For the square footing (1.5mx1.5m) shown in Figure (Q5), Find the total settlement (elastic and consolidation) at the center of the foundation. Use the approximate method to compute the stress increase. 1.5m 3.5m 5.0m Figure (05) 1000 kN 1.5m 1.5m Year 19 kN/m² E, -40 MPa H.- 0.25 Y Y-18 kN/m² 20 kN/m² Y 10 kN/m² E, -50 MPa, μ, = 0.3 C = 0.3, C, = 0.04 de -90 kN/m², e. - 0.8 Rock W.T

For the square footing (1.5mx1.5m) shown in Figure (Q5), Find the total settlement (elastic and consolidation) at the center of the foundation. Use the approximate method to compute the stress increase. 1.5m 3.5m 5.0m Figure (05) 1000 kN 1.5m 1.5m Year 19 kN/m² E, -40 MPa H.- 0.25 Y Y-18 kN/m² 20 kN/m² Y 10 kN/m² E, -50 MPa, μ, = 0.3 C = 0.3, C, = 0.04 de -90 kN/m², e. - 0.8 Rock W.T

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

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.

Question

For the square footing (1.5mx1.5m) shown in
Figure (Q5), Find the total settlement (elastic
and consolidation) at the center of the
foundation. Use the approximate method to
compute the stress increase.
EST
3.5m
was
Figure (Q5)
1000 kN
1.5m 1.5m
Y-18 kN/m²
Year 19 kN/m²
E, -40 MPa
P. - 0.25
You 20 kN/m²
Yw= 10 kN/m'
E, -50 MPa, μ, 0.3
C = 0.3, C, = 0.04
-90 kN/m², e. - 0.8
Rock

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