Modelling mechanical systems with nite elements

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Transcription:

Modelling mechanical systems with nite elements Bastian Kanning February 26, 2010

Outline 1 Introduction 2 Virtual displacements 3 Getting started The ansatz functions 4 Mass and stiness elementwise 5 Putting it all together 2 / 23

Machine tools Figure: Kugler experimental platform 3 / 23

Modelling mechanical systems with nite elements The process Figure: Machine in action 4 / 23

Modelling mechanical systems with nite elements Tool tips Figure: Tool tips and pen 5 / 23

Motion of continua The Motion of continua may be modelled by with some boundary conditions. B = stiness w(t) = displacement µ = mass p(t) = force B 4 2 w(x, t) + µ w(x, t) = p(t), (1) x 4 t2 6 / 23

Main goal Discretize the PDE, and transform it into M d 2 u(t) + Su(t) = p(t), (2) dt2 by using ansatz functions. M ˆ= mass distribution u(t) ˆ= displacement S ˆ= stiness distribution p(t) ˆ= force 7 / 23

Deformation energy With Hooke's law dene the (potential) deformation energy V (w(x, t)) := B L ( 2 ) 2 w(x, t) dx, x 2 with some bending stiness B (the elastic modulus). 8 / 23

Work of mass - and external forces The kinetic energy consist of the work of mass forces: with some mass µ. W m (w(x, t)) := L L µ 2 w(x, t)dx t2 work of external forces: W e (p(x, t)) := p(x, t)dx. 9 / 23

Hamilton's principle Between l 0, l 1, the true evolution w(x, t) of a system is a stationary point of S(w(x, t)) := l1 l 0 V (w(x, t)) + W m (w(x, t)) W e (w(x, t))dx. Setting the derivative to zero yields that δw l δw (x)b 2 0 x w(x,t)dx = 2 l 0 δw(x)p(x,t)dx l 0 δw(x)µ 2 t w(x,t)dx. 2 (3) 10 / 23

Eq. (3) is called the principle of virtual displacements (POVD) and is equivalent to the PDE (1). 11 / 23

Eq. (3) is called the principle of virtual displacements (POVD) and is equivalent to the PDE (1). Virtual displacements δw are arbitrary, not happening in reality, dierentially small, timeless, i.e. without inertia. geometrically possible, 11 / 23

Modelling mechanical systems with nite elements Objective: tool holder Figure: Tool holder and workpiece 12 / 23

From now on: Describing displacements Discretizing the POVD Introducing ansatz functions Calculating stiness and mass matrices 13 / 23

Material properties and DOF strain stiness D torsion stiness B x bending stinesses B y and B z mass distribution µ rotating mass distributions µ mx, µ my and µ mz w x, w y, w z - displacement in x, y- and z-axis direction, resp. β x, β y, β z - cross section torsion and slopes, 14 / 23

How to describe displacements The element displacement vector for the i-th element reads u T i (t) = [w x0, w xl, β x0, β xl, w y0, β y0, w yl, β yl, w z0, β z0, w zl, β zl ] i (t). 15 / 23

How to describe displacements The element displacement vector for the i-th element reads u T i (t) = [w x0, w xl, β x0, β xl, w y0, β y0, w yl, β yl, w z0, β z0, w zl, β zl ] i (t). Where w 01 β 01 0, w li 1 w 0i and β li 1 β 0i, i = 2,..., n. The system displacement vector reads u T (t) = [w 1, w 2, w 3, β 1, β 2, β 3, w 4, w 5, w 6, β 4, β 5, β 6,..., w 6n,... β 6n ](t). 15 / 23

Outline 1 Introduction 2 Virtual displacements 3 Getting started The ansatz functions 4 Mass and stiness elementwise 5 Putting it all together 16 / 23

Ansatz functions For each element we have l δw (x)b 2 0 x w(x,t)dx = 2 l 0 δw(x)p(x,t)dx l 0 δw(x)µ 2 t w(x,t)dx. 2 17 / 23

Ansatz functions For each element we have l δw (x)b 2 0 x w(x,t)dx = 2 l 0 δw(x)p(x,t)dx l 0 δw(x)µ 2 t w(x,t)dx. 2 Transform the integrals into matrices by use of the ansatz functions where ξ = x/l. f 1 (ξ) = 1 3ξ 2 + 2ξ 3, f 2 (ξ) = ξ(1 ξ) 2 l, f 3 (ξ) = 3ξ 2 2ξ 3, f 4 (ξ) = ξ 2 (1 ξ)l, g 1 (ξ) = 1 ξ, g 2 (ξ) = ξ, 17 / 23

With f (x) := [g 1, g 2, f 1,..., f 4 ](x), one elements displacement may now be described by the equations w i (x, t) = f T (x)u i (t) and δw i (x) = δu T i f (x). 18 / 23

With f (x) := [g 1, g 2, f 1,..., f 4 ](x), one elements displacement may now be described by the equations w i (x, t) = f T (x)u i (t) and δw i (x) = δu T i f (x). With these notions we obtain l 0 δw i (x)b 2 x 2 w i(x, t)dx = B l 0 = δu T i B (δu T i l 0 f (x)) (f T (x)u i (t))dx (f (x)f T (x))dx u i (t). 18 / 23

The element stiness matrix It holds f f T = f 1 f 1 f 1 f 4....... f 4 f 1 f 4 f 4, and due to the restriction to constant bending stiness B we can write 12 6l i 12 6l i li B (f f T ) dx i = B 6l i 4l 2 i 6l i 2l 2 i 0 li 12 6l i 12 6l i =: S B y = S Bz. 6l i 2l 2 i 6l i 4l 2 i 19 / 23

On an analogous way we obtain [ 1 1 1 1 and S D = D li ], S T = B [ x li 1 1 1 1 ], the element stiness matrix Ŝ i = S D S T S By. S Bz 20 / 23

The element mass matrix ˆM i = M D M T M By. M Bz consisting of M By,z := µl i 420 156 22l i 54 13l i 22l i 4l 2 i 13l i 3l 2 i 54 13l i 156 22l i 13l i 3l 2 i 22l i 4l 2 i M D = µl i 6 [ 2 1 1 2 + µ my,mz 30l i ], M T = µ mxl i 6 36 3l i 36 3l i 3l i 4l 2 i 3l i l 2 i 36 3l i 36 3l i 3l i l 2 i 3l i 4l 2 i ] [ 2 1 1 2, 21 / 23

Re-order displacements Two transforms of the elementwise matrices are made. Dene matrix T such that u i (t) = T [w x0, w y0, w z0, β x0, β y0, β z0, w xl, w yl, w zl, β xl, β yl, β zl ] T i (t) and thus S i = T T Ŝ i T M i = T T ˆMi T. 22 / 23

The system matrices We nally obtain system matrices by the transformation S := i A T i S i A i and M := i A T i M i A i, where A i is chosen such that u i (t) = A i u(t) and δu i = A i δu. 23 / 23