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2.0 SHALLOW FOUNDATION :
A shallow foundation must :
- be safe against overall shear failure in the soil- not undergo excessive settlement
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
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Vesic (1973) proposed a relationship for the bearing capacityfailure on sands in terms of relative density, Dr depth offoundation, Df and B*, Figure 2.2
Where : LBBLB += 2* and B width, L length of foundation
NOTE : L IS ALWAYS GREATER THAN B
For square; B=L and for circular; B=L=Diameter of foundationand B* = B
Figure 2.2 Modes of foundation failure in sand, (Vesic, 1973)
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Terzaghi suggested for a continuous or strip foundation withfailure surface as in Figure 2.3
Figure 2.3 Bearing capacity failure in soil under rough rigidcontinuous foundation
Soil above the bottom of foundation is surcharge, q = Df The failure zone under the foundation is separated into three
parts namely;- triangular ACD under the foundation- radial shear zones ADF and CDE with curves DE and DF
as arcs of logarithmic spiral- Rankine passive zones AFH and CEG
CAD and ACD are assume to equal friction angle,
Thus ultimate bearing capacity, qu for general shear failure can
be expressed as :
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1819202122232425
15.1216.5617.6918.9220.2721.7523.3625.13
6.046.707.448.269.1910.2311.4012.72
2.593.073.644.315.096.007.088.34
44454647484950
151.95172.28196.22224.55258.28298.71347.50
147.74173.28204.19241.80287.85344.63415.14
261.60325.34407.11512.84650.67831.991072.80
From Kumbhojkar (1993)
And ultimate bearing capacity, qufor local shear failure can beexpressed as :
)..........('3.0''867.0
).........('4.0''867.0
)........('5.0''3
2
foundationcircularBNqNcNq
foundationsquareBNqNcNq
foundationstripBNqNcNq
qcu
qcu
qcu
++=
++=
++=
Where :Nc, Nq, N (see Table 2.2) are reduced bearing capacity
factors can be calculated by using Nc, Nq, N - bearing capacity
factors with
= tan
3
2tan' 1
Table 2.2 Terzaghis Modified Bearing Capacitys Factors
Nc Nq N Nc Nq N01
234567891011121314151617
1819202122232425
5.705.90
6.106.306.516.746.977.227.477.748.028.328.638.969.319.6710.0610.47
10.9011.3611.8512.3712.9213.5114.1414.80
1.001.07
1.141.221.301.391.491.591.701.821.942.082.222.382.552.732.923.13
3.363.613.884.174.484.825.205.60
0.000.005
0.020.040.0550.0740.100.1280.160.200.240.300.350.420.480.570.670.76
0.881.031.121.351.551.741.972.25
2627
28293031323334353637383940414243
44454647484950
15.5316.30
17.1318.0318.9920.0321.1622.3923.7225.1826.7728.5130.4332.5334.8737.4540.3343.54
47.1351.1755.7360.9166.8073.5581.31
6.056.54
7.077.668.319.039.8210.6911.6712.7513.9715.3216.8518.5620.5022.7025.2128.06
31.3435.1139.4844.4550.4657.4165.60
2.592.88
3.293.764.394.835.516.327.228.359.4110.9012.7514.7117.2219.7522.5026.25
30.4036.0041.7049.3059.2571.4585.75
Example 2.1
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Given : A square foundation, 1.5m x 1.5m in plan viewSoil parameters : = 20, c = 15.2 kN/m2, =17.8 kN/m3
Assume : FS = 4, general shear failure condition and Df= 1 mFind : Allowable gross load on the foundation
Solution : ).........(4.0'3.1 foundationsquareBNqNNcq qcu ++=
