Strength Analysis in Geomechanics Part 9 ppsx

5.3 Axisymmetric Problem 149 For the crack and punch we have two other border demands as f  (π)=0, f  (0) = 1 and f  (π)=1,f  (0) = 0 respectively. It is interesting to notice that once again the approximate solution gives at m = 1 the rigorous results. Now we find f  ,f  (see calculations in Appendix J) and according to (5.92) – stresses σ θ , σ r , τ rθ , τ e ., strain ε θ , displacement u r , integral J and factor K as K m+1 =(κ +1)πτ 2 l/2I(m)/f  (π)/ m where I(m) is the same as in Sect. 5.2.1. Condition τ e = constant gives equation forrinform 2r(2τ e ) m+1 GΩ(t)/τ 2 (κ +1)πl=F m+1 /I(m)/f  (π)/ m . (5.100) Diagrams σ θ /σ yi , σ r /σ yi , τ rθ /σ yi are given in Fig. 3.11 by the same lines as for m = 1 but with index 0. From the figure we can see that with the growth of m the distribution of stresses changes very strongly. In the same manner the problem of the punch horizontal movement can be considered. The curves for the stresses can be received by reflection of the previous ones relatively to axis θ = π/2. 5.3 Axisymmetric Problem 5.3.1 Generalization of Boussinesq’s Solution As in Sect. 3.3.2 we suppose for incompressible material (ν =0.5) σ χ = σ θ = τ ρχ = 0, and from the first static equation (2.77) as well as from rheological law (1.29) at α = 0 we have for stress and strain following relations σ ρ =f(χ)/ρ 2 , ε ρ =g(χ)Ω(t)ρ −2m (5.101) whereg=f m . Since for this case in (2.79) ε χ = ε θ we find easily u χ =U(ρ)sinχ, u ρ = ϕ(χ)+g(χ)Ω(t)ρ 1−2m /(1 − 2m). (5.102) Using condition ε ρ = −2ε χ we derive 0.5Ω(t)(3 − 2m)g(χ)ρ 1−2m /(1 − 2m) + ϕ(χ)+U(ρ)cosχ =0. (5.103) Putting u ρ ,u χ from (5.102) into condition γ ρχ = 0 we determine ϕ  (χ)+g  (χ)Ω(t)ρ 1−2m /(1 − 2m) + ρ 2 (sin χ)∂(U(ρ)/ρ)/∂ρ. (5.104) Excluding ϕ  (χ) from (5.103), (5.104) we obtain the expression in which both parts must be equal to the same constant, say n, since each of them depends only on one variable (neglecting t as a parameter) in form Ω(t)g  (χ)/2sinχ = ρ 2m dU(ρ)/dρ = −n 150 5 Ultimate State of Structures at Small Non-Linear Strains P Z Fig. 5.17. Computation of constant n with obvious solutions f(χ)=Ω −µ (C + 2n cos χ) µ , U(ρ)=D− nρ 1−2m /(1 − 2m). (5.105) Since at χ = π/2 we have σ ρ = 0 we must put in the first (5.105) C = 0 and constant n should be found from condition (Fig. 5.17) P=−2 λ  0 σ ρ ρ 2 sin χ cos χdχ. (5.106) Putting here σ ρ from (5.101) we find after calculations σ ρ = −P(µ +2)cos µ χ/2π(1 − cos µ+2 λ)ρ 2 . (5.107) Taking in the second relation (5.105) D = 0 we get the displacement as u χ = −Ω(t)(P(µ +2)/2π(1 − cos 2+µ λ)) m ρ 1−2m (sin χ)/(1 − 2m). (5.108) The most interesting case takes place at λ = π/2 when we receive from expressions (5.107), (5.108) σ ρ = −P(2 + µ)(cos µ χ)/2πρ 2 , u χ = −Ω(t)(P(2 + µ)/2π) m ρ 1−2m (sin χ)/(1 − 2m). (5.109) It is easy to notice that the highest value of σ ρ at ρ = constant is on the line χ = 0. It is not difficult to find that there stress