CH2 Shallowfoundation (SEM2 200910)
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Transcript of CH2 Shallowfoundation (SEM2 200910)
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2.0 SHALLOW FOUNDATION :
2.1
General Concept
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 : L B BL B 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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2.2 Terzaghi’s Bearing Capacity
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 DFas arcs of logarithmic spiral
- Rankine passive zones AFH and CEG
CAD and ACD are assume to equal friction angle, Ø
Thus ultimate bearing capacity, qu for general shear failure can
be expressed as :
)..........(3.03.1
).........(4.03.1
)........(5.0
foundationcircular BN qN cN q
foundation square BN qN cN q
foundation strip BN qN cN q
qcu
qcu
qcu
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Where : c – cohesion of soil - unit weight of soil
q = Df Nc, Nq, N - bearing capacity factors
And
1cot1
24cos2
cot2
tan2/4/32
qc N e
N
245cos2 2
tan2/4/32
e N q
tan1cos2
12
p K N
where p K - passive pressure coefficient
Table 2.1 summarizes values for Nc, Nq, and N
Table 2.1 Terzaghi’s Bearing Capacity’s Factors
Ø Nc Nq N Ø Nc Nq N
01234567891011121314
1516171819202122232425
5.706.006.306.626.977.347.738.158.609.099.6110.1610.7611.4112.11
12.8613.6814.6015.1216.5617.6918.9220.2721.7523.3625.13
1.001.101.221.351.491.641.812.002.212.442.692.983.293.634.02
4.454.925.456.046.707.448.269.1910.2311.4012.72
0.000.010.040.060.100.140.200.270.350.440.560.690.851.041.26
1.521.822.182.593.073.644.315.096.007.088.34
262728293031323334353637383940
41424344454647484950
27.0929.2431.6134.2437.1640.4144.0448.0952.6457.7563.5370.0177.5085.9795.66
106.81119.67134.58151.95172.28196.22224.55258.28298.71347.50
14.2115.9017.8119.9822.4625.2828.5232.2336.5041.4447.1653.8061.5570.6181.27
93.85108.75126.50147.74173.28204.19241.80287.85344.63415.14
9.8411.6013.7016.1819.1322.6526.8731.9438.0445.4154.3665.2778.6195.03115.31
140.51171.99211.56261.60325.34407.11512.84650.67831.991072.80
From Kumbhojkar (1993)
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And ultimate bearing capacity, qu for local shear failure can beexpressed as :
)..........('3.0''867.0
).........('4.0''867.0
)........('5.0''32
foundationcircular BN qN cN q
foundation square BN qN cN q
foundation strip BN qN cN q
qcu
qcu
qcu
Where : N’c, N’q, N’ (see Table 2.2) are reduced bearing
capacity factors can be calculated by using N’c, N’q, N’ -
bearing capacity factors with
tan
3
2tan' 1
Table 2.2 Terzaghi’s Modified Bearing Capacity’s Factors
Ø N’ c N’ q N’ Ø N’ c N’ q N’
012345678
9101112131415161718192021222324
25
5.705.906.106.306.516.746.977.227.47
7.748.028.328.638.969.319.6710.0610.4710.9011.3611.8512.3712.9213.5114.14
14.80
1.001.071.141.221.301.391.491.591.70
1.821.942.082.222.382.552.732.923.133.363.613.884.174.484.825.20
5.60
0.000.0050.020.040.0550.0740.100.1280.16
0.200.240.300.350.420.480.570.670.760.881.031.121.351.551.741.97
2.25
262728293031323334
35363738394041424344454647484950
15.5316.3017.1318.0318.9920.0321.1622.3923.72
25.1826.7728.5130.4332.5334.8737.4540.3343.5447.1351.1755.7360.9166.8073.5581.31
6.056.547.077.668.319.039.8210.6911.67
12.7513.9715.3216.8518.5620.5022.7025.2128.0631.3435.1139.4844.4550.4657.4165.60
2.592.883.293.764.394.835.516.327.22
8.359.4110.9012.7514.7117.2219.7522.5026.2530.4036.0041.7049.3059.2571.4585.75
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Example 2.1
Given : A square foundation, 1.5m x 1.5m in plan viewSoil parameters :
Ø’ = 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 foundation square BN qN N cq qcu
For Ø’ = 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 BN qN cN q qcu 2/52187.3843.13255.349 mkN
Allowable bearing capacity : 2/1304
521mkN
FS
qq uall
Thus total allowable gross load, Q
kN B AqQ all all 5.292)5.15.1(130130 2
Example 2.2
Given : Repeat example 2.1 Assume : Local shear failure conditionSolution :
).........('4.0''867.0 foundation square BN qN cN q qcu
For Ø’ = 20°, (Table 2.2); Nc = 11.85, N
q = 3.88, N
= 1.12
)12.1)(5.1)(8.17)(4.0(88.38.17185.112.15867.0'4.0'''867.0
BN qN N cqqcu
2/3.2370.121.692.156 mkN
Allowable load :2/3.59
4
3.237mkN
FS
qq uall ; kN AqQ all all 4.133)5.15.1(3.59
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2.3
Effect of Water Table on Bearing Capacity
All equations mentioned before are based on the location of
water table well below the foundation; if otherwise, somemodification 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 D D
-
where :- ' - effective unit weight = w sat
- sa t - 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 f Dq
-
'' Bd
Case III : d ≥ B -
water has no effect on the qu
Note : the values of bearing capacity factors used strictly depending onwhether the condition is general or local shear failure.
