undrained shear strength

SCHOOL OF CIVIL & RESOURCE ENGINEERING

GEOMECHANICS

CIVL 2122

PART 2B

COLLAPSE CALCULATIONS USING

UNDRAINED SHEAR STRENGTH

PROFESSOR MARTIN FAHEY SCHOOL OF CIVIL & RESOURCE ENGINEERING

Geomechanics (CIVL 2122) Collapse Calculations

School of Civil & Resource Engineering, The University of Western Australia

Table of Contents

1. Collapse calculations using undrained shear strength....................................................................1 1. Lower bound theorem ....................................................................................................................1 2. Upper bound theorem.....................................................................................................................3 3. Collapse of a footing on clay .........................................................................................................4

3.1 Simple lower bound ...........................................................................................................4 3.2 Simple upper bound ...........................................................................................................5

4. Collapse of a slope (cutting) in clay...............................................................................................6 5. Vane shear test ...............................................................................................................................7

Geomechanics (CIVL 2122) 1 Collapse Calculations


1. Collapse calculations using undrained shear strength

For fine-grained saturated soil (i.e. generally clayey soil), foundation loads are imposed quickly enough for the response to be considered ‘undrained’. We can therefore think of these soils of having (at a particular time) a fixed value of undrained shear strength, su, which is not altered by the application of the load. Generally, the shear strength increases with time after the load is applied, as consolidation occurs and the excess pore pressure dissipates and the effective stress increases. The critical stage with regard to stability is therefore immediately after the load is applied.

This is generally not the case for sands, where the rate of application of the foundation load is not fast enough for excess pore pressures to build up during loading (they drain away as fast as they are generated). Thus, as the foundation load is applied, the sand under the foundation increases in strength, so that the strength changes continually, and what is discussed in this section does not apply.

Applying the load P has no (immediate) effect on strength (because effective stress does

not change) → stability depends on initial in situ strength

Saturated clay soil

P

Applying the load P has no (immediate) effect on strength (because effective stress does

not change) → stability depends on initial in situ strength

Saturated clay soil

P

Applying the load P changes effective stress immediately –

strength changes as the load is applied

Saturated (or dry) sand

P

Applying the load P changes effective stress immediately –

strength changes as the load is applied

Saturated (or dry) sand

P

1. Lower bound theorem

The lower bound theorem allows an estimate of the collapse load to be calculated which, if the rules are followed, we can be sure is either less than, or equal to, the true collapse load (i.e. it’s a ‘lower bound’ to the true collapse load). The rules are:

• Assume (guess) some distribution of stress in the ground due to the load • The distribution does not have to be realistic, but the stresses must be in equilibrium • The maximum shear stress also cannot be greater than the shear strength • The load that is in equilibrium with this stress distribution is a lower bound to the true collapse

load • If the stress distribution guessed happens to be the correct one, the collapse load calculated is

the true collapse load.

Compression and Consolidation 2 Geomechanics (CIVL 2122)


Example: Vertical cutting in clay, with shear strength su. What is the maximum surface stress (q) that can be supported? (Assume the soil is weightless).

Shear strength su

σv

σh

Surface stress = q

Uns

uppo

rted

cut

Shear strength su

σv

σh

Surface stress = q

Uns

uppo

rted

cut

σvσh

τ

τ = suMohr Circle cannot go outside these limits

σv = diameter = 2×su

σσvσh

τ

τ = suMohr Circle cannot go outside these limits

σv = diameter = 2×su

σ

• For equilibrium (with zero stress on the face of the cutting), σh has to be zero.

• The most that σv can be is 2×su (to give a Mohr circle with radius = τ = su).

• For equilibrium, σv = q

• Therefore, a lower bound estimate of the collapse stress is: q = 2×su



2. Upper bound theorem

The upper bound theorem allows an estimate of the collapse load to be calculated which, if the rules are followed, we can be sure is either greater than, or equal to, the true collapse load (i.e. it’s an ‘upper bound’ to the true collapse load). The rules are:

• Assume (guess) some failure mechanism – sliding of rigid plastic blocks • The mechanism must be feasible (possible) – e.g. one block cannot occupy the same volume as

another block • Calculate the work done by the external forces in an increment of movement, and the work

dissipated in plastic work internally. These must be equal. • This gives a collapse load that is an upper bound to the true collapse load • If the failure mechanism guessed happens to be the correct one, the collapse load calculated is

the true collapse load.

