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1.050 Engineering Mechanics I Review session 1 1.050 – 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. Fracture mechanics 11. Plasticity (permanent deformation) Lectures 1-3 Sept. Lectures 4-15 Sept./Oct. Lectures 16-19 Oct. Lectures 20-32 Oct./Nov. Lectures 33-37 Dec. 2 1Notes regarding final exam • Please contact me or stop by at any time for any questions • The final will be comprehensive and cover all material discussed in 1.050. Note that the last two p-sets will be important for the final. – To get an idea about the style of the final, work out old finals and the practicefinal – There will be 2-3 problems with several questions each (e.g. beam problem/truss problem, continuum problem) – We will post old final exams from 2005 and 2006 today – We will post an additional, new practice final exam on or around Wednesdaynext week – Another list of variables and concepts will be posted next week – Stay calm, read carefully, and practice time management 3 Stress, strain and elasticity -concepts 4 25 Overview: 3D linear elasticity 0div =+ g rρσ )(xrσStress tensor Basis: Physical laws (Newton’s laws) BCs on boundary of domain Ω nnT d T d rrr r ⋅=Ω∂σ)(: :Ω nnT rrr ⋅= σ)( jiij σσ = )(xrεStrain tensor Basis: Geometry BCs on boundary of domain Ω ξξξ rr r =Ω∂d d : Linear deformation theory 1Grad <<ξ r ( )( )Tξξε rr gradgrad2 1 +=Statically admissible (S.A.)Kinematically admissible (K.A.) Elasticity εσ :c= klijklij c εσ = εεσ GGK v 21 3 2 +⎟ ⎠ ⎞⎜ ⎝ ⎛ −= Basis: Thermodynamics Isotropic solid ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ = i j j i ij xx ξξε 2 1 Isotropic elasticity σ =⎜⎛ K − 2 G ⎟⎞εv1+ 2Gε =⎜⎛ K − 2 G ⎟⎞(ε11 +ε22 +ε33 )1+ 2Gε ⎝ 3 ⎠ ⎝ 3 ⎠ σ σ σ 11 = ⎜⎛ K − 2 G⎟⎞(ε11 +ε22 +ε33 )+ 2Gε11⎝ 3 ⎠ = ⎛⎜ K − 2 G⎞⎟(ε +ε +ε )+ 2Gε ⎝ 3 ⎠ Linear isotropic elasticity 22 11 22 33 22 33 = ⎜⎝⎛ K − 23 G ⎟⎠⎞(ε11 +ε22 +ε33 )+ 2Gε33 Written out for individual σ12 = 2Gε12 stress tensor coefficients σ23 = 2Gε23 σ13 = 2Gε13 Linear isotropic elasticity σ11 = ⎜⎛ K + 4 G ⎟⎞ε11 + ⎜⎛ K − 2 G ⎟⎞ε22 + ⎜⎛ K − 2 G ⎟⎞ε33 Written out for individual ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ σ 22 = ⎜⎛ K − 2 G ⎟⎞ε11 + ⎜⎛ K + 4 G ⎟⎞ε22 + ⎜⎛ K − 2 G ⎟⎞ε33 stress tensor coefficients, ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ collect terms that multiply σ 33 = ⎛⎜ K − 2 G ⎞⎟ε11 + ⎛⎜ K − 2 G ⎞⎟ε22 + ⎛⎜ K + 4 G ⎞⎟ε33 strain tensor coefficients ⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ 4σ12 = 2Gε12 c1111 = c2222 = c3333 = K + G3 σ 23 = 2Gε23 2 c1122 = c1133 = c2233 = K − G σ13 = 2Gε13 3 6 c1212 = c2323 = c1313 = 2G 3Variable Definition Notes & comments ν GK GK + − = 3 23 2 1νxxzzyy νεεε = −= Poisson’s ratio (lateral contraction under uniaxial tension) E GK KGE + = 3 9 xxxx Eεσ = Young’s modulus (relates stresses and strains under uniaxial tension) x z F F Æ σ= F/A x Æ ε= x/L 7 Uniaxial beam deformation Solving problems with strength approach Use conditions for S.A. plus strength criterion (S.C.) 8 4Variable Definition Notes & comments Two pillars of stress-strength approach At any point, must be: (1) Statically admissible (S.A.) and (2) Strength compatible (S.C.) σ 9 • Equilibrium conditions “only” specify statically admissible stress field, without worrying about if the stresses can actually be sustained by the material – S.A. From EQ condition for a REV we can integrate up (upscale) to the structural scale Examples: Many integrations in homework and in class; Hoover dam etc. • Strength compatibility adds the condition that in addition to S.A., the stress field must be compatible with the strength capacity of the material – S.C. In other words, at no point in the domain can the stress vector exceed the strength capacity of the materialExamples: Sand pile, foundation etc. – Mohr circle Variable Definition Notes & comments DS Strength domain for beams Moment capacity for beams For rectangular cross-section b,h N = N0 = bhσ0N = N x limx Strength capacity for beams lim 0 M Ny xf (My , Nx ) = + −1 ≤ 0M0 Nx 2M N⎛ ⎞y xf (My , Nx ) = + ⎜⎜ ⎟⎟ −1 ≤ 0M0 ⎝ Nx ⎠ M-N interaction (linear) f (My , Nx ) ≤ 0 M-N interaction (actual); convexity 10 1 1 5Variable Definition Notes & comments Safe strength domain : load bearing capacity of i-th load case Linear combination is safe (convexity) 11 Mohr circle Display 3D strain tensor in 2Dprojection – enables us to ‘see’ largest shear stresses, largest normal stresses… Thereby facilitates application ofstrength criterion 12 6Variable Definition Notes & comments Strength domain (general Dk definition) Equivalent to condition for S.C. cDk,Tresca Tresca criterion Max. shear stress ∀n v : Dk ,Tension−cutoff Max. tensile stress c 0( ≤Tf ) −= cσv Tension cutoff criterion 13 Variable Definition Notes & comments ,Mohr Coulomb−Dk =µ σ τ c cohesion c=0 dry sand Max. shear stress function of σ Mohr-Coulomb σ τ 14 Angle of repose 7Variable Definition Notes & comments S = ∫ S dSS Cross-sectional area I = ∫ S z dSI 2 Second order area moment EI y z y EI dx dEIM ϑξ == − 2 02 Beam bending stiffness (relates bending moment and curvature) ES f dx d xx = −2 02ξ Governing differential equation, axial forces EI f dx d zz = 4 04ξ Governing differential equation, shear forces • Step 1: Write down BCs (stress BCs and displacement BCs), analyze the problem to be solved (read carefully!) • Step 2: Write governing equations for • Step 3: Solve governing equations (e.g. byintegration), results in expression with unknownintegration constants • Step 4: Apply BCs (determine integration constants) ..., xz ξξ 15 Solution procedure to solve beam elasticity problems Energy approach Approximate solution or


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