By Frank Rieg, Reinhard Hackenschmidt, Bettina Alber-Laukant

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**Sample text**

On account of the vectorial character we consider the scalar product, which corresponds to the square of the length dL2 = dX T dX In the deformed state, with the coordinates of the terminal points x(X) and x(X + dX), the square of the length can be calculated as follows dL2 = dx T dx For further approach we need the Taylor expansion of the coordinate x (X + dX) which describes the deformed line fragment or later the deformed component. x ∂x x(X + dX) = x(X) + dX = y + ∂X z ∂x ∂X ∂y ∂X ∂z ∂X ∂x ∂Y ∂y ∂Y ∂z ∂Y ∂x ∂Z ∂y ∂Z ∂z ∂Z dX dY dZ 54 3 Some Elasticity Theory For the length of the deformed line fragment we receive a term, which depends on the length of the undeformed line fragment.

At first we have to prove, of course, the static load capacity as we did before by applying the finite element analysis of the elastostatic mechanics. When this has happened, a so-called modal analysis should be carried out to check whether the so-called natural frequency of the component is in the frequency band of the oscillating force. If this is the case, look ahead! Natural frequencies characteristically operate autonomous. Neglecting damping effects, by elastic deformation and forces of gravity as a result of the inertia, the forces are exactly in balance, a continuous stimulation is not necessary for the component to proceed the movement.

1 γxy 2 1 γxz 2 Now we need the strain-displacement relations. Also the shear stresses are called σij with σij, i=j normal stresses σij, i≠j shear stresses Summarizing the 6 equations from above into matrix form, we receive in usual notation: εxx 1 −ν −ν 0 0 0 σxx −ν 1 −ν σyy εyy 0 0 0 σzz εzz 1 −ν −ν 1 0 0 0 = σxy εxy E 0 0 0 2 (1 + ν) 0 0 εyz 0 0 0 σyz 0 2 (1 + ν) 0 0 0 0 0 0 2 (1 + ν) εzx σzx 40 3 Some Elasticity Theory The same in FEA index notation: ε11 1 −ν −ν 0 −ν 1 −ν ε22 0 ε33 1 −ν −ν 1 0 = ε12 E 0 0 0 2 (1 + ν) ε23 0 0 0 0 0 0 0 0 ε31 0 0 0 0 2 (1 + ν) 0 0 0 0 0 0 2 (1 + ν) or in symbolic matrix notation: ε= σ11 σ22 σ33 σ12 σ23 σ31 1 Mσ E or in index notation: εik ik = 1 · E Mij σjk j To solve these equations for σ, we can write quite formally: σ = M−1 ε E We have to calculate the inverse of M in a very elaborate process.