For = 20, (Table 2.1); Nc = 17.69, Nq = 7.44, N= 3.64
Thus( ) ( ) ( ) ( ) )64.3)(5.1)(8.17)(4.0(44.78.17169.172.153.14.03.1 ++=++=
BNqNcNq
qcu
2/52187.3843.13255.349 mkN=++=
Allowable bearing capacity : 2/1304
521mkN
FS
qq uall ===
Thus total allowable gross load, Q
kNBAqQ allall 5.292)5.15.1(1301302 ====
Example 2.2
Given : Repeat example 2.1Assume : Local shear failure conditionSolution :
).........('4.0''867.0 foundationsquareBNqNcNq qcu ++=
For = 20, (Table 2.2); Nc = 11.85, Nq = 3.88, N= 1.12( ) ( ) ( ) ( ) )12.1)(5.1)(8.17)(4.0(88.38.17185.112.15867.0'4.0'''867.0 ++=++=
BNqNNcq
qcu
2/3.2370.121.692.156 mkN=++=
Allowable load :2/3.59
4
3.237mkN
FS
qq uall === ; kNAqQ allall 4.133)5.15.1(3.59 ===
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All equations mentioned before are based on the location ofwater table well below the foundation; if otherwise, some
modification should be made according to the location of thewater table, see Figure 2.4
Figure 2.4 Modification of bearing capacity for water table
Case I : 0 D1 Df
- q(effective surcharge) = '21 DD +
- where :- '- effective unit weight = wsat - sat - saturated unit weight of soil
- w - unit weight of water = 9.81kN/m3 or 62.4 lb/ft3
- '= in the last term of the equation Case II : 0 d B
- the value fDq =
- ( )'' +==B
d
Case III : d B- water has no effect on the qu
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FS
qq uall = , where :
- 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
FS
qqq unetall
=)( , and :
f
unetu
Dq
qqq
=
=)( ;
Where : qall(net) net allowable bearing capacityqu(net) net ultimate bearing capacity
Procedure for FSshear
a. Find developed cohesion,cd and angle of friction,d;
==
shear
d
shear
dFS
andFS
cc
tantan............. 1
)..........(3.03.1
).........(4.03.1
)........(5.0
foundationcircularBNqNcNq
foundationsquareBNqNcNq
foundationstripBNqNcNq
qcu
qcu
qcu
++=
++=
++=
b. Terzaghis equations become (with cd and d):
)..........(3.03.1
).........(4.03.1
)........(5.0
foundationcircularBNqNNcq
foundationsquareBNqNNcq
foundationstripBNqNNcq
qcdu
qcdu
qcdu
++=
++=
++=
With :Nc, Nq, N - bearing capacity factors for d
c. Thus, the net allowable bearing capacity :
( ) BNNqNcqqq qcdallnetall2
11)( ++==
Example 2.3
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UsingFS
qqq unetall
=)( ; and FS = 5; find net allowable load for the
foundation in example 2.1 with qu = 521 kN/m2
With qu = 521 kN/m
2
; q = 1(17.8) = 17.8 kN/m
2
2
)( /64.1005
8.17521mkN
FS
qqq unetall =
=
=
Hence Qall(net) = 100.64(1.5x1.5) = 226.4 kN
Example 2.4
Using Example 3.1, and Terzaghis equation
).........(4.03.1 foundationsquareBNqNcNq qcu ++= with FSshear = 1.5;
Find net allowable load for the foundation
For c=15.2 kN/m2, = 20 and
==
shear
d
shear
dFS
andFS
cc
tantan............. 1
cd =2/13.10
5.1
2.15mkN
FS
c
shear
==
d = tan-1[shearFS
tan] = tan-1[
5.1
20tan] = 13.64
With : BNNqNcq qcdnetall 4.013.1)( ++=
From Table 2.1 : =13.6 ; 2.1N ; 8.3qN ; 12cN (estimation)
Hence :
( ) ( ) ( ) ( ) ( ) ( )2
)(
/2202.128.490.158
2.15.18.174.018.38.171213.103.1
mkN
q netall
=++=
++=
( ) ( ) kNQ netall 4955.15.1220)( ==
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The need to address for rectangular shape foundation where :(0 1
10
tan2
245tan eNq
+=
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( )
