Stability of Slopes

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2/18/2016 1 3010 Foundation Engineering(3-1-0-4) 1 Course instructor Dr. Trudeep N. Dave Institute of Infrastructure Technology Research and Management E-mail: [email protected] Class timings: Monday: 11:00 to 12:00 Tuesday: 10.00 to 11.00 Thursday: 11.00 to 12.00 La Conchita, Santa Barbara, California, Spring 1995

Transcript of Stability of Slopes

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3010 Foundation Engineering(3-1-0-4)

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Course instructor

Dr. Trudeep N. DaveInstitute of Infrastructure Technology Research and Management

E-mail: [email protected]

Class timings:Monday: 11:00 to 12:00 Tuesday: 10.00 to 11.00Thursday: 11.00 to 12.00

La Conchita, Santa Barbara, California, Spring 1995

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Introduction – Modes of Slope Failure

• An exposed ground surface that stands at an angle with thehorizontal is called an unrestrained slope.

• The slope can be natural or man-made. It can fail in various modes.

Types of slope failure (Cruden and Varnes, 1996)

• Fall – Detachment of soil/rock fragment.

• Topple – Forward rotation of soil/rock about an axis below C.G. ofmass being displaced.

• Slide – Downward movement of soil mass on a surface of rupture.

• Spread – Slide by translation.

• Flow – Downward movement of soil similar to viscous fluid.

FallTopple

Slide

Spread Flow

Why failure of slope occurs

Important factors that cause instability in a slope and lead to failure are

• Due to the action of gravitational forces

• Due to seepage forces within the soil

• Erosion of surface of slopes due to flowing water

• Sudden lowering of water adjacent to slope

• Forces due to earthquake

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Why failure of slope occurs

Analysis of stability of slope consists of

• The determination of the most severely stresses on theinternal surface and the magnitude of the shearing stressesto which it is subjected,

• The determination of the shearing strength along thissurface

Important factors that cause instability in a slope and lead to failure are

• Due to the action of gravitational forces

• Due to seepage forces within the soil

• Erosion of surface of slopes due to flowing water

• Sudden lowering of water adjacent to slope

• Forces due to earthquake

Types of Slopes

• Natural slopes: Due to natural causes

• Man made slopes: Cutting and embankments

• Infinite Slope: If a slope represents the boundary of a semi-infinite soil mass, and the properties for all identical depthsbelow the surface are constant, it is called infinite slope.

• Finite Slope: If the slope is of limited extent, it is calledfinite slope.

Factor of Safety

Factor of Safety Fs = tf / td

Fs = factor of safety with respect to strength

tf = average shear strength of the soil

td = average shear stress developed along the potential failuresurface

• The shear strength of a soil consists of two components,cohesion and friction, and may be written as

c’ = cohesion

f’ = angle of friction

s’ = normal stress on potential failure plane

tf = c’ + s’ tanf’

Factor of Safety (contd…)

• The shear strength of a soil consists of two components, cohesion and friction, and may be written as

c’d = cohesion at potential failure surface

f’d = angle of friction at potential failure surface

td = c’d + s’ tanf’d

Factor of Safety w. r. t. Cohesion

Factor of Safety w. r. t. friction

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When F’c = F’f

When Fs is equal to 1, the slope is in a state of impendingfailure.In general, Fs = 1.5 with respect to strength is acceptable forthe design of a stable slope.

Stability of Infinite Slopes

Shear Strength ofSoil may be given by

Assuming that the pore water pressure is zero, while evaluating thefactor of safety against a possible slope failure along a plane ABlocated at a depth H below the ground surface. The slope failure canoccur by the movement of soil above the plane AB from right to left.

Stability of Infinite Slopes

Shear Strength ofSoil may be given by

Considering slope element abcd, with a unit length perpendicular tothe plane of the section. The forces, F, that act on the faces ab and cdare equal and opposite and may be ignored. The weight of the soilelement is

The weight W can be resolved into two components:1. Force perpendicular to the plane AB = Na = W cos b = gLH cos b2. Force parallel to the plane AB = Ta = W sin b = gLH sin b. Force that tends to cause the slip along the plane.

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Considering slope element abcd, with a unit length perpendicular tothe plane of the section. The forces, F, that act on the faces ab and cdare equal and opposite and may be ignored. The weight of the soilelement is

The weight W can be resolved into two components:1. Force perpendicular to the plane AB = Na = W cos b = gLH cos b2. Force parallel to the plane AB = Ta = W sin b = gLH sin b. Force that tends to cause the slip along the plane.

Thus, the effective normal stress and the shear stress at the base of theslope element can be given, respectively, as

……(1)

(2)……..

The reaction to the weight W is an equal and opposite force R. Thenormal and tangential components of R with respect to the plane AB are

For equilibrium, the resistive shear stress that develops at the base of the element is equal to (Tr)/(Area of base)

The resistive shear stress

….From (1)

Factor of safety with respect to strength

For granular soils, c’ = 0, and the factor of safety, Fs becomes equal to(tan f)/(tan b).

This indicates that in an infinite slope in sand, the value of Fs isindependent of the height H and the slope is stable as long as b < f.’

If a soil possesses cohesion and friction, the depth of the plane alongwhich critical equilibrium occurs may be determined by substitutingFs = 1 and H = Hcr into

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For steady state seepagethrough the soil and theground water tablecoincides with theground surface, as shownin figure

Prob : 1

Part (a)

Part (b)

Finite Slopes

• When the value of Hcr approaches the height of the slope, the slopegenerally may be considered finite.

