1 / 41

Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION. Course : S0484/Foundation Engineering Year : 2007 Version : 1/0. SHALLOW FOUNDATION. Topic: General Terzaghi Model Meyerhoff Model Brinch Hansen Model Influence of multi layer soil Influence of ground water elevation

jana
Download Presentation

Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Session 5 – 6 BEARING CAPACITY OF SHALLOW FOUNDATION Course : S0484/Foundation Engineering Year : 2007 Version : 1/0

  2. SHALLOW FOUNDATION Topic: • General • Terzaghi Model • Meyerhoff Model • Brinch Hansen Model • Influence of multi layer soil • Influence of ground water elevation • Shallow Foundation Bearing by N-SPT value

  3. TYPES OF SHALLOW FOUNDATION

  4. TYPES OF SHALLOW FOUNDATION

  5. TERZAGHI MODEL Assumptions: • Subsoil below foundation structure is homogenous • Shallow foundation Df < B • Continuous, or strip, footing : 2D case • Rough base • Equivalent surcharge

  6. TERZAGHI MODEL • FAILURE ZONES: • ACD : TRIANGULAR ZONES • ADF & CDE : RADIAL SHEAR ZONES • AFH & CEG : RANKINE PASSIVE ZONES

  7. STRIP FOUNDATION qult = c.Nc + q.Nq + 0.5..B.N SQUARE FOUNDATION qult = 1.3.c.Nc + q.Nq + 0.4..B.N CIRCULAR FOUNDATION qult = 1.3.c.Nc + q.Nq + 0.3..B.N TERZAGHI MODEL (GENERAL FAILURE) • Where: • c = cohesion of soil • q =  . Df ; Df = the thickness of foundation embedded on subsoil • = unit weight of soil B = foundation width Nc, Nq, N = bearing capacity factors

  8. BEARING CAPACITY FACTORS GENERAL FAILURE

  9. BEARING CAPACITY FACTORS GENERAL FAILURE

  10. TERZAGHI MODEL (LOCAL FAILURE) • STRIP FOUNDATION qult = 2/3.c.Nc’ + q.Nq’ + 0.5..B.N’ • SQUARE FOUNDATION qult = 0.867.c.Nc’ + q.Nq’ + 0.4..B.N’ • CIRCULAR FOUNDATION qult = 0.867.c.Nc’ + q.Nq’ + 0.3..B.N’ • Where: • c = cohesion of soil • q =  . Df ; Df = the thickness of foundation embedded on subsoil • = unit weight of soil B = foundation width Nc, Nq, N = bearing capacity factors ’ = tan-1 (2/3. tan)

  11. BEARING CAPACITY FACTORS LOCAL FAILURE

  12. BEARING CAPACITY FACTORS

  13. GROUND WATER INFLUENCE

  14. GROUND WATER INFLUENCE • CASE 1 0  D1 < Df q = D1.dry + D2 . ’ • CASE 2 0  d  B  q = dry.Df the value of  in third part of equation is replaced with  = ’ + (d/B).(dry - ’)

  15. FACTOR OF SAFETY Where: qu = gross ultimate bearing capacity of shallow foundation qall = gross allowable bearing capacity of shallow foundation qnet(u) = net ultimate bearing capacity of shallow foundation qall = net allowable bearing capacity of shallow foundation FS = Factor of Safety (FS  3)

  16. NET ALLOWABLE BEARING CAPACITY PROCEDURE: • Find the developed cohesion and the angle of friction • Calculate the gross allowable bearing capacity (qall) according to terzaghi equation with cd and d as the shear strength parameters of the soil • Find the net allowable bearing capacity (qall(net)) FSshear = 1.4 – 1.6 Ex.: qall = cd.Nc + q.Nq + ½ .B.N Where Nc, Nq, N = bearing capacity factor for the friction angle, d qall(net) = qall - q

  17. EXAMPLE – PROBLEM A square foundation is 5 ft x 5 ft in plan. The soil supporting the foundation has a friction angle of  = 20o and c = 320 lb/ft2. The unit weight of soil, , is 115 lb/ft3. Assume that the depth of the foundation (Df) is 3 ft and the general shear failure occurs in the soil. Determine: - the allowable gross load on the foundation with a factor of safety (FS) of 4. - the net allowable load for the foundation with FSshear = 1.5

