CE443FoundationDesignDepartmentofCivilandEnvironmentalEngineering,

NewJerseyInstituteofTechnologySpring2015

BearingStressesandElasticSettlement

Settlement Allowablebearingcapacitymaybecontrolledby: Bearingcapacity Settlement

Thestatisticalaccumulationofmovementsinthedirectionofinterestisdefinedassettlement(s).

Where:=strain=q/Esq=changeinstress=f(H,load)Es=modulusH=depthofinfluence

SettlementinSoilsSettlementisadisplacementcausedbyachangeinstress.Insoilswedefinethreedifferenttypes:1. Elastic(immediate)settlement(Se):

Resultsfromlateralmovementsofthesoilinresponsetochangesineffectivestress,anditoccurswithoutnetvolumechange.

Clay Elastic Sand ElasticandPlastic

2. PrimaryConsolidationsettlement(Sc):Mostimportant. Soilissubjectedtoanincreaseineffectivestressandtheindividualparticlesrearrangeintoatighterpacking.

3. SecondaryConsolidationsettlement(Ss):Primarilyduetoparticlereorientation,creep,anddecompositionoforganicmaterials.Itoccursataconstanteffectivestress.

.

Stresses in an elastic medium caused by a point load

IncreaseinStressesduetoFoundationLoads

MethodsofStressCalculation: Approximate

2:1Method

Boussinesq Method AccurateMethods

Chartmethod M&Nmethod

ChangeinStressesduetoFooting

The most common and the method to be used in this class (unless specified) is called 2:1 method.

Stress below foundation spread at a 2:1 ratio. Other areas the increase is zero.

Depending on the soil, one can use other angles.

Example

A 6mX12m footing is loaded with 2kPa load. Find the stress at the corner of the footing at 6m depth.

Hence,

q = 2 kPaX6X12/(6+6)/(12+6)

=0.67kPa

Boussinesq Method

Boussinesq(1885)developedrelationshipstodeterminenormalandshearstressesatanypointinsideahomogeneous,elasticandisotropicmedium.

Stressduetoconcentratedload:

3

2 1

Note:

Example

Equivalent point load at the center of the footing =144kN

z=6m, r=6.7m. Hence,

Notice that R = sqrt(3^2 + 6^2)

.

= 0.252 kPa

A 6mX12m footing is loaded with 2kPa load. Find the stress at the corner of the footing at 6m depth.

Boussinesq Method Stressduetocircularlyloadedarea

Stressesunderthecenter:

11

1 2

Stressesatadistance(r)fromthecenter:Usetable5.1

Boussinesq Method

StressbelowrectangularareaUndercornerofarectangulararea:

3

2

Thus,

For calculatem andn anduseTable5.2

;

Boussinesq MethodStressbelowcornerofrectangulararea

Boussinesq MethodStressbelowcornerofrectangulararea(cont.)

Boussinesq MethodStressbelowrectangularareaUnderanypointunderarectangulararea:1. Divideintosmallerrectangles

eachwithonecornerabovethepointofinterest.

2. Calculateindividuallythestressincreaseduetoeachrectangle.

3. Addorsubtracteachoneofthe

stressincreases.

ChartMethodNewmark(1942)proposedusinginfluencechartsfromwhichverticalstressescanbecomputed.Method: Dividethedimensionsofthefootings(L&B)bythedepthofinterest(H)toobtainB/HandL/Hvalues.

Drawtoscalethefootingusingthescaleinthechart,i.e.,ifB/His1.5thewidthwouldbe1.5timeslengthshowninthechart.

Placetheabovedrawingontopofthechartwiththepointwherestressneededtobecalculatedatthecenterofthechart.

Countthenumberofsegments(M)ofthechartunderthebuilding.

Stressatthepointwillbeequalto: Newmark Chart

ExampleA 6mX12m footing is loaded with 2kPa load. Find the stress at the corner of the footing at 6m depth.

B/H=1

L/H=2

Thus, M=35.5

Hence,

q = 0.005*35.5*2 kPa = 0.355 kPa

ExampleA 6mX12m footing is loaded with 2kPa load. Find the stress at the center of the footing at 6m depth.

