CE 443 - 5 Bearing Stresses and Elastic Settlement - 2.pdf
Transcript of CE 443 - 5 Bearing Stresses and Elastic Settlement - 2.pdf
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CE443FoundationDesignDepartmentofCivilandEnvironmentalEngineering,
NewJerseyInstituteofTechnologySpring2015
BearingStressesandElasticSettlement
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Settlement Allowablebearingcapacitymaybecontrolledby: Bearingcapacity Settlement
Thestatisticalaccumulationofmovementsinthedirectionofinterestisdefinedassettlement(s).
Where:=strain=q/Esq=changeinstress=f(H,load)Es=modulusH=depthofinfluence
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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.
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Stresses in an elastic medium caused by a point load
IncreaseinStressesduetoFoundationLoads
MethodsofStressCalculation: Approximate
2:1Method
Boussinesq Method AccurateMethods
Chartmethod M&Nmethod
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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.
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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
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Boussinesq Method
Boussinesq(1885)developedrelationshipstodeterminenormalandshearstressesatanypointinsideahomogeneous,elasticandisotropicmedium.
Stressduetoconcentratedload:
3
2 1
Note:
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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)
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= 0.252 kPa
A 6mX12m footing is loaded with 2kPa load. Find the stress at the corner of the footing at 6m depth.
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Boussinesq Method Stressduetocircularlyloadedarea
Stressesunderthecenter:
11
1 2
Stressesatadistance(r)fromthecenter:Usetable5.1
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Boussinesq Method
StressbelowrectangularareaUndercornerofarectangulararea:
3
2
Thus,
For calculatem andn anduseTable5.2
;
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Boussinesq MethodStressbelowcornerofrectangulararea
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Boussinesq MethodStressbelowcornerofrectangulararea(cont.)
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Boussinesq MethodStressbelowrectangularareaUnderanypointunderarectangulararea:1. Divideintosmallerrectangles
eachwithonecornerabovethepointofinterest.
2. Calculateindividuallythestressincreaseduetoeachrectangle.
3. Addorsubtracteachoneofthe
stressincreases.
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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
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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
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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
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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
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Newmarks MandNMethod
FromTable
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Newmarks MandNMethod
I
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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
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Newmarks MandNMethod:OtherShapes
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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
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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
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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
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PressureBulbs
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ElasticSettlementforSoils
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SettlementandStressDistribution
Flexible Footing
Rigid Footing
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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
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A1andA2Values:
Janbus MethodForElasticSettlementofSaturatedClays(cont.)
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se=Bqo(1-s2)av /Es (average for a flexible footing)
se=Bqo(1-s2)r /Es (for a rigid footing)
ElasticSettlement
Usingthetheoryofelasticitywecancalculateelasticsettlementinsoil.
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ElasticSettlement: values
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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
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ElasticSettlement:IF
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ElasticSettlement:I1 andI2
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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
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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
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Thank you!