Chapter 12, Problem 12.4P

(a)

To determine

Find the drained angle of friction for the silty clay soil.

Answer to Problem 12.4P

The drained angle of friction for the silty clay soil is $\underset{\_}{19.2°}$.

Explanation of Solution

Given information:

The diameter (d) of clay sample is 63.5 mm.

The height (h) of the clay sample is 32 mm.

Calculation:

Calculate the area of specimen (A) as follows:

$A=\frac{\pi d^2}{4}$

Substitute 63.5 mm for d.

$\begin{array}{c}A=\frac{\pi {\left(63.5\right)}^2}{4}\\ =\frac{12,661.26}{4}\\ =3,165.32{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =0.00316{\text{m}}^{\text{2}}\end{array}$

Find the normal stress $\left({\sigma }^{\prime }\right)$ for Test No. 1 using the formula as follows:

${\sigma }^{\prime }=\frac{N}{A}$

Here, N is the normal force and A is the area of specimen.

Substitute $84\text{N}$ for N  and $0.00316{\text{m}}^{\text{2}}$ for A.

$\begin{array}{c}{\sigma }^{\prime }=\frac{84}{0.00316}\\ =26,582.27\text{N}/{\text{m}}^{\text{2}}\left(\frac{1\text{kN}}{1,000\text{N}}\right)\\ =26.5\text{8}\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Similarly calculate the normal stress $\left({\sigma }^{\prime }\right)$ for remaining Tests as shown in Table 1.

Test no	Normal force, N(N)	Normal stress $\left({\sigma }^{\prime }\right)$$\left(\text{kN}/{\text{m}}^{\text{2}}\right)$
1	84	26.58
2	168	53.16
3	254	80.38
4	360	113.92

Table 1

Find the shear strength $\left({\tau }_f\right)$ of the failure for Test No. 1 as follows:

${\tau }_f=\frac{S}{A}$

Substitute $28.9\text{N}$ for S and $0.00316{\text{m}}^{\text{2}}$ for A.

$\begin{array}{c}{\tau }_f=\frac{28.9}{0.00316}\\ =9,145.5\text{N}/{\text{m}}^{\text{2}}\left(\frac{1\text{kN}}{1,000\text{N}}\right)\\ =9.14\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the angle of friction $\left(\varphi \right)$ for Test No. 1 as follows:

$\varphi ={\tan }^{-1}\left(\frac{{\tau }_f}{{\sigma }^{\prime }}\right)$

Substitute $26.5\text{8}\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }^{\prime }$ and $9.14\text{kN}/{\text{m}}^{\text{2}}$ for ${\tau }_f$.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{9.14}{26.58}\right)\\ =18.976°\end{array}$

Similarly calculate the angle of friction $\left(\varphi \right)$ and shear strength $\left({\tau }_f\right)$ for remaining tests as shown in Table 2.

Test no	Shear force, S (N)	Shear strength, ${\tau }_f$$\left(\text{kN}/{\text{m}}^{\text{2}}\right)$	Angle of friction $\left(\varphi °\right)$
1	28.9	9.14	18.97
2	59.6	18.86	19.53
3	89.1	28.19	19.33
4	125.3	39.65	19.19

Table 2

Calculate the drained angle of friction $\left(\varphi \right)$ for the silty clay soil using Table 2.

The average value of angle of friction of all tests is the drained angle of friction.

Refer to the Table 2,

Find the drained angle of friction as follows:

$\begin{array}{c}\varphi =\frac{18.97°+19.53°+19.33°+19.19°}{4}\\ =\frac{77.02°}{4}\\ =19.2°\end{array}$

Thus, the drained angle of friction for the silty clay soil is $\underset{\_}{19.2°}$.

(b)

To determine

Find the shear strength of the clay in the field at location A.

Answer to Problem 12.4P

The shear strength of the clay in the field at location A is $\underset{\_}{30.92\text{kN}/{\text{m}}^{\text{2}}}$.

Explanation of Solution

Given information:

The specific gravity $\left(G_s\right)$ of silty sand is 2.69.

The specific gravity $\left(G_s\right)$ of silty clay is 2.72.

