MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
3-BEARING CAPACITY OF SOILS
INTRODUCTION
The soil must be capable of carrying the loads from any engineered structure placed upon it
without a shear failure and with the resulting settlements being tolerable for that structure.
The evaluation, of the limiting shear resistance or ultimate bearing capacity qult of the soil
under a foundation load, depends on the soil strength perimeters.
The recommendation for the allowable bearing capacity qa to be used for design is based on
the minimum of either
1. Limiting the settlement to a tolerable amount.
2. The ultimate bearing capacity, which considers soil strength, where qa= qult / F. Where F
is the factor of safety, which typically is taken as 3 for the bearing capacity of shallow
foundations.
Probably, one of two potential mode of failure may occur when a footing loaded to produce
the ultimate bearing capacity of the soil.
a. Rotate about some center of rotation (probably along the vertical line oa) with shear
resistance developed along the perimeter of the slip zone shown as a circle
b. Punch into the ground as the wedge agb.
a. Footing on = 0 Soil
Upper Bound Solution
∑
=0↷
×
2
+
×
−
×
2
=0
⎯⎯⎯
=
+
Lower Bound Solution
At point o, left side element 1 and adjacent right side element 2, the horizontal stresses
should be equal
3, 1= 1, 2
45 −
∅
−2
2
For =0, tan 45°=1 then
45 −
∅
=
2
45 +
∅
+2
2
=
+
For the case of =0, i.e. foundation on ground surface, the average value is
2015 - 2016
45 +
∅
2
= .
pg. 19
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
b. Footing on φ-c Soil
2015 - 2016
pg. 20
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
BEARING CAPACITY EQUATIONS
Terzaghi Bearing-Capacity Equation
One of the early sets of bearing-capacity equations was proposed by Terzaghi (1943) for
Shallow (D ≤ ) strip footing. Terzaghi's equations were produced from a slightly modified
bearing-capacity theory developed by Prandtl (1920) but Terzaghi used shape factors.
=
+
+ .
Where
C= cohesion of soil
= angle of internal friction of soil
B= least lateral dimension of footing (diameter for round footing).
= unit weight of soil (use buoyant (submerged) unit weight for soil below water table.
= D
a
.
°
=
=
cos (45 + ∅/2)
0
5.7
1.0
0.0
=
−1
∅
Shape Factor
∅
=
2
∅
(
− 1)
Strip
1
Round
1.3
Square
1.3
Rectangle
1+ 0.3B/L
1
0.6
0.8
1- 0.2B/L
5
10
15
20
25
30
34
35
40
45
48
50
7.3
9.6
12.9
17.7
25.1
37.2
52.6
57.8
95.7
172.3
258.3
347.5
1.6
2.7
4.4
7.4
12.7
22.5
36.5
41.4
81.3
173.3
287.9
415.1
0.5
1.2
2.5
5.0
9.7
19.7
36.0
42.4
100.4
297.4
780.1
1153.2
Conclusions
1. The ultimate bearing capacity increases with the depth of footing.
2. The ultimate baring capacity of cohesive soil ( =0)is independent of footing size; i.e.,
at ground level qult = 5.7 c
3. The ultimate bearing capacity of a cohesionless soil(c=0) is directly dependent on the
footing size, but the depth of footing is more important than size.
Meyerhof’s Bearing-Capacity Equation
Meyerhof (1951, 1963) proposed a bearing-capacity equation similar to that of Terzaghi but
included a shape factor Sq with the depth term Nq. He also included depth factors di, and
inclination factors ii, for cases where the footing load is inclined from the vertical.
Up to a depth of D « B ≈ the Meyerhof qult is not greatly different from the Terzaghi value.
The difference becomes more pronounced at larger D/B ratios.
=
=
2015 - 2016
+
+
+ .
+ .
