1. 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

2. 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

3. TYPES OF SHALLOW FOUNDATION

4. TYPES OF SHALLOW FOUNDATION

5. 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

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

7. TERZAGHI MODEL (GENERAL FAILURE)
STRIP FOUNDATION
qult = c.Nc + q.Nq .B.N
SQUARE FOUNDATION
qult = 1.3.c.Nc + q.Nq .B.N
CIRCULAR FOUNDATION
qult = 1.3.c.Nc + q.Nq .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

8. BEARING CAPACITY FACTORS
GENERAL FAILURE

9. BEARING CAPACITY FACTORS
GENERAL FAILURE

10. TERZAGHI MODEL (LOCAL FAILURE)
STRIP FOUNDATION
qult = 2/3.c.Nc’ + q.Nq’ .B.N’
SQUARE FOUNDATION
qult = c.Nc’ + q.Nq’ .B.N’
CIRCULAR FOUNDATION
qult = c.Nc’ + q.Nq’ .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)

11. BEARING CAPACITY FACTORS
LOCAL FAILURE

12. BEARING CAPACITY FACTORS

13. GROUND WATER INFLUENCE

14. 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 - ’)

15. 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)

16. 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

17. 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

18. EXAMPLE – SOLUTION
Foundation Type: Square Foundation

19. EXAMPLE – SOLUTION

20. GENERAL BEARING CAPACITY EQUATION
Meyerhof’s Theory Df

21. BEARING CAPACITY FACTOR

22. SHAPE, DEPTH AND INCLINATION FACTOR

23. 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.

24. 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)

25. ECCENTRICALLY LOADED FOUNDATIONS

26. ECCENTRICALLY LOADED FOUNDATIONS

27. 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

28. 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

29. EXAMPLE – SOLUTION
With c = 0, the bearing capacity equation becomes

30. TWO-WAY ECCENTRICITY

31. TWO-WAY ECCENTRICITY – CASE 1

32. TWO-WAY ECCENTRICITY – CASE 2

33. TWO-WAY ECCENTRICITY – CASE 3

34. TWO-WAY ECCENTRICITY – CASE 4

35. BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL UNDERLAIN BY WEAKER SOIL

36. BEARING CAPACITY OF LAYERED SOILS

37. BEARING CAPACITY OF LAYERED SOILS
Rectangular Foundation

38. 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

39. 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

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

41. SOLUTION