σ ρ at m = 1 is 1.5 times more than at µ = 0. The biggest value of u χ is at χ = π/2 but its dependence on m is more complex. However the second relation (5.109) allows to calculate the displacements in some distance from the structure loaded by forces with a resultant P. 5.3 Axisymmetric Problem 151 0 0.2 0.4 2 4 z/a Fig. 5.18. Comparison of stress distribution To appreciate a practical meaning of the results we compare for m = 1 the distribution of stress σ z on axis z for the concentrated force P = qa 2 and for the circular punch of radius a when we have from (5.109) and (3.122) respectively /σ z //q = 3(a/z) 2 /2π, σ z /q=1−(1 + (a/z) 2 ) −3/2 . (5.110) From Fig. 5.18 where by solid and broken lines diagrams /σ z /(z) are shown we can see that at z/a > 3 the simplest solution for concentrated force can be used. Since at µ < 1 a distribution of stresses becomes more even we can expect better coincidence of similar curves with the growth of a non-linearity. It is interesting to notice that according to Figs. 5.18, 5.4 vertical stress in axisymmetric problem is approximately twice less than in the plane one. This explains higher load-bearing capacity of compact foundations. 5.3.2 Flow of Material within Cone Common Equations We solve this problem at the same suppositions as that in Sect. 4.3.3 From (2.79) at u χ = 0 we compute ε ρ = −2U/ρ 3 , ε θ = ε χ =U/ρ 3 , γ ρχ ≡ γ =dU/ρ 3 dχ, γ m =g/ρ 3 (5.111) where U = U(χ)and g(χ)=  9U 2 +U 2 . Similar to (5.18) and (5.68) we use representations ε χ = −g(cos 2ψ)/3ρ 3 , γ ρχ = g(sin 2ψ)/ρ 3 (5.112) 152 5 Ultimate State of Structures at Small Non-Linear Strains putting which into the first law (2.82) we have equations (g cos 2ψ)  +3gsin2ψ =0, dg/gdχ = 2(dψ/dχ −3/2) tan 2ψ. (5.113) The latter gives boundary condition dψ/dχ = 3/2 at ψ = π/4. From expres- sions for strains above we can also find dln/U//dχ = −3 tan 2ψ, g(χ)=−3U/ cos 2ψ, (5.114) From (5.17) and (5.112) we derive representations τ ρχ = τ = ω(t)ρ −3µ g µ sin 2ψ, σ ρ σ χ = ω(t)(C + ρ −3µ (K +2 −1 x2g µ (cos 2ψ)/3)) (5.115) where C is a constant and function K(θ) can be found from the first static equation (2.77) as follows 3µK=(g µ sin 2ψ)  +cotχ(g µ sin 2ψ) + 4(1 − µ)g µ cos 2ψ. (5.116) Putting (3.115) into (2.78) we derive (g µ sin 2ψ)  +(g µ sin 2ψ)  cot χ +(9µ(1 − µ) − 1/ sin 2 χ)g µ sin 2ψ + 2(2 − 3µ)(g µ cos 2ψ)  =0. (5.117) Combining (5.113), (5.117) we have at Ψ according to (5.49) two differential equations Θ=dχ/dψ, (5.118) (cot 2ψ)dΘ/dχ − 2(µ − 1+2µ/Ψ) + Θ((6µ 2 +3µ +4(1− µ)cos 2 2ψ)/Ψ − cot χ cot 2ψ) − 3Θ 2 (3µ 2 − (µ + µ tan 2ψ cot χ +1/3sin 2 χ)cos 2 2ψ)/2Ψ = 0 (5.119) the second of which should be solved at different Θ o = Θ. Then we integrate (3.118) at border demand χ(0) = 0. The searched function must also satisfy condition χ = λ at ψ = π/4. Now we receive from (5.113), (5.114), (2.65) U(χ), g(χ)andτ e . Putting stress σ ρ from (5.115) into integral static equation (4.105) we find at −q ∗ = σ ρ (a, λ)=σ χ (a, λ) expression for max τ e as max τ e =3µq ∗ maxg µ (χ)/((g µ (θ)sin2ψ)     χ=λ − g µ (λ)cotλ −2J 3 / sin 2 λ) where J 3 = λ  0 g µ (θ)(sin 2ψ sin 2 χ +2cos2ψ sin 2χ)dχ. Then the criteria max τ e = τ u and dγ m /dt →∞must be used as before. For the latter we have ε ∗ =1/α, Ω(t ∗ )=(αe2 max τ e ) −m . 