Tham khảo tài liệu 'dohrmann episode 1 part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Finally the nodal force vector fh associated with hourglass stiffness is obtained by differentiating Uh with respect to d. The result is fh eV1 3Gh I - T - T d 20 It follows from Eq. 20 that fh is orthogonal to 4 q. In other words hourglass stiffness does not cause any restoring forces if the nodal displacements are consistent with a linear displacement field the desired result. We note that the hourglass control given by Eq. 20 is also applicable to other uniform strain elements such as the eight-node hexahedron. The development thus far has been focused solely on 3D elements. Corresponding results for 2D elements are obtained simply by redefining q d w and as r -17 q Cy Yxy rx ry rXy 21 d ui V1 u2 v2 . un vn T 22 w diag wi wi w2 w2 . wn wn 23 and _ _ 71 0 yỵ 1 0 yi 0 7 1 0 0 1 71 24 n 0 n 1 0 yn 0 yn 0 0 1 xn In the finite element method equivalent nodal forces for surface tractions are commonly obtained by integrating the product of the shape functions and the tractions over the loaded area. This procedure cannot be used with the least squares approach because shape functions are never introduced. Two alternative options are available for determining equivalent nodal loads. The first involves subjecting a collection of elements to a constant state of stress. Equivalent nodal forces can then be determined from the calculated reaction forces. A second method presented in the Appendix makes use of a mean quadrature formulation that is equivalent to the least squares approach under certain conditions. The six-node triangle is defined to have three vertex nodes and three mid-edge nodes as shown in Figure la. The nodal weights for the element are chosen as wi . We 1 a 1 a 1 a 4a 4a 4a 25 5 where Oí G 0 1 is a scalar weighting parameter. When a 1 5 the weighting for each node is identical. Consider a surface traction of constant value applied to the edge shared by nodes 1 2 and 4. The equivalent nodal forces are given by 1 1 - a F 2 A 1 - a F 2 26 4 aF 27 where F is .
