1. Consider Figure FTR 1 (Taken from Geotechnical Engineer- ing Design, Ming Xiao, strictly for educational purposes) shown. If the specific gravity of the soils are respectively from top to bottom 2.6, 2.7 and 2.73, estimate the effec- tive stresses at the mid thickness of each layer. 2. Still considering Figure FTR 1, estimate the additional stress- es induced by the footing on the mid thickness of each layer using both 2-1 approximation and graphical method as proposed by Boussinesq. 3. A soil layer consists of 4 meters of sand underlying 5 meters of clay. The water table is naturally located at the rock bed under the soil strata but rose by 6 meters due to a recent storm. Further, capillary action is expected to de- velop due to the nature of the soil in question. Compute the effective stresses at the end of expected capillary ac- tion, at the end of the water table, at the interface of the Q = 900 kN %3D Ground surface D, = 1 m Ground water table L× B = 3m × 2m €o= 0,5, c̟ = 0.35, c, = 0.06, oʻ, = 75 kN/m² Hj = 1m %3D H= 9 m eo = 0.48, c̟ = 0.33, c,=0.05, ơʻ, = 100 kN/m² Hz =2m €o = 0,45, c̟ = 0.30, c, = 0.05, oʻ, = 180 kN/m² H3 = 6m clay and sand and finally at the bottom of the soil strata. Pertinent soil properties are as follows: Figure FTR 1 Average S, Capil- lary Zone G, D10 Clay 2.71 0.98 0.0125 25 55% Sand 2.74 0.93 N.A. N.A. N.A.
1. Consider Figure FTR 1 (Taken from Geotechnical Engineer- ing Design, Ming Xiao, strictly for educational purposes) shown. If the specific gravity of the soils are respectively from top to bottom 2.6, 2.7 and 2.73, estimate the effec- tive stresses at the mid thickness of each layer. 2. Still considering Figure FTR 1, estimate the additional stress- es induced by the footing on the mid thickness of each layer using both 2-1 approximation and graphical method as proposed by Boussinesq. 3. A soil layer consists of 4 meters of sand underlying 5 meters of clay. The water table is naturally located at the rock bed under the soil strata but rose by 6 meters due to a recent storm. Further, capillary action is expected to de- velop due to the nature of the soil in question. Compute the effective stresses at the end of expected capillary ac- tion, at the end of the water table, at the interface of the Q = 900 kN %3D Ground surface D, = 1 m Ground water table L× B = 3m × 2m €o= 0,5, c̟ = 0.35, c, = 0.06, oʻ, = 75 kN/m² Hj = 1m %3D H= 9 m eo = 0.48, c̟ = 0.33, c,=0.05, ơʻ, = 100 kN/m² Hz =2m €o = 0,45, c̟ = 0.30, c, = 0.05, oʻ, = 180 kN/m² H3 = 6m clay and sand and finally at the bottom of the soil strata. Pertinent soil properties are as follows: Figure FTR 1 Average S, Capil- lary Zone G, D10 Clay 2.71 0.98 0.0125 25 55% Sand 2.74 0.93 N.A. N.A. N.A.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
answer QUESTION No. 2
Transcribed Image Text:1. Consider Figure FTR 1 (Taken from Geotechnical Engineer-
ing Design, Ming Xiao, strictly for educational purposes)
shown. If the specific gravity of the soils are respectively
from top to bottom 2.6, 2.7 and 2.73, estimate the effec-
tive stresses at the mid thickness of each layer.
2. Still considering Figure FTR 1, estimate the additional stress-
es induced by the footing on the mid thickness of each
layer using both 2-1 approximation and graphical method
as proposed by Boussinesq.
3. A soil layer consists of 4 meters of sand underlying 5 meters
of clay. The water table is naturally located at the rock
bed under the soil strata but rose by 6 meters due to a
recent storm. Further, capillary action is expected to de-
velop due to the nature of the soil in question. Compute
the effective stresses at the end of expected capillary ac-
Q = 900 kN
Ground surface
D; = 1 m
Ground water table
LxB = 3m x 2m
eo = 0.5, c. = 0.35, c, = 0.06, oʻ. = 75 kN/m²
H = 1m
H=9 m
H2 = 2m
eo = 0.48, c̟ = 0.33, c, = 0.05, ơʻ, = 100 kN/m²
eo = 0,45, c = 0.30, c, = 0.05, oʻ, = 180 kN/m2
H3 = 6 m
tion, at the end of the water table, at the interface of the
clay and sand and finally at the bottom of the soil strata.
Pertinent soil properties are as follows:
Figure FTR 1
Average S, Capil-
lary Zone
G,
e
D10
Clay
2.71
0.98
0.0125
25
55%
Sand
2.74
0,93
N.A.
N.A.
N.A.
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