Vertical load
Inclined load
pg. 21
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
=
=
=
(
)
45 +
−1
∅
− 1 tan(1.4∅)
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
2
Shape, depth and inclination factors
Hansen's Bearing-Capacity Method
Hansen (1970) proposed the general bearing-capacity case and N factor equations. This
equation is readily seen to be a further extension of the earlier Meyerhof work. The
extensions include base factors for situations in which the footing is tilted from the
horizontal bi and for the possibility of a slope of the ground supporting the footing to give
ground factors gi. The Hansen equation implicitly allows any D/B and thus can be used for
both shallow and deep foundation.
=
+
When =0
=
(
= .
)
( + ′ + ′ − ′ −
45 +
2
=
−1
∅
= 1.5
− 1 tan(∅)
su =undrained =0 ultimate shear strength
++ .
−
)+
Vesic's Bearing-Capacity Equations
The Vesic (1973, 1975) procedure is essentially the same as the method of Hansen (1961)
with select changes. The Nc and Nq terms are those of Hansen but N is slightly different.
There are also differences in the ii, bi, gi terms. The Vesic equation is somewhat easier to use
than Hansen's because Hansen uses the / terms in computing shape factors si whereas Vesic
does not.
=
+
++ .
=
(
)
45 +
=
−1
∅
= 2 + 1 tan(∅)
2015 - 2016
2
pg. 22
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
2015 - 2016
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
pg. 23
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
Which Equations to Use
The bearing capacity can be obtained—often using empirical SPT or CPT data directly—to a
sufficient precision for most projects.
The Terzaghi equations, being the first proposed, have been very widely used because of
their greater ease of use. They are only suitable for a concentrically loaded footing on
horizontal ground. Both the Meyerhof and Hansen methods are widely used. The Vesic
method has not been much used
Use
Terzaghi
Hansen, Meyerhof, Vesic
Hansen, Vesic
2015 - 2016
Best for
Very cohesive soils where D/B < 1 or for a quick estimate of
quit to compare with other methods. Don’t use for footings
with moments and/or horizontal forces or for tilted bases
and/or sloping ground.
Any situation that applies, depending on preference or
familiarity with a particular method.
When base is tilted; when footing is on a slope or when D/B
> 1.
pg. 24
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
ADDITIONAL CONSIDERTIONS
1. A reduction factor is applicable for the third term of B effect as follows;
r = 1 - 0.25
( )
for B≥
2. Which φ
The soil element beneath the centerline of a strip footing is subjected to plane strain
loading, and therefore, the plain strain friction angle must be used in calculating its bearing
capacity, the plane strain friction angle can be obtained from a plane strain compression
test. The loading condition of a soil element along the vertical centerline of a square or
circular footing more closely resembles axisymmetric loading than plane strain loading, thus
requiring a triaxial friction loading, which can be determined from a consolidated drained or
undrained triaxial compression test.
Meyerhof proposed the corrected friction angle for use with rectangular footing as:
∅
= (1.1 − 0.1 )∅
Example: Compute the allowable bearing
pressure using the Terzaghi equation for
the footing and soil parameters shown.
Use a safety factor of 3 to obtain qa?
Critical Thinking:
F=3 is recommended for cohesive soil
Solution:
Find the bearing capacity. (B not known but D is)
From equations and tables
Nc =17.7
Nq=7.4
Nϒ=5.0
Sc =1.3
Sϒ = 0.8
=
+
+ .
= 20 x 17.7 x 1.3 + 17.30 x 1.2 x 7.4 + 0.5 x B x 17.3 x 5.0 x 0.8 = 613.8 +34.6 B kPa
qa= qult / F.
qa= 613.8 +34.6B)/3 = 205 +11.5B
Likely B ranges from 1.5 to 3m, with B=3m
qa = 205 +11.5x 1.5 = 220 kPa
qa = 205 +11.5x 3 x 0.95 = 240 kPa
2015 - 2016
r = 1 - 0.25 log( ) = 0.95
consider qa =220 kPa
pg. 25
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
EFFECT OF WATER TABLE ON BEARING CAPACITY
The effective unit weight of the soil is used in the bearing-capacity equations for computing
the ultimate capacity, in calculation for q in the qNq term and in the term 0.5yBNy.
1. When the water table is below the wedge zone [depth approximately 0.5B tan (45 + /2)],
the water table effects can be ignored for computing the bearing capacity.
