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

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Numerical Problem and solution on Bearing Capacity ( Terzaghi and Meyerhof Theory )

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- 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)

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