1. BEARING CAPACITY OF SHALLOW FOUNDATIONS of Shallow Foundation

2. A foundation is required for distributing the loads of the superstructure on a large area
The design of foundations of structures such as buildings, bridges, dams, etc. generally requires a knowledge of such factors:
The load of the superstructure
The requirements of the local building code
The behavior of stress –strain of soils
The geological conditions

3. To perform satisfactorily, shallow foundations should be designed with two main characteristics:
They have to be safe against overall shear failure in the soil that supports them.
They cannot undrgo excessive settlement (settlement is within acceptable level)

4. Foundation Engineering is a clever combination of soil mechanics, engineering geology and proper judgment derived from past experience. To a certain extent, it may be called an ART.

5. Geotechnical (Foundation) Engineer
The aim of
Geotechnical (Foundation) Engineer
is to determine
The
most economical
Foundation type

6. Shallow Foundation Deep Foundation
The foundation types for structures have two main categories:
Shallow Foundation
Deep Foundation

7. Shallow Foundation
When Df / B ≤ 3 to 4
Ground Surface

9. Deep Foundation
bed rock
weak soil
When Df / B < 3 to 4

11. Ultimate Bearing Capacity
The load per unit area of the foundation at which shear failure in soil occurs

12. Modes of Failure of Soil under a Foundation
General Concept
Modes of Failure of Soil
under a Foundation

13. Modes of shear Failure :
Vesic (1973) classified shear failure of soil under a foundation base into three categories depending on the type of soil & location of foundation. 1) General Shear failure. 2) Local Shear failure. 3) Punching Shear failure

14. General Shear failure
The load - Settlement curve in case of footing resting on surface of dense sand or stiff clays shows pronounced peak & failure occurs at very small strain.
A loaded base on such soils sinks or tilts suddenly in to the ground showing a surface heave of adjoining soil
The shearing strength is fully mobilized all along the slip surface & hence failure planes are well defined.
The failure occurs at very small vertical strains accompanied by large lateral strains.

15. LOCAL SHEAR FAILURE
When load is equal to a certain value qu(1),The foundation movement is accompanied by sudden jerks.
The failure surface gradually extend out wards from the foundation.
The failure starts at localized spot beneath the foundation & migrates out ward part by part gradually leading to ultimate failure.
The shear strength of soil is not fully mobilized along planes & hence failure planes are not defined clearly.
The failure occurs at large vertical strain & very small lateral strains

16. Punching Share failure
The loaded base sinks into soil like a punch.
The failure surface do not extend up to the ground surface.
No heave is observed.
Large vertical strains are involved with practically no lateral deformation.
Failure planes are difficult to locate.

17. dr. isam jardaneh / foundation engineering / 2010

18. dr. isam jardaneh / foundation engineering / 2010

19. Terzaghi’s Bearing Capacity Theory for General Shear Failure
Terzaghi (1943) analyzed a shallow continuous footing by making some assumptions

21. The failure zones do not extend above the horizontal plane passing through base of footing
The failure occurs when the down ward pressure exerted by loads on the soil adjoining the inclined surfaces on soil wedge is equal to upward pressure.
Downward forces are due to the load (=qu× B) & the weight of soil wedge (1/4 γB2 tanØ)
Upward forces are the vertical components of resultant passive pressure (Pp) & the cohesion (c’) acting along the inclined surfaces.

22. The failure area in the soil under the foundation can be divided into three major zones, they are:

24. For equilibrium: ΣFv = 0 1/4 γ B2tan ø + quxB = 2Pp +2C’ × Li sinø’
For equilibrium: ΣFv = 0  1/4 γ B2tan ø + quxB = 2Pp +2C’ × Li sinø’ where Li = length of inclined surface CB ( = B/2 /cosø’)   Therefore, qu× B = 2Pp + BC’ tanø’ - ¼ γ B2tanø’ – (1) The resultant passive pressure (Pp) on the surface CB & CA constitutes three components i.e. (Pp)r, (Pp)c & (Pp) q, Thus, Pp = (Pp)r + (Pp)c + (Pp)q

25. qu× B= 2[ (Pp)r +(Pp)c +(Pp)q ]+ Bc’tanø’-¼ γ B2 tanø’    Substituting; 2 (Pp)r - ¼rB2tanø1 = B × ½ γ BNr (Pp)q = B × γ D Nq & 2 (Pp)c + Bc1 tanø1 = B × C1 Nc;  We get, qu =C’Nc + γ Df Nq γ B N γ  This is Terzaghi’s Bearing capacity equation for determining ultimate bearing capacity of strip footing. Where Nc, Nq & Nγ are Terzaghi’s bearing capacity factors & depends on angle of shearing resistance (ø)

26. Ultimate Bearing Capacity Equation for Continuous Footing (General Shear Failure)

30. Terzaghi’s Bearing Capacity Theory for Local Shear Failure

33. Effect of water table on Bearing Capacity
The equation for ultimate bearing capacity by Terzaghi has been developed based on assumption that water table is located at a great depth.
If the water table is located close to foundation; the equation needs modification.

