chapter 1 shallow foundation- ultimate bearing capacity
TRANSCRIPT
1. Shallow Foundation: Ultimate Bearing Capacity
1.1. Terzaghi’s Bearing Capacity Theory
Ultimate bearing capacity (qu) of general shear failure:
Continuous or strip foundation: γγBNqNNcq qcu 5.0' ++=
Square foundation: γγBNqNNcq qcu 4.0'3.1 ++=
Circular foundation: γγBNqNNcq qcu 3.0'3.1 ++=
)1('cot
2
'
4cos2
'cot2
'tan)2/'4/3(2
−=
+
=−
qc Ne
N φφπ
φφφπ
+
=−
2
'45cos2 2
'tan)2/'4/3(2
φ
φφπe
N q
'tan1'cos2
12
φφ
γ
γ
−=
pKN
Table 1: Terzaghi’s Bearing Capacity Factors
φ’ Nc Nq Nγ
0 5.70 1.00 0.00 5 7.34 1.64 0.14 10 9.61 2.69 0.56 15 12.86 4.45 1.52 20 17.69 7.44 3.64 25 25.13 12.72 8.34 30 37.16 22.46 19.13 35 57.75 41.44 45.41 40 95.66 81.27 115.31 45 172.28 173.28 325.34
50 347,50 415.14 1072.80
Figure 1. Terzaghi’s Bearing Capacity Factors
0
5
10
15
20
25
30
35
40
45
50
55
60
0 5 10 15 20 25 30 35 40
Ultimate bearing capacity (qu) of local shear failure:
Continuous or strip foundation: γγ '5.0'''3
2BNqNNcq qcu ++=
Square foundation: γγ '4.0'''867.0 BNqNNcq qcu ++=
Circular foundation: γγ '3.0'''867.0 BNqNNcq qcu ++=
N’c, N’q, and N’γ, the modified bearing capacity factors, can be calculated by using the
bearing capacity factor equations (for Nc, Nq, and Nγ respestively)by replacing φ’ by
)'tan(tan'321 φφ −= .
φφφφ’ ( °°°° )
Nc
; N
q ;
Nγγ γγ
Nc
Nq
Nγ
Factor of Safety (FS):
Based on gross allowable bearing capacity:
all
u
q
qFS =
Based on net ultimate bearing capacity:
)(netall
u
q
qqFS
−= ; fDq γ=
Water Table:
Case I: 0 ≤ D1 ≤ Df ; )(21 wsatDDq γγγ −+= ; wsat γγγ −=′
Case II: 0 ≤ d ≤ B ; fDq γ= ; )( γγγγ ′−+′=B
d
Case III: when water table is located that d ≥ B, the water will have no effect on the
ultimate bearing capacity.
1.2. The General Bearing Capacity
General bearing capacity equation: idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq γγγγγ21++′=
Bearing Capacity Factors:
φπφ ′′+= tan)
245(tan 2 eN q
φ ′−= cot)1( qc NN
φγ′+= tan)1(2 qNN
Table 2: Bearing Capacity Factors (General Bearing Capacity Equation)
φ’ Nc Nq Nγ
0 5.7 1 0
5 6.49 1.57 0.45
10 8.35 2.47 1.22
15 10.98 3.94 2.65
20 14.83 6.4 5.39
25 20.72 10.66 10.88
30 30.14 18.4 22.4
35 46.12 33.3 48.03
40 75.31 64.2 109.41
45 133.88 134.88 271.76
50 266.89 319.07 762.89
Figure 2. Bearing Capacity Factors (General Bearing Capacity Equation)
0
5
10
15
20
25
30
35
40
45
50
55
60
0 5 10 15 20 25 30 35 40
Shape Factors:
L > B
+=
c
q
csN
N
L
BF 1
φ ′
+= tan1
L
BFqs
−=
L
BF s 4.01γ
Depth Factors:
For Df / B ≤≤≤≤ 1 For Df / B > 1
+=
B
DF
f
cd 4.01
B
DF
f
qd
2)sin1(tan21 φφ ′−′+=
1=dFγ
+= −
B
DF
f
cd
1tan)4.0(1
B
DF
f
qd
12 tan)sin1(tan21 −′−′+= φφ
1=dFγ
Inclination Factors:
β = inclination of the load on the foundation with respect to the vertical.
