Chapter 12, Problem 12.1P

### Principles of Geotechnical Enginee...

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

Chapter
Section

### Principles of Geotechnical Enginee...

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

# 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 angle of friction, ϕ.

The angle of friction, ϕ is 24.9°_.

Explanation

Given information:

The diameter (d) of specimen is 71 mm.

The height (h) of specimen is 25 mm.

Shear force (Vu) at the failure is 276 N.

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

Calculation:

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

τf=c+σtanϕ (1)

Here, c is the cohesion, τf is shear strength of the dry sand, and ϕ is angle of friction.

Find the shear strength (τf) of the dry sand as follows:

τf=SA=Sπd24

Here, S is the shear force and A is area of specimen.

Substitute 276 N for shear force and 71 mm for d.

τf=276π×(71mm×1m1,000mm)24=69.71×103N/m2×1kN103N=69.71kN/m2

Find the angle of friction (ϕ) as follows:

Substitute 69.71kN/m2 for τf, 150kN/m2 for σ, and 0 for c in Equation (1).

The value of c for sand and inorganic silt is 0.

69.71=0+150tan(ϕ)69.71=150tan(ϕ)ϕ=tan1(69.71150)ϕ=24.9°

Thus, the angle of friction, ϕ is 24.9°_.

(b)

To determine

Find the shear force (S) required to cause failure.

The shear force required (S) to cause failure is 367.6N_.

Explanation

Given information:

The normal stress (σ) is 200kN/m2.

Calculation:

Refer part (a).

Substitute 0 for c, 200kN/m2 for σ, and 24.9° for ϕ in Equation (1).

τf=0+200tan24.9°=92.84kN/m2

Find the shear force is required to cause failure as follows:

S=τf×As=τf×(πd24)

Substitute 92.84kN/m2 for τf and 71 mm for d.

S=92.84×(π×(71mm×1m1000mm)24)=92.84×3.9592×103=0.3676kN×1000N1kN=367.6kN

Thus, the shear force (S) required to cause failure is 367.6N_.

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