Soil Settlement By Kamal Tawfiq Ph D P

Soil Settlement By Kamal Tawfiq, Ph. D. , P. E. , F. ASCE Fall 2013

Soil Settlement: Total Soil Settlement = Elastic Settlement + Consolidation Settlement Stotal = Se + Sc Elastic Settlement or Immediate Settlement depends on Elastic Settlement { { Load Type (Rigid; Flexible) Settlement Location (Center or Corner) Theory of Elasticity Time Depended Elastic Settlement (Schmertman & Hartman Method (1978) Elastic settlement occurs in sandy, silty, and clayey soils. By: Kamal Tawfiq, Ph. D. , P. E.

By: Kamal Tawfiq, Ph. D. , P. E. Consolidation Settlement (Time Dependent Settlement) * Consolidation settlement occurs in cohesive soils due to the expulsion of the water from the voids. * Because of the soil permeability the rate of settlement may varied from soil to another. * Also the variation in the rate of consolidation settlement depends on the boundary conditions. SConsolidation = Sprimary + Ssecondary Primary Consolidation Secondary Consolidation Volume change is due to reduction in pore water pressure Volume change is due to the rearrangement of the soil particles (No pore water pressure change, Δu = 0, occurs after the primary consolidation) Water Table (W. T. ) Expulsion of the water Water Voids Solids When the water in the voids starts to flow out of the soil matrix due to consolidation of the clay layer. Consequently, the excess pore water pressure (Du) will reduce, and the void ratio (e) of the soil matrix will reduce too.

Elastic Settlement Bqo Se = Es Bqo Se = Where α = Es 1 p 2 (1 - μs) α 2 2 (1 - μs) α (corner of the flexible foundation) (center of the flexible foundation) [ ln ( √ 1 + m 2 + m / √ 1 + m 2 - m ) + m. ln ( √ 1 + m 2 + 1 / √ 1 + m 2 - 1 ) m = B/L B = width of foundation L = length of foundation By: Kamal Tawfiq, Ph. D. , P. E.

Se = Bqo (1 - μs) α Es 3. 0 2. 5 α αav αr α, αav, αr 2. 0 1. 5 For circular foundation α=1 αav = 0. 85 αr = 0. 88 1. 0 3. 0 1 2 3 4 5 6 7 8 9 10 L/B Values of α, αav, and αr By: Kamal Tawfiq, Ph. D. , P. E.

Elastic Settlement of Foundation on Saturated Clay Janbu, Bjerrum, and Kjaernsli (1956) proposed an equation for evaluation of the average elastic settlement of flexible foundations on saturated clay soils (Poisson’s ratio, μs = 0. 5). Referring to Figure 1 for notations, this equation can be written as Se = A 1 A 2 qo. B/Es where A 1 is a function H/B and L/B, and is a function of Df/B. Christian and Carrier (1978) have modified the values of A 1 and A 2 to some extent, and these are presented in Figure 2. 2. 0 L/B = ∞ 1. 0 L/B = 10 1. 5 A 2 0. 9 5 A 1 1. 0 2 Square Circle 0. 5 0. 8 0 5 10 Df/B 15 20 0 0. 1 1 10 1000 H /B Values of A 1 and A 2 for elastic settlement calculation (after Christian and Carrier, 1978) By: Kamal Tawfiq, Ph. D. , P. E.

Dp Dp Dp DH = Cc H P + DP log ( o ) P 0 1 + e. O Dp

Dp 2 Dp Dp DH = CS H 1 + e. O log ( PC C CH P + DP 2 )+ log ( o ) Po PC 1 + e. O 2

Dp 2 Dp Dp 2 DH = CS H 1 + e. O log ( Po + DP 2 ) Po

Dr. Kamal Tawfiq - 2010 Example: 1. 2. 3. 4. Figure 1 gsand = 96 pcf Soil sample was obtained from the clay layer Conduct consolidation test [9 load increments ] Plot e vs. log (p) (Figure 2) Determine Compression Index (Cc ) & Swelling Index (Cs) 0. 9 Sand Void Ratio (e) 0. 7 Soil Sample eo = 0. 795 Cc = 0. 72 In the lab and before the consolidation test the stresses on the sample = 0. 6 During testing, the geostatic stress is gradually recovered 0. 4 0. 3 100 Determine Po 16 ft wc = 0. 3 In the ground, the sample was subjected to geostatic stresses. 0. 5 5. Clay gclay = 110 pcf 4 ft Dp 1 Dp 2 0. 8 In the lab the stresses are added to the soil sample 3 ft W. T. 1 In the lab and after removing the soil sample from the ground, the stresses on the soil sample = 0 Dp 1 Dp Stress Dp 7 Increments Dp Dp 9 G. S. Po = 3. (96) + 4. (96 -62. 4) + 8. (110 -62. 4) = 803. 2 lb/ft 2 Cs = 0. 1 Cc Cs Dp 9 Po 1000 Log (p) 10000 Figure 2

Dr. Kamal Tawfiq - 2010 Tangent to point 1 Example: 6. G. S. gsand = 96 pcf Using Casagrande’s Method to determine Pc 3 ft W. T. Sand gclay = 110 pcf Pc = 800 lb/ft 2 4 ft Clay Po 16 ft wc = 0. 3 Overconsolidation Ratio Pc Po 1 =1 The soil is Normally Consolidated N. C. soil 0. 9 Dp 1 0. 8 Void Ratio (e) OCR = Point of maximum curvature 6 X X 1 0. 7 2 5 0. 6 4 3 Tan gen t to poi nt 1 0. 5 0. 4 7 0. 3 100 Po = Pc 1000 Log (p) 10000

