Energy Bounds in Linear Elasticity

Transcription

1 Chapter 1 Energy Bounds in Linear Elasticity Focus of this Recitation: Thefocusofthisrecitationistoputtoworktheenergyboundsoflinearelasticity for the derivation of upper bound and lower bound energy solutions of the form: ) E com σ, ) T E com σ, T S.A. K.A. < = < E }{{ } ) E pot ε, ) pot ε, ξ ) 1.1) ξ }{{ } ) Ψ σ ) W T Ψε ) W ξ The main idea of these energy bounds is that the actual energy change that occurs in a material structural system can be approximated by means of the complementary energy E com σ, T ) and of the potential energy E pot ε, ξ ), which both realize a minimum for the solution: The complementary energy realizes a minimum among all possible statically admissible stress fieldss.a.): ) ) E com σ, T = min E σ com σ, T S.A. ) <E com σ, T 1.) S.A. The potential energy realizes a minimum among all possible kinematically admissible displacement fieldsk.a.): E pot ε, ) ξ = min E pot ε, ξ ) K.A. <E pot ε, ξ ) 1.3) ξ K.A. 18

2 1.1. REVIEW Review The two energy bounds defined by 1.1) provide a means toestimate the actual energy change in a material structural system Target For this actual exact solution, the solution or target energy can be evaluated from the sole knowledge of the external works, which is known as Clapeyron s formula: E com σ, T ) = 1 ) )) ) W T W ξ = E pot ξ,ε 1.4) ) wherew T istheworkbyprescribeddisplacements ξ d : ) W T = d ξ Tda 1.5) Ω ξ d ) andw ξ istheworkbyprescribedforcesρ g dω; T d da): ) W ξ = Ω ξ ρ g dω+ Ω T d ξ T d da 1.6) We will refer to this actual energy solution as the target solution Complementary Energy Approach = Stress Approach The complementary energy approach is a "stress approach", in which the complementaryenergye com σ, T ) is evaluated with statically admissible stress fieldss.a.): E com σ, T ) ) =Ψ σ ) W T 1.7)

3 130 CHAPTER 1. ENERGY BOUNDS IN LINEAR ELASTICITY whereψ σ )isthecomplementaryfreeenergy: ) Isotropic: Ψ σ 1 σ )= m Ω K +s dω G with: 1.8) σ m= 1 3 trσ ; s = 1 σ :σ 3σ m) ) andw T istheworkbyprescribeddisplacements ξ d definedby1.5). This stress approach involves the following steps: 1. Evaluate the target complementary energy solution from Clapeyron s formula1.4)[careful with sign!].. Chooseastressfieldσ andcheckthatitisstaticallyadmissible;thatis: on Ω T d : ons : T d =σ n T n)+ T n)=0 1.9) inω : divσ +ρ g =0 3. Determine the complementary energy from Eqs. 1.7),1.8) and1.5). 4. Minimize this complementary energy expression w.r.t. any"stress-degreeof-freedom" to obtain the lowest possible complementary energy for the chosenstressfieldσ. 5. Compare with Target : E com σ, T ) S.A. <E com σ, T ) 1.10) Potential Energy Approach = Displacement Approach The potential energy approach is a"displacement approach", in which the potentialenergye pot ε, ξ ) isevaluatedwithkinematicallyadmissibledisplacementfields ξ : E pot ε, ξ ) ) =Ψε ) W ξ 1.11)

4 1.. EXAMPLE 1: FROM LECTURE#8) 131 whereψε )isthehelmholtz)freeenergy: Isotropic: Ψε 1 )= Kǫ Ω v +Gǫ d) dω with: 1.1) ǫ v=trε ; ǫ d = ε :ε 1 ) 3 ǫ v andw ξ ) istheworkbyprescribedforcesdefinedby1.6). This displacement approach involves the following steps: 1. Evaluate the target potential energy solution from Clapeyron s formula 1.4).. Choose a displacement field ξ x) and check that it is kinematically admissible; that is on Ω ξ d: ξ d = ξ x) 1.13) Calculate the strain tensor: ε = 1 grad ξ +grad ) ξ ) T 1.14) 3. Determine the potential energy from Eqs. 1.11),1.1) and1.6). 4. Minimize this potential energy expression w.r.t. any"displacement-degreeof-freedom"inordertoobtainforthechosen ξ x)thesmallestpossible value of the potential energy. 5. Compare with Target : E pot ε, ξ ) K.A. <E pot ε, ξ ) 1.15) 1. Example 1: from lecture#8) 1..1 Problem Formulation We consider a homogeneous isotropic elastic soil layer of height h, subjected at itssurfacez=0toasurfacepressurep. Thefocusofthisexerciseistoestimate the surface settlement s. Using the energy approach, determine the following:

