Mechanical engineering textbooks solve the classical stress problem of a steel plate perforated by a circular hole assuming the steel plate to be uni- formly stressed by a unidirectional field of tensile stresses a. The stresses around the circular hole are calculated using Airy functions.
It is a simple matter to reverse this problem and calculate the stresses about a circular gallery excavated in rock. When loaded by its own weight only, the compression stress field in the rock is a uniform potential vertical field.
Assuming a horizontal component of forces also, the two potential fields can be added.
This problem of rock mechanics was first solved by Schmidt (1926), and in 1938 the Chilean geologist, Fenner, published a similar solution. Terzaghi
& Richart (1952) have explored the problem in more detail in a classic paper, assuming different ratios for horizontal and vertical residual stress conditions and different cavity shapes; circular or elliptical galleries, spherical cavities, etc. Mining engineers tackled the problem of rectangular mining galleries.
An Airy function F of two variables (x, y) must be established to meet a general condition:
2d*F(x,y)
dx2dy2 8 /
and additionally to satisfy the boundary conditions of the problem. The stresses aX9 oy, and rxy in the (x, y) directions at the depth y are then func- tions of x and y (fig. 5.3):
32F 32F J d2F
d
In these equations the field of external forces is accepted as the vertical weight component of the rock in the y direction and it explains the term yy9 where y is the specific weight of the rock.
Fig. 5.3 Airy function of two variables.
In the case of a problem with axial symmetry like that of a circular hole in a steel plate or a circular gallery in rock, the variables to be considered are obviously r/x and the angle 0 (fig. 5.4), where x is now measured in the radial direction and the stresses to be observed at any point P inside the rock are ar in the radial direction at in a circumferential direction and the shear stresses rrt in the same direction as the gallery.
Fig. 5.4 Stresses about a circular gallery.
Strain and stress about cavities: theory 91 Assuming the field of external forces to be a potential field of vertical uniform loads p with no horizontal field of forces, the detailed calculation (Terzaghi & Richart, 1952) shows that
where the radius of the circular gallery is r and x is the distance from the centre O of the gallery to point P in the rock (JC > r). The angle 0 is zero in the vertical direction, upwards directed (fig. 5.4). The boundary conditions are obviously: for x = r, <rr = 0 and r = 0 all along the gallery. For x = oo, ar = p in the vertical direction (0 = 0). For x = oo, <rr = 0 in the horizontal direction (0 = 90°). Therefore, along the vertical axis for 0 = 0 and cos 20 = 1
5r2 3r*
which yields, as expected:
for JC = r or = 0, for JC = oo crr =/?.
Along the horizontal axis, for 0 = 90° and cos 20 = — 1:
and so
for x = r af = 0, for x = oo, <rr = 0, as required.
The tangential or circumferential stresses at are as follows:
p
and for
ort=|(l-3)=-/>.
For
which yields for x = r9at = 3/?. The ar and at vary with x (figs 5.5a and 5.5b).
The circumferential stress at along a horizontal diameter varies from at = 3/? in immediate vicinity of the tunnel bore to at = p at great distance; at a distance x = 2r, ert is still at = 1-22/?. The high circumferential stress for x = r may cause an overstraining of the tunnel wall and subsequent rock
failure.
The stresses along a vertical diameter are even more critical. Tensile stress is ot = —p in a circumferential direction near the tunnel bore. These stresses at remain negative until the point 1 — 3r2/x2 = 0 or x = ry/Z = l-73r is reached when the at become positive. Along a horizontal diameter, the radial stresses o> start from o> = 0 along the tunnel wall and increase to
1.22r
t! U11, timti
Fig. 5.5 (a) Diagram of radial stresses, ar; (b) diagram of circumferential stresses, at. a maximum ar = 0-375/? for x = ry/2 and then decrease. Along a vertical diameter, the ar start from ar = 0 to pass through slightly negative values.
There is a value o> = 0 for x = l-22r. These negative values are low: for example, for x = 1-1, ar = —0-04/? only. There is an obvious danger that these negative values may overstrain the tunnel soffit where tensile strength may be low after rock blasting.
The solution suggested by Heim's hypothesis, where horizontal residual stresses ah are equal to the vertical residual stresses av (ah = av = /?), is obtained by superimposing two solutions as before, but at right angles.
Because of the symmetry of the solutions (oh = av)9 ar and at do not depend any more on the angle <f> and for any value 0.
and
ar = p{\ — r2jx2) with aT = 0 for x = r,
at = p{\ + r2jx2) with at = 2/7 for x = r.
A cylinder of uniformly strained rock with maximum circumferential stresses at = 2/7 is formed round the excavated gallery.
If the unlined tunnel were filled with water at a pressure p, equal to the weight of the rock overburden, the pressures in the rock masses would revert
Strain and stress about cavities: theory 93 to a uniform residual stress value p and be balanced, provided that av = oh. If the unlined tunnel were bored in a rock mass, subjected only to a field of vertical forces av =/? with no horizontal field ah = 0, and then filled with water at pressure p, severe stresses near the tunnel soffit would reach at = —2p and fissures would most probably develop there. Tunnel designers must know the residual rock stresses in the rock mass through which a tunnel has to be excavated, as the final stressed state depends on the combina- tion of hydrostatic pressure and residual stresses.
In many cases the horizontal component of the residual stresses is ah = kav
with 0 < k < 1 (fig. 5.6). Tables have been calculated and stress diagrams published for characteristic values of A: (Terzaghi & Richart, 1952).
i
<Jt = —p
<*h = 0
(a)
Fig. 5.6 (a) av = p, ah = 0, k = 0; for 6 = 0, ot = -p and for 0 = 90°, <rt = 3/?.
(6) av =p, ah = kov, k < 1; for 0 = 0, cr, > - / ? ; and for 0 = 90°, at < 3p.
(c) av = pfoh =z kav, k = 1; for any value of 0, at = 2p.
The theories developed in this chapter may be used for estimating borehole diameter deformations under radial pressure.
According to the method of Merril & Peterson, the radial deformation of a circular hole in ideal rock is used to calculate the plane stress and strain around the hole. For uniaxial load, plane stress:
dor
u = —?(l + 2cos20);
Jb
for uniaxial load, plane strain:
u = — (1 - v2){\ + 2 cos 20);
for biaxial load, plane stress:
w = •=; [(** + <*v) + 2(ax - ay) cos 20];
E
and for biaxial load, plane strain:
2(ax - oy) cos 20];
where u is radial displacement, d is diameter of hole, ax, oy are mutually perpendicular applied stresses, 0 is angle clockwise^from ax and v is Poisson's ratio. Furthermore when ul9 u2, u3 = borehole deformations at 60° angles,
Ox + E
3d1 u2 + w3).
5.4.2 Extensions of the theory
(a) When rock is being overstressed and overstrained around the cavity the theory of elastic deformation no longer applies. Plastic deformations caused by overstraining and loosening of the rock penetrates into the rock mass to a depth RL which can be estimated with the method of Kastner (1962, see section 10.10).
(b) Overstrained rock should be supported. A simplified theory of rock support is developed in sections 10.3.3 and 10.3.4.
(c) Problems concerning some special boundary conditions are often complicated. These are summarized in section 10.3.5.