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critical depth inf

Calculates the critical depth for failure of an infinite slope

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Critical Depth Inf
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Interface

C++

Critical Depth Inf

doublecritical_depth_inf(	double	`slope`
	double	`water_depth`
	double	`soil_density`
	double	`soil_c`
	double	`soil_phi`
	double	`F` `= 1`	)

This function calculates the critical depth, $\inline Z_c$ along which failure of an infinite slope is most likely to occur. The term <em>infinite slope slide</em> is commonly used to describe a plane translational movement at a shallow depth, which is also parrallel to the surface. The effect of curvature at the extreme ends of the slope are ignored, which leads to a conservative results when dealing with either straight or concave slopes but overestimates the stability of convex curves and corners.

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The basis for the solution is balancing the component of weight in the slipe direction, T, against the frictional resistance of the soil, R. i.e.

$T=\gamma z b \cos \beta_c \sin \beta_c$

(2)

$R=\tau' b$

(3)

The most challenging aspect to the use of this function is understanding the properties of the soil within your slope. Ideally laboratory triaxial tests should be performed. Alternatively you reply on standard values for different soil types and combine this with an your knowledge of the how the slope is formed. As such through the appropriate selection of soil_c and soil_phi, this function may be used in a number of undrained or drained conditions:

In quick undrained conditions (for example shortly after forming a slope) it is usual to only consider the undrained cohesion (undrained shear strength) of the soil, which implies that no drainage of water from between the soil granual pores (porewater) has been allowed to occur. In these circumstances any increase in prore pressure will equal any increase in total stress and therefore no increase in effective stress occurs. In such a circumstance soil_c= $\inline c_u$ and soil_phi=0, where $\inline c_u$ is the undrained cohesive strength of the soil. As such overall shear strength of the soil is

$\tau_f = c_u$

(4)

Note the depth to water is not used in these calculations.

In consolidated-undrained conditions, some disipation of porewater is assumed so you should be looking for values of soil_c= $\inline c_{cu}$ and soil_phi= $\inline \phi_{cu}$ , where the subscript 'cu' refers to the consolidated-undrained conditions of the soil. Laboratory test are almost certainly need to find these values, with the eventual shear strength of the soil described by

$\tau_f = c_{cu} + \sigma_n \tan \phi_{cu}$

(5)

It is important in these types of conditions, that depth of the water table, H, is set to -1, thereby preventing the hydrostatic pressure of any water from reducing the total normal stress, $\inline \sigma_n$ .

In drained conditions, the consolidation of the soil is assumed to be complete, so although water maybe within the slope, there is no excess water pore pressures associated with past loads - only hydrostatic pressures. For normally consolidated soils, $\inline soil_c=0$ and $\inline soil_phi=\phi'$ , where $\inline \phi'$ is the effective angle of internal friction which describes the proportion of the effective normal stress ( $\inline \sigma_n'$ ) that contributes to the shear strength. In cases where the soil is consolidated by force there maybe some apparent cohesion, thus we can also define $\inline soil_c= c'$ .

The overall effective shear strength is then defined as

$\tau_f = c' + \sigma_n' \tan \phi'$

(6)

where $\inline \sigma_n'$ is the effective normal stress equal to

$\sigma_n' = \left( \gamma z b - \gamma_w (z-h) b \right) \cos^2 \beta_c$

(7)

Standard values for drain soil strength are: <table border="1"> <tr><td><b>Soil Type</b></td><td> $\inline \phi' (degrees)$ </td></tr> <tr><td>High plasticity clay</td><td>23</td></tr> <tr><td>Low plasticity clay</td><td>29</td></tr> <tr><td>Quartz sand</td><td>32</td></tr> <tr><td>Carbonate sand</td><td>40</td></tr> </table>

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Understanding The Results

The results from this function can appear confussing, given certain input combinations. When the soil has no base shear strength, and the soil is either totally dry or totally saturated, i.e. soil_c=0 and H=0 or H=-1, then this function will return a critical depth of zero. This means that the slope can fail at any depth and no one height is any more stable than another. Logically this is correct, since as you decend deeper into the soil the slip activating force T will grow linearly, mean while the resistant force R will also grow linearly and the ratio between T and R will remain constant.

Example 1

The soil in a long slope has an undrained shear strength of 50kN/m2 and a unit density of 2000kg/m3. Estimate the depth at which a shear slip may develop for slopes between 10 degrees and 30 degrees.

#include <stdio.h>
#include <codecogs/engineering/geotechnics/critical_depth_inf.h>
int main()
{
  double soil_density=2000;  // kg/m3
  double soil_c=50000;       // undrained strength of soil of 50kN/m2
  for(double slope=10;slope<30;slope++)  // iterate over possible slopes
  {
    double zc=Engineering::Soil::critical_depth_inf(slope, 2, soil_density, soil_c, 0);
    printf("\n slope=%lf  critical depth=%lf",slope, zc);
  }
  return 0;
}

Output:

slope=10.000000  critical depth=14.907330
 slope=11.000000  critical depth=13.610564
 slope=12.000000  critical depth=12.535401
 slope=13.000000  critical depth=11.630800
 slope=14.000000  critical depth=10.860311
 slope=15.000000  critical depth=10.197214
 slope=16.000000  critical depth=9.621479
 slope=17.000000  critical depth=9.117796
 slope=18.000000  critical depth=8.674268
 slope=19.000000  critical depth=8.281511
 slope=20.000000  critical depth=7.932024
 slope=21.000000  critical depth=7.619749
 slope=22.000000  critical depth=7.339733
 slope=23.000000  critical depth=7.087898
 slope=24.000000  critical depth=6.860853
 slope=25.000000  critical depth=6.655759
 slope=26.000000  critical depth=6.470225
 slope=27.000000  critical depth=6.302225
 slope=28.000000  critical depth=6.150031
 slope=29.000000  critical depth=6.012167

Example 2

You wish to create a long slope within a quartz sand, which you can assume to quickly become fully drained during construction. The water table is estimated to run parrallel to the surface at a depth of 2m. Estimate the depth at which failure is likely to occur on a 30 degree slope, assuming a factor of safety of 1.5 (i.e. you what a conservatively high estimate of a possible slip plane).

#include <stdio.h>
#include <codecogs/engineering/soil/critical_depth_inf.h>
int main()
{
  double soil_density=3200;  // kg/m3
  double soil_phi=32;        // 32 degrees from table
  double slope=30;           // required slope angle
  double zc=Engineering::Soil::critical_depth_inf(slope, 2, soil_density, 0, soil_phi);
  printf("\n critical depth=%lf",slope, zc);
  return 0;
}

Output:

critical depth=0.387849

Note

The density of water is assumed to be $\inline \gamma_w=1000 kg/m^3$ .

Parameters

slope	The angle $\inline \beta_c$ of the slope. [degrees]
water_depth	The distance H beneath the surface at which the soil is fully saturated. Use H=-1 for unsaturated or undrained conditions and H=0 for fully saturated everywhere. [m]
soil_density	The density of soil, $\inline \gamma$ . [ $\inline kg/m^3$ ]
soil_c	The natural cohesion (base shear strength) of the soil. In undrained conditions this is often written as $\inline c_u$ , and in drained conditions it becomes the effective base shear strength usually written as $\inline c'$ . In naturally drained soils, $\inline c'$ is often zero.
soil_phi	The angle of effective internal friction, $\inline \phi'$ . [degrees]
F	Default value = 1

Source Code

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