The unit weight $\left({\gamma }_w\right)$ of water is $9.81\text{kN}/{\text{m}}^{\text{3}}$.

The void ratio (e) is 0.72.

The water content (w) of silty clay is $22\%$.

The depth $\left(h_1\right)$ of silty sand above water table is $2\text{m}$.

The depth $\left(h_2\right)$ of silty sand below water table is $2.2\text{m}$.

The depth $\left(h_3\right)$ of silty clay is $3.5\text{m}$.

Calculation:

Calculate the dry unit weight $\left({\gamma }_{\text{d}}\right)$ for silty sand using the relation as follows:

${\gamma }_{\text{d}}=\frac{G_s{\gamma }_w}{1+e}$

Substitute 2.69 for $G_s$, $9.81\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_w$, and 0.72 for e.

$\begin{array}{c}{\gamma }_{\text{d}}=\frac{2.69\times 9.81}{1+0.72}\\ =\frac{26.3889}{1.72}\\ =15.342\text{kN}/{\text{m}}^{\text{3}}\end{array}$

Calculate the saturated unit weight $\left[{\left({\gamma }_{\text{sat}}\right)}_{\text{s}}\right]$ for silty sand using the relation as follows:

${\left({\gamma }_{\text{sat}}\right)}_{\text{s}}=\frac{\left(G_s+e\right){\gamma }_w}{\left(1+e\right)}$

Substitute 0.72 for e, $9.81\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_w$, and 2.69 for $G_s$.

$\begin{array}{c}{\left({\gamma }_{\text{sat}}\right)}_{\text{s}}=\frac{\left(2.69+0.72\right)9.81}{\left(1+0.72\right)}\\ =\frac{33.452}{1.72}\\ =19.448\text{kN}/{\text{m}}^3\end{array}$

Calculate the saturated unit weight $\left[{\left({\gamma }_{\text{sat}}\right)}_{\text{c}}\right]$ for silty clay using the relation as follows:

${\left({\gamma }_{\text{sat}}\right)}_{\text{c}}=\frac{\left(1+w\right)G_s{\gamma }_w}{\left(1+w{G}_s\right)}$

Substitute $22\%$ for w, 2.72 for $G_s$, and $9.81\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_w$.

$\begin{array}{c}{\left({\gamma }_{\text{sat}}\right)}_{\text{c}}=\frac{\left(1+0.22\right)2.72\times 9.81}{\left(1+0.22\times 2.72\right)}\\ =\frac{32.55}{1.5984}\\ =20.36\text{kN}/{\text{m}}^{\text{3}}\end{array}$

Determine the normal stress $\left({{\sigma }^{\prime }}_A\right)$ at point A as shown below:

${{\sigma }^{\prime }}_A=h_1\times {\gamma }_d+h_2\times \left[{\left({\gamma }_{\text{sat}}\right)}_{\text{s}}-{\gamma }_{\text{w}}\right]+h_3\times \left[{\left({\gamma }_{\text{sat}}\right)}_{\text{c}}-{\gamma }_{\text{w}}\right]$

Substitute $2\text{m}$ for $h_1$, $15.342\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_d$, $2.2\text{m}$ for $h_2$, $19.448\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{s}}$ $9.81\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_w$, $3.5\text{m}$ for $h_3$, $20.36\text{kN}/{\text{m}}^{\text{3}}$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$.

$\begin{array}{c}{{\sigma }^{\prime }}_A=2\times 15.34+2.2\times \left(19.45-9.81\right)+3.5\times \left(20.36-9.81\right)\\ =30.68+21.208+36.925\\ =88.81\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the shear strength $\left({\tau }_f\right)$ of the clay in the field at location A as follows:

${\tau }_f={{\sigma }^{\prime }}_A\tan {\varphi }^{\prime }$

Here, ${\varphi }^{\prime }$ is the angle of friction.

Substitute $88.81\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_A$ and $19.2°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{\tau }_f=88.81\times \tan 19.2°\\ =88.81\times 0.348\\ =30.92\text{kN}/{\text{m}}^{\text{2}}\end{array}$