Problems on bearing capacity of soil

1. Problems on Bearing Capacity

2. Example # 01
• A square footing 2.5 m by 2.5 m is built in a
homogenous bed of sand density 2.0 t/m3 and
having an angle of shearing 38o . The depth of
a base of the footing is 1.5 m below the
ground surface. Calculate the safe load that
can be carried by a footing with a factor of
safety 3 against complete shear failure. Use
Terzaghi’s analysis.

3. Solution:
Given: B = 2.5 m D = 1.5 m γ = 2 t/m3
φ = 38o which is greater than 36o so general shear
failure will occur.
Values of bearing capacity factors from Fig.
Nq = 47 Nγ = 64
Since c = 0 ∴ qf = γ D Nq + 0.4 γ B Nγ
qnf = γ D (Nq -1) + 0.4 γ B Nγ
= 2 x 1.5 (47-1) + 0.4 x 2 x 2.5 x 64 =
138 + 128 = 366 t/m2
qs = qnf /F + γD = 366/3 + 2 x 1.5 = 122 + 3
= 125 t/m2
∴ Maximum safe load = B2 x qs = (2.5)2 x 125 = 780 t.

4. Example # 02
• A strip footing 1 m wide at its base is located
at a depth of o.8 m below the ground surface.
The properties of the foundation soil are : γ =
1.8 t/m3 and φ = 20o and c = 3 t/m2 .
Determine the safe bearing capacity, using a
factor of safety 3 . Use Terzaghi’s analysis.
Assume soil’s local shear failure.

5. Solution:
qf = 2/3 c Nc ′+ γ D Nq ′ + 0.5 γ B Nγ′
For φ = 20o From table: Nc ′= 9.5, Nq ′= 3.5
Nγ = 1.7
qf = (2/3 x 3 x 9.5) + (1.8 x 0.8 x 3.5) + 0.5
(1.8 x 1 x 1.7) = 19 + 5.04 + 0.9 = 24.94 t/m2
qnf = qf - γ D = 24.9- 1.8 x 0.8 = 23.5 t/m2
qs = qnf /F + γ D = 23.5/3 + (1.8 x 0.8) =
7.83 + 1.44 = 9.27 ≈ 9.3 t/m2

6. • Solve above example if the water table is
located at a depth of 1.5 m below the ground
level.
• Rw1 = 1 Rw2 =?
• Zw2 = depth of water table below the base of
footing.
• = 1.5 – 0.8 = 0.7
∴Rw2 = 0.5 (1+Zw2 /B ) = 0.5 (1+ 0.7/1) = 0.85
qf = 2/3 c Nc ′+ γ D Nq ′Rw1 + 0.5 γ B Nγ′Rw2
= 2/3 x 3 x 9.5 + (1.8 x 0.8 x 3.5 x 1) + o.5 x 1.8
x 1 x 1.7 x 0.85 = 19 + 5.04 + 0.76 = 24.8 t/m2

7. • qnf = qf - γ D = 24.8 – 1.44 = 23.36
• qs = qnf /F + γ D = 23.36/3 + 1.44 = 7.76 + 1.44
9.2 t/m2

8. Example # 03
• An R.C column having square in shape is to rest
1.5 m below ground level. The total load to be
transmitted including the weight of the column is
200 tons. As the area is subjected to frequent
flooding, the friction of the footing along the
sides is to be neglected and a factor of safety 2.5
is to be allowed. If the saturated density of the
sand be 2.4 g/cc, angle of internal friction 33o and
value of Nγ = 33, Nq = 32, find the suitable size of
the footing for the above condition.

9. Solution:
Assume the size of footing as 2m x 2m.
B = 2 m and D = 1.5 m ∴ B>D, foundation is shallow.
qf =1.3 cNc + γ D Nq + 0.4 γ B Nγ
Now c = 0 , Nq = 32, Nγ = 33 γ = 2.4 g/cc (2.4 t/m3 ) B = 2.
qf = 0 + 2.4 x 1.5 x 32 + 0.4 x 2.4 x 2 x 33.
= 115.2 + 63.4 = 178.6
qs = 178.6/2.5 = 71.4 (F.S = 2.5)
Now area of footing 2x 2.
∴ Total load = 4 x 71.4 = 285.6 tones
It is higher than the required 200 tones.
∴ it is safe. We can make it more economical by reducing
the section and retrial may be made.

