CE 366 – BEARING CAPACITY (Problems & Solutions)
P1
Question:
An excavation will be made for a ten storey 15x25 m building. Temporary support of
earth pressure and water pressure will be made by deep secant cantilever pile wall. The
gross pressure due to dead and live loads of the structure and weight of the raft is 130 kPa
(assume that it is uniform).
this level can only be fill (placed after
attained after a construction is
water proofing will relatively long time fully completed)
be provided
10 storey building (15x25m)
2 1m
original GWT 4m
position
1 sat = 20 kN/m3
moist =18 kN/m3
medium medium
dense dense GWT is lowered 4m
sand sand
medium stiff clay 2m
sat = 21 kN/m3
a) What is net foundation pressure at the end of construction but before the void space
between the pile wall and the building has been filled, and there is no water inside the
foundation pit yet (water level at the base level) (GWT position 1).
b) What is net foundation pressure long after the completion of the building, i.e. water
level is inside the pile wall and the backfill between the building and the pile wall is
placed (GWT position 2). What is the factor of safety against uplift?
1
Solution:
a) qnet = final effective stress - initial effective stress
at foundation level at foundation level
1m moist = 18 kN/m3
5m
sat = 20 kN/m3
o‘ o‘ = 18x1 + 4x(20-9.8) = 58.8 kPa
( gross pressure – uplift pressure) = final effective stress at foundation level, f ’
gross pressure =130 kPa (given)
uplift pressure = 0 kPa (Since GWT is at foundation level (1), it has no effect on
structure load)
f’ =130 –0 = 130 kPa
qnet =130 – 58.8
= 71.2 kPa
b) f ’ = 130 – 4x9.8 = 90.8 kPa
uplift pressure
o’ = 58.8 kPa (same as above)
qnet = 90.8 – 58.8
= 32.0 kPa
OR
qnet=qgross-satD =130-(18x1+4x20)
=32.0 kPa
Factor of safety against uplift is:
(FS)uplift = weight of structure / uplift
= (130x15x25) / (4x9.8x15x25)
= 3.3
2
P2
Question:
Calculate the FS against uplift and calculate effective stress at the base level for water
level at (1) and (2) for the canal structure given below. Note that the canal is very long
into the page.
0.75 3.50 0.75 concrete = 24 kN/m3
5.0 m
(2) (2)
3.0
2.85
ground very long
level concrete pit
1.0
(1) (1) 3.0 m
waterproof 2.0
membrane
1.0
Solution:
water table at (1)
Factor of Safety against uplift = (2x6x0.75 + 5x1)x24 / (3x5)x9.8
weight of pit uplift
= 336 / 147
= 2.28
Base pressure = 336 / 5 = 67.2 kN/m2 due to weight of structure.(per meter of canal)
147 / 5 = 29.4 kN/m2 is supported by groundwater
67.2 – 29.4 = 37.8 kN/m2 is supported by soil (effective stress at the base)
base pressure 29.4 kPa : supported by
due to 67.2 kPa groundwater (uplift)
structure 37.8 kPa : supported by
soil
3
water table at (2)
FS = 336 / (6.85x5x9.8)
= 1.0 < 1.5 NOT OKEY
⇒ base pressure = 67.2 kPa is supported by ground water
uplift = weight of structure
Soil does not carry any load, structure tends to float
4
P3
Question:
A residential block will be constructed on a clay deposit. The building will rest on a mat
foundation at 2m depth and has 20mx20m dimensions in plan.
The clay deposit is 26m deep and overlies limestone. The groundwater level is at 2m
depth. The bulk unit weights are 18 and 20 kN/m3 above and below water table
respectively.
The clay has c’=5 kN/m2, ’=200, cu=48 kN/m2, u=0. The coefficient of volume
compressibility is 1.00x10-4 m2/kN at the ground surface and decreases with depth at a
rate of 0.02x10-4 m2/kN per meter. Use Eu/cu = constant = 1250 and Is = 1.2
a) Calculate ultimate bearing capacity of the foundation in the short term?
b) For the foundation described above what is the (gross) allowable bearing capacity?
NOTE: For u=0 case use Skempton values, use a safety factor of 3.00 against shear
failure of the foundation. Use sublayers. Maximum allowable total settlement of the
building is 15 cm.
