BFC 4043
SHALLOW FOUNDATION :
2.0
A
shallow foundation foundation must : - be safe safe agains againstt overa overallll shea shearr failu failure re in the the soil soil - not not unde undergo rgo exc exces essi sive ve set settl tlem emen entt
Nature of bearing capacity failure are : (as shown in Figure 2.1) -
general shear failure (for stiff clay or dense sand ) local shear failure (for medium dense sand or clayey soil ) punching shear failure (loose sand or soft clay )
Figure 2.1 Nature of bearing capacity failure : (a) general shear (b) local shear (c) punching shear
BFC 4043
Vesic (1973) (1973) proposed a relationship relationship for the the bearing capacity capacity failure on sands in terms of relative density, Dr depth of foundation, Df and B*, Figure 2.2 Where :
B* =
2 BL B + L
and B – width, L – length of foundation
NOTE : L IS ALWAYS ALWAYS GREATER THAN THAN B
For square; square; B=L and for circul circular; ar; B=L=Diam B=L=Diamete eterr of foundat foundation ion and B* = B
Figure 2.2 Modes of foundation failure in sand, (Vesic, 1973)
BFC 4043
Vesic (1973) (1973) proposed a relationship relationship for the the bearing capacity capacity failure on sands in terms of relative density, Dr depth of foundation, Df and B*, Figure 2.2 Where :
B* =
2 BL B + L
and B – width, L – length of foundation
NOTE : L IS ALWAYS ALWAYS GREATER THAN THAN B
For square; square; B=L and for circul circular; ar; B=L=Diam B=L=Diamete eterr of foundat foundation ion and B* = B
Figure 2.2 Modes of foundation failure in sand, (Vesic, 1973)
BFC 4043
Terzaghi suggested for a continuous continuous or strip foundation with failure surface as in Figure 2.3
Figure 2.3 Bearing capacity failure in soil under rough rigid continuous foundation
Soil above the bottom of foundation is surcharge , q = γ Df The failure zone under the foundation is separated into three parts namely; - tria triang ngul ular ar ACD ACD unde underr the the foun founda dati tion on - radial radial she shear ar zone zoness ADF and and CDE CDE with with curve curvess DE and and DF as arcs of logarithmic spiral - Rank Rankin ine e pas passi sive ve zon zones es AFH AFH and and CE CEG G ∠ CAD and ∠ ACD are assume to equal equal friction angle, angle, Ø Thus ultimate bearing capacity, q u for general shear failure can be expressed as :
BFC 4043
qu qu qu
BN γ .......( strip. foundation) = cN c + qN q +0.5γ BN γ ........( square. foundation) = 1.3cN c + qN q + 0.4γ BN γ .........(circular . foundation) = 1.3cN c + qN q + 0.3γ
Where :
c – cohesion of soil γ - unit weight of soil q = γ Df Nc, Nq, N γ - bearing capacity factors
And N c
N q
e 2( 3π / 4−φ / 2 ) tan φ = cot φ − 1 = cot φ ( N q − 1) π φ 2 cos 2 + 4 2 e 2( 3π / 4−φ / 2 ) tan φ
=
2 cos 2 45 + N γ
φ
2
1 K tan φ = p2γ − 1 2 cos φ
where K pγ - passive pressure coefficient
Table 2.1 summarizes values for Nc, Nq, and N γ Table 2.1 Terzaghi’s Bearing Capacity’s Factors
Ø
Nc
Nq
N γ
Ø
Nc
Nq
N γ
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
5.70 6.00 6.30 6.62 6.97 7.34 7.73 8.15 8.60 9.09 9.61 10.16 10.76 11.41 12.11 12.86 13.68 14.60
1.00 1.10 1.22 1.35 1.49 1.64 1.81 2.00 2.21 2.44 2.69 2.98 3.29 3.63 4.02 4.45 4.92 5.45
0.00 0.01 0.04 0.06 0.10 0.14 0.20 0.27 0.35 0.44 0.56 0.69 0.85 1.04 1.26 1.52 1.82 2.18
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
27.09 29.24 31.61 34.24 37.16 40.41 44.04 48.09 52.64 57.75 63.53 70.01 77.50 85.97 95.66 106.81 119.67 134.58
14.21 15.90 17.81 19.98 22.46 25.28 28.52 32.23 36.50 41.44 47.16 53.80 61.55 70.61 81.27 93.85 108.75 126.50
9.84 11.60 13.70 16.18 19.13 22.65 26.87 31.94 38.04 45.41 54.36 65.27 78.61 95.03 115.31 140.51 171.99 211.56
BFC 4043
18 19 20 21 22 23 24 25 From Kumbhojkar
15.12 16.56 17.69 18.92 20.27 21.75 23.36 25.13 (1993)
6.04 6.70 7.44 8.26 9.19 10.23 11.40 12.72
2.59 3.07 3.64 4.31 5.09 6.00 7.08 8.34
44 45 46 47 48 49 50
