Symmetries and invariant solutions of mathematical models of plastic ow during linear friction welding

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1 Symmetries and invariant solutions of mathematical models of plastic ow during linear friction welding Artur Araslanov Ufa State Aviation Technical University, Ufa, Russia. Laboratory of Group Analysis of Mathematical Models in Natural and Engineereing Sciences, Ufa, Russia

2 Introduction Linear friction welding (LFW) is a type of welding in which the heating and melting of the metal is done by the work of friction forces during a linear oscillating motion of welded samples relative to each other.

3 Stages of linear friction welding

4 The problem The problem is to nd deformation velocities of the plastic layer during the 3rd phase of linear friction welding (LFW) during which the metal in a contact zone starts to melt and goes into plastic state.

5 Main equations continuity equation motion equation W = 0, (1) ρ W t + ρ( W ) W = p + S + ρf, (2) heat transfer equation cρt t + ( W )T = (λ T ) + Φ, (3) where W velocity of plastic ow of material, ρ density, p pressure, S stress deviator tensor, F mass forces vector, t time, T temperature, c = c(t ) heat capacity, λ = λ(t ) thermal conductivity coecient, Φ dissipative function.

6 Supporting equations The relation between stress deviator tensor and strain rate deviator tensor is S ij = 2σ u 3 ε u E ij, (4) where σ u = 3 2 S ijs ij (5) stress deviator tensor's intensity, 2 ε u = 3 E ije ij (6) strain rate deviator tensor's intensity. Dissipative function is Φ = σ u ε u. (7)

7 Models of material Depending on the relation between σ u and ε u there are dierent material models, simplest of which are: 1. σ u = σ s perfectly plastic material (σ s - yield strength); 2. σ u = µ ε u viscous liquid (µ - viscosity coecient); 3. σ u = σ s + µ ε u viscoplastic material.

8 Construction of a model Consider the 2D model of 2 bars as the most simple one. Then the deformation velocity has 2 components: W = (u, v). Assume the horizontal oscillation velocity of the top bar be U(t) = U 0 (t) sin ωt. Denote the vertical (sinking) velocity with V (t). Consider only 1 semicycle of oscillations since the rest is symmetric. Introduce Cartesian coordinate system with the origin located in the geometric center of contact zone, Ox axis co-directed with the oscillation velocity and Oy axis directed normally to the plane of contact zone.

9 Assumptions and boundary conditions Due to the thinness of plastic layer we can assume (similarly to the Prandtl problem) that the velocity component v and tangential stresses S xy does not depend on x. Then we acquire next expressions for velocities: u = u 0 (y, t) + v 1 (t)x, v = v 1 (t)y. (8) where u 0 (t, y), v 1 (t) are functions to be found. Assume that on the top edge of plastic layer the constant pressure q is given.

10 2D model of LFW

11 Perfectly plastic ow. The model Equation system (1)(3) for the case of perfectly plastic material becomes next: u x + v y = 0, ( ρ[u t + U u ) ( ) x u y + v x + uu x + vu y ] = p x + 2 τ s + τ s, H x H y ( ρ[v t + V v ) ( ) y u y + v x + uv x + vv y ] = p y + 2 τ s + τ s, H y H x cρt t + (u + U)T x + (v + V )T y = (λt x ) x + (λt y ) y + τ s H, (9) where τ s = σs 3 maximal tangential stress during plastic ow, H 2E ij E ij = 2(u 2 x + v 2 y ) + (u y + v x ) 2. (10)

12 Perfectly plastic ow. The solution Using the boundary conditions we can nd the function v 1 (t): v 1 (t) = 1 t + C, (11) where C constant being calculated using boundary conditions. Using (8) we get the vertical component of deformation velocity: v(y, t) = y t + C. (12)

13 Perfectly plastic ow. The solution For the function u 0 (y, t) near the edges of plastic layer we get the equation ρ(u 0t + U v 1 u 0 + v 1 yu 0y ) = 4τ s v 2 1 u 0yy u 3 0y + g 1 (t), (13) where g 1 (t) - function being dened using boundary conditions.

