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
Discussion:
13 comments Page 1 of 2.
Shakti said:
4 years ago
How you will get the coefficient?
@Nandish.
The Unit will be m2/t, right?
@Nandish.
The Unit will be m2/t, right?
Sanku said:
5 years ago
Given Answer is correct. I agree.
Sagar nadavadekar said:
7 years ago
Av= -(eo-e)/po-p.
= -(1.25-1.5)/50-25,
=0.01.
= -(1.25-1.5)/50-25,
=0.01.
(1)
Deepak Baswal said:
7 years ago
.01 is the correct answer.
(1)
Prachi said:
7 years ago
The Answer is correct. The Coefficient of compressibility is asked that is e/σ= 0.25/25.
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).
The Coefficient of compressibility is asked, log should be used in the denominator if coefficient of compression (also known as compression index).
(2)
Priya said:
8 years ago
Yes, I agree @Nandish.
(1)
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)).
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)
Rameshwar said:
8 years ago
CC = change in e/log10(initial stress+increased stress/initial stress).
Cc = 0.25/log(50/25) = 0.83.
Cc = 0.25/log(50/25) = 0.83.
Mahesh said:
8 years ago
.830 is the correct answer.
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