+=
B
DF
f
cd
1tan4.01 , ( )
+=
B
DF
f
qd
12 tansin1tan21
1=dF
NOTE : tan-1(Df/B) is in
- inclination2
901
==
qici FF
2
1
=
iF
Where : inclination of load from vertical
For undrained condition ( = 0)
qFFNcq cdcscuu +=
cdcscuuunet FFNcqqq ==)(
Skemptons :
+
+=
L
B
B
Dcq
f
unet 2.012.015)(
Table 2.3Vesics Bearing Capacity Factors for General Equation (1973)
Nc Nq N Nq/ Nc Tan Nc Nq N Nq/ Nc Tan
012345678910
1112131415161718192021
5.145.385.635.906.196.496.817.167.537.928.35
8.809.289.8110.3710.9811.6312.3413.1013.9314.8315.82
1.001.091.201.311.431.571.721.882.062.252.47
2.712.973.263.593.944.344.775.265.806.407.07
0.000.070.150.240.340.450.570.710.861.031.22
1.441.691.972.292.653.063.534.074.685.396.20
0.200.200.210.220.230.240.250.260.270.280.30
0.310.320.330.350.360.370.390.400.420.430.45
0.000.020.030.050.070.090.110.120.140.160.18
0.190.210.230.250.270.290.310.320.340.360.38
2627282930313233343536
3738394041424344454647
22.2523.9425.8027.8630.1432.6735.4938.6442.1646.1250.59
55.6361.3567.8775.3183.8693.71105.11118.37133.88152.10173.64
11.8513.2014.7216.4418.4020.6723.1826.0929.4433.3037.75
42.9248.9355.9664.2073.9085.3899.02115.31134.88158.51187.21
12.5414.4716.7219.3422.4025.9930.2235.1941.0648.0356.31
66.1978.0392.25109.41130.22155.55186.54224.64271.76330.35403.67
0.530.550.570.590.610.630.650.680.700.720.75
0.770.800.820.850.880.910.940.971.011.041.08
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22232425
16.8818.0519.3220.72
7.828.669.6010.66
7.138.209.4410.88
0.460.480.500.51
0.400.420.450.47
484950
199.26229.93266.89
222.31265.51319.07
496.01613.16762.89
1.121.151.20
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
)'2
1( idsqiqdqsqu FFFBNFFFqNq += ;
( ) ( )
2
/6.12187.0 mkNq ==
From Table 2.3 : For =30: Nq = 18.4, N = 22.4, Nq/ Nc = 0.61,Tan = 0.58
( ) 58.158.01tan1 =+=+=B
B
L
BFqs ;
6.04.014.01 =
==B
B
L
BFs
( ) ( ) ( )BBB
DF
f
qd
202.01
7.030sin158.021sin1tan21
22 +=+=+= ;
1=dF
605.090
201
901
22
=
=
== qici FF ;
11.030
2011
22
=
=
=
iF
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So
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) BBB
B
FFFBNFFFqNq idsqiqdqsqu
3.1368.442.22111.016.04.221821605.0202.0158.14.186.12
)'2
1(
++= + +=
+=
BBB
setthusq
q uall 43.489.14
73.73150
:3 2
++==
By trial and error : B=1.3m
Eccentrically loaded foundations give non-uniform distributionof pressure, Figure 2.6
Figure 2.6 Eccentrically loaded foundations
Eccentricity,Q
Me =
qmax and qmin is given by :
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+=
B
e
BL
61max and
=
B
e
BL
61min
if e > B/6, and qmin becomes negative then :
( )eBL
23
4max
=
Factor of safety against bearing capacity failure; effective areamethod, 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, qu :
idsqiqdqsqcicdcscu FFFNBFFFqNFFFcNq '2
1' ++=
- use L and B to find sqscs FandFF ..,
- use B to find dqdcdFandFF
..,
c. Total ultimate load, ( )''''' LBqAqQ uuult == ; where A effective
area
d. Factor of safety,Q
QFS ult=
e. Check FS against qmax ;max
'
q
qFS u=
Example 2.6
Given : A square foundation as shown in Figure 2.7. Using generalbearing capacity factors, (table 2.3)
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Figure 2.7Find : Ultimate load, Qult,
assume one way load eccentricity, e = 0.15m
Solution : with c = 0; idsqiqdqsqu FFFNBFFFqNq '21
