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Continuum Mechanics Continuum mechanics Before we can discuss the detailed physics relevant in micro- and nanoscale machines, we need to have an understanding of the mechanical properties of bulk solids. Continuum mechanics is based on the familiar idea that if we coarse-grain the material, we can effectively understand its gross properties by a small number of well-defined parameters, rather than following each atom. Kobe U. l w h Continuous solid Hydrogen molecule Short carbon nanotube N ~ 1022 N ~ 4 N ~ 105 Elasticity theory Schroedinger eqn. Molecular dynamics stresses, strains, Energy levels Normal modes normal modes Definitions Stresses are forces per unit area. • Because forces are directed and areas are also directed, stresses are tensorial. • Usually deal with individual tensorial components (scalars). FN z w FN l σ xx = x wh h y Normal stress First letter indicates direction FS σ zy= − of surface normal; second lw FS indicates direction of force, so F σ = + S yz lw 3 = ith component of force Shear stresses − ∑σijdA j j=1 acting on a surface dA=dA n, Definitions Define u(r) to be the local displacement of a medium from an undeformed state. Two points that started out a distance dr apart end up separated by dr'()= r d+ r d⋅ ∇ u Intuitive idea: stresses result from nonuniform translations. Looking at one component, ∂u 1 ⎛ ∂u ∂u ⎞ 1 ⎛ ∂u ∂u ⎞ i ≡ ⎜ i + j ⎟ + ⎜ i − j ⎟ ⎜ ⎟ ⎜ ⎟ ∂rj 2 ⎝ ∂rj ∂ri ⎠ 2 ⎝ ∂rj ∂ri ⎠ strain tensor uniform rigid rotation ~ ≡ ε = εijkφ l δV Can see from definitions that Tr= ε~ ∇u ⋅ = V Stresses and strains ⎛ ∂u ∂u ⎞ s law as:Can write Hooke’ σ= − λδ ∇u ⋅⎜ − γ i + j ⎟ ij ij ⎜ ⎟ ⎝ ∂rj ∂ri ⎠ ~ ⎛ 1 ~⎞ Rearranging,σ=ij − δKij ε Tr −2 γ⎜ ij ε −ijTr δ⎟ ε ⎝ 3 ⎠ 2 K≡λ + γ 3 of the stress:a function Often we the strain as want to find 1 ~ 1 ⎛ 1 ~⎞ εij= − δijTr − σ σ⎜ ij − δ ijTr ⎟ σ 9K 2γ ⎝ 3 ⎠ Consider three specific cases: • pressureUniform hydrostatic • Axial stress • Uniform shear Stresses and strains 1 ~ 1 ⎛ 1 ~⎞ eessurUniform pr p εij= − δijTr − σ σ⎜ ij − δijTr ⎟ σ 9K 2γ ⎝ 3 ⎠ p ε= − δ ij 3K ij a cube.remains cube f-diagonal terms: No of essAxial str σzz = pcomponents zero.other , all ⎛ 1 1 ⎞ ε=zz −⎜ + ⎟ p 1 ⎛3K − γ 2 ⎞ ⎝ 9K 3γ ⎠ εxx= ε yy = −⎜ ⎟ε zz 2 ⎝ 3K + γ ⎠ ⎛ 9Kγ ⎞ σ zz=⎜ − ⎟ε zz ⎝ 3K + γ ⎠ o, Poisson ratiν s modulus, oung’YE Difference between fluids and solids Uniform shear σxy = σyx = -f f f ε= ε = δθ = xy yx 2γ γ shear modulus Fluids have no shear modulus! They deform continuously under shear. Numbers and reality To get some sense of sizes of things: Stainless steel: E ~ 200 GPa, γ ~ 76 Gpa Silicon: E ~ 156 GPa, γ ~ 65 Gpa Silver: E ~ 75 GPa, γ ~ 27 Gpa All these are assuming linear elasticity. What really happens: σ brittle ultimate tensile strength failure critical yield stress ductile ε Bending Let’s see why I-beams are stiffer