4. Shallow Foundations
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The document discusses shallow foundations, which are defined as those with a depth less than 3 meters or less than their breadth. It outlines types of shallow foundations, including pad, strip, and raft foundations, along with key design criteria such as adequate depth, limiting settlement, and safety against shear failure. Additionally, it details the mechanics of shear failure modes and includes several examples of calculating net ultimate bearing capacity for various foundation scenarios.
4. Shallow Foundations
- 1. 4.BEARING CAPACITY OF SHALLOW FOUNDATIONS Dr MWAJUMA I. LINGWANDA The most carefully designed structures can fail if they are not supported by suitable foundations
- 2. Definition of Shallow foundations ▪ Shallow foundations are also called spread foundations ▪ But how shallow is shallow? ▪ The common definition of shallow foundations refers to the case where the founding depth is less than the breadth ▪ i.e. when 𝑫 < 𝑩 the foundation is shallow ▪ What about wide raft foundations? ▪ The above definition is not acceptable for wide rafts ▪ Therefore, a foundation is shallow if the depth is less than 3 m or less than the breadth of the foundation CEB 3319 FOUNDATION ENGINERING 2
- 3. Types of shallow foundations • Pad foundations These are usually provided to support structural columns. They may be circular, square or rectangular slab section • Strip foundations A strip foundation has a length >> than breadth. They are normally provided for load bearing walls or closely spaced columns where it is not economical to deal with individual pads CEB 3319 FOUNDATION ENGINERING 3
- 4. Types of shallow foundations cont.. • Raft/Mat foundations Raft foundations are required on soils of low bearing capacity or where structural columns are closely spaced in both directions such that individual pads would nearly touch each other. CEB 3319 FOUNDATION ENGINERING 4
- 5. Foundation design criteria Three main design criteria must be considered for a shallow foundation 1. Adequate depth ✓ The founding depth i.e. the depth from the surface to its underside should be sufficient to prevent effects due to changes of ground surface ✓ Sufficient depth to prevent overturning moments when the structure is subjected to horizontal forces 2. Limiting settlement ✓ The amount of tolerable total and differential settlements depends on functional and user requirements and also economic factors ✓ Most settlement damage may be classified as architectural damage as they normally affect finishes. CEB 3319 FOUNDATION ENGINERING 5
