1.050 Engineering Mechanics Lecture 22: Isotropic elasticity1.050 – Content overview 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) 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.1.050 – Content overview I. Dimensional analysis II. Stresses and strength III. Deformation and strain IV. Elasticity Lecture 20: Introduction to elasticity (thermodynamics) Lecture 21: Generalization to 3D continuum elasticity Lecture 22: Special case: isotropic elasticity Lecture 23: Applications and examples … V. How things fail – and how to avoid itImportant concepts: Isotropic elasticity • Isotropic elasticity = elastic properties do not depend on direction • In terms of the free energy change, this means that the change ofthe free energy does not depend on the direction of deformation • Rather, it depends on quantities that are independent on thedirection of deformation (i.e., independent of coordinate system) • Idea: Use invariants of strain tensor to calculate free energy change – Volume change – Shape change (shear deformation) • Note: Invariants are defined as properties of strain tensor that areindependent of coordinate system (C.S.)Important mathematical tools Trace of a tensor ()= :1 =ε11 +ε22 +ε33 = dΩ− dΩ Relates to the chain of tr ε ε d 0 volume of REV dΩ Independent of C.S. –0 trace of a tensor is an invariant ε= 1 (ε:εT )= 1 ∑∑εij 2 ‘Magnitude’ of a tensor (2nd order norm)2 2 i j Note: Analogy to the ‘magnitude’ of a tensor is the norm of a first order tensor (=vector), that is, its lengthOverview: Approach Step 1: Calculate change in volume εv =tr(ε)=ε:1 Step 2: Calculate magnitude of angle change Define strain deviator tensor = tensor that describes deformation without the volume change (trace of strain deviator tensor is zero!) =⎜⎛ε−1 tr()⎟⎞ tr e =ε11 +ε22 +ε33 −1 (3(11 +ε22 +ε33 =e ε1 () ε)) 0 ⎝ 3 ⎠ 3 εd = 2 e = 21 (e : eT )= 21 ∑∑eij 2 2 2 i j Step 3: Define two coefficients to link energy change with deformation (“spring model”): Ψ = 2 1 2 vKε+2 1 2 dGε Bulk modulus Shear modulusNote The approach that the free energy under deformation depends only on volume change and overall angle change is not derived from physical principles Rather, it is an assumption, which is made to ‘model’ the behavior of a solid (modeling is finding a mathematical representation of a physical phenomenon) Generally, models must be validated, for instance through experiments Alternative approach: Calculation of from ‘first principles’ – by explicitly considering the atomistic scale of atomic, molecular etc. interactions Spring 2008: 1.021J Introduction to Modeling and Simulation (Buehler, Radovitzky, Marzari) – continuum methods, particle methods, quantum