5. Bearing Capacity ofShallow Footings

5. Bearing Capacity ofShallow Footings

CIV4249: Foundation Engineering

Monash University

CIV4249: Foundation Engineering

Monash University

Bearing CapacityBearing Capacity

• Ultimate or serviceability limit state?• “What is the maximum pressure which the

soils can withstand for a given foundation before the soil will fail?”

• Design for less but how much less?• Uncertainty with respect to:

– Loads– Capacity

Limit State DesignLimit State Design• Limit state design equation:

y F < f R

• F = action (kN or kPa)• y = load factor

– (AS1170) - Loading Code– Dead Load = 1.25– Live Load = 1.50– Hydrostatic = 1.00

• Typical value– 2/3 dead + 1/3 live– y = 1.33

• R = capacity (kN or kPa)• f = capacity redn factor

– (AS2159) - Piling Code– Static test = 0.70 to 0.90– CPT design = 0.45 to 0.65– SPT design = 0.40 to 0.55

• Why a range?– variability in site conditions

and in quality or quantity of exploration

Factor of SafetyFactor of Safety

• Working or Allowable stress method is currently used in practice

• No Australian Standard• By convention Factor of Safety = 2.5 to 3.0• qallow = qult ¸ FoS• I want you to apply limit state design principles• Equivalent “Factor of safety” =

y / f• For y = 1.33 implies f =

0.44 to 0.53

Geotechnical DesignGeotechnical Design

• Generally working with stresses• On LHS concerned only with that applied

stress which acts to cause rupture• On RHS concerned with the available

strength which acts to prevent rupture

y F < f R

• Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m• What is the applied stress in these two situations?

Applied Stress, FApplied Stress, F

1.2m1.0m

Net Applied Stress, FNet Applied Stress, F

• Fwall = 120 kN/m : Wwall = 20 kN/m : Wfoot = 10 kN/m : g = 20 kN/m3

• What is the net applied stress in these two situations?

1.2m1.0m

qnet = 125 kPa qnet = 105 kPa

Net Applied Stress RuleNet Applied Stress Rule

• For bearing capacity:

qnet applied = s 'below - s 'beside

ALWAYS WORK WITH NET APPLIED STRESSES

NEVER WORK WITH GROSS APPLIED STRESSES

Available Strength, RAvailable Strength, R

• Methods that can be used to determine available strength:1. Historical / experience :

• Building Codes may specify allowable values in particular formations

2. Field loading tests• Plate loading tests for very large projects

3. Analytical solutions• Upper and lower bound solutions for special cases

4. Approximate solutions• Solutions for general cases

Field (Plate) Loading TestsField (Plate) Loading Tests

0.3m

1.2m

+ -• Testing footing under

actual soil conditions• Measure load-deflection

behaviour

• Expensive mobilization and testing

• Need to apply scaling laws• Different zone of influence• Affected by fabric -

– fissuring, partings etc.

Analytical SolutionsAnalytical Solutions

• The failure of real soils with weight, cohesion and friction is a complex phenomenon, not amenable to simple theoretical solutions.

• If simplifying assumptions are made, it is possible to develop particular analytical solutions.

• These analytical solutions must be based either on principles of equilibrium or kinematic admissibility.

• “If an equilibrium distribution of stresses can be found which balances the applied load, and nowhere violates the yield criterion, the soil mass will not fail or will be just at the point of failure” - i.e. it will be a lower-bound estimate of capacity.

Lower Bound SolutionLower Bound Solution

1 2Weightlesssoil f = 0

qu

0 2c 2c 4c

= 4c

Upper Bound SolutionUpper Bound Solution

• “If a solution is kinematically admissible and simultaneously satisfies equilibrium considerations, failure must result - i.e. it will be an upper-bound estimate of capacity.”

