16.21 Techniques of Structural Analysis and sig Spring 2003 Unit #4- Thermodynamics Principles First Law of Thermodynamics (K+U)=P+H where ·K: kinetic energy ·U: internal energy P. Power of external forces hear exchange per unit time 2/y at at2 pUdv, U: Internal energy density du V+/t ods
� � � 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #4 - Thermodynamics Principles First Law of Thermodynamics d dt � K + U � = P + H (1) where: • K: kinetic energy • U: internal energy • P: Power of external forces • H: hear exchange per unit time 1 ∂u ∂u 1 ∂ui ∂ui K = ρ · dV = ρ dV (2) 2 V ∂t ∂t 2 V ∂t ∂t U = ρU�dV, U�: Internal energy density (3) V � � ∂u ∂u P = f · dV + t · dS (4) V ∂t S ∂t 1
n components Replacing ti=njoji in this expression P=find+n;ojiods Using Gauss'Theorem au 0 au d duil at aui(why?) a du at a x a au at ax 2 at at Notation: Time derivative =i, Examples
˙ ˙ � � � � � � �� � In components: ∂ui ∂ui P = fi dV + ti dS (5) V ∂t S ∂t Replacing ti = njσji in this expression: ∂ui ∂ui P = fi dV + njσji dS (6) V ∂t S ∂t Using Gauss’ Theorem: ∂ui � ∂ � ∂ui � P = fi dV + σji dV ∂t V ∂xj ∂t � V ��∂σji � ∂ui ∂ ∂ui � = + fi + σji dV V ∂t ∂xj ∂t � ∂xj �� � � �� � ∂2ui ∂ ∂ui ρ ∂t2 (why?) ∂t ∂xj (7) � � ∂2ui ∂ui ∂ ∂ui � = ρ V � ∂t2 ��∂t � + σji ∂t ∂xj dV 1 ∂ �∂ui �2 ∂ σji �ji 2 ∂t ∂t ∂t Notation: Time derivatives: ∂( ) ∂t = ( ˙ ) Examples: ∂ui • ∂t = u˙i, ∂u = u ∂t ∂2ui • ∂t2 = u¨i ∂�ij • = �ij ∂t Spatial derivatives: ∂( ) = ( ),i ∂xi Examples: 2
O With this notation, the power of the external forces can be rewritten as au: au dv+/oieiidT deformation power where the"pdv"inside the first integral was included inside the time deriva- tive since it is a constant due to conservation of mass. We conclude that part of the power of the external forces goes into changing the kinetic energy of the material and the rest into deforming the material. We call the latter the deformation power and it represents the rate at which the stresses do work on the deforming material Replacing in the first law, equation(1) dt (k+u)=dt(k)+lojendv+h After canceling the kinetic energy from both sides, the first law expresses the fact that the internal energy of a deforming material can be changed either by heating or by deforming the material dt dt jiejidv+H In the isothermal case(H=0) aU 0-o)d=0 in local fo In ideal elasticity, we assume that all the work of deformation is converted into internal energy, i.e., the internal energy density is a state function of the deformation: U=UC 3
˙ ˙ ˙ ˙ � � �� � � � • ∂ ∂ σ x j j i = σji,j With this notation, the power of the external forces can be rewritten as: d 1 ∂ui ∂ui P = ρ dV + σji�jidV dt � V 2 ∂t ∂t � � V �� � (8) K deformation power where the “ρdV ” inside the first integral was included inside the time derivative since it is a constant due to conservation of mass. We conclude that part of the power of the external forces goes into changing the kinetic energy of the material and the rest into deforming the material. We call the latter the deformation power and it represents the rate at which the stresses do work on the deforming material. Replacing in the first law, equation (1): d � � d � � K + U = K + σji�jidV + H (9) dt dt V After canceling the kinetic energy from both sides, the first law expresses the fact that the internal energy of a deforming material can be changed either by heating or by deforming the material: dU dt = d dt � V ρU�dV = � V σji�˙jidV + H (10) In the isothermal case (H = 0): � � ∂ � � U ρ − σij �ij dV = 0 (11) V ∂t or, in local form: ∂U� ρ = σij �ij (12) ∂t In ideal elasticity, we assume that all the work of deformation is converted into internal energy, i.e., the internal energy density is a state function of the deformation: U = U(�ij ) (13) 3
at ae 14) Replace in first law, equation(12 PaEij-Oijei 15) 圆 i.e., the stresses derive from a potential
˙ ˙ ˙ Then: ∂U� ∂U� = �ij (14) ∂t ∂�ij Replace in first law, equation (12: ∂U� ρ �ij = σij �ij ⇒ (15) ∂�ij ∂U� ρ = σij (16) ∂�ij i.e., the stresses derive from a potential. 4