Question

An infinite slope with slope angle \(\beta = 22^{\circ}\) consists of soil with the following properties:
Unit weight \(\gamma = 15.72\) kN/m\(^3\)
Cohesion \(c' = 12\) kPa
Angle of internal friction \(\phi' = 15^{\circ}\)
The critical height of the slope (in m) is (rounded off to two decimal places).

Hint:

Set the factor of safety, shear strength divided by mobilized shear stress on a plane parallel to the slope, equal to 1, and solve for the depth.

Answer 1

Correct Answer: 6.53

Step 1: Understanding the Question.
For an infinite slope made of soil with both cohesion and friction, the critical height is the maximum vertical depth of a potential failure plane (parallel to the slope) at which the factor of safety against sliding just equals 1.

Step 2: Find the normal and shear stress on a plane at depth \(H\), parallel to the slope.
On a plane parallel to the slope surface, at vertical depth \(H\), the normal and shear stresses caused by the weight of the soil above are:
\[ \sigma = \gamma H\cos^2\beta, \qquad \tau = \gamma H\sin\beta\cos\beta \]
where \(\gamma\) is the unit weight of the soil and \(\beta\) is the slope angle.

Step 3: Apply the Mohr-Coulomb shear strength criterion.
The shear strength available on that plane is
\[ \tau_f = c' + \sigma\tan\phi' = c' + \gamma H\cos^2\beta\tan\phi' \]
At the critical height, the factor of safety, strength divided by the actual shear stress, equals 1, so the available strength exactly equals the mobilized shear stress:
\[ \gamma H_c\sin\beta\cos\beta = c' + \gamma H_c\cos^2\beta\tan\phi' \]

Step 4: Solve for the critical height \(H_c\).
\[ \gamma H_c\cos\beta(\sin\beta-\cos\beta\tan\phi') = c' \]
\[ H_c = \frac{c'}{\gamma\cos\beta(\sin\beta-\cos\beta\tan\phi')} \]

Step 5: Substitute the given values.
\(\gamma = 15.72\ \text{kN/m}^3\), \(c' = 12\) kPa, \(\beta = 22^{\circ}\), \(\phi' = 15^{\circ}\).
\(\cos22^{\circ} = 0.9272\), \(\sin22^{\circ} = 0.3746\), \(\tan15^{\circ} = 0.2679\).
\[ \sin\beta-\cos\beta\tan\phi' = 0.3746-(0.9272\times0.2679) = 0.3746-0.2484 = 0.1262 \]
\[ H_c = \frac{12}{15.72\times0.9272\times0.1262} = \frac{12}{1.839} \approx 6.53\ \text{m} \]

Final Answer:
Rounded off to two decimal places, the critical height of the slope is approximately \(6.53\) m.
\[ \boxed{H_c \approx 6.53\ \text{m}} \]