Weightlesssoil f = 0

qu

r

O

c

c

qu r. r/2 = p r.c.r

= 2pc

e.g. slope stability - optimize failure surface; choose FoS

Other classic analytical solutions for weightless soils:

Other classic analytical solutions for weightless soils:• Solutions with f = 0 :

– Prandtl smooth punch : qu = 5.14c

– Prandtl rough punch : qu = 5.7c

• Solutions with f ¹ 0 :– Rough punch

passive active

log spiral

Solutions for real soilsSolutions for real soils

• There is no rigorous mathematical solution for a soil which contains cohesion, c, and angle of friction, f, and weight, g.

• Empirical or numerical approaches must be used to provide methods of estimating bearing capacity in practical situations.

• Numerical approaches include finite element and boundary element methods and would rarely be used in practice*

Terzaghi Approximate AnalysisTerzaghi Approximate Analysis

• Solution for soil with c, f, g and D > 0• Solution is based on superposition of 3

separate analytical cases:– Soil with f and g but c = D = 0 : qu = Ng.f(g)

– Soil with f and D but c = g = 0 : qu = Nq f(D)

– Soil with f and c but g = D = 0 : qu = Nc f(c)

• Each case has a different failure surface, so superposition is not theoretically valid.

Terzaghi Bearing EquationTerzaghi Bearing Equation

Solution for c and only soil

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N

Solution for D and only soil

Solution for and only soil

Terzaghi Bearing EquationTerzaghi Bearing Equation

B

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N

p'o = 'o D

Generalized soil strength : c, (drainage as applicable)

Soil unit weight : ' (total oreffective as applicable)

Overburden

Failure Zone (depth 2B)

Adopt weighted average values !

Terzaghi Bearing EquationTerzaghi Bearing Equation

– applies to strip footing– Nc, Nq and N are functions of f, and are usually

given in graphical form – c, f and g' refer to soil properties in the failure

zone below the footing– p'o is the effective overburden pressure at the

founding level– shear strength contribution above footing level

is ignored : conservative for deeper footings

qu nett = c.Nc + p'o (Nq - 1) + 0.5B'N

Application to other than strip footings

Application to other than strip footings

• Strip footings represent a plane-strain case

• What is different for a rectangular footing?• Correction factors applied - e.g. Schultz:

– Nc multiplier is (1+ 0.2B/L)

Example #1Example #1

Stiff Clay : cu = 75 kPa fu = 0o g = 18 kN/m3

1.0

1.7 x 2.3

• Shape Factor =• c =• Nc =

• Nq =

• Ng =

• Qu nett =

• / Qu nett =

• (1 + 0.2*1.7/2.3) = 1.148• 75 kPa• 5.7• 1.0• undefined• 1.148*75*5.7*1.7*2.3 = 1919 kN• 0.45 * 1919/1.33 = 649 kN = 166 kPa

Example #2Example #2

Medium Sand : c = 0 kPa f' = 35o g = 20 kN/m3

1.0

1.7 x 2.3

• c =

• p'o =

• Nq =

• g' =• Ng =

• Qu nett =

• /f y Qu nett =

• 0 kPa

• 1.0*20 = 20 kPa• 40

• 10.2kN/m3

• 40• [20*(40-1)+0.5*0.852*1.7*10.2*40]*1.7*2.3 = 4205 kN• 0.45 * 4205/1.33 = 1422 kN = 364 kPa

qmin

qmax

e

P

e < B/6 :

qmin = P (1-6e/B)/BL

qmax = P (1+6e/B)/BL

rigid

Footings with eccentric loadsFootings with eccentric loads

qmin = 0qmax = 4P . 3L(B-2e)

qmin

qmax

e

P

e > B/6 : rigid

Footings with eccentric loadsFootings with eccentric loads

Meyerhof Method for eccentric loads

Meyerhof Method for eccentric loads

P

e

L

2e L' = L- 2e

B

Meyerhof Method for eccentric loads

Meyerhof Method for eccentric loads

0.00

0.50

1.00

1.50

2.00

2.50

0.00 0.05 0.10 0.15 0.20 e/B

PB

/L

q(min)