5 13 CHAPTER 1. ENERGY BOUNDS IN LINEAR ELASTICITY z p Figure 1.1: Soild layer subjected to a surface pressure. i) Target : Determine the potential energy and the complementary energy of the solution. ii) Lower Bound Approach: Determine an upper bound of the surface settlement by using the complementary energy approach, with a diagonal stress fieldoftheform: σ =a e x e x + e y e y )+b e z e z 1.16) wherea, b) are two stress constants to be determined. iii) Upper Bound Approach: Determine a lower bound of the displacement by using the potential energy approach, with a displacement field of the form: ξ =s 1 h) x e z 1.17) iv) Conclude this exercise by comparing the two displacement bounds so obtained. 1.. Target For the Target, we use Clapeyron s formula1.4) to evaluate the potential energy from the external work. In this problem, the work provided

6 1.. EXAMPLE 1: FROM LECTURE#8) 133 ) frotheoutsideisonlyduetothesurfacepressureatz=0;thusw T =0 and: ) E com σ, T = 1 ) W ξ = 1 ) psa = E pot ξ,ε whereaisthesurfacearea. 1.18) 1..3 Lower Bound Energy Approach = Upper Displacement Bound Thestressfieldσ mustbestaticallyadmissible. Fora,bconstant,theconditionσ needstosatisfyisthestressboundarycondition: z=0: T d =p e z σ e z 1.19) b= p The second stress parameter a remains undetermined. It is a"stress degree of freedom", and will be determined from minimizing the complementary energy. To calculate the complementary energy, we determine the stress invariants: σ m= 1 3 trσ = a p 1.0) 3 s = 1 σ :σ 3σ 1 m) = trσ σ ) 3σ m) = 1 ) ) a p a +p 3 3 = 3 ap+1 3 a p 1.1) [Note that σ m and s are constant in this problem, which facilitates the evaluation of subsequent integrals].

7 134 CHAPTER 1. ENERGY BOUNDS IN LINEAR ELASTICITY Thus, noting that W = 0 no work by prescribed displacements), we obtain from1.7) and1.8): E com σ, T ) ) = Ψ σ 1 σ )= m Ω K +s dω G = Ah 1 a 1 p) ap+ ) a p 1.) K G InordertoobtainthepossiblysmallestvalueofE com σ, T ) we minimize this function w.r.t. the"stress-degree-of-freedom", a: E com a) a = Ah 9GK a4g+3k) pg 3K))=0 1.3) a= G 3K 4G+3K p Resubstitution in the complementary energy expression yields: min a E com a)= 3 Ah p 4G+3K Finally, a comparison with the target solution yields: E com σ, T )= 1 psa< 3 Ah p 4G+3K =E com σ, T ) 1.4) 1.5) p s<3h 4G+3K Thus, the lower energy bound approach based on statically admissible stress fields yields an upper bound for the displacement Upper Bound Energy Approach = Lower Displacement Bound Thedisplacementfield ξ mustbekinematicallyadmissible,whichisthecase here, since: z=h: ξ d =0 s e z 1 z ) 1.6) h z=h

8 1.. EXAMPLE 1: FROM LECTURE#8) 135 In this displacement approach, s is a "displacement degree of freedom", which will be determined from minimizing the potential energy. To determine the potential energy1.11), we note that the external work by prescribed forces1.6) is: W ) ξ = ξ T d da=ap ξ e z =Aps 1.7) Ω T d To determine the free energy, we determine the strain invariants: ε = ε zz e z e z = s e z e z 1.8) h ǫ v = trε = s 1.9) h ǫ d = ε :ε 1 ) 3 ǫ v = 4 ) s 1.30) 3 h Hence,notingthatǫ v andǫ d areconstants: Ψε )=Ah 1 ) Kǫ v +Gǫ 1 d =Ah s h ) K+G 4 ) 3 The potential energy thus is: E pot ε, ξ ) ) = Ψε ) W ξ [ = A h 1 ) s ] K+G 4 ) ps h ) 1.3) InordertoobtainthesmallestvalueofE pot ε, ξ ) weminimizetheexpression w.r.t. the"displacement-degree-of-freedom": E pot s ) s =A s h ) s = h ) K+G 4 ) ] p = ) p ) K+G 4 3