151.95 172.28 196.22 224.55 258.28 298.71 347.50
147.74 173.28 204.19 241.80 287.85 344.63 415.14
261.60 325.34 407.11 512.84 650.67 831.99 1072.80
And ultimate bearing capacity, qu for local shear failure can be expressed as : 2
qu
=
cN ' c + qN ' q +0.5γ BN 'γ .......( strip. foundation) 3 qu = 0.867cN ' c + qN ' q +0.4γ BN 'γ ........( square. foundation) qu
= 0.867cN 'c + qN ' q +0.3γ BN 'γ .........(circular . foundation)
Where : N’c, N’q, N’ γ (see Table 2.2) are reduced bearing capacity factors can be calculated by using N’c, N’q, N’ γ - bearing capacity −1 2 ' tan = φ tan φ factors with 3 Table 2.2 Terzaghi’s Modified Bearing Capacity’s Factors Ø
N’ c
N’ q
N’ γ
Ø
N’ c
N’ q
N’ γ
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
5.70 5.90 6.10 6.30 6.51 6.74 6.97 7.22 7.47 7.74 8.02 8.32 8.63 8.96 9.31 9.67 10.06 10.47 10.90 11.36 11.85 12.37 12.92 13.51 14.14 14.80
1.00 1.07 1.14 1.22 1.30 1.39 1.49 1.59 1.70 1.82 1.94 2.08 2.22 2.38 2.55 2.73 2.92 3.13 3.36 3.61 3.88 4.17 4.48 4.82 5.20 5.60
0.00 0.005 0.02 0.04 0.055 0.074 0.10 0.128 0.16 0.20 0.24 0.30 0.35 0.42 0.48 0.57 0.67 0.76 0.88 1.03 1.12 1.35 1.55 1.74 1.97 2.25
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
15.53 16.30 17.13 18.03 18.99 20.03 21.16 22.39 23.72 25.18 26.77 28.51 30.43 32.53 34.87 37.45 40.33 43.54 47.13 51.17 55.73 60.91 66.80 73.55 81.31
6.05 6.54 7.07 7.66 8.31 9.03 9.82 10.69 11.67 12.75 13.97 15.32 16.85 18.56 20.50 22.70 25.21 28.06 31.34 35.11 39.48 44.45 50.46 57.41 65.60
2.59 2.88 3.29 3.76 4.39 4.83 5.51 6.32 7.22 8.35 9.41 10.90 12.75 14.71 17.22 19.75 22.50 26.25 30.40 36.00 41.70 49.30 59.25 71.45 85.75
Example 2.1
BFC 4043
Given
:
Assume Find
: :
A square foundation, 1.5m x 1.5m in plan view Soil parameters : Ø’ = 20°, c’ = 15.2 kN/m2, γ =17.8 kN/m3 FS = 4, general shear failure condition and Df = 1 m Allowable gross load on the foundation
Solution
:
qu
BN γ ........( square. foundation) = 1.3c' N c + qN q + 0.4γ
For Ø’ = 20°, (Table 2.1); N c = 17.69, Nq = 7.44, N γ = 3.64 Thus qu = 1.3cN c + qN q + 0.4γ B N γ = 1.3(15.2) ( 17.69) + ( 1 × 17.8) ( 7.44)
+
(0.4)(17.8)(1.5)(3.64)
= 349.55 + 132.43 + 38.87 = 521kN / m2 Allowable bearing capacity :
qall =
qu FS
=
521 4
= 130kN / m 2
Thus total allowable gross load, Q Qall = qall × A = 130 B 2
= 130(1.5 × 1.5) = 292.5kN
Example 2.2 Given Assume Solution
: : :
Repeat example 2.1 Local shear failure condition
BN 'γ ........( square. foundation) = 0.867cN 'c + qN 'q +0.4γ For Ø’ = 20°, (Table 2.2); N c = 11.85, Nq = 3.88, N γ = 1.12 qu
qu
=
0.867c ' N 'c + qN 'q + 0.4γ B N 'γ = 0.867( 15.2) ( 11.85) + ( 1 × 17.8) ( 3.88)
+
(0.4)(17.8)(1.5)(1.12)
= 156.2 + 69.1 + 12.0 = 237.3kN / m2 Allowable load : q 237.3 qall = u = = 59.3kN / m2 ; FS
4
Qall = qall × A = 59.3(1.5 × 1.5) = 133.4kN
BFC 4043
All equations mentioned before are based on the location of water table well below the foundation; if otherwise, some modification should be made according to the location of the water table, see Figure 2.4
Figure 2.4 Modification of bearing capacity for water table
Case I : 0 ≤ D1 ≤ Df
q’(effective surcharge) = D1 + D2γ ' - where : - γ ' - effective unit weight = γ sat − γ w - γ sat - saturated unit weight of soil 3 3 - γ w - unit weight of water = 9.81kN/m or 62.4 lb/ft - γ = γ ' in the last term of the equation Case II : 0 ≤ d ≤ B -
-
the value
q
-
γ = γ = γ '+
d B
D f = γ
( γ − γ ')
Case III : d ≥ B - water has no effect on the qu
BFC 4043
q all
=
qu
, where :
FS
qall - gross allowable load-bearing capacity, - qu – gross ultimate bearing capacity, - FS – factor of safety -
Values
of FS against bearing capacity failure is 2.5 to 3.0. Net stress increase on soil = net ultimate bearing capacity/FS
q all ( net )
=
qu
−q
FS
,
and :
qu ( net )
= qu − q ;
q = γ D f
Where : qall(net) – net allowable bearing capacity qu(net) – net ultimate bearing capacity
Procedure for FSshear a.
