MHA042 - Material mechanics: Duggafrågor

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MHA042 - Material mechanics: Duggafrågor 1) For a static uniaxial bar problem at isothermal (Θ const.) conditions, state principle of energy conservation (first law of thermodynamics). On the basis of this relationship, derive the relation e Σ Ε, where e is the internal energy density, for the considered bar problem.

2 theory_questions_2008.nb Additional questions for the written exam 1) a) Show, by careful argumentation, that a viscous dissipative material based on the rate equation k f Ε, k responds like an elastic material for very rapid loading represented by the step function (Ε ). b) On the basis of the first discussion in a), what can one say about the model response of the viscoelastic Maxwell model? Response: Given the evolution law k f Ε, k, where f Ε, k contains damping characteristics, we may always integrate the internal variable as t k t f Ε, k t 0 The linearized response pertinent to a step loading Ε Ε 0 H S t t then becomes dk t f Ε, k t 0 as t 0 if f Ε, k is bounded Hence, for a step load the Ε Ε 0 H S t t, dk 0 at t t, whereby the response becomes (initially) elastic, e.g. for a Maxwell type model we obtain dσ dε fdt dε dε Ε 0 S dt Ε 0 at t t Ε 0. In the context of Maxwell visco-elasticity with a step load we have Σ k f Ε, k Μ Σ dt dk Ε 0 H S t t k 0 as dt 0 t bounded Σ Ε k, Ε Ε 0 H S t t Ε 0 H S t t k Response: The viscoelastic Maxwell model is characterized by the constitutive equations: Σ k f Ε, k Μ ; Σ Ε k k Μ dk Ε k dt Ε k Ε k The stress response for the step load Ε Ε 0 H S t t then becomes (for t 0) dσ dε fdt Σ Ε 0 Ε 0 k t with k Ε 0 k 0 Change of time scale s Βt, i.e. s Β, leads to t

theory_questions_2008.nb 3 dk Ε k dt t Ε k 1 Β ds The response depends on the time scale Β, whereby it is rate dependent! Let us also consider the strain controlled situation corresponding to monotonic loading Ε a t with Ε a 0. We then find that dk a t k dt dσ The solution is obtained as a dt a t k dt Σ a t t a t k t with k a t k 0 Σ 1 t a t t 2) Consider the class of non-viscous materials governed by the rate equation k f Ε, k Ε. a) What are the practical consequences of non-viscous (or rate independent response)? What can one say about the response at very rapid loading? b) On the basis of the of the discussion in a), what can one say about the model response of the elastic-plastic linear isotropic hardening (Maxwell) model? Response: It may be noted that the increment dk depends only on the strain increment dε independent of the time scale, i.e. dk dt f Ε, k dε dt dk f Ε, k dε Hence, if we would have chosen another time scale say s Βt, i.e. s Β 0 we obtain the same result, i.e. dk ds dε ds f Ε, k ds dt ds dt dk dε ds dk dε f Ε, k 0 f Ε, k dk f Ε, k dε ds ds dt ds ds Hence, the response is rate-independent, or indifferent to the change of clock. At rapid loading defined by the step Ε Ε 0 H S t t one obtains the instantaneous response k f Ε, k Ε f Ε t, k t Ε0 S t t dk f Ε t, k t Ε 0 S t t dt f Ε t, k t Ε 0

4 theory_questions_2008.nb Response: The constitutive equations of the elastoplastic linear hardening model are characterized by: Flow rules: State equations: p Ε Λ Σ k Λ Loading condition: Ν Σ Λ Ν Σ Ε Ε p ; Σ Ε Ε p Ε Λ Ν Κ H k; Κ H k Λ H Λ 0, Σ, Κ 0, Λ 0 In case of continued loading for stress state {Σ, Κ} that subsequent stress states has to satisfy the yield condition: Σ dσ, Κ dκ Σ, Κ 0 dσ Σ dκ 0 0 consistency condition The consistency condition could also the obtained from the linearized version of Λ 0 leading to Λ Λ Λ Λ 0 0 0 Hence, one obtains Ν Σ Κ Ν Ε Λ Ν leading to (bilinear) elastoplastic response Λ H Ν Ε Λ ΝΝ H h H 0 2 0 Λ 1 h Ν Ε 1 h Ν Ε Σ ep Ε with ep H H L : Ν Ε 0 U : Ν Ε 0 In plastic loading we have Ν Ε 0 Σ Ε Λ Ν Ε h Ε Ν 2 h = + H > 0 is named the "plastic modulus" 2 h Ε h H Ε ep Ε We thus find that Σ ep Ε whereby the response of the considered elastic-plastic is rate-independent, cf. question 16! 3) On the basis of the of the discussion in 1) what can one say about the model response of the visco-plastic linear isotropic hardening model? Consider the response in a first case when the relaxation time is finite and in a second time when 0. Response: The constitutive equations of the visco - plastic linear hardening model of Bingham type are characterized by : Flow rules:

