Craig's Soil Mechanics 7th Edition
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Craig's Soil Mechanics 7th Edition

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realized that the same result would be obtained by the direct applica-
tion of the effective stress equation, the total vertical stress at A being the weight of
saturated soil and water, per unit area, above A. Thus

�A ¼ 11�sat þ 4�w ¼ 220þ 39 ¼ 259 kN=m2
uA ¼ 122 kN=m2
�0A ¼ �A � uA ¼ 259� 122 ¼ 137 kN=m2

The only difference in concept is that the boundary water force per unit area on top
of the column of saturated soil AD contributes to the total vertical stress at A.
Similarly at B

�B ¼ 6�sat þ 1�w ¼ 120þ 9:8 ¼ 130 kN=m2

hB ¼ 2:4
12

� 8 ¼ 1:6m
zB ¼ �7:0m
uB ¼ �wðhB � zBÞ ¼ 9:8ð1:6þ 7:0Þ ¼ 84 kN=m2
�0B ¼ �B � uB ¼ 130� 84 ¼ 46 kN=m2

2 Now consider the combination of effective weight and seepage force. The direction
of seepage alters over the depth of the column of soil AD as illustrated in Figure 3.8(b),
the direction of seepage for any section of the column being determined from the flow
net; the effective weight of the column must be combined with the vertical components
of seepage force. More conveniently, the effective stress at A can be calculated using
the algebraic sum of the buoyant unit weight of the soil and the average value of the
vertical component of seepage pressure between A and D.
Between any two equipotentials the hydraulic gradient is �h/�s (Equation 2.17).

Hence, if � is the angle between the direction of flow and the horizontal, the vertical
component of seepage pressure ( j sin �) is

�h

�s
�w sin � ¼ �h

�z
�w

where �z (¼�s/ sin �) is the vertical distance between the same equipotentials. The
calculation is as follows.

Number of equipotential drops between D and A ¼ 3:8.
Loss in total head between D and A ¼ 3:8

12
� 8 ¼ 2:5m.

Average value of vertical component of seepage pressure between D and A, acting
in the same direction as gravity

¼ 2:5
11

� 9:8 ¼ 2:3 kN=m3

Influence of seepage on effective stress 87

Buoyant unit weight of soil, �0 ¼ 20� 9:8 ¼ 10:2 kN/m3.
For column AD, of unit area, resultant body force

¼ 11ð10:2þ 2:3Þ ¼ 137 kN
i.e. effective vertical stress at A ¼ 137 kN/m2.

The calculation is now given for point B.

Loss in total head between B and C ¼ 2:4
12

� 8 ¼ 1:6m.
Average value of vertical component of seepage pressure between B and C, acting
in the opposite direction to gravity

¼ 1:6
6
� 9:8 ¼ 2:6 kN=m3

Hence, �0B ¼ 6(10:2� 2:6) ¼ 46 kN/m2.

Example 3.4

Using the flow net in Figure 3.8(a), determine the factor of safety against failure by
heaving adjacent to the downstream face of the piling. The saturated unit weight of the
soil is 20 kN/m3.
The stability of the soil mass EFGH in Figure 3.8(a), 6m by 3m in section, will be

analyzed.

By inspection of the flow net, the average value of total head on the base GH is
given by

hm ¼ 3:5
12

� 8 ¼ 2:3m

The average hydraulic gradient between GH and the soil surface EF is

im ¼ 2:3
6
¼ 0:39

Critical hydraulic gradient, ic ¼ �
0

�w
¼ 10:2

9:8
¼ 1:04

Factor of safety, F ¼ ic
im
¼ 1:04
0:39

¼ 2:7

PROBLEMS

3.1 A river is 2m deep. The river bed consists of a depth of sand of saturated unit
weight 20kN/m3. What is the effective vertical stress 5m below the top of the sand?

3.2 The North Sea is 200m deep. The sea bed consists of a depth of sand of satur-
ated unit weight 20 kN/m3. What is the effective vertical stress 5m below the top
of the sand?

