A 5 m deep deposit of sand and silt containing organic layers is to be compacted using explosives placed in the boreholes located 8 m apart on a square grid. It is intended to create pore pressures which are at least equal to 80% of the effective overburden pressure. Estimate how many kg of TNT have to be distributed in each borehole in order to achieve this degree of liquefaction. How much is that per cubic meter of compacted soil?

kg of TNT = 10 kg

per cubic meter of compacted soil = 0.5%

Explanation:

Given Data

(1) Depth of sand to be compacted- 5m

(2) Explosives placed in boreholes located at 8m in square grid.

Requirement - To create atleast 80% pore pressure of the effective overburden pressure.

Upper bound value

RUG-RPP = 14 x (kg)1/3

Lower bound value        RUG-RPP = 3 x (kg)1/3

Hence, for 8m distance for upper bound value of TNT in kg, the equation can be written as

8 = 14 x (kg)1/3

Hence, . TNT weight required for this is 0.186 kg.

Hence, for 8m distance for lower bound value of TNT in kg, the equation can be written as

8 = 3 x (kg)1/3.

Hence, . TNT weight required for this is 18.96 kg.

Now, to achieve atleast 80% pore water pressure the weight of TNT required is 0.186 kg as per upper bound and 18.96 kg as per lower bound is required.

It is advised to take average of these two to achieve the required values of pore water pressure generation i.e 10 kg of TNT is required.

Considering unit weight of compacted sand as 20 kN/m3, the amount of TNT is 10 kg which is equivalent to 98.1 N. This is approximately 0.5% of the compacted soil.

this is true

explanation:

i looked it up

answer: 500 wph.

explanation: