Box 2.12Calculating settlement. Worked exampleA large landfill is planned on 10 m of normally consoli- midpoint can be taken as representative of the wholedated clays with an underlying rock stratum. The watertable is at ground surface. Clay samples extracted fromthe stratum at an intermediate point at a depth of 5 mprovide the following soil properties: saturated unit weight,Kat = 20 kN/m3, void ratio, e, = 0.8, and compression index,ç = 0.15. Determine the settlement of the clay layer ifstratum:Final void ratio:a+ Aơep -e =c log50+ 80the increase in vertical stress due to the landfill load is= 0.8 -e =0.15 log-50= e- 0.74Aơ = 80 kPa.Vertical unit strain:Solution:eo - erHo1+eo0.8-0.74The hypothesis of uniform loading over an infinite lateralextent, allows one-dimensional conditions to be assumed.The figure shows the effective stress distribution corre-sponding to the initial and final situations (unit weight ofwater is , = 10 kN/m³).Taking that the increment in effective stress isconstant throughout the thickness of the day layer, theAH= 0.0331+0.8Total settlement (clay thickness Ho = 1,000 cm):AH0.033 == AH = 33 cm1000AG = 80 kPaAo = 80 kPa5 m5 m50 kPa130 kPaA CI5 m105 m180 kPa100 kPaStudy example 2. 8 and 2.11 before answering the given problem.Based on example 2.12, what would be the settlement of the clay layer if the following parameters werechanged:- The porosity is 0.5 and the saturated unit weight of the soil is 18 kN/m3 Воx 2.8Воx 2.11Piping conditions. Worked exampleCalculating the degree of overconsolidation. Worked exampleIn a normally consolidated clay deposit the water table isat the surface. An erosion process lowers the ground sur- for different depths before and after erosicn, as well as theface by 3 m. Assuming that the water table coincides at all degree of overconsolidation. The required OCR-depth ratiotimes with the ground surface, find the degree of overcon- after erosion is shown in the figure.solidation induced by the erosion process.(For clays take Kat = 21 kN/m?, and for pore waterX- 9.81 kN/m).The attached table shows effective vertical stressesThe stratigraphic column under the horizontal surface of awide valley is compased of 3 m of coarse gravels lying on12 m of clay deposits. Below the clays is a layer of highlypermeable fractured sandstone. The water table in thegravel layer lies 0.6 m below ground surface. In contrast,in the sandstone layer the water is under artesian condi-tions, with a piezometric height of 6 m above the surfaceof the ground. The apparent unit weights of the differentsoil strata are:the water would rise up to 6 m above the surfaceof the valley; i.e.:- 21m uc = 21-9.81 = 206.01 kPaYwOCR1.000+1.502.002.503.003.504.00Total vertical stress at Cis:c=21.6 -2Solution:Gravels (above the water table): yd = 16.8 kN/m?Saturated gravels (below the water table): =20.8 kN/mClay (saturated): = 21.6 kNmAssuming the ground the ground the ground theground has no strength but only weight, piping willAs the water table is always at the surface, the totalstresses, pore pressures, and effective vertical stresses canbe represented by the following expressions:10-occur when:15-- Kat20A large dry excavation has been projected in thevalley, for which the water level has to be drawdown atthe base of the excavation. Determine the depth at whichconditions for piping would be reached if:Artesian conditions are maintained in theso that equalling the two previous expressions willgive:o,- (Kat - K 225-where z is the depth measured from the surface at anygiven time.206.01=9.54 m d=15-9.54 = 5.46 m3021.6a)sandstones.b)Using the same operation as in the previousexample:Initial depth (m)Final depth (m)(kPa)(kPa)OCRb)Drainage wells are bored to lower the piezometricheight in the sandstones by 6 m (unit weight ofwater y, = 9.81 kN/m).444.7611.194.00255.9522.382.50-15m uc =15-9.81- 147.15 kPaTw67.1433.572.00Solution:7478.3344.761.75c=21.6 z89.5255.951.60a)The artesian conditions in the sandstone layer mean147.15=6.81m d=15-6.81=8.19m134.28that if a piezometer is installed at e.g. point C,21.612100.711.331613179.04145.471.232017223.8190.231.186 m2421268.56234.991.142825313.32279.751.120.6 m $24mImGravels3 m3229358.08324.511.1015 m12 m12 mMoment (2): OCR (2)-1 (NC)Moment (3): OCR (3)-ClaysClaysMoment (3): OCR (3) -1 (NC)Moment (2): OCR (2)->OCR(3) > 1; a-aPorous sandstonesAs can be seen, the overconsolidation ratio is equal to 1for normally consolidated (NC) states and greater than 1 forb)Moment (4): OCR (4) - -1 (Noverconsalidated states.(ug yıdag

