9.5. Examples

Example 1: For the soil deposit shown below, draw the total stress, pore water pressure and effective stress diagrams. The water table is at ground level. Solution: Total stress At - 4m, = 1.92 x 4 = 7.68 At -11m, = 7.68 + 2.1 x 7 = 22.38 Pore water pressure At - 4 m, u = 1 x 4 = 4 At -11 m, u = 1 x 11 = 11 Effective stress At - 4 m , = 7.68 - 4 = 3.68 At -11m , = 22.38 - 11 = 11.38

In this example, we calculate total stress, pore water pressure, and effective stress at two different depths: -4 meters and -11 meters. Total stress is the weight of the soil above that point, while pore water pressure is the hydraulic pressure exerted by water in the pores of the soil.

• At -4 m, we calculate total stress by multiplying the soil density (1.92 T/m³) by the depth (4 m), resulting in 7.68 tons.
• At -11 m, we add the stress from the first segment (7.68) to the weight of the additional soil (2.1 T/m³ × 7 m) to get a total stress of 22.38 tons.

Pore water pressure at -4 m is found by multiplying the water density (1 T/m³) by the depth (4 m), resulting in 4 tons. At -11 m, it is 11 tons.

To find effective stress, we subtract pore water pressure from total stress. Thus, at -4 m, the effective stress is 3.68 tons and at -11 m, it is 11.38 tons.

Think of a sponge soaked in water. The water inside the sponge represents pore water pressure which pushes against the sponge material. The weight of the sponge, which represents the total stress, is pushing down—if you were to squeeze the sponge, you'd notice that the force you need to apply (effective stress) depends on how much water is in it (pore water pressure). If it were fully soaked, it would be hard to push down, similar to how effective stress is calculated in soil mechanics.

Example 2: An excavation was made in a clay stratum having = 2 T/m3. When the depth was 7.5 m, the bottom of the excavation cracked and the pit was filled by a mixture of sand and water. The thickness of the clay layer was 10.5 m, and below it was a layer of pervious water-bearing sand. How much was the artesian pressure in the sand layer? Solution: When the depth of excavation was 7.5 m, at the interface of the CLAY and SAND layers, the effective stress was equal to zero. Downward pressure due to weight of clay = Upward pressure due to artesian pressure (10.5 - 7.5) = h, where h = artesian pressure head 3 x 2 = 1 x h h = 6 m = 0.6 kg/cm2 or 6 T/m2 artesian pressure

In this example, we assess a situation where an excavation has been made into clay, with water-bearing sand beneath it. When the excavation was 7.5 m deep, the effective stress at the interface of the clay and sand was zero, indicating that the upward artesian pressure from the sand was balancing the downward weight of the clay.

We calculate the downward pressure by using the formula for pressure due to a column of soil. The pressure contributed by the clay above is calculated as the depth of the clay column (10.5 m - 7.5 m = 3 m) multiplied by the density of the clay (2 T/m³), resulting in a downward pressure of 6 T/m².

To find the artesian pressure, we realize it must equal the downward pressure, which allows us to set them equal and solve for the artesian pressure head (h), concluding it is also 6 m.

Imagine a syringe filled with water and sealed at one end. If you push the plunger, the pressure inside the syringe increases; this pressure can be compared to artesian pressure. In our example, just like the water pushes back against the plunger, the artesian pressure from the sand pushes up against the weight of the clay. When the effective stress is zero, it’s like the water in the syringe has completely supported all the weight above it, allowing it to float freely.

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