# A drained triaxial compression test on a saturated clay yielded the effective shear strength parameters as c' = 15 kPa and ϕ' = 22°. Consolidated Undr

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A drained triaxial compression test on a saturated clay yielded the effective shear strength parameters as c' = 15 kPa and ϕ' = 22°. Consolidated Undrained triaxial test on an identical sample of this clay at a cell pressure of 200 kPa developed a pore water pressure of 150 kPa at failure. The deviator stress (expressed in kPa) at failure is _________

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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