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1 CST ELEMENT Constant Strain Triangular Element Decompose two-dimensional domain by a set of triangles. Each triangular element is composed by three corner nodes. Each element shares its edge and two corner nodes with an adjacent element Counter-clockwise or clockwise node numbering Each node has two DOFs: u and v displacements interpolation using the shape functions and nodal displacements. Displacement is linear because three nodal data are available. Stress & strain are constant. u 1 v 1 u 2 v 2 u 3 v 3 3 1 2 x y
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1 CST ELEMENT Constant Strain Triangular Element –Decompose two-dimensional domain by a set of triangles. –Each triangular element is composed by three.

Dec 17, 2015

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Myra Walton
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  • Slide 1
  • Slide 2
  • 1 CST ELEMENT Constant Strain Triangular Element Decompose two-dimensional domain by a set of triangles. Each triangular element is composed by three corner nodes. Each element shares its edge and two corner nodes with an adjacent element Counter-clockwise or clockwise node numbering Each node has two DOFs: u and v displacements interpolation using the shape functions and nodal displacements. Displacement is linear because three nodal data are available. Stress & strain are constant. u1u1 v1v1 u2u2 v2v2 u3u3 v3v3 3 1 2 x y
  • Slide 3
  • 2 DISPLACEMENT INTERPOLATION Assumed form for displacements Components u(x,y) and v(x,y) are separately interpolated. u(x,y) is interpolated in terms of u 1, u 2, and u 3, and v(x,y) in terms of v 1, v 2, and v 3. interpolation function must be a three term polynomial in x and y. Since we must have rigid body displacements and constant strain terms in the interpolation function, the displacement interpolation must be of the form The goal is to calculate i and i, i = 1, 2, 3, in terms of nodal displacements.
  • Slide 4
  • 3 CST ELEMENT cont. Equations for coefficients x-displacement: Evaluate displacement at each node In matrix notation When is the coefficient matrix singular? u1u1 v1v1 u2u2 v2v2 u3u3 v3v3 3 1 2
  • Slide 5
  • 4 SOLUTION Explicit solution: where Area:
  • Slide 6
  • 5 INTERPOLATION FUNCTION Insert to the interpolation equation N 1 (x,y) N 2 (x,y) N 3 (x,y)
  • Slide 7
  • 6 SIMILARLY FOR V Displacement Interpolation A similar procedure can be applied for y-displacement v(x, y). N 1, N 2, and N 3 are linear functions of x- and y-coordinates. Interpolated displacement changes linearly along the each coordinate direction. Tent functions Shape Function
  • Slide 8
  • 7 MATRIX EQUATION Displacement Interpolation [N]: 26 matrix, {q}: 61 vector. For a given point (x,y) within element, calculate [N] and multiply it with {q} to evaluate displacement at the point (x,y).
  • Slide 9
  • 8 STRAINS Strain Interpolation differentiating the displacement in x- and y-directions. differentiating shape function [N] because {q} is constant. Strains are constant inside!
  • Slide 10
  • 9 B-MATRIX FOR CST ELEMENT Strain Interpolation [B] matrix is a constant matrix and depends only on the coordinates of the three nodes of the triangular element. the strains will be constant over a given element
  • Slide 11
  • 10 PROPERTIES OF CST ELEMENT Since displacement is linear in x and y, the triangular element deforms into another triangle when forces are applied. a straight line drawn within an element before deformation becomes another straight line after deformation. Consider a local coordinate such that = 0 at Node 1 and = a at Node 2. Displacement on the edge 1-2: Since the variation of displacement is linear, the displacements should depend only on u 1 and u 2, and not on u 3. 3 1 2 a
  • Slide 12
  • 11 PROPERTIES OF CST ELEMENT cont. Property of CST Element Inter-element Displacement Compatibility Displacements at any point in an element can be computed from nodal displacements of that particular element and the shape functions. Consider a point on a common edge of two adjacent elements, which can be considered as belonging to either of the elements. Because displacements on the edge depend only on edge nodes, both element will produce the same displacement for a point on a common edge. 3 1 2 x a Element 1 Element 2 3 1 2
  • Slide 13
  • 12 EXAMPLE - Interpolation nodal displacements {u 1, v 1, u 2, v 2, u 3, v 3, u 4, v 4 } = {0.1, 0, 0.1, 0, 0.1, 0, 0.1, 0} Element 1: Nodes 1-2-4 1 2 1 2 3 4 (0,0) (1,0) (0,1) (1,1) x y Calculate strains directly
  • Slide 14
  • 13 EXAMPLE Interpolation cont. Element 2: Nodes 2-3-4 1 2 1 2 3 4 (0,0) (1,0) (0,1) (1,1) x y Strains are discontinuous along the element boundary