1.050 Engineering Mechanics Lecture 24: Beam elasticity – derivation of governing equation 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. 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. 2 11.050 – 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 Lecture 24: Beam elasticity Lecture 25: Applications and examples (beam elasticity) Lecture 26: … cont’d and closure … V. How things fail – and how to avoid it 3 Goal of this lecture • Derive differential equations that can be solved to determine stress, strain and displacement fields in beam • Consider 2D beam geometry: z x + boundary conditions (force, clamped, moments…) • Approach: Utilize beam stress model, strain model for beams and combine with isotropic elasticity 4 2Stress ()=⎜⎛σ 0 xx 00 σxz ⎟⎞ σij ⎜ 0 ⎟ ⎜σ 0 0 ⎟ ⎝ xz ⎠ Shape of stress tensor for 2D beam problem N =∫σxxdS Qz =∫σxzdS SS My =∫zσxxdS S dM y = Qz d 2 My = − fzdx dx2 dN = − fdx x Isotropic elasticity: σ=⎜K G ⎟ε1+ 2Gε 5 ⎝ 3 ⎠v Strain Navier-Bernouilli beam model ε =ε0 +ϑ0 zxx xx y 0 d 2ξz 0 ϑy = − 2 Curvaturedx dξ0 ε0 = x Axial strain xx dx Thus: dξ0 d 2ξ0 x zεxx = − 2 zdx dx Strain completely determined from displacement of beam reference axis ⎛−2 ⎞ Derivation of beam constitutive equation in 3-step approach Section number below corresponds to section numbering used in class Step 1: Consider continuum scale alone (derive a relation between stress and strain for the particular shape of the stress tensor in beam geometry) 2.1) Step 2: Link continuum scale with section scale (use reduction formulas) 2.2) Step 3: Link section scale to structural scale (beam EQ equations) 2.3) 6 3Overview Continuum scale Section scale Structural scale εσ ,yz MQN ,, )(),(),( ),(),(),( xxx xMxQxN zxy yz ξξω Reminder: Rotation (slope) 0 zξ slope z x Curvature (=first derivative of rotation) ω0 y = − dξz 0 ϑy 0 = − d 2ξ 2 z 0 = dωy 0 7dx dx dx x F ⎟⎠⎜⎝ 000 2.1) Step 1 (continuum scale) ⎛σ xx 0 0⎞ z Consider a beam in uniaxial tension: ()σij = ⎜⎜ 0 0 0⎟⎟ σ xx = ⎜⎛ K − 2 G ⎟⎞(εxx +ε yy +εzz )+ 2Gε xx (1) ⎝ 3 ⎠ 3 unknowns, 2 (2)σ yy = ⎜⎛ K − 2 G ⎟⎞(εxx +ε yy +εzz )+ 2Gε yy = ! 0equations; can ⎝ 3 ⎠eliminate one variable and obtain σ zz = ⎝⎜⎛ K − 23 G ⎠⎟⎞(εxx +ε yy +εzz )+ 2Gεzz = ! 0 (3) relation between 2 remaining ones Eqns. (2) and (3) provide relation between εxx and ε yy ,εzz : 1 3K − 2Gε =ε = − ε = −νεyy zz xx xx2 3K + G 8 =:ν Poisson’s ratio 410Physical meaning “Poisson’s effect” • The ‘Poisson effect’ refers to the fact that beams contract in the lateral directions when subjected to tensile strain ε =ε =−νεyy zz xx d1 d2 d2 =(1+εyy )d1 =(1−νεxx ) 9 9KGFrom eq. (1) (with Poisson relation): σ= εxx xx3K + G =: E Young’s modulus xxxx Eεσ = This result can be generalized: In bending, the shape of the stress tensor is identical, for any point in the cross-section (albeit the component σzztypically varies with the coordinate z) Thus, the same conditions for the lateral strains applies ⎛σxx (z) 0 0⎞ ()=⎜ 0 0 ⎟ Therefore: We can use the same formulas! σij ⎜ 0⎟ ⎜ ⎟⎝ 0 0 0⎠ 5112.2) Step 2 (link to section scale) Now: Plug in relationσ xx = Eεxx into reduction formulas Consider that εxx = ddx ξx 0 − ddx 2ξ 2 z 0 z and thus σ xx = E⎜⎜⎝⎛ ddx ξx 0 − ddx 2ξ 2 z 0 z ⎟⎟⎠⎞ Results in: Assume: E constant over S =0 ∫ S ⎜⎜⎝⎛ ddx ξx 0 ddx 2ξ 2 z 0 ⎟⎟⎠⎞ N = E ddx ξx 0 ∫ S dS − E ξ dx d2 2 z 0 ∫ S zdSN = E − z dS =0 My =∫ SE⎜⎜⎝⎛ ddx ξx 0 z − ddx 2ξ 2 z 0 z2 ⎟⎟⎠⎞dS My = E dx dξx 0 ∫ S zdS − E ddx 2ξ 2 z 0 ∫ Sz2dS = I Finally: N = ES dξx 0 My = −EI d 2ξ 2 z 0 Area dx dx moment of inertia 2.3) Step 4 (link to structural scale) Beam EQ equations: Beam constitutive equations: d 4ξ 0 zM = −EI = − fy 4 zdxd 2 My = − fzdx2 d 4ξ 0 fz z= with: dx4 EI z 0dN = − fx My = −EI ddx 2ξ 20 d 2ξx = − fx 2dx dξ 0 dx ES xN = ES dx 12 6Beam bending elasticity Governed by this differential equation: d 4ξ 0 fz z= dx4 EI Integration provides solution for displacement Solve integration constants by applying BCs Note: E = material parameter (Young’s modulus) I = geometry parameter (property of cross-section) f= distributed shear force z How to solve? Lecture 25 13