Geotech FE Review

8/16/2019 Geotech FE Review

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FE Review

Geotechnical EngineeringJanuary, 2016


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Soil Properties and

Classification


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Definition of Soil

An assemblage (or system) of particles

that vary in size, shape and chemical

composition


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Particle shapes

• Cohesionless soil (sand and gravel) – Spheroid shape (tiny pebbles)

• Cohesive soil (silt and clay)

– Flat platelets (confetti of plastic wrap)


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Size of Particles

• Gravel 4.75mm to 75mm

• Sand 0.075mm to 4.75mm

• Silt and clay smaller than 0.075mm

• 4.75mm = No. 4 sieve (4 openings/inch)

• 0.075mm = No. 200 sieve (200 openings/in.)


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Grain Size (sieve) Analysis


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Quantifying Gradation

• Uniformity Coefficient

– Cu = D60/D10

• Coefficient of Gradation (curvature)

– Cc = D302/ (D60 x D10) Where,

– D60 = Diameter at which 60% of grains are passing




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Principles of Foundation Engineering


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Soils with Fines (-200 sieve)

Moisture line (increasing)

Solid

State

Non-plastic

State

Plastic

State

Liquid

State

SL,(ωs) PL,(ωp) LL,(ωl)

SL = Shrinkage Limit

PL = Plastic LimitLL = Liquid Limit

PI (Ip) = Plasticity Index = LL - PL

Atterberg Limits


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Unified Soil Classification System

Primary Descriptions

G = Gravel, S = Sand, C = Clay, M = Silt

Secondary DescriptionsP = Poorly Graded

W = Well Graded

L = Low Plasticity

H = High PlasticityO = Organic soil

Pt = Highly organic soil (peat)

i i l f d i i i


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A soil has 3 phases

• Solids

• Voids – Filled with air and/or

water


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Idealized Phase Diagram

Air

Water

Solids

Vt

Va

Vv

Vs

VwWt

Ws

Ww


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Soil Properties

• Void Ratio: e = Vv/Vs

• Porosity: = Vv/Vt

• Moisture Content: ω = Ww/Ws• Deg. of Saturation: S = (Vw/Vv) X 100


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Helpful Relationships

Specific Gravity: Gs = solid mineral/ w

= e/(1+e)e = /(1- )

(Gs) (ω) = (S) (e) (“S” as decimal)


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Unit Weight:

• t = Wt/Vt (lbs./ft.3), sometimes called

density

• Consider the 1 cubic foot box


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Unit Weights

Dry unit weight: d = Ws/Vt

Moist unit weight: m = (Ws + Ww)/VtSaturated unit weight: sat = (Ws + Ww)/Vt

@ S = 100%

Buoyant (effective) unit weight: b = sat - w ( w =62.4pcf)

Relative Density: Dr = [(emax-e)/(emax-emin)] X100

d = (( w) (Gs))/(1+e)

m = (( w) (Gs)(1+ ω))/(1+e)


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Verifying Compaction

Determine the in-place unit weight (dry density) and

compare it to a sample compacted under consistentlaboratory conditions

Proctor Test

• Start with dry soil samples• Add some water and compact the soil in a mold

of known value with a consistent amount of

energy and measure the dry unit weight• Repeat the process incrementally adding water

and plot the dry density vs. moisture content


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MDD = Maximum Dry Density = Dry Unit

Weight at the peak of the curve

OMC = Optimum Moisture Content =

moisture content corresponding to MDD


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In-place Density Determination

• Methods

– Drive Sleeve Method

– Sand Cone Method

– Nuclear Gauge

• Procedure

– Measure the in-place total density and moisture

– Calculate the in-place dry density

– Compare the in-place dry density with the MDD

– Determine if specified % density has been met

(90%, 95% etc.)

I l D i R


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In-place Density Report


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Permeability and Groundwater

Flow


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Flow Net Rules


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Flow Velocity Through Porous Media

Face Velocity; v = Q/A = ki (also known as Darcy Velocity, Hydraulic Velocity or Specific Discharge

Seepage Velocity; vs = v/n (n = porosity)


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Shallow Foundations 1

Bearing Capacity


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The Purpose of Foundations

P P

B

qs = soil pressure (lbs/ft2)

P = foundation load (lbs)

I di id l C l F i C i W ll F i


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Individual Column Footing

P

B

qs = soil pressure = P/Af

P = Concentrated foundation load,

point load (Kilopounds = 1000

pounds, Kips)

Af = Area of footing (ft2, B x L)

Continuous Wall Footing

W

B

qs = soil pressure = W/B

W = Continuous foundation load,

wall/strip load (Kips/lineal foot)

B = Width of footing (ft)

W = 9 kip/lin. ft. = 9 K/ft

B = 3 ft

qs = 9 K/ft ÷ 3 ft = 3 K/sq. ft.

P = 100 K

B = 5 ft, L = 5ft, Af = 5’ x 5’ = 25 sq. ft.

qs = 100 K ÷ 25 sq. ft. = 4 K/sq. ft.


