A pressuremeter test was carried out at a site at a depth of 7 m below the ground surface. The water table level was at a depth of 1.5 m. The average unit weight of saturated soil is 17.3 kN/m^3. The corrected pressuremeter curve is given in Fig. Ex. 9.11 and the depleted volume of the probe is Vc

Question

A pressuremeter test was carried out at a site at a depth of 7 m below the ground surface. The water table level was at a depth of 1.5 m. The average unit weight of saturated soil is 17.3 $kN / m ^{3}$ . The corrected pressuremeter curve is given in Fig. Ex. 9.11 and the depleted volume of the probe is $V_{c}=535 cm ^{3}$ . Determine the following

(a) The coefficient of earth pressure for the at-rest condition

(b) The Menard pressuremeter modulus $E_{m}$

(c) The undrained shear strength $c_{u} \text {. Assume that } p_{\text {oh }}=p_{\text {om }}$ in this case

(a) The coefficient of earth pressure for the at-rest condition

(b) The Menard pressuremeter modulus [latex]E_{m}[/latex]

(c) The undrained shear strength [latex]c_{u} ext {. Assume that } p_{ ext {oh }}=p_{ ext {om }}[/latex] in this case

Accepted Answer

From Fig. Ex 9.11, [latex]p_{o h}=p_{o m}=105[/latex] kPa

The effective overburden pressure is

[latex]p_{0}^{\prime}=17.3 imes 7-5.5 imes 9.81=67.2[/latex] kPa

The effective horizontal pressure is

[latex]p_{0 h}^{\prime}=105-5.5 imes 9.81=51.0[/latex] kPa

(a) From Eq (9.21)

[latex]p_{o h}=(\gamma z-u) K_{0}+u[/latex] (9.21)

[latex]K_{0}=\frac{p_{o h}^{\prime}}{p_{o}^{\prime}}=\frac{51.0}{67.2}=0.76[/latex]

(b) From Eq (9.25)

[latex]E_{m}=2.66 V_{m} \frac{\Delta p}{\Delta u}[/latex] (9.25)

[latex]E_{m}=2.66 V_{m} \frac{\Delta p}{\Delta u}[/latex]

From Fig. Ex 9.11

[latex]\begin{array}{ll}v_{f}=200 cm ^{3} & p_{f}=530 kPa \v_{o}=160 cm ^{3} & p_{o m}=105 kPa\end{array}[/latex]

From Eq (9.24) [latex]V_{m}=535+\frac{200+160}{2}=715 cm ^{3}[/latex]

[latex]V=V_{m}=V_{c}+\frac{v_{0}+v_{f}}{2}[/latex] (9.24)

[latex]\frac{\Delta p}{\Delta v}=\frac{530-105}{200-160}=10.625[/latex]

Now [latex]E_{m}=2.66 imes 715 imes 10.625=20,208[/latex] kpa

(c) From Eq (9.26)

[latex]c_{u}=\frac{\bar{p}_{l}}{9}[/latex] (9.26)

[latex]c_{u}=\frac{\overline{p_{l}}}{9}=\frac{p_{l}-p_{o h}}{9}[/latex]

From Fig Ex 9.11

[latex]\overline{p_{l}}=950-105=845 kPa[/latex]

Therefore [latex]c_{u}=\frac{845}{9}=94 kPa[/latex]

From Eq (9.27)

[latex]c_{u}=\frac{\bar{p}_{l}}{10}+25 kPa[/latex] (9.27)

[latex]c_{u}=\frac{\overline{p_{l}}}{10}+25=\frac{845}{10}+25=109.5[/latex] kpa

Question 9.11: A pressuremeter test was carried out at a site at a depth of...

Related Answered Questions