than rods of the same cross- sectional area. compression y There is a value of y for which the material is in neither tension nor compression. tension That y position is the “neutral axis”. Set up coordinates so that y = 0 there. Suppose there’s some externally applied torque T that must be balanced in equilibrium. We know the total force must be zero, too - that’s what sets the position of the neutral axis. Assume beam has some width w(y). Bending For y = 0 as the neutral axis, assuming strain linear in y, compression F= w()() y(σ dy) y y x ∫ xx y2 w= ()() y dy⋅ k y ∫y1 Since this must = 0, we find that tension the y = 0 axis must be at the centroid of the cross-section in the y-direction. Now compute the moment (torque) for this case: M z()()()= w y(σ ydy) y ∫ xx The moment that is generated y2 w= y()() ydy⋅ k yelastically by this kind of bending is ∫y1 proportional to the areal moment of inertia around the neutral axis! Bending neutral coordinates, Again, for arbitrary axis is such that yw() y dy y = ∫ w∫ () y dy Areal moment of inertia about the neutral axis is then just I= y()() y − 2 w y dy Examples: ∫ 4 h bh3 πa I = I = 12 4 b radius a f in I-beams are flexure because their area stifis concentrated far from their neutral axis! δθ/2 Bending δθ Actually computing the shape of the beam, ()y ⋅δθ σ ()y= E zz δz δz M z()()()= w yσ ydy y ∫ ( zz ) y2 ⎛ ∂θ ⎞ w= y() ydy⎜ Ey⎟ ∫y1 ⎝ ∂z ⎠ ∂θ means Bigger areal moment of inertia = E I ∂z for deflection rate of angular smaller given bending torque. ∂u Note that θ()z ≈ y ∂z ∂2u M()() z= EI zy Bernoulli-Euler equation ∂z 2 Bending p o find the beam profile Tuy(z we ,) generally also worry about shear forces in the beam.must M M+dM Analyzing the forces and on moments a V V+dV ,orderof beam to first segment e littl ∂M Vdz= − dM =V − ∂z dV= − pdz ∂V = −p ∂z equation:ferential difwe a 4th order get and plugging in, ferentiating Dif ∂ 2 M ∂ 2 ⎛ ∂ 2u ⎞ =p → ⎜ EI y ⎟ =p() z 2 2 ⎜ 2 ⎟ ∂z ∂z ⎝ ∂z ⎠ Note that EI of a beam.the “flexural rigidity” called is often Boundary conditions and beam shapes For the fixed end of a beam, ∂u (u 0= z ) = 0 , y = 0 y ∂z z=0 For a hinged beam, ∂ 2u ( 0u )= z 0 , = M = z( = 0 →)y 0 = 0 y ∂z 2 z=0 Torsion An analogous thing happens in torsion. The shear stress at some position x, y is linearly proportional to its radial position - in a beam with a uniform twist, there is no shear strain on the axis. The shear stress from that little patch of area contributes an amount of torque proportional to that stress times that radial distance. Total torque supplied by twisting the beam is proportional to the polar areal moment of inertia: J= ∫ ρ 2 dA πd 4 For a circular rod of diameter d: J = 32 For a thin tube of radius a and wall thickness t: J= 2π a3 t Torsion Actually computing the twist of a torsion member: ρδφ φ(z) At some radius ρ, the shear strain = δz