5.3 Axisymmetric Problem 153 Some Particular Cases At µ = 0 we have the solution of Sect. 4.3.3. If µ = 1 we compute from (5.113), (5.117) equation (g sin 2ψ)  +(gsin2ψ)  cot χ +(6− 1/ sin 2 χ)g sin 2ψ =0 with obvious solution gsin2ψ =2Dsin2χ (5.120) where D is a constant. Then from (5.112) gcos2ψ =3D(cos2χ −cos 2λ). (5.121) From (5.120), (5.121) we receive tan 2ψ = 2(sin 2χ)/3(cos 2χ −cos 2λ) Diagrams ψ(χ) at different λ according to this relation are drawn in Fig. 5.19. Similar to the general case we have ultimate condition as max τ e =q ∗ x 1 2/3tanλ (λ > < 33.7 ◦ ) (5.122) At µ =2/3 we calculate from (5.117) equation (g 2/3 sin 2ψ)  +(g 2/3 sin 2ψ)  cot χ +(2− 1/ sin 2 χ)g 2/3 sin 2ψ =0 with obvious solution solution g 2/3 sin 2ψ =Hsinχ (5.123) where H is a constant. Putting (5.123) into (5.113) we derive differential equation of the first order 0 15 30 30 60 Fig. 5.19. Dependence ψ(χ)atµ =1 154 5 Ultimate State of Structures at Small Non-Linear Strains 0 60 120 2.3 3.051.280.58 1 15 30 4810 χ o ψ o Fig. 5.20. Diagrams ψ(χ)atµ =2/3 and different Θ o dχ/dψ = 2(2 + 3 cot 2 2ψ)/3(2 + cot χ cot 2ψ) (5.124) that should be integrated at different Θ(0) = Θ o . Diagrams χ(ψ)atΘ o -values in the curve’s middles and λ at their tops are given in Fig. 5.20. Putting σ ρ from (5.113) into (4.105)we find C and from condition dτ e /dχ = 0 with consideration of (5.124) – equality tan 2ψ = 3 tan χ which gives to max τ e (it increases with a growth of χ) value max τ e =q ∗ sin 2 λ  cos 2 λ +9sin 2 λ/(3 cos λ −cos 3 λ − 2 − 12J 4 ). Here as before −q ∗ = σ ρ (a, λ)=σ χ (a, λ)and J 4 = λ  0 (sin 2 χ cos χ)(tan 2ψ) −1 dχ. Diagrams J 4 (λ) and max τ e (λ) are shown in Figs. 5.21, 5.22 respectively. The broken line in the latter picture refers to the case µ = 1 (computations for µ =1/3 see in Appendix K-interrupted by points curve in the figure) and pointed line refers to solution (4.106). 5.3.3 Cone Penetration and Load-Bearing Capacity of Circular Pile Here common relations (5.111). . . (5.119) are valid. We put stresses according to (5.115) into integral static equations (4.107). to detail the constants and according to (2.65) we compute max τ e at a = ρ as 5.3 Axisymmetric Problem 155 0 0.4 0.8 1.2 J 4 45 90 135 Fig. 5.21. Diagram J 4 (λ)forµ =2/3 0 1 2 30 60 Fig. 5.22. Diagram max τ e (λ) max τ e =3µ(P/π − p ∗ (a + 1) 2 sin 2 λ)(maxg µ (χ))/a(a(2(g µ (χ)sin2ψ)     χ=λ sin 2 λ +g µ (λ)(1 + 3µ)sin2λ((1 + l/a) 2−3µ − 1)/(2 − 3µ) − l(g µ (λ)sin2λ +2J 5 )(2 + l/a)) (5.125) where p ∗ is the strength of soil in a massif at compression and J 5 = λ  0 g µ (χ)((1 + 3µ)sin2ψ sin 2 χ +2cos2ψ sin 2χ)dχ. At λ → π,a→∞we find for a circular pile max τ e =(P/π − p ∗ b 2 )maxg µ (χ)/l(bg µ (λ)+J 5 (λ))(2 + l/a). (5.126) In the same manner we consider the particular cases and consequently for µ = 0 we receive from (5.117), equation (4.103) and hence the solution of Sect. 4.3.4. 