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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.
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,