shallow foundations 2006

7/29/2019 Shallow Foundations 2006

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DIVISION OF INFORMATION TECHNOLOGY, ENGINEERING AND THE ENVIRONMENT 1

Bearing Capacity- shallow foundation systems

D. A. Cameron

Rock and Soil Mechanics 2006


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TERMS

Thefoundation of a structure is the earth

upon which the structure is supported

structural loads are transferred to the soil

via the footing

Reinforced (?) concrete strip footings & pads

take column (point) loads & distributed

loading (line loads)


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Foundation requirements

Capacity = ultimate

bearing capacity

qu Soil mass at point of

failing

SAFE bearing

capacity

qsafe FoS applied to qu

ALLOWABLE bearing

pressure

qa Allowablesettlementmay dictate

1. Safe bearing capacity orstrength2. Allowable bearing pressure


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Presumptive qsafe values?

NoWTwithin depth, D = B, footing breadth

Dense gravel > 600 kPa

Loose gravel < 200 kPa

Stiff clays 150 - 300 kPa

Soft clays < 75 kPa


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Modes of Failureof ShallowFoundations

General shear

Punching shear

Local shear

Fairly

incompressible soil

Soft, very

compressible soil

Intermediate soil

condition


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Bearing Capacity Analysis

Solution is usually based on:

a) General bearing capacity failure

b) Strip footing - width, B, length L = ,founded at depth, D

c) Soil - homogeneous & isotropic,

unit weight,

d) Soil Strength - parameters c and


7/62DIVISION OF INFORMATION TECHNOLOGY, ENGINEERING AND THE ENVIRONMENT 7

SOIL

Strain

Stres

s

SURFACE FOOTING, B x L =

qaverage = Q/B per m length

Soil strength,cand

Force, Q

BRigid-plasticsoil behaviour



Choice of Strengths

SAND Clean sand

c = 0 = c

, ,

not affected by

inundation

BUT lower qu!

CLAY Saturated, NC

undrained loading?

u = 0

usually more critical

drained loading?

c ,



B

Q

The captured, driven soil wedge

Circular for u = 0

MECHANISM


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Simple upper bound foru = 0 soils

Centre ofrotation?

B

Radius, B

qu


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Solution

Take moments about the centre

Disturbing moment = restoring moment

quB(0.5B) = (cuB)B

qu = (2)cu

qu = 6.3 cu


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Simple lower bound foru = 0 soils

B45 45

Active

wedge

Passive

wedge

Ha Hp

Ka = Kp = 1


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Solution

Shear stress

Active statePassive state

Principal stress

= HaHp qu

cu

0

qu = 4cu


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The Correct Answer

Lower bound estimate (safe)

qu = 4cu

The correct answer:

qu = 5.14cu

Upper bound estimate (unsafe)

qu = 6.3cu


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The Accepted Failure Mechanism

1. Active wedge under footing

wedge at (45 + /2)

2. Rotational zone ofplasticity

(radial shear zone) log spiral

shear stresses on either side of the radialshear zone = fn(spiral)

3. Passive wedges to either side of footing wedges at (45 - /2)

resist soil rotation


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45 - /2

Mechanismthe sameboth sides

Rotational or radial shear

zone soil plastic

Log spiral

A

P

45 + /2


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Rotational or radial shear

zone soil is plastic

Log spiral

P

45 + /2

P

45 - /2


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Log Spiraldictates change of shear in theradial shear zone

tan2

2

1 e

2

1

Normal forces offset by friction,

no moments about centre


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Comment

Details of the geometry of thefailure mechanism are really only of

any practical purpose for

consideration of the influence ofadjacent or underground works, or

the influence of soil profile changes

N.B. = 0 soils have smallest

mechanism


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The General Bearing

Capacity Equation

Ultimate bearing capacity for a strip footing,

subjected to vertical load

other situations handled by empiricalexpressions

qocuu N2

BNqNc1B

Qq


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Considers 3 cases:

1. Surface footing, soil, with cohesion, but no friction

2. Surface footing, soil friction and soil

3. Burial (surcharge & extra shear resistance)

Contributions are simply summed

an approximation,

backed up by experience and experimentation

FORMULATIONLimit Equilibrium Analysis


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Bearing Capacity Factors

Nc, Nq & N are the bearing

capacity factors corresponding to:

1. Cohesion case

2. Surcharge or burial case

3. Self weight of soil case

Nc, NqandNare all functions of


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The Bearing Capacity Factors, Nc, Nq

0

20

40

60

80

100

120

140

160

0 10 20 30 40 50

Angle of Friction (kPa)

Bearing

CapacityFactor

Nc

Nq


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0

5

10

15

20

25

30

0 5 10 15 20 25 30


Bearing

Cap

acityFactor

Nc Nq

5.14

1.0

NC clay,saturated,short termloading


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The Bearing Capacity Factor, N

