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

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/m3kN / m ^{3}. The corrected pressuremeter curve is given in Fig. Ex. 9.11 and the depleted volume of the probe is Vc=535cm3V_{c}=535 cm ^{3}. Determine the following

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

(b) The Menard pressuremeter modulus EmE_{m}

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

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From Fig. Ex 9.11, poh=pom=105p_{o h}=p_{o m}=105 kPa

 

The effective overburden pressure is

 

p0=17.3×75.5×9.81=67.2p_{0}^{\prime}=17.3 \times 7-5.5 \times 9.81=67.2 kPa

 

The effective horizontal pressure is

 

p0h=1055.5×9.81=51.0p_{0 h}^{\prime}=105-5.5 \times 9.81=51.0 kPa

 

(a) From Eq (9.21)

 

poh=(γzu)K0+up_{o h}=(\gamma z-u) K_{0}+u (9.21)

 

K0=pohpo=51.067.2=0.76K_{0}=\frac{p_{o h}^{\prime}}{p_{o}^{\prime}}=\frac{51.0}{67.2}=0.76

 

(b) From Eq (9.25)

 

Em=2.66VmΔpΔνE_{m}=2.66 V_{m} \frac{\Delta p}{\Delta \nu} (9.25)

 

Em=2.66VmΔpΔνE_{m}=2.66 V_{m} \frac{\Delta p}{\Delta \nu}

 

From Fig. Ex 9.11

 

vf=200cm3pf=530kPavo=160cm3pom=105kPa\begin{array}{ll}v_{f}=200 cm ^{3} & p_{f}=530 kPa \\v_{o}=160 cm ^{3} & p_{o m}=105 kPa\end{array}

 

From Eq (9.24) Vm=535+200+1602=715cm3V_{m}=535+\frac{200+160}{2}=715 cm ^{3}

 

V=Vm=Vc+v0+vf2V=V_{m}=V_{c}+\frac{v_{0}+v_{f}}{2} (9.24)

 

ΔpΔv=530105200160=10.625\frac{\Delta p}{\Delta v}=\frac{530-105}{200-160}=10.625

 

Now Em=2.66×715×10.625=20,208E_{m}=2.66 \times 715 \times 10.625=20,208 kpa

 

(c) From Eq (9.26)

 

cu=pˉl9c_{u}=\frac{\bar{p}_{l}}{9} (9.26)

 

cu=pl9=plpoh9c_{u}=\frac{\overline{p_{l}}}{9}=\frac{p_{l}-p_{o h}}{9}

 

From Fig Ex 9.11

 

pl=950105=845kPa\overline{p_{l}}=950-105=845 kPa

 

Therefore cu=8459=94kPac_{u}=\frac{845}{9}=94 kPa

 

From Eq (9.27)

 

cu=pˉl10+25kPac_{u}=\frac{\bar{p}_{l}}{10}+25 kPa (9.27)

 

cu=pl10+25=84510+25=109.5c_{u}=\frac{\overline{p_{l}}}{10}+25=\frac{845}{10}+25=109.5 kpa

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