session 5 – 6 bearing capacity of shallow foundation course: s0484/foundation engineering year:...
TRANSCRIPT
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• Shallow Foundation Bearing by N-SPT
value
TYPES OF SHALLOW FOUNDATION
TYPES OF SHALLOW FOUNDATION
TERZAGHI MODEL
Assumptions:
• Subsoil below foundation structure is homogenous
• Shallow foundation Df < B
• Continuous, or strip, footing : 2D case
• Rough base
• Equivalent surcharge
TERZAGHI MODEL
FAILURE ZONES:1. ACD : TRIANGULAR ZONES2. ADF & CDE : RADIAL SHEAR ZONES3. AFH & CEG : RANKINE PASSIVE ZONES
• 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
tan1cos2
1
24cos.2
1
24cos.2
cot
2
2
tan2/4/32
2
tan2/4/32
pyKN
eNq
eNc
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 (GENERAL FAILURE)
BEARING CAPACITY FACTORS
GENERAL FAILURE
BEARING CAPACITY FACTORS
GENERAL 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’
'tan1'cos2
1
2'
4cos.2
1
2'
4cos.2
'cot
2
2
'tan2/'4/32
2
'tan2/'4/32
pyKN
eNq
eNc
’ = tan-1 (2/3. tan)
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
BEARING CAPACITY FACTORS
LOCAL FAILURE
BEARING CAPACITY FACTORS
GROUND WATER INFLUENCE
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
FS
FS
unetnetall
uall
)()(
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)
f
uunet
Dq
qqq
.
)(
NET ALLOWABLE BEARING CAPACITY
PROCEDURE:1. Find the developed cohesion and the angle of friction
2. Calculate the gross allowable bearing capacity (qall) according to terzaghi equation with cd and d as the shear strength parameters of the soil
3. Find the net allowable bearing capacity (qall(net))
sheard FS
cc
sheard FS
tantan 1
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
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
Foundation Type: Square Foundation
EXAMPLE – SOLUTION
GENERAL BEARING CAPACITY EQUATION
idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq .....).5.0(........
Df
Meyerhof’s Theory
BEARING CAPACITY FACTOR
tan)1(2
cot1
245tan tan.2
NqN
NqNc
eNq
SHAPE, DEPTH AND INCLINATION FACTOR
EXAMPLE 2
2 m
GWL
dry = 13 kN/m3
sat = 18 kN/m3
c = 1 kg/cm2
= 20o
P = 73 ton
Tank
Foundation
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
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)
dry = 13 kN/m3B = 4m
SQUARE FOUNDATION
ECCENTRICALLY LOADED FOUNDATIONS
ECCENTRICALLY LOADED FOUNDATIONS
ONE WAY ECCENTRICITY
Meyerhof’s step by step procedure:• Determine the effective dimensions of the foundation as :
B’ = effective width = B – 2eL’ = 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, Fs use effective length and effective widthto determine Fcd, Fqd, Fd use B
• The total ultimate load that the foundation can sustain isQult = qu’.B’.L’ ; where B’xL’ = A’ (effective area)
• The factor of safety against bearing capacity failure isFS = Qult/Q
• Check the factor of safety against qmax, or, FS = qu’/qmax
idsqiqdqscicdcsu FFFNBFFFNqqFFFNccq ......5,0........
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 – SOLUTION
With c = 0, the bearing capacity equation becomes
TWO-WAY ECCENTRICITY
TWO-WAY ECCENTRICITY – CASE 1
TWO-WAY ECCENTRICITY – CASE 2
TWO-WAY ECCENTRICITY – CASE 3
TWO-WAY ECCENTRICITY – CASE 4
BEARING CAPACITY OF LAYERED SOILS
STRONGER SOIL UNDERLAIN BY WEAKER SOIL
BEARING CAPACITY OF LAYERED SOILS
HB
K
H
DH
B
Hcqq sfabu 1
121
tan21
2
)2(2)2(22
)1(1)1(11
2
12
1
BNNcq
BNNcq
c
c
)2()2(2)2()2(1)2()2(2 2
1sqsqfcscb FBNFNHDFNcq
BEARING CAPACITY OF LAYERED SOILS
tsfa
bu qHB
K
H
DH
B
Hcqq
1
121
tan21
2
tsfa
bu qHB
K
H
D
L
BH
B
Hc
L
Bqq
1
121
tan211
21
Rectangular Foundation
)1()1(1)1()1(1)1()1(1
)2()2(2)2()2(1)2()2(2
2
12
1
sqsqfcsct
sqsqfcscb
FBNFNDFNcq
FBNFNHDFNcq
)2(2)2(22
)1(1)1(11
2
12
1
BNNcq
BNNcq
c
c
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
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
Correction of standard penetration number
(Liao and Whitman relationship)
SOLUTION