2. When the water table lies within the wedge zone, if B is known, computing the average
effective unit weight to be used in the 0.5 BN term will be as follows
′
= (2 − )
+
( − )
Where
H
= 0.5 B tan (45 + /2)
dw = depth of water table below base of footing
= wet unit weight of soil in depth dw
’ = submerged unit weight of soil below water table
=
−
Example: for the square footing shown,
what is the allowable bearing capacity?
use F=2, Assume the soil above water table
is wet with normal water content wN =10%
and Gs=2.68
Critical Thinking:
F=2 is recommended for cohesionless soil
Effective unit weight should be calculated
Solution:
= (1.1 -0.1B/L)
tri
=35°
H = 0.5 B tan (45 + /2) = 0.5x2.5xtan (45+35/2) = 2.4 m
dw = 1.95-1.1= 0.85
∴ Water table lies within the wedge zone
18.10
=
=
= 16.45
/
1+
1 + 0.1
16.45
=
=
= 0.626
(9.81) 2.68(9.81)
=1−
= 1 − 0.626 = 0.374
The saturated unit weight is the dry weight +weight of water in voids
= 16.45 + 0.374(9.81) = 20.12 /
′
= (2 − )
+
( − )
= (2 2.4 − 0.85)
2015 - 2016
0.85
20.12 − 9.81
18.1 +
(2.4 − 0.85) = 14.85
2.4
2.4
/
pg. 26
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
To calculate q in the qNq term use
q = 18.1 1.1 = 19.91
/
In the 0.5 BN term use
= 14.85
/
1. Bearing Capacity by Terzaghi’s equations
=
+
= 35 Nc = 57.8
Nq = 41.4
r = 1 - 0.25 log(
+ .
.
Nϒ = 42.4
) = 0.975
= 0 + 19.91 x 41.4 + 0.5x2.5x14.85x42.4x0.8x0.975 = 1438 kN/m2
qa =qult /2 =1438/2=719 kN/m2
2. Bearing Capacity by Meyerhof’s equations
=
+
= 35 Nc = 46.1 Nq = 33.3
Nϒ = 37.1
+ .
Sq = s =1+0.1Kp B/L =1+0.1x3.7x1=1.37
d q= d =1+0.1√3.7 x 1=1.08
qult = 19.91x33.3x1.37x1.08 + 0.5x2.5x14.85x37.1x1.37x1.08x0.975=2000 kN/m2
qa =qult /2 =2000/2=1000 kN/m2
3. Bearing Capacity by Hansen’s equations
qa =qult /2 =1524/2=762 kN/m2
BEARING CAPACITY FOR FOOTINGS ON LAYERED SOILS
There are three general cases of the footing on a layered soil as follows:
Case 1. Footing on layered clays (all = 0).
a. Top layer weaker than lower layer (c1 < c2)
b. Top layer stronger than lower layer (c1> c2)
CR= strength ratio
If
=CR ≤ 0.7
1.5
=
+ 5.14
≤ 5.14
=
3
When 0.7≤CR ≤ 1
IF
=CR ≥ 1
= 4.14 +
2015 - 2016
+ 6.05
≤ 6.05
Reduce the aforesaid
0.5
&
by 10%
= 4.14 +
1.1
pg. 27
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
= 5.05 +
0.33
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
= 5.05 +
0.66
&
×
×2
+
Notice: When the top layer is very soft with a small d 1/B ratio, one should give consideration
either to placing the footing deeper onto the stiff clay or to using some kind of soil
improvement method
=
Case 2. Footing on layered -c soils with a, b same as case 1.
A modified angle of friction and soil cohesion may be used to determine the bearing
capacity when the depth of the top layer( ) under the base of foundation is less than(H).
Calculate H ; H = 0.5 B tan (45 + /2)
IF H > d1 then compute
∅ + ( − )∅
+( − )
∅ =
=
For case 1 and 2 ,if the top layer is soft the calculated bearing capacity should not be more
than
qult= 4c1 +q
Case3. Footing on layered sand and clay soils.
a. Sand overlying clay
b. Clay overlying sand
IF H > d1 then
1. Find qult based on top-stratum soil parameters using one of bearing capacity equations.
2. Assume a punching failure bounded by the base perimeter of dimensions B x L. Here
include the q contribution from d1, and compute q’ult of the lower stratum using the base
dimension B.
