SHALLOW FOUNDATION Session 5 – 10 Course : S0825/Foundation Engineering Year : 2009 SHALLOW FOUNDATION Session 5 – 10

SHALLOW FOUNDATION Topic: BEARING CAPACITY SETTLEMENT General Terzaghi Model Meyerhoff Model Influence of ground water elevation Influence of multi layer soil Shallow Foundation Bearing by N-SPT value SETTLEMENT Immediate Settlement Consolidation Settlement Bina Nusantara

SHALLOW FOUNDATION SESSION 5 – 6 Bina Nusantara

TYPES OF SHALLOW FOUNDATION Bina Nusantara

TYPES OF SHALLOW FOUNDATION Bina Nusantara

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

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

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 Bina Nusantara

BEARING CAPACITY FACTORS GENERAL FAILURE Bina Nusantara

BEARING CAPACITY FACTORS GENERAL FAILURE Bina Nusantara

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 Bina Nusantara ’ = tan-1 (2/3. tan)

BEARING CAPACITY FACTORS LOCAL FAILURE Bina Nusantara

BEARING CAPACITY FACTORS Bina Nusantara

GROUND WATER INFLUENCE Bina Nusantara

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 - ’) Bina Nusantara

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) Bina Nusantara

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 Bina Nusantara

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 Bina Nusantara

EXAMPLE – SOLUTION Foundation Type: Square Foundation Bina Nusantara

EXAMPLE – SOLUTION Bina Nusantara

SHALLOW FOUNDATION SESSION 7 – 8 Bina Nusantara

GENERAL BEARING CAPACITY EQUATION Meyerhof’s Theory Df Bina Nusantara

BEARING CAPACITY FACTOR Bina Nusantara

SHAPE, DEPTH AND INCLINATION FACTOR Bina Nusantara

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. Bina Nusantara

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

ECCENTRICALLY LOADED FOUNDATIONS Bina Nusantara

ECCENTRICALLY LOADED FOUNDATIONS Bina Nusantara

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 Bina Nusantara

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 Bina Nusantara

EXAMPLE – SOLUTION With c = 0, the bearing capacity equation becomes Bina Nusantara

TWO-WAY ECCENTRICITY Bina Nusantara

TWO-WAY ECCENTRICITY – CASE 1 Bina Nusantara

TWO-WAY ECCENTRICITY – CASE 2 Bina Nusantara

TWO-WAY ECCENTRICITY – CASE 3 Bina Nusantara

TWO-WAY ECCENTRICITY – CASE 4 Bina Nusantara

BEARING CAPACITY OF LAYERED SOILS STRONGER SOIL UNDERLAIN BY WEAKER SOIL Bina Nusantara

BEARING CAPACITY OF LAYERED SOILS Bina Nusantara

BEARING CAPACITY OF LAYERED SOILS Rectangular Foundation Bina Nusantara

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 Bina Nusantara

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 Bina Nusantara

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

SOLUTION Bina Nusantara

SHALLOW FOUNDATION SESSION 9 – 10 Bina Nusantara

GENERAL The settlement of shallow foundation may be divided into 3 broad categories: Immediate settlement, which is caused by the elastic deformation of dry soil and of moist and saturated soils without any change in the moisture content. Immediate settlement are generally based on equations derived from the elasticity theory Primary consolidation settlement, which is the result of a volume change in saturated cohesive soils because of expulsion of the water that occupies the void spaces. Secondary consolidation settlement, which is observed in saturated cohesive soils and is the result of the plastic adjustment of soil particles. This course will focus at immediate and primary consolidation settlement only. Bina Nusantara

IMMEDIATE SETTLEMENT Bina Nusantara

IMMEDIATE SETTLEMENT General Equation (Harr, 1966) ; ; H =  Flexibel Foundation At the corner of foundation At the center of foundation Average Rigid Foundation ; ; H =  Es = Modulus of elasticity of soil B = Foundation width L = Foundation length Bina Nusantara

