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1.050 – Content overview I. Dimensional analysis 1. On monsters, mice and mushrooms Lectures 1-31.050 Engineering Mechanics I 2. Similarity relations: Important engineering tools Sept. II. Stresses and strength 3. Stresses and equilibrium Lectures 4-154. Strength models (how to design structures,foundations.. against mechanical failure) Sept./Oct. Lecture 32 III. Deformation and strain Energy bounds in beam structures (cont’d) -5. How strain gages work? How to solve problems 6. How to measure deformation in a 3D Lectures 16-19 structure/material? Oct. IV. Elasticity 7. Elasticity model – link stresses and deformation Lectures 20-328. Variational methods in elasticity Oct./Nov. V. How things fail – and how to avoid it 9. Elastic instabilities 10. Plasticity (permanent deformation) Lectures 33-37 1 211. Fracture mechanics Dec. 1.050 – Content overview Review: 3D isotropic elasticity I. Dimensional analysis ⎧max(−εcom(σ '))⎫ II. Stresses and strength σ 'S.A. r III. Deformation and strain −εcom(σ ') ≤ ⎨⎪⎪is equal to ⎬⎪⎪≤ εpot(ξ ') IV. Elasticity σ 'S.A. ⎪ rmin εpot(ξ r ') ⎪ξ r ' K.A. … ⎪⎩ ξ ' K.A. ⎪⎭Lecture 27: Introduction: Energy bounds in linear elasticity (1D system) Lecture 28: Introduction: Energy bounds in linear elasticity (1D system), cont’d Lecture 29: 1D examples Lower bound Solution Upper bound Lecture 30: Generalization to 3D Complementary energy Potential energyLecture 31: Energy bounds in beam structures Lecture 32: Energy bounds in beam structures (cont’d): How to solve problems approach approach V. How things fail – and how to avoid it r Lecture 33 (Mon): Buckling (loss of convexity) ε (σ ') =ψ *(σ ') − W *(T ')Lecture 34 (Wed): Fracture mechanics I (and surprise!) com Lecture 35 (Fri): Fracture mechanics II r r Lecture 36 (Mon): Plastic yield εpot(ξ ') =ψ (ε ') − W (ξ ')Lecture 37 (Wed): Wrap-up plastic yield and closure 3 4 17Beam structures (2D) Complementary free energy ψ* = ⎢⎡1 N 2 +1 My 2 ⎥⎤dx∫ 2 ES 2 EI x=0..l ⎣ ⎦ Free energy ⎡1 021 02 ⎤ψ= ES() ()dx∫⎢εxx + EI ϑy ⎥x=0..l ⎣2 2 ⎦ Note: For 2D, the only contributions are axial forces & moments and axial strains and curvatures (general 3D case see manuscript page 263 and following) 5 Clapeyron’s formulas ψ=ψ* = 1(W * +W )2 εpot = 1(W * −W )2 εcom = 1(W −W *)2 Significance: Calculate solution potential/complementary energy (“target”) from BCs Beam structures External work by prescribed displacements * rd r d d d dW =∑[ξ(xi ) ⋅ R +ωy (xi )My,R ]=∑[ξx (xi )Rx +ξz (xi )Rz +ωy (xi )My,R ] i i External work by prescribed force densities/forces/moments rr rr0 d 0 d dW =∫ξ ⋅ f (x)dx +∑[ξ ⋅ F (xi ) +ωyM y (xi )] x=0..li 0 d 0 d 0 d 0 d d= ∫[ξx fx (xi ) +ξz fz (xi )]dx +∑[ξx Fx (xi ) +ξz Fz (xi ) +ωyM y (xi )] x=0..li 6 Beam elasticity ⎧ max (−ε (F r ', M '))⎫ N ',My ' S.A. com S y −εcom(F r S ', My ') ≤⎨⎪⎪is equal to ⎬⎪⎪≤εpot(r ξ r ',ωy ') N ',M y 'S.A. ⎪⎪ min ε (ξ v ',ω') ⎪⎪ ξ'K.A. ⎩ξ r ' K.A. pot y ⎭ Lower bound Solution Upper bound r Complementary energy FS', My ' Potential energy approach that provide approach “Stress approach” absolute max “Displacement of −εcom approach) rWork with unknown but ξ',ωy ' S.A. moments and Work with unknown forces that provide but K.A. displacementsabsolute min 8 of εpot 210Step-by-step solution approach Example Use complementary energy approach! • Step 1: Express target solution (Clapeyron’s formulas) – calculatecomplementary energy AT solution P • Step 2: Determine reaction forces and reaction moments • Step 3: Determine force and moment distribution, as a function of reaction forces and reaction moments (need My and N) l/2 δ l/2 • Step 4: Express complementary energy as function of reaction forces and δ = unknown displacement at point of reaction moments (integrate) load application • Step 5: Minimize complementary energy (take partial derivatives w.r.t. all Structure is statically indeterminate to degree 1unknown reaction forces and reaction moments and set to zero); result: set of unknown reaction forces and moments that minimize the complementary energy Can not be solved by relying on static equilibrium only (too many unknown forces, ‘hyperstatic’).• Step 6: Calculate complementary energy at the minimum (based on resultingforces and moments obtained in step 5) • Step 7: Make comparison with target solution = find solution displacement 9 Goal: Solve problem using complementary energy approach Example Example Step 1: Target solution ε= 1 Pδ Step 3: Determine force and moment distribution Concept of com 2 (as a function of hyperstatic force R’): superposition often helpful Step 2: Determine hyperstatic forces and moments (here: R’) ⎧ Pl 2x x P My (x) = ⎪⎨ 2 (1 − l ) − R' l(1 − l ) 0 ≤ x ≤ l / 2 P ⎩⎪ − R' l(1 − xl ) l /2 < x ≤ l My (x) + Note: Only need expression for N and My R’ l/2 l/2 M (x)y Hyperstatic force ⎣2 ES 2 EI 11 εcom (R') = 1 ⎛⎜⎜l3 R'2 − 5 l 3 R' P + 1 l3P2 ⎞⎟⎟12 δ Step 4: Express complementary energy 2N =0 W =0 No contributionR’ εcom =ψ * − W * =∫ ⎡⎢ 1 + 1 My 2 ⎤⎥dx − * from prescribed displacementsx=0..l ⎦ 2EI ⎝ 3 24 24 ⎠ 3Example Step 5: Find min of ε (R') ∂εcom (R') com = 0 ∂R' 1 ⎛ 2l3 R' 5 ⎞ ⎜⎜ − l3 P⎟⎟ = 0 2EI ⎝ 3 24 ⎠ 5R' = P 16 Step 6: Minimum complementary energy εcom (R' = 5 P) = 7 l3 P 16 1536EI 13 Example Step 7: Compare with target solution εcom = 1 Pδ ≤εcom (R' = 5 P) = 7 l 3P 2 16 1536EI 73δ≤ l P768EI δ= 7 l3P represents a minimum of the complementary energy 768EI Is it a global minimum, that is, the solution? 1. My’ is S.A. 2. R’ is the only hyperstatic reaction force (in other words, the only source of additional moments) 3. Therefore, the minimum is actually a global minimum, and therefore, it is the solution 14 Generalization (important) • For any homogeneous beam problem, the minimization of the complementary energy with respect to all hyperstatic forces and moments Xi ={Ri , My,R;i } yields the solution of the linear elastic beam problem: ∂∂ Xi (εcom( Xi ))= ! 0 1*(W − W )≡ minε ( X )2 Xi com i 15


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