Question

During the first stage of triaxial test wher
the cell pressure is increased from 0.10
N/mm² to 0.26 N/mm², the pore water
pressure increases from 0.07 N/mm² to
0.15 N/mm². Skempton's pore pressure
parameter B is​

Answer 1

Answer: 0.5

Step-by-step explanation:

Skempton's pore pressure  parameter B: It can be defined as the ratio of change in the pore water pressure with the change in the cell pressure.

Initial Cell Pressure (σ3) = 0.10 $\frac{N}{mm^2}$

Final Cell Pressure (σ3)= 0.26 $\frac{N}{mm^2}$

∴ Difference in Cell Pressure (∆σ3) = 0.16 $\frac{N}{mm^2}$

Initial pore water pressure(U) = 0.07 $\frac{N}{mm^2}$

Final pore water pressure(U) = 0.15 $\frac{N}{mm^2}$

∴ ∆U = 0.08 $\frac{N}{mm^2}$

Now, we know that

∆U = B x ∆σ3

0.0 8 = B × 0.16  

B = 0.5

∴Skempton's pore pressure  parameter B is​ 0.5

Answer 2

Answer:

Step-by-step explanation:

Parameter B is given by

B = change in pore pressure (u)

           change in pressure

B = 0.15/0.26

B = 0.5