2 Direct shear tests were P size of the specimen was results were as given in Test no. Normal force 90 2. 135 3. 315 450 Find the shear stress par 3, Drive the general formula the horizonta frictional angel, and proof, 1-sin 14 sine 1-sin -20 e fror 1+ sin e

Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Braja M. Das, Khaled Sobhan
Chapter12: Shear Strength Of Soil
Section: Chapter Questions
Problem 12.7P
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5,The following result were obtained from a series of undrained triaxial test carried out on undisturbed samples of a
compacted soil: Each sample is originally 76mm Long and 38mm in diameter, experienced a vertical deformation 2 mm
Draw the strength envelop & Determine coulomb equation for the shear strength of the soil in terms of total stresses,
Cell pressure (kPa) Additional axial load at Failure (N)
100
350
250
400
400
500
Transcribed Image Text:5,The following result were obtained from a series of undrained triaxial test carried out on undisturbed samples of a compacted soil: Each sample is originally 76mm Long and 38mm in diameter, experienced a vertical deformation 2 mm Draw the strength envelop & Determine coulomb equation for the shear strength of the soil in terms of total stresses, Cell pressure (kPa) Additional axial load at Failure (N) 100 350 250 400 400 500
Direct shear tests were performed on a dry. sandy soil. The
size of the specimen was 50 mm x 50 mm x 20 mm. Test
results were as given in the table.
Test no.
Normal force (N)
Shear force at failure (N)
90
54
2.
135
82.35
315
189.5
4.
450
270.5
Find the shear stress parameters.
3, Drive the general formula the horizontal stress as the function of vertical stress, cohesion and internal
frictional
angel,
and
proof,
0 =45° +/2
o, -o, tan (45 ++ 2c tan(45 +
-) and
1-sin
- sin o
-2c
1+ sino
from fig below.
1+sino
Transcribed Image Text:Direct shear tests were performed on a dry. sandy soil. The size of the specimen was 50 mm x 50 mm x 20 mm. Test results were as given in the table. Test no. Normal force (N) Shear force at failure (N) 90 54 2. 135 82.35 315 189.5 4. 450 270.5 Find the shear stress parameters. 3, Drive the general formula the horizontal stress as the function of vertical stress, cohesion and internal frictional angel, and proof, 0 =45° +/2 o, -o, tan (45 ++ 2c tan(45 + -) and 1-sin - sin o -2c 1+ sino from fig below. 1+sino
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