7.12Proble ms
Example 7.1:A concrete pile of 40 cm diameter
was driven into sand of loose to medium density to
a depth of 12m. The following properties are
known: average unit weight of soil along the
length of the pile, γ = 18 kN/m3 , average  = 35°,
average Ks = 1.0 and δ= 0.7  .
Calculate:
(a) the ultimate bearing capacity of the pile, and
(b) the allowable load with Fs = 2.
Assume
i) the water table is at great depth. Use Berezantsev's method.
Fig Ex.7.1
ii) the water table is at the ground surface and γ sat = 19 kN/m3
Solution:
i) Qu  Qb  Q f  q0 ' Ab N q  qo ' As K s tan 
q0  L  216 kN / m 2 , q0 '  0.5L  108kN / m 2
Ab 
  0.4 2
 0.125 m 2 , As  2rL  15.08m 2 and δ= 0.7  =24.50 .
4
L
 30,   35 0 , N q  48
D
Qu  Qb  Q f  216  0.125  48  108  15.8  1  0.4557  2073 .606 kN
Qallowable 
2073 .606
 1036 .80 kN
2
ii) If GWT is at the ground surface & γsat = 19 kN/m3
 b  19  9.81  9.19 kN / m 3
q0  9.19  12  110 .28kN / m 2 , q0 '  0.5  110 .28  55.14 kN / m 2
Qu  110.28  0.125  48  55.14 15.08  0.4557  1040 .599 kN
Qallowable 
1040 .599
 520 .299 kN
2
Example 7.2:A concrete pile 40 cm in diameter and 15 meter long is driven into a homogeneous
mass of clay soil of medium consistency. The water table is at the ground surface. The unit
cohesion of the soil under undrained condition is 50 kN/m2 and the adhesion factor α= 0.75.
Compute Qu and Qa with F, = 2.5.
Solution:
Qu  Qb  Q f  cb N c Ab  S sCu
Here, Cb= Cu, Nc= 9, Ab= 0.1256 m2 , As=18.84 m2
 Qu  50  9  0.1256  18.84  0.7  50  63.114 kN
Qallowable 
63.114
 25.245 kN
2.5
Example 7.3:A concrete pile of 35 cm diameter is driven into a homogeneous mass of cohesion
less soil. The pile carries a safe load of 600kN.A static cone penetration test conducted at the site
indicates an average value of qc = 35 kg/cm2 along the pile and 100 kg/cm2 below the pile tip.
Compute the length of the pile with Fs = 2.5.
Solution:
qb  pile  q p cone
Q p =100 kg/cm2 =10,000 kN/m2
Qb  (  0.35 2  10000 )  962 .112 kN
Let L be the required length of pile
Avg. Qc  35kg / cm 2 , fs=17.16 kN/m2
Q f  17.16  3.14  0.35  L  18.85 LkN
Given Q a= 600 kN and Factor of Safety is 2.5, so, Q u=1500 kN.
Now, 1500=962.112+18.85L
 L=28.53 m=29m
So, the required length of the pile to carry a safe load of 600 kN will be 29 meters.
Example 7.4:A concrete pile of 45 cm diameter was driven into sand of loose to medium density
to a depth of 15m. The following properties are known:
(a) Average unit weight of soil along the length of the pile,
= 17.5 kN/m3 , average ∅ = 30°,
(b) Average K s  1 and, δ = 0.75∅.
Calculate
(a) the ultimate bearing capacity of the pile, and (b) the allowable load with
= 2.5. Assume
the water table is at great depth. Use Berezantsev's method.
Solution:
Qu  Qb  Q f  q0 ' Ab N q  qo ' As K s tan 
where, q0 '   L  17.5  15  262.5kN / m 2
q0 ' 
Ab 
1
262 .5
L 
 131 .25kN / m 2
2
2

4
 0.45 2  0.159 m 2
, As  3.14  0.45  15  21.195 m 2
  0.75  0.75  30  22.50 , tan = 0.4142
For,
L
15

 33 .3 and   30 0 , N q  16.5
d 0.45
Substituting the known values, we have
Qu  Qb  Q f  262 .5  0.159  16.5  131 .25  21.195  1  0.4142  689  1152  1841kN
Qa 
1841
 736 kN
2.5
Example 7.5
Assume in Ex. 7.4 that the water table is at the ground surface and γ sat = 18.5 kN/m3 . All the
other data remain the same. Calculate Qu and Qa .