b.
qu qu qu
Find developed cohesion,cd and angle of friction,Ød; tan φ c c d = .......and ......φ d = tan −1 FS shear FS shear qu = cN c + qN q +0.5γ BN γ .......( strip. foundation) qu = 1.3cN c + qN q + 0.4γ BN γ ........( square. foundation) qu = 1.3cN c + qN q + 0.3γ BN γ .........(circular . foundation) Terzaghi’s equations become (with cd and Ød): BN γ .......( strip. foundation) = cd N c + qN q +0.5γ BN γ ........( square. foundation) = 1.3cd N c + qN q + 0.4γ = 1.3cd N c + qN q + 0.3γ BN γ .........(circular . foundation)
With : Nc, Nq, N γ - bearing capacity factors for Ø d c. Thus, the net allowable bearing capacity : 1 q all ( net ) = q all − q = c d N c + q ( N q − 1) + γ BN γ 2
Example 2.3
BFC 4043
Using
q all ( net ) =
qu − q FS
; and FS = 5; find net allowable load for the
foundation in example 2.1 with q u = 521 kN/m2 With qu = 521 kN/m2; q = 1(17.8) = 17.8 kN/m 2 q all ( net )
=
qu
−q
FS
Hence
=
521 − 17.8 5
= 100.64kN / m 2
Qall(net) = 100.64(1.5x1.5) = 226.4 kN
Example 2.4 Using Example 3.1, and Terzaghi’s equation qu
BN γ ........( square. foundation) with FSshear = 1.5; = 1.3cN c + qN q + 0.4γ
Find net allowable load for the foundation For c=15.2 kN/m2, Ø = 20° and c
=
cd = FS
shear
-1
15.2 1.5
tan φ
Ød = tan [ FS
shear
With :
q all ( net )
=
c FS shear
.......and ......φ d
tan φ = tan −1 FS shear
= 10.13kN / m 2 ] = tan-1[
tan 20 1.5
] = 13.64°
BN γ = 1.3c d N c + q N q − 1 + 0.4γ
From Table 2.1 : Ø=13.6° ; Hence :
c d
N γ
≈ 1.2 ; N q ≈ 3.8 ; N c ≈ 12 (estimation)
= 1.3(10.13)(12 ) + 17.8( 3.8 − 1) + 0.4(17.8 )(1.5 )(1.2 ) = 158.0 + 49.8 + 12.2 = 220kN / m 2 Qall ( net ) = 220(1.5)(1.5) = 495kN q all ( net )
BFC 4043
The need to address for rectangular shape foundation where : (0
Where : c – cohesion q – effective stress at the level of the bottom of foundation γ - unit weight of soil B – width (or diameter) of foundation Nc, Nq, N γ - bearing capacity factors F cs , F qs , F γ s - shape factors F cd , F qd , F γ d - depth factors F ci , F qi , F γ i - load inclination factors Values
- bearing capacity factors : N q
N c = N q − 1 cot φ φ = tan 2 45 + e π tan φ 2 - shape, depth and inclination factors : - shape F cs
= 1+
B N q L N c
,
F qs
= 1+
B L
tan φ ,
F γ s
N γ
= 2 N q + 1 tan φ
B
= 1 − 0.4
L
Where : L – length of the foundation and (L>B) - depth if D /B ≤1 f F cd
D f
= 1 + 0.4
B
if D /B >1 f
,
F qd
= 1 + 2 tan φ (1 − sin φ ) 2
D f B
,
F γ d
=1
BFC 4043
D 2 −1 D f ( ) F 1 2 tan 1 sin tan = 1 + ( 0.4) tan −1 f = + − φ φ , qd B B F γ d = 1 F cd
NOTE : tan-1(D /B) is in f - inclination 2 2 β β ° F ci = F qi = 1 − F γ i = 1 − 90° φ Where : β – inclination of load from vertical
For undrained condition (Ø = 0) = cu N c F cs F cd + q q net ( u ) = q u − q = cu N c F cs F cd qu
Skempton’s : q net ( u )
D B 1 0 . 2 = 5c1 + 0.2 f + B L
Table 2.3 Vesic’s Bearing Capacity Factors for General Equation (1973) Ø
Nc
Nq
N γ
Nq / Nc
Tan Ø
Ø
Nc
Nq
N γ
Nq / Nc
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
