loads_etc - To show that the nodal forces are...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Work equivalent (consistent) normal loads Mechanical loads : concentrated loads, surface traction, body forces. Normal surface traction on a side of a plane element whose sides remain straight (q is force/length): Work-equivalent nodal forces:
Background image of page 1

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

View Full DocumentRight Arrow Icon
2
Background image of page 2
3 Distributed Shear Traction Shear traction on a side of a plane element whose sides remain straight (q is force/length): In (b), a Q4 element and two CSTs share the top midnode so that the nodal loads from Q4 and the right CST are combined.
Background image of page 3

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

View Full DocumentRight Arrow Icon
4 Quadratic Normal Surface Traction Quadratic normal surface traction on a side of a plane element whose sides may deform quadratically: Work equivalent nodal loads lead to greater accuracy than lumped loads. But….
Background image of page 4
5 Uniform Body Force Work-equivalent nodal forces corresponding to weight as a body force (rectangular quadrilaterals for work-equivalence): LST has no vertex loads and vertex loads of Q8 are upwards! The resultant in all cases is W, the weight of the element.
Background image of page 5

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

View Full DocumentRight Arrow Icon
6 Equivalence of nodal forces and weight
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: To show that the nodal forces are work-equivalent to the element weight for a Q4 element of unit thickness: total work of the nodal forces=(v 1 +v 2 +v 3 +v 4 )W/4 total work of the body force: By performing the indicated integration, the two work expressions can be shown to be equal. dxdy v y x N dxdy A W y x v i i i A W = = 4 1 ) , ( ) , ( ( ) i i i A W v dxdy y x N ) , ( 4 1 = = 7 Connecting beam and plane elements Since all of the previous plane elements have translational dof only, no moment can be applied to their nodes. Therefore the connection (a) of a beam and a plane elements cannot transmit a moment and the beam element can freely rotate. (Singular K !) A solution is in (b) where beam is extended. Rotational dof at A, B and C are associated with the beam elements only. A plane element with drilling dof would also work but is not recommended....
View Full Document

This note was uploaded on 06/06/2011 for the course EAS 4240 taught by Professor Peterifju during the Spring '08 term at University of Florida.

Page1 / 7

loads_etc - To show that the nodal forces are...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online