Solution:
Water table at the ground surface  sat  18.5kN / m 3
 b   sat   w  18.5  9.81  8.69kN / m 3
q0 '   b L  8.69  15  130 .35kN / m 2
q0 ' 
1
130 .35
bL 
 65.18kN / m 2
2
2
Substituting the known values
Qu  130.35  0.159 16.5  65.18  21.195 1 0.4142  342  572  914 kN
Qa 
914
 366 kN
2.5
It may be noted here that the presence of a water table at the ground surface in cohesionless soil
reduces the ultimate load capacity of pile by about 50 percent.
Example 7.6:Determine Q b, Qf, Q u and Q a by using the SPT value for ∅ = 30°
Fig Ex.7.6: Terzaghi's bearing capacity factors which take care of mixed state of local and
general shear failures in sand (Peck et al., 1974)
Solution:
From Fig Ex.7.6, N cor = 10 for ∅ = 30°.
Qu  Qb  Q f  40 N cor
L
Ab  2 N cor As
d
where, Qb  Qb1  400 N cor Ab
Given: L = 15 m, d = 0.45 m, Ab = 0.159 m2 , As = 21.2 m2
Qb  40  10 
15
 0.159  2120 kN
0.45
Qb1  400 10  0.159  636 kN
Since Qb> Q b1 , use Q b1
Q f  2  10  21.2  424 kN
Now Qu  636  424  1060 kN
Qa 
1060
 424 kN
2.5
Example 7.7
A concrete pile 0.4m in diameter and 15m long is driven into a homogeneous mass of clay soil of
medium consistency. The water table is at the ground surface. The unit cohesion of the soil under
undrained condition is 50 kN/m2 and the adhesion factor α = 0.75. Compute Q u and Q a with Fs=
2.5.
Solution:
Given: L = 15m, d = 0.4m, c u = 50 kN/m2 , α = 0.75
Qu  Qb  Q f  cb N c Ab  As cu
where, cb  cu  50 kN/m2 , Nc =9;Ab= 0.126 m2 ; As = 18.84 m2
Substituting the known values, we have
Qu  50  9  0.126  18.84  0.75  50
= 56.70+706.50 = 763.20 kN
Qa 
763 .20
 305 .28 kN
2.5
Example 7.8:
A concrete pile of 45 cm diameter is driven through a system of layered cohesive soils. The
length of the pile is 16m. The following data are available. The water table is close to the ground
surface.
Top layer 1: Soft clay, thickness = 8 m, unit cohesion
u
= 30 kN/m2 and adhesion factor α =
0.90.
Layer 2: Medium stiff, thickness = 6 m, unit cohesion
u
= 50 kN/m2 and α = 0.75.
Layer 3: Stiff stratum extends to a great depth, unit cohesion
u
= 105 kN/m2 and α = 0.50.
Compute Qu and Qa with Fs = 2.5.
Solution:
Here, the pile is driven through clay soils of different consistencies.
The equations for Qu expressed as
Qu  9cb Ab  POL cu L
Here, cb = uof layer 3, P = 1.413 m, Ab = 0.159 m2
Substituting the known values, we have
Qu  9 105  0.159  1.4130.9  30  8  0.75  50  6  0.5 105  2  150.25  771.5  921.75kN
Qa 
921 .75
 369 kN
2.5
Example 7.9:
A precast concrete pile of size 0.45 x 0.45 m is driven into stiff clay. The unconfined
compressive strength of the clay is 200 kN/m2 . Determine the length of pile required to carry a
safe working load of 400 kN with Fs= 2.5.
Fig Ex.7.9: Adhesion factor a for piles with penetration lengths less than 50 m in clay.
