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
\({\rm{\sigma }}{{\rm{'}}_1} = {\rm{\sigma }}{{\rm{'}}_3} \times {\tan ^2}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right) + 2 \times {\rm{C'}} \times \sqrt {{\rm{tan\;}}\left( {45 + \frac{{\phi {\rm{'}}}}{2}} \right)} \)
σ3 = σc = cell pressure, σd = deviator stress
σ1 = principal stress = σc + σd
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°
σc = σ3 = 200 kPa, U = 150 kPa
σ’3 = σ3 – 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,
σ1 = σc + σd
∴ σd = σ’1 - σ’3 = 154.376 – 50 = 104.376 kPa
σd = deviator stress = 104.376 kPa