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2.4 Factor of Safety, FS
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 unet all
)( , and :
f
unet u
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,Ød;
shear
d
shear
d FS
and FS
cc
tantan............. 1
)..........(3.03.1
).........(4.03.1
)........(5.0
foundationcircular BN qN cN q
foundation square BN qN cN q
foundation strip BN qN cN q
qcu
qcu
qcu
b. Terzaghi’s equations become (with cd and Ød):
)..........(3.03.1
).........(4.03.1
)........(5.0
foundationcircular BN qN N cq
foundation square BN qN N cq
foundation strip BN qN N cq
qcd u
qcd u
qcd u
With : Nc, Nq, N - bearing capacity factors for Ød
c. Thus, the net allowable bearing capacity :
BN N q N cqqq qcd all net all 2
11)(
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Example 2.3
Using FS
qqq unet all
)( ; and FS = 5; find net allowable load for the
foundation in example 2.1 with qu = 521 kN/m
2
With qu = 521 kN/m2; q = 1(17.8) = 17.8 kN/m2
2
)( /64.1005
8.17521mkN
FS
qqq unet all
Hence Qall(net) = 100.64(1.5x1.5) = 226.4 kN
Example 2.4
Using Example 3.1, and Terzaghi’s equation
).........(4.03.1 foundation square BN qN cN q qcu with FSshear = 1.5;
Find net allowable load for the foundation
For c=15.2 kN/m2, Ø = 20° and
shear
d
shear
d
FS
and
FS
cc
tantan............. 1
cd =2/13.10
5.1
2.15mkN
FS
c
shear
Ød = tan-1[
shear FS
tan] = tan-1[
5.1
20tan] = 13.64°
With : BN N q N cq qcd net all 4.013.1)(
From Table 2.1 : Ø=13.6° ; 2.1 N ; 8.3q N ; 12c N (estimation)
Hence :
2
)(
/2202.128.490.158
2.15.18.174.018.38.171213.103.1
mkN
q net all
kN Q net all 4955.15.1220)(
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2.5 The General Bearing Capacity Equation
The need to address for rectangular shape foundation where :(0 1
tan2
245tan e N q
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B
D F
f
cd
1tan4.01
B
D F
f
qd
12 tansin1tan21 1d F
NOTE : tan-1(Df /B) is in radian
-
inclination2
901
qici F F
2
1
i F
Where : β – inclination of load from vertical
For undrained condition (Ø = 0)
q F F N cq cd cscuu
cd cscuuunet F F N cqqq )(
Skempton’s :
L
B
B
Dcq f
unet 2.012.015)(
Table 2.3 Vesic’s Bearing Capacity Factors for General Equation (1973)
Ø Nc Nq N Nq / Nc Tan Ø Ø Nc Nq N Nq / Nc Tan Ø012345678910111213141516171819202122232425
5.145.385.635.906.196.496.817.167.537.928.358.809.289.8110.3710.9811.6312.3413.1013.9314.8315.8216.8818.0519.3220.72
1.001.091.201.311.431.571.721.882.062.252.472.712.973.263.593.944.344.775.265.806.407.077.828.669.6010.66
0.000.070.150.240.340.450.570.710.861.031.221.441.691.972.292.653.063.534.074.685.396.207.138.209.4410.88
0.200.200.210.220.230.240.250.260.270.280.300.310.320.330.350.360.370.390.400.420.430.450.460.480.500.51
0.000.020.030.050.070.090.110.120.140.160.180.190.210.230.250.270.290.310.320.340.360.380.400.420.450.47
26272829303132333435363738394041424344454647484950
22.2523.9425.8027.8630.1432.6735.4938.6442.1646.1250.5955.6361.3567.8775.3183.8693.71105.11118.37133.88152.10173.64199.26229.93266.89
11.8513.2014.7216.4418.4020.6723.1826.0929.4433.3037.7542.9248.9355.9664.2073.9085.3899.02115.31134.88158.51187.21222.31265.51319.07
