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Yield Criteria for Non-Metals
A soil is found to obey the Mohr-Coulomb failure criterion:

   τ=τ*−σn tan φ

where τ and σn are the shear stress and the normal tensile stress on the plane on which failure takes place, τ*=18 kPa and φ=34.6°. It is found to slip when subjected to a shear stress of 50 kPa along the slip plane. Determine the normal stress on the slip plane and the two principal stresses.
Hint:

Use the Coulomb criterion construction to find the radius of a Mohr's circle.
σ τ σ3 σc σ1 0 φ σn R τ τ φ φ=34.6 R -σn=-46.39 τ° = 18σ τ=50 R=60.74σ -20.14 kPa -80.88 kPa -141.62 kPa
The diagram above depicts the Mohr-Coulomb failure criterion, where σc denotes the centre of the Mohr's circle, σ1 and σ3 are the two maximum and minimum principal stresses.

(The third principal stress must be between σ1 and σ3)
From the equation:

\[\tau = \tau* - \sigma_{n}\text{tan}\phi \]

it follows that if τ=50 kPa, τ*=18 kPa and φ=34.6°, \[\sigma_{n} = \frac{\tau*-\tau}{\text{tan}\phi} = \frac{-32}{\text{tan}34.6}= -46.39 \text{kPa}\] is the normal stress on the slip plane. This is a compressive stress.
From simple geometry, the radius, R, of the Mohr's circle is given by: \[\frac{\tau}{R} = cos\phi\] and so r = 60.74 kPa.

Also, \(\frac{\left|\sigma_{c} - \sigma_{n}\right|}{\tau} = tan\phi\) and so,

\(\left|\sigma_{c} - \sigma_{n}\right|\) = 34.49 kPa

Hence σc = -80.88 kPa.
It follows that:

σ1 = σc - R= -141.62 kPa

and

σ3 = σc + R = -20.14 kPa.