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Rock Mechanics.pdf - Mining and Blasting

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Figure 11.6 Calculated required<br />

support line for the sidewalls in a sample<br />

problem.<br />

ROCK SUPPORT AND REINFORCEMENT<br />

Table 11.1 Required support line calculations for sample problem.<br />

pi(MPa) 10 4 2 1.222 1.0 0.5 0.2 0.1<br />

re(m) — — — — 3.316 4.690 7.415 10.487<br />

i(m) 0 0.015 0.020 0.022 0.027 0.063 0.228 0.632<br />

(re − a)(MPa) 0 0 0 0 0.008 0.042 0.110 0.187<br />

proof = pi + (re − a)(MPa) 10 4 2 1.222 1.008 0.542 0.310 0.287<br />

pfloor = pi − (re − a)(MPa) 10 4 2 1.222 0.992 0.458 0.090 (−0.087)<br />

To determine the ground characteristic or required support curve, substitute successive<br />

values of pi in equation 7.15 to obtain a series of values of re which are then<br />

substituted into equation 11.7 to obtain the corresponding values of i =−ui. The<br />

results so obtained are tabulated in Table 11.1 <strong>and</strong> plotted in Figure 11.6. The critical<br />

support pressure below which a fractured zone will develop is found by putting re = a<br />

in equation 7.15 which gives pi cr = 1.222 MPa. In order to restrict radial displacements<br />

to values of i, calculated for sidewall support pressures of pi, roof <strong>and</strong> floor<br />

pressures of pi + (re − a) <strong>and</strong> pi − (re − a) will be required.<br />

The complete solution of a rock–support interaction problem requires determination<br />

of the support reaction or available support line in addition to the ground<br />

characteristic or required support line considered so far. Using methods introduced<br />

by Daemen (1975), Hoek <strong>and</strong> Brown (1980) have presented methods of calculating<br />

support reaction lines for concrete or shotcrete linings, blocked steel sets <strong>and</strong><br />

ungrouted rock bolts or cables. Details of these calculations are given in Appendix C.<br />

Figure 11.7 shows the results of a set of calculations carried out for a sample<br />

problem using the material model of Figure 11.5. A 5.33 m radius access tunnel is<br />

driven in a fair quality gneiss at a depth of 120 m where the in situ state of stress<br />

is hydrostatic with p = 3.3 MPa. The properties of the rock mass are c = 69 MPa,<br />

m = 0.5, s = 0.0001, E = 1.38 GPa, = 0.2, f = 4.2, mr = 0.1, sr = 0 <strong>and</strong> r =<br />

20 kN m −3 . In this problem, the self-weight of the fractured rock around the tunnel<br />

has an important influence on radial displacements, as shown in Figure 11.7.<br />

The support reaction or available support line for 8I23steel sets spaced at 1.5 m<br />

centres with good blocking was calculated using the following input data: W =<br />

0.1059 m, X = 0.2023 m, As = 0.0043 m 2 , Is = 2.67 × 10 −5 m 4 , Es = 207 GPa,<br />

ys = 245 MPa, S = 1.5 m, = 11.25 ◦ , tB = 0.25 m, EB = 10.0 GPa <strong>and</strong> i0 =<br />

0.075 m. The available support provided by these steel sets is shown by line 1 in<br />

Figure 11.7 which indicates that the maximum available support pressure of about<br />

0.16 MPa is quite adequate to stabilise the tunnel. However, because the set spacing<br />

of 1.5 m is quite large compared with the likely block size in the fractured rock, it<br />

will be necessary to provide a means of preventing unravelling of the rock between<br />

the sets.<br />

The importance of correct blocking of steel sets can be demonstrated by changing<br />

the block spacing <strong>and</strong> block stiffness. Line 2 in Figure 11.7 shows the available<br />

support line calculated with = 20 ◦ <strong>and</strong> EB = 500 MPa. The support capacity has<br />

now dropped below a critical level, <strong>and</strong> is not adequate to stabilise the tunnel roof.<br />

Since it has already been recognised that some other support in addition to steel<br />

sets will be required, the use of shotcrete suggests itself. Line 3 in Figure 11.7 is<br />

the available support curve for a 50 mm thick shotcrete layer calculated using the<br />

following data: Ec = 20.7 GPa,c = 0.25, tc = 0.050 m, cc = 34.5 MPa. Because<br />

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