Ignoring the given Es-data and considering that the supporting soil is clean, normally consolidated sand with thefollowing available data:A rigid shallow foundation 1x 2m is shown in the scheme. Cakulate the elastic setlement at the centre of thefoundation.q=150Pa1m10000Sand-032800012000wwwScheme 1.2ImSand2,m0-25 725-651165-10.014Scheme 13Assume the rodk layer exists at depth z- 10m.C ExpandTranscribed Image TextA rigid shallow foundation 1x 2 m is shown in the scheme. Calculate the elastic settlement at the centre of the foundation. q=150 kPa im E, (kPa) 1m. 110000 Sand Hs=0.3 2 8000 34 12000 Rock Scheme 3.21m g=150 kPa z, m Sand Hs =0.3 0-2.5 7 2.5-6.5 11 6.5- 10.0 14 Scheme 3.3 Assume the rock layer exists at depth z= 10 m. (u) z

The elastic settlement at the center of footing: Si=qB1-μ2IfEs As the given data is: q=150 kPa…

Answered: Ignoring the given Es-data and…

Ignoring the given Es-data and considering that the supporting soil is clean, normally consolidated sand with the following available data:

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter6: Shallow Foundations: Ultimate Bearing Capacity

Section: Chapter Questions

Problem 6.21P: Consider a continuous foundation of width B = 1.4 m on a sand deposit with c = 0, = 38, and = 17.5...

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Foundation construction

Question

Ignoring the given Es-data and considering that the supporting soil is clean, normally consolidated sand with the
following available data:
A rigid shallow foundation 1x 2m is shown in the scheme. Cakulate the elastic setlement at the centre of the
foundation.
q=150Pa
1m
10000
Sand
-03
2
8000
12000
www
Scheme 1.2
Im
Sand
2,m
0-25 7
25-65
11
65-10.0
14
Scheme 13
Assume the rodk layer exists at depth z- 10m.
C Expand
Transcribed Image Text
A rigid shallow foundation 1x 2 m is shown in the scheme. Calculate the elastic settlement at the centre of the foundation. q=
150 kPa im E, (kPa) 1m. 110000 Sand Hs=0.3 2 8000 34 12000 Rock Scheme 3.21m g=150 kPa z, m Sand Hs =0.3 0-2.5 7 2.5-
6.5 11 6.5- 10.0 14 Scheme 3.3 Assume the rock layer exists at depth z= 10 m. (u) z

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