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- 1. GEOTECHNICAL ENGINEERING - II Engr. Nauman Ijaz SETTLEMENT ANALYSIS Chapter # 03 UNIVERSITY OF SOUTH ASIA
- 2. SETTLEMENT When a soil deposit is loaded, deformation will occur due to change in stress. The total vertical downward deformation at the surface resulting from the load is called Settlement. Similarly when load is decreased (e.g during excavation) the deformation may vertically upward and is known as swelling. Estimate of settlement and swelling are made using identical procedures.
- 3. TYPES OF SETTLEMENT a) b) Types with respect to Permanency. Permanent settlement Temporary Settlement. Types with respect to Mode of Occurrence. a) b) c) a) b) Primary consolidation settlement (Sc) Secondary consolidation settlement (Ss) Immediate settlement (Si) Types with respect to Uniformity. Uniform settlement Differential settlement
- 4. Types with respect to Permanency Permanent Settlement (Irreversible settlement): • • • • • This type of settlement is caused due to distortion brought about by sliding and rolling of particles under the action of applied stresses. The sliding and rolling will reduce the voids resulting in reduction of volume of soil deposit. The increased in stresses may also crush the soil particles while alter the material and produce some settlement. This type of settlement is permanent and undergoes insignificant recovery upon removal of load. Settlement due to consolidation (both primary and Secondary) generally falls under this category.
- 5. TEMPORARY SETTLEMENT Settlement due to elastic compression of soil are usually reversible and recover a major part upon load release. Immediate settlement falls under this category. This settlement is generally small in soils.
- 6. TYPES WITH RESPECT TO MODE OF OCCURRENCE: PRIMARY CONSOLIDATION SETTLEMENT (Sc): These settlements are time dependent or long term settlements and completion time varies from 1- 5 years or more. This is also known as Primary consolidation (i.e the settlement caused due to expulsion of water from the pores of saturated fine grained soils (clays). This type of settlement is predominant in saturated inorganic fine grained soil (clays).
- 7. SECONDARY CONSOLIDATION This is the consolidation under constant effective stress causing no drainage. This is very predominant in certain Organic soils, but insignificant for inorganic soil. This is similar to creep in concrete.
- 8. IMMEDIATE SETTLMENT This type of settlement is predominant in coarse grained soils of high permeability and in unsaturated fine grained soils of low permeability. The completion time is usually few days (say about 7 days). Usually this type of settlement is completed during construction period and is called build in settlement. Also known as short term settlement.
- 9. Settlement produced due to inadequate shear strength of the soil mass is caused due to bearing capacity failure of soil. Settlement due to lateral expulsion of soils from underneath the foundation is an example of this category. This settlement can not be estimated using present knowledge of soil mechanics but can be controlled easily by controlling bearing capacity. So the total settlement; St = Si + Sc + Ss
- 10. TYPES WITH RESPECT TO UNIFORMITY UNIFORM SETTLEMENT: When all the points settle with an equal amount, the settlement is known as uniform settlement. This type of settlement is possibly only under relatively rigid foundation loaded with uniform pressure and resting on uniform soil deposit, which is a very rare possibility. This type of settlement may not endanger the structure stability but generally affects the utility of the structure by jamming doors/windows, damaging the utility lines ( sewer, water supply mains etc)
- 11. DIFFERENTIAL SETTLEMENT When different parts of a structure settle by different magnitude, the settlement is called differential settlement. This is very important as it may endanger the structural stability and may cause catastrophic failure. If soil is granular, then differential settlement will be 2/3 of the total maximum settlement. In case of cohesive soil, possible differential settlement is about 1/3 of the maximum settlement.
- 12. SETTLEMENTS OF FOUNDATIONS NO SETTLEMENT * TOTAL SETTLEMENT * DIFFERENTIAL SETTLEMENT Uniform settlement is usually of little consequence in a building, but differential settlement can cause severe structural damage
- 13. DIFFERENTIAL SETTLEMENT
- 14. DIAGONAL CRACKS IN BRICK WORK DUE TO DIFFERENTIAL SETTLEMENT
- 15. LEANING TOWER OF PISA TOWER
- 16. PLASTIC AND ELASTIC DEFORMATION All materials deform when subjected to an applied load. If all this deformation is retained when load is released, it is said to have experienced Plastic deformation. Conversely if the material returns to its original size and shape when the load is released, it is said to have experienced elastic deformation. Soil exhibits both plastic and elastic deformation.