3. Compare qult to q'ult and use the smaller value.
Example: A footing of B = 3, L = 6 m is to be
placed on a two-layer clay deposit as shown.
Estimate the ultimate bearing
Capacity using Hansen’s equation?
Critical Thinking:
Footing on layered caly soil =0
2015 - 2016
pg. 28
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
Solution:
Check H ; H = 0.5 B tan (45 + /2)
H=0.5 x 3 tan 45°= 1.5 m > (d1=1.22 m)
115
= 1.5 > 1
77
Nc1= 4.14 +0.5x3/1.22= 5.37
=
=
=
×
+
×2=
N c2= 4.14 + 1.1x3/1.22= 6.84
5.37 × 6.84
×2 =6
5.37 + 6.84
= .
( + ′ + ′ − ′ −
−
)+
qult= 6 x 77 (1 + (0.2x3/6) + (0.4x1.83/3) ) + 17.26 x1.83 x1x1 = 652 kPa
BEARING CAPACITY FOR FOOTINGS WITH ECCENRIC LOADING
Two methods can be followed to solve this case
Method 1. Use either the Hansen or Vesic bearing-capacity equation with the following
adjustments:
a. Use B' in the yBNy term.
b. Use B' and L' in computing the shape factors.
c. Use actual B and L for all depth factors.
qa = qult/F
and Pa = qaB'L'
Method 2. Use the Meyerhof general bearing-capacity equation and a reduction factor Re
used as
quIt, des = quIt, comp x Re
Re = 1 - 2e/B
(cohesive soil)
Re = 1 - √e/B
(cohesionless soil and for 0 < e/B < 0.3)
Alternatively, one may directly use the Meyerhof equation with B' and L’ used to compute
the shape and depth factors and B’ used in the 0.5 B'N term.
2015 - 2016
pg. 29
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
Example: A square footing is 1.8 X 1.8 m
with a 0.4 X 0.4 m square column. It is
loaded with an axial load of 1800 kN
and Mx = 450 kN m; My = 360 kN m.
Undrained triaxial tests (soil not
saturated) give = 36° and c = 20 kPa.
The footing depth D = 1.8 m; the soil
unit weight y = 18.00 kN/m3; the
water table is at a depth of 6.1 m
from the ground surface. What is
the allowable soil pressure, if F = 3.0,
using the Hansen bearing-capacity
equation with B', L', Meyerhof's
equation; and the reduction factor Re?
Solution:
ex = My/V= 360/1800 = 0.2 < B/6
ey = Mx/V = 450/1800 = 0.25 < L/6
Bmin = 4ex +wx = 4x0.2+0.4 = 1.2 <1.8
Lmin = 4ey +wy =4x0.25+0.4= 1.4 <1.8
B’ = B - 2ex = 1.8-2x0.2 = 1.4m
L’ = L - 2ey = 1.8-2x0.25 = 1.3m
Hansen’s bearing-capacity equation
Nc = 51 Nq= 38
N = 40
Compute shape factors using B’, L’
sc= 1+ (Nq/Nc) (B’/L’)= 1.69
sq = 1+ ( B’/L’) sin =1.55
s = 1-0.4B’/L’=0.6
Compute depth factors using B , L
dc= 1+0.4D/B = 1.4
dq = 1+2tan (1-sin )2D/B=1.25
d =1
=
+
++ .