' +=
Where :
q = 0.7(18) = 12.6 kN/m2
for = 30, from Table 2.3 : Nq=18.4 and N=22.4B = 1.5 2(0.15) = 1.2mL = 1.5m
Thus values for general bering capacity equations : (using B and L)
462.130tan5.1
2.11tan
'
'1 =
+=+=
L
BFqs
( )( ) ( )
135.15.1
7.0289.01sin1tan21
2 =+=+=B
DF
f
qd
68.05.1
2.14.01
'
'4.01 =
=
=L
BFs
1=dF
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2
21
/2.5495.1647.384
168.04.222.118135.1462.14.186.12'
mkN
q u
=+=
+=
15
Sand :
0
30
/18 3
==
=
c
mkN
1.5m x 1.5 m
0.7 m
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Qult = qu 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 auniform deposit of stiff clay and must support a wall that imposes aloading of 152 kN/m of wall length. Use general bearing capacity
factors.
Find : The width of footing with FS of 3.
Figure 2.8Solution :
22
/9.722
/8.145
2;
........(5.0
mkNmkNq
cwith
foundationstripBNqNcNq
u
qcu
===
++=
And =0; from the Table 2.3 Nc = 5.14, Nq = 1.0 and N
=0
mmkN
mkNdthofwallrequiredwi
mkNmkN
q
mkNBmkNmmkNmkNq
all
ult
15.1/4.132
/0.152
/4.1323
/3.397
/3.397)0)()(/82.18(5.0)0.1)(2.1)(/82.18()14.5)(/9.72(
2
22
2332
==
==
=++=
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B required is 1.5 meter to be conservative
2.8
Example 2.8 :Soil deposit has the following ;=20.44 kN/m3, =30, c=38.3kN/m2
Square footing located 1.52 m below surface, carries 2670 kN andgroundwater is negligible. Use Terzaghis values, (Table 2.1).
Find : The right dimension B. Use Terzaghis equation).........(4.03.1 foundationsquareBNqNcNq qcu ++=
With =30; Nc=37.16, Nq=22.46, N
=19.13
Assume B=3 m;
mBBmmkN
kNwallofwidthrequired
mkNmkN
q
mkNmkN
mmkNmmkNmkNq
all
ult
63.165.2/7.1005
2670
/7.10053
/3017
/3017/4696981850
)13.19)(3)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1
22
2
22
22
332
====
==
=++=
++=
Assume B=1 m;
17
2670 kN
= 20.44 kN/m3
1.52m =30
c = 38.3 kN/m2Figure 2.9
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mBBmmkN
kNwallofwidthrequired
mkNmkN
q
mkNmkN
mmkNmmkNmkNq
all
ult
65.172.2/980
2670
/9803
/2939
/2939/3916981850
)13.19)(1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1
222
22
22
332
====
==
=++=
++=
Assume B=2m;
mBBmmkN
kNwallofwidthrequired
mkNmkN
q
mkNmkN
mmkNmmkNmkNq
all
ult
67.180.2/954
2670
/9543
/2861
/2861/3136981850
)13.19)(2)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1
22
2
22
22
332
====
==
=++=
++=
Assume B=1.8m;
mBBmmkN
kNwallofwidthrequired
mkNmkN
q
mkNmkN
mmkNmmkNmkNq
all
ult
68.183.2/943
2670
/9433
/2830
/2830/2826981850
)13.19)(8.1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1
22
2
22
22
332
====
==
=++=
++=
Assume B=1.7m;
mBBmmkN
kNwallofwidthrequired
mkNmkN
q
mkNmkNmmkNmmkNmkNq
all
ult
7.185.2/938
2670
/9383
/2814
/2814/2666981850)13.19)(7.1)(/44.20(4.0)46.22)(52.1)(/44.20()16.37)(/3.38(3.1
22
2
22
22
332
====
==
=++=++=
Therefore use 1.7m x 1.7m
2.9
Can be computed by using flexural formula of :
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y
y
x
x
I
xM
I
yM
A
Qq =
Where :
q contact pressureQ total axial vertical loadA area of footing
Mx, My total moment about respective x and y axesIx, Iy moment of inertia about respective x and y axesx, 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 loadof 222.4kN. Estimate soil contact pressure and FS against bearingcapacity.
