1.050 Engineering Mechanics I Lecture 28 Introduction: Energy bounds in linear elasticity (cont’d) 1 1.050 – Content overview I. Dimensional analysis II. Stresses and strength III. Deformation and strain IV. Elasticity … Lecture 23: Applications and examples Lecture 24: Beam elasticity Lecture 25: Applications and examples (beam elasticity) Lecture 26: … cont’d and closure Lecture 27: Introduction: Energy bounds in linear elasticity (1D system) Lecture 28: Introduction: Energy bounds in linear elasticity (1D system), cont’d Lecture 29: Generalization to 3D, examples … V. How things fail – and how to avoid it Lectures 32 to 37 3 1.050 – Content overview I. Dimensional analysis 1. On monsters, mice and mushrooms Lectures 1-3 2. Similarity relations: Important engineering tools Sept. II. Stresses and strength 3. Stresses and equilibrium Lectures 4-15 4. Strength models (how to design structures, foundations.. against mechanical failure) Sept./Oct. III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D Lectures 16-19 structure/material? Oct. IV. Elasticity 7. Elasticity model – link stresses and deformation Lectures 20-31 8. Variational methods in elasticity Oct./Nov. V. How things fail – and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) Lectures 32-37 11. Fracture mechanics Dec. 2 Outline and goals Use concept of concept of convexity to derive conditions that specify the solutions to elasticity problems Obtain two approaches: Approach 1: Based on minimizing the potential energy Approach 2: Based on minimizing the complementary energy Last part: Combine the two approaches: Upper/lower bound 4 15Reminder: convexity of a function f (x) ∂f | (b − a) ≤ f (b) − f (a)∂xx=a secant f (x) tangent x ab Total external work vr vr Wd =ξ ⋅ Fd +ξ d ⋅ R Work done by prescribed forces Displacement s unknown Work done by prescribed displacements, force unknown 6Free energy and complementary free energy functions are convex! ii ψψ , * Total internal work Ni * iψComplementary free energy State equations Combining it… vr vr ! Wd =ξ ⋅ Fd +ξ d ⋅ R =ψ +ψ * ∂ψNi = ∂δi ∂ψ * ψi δi =∂NiFree energy δi ∑δiNi =ψ *(Ni ) +ψ (δi ) i 7 * d d−(ψ −ξ v ⋅ R r)= ! ψ −ξ v ⋅ F r Complementary Potential energy energy =: ε=: εcom pot Solution to elasticity problem −εcom =εpot 8 211Example system: 1D truss structure Rigid boundary 1 32 N1 N3N2 Rigid bar δ1 δ2 δ3 P ξ 0 9 Minimum potential energy approach (1) ξ0' P =∑Niδi ' (2) ξ0 P =∑Niδi (1)-(2) P(ξ0' −ξ0) =∑Ni (δi ' −δi ) =∑ ∂ψ (δi ' −δi ) i i ∂δi ∂ψNi = ∂ψ ' ∂ ' δi Convexity: (δi −δi ) ≤ψ (δi ) −ψ (δi )∂δi P(ξ0' −ξ0) ≤ψ (δi ') −ψ (δi ) ),()()(),( ' 0 ' pot ' 0 ' 00pot ξδεξδψξδψξδε iiii PP =−≤−= Minimum potential energy approach Consider two kinematically admissible (K.A.) displacement fields (1) Approximation δ1 ‘ =δ2 ‘ =δ3 ‘ =ξ0 ‘ to solution (K.A.) 1 N1 (2) δ1δ23 2N3N1 δ2 P δ3 δ2 =δ1 + 2 (ξ0 −δ1 )Actual solution 3ξ0 Prescribed force δ3 =δ1 + 4 (ξ0 −δ1 )3Unknown displacement 10 Minimum potential energy approach εpot(δi ,ξ0) =ψ (δi ) − Pξ0 ≤ψ (δi ') − Pξ0' =εpot(δi ',ξ0') Potential energy of actual solution is always smaller than the solution to any other displacement field Therefore, the actual solution realizes a minimum of the potential energy: εpot(δi ,ξi ) = min εpot(δi ',ξi ')'δi K.A. To find a solution, minimize the potential energy for a selected choice of kinematically admissible displacement fields 12We have not invoked the EQ conditions! 31416Minimum complementary energy approach Conditions for statically admissible (S.A.) Consider two statically admissible force fields N1 + N2 + N3 = R (1) 3N1 + N2 − N3 = 0 N1' N2' N3' Approximation N1', N2' , N3' to solution Still S.A. 1 R' N1 (2) δ23 2N3N1 δ2 R δ3 N1 N2 N3 N1 = 1/12R Actual ξ0 d N2 = 1/3Rsolution Prescribed (obtained in displacement lecture 20) N3 = 7/12R Unknown force R 13 Minimum complementary energy approach (1) ξ0 dR' =∑Ni 'δi (2) ξ0 dR =∑Niδi (1)-(2) ξ0 d (R' − R) =∑δi (Ni ' − Ni ) =∑∂ψ *(Ni )(Ni ' − Ni ) i ∂Ni ∂ψ * i δi = ∂N Convexity: ∂∂ψ Ni *(Ni ' − Ni ) ≤ψ *( iNi ') −ψ *(Ni ) ξ0 d (R' − R) ≤ψ *(Ni ') −ψ *(Ni ) ),()()(),( '' com ' 0 '* 0 * com RNRNRNRN i d i d ii εξψξψε =−≤−= Minimum complementary energy approach ),()()(),( '' com ' 0 '* 0 * com RNRNRNRN i d i d ii εξψξψε =−≤−= 15 Complementary energy of actual solution is always smaller than the solution to any other displacement field Therefore, the actual solution realizes a minimum of the complementary energy: ),(min),( '' comS.A. com ' RNRN iNi i εε = To find a solution, minimize the complementary energy for a selected choice of statically admissible force fields We have not invoked the kinematics of the problem! Combine: Upper/lower bound Recall that the solution to elasticity problem −εcom =εpot Therefore ' '−εcom(Ni , R) = max(−εcom(Ni , R )) (change sign)' Ni S.A. ε (δ ,ξ ) = min ε (δ ',ξ ')pot i i δi K.A. pot i i' ' '⎧max(−ε (N , R ))⎫ ' com i i −εcom(Ni ', R') ≤ ⎪⎪⎨ N S.A.is equal to ⎪⎪⎬ ≤ εpot(δi ',ξi ') ' '⎪ min ε (δi ,ξi ) ⎪ Lower bound ⎪⎩ δi ' K.A. pot ⎪⎭ Upper bound At the solution to the elasticity problem, the upper and lower bound coincide 417Approach to approximate/numerical solution of elasticity problems • Minimum potential energy approach:Select a guess for a displacement field; the only condition that must be satisfied is that it is kinematically admissible. In a numerical solution, this displacement field istypically a function of some unknown parameters (a1,a2,…) – Express the potential energy as a function of the unknown parameters a1,a2,… – Minimize the potential energy by finding the appropriate set of parameters(a1,a2,…) for the minimum – generally yields approximate solution – The actual solution is given by the displacement field that yields a total minimum of the potential energy. Otherwise, an approximate solution is obtained • Minimum complementary energy approach:Select a guess for a force field; the only condition that must be satisfied is that it is statically admissible. In a numerical solution, this force