- 6. Foundation design criteria cont.. 3. Factor of safety against shear failure ✓ In foundations, shear failure occurs when soil divides into separate zones which move partially or fully along slip surfaces ✓ At this point, a plastic yielding condition has developed whereby the shear stress along the slip surface has reached a limiting value(ultimate limit state) ✓ The principal criteria for design will be satisfying a given factor of safety against shear failure CEB 3319 FOUNDATION ENGINERING 6
- 7. Three principal modes of shear failure 1. General shear failure ➢ This mode of failure occurs in dense or overconsolidated soils of low compressibility ➢ A clearly defined plastic yield slip forms under the footing and develop outwards to the surface ➢ Heaving of the ground occurs on both sides of the footing ➢ Failure is sudden and often accompanied by severe tilting on one side CEB 3319 FOUNDATION ENGINERING 7
- 8. Three principal modes of shear failure cont.. 2. Local shear failure ➢ Occurs in compressible soils ➢ A significant vertical movement may take place before any noticeable heaving ➢ Shear planes develop but fail to extend to the ground surface ➢ Very little tilting may take place ➢ The settlement which occurs usually is the principal design criteria CEB 3319 FOUNDATION ENGINERING 8
- 9. Three principal modes of shear failure cont.. 3. Punching shear failure ➢ Occurs in weak compressible soils ➢ Considerable vertical movement take place ➢ Slip surfaces restricted to vertical planes adjacent to the sides of the footing ➢ Heaving at the surface is usually absent and may be replaced by a drag down ➢ Normally there will be no tilting ➢ This kind of failure may be observed in highly compressible clays and in loose sands CEB 3319 FOUNDATION ENGINERING 9
- 10. Important Definitions • Total foundation pressure (𝒒) or gross loading intensity Is the intensity of total pressure on the ground beneath the foundation after the structure has been erected and fully loaded • The net bearing pressure (𝒒 𝒏) or the net foundation pressure Is the net change in total stress experienced by the soil at the base of the foundation Let 𝝈 𝟎 = the stress removed due to excavation (𝝈 𝟎 = 𝜸𝑫) 𝒒 = Total foundation pressure/ contact pressure at the base of the foundation Then, 𝒒 𝒏 = 𝒒 − 𝝈 𝟎 CEB 3319 FOUNDATION ENGINERING 10
- 11. Important Definitions cont.. • The net effective bearing pressure (𝒒 𝒏 ´ ) or the net effective foundation pressure The net foundation pressure expressed in terms of effective stress. When the water table is above the base of the foundation the net effective bearing pressure is given by; 𝒒 𝒏 ´ = 𝒒 − 𝝈 𝟎 ´ Where 𝝈 𝟎 ´ = 𝜸𝑫 − 𝜸 𝒘 𝒉 𝒘 CEB 3319 FOUNDATION ENGINERING 11
