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A conventional drained triaxial compression test was conducted on a normally consolidated clay sample under an effective confining pressure of 200 kPa. The deviator stress at failure was found to be 400 kPa. An identical specimen of the same clay sample is isotropically consolidated to a confining pressure of 200 kPa and subjected to a standard undrained triaxial compression test. If the deviator stress at failure is 150 kPa, the pore pressure developed (in kPa, up to one decimal place) is _______.

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For normally consolidated clay in a drained Triaxial Test, c’ = 0

\({{\sigma }_{1}}'=~{{\sigma }_{3}}'{{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing '}{2} \right] \right)}^{2}}\)

\(\sigma _{3}^{'}+{{\sigma }_{d}}'=~{{\sigma }_{3}}'{{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing '}{2} \right] \right)}^{2}}~\)…….. u = 0

In the drained test:

\(200+400=~200{{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing '}{2} \right] \right)}^{2}}\)

ϕ' = 30

For standard undrained Triaxial test,

\({{\sigma }_{1}}'=~{{\sigma }_{3}}'{{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing '}{2} \right] \right)}^{2}}\) ……. In terms of effective stress.

\(({{\text{ }\!\!\sigma\!\!\text{ }}_{1}}-\text{u})=\text{ }\!\!~\!\!\text{ }({{\text{ }\!\!\sigma\!\!\text{ }}_{3}}-\text{u}){{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing \text{ }\!\!'\!\!\text{ }}{2} \right] \right)}^{2}}\)

\(({{\text{ }\!\!\sigma\!\!\text{ }}_{3}}+{{\text{ }\!\!\sigma\!\!\text{ }}_{\text{d}}}-\text{u})=\text{ }\!\!~\!\!\text{ }({{\text{ }\!\!\sigma\!\!\text{ }}_{3}}-\text{u}){{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing \text{ }\!\!'\!\!\text{ }}{2} \right] \right)}^{2}}\)

\(\left( 200+150-\text{u} \right)=\left( 200-\text{u} \right){{\left( \text{tan }\!\!~\!\!\text{ }\left[ 45+\frac{\varnothing \text{ }\!\!'\!\!\text{ }}{2} \right] \right)}^{2}}\)

u = 125 KPa

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