  18. EXAMPLE – SOLUTION Foundation Type: Square Foundation

  19. EXAMPLE – SOLUTION

  20. GENERAL BEARING CAPACITY EQUATION Meyerhof’s Theory Df

  21. BEARING CAPACITY FACTOR

  22. SHAPE, DEPTH AND INCLINATION FACTOR

  23. P = 73 ton Tank dry = 13 kN/m3 sat = 18 kN/m3 c = 1 kg/cm2  = 20o 2 m Foundation GWL EXAMPLE 2 Determine the size (diameter) circle foundation of tank structure as shown in the following picture With P is the load of tank, neglected the weight of foundation and use factor of safety, FS = 3.5.

  24. EXAMPLE 3 SQUARE FOUNDATION B = 4m dry = 13 kN/m3 • DETERMINE THE FACTOR OF SAFETY FOR: • CASE 1 : GWL LOCATED AT 0.3m (MEASURED FROM THE SURFACE OF SOIL) • CASE 2 : GWL LOCATED AT 1.5m (MEASURED FROM THE SURFACE OF SOIL)

  25. ECCENTRICALLY LOADED FOUNDATIONS

  26. ECCENTRICALLY LOADED FOUNDATIONS

  27. ONE WAY ECCENTRICITY Meyerhof’s step by step procedure: • Determine the effective dimensions of the foundation as : B’ = effective width = B – 2e L’ = effective length = L Note: • if the eccentricity were in the direction of the length of the foundation, the value of L’ would be equal to L-2e and the value of B’ would be B. • The smaller of the two dimensions (L’ and B’) is the effective width of the foundation • Determine the ultimate bearing capacity to determine Fcs, Fqs, Fs use effective length and effective width to determine Fcd, Fqd, Fd use B • The total ultimate load that the foundation can sustain is Qult = qu’.B’.L’ ; where B’xL’ = A’ (effective area) • The factor of safety against bearing capacity failure is FS = Qult/Q • Check the factor of safety against qmax, or, FS = qu’/qmax

  28. EXAMPLE – PROBLEM A Square foundation is shown in the following figure. Assume that the one- way load eccentricity e = 0.15m. Determine the ultimate load, Qult

  29. EXAMPLE – SOLUTION With c = 0, the bearing capacity equation becomes

  30. TWO-WAY ECCENTRICITY

  31. TWO-WAY ECCENTRICITY – CASE 1

  32. TWO-WAY ECCENTRICITY – CASE 2

  33. TWO-WAY ECCENTRICITY – CASE 3

  34. TWO-WAY ECCENTRICITY – CASE 4

  35. BEARING CAPACITY OF LAYERED SOILS STRONGER SOIL UNDERLAIN BY WEAKER SOIL

  36. BEARING CAPACITY OF LAYERED SOILS

  37. BEARING CAPACITY OF LAYERED SOILS Rectangular Foundation

  38. BEARING CAPACITY OF LAYERED SOILS SPECIAL CASES • TOP LAYER IS STRONG SAND AND BOTTOM LAYER IS SATURATED SOFT CLAY (2 = 0) • TOP LAYER IS STRONGER SAND AND BOTTOM LAYER IS WEAKER SAND (c1 = 0 , c2 = 0) • TOP LAYER IS STRONGER SATURATED CLAY (1 = 0) AND BOTTOM LAYER IS WEAKER SATURATED CLAY (2 = 0) Find the formula for the above special cases

  39. BEARING CAPACITY FROM N-SPT VALUE A square foundation BxB has to be constructed as shown in the following figure. Assume that  = 105 lb/ft3, sat = 118 lb/ft3, Df = 4 ft and D1 = 2 ft. The gross allowable load, Qall, with FS = 3 is 150,000 lb. The field standard penetration resistance, NF values are as follow: Determine the size of the foundation

  40. SOLUTION Correction of standard penetration number (Liao and Whitman relationship)

  41. SOLUTION

More Related