88 Effective stress

3.3 A layer of clay 4m thick lies between two layers of sand each 4m thick, the top
of the upper layer of sand being ground level. The water table is 2m below
ground level but the lower layer of sand is under artesian pressure, the piezo-
metric surface being 4m above ground level. The saturated unit weight of the
clay is 20 kN/m3 and that of the sand 19 kN/m3; above the water table the unit
weight of the sand is 16.5 kN/m3. Calculate the effective vertical stresses at the
top and bottom of the clay layer.

3.4 In a deposit of fine sand the water table is 3.5m below the surface but sand to
a height of 1.0m above the water table is saturated by capillary water; above this
height the sand may be assumed to be dry. The saturated and dry unit weights,
respectively, are 20 and 16 kN/m3. Calculate the effective vertical stress in the
sand 8m below the surface.

3.5 A layer of sand extends from ground level to a depth of 9m and overlies a layer
of clay, of very low permeability, 6m thick. The water table is 6m below the
surface of the sand. The saturated unit weight of the sand is 19 kN/m3 and that of
the clay 20 kN/m3; the unit weight of the sand above the water table is 16 kN/m3.
Over a short period of time the water table rises by 3m and is expected to remain
permanently at this new level. Determine the effective vertical stress at depths of
8 and 12m below ground level (a) immediately after the rise of the water table
and (b) several years after the rise of the water table.

3.6 An element of soil with sides horizontal and vertical measures 1m in each
direction. Water is seeping through the element in a direction inclined upwards at
30� above the horizontal under a hydraulic gradient of 0.35. The saturated unit
weight of the soil is 21 kN/m3. Draw a force diagram to scale showing the
following: total and effective weights, resultant boundary water force, seepage
force. What is the magnitude and direction of the resultant body force?

3.7 For the seepage situations shown in Figure 3.9, determine the effective nor-
mal stress on plane XX in each case (a) by considering pore water pressure

Figure 3.9

Problems 89

and (b) by considering seepage pressure. The saturated unit weight of the soil
is 20 kN/m3.

3.8 The section through a long cofferdam is shown in Figure 2.24, the saturated unit
weight of the soil being 20 kN/m3. Determine the factor of safety against ‘boiling’
at the surface AB and the values of effective vertical stress at C and D.

3.9 The section through part of a cofferdam is shown in Figure 2.25, the saturated
unit weight of the soil being 19.5 kN/m3. Determine the factor of safety against a
heave failure in the excavation adjacent to the sheet piling. What depth of filter
(unit weight 21 kN/m3) would be required to ensure a factor of safety of 3.0?

REFERENCES

1 Skempton, A.W. (1961) Effective stress in soils, concrete and rocks, in Proceedings of

Conference on Pore Pressure and Suction in Soils, Butterworths, London, pp. 4–16.

2 Taylor, D.W. (1948) Fundamentals of Soil Mechanics, John Wiley & Sons, New York.

3 Terzaghi, K. (1943) Theoretical Soil Mechanics, John Wiley & Sons, New York.

90 Effective stress

Chapter 4

Shear strength

4.1 SHEAR FAILURE

This chapter is concerned with the resistance of a soil to failure in shear, a knowledge
of which is required in analysis of the stability of soil masses. If at a point on any plane
within a soil mass the shear stress becomes equal to the shear strength of the soil then
failure will occur at that point. Originally, prior to the postulation of the principle of
effective stress, the shear strength (�f ) of a soil at a point on a particular plane was
expressed by Coulomb as a linear function of the normal stress at failure (�f ) on the
plane at the same point:

�f ¼ cþ �f tan� ð4:1Þ

where c and � are the shear strength parameters referred to as the cohesion intercept
and the angle of shearing resistance, respectively. However, in accordance with the
principle that shear stress in a soil can be resisted only by the skeleton of solid
particles, shear strength should be expressed as a function of effective normal stress
at failure (�0f ), the shear strength parameters being denoted c

0 and �0:

�f ¼ c0 þ �0f tan�0 ð4:2Þ

Failure will thus occur at any point in the soil where a critical combination of shear
stress and effective normal stress develops. It should be appreciated that c0 and �0 are
simply mathematical constants defining