A cylindrical soil specimen of saturated clay, 3.50 cm diameter a

A cylindrical soil specimen of saturated clay, 3.50 cm diameter a
| A cylindrical soil specimen of saturated clay, 3.50 cm diameter and 8 cm length, testing machine. The specimen failed under a vertical load of 50 kg together with an accompanying additional deformation of 8 mm. What is the unconfined compressive strength of this clay?

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

\(\varepsilon = \frac{{\Delta L}}{{{L_o}}} = \frac{8}{{80}} = 0.10\)

\({A_o} = \frac{\pi }{4} \times d_o^2 = \frac{\pi }{4} \times {3.5^2} = 9.62\;c{m^2}\)

\(\therefore {A_f} = \frac{{{A_o}}}{{1 - \varepsilon }} = \frac{{9.62}}{{1 - 0.10}} = 10.69\;c{m^2}\)

\(\therefore {q_u} = \frac{P}{{{A_f}}} = \frac{{50}}{{10.69}} = 4.67\;kg/c{m^2}\)