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

2. 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
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3. SHALLOW FOUNDATION
SESSION 5 – 6
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4. TYPES OF SHALLOW FOUNDATION
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5. TYPES OF SHALLOW FOUNDATION
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6. TERZAGHI MODEL Assumptions:
Subsoil below foundation structure is homogenous
Shallow foundation Df < B
Continuous, or strip, footing : 2D case
Rough base
Equivalent surcharge
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7. TERZAGHI MODEL FAILURE ZONES: ACD : TRIANGULAR ZONES
ADF & CDE : RADIAL SHEAR ZONES
AFH & CEG : RANKINE PASSIVE ZONES
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8. 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
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9. BEARING CAPACITY FACTORS
GENERAL FAILURE
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10. BEARING CAPACITY FACTORS
GENERAL FAILURE
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11. 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
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’ = tan-1 (2/3. tan)

12. BEARING CAPACITY FACTORS
LOCAL FAILURE
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13. BEARING CAPACITY FACTORS
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14. GROUND WATER INFLUENCE
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15. 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 - ’)
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16. 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)
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17. 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
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18. 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
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19. EXAMPLE – SOLUTION
Foundation Type: Square Foundation
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20. EXAMPLE – SOLUTION
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21. SHALLOW FOUNDATION
SESSION 7 – 8
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22. GENERAL BEARING CAPACITY EQUATION
Meyerhof’s Theory Df
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23. BEARING CAPACITY FACTOR
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24. SHAPE, DEPTH AND INCLINATION FACTOR
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25. 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.
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26. 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)
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27. ECCENTRICALLY LOADED FOUNDATIONS
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28. ECCENTRICALLY LOADED FOUNDATIONS
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29. 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
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30. 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
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31. EXAMPLE – SOLUTION With c = 0, the bearing capacity equation becomes
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32. TWO-WAY ECCENTRICITY
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33. TWO-WAY ECCENTRICITY – CASE 1
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34. TWO-WAY ECCENTRICITY – CASE 2
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35. TWO-WAY ECCENTRICITY – CASE 3
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36. TWO-WAY ECCENTRICITY – CASE 4
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37. BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL UNDERLAIN BY WEAKER SOIL
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38. BEARING CAPACITY OF LAYERED SOILS
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39. BEARING CAPACITY OF LAYERED SOILS
Rectangular Foundation
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40. 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
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41. 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
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42. SOLUTION Correction of standard penetration number
(Liao and Whitman relationship)
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43. SOLUTION
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44. SHALLOW FOUNDATION
SESSION 9 – 10
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45. 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.
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46. IMMEDIATE SETTLEMENT
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47. 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
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48. IMMEDIATE SETTLEMENT
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49. 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
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50. IMMEDIATE SETTLEMENT
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51. IMMEDIATE SETTLEMENT
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52. 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
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53. EXAMPLE Solution: Part a. Part b. For L/B = 2/1 = 2    1.53, so
For L’/B’ = 2, and H/B’ = 10  F1  and F2  0.033, so
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54. IMMEDIATE SETTLEMENT General Equation (Bowles, 1982) and F1 time by 4
Es = Modulus of elasticity of soil
H = effective layer thickness, ex B below foundation
At the center of Foundation
and F1 time by 4
At the corner of Foundation
and F1 time by 1
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55. IMMEDIATE SETTLEMENT
For saturated clay soil
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56. 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
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57. IMMEDIATE SETTLEMENT Circle Foundation or L/B =1 z = 0  Iz = 0.1
Young Modulus
Circle Foundation or L/B =1
z =  Iz = 0.1
z = z1 = 0,5 B  Iz = 0.5
z = z2 = 2B  Iz = 0.0
Foundation with L/B ≥ 10
z =  Iz = 0.2
z = z1 = B  Iz = 0.5
z = z2 = 4B  Iz = 0.0
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58. 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)
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59. EXAMPLE
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60. 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
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61. CONSOLIDATION SETTLEMENT
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62. CONSOLIDATION SETTLEMENT
Normal Consolidation
Over consolidation or or
po + p < pc
po < pc < po+p
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63. 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
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64. 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.
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65. EXAMPLE po = (2.5)(16.5) + (0.50)(17.5-10) +(1.25)(16-10) = 52.5 kN/m2
2:1 method
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66. ALLOWABLE SETTLEMENT
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