- 17. ANGULAR DISTORTION Angular distortion between two points under a structure is equal to the differential settlement between the points divided by the distance between them. Angular distortion is also known as Relative Rotation. Differential settlement = ∆S = Smax – Smin Angular Distortioin = ∆S /L
- 18. LIMITS OF ALLOWABLE AND TOTAL DIFFERENTIAL SETLLEMENT Tolerable differential settlement of buildings, in inches, recommended maximum values in parentheses. CRITERION ISOLATED FOUNDATION ANGULAR DISTORTION RAFT 1/300 GREATEST DIFFERENTIAL SETTLEMENT 1³/4 (11/2) (CLAYS) 1³/4 (1) (SANDS) MAXIMUM SETTLEMENT CLAYS 3 (21/2) 3- 5 (21/2 - 4) SANDS 2 (11/2) 2-3 (11/2 - 21/2)
- 19. For normal structures with isolated foundation, total settlements up to 50mm and differential settlement between adjacent columns up to 20mm is acceptable. According to Euro Code 7, the maximum acceptable relative rotation for open frames and load bearing or continuous brick wall are likely to range from about 1/2000 to about 1/300 to prevent the occurrence of a serviceability limit state in the structure. According to Bjerrum safe limit to avoid cracking in the pannel walls of frame structure ( partition wall) = 1/500
- 20. Large total and differential settlement may be acceptable provided the relative rotations remains within acceptable limits and provided total settlement do not cause problems with the services entering the structure or cause titling. Average maximum settlement of Structures on Perma frost. (a layer of soil beneath the surface that remains frozen through out the year) Structure Average Max Settlement (mm) Reinforced Concrete 150 at 40mm/year Masonry ,Precast Concrete 200 at 60mm/year Steel frames 250 at 80mm/year Timber 400 at 129mm/year
- 21. CAUSES OF SETTLEMENT Following are the major causes of settlement; Changes in Stress due: a) Applied structural load or excavation . b) Movement of ground water table . c) Vibrations due to Machines, Earth quake. Desiccation due to surface drying and/or plants life. (Desiccation = Removal of water from soil Loss (evaporation) of water / effective stress(inter-granular stresses) increase /Mass shrinkage will start ) ( Reason= High fines content, Volume of water is the direct function of shrinkage)
- 22. Changes due to structure of soil. Adjacent excavation. Mining Subsidence. Swelling and Shrinkage. Lateral expulsion of soil. Land slides.
- 23. Remedial Measures a) b) Philosophy of remedial measures is to; Reduce or eliminate settlement. Design structure to with stand the Settlement. To reduce or eliminate stresses following considerations can be followed; Reduce Contact pressure. Reduce Compressibility of soil deposits using various ground improvement techniques ( Stabilization, pre-compression, vibro-flotation etc). Remove soft compressible material such as peat, muck. etc
- 24. Built slowly on cohesive soils to avoid lateral expulsion of soil mass, and to give time for pore pressure dissipation. Consider using deep foundation (piles or piers). Provide lateral restraint against lateral expulsion. To achieve uniform settlement one can resolve to; Design of footing for uniform pressure. Use artificial cushion underneath the less settling foundation parts of the structure.
- 25. Built different parts of foundation of different weight on different soil at different depths. Built the heavier parts of structure first (such as towers) and lighter parts later.
- 26. CONSOLIDATION (OEDOMETER) TEST This test is performed to determine the magnitude and rate of change in volume of a laterally confined soil specimen undergoes when subjected to different vertical pressure To compute the consolidation settlement in a soil we need to know stress- strain properties. (i.e relationship between( σZ & εz ). This normally involves bringing the soil sample to the laboratory, subjecting it to a series of loads and measuring corresponding settlements.
- 27. This test is known as consolidation test or Oedometer test. We mostly interested in engineering properties of natural soils as they exist in the field, so consolidation tests are usually performed on high quality Undisturbed samples. It is also important for samples that were saturated in the field to remain so during storage and testing. If the sample is allowed to dry, a process we call Desiccation, negative pore pressure will develop and may cause irreversible changes in the in the soil.