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Pressure from Point Loads (Boussinesq’s Equation)


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Pressure from Applied Distributed Loads

The “60 degree Method”



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Applied Stress from Uniform

Circular Load

0.28



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Applied Stress from Uniform

Infinite and Square Footings


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Reaction of Soil Under Load

1. Shear Failure (bearing capacity)

– Based on soil strength

2. Settlement (compression/consolidation)

– Based on soil compressibility

– Stress- strain

h h l l


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Strength Parameters – Cohesionless Soils

• Cohesionless soil – strength based onintergranular friction

• Particles must slip over each other (two pieces

of coarse sandpaper)

h h i il


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Strength Parameters – Cohesive Soils

• Strength based on inter-granular attraction

(stickiness), like two pieces of plastic wrap

• The strength parameter for cohesive soil is

Cohesion, “C”, lbs/ft 2

• Cohesion is determined in the laboratory

(unconfined compressive strength)

• C = ½ q; q = unconfined compressive strength

• Soils can have strength from both C and Φ

S = C + σ tanΦ


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Bearing Capacity Failure

• Rotational failure

• Soil along the failure surface resists failure

P

B

B


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Types of Bearing Capacity Shear

• General Shear – Sudden failure, medium dense soil

• Local shear

– Not as sudden, “irregular”, loose soil

• Punching shear

– Not rotational, but vertical

– Very loose soil



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Rights Reserved. 3 - 54

Terzaghi – Meyerhoff Bearing Capacity


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Terzaghi Meyerhoff Bearing Capacity

Equation

• To determine the ultimate bearing capacity (continuous footing,

general shear)

• qu = CNc + qNq + 0.4 BN where;

C = Cohesion

q = Soil stress at footing level = γDf ;

Df = depth of footing, γ = unit weight of soilNc, Nq, N γ = Bearing capacity factors based on Φ

Equation is similar for square footings

• For individual column footings (square or round), apply shape

factors to Ncand N

γ



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Water Table EffectsCase 2


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Case 1

If 0 < D1 < Df . . . make

2 modifications to equation

• q = D1γm + D2γb• In 3rd factor, γ = γb

Case 3

If d ≥ B . . . no modifications

Water table is at a depth greater

than or equal to the footing width.

No effect on failure zone

Case 2

If 0 < d < B . . . make

1 modification to

equation

• In 3rd factor, γ = γ

• γ = γb + d/B x (γm - γb)


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Consolidation and Foundation

Settlement


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Why do Footings Settle?

1. Loaded foundation causes increase in stress

within underlying soils (∆pavg)

2. Soil reacts to increased stress by compressing

3. As soil compresses, the ground above

deflects and the footing settles


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Two Modes of Settlement

1. Consolidation (long term) settlement

– Applicable to cohesive soils (clay and silt)

– Primary mechanism is dissipation of pore water

pressure (“plastic” settlement)

– Void ratio decreases as pressure increases

– Since soil has very low permeability, water

escapes very slowly, thus compression (andsettlement) occurs very slowly (long term)


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Consolidation States

• Normally consolidated

– Soil element is currently at the highest pressure

that it has ever been at (p0 = pc, where pc is pre-

consolidation pressure)• Over consolidated

– Soil element has, in the past, experienced a higher

pressure than it has currently (pc > p0)



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(continued)

2. “Immediate” (short term) settlement

– Applicable to cohesionless soils (sand)

– Primary mechanism is reorganization of soil

particles

– High permeability, soil drains quickly; little or no

soil water pressure

– Soil is modeled as an elastic medium (spring) withan elastic soil modulus (Es)

C lid ti T t


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Consolidation Test


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Consolidation Equation


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Consolidation Equation

Over-consolidated where p0 + ∆pavg < pc

Sc = CR Hc p0 + ∆pavg (eq. 5.82)

1+e0 p0

Over-consolidated where p0 < pc < p0 + ∆pavg Sc = CR Hc pc Cc Hc p0 + ∆pavg

1+e0 p0 1+e0 pc

CR = Swelling or Recompression Indexpc = Pre-consolidation pressure

log

log + log


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Virgin Curve

Approximation

Recompression Curve

Approximation


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Combined Stress and Mohr’s

Circle


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Lateral Earth Pressure and

Retaining Walls


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P

B

Vertical Stress (Pressure) Lateral Pressure

Lateral Earth Pressure; Typical Applications


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Lateral Earth Pressure; Typical Applications

Basements Sea Walls

Retaining Walls

Three Soil States

1. “At Rest” state

2. Active state

3. Passive state

“At Rest” State


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At Rest StateSoil is not moving . . . “at rest”

σv (pv)

σh (ph)

σh = σvK0

K0 = Coefficient of Earth

pressure at rest

For cohesionless soil;

K0 = 1 – sin Φ

Rigid Pipe

10’γm = 110 lbs/ft3

Φ = 30o

Calc. lateral earth pressure on pipeK0 = 1 – sin Φ =1- sin 30

o = 0.50

σv = pv = γmd = (110pcf)(10’) = 1100 psf

σh = ph = σv K0 = (1100psf)(0.5)