- 12. Important definitions cont.. • The ultimate bearing capacity (𝒒 𝒇) Is the value of the loading intensity at which the supporting ground fails in shear • The net ultimate bearing capacity (𝒒 𝒏𝒇) • Is the value of net loading intensity at which the supporting ground fails in shear • LOAD – SETTLEMENT CURVE CEB 3319 FOUNDATION ENGINERING 12 Settlement Load 0
- 13. Important definitions cont.. • The factor of safety against shear failure is defined as the ratio between the net ultimate bearing capacity and the net bearing pressure 𝑭 = 𝒒 𝒏𝒇 ´ 𝒒 𝒏 ´ = 𝒒 𝒇 − 𝝈 𝒐 ´ 𝒒 − 𝝈 𝒐 ´ For building foundations, the recommended factor of safety is between 3 and 2.5 • Presumed bearing pressure Is the net loading intensity considered appropriate to the particular type of ground for preliminary design purposes. The value is based on local experience or by calculation from strength tests using a factor of safety against shear failure CEB 3319 FOUNDATION ENGINERING 13
- 14. Important definitions cont.. • The allowable bearing pressure (𝒒 𝒂) Is the maximum allowable net loading intensity of a ground in a given case, taking into consideration the bearing capacity and the estimated amount and rate of settlement that will occur. Considering shear failure, allowable design bearing capacity is given by 𝒒 𝒂 = 𝒒 𝒇 − 𝝈 𝒐 ´ 𝑭 + 𝝈 𝒐 ´ CEB 3319 FOUNDATION ENGINERING 14
- 15. Simple bearing capacity formula • Assumptions i. The footing experiences a bearing capacity failure ii. Failure occurs along a circular surface iii. Undrained conditions prevail, 𝒄 𝒖 > 𝟎, 𝝓 𝒖 = 𝟎 iv. The shear strength of the soil between the ground and depth D is neglected v. The soil between the ground and D is considered as a surcharge and produces a vertical stress = 𝜸𝑫 • Taking moments about point O, • 𝒒 𝒇 𝑩. 𝑩 𝟐 = 𝒄 𝒖 𝝅𝑩. 𝑩 + 𝜸𝑫𝑩. 𝑩 𝟐 • 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 + 𝜸𝑫 • 𝒒 𝒇 ≈ 𝟔. 𝟐𝟖𝒄 𝒖 + 𝜸𝑫 CEB 3319 FOUNDATION ENGINERING 15 Consider a long footing with cross section shown below
- 16. Class exercise With similar assumptions regarding ground conditions as the previous foundation, derive a simple bearing capacity formula for the following cases I. The foundation rests on the ground level II. The foundation rests at a depth D and a long and wide surcharge load due to soil compacted on each side at the ground level to a height H. The unit weight of the compacted soil being 𝜸 𝟐 (i) 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 (ii) 𝒒 𝒇 = 𝟐𝝅𝒄 𝒖 + 𝜸𝑫 + 𝜸 𝟐 𝑯 CEB 3319 FOUNDATION ENGINERING 16