- 28. Consolidation Apparatus
- 29. The test begins by applying vertical normal load P. It produces a vertical effective stress of; σ’Z = P/A - U Where; σ’Z = Vertical effective stress. P A U = Applied Load. = Cross sectional area of soil specimen. = pore water pressure inside the soil specimen. The water bath barely covers the specimen , so the pore water pressure is very small as compared to the vertical stress and thus may be ignored; σ’Z = P/A
- 30. The vertical strain εz is noted by monitoring the dial gage, for each corresponding increase in load. εz = Change in Dial Gage Reading Initial height of the sample Increase the load to some higher value and allow the soil to consolidate again, thus obtaining a second value of (σZ ,εz). This process will continues until we have reached the desired peak vertical stress; from this loading sequence we obtain the loading curve ABC. We then incrementally unload the sample and allow it to rebound thus producing unloading curve CD. Shown in the figure presented in the next slide. Data is plotted on logarithmic scale.
- 31. AB representing the Recompression Curve BC representing the Virgin Curve ,CD representing the Rebound Curve
- 32. SIGNIFICANCE The consolidation properties determine from the consolidation test are used to estimate the magnitude of both primary and secondary settlement. The consolidation parameters we find out; 1. Compression Index (Cc) . 2. Recompression Index (Cr) . 3. Coefficient of Volume change (mv) . 4. Pre-consolidation pressure . 5. 6. e – field Density Coefficient of Consolidation (Cv) . .
- 33. The first 4 parameters can be determined from e ~ σ’Z graph. Field density is determined by; γd = Gs γw 1+e
- 34. Cc = Compression Index σ’Z0 = Initial vertical effective stress σ’Zf = Final vertical effective stress Cc = ∆e/ log(σ’Zf / σ’Z0)
- 35. Cr = σ’C = Recompression Index Pre- consolidation pressure Pre-consolidation pressure is the maximum pressure that the soil has been subjected in the past. It varies with depth and is used to identify over consolidation of specimen.
- 36. NORMALLY CONSOLIDATED (NCC) A soil is said to be normally consolidated when; σ’C = σ’ 0Z σ’C = σ’ 0Z Pre-consolidation Pressure. = Present effective overburden pressure.
- 37. OVER CONSOLIDATED (OCC) The soil is said to be over consolidated when; σ’C > > > σ’0Z 1. 2. 3. 4. This shows that soil has been subjected to some over burden pressure in the past which has been removed. This over burden pressure may be due to; Snow loading Past Structure which now has been removed. Level of ground is lowered. Lowering of the ground water table.(In the past ground water table is high)
- 38. OVER CONSOLIDATED RATIO (OCR) Over consolidated ratio is defined as ratio of pre-consolidation pressure to present overburden pressure. OCR = σ’C / σ’ 0Z Greater the value of OCR the more the soil is consolidated
- 39. HOW TO ACCESS CONSOLIDATION SETTLEMENT:
- 40. εf As; we have; = εL ∆H/H = ∆h/h ∆V/Vo = ∆e/ (1+eo) ∆H/H = ∆e/ (1+eo) ∆H = H × ∆e/ (1+eo) Cc = ∆e / log(σ’Zf / σ’Z0) ∆e = Cc H log(σ’Zf / σ’Z0) ∆H = Cc H log(σ’Zf / σ’Z0)…….(A) (This is the equation for NCC soil) 1+eo σ’Z0 = Initial Pressure or present Over burden pressure. σ’Zf = Final Pressure.
