1.050 – Content overview 1.050 Engineering Mechanics I Lecture 29 Energy bounds in 1D systems Examples and applications 1.050 – Content overview I. Dimensional analysis II. Stresses and strength III. Deformation and strain I. Dimensional analysis 1. On monsters, mice and mushrooms 2. Similarity relations: Important engineering tools II. Stresses and strength 3. Stresses and equilibrium 4. Strength models (how to design structures, foundations.. against mechanical failure) III. Deformation and strain 5. How strain gages work? 6. How to measure deformation in a 3D structure/material? IV. Elasticity 7. Elasticity model – link stresses and deformation 8. Variational methods in elasticity V. How things fail – and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) 1 11. Fracture mechanics Lectures 1-3 Sept. Lectures 4-15 Sept./Oct. Lectures 16-19 Oct. Lectures 20-31 Oct./Nov. Lectures 32-37 Dec. 2 Example system: 1D truss structure Rigid boundary 1 32 IV. Elasticity … Lecture 23: Applications and examples N1 N3N2Lecture 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 Rigid bar δ1 δ2 δ3Lecture 29: 1D examples Lecture 30: Generalization to 3D P … V. How things fail – and how to avoid it Lectures 32 to 37 ξ0 3 4 11 Minimum potential energy approach Conditions for Consider two kinematically admissible (K.A.) kinematically admissible displacement fields (K.A.): Deformation must be compatible w/ (1) rigid bar Approximation δ1 ‘ =δ2 ‘ =δ3 ‘ =ξ0 ‘ to solution (K.A.) N1 (2) δ1δ23 2N3N1 δ2 P δ3 δ2 =δ1 + 2 (ξ0 −δ1 )Actual solution 3ξ 0 Prescribed force δ3 =δ1 + 4 (ξ0 −δ1 )3Unknown displacement 5 Minimum potential energy approach (1) Approximation δ1 ‘ =δ2 ‘ = ‘ δ3 ‘ =ξ0 ‘ εpot(ξ0' ) = − 1 P2 to solution 6k(K.A.) is larger than (2) δ1δ2 =δ1 + 2 (ξ0 −δ1 )Actual solution 3 δ3 =δ1 + 4 (ξ0 −δ1 )3 2 10pot 48 11)(ξε , Pk−=δ 7 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 We have not invoked the EQ conditions! 6 Minimum 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' Approximation N1', N2' to solution Still S.A. 1 R' ξ 0 d 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) N = 7/12R Unknown force R ξ 0 d 3 8 2121 Minimum complementary energy approach ),()()(),( '' com ' 0 '* 0 * com RNRNRNRN i d i d ii εξψξψε =−≤−= 9 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! Minimum complementary energy approach (1)N1' N2' Approximation N1', N2' to solution Still S.A. εcom(R') = − 1 k(ξ0 d )2 5 R' ξ 0 d (2) is larger thanN1 N2 N3 N1 = 1/12R Actual solution N2 = 1/ 3R (obtained in lecture 20) R ξ0 dN3 = 7/12R ( )2 01com 11 12(Rε ), dkN ξ−= 10 Combine: Upper/lower bound Treat this BC with complementary energy approach N1 εcom(N1) =ψ *(Ni ) −0ξ d R = 0 δ23 2N3N1 δ2 P δ3 ε (N ) = 1 (12N 2 − 2PN + P2 )ξ 0 com 1 1 14kPrescribed force Unknown displacement Find min: ∂εcom(N1) = 0 N = 1 P εcom = 11 P2 ∂N1 112 48k 11 Combine: Upper/lower bound ' '⎧max(−ε (N , R ))⎫ ' com i i ' ' ⎪⎪N S.A. ⎪⎪' '−εcom(Ni , R ) ≤⎨ is equal to ⎬≤εpot(δi ,ξi ) ' '⎪ min ε (δi ,ξi ) ⎪ Lower bound ⎪⎩ δi ' K.A. pot ⎪⎭ Upper bound ε= 11 P2 −ε = − 11 P2 com com48k 48k ε pot(ξ0,δ1) = − 11 P2 48k At the solution to the elasticity problem, the upper and lower bound coincide 3Another example Approximate solution for coupled beam-truss problem 13 Rigid boundary Step-by-step approach • Step 1: Determine K.A. displacement field (for approximation, find appropriate assumed displacement field) • Step 2: Express work balance – find εpot / εcom • Step 3: Find min of εpot / εcom • Step 4: Determine displacement field, forces etc. • Solution is approximation to actual solution 14 Minimum potential energy approach Step 1: Assume K.A. displacement field 21 32 4 ξz (x;α, β ) = β +α⎜⎛ x ⎟⎞ ⎝ 3L ⎠N1 N3 N4N2 (approximation of the actual solution…) L L L δ1 Elastic beam x δ2 δ3 δ4 δ1 δ2 δδ4P β ξ == )0(xz ξ (xz αβ +== )3L 3 ξz ξ0 15 16 418Minimum potential energy approach Displacement of the four truss members δ1 =ξz (x =0) =β δ2 =ξz (x = L) =β+α 92⎛x ⎞ z(*) δ3 =ξz (x =2L) =β+4 αξ(x;α, β) =β+α⎝⎜3L ⎠⎟ 9 δ4 =ξz (x =3L) =β+α ξ0 =δ4 17 Minimum potential energy approach Step 2: Page 215 in manuscript Total free energy of a beam: (chapter 5) 3L h / 2 b /2 10 02ψB =∫∫ ∫ E(εxx +ϑyz)dydzdx2 x=0 z=−h / 2 y=−b / 2 with: ε0 = 0 (no displacement in the x-direction) xx ϑ0 = −∂2ξz = −2α (curvature can be calculated from y ∂x29L2 the assumed displacement field) ψB = E 4α24 3L h / 2 b /2 z2dydzdx ψB (α, β) = bh3E 3 α2∫∫ ∫ 2 81L 21:= Bk162L x=0 z=−h /2 y=−b /2 “spring constant” Minimum potential energy approach ψB (α) = 1 kBα2 2 Total free energy: 12Sum of free energies of four trusses… ψi = kδi2 ψ(α, β) =ψB (α) +∑ψi (α, β) i=1..4 ψ(α, β) = 12 kBα2 +12 k⎜⎝⎜⎛β2 +⎝⎜⎛β+α 9 ⎠⎟⎞2 +⎝⎜⎛β+49 α ⎠⎟⎞2 +(β+α)2 ⎟⎠⎟⎞ External work W = F (α+β) 19 Minimum potential energy approach εpot(α, β) = 21 kBα2 +21 k⎜⎝⎜⎛β2 +⎜⎝⎛β+α 9 ⎟⎠⎞2 +⎜⎝⎛β+49 α⎟⎠⎞2