theory_questions_2008.nb 5 State equations: p Ε Λ Σ k Λ Ν Σ Λ Ν Σ Ε Ε p ; Σ Ε Λ Ν Κ H k; Κ Λ H Λ 1 Η 1 1) Consider the relaxation time as finite Σ Ε Λ Ν Ε 1 0 dt Ν dσ dε which may be integrated for the step load Ε Ε 0 H S t t (for t 0) as Σ Ε 0 1 0 t Ν Ν t Ε 0 1 Σ Σ y Κ t 0 t We thus conclude that the model response consists of an initial elastic response Ε 0 followed by stress relaxation. Also note that dσ dε dt Ν dε 1 Β ds Ν Response depends on the time scale whereby it is rate-dependent. 2) Consider the limiting case 0 Σ Ε Λ Ν Κ Λ H The condition 0 may be formulated as Lim 0 t Λ 0 We obtain the loading conditions pertinent to elastoplastic behavior, i.e. 0, Λ 0, Λ 0 and 0 Response is identical to that of linear isotropic hardening, i.e. one obtains the linearized response Λ 1 h Ν Ε 1 h Ν Ε Σ ep Ε with ep H H L : Ν Ε 0 U : Ν Ε 0 We thus find that the response becomes rate independent when 0. 4) Define the difference between the hardening modulus, H and the elastic-plastic tangent stiffness modulus, ep. Illustrate in a stress-strain response curve. Derive ep for the case of linear mixed hardening elastoplasticity. Response: The constitutive equations of the elastoplastic mixed linear hardening model are characterized by: Flow rules:

6 theory_questions_2008.nb p Ε Λ Σ Λ Ν; Ν Sign Σred k Λ Λ a Λ Α -Λ Ν Constitutive (rate) equations: Loading conditions Combination yields Σ Ε - Ε p Ε - Λ Ν Κ r H k r H Λ Α 1 - r H a 1 - r H Λ Ν Λ 0, Σ, Α, Κ 0, Λ 0, linearized loading : 0 Σ Σ Κ Α Α Ν Σ Κ Ν Α Ν Ε - Λ Ν r H Λ Ν 1 - r H Λ Ν Ν Ε - Ν 2 r H 1 - r H Ν 2 0 leading to the plastic multiplier as Λ Ν Ε with h H h The tangent relation is (just like in isotropic hardening plasticity) given by: Σ Ε - Λ Ν Ε - Ε h ep Ε with ep H h L : Ν Ε 0 : Ν Ε 0 5) In the derivation of plasticity theory, it is assumed that h def H 0 always holds. Why is this a necessary condition? Hint: We would like the value of Ν Ε to be the quantity that uniquely determines whether the change of state represents plastic loading (L) or elastic unloading (U). Response: From the linear hardening as well as for the kinematic hardening elastoplastic response we have that

theory_questions_2008.nb 7 Λ 1 h Ν Ε Σ ep Ε with ep H H L : Ν Ε 0 U : Ν Ε 0 In this development, we must have h > 0 to maintain control of the loading conditions L : Ν Ε 0 and U : Ν Ε 0. 6) Formulate the equivalence between the weak and strong representations of equilibrium for a solid subjected to quasistatic forces. 7) stablish the free energy and derive the stress response for elastic material behavior when the free energy is defined by Ψ Ε. stablish also elastic modulus stiffness tensor and the stress response at the restriction to isotropic elasticity specified in terms of the shear modulus G bulk modulus K. 8) Characterize the generic format of the constitutive equations of associated isotropic linear hardening plasticity for the multiaxial situation. Discuss the resulting continuum tangent stiffness behavior along with the condition for plastic loading. 9) Derive the linearized response (or algorithmic tangential behavior) in plastic loading of the integrated constitutive response of the "J 2 -hardening" plasticity model (involving isotropic hardening).

8 theory_questions_2008.nb 10) (a) Show that Σ and its deviator Σ dev have the same eigendirections and that the principal values are related as