Solution:
2m
20x20
z d=18kN/m3
26m
sat=20kN/m3
c’=5kPa ’=20
cu=48 kPa u
limestone
Skempton expression for u is : qf = cuNc + sat D (total stress analysis)
qnf = cuNc
5
Short Term :
D 2
0.1 N c square 6.4 (Skempton Chart, page 73 Fig.4.6 in Lecture Notes)
B 20
q f 48x6.4 18x 2 343.2 kPa
q nf q f D c u N c 307.2 kPa
Settlement Check :
St = Si + Sc
IMMEDIATE SETTLEMENT IN CLAY, Si:
qB
Si (1 2 )I s where q q net (net foundation presure) q nf 307.2 102.4 kPa
E FS 3
Note that in clay for UNDRAINED CASE 0.5
undrained mod ulus, E u 60 000 kPa
Is 1.2 (given)
102.4x20
Si (1 0.52 )x1.2 0.031m 31mm
60x103
CONSOLIDATION SETTLEMENT IN CLAY, Sc:
2m
20x20
z . mid-point of sublayer 1, z =6m
1
H1=12 m
26m
. mid-point of sublayer 2, z =18m H2=12 m
2
limestone
6
Vertical Stress due to qnet should be determined at the mid-point of each sublayer
10m
Soed = mv
10m =4qIr ; q=qnet=102.4 kPa
mv = [1-0.2(2+z)]x10-4
Layer no z m=n=10/z Ir mv(m2/kN)
1 6 1.67 0.2 81.9 0.84x10-4
2 18 0.55 0.093 38.1 0.6x10-4
Soed= ( 0.84x10-4x81.9x12)+( 0.6x10-4x38.1x12)=0.110m=110mm
St = 31+110 141mm<150mm (allowable) OK.
GENERALLY IN CLAY SHEAR FAILURE CONTROLS THE DESIGN,
SETTLEMENT IS NOT CRITICAL. BUT IT SHOULD BE CHECKED ALSO
(qall)net = 102.4 kPa
(qall)gross= 102.4+2x18 = 138 kN/m2
7
P4
Question:
A footing of 4mx4m carries a uniform gross pressure of 300 kN/m2 at a depth of 1.5m in
a sand. The saturated unit weight of the sand is 20 kN/m3 and the unit weight above the
water table is 17 kN/m3. The shear strength parameters are c’=0, ’=320. Determine the
factor of safety with respect to shear failure for the following cases;
a) The water table is at ground surface
b) The water table is 1.5m below the surface
Solution:
(q ult ) net q nf q D q f D
FS ult
q net q n q gross D qn
For square footing:
q f q ult 0.4BN 1.2cN c DN q
c' 0 and ' 320 N 26 , N q 29 (see page 69 Figure 4.3 in Lecture Notes)
jıjıjıjı
D=1.5m
B=4.0m
8
a)
q f 0.4B' N ' DN q 0.4x4x(20 10)x 26 (20 10)x1.5x 29 851kPa
q nf q f ' D 851 (20 10)x1.5 836 kPa
q gross 300 kPa
i. q net 300 20x1.5 270 kPa OR
ii. q net (300 1.5x10) 1.5(20 10) 270 kPa
836
FS 3.1
270
b)
q f 0.4B' N d DN q 0.4x4x(20 10)x26 17x1.5x29 1156 kPa
q nf q f D 1156 17x1.5 1130 kPa
q gross 300 kPa
q net 300 17x1.5 275 kPa
1130
FS 4.1
275
9
P5 FOOTING ON SAND
Question:
The column loads, wall loads and the pertinent soil data for a proposed structure is given
below. Design the square column and wall footings for a permissible settlement of 30
mm, using Peck & Hanson & Thornburn charts. Make a reasonable assumption to
obtain an average N value below the footing.
