5003 Lectures - Faculty of Engineering and Applied Science

E5003 - Ship Structures I 116
© C.G. Daley
A rotation at #6 require a similar set of forces, so that we can also write;
4EI
2EI
k66
= , k36
=
L L
6EI
−6EI
k 26 =
2 , k56
=
2 , k 16
L L
= k 46 = 0
We can collect all these terms in the matrix;
⎡ AE
⎢
L
⎢
⎢ 0
⎢
⎢
⎢ 0
K = ⎢
⎢
− AE
⎢ L
⎢
⎢ 0
⎢
⎢ 0
⎢⎣
0
12EI
L
6EI
L
−12EI
L
6EI
L
3
2
0
3
2
0
6EI
2
L
4EI
L
0
− 6EI
2
L
2EI
L
−
AE
L
0
0
AE
L
0
0
0
−12EI
L
− 6EI
L
0
12EI
L
− 6EI
L
3
2
3
2
⎤
0 ⎥
6EI
⎥
⎥
2
L ⎥
2EI
⎥
L
⎥
⎥
0 ⎥
⎥
− 6EI
⎥
2
L
⎥
4EI
⎥
⎥
L ⎥⎦
Note that the matrix is symmetrical. This means that terms such as k35
(moment at #3 due to displacement at #5) is equal to k35 (force at #5 due to
rotation at #3). This may seem quite odd that these two items would be equal.
We will examine this in the next Chapter.
The standard 3D degrees of freedom for a beam are shown below;
The K matrix for a 3D beam is a 12x12 (144 terms).
3D beam = 12 degrees of freedom