Given :1.52m by 1.52m square footing; P=222.4kN; soil =18.85kN/m3
concrete =24 kN/m3; qu = 143.64 kN/m2
Find :a. Soil contact pressureb. FS against bearing capacity pressure
19
0.14m20.91m
1.22m
0.31m1.52m
Figure 2.10
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Solution :
a.y
y
x
x
I
xM
I
yM
A
Qq = ; Mx=My=0; since load on centroid
Total load calculation, Q :
Column load, P = 222.4kNWeight of footing base= (1.52m)(1.52m)0.31m(24kN/m3) = 17.19 kNWeight of footing pedestal= (0.14m)(0.14m)(0.91m)(24kN/m3) = 0.43 kNWeight of backfill soil
= [(1.52m)(1.52m)-(0.14m)(0.14m)](0.91m) x 18.85kN/m3
= 39.3kN
Q = 222.4 + 17.19 + 0.43 + 39.3 = 279.32kNArea, A = 1.52mx1.52m = 2.31m2
Soil contact pressure or Stress, q = Q/A = 120.9 kN/m2
b.2
2
/82.712
/64.143
2
4.02.1
mkNmkNq
c
BNNDcNq
u
qfcult
===
++=
Assuming cohesive soil has : =0 and c>0; thus :Nc=5.14, Nq=1.0, N=0, Df=1.22m
85.39.120
98.465
/98.465
0)0.1)(22.1(85.18)14.5)(82.71(2.14.02.1
2
===
=
++=++=
q
qFS
mkN
BNNDcNq
ult
qfcult
Since FS > 3.0; thus ok.
20
Example 2.10
Draw soil contact pressure for
footing in Figure 2.11
Conversion to SI unitP=222.4 kN;H=88.96 kN;M=81.35kN.m;W=88.96 kNDf=1.22m;B=2.29m (7.5ft);
L=1.52m (5ft)
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Given : 2.29m by 1.52m rectangular footingFind : Contact pressure and soil pressure diagram
Solution :
Using flexural formula;y
y
x
x
I
xM
I
yM
A
Qq =
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; 43
52.112
)29.2(52.1m
mmIy ==
22
22
42
/53.53...../47.232
/143/47.8952.1
)145.1)(88.189(
48.3
36.311
mkNqandmkNq
mkNmkNm
mmkN
m
kNq
leftright =+=
=
=
Take V = 0 and Mc = 0 will produce :
V = 0 : )........(36.311)52.1(2
......0))((2
AkNmqd
andWPLdq
=
=
Mc = 0 : see Figure 2.12 (b) and (c)
03
))((2
))(( =
+
dxLd
qSHM
21
Figure 2.11
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( ) ( )
2/46.254.,36.311)52.1)(61.1(2
:)(int.