12.5414.4716.7219.3422.4025.9930.2235.1941.0648.0356.3166.1978.0392.25109.41130.22155.55186.54224.64271.76330.35403.67496.01613.16762.89
0.530.550.570.590.610.630.650.680.700.720.750.770.800.820.850.880.910.940.971.011.041.081.121.151.20
0.490.510.530.550.580.600.620.650.670.700.730.750.780.810.840.870.900.930.971.001.041.071.111.151.19
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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( id sqiqd qsqu F F F BN F F F qN q ;
2/6.12187.0 mkN q
From Table 2.3 : For Ø’ =30°: Nq = 18.4, N = 22.4, Nq / Nc = 0.61,Tan Ø = 0.58
58.158.01tan1 B
B
L
B F qs ; 6.04.014.01
B
B
L
B F s
B B B
D F
f
qd
202.01
7.030sin158.021sin1tan21
22 ; 1d F
605.090
201
901
22
qici F F ; 11.0
30
2011
22
i F
So
B B
B B
F F F BN F F F qN q id sqiqd qsqu
3.1368.44
2.22111.016.04.22182
1605.0
202.0158.14.186.12
)'2
1(
B B B
set thusq
q uall 43.489.14
73.73150:
3 2
By trial and error : B=1.3m
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2.6
Eccentrically Loaded Foundations
Eccentrically loaded foundations give non-uniform distribution
of pressure, Figure 2.6
Figure 2.6 Eccentrically loaded foundations
Eccentricity,Q
M e
qmax and qmin is given by :
B
e
BL
Qq
61max and
B
e
BL
Qq
61min
if e > B/6, and qmin becomes negative then :
e B LQ
q23
4max
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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 :
id sqiqd qsqcicd cscu F F F N B F F F qN F F F cN q '2
1'
- use L’ and B’ to find sqscs F and F F ..,
-
use B to find d qd cd F and F F ..,
c. Total ultimate load, ''' '' L Bq AqQ uuult ; where A’ – effective
area
d.
Factor of safety,Q
Q FS ult
e.
Check FS against qmax ;max
'
q
q FS u
Example 2.6
Given : A square foundation as shown in Figure 2.7. Using generalbearing capacity factors, (table 2.3)
Figure 2.7
Sand :
0
30/18 3
c
mkN
1.5m x 1.5 m
0.7 m
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Find : Ultimate load, Qult,assume one way load eccentricity, e = 0.15m
Solution : with c = 0;id sqiqd qsqu F F F N B F F F qN q '
2
1'
Where :
q = 0.7(18) = 12.6 kN/m2
for Ø = 30°, from Table 2.3 : Nq=18.4 and N =22.4
B’ = 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
B F qs
135.15.1
7.0289.01sin1tan21
2
B
D F
f
qd
68.05.1
2.14.01'
'4.01
L
B F s
1d F
2
21
/2.5495.1647.384
168.04.222.118135.1462.14.186.12'
mkN
q u
Qult = q’ u X A’ = 549.2 X (1.5X1.2) = 988kN
Qall = 988/3 = 330kN with FS=3
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2.7 Load on strip footing
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 capacityfactors.
Find : The width of footing with FS of 3.
Figure 2.8Solution :
2
2
/9.722
/8.1452
;
........(5.0
mkN mkN qcwith
foundation strip BN qN cN q
u
qcu
And Ø=0°; from the Table 2.3 Nc = 5.14, Nq = 1.0 and N =0
m
mkN
mkN dthofwall requiredwi
mkN mkN
q
mkN BmkN mmkN mkN q
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
B required is 1.5 meter to be conservative
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2.8 Dimension of loaded square pad footing
Example 2.8 :Soil deposit has the following ; =20.44 kN/m3, Ø=30°, c=38.3kN/m2
Square footing located 1.52 m below surface, carries 2670 kN andgroundwater is negligible. Use Terzaghi’s values, (Table 2.1).
