Numerical Problem and solution on Bearing Capacity ( Terzaghi and Meyerhof Theory ) http://usefulsearch.org (user friendly site for new internet user)

1. Topic: Bearing Capacity Of Soil
Using Terzaghi’s Equation
Using Meyerhof’s equation

2. Terzaghi's Bearing Capacity
Equations Based Problems

3.  STRIP FOOTING
 1. A strip footing of width 1 m is resting on a soft clay strata at a depth
of 1 m below the ground surface. The angle of internal friction is zero,
and cohesion c = 20 kN/m2. The water table is at a great depth.
Calculate the allowable bearing capac1ty of soil using Terzaghi’s
equation. ( γ=12 KN/m3)
 Given:-
 Soil type:- Cohesive Soil.
 Cohesion:- 20 kN/m2
 φ = 0
 Unit weight of soil:- 12 kN/m3
 1 m wide strip footing, bottom of footing at 1 m below ground level.
 Factor of safety:- 3

4. SOLUTION 1:-
 From Table, Nc = 5.71, Nq = 1.0, Nγ = 0.0
 Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for strip footing.
 Qu = cNc + γDNq + 0.5γBNγ
 Qu = 20x5.71 + 12x1.0x1.0 + 0.5x12x1x0 = 126.25 kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 126.2/3 = 42.06 kN/m2 (ANSWER)

5.  RECTANGULAR FOOTING
 2. A rectangular footing 1mx2m is placed at a depth 2m below the
ground surface on a c- φ soil. Calculate the safe bearing capacity using
a factor of safety 2.5. The soil has following parameters : γ=12 KN/m3,
c=10 KN/m2, φ=20
 Given:-
 Cohesion:- 10 kN/m2
 φ = 20
 Unit weight of soil:- 12 kN/m3
 1x2 m rectangular footing, bottom of footing at 2 m below ground level.
 Factor of safety:- 2.5

6.  Solution 2:-
 From Table, Nc = 17.7, Nq = 7.4, Nγ = 5.0 for φ = 20
 Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for square footing.
 Qu = (1+0.3B/L)cNc + γDNq + 0.5(1-0.2B/L)γBNγ
 Qu = 1.15x10x17.7 + 12x2x7.4 + 0.45x12x1x5 = 408.15 kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 408.15/2.5 = 163.26 kN/m2 (ANSWER)

7.  SQUARE FOOTING
 3.If a square footing of size4mx4m is resting on the surface deposit of
clay with a cohesion value 120 kN/m2 at a depth 3m below ground level.
Determine the ultimate bearing capacity of the footing. (Unit weight of
soil= 18 kN/m3)
 Given:-
 Soil type:- Clay.
 Cohesion:- 120 kN/m2
 φ = 0
 Unit weight of soil:- 18 kN/m3
 4 m square footing, bottom of footing at 3 m below ground level.
 Factor of safety:- 3

8.  Solution 3:-
 From Table, Nc = 5.7, Nq = 1.0, Nγ = 0 for φ = 0
 Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for square footing.
 Qu = 1.3cNc + γDNq + 0.4γBNγ
 Qu = 1.3x120x5.7 + 18x3x1 + 0.4x18x4x0 = 943.2 kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 943.2/3 = 314.4 kN/m2(ANSWER)

9.  CIRCULAR FOOTING
 4.Determine the allowable bearing capacity of a circular footing of 4m
diameter and bottom of footing at 5m below ground level provided
with a factor of safety 3.5.The foundation soil has c=20 kN/m2, φ=30
and unit weight 12 kN/m3. using Terzaghi’s analysis.
 Given:-
 Soil type:- Sandy Clay.
 Cohesion:- 20 kN/m2
 φ = 30
 Unit weight of soil:- 12 kN/m3
 4 m diameter circular footing for a circular tank, bottom of footing at 5
m below ground level.
 Factor of safety:- 3.5

10.  Solution 4:-
 From Table, Nc = 37.2, Nq = 22.5, Nγ = 19.7 for φ = 30
 Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for circular footing.
 Qu = 1.3cNc + γDNq + 0.3γBNγ
 Qu = 1.3x20x37.2 + 12x5x22.5 + 0.3x12x4x19.7 = 2600.88 kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 2600.88/3.5 = 743.1 kN/m2

11.  EFFECT OF WATER TABLE
 5. A strip footing on 3m wide has a depth of 2 m in sand. The saturated
unit weight of sand is 18.5 kN/m3 and unit weight above water table is
16kN/m3. the shear strength parameters are c=0 and φ = 30. Determine
allowable soil bearing capacity using Terzaghi’s equations.
a) Water table is at 4m ground level.
 Given:-
 Soil type:- Cohesionless Soil.
 Cohesion:- 0 (Negligible)
 Unit weight of dry soil, γ1:- 16 kN/m3
 Unit weight of saturated soil, γ2:- 18.5 kN/m3
 3 m wide strip footing, bottom of footing at 2 m below ground level.
 Water Table is at 4 m below ground level.
 Factor of safety:- 3

