Engineering Rock Mechanics

Question and answers: design of underground excavations 381

3.50 -.

Proof = Pi + (re - r)Yrock Pfloor = Pi - (re - r)l'rock.

We are told that f = 1.4, d = 3 (because ul = 3q = do3) and
I/ro& = 25 kN/m2. Sample calculations are shown in the table, with
the three ground characteristics shown in the diagram below.

' elastic

9.00

2.00 4.000

1.80 4.216

1.60 4.472

1.40 4.781

1.20 5.164

1.00 5.657

0.80 6.325

0.60 7.303

0.40 8.944

0.20 12.649

2- 2.50 ~~

e

2

G 150 -.

F

1 00 --

0.50 ~~

000 7

z 200 --

Q

0

0.0000

0.0067

0.0074

0.0084

0.0096

0.0114

0.0139

0.0178

0.0247

0.0394

0.0892

limit of elastic

/ behaviour

side wall

floor

---_ ----___. - - - - - - - 1.

.-

2.00 2.00

1.81 1.79

1.61 1.59

1.42 1.38

1.23 1.17

1.04 0.96

0.86 0.74

0.68 0.52

0.52 0.28

0.42 -0.02

The diagram shows how the floor stabilizes without a support pres-
sure after a radial displacement of about 85 mm. However, the roof
and sidewalls never stabilize and will always require supporting. It is
also interesting to see from the table of results how the thickness of the
zone of fractured rock increases with reducing support pressure. If an
engineering requirement is to minimize the size of the fractured zone,
then the support pressure needs to be close to 2 MPa.

420.5 For the design of part of a large underground civil defence fa-
cility in a rock mass, there are two competing excavation geometries,
as shown in the sketches of the vertical cross-sections given below.
Both geometries consist of excavated rooms separated by rock pil-
lars.