__Concept:__

**Equation of plastic equilibrium is given as follows**

\({{\rm{\sigma }}_1} = {{\rm{\sigma }}_3} \times {{\rm{N}}_\phi } + 2 \times {\rm{C}} \times \sqrt {{{\rm{N}}_\phi }} \)

\({{\rm{N}}_\phi } = {\rm{Flow\;value}} = {\tan ^2}\left( {45 + \frac{\phi }{2}} \right)\)

When effective stress is considered

\({\rm{\sigma }}{{\rm{'}}_1} = {\rm{\sigma }}{{\rm{'}}_3} \times {\tan ^2}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right) + 2 \times {\rm{C'}} \times \sqrt {{\rm{tan\;}}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right)} \)

σ_{3} = σ_{c} = cell pressure, σ_{d} = deviator stress

σ_{1 }= principal stress = σ_{c }+ σ_{d}

C = cohesion of the soil, ϕ = angle of internal friction,

U = pore water pressure

σ’_{1} and σ’_{3} are effective stress parameters

C’ and ϕ’ are effective or drained shear strength parameters

__Calculation:__

C’ = 15 kPa, ϕ’ = 22°

σ_{c} = σ_{3} = 200 kPa, U = 150 kPa

σ’_{3} = σ_{3} – U = 200 – 150 = 50 KPa

\({\rm{\sigma }}{{\rm{'}}_1} = {\rm{\sigma }}{{\rm{'}}_3} \times {\tan ^2}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right) + 2 \times {\rm{C'}} \times \sqrt {{\rm{tan\;}}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right)} \)

\({\rm{\sigma }}{{\rm{'}}_1} = 50 \times {\tan ^2}\left( {45 + \frac{{22^\circ }}{2}} \right) + 2 \times 15 \times \sqrt {{\rm{tan\;}}\left( {45 + \frac{{22^\circ }}{2}} \right)} \)

σ’_{1} = 154.376 kPa

we know that,

σ_{1 }= σ_{c }+ σ_{d}

∴ σ_{d} = σ’_{1 }- σ’_{3 } = 154.376 – 50 = 104.376 kPa

**σ**_{d} = deviator stress = 104.376 kPa