= 20x51x1.69x1.4 + 18x1.8x38x1.55x1.25 + 0.5x1.4x18x40x0.6x1= 5100 kPa
qa = 5100/3 =1700kpa
The actual soil pressure is 1800/ (1.4x1.3) = 989 kPa
2015 - 2016
pg. 30
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
Meyerhof’s equation
Kp =tan2 (45+ /2) =3.85
Nc =51 Nq=38 N =44
sc = 1+ (0.2Kp B/L) = 1.77
sq = s = 1+ (0.1Kp B/L) =1.39
dc = 1+ (0.2√Kp D/B) = 1.39
dq = d = 1+ (0.1√Kp D/B) =1.20
qult = 20x51x1.77x1.39 + 18x1.8x38x1.39x1.2 + 0.5x18x1.8x44x1.39x1.2 =5752 kPa
Re=1-(2e/B) cohesive soil
Rex=1-(2x0.2/1.8) = 0.78
Rey= 1-(2x0.25/1.8) = 0.72
qult= 5752x0.78x0.72 = 3230 kPa
qa= 3230/3 =1073 kPa
The actual soil pressure is 1800/ (1.8x1.8) = 555 kPa
Re= 1- √(e/B) cohesionless soil (This can be used since the cohesion of the soil is low (20)
Rex = 1- √(e/B) = 1 - √0.2/1.8 = 0.67
Rey= 1- √(e/B)= 1 - √0.25/1.8 = 0.63
qult= 5752x0.67x0.63=2428 kPa
qa=2428/3 =809 kPa
BEARING CAPACITY FOR FOOTINGS WITH INCLINED LOAD
Inclined loads are produced when the footing is loaded with both a vertical V and a
horizontal component(s) Hi of loading (refer to Table 4-5 and its figure). This loading is
common for many industrial process footings where horizontal wind loads are in
combination with the gravity loads. In any case the load inclination results in a bearing
capacity reduction over that of a foundation subjected to a vertical load only.
Safety Factor Against Sliding
The footing must be stable against both sliding and bearing in case it subjected to a
horizontal load component, for sliding the general equation for the factor of safety is;
+ ∁
+
=
Example: The data obtained by the
Load test is Vult= 1060 kN
HL ult= 382 kN
Find the ultimate bearing capacity by
Hansen and Vesic’s equations?
Critical Thinking:
Change triaxial to
2015 - 2016
plane strain
pg. 31
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
Solution:
plane strain =
(1.1 – 0.1B/L)
triaxial
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
=(1.1– 0.1x 0.5/2)43° = 47°
Hansen’s Bearing Capacity Equation
=
From Equations ;
Nq = 187 N = 300
+
++ .
dqB =1+2 tan (1-sin )2 D/B = 1.155
dqL =1+2 tan (1-sin )2 D/L = 1.04
d B = d L =1
iqB = (1-(0.5HB / (V+Af Ca Cot )))2.5 = 1
i B = (1-(0.7HB / (V+ Af Ca Cot )))3.5 = 1
Horizontal force parallel to B =0
iqL = (1-(0.5HL / (V+Af Ca Cot )))2.5 = (1 – 0.5x 382/1060)2.5 = 0.608
i L = (1-(0.7HL / (V+Af Ca Cot )))3.5 =
= 0.361
SqB =1+ sin B’ iqB / L’ = 1.18
SqL =1+ sin L’ iqL /B’ = 2.78
S B =1 -0.4 B’ i B /( L’ i L) = 0.723
S L =1- 0.4 L’ i L /( B’ i B) = 0.422 Use 0.6
both should be equal or more than 0.6
Substitute the factors obtained for each direction in Hansen’s equation and take the
smaller ultimate bearing capacity
qult = 9.43 x 0.5x187x1.18x1.16x1 +0.5x0.5x9.43x300x0.732x1x1 =1700 kPa
qult = 9.43 x 0.5x187x2.78x1.04x0.608 +0.5x2.0x9.43x300x0.6x1x0.361 =2150 kPa
Vesic’s Bearing Capacity Equation
=
+
++ .