depth 1 2 3 4 5 6 7 8 9 10 11 12
N 8 14 11 16 18 11 9 13 18 20 50/11 50/7
280 kN/m 900 kN 900 kN 280 kN/m
3m 3m 3m
Df =1m
Dw
wall column = 18 kN/m3 1.5m
GWT
SAND = 21 kN/m3
w = 10 kN/m3
Footing on Cohesionless Soils:
Assumptions:
significant depth: 0.5 B above, 2 B below the footing
weight of excavated soil weight of (footing + column) in the soil
column load / area qnet
footings to be designed for the largest qnet (i.e. column ftg)
10
Solution:
NOTE: For Peck-Hanson-Thorburn, N values should be corrected for overburden stress
Depth Nfield o ‘ CN N1
1 8 18 2.0 16
2 14 36 1.63 23
3 11 50.5 1.38 15
4 16 61.5 1.25 20
5 18 72.5 1.15 21
6 11 83.5 1.07 12
7 9 94.5 1.01 9
8 13 105.5 0.95 12
9 18 116.5 0.91 16
10 20 127.5 0.87 17
11 50/11 -
12 50/7 -
CN (overburden correction) values are calculated by using eq.2.3 (page 31) in Lecture
Notes. (CN = 9.78x(1/σvı)0.5≤2)
Square column footings Peck & Hanson & Thornburn charts:Fig 4.8 in Lecture Notes
⇒ assume B=3.0 m
⇒ To obtain the average N value to be used in the calculations
Consider 0.5B=0.5x3=1.5m above
2.0B=2.0x3=6.0m below the foundation level
Depth N1
0.5B=1.5m
1 16
2 23
3 15
4 20 2.0B=6.0m
5 21
6 12
7 9
8 12
9 16
10 17
N1,av = (16+23+15+20+21+12+9) / 7 = 17
11
Cw = 0.5 + 0.5x[2.5/(1+3)] = 0.81
(qn )all=11x N1,av x cw (kN/m2) for 25 mm settlement (page 78 in Lecture
Notes) (qn )all=11x17x0.81 = 151 kPa
S all (mm)
(qn ) all (qn ) all x
25
qall = 151x(30/25) = 181 kPa
qnet = 900/(3x3) = 100 kPa
181 >>100 overdesign
⇒ assume B = 2.0 m
Depth Ncor
0.5B=1.0m
1 16
2 23
3 15
4 20 2.0B=4.0m
5 21
6 12
7 9
8 12
9 16
10 17
Nav = (16+23+15+20+21) / 5 = 19
Cw = 0.5 + 0.5x[2.5/(1+2)] = 0.92
(qn )all=11x19x0.92 = 192 kPa
qall = 192x(30/25) = 230 kPa
qnet = 900/(2x2) = 225 kPa
230 225 OK
B = 2.0 m
12
Wall footings
⇒ Use qnet = 225 kPa
280
B 1.25m
225
Check B value
Nav = (16+23+15) / 3 = 18
Cw = 0.5 + 0.5x[2.5/(1+1.25)] ≤ 1.0 Cw = 1.0
(qn )all=11x18 = 198 kPa
qall = 198x(30/25) = 238 kPa
238 > 225 OK
13
P6 FOOTING ON CLAY
Question:
A public building consists of a high central tower which is supported by four widely
spaced columns. Each column carry a combined dead load and representative sustained
load of 2500 kN inclusive of the substructure (gross load). Trial borings showed that
there is a 7.6m of stiff fissured Ankara clay (cu=85 kPa, Eu = 30 MN/m2 and
mv = 1x10-4 m2/kN) followed by dense sand which is relatively incompressible.
Determine the required foundation width (assume square foundation) and allowable
bearing pressure for the tower footings.
Assume wet = sat = 18.6 kN/m3 (above and below GWT)
w = 10 kN/m3
Consider immediate and consolidation settlements. Divide the clay layer into 4 equal
sublayers.
The foundation depth can be taken as 2m.
⇒ D=2.0m, cu = 85 kPa, Skempton-Bjerrum factor:µ=0.5, Dw = 1.2 m, F.S. = 2.5
Solution:
Assume B=2.0m
Df/B=1 ⇒ Nc = 7.7 (Skempton)
qnf = (qult)net = cuNc = 85x7.7 = 654.5 kPa
for FS=2.5 (qnet)safe = 654.5/2.5 = 261.8 kPa
2500kN
qnet = 2500/(2x2) – 2x18.6 = 587.8 kPa
OR
1.2
2m qnet = (2500/(2x2) – 0.8x10)-(1.2x18.6+0.8x8.6)
= 587.5 kPa
(qnet)safe << qnet NOT ACCEPTED
14
Assume B=3.0m
Df/B=0.67 ⇒ Nc = 7.4 (Skempton)
qnf = (qult)net = cuNc = 85x7.4 =629 kPa
for FS=2.5 (qnet)safe = 629/2.5 = 251.6 kPa
qnet = 2500/3x3 – 2x18.6 = 241 kPa
(qnet)safe qnet OK
B=3.0m
Settlements
B=3.0m Eu = 30000 kPa Df=2.0m
Compressible layer thickness H=7.6-2=5.6m
H D ⇒ 0 0.95
1.87 0.67 1 0.57
B B
241x3
S i 0.57x0.95x 0.013m 13mm
30000
2500kN
1.2
2m
1.4
P1
1.4 P2
1.4 P3
1.4 P4
SAND
Sand is relatively incompressible
(also = 2B)
15
qnet=241 kPa
q net BL
P
( B z )( L z )
(Use 2:1 approximation)
Layer no Thickness,H (m) ΔP
1 1.4 158
2 1.4 83.4
3 1.4 51.3
4 1.4 34.8
Note that:
⇒ P= vertical stress due to qnet at the mid-point of each sublayer
Soed=mv..H
Soed=1x10-4x1.4x(158+83.4+51.3+34.8)=4.585x10-2m=45.85mm
Apply Skempton-Bjerrum factor =0.5
Sc = Soed 45.85x0.5 22.9mm
Stotal = SI + Sc = 13+22.9 = 35.9mm
16
P7 RAFT FOUNDATION ON DEEP CLAY LAYER
Question:
A 16-storey apartment block is to be constructed at a site. The soil profile consists of a
deep clay layer. The ground water table is at 4m depth. The base of the raft under the
building is 8m deep from the ground surface. The profile and the soil properties are
shown in the figure below.
