   Chapter 12, Problem 12.2P Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Solutions

Chapter
Section Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

Consider the specimen in Problem 12.1.a. If a direct shear test is conducted with a normal force of 675 N, what would be the principal stresses at failure?b. What would be the inclination of the major principal plane with the horizontal?12.1 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, ϕ ′ .b. For a normal stress of 200 kN/m2, what shear force is required to cause failure?

(a)

To determine

Find the principle stress at failure of the specimen.

Explanation

Given information:

The normal force (N) is 675 N.

The diameter (d) of the specimen is 71 mm.

The height of the specimen is 25 mm.

The effective angle of friction ϕ is 24.9°.

The normal stress (σ) on the failure plane is 150kN/m2.

Calculation:

Show the expression to find normal stress on failure plane (σ) using equation as follows:

σ=NA=Nπd24 (1)

Here, A is area of specimen.

Substitute 675N for normal force and 71mm for d in Equation (1).

σ=675Nπ×7124=0.1705N/mm2×(1kN1,000N)×1(1m1,000mm)2=170.5kN/m2

Show the expression of Mohr’s coulomb failure as follows:

τf=c+σtanϕ (2)

Here, c is the cohesion, τf is shear strength at failure, and ϕ is effective angle of friction.

Consider the shear test is conducted for dry silt sand. Hence, the cohesion (c) is 0.

Substitute 0 for c, 170.5kN/m2 for σ, and 24

(b)

To determine

Find the inclination of the major principle plane with the horizontal.

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