Belytschko T. - Introduction (779635), страница 28
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The total force on the body is given byf (t ) = ∫ ρb( x,t ) dΩ+ ∫ t(x, t)dΓΩ(3.5.26)ΓThe linear momentum of the body is given byp(t ) = ∫ ρv(x,t )dΩ(3.5.27)Ωwhere ρv is the linear momentum per unit volume.3-44T. Belytschko, Continuum Mechanics, December 16, 199845Newton’s second law of motion for a continuum states that the materialtime derivative of the linear momentum equals the net force. Using (3.5.26) and(3.5.27), this givesDp=f⇒DtDDt∫ ρvdΩ= ∫ ρbdΩ + ∫ tdΓΩΩ(3.5.28)ΓWe now convert the first and third integrals in the above to obtain a single domainintegral so Eq. (3.5.1) can be applied.
Reynold’s Transport Theorem applied tothe first integral in the above givesDDtD ( ρv ) + div(v) ρv dΩ = (ρ Dv + v() ∫ Dt Dt + ρdiv(v))dΩ∫ ρv dΩ= ∫ ( DtDρΩ(3.5.29)ΩΩwhere the second equality is obtained by using the product rule of derivatives forthe first term of the integrand and rearranging terms.The term multiplying the velocity in the RHS of the above can berecognized as the continuity equation, which vanishes, givingDDt∫ ρvdΩ = ∫ ρ DvDt dΩΩ(3.5.30)ΩTo convert the last term in Eq. (3.5.28) to a domain integral, we invoke Cauchy’srelation and Gauss’s theorem in sequence, giving∫ t dΓ= ∫ n⋅σ dΓ= ∫ ∇⋅ σdΩΓΓΩor∂σ ij∫ t jdΓ = ∫ niσ ij dΓ = ∫ ∂xi dΩΓΓ(3.5.31)ΩNote that since the normal is to the left on the boundary integral, the divergence isto the left and contracts with the first index on the stress tensor.
When thedivergence operator acts on the first index of the stress tensor it is called the leftdivergence operator and is placed to the left of operand. When it acts on thesecond index, it is placed to the right and call the right divergence. Since theCauchy stress is symmetric, the left and right divergence operators have the sameeffect. However, in contrast to linear continuum mechanics, in nonlinearcontinuum mechanics it is important to become accustomed to placing thedivergence operator where it belongs because some stress tensors, such as thenominal stress, are not symmetric. When the stress is not symmetric, the left andright divergence operators lead to different results.
When Gauss’s theorem isused, the divergence on the stress tensor is on the same side as the normal inCauchy’s relation. In this book we will use the convention that the normal anddivergence are always placed on the left.Substituting (3.5.30) and (3.5.31) into (3.5.28) gives∫ ( ρ DvDt −ρb −∇⋅ σ )dΩ= 0(3.5.32)Ω3-45T. Belytschko, Continuum Mechanics, December 16, 199846Therefore, if the integrand is C -1, since (3.5.32) holds for an arbitrary domain,applying (3.5.1) yieldsρ DvDt = ∇⋅ σ + ρb ≡ divσ + ρborDv ∂σρ Dti = ∂x ji + ρbij(3.5.33)This is called the momentum equation or the equation of motion; it is also calledthe balance of linear momentum equation.
The LHS term represents the changein momentum, since it is a product of the acceleration and the density; it is alsocalled the inertial term. The first term on the RHS is the net resultant internalforce per unit volume due to divergence of the stress field.This form of the momentum equation is applicable to both Lagrangian andEulerian descriptions.
In a Lagrangian description, the dependent variables areassumed to be functions of the Lagrangian coordinates X and time t, so themomentum equation isρ( X,t )∂v(X ,t)∂t()= divσ φ−1 (x,t ),t +ρ ( X,t )b(X ,t )(3.5.34)Note that the stress must be expressed as a function of the Eulerian coordinatesthrough the motion φ −1 (X,t ) so that the spatial divergence of the stress field canbe evaluated; the total derivative of the velocity with respect to time in (3.5.33)becomes a partial derivative with respect to time when the independent variablesare changed from the Eulerian coordinates x to the Lagrangian coordinates X.In an Eulerian description, the material derivative of the velocity is writtenout by (3.2.9) and all variables are considered functions of the Euleriancoordinates.
