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?

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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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
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?
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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