Find : The right dimension B. Use Terzaghi’s equation).........(4.03.1 foundation square BN qN cN q qcu
With Ø=30°; Nc=37.16, Nq=22.46, N =19.13
Assume B=3 m;
m B BmmkN
kN wall of widthrequired
mkN mkN
q
mkN mkN
mmkN mmkN mkN q
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;
m B BmmkN
kN wall of widthrequired
mkN mkN
q
mkN mkN
mmkN mmkN mkN q
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
22
2
22
22
332
Assume B=2m;
2670 kN
γ = 20.44 kN/m3
1.52m Ø=30˚
c = 38.3 kN/m2
Figure 2.9
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m B BmmkN
kN wall of widthrequired
mkN mkN
q
mkN mkN
mmkN mmkN mkN q
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
222
22
22
332
Assume B=1.8m;
m B BmmkN
kN wall of widthrequired
mkN mkN
q
mkN mkN
mmkN mmkN mkN q
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;
m B BmmkN
kN wall of widthrequired
mkN mkN
q
mkN mkN
mmkN mmkN mkN q
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
Contact Pressure and stability check.
Can be computed by using flexural formula of :
y
y
x
x
I
x M
I
y M
A
Qq
Where :
q – contact pressureQ – total axial vertical load
A – area of footingMx, My – total moment about respective x and y axesIx, Iy – moment of inertia about respective x and y axes
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x, y – distance from centroid to the outer most point atwhich the contact pressure is computed along respective x andy 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; so il =18.85kN/m
3
concrete =24 kN/m3; qu = 143.64 kN/m
2
Find :a.
Soil contact pressureb. FS against bearing capacity pressure
Solution :
a. y
y
x
x
I
x M
I
y M
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
222.4KN
0.14m20.91m
1.22m
0.31m1.52m
Figure 2.10
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= [(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 = 279.32kN
Area, 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
mkN mkN q
c
BN N DcN q
u
q f cult
Assuming cohesive soil has : Ø=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
q FS
mkN
BN N DcN q
ult
q f cult
Since FS > 3.0; thus ok.
Example 2.10
Draw soil contact pressure forfooting 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)
Figure 2.11
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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 x M
I y M
AQq
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
mm I y
22
22
42
/53.53...../47.232
/143/47.8952.1
)145.1)(88.189(
48.3
36.311
mkN qand mkN q
mkN mkN m
mmkN
m
kN q
left right
Take ΣV = 0 and ΣMc = 0 will produce :
ΣV = 0 : )........(36.311)52.1(2
......0))((2
AkN mqd
and W P Ld q
ΣMc = 0 : see Figure 2.12 (b) and (c)
03
))((2
))((
d x Ld
qS H M
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
mkN qkN mmq
Ao substituemd d
Bd m
mkN mkN
Figure 2.12 (a) and (b)
1.61m
254.46kN/m2
2.29m2.29m
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Example 2.11 Checking stability on shallow foundation
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/ft
2.
Find :1.
Contact pressure and soil pressure diagram.2.
Shear and moment at section A-A (in the Figure E3.14)3.
FS against sliding if coefficient of friction, δ = 0.40
4.
FS against overturning.
Solution :
1. y
y
x
x
I
x M
I
y M
A
Qq
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
22
42 /50.0/24.2
108
)3(.18
36
5.80 ft kip ft kips
ft
ft ft kip
ft
kips
I
x M
I
y M
A
Qq
y
y
x
x
So : qright = 2.74 kips/ft2 < 3.0 kips/ft2 ; OK
qleft = 1.74 kips/ft2 < 3.0 kips/ft2 ; OK
Figure 2.13
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2. ΔFDG and ΔEDH are similar triangles; so
ft kips ft kips ft
kips A Aat Moment
kipskipskips
ft ft kip ft ft ft kips ft A Aat Shear
ft kip DE ft
ft DE ft FG
ft
ft ft
EH ft kips DF FG
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
forces Horizontal eandsoil betweenbas frictionof t coefficienload vertical Total sliding against FS
4. 4.13)5.4(4
)2/6(5.80
.