Equation (3.5.33) becomes ∂v (x,t )ρ( x, t) ∂t +(v(x, t) ⋅grad )v(x,t ) = divσ(x,t ) + ρ (x, t )b(x ,t )(3.5.35) ∂v ∂σ jior ρ i +vi , j v j =+ ρbi ∂t ∂x jAs can be seen from the above, when the independent variables are all explicitlywritten out the equations are quite awkward, so we will usually drop theindependent variables. The independent variables are specified wherever thedependent variables are first defined, when they first appear in a section orchapter, or when they are changed. So if the independent variables are not clear,the reader should look back to where the independent variables were lastspecified.In computational fluid dynamics, the momentum equation is sometimesused without the changes made by Eqs.
(3.5.13-3.5.30). The resulting equation isD( ρv ) ∂ ( ρv )Dt ≡ ∂t +v ⋅grad ( ρv ) = divσ + ρb3-46(3.5.36)T. Belytschko, Continuum Mechanics, December 16, 199847This is called the conservative form of the momentum equation with consideredρv as one of the unknowns. Treating the equation in this form leads to betterobservance of momentum conservation.3.5.7Equilibrium Equation.
In many problems, the loads are appliedslowly and the inertial forces are very small and can be neglected. In that case,the acceleration in the momentum equation (3.5.35) can be dropped and we have∂σ ji+ ρbi = 0∂x j∇⋅ σ + ρb = 0 or(3.5.37)The above equation is called the equilibrium equation. Problems to which theequilibrium equation is applicable are often called static problems. Theequilibrium equation should be carefully distinguished from the momentumequation: equilibrium processes are static and do not include acceleration. Themomentum and equilibrium equations are tensor equations, and the tensor forms(3.5.33) and (3.5.37) represent nSD scalar equations.3.5.8Reynold's Theorem for a Density-Weighted Integrand.Equation (3.5.30) is a special case of a general result: the material time derivativeof an integral in which the integrand is a product of the density and the function fis given byDDt∫ ρf dΩ= ∫ ρ Dt dΩDfΩ(3.5.38)ΩThis holds for a tensor of any order and is a consequence of Reynold's theoremand mass conservation; thus, it can be called another form of Reynold's theorem.It can be verified by repeating the steps in Eqs.
(3.5.29) to (3.5.30) with a tensorof any order.3.5.9Conservation of Angular Momentum. The conservation ofangular momentum provides additional equations which govern the stress tensors.The integral form of the conservation of angular momentum is obtained by takingthe cross-product of each term in the corresponding linear momentum principlewith the position vector x, givingDDt∫ x × ρvdΩ = ∫ x × ρbdΩ + ∫ x × tdΓΩΩ(3.5.39)ΓWe will leave the derivation of the conditions which follow from (3.5.39) as anexercise and only state them:σ = σTorσ ij = σ ji(3.5.40)In other words, conservation of angular momentum requires that the Cauchystress be a symmetric tensor. Therefore, the Cauchy stress tensor represents 3distinct dependent variables in two-dimensional problems, 6 in three-dimensionalproblems.
The conservation of angular momentum does not result in anyadditional partial differential equations when the Cauchy stress is used.3-47T. Belytschko, Continuum Mechanics, December 16, 1998483.5.10Conservation of Energy. We consider thermomechanicalprocesses where the only sources of energy are mechanical work and heat.
Theprinciple of conservation of energy, i.e. the energy balance principle, states thatthe rate of change of total energy is equal to the work done by the body forces andsurface tractions plus the heat energy delivered to the body by the heat flux andother sources of heat. The internal energy per unit volume is denoted byρwint where w int is the internal energy per unit mass. The heat flux per unit areais denoted by a vector q, in units of power per area and the heat source per unitvolume is denoted by ρs . The conservation of energy then requires that the rateof change of the total energy in the body, which includes both internal energy andkinetic energy, equal the power of the applied forces and the energy added to thebody by heat conduction and any heat sources.The rate of change of the total energy in the body is given byD ρwint dΩ, P kin = DP tot = P int +P kin , P int = Dt∫DtΩ∫ 21 ρv⋅v dΩ(3.5.41)Ωwhere P int denotes the rate of change of internal energy and P kin the rate ofchange of the kinetic energy.