9 136 CHAPTER 1. ENERGY BOUNDS IN LINEAR ELASTICITY The smallest potential energy for the chosen displacement field is thus: mine pot s )= 3 p ha s 4G+3K Finally, a comparison with the target solution yields: 1.34) ) E pot ξ,ε = 1 psa< 3 p ha 4G+3K =min E pot s ) s 1.35) s s ph = ) K+G Comparison: Lower Upper Bound Energy and Displacements Letusbringthedifferentenergyboundstogetherinthesenseofthefundamental relation1.1): mine com a)= 3 a Ah p 4G+3K ) E com σ, T S.A. < = 1 psa = E pot ε, ) ξ K.A. < min s E pot s )= 3 p ha 4G+3K S.A. < K.A. < 1.36) From a comparison of the energies, we readily realize that the lower energy bound min a E com a) is equaltothe upper bound K.A. < min s E pot s ). This means that the chosen stress field and displacement field are actual the solution of the problem. This is also found by comparing the displacement bounds. Since the upper energy bound provides lower settlement boundi.e. it overestimates the rigidity), and since the lower energy bound provides an upper settlement bound, the very fact that the settlements obtained with the two different approaches coincide, means that this is the actual solution of the problem: p 3h 4G+3K S.A. s K.A. s = ph K+G 4 3 ) 1.37)

10 1.3. EXAMPLE : DEEP TUNNELING THROUGH HETEROGENEOUS ROCK Example : Deep Tunneling Through Heterogeneous Rock What s the Meaning of Heterogeneous XYZ The previous example was restricted to homogeneous materials, meaning thattheelasticpropertiesdonotvaryfromonepointtotheother. However, the same approach also applied to heterogeneous materials and structures, for which the elastic properties vary from one point to another: K=K x);g=g x) 1.38) The problem is quite complex, and there are often no closed form solutions possible for such heterogeneous problems. The energy bounds then provide a unique opportunity to develop upper and lower energy bounds. In this case, the spatial variability of the elastic properties must be considered in the evaluation of the free energy integral1.1) and the complementary free energy integral1.8): Ψε ) = Ψ σ ) = Ω Ω Problem Formulation K x) ǫ v +G x) ǫ σ m K x) + s G x) d) dω 1.39) ) dω 1.40) WeconsiderthetunnelingofacylinderofradiusRisathighdepthH R) throughamountainthat is composedof two rocks separatedalong a plane Γ Fig. 1.). Giventhehighdepth,thetworockscanbemodeledastwosemiinfinite domains which are perfectly connected along the separation plane Γ. The tunnel is driven in the z-direction, normal to the separation plane Γ, which separates the tunnel section into two half-circles. Throughout this exercise, we willusecylindercoordinatesr,θ,z),withzbeingthetunnelaxis. We have seen a similar problem in a previous recitation. However, the main difference here is that the surrounding rock is not homogeneousconstant elasticity properties), but composed of two different rocks having different elastic properties: { K1,G Kθ),Gθ))= 1 ); 0<θ<π 1.41) K,G ); π<θ<π

11 138 CHAPTER 1. ENERGY BOUNDS IN LINEAR ELASTICITY H >> R r r K,G 1 1 θ Γ K 1,G 1 R θ Γ K,G K,G σ = 01 0 p δ Figure 1.: Deep tunneling through heterogeneous rock: two infinite halfspaces separated along a plane Γ.

12 1.3. EXAMPLE : DEEP TUNNELING THROUGH HETEROGENEOUS ROCK139 In contrast to the homogeneous deep tunneling problem, there is no closed form solution of this elasticity problem. By putting the energy bounds to work, we want to provide a back-of-the-enveloppe estimate of the on-average elastic radial displacementpositive inward) that occurs at the tunnel wall; that is an upper and a lower bound of the displacementfig. 1.): where n= e r. δ= 1 π π 0 ξr=r,θ) ndθ 1.4) 1. Target : By using Clapeyron s formula, show that the energy of the actual solution is equal to: E pot ξ,ε ) = = p 0 Rπδ 1.43) wherep 0 =ρ 0 gh isthehydrostaticgeological)pressureprevailinginthe rock before tunneling.. Upper Bound Approach: Consider a radial displacement field of the form: ξ =ur) er = C r e r;i=1, 1.44) where C is a"displacement-degree-of-freedom". Using the potential energyapproach,determinealowerboundofthedisplacementδ. Whyis ξ not the displacement solution of the heterogeneous problem? 3. Lower Bound Approach: Consider a radial stress field of the form: σ σ 0=p 0 R r) e r e r + e θ e θ ) 1.45) whereσ 0= p 0 1isthegeostaticstressstate. Usingthecomplementary energy approach, determine an upper bound of the displacement δ. Why isσ notthesolutionoftheheterogeneousproblem? [The solution is developed in the lecture notes, pp ]