Math Problem Statement

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.

Solution

Given data### Given

  • Slope angle B=25B = 25^\circ
  • Height of soil layer h=12 feeth = 12 \text{ feet}
  • Moist unit weight γm=115 pcf\gamma_m = 115 \text{ pcf}
  • Saturated unit weight γs=125 pcf\gamma_s = 125 \text{ pcf}
  • Water table depth from the bedrock = 4 feet4 \text{ feet}
  • Cohesion c=200 psfc = 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 FsF_s for an infinite slope with no cohesion is given by:

Fs=tanϕtanBF_s = \frac{\tan \phi}{\tan B}

Where:

  • ϕ\phi = friction angle of the soil
  • BB = slope angle = 2525^\circ

For failure, the factor of safety Fs=1F_s = 1, so:

1=tanϕtan251 = \frac{\tan \phi}{\tan 25^\circ}

Solving for ϕ\phi:

ϕ=tan1(tan25)\phi = \tan^{-1} (\tan 25^\circ)

Thus,

ϕ=25\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:

Fs=c+(γmγwhw)hcos2BtanϕγmhsinBcosBF_s = \frac{c + (\gamma_m - \gamma_w h_w) h \cos^2 B \tan \phi}{\gamma_m h \sin B \cos B}

Where:

  • γm\gamma_m = moist unit weight = 115pcf115 \, \text{pcf}
  • γw\gamma_w = unit weight of water = 62.4pcf62.4 \, \text{pcf}
  • hwh_w = height of water table = 4ft4 \, \text{ft}
  • hh = total height of soil = 12ft12 \, \text{ft}
  • BB = slope angle = 2525^\circ
  • cc = cohesion = 200psf200 \, \text{psf}

Let me compute the required friction angle ϕ\phi when cohesion is involved.The friction angle ϕ\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.

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Math Problem Analysis

Mathematical Concepts

Geotechnical Engineering
Slope Stability Analysis
Friction Angle
Cohesion

Formulas

Factor of safety with no cohesion: Fs = tan(φ) / tan(B)
Factor of safety with cohesion: Fs = (c + (γm - γw * hw) * h * cos²(B) * tan(φ)) / (γm * h * sin(B) * cos(B))

Theorems

Infinite Slope Stability Theory

Suitable Grade Level

Undergraduate Civil Engineering