2
901
−==
o
oβqici FF
2
1
′−=
φ
βγiF
Problem:
A square foundation (B x B) has to be constructed as shown below. Assume that γ = 16.5
kN/m3, γsat = 18.5 kN/m
3, Df = 1.2 m, and D1 = 0.6 m. The gross allowable load, Qall,
with FS=3 is 675 kN. The standard penetration resistance, N60 values are as follows:
Depth (m) N60 (blow/m)
1.5
3.0
4.5
6.0
7.5
4
6
6
10
5
Determine the size of the footing!
Solution:
5.0
0
60601'
)(
=
σap
NN
20'
20
5.05.0
0
60 +
=′
σφ ap
N
pa (atmospheric pressure) ≈ 95 kN/m2. Now the following table can be prepared:
Depth (m) N60 σ’0 (kN/m2) φ’(deg)
1.5
3.0
4.5
6.0
7.5
4
6
6
10
5
0.6x16.5+0.9(18.5-9.8)=17.73
17.73+1.5(18.5-9.8)=30.78
30.78+1.5(18.5-9.8)=43.83
43.83+1.5(18.5-9.8)=56.88
56.88+1.5(18.5-9.8)=69.93
33.6
34.5
33.3
36.0
30.8
Average φ’=33.64°≈ 34°
Next we have
qall = Qall / B2 = 675 / B
2 kN/m
2
with c’ = 0, we obtain
qall = FS
qu = 3
1
+ dsqdqsq FFBNFFqN γγγγ '
2
1
For φ’=34° from table 3.3., Nq=29.44 and Nγ=41.06.
Hence,
Fqs = 1+L
Btan φ’ = 1+tan 34 = 1.67
Fγs = 1 – 0.4(L
B) = 1- 0.4 = 0.6
Fqd = 1 + 2 tan φ’(1-sinφ’)2
B
D f= 1 + 2 tan 34 (1-sin 34)
2
B
4= 1 +
B
05.1
Fγd = 1
And
q = (0.6)(16.5) + 0.6(18.5-9.8) = 15.12 kN/m2
so
qall = 3
1
−
+
+ )1)(6.0)(06.41)()(8.95.18(
2
105.11)67.1)(44.29)(12.15( B
B
= 247.79 + B
18.260+ 107.16B
Substitution,
675 / B2 = 247.79 +
B
18.260+ 107.16B
By trial and error, we find that B ≈ 1.025 m
1.3. Eccentrically Loaded Foundation
Figure 3. Eccentrically Load
LB
M
BL
2max
6+= ;
LB
M
BL
2min
6−=
Q
Me =
For e<B/6:
+=
B
e
BL
61max ;
−=
B
e
BL
61min
For e>B/6:
)2(3
4max
eBL
−= ; negativeq =min
Ultimate Bearing Capacity under Eccentric Loading – Meyerhof’s Theory:
Step 1: Determine the effective dimension of the foundation:
B’ = effective width = B-2e
L’ = effective length = L
Step 2: Use general equation for ultimate bearing capacity:
idsqiqdqsqcicdcscu FFFBNFFFqNFFFNcq γγγγγ21++′=
Step 3: Determine the total ultimate load
'' AqQ uult =
A’= effective area = (B’)(L’)
Step 4: Determine factor of safety against bearing capacity failure
Q
QFS ult=
Step 5: Check the factor of safety against qmax;
max
'
q
qFS u=
Problem:
A continuous foundation is shown below. If the load eccentricity is 0.15 m, determine the
ultimate load, Qult, per unit length of the foundation. Use Meyerhof’s effective area.
Solution:
For c’=0,
idsqiqdqsqu FFFBNFFFqNq γγγγγ210 ++=
where,
q = (17.5)(1.8) = 31.5 kN/m2
For φ’ = 35°, from Table.1 Terzaghi’s Bearing Capacity Factors, Nq = 33.3 and Nγ =
48.03
And B’ = 2 – (2)(0.15) = 1.70 m
Because the foundation is a strip foundataion, B’ / L’ is zero.
Hence,
1==sqs FF γ
1==iqi FF γ
Fqd = 1 + 2 tan φ’(1-sinφ’)2
B
D f= 1 + 2 tan 35 (1-sin 35)
2
B
4
= 1 + 0.255
2
8.1
= 1.2295
Fγd = 1
And
)1)(1)(1)(03.48)(70.1)(5.17()1)(2295.1)(1)(3.33)(5.31(021++=uq
= 2004.13 kN/m2
Thus,
Qult = (B’)(1)(qu’) = (1.70)(1)(2004.13) = 3407.021 kN/m = 340.7 ton/m