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 mmNormal stress: 150 kN/m2Shear force at failure: 276 Na. 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.²Determine the shear force required to cause failure.3. Refer to Figure 12.54. Shear strength parameters are needed for the design of afoundation placed at a depth of 2 m in the silty sand layer. Soils collected from thissand were compacted in the direct shear mold (diameter = 63.5 mm) at the same dryunit weight as the field and subjected to four direct shear tests.Results are as follows:Testno.1234Normalforce(N)2004008001600Shear forceat failure(N)105205414830 2 m2.2 m5.25 m3.5 mA2.6 mG.W.TSilty sande = 0.72; G = 2.69Silty clayw = 22%; G₂ = 2.72Poorly graded sandYsat = 19.8 kN/m³a. Determine the shear strength parameter ' for the soil.b. Determine the shear strength at the bottom of the silty sand layer.Part 2 - concept understanding1. 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 loadingcondition? What is basis for this criterion?3. Why does a sample need to be saturated to interpret the results from triaxialundrained tests?

This question is related to the Soil Strength from Geotechnical Engineering. In this question , ask…

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?