CH.1. DESCRIPTION OF MOTION Multimedia Course on Continuum Mechanics Overview 1.1. Definition of the Continuous Medium 1.1.1. Concept of Continuum Lecture 1 1.1.2. Continuous Medium or Continuum 1.2. Equations of Motion 1.2.1 Configurations of the Continuous Medium 1.2.2. Material and Spatial Coordinates Lecture 2 1.2.3. Equation of Motion and Inverse Equation of Motion 1.2.4. Mathematical Restrictions 1.2.5. Example Lecture 3 1.3. Descriptions of Motion 1.3.1. Material or Lagrangian Description Lecture 4 1.3.2. Spatial or Eulerian Description 1.3.3. Example Lecture 5 2 Overview (cont’d) 1.4. Time Derivatives Lecture 6 1.4.1. Material and Local Derivatives 1.4.2. Convective Rate of Change Lecture 7 1.4.3. Example Lecture 8 1.5. Velocity and Acceleration 1.5.1. Velocity Lecture 9 1.5.2. Acceleration 1.5.3. Example 1.6. Stationarity and Uniformity 1.6.1. Stationary Properties Lecture 10 1.6.2. Uniform Properties 3 Overview (cont’d) 1.7. Trajectory or Pathline 1.7.1. Equation of the Trajectories 1.7.2. Example 1.8. Streamlines Lecture 11 1.8.1. Equation of the Streamlines 1.8.2. Trajectories and Streamlines 1.8.3. Example 1.8.4. Streamtubes 1.9. Control and Material Surfaces 1.9.1. Control Surface 1.9.2. Material Surface Lecture 12 1.9.3. Control Volume 1.9.4. Material Volume 4 1.1 Definition of the Continuous Medium Ch.1. Description of Motion 5 The Concept of Continuum Microscopic scale: Matter is made of atoms which may be grouped in molecules. Matter has gaps and spaces. Macroscopic scale: Atomic and molecular discontinuities are disregarded. Matter is assumed to be continuous. 6 Continuous Medium or Continuum Matter is studied at a macroscopic scale: it completely fills the space, there exist no gaps or empty spaces. Assumption that the medium and is made of infinite particles (of infinitesimal size) whose properties are describable by continuous functions with continuous derivatives. 7 Exceptions to the Continuous Medium Exceptions will exist where the theory will not account for all the observed properties of matter. E.g.: fatigue cracks. In occasions, continuum theory can be used in combination with empirical information or information derived from a physical theory based on the molecular nature of material. The existence of areas in which the theory is not applicable does not destroy its usefulness in other areas. 8 Continuum Mechanics Study of the mechanical behavior of a continuous medium when subjected to forces or displacements, and the subsequent effects of this medium on its environment. It divides into: General Principles: assumptions and consequences applicable to all continuous media. Constitutive Equations: define the mechanical behavior of a particular idealized material. 9 1.2 Equations of Motion Ch.1. Description of Motion 10 Material and Spatial points, Configuration A continuous medium is formed by an infinite number of particles which occupy different positions in space during their movement over time. MATERIAL POINTS: particles SPATIAL POINTS: fixed spots in space The CONFIGURATION Ωt of a continuous medium at a given time (t) is the locus of the positions occupied by the material points of the continuous medium at the given time. 11 Configurations of the Continuous Medium Ω0: non-deformed (or reference) Ω or Ωt: deformed (or present) configuration, at reference time t0. configuration, at present time t. Γ0 : non-deformed boundary. Γ or Γt : deformed boundary. X : Position vector of a particle at reference time. x : Position vector of the same particle at present time. ϕ ()X,t Γ t0 = 0 → reference time tT∈→[]0, current time Γ0 Initial, reference Ω or undeformed configuration X Ω0 x Present or deformed configuration 12 Material and Spatial Coordinates The position vector of a given particle can be expressed in: Non-deformed or Reference Configuration XX 1 X =XY = ≡ material coordinates (capital letter) [] 2 XZ3 Deformed or Present Configuration xx 1 []x =xy2 = ≡ spatial coordinates (small letter) xz3 13 Equations of Motion The motion of a given particle is described by the evolution of its spatial coordinates (or its position vector) over time. xx= ϕ (particle label,,tt)(= particle label ) xii= ϕ (particle label, ti){}∈ 1, 2, 3 φ (particle label, t) is the motion that takes the body from a reference configuration to the current one. The Canonical Form of the Equations of Motion is obtained when the “particle label” is taken as its