Dr. Kamal Tawfiq - 2010 Casagrande’s Method to Determine Preconsolidation Pressure (Pc) 1 Normally Consolidated Soil Point of maximum curvature 1 0. 9 ction nterse 6 I X X 1 4 div ide th 0. 7 3 ten e ang Tan gen he dt 0. 6 5 2 Horizontal line Ex str Void Ratio (e) 0. 8 of 4 & t to le bet ween 2&3 poi nt 1 h aig ne t li 0. 5 0. 4 7 0. 3 Po = Pc 1000 100 Log (p) Overconsolidation Ratio OCR = Pc Po =1 The soil is Normally Consolidated (N. C. ) soil 10000

Dr. Kamal Tawfiq - 2010 2 Casagrande’s Method to Determine Pc Overconsolidated Soil Point of maximum curvature ction nterse 6 I 5 2 Horizontal line X X 1 4 div ide th 3 5 Void Ratio (e) of 4 & e ang gen he dt ten Ex Tan ne t li igh str 7 Po Overconsolidation Ratio OCR = Pc Po The soil is oversonsolidated (O. C. ) soil >1 Pc Log (p) t to le bet poi nt 1 ween 2&3

Building qdesign Example: gsand = 96 pcf A 150’ x 100’ building will be constructed at the site. The vertical stress due to the addition of the building qdesign =1000 lb/ft 2 Dr. Kamal Tawfiq - 2010 G. S. W. T. Sand Clay The weight of the building Qdesign will be transferred to the mid height of the clay layer 3 ft 4 ft D P 1 Po Po 16 ft eo = wc. Gs = 0. 3 x 2. 65 = 0. 795 1 Qdesign = 15, 000 lb The added stress at 15’ from the ground surface is 0. 9 Dp 1 Dp = 15, 000 lb (150+15) x (100+15) DP = 790. 51 lb/ft 2 Void Ratio (e) 0. 8 0. 7 0. 6 0. 5 DP + Po = 790. 51 + 803 = 1593. 51 lb/ft 2 0. 4 0. 3 100 Po 1000 Log (p) 10000

Building Example: qdesign DP + Po = 790. 51 + 803 = 1593. 51 lb/ft 2 gsand = 96 pcf Consolidation Settlement DH = Cc H 1 + e. O log ( Dr. Kamal Tawfiq - 2010 G. S. W. T. Sand Clay Po + DP ) Po 0. 72 x 16 1593. 51 log ( ) 1 + 0. 795 803 DH = 1. 9 ft 3 ft 4 ft D P 1 Po Po 16 ft eo = wc. Gs = 0. 3 x 2. 65 = 0. 795 1 0. 9 Dp 1 Void Ratio (e) 0. 8 D P 1 0. 7 0. 6 0. 5 0. 4 0. 3 100 Po 1000 Po + DP Log (p) 10000

Demolished When the building was removed, the soil has become an overconsolidated clay. qdesign gsand = 96 pcf The rebound has taken place through swelling from pint 1 to point 2 Dr. Kamal Tawfiq - 2010 G. S. W. T. Sand Clay 3 ft 4 ft Po Po 16 ft eo = wc. Gs = 0. 3 x 2. 65 = 0. 795 1 0. 9 Dp 1 Void Ratio (e) 0. 8 D P 1 0. 7 2 0. 6 1 0. 5 0. 4 0. 3 100 Po 1000 Po + DP Log (p) 10000

Constructing a new building Scenario #1 The soil now is overconsolidated Soil: qdesign gsand = 96 pcf DH = G. S. W. T. Sand The new building is heavier in weight CS H 1 + e. O Dr. Kamal Tawfiq - 2010 Clay P C CH P + DP log ( C ) + log ( o ) Po PC 1 + e. O 3 ft 4 ft D P 2 Po Po 16 ft eo = wc. Gs = 0. 3 x 2. 65 = 0. 795 1 eo = 0. 61 0. 9 DH = 0. 1 x 16 1 + 0. 61 log ( 1593. 51 803 + 0. 72 x 16 log ( 2100 1 + 0. 61 1593. 51 = ) D P 2 0. 8 Void Ratio (e) Assume Po + Dp 2 = 2100 psf Dp 1 D P 1 0. 7 0. 6 CS 0. 5 CC ) 0. 4 0. 3 100 Po 1000 Pc Log (p) Po + DP New Building 10000

Dr. Kamal Tawfiq - 2010 Scenario # 2 The soil now is overconsolidated Soil: Constructing a new building qdesign gsand = 96 pcf G. S. W. T. Sand Clay The new building is lighter in weight 3 ft 4 ft D P 2 Po Po 16 ft eo = wc. Gs = 0. 3 x 2. 65 = 0. 795 CS H P + DP log ( o ) P 0 1 + e. O DH = 1 0. 9 eo = 0. 61 = log ( 1600 1593. 51 ) Void Ratio (e) 0. 1 x 16 1 + 0. 61 D P 2 0. 8 Assume Po + Dp 2 = 1600 psf DH = Dp 1 D P 1 0. 7 0. 6 CS 0. 5 0. 4 0. 3 100 Po 1000 Po + DP Log (p) 10000 New Building