"please resolve the question but using method 2:1"note / the value load (q) is KN/m2method 2:1 q/(B+z) why the q * area"note in the picture solve the question but please solve in method 2:1" A circular foundation 2m in diameter is shown in the figure below. A normallyconsolidated clay layer 5m thick is located below the foundation. Determine the consolidationsettlement of the clay. method 2:1(1) As one layer of clay of 5m thick:- Divide the clay layer into (5) sub-layers each of 1m thick:(2) Calculation of increase of stress below the center of each sub-layer Ao (1):(3) Weighted average pressure increase (Simpson's rule):Circular foundationdiameter B - 2mG.S.1.0mSandq = 150 kN/m²y = 17 kNhm²0.5mW.T.Sand0.5mNormally consolidated clayY sat. = 18.5 kN/m²C =0.16, e, =0.855.0mSolution :(1) As one layer of clay of 5m thick:At the center of clay: 0,=1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²For circular loaded area, the increase of stress below the center is given by:1Ao = q{1-where: b=the radius of the circular foundation,[(b/z)² +1]3/21At mid-depth of the clay layer: z = 3.5m; Ao =150{1-=16.66 kN/m²(1/3.5)² +1]3/2Ae =C, log10o, + Ao- = 0.16.log1051.82 +16.66= 0.019451.82Де-H; =1+ 0.850.0194-(5)(1000) =52.4 mm1+e,(2) Divide the clay layer into (5) sub-layers each of 1m thick:• Calculation of effective stress at the middle of each sub-layer o'oi):For 1*. Layer: oom=1.5(17) +0.5(19-9.81) + 0.5(18.5-9.81) = 35.44 kN/m²For 2nd. Layer: oo(2)= 35.44 +1.0(18.5-9.81) = 35.44 + 8.69 = 43.13 kN/m²For 3rd. Layer: oo(3) = 43.13 + 8.69 = 51.81 kN/m²For 4. Layer: (4)= 51.81 + 8.69 = 60.51 kN/m²For 5". Layer: (5)= 60.51 + 8.69 = 69.20 kN/m? • Calculation of increase of stress below the center of each sub-layer Aoi):1For 1“. Layer. Aoq) =150'1--|= 63.59 kN/m²[(1/1.5)² + 13/21For 2nd. Layer: Ao 2) =1= 29.93 kN/m²[(1/2.5)² + 1]3/21For 3rd Layer: Ao) =150{1.=16.66 kN/m²[(1/3.5)² +1]3/21For 4. Layer: Ao(4) =150- =10.46 kN/m²[(1/4.5)² +1]3/21For 5th. Layer: Ao(5) =150= 7.14 kN/m²[(1/5.5)² +1]3/2Ae 6)AH;Ae (1)- AH;1+e,Layer no.kN/m2kN/m²m35.4443.1351.8263.5929.930.07270.03660.03930.01980.01050.00600.00371213.116.660.01944160.5110.460.0111169.207.140.00682Σ -00793oʻoi) + Ao) .Ae 6) = C, log10Cc =0.16, e,=0.85, Sc= 0.0793 m= 79.3 mm.(3) Weighted average pressure increase (Simpson's rule):At the center of clay: o% =1.5(17) + 0.5(19-9.81) + 2.5(18.5-9.81)= 51.82 kN/m²1At z = 1.0m from the base of foundation: Ao =150= 75 kN/m²[(1/1)² +1}3/2]1At z = 3.5m from the base of foundation: Ao =150{1-=16.66 kN/m?[(1/3.5)² + 1}3/2 J1At z = 6.0m from the base of foundation: Ao =150{1-= 6.04 kN/m²[(1/6)? +1]3/2 ](Ao; + 4Aom + Ao) ==[75 + 4(16.66) + 6.04] = 24.61 kN/m²o, +Ao -0.16.log10.. Aoavg.51.82 + 24.61Ae =C, log10= 0.02751.82Де-Hi1+e.0.027-(5)(1000) =72.9 mm1+ 0.85

Effective stress: Effective stress is defined as the difference between the total stress and the…

A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: - Divide the clay layer into (5) sub-layers each of 1m thick: (2) Calculation of increase of stress below the center of each sub-layer Ao- (): (3) Weighted average pressure increase (Simpson's rule): Circular foundation diameter B - 2m G.S. 1.0m Sand q = 150 kN/m² y = 17 kN/hm² 0. 5m W.T. Sand 19 kNim!. 0.5m Normally consolidated clay Y sat = 18.5 kN/m² Cc =0.16, e, =0.85 5.0m Solution :

A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: - Divide the clay layer into (5) sub-layers each of 1m thick: (2) Calculation of increase of stress below the center of each sub-layer Ao- (): (3) Weighted average pressure increase (Simpson's rule): Circular foundation diameter B - 2m G.S. 1.0m Sand q = 150 kN/m² y = 17 kN/hm² 0. 5m W.T. Sand 19 kNim!. 0.5m Normally consolidated clay Y sat = 18.5 kN/m² Cc =0.16, e, =0.85 5.0m Solution :

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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