hw3_f03_solns

# hw3_f03_solns - EMA 405 Fall 2003 Solutions for HW#3 1 The...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EMA 405, Fall 2003 Solutions for HW #3 1. The Q8 element has nodes located at each corner and at the middle of each side. The interpolation functions, the N i ( x,y ), describe shapes that are parabolic along the two element sides that include the nonzero node for the corner-centered N i ( x,y ) and shapes that are parabolic along only one side and zero on all others for the side-centered N i ( x,y ). For a square region (or in element coordinates), the shapes would appear the same if we rotated the arrangement by any multiple of π /2, so there are only each corner-centered shape and each side-centered shape have the same weight. This simplifies the computation of work equivalent forces for a uniform body force, because it means that we only have to compute two integrals. In a manner similar to applying pressure on the boundary of an element, the work equivalent loads F i for a body force w satisfy ( ) ∫ ∑ ∫ ∑ ⋅ = ⋅ = ⋅ i i i i i i y x N dVol dVol , FEA U w U w U F for all possible nodal displacements, U i . [The above relation states that the sum of the work done by the work equivalent loads is equivalent to the work done by the distributed body force, given the form of the finite element analysis solution, U FEA .] To hold for all possible displacements, the coefficients of the U i on each side of the equation must be equivalent after exchanging the order of summation and integration on the rhs: ( ) ∫ = y x N dVol i i , w F Note that the integration in the z- direction is trivial, because nothing depends on z in these 2D planar analyses. In addition, with the body force being uniform, we can move w outside the integral. Using node 3 (top right corner) as our representative corner shape function and b as the base or z-dimension of the structure (just 1 m, 1 ft., etc. for plane stress analysis), the work equivalent force at a corner node is ( ) 3 3 4 3 2 2 4 ) 1 ( 4 )] 1 ( ) 1 ( 1 )[ 1 )( 1 ( 4 1 , 2 1 1 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 3 3 w w w w w F b x x dx b xy xy y x y x dxdy b y x y x dxdy b y x N dxdy b − = − + = − + + + + = − − − − + + = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ − − − − − − − 1/9 The total force per element from the body force is the element volume times the constant volume force, F tot =4 b w , so the work equivalent force on a corner node can be expressed as 12 tot corner F F − = . This is in the opposite direction of the applied body force, because the corner shape functions are negative over a substantial fraction of the element. For a representative work-equivalent load calculation for a mid-side node, consider node 6. ( ) 3 4 3 4 3 4 2 ) 1 )( 1 ( 2 1 , 1 1 2 1 1 1 1 1 1 1 1 6 6 w w w w F b x dx b y x dxdy b y x N dxdy b + = − = − + = = ∫ ∫ ∫ ∫ ∫ − − − − − Thus, for the mid-side nodes, we have 3 tot side F F + = Note that the sum of the work equivalent loads for the four sides and four corners is equal to the integrated body force over the element, as it must be....
View Full Document

## This note was uploaded on 07/29/2008 for the course EMA 405 taught by Professor Witt during the Spring '04 term at University of Wisconsin.

### Page1 / 9

hw3_f03_solns - EMA 405 Fall 2003 Solutions for HW#3 1 The...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online