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ENCE 4610ENCE 4610Foundation Analysis and DesignFoundation Analysis and Design

Lecture 3Bearing Capacity of Shallow Foundations:

Eccentric Loading of FoundationsEffect of Water Table and Layered Soils

BrinchBrinch--HansenHansen’’ss Complete Complete FormulaFormula

Similar in basic format to Terzaghi's Method, but takes into account a larger number of factorsSome variations in the way it is implemented

Shape FactorsShape Factors

From Verruijt; Fellenius' factors slightly different

Load Inclination FactorsVerruijt

The problem with load inclination factors such as this is that they require knowing the load on the foundation before it is analysed for bearing capacity. In some cases this is OK, in some it is not.

Fellenius

For this course, the preferred values are those of FelleniusOnly problem is that i

γis

singular for purely cohesive soilsFellenius also gives values for AASHTO code

Allowable Bearing Allowable Bearing CapacityCapacity

Most foundations designed by ASD for geotechnical strength (LRFD to be discussed later)

Foundation is then designed so that the allowable bearing pressure is not exceeded

Fr=

Fq=q=r uult

aa

Factors when considering selection of a factor of safety

Bearing Capacity Example

q = (2)(121) = 242 psfφ = 31 degreesShape Factors

sc = 1.2sq = 1+sin(31) = 1.515sγ

= 0.7

Inclination factors unity for vertical loadUnit load on foundation = (76)/((5)(5)) = 4.75 ksf

Bearing Capacity Example

p = (1)(1.2)(0)(32.671) + (1)(1.515)(0.242)(20.631) + (1)(0.7)(0.5)(0.121)(5)(23.591) = 12.55 ksf

FS = (12.55-.242)/4.75Effective stress at the base

of the foundation is customarily included in this way

Concentric vs. Vertical Loading

Expressing Load Eccentricity and Inclination

Load Divided by Inclination Angle

Total Load QTotal Vertical Load Qv

Total Horizontal Load Qh

Angle of Inclination α = arctan(Qh/Qv)Load can be concentric or eccentric

Load and EccentricityTotal Vertical Load QEccentricity from centroid of foundation eHorizontal Load (if any) not included

Moment and Eccentric Load

Total eccentric vertical Load Q with eccentricity eReplace with concentric vertical load Q and eccentric moment M=Qe

Continuous Foundations

Moments, loads expressed as per unit length of foundation, thus Q/b or M/b

OneOne--Way LoadingWay LoadingOne-way loading is loading along one of the centre axes of the foundationThree cases to considerResultant loads outside the “middle third” result in foundation lift-off and are thus not permitted at all

Equations for OneEquations for One--Way Pressures with Way Pressures with

Eccentric/Moment LoadsEccentric/Moment Loads

If q at any point is less than zero, resultant is outside the middle third

Wf is foundation weight

Example of One Way Example of One Way EccentricityEccentricity

• Giveno Continuous Foundation as

showno Groundwater table at great

deptho Weight of foundation

(concrete) not included in load shown

• Findo Whether resultant force acts

in middle thirdo Minimum and maximum

bearing pressures

Example of One Way Example of One Way EccentricityEccentricity

Compute Weight of FoundationWf/b = (5)(1.5)(150) = 1125 lb/ft

Compute eccentricity

Thus, eccentricity is within the “middle third” of the foundation and foundation can be analysed further without enlargement at this point

e= (M /b)Q /b = 8000

12000+1125 = 0.61 ft .

B6

= 56

= 0.833 ft .>0 .61 ft .

Example of One Way Example of One Way EccentricityEccentricity

Compute minimum and maximum bearing pressures

Two-Way EccentricityEccentricity in both “B”

and “L” directions produces a planar distribution of stress

Kern of Stability– Foundation stable

against overturn only if resultant falls in the kern in the centre of the foundation

– Resultant in the kern if6 eB

B +6eL

L ≤ 1

eB, eL = eccentricity in B, L directions

Bearing Pressure at CornersBearing Pressure at CornersTwoTwo--Way EccentricityWay Eccentricity

• Helpful hint to prevent confusion of eccentricity of finite vs. infinite (continuous) foundations

o Always use one-way eccentricity equations for continuous foundations

o Always use two-way eccentricity equations for finite foundations

o Two-way equations will reduce to one-way equations if one of the eccentricities (eB, eL) is zero