156 5 Ultimate State of Structures at Small Non-Linear Strains At µ =1wehave max τ e =4.5(P/π − p∗(a + 1) 2 sin 2 λ)x 1 2/3tanλ /la(2(5 − 6sin 2 λ)/(1 + l/a) − (2 + 3 sin 2 λ)(2 + l/a)) λ<146 ◦ λ<146 ◦ and for the pile the yielding (the first ultimate) load is P yi = πb(2τ yi l+p ∗ b). (5.127) We can see that this result has obvious structure and coincides with approxi- mate relation (4.110) for ideal plasticity. Similarly we compute for µ =2/3 max τ e =(P/π − p ∗ (a + 1) 2 sin 2 λ)  cos 2 λ +9sin 2 λ/6a(2a cos λ sin 2 λ ln(1 + l/a) − l(2 + l/a)(1 − cos λ +2J 4 )). Here J 4 is given in Sect. 5.3.2. For the circular pile this relation predicts big values of ultimate load and so we can take in the safety side P u = πb(p ∗ b+2τ u l). (5.128) 5.3.4 Fracture of Thick-Walled Elements Due to Damage Stretched Plate with Hole We consider plate of thickness h with axes r, θ, z (Fig. 5.23) and use the Tresca-Saint-Venant hypotheses. Since here σ θ > σ r > σ z =0wehaveε r =0 and from (2.32) at α =0,σ eq =2τ e = σ θ ε ≡ ε θ = 3Ω(t)σ θ m /4 σ θ =(4/3Ω) µ ε µ (5.129) where ε =u/r and radial displacement u depends only on t. Putting (5.129) into the static equation of this task hσ θ =d(hrσ r )/dr, (5.130) integrating it at h = constant as well as at boundary conditions σ r (a) = 0, σ r (b) = p and excluding factor (4u/3Ω) µ we receive with the help of (5.129) σ θ =p(1−µ)(b/r) µ /(1 − β µ−1 ) (5.131) where β = b/a. Putting σ θ into (2.66) we find for the dangerous (internal) surface e −αε ε =3β(1 − µ) m Ω(t)p m (1 − β µ−1 ) m /4. (5.132) Applying to (5.132) criterion dε/dt→∞we have 5.3 Axisymmetric Problem 157 a b r u p Fig. 5.23. Stretched plate ε ∗ =1/α, p m Ω(t ∗ ) = 4(1 −β µ−1 ) m /3β(1 − µ) m αe. (5.133) When the influence of time is negligible we compute from (5.133) at Ω = constant critical load p ∗ =(4/3) µ (1 − β µ−1 )/(1 − µ)(αβΩe) µ . (5.134) At small µ that value must be compared to ultimate load p u which follows from (5.131) at µ → 0as p u = σ yi (1 − 1/β) where σ yi is a yielding point at an axial tension or compression and the small- est value should be taken. At m near to unity we must compare p ∗ with yielding load which follows from (5.131) at m = 1 in form: p yi =(σ yi /β)lnβ and the consequent choice should be made. Sphere For a sphere under internal q and external p pressures (Fig. 3.23) we denote the radial displacement also as u and according to relations (2.80) we compute ε θ =u/ρ, ε ρ = du/dρ and from the constant volume demand (2.81) we find u=C/ρ 2 , ε ρ = −2C/ρ 3 , ε θ =C/ρ 3 =u/ρ (5.135) where constant C is to be established from boundary conditions. Now from (2.32) at α =0andσ eq = σ θ − σ ρ we deduce 158 5 Ultimate State of Structures at Small Non-Linear Strains ε θ = Ω(t)(σ θ − σ ρ ) m /2 or