Course Year Version S 0484Foundation Engineering 2007 10

Course Year Version : S 0484/Foundation Engineering : 2007 : 1/0 Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

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

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

TERZAGHI MODEL FAILURE ZONES: 1. ACD : TRIANGULAR ZONES 2. ADF & CDE : RADIAL SHEAR ZONES 3. 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 D 1 < Df q = D 1. dry + D 2. ’ • 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: 1. Find the developed cohesion and the angle of friction FSshear = 1. 4 – 1. 6 2. Calculate the gross allowable bearing capacity (qall) according to terzaghi equation with cd and d as the shear strength parameters of the soil Ex. : qall = cd. Nc + q. Nq + ½ . B. N Where Nc, Nq, N = bearing capacity factor for the friction angle, d 3. Find the net allowable bearing capacity (qall(net)) 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 = 20 o and c = 320 lb/ft 2. The unit weight of soil, , is 115 lb/ft 3. 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 P = 73 ton dry = 13 k. N/m 3 sat = 18 k. N/m 3 c = 1 kg/cm 2 = 20 o Tank 2 m Foundation GWL With P is the load of tank, neglected the weight of foundation and use factor of safety, FS = 3. 5.

EXAMPLE 3 SQUARE FOUNDATION B = 4 m dry = 13 k. N/m 3 DETERMINE THE FACTOR OF SAFETY FOR: -CASE 1 : GWL LOCATED AT 0. 3 m (MEASURED FROM THE SURFACE OF SOIL) -CASE 2 : GWL LOCATED AT 1. 5 m (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 – 2 e 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-2 e 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’x. L’ = 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. 15 m. 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 BOTTOM LAYER IS SATURATED SOFT CLAY ( 2 = 0) – TOP LAYER IS STRONGER SAND BOTTOM LAYER IS WEAKER SAND (c 1 = 0 , c 2 = 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 Bx. B has to be constructed as shown in the following figure. Assume that = 105 lb/ft 3, sat = 118 lb/ft 3, Df = 4 ft and D 1 = 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

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

SOLUTION