Lecture notes, lectures 1-5 PDF

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Part 1

Review of Soil Mechanics

1

Phase Relationship Volume

Weight Wa=0

Air

Va

Total unit weight g = W / V Moisture content w = Ww / Ws

Vv W

Ww

Water

Vw

V

Void ratio

e = Vv / Vs Sre = wGs

Ws

Solids

Vs

Specific gravity

Gs ~ 2.7

4

Unit Weights Total unit weight: g, gmoist, gsat (kN/m3) Fill

~ 20

Marine clay ~ 16 Stiff clay

~ 18

Hard clay

~ 20

Concrete

~ 24

Water

gw = 9.81 ~ 10

Buoyant unit weight gb gb = g - gw

5

Relative Density - Dr (%) Compactness

Dr (%)

or compactness of coarse-

Very loose

0 – 15

grained soils such as sand

Loose

15 – 35

Medium dense

35 – 65

Dense

65 – 85

Very dense

85 - 100

This a measure of consistency

and gravel. emax – e Dr = --------------- x 100% emax – emin

Dr= 0%

emax

Dr = ?

e

Dr= 100%

emin 6

Atterberg Limits Liquid Limit (wL) or (LL) WL

Plastic Limit (wP) or (PL) IP WP

Plasticity Index (IP) or (PI) IP = wL – wP or

PI = LL - PL

7

Particle Size Boulder

> 350 mm

Cobble

75 – 350 mm

~ durian

Gravel

4.75 – 75 mm

~ lime

Sand

0.075 – 4.75 mm

Silt/Clay

< 0.075 mm

> basketball

~ coarse grain sugar

< powdered sugar

8

W – well graded

USCS Soil Classification System

G – gravel S – sand M – silt C – clay O - organic

P – poorly graded

M – silty C – clayey L – low plasticity H – high plasticity

Symbols adopted

G – GW, GP, GM, GC, GM-GC S – SW, SP, SM, SC, SM-SC M – ML, MH C – CL, CH O – OL, OH 9

10

U-line A-line

Plasticity Chart for Classification between Clay and Silt 11

Bernoulli's equation total head = pressure head + velocity head + elevation head

hT =

hp

hT = total head

+ hv hp = u/gw

u = water pressure

+

Z

hv = v2/2g

v = velocity

Z = vertical distance with respect to a datum Since v is small in soils Some books use h and hz instead of hT and Z: h = hp + hz

hT = hp + Z Pressure head

12

Elevation head

Darcy's Law Hydraulic gradient

Dh

i = Dh/L Define discharge velocity:

Rate of flow q:

v = q/A

q = kiA

L

v = ki flow q area A

Assumption: Steady-state flow

Find the total head at point D

Example 1 overflow

20 cm

C 25 cm

D 30 cm

10 cm

soil

E 14

Datum 20 cm

C in the "entry" point for water flowing into the soil ZC = 35 cm, hCp = 20 cm, hCt = 55 cm

hCp

20 cm

C ZC

25 cm

D 30 cm

10 cm

soil

E 15

Datum 20 cm

E in the "exit" point for water flowing out of the soil ZE = -20 cm,

hEp = 20 cm,

hEt = 0 cm

20 cm

C 25 cm

D 30 cm

10 cm

hEp

soil

E 16

Datum 20 cm ZE

Point

Z, cm

Hp, cm

hT, cm

C

35

20

55

E

-20

20

0

D

10

20

30

C

55 cm

30 cm

D

30 cm

17

55 cm

E

Flow net and seepage force

Pore Pressure

Du Du Sand

zw

Sand

zw

Sand

zw

Clay

Clay

Clay

Sand

Sand

Sand

u = uh = gw zw

u = uh+Du = gw zw

u = uh- Du = gw zw 19

Total Stress sz Effective stress principle

Sand

g1

z1

g2

z2

sz = s'z + u sz = S gi zi sz = g1 z1 + g2 z2 + g3 z3 + g4 z4

Clay

g3

z3

g4

z4

(Regardless of the location of GWT below ground surface)

Sand

20

Effect Stress s'z Under all conditions g1

g2

Sand

z1

z2

zw Clay

Sand

s'z = sz - u s'z = S gi zi - gwzw s'z = g1 z1 + g2 z2 + g3 z3 + g4 z4 - gw zw Under hydrostatic condition

g3

z3

g4

z4

s'z = S g'i zi s'z = g1 z1 + gb2 z2 + gb3 z3 + gb4 z4 where gbi = gi - gw 21

Compression of a Clay Layer 1. Obtain a soil sample from the ground.

