Page 604 - Rock Mechanics For Underground Mining

ANSWERS TO PROBLEMS

2 (a) 0.087 – 0.100 m; 0.083 – 0.104 m.
(b) 26 m.
◦
348 .
4 038/20.
◦
571 measured in the plane normal to the line of intersection.
6 275/50.
7 312/17.
8 RMR = 69.

Chapter 4

1 E t = 14.3GPa, = 0.23.
2 1 = 133.3MPa; 1 = 85 MPa.
3 (a) 1 = 133 + 3.0 3 MPa.
0.75
1 3
4 = 1.0 + 4.75 is one of several possible solutions.
100 100
◦
5 s = 15.0 + n tan 27.5 MPa.
7 (a) b = 30 , i = 15 .
◦
◦
(b) Peak strength envelope bilinear with c = 0, = 45 for 0 < n < 1MPa,
◦
and c = 0.5MPa, = 26.5 for 1 < n < 5 MPa. Residual strength
◦
envelope bilinear with c r = 0, r = 30 for 0 < n < 1MPa,
◦
and c r = 0.1MPa, r = 26 for 1 < n < 5MPa.
◦
8 u = 1.247 MPa; u = 0.642 MPa.

9 = 0.346 + 3.0 MPa.

1 3
Chapter 5
i
i
1 (b) i = 1.35 MPa, = 1.35 MPa, = 0.86 MPa.
xx zz zx
(c) xx = 3.51 MPa, zz = 4.59 MPa, zx = 0.86 MPa.
1 = 5.07 MPa, 2 = 3.03 Mpa, 1 = 61.1 ◦
2 p zz = 20.80 MPa, fissure water pressure u = 6.86 MPa.
13.69 < p xx < 43.29 MPa.

3 (a) Plane principal stresses are vertical and horizontal.
(b) 1 = 20.0 MPa (vertical), 2 = 15.0 MPa (horizontal),
3 = 12.0 MPa (parallel to tunnel axis).

4 All stresses are in MPa.
(b) p ll = 10.22, p mm = 17.09, p nn = 10.68, p lm =−1.30, p mn =−4.86,
p nl = 1.12.
(c) p xx = 15.59, p yy = 11.06, p zz = 11.35, p xy = 3.05, p yz = 3.12,
p zx = 3.99.
586