61.1079.10351.35653.10835.81
.....032
29.236.31122.196.88.35.81
mkNqkNmmq
Aosubstitue
mdd
Bdm
mkNmkN
==
==++
=
+
Example 2.11
22
Figure 2.12 (a) and (b)
1.61m
254.46kN/m2
2.29m9m
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Given : A 6 ft x 6 ft footing as shown; load P=60kips; weight ofconcrete footing including pedestal + base pad, W1=9.3kips; backfill,W2=11.2kips; horizontal load = 4kips; qall for soil = 3.0 kips/ft2.
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, = 0.40FS against overturning.
Solution :
1.y
y
x
x
I
xM
I
yM
A
Qq =
Q=P+W1+W2=60+9.3+11.2=80.5kipsA=6ftx6ft=36ft2
My=4kipsx4.5ft=18kip-ft (about point C)x=6ft/2=3ftIy=6ft(6ft)3/12=108ft4; Mx=0; Mxy/Ix=0
22
42/50.0/24.2
108
)3(.18
36
5.80ftkipftkips
ft
ftftkip
ft
kips
I
xM
I
yM
A
y
y
x
x ===
So : qright = 2.74 kips/ft2 < 3.0 kips/ft2 ; OKqleft = 1.74 kips/ft2 < 3.0 kips/ft2 ; OK
23
gure 2.13
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2. FDG and EDH are similar triangles; so
( ) ftkipsftkipsft
kipsAAatMoment
kipskipskips
ftftkipftftftkipsftAAatShear
ftkipDEft
ftDEftFG
ft
ftft
EHftkipsDFFG
EH
DF
DE
.7.3925.253.22
25.293.31:...
46.3453.293.31
)6)(/375.0)(25.2()6)(/375.074.2(25.2:...
/375.0;..6
25.2
0.1;...6
25.22
5.1
2
6
;.../0.174.174.2;.....
32
2
212
2
2
=+
=
=+=
+=
===
=====
3.( )
05.84
)40.0(2.113.960
.........
=++
=
=
kips
kipskipskips
forcesHorizontaleandsoilbetweenbasfrictionoftcoefficienloadverticalTotalslidingagainstFS
4. 4.13)5.4(4
)2/6(5.80
.
.Re.. ===
ftkips
ftkips
momentTurning
momentsistinggoverturninagainstFS
24
Pressure diagram
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Foundation settlement under load can be classified according to twomajor types :
(a) immediate or elastic settlement, Se
(b) consolidation settlement, Sc
Elastic settlement, Se takes place immediately during or afterconstruction of structure.
Consolidation settlement, Sc is time dependent comprises of twophases; namely, primary and secondary consolidation settlement.
Elastic settlement of foundations on saturated clay is given by Janbuet al., (1956) using the equation :
s
eE
BqAAS 021=
where :
A1 is a function of H/B and L/B and A2 is a function of Df/B
All parameters of H, B and Df (with L into the paper) are asshown in Figure 2.14.
25
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Parameters
A2 Versus Df/B
A1 Versus H/B and L/B
26
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Schmertmann, (1978) proposed that the elastic settlement in sandy
soil as :
( ) =2
0
21
z
s
ze z
E
IqqCCS
where :
Iz strain influence factor
C1 correction factor due to depth =( )
q5.01
C2 correction factor due to soil creep =
+
1.0log2.01
yearsintime
q - stress at the level of foundation (due to loading + selfweight of footing + weight of soil above footing)
fDq =
Calculation of elastic settlement using strain influencefactor
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The variation of Iz with depth below the footing for square orcircular are as below :
Iz = 0.1 at z = 0Iz = 0.5 at z = z1 = 0.5BIz = 0 at z = z2 = 2B
Footing with L/B 10 (rectangular footing) :
Iz = 0.2 at z = 0Iz = 0.5 at z = z1 = BIz = 0 at z = z2 = 4B
Elastic parameters such as Es and s in Table 2.4 can be used if thereal laboratory test results not available.