material coordinates not x=ϕ ()() X,,tt = xX T particle label ≡≡{XXX123} X xii= ϕ (){} XX123, , Xt , i∈ 1, 2, 3 14 Inverse Equations of Motion The inverse equations of motion give the material coordinates as a function of the spatial ones. ϕ (X,t) Γ Γ0 Ω −1 X ϕ (x,t) Ω0 x not − X=ϕ 1 ( x,,tt) = Xx( ) −1 Xii= ϕ ( xxxt123, , ,) i∈{ 1, 2, 3} 15 Mathematical restrictions for φ and φ-1 defining a “physical” motion Consistency condition ϕ ( XX ,0 ) = , as X is the position vector for t=0 Continuity condition 1 ϕ ∈ C , φ is continuous with continuous derivatives Biunivocity condition φ is biunivocal to guarantee that two particles do not occupy simultaneously the same spot in space and that a particle does not occupy simultaneously more than one spot in space. Mathematically: the “Jacobian” of the motion’s equations should be different from zero: ∂ϕ (X,t) ∂ϕ J = =deti ≠ 0 ∂∂ XXj The “Jacobian” of the equations of motion should be “strictly positive” ∂ϕ (X,t) ∂ϕ J = =deti > 0 density is always positive ∂∂ XXj (to be proven) 16 Example The spatial description of the motion of a continuous medium is given by: 22tt x11= X e x= Xe −−22tt xX( ,t) ≡ x22 = X e ≡=y Ye =+=+22tt x355 X 13 t X e z Xt Ze Find the inverse equations of motion. 17 2t x11= Xe −2t xX( ,t) ≡= x22 Xe Example - Solution 2t x3=5 Xt 13 + Xe Check the mathematical restrictions: ϕ = Consistency Condition ( XX ,0 ) ? 20⋅ Xe11 X −⋅20 xX( ,0t=) = Xe22 = X = X ⋅+ 20⋅ 50X13 Xe X 3 1 Continuity Condition ϕ ∈ C ? Biunivocity Condition ? ∂∂∂xxx111 ∂∂∂XXX 123e2t 00 ∂x ∂∂∂xxx −− J=i =222=00e2t =⋅ ee 2 t 22 tt ⋅=≠ e e 2 t0 ∀ t ∂X ∂∂∂ XXX j 12350te2t ∂∂∂xxx333 ∂∂∂XXX123 ∂ϕ (X,t) Density positive ? J = > 0 2t ∂X Je= > 0 18 2t x11= Xe −2t xX( ,t) ≡= x22 Xe Example - Solution 2t x3=5 Xt 13 + Xe Calculate the inverse equations: x x= Xe22tt ⇒== X1 xe− 11 1e2t 1 x x= Xe−22tt ⇒== X2 xe 22 2e−2t 2 x− 5 Xt =+2t ⇒=31 =−−−22tt =− − 2 t − 4 t x35 Xt 13 X e X3 2t ( x355( xe 1) t) e xe3 txe 1 e −2t X11= xe −12 t Xx≡==ϕ ( ,t) X22 xe −−24tt X33= x e − 5 tx 1 e 19 1.3 Descriptions of Motion Ch.1. Description of Motion 20 Descriptions of Motion The mathematical description of the particle properties can be done in two ways: Material (Lagrangian) Description Spatial (Eulerian) Description 21 Material or Lagrangian Description The physical properties are described in terms of the material coordinates and time. It focuses on what is occurring at a fixed material point (a particle, labeled by its material coordinates) as time progresses. Normally used in solid mechanics. 22 Spatial or Eulerian Description The physical properties are described in terms of the spatial coordinates and time. It focuses on what is occurring at a fixed point in space (a spatial point labeled by its spatial coordinates) as time progresses. Normally used in fluid mechanics. 23 Example The equation of motion of a continuous medium is: x= X − Yt x= xX( ,t) ≡=+ y Xt Y z=−+ Xt Z Find the spatial description of the property whose material description is: XYZ++ ρ ( X,Y,Z,t ) = 1+ t 2 24 x= X − Yt x= xX( ,t) ≡=+ y Xt Y Example - Solution z=−+ Xt Z Check the mathematical restrictions: Consistency Condition φ ( X , 0 ) = X ? XY−⋅0 X xX( ,0t=) = XY ⋅+ 0 = Y =X XZ⋅+0 Z 1 Continuity Condition φ ∈ C ? Biunivocity Condition ? ∂∂∂xxx ∂∂∂ XYZ10−t ∂x ∂∂∂yyy J=i = =t 1 0 = 111 ⋅⋅−⋅− 1( tt) ⋅ = 1 + t2 ≠ 0 ∀ t ∂X ∂∂∂ XYZ j −t 01 ∂∂∂zzz ∂∂∂XYZ ∂φ(X,t) Any diff. Vol. must be positive J = > 0 ? ∂X Jt=+>102 25 x= X − Yt x= xX( ,t) ≡=+ y Xt Y Example - Solution z=−+ Xt Z Calculate the inverse equations: x=−⇒ X Yt X =+ x Yt y−− Y y xt ⇒ + = ⇒2 +=− ⇒ = yY− x Yt Yt Y y xt Y 2 y=+⇒= Xt Y X tt1+ t y− xt x + xt22 +− yt xt x + yt X=+=+ x Yt x t = = 1+t2 11 ++ tt 22 22 x+ yt z + zt ++ xt yt z=−+ Xt Z ⇒ Z =+=+ z Xt z t = 11++tt22 x+ yt X = + 2 1 t y− xt ≡==ϕ −1 Xx( ,tY) 2 1+ t z+ zt22 ++ xt yt Z = 1+ t 2 26 Example - Solution Calculate the property in its spatial description: x+ yt X = + 2 1 t y− xt ≡==ϕ −1 Xx( ,tY) 2 1+ t z+ zt22 ++ xt yt Z = 1+ t 2 x+ yt y − xt z + zt22 ++ xt yt ++ XYZ+ + 11++tt22 1 + t 2 x++ y yt + yt22 ++ z zt ρ = = = ( X,Y,Z,t ) 22 2 11++tt (1+ t 2 ) 22 XYZ++ xy+(11 ++ tt) + z( + t) ρρ=⇒= ( X,Y,Z,t) 2 2 ( x, y,z,t) 1+ t (1+ t 2 ) 27 1.4 Time Derivatives Ch.1. Description of Motion 28 Material and Local Derivatives The time derivative of a given property can be defined based on the: Material Description Γ(X,t) TOTAL or MATERIAL DERIVATIVE Variation of the property w.r.t.