Ds Ds'z DH H

s'zo

2. Consolidate the sample to the insitu s'zo. s'zo

Ds Ds'z Dh

A clay layer is subjected to an applied pressure (Ds Ds Ds'z). How can we determine the amount of compression (DH)?

h

3. Apply additional pressure Ds Ds'z. 4. Measure compression Dh due to Ds Ds'z. DH = (D Dh / h) ∙ H

or

DH = Dez ∙ H

From the test , we obtained mv = Dez / Ds Ds'z Ds'z ∙ H Therefore, DH = mv Ds mv = coefficient of volume compressibility 22

Consolidation Test

ez

e

clay

s'z 23

Compressibility Parameters e-log p’ curve yields Compression Index Cc and Recompression Index Cr e-log p’ curve yields Compression Ratio Cec and Recompression Ratio Cer

Cec = Cc / ( 1 + e)

and

Cer = Cr / ( 1 + e) s'c

Cer Cr

ez

Cec

e Cc

e-log p' curve s'z (Log Scale)

s'z (Log Scale)

24

Preconsolidation Pressure s'c and Overconsolidation Ratio OCR

s'c e

s'c = preconsolidation pressure or maximum past pressure the soil ever experienced OCR = overconsolidation ratio OCR = s'c / s'zo s'zo = effective overburden pressure calculated based on the existing soil profile

log s'z or log p' 25

Settlement in NC Clay Ds Ds'z

s'c = s s'zo dc

H

e

s'zf = s'zo + Ds'z

s'zo

log s'z

26

Settlement in OC Clay – Case 1 s'zo

Ds Ds'z

s'zf = s'zo + Ds'z

dc

s'c

e H

s'zo

log s'z

27

Settlement in OC Clay – Case 2 s'zo

Ds Ds'z dc H

e

s'c s'zf = s'zo + Ds'z

s'zo

log s'z

28

Consolidation theory • Tergaghi’s 1D consolidation theory:

u e  2u e = cv 2 t z

where

z   Due  = f  T , 1 ui  H dr 

cv =

k g w mv

T=

cv t H dr2

where

• Degree of consolidation: U avg =

d c (t ) 100% (d c )ult 29

Summary • Physical properties of soil • Darcy’s law; calculation of total head and pore water pressure; • Effective stress calcualtions • Compressibility and consolidation properties of soil – e – log p’ curve, settlement calculation; rate of consolidation. 30

Part 2

Soil as a Soil as a Continuum

1

Topics • • • •

Stresses and strains in soil Stress-strain relationships Failure of soil Mohr circle

2

Learning outcomes • Understand how the stress-strain behaviour of soil may be adequately described. Understand the limitations of a linear elastic model for soil. • Understand how failure of soil is defined and the use of Mohr circle to analyse stresses acting on a soil element.

3

Stresses and strains in soil

4

Stresses in 2D

5

Strains in soil Disp. In x: u

Equation of compatibility in 2D:

Disp. In z: w 6

Typical stress-strain relationships for soil Fig. 5.3 (a) Typical stress–strain relationship for soil, (b) (b) elastic–perfectly plastic model, (c) rigid–perfectly plastic model, and (d) strain hardening and strain softening elastic–plastic models.

7

Linear Elasticity The stress-strain relationship for CHILE (continuous, homogeneous, isotropic, linear elastic materials) type of material is Hooke’s Law:

(5.5a)

(5.4)

(5.5b)

Here: E is the Young’s Modulus,  is the Poisson’s Ratio, and G is the Shear Modulus. Only two constants are independent.

(5.6) 8

Linear Elastic Stress-Strain Relationship Using Eqs. (5.4), (5.5a&b) and (5.6), we can calculate strains using stresses. The stress-strain relationship in matrix form is:

(5.7)

(5.29)

9

Non-Linear Elasticity We know the stress-strain relationship is not linear. Thus, the Shear modulus G is not a constant, but decreases with increasing shear strain and increases with increasing effective stresses) as shown typically in Fig. 5.4a.

