1.9 shear strength - national chiao tung...

CE 416.3 Class Notes I.R. Fleming 9

Chapter 1 (cont) Soil Mechanics Review Part B

1.9 Shear Strength Introduction

• Soil strength is measured in terms of shear resistance

• Shear resistance is developed on the soil particle contacts

• Failure occurs in a material when the normal stress and the shear stress reach some limiting combination

Figure 1.5 Shear Strength Model.

Shear Strength Formula

• Shear strength formula is based on the Mohr-Coulomb failure criteria, tanφσ'cτ += ′

Where: τ = shear strength c = cohesion φ′ = angle of shearing resistance

Soil ParticlesVoids (Water and/or Air)

Shear Resistance

Normal Load

WaterPressure

CE 416.3 Class Notes I.R. Fleming 0

Shear Strength Testing • Direct shear test – simple, inexpensive, limited configurations

Figure 1.6 Direct Shear Test.

• Triaxial test – may be complex, expensive, several configurations

Figure 1.7 Triaxial Test.

CE 416.3 Class Notes I.R. Fleming Page 21

Undrained Loading (φ = 0 Concept)

• Total stress change is the same as the pore water pressure increase in undrained loading, i.e. no change in effective stress

• Changes in total stress do not change the shear strength in undrained loading

Unconfined Compression Test

• A special type of unconsolidated-undrained triaxial test in which the confining pressure, σ3, is set to zero

• The axial stress at failure is referred to the unconfined compressive strength, Qu (not to be confused with qu)

• The unconfined shear strength, cu, may be defined as, 2Qc u

u =

Figure 1.8 Unconfined Compression Test.


Effect of Shear Strength on Clays and Sands

• Volume changes occur in drained loading • Pore water pressures develop in undrained loading which effect

the effective stress and in turn the shear strength σ

τ

Loose Arrangement- Loose sands- NC clays- Tries to decrease volume- Positive pore water pressure- Decreases effective stress

σ

τ

Dense Arrangement- Dense sands- OC clays- Tries to increase volume- Negative pore water pressure- Increases effective stress

Volume Change (Drained)

∆V

+

-

ε

Loose or NC

Dense or OC

Pore Pressure (Undrained)

∆u

+

-

ε

Loose or NC

Dense or OC

Figure 1.9 Effect of Shear Strength on Clays and Sands.


1.10 Vertical Stress Increase with Depth

• Allowable settlement, usually set by building codes, may control the allowable bearing capacity

• The vertical stress increase with depth must be determined to calculate the amount of settlement that a foundation may undergo

Stress due to a Point Load

• In 1885, Boussinesq developed a mathematical relationship for vertical stress increase with depth inside a homogenous, elastic and isotropic material from point loads as follows:

( ) 25

22z

r1z2

P3

+⋅

⋅=

π

∆σ

• For this solution, material properties such as Poisson’s ratio and modulus of elasticity do not influence the stress increase with depth, i.e. stress increase with depth is a function of geometry only.

Figure 1.10 Stress Increase from a Point Load.


For r/z >1, the Boussinesq solution reduces to approximate expressions:

Stress under a continuous strip load:

Stress due to a Circular Load • The Boussinesq Equation as stated above may be used to derive

a relationship for stress increase below the center of the footing from a flexible circular loaded area:

( )

+−=

− 23

2z2

B11oq∆σ

α δ


Figure 1.11 Stress Increase from a Circular Load.

• The stress increase may also be calculated at a distance, r, from the center of the flexible circular footing using Table 1.6

Table 1.6 Variation of ∆σ/qo for Circular Areas.


Figure 1.12 Variation of stress increase under a circular load


Stress due to a Rectangular Load • The Boussinesq Equation may also be used to derive a

relationship for stress increase below the corner of the footing from a flexible rectangular loaded area:

Figure 1.13 Stress Increase from a Rectangular Load.

The incremental stress may also be expressed using Influence Factors, Iq∆σ o= Where: I = Influence Factor

• The influence factors are complex to calculate but may be graphed in terms of m and n, where:

zBm = &

zLn =


Figure 1.14 Variation of Influence Factor for Stress Increase from a

Rectangular Load.


• The stress increase may also be calculated at other points within the footing by dividing the area into four rectangles, calculating the stress increase from the corner of each rectangle and summing the four values, (SUPERPOSITION)

• The stress increase may also be calculated at the center of the footing using Table 1.7 where:

BLm1 =

=

2Bzn1

coIq∆σ =

Table 1.7 Variation of Ic for Rectangular Areas.


Newmark’s Influence Chart

• The increase in stress under any type of a flexible loaded area can be easily determined using Newmark’s Influence Chart, developed in 1942.