with consideration of (5.135) σ θ − σ ρ = (2C/Ωρ 3 ) µ . (5.136) Putting (5.136) into static equation (2.80) we get after integration at border demands σ ρ (b) = −p, σ ρ (a) = −q and exclusion of constants σ θ − σ ρ = 3(q − p)µ(b/ρ) 3µ /2(β 3µ − 1). (5.137) Now we use constitutive law (2.32) which for our structure is e −αε ε =0.5Ω(t)(σ θ − σ ρ ) m (5.138) where ε = ε θ . Using here σ θ − σ ρ from (5.136) and criterion dε/dt→∞we deduce ε ∗ =1/α, (q −p) m Ω(t) = 2(2m/3) m (1 − β −3µ ) m /αe. (5.139) When the influence of time is not high critical difference of the pressures at Ω = constant can be got (q − p) ∗ =2 1+µ m(1 − β −3µ )/3(αΩe) µ . (5.140) At small µ this value should be compared with (q − p) u according to (4.97) and the smaller one must be taken. Similar choice have to be fulfilled between (q − p) ∗ and (q − p) yi given by (4.94) at µ near unity. Cylinder In an analogous way the fracture of a thick-walled tube can be studied. From (2.32) at α =0,ε x =0,ε θ ≡ ε and σ eq = σ θ − σ r we have ε =(3/4)Ω(t)(σ θ − σ r ) m (5.141) and providing the procedure above for the disk and the sphere we find /17, 27/ σ θ − σ r =2µ(q − p)(b/r) 2µ /(β 2µ − 1). (5.142) Equation (2.32) for this structure is e −αε ε = 3Ω(t)(σ θ − σ r ) m /4. (5.143) [...]... Bar in tension under hydrostatic pressure 6.1.2 Fracture Time of Axisymmetrically Stretched Plate In order to integrate differential equation (5.130) in the range of finite strains we take according to the condition of Tresca-Saint-Venant in Sect 5.3.4 r = ro + u(t) We replace strains and displacements by their rates and rewrite 6.1 Use of Hoff’s Method 163 (5.130) with consideration of (5.1 29) at B instead... together with (6.11) into (6.10), separating the variables and integrating as before we have finally 1 (J6 )m (β − 1)−2 dβ B(q − p) t∞ = (4/3)(βo − 1) cos λo m βo The integrals in (6.12) should be calculated as a rule approximately (6.12) 166 6 Ultimate State of Structures at Finite Strains 6.1.4 Final Notes Although the method in this sub-chapter uses somewhat unrealistic supposition of an in nite elongation... fracture time is near to test data An analysis shows that the reason of it lays in the nonlinearity of equations linking the rate of strains with stresses Because of that the approach is widely used for the prediction of the failure moment of structures For example in /17/ a row of elements are considered Among them a grating of two bars, thin-walled sphere and tube under internal pressure, a long membrane... and introduce ratio β = cos ψ/ cos λ Then λ is function of β as cos λ = ((βo − 1)/(β − 1)) cos λo and integral J6 in (5.146) is also a function of β Now we find dβ/dt = (β − 1)(tan λ)dλ/dt (6.10) and since ε(λ) = ln(sin λ/ sin λo ) (Fig 3.24) then dλ/dt = tan λdε(λ)/dt and from (5.141) with dε/dt, B instead of ε, Ω and (5.145) we derive dε(λ)/dt = (3B/4)(q − p)m (cos λ)/(J6 )m sin2 λ (6.11) Putting... q/po and according to (1.42) p = po eε The integration of (6.1) in limits 0 ≤ ε ≤ ∞, 0 ≤ t ≤ t∞ gives m−1 m (−1)i (1 − i/(1 + κo )i )/i!i(m − 1 − i)!)