14 Perfectly plastic ow. Asymptotics Having found admitted generators of transformed equation (13) and corresponding invariants we get asymptotics near the edge of plastic layer: u 0 (y, t) = 3 φ1 (t) y 1/3 + ψ 1 (t), u 0 (y, t) = y 2 2 t + C + ψ 2(t), u 0 (y, t) = 2 y ln (t + C) + ψ 3 (t). where φ 1, ψ 1, ψ 2, ψ 3 functions satisfying known equations.

15 Viscous ow. The model Equation system (1)(3) for a case of viscous material becomes next: u x + v y = 0, ρ[u t + U + uu x + vu y ] = p x + 2(µu x ) x + (µ(u y + v x )) y, ρ[v t + V + uv x + vv y ] = p y + 2(µv y ) y + (µ(u y + v x )) y, cρt t + (u + U)T x + (v + V )T y = (λt x ) x + (λt y ) y + µh 2, (14) where µ = µ(t ) viscosity coecient.

16 Viscous ow. The solution Using boundary conditions we can nd the function v 1 (t): v 1 (t) = 1 t + C, (15) where C constant being dened using boundary conditions. Using (8) we get the vertical component of deformation velocity: v(y, t) = y t + C. (16)

17 Viscous ow. The solution For a function u 0 (y, t) we get the equation g 1 (t) = ρ(u 0t + U v 1 u 0 + v 1 yu 0y ) µu 0yy, (17) where g 1 (t) - function being dened using boundary conditions.

18 Viscous ow. The solution Having found admitted generators of transformed equation (17) and corresponding invariants we can nd particular solutions: C u 0 (y, t) = ψ 1 (t) + φ 1 (t) e y2 2 dy. 2at(t + C) u 0 (y, t) = ψ 2 (t) + φ 2 (t) e y 2 4a(t+C) dy. u 0 (y, t) = ψ 3 (t) + φ 3 (t) t + C e y 2 4a(t+C) dy. u 0 (y, t) = ψ 4 (t) + u 0 (y, t) = ψ 5 (t) + ( ) ] y t [φ 4 (α 4 (t)y 2 + β 4 (t)) dy. [ ( y ) ] φ 5 (α 5 (t)y 2 + β 5 (t)) dy. t where φ 1, ψ 1, φ 2, ψ 2, φ 3, ψ 3, φ 4, ψ 4, φ 5, ψ 5 functions satisfying known equations.

19 Viscoplastic ow. The model Equation system (1)(3) for a case of visco-plastic material becomes next: u x + v y = 0, (( ρ[u t + U τs ) ) + uu x + vu y ] = p x + 2 H + µ u x + x (( τs ) ) + H + µ (u y + v x ), y (( ρ[v t + V τs ) ) + uv x + vv y ] = p y + 2 H + µ v y + y (( τs ) ) + H + µ (u y + v x ), y cρt t + (u + U)T x + (v + V )T y = (λt x ) x + (λt y ) y + ( τs ) + H + µ H 2. (18)

20 Viscoplastic ow. The solution Using boundary conditions we can nd the function v 1 (t): v 1 (t) = 1 t + C, (19) where C constant being dened using boundary conditions. Using (8) we get the vertical component of deformation velocity: v(y, t) = y t + C. (20)

21 Viscoplastic ow. The solution For a function u 0 (y, t) near the edge of plastic layer we get the equation ρ(u 0t + U v 1 u 0 + v 1 yu 0y ) = 4τ s v 2 1 u 0yy u 3 0y + µu 0yy + g 1 (t), (21) where g 1 (t) function being dened using boundary conditions.

22 Viscoplastic ow. Asymptotics Having found admitted generators of trasformed equation (21) and corresponding invariants we can get the asymptotic near the edge of the plastic layer: ( ) y u 0 (y, t) = (t + C) 2/3 2 φ 1 dy. t + C where φ 1 function satisfying the known equation.

23 Conclusion As a result of the research particular solutions of problem of nding deformation velocities of plastic layer during LFW were found for 3 dierent material models. For a viscoplastic model, which represent the real process the most accurate, only one solution on the edge was found. Acquired results can be used in further researches of linear friction welding process.

24 Bibliography Kachanov L.M. Placticity theory basics. M.: Nauka, p. Ilyushin A.A. Proceedings. Tome 2. Plasticity. M.: Fizmatlit, p. Kiiko I.A. Viscoplastic ow of materials. M.: MGU, p. Ibrahimov N.H. ABCs of group analysis. M.: Znanie, p.

25 Thank you for attention.

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