Figure 5.3a

Figure 5.4a

10

Shear Failure in Soils • Soil is a particulate material, so shear failure occurs when stresses between the particles are such that they slide or roll past each other. Hence, the shear strength of soil is mainly controlled by friction. 11

Volume Change

Normal Stress Shear Stress

Shear Failure in Soils

Sliding Block

Soil

Friction is proportional to the normal force N

For soil, t increases with sn

f = m N = N tan f

t = sn tanf f

12

Failure Equation

f = m N = N tan f

t = sn tanf f

Figure 5.5 Friction strength along a plane; (b) strength of an assembly of particles along a plane of slip 13

Failure State σv’

Failure occurs when the Mohr circle tangent to the failure envelope, e.g., circle B.

σh’

σh’

σv’

Mohr’s circle review Normal and shear stresses on a plane

15

Mohr’s circle review Sign conventions – normal stress are (+) – shear stresses causing about center of element are (+) +

+ +

+ +

+ +

+

16

Mohr’s circle review:

Normal and shear stresses on a plane Sign conventions 

–

of plane is (+)

counterclockwise

+ +

17

Mohr’s circle review:

Normal and shear stresses on a plane Areas of sides: Define AEF=A,

then

ABF=Asin( ), AEB=Acos()

18

Mohr’s circle review:

Normal and shear stresses on a plane Sum forces acting on the wedge in the N direction: N



T

19

Sum forces in N direction FN  s n A  s x s  ( A  s)  t xy c  ( A  s)  t xy s  ( A c)  s y c  ( A c)  0 or

s n  s x s 2  s y c 2  2t xy s  c where s  sin( ), c  cos( ) Using the trigonometric identities 1 1 sin 2 ( )  (1  cos(2 )), cos 2 ( )  (1  cos(2 )), and 2 2 2sin( )cos( )  sin(2 ) in the above gives s y  sx s y  sx cos(2 )  t xy sin(2 ) (1) sn   2 2

20

Sum forces in T direction FT  t n A  s x c  ( A  s)  t xy s  ( A  s)  t xy c  ( A c)  s y s  ( A c)  0 or

t n  (s y  s x ) s  c  t xy ( s 2  c 2 ) where s  sin( ), c  cos( ) Using the trig. identities cos 2 ( )  sin 2 ( )  cos(2 ) and 2sin( )cos( )  sin(2 ) s sx tn  y sin(2 )  t xy cos(2 ) (2) 2

gives

21

Mohr’s circle

Squaring and adding Equations (1) and (2) gives 2

2

s y s x   2  s y s x  2 s n   t n     t xy 2 2     or

(3)

s n  a 2 t n2  R 2

This is the equation of a circle in the ( s n ,tn) plane having a center at (a ,0) and radius R, where center a 

s y s x 2

2

,

 s y s x  2  radius R=  t  xy 2   22

Mohr’s circle When the shear stress tn =0, Equation (3) gives sn 

s y  sx 2

2

 s y  sx  2     t xy 2  

(4)

which gives the two values of principal stress, s 1 and s 3 acting on the principal planes

23

Mohr’s circle The stresses (sn ,tn ) on plane EF can be found by rotating from point M by an angle 2 in the CCW direction to reach point Q. sn t n

(plane AD) (plane EF)

(plane AB)

24

Mohr’s circle The stresses (sn ,tn ) on plane EF can be found by rotating from point M by an angle 2 in the CCW direction to reach point Q. sn t n

(plane AD) (plane EF)

(plane AB)

25

Example with principal stresses given

sn t n

Note: this is a special case, where 2 is measured from the normal-stress axis only because s1 acts on the horizontal plane. The previous figure shows the general case. 26

Pole method 1. Identify a point on Mohr’s circle corresponding to known normal and shear stresses on a plane (points M or R in the following figures). 2. Draw a line through that point parallel to the plane on which the known stresses act. 3. The intersection of this line with Mohr’s circle is the Pole or origin of planes (point P). 4. Stresses on any new plane are then given by point Q, which is the intersection point of a line drawn parallel to the new plane and passing through the pole P. 27

Pole method

sn t n

Step 1: Draw a horizontal line going through M (because plane AB is horizontal) 28

Pole method

sn t n

Step 2: Draw a vertical line through point R ( because plane BC is vertical). The intercepting point P is the pole. 29

Pole method

sn t n

Step 3: Draw a line through the pole P parallel to plane EF to find point Q, which gives the stresses on plane EF. sn and tn can be read graphically or determined by trigonometry 30

Example 1: pole method tn

sn

10

31

Summary • The stress-strain behaviour of soil is nonlinear and a linear elastic model is an over-simplification. • The shear strength of soil is mainly controlled by friction. • Mohr circle is used to analyse stresses acting on a soil element.