10. Example # 04
• Size of an isolated footing is to be limited to
1.5 m square. Calculate the depth at which
the footing should be placed to take a load of
200 tonnes with a factor of safety 3.The soil is
having angle of internal friction φ = 30o , γ =
2.1 g/cc, weight of footing 5% of the external
load. Nq = 22, Nγ = 20.

11. Solution:
A soil is sandy so no cohesion
Allowable bearing capacity for square footing:
qs = 1/F { 1.3 cNc + γD (Nq – 1) + 0.4 γ B Nγ } + γD
Now total load = 200 + 5/100 x 200 = 210 tonnes
qs = 210/1.5 x 1.5.
Put values in the above eq.
210/1.5 x 1.5 = 1/3 { 2.1 x D (22-1) + 0.4 x 2.1 x 1.5 x 20} + 2.1
x D
= 1/3 {2.1 x 21 x D + 25.2} 2.1 D
= 14.7D + 8.4 + 2.1D
16.8D = 93.4 – 8.4 = 85
∴ D = 85/16.8 = 5.6 meters
∴ The footing will be placed at 5.6 m below the ground level

12. Example # 05
A 3.0 m strip footing rests 2.5 m below
ground level over sandy clay having unit
weight of 2250 kg/m3 . Quick tests in a shear
box apparatus gave shear strength of 0.37 and
0.50 kg/cm2 for normal stress of 0.75 and 1.5
kg/cm2 respectively. Find out cohesion and
angle of internal friction. From Terzaghi’s
formula, calculate the ultimate load per meter
run of the foundation.

13. Solution:
• As per Coulomb’s eq. τ = c + σ tan φ -------(A)
• c = ? And φ = ?
• Two equations can determine c and φ.
• Given: τ1 = 0.37 kg/cm2 τ2 = 0.5 kg/cm2
• σ1= 0.75 kg/cm2 σ2= 1.5 kg/cm2
Put values in eq. (A)
0.37 = c + 0.75 tanφ. ----------(i)
0.5 = c + 1.5 tanφ.------------(ii)
Solve simultaneously.
c = 0.25 kg/cm2 tan φ = 0.1733 ∴ φ = 9.83o ≈ 10o .
Assume local shear failure.
∴ Nc ′ = 8 , Nq ′= 1.9 Nγ ′ = 0.5
qf = 2/3 c Nc ′ + γD Nq ′ + 0.5 γ B Nγ ′

14. • qf = 2/3 x 0.25 x 8 + 2250/109 x 2.5 x 102 x1.9
+ 0.5 x 2250/109 x 3.0x102 x0.5
• = 1.333 + 0.01069 + 0.001688 = 1.345
kg/cm2 = 13.45 t/m2
• Ultimate load per meter run of foundation =
13.45 x 3 = 40.37 Tonnes/meter length.

15. Example # 06
• Compute the allowable bearing pressure using
the Terzaghi’s equation for the footing and
soil parameters shown in Fig. Use factor of
safety 3 to obtain qa . The soil data are
obtained from a series of U-triaxial tests. Is
the soil saturated ?
γ= 110 pcf
φ = 20o
C = 300 psfB
D = 4 ft

16. Solution:
• Since a U –test gives a φ angle. The soil is not saturated.
• For φ = 20o , Assuming square foundation general shear failure.
• qult =1.3 cNc + q Nq + 0.4 γ B Nγ
• Now for φ = 20o , Nc = 17.7, Nq = 7.4 Nγ = 5.0
• Put values.
• qult =1.3 x 300/1000 x 17.7 + 4 x 110/1000 x 7.4 + 0.4 x 110/1000 x
B x 5.0 = (10.2 + 0.22B) ksf
• Qa = qult /F.S = 10.2 0.22B /3 = (3.4 + 0.07B) ksf
• Normally B varies between 5 to 10 ft.
• For B = 5ft , 3.4 + 0.07 x 5 = 3.75 ksf
• = 10 ft, 3.4 +0.07 x 10 = 4.10 ksf
• ∴ Recommended is : 3.5 ksf

17. Example # 07
• The results of full scale tests conducted by
H.Muhs in Berlin reported by J.B.Hansen. The
pertinent data associated with this test as
reported by Hansen are footing dimension L =
2 m, B 0.5 m , D = 0.5 m dense sand γ = 0.95
t/m3 , c = 0 . Failure load Q 190 Ton. φ = 47o .
• Compare the results with Terzaghi Analysis
and Hansen.

