Engineering Rock Mechanics

Questions and answers: underground excavation instabilify mechanisms^35 1

h = 5 (123 tan 823 + 123 tan 835 + 152 tan 852)

=

= i(35.7 + 36.9 + 36.4) = 36.3 mm = 7.27 m

tan70 +49, tan37 +45 - tan39)

W = iyAih = i x 22 x 49.06 x 7.27 = 2615.6 kN, and

p = fyh = 4 x 22 x 7.27 = 53.3 kN/m2.

Angles taken from the hemispher-

ical projection:

0% = - =^1127 -^1601 =^33

845 = la4 - a51 = 1160 - 3351 = 175

053 = la5 - a31 = (335 - 1271 = 208

834 = 20

845 = 2

853 = 37

Block 345

Scale drawing of block (original scale 5 mm: 1 m):

‘34=57 , *,:...:y

* I...,. ’../ ,+ L3=14mm

~. 1

. , .... ‘ ...’
L45=83mm ~. :,:,:~.~..

Af = 4 (il3l41 sin 034 I + $4Z5 1 sin 045 I + $513 1 sin 053 I)

= i(i

= i(286.0 + 284.3 + 285.9) = 285.4 mm2 = 11.42 m2

14.75. I sin 331 + .75 .87 -^1 sin^1751 +^4 .87 14. I sin 2081)

h = $34 tan 834 + 134 tan 845 + 153 tan 853)

= &(5 tan20 + 83 - tan 2 + 9. tan371

= $( 1.82 + 2.90 + 6.78) = 3.83 mm E 0.77 m