Numerical problem bearing capacity terzaghi , group pile capacity (usefulsearch.org) (useful search)

1. A strip footing, 1m wide at its base is located at a depth of 0.8m below the
ground surface. The properties of the foundation soil are:
ϒ=18kN/m³, c=30kN/m² and φ=20°. Determine the safe bearing capacity using a
factor of safety of 3. use Terzaghi’s analysis. Assume the soil fails by local shear.
Solution :
The bearing capacity is given by
qf= 2/3 c NC̕ + ϒ D Nq̕ +0.5 ϒB Nᵧ̕
For φ= 20°, the bearing capacity factors are taken from table as;
Nc̕ =11.8; Nq̕ = 3.9 ; Nᵧ̕ = 1.7
qf= (2/3 × 30× 11.8) +( 18× 0.8 ×3.9)+ (0.5× 18 ×1 ×1.7)
=236 + 56.2 +15.3 =307.5 kN/m²
qnf = qf – ϒD =307.5 −18 × 0.8 = 293.1 kN/m²
qf= qnf/F +ϒD =293.1/3 +(18×0.8) = 112.1 kN/m².

2. A rectangular footing 2m×3m rests on a c-φ soil, with its base at 1.5m below the
ground surface. Calculate the safe bearing capacity using a factor of safety of 3 on
(1) net ultimate bearing capacity and (2) ultimate bearing capacity. The soil has
the following parameters ϒ=18kN/m³ ; c= 10 kN/m² ; φ= 30°. Use terzaghi’s
analysis.
Solution: Take σ̅ = ϒD, We get
qf =c Nc (1+ 0.3 B/L) + ϒ DNq + 0.5 ϒBNᵧ (1−0.2 B/L)
For φ= 30° we get
Nc = 37.2 ; Nq = 22.5 and Nᵧ =19.7
qf= 10× 37.2 (1+0.3×2/3) +18×1.5×22.5 + 0.5×18×2×19.7×(1−0.2×2/3)
or qf= 446.4 + 607.5 +307.3 = 1361.2 kN/m²
(1) qnf = qf -ϒ D =1361.2 – 18×1.5 =1334.2 kN/m²
qs= qnf/F + ϒD =1334.2/3 + 18 ×1.5 =471.7 kN/m²
(2) Applying F.S on qf , we get
qs = qf/3 =1361.2/3 = 453.7 kN/m²

3. A soft, normally consolidated clay layer is 6m thick with a natural water content of 30
percent. The clay has a saturated unit weight of 17.4kN/m³, a specific gravity of 2.67 and
a liquid limit of 40 per cent. The ground water level is at the surface of the clay.
Determine the settlement of the foundation if the foundation load will subject the
centre of the clay layer to a vertical stress increase of 8kN/m².
Solution :
σo= effective initial overburden pressure, measured at the centre of the layer
= ϒ’ Z =(17.4 – 9.81) (1/2 × 6) =22.77 kN/m²
∆σ = vertical stress increment, due to foundation load = 8 kN/m²
Cc = compression index= 0.009(wl – 10) = 0.009 (40 - 10) = 0.27
eo= initial voids ratio = wG = 0.3 ×2.67 =0.801
sc= C × Cc/ 1+eo Hlog × σo+ ∆σ/σo
Taking C=1, we get Sc= 0.2/1+0.801 ×6 log 22.77 + 8/ 22.77
=0.087m= 87mm

4. In a 16 pile group, the pile diameter is 45 cm and centre to centre spacing of the square
group is 1.5m. If c=50 kN/m², determine whether the failure would occur with the pile
acting individually, or as a group? Neglect bearing at the tip of the pile. All the piles are
10 m long. Take m=0.7 for shear mobilisation around each pile.
Solution : n= 16; d= 45 cm; L= 10 m
width of group = B=(150×3) + 45= 495 cm= 4.95m
(1) For the group
Qug = c × perimeter × length = c× 4B ×L =50×4×4.95×10 = 9900kN
(2)For the piles acting individually :
Qug = n Qup = n{ mcAp}
Ap= π Dl =π × 0.45× 10
Qug= 16 × 0.7× 50×π×0.45 ×10 = 7917 kN
Which is less than the load carried by the group action. Hence the foundation will fail
by the piles acting individually , and the load at failure would be 7917 kN.

5. A rectangular footing 2m*3m carries a column load of 600 kN at a depth of 1 m. The
footing rests on a c-φ soil strata 6m thick , having poisson’s ratio of 0.25 and young’s
modulus of elasticity as 20000 kN/m². Calculate the immediate elastic settlement of the
footing.
Solution :
Si = qB (1-µ²/Es) Iw
Here q= intensity of contact pressure = 600/2×3 =100 kN/m²
B= least lateral dimension of footing = 2 m
µ= Poisson’s ratio = 0.25 ; Es= 20000 kN/m²
Iw= Influence factor = 1.06 for rigid rectangular footing having L/B =1.
Si = 100 × 2× [1-(0.25)²/20000] ×1.06 = 9.94 ×10ˉ³m
= 9.94mm

6. A footing is to be constructed at a depth of 1.2 m in a uniform deposit of stiff clay having
unconfined compressive strength of 145.8 kN/m² and density of 18.82kN/m³. The
footing will support a wall that imposes a loading of 152 kN/m of wall length. Calculate
the width of the footing using a factor of safety of 3.0?
Solution :
qu= 145.8 kN/m² ; hence c=145.8/2 =72.9 kN/m²
For strip footing qnf= cNc + σ̅ (Nq – 1) + ½ BϒNr
For clay φ= 0. Hence Nc = 5.14 , Nq= 1 and Nᵧ= 0
qnf = 5.14 c = 5.14× 72.9 = 374.706kN/m²
qs = qnf/3 = 374.706 = 124.9 kN/m²
B= P/qs = 152/ 124.9 = 1.22m

7. A raft foundation 10 m wide and 12 m long is to be constructed in a clayey soil having a
shear strength of 12 kN/m². Unit weight of soil is 16kN/m³. if the ground surface carries
a surcharge of 20kN/m² calculate the max depth of foundation to ensure a factor of
safety of 1.2 against base failure. Nc for clay is 5.7.
Bearing capacity of soil for rectangular footing in cohesive soil is given by :
Qf= c Nc(1 + 0.3 B/L)+σ̅= cNc (1+0.3 B/L)+(ϒD+q)
Qf= 12×5.7(1+.3 × 10/12) +16D + 20 =105.5 + 16D
base failure will occur when Qf is equal to zero.
D= -105.5/16 =(-)6.59 (minus sign indicates that it is excavation
Critical depth = 6.59 and Safe depth = 6.59/1.2=5.49m