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Rock Mechanics.pdf - Mining and Blasting

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Answers to problems<br />

Chapter 2<br />

1 (a) 200 kN, with clockwise moment.<br />

(b) xx =−15 MPa, yy = 16 MPa, xy = 10 MPa.<br />

(c) ll = 1.41 MPa, mm =−0.41 MPa, lm = 18.42 MPa.<br />

(d) 1 = 8.95 MPa, 2 =−17.95 MPa, = 73.6 ◦ .<br />

(e) = 0 ◦ ; tx = 15 MN m −2 , ty =−10 MN m −2 .<br />

= 60 ◦ ; tx =−1.16 MN m −2 , ty =−18.86 MN m −2 , R = 18.90 MPa<br />

= 90 ◦ ; tx =−10 MN m −2 , ty =−16 MN m −2<br />

2 (a) xx = 7.825, yy = 6.308, zz = 7.866;<br />

xy = 1.422, yz = 0.012, zx =−1.857.<br />

(b) mm = 9.824; nl = 0.079.<br />

(c) I1 = 22.0, I2 = 155.0, I3 = 350.0;.<br />

1 = 10.0 (1 = 43.4 ◦ , 1 = 73.9 ◦ , 1 = 129.05 ◦ );<br />

2 = 7.0 (2 = 108.5 ◦ , 2 = 131.8 ◦ , 2 = 132.4 ◦ );<br />

3 = 5.0 (3 = 51.8 ◦ , 3 = 132.4 ◦ , 3 = 66.2 ◦ ).<br />

4 (a) εxx = 1<br />

2G<br />

<br />

− y<br />

<br />

.<br />

r 2 + 2x 2y r 4<br />

εyy = 1<br />

<br />

(1 − 4)<br />

2G<br />

y 2y3<br />

+<br />

r 2 r 4<br />

<br />

,<br />

xy = 1<br />

<br />

2(1 − 2)<br />

2G<br />

x 4xy2<br />

+<br />

r 2 r 4<br />

<br />

.<br />

(b) xx =−(1 − 2) y<br />

r 2 + 2x 2y r 4<br />

.<br />

yy = (1 − 2) y 2y3<br />

+ ,<br />

r 2 r 4<br />

xy = (1 − 2) x 2xy2<br />

+<br />

r 2 r 4<br />

.<br />

5 (a) 1 = 25.0, 2 = 7.1, 1 = 58 ◦ , 2 = 148 ◦ .<br />

n = 21.0, = 7.2.<br />

(b) ll = 20.9, lm = 7.46.<br />

(c) 1 = 24.94, 2 = 7.10, 1 = 58.3 ◦ , 2 = 148.3 ◦ .<br />

Chapter 3<br />

1 (a) ¯x0A = 0.369 m; ¯x0B = 0.274 m.<br />

(b) ¯x = 0.288 m.<br />

(c) RQD = 95.2.<br />

585

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