Question

Weathered residual soil and the underlying bedrock layer have a slope of B = 25 degrees. The soil layer has a vertical height of 12 feet, a moist unit weight of 115 pef and a saturated unit weight of 125 pef. The water table is 4 feet above the bedrock.

a. Determine the friction angle (in degree )at failure with no cohesion.

b. Determine the friction angle (in degree) with cohesion equal to 200 psf.

### Given data### Given 

- Slope angle \( B = 25^\circ \)
- Height of soil layer \( h = 12 \text{ feet} \)
- Moist unit weight \( \gamma_m = 115 \text{ pcf} \)
- Saturated unit weight \( \gamma_s = 125 \text{ pcf} \)
- Water table depth from the bedrock = \( 4 \text{ feet} \)
- Cohesion \( c = 200 \text{ psf} \) for part (b), and no cohesion for part (a).

### Part (a): Friction angle at failure with no cohesion

We'll use **infinite slope stability analysis** for a purely frictional soil (i.e., with no cohesion):

The factor of safety \( F_s \) for an infinite slope with no cohesion is given by:

\[
F_s = \frac{\tan \phi}{\tan B}
\]

Where:
- \( \phi \) = friction angle of the soil
- \( B \) = slope angle = \( 25^\circ \)

For failure, the factor of safety \( F_s = 1 \), so:

\[
1 = \frac{\tan \phi}{\tan 25^\circ}
\]

Solving for \( \phi \):

\[
\phi = \tan^{-1} (\tan 25^\circ)
\]

Thus,

\[
\phi = 25^\circ
\]

So, the friction angle at failure with no cohesion is **25 degrees**.

---

### Part (b): Friction angle with cohesion equal to 200 psf

For the case with cohesion, the factor of safety is given by:

\[
F_s = \frac{c + (\gamma_m - \gamma_w h_w) h \cos^2 B \tan \phi}{\gamma_m h \sin B \cos B}
\]

Where:
- \( \gamma_m \) = moist unit weight = \( 115 \, \text{pcf} \)
- \( \gamma_w \) = unit weight of water = \( 62.4 \, \text{pcf} \)
- \( h_w \) = height of water table = \( 4 \, \text{ft} \)
- \( h \) = total height of soil = \( 12 \, \text{ft} \)
- \( B \) = slope angle = \( 25^\circ \)
- \( c \) = cohesion = \( 200 \, \text{psf} \)

Let me compute the required friction angle \( \phi \) when cohesion is involved.The friction angle \( \phi \) with cohesion equal to 200 psf is approximately **54.8 degrees**.

### Summary:
- (a) The friction angle at failure with no cohesion is **25 degrees**.
- (b) The friction angle at failure with cohesion of 200 psf is approximately **54.8 degrees**.

Would you like further details on these calculations, or do you have any specific questions?

Here are 5 related questions you could explore:
1. How does the depth of the water table influence the stability of slopes?
2. What happens to the friction angle if the slope angle is increased?
3. How does adding cohesion affect slope stability?
4. Can you explain the concept of "infinite slope" in soil mechanics?
5. How would the presence of pore pressure influence the analysis?

**Tip:** When cohesion is present, it significantly increases the soil's ability to resist shear failure, especially on steeper slopes.

Weathered residual soil and the underlying bedrock layer have a slope of B = 25 degrees. The soil layer has a vertical height of 12 feet, a moist unit weight of 115 pef and a saturated unit weight of 125 pef. The water table is 4 feet above the bedrock. a. Determine the friction angle (in degree) at failure with no cohesion. b. Determine the friction angle (in degree) with cohesion equal to 200 psf.

Learn how to calculate the friction angle at failure for weathered residual soil with a slope of 25 degrees and a water table 4 feet above the bedrock. Explore the effect of no cohesion and the impact of 200 psf cohesion on slope stability, using geotechnical engineering principles and infinite slope theory.

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