.Re..
ft kips
ft kips
moment Turning
moment sisting g overturninagainst FS
Pressure diagram
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2.10 Settlement of shallow foundationFoundation 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.
2.10.1 Elastic settlement of foundations on saturated clay
Elastic settlement of foundations on saturated clay is given by Janbuet al., (1956) using the equation :
s
e E
Bq A AS 021
where :
A 1 is a function of H/B and L/B and A 2 is a function of Df /B All parameters of H, B and Df (with L into the paper) are asshown in Figure 2.14.
Figure 2.14 : Parameters
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Figure 2.15 : A 2 Versus Df /B
Figure 2.16 : A 1 Versus H/B and L/B
2.10.2 Elastic settlement of foundations on sandy soil: use ofstrain influence factor
Schmertmann, (1978) proposed that the elastic settlement in sandysoil as :
2
0
21
z
s
z e z
E
I qqC C S
where :
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Iz – strain influence factor
C1 – correction factor due to depth =
qq
q5.01
C2 – correction factor due to soil creep =
1.0
log2.01 yearsintime
q - stress at the level of foundation (due to loading + self
weight of footing + weight of soil above footing)
f Dq
Figure 2.17 : Calculation of elastic settlement using strain influencefactor
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 = 0
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Iz = 0.5 at z = z1 = BIz = 0 at z = z2 = 4B
2.10.3 Range of material parameters
Elastic parameters such as E s and μ s in Table 2.4 can be used if thereal laboratory test results not available.
Table 2.4 : Elastic parameters of various soils
Type of soil Modulus of Elasticity,Es
(MN/m2
)
Poisson’s ratio, μs
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
2.10.4 Consolidation settlement
(a) Primary consolidation, Sc
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 :
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00
log1 p
p
e
H C S cc
where : Cc – compression index (given)
H – thickness of clay layere0 – 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
Figure 2.18 : Chart for determining stresses below corners of rigidand isotropic.
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Example 2.7
Given :
Figure 2.19
A foundation to be constructed as in Figure 2.19. The base of thefoundation 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
20
/3.2355.19519881.983.1919820083.1963
5400
,
mkN mm
kN
excavationof weight pressureload qincrease stress Net
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By dividing the base into 4 equal size of 1.5m by 3.0m :
mz = 1.5m nz = 3.0m
mmm
m z 7.62
6.1850.1925.195
224.07.6
5.1
m
mm ; 448.0
7.6
0.3
m
mn
From Figure 2.18, the influence coefficient is 0.04
Therefore ; 22 /6.37/3.23504.04 mkN mkN p
Final pressure, p = p0 + Δp = 123.1 + 37.6 = 160.7 kN/m2.
Therefore; mmkN
mkN m
p
p
e
H C S cc 212.0
/1.123
/7.160log
38.11
4.668.0log
1 2
2
00
(b) secondary consolidation
Secondary settlement, Ss is computed from the following calculation(U.S. Department of the Navy, 1971)
p
s s
t
t H C S log
where :
Ss – secondary compression settlement
Cα – coefficient of secondary compression, can be determinedfrom Figure
3.26H – thickness of clay layer that is consideredts – time for which settlement is requiredtp – time to completion of primary consolidation
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Figure 2.20 : Value of Cα
2.11 Allowable bearing pressure in sand based onsettlement consideration.
Bowles (1977) proposed a correlation of the net allowablebearing pressure for foundations with SPT (N-values).
The following equations are used :
m B for S
F N mkN q d all net 22.14.25
16.19)/( 2)(
And
m B for S
F B
B N mkN q d all net 22.1
4.2528.3
128.398.11)/(
2
2
)(
Where :
Fd – depth factor = 33.133.01
B
D f
S – tolerable settlement, in mm.
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Example 2.8
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
d d all net F B
B F
B
BmkN q
22
2
)(28.3
128.38.119
4.25
4.25
28.3
128.31098.11)/(
33.133.01
B
D F
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
unet u
Dq
qqq
)( ; id sqiqd qsqcicd cscu F F F BN F F F qN F F F cN q
2
1
q F F F BN F F F qN qqq id sqiqd qsqult net u
2
1
For N=10; friction angle of Ø=34˚ is considered (from table onSI)
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With no inclination so Fqi=Fγi=1.0From table 2.3 Nq=29.44, Nγ=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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