Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Subject is soils mechanics
An acrylic container having a volume of 0.30 ft is filled with dry sand under relatively
loose condition. The dry weight of the sand in the container is 31 lb. Water is then
carefully added to the container so as not to disturb the condition of the sand. When the
container is filled with water, the sand becomes fully saturated, and the combined weight
of solids plus water is 38.2 lb. From these data, determine all of the following:
(A) Void ratio of the sand in the container. (Show all pertinent volume-weight
relationships and calculations.)
(B) Specific gravity of the soil solids. (Show all pertinent volume-weight relationships
and calculations.)
(C) Saturated unit weight, and dry unit weight, of the sand, in lb/ft and kN/m³. (Show
all pertinent volume-weight relationships and calculations.)
Conceptual Illustrations: Exercise 5
Dry Sand
V = 0.30 ft³
W = 31 lb
Saturated Sand
V = 0.30 ft³
W=W₁ + W₁ = 38.2 lb
S
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Transcribed Image Text:An acrylic container having a volume of 0.30 ft is filled with dry sand under relatively loose condition. The dry weight of the sand in the container is 31 lb. Water is then carefully added to the container so as not to disturb the condition of the sand. When the container is filled with water, the sand becomes fully saturated, and the combined weight of solids plus water is 38.2 lb. From these data, determine all of the following: (A) Void ratio of the sand in the container. (Show all pertinent volume-weight relationships and calculations.) (B) Specific gravity of the soil solids. (Show all pertinent volume-weight relationships and calculations.) (C) Saturated unit weight, and dry unit weight, of the sand, in lb/ft and kN/m³. (Show all pertinent volume-weight relationships and calculations.) Conceptual Illustrations: Exercise 5 Dry Sand V = 0.30 ft³ W = 31 lb Saturated Sand V = 0.30 ft³ W=W₁ + W₁ = 38.2 lb S
Expert Solution
Check Mark
Step 1

Given,

Volume (V) = 0.30 ft3 

Dry weight = 31 lb 

Saturated weight = 38.2 lb

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