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2.001 - MECHANICS AND MATERIALS I Lecture #26 12/11/2006 Prof. Carol Livermore Energy Methods 3 Basic Ingredients of Mechanics: 1. Equilibrium 2. Constitutive Relations σ − . . . Stress-Strain F − δ Force-Deformation 3. Compatibility Dealing with geometric considerations Castigliano’s Theorem Start with work: F · ds W = Conservative Forces: EX: Gravity 1 u = mgh ⇒ Potential Energy, Path Independent h W = F · ds = mgdz = mgh 0 Note: Elastic systems are conservative. Plastic deformation is not conservative. EX: Springs Energy = 1/2kδ 2 W = δ δ P dδ = kδdδ = 0 0 Stored Energy = Work Put In U =W What if the spring were stretched a little further? 2 1 2 kδ 2 dU = P dδ dU P = dδ Complementary Energy (U ∗ ) ∗ U = δdP Note: U = U ∗ due to linearity U= U ∗ if this is non-linear What if the load were stretched a little further? 3 dU ∗ = δdP dU ∗ δ= dP Recall, for linear elastic material δ= dU dP Castigliano’s Theorem: Express complementary energy in terms of the loads Add up all the complementary energy stroed in all the pieces Find displacement of given point from derivative of U ∗ with respect to P at that point in that direction Example: Linear Spring U∗ = δdP = k2 δ2 1 P P2 dP = = = kδ 2 k 2k 2k 2 Example: Springs in series F1 = F2 = P U= Ui = i F12 F2 P2 P2 + 2 = + 2k2 2k1 2k2 2k1 Use Castigliano’s Theorem δ= P P U∗ = + dP k1 k2 4 Find δ1 using Castigliano’s Theorem Put a ﬁctitous Pf to coinside with δ1 Fx = 0 −F1 + Pf + P = 0 ⇒ F1 = Pf + P −F2 + P = 0 ⇒ F2 = P U= F2 (Pf + P )2 P2 F12 + 2 = + 2k1 2k2 2k1 2k2 δ1 = ∂U Pf + P = ∂Pf k1 Recall Pf =0, so: δ1 = δ= P k1 dU Pf + P P = + dP k1 k2 Pf = 0 So: δ= P P + k2 k1 Note: This was done without doing compatibility explicitly. Example Truss 5 U= Ui = i 2 FAB F2 + BC 2kAB 2kBC Equilibrium of B Fx = 0 FAB − FBC sin θ = 0 Fy = 0 −P − FBC cos θ = 0 P = −FBC cos θ −P cos θ = P tan θ FBC = FAB U= (P tan θ)2 + 2kAB −uB y = P cos θ 2 1 2kBC dU P tan2 θ P = + dP kAB cos2 θkBC Strain Energy Density W = F dδ u= σd This is useful in beam bending. 6 Stored Energy in A Beaam (σxx , xx ) Total Energy Stored U= dV V 0 σxx dxx Recall: For a beam: σxx = Exx For one material: σxx = −M y I Beam: U = dV σxx dxx σxx σxx dσxx = dV E V 0 2 σxx = dV 2E V V For one material: U= V M (x)2 2 y dV 2EI 2 So: U= 1 2EI M (x)2 dx For special case of one material beam. a 7 Example: M (x) = −P (L − x) L P 2 L3 1 U= P 2 (L − x)2 dx = 2EI 0 6EI δtip = dU P L3 = dP 3EI 8