﻿ Craig's Soil Mechanics 7th Edition - Mecânica dos Solos 2 - 38
458 pág.

# Craig's Soil Mechanics 7th Edition

Disciplina:Mecânica dos Solos 21.021 materiais5.709 seguidores
Pré-visualização50 páginas
```at any point below an area of any shape
carrying a uniform pressure q. The chart (Figure 5.11) consists of influence areas, the
boundaries of which are two radial lines and two circular arcs. The loaded area is
drawn on tracing paper to a scale such that the length of the scale line on the chart
represents the depth z at which the vertical stress is required. The position of the
loaded area on the chart is such that the point at which the vertical stress is required is
at the centre of the chart. For the chart shown in Figure 5.11 the influence value is
0.005, i.e. each influence area represents a vertical stress of 0.005q. Hence, if the

Figure 5.11 Newmark’s influence chart for vertical stress. Influence value per unit pressure ¼
0:005. (Reproduced from N.M. Newmark (1942) Influence Charts for Computation of
Stresses in Elastic Foundations, University of Illinois, Bulletin No. 338, by permission
of Professor Newmark.)

152 Stresses and displacements

number of influence areas covered by the scale drawing of the loaded area is N, the
required vertical stress is given by

�z ¼ 0:005Nq

Example 5.1

A load of 1500 kN is carried on a foundation 2m square at a shallow depth in a soil
mass. Determine the vertical stress at a point 5m below the centre of the foundation
(a) assuming that the load is uniformly distributed over the foundation and (b)
assuming that the load acts as a point load at the centre of the foundation.

(a) Uniform pressure,

q ¼ 1500
22

¼ 375 kN=m2

The area must be considered as four quarters to enable Figure 5.10 to be used. In this case

mz ¼ nz ¼ 1m

Then, for z ¼ 5m

m ¼ n ¼ 0:2

From Figure 5.10,

Ir ¼ 0:018

Hence,

�z ¼ 4qIr ¼ 4� 375� 0:018 ¼ 27 kN=m2

(b) From Table 5.1, Ip ¼ 0:478 since r/z ¼ 0 vertically below a point load.

Hence,

�z ¼ Q
z2
Ip ¼ 1500

52
� 0:478 ¼ 29 kN=m2

The point load assumption should not be used if the depth to the point X (Figure
5.5(a)) is less than three times the larger dimension of the foundation.

Example 5.2

A rectangular foundation 6� 3m carries a uniform pressure of 300 kN/m2 near the
surface of a soil mass. Determine the vertical stress at a depth of 3m below a point (A)

Stresses from elastic theory 153

on the centre line 1.5m outside a long edge of the foundation (a) using influence
factors and (b) using Newmark’s influence chart.

(a) Using the principle of superposition the problem is dealt with in the manner shown
in Figure 5.12. For the two rectangles (1) carrying a positive pressure of 300 kN/m2,
m ¼ 1:00 and n ¼ 1:50, and therefore

Ir ¼ 0:193

For the two rectangles (2) carrying a negative pressure of 300 kN/m2, m ¼ 1:00 and
n ¼ 0:50, and therefore

Ir ¼ 0:120

Hence,

�z ¼ ð2� 300� 0:193Þ � ð2� 300� 0:120Þ
¼ 44 kN=m2

(b) Using Newmark’s influence chart (Figure 5.11) the scale line represents 3m, fixing
the scale to which the rectangular area must be drawn. The area is positioned such that
the point A is at the centre of the chart. The number of influence areas covered by the
rectangle is approximately 30 (i.e. N ¼ 30), hence

�z ¼ 0:005� 30� 300
¼ 45 kN=m2

Example 5.3

A strip footing 2m wide carries a uniform pressure of 250 kN/m2 on the surface of
a deposit of sand. The water table is at the surface. The saturated unit weight of the
sand is 20 kN/m3 and K0 ¼ 0:40. Determine the effective vertical and horizontal
stresses at a point 3m below the centre of the footing before and after the application
of the pressure.

Figure 5.12 Example 5.2.