- 17. Terzaghi´s bearing capacity formula • The formula was developed for long foundations (strip foundations) then extended to square and circular foundations • A general shear failure is considered CEB 3319 FOUNDATION ENGINERING 17
- 18. Terzaghi´s bearing capacity formula cont.. • Assumptions ✓ The depth of the foundation is less than or equal to its width (𝑫 ≤ 𝑩) ✓ No sliding occurs between the foundation and the soil ✓ The soil is homogeneous semi infinite mass ✓ The shear strength of the soil is presented by both c and ϕ ✓ Failure is by general shear mode ✓ The soil between the ground surface and a depth D has no shear strength i.e. it serves only as a surcharge load CEB 3319 FOUNDATION ENGINERING 18
- 19. Terzaghi´s bearing capacity formula cont.. • Terzaghi proposed a three-term expression for ultimate bearing capacity of a long strip footing of breadth B as 𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝝈 𝒐 𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 • 𝒒 𝒇 = ultimate bearing capacity • 𝒄 = cohesion of soil beneath the foundation • 𝝈 𝟎 = overburden pressure at the foundation level • 𝑩 = breadth of the foundation • 𝜸 = unit weight of the soil beneath the foundation • 𝑵 𝒄, 𝑵 𝒒 and 𝑵 𝜸 areTerzaghi´s bearing capacity factors which are function of ϕ only 19
- 20. Terzaghi´s bearing capacity formula cont.. • 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓 . 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝝓 𝟐 • 𝑵 𝒄 = 𝑵 𝒒−𝟏 𝒕𝒂𝒏𝝓 for ϕ > 0 • 𝑵 𝒄 = 𝟓. 𝟕 for ϕ =0 • 𝑵 𝜸 ≈ 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 • Under undrained conditions, ϕ =0 and c = cu • 𝑵 𝒒 = 𝟏 • 𝑵 𝒄 = 𝟓. 𝟕 • 𝑵 𝜸 = 𝟎 • 𝒒 𝒇 = 𝟓. 𝟕𝒄 𝒖 + 𝝈 𝟎 CEB 3319 FOUNDATION ENGINERING 20 Note that Because of the shape of the failure surface, the values of 𝑐 and 𝜙 only need to represent the soil between the bottom of the footing and a depth B below the bottom. The soils between the ground surface and a depth D are treated simply as overburden.
- 21. Bearing capacity factors in chart 21
- 22. Example 1 A strip footing is 2.8 m wide and is founded at a depth of 2.4 m in a soil having the following properties: 𝒄´ = 𝟏𝟐𝒌𝑵/𝒎 𝟐 𝝓´ = 𝟐𝟎 𝒐 𝜸 = 𝟏𝟗𝒌𝑵/𝒎 𝟑 . Determine the net ultimate bearing capacity. CEB 3319 FOUNDATION ENGINERING 22
- 23. Solution to example 1 Sketch the problem and indicate the known parameters (soil properties and dimensions) 𝒒´ 𝒏 = 𝒒´ 𝒇 − 𝝈´ 𝒐 𝒒´ 𝒏 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸´𝑵 𝜸 − 𝜸´𝑫 determine the bearing capacity coefficients 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓 . 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝝓 𝟐 = 𝒆 𝝅𝒕𝒂𝒏𝟐𝟎 × 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝟐𝟎 𝟐 = 𝟔. 𝟑𝟗𝟗 CEB 3319 FOUNDATION ENGINERING 23
