By Andreas Öchsner
This booklet introduces readers to fashionable computational mechanics according to the finite point strategy. It is helping scholars reach mechanics classes by means of exhibiting them the best way to observe the basic wisdom they received within the first years in their engineering schooling to extra complicated topics.
In order to deepen readers’ knowing of the derived equations and theories, every one bankruptcy additionally comprises supplementary difficulties. those difficulties begin with basic wisdom questions about the idea provided within the bankruptcy, via calculation difficulties. In overall over eighty such calculation difficulties are supplied, in addition to short suggestions for each.
This e-book is principally designed to fulfill the wishes of Australian scholars, reviewing the math coated of their first years at collage. The 13-week direction contains 3 hours of lectures and hours of tutorials in keeping with week.
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Extra resources for Computational Statics and Dynamics: An Introduction Based on the Finite Element Method
25) Linear Element Formulation Let us consider in the following a rod element which is composed of two nodes as schematically shown in Fig. 7. e. a displacement in the direction of the principal axis (cf. Fig. 7a) and each node can be only loaded by a single force acting in x-direction (cf. Fig. 7b). Since there are only two nodes with two unknowns, the equation for the unknown displacement in the element and its virtual counterpart (cf. Eqs. 3 Finite Element Solution 19 Fig. 7 Definition of the one-dimensional linear rod element: a deformations; b external loads.
4 Transformation between Cartesian (x) and natural coordinates (ξ) Configuration Transformation 2x − 1, L dξ 2 = . dx L ξ= 2x , L 2 dξ = . dx L ξ= 2 (X − X 1 ) − 1, X2 − X1 dξ 2 = . dX L ξ= composed of constant values, it is sufficient to consider a one-point integration rule (ξ = 0, w = 2) to achieve the analytical result8 as: Ke = EA 2L 1 −1 −1 1 ξ=0 × 2 = w EA 1 −1 . 4 can be further generalized. 41) where x1 and x2 are the coordinates of the start and end node in the elemental Cartesian coordinate system.
A displacement in x-direction and each node can be only loaded by a single force acting along the x-axis. e. at x = L2 , of the element. Since there are now three nodes with three unknowns, the equation for the unknown displacement in the element and its virtual counterpart (cf. Eqs. ; displacement BC u at node 2 FE gives exact nodal values but only approximate solution between nodes Fig. 9 Definition of the one-dimensional quadratic rod element: a deformations; b external loads. 54) and ⎡ W (x) = δuTp N(x) = δu 1 δu 2 δu 3 ⎤ N1 × ⎣ N2 ⎦ .