Elio Sacco. Dipartimento di Ingegneria Civile e Meccanica Università di Cassino e LM

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Primrose Gordon
5 years ago
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1 Elio Sacco Diartimento di Ingegneria Civile e Meccanica Università di Cassino e LM

$if ε < 0 = E ( ε δ) with δ = ε if ε 0 ε e δ elastic strain fracture$

2 Uni-axial resonse E ε if ε < 0 = 0 if ε 0 stress-strain t ε = 0 = c 0 if ε < 0 = E ( ε δ) with δ = ε if ε 0 ε e δ elastic strain fracture strain

3 Uni-axial resonse stress-strain ε A free internal energy exists 1 E ε if ε < 0 ϕ( ε) = 0 if ε if ε < 0 ϕεδ (, ) = E ( ε δ) with δ= ε if ε 0 internal energy ϕ ε ϕ 0 if ε < 0 = = E ( ε δ) with δ = ε ε if ε 0 For any ossible closed cycle the material is not able to dissiate energy; the whole energy sent to deform the material is comletely recovered during the unloading. 3

4 D no tension material K stress admissibility conditions = I1 0 = I 0 1 cone of admissible stresses { : 0& 0} K = I I 1 K * δ total strain ε= ε + δ e normality rule for fracture strain ( ) T * * δ 0 K 4

5 D no tension material K stress admissibility conditions 1 normality rule for fracture strain setting setting it results ( ) T * * δ 0 K * = it results T 0 δ 0 * T = it results δ 0 ( ) ( ) T δ= * * * * as δ + δ 0 0, 0 with δ 1 1+ δ = 0 so that setting δ + δ δ δ = δ δ 0 0, 0 K * * * * * * * = 0 δ 0 and setting * 1 * 1 δ = 0 δ 0 δδ KK 5

6 D no tension material determination of the fracture strain isotroic constitutive law (lane strain/stress) ( - ) ( - ) ( - ) ( - ) = α ε δ + β ε δ = β ε δ + α ε δ 1 1 normality rule for fracture strain δ + δ = with δ1 0 δ K * δ K 1 Four cases can arise: case 1 1 δ1 δ case 1 δ1 δ case 3 1 δ1 δ case 4 δ δ < 0 < 0 = 0 = 0 < 0 = 0 = 0 0 = 0 < 0 0 = 0 = 0 =

7 Case 1 < 0 < 0 δ = 0 δ = the rincial strain comonents satisfy the condition αε1+ βε βε + αε 1 Case < 0 = 0 δ = 0 δ 1 1 ( ) = αε + β ε -δ = β ε1+ α ε δ = δ = α ε + βε1 α ( - ) 0 ( ) the rincial strain comonents satisfy the condition β 1 = α- ε1 α ε1 βε + αε 1 7

8 Case 3 = 0 < 0 δ δ = = α ε1 δ1 + β ε = δ1 = α ε1+ βε α = β ε -δ + αε ( - ) 0 ( ) ( ) 1 1 the rincial strain comonents satisfy the condition β = α- ε α α ε1+ βε ε Case 4 = 0 = 0 δ δ 1 1 ( ) ( ) ( ) ( ) 1 = α ε1 δ1 + β ε δ = 0 δ1 = ε1 = β ε δ + α ε δ = 0 δ = ε 1 1 the rincial strain comonents satisfy the condition ε1 ε 8

9 Structural roblem b, stress field equilibrated with the body and surface forces ε u comatible strain and dislacement fields 0 0 admissible condition for the stress 1 Ω f Ω δ 0 δ 0 1 δ ositiveness of fracture strain T 1 1+ δ = δ 0 Ω u PLV: let εε ee = 00, so that uu is comatible with the only fracture field δδ. It results: T T T δ= ub+ u 0 Ω Ω Ω f If the virtual work of the loading forces for dislacements comatible with ure fracture strains is not ositive, the loading condition is admissible. 9

10 Some consequence of the NTM model if the load is admissible, a solution exists the solution is unique in terms of stresses the solution is not unique in terms of dislacements the material model is nonlinear elastic monotone unloading follows the same loading ath of the monotone loading the dissiation is always zero internal friction angle = π/ τ 10

11 The roblem does not admit solution The roblem admit solution = = Uniqueness of the solution in terms of stresses NOT unique in terms of dislacements 11

12 A secial roblem solution fracture fracture 1

13 solution N M 1 N = h h h M = N = 6 1 M h χ = = EI 1 EI v v v z z 4 Eh ( 0) = 0, ( ) = 0 = ( ) h v '' = χ = 1 EI h 4 EI v= z + cz 1 + c Note that when h tends to zero then the deflection tends to infinity fracture 13

Elio Sacco. Dipartimento di Ingegneria Civile e Meccanica Università di Cassino e LM

Elio Sacco. Dipartimento di Ingegneria Civile e Meccanica Università di Cassino e LM Elio Sacco Diartimento di Ingegneria Civile e Meccanica Università di Cassino e LM 1D erect elastolastic resonse OA elastic resonse AB nonlinear resonse, constant stress BC elastic nloading CB elastic