Session 5 6 BEARING CAPACITY OF SHALLOW FOUNDATIONCourse: S0484/Foundation EngineeringYear: 2007Version: 1/0

SHALLOW FOUNDATIONTopic:GeneralTerzaghi ModelMeyerhoff ModelBrinch Hansen ModelInfluence of multi layer soilInfluence of ground water elevationShallow Foundation Bearing by N-SPT value

TYPES OF SHALLOW FOUNDATION

TYPES OF SHALLOW FOUNDATION

TERZAGHI MODELAssumptions:Subsoil below foundation structure is homogenousShallow foundation Df < BContinuous, or strip, footing : 2D caseRough base Equivalent surcharge

TERZAGHI MODELFAILURE ZONES:ACD : TRIANGULAR ZONESADF & CDE : RADIAL SHEAR ZONESAFH & CEG : RANKINE PASSIVE ZONES

TERZAGHI MODEL (GENERAL FAILURE)STRIP FOUNDATIONqult = c.Nc + q.Nq + 0.5..B.NSQUARE FOUNDATIONqult = 1.3.c.Nc + q.Nq + 0.4..B.NCIRCULAR FOUNDATIONqult = 1.3.c.Nc + q.Nq + 0.3..B.N

Where:c = cohesion of soilq = . Df ; Df = the thickness of foundation embedded on subsoil = unit weight of soilB = foundation widthNc, Nq, N = bearing capacity factors

BEARING CAPACITY FACTORSGENERAL FAILURE

BEARING CAPACITY FACTORSGENERAL FAILURE

TERZAGHI MODEL (LOCAL FAILURE)STRIP FOUNDATIONqult = 2/3.c.Nc + q.Nq + 0.5..B.NSQUARE FOUNDATIONqult = 0.867.c.Nc + q.Nq + 0.4..B.NCIRCULAR FOUNDATIONqult = 0.867.c.Nc + q.Nq + 0.3..B.N

= tan-1 (2/3. tan)Where:c = cohesion of soilq = . Df ; Df = the thickness of foundation embedded on subsoil = unit weight of soilB = foundation widthNc, Nq, N = bearing capacity factors

BEARING CAPACITY FACTORSLOCAL FAILURE

BEARING CAPACITY FACTORS

GROUND WATER INFLUENCE

GROUND WATER INFLUENCECASE 10 D1 < Df q = D1.dry + D2 .

CASE 20 d B q = dry.Dfthe value of in third part of equation is replaced with = + (d/B).(dry - )

FACTOR OF SAFETYWhere:qu = gross ultimate bearing capacity of shallow foundationqall = gross allowable bearing capacity of shallow foundationqnet(u) = net ultimate bearing capacity of shallow foundationqall = net allowable bearing capacity of shallow foundationFS = Factor of Safety (FS 3)

NET ALLOWABLE BEARING CAPACITYPROCEDURE:Find the developed cohesion and the angle of friction

Calculate the gross allowable bearing capacity (qall) according to terzaghi equation with cd and d as the shear strength parameters of the soil

Find the net allowable bearing capacity (qall(net)) FSshear = 1.4 1.6Ex.: qall = cd.Nc + q.Nq + .B.NWhere Nc, Nq, N = bearing capacity factor for the friction angle, dqall(net) = qall - q

EXAMPLE PROBLEM A square foundation is 5 ft x 5 ft in plan. The soil supporting the foundation has a friction angle of = 20o and c = 320 lb/ft2. The unit weight of soil, , is 115 lb/ft3. Assume that the depth of the foundation (Df) is 3 ft and the general shear failure occurs in the soil. Determine:- the allowable gross load on the foundation with a factor of safety (FS) of 4.- the net allowable load for the foundation with FSshear = 1.5

EXAMPLE SOLUTION Foundation Type: Square Foundation

EXAMPLE SOLUTION

GENERAL BEARING CAPACITY EQUATIONDfMeyerhofs Theory

BEARING CAPACITY FACTOR

SHAPE, DEPTH AND INCLINATION FACTOR

EXAMPLE 2Determine the size (diameter) circle foundation of tank structure as shown in the following pictureWith P is the load of tank, neglected the weight of foundation and use factor of safety, FS = 3.5.

EXAMPLE 3DETERMINE THE FACTOR OF SAFETY FOR:CASE 1 : GWL LOCATED AT 0.3m (MEASURED FROM THE SURFACE OF SOIL)CASE 2 : GWL LOCATED AT 1.5m (MEASURED FROM THE SURFACE OF SOIL)dry = 13 kN/m3B = 4mSQUARE FOUNDATION

ECCENTRICALLY LOADED FOUNDATIONS

ECCENTRICALLY LOADED FOUNDATIONS

ONE WAY ECCENTRICITYMeyerhofs step by step procedure:Determine the effective dimensions of the foundation as :B = effective width = B 2eL = effective length = LNote:if the eccentricity were in the direction of the length of the foundation, the value of L would be equal to L-2e and the value of B would be B.The smaller of the two dimensions (L and B) is the effective width of the foundationDetermine the ultimate bearing capacity

to determine Fcs, Fqs, Fs use effective length and effective widthto determine Fcd, Fqd, Fd use BThe total ultimate load that the foundation can sustain isQult = qu.B.L ; where BxL = A (effective area)The factor of safety against bearing capacity failure isFS = Qult/QCheck the factor of safety against qmax, or, FS = qu/qmax

EXAMPLE PROBLEM A Square foundation is shown in the following figure. Assume that the one- way load eccentricity e = 0.15m. Determine the ultimate load, Qult

EXAMPLE SOLUTIONWith c = 0, the bearing capacity equation becomes

TWO-WAY ECCENTRICITY

TWO-WAY ECCENTRICITY CASE 1

TWO-WAY ECCENTRICITY CASE 2

TWO-WAY ECCENTRICITY CASE 3

TWO-WAY ECCENTRICITY CASE 4

BEARING CAPACITY OF LAYERED SOILSSTRONGER SOIL UNDERLAIN BY WEAKER SOIL

BEARING CAPACITY OF LAYERED SOILS

BEARING CAPACITY OF LAYERED SOILSRectangular Foundation

BEARING CAPACITY OF LAYERED SOILSSPECIAL CASESTOP LAYER IS STRONG SAND AND BOTTOM LAYER IS SATURATED SOFT CLAY (2 = 0)TOP LAYER IS STRONGER SAND AND BOTTOM LAYER IS WEAKER SAND (c1 = 0 , c2 = 0)TOP LAYER IS STRONGER SATURATED CLAY (1 = 0) AND BOTTOM LAYER IS WEAKER SATURATED CLAY (2 = 0)Find the formula for the above special cases

BEARING CAPACITY FROM N-SPT VALUEA square foundation BxB has to be constructed as shown in the following figure. Assume that = 105 lb/ft3, sat = 118 lb/ft3, Df = 4 ft and D1 = 2 ft. The gross allowable load, Qall, with FS = 3 is 150,000 lb. The field standard penetration resistance, NF values are as follow: Determine the size of the foundation

SOLUTIONCorrection of standard penetration number(Liao and Whitman relationship)

SOLUTION