Chapter 12, Problem 12.10P

(a)

To determine

Find the undrained cohesion of the clay using Bjerrum’s $\lambda$ relationship.

Answer to Problem 12.10P

The undrained cohesion of the clay using Bjerrum’s $\lambda$ relationship is $\underset{\_}{18.9{\text{kN/m}}^2}$.

Explanation of Solution

Given information:

The height of the vane (h) is 127 mm.

The width of the vane (d) is 63.5 mm.

The torque (T) is $20\text{N}\cdot \text{m}$.

The liquid limit (LL) is 50.

The plastic limit (PL) is 18.

Calculation:

Find the ratio of height (h) to diameter (d):

$\text{Height}\text{to}\text{diameter}\text{ratio}=\frac{h}{d}$

Substitute 127 mm for h and 63.5 mm for d.

$\begin{array}{c}\text{Height}\text{to}\text{diameter}\text{ratio}=\frac{127}{63.5}\\ =2\end{array}$

Find the value of K using the formula.

$K=\frac{7\pi d^3}{6}$

Substitute 63.5 mm for d.

$\begin{array}{c}K=\frac{7\pi \times {\left(63.5\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^3}{6}\\ =0.000938{\text{m}}^{\text{3}}\end{array}$

Find undrained shear strength $\left(c_{u\left(VST\right)}\right)$ of vane shear test using the formula:

$c_{u\left(VST\right)}=\frac{T}{K}$

Substitute $20\text{N}\cdot \text{m}$ for T and $0.000938{\text{m}}^{\text{3}}$ for K.

$\begin{array}{c}{c}_{u\left(VST\right)}=\frac{20}{0.000938}\\ =\left(\frac{21,322{\text{N/m}}^2}{1,000}\right){\text{kN/m}}^2\\ =21.3{\text{kN/m}}^2\end{array}$

Find the plasticity index (PI) using the relation:

$PI=LL-PL$

Substitute 50 for LL and 18 for PL.

$\begin{array}{c}PI=50-18\\ =32\end{array}$

Find the correction factor $\lambda$ using the formula:

$\lambda =1.7-0.54\log \left(PI\right)$

Substitute 32 for PI.

$\begin{array}{c}\lambda =1.7-0.54\log \left(32\right)\\ =0.887\end{array}$

Find the corrected undrained shear strength $\left(c_{u\left(Corrected\right)}\right)$ using the relation:

$c_{u\left(\text{Corrected}\right)}=c_{u\left(VST\right)}\times \lambda$

Substitute $21.3{\text{kN/m}}^2$ for $c_{u\left(VST\right)}$ and 0.887 for $\lambda$.

$\begin{array}{c}{c}_{u\left(\text{Corrected}\right)}=21.3\times 0.887\\ =18.9{\text{kN/m}}^2\end{array}$

Thus, the undrained cohesion of the clay by Bjerrum’s $\lambda$ relationship is $\underset{\_}{18.9{\text{kN/m}}^2}$.

(b)

To determine

Find the undrained cohesion of the clay using Morris and Williams’ $\lambda$ and PI relationship.

Answer to Problem 12.10P

The undrained cohesion of the clay using Morris and Williams’ $\lambda$ and PI relationship is $\underset{\_}{14.08{\text{kN/m}}^2}$.

Explanation of Solution

Refer to part (a);

The value of PI is 32.

Find the correction factor using the formula:

$\lambda =1.18e^{-0.08\left(PI\right)}+0.57$

Substitute 32 for PI.

$\begin{array}{c}\lambda =1.18e^{-0.08\left(32\right)}+0.57\\ =0.661\end{array}$

Find the corrected undrained shear strength $\left(c_{u\left(Corrected\right)}\right)$ using the relation:

$c_{u\left(\text{Corrected}\right)}=c_{u\left(VST\right)}\times \lambda$

Refer part (a), the undrained shear strength $\left(c_{u\left(VST\right)}\right)$ is $21.3{\text{kN/m}}^2$.

Substitute $21.3{\text{kN/m}}^2$ for $c_{u\left(VST\right)}$ and 0.661 for $\lambda$.

$\begin{array}{c}{c}_{u\left(\text{Corrected}\right)}=21.3\times 0.661\\ =14.08{\text{kN/m}}^2\end{array}$

Thus, the undrained cohesion of the clay by Morris and Williams’ $\lambda$ and PI relationship is $\underset{\_}{14.08{\text{kN/m}}^2}$.

(c)

To determine

Find the undrained cohesion of the clay using Morris and Williams’ $\lambda$ and LL relationship.

Answer to Problem 12.10P

The undrained cohesion of the clay using Morris and Williams’ $\lambda$ and LL relationship is $\underset{\_}{39.5°}$.

Explanation of Solution

Given information:

The liquid limit (LL) is 50.

Calculation:

Find the correction factor using the formula:

$\lambda =7.01e^{-0.08\left(LL\right)}+0.57$

Substitute 50 for LL.

$\begin{array}{c}\lambda =7.01e^{-0.08\left(50\right)}+0.57\\ =0.698\end{array}$

Find the corrected undrained shear strength $\left(c_{u\left(Corrected\right)}\right)$ using the relation:

$c_{u\left(\text{Corrected}\right)}=c_{u\left(VST\right)}\times \lambda$

Refer part (a), the undrained shear strength $\left(c_{u\left(VST\right)}\right)$ is $21.3{\text{kN/m}}^2$.

Substitute $21.3{\text{kN/m}}^2$ for $c_{u\left(VST\right)}$ and 0.698 for $\lambda$.

$\begin{array}{c}{c}_{u\left(\text{Corrected}\right)}=21.3\times 0.698\\ =14.88{\text{kN/m}}^2\end{array}$

Thus, the undrained cohesion of the clay by Morris and Williams’ $\lambda$ and LL relationship is $\underset{\_}{14.88{\text{kN/m}}^2}$.