- 24. Solution to example 1 cont.. 𝑵 𝒄 = 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟔. 𝟑𝟗𝟗 − 𝟏 𝒕𝒂𝒏𝟐𝟎 = 𝟏𝟒. 𝟖𝟑𝟓 𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟔. 𝟑𝟗𝟗 − 𝟏 𝒕𝒂𝒏𝟐𝟎 = 𝟑. 𝟓𝟑𝟕 Substituting to the net bearing capacity equation 𝒒´ 𝒏 = (𝟏𝟐 × 𝟏𝟒. 𝟖𝟑𝟓) + (𝟏𝟗 × 𝟐. 𝟒 × 𝟔. 𝟑𝟗𝟗) + ( 𝟏 𝟐 × 𝟐. 𝟖 × 𝟏𝟗 × 𝟑. 𝟓𝟑𝟕) − 𝟏𝟗 × 𝟐. 𝟒 𝒒´ 𝒏 = 𝟓𝟗𝟎. 𝟖𝟓 − 𝟒𝟓. 𝟔 𝒒´ 𝒏 = 𝟓𝟒𝟓. 𝟐𝟓 ൗ𝒌𝑵 𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 24
- 25. Example 2 Determine the net ultimate bearing capacity of a strip footing of breadth 2.5 m which is to be founded 1.25 m below the surface of a thick layer of clay having 𝒄 𝒖 = 𝟖𝟖𝒌𝑵/𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 25
- 26. Solution to example 𝒒 𝒏 = 𝒒 𝒇 − 𝝈 𝒐 𝒒 𝒏 = 𝒄 𝒖 𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 − 𝜸𝑫 Bearing capacity coefficients for 𝜑 = 0; 𝑵 𝒒 = 𝟏 𝑵 𝒄 = 𝟓. 𝟕 𝑵 𝜸 = 𝟎 Substituting the coefficients we have 𝒒 𝒏 = 𝟓. 𝟕𝒄 + 𝜸𝑫 − 𝜸𝑫 𝒒 𝒏 = 𝟓. 𝟕𝒄 𝒒 𝒏 = 𝟓. 𝟕 × 𝟖𝟖 𝒒 𝒏 = 𝟓𝟎𝟏. 𝟔 ൗ𝒌𝑵 𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 26
- 27. Example 3 A strip foundation presented in the figure below is required to transmit an inclusive load of 650kN/m to the supporting soil. Adopting a factor of safety of 3, determine the breadth of the footing required. CEB 3319 FOUNDATION ENGINERING 27
- 28. Solution to example 3 The supported inclusive load is an allowable load with a factor of safety F = 3 𝒒 𝒂 = 𝒒 𝒇 − 𝝈 𝒐 ´ 𝑭 + 𝝈 𝒐 ´ 𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 Since the soil has no cohesion, the ultimate bearing capacity equation reduces to 𝒒 𝒇 = 𝜸𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 Bearing capacity coefficients 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓. 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝝓 𝟐 = 𝒆 𝝅𝒕𝒂𝒏𝟑𝟎. 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝟑𝟎 𝟐 = 𝟏𝟖. 𝟒𝟎𝟐 𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟏𝟖. 𝟒𝟎𝟐 − 𝟏 𝒕𝒂𝒏𝟑𝟎 = 𝟏𝟖. 𝟎𝟖𝟓 28
- 29. Solution to example 3 cont.. 𝒒 𝒇 = (𝟏𝟗 × 𝟏. 𝟓 × 𝟏𝟖. 𝟒𝟎𝟐) + ( 𝟏 𝟐 × 𝑩 × 𝟐𝟎 × 𝟏𝟖. 𝟎𝟖𝟓) 𝒒 𝒇 = 𝟓𝟐𝟒. 𝟒𝟔 + 𝟏𝟖𝟎. 𝟖𝟓𝑩 Substituting 𝑞 𝑓 into 𝑞 𝑎 equation..The value of 𝑞 𝑎 given is converted to a stress in ൗ𝑘𝑁 𝑚2 𝟔𝟓𝟎 𝑩 = 𝟓𝟐𝟒. 𝟒𝟔 + 𝟏𝟖𝟎. 𝟖𝟓𝑩 − 𝟏𝟗 × 𝟏. 𝟓 𝟑 + 𝟏𝟗 × 𝟏. 𝟓 1950 = 180.85𝐵2 + 495.69𝐵 + 85.5𝐵 𝟏𝟖𝟎. 𝟖𝟓𝑩 𝟐 + 𝟓𝟖𝟏. 𝟏𝟗𝑩 − 𝟏𝟗𝟓𝟎 = 𝟎 𝑩 = 𝟐. 𝟎𝟓 𝒎 CEB 3319 FOUNDATION ENGINERING 29
- 30. Effect of water table • POSITION I: 𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫 𝟏 + 𝜸´𝑫 𝟐 𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸´𝑵 𝜸 • POSITION II: (water table near the base of the footing) 𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸´𝑵 𝜸 • POSITION III: (water table far from the base of the footing) 𝒒 𝒇 = 𝒄𝑵 𝒄 + 𝜸𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 30
