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 rectangular element will add one node. he extra node allows us to include one more term in the linear multivariate u and v functions. uxy ˆ(, ) a1 ax 2 ay 3 axy 4 vxy ˆ(, ) a axayaxy 5 6 7 8 New linear terms that can be added to the displacement functions of the CS element.
he solution process for the a i coefficients follows that of the CS element. Note that the element is aligned with the reference frame. his simplifies the integration process on the next page. u a1 a2xa3ya4xy u a5 a6xa7ya8xy uˆ a a x a y a x y uˆ a a x a y a x y, u uˆ a a x a y a xy uˆ aax ay axy 1 1 2 1 3 1 4 1 1 2 1 2 2 3 2 4 2 2 1 2 vˆ1 a5 a6x1 a7y1 a8x ˆ 1y1 v2 a5 a6x2 a7y2 a8x2y2 3 1 2 3 3 3 4 3 3 4 1 2 4 3 4 4 4 4 3, u4 vˆ 3 a5 a6x3 a7y3 a ˆ 8x3y3 v4 a5 a6x4 a7y4 a8x4y4 u
Lecture 21: Isoparametric Formulation of Plane Elements. he stiffness matrix is obtained by forming the first integral expression of the stationary potential energy principle. Bd h b 0 ˆ t ˆ B DB dxdyd t N b da N S da f h b ( e) ( e) d d d d d 1x 2x 4x 4y A he bounds make the integration a little easier. But the element must remain horizontal in the chosen frame for these bounds to remain nice. ( yh) 0 ( h y) 0 ( h y) 0 ( h y) 0 1 B 0 ( x b) 0 ( b x) 0 ( b x) 0 ( b x) 4bh ( x b) ( y h) ( b x) ( h y) ( b x) ( h y) ( bx) ( h y) S he extra term in the u and v approximations does cause the B matrix to vary over the element domain.
he Rectangular Plane Stress Element requires an element frame be used to simplify the integration process. For planar elements we don t want to apply rotation transforms. We would have to transform four 2x1 vectors. he transform,, would be an 8x8 matrix. 10.2: Isoparametric Formulation of the Plane Quadrilateral Element Stiffness Matrix (and element equations). Use natural coordinates (s and t) to define the variation of x, y, u, and v over the region. In addition to the regular first 4 Logan steps we must come up with a mapping between (x,y) and (s,t). We will build on the isoparametric procedure used for the bar element.
he planar element as seen in the s,t plane MAPPING functions. he planar element as seen in the x,y plane. As in 10.2. we are using straight edges to form the element. We will have only linear terms in our multivariate displacement
Step 1: Set the element type. We want to have linear edges to the element in the deformed configuration. We expect to have linear shape functions. Form a linear definition for the variation of the x and y coordinates. x a a sa t a st 1 2 3 4 y a a sa t a st 5 6 7 8 8 unknowns: require 8 equations to specify a unique set of polynomial coefficients. Linear multivariate polynomial
a Can set up 8 scalar equations based on the known nominal vertex coordinates. 1 a 1 s t st 0 0 0 0 x a 0 0 0 0 1 s t st y a 1 s t s t 0 0 0 0 x a 0 0 0 0 1 s t s t y a 1 s t s t 0 0 0 0 x a 0 0 0 0 1 s t s t y a 1 s t s t 0 0 0 0 x a 0 0 0 0 1 s t s t y 1 1 1 1 1 1 2 1 1 1 1 1 3 2 2 2 2 2 4 2 2 2 2 2 5 3 3 3 3 3 6 3 3 3 3 3 7 4 4 4 4 4 8 4 4 4 4 4 Solve for the a coefficient values and backsubstitute into the original linear poly
he mapping between the real world Cartesian frame and the natural coordinate system is defined by the shape functions. x y N N N N x y x y 1 2 3 4 28 1 1 2 4 18 he mapping N j N 0 j 0 N j N N 1 s1t 1s1t N 4 4 1 2 1 s1t 1s1t N 4 4 3 4
he shape functions ensure an interpolation of the node points by the edges.
Step 2: Select a displacement function. Premise of the isoparametric formulation is that we apply the same approximation to the displacement field, {uv} d d d 1x 1y 2x u d2 y 1 2 3 428 v d3x N N N N N d d 3 y d 4x 4 y displacement vector at node 4.
