65 \times 0.55}} \times 10=16.71\;kN/{m^3}\]

[Se = Gsw and for fully saturated soil S =1]

The saturated unit weight of the clay-II is

\[{\gamma _{sat(clay - II)}}={{{G_s} + e} \over {1 + e}}{\gamma _w}={{{G_s} + {G_s}w} \over {1 + {G_s}w}}{\gamma _w}={{2.7 + 2.7 \times 0.6} \over {1 + 2.7 \times 0.6}} \times 10=16.49\;kN/{m^3}\]

At elevation -2.5 m

Total stress (Ïƒ) = 17.72 Â´ 2.5 = 44.3 kN/m^{2}

Pore water pressure (u) = 0

Effective stress () = 44.3 - 0 = 44.3 kN/m^{2}

At elevation -6 m

Total stress (Ïƒ) = 17.72 Â´ 2.5 + 19.7 Â´ 3.5 = 113.25 kN/m^{2}

Pore water pressure (u) = 10 Â´ 3.5 = 35 kN/m^{2}

Effective stress ( \[\sigma '\] ) = 113.25 - 35 = 78.25 kN/m^{2}

At elevation -10 m

Total stress (Ïƒ) = 17.72 Â´ 2.5 + 19.7 Â´ 3.5 + 16.71 Â´ 4 = 180.1 kN/m^{2}

Pore water pressure (u) = 10 Â´ 7.5 = 75 kN/m2

Effective stress ( \[\sigma '\] ) = 180.1 - 75 = 105.1 kN/m^{2}

At elevation -14 m

Total stress (Ïƒ) = 17.72 Â´ 2.5 + 19.7 Â´ 3.5 + 16.71 Â´ 4 + 16.49 Â´ 4 = 246.1 kN/m^{2}

Pore water pressure (u) = 10 Â´ 11.5 = 115 kN/m^{2}

Effective stress ( \[\sigma '\] ) = 246.1 - 115 = 131.1 kN/m^{2}

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