0

50

100

150

200

250

0 10 20 30 40 50


Bearing

CapacityFactor,

N

rough

smooth

Hansen


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0

2

4

6

8

10

12

14

16

0 5 10 15 20 25 30


B

earing

Capac

ityFactor,

N

rough

smooth

0.0


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qocu N

2

BNqNcq

Cohesion term

Surface

surcharge term

Soil self-weight term

The Bearing Capacity Equation

- for a long strip footing


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Variations in the Bearing Capacity

Factors between methods

1) Hansen [or Brinch-Hansen] analyses

generally accepted as most accurate

2) Terzaghi, the pioneer, misconstrued Nc

3) Meyerhofthe 2nd best of these 3


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Effects of Soil Properties

Bearing capacity = fn(, , B, c and qo)

has the greatest influence

Both the 2nd and 3rd terms in the equation

depend on

if water is above the bottom of the footing,the surcharge weight is also affected


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THE SURCHARGE TERM- most footings are buried

1

qo = 1D

D


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Effect of Footing Size

Last term with N has B orfooting breadth

Wider footing, greater bearing capacity

BUT for = 0 soil, B has little effect

N = 0! for = 0


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Factor of Safety, qu to qsafe?

A Factor of Safety of 3?

oqoc

safe qSofF

NB5.01)(NqNcq

= [nett ultimate bearing capacity (FoS)] + surcharge

= [qu nett (FoS)] + qo


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The General Bearing Capacity Eqn.

Considers soil rigidity, (r)

footing shape, (s)

depth of embedment, (d)

Inclined load, (i) base inclination, (b)

Ground inclination, (g)

qoqqqqqqcccccccu N2B

gbidsrNqgbidsrNcgbidsrq

Note: shape factors not used with inclination factors?


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Effect of depth, D ( u = 0)

D/B Brinch -Hansen

0 5.14cu + qo

1 7.2cu + qo

8.4cu + qo


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Effect of depth, D ( u = 0)

D = 0qu surface

D = B

D >> B

1.4qu surface

1.63qu surface


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The Influence of a WT

If zw

within 1.5D, assume at underside

of footing and use in self weight termBuoyancy if too high?

zw1


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Footing Shapeshape factors B=L

L/B 1 2 5sc 1.19 1.10 1.04

= 0 sq 1.00 1.00 1.00

s 0.6 0.8 0.92

sc 1.61 1.31 1.12

= 30 sq 1.58 1.29 1.12

s 0.6 0.8 0.92

sc 2.01 1.50 1.20= 45 sq 2.00 1.50 1.20

s 0.6 0.8 0.92


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Breduced = B'

Plan

l

Q

e

Eccentricity of Loading

e2BBBeffective


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EXAMPLE 8.7From Barnes, (changed slightly)

A long, reinforced concrete, retaining wall is to be founded at 1.5 m

depth below ground level in a clay with the water table at 1.5 m

below ground level. The width of the footing is 2.5 m and the base

is 1.5 m below ground level. The thickness of the footing is 0.5 m.

The top 1 m of excavation is to be backfilled.

A vertical line load of 90 kN/m is located 0.45 m off the centreline of

the footing.

If c = 4 kPa, = 23 and = 22 kN/m3; cu = 40 kPa and u = 0 ;

concrete = 25 kN/m3 and backfill = 21 kN/m3, then find the factor ofsafety against bearing capacity failure for both short term and long

term conditions

DO NOT ignore depth factors


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Problem from Notes

A column carries 900 kN.

The foundation soil is dry sand, 18 kN/m3, = 40.A minimum factor of safety of 2.5 is required.

FIND the size of :

a) A square footing if it is placed at the ground surfaceb) A rectangular surface footing L/B = 2

c) A square footing if it is placed 1 m below the surface.

d) A square footing, 1 m below the surface - the water

table is expected to rise to the underside of thefooting. Below the water table, the unit weight is 21kN/m3


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ANSWER (a)

e = 0, so, q = 900/B2 kPa

qu req = 2.5 x 900/B2 = 2250/B2

BUT for dry sand, AND a surface footingqu = 0.5(s)BN

2250/B2 = 0.5(s)18BN

250 = (s)B3N = 0.6 x 85.8 B3

SOLVE: B = 1.69 m


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ANSWER (c)

q = 900/B2 kPa

qu req = 2.5 x 900/B2 - sqdqqo = 2250/B

2 - 18sqdq

BUT for dry sand, AND a square footing at 1 m,

qu = sqdqqoNq + 0.5(s)(d)BN

2250/B2 18(1.84)1.16= 1.84 (1.16)(18)64.2 +

0.5(0.6)(1)18B(85.8)