The dimensions of the building and the raft are the same (15mx30m). Total weight of the
building (dead+live+raft) is 90 000 kN.
Find the net foundation pressure and check the factor of safety against bearing capacity
and calculate the total settlement of the building.
No secondary settlements are expected. Take the Skempton-Bjerrum correction factor =
0.75. Consider the compressions of the soil within 20m distance from the foundation
level. The G.W.T. is at the “Stage 2” level prior to construction, lowered to “Stage1” level
during the construction and rises back to “Stage 2” level in the long term.
B=15m Original GWT
position
=18 kN/m3 4m
Stage 2
3
sat=20kN/m RAFT 4m
Stage 1
Medium stiff clay 5m
mv1=0.025x10-2
I
cu=40kN/m2 Eu=20MPa
II 5m
Medium stiff clay III 5m
mv2=0.015x10-2
cu=45kN/m2 Eu=20MPa IV 5m
17
Solution:
Total weight of the building (dead+live+raft)=Qgross=90000 kN
qgross= 90000/(15x30) = 200 kPa
Stage 1 (GWT is lowered to the foundation level)
Uplift = 0
o’=4x18+4(20-9.8) = 112.8 kPa
qnet=(200-0)-112.8 = 87.2 kPa (net foundation pressure)
Stage 2 (GWT is raised to its original position)
Uplift= 4x9.8= 39.2 kPa
o’=4x18+4(20-9.8) = 112.8 kPa
qnet=(200-39.2)-112.8 = 48 kPa
qnet= 87.2 kPa is MORE CRITICAL
Net bearing capacity of the foundation : qnf = qf - D=cuNc+D-D=cuNc
cu = 40 kPa
Df/B=8/15=0.53 (Nc)square=7.1
(Nc)rect.=(Nc)square (0.84+0.16B/L) = 7.1(0.84+0.16x15/30) = 6.5
qnf = 6.5x40 = 260 kPa
q nf 260
Safety factor against shear FS 3.0 OK
q net 87.2
Settlement Analysis:
Total settlement= St= SI + Sc
Consider the compressions of the soil within 20m distance from the foundation level.
qB 87.2x15
Initial settlement : Si 0 1 0.95x0.5x 3.1cm
Eu 20000
18
Consolidation Settlement : Sc = mv H
For consolidation settlement; consider 5m thick sublayers.
15
7.5 = 4qIr
qnet = 48 kPa sınce consolıdatıon is a
LONG TERM situation
n=B/z m=L/z Ir = 4qIr mv (m2/kN)
7.5/2.5 15/2.5 0.245 47 0.025x10-2
7.5/7.5 15/7.5 0.2 38.4 0.025x10-2
7.5/12.5 15/12.5 0.145 27.8 0.015x10-2
7.5/17.5 15/17.5 0.102 19.6 0.015x10-2
Sc = 0.025x10-2x47x5 + 0.025x10-2x38.4x5 + 0.015x10-2x27.8x5 + 0.015x10-2x19.6x5
Sc = 0.142m=14.2cm
⇒ =0.75 (Skempton-Bjerrum)
Sc = 14.2x0.75= 10.7cm
St = 3.1+10.7 = 13.8 cm
19