/(κo )m B(po ) t∞ = 2(ln(1 + κo ) + (m − 1)! i=1 From Fig 6.2 where for some κo curves tu (µ) according to the latter expression are given by broken lines we can see that tu ≡ t∞ diminishes with an increase of hydrostatic component In a similar way... for a bar in tension by stresses q under hydrostatic pressure p (Fig 6.3) In this case /17/ σ1 = q, σ2 = σ3 = −p and hence in (2.31) S1 = 2(p + q)/3, σeq = p + q Comparing this data to the previous ones we can see that rate dε/dt in the latter problem is twice of that for the plate Hence t∞ for the bar is one half of that in the plate case 162 6 Ultimate State of Structures at Finite Strains q p q... example at m = 2 and the shown meanings of λ, ψ its value is 0 .9 In order to appreciate the moment of fracture we put (5.145) into (5.143) and use criterion dε/dt → ∞ when we have for a dangerous (internal) surface ε∗ = 1/α, Ω(t∗ )(q − p)m = 4(J6 )m (sin2 ψ)/3eα cos ψ (5.147) If the in uence of time is negligible we derive from (5.147) at Ω = constant (q − p)∗ = (4/3)µ J6 (sin2 ψ/αeΩ cos ψ)µ (5.148) Once... 6 Ultimate State of Structures at Finite Strains β)4 /16β), Bp2 t∞ = (16/3) ln((1 + Bp2 t∞ = 16((3(1 + β2 )/2 + 2β) − 2(1 + β) β + 2(ln((1 + 2 + β ln((1 + 1/ β)/2) 2 β)/2))/3β 6.1.3 Thick-Walled Elements Under Internal and External Pressures We begin with a sphere and replace εθ , u in (5.135) by their rates dεθ /dt, V Then we suppose in (5.136) Ω(t) = Bt According to definition β = b/a we have dβ/dt... Axisymmetric Problem 1 59 Using here expression (5.142) at r = a and the criterion dε/dt → ∞ we derive ε∗ = 1/α, (q − p)m Ω(t∗ ) = 4(m/2)m (1 − β−2µ )m /3αe (5.144) When in uence of time is negligible we can find as before critical difference of pressures as (q − p)∗ = (4/3)µ m(1 − β−2µ )/2(αΩe)µ and again for µ near to zero this value must be compared with (q − p)u according to (4 .99 ) and smaller one have... relations for strains (3.117) and stresses (3.116) as well as law (5.141) at σr → σχ we find σθ − σχ = (q − p) sinµ χ/J6 sin2µ χ where (5.145) λ (cos1+µ χ/ sin2µ+1 χ)dχ J6 = (5.146) ψ Ω * (q−p)m 2 2 3 0.5 1 01 4 Fig 5.24 Dependence of t∗ on β and m β 160 5 Ultimate State of Structures at Small Non-Linear Strains The computations for m = 1 when J6 = A/2 from Sect 3.3.1 at λ = π/3, ψ = π/6 show that integral J6 . follows 3µK=(g µ sin 2ψ)  +cotχ(g µ sin 2ψ) + 4(1 − µ)g µ cos 2ψ. (5.116) Putting (3.115) into (2.78) we derive (g µ sin 2ψ)  +(g µ sin 2ψ)  cot χ + (9 (1 − µ) − 1/ sin 2 χ)g µ sin 2ψ + 2(2 −. equation (g 2/3 sin 2ψ)  +(g 2/3 sin 2ψ)  cot χ +(2− 1/ sin 2 χ)g 2/3 sin 2ψ =0 with obvious solution solution g 2/3 sin 2ψ =Hsinχ (5.123) where H is a constant. Putting (5.123) into (5.113) we. shown in Figs. 5.21, 5.22 respectively. The broken line in the latter picture refers to the case µ = 1 (computations for µ =1/3 see in Appendix K-interrupted by points curve in the figure) and pointed