18. Solution:
• Terzaghi’s eq. As L/B = 2/0.5 =4, use strip footing
formula.
• qu = c Nc + γ D Nq + 1/2 γ B Nγ.
• For φ = 47o , Nq = 246 , Nγ = 585
• = 0 + 0.95 x 0.5 x 246 + ½ x 0.95 x 0.5 x 585
• =116.85 + 138.9 = 255.78 t/m2. More than
actual.
• Test result = Q/A =190/2 x 0.5 = 190 T/m2
same if checked by Hansen eq. = 180.6 ton/m2

19. Example # 08
• A given series of values for qu in the of interest
from SPT samples from a boring log gives
average 200 kpa. Estimate the allowable
bearing capacity for square footing by
Terzaghi’s formula. Use F.S =3.

20. Solution:
• Now c = qu /2 = 200/2 = 100 kPa.
• Terzaghi’s eq. for square footing- φ = 0
• qu = 1.3 c Nc = 1.3 x 100 x 5.7 = 741 kPa
• qa= 741/3 = 247 kPa
• Normally qu from SPT is not reliable. UDS are
taken.
• Or in case of SPT qu result F.S = 4 to 5.
• Generally qa = qu =unconfined compressive
strength = c Nc/F.S = qux 5.73 /2 x3 = qu

21. Example # 09
• A 30 cm square bearing plate settles by 1.5
cm in a plate loading test on a cohesion less
soil when the intensity of loading is 2 kg/cm2.
What will be the settlement of a prototype
shallow footing 1.0 m square under the same
intensity of loading.

22. Solution:
• Given Bp = 30 cm B = 100 cm
• ρp= 1.5 cm.
• ρ = 1.5 [ 100 (30 + 30.5)/30 (100 +
30.5)]2 = 3.6 cm

23. Example # 10
• The results of a plate load test in a sandy soil
are given below. The size of the plate is 0.305
x .305 m. determine the size of square footing
of column foundation that should carry a load
of 2500 kN with a maximum settlement of 25
mm.
Load/unit area
kN/m2
200 400 600 700
Settlement
(mm)
5 12.5 28 60

24. Solution:
Given:
Size of the plate 0.305 x 0.305 m.
Load v/s settlement , draw a curve or plot the data.
Load on footing------2500 kN
Max. settlement = 25mm 200 400 800 kN/m2
600
10
30
50
(mm)
80
Load
Settlement
70

25. Now settlement of footing:
Now by trial and error
Procedure assume the
size of footing and find the
amount of settlement
where the trial agrees
with the conditions
given will be the size
of footing.
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22
22
12.3
12.3
48.30
48.30
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p
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F
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B
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B
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Or
B
B
B
B
SS
(Size in m)

26. Qo , kN Assumed
width, BF (cm)
qo
Qo /BF
2 ,
kN/m2
Sp
correspondin
g to Qo (mm)
SF from above
Eq. (mm)
2500 400 156.25 4 13.8
2500 300 277.8 8 26.35
2500 320 244.10 6.8 22.70
2500 310 260.10 7.2 23.86

27. Recommended 3.1 x 3.1 m size of footing
Qo , kN Assumed
width, BF (cm)
qo
Qo /BF
2 ,
kN/m2
Sp
correspondin
g to Qo (mm)
SF from above
Eq. (mm)
2500 400 156.25 4 13.8
2500 300 277.8 8 26.35
2500 320 244.10 6.8 22.70
2500 310 260.10 7.2 23.86

28. Example # 11
• The results of two plate load tests are given in
the following table:
• Determine the size of square footing to carry a
load of 715 kN with tolerable settlement as
20mm. Solve for foundation.
Plate dia. (B) (m) Total Load, Q (kN) Settlement (mm)
0.305 32.2 20
0.610 71.8 20

