Lecture 20: Isoparametric Formulations.

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1 Chapter #0 Isoparametric Formulation. Isoparametric formulations help us solve two problems. Help simplify the definition of the approimate displacement field for more comple planar elements (4-sided elements, elements with curved edges, ). Significantly reduce the integration process by ensuring that we always have integrals in terms of natural coordinates that are taken over fied bounds. We start by applying the isoparametric formulation to a bar element. Move on to derivation of the stiffness matri and ee sfor a plane element that can be used in plane stress and plane strain problems. The derivations will show the set up of the integral epression. We will apply Guassian quadrature to evaluate these integrals.

2 0. Isoparametric Formulation of the Bar Element Euations. Step : Set the element type. Step now includes the definition of a natural (or curvilinear coordinate), s. We must create a mapping between the curvilinear coordinate s and the cartesian coordinate. This mapping will match the element boundaries/keypoints/nodes between the natural coordinate plane and the Cartesian coordinate system. Element nodes Element centroid

3 The mapping procedure is very similar to the creation of shape functions as in Logan Ch. s, 3, 4, 5, and 6. Set the order of the approimation: = a + as Apply some kinematic constraints on the approimation. = a + a( ) = a + a ( + ) a = a = ( s) + ( + s) + a = a

4 Defining the shape functions that are the core of the mapping between s and [ ] N N = ( s) ( + s) The shape functions define the variation of a quantity (now a coordinate ) over some domain of interest (now a natural coordinate s). We can apply the same shape functions to define the variation of other values over the s.. + domain ( ) N N = =

5 Step : Select a displacement function: u dˆ u = [ N N] = [ N N] u ˆ d

6 Step 3: Define the stress strain relationships. Regardless of what natural coordinates we introduce, we do not change the governing physics of the structural problem: du ε = d du du ds = d d ds Governing physics/constitutive equations. How we realize the differential term(s) in the constitutive equations. du u u = ds d L = = ds du ( u u) d ˆ ε = = = B ; B = d ( L) dˆ L L Again, we use the B notation to refer to a matri that relates displacements to strains.

7 Step 4: Derive the element equations: 0 T ˆ T T ˆT = ˆ B DB dv d + N b dv + N TS ds + f d V( e) V( e) S( e) For line elements the integrals reduces to integration in -D:. (uniform cross sectional area assumed in Step.) 0 L L L T ˆ T T ˆT = A ˆ B DB d d + A N b d+ P N TS d+ f d = 0 = 0 = 0 3 P Perimeter around the bar cross section.

8 Consider the first term in the functional : (the strain energy term that produces the stiffness matri) L s=+ T T kˆ = A B DB d = A B DB d = 0 s= d L = = ds L Jacobian d = ds matri. s=+ ˆ T L k = A B DB Jds J = s=

9 Some notes on the ever-present Jacobian matri: Main Entry: Ja co bi an Pronunciation: j&-'ko-be-&n, yä- Function: noun Etymology: K. G. J. Jacobi died 85 German mathematician : a determinant defined for a finite number of functions of the same number of variables in which each row consists of the first partial derivatives of the same function with respect to each of the variables Main Entry: de ter mi nant Pronunciation: di-'t&r-m&-n&nt Function: noun :an element that identifies or determines the nature of something or that fies or conditions an outcome

10 Some more notes on the ever-present Jacobian matri: Referred to as a Jacobian Matri or just Jacobian. Any matri that defines the conversion from one rate to another. Like a matri of currency conversions. J in this case is a scalar entity: J d change in the cartesian coordinate ' ' = ds change in the natural coordinate ' s ' Robotics: defines how a collection of joint rates maps over a translational and rotational velocity of an end effector. Basis for kinematic and dynamic robotics problems. Dynamics: how do changes in the state of a body affect the dynamic equilibrium of that body. How we measure the stiffness of a set of dynamic equations. Has consequence in how we apply numerical integrators to solve dynamics problems.

11 Completing the evaluation of k. s=+ ˆ T k = A B DB Jds s= s=+ AL L = E ds L L s= L AE = L So what s the big deal about introducing s?

12 Look at the second term in the original functional L ˆ B T f = A N b d = 0 s s=+ L = A b ds + s s= Still pretty easy to evaluate but we saw for the LST how the integrals get a lot more complicated. But now we know the limits of the integration and the shape functions are well defined over the - to + interval

13 0.4. Guassian Quadrature (Numerical Integration). Numerical integration methods provide a value for an integral I: I = y( ) d If we turn to numerical approaches it is because the integrand is to comple to get a useful epression for I. Numerical methods sample the integrand, y(), and approimate the integral, I, with a weighted summation of the sample values. The choice of the weighting factors is what defines the particular integration scheme. Numerical methods effectively replace the original function, y(), with a polynomial approimate.

14 y( i ) = value of the function at the sample point i. W i = weighting applied to this sample. I = y( ) d W y( ) n i= i i Guassian Quadrature uses n sample values to create a polynomial of order n- which replaces the original function. For MECH 40 we are integrating over domains: s.. + ( ) and/or (.. ) t +

15 Guassian quadrature: Uses a symmetric distribution of sample points over the - to + s domain. Symmetric points carry the same weight. Any number of points can be chosen. One-point Guassian quadrature, Two-point Guassian quadrature, If an odd number of points is chosen then one of the points is the element centroid. If we consider how the two-point quadrature formula is obtained then we can repeat this procedure to obtain quadrature formulas that are not tabulated.

17 Using the two point formula we could quickly discretize the body fied load we looked at earlier: s L s=+ ˆ B T L f = A N b d = A b ds s = 0 s= + s s AL AL = (.000) b + (.000) b + s + s Note: we would have to evaluate b at the points: = ( ) ( ) + + and = ( ) + ( ) s= s=

Lecture 21: Isoparametric Formulation of Plane Elements.

Lecture 21: Isoparametric Formulation of Plane Elements. 6.6. Rectangular Plane Stress/Strain Element. he CS element uses a triangular shape. he 3 nodes of the CS element allow us to employ linear multivariate approximations to the u and v displacements. he