= 550 psf

Lateral Stresses on Walls


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Lateral Stresses on Walls

Unyielding, Smooth Wall

d

z

Vertical and horizontal stresses

increase with depth

@ z=0, σv = (γm)(0) ∴ σh = (K0)(0) = 0

@ z=d/2, σv = (γm)(d/2) ∴ σh = (K0)(γm)(d/2)

@ z=d, σv = (γm)(d) ∴ σh = (K0)(γm)(d)

Lateral Stress Diagram


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a e a S ess ag a

h P

h/3σh @ z=h

b

Horizontal stress varies linearly with

depth creating a triangle stress

diagram

The total force “P” (per lineal foot of

wall) equals the area of the

diagram; P = ½ (b)(h)

Where; h = depth of the wall

b = σh at bottom of wall

b = (K0)(γm)(h)

∴ P = ½ (K0)(γm)(h)(h)

= ½ (K0)(γm)(h2)

P acts at the centroid of the stress

Diagram (1/3)(h)

Example


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p

10’P

h/3

σh @ 10’

γm = 110lbs/ft3

γsat = 120 lbs/ft3γb = 57.6 lbs/ft

3

K0 = 0.45

Calc. lateral soil force against the

unyielding wall

σh @ 10’ = (σv @ 10’)(K0)

= (γm)(10ft)(K0)

= (110 pcf)(10’)(0.45) = 495 psf

P = ½ bh (area of diagram)

P = ½(495 psf)(10 ft) = 2,475 lbs

(per lineal foot of wall)

A Wrinkle in the Example (water table @ 3 ft.)


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10’

3’

σh @ 0’ = σv @ 0’ = 0

σh @ 3’ = σv @ 3’ (K0) = 110 pcf(3’)(0.45) = 149 psf

σh @ 10’ = σv @ 10’ (K0) = {(3)(110)+(57.6)(7’)}(.45) = 330

psf

σh @ 3’ = 149 psf

σh @ 10’ = 330 psf

Water pressure

µ10’ = 7’(62.4 pcf)

= 437 psf

Total lateral force, Ptot = area of both stress diagrams


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tot

149 psf

330 psf 437 psf

P1

P2P3

P1 = ½ (149 psf)(3’) = 224 lbs/ft of wall (fow)

P2 = ½ (149 psf + 330 psf)(7’) = 1676 lbs/fow

Total lateral soil force = P1 + P2 = 224 + 1676 = 1900 lbs/fow

P3 = ½(437 psf)(7’) = 1530 lbs/fow

Total lateral force = P1 + P2 + P3 = 1900 + 1530 = 3430 lbs/fow

Compare with no WT case, Ptot = 2475 lbs/fow (nearly 40% higher)

3’

7’ 7’

Common Stress States for Wall Design


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g

Active Passive

Wall moves away from soil

(soil is a causing force)Wall moves toward soil

(soil is a resisting force)

Soil Soil

Active State


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Active State

Failure Wedge

45o + Φ/2

Ka = coefficient of active earth pressure = tan2 (45o - Φ/2)

Calculate lateral pressure the same as for “At Rest” state,

but substitute Ka into the analysis

Passive State


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Passive State

Failure Wedge

45o - Φ/2

Kp = coefficient of passive earth pressure = tan2 (45o + Φ/2)

Calculate lateral pressure the same as for “At Rest” and

Active states, but substitute Kp into the analysis

Example


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p

Generally, active pressure will work to cause failure

while passive pressure works to prevent failure

H1 = 10ft

H2 = 3ft

γm = 110 lbs/ft2

Φ = 30 degrees

Will wall translate to the left

(failure)

Pp > Pa not to fail

Ka = tan2 (45o – Φ/2) = 0.33

Kp = tan2 (45o + Φ/2) = 3.0

Pressure Diagrams


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Pa

Pp

σp @ 3’ = σv @ 3’ (Kp)

= 110 pcf(3’)(3.0) = 990 psf

Pp = ½ (990)(3) = 1485 lbs/fow

σa @ 10’ = σv @ 10’ (Ka)

= 110 pcf(10’)(0.33) = 363 psf

Pa = ½ (363)(10) = 1815 lbs/fow

Pa > Pp

∴ Causing force is greater then resisting force and wall will fail


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Slope Stability


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R shear = SL = cL

R friction = W (cos α) (tan ϕ)Tff = R total = cL + W (cos α) (tan ϕ)

Tmob = W (sin α)

R shear = SL = cL

R friction = W (cos α) (tan ϕ)

T = R = cL + W (cos α) (tan ϕ)


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Tff = R total = cL + W (cos α) (tan ϕ)

Tmob = W (sin α)


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R shear = SL = cL

R friction = W (cos α) (tan ϕ)

Tff = R total = cL + W (cos α) (tan ϕ)

Tmob = W (sin α)

Acknowledgements


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Acknowledgements

• Some information presented in this review hasbeen obtained from sources noted within and

also from;

– Foundation Engineering, Braja M. Das, 7th Edition – Test Masters Review

– “Learn Civil Engineering”

• All information included herein is used strictlyfor academic purposes

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