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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

Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

Course : S0484/Foundation Engineering

Year : 2007

Version : 1/0

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
terzaghi model
TERZAGHI MODEL

Assumptions:

  • Subsoil below foundation structure is homogenous
  • Shallow foundation Df < B
  • Continuous, or strip, footing : 2D case
  • Rough base
  • Equivalent surcharge
terzaghi model6
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

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
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 influence14
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
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
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
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
EXAMPLE – SOLUTION

Foundation Type: Square Foundation

example 2

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
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
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 problem28
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 solution29
EXAMPLE – SOLUTION

With c = 0, the bearing capacity equation becomes

bearing capacity of layered soils
BEARING CAPACITY OF LAYERED SOILS

STRONGER SOIL UNDERLAIN BY WEAKER SOIL

bearing capacity of layered soils37
BEARING CAPACITY OF LAYERED SOILS

Rectangular Foundation

bearing capacity of layered soils38
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
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
SOLUTION

Correction of standard penetration number

(Liao and Whitman relationship)