Strength Analysis in Geomechanics Part 2 ppt

... elastic. 22 1 Introduction: Main Ideas C A R H a) b) R H Q Q B E h nj Ψ + j Ψ Ψ Fig. 1 .20 . Retaining wall sliding plane AB forms friction angle ϕ with normal n to AB. It means that R is inclined ... The 1.4 Main Properties of Soils 11 Ot 1 2 3 S Fig. 1.9. Combined in uence of time and loading Combined In uence of Time and Loading At small loads F the settling grows slowly (curve 1 in...

Ngày tải lên: 10/08/2014, 12:21

... tan λ: (4a cos υ +sin2υ) tan 2 λ 2( a 2 +cos2υ +2asinυ) tan λ −sin 2 =0. (4.56) Now we find the ultimate load according to the field of slip lines in Fig. 4 .22 . From Fig. 1 .22 we have for a cohesive ... ODCD’ σ x = σ yi π /2, τ xy =0, σ y = σ yi (1 + π /2) and according to (2. 72) we compute σ r = σ yi (π /2+ sin 2 θ), σ θ = σ yi (π /2 + cos 2 θ), τ rθ = τ yi sin2θ. (4.40) In...

Ngày tải lên: 10/08/2014, 12:21

... 978-3-540-44718-4 ISBN 978-3-540-370 52- 9 ISBN 978-3-540-3 726 1-8 (Continued after index) Elsoufiev, S.A. Strength Analysis in Geomechanics, 20 07 20 07 Vibration of Strongly Nonlinear Discontinuous Awrejcewicz, ... 20 3 Appendix H 20 7 Appendix I 20 9 Appendix J 21 1 Appendix K 21 3 Appendix L 21 5 Appendix M 21 7 Appendix N 22 3 Index 22 9 Series Editors: f¨ur Springe...

Ngày tải lên: 10/08/2014, 12:21

... of constants as C o =0.5p(sin 2 2 cos 2 ) −1 , C 1 = −p /2 (3 .24 ) and according to (2. 65) τ e =0.5p  1 −2cos2θcos 2 +cos 2 2λ/(sin 2 2 cos 2 ). (3 .25 ) The analysis of (3 .25 ) shows that at λ = ... 2 )+C o (2( 1 2 )θ cos 2 +sin2θ))/2G, (3 .27 ) u θ =r(pθ(1 2 )+C o (cos 2 2( (1 2 )θ 2 + 2( 1 − ν)lnr)cos2λ)/2G. For incompressible material (ν =0.5) relations (3.36), (3 .27...

Ngày tải lên: 10/08/2014, 12:21

... according to the asymptotic approach and (3.1 02) , (3.103) as σ r = −(K 2 / √ 2 r) (2 −3cos 2 (θ /2) ) sin(θ /2) , σ θ = −3(K 2 / √ 2 r) cos 2 (θ /2) sin(θ /2) , τ rθ =(K 2 / √ 2rπ)(1 −3sin 2 (θ /2) ) cos(θ /2) , ... cos(θ /2) , τ rθ =(Q/π √ 2rl)(3 sin 2 (θ /2) 2) sin(θ /2) , τ e =(Q /2 √ 2rl)  1+3cos 2 θ, (3.110) u r =u 1 − (Q/πG)  r/2l((κ +1) /2 − sin 2 (θ /2) ) co...

Ngày tải lên: 10/08/2014, 12:21

... −p=0.5σ yi (1 + 2 ln(sin υ/ sin ψ − sin 2 υ(cos υ/ sin 2 υ + ln(tan(λ /2) / tan(υ /2) )), (q − p) u = σ yi ln(sin λ/ sin ψ). (4.100) 128 5 Ultimate State of Structures at Small Non-Linear Strains solution ... b σ r = −p+c 2 σ yi (1 − b 2 /r 2 )/2b 2 , σ θ = −p+c 2 σ yi (1 + b 2 /r 2 )/2b 2 . The dependence of q − p on c and its ultimate value at c = b are q − p=0.5σ yi...

Ngày tải lên: 10/08/2014, 12:21

... law M= 2 λ  0 τ(θ)r 2 dθ (5. 72) we compute σ r = 2( M/r 2 B 6 )sin2θ, τ =(M/r 2 B 6 )(cos 2 −cos 2 ) (5.73) where B 6 =sin2λ 2 cos 2 . (5.74) Now (2. 65) gives τ e = M(1 + cos 2 2λ −2cos2θ cos 2 ) 0.5 /r 2 B 6 . ... a/l) sin λ)  cos 2 λ +4sin 2 λ /2( (a/l) ln(l/a+1)sin2λ − (λ +J 1 )). (5. 62) For the cases a = 0 and a →∞, λ → π we have respectively max τ e =(P/2l −p ∗...

Ngày tải lên: 10/08/2014, 12:21

... equation (g 2/ 3 sin 2 )  +(g 2/ 3 sin 2 )  cot χ + (2 1/ sin 2 χ)g 2/ 3 sin 2 =0 with obvious solution solution g 2/ 3 sin 2 =Hsinχ (5. 123 ) where H is a constant. Putting (5. 123 ) into (5.113) ... obvious solution gsin2ψ =2Dsin2χ (5. 120 ) where D is a constant. Then from (5.1 12) gcos2ψ =3D(cos2χ −cos 2 ). (5. 121 ) From (5. 120 ), (5. 121 ) we receive tan 2 = 2( sin 2...

Ngày tải lên: 10/08/2014, 12:21

Ngày tải lên: 15/12/2013, 06:15

Ngày tải lên: 26/01/2014, 16:20