38. Find the allowable gross load on the given foundation
Example
Find the allowable gross load on the given foundation
For soil
γ= 18KN/m3
C= 16KN/m2
Ф= 20 2m
1.5x1.5m
Assume:
General shear failure
Factor of safety 3.5

39. qu= 1.3{(16)(17.69)}+{(2x18)(7.44)}+{(0.4)(18)(1.5)(3.64)}=
For square footing
qu= 1.3 C Nc + q Nq + 0.4γBNγ
For Ф = 20 Nc = 17.69
Nq = 7.44
Nγ = 3.64
qu= 1.3{(16)(17.69)}+{(2x18)(7.44)}+{(0.4)(18)(1.5)(3.64)}=
= KN/m2
qall = qu / FS = /3.5 = KN/m2

40. The General Bearing Capacity Equation
Terzaghi’s bearing capacity equations have the following shortcomings:
They don not address the case of rectangular foundations.
They do not take into account the shearing resistance along the failure surface in the soil above the bottom of foundation.
They do not take in account inclined load.

41. Meyerhof (1963) suggested the general form of general capacity equation:

42. Bearing Capacity Factors for General Bearing Capacity Equation

45. ᵝ

46. Effect of Soil Compressibility

47. Gs = Es/2(1+μs

50. Find B Example =20 For the soil γ = 16KN/m3 1.0m C =0 Ф = 30 1.5m
For the foundations
square footing
Factor of safety 4
Q = 150KN
Find B ᵝ
=20
1.0m
1.5m
0.5m B

51. qu=CNcFcsFcdFci+qNqFqsFqdFqi+0.5γBNγFγsFγdFγi
Because C=0
qu=qNqFqsFqdFqi+0.5γBNγFγsFγdFγi
For Ф=30
Nq= 18.4
Nγ=22.4
Fqs= 1+(B/L)tanФ= = 1.577
Fγs=1-0.4(B/L)=0.6
Fqd= 1+2tanФ(1-sinФ)2 Df/B= /B
Fγd= 1

52. qu=qNqFqsFqdFqi+0.5γBNγFγsFγdFγi
q = 1x ( ) =
qu=qNqFqsFqdFqi+0.5γBNγFγsFγdFγi
qu =(19.095)(18.4)(1.577){1+(0.433/B)}(0.605)+(0.5){( )}(B)(22.4)(0.6)(1)(0.11) = /B B
qall = qu/FS = /B+1.144B
Q=qall*B qall = 150/B2
150/B2 = /B+1.144B
B= 1.13m

53. Bearing Capacity of Eccentrically Loaded Foundation

55. Ultimate Bearing Capacity under Eccentric Loading Meyerhof’s Theory
In 1953, Meyerhof proposed a theory that is generally referred to as the effective area method.
The following is a step – by – step procedure for determining the ultimate load that the soil can support:

57. Step 4. the factor of safety FS= Qult/Qall

58. Find Qult e Example For the soil γ= 18 C=0 Ф= 30 For foundation
e= 0.15m e
Find Qult

59. qu=CNcFcsFcdFci+qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
qu=qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
For C=0
Step 1 B’ = B-2e = 1.8 – (2x0.15) = 1.5m
L’ = 1.8m
For Ф = Nq= 18.4
Nγ=22.4
Fqs= 1+ (B’/L)tanФ = 1.48
Fγs=1-0.4 (B’/L) = 0.66
Fqd= 1+2tanФ(1-sinФ)2 Df/B= =1.24
Fγd= 1
q= 1.5 x 18 = 27KN/m2
Fqi= 1
Fγi= 1
Use B not B’

60. qu=qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
= KN/m2
Qu = qu x A’
Qu = x ( 1.8 x 1.5 ) = KN
Step 2
Step 3

61. Homework