TwoTwo--Way Eccentricity Way Eccentricity ExampleExample

Given◦ Grain silo design as shown◦ Each silo has an empty

weight of 29 MN; can hold up to 110 MN of grain

◦ Weight of mat = 60 MN◦ Silos can be loaded

independently of each other

Find◦ Whether or not eccentricity

will be met with the various loading conditions possible

◦ Eccentricity can be one-way or two-way

12 m

12 m

TwoTwo--Way Eccentricity Way Eccentricity ExampleExample

One-Way Eccentricity

Largest Loading: two adjacent silos full and the rest emptyQ = (4)(29) + 2(110) + 60 = 396 MNM = (2)(110)(12) = 2640 MN-m

e= MQ

e= 2640396

e=6 .67 mB6

= 506

= 8 .33m>6. 67m

Eccentricity OK for one-way eccentricity

TwoTwo--Way Eccentricity Way Eccentricity ExampleExample

Two-Way EccentricityLargest Loading: one silo full and the rest emptyP = (4)(29) + 110 + 60 = 286 MNMB = ML = (110)(12) = 1320 MN-m

eB=eL= MQ

= 1320286

= 4 .62m

6 eB

B+

6eL

L= 2((6)(4 .62)

50 )= 1.11>1

Not acceptable

TwoTwo--Way Eccentricity Way Eccentricity ExampleExample

Two-Way EccentricitySolution to Eccentricity Problem: increase the size of the mat

Necessary to also take other considerations into account (bearing failure, settlement, etc.)

6 eB

B+

6eL

L= 2((6)(4 .62)

B )= 1

B=L=55. 4 m

Equivalent Equivalent Footing Footing

ProcedureProcedure

(NAVFAC DM 7.02)

Structural Geotechnical

Equivalent Footing Using TwoEquivalent Footing Using Two--Way Eccentricity ExampleWay Eccentricity Example

Largest Loading: one silo full and the rest empty

Result of Two-Way Eccentricity AnalysiseB = eL = 4.62 mB = L = 55.4 m (expanded foundation)Equivalent Footing DimensionsB’ = B – 2eB = 55.4 – (2)(4.62)B’ = 45.8 m = L’ (as B = L and eB = eL)

Equivalent Footing Using TwoEquivalent Footing Using Two--Way Eccentricity ExampleWay Eccentricity Example

One-Way Eccentricity

Largest Loading: two adjacent silos full and the rest emptyB = L = 55.4 m (expanded foundation)eB = 6.67meL = 0 mB’ = B’-2eB = 55.4 –(2)(6.67) = 42.1 mL = L’ = 55.4 m B

6= 50

6= 8 .33m>6. 67m

Groundwater and Groundwater and Layered Soil EffectsLayered Soil Effects

• Shallow groundwater affects shear strength in two ways:

o Reduces apparent cohesion that takes place when soils are not saturated; may necessitate reducing the cohesion measured in the laboratory

o Pore water pressure increases; reduces both effective stress and shear strength in the soil (same problem as is experienced with unsupported slopes)

• Layered Soils are virtually unavoidable in real geotechnical situations

• Softer layers below the surface can and do significantly affect both the bearing capacity and settlement of foundations

Solution for Groundwater and Solution for Groundwater and Layered Soil Bearing CapacityLayered Soil Bearing Capacity

• Three ways to analyze layered soil profiles:1. Use the lowest of values of

shear strength, friction angle and unit weight below the foundation. Simplest but most conservative.

2. Use weighted average of these parameters based on relative thicknesses below the foundation. Best balance of conservatism and computational effort.1. Use width of foundation B as

depth for weighted average3. Consider series of trial

surfaces beneath the footing and evaluate the stresses on each surface (similar to slope failure analysis.) Most accurate but calculations are tedious; use only when quality of soil data justify the effort

• Weighted average is, overall, the best way of handling both of these situations

• Groundwater creates additional soil layer

• Valid unless soil strengths have major variations

Example with Layered Example with Layered Soils and GroundwaterSoils and Groundwater

• Giveno Square spread footing as

shown

• Findo Check adequacy

against bearing capacity failure using weighted average method, Brinch-Hansen Formula and FS = 3

2.5 m

Layered Soil ExampleLayered Soil Example• Weighted Values of Soil

Parameterso c' = (0.28+0.33)(5 kPa) +

(0.39)(0) = 3 kPao φ' = (0.28+.33)(32) + (0.39)(38)

= 34ºo γ = (0.33)(18.2) + (0.28)(18.2-

9.8) +(0.39)(20.1-9.8) = 12.4 kN/m3

• Note that submerged unit weights were used for submerged layer

• This avoids the need to then compute an average buoyant weight

• In the event a layer is not submerged, the moist unit weight can be used for that layer and a buoyant weight for the submerged layers

• Failure zone exists for a distance B (1.8 m) below the foundation, i.e., from 1.9 m to 2.7 m below the ground surface (an assumption of the method)

o Unsubmerged silty sand layer is 0.6 m deep

o Submerged silty sand layer is 0.5 m deep

o Fine-to-medium sand layer is 0.7 m in the failure zone

• Weighting factorso Unsubmerged silty sand layer:

0.6/1.8 = 0.33o Submerged silty sand layer:

0.5/1.8 = 0.28o Fine-to-medium sand layer:

0.7/1.8 = 0.39

Questions?