2250/B2

38.4 = 2466 + 463.3B

0 = 0.206B3 + 1.11B2 - 1

SOLVE: B = 0.88 m assumed B = 1 m for depth factors


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ANSWER (b)

q = 900/B2 kPa

qu req = 2.5 x 900/B2 = 2250/B2 - surface footing

BUT for dry sand, and L/B = 2,

qu = 0.5(s)BN

2250/B2 = 0.5(0.8)18B(85.8)

3.642 = B3

SOLVE: B = 1.54 m


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ANSWER (d)

q = 900/B2 kPa

qu req = 2.5 x 900/B2 - sqdqqo = 2250/B

2 - 18sqdq

BUT for wet sand, AND a square footing at 1 m,

qu = sqdqqoNq + 0.5(s)(d) BN

2250/B2 18(1.84)1.16= 1.84 (1.16)(18)64.2 +

0.5(0.6)(1)(11.2)B(85.8)

2250/B2

38.4 = 2466 + 288.2B

0 = 0.128B3 + 1.11B2 - 1

SOLVE: B = 0.90 m assumed B = 1 m for depth factors


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Q

DB

a

Changed shearmechanism

Shallower,

longer?

InclinedLoading

qo D

c , , Q/(BL) =qu/(cosa)


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Inclined Load correction factors(Meyerhof: guide onlyuse Hansen)

FACTOR DEPTH OF

FOOTING

INCLINATION OF LOAD FROM

VERTICAL

D 0 10 20 30

N 0 1.0 0.5 0.2 0

N B 1.0 0.6 0.4 0.25

Nc 0 to B 1.0 0.8 0.6 0.4


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DEFINITION of nfor inclined loading

Inclination of Load

correction factors, ic, iq, i

B

L Q

PLAN

n

Q

B

ELEVATION

a


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Inclined Loading

Has a tremendous influence

Cannot use shape factors (strip solution)

The solution is always for the VERTICAL

component only of the force, Q

Q = [N2 + T2] and qu = N/A

MUST design against sliding arising from

the HORIZONTAL component of force

i.e. T = N(tan )


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CLASS EXAMPLE

A footing 2 m wide by 4 m long is to be placed on adense layer of sand, overlain by poorly compacted fill

(fill = 16 kN/m3) to a depth of 2 m. The sand has the

properties c = 0 kPa, = 38 and = 20 kN/m3.

There is no water table near the footing.

Check the ultimate bearing capacity of the footing for

a vertical central load

a central load inclined at 10 to the vertical (in plan,the load is parallel to the short side or breadth, B, of

the footing)


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Non-Homogeneous Soil

- soil profiles

Approximations can apply, SO LONG AS the

general mechanism of instability remains

NOT THE CASE for very soft clay, in

thin layer, over hard soil

Toothpaste tube?

NOR strong thin layer over weak layer

Punching shear


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Non-Homogeneous Soil?

Stronger layer

Weaker Layer

21

Load spreadingtechnique

Load, QI l l


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Footing, B x L

Depth to layer, DL

New footing area

(B+DL)(L+DL)

Pressure, qL = Q/area

Check qa2

Ensure lower layer not

overstressed by qL

- reduce qa as required

Ignore lower layerand check surface qa


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Example

Returning to the previous question, had

the footing been a surface footing and

the poorly compacted fill had the

properties c = 5 kPa, = 25, compute

the ultimate bearing capacity under

vertical loading .


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Alternative treatment ifu = 0 soils

Layer 1,

cu1

Layer 2 ,

cu2

Centre of

rotation?

B

Radius, B?

O h F


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Other Factors

a) Adjacent footingssuitably spacedto reduce interaction (refer to failure

mechanisms)

b) Rate of loading appropriate

strength parameters?

c) Inclined slopes adjacent to footings

influence on slope stability?


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SUMMARY

Bearing capacity is determined by limit

equilibrium methods

Complexity requires semi-empirical

solutions

Endnote: some new developments willimprove these methods


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GROUND COVERED

1) types of failure

2) general bearing capacity analysis

3) correction factors for depth,shape, inclination

4) influence of eccentric load

5) dealing with soil profiles


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BEARING CAPACITY ON ROCKS

Rock can be stronger than a concrete footing

BUT consideration of rock mass strength may

be necessary

closed joints frequently spaced, can treat as

Bell-Terzaghi bearing problem

ALSO defects and their orientation may impact

upon performance

R k M ??


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Rock Mass??

Close, but open vertical joints

B

S

Unconfined compressivestrength of rock rules!

R k M ??


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Rock Mass??

Widely spaced joints

B

S

Tension splitting of rock slab?Bishnois theoretical solution

Refer: Sowers G F, Introductory Soil Mechanicsand Foundations

Oth C id ti


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Other Considerations

Rigid

Weak

Flexural failure

Punching failure

H

Weak

Rigid, but thinner

H


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KEY POINT

Mechanisms of bearing capacity

failure of rock masses are usually

quite different from soil

shallow foundations 2006

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