- 31. Example 4 A strip footing of breadth 1.75 m is required to transmit a net load of 650kN/m to a compacted sand of the following properties: : 𝒄´ = 𝟎 𝝓´ = 𝟑𝟖 𝒐 𝜸 𝒔𝒂𝒕 = 𝟐𝟎. 𝟓𝒌𝑵/𝒎 𝟑. Assume that the water table may rise to the surface and adopting a factor of safety of 3, determine the required depth of the footing. CEB 3319 FOUNDATION ENGINERING 31
- 32. Solution to example 4 From the definition of factor of safety 𝑭 = 𝒒 𝒏𝒇 ´ 𝒒 𝒏 ´ 𝒒´ 𝒏𝒇 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸´𝑵 𝜸 − 𝜸´𝑫 Bearing capacity coefficients 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝟑𝟖 . 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝟑𝟖 𝟐 = 𝟒𝟖. 𝟗 𝑵 𝜸 = 𝟏. 𝟖 𝟒𝟖. 𝟗 − 𝟏 𝒕𝒂𝒏𝟑𝟖 = 𝟔𝟕. 𝟒 Substituting the known variables and observing the effect of water table CEB 3319 FOUNDATION ENGINERING 32
- 33. Solution to example 4 cont.. 𝒒´ 𝒏𝒇 = 𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏 × 𝑫 × 𝟒𝟖. 𝟗 + 𝟏 𝟐 × 𝟏. 𝟕𝟓 × 𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏 × 𝟔𝟕. 𝟒 − (𝟐𝟎. 𝟓 − 𝟗. 𝟖𝟏)𝑫 𝒒´ 𝒏𝒇 = 𝟓𝟏𝟐. 𝟎𝟏𝟓𝑫 + 𝟔𝟑𝟎. 𝟒𝟒 Recall that 𝑭 = 𝒒 𝒏𝒇 ´ 𝒒 𝒏 ´ Substituting we have 𝟑 = 𝟓𝟏𝟐. 𝟎𝟏𝟓𝑫 + 𝟔𝟑𝟎. 𝟒𝟒 ൗ𝟔𝟓𝟎 𝟏. 𝟕𝟓 𝑫 = 𝟎. 𝟗𝟒 𝒎 ~𝟏 𝒎 33
- 34. Example 5 Determine the design load on a strip footing indicated in the figure below. The soil characteristic are 𝒄´ = 𝟏𝟎𝒌𝑵/𝒎 𝟐 𝝓´ = 𝟐𝟖 𝒐 𝜸 = 𝟏𝟖𝒌𝑵/𝒎 𝟑 𝜸 𝒔𝒂𝒕 = 𝟏𝟗. 𝟓 ൗ𝒌𝑵 𝒎 𝟑 CEB 3319 FOUNDATION ENGINERING 34
- 35. Solution to example 5 𝒒´ 𝒏𝒇 = 𝒄´𝑵 𝒄 + 𝜸´𝑫𝑵 𝒒 + 𝟏 𝟐 𝑩𝜸´𝑵 𝜸 − 𝜸´𝑫 Bearing capacity factors 𝑵 𝒒 = 𝒆 𝝅𝒕𝒂𝒏𝝓 . 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝝓 𝟐 = 𝒆 𝝅𝒕𝒂𝒏𝟐𝟖 × 𝒕𝒂𝒏 𝟐 𝟒𝟓 + 𝟐𝟖 𝟐 = 𝟖. 𝟖𝟒 𝑵 𝒄 = 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟖. 𝟖𝟒 − 𝟏 𝒕𝒂𝒏𝟐𝟖 = 𝟏𝟒. 𝟕𝟒 𝑵 𝜸 = 𝟏. 𝟖 𝑵 𝒒 − 𝟏 𝒕𝒂𝒏𝝓 = 𝟏. 𝟖 𝟖. 𝟖𝟒 − 𝟏 𝒕𝒂𝒏𝟐𝟖 = 𝟕. 𝟓𝟎 𝒒´ 𝒏𝒇 = 𝟏𝟎 × 𝟏𝟒. 𝟕𝟒 + 𝟏𝟖 × 𝟏. 𝟓 + (𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏) × 𝟏 × 𝟖. 𝟖𝟒 + 𝟎. 𝟓 × 𝟑 × 𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏 × 𝟕. 𝟓𝟎 − 𝟏. 𝟓 × 𝟏𝟖 + (𝟏𝟗. 𝟓 − 𝟗. 𝟖𝟏) × 𝟏 𝒒´ 𝒏𝒇 = 𝟓𝟒𝟒. 𝟎𝟔 ൗ𝒌𝑵 𝒎 𝟐 or 𝒒´ 𝒏𝒇 = 𝟏𝟔𝟑𝟐. 𝟏𝟖 Τ𝒌𝑵 𝒎 35
- 36. Generalized formula for bearing capacity • When the assumptions stated by Terzaghi are not fulfilled, the following expression should be used for ultimate bearing capacity 𝒒 𝒇 = 𝒄𝑵 𝒄 𝒔 𝒄 𝒅 𝒄 𝒍 𝒄 + 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 𝒅 𝒒 𝒍 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 𝒔 𝜸 𝒅 𝜸 𝒍 𝜸 𝒔 𝒄, 𝒔 𝒒, 𝒔 𝜸 are shape factors, applied for non continuous footing 𝒅 𝒄, 𝒅 𝒒, 𝒅 𝜸 are depth factors, applied when D > B 𝒍 𝒄, 𝒍 𝒒, 𝒍 𝜸 are load inclination factors, applied when the load is inclined CEB 3319 FOUNDATION ENGINERING 36
