土力学课件

University of Dundee

Division of Civil Engineering

EG22005: Geomechanics

Tutorial Sheet 5 – Shear strength

1.Three shear box tests are conducted on identical samples of soil. Each test is conducted

with a different mass, N, providing the normal load and the shear force at the point of

failure, F is recorded. The cross-sectional area of the box is circular with diameter

100 mm and the tests are conducted with u = 0. The results of the tests are:

Test N, kg F kN

1 25 0.259

2 75 0.542

3 100 0.684

(a) Re-express the data in terms of stresses, σ'n and τ.

(b)Plot the failure states for these tests in σ'n and τ space.

(c) Draw the failure envelope and calculate the values of the effective stress parameters

c' and φ'.

First convert mass N to a force by multiplying by gravitational acceleration g (= 9.8 m/s2 today), then convert both the normal and shear forces into stresses by dividing them by the shear box plan area.

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By adding a trendline to this data, it should intercept the shear stress axis at about 15 kPa and have a slope of about 300.

2. A military tank has mass 60 tonnes and is given motion by two tracks with a total area of

10 m2.

Calculate the horizontal traction that the tank can provide if it is driving on:-

(a) clay of c u = 20 kPa

(b) cohesionless sand of ’ = 300

The sand in part (b) is found to have an additional dilation of 80 under the conditions expected. How much traction can the tank now develop?

[200 kN, 339 kN, 459 kN]

Assume that the tank has sufficient grip that it fails the soil when it slips, rather than at the interface.

If the soil is clay and fails when stresses exceed 20 kPa, then the tank can provide traction of c u.A = 20 kPa x 10 m2 = 200 kN on this soil before it fails.

For the sand, we need to calculate the effective stress level.

Tank has weioght force = 60 x 103 kg × 9.8 m/s2 = 588 kN

T his represents a normal stress σ = force/area = 58.8 kPa

Strength is proportional to this τ = σ’ tan (phi’) = 58.8 kPa × tan (300) = 33.9 kPa

This represents a force of τ.A = 339 kN.

If we have dilation of the soil too, this gives a peak strength to the soil of

τ = σ’ tan (phi’ + psi) = 58.8 kPa × tan (300 + 80) = 45.9 kPa

or a force of 459 kN.

(although of course as this is a peak force then if this is exceeded then the soil will be unable to sustain this level of traction, and the tank will slip badly until the applied force drops to the previous level of 339 kN – remember the shear box)