Question

What will be the value of the coefficient of volume change if, in a consolidation test, the void ratio decreased from $0.70$ to $0.65$ when the load was changed from $50 \, \text{kN/m}^2$ to $100 \, \text{kN/m}^2$?

• A. $5.88 \times 10^{-4} \, \text{m}^2/\text{kN}$
• B. $2.88 \times 10^{-4} \, \text{m}^2/\text{kN}$
• C. $3.12 \times 10^{-4} \, \text{m}^2/\text{kN}$
• D. $4.05 \times 10^{-4} \, \text{m}^2/\text{kN}$

Hint:

Always verify changes in void ratio and stress are substituted accurately into the formula $m_v = \frac{\Delta e}{\Delta \sigma}$ to avoid calculation errors.

Answer 1

The Correct Option is A

The coefficient of volume change ($m_v$) is calculated using the formula:
\[m_v = \frac{\Delta e}{\Delta \sigma \times (1 + e)}\]
Here:
$\Delta e =$ change in void ratio,
$\Delta \sigma =$ change in stress,
$e =$ initial void ratio.
Input values are:
\[e = 0.70, \quad \Delta e = 0.70 - 0.65 = 0.05, \quad \Delta \sigma = 100 - 50 = 50 \, \text{kN/m}^2\]
Substituting these values into the formula gives:
\[m_v = \frac{0.05}{50 \times (1 + 0.70)}\]
Simplifying this expression:
\[m_v = \frac{0.05}{50 \times 1.70}\]
The calculation yields:
\[m_v = \frac{0.05}{85} = 0.000588 \, \text{m}^2/\text{kN}\]
In scientific notation, this is:
\[m_v = 5.88 \times 10^{-4} \, \text{m}^2/\text{kN}\]
The final answer is $5.88 \times 10^{-4} \, \text{m}^2/\text{kN}$ (Option A).