Example: Same problem as above. Assume mechanism of failure: Rigid block ABC (at angle 45º) slides into excavation. An increment of vertical movement δ implies an equal increment of horizontal movement δ, and a movement along the sliding plane AB = δ.2 .

Shear strength su

b

bSurface stress = q

A

B

C

δ = vertical movement of top surface

δ

δ.2 AB Length =

Movement vectors: Assume δ is increment of vertical movement

δ.245º

Shear strength su

b

bSurface stress = q

A

B

C

δ = vertical movement of top surface

δ

δ.2 AB Length =

Movement vectors: Assume δ is increment of vertical movement

δ.245º

Calculations are for a unit thickness ‘back into the page’.

Vertical force acting on block = q.b

{ {

{

u

uie

untdisplacemeForce

ui

ntdisplacemeforcee

s.2q.b2.s.b.qWWBut

.b2.s.2b.2.sWplane) failure on shear work (plastic work Internal

b.qWforce by this done work External

=δ=δ⇒=

δ=δ×=

δ×=

43421

In this (rare) case, the upper bound solution = lower bound solution. Therefore, this is the exact solution.

b



3. Collapse of a footing on clay

3.1 Simple lower bound Guess simple stress distribution: Three regions, as shown

Shear strength suUnit weight γ

Stress on strip footing = q

σv

σh

h σv

σh

σv

σh

Region 1 Region 2 Region 1

Stress discontinuities(frictionless in this example)

Strip footing: A footing that is very long ‘back into the page’

σh

σv

σv

σv

σh

σv1 2

σh in 1 must be equal to σh in 2, for equilibrium.

But σv in 1 does not have to be equal to σv in 2.

Shear strength suUnit weight γ


σv

σh

σv

σh

h σv

σh

σv

σh

σv

σh

σv

σh

Region 1 Region 2 Region 1

Stress discontinuities(frictionless in this example)

Strip footing: A footing that is very long ‘back into the page’

σh

σv

σv

σhσh

σv

σv

σv

σh

σv

σv

σhσh

σv1 2

σh in 1 must be equal to σh in 2, for equilibrium.

But σv in 1 does not have to be equal to σv in 2.

Region 1 Region 2

σv1 = γ.h σv2 = γ.h + q

σh1 = ? must be equal σh2 = ?

Major principal stress is horizontal Major principal stress is vertical There has to be equilibrium across the discontinuity, which means that σh must be the same on both sides. Draw Mohr Circles of Stress for both sides of the discontinuity, with σh1 = σh2.

τ

σ

σv2σh1τ = su

σh2σv1

τ = su

Region 1 Region 2

τ

σ

σv2σh1τ = su

σh2σv1

τ = su

Region 1 Region 2

σv1 + 2 × diameters = σv2 ⇒ γh + 2×(2×su) = σv2 = q + γh

⇒ q = 4.su Lower bound estimate of collapse load q.

Even though the stress distribution is very unrealistic, we’ve followed the rules about equilibrium, and not exceeding the shear strength. Therefore, this is a guaranteed lower bound. A better (higher) lower bound would be obtained if we used a better (more realistic) stress distribution.



3.2 Simple upper bound

Shearing resistance τ = su


dθB.dθ

B

Centre of rotation

Mechanism: Assume circular failure surface, with centre of rotation at edge of footing.

Length of failure surface (arc) = π.B

Displacement of any point on the arc = B.dθ

Average displacement of the footing = (B.dθ)/2Shearing resistance τ = su


dθB.dθ

B

Centre of rotation

Mechanism: Assume circular failure surface, with centre of rotation at edge of footing.

Length of failure surface (arc) = π.B

Displacement of any point on the arc = B.dθ

Average displacement of the footing = (B.dθ)/2

{{

{

solution) bound(Upper

:)shear work (plastic work Internal

:footing theon done work External

ntdisplacemeforce

s28.6s2q

d..B.sd2

BqWW

d..B.sd.B.B.sW

d2

Bqd2B.B.qW

uu

2u

2

ie

2uui

2

ntdisplacemeforce

e

=π=⇒

θπ=θ⇒=

θπ=θ×π=

θ=θ=

321

To get a better (i.e. lower) upper bound, search for a mechanism that will give a lower value of the collapse bearing pressure. So, for this example, we have:

Lower bound q = 4.su Upper bound q = 6.28 su.

Therefore, we have bounded the true value: 4.su < q < 6.28.su

If we were not happy to accept this range of uncertainty on the true collapse bearing pressure, we would try to improve both solutions, to make the bounds closer together.