154 Stresses and displacements

�0z ¼ 3�0 ¼ 3� 10:2 ¼ 30:6 kN=m2

�0x ¼ K0�0z ¼ 0:40� 30:6 ¼ 12:2 kN=m2

After loading: Referring to Figure 5.7(a), for a point 3m below the centre of the
footing,

¼ 2 tan�1 1
3

� �

sin
¼ 0:600

� ¼ �
2

; cosð
þ 2�Þ ¼ 1

The increases in total stress due to the applied pressure are:

��z ¼ q

ð
þ sin
Þ ¼ 250

ð0:643þ 0:600Þ ¼ 99:0 kN=m2

��x ¼ q

ð
� sin
Þ ¼ 250

ð0:643� 0:600Þ ¼ 3:4 kN=m2

Hence,

�0z ¼ 30:6þ 99:0 ¼ 129:6 kN=m2

�0x ¼ 12:2þ 3:4 ¼ 15:6 kN=m2

5.3 DISPLACEMENTS FROM ELASTIC THEORY

The vertical displacement (si) under an area carrying a uniform pressure q on the
surface of a semi-infinite, homogeneous, isotropic mass, with a linear stress–strain
relationship, can be expressed as

si ¼ qB
E
ð1� v2ÞIs ð5:27Þ

where Is is an influence factor depending on the shape of the loaded area. In the
case of a rectangular area, B is the lesser dimension (the greater dimension being L)
and in the case of a circular area, B is the diameter. The loaded area is assumed to
be flexible. Values of influence factors are given in Table 5.2 for displacements
under the centre and a corner (the edge in the case of a circle) of the area and for
the average displacement under the area as a whole. According to Equation 5.27,

Displacements from elastic theory 155

vertical displacement increases in direct proportion to both the pressure and the
width of the loaded area. The distribution of vertical displacement is of the form
shown in Figure 5.13(a), extending beyond the edges of the area. The contact
pressure between the loaded area and the supporting mass is uniform. It should
be noted that, unlike the expressions for vertical stress (�v) given in Section 5.2, the
expression for vertical displacement is dependent on the values of elastic modulus
(E ) and Poisson’s ratio (�) for the soil in question. Because of the uncertainties
involved in obtaining elastic parameters, values of vertical displacement calculated
from elastic theory, therefore, are less reliable than values of vertical stress.
In the case of an extensive, homogeneous deposit of saturated clay, it is a reasonable

approximation to assume that E is constant throughout the deposit and the distribu-
tion of Figure 5.13(a) applies. In the case of sands, however, the value of E varies with
confining pressure and, therefore, will increase with depth and vary across the width of
the loaded area, being greater under the centre of the area than at the edges. As a
result, the distribution of vertical displacement will be of the form shown in Figure
5.13(b); the contact pressure will again be uniform if the area is flexible. Due to the
variation of E, and to heterogeneity, elastic theory is little used in practice in the case
of sands.
If the loaded area is rigid the vertical displacement will be uniform across the width

of the area and its magnitude will be only slightly less than the average displacement
under a corresponding flexible area. For example, the value of Is for a rigid circular
area is
/4, this value being used in the calculation of E from the results of in-situ plate
loading tests. The contact pressure under a rigid area is not uniform; for a circular area
the forms of the distributions of contact pressure on clay and sand, respectively, are
shown in Figures 5.14(a) and (b).
In most cases in practice the soil deposit will be of limited thickness and will be

underlain by a hard stratum. Christian and Carrier [3] proposed the use of results by

Table 5.2 Influence factors for vertical displacement under
flexible area carrying uniform pressure

Shape of area Is

Centre Corner Average

Square 1.12 0.56 0.95
Rectangle, L/B ¼ 2 1.52 0.76 1.30
Rectangle, L/B ¼ 5 2.10 1.05 1.83
Circle 1.00 0.64 0.85

Figure 5.13 Distributions of vertical displacement: (a) clay and (b) sand.

156 Stresses and displacements

Giroud [7] and by Burland [2] in such cases. The average vertical displacement under a
flexible area carrying a uniform pressure q is given by

si ¼ �0�1 qB
E

ð5:28Þ

where �0 depends on the depth of embedment and �1 depends on the layer thickness
and the shape of the loaded area. Values of the coefficients �0 and �1 for Poisson’s
ratio equal to 0.5 are given in Figure 5.15. The principle```