• For simplicity, when analyzing the stability of a finite slope in ahomogeneous soil, we need to make an assumption about thegeneral shape of the surface of potential failure.

• Swedish geotechnical commission recommended that the actualsurface of sliding may be approximated to be circularly cylindrical.

• Most conventional stability analyses of slopes have been made byassuming that the curve of potential sliding is an arc of a circle.

• In many circumstances (for example, zoned dams and foundations onweak strata), stability analysis using plane failure of sliding is moreappropriate and yields excellent results.

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Analysis of Finite Slopes with Circular Failure Surfaces

Modes of failure of finite slope: (a) slope failure; (b) shallow slope failure; (c) base failure

• Mass Procedure: Mass of the soil above the surface of sliding istaken as unit. This procedure is useful when the soil that forms theslope is assumed to be homogeneous, although this is not the casein most natural slopes.

• Method of slices: In this procedure,

1. The soil above the surface of sliding is divided into a number ofvertical parallel slices.

2. The stability of each slice is calculated separately.

3. This is a versatile technique in which the nonhomogeneity of the soilsand pore water pressure can be taken into consideration.

4. It also accounts for the variation of the normal stress along thepotential failure surface.

Types of Stability Analysis Procedures

Fellenius (Ordinary) Method of Slices

• AC is an arc of a circle representing the trial failure surface.

• The soil above the trial failure surface is divided into several verticalslices. The width of each slice need not be the same.

• Considering a unit length perpendicular to the cross section shown,the forces that act on a typical slice (nth slice) are shown in Figure.

• Wn is the weight of the slice. The forces Nr and Tr, respectively, arethe normal and tangential components of the reaction R.

• Pn and Pn+1 are the normal forces that act on the sides of the slice.

• The shearing forces that act on the sides of the slice are Tn and Tn+1.

• For simplicity, the pore water pressure is assumed to be zero.

Trial failure surface

forces acting on nth slice

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• For (force)Equilibrium consideration

• The resisting shear force can be expressed as

• The normal stress s ‘ can be equated as

• For equilibrium of the trial wedge ABC, the moment of the drivingforce about O equals the moment of the resisting force about O, or

• It can be represented as

Factor of safety = MR/MD

• The value of an is positive when the slope of the arc is in the samequadrant as the ground slope. To find the minimum factor ofsafety—that is, the factor of safety for the critical circle—one mustmake several trials by changing the center of the trial circle.

• For convenience, a slope in a homogeneous soil. However, themethod of slices can be extended to slopes with layered soil, asshown in Figure.

• The general procedure of stability analysis is the same. However, forthe factor of safety calculation, the values of f’ and C’ will not be thesame for all slices. For example, for slice No. 3 we have to use afriction angle of f’ = f’3 and cohesion C’ = C’3. Similarly, for slice no.2.

• Total shear strength parameters

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The stability of slopes of an earth dam is testedunder the following conditions:

1. Stability of downstream slope during steadyseepage

2. Stability of upstream slope during suddendrawdown

3. Stability of upstream and downstream slopesduring and immediately after construction.

Stability of Slopes of Earth Dam Stability of Downstream Slope During Steady Seepage

• Critical condition of D/S slope occurs when the reservoiris full and percolation is at its maximum rate. Thedirection of seepage forces tend to decrease the stability.

• In other words, the Pore Water Pressure acting on thesoil mass below the saturation line reduces the effectivestress responsible for mobilizing shearing resistance.

• The Normal components N’ are calculated on the basisof Buoyant Unit Weight g’ of the dam, while Tcomponents are to be calculated on the basis ofsaturated unit weight.

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Stability of Upstream Slope During Sudden Drawdown

• For U/S slope, the critical condition is when the reservoiris suddenly emptied without allowing any appreciablechange in the water level within the saturated mass ofsoil.

• With complete drawdown, hydrostatic force acting alongthe U/S slope at the time of full reservoir is emptied. Theweight of water within the soil now tends to help a slidingfailure without the hydrostatic pressure on the slope tocounteract it.

• The Normal components N’ are calculated on the basis ofBuoyant Unit Weight g’ of impervious U/S slope while Tcomponents are to be calculated on the basis of saturatedunit weight.

Stability of U/S and D/S Slopes During or Immediatelyafter Construction• When a dam is built of relatively impervious compressible

soil, excess pore pressure develops in the air and waterentrapped in the pore space.

• As the pore pressure greatly affects the shear strength ofsoil, it is essential to know its magnitude for stabilityanalysis.

• Plot between effective stress s’ and percent consolidation∆ is obtained from consolidation test.

• The construction pore pressure is estimated using Hilf’sequation : u = (Pa * ∆)/ (Va + hc * gw * ∆)

• Knowing effective stress and pore pressure u, the totalstress s is obtained by s = s’ + u

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Bishop’s Simplified Method of Slices

• A more refined solution to the ordinary method of slices.• The effect of forces on the sides of each slice are accounted for to some degree.

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Summing the forces in the vertical direction gives

Taking the moment about O gives

Note that the term Fs is present on both sides of above Equation Hence,we must adopt a trial-and-error procedure to find the value of Fs.

Bishop’s simplified method is probably the most widely used.