• The graphical form is based on equations of R/z

• The method consists of concentric circles drawn to scale, each square contributes a fraction of the stress

• Most charts contain 200 circles and each square contributes 1/200 (or 0.005) units of stress (influence value, IV)

• Follow the 5 steps to determine the stress increase:

1. Determine the depth, z, where you wish to calculate the stress increase

2. Adopt a scale of z=AB

3. Draw the footing to scale and place the point of interest over the center of the chart

4. Count the number of elements that fall inside the footing, N

5. Calculate the stress increase as: ( ) ( )NIVq∆σ o ⋅=

Example 3.3

A flexible rectangular footing, 5 m by 2.5 m, is located on the ground surface and loaded with qo=250 kPa, determine the stress increase caused by this loading at 2.5 m outside the edge of the footing at a depth of 5 m below grade


Figure 1.15 Newmark’s Chart.


Simplified Methods

• page 215 of your text contains a number of approximate solutions for various loaded areas.

2:1 Method

• The 2:1 method is an approximate method of calculating the apparent “dissipation” of stress with depth by averaging the stress increment onto an increasingly bigger loaded area based on 2V:1H.

• This method assumes that the stress increment is constant across the area (B+z)·(L+z) and equals zero outside this area.

• The method employs simple geometry of an increase in stress proportional to a slope of 2 vertical to 1 horizontal

Figure 1.16 2:1 Method.

• According to the method, the increase in stress is calculated as

follows: ( ) ( )zLzBBLq∆σ o

+⋅+=


1.7 Consolidation Introduction

• Settlement – total amount of settlement • Consolidation – time dependent settlement • Consolidation occurs during the drainage of pore water caused by

excess pore water pressure

Figure 1.16 Spring Analogy for the Consolidation Process.

Settlement Calculations • Settlement is calculated using the change in void ratio

Figure 1.17 Settlement from the Phase Diagram.


• Settlement, oo

He1

∆eS+

=

Where: ∆e = change in void ratio eo = initial void ratio Ho = initial height • Change in void ratio, ( )o'f

'c logσlogσC∆e −=

Where: Cc = compression index • Settlement is dependent on the preconsolidation pressure, p’c or σ’c

Figure 1.18 Settlement from Consolidation Curves.

• Note: the slope of the consolidation curve changes after the preconsolidation pressure which effects the change in void ratio and in turn the total settlement


Example 1.1

Calculate the total settlement by drawing down the water table from surface to a depth of 3 m, i.e. to the upper sand/clay interface


Consolidation Calculations

• Consolidation is calculated using Terzaghi’s one dimensional consolidation theory

• Need to determine the rate of dissipation of excess pore water pressures

Figure 1.19 Drainage of Excess Pore Water Pressure.

• Coefficient of consolidation, wv

v γmKc =

Where: K = hydraulic conductivity

mv = volume coefficient of compressibility

γw = unit weight of water

• Time factor, 2dr

v HtcT =

Where: t = time

Hdr = length of drainage path


• Consolidation ratio, max

t

SSU =

Where: St = settlement at time, t

Smax = maximum settlement

Figure 1.20 Plot of Time Factor versus Consolidation Ratio.


Example 1.2

Calculate the total settlement by placing the 120 kPa load on the clay surface


Calculate the time for 20 cm of consolidation


1.6 Lateral Earth Pressure

Figure 1.21 Active Earth Pressure

σ΄x = KA σ΄z - 2c√KA

Φ+Φ−

=sin1sin1

AK

The Active Case – wall moves away from the soil


Figure 1.22 Active Failure Wedge and Pressure Distribution


Figure 1.23 Active and Passive Earth Pressures


Figure 1.24 Passive Earth Pressure

σ΄x = KP σ΄z + 2c√KP

Φ−Φ+

=sin1sin1

PK


Figure 1.25 Active Failure Wedge and Pressure Distribution


Earth Pressure “at Rest”

KO= (1-sinΦ́) OCRsinΦ́ Consider the physical meaning of this

Example: for a clayey till with γB = 20 kN/m3, Φ ́=33°, c= 5 kPa that extends from surface to depth with WT at 3 m BGS, what is the horizontal effective stress at 3 and 15 m depth if the preconsolidation pressure σ´c (or p´c) is 600 and 800 kPa at 3 and 15 m respectively ?

a) at 3 m BGS σ´z = 60 kPa so OCR= 600/60 = 10

σ´x = (60 kPa) · (1-sin33°) · 10sin33° = 96 kPa

b) at 15 m depth σ´z = 15·20–9.8·12 = 182 kPa so OCR= 800/182 = 4.4

σ´x = (182 kPa) · (1-sin33°) · 4.4sin33° = 185 kPa

1.9 shear strength - national chiao tung...

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