B/H=1

L/H=2

Thus, M=96

Hence,

q = 0.005*96*2 kPa = 0.96 kPa

ExampleA 6mX12m footing is loaded with 2kPa load. Find the stress due to the footing at a location 6m away from the out side edges of the footing at 6m depth.

B/H=1

L/H=2

Location / H = 1

Thus, M=2

Hence,

q = 0.005*2*2 kPa = 0.02 kPa

Newmarks MandNMethod

FromTable

Newmarks MandNMethod

I

Example

A 6mX12m footing is loaded with 2kPa load find the stress at the corner of the footing at 6m depth.

M=B/H=1

N=L/H=2

Thus, I=0.200

Hence,

Dq = 0.200*2 kPa = 0.4 kPa

Newmarks MandNMethod:OtherShapes

Example

A 6mX12m footing is loaded with 2kPa load. Find the stress at the center of the footing at 6m depth.

M=3/6=0.5

N=6/6=1

Hence, I=0.120

Thus,

q = (0.12*2 kPa)*4 = 0.96 kPa

ExampleA 6mX12m footing is loaded with 2kPa load (solid green). Find the stress due to the footing at a location 6m away from the out side edges of the footing (black frame) as shown at 6m depth.

Stress calculated based on principle of superposition=

Black frame-blue -yellow+red

O

12m

6m

18m

12m

Example

For z = 6m

M and N for black frame

M=2, N=3 hence I=0.238

M and N for blue

M=1, N=3 hence I=0.203

M and N for yellow

M=1, N=2 hence I=0.200

M and N for red

M=1, N=1 hence I=0.175

Hence q=(0.238-0.203-0.200+0.175)*2 kPa = 0.02 kPa

12m

18m

6m

6m

PressureBulbs

ElasticSettlementforSoils

SettlementandStressDistribution

Flexible Footing

Rigid Footing

Janbus MethodForElasticSettlementofSaturatedClays

Janbu (1956) proposed an equation for calculating the average settlement of flexible footings on a saturated clay with s=0.5

Se = A1A2Bqo/Es

A1andA2Values:

Janbus MethodForElasticSettlementofSaturatedClays(cont.)

se=Bqo(1-s2)av /Es (average for a flexible footing)

se=Bqo(1-s2)r /Es (for a rigid footing)

ElasticSettlement

Usingthetheoryofelasticitywecancalculateelasticsettlementinsoil.

ElasticSettlement: values

ElasticSettlement

For Center of Footing:

=4, N=H/(B/2) and M=L/B, B = B/2

For Corner of Footing:

=1, N=H/B and M=L/B, B=B

With Shape and Depth Factors

H1 1 2

1

ElasticSettlement:IF

ElasticSettlement:I1 andI2

ElasticPropertiesofSoils

Soil Type Poisson's RatioSoft Clay 0.15-0.25Medium Clay 0.2-0.5Silty Sand 0.2-0.4Loose Sand 0.2-0.4Medium Sand 0.25-0.4Dense Sand 0.3-0.45

Soil Type Modulus in MN/m2 Modulus in psi Loose sand 10.5-24 1500-3500 Medium Dense sand 17-27 2500-4000 Dense Sand 34-55 5000-8000 Silty Sand 10-17 1500-2500 Sand/Gravel 69-172 10000-25000 Soft Clay 5-20 600-3000 Medium Stiff Clay 20-40 3000-6000 Stiff Clay 40-95 6000-14000

ExampleA 6mX6m footing located at 3m depth is loaded with 200kPa load. Find the settlement at the corner of the footing, if it is in a medium stiff soil of 18m depth. You may assume Es=10000kPa and Poissons ratio of 0.3

M=L/B=1, N=H/B=3 D/B=0.5 Hence IF=0.775 , I1=0.363 I2=0.048

Hence s=200*6(1-0.3*0.3)*

[0.363+(1-2*.3)/(1-.3)*0.048]0.775/10000

s=33 mm

Alternate

Alpha = 0.85

S =(6*200/10000) *(1-0.3^2)0.85/2= 46mm

Thank you!