The rate of the work by the body forces in thedomain and the tractions on the surface is∫∫∫∫P ext = v ⋅ρb dΩ+ v ⋅t dΓ= vi ρbi dΩ+ vit i dΓΩΓΩ(3.5.42)ΓThe power supplied by heat sources s and the heat flux q is∫∫∫∫P heat = ρs dΩ− n⋅q dΓ = ρsdΩ − ni qidΓΩΓΩ(3.5.43)Γwhere the sign of the heat flux term is negative since positive heat flow is out ofthe body.The statement of the conservation of energy is writtenP tot = P ext + P heat(3.5.44)i.e.
the rate of change of the total energy in the body (consisting of the internaland kinetic energies) is equal to the rate of work by the external forces and rate ofwork provided by heat flux and energy sources. This is known as the first law ofthermodynamics. The disposition of the internal work depends on the material. Inan elastic material, it is stored as elastic internal energy and fully recoverableupon unloading.
In an elastic-plastic material, some of it is converted to heat,whereas some of the energy is irretrievably dissipated in changes of the internalstructure of the material.Substituting Eqs. (3.5.41) to (3.5.43) into (3.5.44) gives the full statementof the conservation of energy3-48T. Belytschko, Continuum Mechanics, December 16, 1998DDt49∫ ( ρw int + 12 ρv⋅ v )dΩ = ∫ v⋅ ρbdΩ+ ∫ v ⋅tdΓ + ∫ ρsdΩ− ∫ n⋅q dΓΩΩΓΩ(3.5.45)ΓWe will now derive the equation which emerges from the above integral statementusing the same procedure as before: we use Reynolds’s theorem to bring the totalderivative inside the integral and convert all surface integrals to domain integrals.Using Reynold’s Theorem, (3.5.38) on the first integral in Eq. (3.5.45) givesDDt∫ ( ρw)int(v⋅v ) + 12 ρv⋅ v dΩ= ∫ ρ Dw+ 12 ρ DDtDt dΩintΩΩ=∫Ω()intρ DwDvDt + ρv ⋅ Dt dΩ(3.5.46)We will use commas in the following to denote spatial derivatives.
ApplyingCauchy’s law (3.4.1) and Gauss’s theorem (3.5.12) to the traction boundaryintegrals on the RHS of (3.5.45) yields:∫ v⋅tdΓ= ∫ n⋅σ ⋅vdΓ = ∫ (vi σ ji ), j dΩ= ∫ (vi, j σ ji + vi σ ji , j )dΩΓΓΩΩ()= ∫ D jiσ ji +Wjiσ ji + viσ ji, j dΩ using (3.3.9)Ω(symmetry of σ andskew symmetry of W)= ∫ D jiσ ji +viσ ji , j dΩΩ=∫ ( D :σ + (∇⋅ σ ) ⋅v) dΩ(3.5.47)ΩInserting these results into (3.5.44) or (3.5.45), application of Gauss’s theorem tothe heat flux integral and rearrangement of terms yields∫ (ρ DwDtintΩ())− D :σ +∇⋅ q − ρs + v⋅ ρ DvDt −∇⋅ σ − ρb dΩ = 0(3.5.48)The last term in the integral can be recognized as the momentum equation, Eq.(3.5.33), so it vanishes.
Then invoking the arbitrariness of the domain gives:intρ DwDt = D :σ −∇⋅ q + ρs(3.5.49)When the heat flux and heat sources vanish, i.e. in a purely mechanicalprocess, the energy equation becomesintρ DwDt = D :σ = σ: D = σ ij Dij(3.5.50)3-49T. Belytschko, Continuum Mechanics, December 16, 199850The above defines the rate of internal energy or internal power in terms of themeasures of stress and strain. It shows that the internal power is given by thecontraction of the rate-of-deformation and the Cauchy stress.
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