- 37. Generalized formula for bearing capacity cont.. Some proposed factors Shape factors: 𝒔 𝒄 = 𝒔 𝒒 = 𝟏 + 𝟎. 𝟐 𝑩 𝑳 𝒔 𝜸 = 𝟏 − 𝟎. 𝟒 𝑩 𝑳 for rectangular footings 𝒔 𝒄 = 𝟏. 𝟑 𝒔 𝒒 = 𝟏. 𝟐 𝒔 𝜸 = 0.8 for square footings Depth factors: 𝒅 𝒄= 𝟏 + 𝟎. 𝟐 𝑫 𝑩 𝒅 𝒒 = 𝟏 + 𝟎. 𝟏 𝑫 𝑩 𝒅 𝜸= 𝟏 CEB 3319 FOUNDATION ENGINERING 37
- 38. Example 6 A rectangular footing of breadth 6 m and length 15 m is founded at a depth of 4.5 m in a fine soil having the following properties: 𝒄 𝒖 = 𝟒𝟎 𝒌𝑵/𝒎 𝟐 𝜸 = 𝟏𝟖𝒌𝑵/𝒎 𝟑 Determine the ultimate bearing capacity of the footing CEB 3319 FOUNDATION ENGINERING 38
- 39. Solution to example 6 • Determine 𝒒 𝒇 𝒒 𝒇 = 𝒄𝑵 𝒄 𝒔 𝒄 + 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 𝒔 𝜸 Bearing capacity coefficients for 𝜑 = 0; 𝑵 𝒒 = 𝟏 𝑵 𝒄 = 𝟓. 𝟕 𝑵 𝜸 = 𝟎 𝒒 𝒇 = 𝟓. 𝟕𝒄 𝒖 𝒔 𝒄 + 𝜸𝑫𝒔 𝒒 𝒔 𝒄 = 𝒔 𝒒 = 𝟏 + 𝟎. 𝟐 𝑩 𝑳 = 𝟏 + 𝟎. 𝟐 𝟔 𝟏𝟓 = 𝟏. 𝟎𝟖 𝒒 𝒇 = 𝟓. 𝟕 × 𝟒𝟎 × 𝟏. 𝟎𝟖 + 𝟏𝟖 × 𝟒. 𝟓 × 𝟏. 𝟎𝟖 𝒒 𝒇 = 𝟑𝟑𝟑. 𝟕𝟐 ൗ𝒌𝑵 𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 39
- 40. Example 7 A footing 2.25 by 2.25m is located at a depth of 1.5 m in a sand, the shear strength parameters to be used in design being 𝒄´ = 𝟎, and 𝝓´= 𝟑𝟖 𝒐. Determine the ultimate bearing capacity (a) if the water table is well below foundation level and (b) if the water table is at the surface. (qf = 1515.04 kN/m2) The unit weight of the sand above the water table is 18 kN/m3; the saturated unit weight is 20 kN/m3. CEB 3319 FOUNDATION ENGINERING 40
- 41. Solution to example 7 (a) Water table well below the foundation base 𝒒 𝒇 = 𝝈 𝟎 𝑵 𝒒 𝒔 𝒒 + 𝟏 𝟐 𝑩𝜸𝑵 𝜸 𝒔 𝜸 𝑁𝑞 = 48.9 𝑁𝛾 = 67.4 𝒔 𝒒 = 𝟏. 𝟐 𝒔 𝜸 = 0.8 𝒒 𝒇 = 𝟏𝟖 × 𝟏. 𝟓 × 𝟒𝟖. 𝟗 + 𝟏 𝟐 × 𝟐. 𝟐𝟓 × 𝟏𝟖 × 𝟔𝟕. 𝟒 𝒒 𝒇 = 𝟐𝟔𝟕𝟔. 𝟐𝟒 ൗ𝒌𝑵 𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 41
- 42. Solution to example 7 cont.. (b)Water table at the surface Bearing capacity as well as shape factors remains the same as in part (a) 𝒒 𝒇 = 𝜸 𝒔𝒂𝒕 − 𝜸 𝒘 𝑫𝑵 𝒒 𝒔 𝒒 + 𝟏 𝟐 𝑩 𝜸 𝒔𝒂𝒕 − 𝜸 𝒘 𝑵 𝜸 𝒔 𝜸 𝑞 𝑓 = 20 − 9.81 × 1.5 × 48.9 + 1 2 × 2.25 × 20 − 9.81 × 67.4 × 0.8 𝑞 𝑓 = 896.92 + 618.12 𝒒 𝒇 = 𝟏𝟓𝟏𝟓. 𝟎𝟒 ൗ𝒌𝑵 𝒎 𝟐 CEB 3319 FOUNDATION ENGINERING 42
- 43. allowable bearing capacity For a given foundation in a given soil, the allowable design value for the applied bearing pressure must satisfy the two criteria listed below: I. An ultimate limit state value (shear strength) Allowable design bearing capacity: 𝒒 𝒂 = 𝒒 𝒇 − 𝝈 𝒐 ´ 𝑭 + 𝝈 𝒐 ´ II. A serviceability limit state (settlement) Allowable design bearing capacity 𝒒 𝒂 = bearing pressure corresponding to a specific limit value 𝑺 𝑳 of undrained or drained settlement 43
