Unit weight of water is 62.4 pcf. How would you find the volumes of water vs soil/solid and how much air their is for soil A when not fully saturated? See images for problem information and equations known. 30%, and 60% finer on grain-size curveVvVve= void ratioVsV-VvV volume of voidV volume of solidV= total volumeVvn= porosity=%3DVVwS= Degree of Saturation=.x 100%VvVw=volume of waterWww=water content=WsWw weight of waterWs = weight of solidsWy= unit weight=-VWsYa = dry unit weight=VWsUnit weight of solids: YsVsWsYsGs= specific gravity=;YwVsYwYw= 62.4 pcf= 9.81 kN/m³= 1 ton/m3%3DYsYd=1+w1+ewGs = SeGSYW%3DYs%DtonUnit weight of water: Yw = 9.81 kN/.= 1m3m362.4 pcf Problem 1. Two undisturbed soil samples, eachhaving a volume of 0.1 ft3, are collected fromdifferent depths of the same soil layer. For sampleA, located above the groundwater table, W= 11 lband w= 9%. Sample B is located below thegroundwater table. If G3= 2.68, determine:a. Void ratio of Ab. Degree of saturation of Ac. Water content of Bd. Total weight of B

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Problem 1. Two undisturbed soil samples, each having a volume of 0.1 ft³, are collected from different depths of the same soil layer. For sample A, located above the groundwater table, W= 11 lb and w= 9%. Sample B is located below the groundwater table. If G3= 2.68, determine: a. Void ratio of A b. Degree of saturation of A c. Water content of B d. Total weight of B

Problem 1. Two undisturbed soil samples, each having a volume of 0.1 ft³, are collected from different depths of the same soil layer. For sample A, located above the groundwater table, W= 11 lb and w= 9%. Sample B is located below the groundwater table. If G3= 2.68, determine: a. Void ratio of A b. Degree of saturation of A c. Water content of B d. Total weight of B

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

Unit Conversion

Standard units are also called SI units. SI units are used for measuring a physical quantity satisfying certain conditions. There are two types of units: fundamental units and derived units. Fundamental units are physical quantities that are independent, that is, they do not depend on other quantities like mass, length, or time. There are plenty of quantities in our lives which need to be measured and compared. Measuring is the comparison of a particular physical quantity with its reference value. A unit is a measure of a fixed physical quantity. A physical quantity is the product of a unit and the numerical value we need to compare with.

Question

100%

Unit weight of water is 62.4 pcf. How would you find the volumes of water vs soil/solid and how much air their is for soil A when not fully saturated? See images for problem information and equations known.

30%, and 60% finer on grain-size curve
Vv
Vv
e= void ratio
Vs
V-Vv
V volume of void
V volume of solid
V= total volume
Vv
n= porosity=
%3D
V
Vw
S= Degree of Saturation=.
x 100%
Vv
Vw=volume of water
Ww
w=water content=
Ws
Ww weight of water
Ws = weight of solids
W
y= unit weight=-
V
Ws
Ya = dry unit weight=
V
Ws
Unit weight of solids: Ys
Vs
Ws
Ys
Gs= specific gravity=;
Yw
VsYw
Yw= 62.4 pcf= 9.81 kN/m³= 1 ton/m3
%3D
Ys
Yd=
1+w
1+e
wGs = Se
GSYW
%3D
Ys
%D
ton
Unit weight of water: Yw = 9.81 kN/.
= 1
m3
m3
62.4 pcf

Problem 1. Two undisturbed soil samples, each
having a volume of 0.1 ft3, are collected from
different depths of the same soil layer. For sample
A, located above the groundwater table, W= 11 lb
and w= 9%. Sample B is located below the
groundwater table. If G3= 2.68, determine:
a. Void ratio of A
b. Degree of saturation of A
c. Water content of B
d. Total weight of B

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