Successfully reported this slideshow.
Upcoming SlideShare
×

# Numerical Problem and solution on Bearing Capacity ( Terzaghi and Meyerhof Theory ) http://usefulsearch.org

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

• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

### Numerical Problem and solution on Bearing Capacity ( Terzaghi and Meyerhof Theory ) http://usefulsearch.org

1. 1. Topic: Bearing Capacity Of Soil Using Terzaghi’s Equation Using Meyerhof’s equation
2. 2. Terzaghi's Bearing Capacity Equations Based Problems
3. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 13. MEYERHOF's Bearing Capacity Equations Based Problems
14. 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. 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. 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. 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. 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. 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)