Example 5 A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R. R a = 3.6 m/s 0.5 m Oil 2.0 m (S.G. 0.9) 35° width = 2 m 10 m

Example 5 A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R. R a = 3.6 m/s 0.5 m Oil 2.0 m (S.G. 0.9) 35° width = 2 m 10 m

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter8: Stresses In A Soil Mass

Section: Chapter Questions

Problem 8.14P

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Concept explainers

Principles of Seepage

In hydrology, when water flows down in a pore or crack of unsaturated soil due to gravity, the deep layer of soil is called seepage. It is the slow movement of water in the soil layer. Seepage is mainly found near hydraulic structures and is responsible for the stability of earthen-type hydraulic structures.

Question

Example 5
A closed tank shown in the given figure, is given an acceleration of 3.6 m/s, the acceleration vector
being inclined at an angle of 35° to the horizontal. Calculate the pressure at point M, N and R.
R
a = 3.6 m/s
0.5 m
2.0 m
Oil
(S.G. = 0.9)
35°
width = 2 m
10 m

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