16.21 Techniques of Structural Analysis and Design Spring 2003 Section 2-Energy and variational Principles Unit #7-Concepts of work and energy Work igure 1: Work of a force on a moving particle
16.21 Techniques of Structural Analysis and Design Spring 2003 Section 2 - Energy and Variational Principles Unit #7 - Concepts of work and energy Work Figure 1: Work of a force on a moving particle 1
Work done by a force dW=f·du=fu2=|f‖ldu‖cos(fu) dw f. du · Work done by a moment: dW=M. de= M: 0 dw Extend definition to material bodies: total work is the addition of the work done on all particles by forces distributed over the volume W by forces distributed over the surface t·udS by concentrated forces W=∑f·u(x) nother classification Work done by external forces: we will assume that external forces dont change during the motion or deformation, i. e, they are independent of the displacements. This will lead to the potential character of the external work and to the definition of the potential of the erternal force as the negative of the work done by the external forces Work done by internal forces: the internal forces do depend on the deformation In general, the work done by external forces and the work done by nternal forces don't match(we saw that part of the work changes the kinetic energy of the material)
� � � • Work done by a force: dW = f · du = fiui = �f� �du� cos(fu� ) (1) � B � B WAB = dW = f · du (2) A A • Work done by a moment: dW = M · dθ = Miθi (3) � B � B WAB = dW = M · dθ (4) A A • Extend definition to material bodies: total work is the addition of the work done on all particles: – by forces distributed over the volume: W = f · udV V – by forces distributed over the surface: W = t · udS S – by concentrated forces: n W = fi · u(xi) i=1 Another classification: • Work done by external forces: we will assume that external forces don’t change during the motion or deformation, i.e., they are independent of the displacements. This will lead to the potential character of the external work and to the definition of the potential of the external forces as the negative of the work done by the external forces. • Work done by internal forces: the internal forces do depend on the deformation. In general, the work done by external forces and the work done by internal forces don’t match (we saw that part of the work changes the kinetic energy of the material). 2
收 F=g E Figure 2: Spring loaded with a constant force Example: Consider the following spring loaded with a constant force WE= Fo, F doesn't change when u goes from 0 to O Fs(u)du, Fs: force on spring kudu WE≠W Remarks 3
Figure 2: Spring loaded with a constant force Example: Consider the following spring loaded with a constant force: WE = Fδ, F doesn’t change when u goes from 0 to δ (5) = mgδ (6) WI = � δ 0 Fs(u)du, FS : force on spring (7) = � δ 0 kudu = 1 2 kδ2 (8) ⇒ WE �= WI (9) Remarks: 3
WE=WI would imply 8= 219, which contradicts equilibrium:8 rium. How can you explain this? teached the system is not in equilib- before the final displacement d is Strain energy and strain energy density Figure 3: Strain energy density Strain energy and strain energy density(see also unit on first law of thermodynamics U=UdV 10) From first law at
˙ � � • WE = WI would imply δ = 2mg k , which contradicts equilibrium: δ = mg k , • before the final displacement δ is reached the system is not in equilibrium. How can you explain this? Strain energy and strain energy density Figure 3: Strain energy density Strain energy and strain energy density (see also unit on first law of thermodynamics): U = UdV (10) V From first law: ∂U� = σij �ij ∂t 4
au Jiideii I. not necessarily linear elastic Ciiklekde 11) Complementary strain energy and complementary strain energy density CA Figure 4: Complementary strain energy density 5
˙ ˙ � � � � � � ∂U� �ij � �� � ∂�ij = σij �ij ???? ∂U� σij = ∂�ij � �ij 0 U = σijd�ij , not necessarily linear elastic Linear case: U = �ij 0 Cijkl�kld�ij = 1 Cijkl�kl�ij = 2 1 σij �ij (11) 2 Complementary strain energy and complementary strain energy density Figure 4: Complementary strain energy density V Uc = UcdV (12) � σij 0 Uc = �ijdσij (13) 5
Linear case: Eii= SiikOk, where Siik=C >Uc=U for a linear elastic material Note: In the literature, you will find the following notation used indis- tinctively o represent strain energy density U to represent complementary strain energy density U: to represent strain energy density U, Uc: to represent complementary strain energy density Example: Compute the strain energy density, strain energy, and their complementary counterparts for the linear elastic bar loaded axially shown in the figure P E∈1de1 E 6
� � � � � Linear case: �ij = Sijklσkl, where Sijkl = C−1 ijkl σij 1 Uc = Sijklσkldσij = Sijklσklσij = 0 2 1 �ijσij (14) 2 ⇒ Uc = U for a linear elastic material (15) Note: In the literature, you will find the following notation used indistinctively: U0, U� : to represent strain energy density U0 ∗ , U�c : to represent complementary strain energy density U : to represent strain energy density U∗ , Uc : to represent complementary strain energy density Example: Compute the strain energy density, strain energy, and their complementary counterparts for the linear elastic bar loaded axially shown in the figure: � �0 � −ν�0 U = σ11d�11 + σ22d�22 + . . . 0 � �0 0 1 = E�11d�11 = E� 2 2 0 0 6
From equilibrium we know: oo From the constitutive law:o=曾=是 1P2 U 2EA2 ALP. PZL 2EA2 2EA g P4 E 2 E 2EA U.dv ALP P2L 2EA22EAO! Potential Energy Capacity of the system(material body external forces)to return work II=U+V,V: potential of external loads ds-fuidv fuds-/元udv This expression applies to linear elastic materials(why?
� � � � P From equilibrium we know: σ0 = A . From the constitutive law: �0 = σ0 = P E AE � 1 P2 ⇒ U = � 2 EA2 � ALP2 P2L U = UdV = = V 2EA2 2EA � σ0 � 0 Uc = �11dσ11 + 0 0 � σ0 σ11 1 σ2 � �σ0 = 0 E dσ11 = 2E 11� 0 = � ALP2 Uc = UcdV = �22dσ22 + . . . σ2 P2 0 = = U�!! 2E 2EA2 P2L = = U!! V 2EA2 2EA Potential Energy Capacity of the system (material body + external forces) to return work Π = U + V , V : potential of external loads (16) V = − t ¯iuidS − f ¯ iuidV (17) S V Π = � V 1 2 σij �ijdV − � S t ¯iuidS − � V ¯fiuidV (18) This expression applies to linear elastic materials (why?). 7