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Energy principles and variational formulations in .NET Drawing Denso QR Bar Code in .NET Energy principles and variational formulations

Energy principles and variational formulations generate, create none none in none projects .NET Framework 2.0 We verify the none none traction-free surface condition for the stress eld and note from the series expansion of the traction vector that this condition is not satis ed unless the thickness of the beam is small compared to its length: a/L 1.. << Tenso r2Analysis.m << Displacement.m SetCoordinates[Cartesian[x, y, z]] domain = {{x, 0, L}, {y, -a/2, a/2}, {z, -b/2, b/2}} w = A ( Cos[Pi*x/L] - 1) + B ( Cos[2*Pi*x/L] - 1) u = {-y D[w, x], w, 0} u /.

x -> 0 (eps = Strain[u]) // MatrixForm (sig = EE Stress[eps]) // MatrixForm normal = {0, -1, 0} traction = sig . normal /. y -> -a/2.

Series[ traction /. a -> aoL L,{aoL ,0,1}] /. aoL -> a/L We can now pro ceed to compute the mass matrix and the stiffness matrix. They can be obtained from the equalities [ ]T [M][ ] = [ ]T [K][ ] = u u 2 dv u u [u ] : C : [u ] dv,. where [ ] = none for none [A B]T . To compute the integrals, we Apply (@@) the Integrate command to the integrand Prepend ed to the integration domain. The results show that we can isolate the two constants in order to simplify the computations: cm = a b L ck = Pi 4 a3 b ( 1 + nu)E .

24 L3 (1 2 )(1 + ). M = Integrate @@ Prepend[domain, rho u . u] cf = CoefficientList[M, {A, B}] MM = {{cf[[1,3]], cf[[2,2]]/2}, {cf[[2,2]]/2, cf[[3,1]]}} K = Integrate @@ Prepend[domain, Flatten[eps] . Flatten[sig]] cf = CoefficientList[M, {A, B}].

7.5 Extremal properties of free vibrations KK = {{cf[[1,3 ]], cf[[2,2]]/2}, {cf[[2,2]]/2, cf[[3,1]]}}. cm = a b L rho none none Mc = Simplify[MM/cm] ck = a 3 b EE (-1 + nu) Pi 4 / ( 24 L 3 (-1 + nu + 2 nu 2) ) Kc = Simplify[KK/ck]. As [K] = ck[Kc ] and [M] = cm[Mc ], the cyclic eigenfrequency can be computed from the solution of the equation: det ([Kc ] [Mc ]) = 0. elsol = Solve[ none for none Det[ Kc - el Mc] == 0, el]. ck . cm In the case a/ none for none L 1 we can obtain a simple expression for and implicitly for the cyclic eigenfrequency and also compute the eigenmode in this limiting case. A series expansion of the matrices provides a linear system of equations that must be satis ed by the eigenmode. The two approximate eigenvalues and eigenmodes in the case a/L 1 are de ned by 1 2281) = (51 5 and 143 + 3 2281 A = B.

51 2281 The proportionality between A and B indicates that the eigenvector de nes only a direction in the space of kinematically admissible elds, without de ning its amplitude.. ellim = Simpli fy[ Map[(Series[#/.a->aoL,{aoL,0,1}] &), elsol], L>0 ] N[ellim] Mclim = Normal[ Series[ Mc /. a -> aoL , {aoL, 0, 1}] ] Kclim = Kc.

system = Threa none none d[ (Kclim - el Mclim). {A, B} == 0 Map[ (Solve[ # , {A, B} ][[1, 1]] & ) , system ]. ] /. ellim Energy principles and variational formulations y x x Figure 7.6. Two con gurations of a bar composed of two different elastic materials. SUMMARY Variational pr none none inciples in elastostatics and vibration analysis were introduced in this chapter, leading to the development of approximate methods of solution that are illustrated in the form of Mathematica code.. EXERCISES 1. Extension o none for none f bimaterial rods Consider two bars composed of two elastic isotropic materials as illustrated in Figure 7.6.

The bars occupy the domain = [ L, L] [ L, L] [ H, H] and the elastic moduli will be denoted by (Ei , i ) and ( i , i ) for i = 1, 2. We shall subject both bars to tension along the axis Oz by applying uniform vertical displacement: uz = at both ends z = H..

(a) Write down conditions of continuity for tractions and displacement vectors at each of the interfaces. (b) Show that for bar (a) an exact solution can be constructed. 2.

Bending of a plate Consider a parallelepipedal plate that in its initial con guration occupies the domain [0, L] [ e, e] [ H, H]. The plate is made of an isotropic homogenous elastic material de ned by the constants (E, ). The prescribed boundary displacements are u (0, y, z) = 0, that is, the face x = 0 is encastre.

The displacement component along e y is given by uy (L, y, z) = d on the face x = L. On the faces z = H, the displacement component along e z is given by uy (x, y, H) = 0. (a) Under the assumptation that all other necessary components of the boundary conditions are such that the surfaces are traction-free, complete the boundary conditions such that the problem is well-posed.

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