A cylindrical soil specimen of saturated clay, 3.50 cm diameter a
![A cylindrical soil specimen of saturated clay, 3.50 cm diameter a](/img/relate-questions.png)
A. 4.67 kg/cm<sup>2</sup>
B. 5.0 kg/cm<sup>2</sup>
C. 5.5 kg/cm<sup>2</sup>
D. 6.0 kg/cm<sup>2</sup>
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
Unconfined compression Test:
The primary purpose of Unconfined Compression Test is to determine the unconfined compressive strength, which is then used to calculate the unconsolidated undrained shear strength of the clay under unconfined conditions.
As per IS2720 (Part10):1991 Clause:7, The unconfined compressive strength of clay can be calculated from the following relationship,
\({q_u} = \frac{P}{{{A_f}}}\)
Where, P = The compressive force ,
\({A_f}\) = Average cross-sectional area at a particular strain \(= \frac{{{A_o}}}{{1 - \varepsilon }}\)
Here, Ao = Intial average cross-section area of the specimen
ε = axial strain \(= \frac{{\Delta L}}{{{L_o}}}\)
∆L = Change in the length of specimen
Lo = intial length of the specimen
Calculation:
∆L = 8 mm, Lo = 8 cm, Initial diameter of specimen = 3.50 cm, Vertical load at failure P = 50 kg
\(\varepsilon = \frac{{\Delta L}}{{{L_o}}} = \frac{8}{{80}} = 0.10\)
\({A_o} = \frac{\pi }{4} \times d_o^2 = \frac{\pi }{4} \times {3.5^2} = 9.62\;c{m^2}\)
\(\therefore {A_f} = \frac{{{A_o}}}{{1 - \varepsilon }} = \frac{{9.62}}{{1 - 0.10}} = 10.69\;c{m^2}\)
\(\therefore {q_u} = \frac{P}{{{A_f}}} = \frac{{50}}{{10.69}} = 4.67\;kg/c{m^2}\)