- 41. FOR OCC For over consolidated soil we have two cases; CASE # 01 When ; σ’Z0 < ∆H = Cr σ’zf < σ’C H log(σ’Zf / σ’Z0)…(B) 1 + eo Cr = Coefficient of Recompression
- 42. Cr = Coefficient of Recompression CASE # 02 When; σ’Z0 < σ’C < σ’zf ∆H = Cr H log(σ’c / σ’Z0 ) + Cc H log(σ’Zf / σ’Z0) 1+eo 1+eo
- 43. OVER CONSOLIDATION MARGIN The σ’C values from the laboratory represent the pre-consolidation stress at the sample depth. However we have to compute σ’C at other depths. To do so compute the overconsolidation margin σ’m, using σ’z0 at the sample depth and the following equation; σ’m = σ’C - σ’z0
- 44. EXAMPLE A 3m deep compacted fill is to be placed over the profile shown in the figure. A consolidated test on a sample from point A produced the following results; Cc = 0.40 Cr = 0.08 eo = 1.10 σ’ = 70.0 Kpa Compute the ultimate consolidation settlement due to weight of this fill. C
- 45. Dr = 40% Cc/1+e0 = 0.008
- 46. σ’Zf σ’Zf σ’Zf = σ’Z0 + γfillHfill = σ’Z0 + (19.2KN/m³) (3.0m) = σ’Z0 + 57.6Kpa Compute the initial vertical stress at sample location. σ’Z0 = Σ γH - U σ’Z0 = (18.5KN/m³) (1.5m) + (19.5KN/m³) (2.0m) + (16.0KN/m³) (4.0m) – (9.8KN/m³) (6.0m) σ’Z0 = 72.0 Kpa. At the sample σ’C ͌ σ’Z0 Clay is Normally consolidated Cc/ (1+e0) = 0.40/(1+1.10) = 0.190 For sand ; Cc/ (1+e0) = 0.008 (this value is taken from table corresponding to the value of Dr)
- 47. At Mid Of layer Layer H(m) σ’Z0 σ’Zf (Kpa) Cc/ (1+e0) δc (mm) (Kpa) 1 1.5 13.9 71.5 0.008 8 2 2.0 37.4 95.0 0.008 6 3 3.0 56.4 114.0 0.190 174 4 3.0 75.0 132.6 0.190 141 5 4.0 96.7 154.3 0.190 154 δc, ult 483mm Using the equation for NCC δc = 480mm Round off ∆H = Cc H log(σ’Zf / σ’Z0)…(B) 1 + eo
- 48. EXAMPLE #02 An 8.5m deep compacted fill is to be placed over the soil profile shown in the figure. Consolidation test on samples from point A and B produced the following results Sample A Sample B Cc 0.25 0.20 Cr 0.08 0.06 e0 0.66 0.45 σ’C 101kpa 510kpa
- 49. σ’Zf σ’Zf σ’Zf = σ’Z0 + γfillHfill = σ’Z0 + (20.3KN/m³) (8.5m) = σ’Z0 + 172.6 Kpa Compute the initial vertical stress at sample location. σ’Z0 = Σ γH - U σ’Z0 = (18.3KN/m³) (2.0m) + (19.0KN/m³) (2.0m) – (9.8KN/m³) (2.0m) σ’Z0 = 55.0 Kpa. σ’Zf = σ’Z0 + 172.6 Kpa σ’Zf = 55.0 Kpa + 172.6 Kpa = 227.6Kpa σ’Z0 < σ’C < σ’zf 55kpa < 101Kpa < 227.6 Kpa (Over consolidated Case#2) We will use this equation ∆H = Cr H log(σ’c / σ’Z0 ) + 1+eo Cc H log(σ’Zf / σ’Z0)} 1+eo
- 50. σ’m σ’m σ’c - σ’Z0 = 101 – 55 = 46Kpa = Therefore σ’c at any depth in the stiff silty clay stratum is equal to σ’z0 + 46Kpa. For sample B: σ’Z0 = Σ γH - U σ’Z0 = (18.3KN/m³) (2.0m) + (19.0KN/m³) (7.0m) + (19.5KN/m³) (10m) – (9.8KN/m³) (17m) σ’Z0 = 198.0 Kpa. σ’Zf = σ’Z0 + 172.6 Kpa σ’Zf = 198.0Kpa + 172.6 Kpa σ’Zf = 370.6 Kpa σ’Z0 < σ’zf < σ’ C (Over consolidated Case#1) 198kpa < 370Kpa < 510 Kpa ∆H = Cr H log(σ’Zf / σ’Z0) 1 + eo
- 51. At Mid Of layer H(m) σ’Z0 σ’c σ’Zf Cr/ Cc/ (Kpa) Layer (Kpa) (Kpa) (1+e0) δc (mm) (1+e0) 1 2.0 18.3 64.3 190.9 0.05 0.15 196 2 3.0 50.4 96.4 223.0 0.05 0.15 206 3 4.0 82.6 128.6 255.2 0.05 0.15 217 4 4.0 120.4 293.0 0.04 0.14 62 5 4.0 159.2 331.8 0.04 0.14 51 6 5.0 202.8 375.4 0.04 0.14 53 7 5.0 251.4 424.0 0.04 0.14 45 δc, ult 830mm
- 52. SETTLEMENT DUE TO SECONDARY CONSOLIDATION Secondary consolidation is also known as creep settlement and it is actually a continuation of the volume change that started during primary consolidation. This settlement takes place at constant effective stress (i.e after all the pore pressure has been dissipated). Co-efficient of secondary consolidation is defined as the vertical strain which occurs during one log cycle of time following completion of primary consolidation and is given by; Cα = ∆H/Ho = (D1 – D2)/Ho where; ∆H = the change in sample height during consolidation following the completion of primary consolidation. Ho = Consolidation sample height under a given pressure. D1 & D2 = dial gage reading along secondary compression curve against any time t1 and t2 where t2 = 10t1
- 53. Settlement due to secondary consolidation is now computed as follow; Sc = (Cα) (Ho) (log tsc/tp) Where; Ho = thickness of compressible stratum. tp & tsc = the time for completion of primary consolidation and time for which secondary consolidation is to be calculated. In sand, settlement caused by secondary compression is negligible, but in peat, it is very significant.
