ISOPARAMETRIC ELEMENT Bruce Irons, 1924-1983 Quadrilateral Shape Most commonly used element (irregular shape) Generalization of rectangular element Use mapping to transform into a square (Reference element). The relationship between (x, y) and (s, t) must be obtained. All formulations are done in the reference element. u1 v1 u2 v2 u4 v4 4 1 2 x y u3 v3 3 4 (-1,1) 1 (-1,-1) 2 (1,-1) 3 (1,1) s t Actual Element (x, y) Reference Element (s, t)

ISOPARAMETRIC MAPPING Definition the same interpolation method is used for displacement and geometry. Procedure Construct the shape functions N1, N2, N3, and N4 at the reference element 4 (-1,1) 1 (-1,-1) 2 (1,-1) 3 (1,1) s t

HOW DOES IT LOOK? Two questions: y t 3 4 4 (-1,1) 3 (1,1) s 1 1 (-1,-1) 2 (1,-1) 2 x Two questions: For a given (x,y), find corresponding (s,t). For a given (s,t), find corresponding (x,y). Linearity of transformation on boundary is important for continuity

THE ISO PART Use the shape functions for interpolating displacement and geometry. For a given value of (s,t) in the parent element, the corresponding point (x,y) in the actual element and displacement at that point can be obtained using the mapping relationship. Displacement interpolation Geometry interpolation

EXAMPLE 6.6 Find mapping point of A in the physical element At point A, (s, t) = (0.5, 0.5) Physical coord x y 4 (0,0) 1 (6,0) 2 (4,4) 3 (2,4) B (1,2) 4 (−1,1) 1 (−1,−1) 2 (1,−1) 3 (1,1) s t A (.5,.5)

EXAMPLE cont. Find mapping point of B in the reference element At point B, (x, y) = (1, 2) Thus, (s, t) = (0, 1) x y 4 (0,0) 1 (6,0) 2 (4,4) 3 (2,4) B (1,2) 4 (−1,1) 1 (−1,−1) 2 (1,−1) 3 (1,1) s t A (.5,.5)

Quiz-like questions Consider the element in Example 6.4. We are given the u displacement at the four nodes as u1=1, u2=3, u3=0, u4=2. Find the equation of u along the 3-4 edge of the quadrilateral as function of x or y. Why is it important that it is linear? What point in the real element corresponds to the center of the reference element (s=t=0)? What is the equation of the line y=x in the s,t plane? Answers in the notes page. x y 4 (0,0) 1 (6,0) 2 (4,4) 3 (2,4) B (1,2) 4 (−1,1) 1 (−1,−1) 2 (1,−1) 3 (1,1) s t A (.5,.5) Fortunately the sides of the reference element correspond to the sides of the real element, so the side 3-4 corresponds to t=1. It is easy to check that when s=t=0 N1=N2=N3=N4=0.25, so this point is mapped into X=(x1+x2+x3+x4)/4, y=(y1+y2+y3+y4)/4 Note that the equations for xB and yB in the previous slide are good for any (x,y). So

JACOBIAN OF MAPPING Shape functions are given in (s,t). But, we want to differentiate w.r.t. (x,y) in order to calculate strain and stress. Use chain rule of differentiation. In Matrix Form Matrix [J] is called the Jacobian matrix of mapping. How to calculate matrix [J]?

JACOBIAN OF MAPPING cont. Derivatives of shape functions w.r.t. (x,y) coordinates: Determinant |J|: Jacobian What happen if |J| = 0 or |J| < 0? shape function derivative cannot be obtained if the |J| = 0 anywhere in the element Mapping relation between (x, y) and (s, t) is not valid if |J| = 0 or |J| < 0 anywhere in the element (–1 ≤ s, t ≤ 1).

MEANING OF NON-POSITIVE JACOBIAN Every point in the reference element should be mapped into the interior of the physical element When an interior point in (s, t) coord. is mapped into an exterior point in the (x, y) coord., Jacobian becomes negative If multiple points in (s, t) coordinates are mapped into a single point in (x, y) coordinates, the Jacobian becomes zero at that point It is important to maintain the element shape so that the Jacobian is positive everywhere in the element Jacobian IS negative when an angle exceeds 180o x y

EXAMPLE 6.7 Jacobian must not be zero anywhere in the domain (-1 ≤ s, t ≤ 1) Nodal Coordinates Iso-Parametric Mapping Jacobian Matrix 1(0, 0) 2(1, 0) 3(2, 2) 4(0, 1) x y

JACOBIAN Jacobian It is clear that |J| > 0 for –1 ≤ s ≤ 1 and –1 ≤ t ≤ 1. Constant t Constant s 4 (-1,1) 1 (-1,-1) 2 (1,-1) 3 (1,1) s t

Quiz-like problem Find the equation of the determinant of the Jacobian if Node 3 of the quadrilateral element of Slide 11 is moved the the midpoint of Node 2 and Node 4, that is to (0.5,0.5), so that the quadrilateral is actually a triangle. We proceed as in Example 6.7