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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:
1. The load of the superstructure2. The requirements of the local building code3. The behavior of stress –strain of soils4. The geological conditions
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)
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.
The foundation types for structures have two main categories:
1.Shallow Foundation
2.Deep Foundation
Ultimate Bearing Capacity
The load per unit area of the foundation at which shear failure in soil occurs
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
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.
General Shear failure
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
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.
Terzaghi’s Bearing Capacity Theoryfor
General Shear Failure
Terzaghi (1943) analyzed a shallow continuous footing by making some assumptions
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.
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
qu× B= 2[ (Pp)r +(Pp)c +(Pp)q ]+ Bc’tanø’-¼ γ B2 tanø’
Substituting; 2 (Pp)r - ¼rB2tanø1 = B × ½ γ BNr2 (Pp)q = B × γ D Nq
& 2 (Pp)c + Bc1 tanø1 = B × C1 Nc; We get,
qu =C’Nc + γ Df Nq + 0.5 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)
EFFECT OF WATER TABLE ON BEARING CAPACITY
The equation for ultimate bearing capacity by Terzaghihas 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.
Example Find the allowable gross load on the given foundation
For soil
γ= 18KN/m3
C= 16KN/m2
Ф= 20
1.5x1.5m
2m
Assume:
General shear failure
Factor of safety 3.5
For square footing
qu= 1.3 C Nc + q Nq + 0.4γBNγ
For Ф = 20 Nc = 17.69Nq = 7.44Nγ = 3.64
qu= 1.3{(16)(17.69)}+{(2x18)(7.44)}+{(0.4)(18)(1.5)(3.64)}=
367.952 + 267.84 + 39.312 = 675.104KN/m2
qall = qu / FS = 675.104/3.5 = 192.89KN/m2
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.
ᵝ
B
1.5m
=20
ExampleFor the soil
γ = 16KN/m3
C =0
Ф = 30For the foundations
square footingFactor of safety 4
Q = 150KN
Find B
1.0m
0.5m
qu=CNcFcsFcdFci+qNqFqsFqdFqi+0.5γBNγFγsFγdFγiBecause C=0
qu=qNqFqsFqdFqi+0.5γBNγFγsFγdFγiFor Ф=30
Nq= 18.4Nγ=22.4Fqs= 1+(B/L)tanФ= 1+0.577 = 1.577Fγs=1-0.4(B/L)=0.6Fqd= 1+2tanФ(1-sinФ)2 Df/B= 1+0.433/BFγd= 1
Fqi= (1- ᵝ/90)2 = (1-20/90)2 = 0.605Fγi= (1- ᵝ / Ф )2 = (1-20/30)2 = 0.11
q = 1x 16 +0.5 (16-9.81 ) = 19.095qu=qNqFqsFqdFqi+0.5γBNγFγsFγdFγi
qu =(19.095)(18.4)(1.577){1+(0.433/B)}(0.605)+(0.5){(16-9.81)}(B)(22.4)(0.6)(1)(0.11) = 335.216+145.148/B + 4.575Bqall = qu/FS = 83.804+36.287/B+1.144B
Q=qall*B2 qall = 150/B2
150/B2 = 83.804+36.287/B+1.144B
B= 1.13m
Ultimate Bearing Capacity under Eccentric Loading Meyerhof’s Theory
In 1953, Meyerhof proposed a theory that is generally referred toas the effective area method.
The following is a step – by – step procedure for determining theultimate load that the soil can support:
qu=CNcFcsFcdFci+qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
For C=0 qu=qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
Step 1 B’ = B-2e = 1.8 – (2x0.15) = 1.5m
L’ = 1.8m
For Ф = 30 Nq= 18.4Nγ=22.4Fqs= 1+ (B’/L)tanФ = 1.48Fγs=1-0.4 (B’/L) = 0.66Fqd= 1+2tanФ(1-sinФ)2 Df/B= 1+0.24=1.24Fγd= 1
q= 1.5 x 18 = 27KN/m2
Fqi= 1Fγi= 1
Use B not B’
qu=qNqFqsFqdFqi+0.5γB’NγFγsFγdFγi
qu =(27)(18.4)(1.48)(1.24)(1) + (0.5)(18)(1.5)(22.4)(0.66)(1)(1) =
911.7 + 199.6 = 1111.3 KN/m2
Qu = qu x A’
Qu = 1111.3 x ( 1.8 x 1.5 ) = 3000.5 KN
Step 2
Step 3
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