(This problem has an exact solution: The true value of the collapse bearing pressure, also called the ultimate bearing pressure qult) is

qult = (2+π).su = 5.14 su

Example calculation: Strip footing is to be designed to carry a load of 150 kN/m (i.e. 150 kN for each metre ‘back into the page’), and is founded on clay with su = 75 kPa. If the Factor of Safety against bearing failure is required to be 3.0, how wide must the footing be?

wide)m 1.2it make(Probably

pressure) bearing allowable"" theis (q all

m16.1129150

qQB

B.qm/kN150QLoad

kPa1293

3863

qq

kPa3867514.5s14.5q

all

ultall

uult

===⇒

==

===

=×==



4. Collapse of a slope (cutting) in clay

Many slope failures have been observed to be circular in shape (just like the failure under a footing) – i.e. rotation about some point.

This is a case where we can use equilibrium (moment equilibrium) to get a solution. This is very similar to the upper bound method, because we also assume a collapse mechanism, as shown below.

The slope shown is 7 m high and has a slope angle of 45º, the circle radius is 15 m, the shear strength of the clay su is 30 kPa, and the angle θ = 57º, and the soil unit weight is 17 kN/m3.

W

Centre of Gravity of sliding mass

x = 6.17 m

O

θ = 57º

A

B

Clay:Shear strength su = 17 kPaUnit weight γ = 17 kN/m3

Rock

15 m

7 m

Area of sliding mass = 40 m2

(i.e. Volume = 40 m3/m ‘into page’)

τ = su

W

Centre of Gravity of sliding mass

x = 6.17 m

O

θ = 57º

A

B

Clay:Shear strength su = 17 kPaUnit weight γ = 17 kN/m3

Rock

15 m

7 m

Area of sliding mass = 40 m2

(i.e. Volume = 40 m3/m ‘into page’)

τ = su

( )

SAFE) therefore 1.0, ( 47.145566715

MMF

m/m.kN455617.6680Mm/kN6801740VolW

m 6.17 x and /m)m 40 (Vol m 40 mass sliding the of area the diagram, the From

x.WM:failure )driving'(' causing Moment

)page' into' slice m per (i.e. m/kN6715M

180571530.R.s

.RL R.L.s M :failure resisting Moment

failure driving'' Momentfailure resisting Moment

MMF Safety, of Factor

D

Rs

D

32D

R

22u

ABABuR

D

Rs

>===

=×==×=γ×=

===

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛π××=θ=

θ==

==

o

o

Other circles (mechanisms) may be less safe, so must search to find the ‘worst’ case (lowest Fs).



R

rdr

R

rdr

5. Vane shear test A shear vane is a device for measuring the undrained shear strength su of (saturated) clayey soils. The version shown here is a small laboratory version, but larger versions, with long rods, are used in the field. This will be covered in the Site Investigation section of this unit.

The vane section is pushed into the soil such that it is completely buried with the soil. The top is then rotated. The top section is connected to the vane stem via a coil spring, so that as the top is rotated, an increasing torque is applied to the vane stem. The torque is measured on the dial. The torque is increased until the vane rotates within the soil, forcing a cylindrical body of soil to rotate with it. The maximum torque applied is indicated on the dial. Since it is the shear strength of the soil that provides the torsional resistance, it is possible to relate the shear strength to the applied torque, as shown below.

This is an example of an equilibrium calculation, where in this case we are comparing an applied torque and a resisting torque.

Calculate the torsional resistance provided by the soil, at the point where the cylinder of soil just starts to rotate. The shear stress in the soil at this stage is τ = su.

Cylindrical part: ( ) u

2

u

cyl

s2

HD2DsDH

armlever stressshear area surface T

π=××π=

××=

Top and bottom surfaces of the cylinder. Assume that the shear stress is uniform over these surfaces, and equal to su.

Element of thickness dr, area = 2π.r.dr

Rsdr.r.2armlever stressshear area dT

u ××π=××=

( )DH3DT6s

: thatso,6

Ds.s2

HDT,Then

6Ds.TT

12Ds.

24Ds.2

3Rs.2drrs.2T

2u

3

uu

2

3

u.bottop

3

u

3

u

3

u

R

0r

2utop

+π=

π+π

=

π=+

π=π=

π=π= ∫=

Applied torque T

Resistance by the soil, along the surfaces of the cylinder of soil defined by the vane

τ = su

T

Applied torque T

Resistance by the soil, along the surfaces of the cylinder of soil defined by the vane

τ = su

T

undrained shear strength

Documents