Step 3: Define the stress and strain relationships. his step is entirely focused on obtaining the B matrix. We need to define the strains that result from our chosen u and v approximates. he B matrix is the key component of Step 4: Note the use of h for thickness of element. ˆ 0 h B DB dad h N b da N ˆ S ds f A( e) A( e) S( e) Elasticity matrix entirely dependent on the boundary conditions (plane stress or plane strain?) he B matrix is entirely dependent on the choice of the approximate displacement field.
Recall the definition of strain. u x What is and? 0 x u Su 0 y v y x u y Differential operator S s 1 2 x const he form of the mapping dictates that x (& u) change with both s and t u u x u y s x s y s u u x u y t x t y t
Can solve for the unknown partials using methods of linear algebra. u u x u y s x s y s u u x u y t x t y t u u x y x s s s u y u u x y x t t t u y J u x yu s s s x u x y u t t t y A matrix that relates rates of change of u in terms of x and y to rates of change in terms of s and t. By definition: a Jacobian matrix
Solution process can be carried out in advance u 1 y u y u 1 y y or ( u) ( u) x J t s s t x J t s s t u 1 x u x u 1 x x or ( u) ( u) y J s t t s y J s t t s We can equate the differential operator to an alternative convenient form. 1 y y x J t s s t 1 x x y J s t t s
he strain becomes (in either the plane stress or plane strain case). x y S xy Nd 0 Nd SN Bd d 1 J y y 0 t s s t x x s t t s x x y y s t t s t s s t Note the difference in notation relative to Logan pg.# 497 eq. (10.2.15)
he B matrix becomes: 1 B38 B B B B J 1 2 3 4 Page #498 for the analytical form of B i (eq. 10.2.18 to 10.2.21) he determinant of the Jacobian is given by (eq. 10.2.22): J x1 0 1t t s s1y1 1 x t 1 0 s 1 s t y x 1s st t 1 0 y 2 2 8 x 3 st s1 0 t 1 y3 4 4
1 B B B B B J 1 2 3 4 Ni Ni a b 0 s t Ni Ni Bi 0 c d t s Ni Ni Ni Ni c d a b t s s t 1 a y1( s1) y2( s1) y3( s1) y4(1 s) 4 1 b y1( t1) y2(1 t) y3( t1) y4(1 t) 4 1 c x1( t1) x2(1 t) x3(1 t) x4(1 t) 4 1 d x1 ( s1) x2( s1) x3( s1) x4(1 s) 4
Step 4: Derive the element equations (the stiffness matrix is of primary interest). 0 ˆ h B DB da d h N b da N ˆ S ds f A A S 1 2 3 4 Looking at term 1 first: k h B DB dxdy h B D B J dtds 1 1 38 33 38 11 88 A s1t1 Plane Stress or Plane Strain?
Lecture 21: Isoparametric Formulation of Plane Elements. 1 1 1 1 1 1 1 1 1 1 1,,, i i s t n j j s n n i j i j n n i j i j j i j st st st s k h B DB J dtds h W B DB J ds h W W B DB J h WW B DB J
o evaluate the integral we have to evaluate the integrand at four places in the s,t plane. I WW f ( s, t ) WW f ( s, t ) WW f ( s, t ) WW f ( s, t ) 1 1 1 1 1 2 1 2 2 1 2 1 2 2 2 2 Substituting the value of W 1 and W 2 : I f( s, t ) f( s, t ) f( s, t ) f( s, t ) 1 1 1 2 2 1 2 2
he second term: 1 1 B f h N b da h N b J ds dt A s1t1 28 21 11 he third term: S f N S da S 1. In a plane stress plane strain problem we only have surface tractions applied around the outer edges. 2. Each edge has a depth h into the plane of the problem. 3. Each straight edge is easily defined using the natural s and t coordinates.
Lecture 21: Isoparametric Formulation of Plane Elements S f N S da S 1 S f h N28 S J dt 21 11 t1 s1 h N J dt 1 28 S 21 11 t1 s1 h N J ds 1 28 S 21 11 s1 t1 h N J ds 1 28 S 21 11 s1 t1 1. Remember that over the edges the shape functions are linear: is there any need to actually carry out this integration? 2. Do you know the results already for a constant traction? A linearly varying traction?