case 2: Tunnel bore stable, tunnel heading not stable
11.2 Shear and horizontal stresses in rock foundations of
Jimenes-Salas at the Eighth Congress on Large Dams, 1964, Edinburgh, reported on the danger of tensile stresses and tensile fissures in the rock
[325]
326 Rock mechanics and dam foundations
foundations at the heel of dams. Italian authors reported to the same Con- gress on the displacement of dam foundations under full hydrostatic load, which could also be attributed to overstraining of the rock under tensile stress.
11.2.1 Classical approach
Conventional methods are used for estimating the vertical stresses and the horizontal shear stresses on the horizontal concrete foundations of gravity and buttress dams. Similar methods can be used for the foundations at the concrete periphery of arch dams. It is commonly assumed that, when rock masses are capable of withstanding the estimated vertical compression stresses, cr^, investigation is also required to establish shear stress conditions assuming the law of Coulomb (see section 7.1 formulas (11) and (14)):
r = c + a tan <f> < ru l t.
When ax and a2 are the principal stresses in homogeneous rock masses, the law of Coulomb can be written as
= sin <j>.
0i + °2 + 2c cot <j>
When shearing along a weaker line of fracture has to be investigated the formula to be used is
ax cos /? sin (<f> — /?) + ay sin /? cos (</> — (i) + c cos <f> — u sin <f> ^ 0, where u is the interstitial water pressure, and where /? is the angle of the plane of fracture versus the principal stress a± (section 7.1, formula (16)).
This appears to be straightforward, provided a1 and a2 are known in direc- tion and value, but it is not, because the horizontal stress components ah
under the dam foundations, inside the rock masses, are not known, but could be measured.
11.2.2 The horizontal stress component ah and the shear stress T under the foundations of a gravity dam
The following simplified approach gives an idea of the real values of ah in the rock under the foundations of a gravity dam (fig. 11.1).
If we assume that the vertical stresses form a triangular load distribution from av = 0 for x = 0 to av = p for x = B the load at point £ (0 < I < B) is
Pc-Pj
(b)
Fig. 11.1 Simplified method for estimating horizontal and shear stresses in the rock foundation under a dam.
At point P(x, z) at a distance r from o* the stresses caused by the isolated load pz are:
= —
77T cos3 0, ox = -^ sin2 0 cos 0,
77T
rX2 = - ^ sin 0 cos2 0.
77T
Integration from £ = 0 to £ = B yields
2/?z3 r
Jo [ ( * -
= -^- Utan"1
D2 + ^2]2 ^
5z 5z(x - B) x2 + z2 -Bx z2 + (x-l
= ^ [z 2 ^"-^) 2 + Z 1Og ^ t 2 + '/ )2
W5 Jo [ ( x - D2 +
=p ± r ^
TT5 LZ2 + (JC - B
t a n- i
z2 + x22 - Bx\
328 Rock mechanics and dam foundations
Similarly, assuming again a triangular load distribution, an isolated hori- zontal load q^ (fig. 11.16):
2 = B
Fig. 11.2 Horizontal stresses, ah, caused by a vertical triangular load. The same diagram also valid for shear stresses, r, caused by horizontal loads (after Del Campo & Piquer, 1962).
causes at point P, stress:
= -25 sin 0 cos2 0, irr
ol = -^ sin3 0,
7TA*
ja = ^ sin2 6 cos 0,
77T77T
and integration from £ = 0 to £ = 5 yields:
Jo [(x^lY + z*?^
at = ^
Bz
*•
Bz2
, _z* + (x-B?
+ x tan" Bz^ I x2-Bx]'
Figures 11.2 and 11.3 show the horizontal stresses, ah9 and the shear stresses, r, caused by a vertical triangular loading, and fig. 11.4 the horizontal shear
z = 0AB
z = 0.5B-
z= B
Fig. 11.3 Shear stresses, T, caused by a triangular vertical load. The same diagram also valid for vertical stresses, av, caused by horizontal loads (after Del Campo &
Piquer, 1962).
z = 0.15 2 = 0.5B
z = B M II
llllllllll \
^ - U > |
Fig. 11.4 Horizontal stresses, ah, caused by a horizontal triangular load.
stresses, ah9 caused by a horizontal triangular loading. They are also valid for shear stresses, r, and the vertical stresses, aV9 caused by horizontal loads. The diagram shows that ah and r are not linear. Figure 11.4 shows clear negative tensile horizontal stresses, oh, in the region of the heel of the dam. When verti- cal and horizontal (hydrostatic) loads act on the dam the corresponding dia- grams have to be combined. Horizontal tensile stresses still exist under the dam.
From a rock mechanics point of view these tensile stresses are of major importance as it is sometimes doubtful whether rock masses can stand them.
330 Rock mechanics and dam foundations
Knowing av and ah the usual formulae allow the calculation of the principal stresses o1 and <x2, for che cases reservoir full and empty.
The basic assumption that vertical and horizontal loadings are triangular is only an approximation. Modern computing methods can be used for direct stress analysis of dam plus rock foundations. For example, Zienkiewicz (1965, 1968) has used the finite element method and assumes that the rock mass stands tensile stresses or that there is a 'no-tension' solution (fig. 11.5).
Fig. 11.5 Stress distribution at the heel of a gravity dam: (a) initial elastic solution, assuming tensile stresses; (b) final 'no-tension* solution, fissured rock (after Zienkiewicz, 1968).
Foundations of gravity dams, buttress dams, arch dams and also anchored dams, have been analysed by this method, and they always show tensile or fissured regions near the heel of the dam.