5.14 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.35 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.82
1.00 1.09 1.20 1.31 1.43 1.57 1.72 1.88 2.06 2.25 2.47 2.71 2.97 3.26 3.59 3.94 4.34 4.77 5.26 5.80 6.40 7.07
0.00 0.07 0.15 0.24 0.34 0.45 0.57 0.71 0.86 1.03 1.22 1.44 1.69 1.97 2.29 2.65 3.06 3.53 4.07 4.68 5.39 6.20
0.20 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.30 0.31 0.32 0.33 0.35 0.36 0.37 0.39 0.40 0.42 0.43 0.45
0.00 0.02 0.03 0.05 0.07 0.09 0.11 0.12 0.14 0.16 0.18 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.32 0.34 0.36 0.38
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47
22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 55.63 61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.88 152.10 173.64
11.85 13.20 14.72 16.44 18.40 20.67 23.18 26.09 29.44 33.30 37.75 42.92 48.93 55.96 64.20 73.90 85.38 99.02 115.31 134.88 158.51 187.21
12.54 14.47 16.72 19.34 22.40 25.99 30.22 35.19 41.06 48.03 56.31 66.19 78.03 92.25 109.41 130.22 155.55 186.54 224.64 271.76 330.35 403.67
0.53 0.55 0.57 0.59 0.61 0.63 0.65 0.68 0.70 0.72 0.75 0.77 0.80 0.82 0.85 0.88 0.91 0.94 0.97 1.01 1.04 1.08
Tan Ø
BFC 4043
22 23 24 25
16.88 18.05 19.32 20.72
7.82 8.66 9.60 10.66
7.13 8.20 9.44 10.88
0.46 0.48 0.50 0.51
0.40 0.42 0.45 0.47
48 49 50
199.26 229.93 266.89
222.31 265.51 319.07
496.01 613.16 762.89
1.12 1.15 1.20
Example 2.5
Figure 2.5 Given
: A square foundation (B x B), Figure 2.5, Q=150 kN. Df = 0.7m, load is inclined at 20˚ from vertical, FS = 3. Use general bearing capacity factors
Find
: The width of foundation B qu
= (qN q F qs F qd F qi + 1 γ ' BN γ F γ s F γ d F γ i ) ;
2 q = ( 0.7 )(18 ) = 12.6kN / m 2
From Table 2.3 : For Ø’ =30°: N q = 18.4, N γ = 22.4, Nq / Nc = 0.61, Tan Ø = 0.58 B B F qs = 1 + tan φ = 1 + ( 0.58) = 1.58 ; L
F γ s
B
B B = 1 − 0.4 = 1 − 0.4 = 0.6 L B
F qd = 1 + 2 tan φ (1 − sin φ ) F γ d
2
D f B
= 1 + 2( 0.58)(1 − sin 30 ) 2 0.7 = 1 + 0.202 ;
=1
2 2 β ° 20 F ci = F qi = 1 − = 1 − = 0.605 ; 90° 90 2 β 20 2 F γ i = 1 − = 1 − = 0.11 φ 30
B
B
BFC 4043
So qu
= (qN q F qs F qd F qi +
1 2
γ ' BN γ F γ s F γ d F γ i )
0.202 1 44.68 = (12.6)(18.4 )(1.58 ) + 13.3B 1 + ( 0.605 ) + (18 )( B )( 22.4 )(0.6 )(1)(0.11 ) = 221.2 + 2 B B qall =
qu 3
thus set :
150 B 2
= 73.73 +
14.89 B
+ 4.43 B
By trial and error : B=1.3m
Eccentrically loaded foundations give non-uniform distribution of pressure, Figure 2.6
Figure 2.6 Eccentrically loaded foundations Eccentricity,
qmax and qmin is given by :
e=
M Q
BFC 4043
q max
1 + 6e and BL B Q
q min
=
1 − 6e BL B Q
if e > B/6, and qmin becomes negative then : q max
=
=
4Q 3 L( B − 2e )
Factor of safety against bearing capacity failure; effective area method, by Meyerhof (1953) a. Find effective dimensions of the dimensions - the smaller of B’ and L’ is the width - effective width, B’ = B – 2e - effective length, L’ = L - if e is in the direction of L than L’ = L – 2e b.
Find the ultimate bearing capacity, q u : 1 q ' u = cN c F cs F cd F ci + qN q F qs F qd F qi + γ B ' N γ F γ s F γ d F γ i -
c.
d.
e.