IMMEDIATE SETTLEMENT Bina Nusantara

IMMEDIATE SETTLEMENT If Df = 0 and H < , the elastic settlement of foundation can be determined from the following formula: (corner of rigid foundation) (corner of flexible foundation) The variations of F1 and F2 with H/B are given in the graphs of next slide Bina Nusantara

IMMEDIATE SETTLEMENT Bina Nusantara

IMMEDIATE SETTLEMENT Bina Nusantara

EXAMPLE Problem: A foundation is 1 m x 2 m in plan and carries a net load per unit area, qo = 150 kN/m2. Given, for the soil, Es = 10,000 kN/m2, s = 0.3. Assuming the foundation to be flexible, estimate the elastic settlement at the center of the foundation for the following conditions: a. Df = 0 and H =  b. Df = 0 and H = 5 m Bina Nusantara

EXAMPLE Solution: Part a. Part b. For L/B = 2/1 = 2    1.53, so For L’/B’ = 2, and H/B’ = 10  F1  0.638 and F2  0.033, so Bina Nusantara

IMMEDIATE SETTLEMENT General Equation (Bowles, 1982) and F1 time by 4 Es = Modulus of elasticity of soil H = effective layer thickness, ex. 2 - 4B below foundation At the center of Foundation and F1 time by 4 At the corner of Foundation and F1 time by 1 Bina Nusantara

IMMEDIATE SETTLEMENT For saturated clay soil Bina Nusantara

IMMEDIATE SETTLEMENT For sandy soil where: Iz = factor of strain influence C1 = correction factor to thickness of embedment foundation = 1 – 0.5x[q/(q-q)] C2 = correction factor due to soil creep = 1+0,2.log(t/0,1) t = time in years q = stress caused by external load q =  . Df Bina Nusantara

IMMEDIATE SETTLEMENT Circle Foundation or L/B =1 z = 0  Iz = 0.1 Young Modulus Circle Foundation or L/B =1 z = 0  Iz = 0.1 z = z1 = 0,5 B  Iz = 0.5 z = z2 = 2B  Iz = 0.0 Foundation with L/B ≥ 10 z = 0  Iz = 0.2 z = z1 = B  Iz = 0.5 z = z2 = 4B  Iz = 0.0 Bina Nusantara

EXAMPLE A shallow foundation 3 m x 3 m (as shown in the following drawing). The subgrade is sandy soil with Young modulus varies based on N-SPT value (use the following correlation: Es = 766N) Determine the settlement occur in 5 years (use strain influence method) Bina Nusantara

EXAMPLE Bina Nusantara

EXAMPLE Depth (m) z Es (kN/m2) Iz (average) (m3/kN) 0.0 – 1.0 1.0 8000 0.233 0.291 x 10-4 1.0 – 1.5 0.5 10000 0.433 0.217 x 10-4 1.5 – 4.0 2.5 0.361 0.903 x 10-4 4.0 – 6.0 2.0 16000 0.111 0.139 x 10-4  1.55 x 10-4 Bina Nusantara

CONSOLIDATION SETTLEMENT Bina Nusantara

CONSOLIDATION SETTLEMENT Normal Consolidation Over consolidation or or po + p < pc po < pc < po+p Bina Nusantara

CONSOLIDATION SETTLEMENT where: eo = initial void ratio Cc = compression index Cs = swelling index pc = preconsolidation pressure po = average effective pressure on the clay layer before the construction of the foundation =  ’.z p = average increase of pressure on the clay layer caused by the foundation construction and other external load, which can be determine using method of 2:1, Boussinesq, Westergaard or Newmark. Alternatively, the average increase of pressure (p) may be approximated by: pt = the pressure increase at the top of the clay layer pm = the pressure increase at the middle of the clay layer pb = the pressure increase at the bottom of the clay layer Bina Nusantara

EXAMPLE A foundation 1m x 2m in plan is shown in the following figure. Estimate the consolidation settlement of the foundation. Assume the clay is normally consolidated. Bina Nusantara

EXAMPLE po = (2.5)(16.5) + (0.50)(17.5-10) +(1.25)(16-10) = 52.5 kN/m2 2:1 method Bina Nusantara

ALLOWABLE SETTLEMENT Bina Nusantara