Civil Engineering - Soil Mechanics and Foundation Engineering - Discussion

Discussion Forum : Soil Mechanics and Foundation Engineering - Section 3 (Q.No. 30)
30.
The void ratio of a soil sample decreases from 1.50 to 1.25 when the pressure is increased from 25 tonnes/m2 to 50 tonnes/m2, the coefficient of compressibility is
0.01
0.02
0.05
0.001
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Vikash said:   1 decade ago
Coefficient of compressibility = change in e/change in stress.
= (1.50-1.25)/(50-25).
= 0.01.

Pranav said:   10 years ago
Cc = Change in e/log (Final pressure/Initial pressure).

= (1.50-1.25)/log(50/25) = 0.8304.

Yogesh said:   9 years ago
@Pranav.

That is Coefficient of Compression index.

Mahesh said:   8 years ago
.830 is the correct answer.

Rameshwar said:   8 years ago
CC = change in e/log10(initial stress+increased stress/initial stress).

Cc = 0.25/log(50/25) = 0.83.

Nandish said:   8 years ago
e0 = initial void ratio = 1.5.
e = final void ratio = 1.25.
P0 = initial pressure = 25 t/m2.
P = final pressure =50 t / m2.
for co efficient of compressibility.
an = ((eo - e) / (p-po)) = (( 1.50-1.25)/(50-25)) = 0.01.
for compressibility index use;
= (( eo - e)/ (log (p/po)).
(6)

Priya said:   8 years ago
Yes, I agree @Nandish.
(1)

Biki said:   7 years ago
.01 is correct.

The Coefficient of compressibility is asked, log should be used in the denominator if coefficient of compression (also known as compression index).
(2)

Prachi said:   7 years ago
The Answer is correct. The Coefficient of compressibility is asked that is e/σ= 0.25/25.

Deepak Baswal said:   7 years ago
.01 is the correct answer.
(1)


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