1. Which failure criterion is used to interpret the results from a direct shear test? Why? 2. Which failure criterion is used to interpret triaxial test results for long-term loading condition? What is basis for this criterion? 3. Why does a sample need to be saturated to interpret the results from triaxial undrained tests?
1. Which failure criterion is used to interpret the results from a direct shear test? Why? 2. Which failure criterion is used to interpret triaxial test results for long-term loading condition? What is basis for this criterion? 3. Why does a sample need to be saturated to interpret the results from triaxial undrained tests?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Answer Part 2 Concept Understanding
![1. Following data are given for a direct shear test conducted on dry silty sand:
Specimen dimensions: diameter = 71 mm; height = 25 mm
Normal stress: 150 kN/m2
Shear force at failure: 276 N
a. Determine the angle of friction, Ø.
b. For a normal stress of 200 kN/m2, what shear force is required to cause failure?
2. For a dry sand specimen in a direct shear test box, the following are given:
Size of specimen: 2.8 in. x 2.8 in. x 1.25 in. (height)
Angle of friction: 41°
Normal stress: 22 lb/in.²
Determine the shear force required to cause failure.
3. Refer to Figure 12.54. Shear strength parameters are needed for the design of a
foundation placed at a depth of 2 m in the silty sand layer. Soils collected from this
sand were compacted in the direct shear mold (diameter = 63.5 mm) at the same dry
unit weight as the field and subjected to four direct shear tests.
Results are as follows:
Test
no.
1
2
3
4
Normal
force
(N)
200
400
800
1600
Shear force
at failure
(N)
105
205
414
830](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9015a8bf-9890-40c9-90d1-4e5e245b9dae%2F465fcb84-dbf2-4d23-a485-033a3c82c3f0%2Fe1lged_processed.png&w=3840&q=75)
Transcribed Image Text:1. Following data are given for a direct shear test conducted on dry silty sand:
Specimen dimensions: diameter = 71 mm; height = 25 mm
Normal stress: 150 kN/m2
Shear force at failure: 276 N
a. Determine the angle of friction, Ø.
b. For a normal stress of 200 kN/m2, what shear force is required to cause failure?
2. For a dry sand specimen in a direct shear test box, the following are given:
Size of specimen: 2.8 in. x 2.8 in. x 1.25 in. (height)
Angle of friction: 41°
Normal stress: 22 lb/in.²
Determine the shear force required to cause failure.
3. Refer to Figure 12.54. Shear strength parameters are needed for the design of a
foundation placed at a depth of 2 m in the silty sand layer. Soils collected from this
sand were compacted in the direct shear mold (diameter = 63.5 mm) at the same dry
unit weight as the field and subjected to four direct shear tests.
Results are as follows:
Test
no.
1
2
3
4
Normal
force
(N)
200
400
800
1600
Shear force
at failure
(N)
105
205
414
830
![2 m
2.2 m
5.25 m
3.5 m
A
2.6 m
G.W.T
Silty sand
e = 0.72; G = 2.69
Silty clay
w = 22%; G₂ = 2.72
Poorly graded sand
Ysat = 19.8 kN/m³
a. Determine the shear strength parameter ' for the soil.
b. Determine the shear strength at the bottom of the silty sand layer.
Part 2 - concept understanding
1. Which failure criterion is used to interpret the results from a direct shear test? Why?
2. Which failure criterion is used to interpret triaxial test results for long-term loading
condition? What is basis for this criterion?
3. Why does a sample need to be saturated to interpret the results from triaxial
undrained tests?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9015a8bf-9890-40c9-90d1-4e5e245b9dae%2F465fcb84-dbf2-4d23-a485-033a3c82c3f0%2Fhpasbz_processed.png&w=3840&q=75)
Transcribed Image Text:2 m
2.2 m
5.25 m
3.5 m
A
2.6 m
G.W.T
Silty sand
e = 0.72; G = 2.69
Silty clay
w = 22%; G₂ = 2.72
Poorly graded sand
Ysat = 19.8 kN/m³
a. Determine the shear strength parameter ' for the soil.
b. Determine the shear strength at the bottom of the silty sand layer.
Part 2 - concept understanding
1. Which failure criterion is used to interpret the results from a direct shear test? Why?
2. Which failure criterion is used to interpret triaxial test results for long-term loading
condition? What is basis for this criterion?
3. Why does a sample need to be saturated to interpret the results from triaxial
undrained tests?
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