From Equations
Nq = 187 N = 404
Sq= 1 + B/L tan = 1 + (0.5/2) tan 47 = 1.27
S = 1 – 0.4 B/L = 1 – 0.4 (0.5/2) = 0.9 ≥ 0.6 OK
dq = 1 + 2 tan (1 – sin )2 k = 1.16
d =1
m=(2 +(L/B)) / (1+(L/B))
iqL = (1-(HL / (V+Af Ca Cot )))m = (1 – 382/1060)1.2 = 0.585
i L = (1-(HL / (V+Af Ca Cot )))m+1 =
= 0.374
qult = 9.43 x 0.5x187x1.27x1.16x0.585 +0.5x0.5x9.43x404x0.9x1x0.374 =1080 kPa
2015 - 2016
pg. 32
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
BEARING CAPACITY OF FOOTINGS ON SLOPES
Example: A 2 x 2 m square footing has a ground
slope of = 0° and base slope of =10°. Are the
footing dimensions adequate for the given load
with a factor of safety 3?
Use Hansen’s and Vesic’s equations
Critical Thinking:
Assume angle of friction between concrete and
soil ( )= =25°
Base adhesion Ca =0.6 to 1 x C = 25 kPa
Neglect passive earth pressure for D = 0.3 m
Solution:
Factor of safety against sliding must be checked
=
+ ∁
+
=
600 tan 25 + 25 × 2 × 2
= 1.9
200
Hansen’s Bearing Capacity Equation
=
+
From table Nc = 20.7 Nq= 10.7 N = 6.8
++ .
dc =1+ 0.4 D/B = 1.06
dq =1+2 tan (1-sin )2 D/B = 1.05
d =1
iqB = (1-(0.5HB / (V+Af Ca Cot )))3 = 0.675
i B = (1-((0.7- /450)HB / (V+Af Ca Cot )))4 = 0.481
i L= 1
ic = iq – (1-iq)/(Nq-1) = 0.641
ScB =1+ (Nq/Nc)( B’ iqB / L) = 1.329
SqB =1+ sin B’ iqB / L = 1.285
S B =1 -0.4 B’ i B / L i L = 0.808
= 10° = 0.175 rad
bcB = 1 – /147 = 0.93
bqB = exp( -2 tan ) = 0.849
b B = exp( -2.7 tan ) = 0.802
Ground factors gi =1 for =0
qult = 25x20.7x1.329x1.06x0.641x0.93 + 17.5x0.3x10.7x1.285x1.05x0.675x0.848 +
0.5x17.5x2x6.8x0.808x1x0.481x0.802 = 515 kPa
Pa = A x qa = 2x2 x 515/3 = 686 kN > 600
2015 - 2016
OK
pg. 33
MAYSAN UNIVERSITY
COLLEGE of ENGINEERING
CIVIL ENGINEERING DEPARTMENT
FOURTH YEAR
FOUNDATION DESIGN CE402
Dr. RAID AI SHADIDI
BEARING CAPACITY FROM IN SITUE SPT
=
(1 + 0.33 )
=
(
=
(1 + 0.33 )
+ 0.3
≤ 1.2
F1
F2
) (1 + 0.33 )
N55
0.05
0.08
N’70
0.04
0.06
> 1.2
qa = allowable bearing pressure for ∆Ho = 25-mm or 1-in. settlement, kPa ,for any other
value of settlement ∆Hi , qai=∆Hi/∆Ho x qa
Kd = 1 + 0.33 D/B ≤ 1.33
N= is the statistical average value of blows for the footing influence zone of about 0.5B
above footing base to at least 2B below. To convert N’70 x 70 = N55 x 55
Example: Find the allowable bearing pressure for a soil of N70 =24, if the foundation depth
=1m and width=3m for a tolerable settlement of 15 mm?
Solution:
=
24 3 + 0.3
1
(
) (1 + 0.33 ) = 537
0.06
3
3
qa for 15mm settlement= 537 x 15/25 = 322 kPa
BEARING CAPACITY FROM IN SITUE CPT
An approximation for the bearing capacity should be applicable for D/B < 1.5 from an
averaged value of qc (cone point resistance in kg/ cm2) over the depth interval from about
B/2 above to 1.1B below the footing base. For cohesionless soils one may use
= 28 − 0.0052(300 −
= 48 − 0.009(300 −
For cohesive soils one may use
)
)
.
.
= 2 + 0.28
= 5 + 0.34
2015 - 2016
/
/
/
/
pg. 34