I don't know how either. Do you have the answer now? In his book, matrix A is [[A1,0],[0,A1]] with A1 is sub-matrix [[1,1,1,1],[1,-1,1,-1],[1,-1,-1,1],[1,1,-1,-1]] but without explanation.
@@namtang Because each element has 2cm of dimension. So, he put the local coordinate system in the center of the element (see Figure E4.6 in the book). After, he substitutes in the PHI matrix each nodal local coordinate, like this: Frist node: [1, x, y, xy] -> [1, 1, 1, 1] Second node: [1, x, y, xy] -> [1, -1, 1, -1] Thirth node: [1, x, y, xy] -> [1, -1, -1, 1] Forfth node: [1, x, y, xy] -> [1, 1, -1, -1]
MIT HIGH QUALITY EDUCATION!
I Really Like The Video Generalized coordinate finite element models From Your
Great Contribution.
The camera quality kinda screwed me on the H(2) element description relative to H.
can someone tell me how to find nodal displacements? One that Prof. referring at 21.40
how could we say A.a=U how could we find the A
I don't know how either. Do you have the answer now?
In his book, matrix A is [[A1,0],[0,A1]] with A1 is sub-matrix [[1,1,1,1],[1,-1,1,-1],[1,-1,-1,1],[1,1,-1,-1]] but without explanation.
@@namtang Because each element has 2cm of dimension. So, he put the local coordinate system in the center of the element (see Figure E4.6 in the book).
After, he substitutes in the PHI matrix each nodal local coordinate, like this:
Frist node: [1, x, y, xy] -> [1, 1, 1, 1]
Second node: [1, x, y, xy] -> [1, -1, 1, -1]
Thirth node: [1, x, y, xy] -> [1, -1, -1, 1]
Forfth node: [1, x, y, xy] -> [1, 1, -1, -1]
Very informative lecture by the Prof. However, presentation should be improved by utilizing PowerPoint or other computer software.
When he gave the classes probably ms powerpoint even still exist I guess