SETTLEMENT Of SHALLOW FOUNDATION

2 types of settlement Elastic ( Immediate ) settlement Time independent Causes: Elastic deformation of dry soil particles Consolidation settlement Time dependant Causes: Water expulsion

ONE DIMENSIONAL CONSOLIDATION TEST CONSOLIDATION SETTLEMENT To calculate the consolidation settlement we have to perform ONE DIMENSIONAL CONSOLIDATION TEST Consolidometer Audiometer

P4<< P3 << P2 << P1 time P1 P2 P3 P4 deformation P4<< P3 << P2 << P1

I II III

Total settlement S = S1 + S2 + S3 I – Initial compression II – Primary Consolidation III – Secondary Settlement I – Initial compression S1 Causes: imperfect contact surface, loading, particles relocation II – Primary Consolidation S2 Causes: water expulsion III – Secondary Settlement S3 Causes: Particles relocation, Particles deformation, Particles destruction Total settlement S = S1 + S2 + S3

e eo e1 e2 While P2 is greater than P1 e2 is smaller than e1 Log P P1 P2

e Cc Cc Compression Index We can use any curve because the slope is the same Cc The slope from the curve Cc = 0.009 ( LL – 10 ) for undisturbed samples Cc = 0.007 ( LL – 10 ) for remolded samples Log P

Pc Soil History PRECONSOLIDATION PRESSURE We can find Soil History by finding what is called PRECONSOLIDATION PRESSURE Or Max Past Effective Overburden Pressure Pc

e Pc Log P

OCC – Over Consolidated Clay Based on values of Pc Clay soils may be NCC – Normally Consolidated Clay OCC – Over Consolidated Clay

P Cc Pc Po NCC – Normally Consolidated Clay. Where Po is greater than Pc Pc = Max Past Effective Overburden Pressure Po = Max Present Effective Overburden Pressure P The slope of the curve in this area is Cc Cc Pc Po

P Po Pc OCC – Over Consolidated Clay. Where Po is smaller than Pc Pc = Max Past Effective Overburden Pressure Po = Max Present Effective Overburden Pressure P The slope of the curve in this area is Cs Cs Po Pc

Cs NCC Cc OCC Pc

Total settlement S = Si + Sp + Ss INITIAL COMPRESSION ( Si ) PRIMARY CONSOLIDATION SETTLEMENT ( Sp ) SECONDARY SETTLEMENT ( Ss ) ep

Cα= /log( ) Ss = C’α H log( ) C’α = ep Cα t1 t2

Iρ Si = q B INITIAL COMPRESSION ( Si ) Where: q- net applied pressure B- width of footing Poisson’s ratio (tab. 6.6 p 167) E- modulus of elasticity (tab. 6.5 p 167) Iρ- influence factor (tab. 6.4 p 167) Poisson’s ratio (tab. 6.6 p 167) E- modulus of elasticity (tab. 6.5 p 167) Iρ- influence factor (tab. 6.4 p 167)

TIME RATE OF SETTLEMENT

Iρ Iρ < summary S = ) Immediate settlement Si = q B Si = q B NCC Initial compression Si = q B NCC Primary consolidation S = P0 + ΔP Pc OCC Consolidation settlement Pc P0 + ΔP < Ss = C’α H log( ) Secondary settlement

Settlement due to surcharge S = f ( P, soil, Z, X, load application type ) Load application types: point line strip circle square rectangle irregular

A- Elastic or immediate settlement The settlement of a foundation can be divided into two major categories: A- Elastic or immediate settlement B- Consolidation settlement

Immediate or Elastic Settlement Consolidation Settlement Secondary Settlement Time

For the calculation of foundation settlement, it is required to determine the vertical stress increase in the soil mass due to the load applied on the foundation. This chapter is divided into the following three parts: Procedure for calculation of vertical stress increase Elastic settlement calculation Consolidation settlement calculation

dr. isam jardaneh / foundation engineering 61303 / 2010

2 : 1 Method dr. isam jardaneh / foundation engineering 61303 / 2010

Average Vertical Stress Increase Due to a Rectangularly Loaded Area As suggested by Griffiths 1984, to find average pressure increase between z = H1 and z = H2 below the corner of a uniformly loaded rectangular area

Elastic Settlement Elastic settlement Based on The Theory of Elasticity

Elastic Settlement Based on The Theory of Elasticity

dr. isam jardaneh / foundation engineering 61303 / 2010

dr. isam jardaneh / foundation engineering 61303 / 2010

Elastic Settlement of Foundation on Saturated Clay

Settlement of Sandy Soil: Use of Strain Influence Factor

Continuous Footing

Need Interpolation ???

Influence Factor Method Example ¯ q = 160 KN/m² Find Elastic Settlement After 5 years Using Influence Factor Method 1.5m γ = 17.8 KN/mᶟ 3 x 3 m 3x3m E KN/m² 1.0m 8000 10000 3.0m 16000

0.1 1m 0.5m 0.5 2.5m 2m

∑ = 1.55x10¯⁴ Depth m ΔZ Es KN/m² Average Iz ----- ΔZ mᶟ/KN 0.0 – 1.0 8000 0.233 0.291x10¯⁴ 1.0 – 1.5 0.5 10000 0.433 0.217x10¯⁴ 1.5 – 4.0 2.5 0.361 0.903x10¯⁴ 4.0 – 6.0 2.0 16000 0.111 0.139x10¯⁴ ∑ = 1.55x10¯⁴ Iz Es

q C1 = 1-0.5 ( ---- ) = 1-0.5 [-------------] = 0.9 C2 = 1 + 0.2 log (5/0.1) = 1.34 Se = C1 C2 ( q – q ) ∑ (Iz/E)Δz Se = 24.9 mm 17.8x1.5 _ q - q 160 – (17.8x1.5) ‒

Range of Materials Parameters for Computing Settlement

Range of Materials Parameters for Computing Settlement

dr. isam jardaneh / foundation engineering 61303 / 2010

Example 1

Example 2

Problem # 1

Problem # 2

Problem # 3

Problem # 4 Estimate the consolidation settlement of the clay layer shown in Figure below using 2:1 method and trapezoidal rule. Note 1 ton = 2000 lb.