Ysat = 18Kn/m3 YW = 10Kn/m3 H0 = 10 m z= 5 Coefficient of Consolidation CC = 0.15 Porosity =…

Calculating settlement. Worked example ge landfill is planned on 10 m of normally consoli- i clays with an underlying rock stratum. The water is at ground surface. Clay samples extracted from tratum at an intermediate point at a depth of 5 m midpoint can be taken as representative of the whole stratum: Final void ratio: de the following soil properties: saturated unit weight, 20 kN/m³, void ratio, e, =0.8, and compression index, 0.15. Determine the settlement of the clay layer if ncrease in vertical stress due to the landfill load is o + Ao, eo -e= c log- 50+ 80 = 0.8 -e =0.15- log- 50 = e- 0.74 80 kPa. Vertical unit strain: tion: AH 0-el= 0.8-0.74 nypothesis of uniform loading over an infinite lateral at, allows one-dimensional conditions to be assumed. figure shows the effective stress distribution corre- ding to the initial and final situations (unit weight of = 0.033 На 1+eo 1+0.8 Total settlement (clay thickness H, = 1,000 cm):

Calculating settlement. Worked example ge landfill is planned on 10 m of normally consoli- i clays with an underlying rock stratum. The water is at ground surface. Clay samples extracted from tratum at an intermediate point at a depth of 5 m midpoint can be taken as representative of the whole stratum: Final void ratio: de the following soil properties: saturated unit weight, 20 kN/m³, void ratio, e, =0.8, and compression index, 0.15. Determine the settlement of the clay layer if ncrease in vertical stress due to the landfill load is o + Ao, eo -e= c log- 50+ 80 = 0.8 -e =0.15- log- 50 = e- 0.74 80 kPa. Vertical unit strain: tion: AH 0-el= 0.8-0.74 nypothesis of uniform loading over an infinite lateral at, allows one-dimensional conditions to be assumed. figure shows the effective stress distribution corre- ding to the initial and final situations (unit weight of = 0.033 На 1+eo 1+0.8 Total settlement (clay thickness H, = 1,000 cm):

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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In the soil profile shown below, if the water table changes from level WT0 to level WT1 permanently, what is the change in the ground surface elevation in mm in the long term after primary consolidation is completed? Assume that the clay layer is initially normally consolidated. Use a single clay layer in your calculations and positive value for settlement. (Acceptable tolerance = 4%)
Refer to Figure 2, estimate the primary consolidation settlement in the clay layer. Given: Ag = 85 kN/m²; Hi = 2 m; H2 = 4 m; and H3= 6 m. Soil characteristics are as follows: Sand: ydx 14 kN/m³, Ysat 18 kN/m³ Clay: LL = 54; e = 0.98; Ysat 19 kN/m'; Cs =- Cc 150 kN/m? H3 Sand Clay ョ→
The volume of clay sample having a natural water content of 40% is 0.0009 ft' and its wet weight is 0.096 Ib. Calculate the degree of saturation of the sample, if G = 2.75.
Refer to Figure below. Estimate the primary consolidation settlement in the clay layer. Given: Δσ = 150 kN/m2; H1 = 3 m; H2 = 4 m; and H3= 5 m. Soil characteristics are as follows: • Sand: γdry = 15 kN/m3, γsat = 19 kN/m3 • Clay: LL = 65; e = 0.65; γsat = 18 kN/m3; ; Cs =1/6cc σʹc = 200 kN/m2