32

Part 3

Shear Strength of Soil - 1

1

Topics • Mohr-Coulomb Failure Equation • Laboratory shear tests – Direct shear test – Unconfined compression (UC) test – Triaxial tests: • Unconsolidated Undrained (UU) test • Consolidated Undrained (CU) test • Consolidated Drained (CD) test

2

Learning outcomes • Learn the Mohr-Coulomb failure equation in terms of both total stress and effective stresses. • Learn the reasons for different use of strength parameters and the meanings of each strength parameter • Direct shear test an typical soil behaviour in direct shear tests 3

Practical (Stability) Problems

4

Some failure cases related to shear strength of soil

5

Shear Failure in Soils • Soil is a particulate material, so shear failure occurs when stresses between the particles are such that they slide or roll past each other. Hence, the shear strength of soil is mainly controlled by friction. 6

Volume Change

Normal Stress Shear Stress

Shear Failure in Soils

Sliding Block

Soil

Friction is proportional to the normal force N

For soil, t increases with sn

f = m N = N tan f

t = sn tanf f

7

Failure Equation

f = m N = N tan f

t = sn tanf f

Figure 5.5 Friction strength along a plane; (b) strength of an assembly of particles along a plane of slip 8

Mohr-Coulomb Failure Equation • Mohr-Coulomb Failure Equation (Coulomb, 1773; Mohr, 1882):

tf = c + snf tanf (1 (12.2) 2.2) Failure shear stress

cohesion

Friction angle Failure normal stress

It defines the relationship between the shear stress and normal stress along a failure plane. 9

Mohr-coulomb failure envelope M-C failure criteria: t f = c + s  tan f  Coulomb’s linear approximation

Mohr’s curved failure envelope, tf = f(f) 10

Mohr-Coulomb Failure Equation Effective stress analysis (mainly used for sandy soil):

tf = c’ + (snf – u) tanf f’

(12.3)

Total stress (or undrained) analysis (mainly for clayey or silty soil):

tf = cu + snf tanf fu

(12.3a)

shear ar strengt strengthh c’ & f’, or cu (& fu=0) are she pa parameters rameters of soil and need to be det determined ermined by either lab or in in-situ -situ test testss 11

Mohr circle of total stresses & Mohr circle of effective stresses

12

Mohr’s circle and Mohr-coulomb failure envelope Relationship between f and s1 and s3 ?

 = 45o +

f 2

13

Mohr’s circle and Mohr-coulomb failure envelope sin f  =

ad fa

fa = fO + Oa fO = c cot f 

From Mohr's circle:

s  + s 3 Oa = 1 2 s   s 3 ad = 1 2

14

Mohr’s circle and Mohr-coulomb failure envelope s 1  s 3

sin f  =

ad 2 = s1 + s 3 fa + c cot f  2

solving for s 1 :  1 + sin f    cosf    2 + c      1  sin f    1 sin f  

s1 = s 3 

When c’ = 0, then 15

Mohr’s circle and Mohr-coulomb failure envelope One can show that 1 + sin f  2  o f  = tan  45 +   1  sin f 2  and

f  cos f   = tan  45o +  1  sin f  2  Therefore :  

s1 = s 3 tan 2  45o +

f 

 o f   + 2 tan c   45 +  2 2 

16

Example -1 Given the soil profile and strength parameters as shown, check whether point A is at the failure state ( here K = σh’ / σv’).

At point A: sv’ = 17*3+17.5*1.1 – 9.8*1.1= 59.5 kPa sh’ = Ksv’ = 0.54*59.5 = 27.1 kPa

Quick Review • What is M-C Failure Equation

• In the figure below, is soil element A at failure state given c’ = 0 and φ’ = 30o 200 kPa

A

100 kPa 18

Strength tests Laboratory tests for shear strength parameters cu, c and f  1. Direct shear test (DS) 2. Unconfined compression test (UC) 3. Tr...