(Data from Dennis and Olson 1983a, b: Stas and Kulhawy, 1984)
Solution:
The equation for Qu is
Qu  N c cu Ab   cu As
We have
Qu = 2.5 x 400 = 1000 kN
Nc= 9, cu = 100 kN/m2
α = 0.48 from Fig Ex.7.9,
u
= cu = 100 kN/m2 , Ab = 0.159m2
Assume the length of pile = L m
Now, As  4  0.45 L  1.8L
Substituting the known values, we have
1000  9  100  0.159  0.48  100  1.8L
 1000  143 .1  86.4 L
1000  143 .1
L
 9.91m
86.4
Example 7.10:
A reinforced concrete pile of size 30 x 30 cm and 10m long is driven into coarse sand extending
to a great depth. The average total unit weight of the soil is 18 kN/m3 and the average N cor value
is 15. Determine the allowable load on the pile by the static formula. Use F s= 2.5. The water
table is close to the ground surface.
Solution:
In this example only the N-value is given. The corresponding ∅ value can be found from Fig
Ex.7.10a which is equal to 32°. Now from Fig Ex.7.10b,
Fig Ex.7.10a: Terzaghi's bearing capacity factors which take care of mixed state of local
and general s hear failures in sand (Peck et al., 1974)
Fig Ex.7.10b: Berezantsev's bearing capacity factor, N (afte r Tomlinson, 1986)
for ∅ = 32°, and
L 10

 33.33 , the value of N q = 25.
d 0. 3
Ab = 0.09m2 , As = 12m2
δ=0.75x32 = 24°, tan δ = 0.445
The relative density is loose to medium dense.
Ks  1
1
2  1  1.33
3
Now, Qu  q0 ' N q Ab  q0 ' K s tan As
 b   sat   w  18.0  9.81  8.19 kN / m 3
q0 '   b L  8.19  10  81.9kN / m 2
q0 ' 
q0 '
 40.59kN / m 2
2
Substituting the known values, we have
Qu  81.9  25  0.09  40.95 1.33  0.445 12  184  291  475kN
Qa 
475
 190 kN
2.5
Example 7.11:
A concrete pile of 40 cm diameter is driven into a homogeneous mass of cohesionless soil. The
pile carries a safe load of 650 kN. A static cone penetration test conducted at the site indicates an
average value of qc = 40 kg/cm2 along the pile and 120 kg/cm2 below the pile tip. Compute the
length of the pile with Fs = 2.5.
Solution:
qb (pile) = qp (cone)
Given, q = 120 kg/cm2 , therefore,
qb =l20 kg/cm2 = 120 x 100 = 12000 kN/m2
qb is restricted to 11,000 kN/m2
Therefore,
Assume the length of the pile = L m
The average,
fs =
= 40 kg/cm2
= 20 kN/m2
Now, Qf= f sAs = 20 x 3.14 x 0.4 x L = 25.12L kN
Given Qa = 650 kN. With Fv = 2.5, Qu = 650 x 2.5 = 1625 kN.
Now,
1625 = Qb + Qf = 1382 – 25.12L
or, L =
= 9.67 m or say 10 m
The pile has to be driven to a depth of 10 m to carry a safe load of 650 kN with F =2.5.
Fig Ex.7.11
Example 7.12:
The plan of a group of nine piles is shown in fig.4.0 A load of 3250kN. Q, inclined at 10° to the
vertical, acts in a direction parallel to the X-X axis and its point of application has eccentricities,
ex =0.5m, ey =0.7m, Determine the values of vertical load and horizontal load that should be used
to check the suitability of the piles.
Solution:
2m
2m
0.5m
1.5m
1.5m
0.7m
Fig Ex.7.12
Horizontal components of inclined load= 3250  sin 10 0  564 kN
Vertical components of inclined load= 3250  cos10 0  3200 kN
For this pile X=2m and Y=1.5m
X
2
 6  2 2  24 ,
Y
2
 6  1.5 2  13.5
 1 2  0.5 1.5  0.7 
Q p  3200  

  737 kN
24
13.5 
9
Horizontal load per pile =
564
 63kN
9
Hence, piles should be checked that they can each support at vertical load of 737kN and
horizontal load, acting at level of underside of pile cap, of 63kN