So the contribution to the restoring torque from a patch at that radius is ⎛ ρδφ ⎞ ∂φ ρ⎜ γ ⎟dA = γ ()ρ 2dA ⎝ δz ⎠ ∂z Integrating this up, for a given restoring torque T(z), ∂φ ∂φ T() z= γ ρ2dA = γ J() z ∂z ∫ ∂z Usual boundary condition for a torsion member is φ(z=0) = 0. Note that maximum shear stress happens at the outermost radius. For a circular rod of radius a, Ta σ = ,xy max J Torsion torsion So, symmetric to find the angle of twist of a circularly member subjected a to given amount of torque T, ⎛T() z ⎞ φ()z= L = ⎜ ⎟dz ∫ ⎜ γJ() z⎟ ⎝ ⎠ TL on,cross-sectiand uniform torque For a total φ()z= L = γJ For cularnoncir members, things get complicated because of has at corners, etc. For rectangular beams, conditions boundary by conformal mapping:analytically been solved T TL c σ = φ = ,xy max αbc 2 βbc 3 γ b b/c 1.00 1.50 2.00 3.00 6.00 10.0 ynfinitI α .208 .231 .246 .267 .299 .312 .333 β .141 .196 .229 .263 .299 .312 .333 Lagrangian formulation Often we’re interested in the dynamics of a system, not just its statics. Can use Hamiltonian formalism to come up with equations of motion for u(r), the classical deformation field. Need to be able to write down kinetic and potential energy densities. Kinetic energy density is simple: ⎛ 1 2⎞ 3 KE = ⎜ ρu& ⎟d r ∫⎝ 2 ⎠ Potential energy is a little trickier. Keeping track of the work done when straining a differential volume leads to a conclusion that, in a simple 1d case, 2 ⎛ 1 ⎛ ∂u ⎞ ⎞ PE = ⎜ EA⎜ z ⎟ ⎟dz ∫ ⎜ ⎟ Longitudinal deformation ⎝ 2 ⎝ ∂z ⎠ ⎠ 2 ⎛ 1 ⎛ ∂ 2u ⎞ ⎞ PE = ⎜ EI⎜ y ⎟ ⎟dz Lateral deformation (bending) ∫⎜ 2 ⎜ ∂z 2 ⎟ ⎟ ⎝ ⎝ ⎠ ⎠ Lagrangian formulation of motion, we o find equations can write down the Lagrangian, and T 1d lateral For the calculus. variational appropriate do the force on the free end of the beam,with a driving displacement case, 2 L ⎡ ⎛ ⎛ ∂ 2u ⎞ ⎞⎤ ⎢⎛ 1 2 ⎞ ⎜ 1 ⎜ y ⎟ ⎟⎥ L = ⎜ ρA& uy ⎟ − EI dz ∫ ⎢⎝ 2 ⎠ ⎜ 2 ⎜ ∂z 2 ⎟ ⎟⎥ 0 ⎣ ⎝ ⎝ ⎠ ⎠⎦ Need of to look at variation time integral of L: 2 t t ⎧ L ⎡ 2 ⎤ ⎫ 2 2 ⎛ ⎛ ∂ u ⎞ ⎞ ⎪ ⎢⎛ 1 2 ⎞ ⎜ 1 ⎜ y ⎟ ⎟⎥ ⎪ δS= δ L dt = δ⎨ ⎜ A ρ & uy ⎟ − EI dz⎬ + F()(,) tδ y u L t ∫ ∫∫⎢⎝ 2 ⎠ ⎜ 2 ⎜ ∂z 2 ⎟ ⎟⎥ t1 t1 ⎪ 0 ⎪ ⎩ ⎣ ⎝ ⎝ ⎠ ⎠⎦ ⎭ to play the old games that Can remembering do the variations, uy and s possible to , and it’s time derivative can be ytreated independentlit’ ation.ferentithe orders of difinterchange Lagrangian formulation Carrying this through, and setting δS = 0, we at PDEs arrive and a boundary condition: ∂ 2u ∂ 4u ρA y + EI 0= ,y <z0 < L ∂t 2 ∂z 4 ∂ 2u ⎛ ∂u ⎞ ∂ 3u − EI y δ ⎜ zy (,)⎟ = L t + u EI y ( zδ L ,= t) F+ ( ) tδ u (= z , = ) L t 0 ∂z 2 ⎜ ∂z ⎟ ∂z 3 y y z= L ⎝ ⎠ z= L 2 3 ∂ u y ∂ u y EI +F( ) t = 0 2 = 0 3 ∂z ∂z z= L z= L So, we now have a PDE looks like a that clearly kind of wave to our physical in addition conditions, and some boundary equation, constraints at z=0.
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