2 F cs , F qs .and .F γ s
use L’ and B’ to find use B to find F cd , F qd .and .F γ d
Total ultimate load, area
Qult
Factor of safety, FS =
= qu' A' = qu' ( B'× L') ; where A’ – effective
Qult Q qu'
Check FS against qmax ; FS = q
max
Example 2.6 Given : A square foundation as shown in Figure 2.7. Using general bearing capacity factors, (table 2.3)
BFC 4043
Sand :
γ = 18kN / m 3
0.7 m
φ = 30° c
1.5m x 1.5 m
=0
Figure 2.7 Find : Ultimate load, Qult, assume one way load eccentricity, e = 0.15m Solution : with c = 0;
q 'u = qN q F qs F qd F qi
+
1 2
B ' N γ F γ s F γ d F γ i γ
Where : q = 0.7(18) = 12.6 kN/m 2 for Ø = 30°, from Table 2.3 : N q=18.4 and N γ =22.4 B’ = 1.5 – 2(0.15) = 1.2m L’ = 1.5m Thus values for general bering capacity equations : (using B’ and L’) B '
1.2 tan 30° = 1.462 1.5
F qs
= 1+
F qd
= 1 + 2 tan φ (1 − sin φ ) 2
L'
tan φ = 1 +
D f B
= 1+
( 0.289)( 0.7 ) 1.5
= 1.135
B' 1.2 = 1 − 0.4 = 1 − 0.4 = 0.68 L' 1.5 F γ d = 1 F γ s
= 12.6(18.4)(1.462 )(1.135) + 12 (18 )(1.2 )( 22.4 )( 0.68 )(1) = 384.7 + 164.5 = 549.2kN / m 2 q'u
BFC 4043
Qult = q’ u X A’ = 549.2 X (1.5X1.2) = 988kN Qall = 988/3 = 330kN with FS=3
Example 2.7 : Given : The strip footing shown below is to be constructed in a uniform deposit of stiff clay and must support a wall that imposes a loading of 152 kN/m of wall length. Use general bearing capacity factors. Find : The width of footing with FS of 3.
Figure 2.8 Solution : qu
BN γ .......( strip. foundation = cN c + qN q +0.5γ
with; c
=
qu 2
=
145.8kN / m 2 2
= 72.9kN / m 2
And Ø=0°; from the Table 2.3 Nc = 5.14, Nq = 1.0 and N γ =0 qult q all
= (72.9kN / m 2 )(5.14) + (18.82kN / m 3 )(1.2m)(1.0) + 0.5(18.82kN / m 3 )( B)(0) = 397.3kN / m 2 =
397.3kN / m 2 3
= 132.4kN / m 2
∴ requiredwidthofwall =
152.0kN / m 132.4kN / m 2
= 1.15m
BFC 4043
B required is 1.5 meter to be conservative
2.8
Example 2.8 : Soil deposit has the following ; γ =20.44 kN/m3, Ø=30°, c=38.3kN/m 2 Square footing located 1.52 m below surface, carries 2670 kN and groundwater is negligible. Use Terzaghi’s values, (Table 2.1). 2670 kN
γ = 20.44 kN/m 3 Ø=30˚ c = 38.3 kN/m 2
1.52m
Figure 2.9
Find : The right dimension B. Use Terzaghi’s equation qu = 1.3cN c + qN q + 0.4γ BN γ ........( square. foundation) With Ø=30°; Nc=37.16, Nq=22.46, N γ =19.13 Assume B=3 m; q ult = 1.3(38.3kN / m 2 )(37.16) + (20.44kN / m 3 )(1.52m)(22.46) + 0.4(20.44kN / m 3 )(3m)(19.13) 2
= 1850 + 698 + 469kN / m = 3017 kN / m
q all =
3017 kN / m 2 3
= 1005 .7 kN / m
∴ required width of wall =
Assume B=1 m;
2
2
2670 kN 1005 .7 kN / m
2
2
2
= 2.65m = B ∴ B = 1.63m
BFC 4043
q ult = 1.3(38.3kN / m 2 )(37.16) + (20.44kN / m 3 )(1.52m)(22.46) + 0.4(20.44kN / m 3 )(1m)(19.13) 2
= 1850 + 698 + 391kN / m = 2939 kN / m
q all =
2939 kN / m 2 3
= 980kN / m
∴ required width of wall =
2
2
2670 kN 980kN / m
2
2
2
= 2.72m = B ∴ B = 1.65m
Assume B=2m; = 1.3(38.3kN / m 2 )(37.16) + (20.44kN / m 3 )(1.52m)(22.46) + 0.4(20.44kN / m 3 )(2m)(19.13) = 1850 + 698 + 313kN / m 2 = 2861kN / m 2 q ult
q all
=
2861kN / m 2 3
= 954kN / m 2
∴ required width of wall =
2670 kN 954kN / m
2
= 2.80m 2 = B 2 ∴ B = 1.67m
Assume B=1.8m; q ult = 1.3(38.3kN / m 2 )(37.16) + (20.44kN / m 3 )(1.52m)(22.46) + 0.4(20.44kN / m 3 )(1.8m)(19.13) 2
= 1850 + 698 + 282kN / m = 2830 kN / m
q all =
2830 kN / m 2 3
= 943kN / m
∴ required width of wall =
2
2
2670 kN 943kN / m
2
2
2
= 2.83m = B ∴ B = 1.68m