mechanicsStress-strain relation Total stress tensor = sum of contribution from volume change and contribution from shape change: σ=σ +σ v d ∂Ψ vσ= v ∂ε σ =∂Ψd d ∂ε Next step: Carry out differentiationsStress-strain relation Total stress tensor = sum of contribution from volume change and contribution from shape change: σ=σ+σ σ=∂Ψv σ =∂Ψd v dv d∂ε ∂ε 1. Calculation of σ v v v vΨv = 1 Kεv 2 σ v =∂Ψ =∂Ψ : ∂ε = Kεv12 ∂ε ∂ε∂εv ∂Ψ ∂εv ∂(tr(ε)) ∂(ε :1)v = Kε == = 1 v∂ε ∂ε∂ε ∂ε vStress-strain relation 2. Calculation of σd Ψd = 1 Gεd 2 2 d dσ = ∂∂Ψε = ∂∂Ψ e : ∂∂ε e = 2Ge : ⎛⎝⎜1− 131⊗1⎞⎠⎟ = 2Ge − 13 (e :1)⊗1d =0 since: ∂Ψ 1 ∂(2e : eT )∂e 1 tr(e)= 0 d = G = 2Ge = 1− 1⊗1 ∂e 2 ∂e ∂ε 3 Note (definition of εd ): tensor product Note (definition of e ): εd = 2 e = 2 12 (e : eT ) e =ε− 13εv1 =ε− 13(ε:1) ⊗1Complete stress-strain relation 3. Putting it all together: σ=σ +σ v d σ=Kεv1+ 2Ge =Kεv1+ 2G⎛⎜ε−1εv1⎞⎟ ⎝ 3 ⎠ e = ⎛⎜ε−1 tr()⎟⎞ε1 ⎝ 3 ⎠ Deviatoric part of the strain tensor σ= ⎛⎜K −2 G ⎞⎟εv1+ 2Gε Reorganized…⎝ 3 ⎠Complete stress-strain relation σ= ⎝⎜⎛ K − 23 G ⎠⎟⎞εv1+ 2Gε= ⎝⎜⎛ K − 23 G ⎠⎟⎞(ε11 +ε22 +ε33 )1+ 2Gε Writing it out in coefficient form: σ11 = ⎜⎛ K − 2 G ⎟⎞(ε11 +ε22 +ε33 )+ 2Gε11 ⎝ 3 ⎠ σ22 = ⎜⎛ K − 2 G ⎟⎞(ε11 +ε22 +ε33 )+ 2Gε22 ⎝ 3 ⎠ σ33 = ⎜⎛ K − 2 G ⎟⎞(ε11 +ε22 +ε33 )+ 2Gε33 ⎝ 3 ⎠ σ12 = 2Gε12 σ23 = 2Gε23 σ13 = 2Gε13Complete stress-strain relation Rewrite by collecting terms multiplying εii σ11 = ⎜⎛ K + 4 G ⎟⎞ε11 + ⎜⎛ K − 2 G ⎟⎞ε22 + ⎜⎛ K − 2 G ⎟⎞ε33 (1)⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ σ22 = ⎜⎛ K − 2 G ⎟⎞ε11 + ⎜⎛ K + 4 G ⎟⎞ε22 + ⎜⎛ K − 2 G ⎟⎞ε33 (2)⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ σ33 = ⎜⎛ K − 2 G ⎟⎞ε11 + ⎜⎛ K − 2 G ⎟⎞ε22 +⎜⎛ K + 4 G ⎟⎞ε33 (3) ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ … collecting terms multiplying ε12,ε23,ε13 σ12 = 2Gε12 σ23 = 2Gε23 σ13 = 2Gε13Complete stress-strain relation ⎛ 4 ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞σ11 = ⎜⎝ K + 3 G ⎟⎠ε11 + ⎜⎝ K − 3 G ⎟⎠ε22 + ⎜⎝ K − 3 G ⎟⎠ε33 (1) 4 c1111 = K + G3 2 c1122 = K − G = c11333Complete stress-strain relation ⎛ 2 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞σ22 = ⎜⎝ K − 3 G ⎟⎠ε11 + ⎜⎝ K + 3 G ⎟⎠ε22 + ⎜⎝ K − 3 G ⎟⎠ε33 (2) 2 c2211 = K − G = c2233 3 4 c2222 = K + G3Complete stress-strain relation σ33 = ⎜⎛ K − 2 G ⎟⎞ε11 + ⎜⎛ K − 2 G ⎟⎞ε22 +⎜⎛ K + 4 G ⎟⎞ε33 (3)⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ 2 c3311 = K − G = c3322 3 4 c3333 = K + G3Complete stress-strain relation σ12 = 2Gε12 σ23 = 2Gε23 σ13 = 2Gε13 c1212 = 2Gc2323 = 2Gc1313 = 2G All other cijkl are zeroSummary: Expression of elasticity tensor 4 c1111 = c2222 = c3333 = K + G3 2 c1122 = c1133 = c2233 = K − G3 c1212 = c2323 = c1313 =