29. Solution:
• Qo = Am + Pn
Put the values in the above Eq. to find m and n
32.2 = π/4 (0.305)2 m + π(0.305)n.---------(i)
71.8 = π/4 (0.610)2 m + π(0.610)n.---------(ii)
By solving simultaneously we can find:
m = 50.68 kN/m2
n = 29.75 kN/m2.
Now again use the above Eq.with these values to find the size
of square footing.
Load on foundation = 715 kN
A = Bf
2 , P = 4 Bf ; Put values
715 = Bf
2 x 50.68 + 4 Bf x 29.75
∴ 50.68 Bf
2 + 119Bf – 715 =0
∴Bf = 2.8 m Ans.

30. Example # 12
• Plate load test performed on a uniform
deposit of sand and the following
observations were recorded.
• The size of the plate was 30 x30cm. Plot the
load-settlement curve and determine the load
on a footing 1.5m x 1.5m. Can it carry safely if
the settlement is not to exceed 50 mm.
Load t/m2 5 10 20 30 40 50 60
Settlement
(mm)
4.5 8.5 16 31.3 50 74 104

31. Solution:
• Given:
• Size of plate 30 x 30 cm
• Footing size 1.5 x 1.5 m
• Settlement of footing 50 mm
• Plot Load vs settlement

32. 10 20 30 40 50 60
Load t/m2
10
30
50
70
90
Settlement(mm)

33. • Now from load – settlement curve the load corresponding
to this settlement = 20 t/m2.
• ∴ Safe on this footing for 50mm settlement = 1.5 x 1.5 x
20 = 45 tons.
mmS
S
Scalculatetovaluestheputor
B
B
B
B
SS
p
p
p
f
p
p
f
pF
18
48.30150
48.3030
30
150
50
.
48.30
48.30
2
2
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34. Example # 13
• A load test was made with a 35 cm square plate
at a depth of one meter below the ground level
in soil with φ = 0. The water table was located at
a depth of 5m below the ground level. Failure
occurred at a load of 5200 kg. What would be the
ultimate bearing capacity per unit area for a 1.6m
wide continuous footing with its base loaded at
the same depth in the same soil. Unit weight of
soil was 1.9 g/cc above water table. For φ = 0,
• Nc = 5.7, Nq = 1 , Nγ = 0. Assume general shear
failure.

35. Solution:
• For square footing,
qf =1.3 c Nc + γD Nq + 0.4 γ B Nγ
Taking the case of load test,
B = 0.35cm , D = 1.
Since φ = 0 , Nq = 1 and Nγ = 0
qu=1.3 c Nc + γD Nq
5200/.35 x .35 x1000 = 1.3 x c x 5.7 + 100 x 1.9 x1.0.
7.42 c = 42.5 – 1.9 = 40.6
c = 40.6/7.42 = 5.47 tonnes/m2 .
Now for contineous footing strip of width 1.0m
qf = c Nc + γD Nq + 0.5 γ B Nγ
= 5.47 x 5.7 + 1.9 x 1 x 1 + 0
= 31.2 + 1.9 = 33.1 ton/m2 .

36. Example # 14
• At a certain site subsoil soil consists of a thick
layer of soft clay (cu = 20 kN/m2, φu = 0) which is
overlain by a srtiffer clay (cu = 87.5 kN/m2, φu = 0)
of a variable thickness. The ground surface is
horizontal and the water table is at a
considerable depth.
• It is proposed to install widely spaced footings
1.5 m square at a depth of 1 m below ground
level in the upper clay material. Estimate the net
working load on the footings, using a load factor
of 3.

