Consolidation parameters from a virgin compression curve: 16-Civ-A4 Geotechnical Materials and Analysis, December 2019, Q5

Question 5: Consolidation parameters from a virgin compression curve (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two points on the virgin (normally consolidated) $e$–$\log\sigma'$ line, and the coefficient of consolidation.

Given data
Point 1	$e_1 = 0.8$, $\sigma'_1 = 200\ \text{kN/m}^2$
Point 2	$e_2 = 0.5$, $\sigma'_2 = 400\ \text{kN/m}^2$
Coefficient of consolidation	$c_v = 0.003\ \text{cm}^2/\text{s}$
Unit weight of water	$\gamma_w = 9.81\ \text{kN/m}^3$

Find. $m_v$ over 200–400 kN/m²; the permeability $k$ in cm/s; and $\sigma'$ at $e=0.65$.

The two data points define the virgin compression line; the mid-value $e=0.65$ is read back to $\sigma'\approx283\ \text{kN/m}^2$.

Approach. Get the coefficient of compressibility $a_v=-\Delta e/\Delta\sigma'$, convert to $m_v=a_v/(1+\bar e)$, then use the diffusion identity $k=c_v\,m_v\,\gamma_w$; finally interpolate along the straight $e$–$\log\sigma'$ line (compression index $C_c$) for the pressure at $e=0.65$.

(i) Coefficient of volume compressibility. The coefficient of compressibility over the increment is $$a_v=\frac{\Delta e}{\Delta\sigma'}=\frac{0.8-0.5}{400-200}=1.5\times10^{-3}\ \text{m}^2/\text{kN}.$$ Over the stated pressure range the void ratio falls from 0.8 to 0.5, so use the average void ratio $\bar e=(0.8+0.5)/2=0.65$ (the value part (ii) refers to): $$m_v=\frac{a_v}{1+\bar e}=\frac{1.5\times10^{-3}}{1.65}=\boxed{9.09\times10^{-4}\ \text{m}^2/\text{kN}}$$ (Using the initial $e_0=0.8$ instead would give $8.33\times10^{-4}\ \text{m}^2/\text{kN}$; the average void ratio is the standard choice for a stated range, as in Das.)
(ii) Coefficient of permeability. The coefficient of consolidation ties $k$ to $m_v$ through $c_v = k/(m_v\gamma_w)$, so $k=c_v\,m_v\,\gamma_w$. Converting $c_v=0.003\ \text{cm}^2/\text{s}=3.0\times10^{-7}\ \text{m}^2/\text{s}$: $$k=(3.0\times10^{-7})(9.09\times10^{-4})(9.81)=2.68\times10^{-9}\ \text{m/s}=\boxed{2.68\times10^{-7}\ \text{cm/s}}$$ a value typical of a silty clay, evaluated with $m_v$ at the average void ratio $\bar e=0.65$, as the question asks.
(iii) Effective pressure at $e=0.65$. The compression index (slope of the straight $e$–$\log\sigma'$ line) is $$C_c=\frac{e_1-e_2}{\log(\sigma'_2/\sigma'_1)}=\frac{0.3}{\log(400/200)}=0.997.$$ Interpolating from point 1, $e=e_1-C_c\log(\sigma'/\sigma'_1)$, set $e=0.65$: $$\log\frac{\sigma'}{200}=\frac{0.8-0.65}{0.997}=0.1505\;\Rightarrow\;\sigma'=200\,(10^{0.1505})=\boxed{283\ \text{kN/m}^2}$$ Because $e=0.65$ is exactly midway between $e_1$ and $e_2$, $\sigma'$ is the geometric mean $\sqrt{200\times400}=282.8\ \text{kN/m}^2$ — a useful check.

Consolidation parameters
Quantity	Value
Coefficient of compressibility $a_v$	$1.5\times10^{-3}\ \text{m}^2/\text{kN}$
Coefficient of volume compressibility $m_v$ (at $\bar e=0.65$)	$9.09\times10^{-4}\ \text{m}^2/\text{kN}$
Coefficient of permeability $k$	$2.68\times10^{-7}\ \text{cm/s}$
Compression index $C_c$	$0.997$
$\sigma'$ at $e=0.65$	$283\ \text{kN/m}^2$

Topic: Extracting compressibility, permeability and the compression index from a one-dimensional consolidation (oedometer) test.

Key relations: $a_v=-\Delta e/\Delta\sigma'$; $m_v=a_v/(1+\bar e)$; $c_v=k/(m_v\gamma_w)\Rightarrow k=c_v m_v \gamma_w$; $C_c=\Delta e/\Delta\log\sigma'$.

Why this works: $m_v$ is the volumetric strain per unit stress, so it links a stress increment to settlement; the coefficient of consolidation bundles stiffness ($1/m_v$) and drainage ($k$) into one diffusion coefficient, which is why $k$ falls straight out of $c_v$ and $m_v$. On the virgin line $e$ varies linearly with $\log\sigma'$, so interpolation is exact.

Common pitfall: Forgetting the $\text{cm}^2\to\text{m}^2$ ($\times10^{-4}$) conversion on $c_v$, or dividing $a_v$ by $(1+e_0)$ using the wrong void ratio; and interpolating $\sigma'$ linearly instead of on a log scale.

Source: Das & Sobhan, Ch. 11 ($a_v,\ m_v,\ C_c,\ c_v$, and $k=c_v m_v\gamma_w$).

Question 5 of 6: Consolidation parameters from a virgin compression curve

Question 5: Consolidation parameters from a virgin compression curve (15 marks)