A dry sand sample was tested in a triaxial machine with the cell pressure of 200 kPa. If the deviator stress at failure is 400 kPa, then what will be the angle of shearing resistance?

Concept:

From the triaxial test

From the above figure,

\(\sin \phi = \frac{Y}{X}\)

Where,

\(Y = \frac{{{\sigma _1} - {\sigma _3}}}{2}\)

\(X = \frac{{{\sigma _1} + {\sigma _3}}}{2}\)

\(\therefore \sin \phi = \frac{Y}{X} = \frac{{\frac{{{\sigma _1} - {\sigma _3}}}{2}}}{{\frac{{{\sigma _1} + {\sigma _3}}}{2}}} = \frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}\)

\(\phi = {\sin ^{ - 1}}\left( {\frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}} \right)\)

Where,

σd = deviator stress

σ3 = σc = cell pressure

σ1 = σc + σd

Calculation:

σd = 400 kPa

σ3 = σc = 200 kPa

σ1 = σc + σd = 400 + 200 = 600 kPa

\(\phi = {\sin ^{ - 1}}\left( {\frac{{{\sigma _1} - {\sigma _3}}}{{{\sigma _1} + {\sigma _3}}}} \right) = \;{\sin ^{ - 1}}\left( {\frac{{600 - 200}}{{600 + 200}}} \right)\; = \;{\sin ^{ - 1}}\left( {\frac{{400}}{{800}}} \right) = 30^\circ \)