3. Given: Refer to fig. 1 ΔονE 100 kPa H1 = 2.5 m , H2 = 2.5 m , H3 = 3.0 m Sand: e, = 0.64 , Gs = 2.65 Clay: eo = 0.90 , Gs = 2.75 , LL = 55% Estimate the primary consolidation settlement of the clay layer that is normally consolidated.

3. Given: Refer to fig. 1 ΔονE 100 kPa H1 = 2.5 m , H2 = 2.5 m , H3 = 3.0 m Sand: e, = 0.64 , Gs = 2.65 Clay: eo = 0.90 , Gs = 2.75 , LL = 55% Estimate the primary consolidation settlement of the clay layer that is normally consolidated.

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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Concept explainers

Calculation of Settlement

The vertical displacement of the structural foundation due to the loading from the superstructure is known as settlement. It is due to the reduction of air voids present in the soil. The amount of vertical displacement depends on the bearing capacity of the soil on which the foundation is laid. The effect of the settlement depends on the magnitude of settlement, uniformity, and the structural design of the structure. The maximum allowable settlement for residential and office buildings should be less than 18 mm, for industrial buildings, it is 38 mm and for commercial buildings, it is 25 mm.

Question

3. Given:
Refer to fig. 1
Δον 100 kPa
H1 = 2.5 m
, H2 = 2.5 m , H3 = 3.0 m
, Gs = 2.65
, Gs = 2.75
Sand: e, = 0.64
Clay : eo = 0.90
, LL = 55%
Estimate the primary consolidation settlement of the clay layer that is normally
consolidated.

Weight of structure
W.T.
Sand layer
H2
Clay layer
Н
as
Compacted sand-
gravel layer
figure 1

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Step 1: calculating the unit weights of dry sand layer , saturated sand layer and clay layer

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Step 2: effective stress at the middle height of clay layer

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Question attached
11.11 An undisturbed clay has a preconsolidation pressure of 100 kN/m. A laboratory consolidation tests yield the following data: o' (kN/m?) e 0.92 0.77 150 300 Determine the void ratio that corresponds to a pressure of 450 kN/m?. DCengage Learing 2014
Take a look at the figure below. Estimate the primary consolidation settlement in the clay layer. The increase in stress is 85 kN/m2, H1 = 2m, H2 = 4m and H3 = 6m. Soil characteristics are as follows:• Sand: e = 0.65; Gs = 2.66• Clay: LL = 54; e = 0.98; Gs = 2.74; pre-consolidation stress = 150 kN/m2; Cs = Cc/6
Question attached
Q5: Figure below shows a soil profile. The uniformly distributed load on the ground surface is Ao. Estimate the primary consolidation settlement of the clay layer given these values: HI = 1.5 m, H2 = 2 m, H3 = 2.5 m Sand: e = 0.62, Gs 2.62 Clay: e = 0.98, Gs = 2.75, LL = 50 Ao = 110 kN/m2 Sand Groundwater table Sand Clay Void ratio =e Hy Normally consolidated Sand
Given the following data for the normally consolidated clay layer in the field. Compute the primary consolidation settlement. Thickness of clay layer= 3.0m Cc= 0.30 e= 0.70 Overburden pressure= 250 Kpa ΔP= 46 kPa Cα= 0.022
Refer to Figure Q3. Estimate the primary consolidation settlement in the clay layer. Given: Δ? = 85 ??/?2 ?1 = 2 ? ?2 = 4 ? ?3 = 6 ?. Soil characteristics are as follows (use appropriate relations):• Sand: ? = 0.65; ?? = 2.66• Clay: ?? = 54;? = 0.98; ?? = 2.74; ?'c = 150 ??/?2; ?? = 1/6 ?c
‹ample 9.1: Following results were recorded from a consolidation test on a specime turated clay, each load being maintained constant for 24 hours before the addition o ext increment.
Compressibililty: Consolidation and Settlement of Soil 500 kN 4m x 3m 3.5m Dry Sand = 15 kN/m Yary Sand Y = 17 kN/m³ 1.5m Clay = 19 kN/m Ysat e. = 1.0 LL = 50% 4.5m 2. Find the total consolidation settlement of the clay 4 years after the completion of primary consolidation settlement. The time for the completion of the primary consolidation settlement is 3 years. Ca = 0.020 (E; = 18 MPa, Poisson's Ratio = 0.20, I = 1.0)
In general, clays tend to have a more pronounced failure: -True -False
A normally consolidated clay sample, which is fully saturated, was consolidated in thetriaxial cell under a particular all-round pressure of σ3. Upon consolidation, drainage valve isclosed and σ3 is kept constant while deviatoric stress was gradually increased under undrainedcondition until failure is reached. The following results are obtained: ½ (σ1 – σ3) = 1/3 σ3 and ½ (σ1′ + σ3′) = 2/3 σ3a) Determine φ′,b) Express the pore pressure, uf, at failure in terms of cell pressure,c) Determine pore pressure coefficient Af′ at failured) Show that if sample is sheared under drained conditions, following results mayobtained: ½ (σ1 – σ3) = σ3 and ½ (σ1′ + σ3′) = 2σ3