12.  Solution 5:-
 From Table, Nc = 37.2, Nq = 22.5, Nγ = 19.7 for φ = 30
 Effect of Water Table:-
 Rw1 = 0.5(1 + Zw1/D) = 0.5(1 + 2/2) = 1
 Rw2 = 0.5(1 + Zw2/B) = 0.5(1 + 2/3) = 0.83
 Determine ultimate soil bearing capacity using Terzaghi’s bearing
capacity equation for strip footing.
 Qu = cNc + γ1DNqRw1 + 0.5(γ1 + γ2)0.5BNγ Rw2
 Qu = 0x37.2 + 16x2x22.5x1 + 0.5x(16+ 18.5)x0.5x3x19.7x.83 =
1143.08kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 1143.08/3 = 381.02 kN/m2

13. MEYERHOF's Bearing Capacity
Equations Based Problems

14.  Strip footing
 1.Calculate the allowable bearing capacity of a strip footing 2m wide in
plane founded at a depth of 3m below ground level. The load on the
footing acts at an angle of 25. Soil parameters are ( c=50 kN/m2,
F.O.S=3 , unit weight=11 kN/m3)
 Solution:-
 Given:-
 Soil type:- Clayey Sand.
 Cohesion:- 50 kN/m2
 φ = 25
 Unit weight of soil:- 11 kN/m3
 2 m wide strip footing, bottom of footing at 3 m below ground level.
 Factor of safety:- 3

15.  Solution 1:-
 Determine ultimate soil bearing capacity using Meyerhof’s bearing
capacity equation for vertical load.
 Passive pressure coefficient:-
 Kp = tan2(45 + φ/2) = tan2(45 + 25/2) = 2.5

 Shape factors:-
 Sc = 1 + 0.2Kp(B/L)
 Sc = 1 + 0.2x2.5x(0) = 1 (For Strip footing, L = Infinite)
 Sq = Sγ = 1+ 0.1Kp(B/L) = 1+ 0.1x2.5x(0) = 1

 Depth factors:-
 Dc = 1+ 0.2√Kp(B/L) = 1 + 0.2x(2/0)x√2.5 = 1
 Dq = Dγ = 1+ 0.1√Kp(D/B) = 1 + 0.1x√2.5x(3/2) = 1.24

16.  From Table, Nc = 20.7, Nq = 10.7, Nγ = 6.8 for φ = 25
 Qu = cNcScDc + γDNqSqDq + 0.5γBNγSγDγ
 Qu = 50x20.7x1x1 + 11x3x10.7x1x1.24 + 0.5x11x2x6.8x1x1.24 = 1565.6
kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 1565.59/3 = 521.86 kN/m2 (ANSWER)

17.  Rectangular footing
 2.Calculate allowable bearing capacity of a rectangular footing 3mx1m
at a depth of 3m below ground level. Angle of internal friction is
20.other parameters are C=50, unit weight=25 kN/m3.
 Solution:-
 Given:-
 Soil type:- Sandy Clay.
 Cohesion:- 50 kN/m2
 φ = 20
 Unit weight of soil: 25 kN/m3
 3 m by 1 m rectangular footing, bottom of footing at 3 m below ground
level.
 Factor of safety:- 3

18.  Solution 2:-
 Determine ultimate soil bearing capacity using Meyerhof’s bearing
capacity equation for vertical load.
 Passive pressure coefficient:-
 Kp= tan2(45 + φ/2) = tan2(45 + 20/2) = 2
 Shape factors:-
 Sc = 1 + 0.2Kp(B/L) = 1 + 0.2x2x(1/3) = 1.13
 Sq = Sγ = 1 + 0.1Kp(B/L) = 1 + 0.1x2x(1/3) = 1.07
 Depth factors:-
 Dc = 1 + 0.2√Kp(B/L) = 1 + 0.2x√2(1/3) = 1.09
 Dq = Dγ = 1 + 0.1√Kp(D/B) = 1 + 0.1x√2(3/1) = 1.42

19.  From Table, Nc = 14.9, Nq = 6.4, Nγ = 2.9 for φ = 20
 Qu = cNcScDc + γDNqSqDq + 0.5γBNγSγDγ
 Qu = 50x14.9x1.13x1.09 + 25x3x6.4x1.07x1.42 + 0.5x25x1x2.9x1.07x1.42 =
1702.01 kN/m2
 Allowable soil bearing capacity,
 Qa = Qu/F.S. = 1702.01/3 = 567.33 kN/m2(ANSWER)