- 54. FOOTINGS ON SANDS AND GRAVEL In theory the method used to predict settlement of spread footings on clays and silts also could be used for sands and gravels but to use this method we need to compute Cc and Cr in these soils, which is very difficult or impossible because of the difficulties in obtaining undisturbed samples. Because of this limitation we will use a different approach in computing settlement in sands and gravels.
- 55. Settlement in sands and gravel are generally much smaller than those in soft or medium clays. The most common method used is that developed by Schmertmann.
- 56. SCHMERTMANN'S METHOD For Square Footing Σ Σ For Strip Footing q = P/A + γCDf - U q’o =γsoilDf
- 57. SCHMERTMANN'S METHOD C1 = Correction factor for footing depth. . C2 = Correction factor for Creep. t = Time in years for which settlement is required Modulus of Elasticity Es
- 58. Square Strip Modified triangular vertical strain influence factor distribution diagram The strain influence factor accounts for both the distribution of stresses below the footing and nonlinear soil behavior immediately below the footing. The peak value is shown in the graph.
- 59. For Square footing (L/B =1) Iz = 0.1 + (Zc/0.5B) ( Ipeak – 0.1) (For Z < 0.5B) Iz = (2/3) Ipeak (2 – Zc/B) (For Z> 0.5B) For a strip Footing (L/B > 10) Iz = 0.2 + (Zc/B) ( Ipeak – 0.2) (For Z < B) Iz = (1/3) Ipeak (4 – Zc/B) (For Z > B)
- 60. IMPORTANT CONSIDERATIONS Primarily used to estimate immediate settlement of foundations in sand. Specially useful when CPT data are available. Results are compatible with field measurements. Based on analysis of vertical strain distribution with a linear elastic half space subjected to a uniform pressure
- 61. Example A strip footing 2.0 × 23 m is subjected to load 450KN/m. The depth of footing is 2m. There is a deep deposit of sand of unit weight 16KN/m³, the water table is deep below the surface. The variation of cone penetration resistance (qc) with depth (z) is shown in the next figure.
- 62. SOLUTION As it is a strip footing; For strip footing; Σ Net foundation base pressure = qn = q – q’o Total foundation pressure = q = P/A + γCDf – U = (450/2) + 2×23.5 – 0 = 272KN/m² Effective overburden pressure at foundation level = q’o = γsoilDf = 16 × 2 = 32KN/m² Net foundation base pressure = qn = q – q’o = 272 – 32 = 240 KN/m²
- 63. Now we calculate; C1 = 1 – 0.5 (32/240) = 0.0933 Now we calculate C2; C2 = 1 + 0.2 log(5/0.1) = 1.339
- 64. Now we calculate Ipeak; = 0.5 + 0.1 undr Root(240/64) =0.7015 P’o = 2× 16 + 2× 16 = 64KN/m² As we know that for a strip Footing (L/B > 10). Iz = 0.2 + (Zc/B) ( Ipeak – 0.2) (For Z < B) Iz = (1/3) Ipeak (4 – Zc/B) (For Z > B)
- 65. Layer ∆z (m) qc Es=3.5qc Zc Iz (Iz/Es)× ∆z KN/m² 1 1 4000 14000 0.5 2 2 6000 21000 2 3 3 8000 28000 4.5 4 2 10000 35000 7 Σ Σ m.irfan B-15952

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