37. • If the footings are fully loaded at this
pressure, estimate the thickness of stiff clay
required below them in order to ensure that
the load factor against shear failure in the soft
clay is at least 3.
• (Ignore settlement effects and assume a load
dispersion at 30o to the vertical. Take Nc = 7.5
on the lower clay and assume Terzaghi’s
bearing capacity equation to apply)

38. Solution:
• Upper layer
• Lower layer:
• Nc = 7.5
• B = 1.5m square footing.
• F.S = 3
G.L
1 m
Z
Z/√3 Z/√31.5m
30o
cu = 87.5 kN/m2
φu = 0
cu = 20 kN/m2
φu = 0
cu = 87.5 kN/m2
φu = 0
cu = 20 kN/m2
φu = 0

39. • Terzaghi’s Eq. qnf =1.3 c Nc where Nc = 5.7
• For upper layer qnf =1.3 x 87.5 x 5.7 = 648.375
kN/m2 ≈ 648 kN/m2
• Net Safe unit working load = 648/3 = 216
kN/m2
• ∴ Net working load on footing (1.5m)2 = 216 x
1.5 x 1.5 = 486 kN.
• This load is dispersed at 30o angle, Assuming
the depth Z below the footing.

40. • ∴ Pressure on top of lower clay stratum
• Load /Area = 486/(1.5 + 2Z/√3)2
• Now allowable pressure on lower stratum:
• Qf/F = cNc /F = 20 x 7.5/3 =50kN/m2
• Equating these two pressures:
• 50 = 486/ (1.5 + 2 Z/√3)2
• ∴112.5 + 173.2 Z + 66.7 Z2= 486
• Z2 + 2.6 Z – 5.6 = 0
• Z = 1.4 m

41. Example # 15
• Determine the net bearing pressure for a 3m x
3m footing at a depth of 2 m in a medium
dense sand so that the total settlement does
not exceed 25mm. The average SPT blows
below the footing (upto B )N are 28/30 cm.
the average moist density is 1.75 t/m3. The
water table is more than 3 m below the
footing.

42. Solution:
• σo′ at B/2 below footing 1.75(2+1.5) = 6.125
t/m2
• Cn = 0.77 log 200/ σo′ = 0.77 log 200/ 6.125 =
1.165
• Nn = Cn N = 1.165 x 28 =33
• q25 = 0.041 Nn S = 0.041 x 33 x 25 = 33.8 t/m2

43. Example # 16
• A proposed strip footing that is 4 ft(1.2m) wide
will be located 2 ft(0.6m) below ground surface.
The soil type is uniform dense sand that has a
friction angle φ = 35o. The total unit weight of the
soil is equal to 125 pcf (19.7 kN/m3). The ground
water table is well below the bottom of the
footing and will not be a factor in the bearing
capacity analysis. Using a factor of safety of 3,
calculate the allowable bearing pressure using
Figs. 6.5 and 6.6.

44. Solution:
• From Fig. 6.5, for φ = 35o, Nγ=37 and Nq = 33. Using
Eq. 6.1 with c = 0:
• qult= ½ γγγγt B Nγγγγ + γγγγt Df Nq = ½ (125)(4)(37) +
(125)(2)(33) = 17500 psf (840 kPa)
• Using F = 3, qall = qult/3 = 17500/3 = 5800 psf (280
kPa)
• From Fig.6.6, for φ = 35o , Nγ= 40 and Nq = 36 Using
Eq. 6.1 with c = 0:
• qult= ½ γγγγt B Nγγγγ + γγγγt Df Nq = ½ (125)(4)(40) +
(125)(2)(36) = 19000 psf (910 kPa)
• Using F = 3, qall = qult/3 = 19000/3 = 6300 psf (300
kPa)

45. Example # 17
• Use the same data as the above example (No
16) problem, but assume silty sand with a
friction angle φ of 30o. Also assume the
footing will not punch into the soil.

46. Solution:
• From Fig. 6.5, for φ = 30o, Nγ=15 and Nq = 19. Using Eq.
6.1 with c = 0:
• qult= ½ γγγγt B Nγγγγ + γγγγt Df Nq = ½ (125)(4)(15) + (125)(2)(19 =
8500 psf (400 kPa)
• Using F = 3, qall = qult/3 = 8500/3 = 2800 psf (130 kPa)
• From Fig.6.6, for φ = 30o , Nγ= 17 and Nq = 20. Using Eq.
6.1 with c = 0:
• qult= ½ γγγγt B Nγγγγ + γγγγt Df Nq = ½ (125)(4)(17 + (125)(2)(20) =
9250 psf (440 kPa)
• Using F = 3, qall = qult/3 = 9250/3 = 3100 psf (150 kPa)