Assume B=1.7m; = 1.3(38.3kN / m 2 )(37.16) + (20.44kN / m 3 )(1.52m)(22.46) + 0.4(20.44kN / m 3 )(1.7m)(19.13) = 1850 + 698 + 266kN / m 2 = 2814kN / m 2 q ult
q all
=
2814 kN / m 2 3
= 938kN / m 2
∴ required width of wall =
2670 kN 938kN / m
2
= 2.85m 2 = B 2 ∴ B = 1.7m
Therefore use 1.7m x 1.7m
2.9
Can be computed by using flexural formula of :
BFC 4043
q
=
Q M x y A
±
I x
±
M y x I y
Where : q – contact pressure Q – total axial vertical load A – area of footing Mx, My – total moment about respective x and y axes Ix, Iy – moment of inertia about respective x and y axes x, y – distance from centroid to the outer most point at which the contact pressure is computed along respective x and y axes. Example 2.9 A pad footing with dimension of 1.52 x 1.52m acted upon by the load of 222.4kN. Estimate soil contact pressure and FS against bearing capacity.
Figure 2.10
0.91m
0.14m
2
1.22m
1.52m 0.31m
Given : 1.52m by 1.52m square footing; P=222.4kN; γ concrete =24 kN/m3; qu = 143.64 kN/m2 Find : a. Soil contact pressure b. FS against bearing capacity pressure
γ soil =18.85kN/m 3
BFC 4043
Solution : a.
q
=
Q M x y A
±
I x
±
M y x I y
; Mx=My=0; since load on centroid
Total load calculation, Q : Column load, P = 222.4kN Weight of footing base = (1.52m)(1.52m)0.31m(24kN/m 3) = 17.19 kN Weight of footing pedestal = (0.14m)(0.14m)(0.91m)(24kN/m 3) = 0.43 kN Weight of backfill soil = [(1.52m)(1.52m)-(0.14m)(0.14m)](0.91m) x 18.85kN/m 3 = 39.3kN Q = 222.4 + 17.19 + 0.43 + 39.3 = 279.32kN Area, A = 1.52mx1.52m = 2.31m2 Soil contact pressure or Stress, q = Q/A = 120.9 kN/m2 q ult = 1.2cN c + γ D f N q + 0.4γ BN γ
b.
c=
qu 2
=
143.64kN / m 2 2
= 71.82kN / m
2
Assuming cohesive soil has : Ø=0° and c>0; thus : Nc=5.14, Nq=1.0, N γ =0, Df =1.22m q ult
= 1.2cN c + γ D f N q + 0.4γ BN γ = 1.2(71.82)(5.14) + 18.85(1.22)(1.0) + 0
= 4652.10 .98kN / m 2 Example FS =
qult
465.98
= = 3.85 q 120.pressure 9 Draw soil contact for footing in Figure 2.11 Since FS > 3.0; thus ok. Conversion to SI unit P=222.4 kN; H=88.96 kN; M=81.35kN.m; W=88.96 kN Df =1.22m; B=2.29m (7.5ft);
BFC 4043
Figure 2.11 Given : 2.29m by 1.52m rectangular footing Find : Contact pressure and soil pressure diagram Solution : Using flexural formula;
q
=
Q M x y A
±
I x
±
M y x I y
Q = P + W = 222.4 kN + 88.96 kN = 311.36 kN. A = 2.29m x 1.52m = 3.48 m2; Mx=0; My=88.96(1.22)+81.35=189.88kN.m (Moment about point C) x = 2.29/2 = 1.145m; q
=
311.36kN 3.48m
2
±
I y
=
1.52m(2.29m) 3
12 (189.88kN − m)(1.145m) 1.52m
4
q right = +232.47kN / m 2 and .....q left
= 1.52m 4
= 89.47 kN / m 2 ± 143kN / m 2
= −53.53kN / m 2 Take ΣV = 0 and ΣMc = 0 will produce : q qd ΣV = 0 : 2 (d )( L) − P − W = 0...and ... 2 (1.52m) = 311.36kN ........( A) ΣMc = 0 : see Figure 2.12 (b) and (c) q d M + ( H )( S ) − ( d )( L) x − = 0 2 3
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2.29m − d = 0.....( B ) 3 2 81.35 + 108.53 − 356.51 + 103.79d = 0 ∴ d = 1.61m 81.35kN .m + 88.96kN (1.22m ) − 311.36
substitue .
int o
( A) :
q (1.61m)(1.52m) = 311.36kN ., q = 254.46kN / m 2 2
2
254.46kN/m 1.61m 2.29m
2.29m
Figure 2.12 (a) and (b)
Example 2.11
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Figure 2.13
Given : A 6 ft x 6 ft footing as shown; load P=60kips; weight of concrete footing including pedestal + base pad, W 1=9.3kips; backfill, W2=11.2kips; horizontal load = 4kips; qall for soil = 3.0 kips/ft2. 1. 2. 3.