Meyerhof

q(max)

average

2-way eccentricity2-way eccentricity

P

e1

L

2e1 L' = L- 2e1

B e2

2e2

B' =

B-

2e2

Footings with momentsFootings with moments

P

M

eP

e = M Ptreat as equivalent eccentric load

Equivalent footing exampleEquivalent footing exampleLight tower

5.3x5.3 m

Vertical Load = 500 kNEquiv Horizontal Load = 30 kN @ 13m above baseDetermine:

a) Maximum and minimum stresses under the footingb) Equivalent footing dimensions

• Effective eccentricity =• e/B = • min =

• max =

• Effective area =

Light tower

5.3x5.3 m

• 30*13/500 = 0.78m• 0.78/5.3 = 0.147 < 0.166B• 500*(1- 6*0.147)/5.32 = 2.1 kPa• 500*(1+ 6*0.147)/5.32 = 33.5 kPa• 5.3 * (5.3 - 2*0.78) = 5.3 * 3.74m

Equivalent footing exampleEquivalent footing example

Inclined LoadsInclined Loads

• Correction Factors, Fc , Fq and F g empirically determined from experiments

Fc = Fq = (1 - d / 90)2

Fg = (1 - d / f)2

Meyerhof Approx AnalysisMeyerhof Approx Analysis

• differs from Terzaghi analysis particularly for buried footings– soil above footing base provides not only

surcharge but also strength– more realistic i.e. less conservative

qu = cNcscdcic + qNqsqdqiq + 0.5g'BNgsgdgig

• s, d, and i are shape, depth and load inclination factors

Analyses by Hansen, VesicAnalyses by Hansen, Vesic

qu = cNcscdcicgcbc + qNqsqdqiqgqbq + 0.5g'BNgsgdgigggbg

Nc ,Nq ,Ng : Meyerhof bearing capacity factors

sc ,sq ,sg : shape factors

dc ,dq ,dg : depth factors

ic ,iq ,ig : load inclination factors

gc ,gq ,gg : ground inclination factors

bc ,bq ,bg : base inclination factors

Example 4 - Bearing capacity after Hansen

Example 4 - Bearing capacity after Hansen

1.00.6

1.7 x 2.3

Medium sand : f' = 34o g = 20 kN/m3

Grading dense : f' = 40o g = 21.5 kN/m3

1.5

Load inclination = 10oGround inclination = 3.5o

Firm Clay : cu = 40 kPa fu = 0o g = 17 kN/m3

Determine the ultimate bearing capacity (in kN)

• 0.831 (1 = 2)

• 0.674 (2 = 3)

• 0.80• 0.80• 1.00• 1.00• 17*0.6+7*0.4 = 13•

1.5*10.2+1.9*11.7/3.4 = 11.0 kN/m3

• =

• Nc =

• Nq =

• N =

• sq =

• s =

• dq =

• d =

Example 4 - Bearing capacity after Hansen

Example 4 - Bearing capacity after Hansen

• iq =

• i =

• gq =

• g =

• bq =

• b =

• q =• =

Qu = 1.7*2.3*(664+168) = 3250 kN

• 1.5*34+1.9*40/3.4 = 37o

• 0• 42.9• 47.4• 1.445• 0.704• 1.239• 1.00

Stratified Deposits - 1Stratified Deposits - 1

soft clay

stiff clay ordense sand

B

2Btrendingstronger

• For uniform soils, zone of influence typically ~ 2B

• Failure surface will tend to be more shallow

• Ignore strength increase?• Place footing deeper?• Take strength of underlying

stiffer material into account• Approaches based on taking

weighted average strength• See Bowles, Das or other text

Stratified Deposits - 2Stratified Deposits - 2

• Ignoring underlying layer unconservative

• compute load spread and analyze as larger footing with reduced stress on underlying soil

• use parameters of underlying soil in bearing equation

• again, look at texts for different approaches

soft clay

dense sand

TerminologyTerminology

• Ultimate Bearing Pressure, qu

– as computed by any number of methods

• Maximum Safe Bearing Pressure, qs

– qs = qu ¸ FoS

• Allowable Bearing Pressure, qa

– take settlement into consideration : qa £ qs

• Design Pressure, qd

– construction practicalities/ standardization may dictate larger footings : qd £ qa