- 44. allowable bearing capacity cont.. (a) Immediate or undrained settlement 𝒒 𝒂 = 𝑺 𝑳 𝑬 𝒖 𝑩(𝟏 − 𝝂 𝟐)𝑰 𝒑 + 𝝈 𝒐 (b) Consolidation or drained settlement 𝒒 𝒂 = 𝑺 𝑳 𝒎 𝒗(𝑰+𝟏)𝑯 𝒐 + 𝝈´ 𝒐 Where • 𝑞 𝑎 = allowable design bearing capacity • 𝑆 𝐿 = a limit value of undrained or drained stability • 𝐸 𝑢 = undrained elastic modulus of the soil • 𝐵 = least lateral dimension of the foundation 𝜈 = Poisson's ratio for the soil • 𝐼 𝑝 = influence factor for vertical displacement • 𝑚 𝑣 = coefficient of volume compressibility for the soil • 𝐻 𝑜 = original thickness of the compressed layer • 𝐼 = stress influence coefficient CEB 3319 FOUNDATION ENGINERING 44
- 45. allowable bearing capacity cont.. Allowable bearing capacity and settlements of sands can alternatively be determined from results of in-situ tests such as standard penetration test and cone penetration test CEB 3319 FOUNDATION ENGINERING 45
- 46. Design process The following is the usual procedure followed in foundation design 1. Proposing the type, shape and size of foundation using presumed bearing values. Suggested in BS 8004 for example, for medium dense sand presumed bearing values are typically 100 – 300 kN/m2 and for stiff clays between 150 – 300 kN/m2 2. Evaluation of the loading conditions and calculation of expected contact pressure 3. Determination of shear strength and settlement characteristics of the soil using laboratory or field tests 4. Evaluation of bearing capacity and compare with the expected pressures 5. Adjusting the size and depth of the footing to arrive to a design which represents both structural and economic efficiency 46
- 47. Exercises 1. A strip footing 2.4 m wide is founded at a depth of 2.8 m in a soil having the following properties: 𝒄´ = 𝟏𝟐 𝐤𝐏𝐚, and 𝝓´= 𝟐𝟎 𝒐, 𝜸 = 𝟏𝟗𝒌𝑵/𝒎 𝟑 Determine (a)The net ultimate bearing capacity (b) the design bearing capacity adopting a factor of safety of 3 2. A square footing is to be founded to a depth of 2.6 m and will transmit a uniform load of 2.6 MN. Adopting a factor of safety of 3, determine the required size of the footing. Soil properties are: 𝒄 𝒖 = 𝟔𝟒 𝐤𝐍/𝒎 𝟐 𝛄 = 𝟏𝟗 𝐤𝐍/𝒎 𝟑 CEB 3319 FOUNDATION ENGINERING 47
- 48. Exercises 3. A rectangular foundation 10 x 5 m is to be designed, using a factor of safety of 3 to carry a uniform load of 86.6 MN, in fully drained conditions in the following soil: 𝒄´ = 𝟎, and 𝝓´ = 𝟑𝟓 𝒐 , 𝜸 = 𝟐𝟎 𝒌𝑵/𝒎 𝟑 , 𝜸 𝒔𝒂𝒕 = 𝟐𝟐 𝒌𝑵/𝒎 𝟑 a) Determine a suitable founding depth b) Determine the percentage reduction in load carrying capacity that results from the ground water level rising i. To the underside of the footing ii. To the ground surface c) Calculate the factor of safety when the 86.6 MN load is applied and the two conditions in (b) occur CEB 3319 FOUNDATION ENGINERING 48