By William G. Pariseau
SolutionsВ manual forВ Design research in Rock Mechanics (2006) by means of William G. Pariseau containing all, absolutely labored strategies to all routines within the correspondingВ textbook, together with many drawings. Textbook:В Hardback, ISBN 978-0-415-40357-3, Paperback, ISBN 978-0-415-45661-6.
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4) 144 p = 135 psi 36 Solutions Manual to Design Analysis in Rock Mechanics γave − 135 144 γave v S h = SH = 1,425 − 135 1−v 144 Sv = 1,425 39. 2 m thick at z = 396 m at z = 396 m p(top) = 552 kPa Find: Effective and total stresses at center depth of sandstone. 2 2 = sp. wt. 87 kN/m3 p = 552 kPa + γw γw p = 552 + 376 p = 928 kPa at center of Ss . Sv = 434γave − 928 (kPa) v S v = SH = 434γave − 928 (kPa) 1−v Introduction 40. 37 Given: Gravity loading and properties table, complete lateral restraint Surface Find: Sh = SH (Ss bottom) & (Lm top).
641(10−4 ) H H = 3,786 ft 17. 15 ←−−−−−−−−−−−−−−−−− Given: Sketch of the potential slope failure shown in the sketch, Find: (a) Factor-of safety of a cable bolted slope when the water table is drawn down 100 ft Bench height = 40 (vertical bolt spacing) Horizontal bolt spacing = 20 Bolt angle = 5◦ down Bolt loading = 700 kips/hole (b) Factor of safety of the same slope but without bolts when the water table is drawn below the toe (c) Reasons for preferring one over the other. Tension crack Bench ft 37Њ 40Ј b ϭ 40Њ 100Ј WT 320Ј a ϭ 32Њ Failure Surface f ϭ 28Њ c ϭ 1440 psf g ϭ 158 pcf S Not to Scale Sketch for problem with given data.
FS = C, Wn P W1 C1, f1 A1 (N tan φ + C) + W1 tan φ + C1 Ws FS(with berm) = FS(without berm) + W1 tan φ1 + c1 A1 > FS(without berm) Ws The added resistance W1 tan φ1 + C1 , comes without added driving force and thus increases the FS. 50 12. 87 cr = 64,800 psf cj = 1,620 psf φr = 32◦ φj = 25◦ Find: (a) Hmax when WT at crest Wn n (b) Hmax when WT at toe NЈ s Solution: W Ws p b R FS = D D = W sin α R = N tan φ + C N = (Wn − p) C = cA Wn = W cos α a A (b) When depressurized P = 0 γH 2 (cot α − cot β) 2 H A= sin α W = FS = cH W cos α tan φ + sin α W sin α W sin α FS = tan φ + tan α γH 2 tan φ tan α γ 2 ∴ FS − cH sin α 2 (cotan α − cotan β)sin α cotan α − cotan β 1 is minimum when FS is minimum H occurs of FS = 1.