4.
Find : Contact pressure and soil pressure diagram. Shear and moment at section A-A (in the Figure E3.14) FS against sliding if coefficient of friction, δ = 0.40 FS against overturning. Solution : 1.
q
=
Q M x y A
±
I x
±
M y x I y
Q=P+W1+W2=60+9.3+11.2=80.5kips A=6ftx6ft=36ft2 My=4kipsx4.5ft=18kip-ft (about point C) x=6ft/2=3ft Iy=6ft(6ft) 3 /12=108ft4; Mx=0; Mxy/Ix=0 q
=
Q M x y A
±
I x
±
M y x I y
=
80.5kips 2
36 ft
±
18kip. ft (3 ft ) 4
108 ft
So : qright = 2.74 kips/ft2 < 3.0 kips/ft 2 ; OK qleft = 1.74 kips/ft2 < 3.0 kips/ft 2 ; OK
= 2.24kips / ft 2 ± 0.50kip / ft 2
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Pressure diagram
2. ΔFDG and ΔEDH are similar triangles; so
DE DF FG
=
EH FG
;..... DF = 2.74 − 1.74 = 1.0kips / ft 2 ;... EH =
= 6 ft ;...
DE 1.0
=
2.25 ft 6 ft
6 ft 1.5 ft 2
−
2
= 2.25 ft
;.. DE = 0.375kip / ft 2
Shear .at .. A − A := 2.25 ft (2.74 − 0.375kips / ft 2 )(6 ft ) + 12 (2.25 ft )(0.375kip / ft 2 )(6 ft )
= 31.93kips + 2.53kips = 34.46kips 2.25 ft + 2.53kips( 2 × 2.25 ft ) = 39.7 kips. ft 3 2
Moment .at .. A − A := 31.93kips FS .against . sliding =
3. =
Total .vertical .load × coefficient .of . friction.betweenbaseandsoil
∑ Horizontal .. forces
( 60kips + 9.3kips + 11.2kips ) (0.40) 4kips
4. FS .against .overturnin g =
= 8.05
Re sisting .moment Turning .moment
=
80.5kips(6 ft / 2) 4kips (4.5 ft )
= 13.4
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Foundation settlement under load can be classified according to two major types : (a) (b)
immediate or elastic settlement, Se consolidation settlement, Sc
Elastic settlement, Se takes place immediately during or after construction of structure. Consolidation settlement, Sc is time dependent comprises of two phases; namely, primary and secondary consolidation settlement.
Elastic settlement of foundations on saturated clay is given by Janbu et al., (1956) using the equation : S e
= A1 A2
q 0 B E s
where : A1 is a function of H/B and L/B and A2 is a function of D /B f All parameters of H, B and Df (with L into the paper) are as shown in Figure 2.14.
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Parameters
A2 Versus D /B f
A1 Versus H/B and L/B
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Schmertmann, (1978) proposed that the elastic settlement in sandy soil as : z 2
S e
= C 1C 2 (q − q )∑ 0
I z
E s
∆z
where : Iz – strain influence factor q C1 – correction factor due to depth = (q − q ) time in years C2 – correction factor due to soil creep = 1 + 0.2 log 0.1 q - stress at the level of foundation (due to loading + self weight of footing + weight of soil above footing) q = γ D f 1 − 0.5
Calculation of elastic settlement using strain influence factor
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The variation of Iz with depth below the footing for square or circular are as below : Iz = 0.1 Iz = 0.5 Iz = 0
at at at
z=0 z = z1 = 0.5B z = z2 = 2B
Footing with L/B ≥ 10 (rectangular footing) : Iz = 0.2 Iz = 0.5 Iz = 0
at at at
z=0 z = z1 = B z = z2 = 4B
Elastic parameters such as E s and μ s in Table 2.4 can be used if the real laboratory test results not available. Elastic parameters of various soils Type of soil Loose sand Medium dense sand Dense sand Silty sand Sand and gravel Soft clay Medium clay Stiff clay
(a)
Modulus of Elasticity, Es (MN/m2) 10.5 – 24.0 17.25 – 27.60 34.50 – 55.20 10.35 – 17.25 69.00 – 172.50 4.1 – 20.7 20.7 – 41.4 41.4 – 96.6
Primary consolidation, Sc
Poisson’s ratio, μs 0.20 – 0.40 0.25 – 0.40 0.30 – 0.45 0.20 – 0.40 0.15 – 0.35 0.20 – 0.50
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Many methods were developed in estimating the value of consolidation settlement, Sc. Due to simplicity only chart based on Newmarks (1942), Figure 2.18 will be used in estimating the consolidation settlement. Primary consolidation, Sc calculated as : S c
