Explain the solution of Mass of water to be added to reach the full saturation on how they come up with that solution. Thank you Degree of saturation:w Gse15(2.7)S=0.846S = 47.9%S=-Mass of water to be added to reachfull saturation:(Gs+e) Yw1+e(2.7+0.846) (9.81)1+0.846Ysat =YsatYsat 18.84 kN/m³Mass of water to be added =(18.84-16.5)(1000)9.81Mass of water to be added = 238.5 kg/m³ The moist unit weight of a soil is16.5 kN/m3. If the water content is 15% andsp.gr. of soil is 2.7, determine thefollowing:(2Dry unit weightPorosityDegree of saturationMass of water in kg/m³ to beadded to reach full saturationSolution:Dry unit weight:YdryY1 + @16.5Ydry=1+0.15Ydry = 14.35 kN/m³2 Porosity:GsYwYdry 1 + e2.7(9.81)1 + e14.35 =e=0.846n=e1+ en = 0.4580.8461+0.846

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Answered: Explain the solution of Mass of water…

Explain the solution of Mass of water to be added to reach the full saturation on how they come up with that solution. Thank you

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

Fluid Materials

In general terms, anything that flows can be termed as a fluid. In other words, a fluid is any substance that cannot resist shear forces and deforms continuously under the application of external shear forces. Fluid materials differ completely from solid materials such as metals and alloys. For instance, if we consider two infinitesimally small cross-sectional areas of fluid materials and solid materials, and apply shear forces to these elements, both the elements will undergo deformation. Upon releasing the shear force, the element of the solid materials will regain its original orientation, but the element of the fluid materials will keep on deforming continuously under the action of the shear force.

Question

Explain the solution of Mass of water to be added to reach the full saturation on how they come up with that solution. Thank you

Degree of saturation:
w Gs
e
15(2.7)
S=
0.846
S = 47.9%
S=-
Mass of water to be added to reach
full saturation:
(Gs+e) Yw
1+e
(2.7+0.846) (9.81)
1+0.846
Ysat =
Ysat
Ysat 18.84 kN/m³
Mass of water to be added =
(18.84-16.5)(1000)
9.81
Mass of water to be added = 238.5 kg/m³

The moist unit weight of a soil is
16.5 kN/m3. If the water content is 15% and
sp.gr. of soil is 2.7, determine the
following:
(2
Dry unit weight
Porosity
Degree of saturation
Mass of water in kg/m³ to be
added to reach full saturation
Solution:
Dry unit weight:
Ydry
Y
1 + @
16.5
Ydry=1+0.15
Ydry = 14.35 kN/m³
2 Porosity:
GsYw
Ydry 1 + e
2.7(9.81)
1 + e
14.35 =
e=0.846
n=
e
1+ e
n = 0.458
0.846
1+0.846

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