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: Use the factor-of-safety formula for an infinite slope directly.
A standard, ready-made formula for the factor of safety of an infinite slope of $c'$-$\phi'$ soil, at depth $H$ below the surface, is:
\[ FS = \frac{\tan\phi'}{\tan\beta} + \frac{c'}{\gamma H\sin\beta\cos\beta} \]
This is the same physics as resolving stresses on the failure plane, but written as a ready formula so we can plug numbers straight in.

Step 2: Set FS = 1 at the critical height and isolate the cohesion term.
\[ 1 = \frac{\tan\phi'}{\tan\beta} + \frac{c'}{\gamma H_c\sin\beta\cos\beta} \]
\[ \frac{c'}{\gamma H_c\sin\beta\cos\beta} = 1-\frac{\tan\phi'}{\tan\beta} \]

Step 3: Compute the friction ratio.
$\tan\beta = \tan22^\circ = 0.4040$, $\tan\phi' = \tan15^\circ = 0.2679$.
\[ \frac{\tan\phi'}{\tan\beta} = \frac{0.2679}{0.4040} = 0.6632 \]
\[ 1-0.6632 = 0.3368 \]

Step 4: Compute $\sin\beta\cos\beta$ and solve for $H_c$.
$\sin22^\circ = 0.3746$, $\cos22^\circ = 0.9272$, so $\sin\beta\cos\beta = 0.3473$.
\[ H_c = \frac{c'}{\gamma\sin\beta\cos\beta\times0.3368} = \frac{12}{15.72\times0.3473\times0.3368} \]
\[ H_c = \frac{12}{1.839} \approx 6.53\text{ m} \]
This matches the stress-based derivation, which confirms the result.

Final Answer:
\[ \boxed{H_c \approx 6.53\text{ m}} \]