H p = C c log 1 + e0 p0
where :
Cc – compression index (given) H – thickness of clay layer e0 – initial void ratio (given) p = p0 + Δp, final pressure p0 – overburden pressure Δp =4(Ip)q0 – net consolidation pressure at mid-height of
clay layer Ip – Influence factor (from Figure 2.18) q0 – net stress increase
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Chart for determining stresses below corners of rigid and isotropic. Example 2.7 Given :
BFC 4043
A foundation to be constructed as in Figure 2.19. The base of the foundation is 3m by 6m, and it exerts a total load of 5400 kN, which include all self weight. The initial void ratio, e 0 is 1.38 and compression index, Cc is 0.68. Required : Expected primary consolidation settlement of clay layer. Solution : p0 = 19.83(200 - 198) + (19.83 – 9.81)(198 - 192) + (17.1 – 9.81) (192 – 185.6)/2 = 123.1 kN/m 2 Weight of excavation = 19.83(200 - 198) + (19.83 – 9.81)(198 – 195.5) = 64.7kN/m 2 = load pressure − weight of excavation
Net stress increase, q 0
=
5400 kN 3m × 6m
− [ (19.83)( 200 − 198) + (19.83 − 9.81)(198 − 195.5 ) ] = 235.3kN / m 2
By dividing the base into 4 equal size of 1.5m by 3.0m : mz = 1.5m z = 195.5m −
m=
1.5m 6.7 m
nz = 3.0m
192.0m + 185.6m
= 0.224 ;
2 n=
= 6.7m
3.0m 6.7 m
= 0.448
From Figure 2.18, the influence coefficient is 0.04 Therefore ; ∆ p = 4( 0.04) 235.3kN / m 2 = 37.6kN / m 2
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Final pressure, p = p0 + Δp = 123.1 + 37.6 = 160.7 kN/m 2. Therefore;
S c
H p 6.4m 160.7 kN / m 2 = 0.212m log = C c 2 log p = 0.68 1 + 1.38 1 e + 123 . 1 kN / m 0 0
Secondary settlement, Ss is computed from the following calculation (U.S. Department of the Navy, 1971) S s
t = C α H log s t p
where : Ss – secondary compression settlement Cα – coefficient of secondary compression, can be determined from Figure 3.26 H – thickness of clay layer that is considered ts – time for which settlement is required tp – time to completion of primary consolidation
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Value of Cα
• Bowles (1977) proposed a correlation of the net allowable bearing pressure for foundations with SPT (N-values). • The following equations are used : S for B ≤ 1.22m q net ( all ) ( kN / m 2 ) = 19.16( N )( F d ) 25.4 And 2 3.28 B + 1 S for B > 1.22m 2 q net ( all ) ( kN / m ) = 11.98( N ) ( F d ) 3.28 B 25.4 Where : D f Fd – depth factor = 1 + 0.33 B ≤ 1.33 S – tolerable settlement, in mm.
Example 2.8
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Given: A shallow square footing for a column is to be constructed. Design load is 1000 kN. The foundation soil is sand. The SPT numbers from field exploration as shown in the table. Assume that the footing must be 1.5m deep, the tolerable settlement as 25.4mm and the size is > 1.22m. Required : (a) The exact size of the footing (b) safety factor for foundation Solution :
Navg = (7+8+11+11+13+10+9+10+12)/9=10 With S=25.4mm and N=10 2 2 3.28 B + 1 25.4 3.28 B + 1 2 q net ( all ) ( kN / m ) = 11.98(10 ) ( F d ) = 119.8 F d 3 . 28 B 25 . 4 3 . 28 B D f F d = 1 + 0.33 ≤ 1.33 B By trial and error (set the table for calculation)
BFC 4043
From the table it is seen that the appropriate B=2.4
Setting general equation and equation for net ultimate with c=0 (for sandy soil) : qu ( net ) = qu − q q = γ D f q u ( net )
;
qu
= cN c F cs F cd F ci + qN q F qs F qd F qi +
= qult − q = qN q F qs F qd F qi +
1 2
1 2
BN γ F γ s F γ d F γ i γ
BN γ F γ s F γ d F γ i − q γ
For N=10; friction angle of Ø=34˚ is considered (from table on SI)
With no inclination so F qi=Fγi=1.0
BFC 4043
From table 2.3 Nq=29.44, Nγ=41.06
So for a tolerable settlement of 25.4mm, the SF required is calculated as : SF=Qnet(u)/Q= 10,322kN/1000kN = 10.3 which is OK, therefore most design controlled by tolerable criterion.