The 14-72-km-long Fionnay-Riddes tunnel had to cross difficult rocks
10.10 Estimate of required rock support based on rock
10.10.4 Plastic deformation of the fissured rock mass
(a) No rock support (after Kastner, 1962 and Jaeger, 1975/76). In most cases the overstressed rock around the tunnel or cavity gets fissured when excava- tion proceeds. Rock deforms plastically. In plastically deforming rock masses the following general equations, in polar coordinates, can be used (see Duvall
&Obert, 1967, p. 75):
32F idF
~dw2 rUr
Gt = 82F/dr2 - d (l
dr \r dtp
(1)
where r and tp are polar coordinates and F an Airy function, still to be determined. o>, at and r are the stresses in radial and circumferential direction, r being a shear stress.
Assuming circular symmetry (av = ah = p), dFjdxp = 0 and:
idF
O*r = - —
r dr at = d2F/dr2 r = 0.
(2)
In plastically deforming rock, the basic equation of Coulomb (Hook) applies to the whole fissured rock mass and
T = c + a tan <f> (3)
is a condition for all the values (c, r).
The Coulomb straight line (3) is traced on fig. 10.51. All Mohr circles are tangent to it and
with
ot + or + 2a a = c cotan <f>.
(4)
T a r = c + a tan <j>
Fig. 10.51 Circles of Mohr.
For o> = 0, crf = crc (uniaxial compression strength of intact rock).
Therefore
ac + 2d9
2 sin Introducing equation (5) into (4) yields
1 + sin 6
Writing
equation (6) becomes
1 + sin (f>
1 — sin 6
- £<*r — crc = 0.
(5)
(6) (7)
(8) In order to get the Airy function F, we introduce equations (2) into (8) and obtain in succession
1 d F
r dr (9)
(10)
292 Underground excavations
At the rock surface of the circular tunnel (fig. 10.50) r = r0 and ar = 0 (no rock support). Therefore:
Equations (10) yield
and for r > r0
1 dF ^ J dF ^
(Tr = - -7- = 0 and — = 0.
r0 dr dr
c-J- ffc
and from (2) within the plastic zone (r < RL):
(11)
where the subscript/? means 'plastic zone'.
It is important to determine how deep the fissures penetrate into the rock mass. Beyond the limit of the plastic and elastic zones, for r > RL, the elastic rock stresses caused by the residual stresses av = ah=p are (see section 5.4):
For r^RL:
Ore. = i
ff<ei = ,
(12)
(the subscript e referring to the 'elastic zone').
Additionally at the limit r = RL a radial pressure, ar, is transmitted from
the plastic zone to the elastic zone. This radial stress ar causes inside the elastic zone the stresses
The total stresses inside the elastic zone are therefore:
Ore
°te =
R2L Rl
Rl\ n R\
T = Tex + Te2 = 0.
For r = RL9 we equate 'elastic' and 'plastic' stresses.
0RL = Orp = Ore and atp = ate
and finally
oRL can be eliminated by addition of the two equations (15):
and
- i) +
(13)
(14)
(15)
(16)
(16a) Equations (16) and (16a) obviously refer to the particular case where or = oh = p9 k = 1 and /?* = 0 (no rock support).
Some interesting examples on how to use equation (16a) are given in Kastner's and Rabcewicz's publications. From Kastner (1962), (see Jaeger, 1975/76) we take the following examples:
(/) The rock is supposed to be characterized by
ac = 20 kg/cm2, <f> = 30°, sin <f> = 0-5, y = 2-0 t/m3. The case/? = av = 120kg/cm2 considered by Kastner, corresponding to an overburden of 600 m, yields RL = 2-55r0.
294 Underground excavations
The case RL = r0, with ac = 2p9 would correspond to an overburden of only 10 kg/cm2 or 50 m rock for y = 2 t/m3.
(ii) This second case concerns a very high overburden with H = 1200 m, and p = 312 kg/cm2 for a rock specific weight y = 2-6 t/m3. The tunnel is assumed to have a radius r0 = 2-0 m and the rock is very good, characterized by:
<rc = 500 kg/cm2, c = 60 kg/cm2, y = 63°, The RL value would be i?L = 2-026 m and
RL — ro = 2-026 — 2-00 = 0-026 m only.
The circumferential stress at = 2p = 624 kg/cm2: there is a possibility for very thin overstressed rock layers around the tunnel to spall. The case mentioned by Kastner is similar to the Mont Blanc Road tunnel under 2000 m head, where k = 0t4 was assumed. Spalling occurred and was stopped with short rock-bolts.
(in) A case analysed by Rabcewicz concerns a rock characterized by:
ac = 120 kg/cm2, c = 30 kg/cm2, <f> = 36-9° and p = 312 kg/cm2 which yields for r0 = 2-00 m, RL = 3-04 m and RL — r0 = 1-04 m.
The length of the required supporting rock-bolts should be larger than RL — r0. The required bolt pressure required for safe rock support will be estimated in the following paragraphs.
(b) Rock support, pm9 required to avoid visco-plastic rock deformations.
(i) Before analysing the general problem of supported plastic rock masses, it is convenient to investigate which rock support /?£ax = p* is required to avoid any rock fissures and rock plasticity. Such conditions are as represented in fig. 10.52.
Let us first consider a pressure /?* acting on a plane surface. Assuming av = p to be the vertical pressure in the mass, and np the pressure parallel to the rock surface, fig. 10.52 shows that:
and
,„ 1 — sin ^ 1 — sin <f>
Pm = n / 7l + s i n ^ ~ ffcl + sin^
1-tin* _*-*_. ( l g )
Fig. 10.52 Circle of Mohr for ct = np and <rr = /?*.
If we relate p to the rock crushing strength, aC9
p = sac, s = p\ac
* ns ~" 1
(ô) When the rock surface is circular, with radius r0, the pressure ar = p*
on the rock surface causes inside the rock mass tangential rock stress at = —/?*. The total tangential rock stress is therefore at = np — /?* and
Pm =*
or
* _np-ac
Pm y i 1
ns — 1
(19)
(20)
In the case of a circular tunnel excavated in homogeneous rock mass where av = ah =p,n = 2 and/?* = (2s - l)crc/(£ + 1).
(c) Plastic deformation of a fissured rock mass: rock support required, general case (circular symmetry) (fig. 10.50). The groups of equations (1) and (2) are still valid. For
and
— = rQar = rop* =
ar 4 ~~ r0
296 Underground excavations This gives inside the plastic zone:
(21) Similarly:
dt* ~ " ' I - 1
ô At the limit of the overstressed zone, for r = i?L, we get
Adding these two equations and eliminating the unknown aRr stress:
or
an equation relating RLfr0 to/?*. This equation can be transformed and finally yields:
2 ipc + (g - l)p) (24)
U-l)'(oc + a-l)p*) }
which is identical to an equation published - without demonstration - in 1962 by Kashner in his great treatise Statik des Tunnel- und Stollenbaues (see equation (7), p. 175 of Kastner).
For entirely elastic conditions (case b (//)),
yielding
* _ 2/7 — GC __ 2s — 1
Pm - T ^ T - y^rr a<
as found before.
Equation (23) is supposed to supersede the well known equation of Fenner (1938)-Talobre (1957) which in some cases gives improbable results.
Remark. When the rock support can be adjusted at will (for example rock prestressing anchors) the use of equation (23) solves the problem. In most cases (for example concrete lining) the reaction p* from the support depends on relative radial deformations of the rock surface and of the sup- port, a problem which can be solved by iteration.