Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

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Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION. Course : S0484/Foundation Engineering Year : 2007 Version : 1/0. SHALLOW FOUNDATION. Topic: General Terzaghi Model Meyerhoff Model Brinch Hansen Model Influence of multi layer soil Influence of ground water elevation

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### Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

Course : S0484/Foundation Engineering

Year : 2007

Version : 1/0

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
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
• AFH & CEG : RANKINE PASSIVE ZONES
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

TERZAGHI MODEL (GENERAL FAILURE)
• 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

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)

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

P = 73 ton

Tank

dry = 13 kN/m3

sat = 18 kN/m3

c = 1 kg/cm2

 = 20o

2 m

Foundation

GWL

EXAMPLE 2

Determine the size (diameter) circle foundation of tank structure as shown in the following picture

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 = 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)
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

BEARING CAPACITY OF LAYERED SOILS

STRONGER SOIL UNDERLAIN BY WEAKER SOIL

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

SOLUTION

Correction of standard penetration number

(Liao and Whitman relationship)