Elastic parameters of various soils
Type of soil Modulus of Elasticity,Es(MN/m2)
Poissons ratio, s
Loose sand 10.5 24.0 0.20 0.40Medium dense sand 17.25 27.60 0.25 0.40
Dense sand 34.50 55.20 0.30 0.45Silty sand 10.35 17.25 0.20 0.40
Sand and gravel 69.00 172.50 0.15 0.35Soft clay 4.1 20.7
0.20 0.50Medium clay 20.7 41.4Stiff clay 41.4 96.6
(a) Primary consolidation, Sc
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Many methods were developed in estimating the value ofconsolidation settlement, Sc.
Due to simplicity only chart based on Newmarks (1942), Figure 2.18will be used in estimating the consolidation settlement.
Primary consolidation, Sc calculated as :
00
log1 p
p
e
HCS cc
+
=
where : Cc compression index (given)
H thickness of clay layere0 initial void ratio (given)p = p0 + p, final pressure
p0 overburden pressurep =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 rigidand isotropic.
Example 2.7
Given :
30
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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, whichinclude all self weight. The initial void ratio, e0 is 1.38 andcompression 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/m2
Weight of excavation = 19.83(200 - 198) + (19.83 9.81)(198 195.5) = 64.7kN/m2
( ) ( ) ( )( )[ ] 2
0
/3.2355.19519881.983.1919820083.19
63
5400
,
mkN
mm
kN
excavationofweightpressureloadqincreasestressNet
=+
=
=
By dividing the base into 4 equal size of 1.5m by 3.0m :
mz = 1.5m nz = 3.0m
mmm
mz 7.62
6.1850.1925.195 =
+=
224.07.6
5.1
== mm
m ; 448.07.60.3
== mm
n
From Figure 2.18, the influence coefficient is 0.04
Therefore ; ( ) 22 /6.37/3.23504.04 mkNmkNp ==
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Final pressure, p = p0 + p = 123.1 + 37.6 = 160.7 kN/m2.
Therefore; mmkN
mkNm
p
p
e
HCS cc 212.0
/1.123
/7.160log
38.11
4.668.0log
1 2
2
00
=
+=
+
=
Secondary settlement, Ss is computed from the following calculation(U.S. Department of the Navy, 1971)
=
p
ss
t
tHCS log
where :
Ss secondary compression settlementC coefficient of secondary compression, can be determined
from Figure3.26H thickness of clay layer that is consideredts time for which settlement is requiredtp time to completion of primary consolidation
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Value of C
Bowles (1977) proposed a correlation of the net allowable
bearing pressure for foundations with SPT (N-values). The following equations are used :
( )( ) mBforS
FNmkNq dallnet 22.14.25
16.19)/( 2)(
=
And
( ) ( ) mBforS
FB
BNmkNq dallnet 22.1
4.2528.3
128.398.11)/(
2
2
)( >
+=
Where :
Fd depth factor = 33.133.01
+ B
Df
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 SPTnumbers from field exploration as shown in the table.
Assume that the footing must be 1.5m deep, the tolerablesettlement 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=10With S=25.4mm and N=10
( ) ( ) ddallnet FB
BF
B
BmkNq
22
2
)(28.3
128.38.119
4.25
4.25
28.3
128.31098.11)/(
+=
+=
33.133.01
+=
B
DF
f
d
By trial and error (set the table for calculation)
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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) :
f
unetu
Dq
qqq
=
=)(
; idsqiqdqsqcicdcscu FFFBNFFFqNFFFcNq 21++=
( ) qFFFBNFFFqNqqq idsqiqdqsqultnetu +== 2
1
For N=10; friction angle of=34 is considered (from table onSI)
With no inclination so Fqi=Fi=1.0
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From table 2.3 Nq=29.44, N=41.06
So for a tolerable settlement of 25.4mm, the SF required iscalculated as : SF=Qnet(u)/Q= 10,322kN/1000kN = 10.3 whichis OK, therefore most design controlled by tolerable criterion.
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