Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION Course : S0484/Foundation Engineering Year : 2007 Version : 1/0 Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

Influence of multi layer soil Influence of ground water elevation SHALLOW FOUNDATION Topic: General Terzaghi Model Meyerhoff Model Brinch Hansen Model Influence of multi layer soil Influence of ground water elevation Shallow Foundation Bearing by N-SPT value

TYPES OF SHALLOW FOUNDATION

TYPES OF SHALLOW FOUNDATION

Subsoil below foundation structure is homogenous TERZAGHI MODEL Assumptions: Subsoil below foundation structure is homogenous Shallow foundation Df < B Continuous, or strip, footing : 2D case Rough base Equivalent surcharge

TERZAGHI MODEL FAILURE ZONES: ACD : TRIANGULAR ZONES ADF & CDE : RADIAL SHEAR ZONES AFH & CEG : RANKINE PASSIVE ZONES

TERZAGHI MODEL (GENERAL FAILURE) STRIP FOUNDATION qult = c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION qult = 1.3.c.Nc + q.Nq + 0.4..B.N CIRCULAR FOUNDATION qult = 1.3.c.Nc + q.Nq + 0.3..B.N Where: c = cohesion of soil q =  . Df ; Df = the thickness of foundation embedded on subsoil = unit weight of soil B = foundation width Nc, Nq, N = bearing capacity factors

BEARING CAPACITY FACTORS GENERAL FAILURE

BEARING CAPACITY FACTORS GENERAL FAILURE

TERZAGHI MODEL (LOCAL FAILURE) STRIP FOUNDATION qult = 2/3.c.Nc’ + q.Nq’ + 0.5..B.N’ SQUARE FOUNDATION qult = 0.867.c.Nc’ + q.Nq’ + 0.4..B.N’ CIRCULAR FOUNDATION qult = 0.867.c.Nc’ + q.Nq’ + 0.3..B.N’ Where: c = cohesion of soil q =  . Df ; Df = the thickness of foundation embedded on subsoil = unit weight of soil B = foundation width Nc, Nq, N = bearing capacity factors ’ = tan-1 (2/3. tan)

BEARING CAPACITY FACTORS LOCAL FAILURE

BEARING CAPACITY FACTORS

GROUND WATER INFLUENCE

GROUND WATER INFLUENCE CASE 1 0  D1 < Df  q = D1.dry + D2 . ’ CASE 2 0  d  B  q = dry.Df the value of  in third part of equation is replaced with  = ’ + (d/B).(dry - ’)

FACTOR OF SAFETY Where: qu = gross ultimate bearing capacity of shallow foundation qall = gross allowable bearing capacity of shallow foundation qnet(u) = net ultimate bearing capacity of shallow foundation qall = net allowable bearing capacity of shallow foundation FS = Factor of Safety (FS  3)

NET ALLOWABLE BEARING CAPACITY PROCEDURE: 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 – 1.6 Ex.: qall = cd.Nc + q.Nq + ½ .B.N Where Nc, Nq, N = bearing capacity factor for the friction angle, d qall(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 EQUATION Meyerhof’s Theory Df

BEARING CAPACITY FACTOR

SHAPE, DEPTH AND INCLINATION FACTOR

EXAMPLE 2 Determine the size (diameter) circle foundation of tank structure as shown in the following picture 2 m GWL dry = 13 kN/m3 sat = 18 kN/m3 c = 1 kg/cm2  = 20o P = 73 ton Tank Foundation With P is the load of tank, neglected the weight of foundation and use factor of safety, FS = 3.5.

EXAMPLE 3 B = 4m dry = 13 kN/m3 DETERMINE THE FACTOR OF SAFETY FOR: SQUARE FOUNDATION B = 4m dry = 13 kN/m3 DETERMINE 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)

ECCENTRICALLY LOADED FOUNDATIONS

ECCENTRICALLY LOADED FOUNDATIONS

ONE WAY ECCENTRICITY Meyerhof’s step by step procedure: Determine the effective dimensions of the foundation as : B’ = effective width = B – 2e L’ = effective length = L Note: 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 foundation Determine the ultimate bearing capacity to determine Fcs, Fqs, Fs use effective length and effective width to determine Fcd, Fqd, Fd use B The total ultimate load that the foundation can sustain is Qult = qu’.B’.L’ ; where B’xL’ = A’ (effective area) The factor of safety against bearing capacity failure is FS = Qult/Q Check 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 – SOLUTION With 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 SOILS STRONGER SOIL UNDERLAIN BY WEAKER SOIL

BEARING CAPACITY OF LAYERED SOILS

BEARING CAPACITY OF LAYERED SOILS Rectangular Foundation

BEARING CAPACITY OF LAYERED SOILS SPECIAL CASES TOP 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 VALUE A 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

Correction of standard penetration number SOLUTION Correction of standard penetration number (Liao and Whitman relationship)

SOLUTION