5003 Lectures - Faculty of Engineering and Applied Science

E5003 - Ship Structures I 112
© C.G. Daley
⎡k
11 k 12 k 13 k 14 k 15 k 16 ⎤
⎢
⎥
⎢
k 21 k 22 k 23 k 24 k 25 k 26 ⎥
⎢k
⎥
31 k 32 k 33 k 34 k 35 k 36
K = ⎢
⎥
⎢k
41 k 42 k 43 k 44 k 45 k 46 ⎥
⎢k
⎥
51 k 52 k 53 k 54 k 55 k 56
⎢
⎥
⎢⎣
k 61 k 62 k 63 k 64 k 65 k 66 ⎥⎦
We will now derive these 36 terms. Luckily they are not all unique.
Axial Terms
The axial terms are found by asking what set of forces is required to create a
unit displacement at d.o.f. #1 (and only #1);
For axial compression, the deflection under load is;
F1L
F1
δ 1 = = 1⇒
= k11
=
AE δ
1
AE
L
the force at d.o.f. #4 is equal and opposite to the force at #1;
F
4
F 4
= −F1
⇒ = k 41
δ
There are no other forces (at #2, 3, 5, 6), so we have;
1
F 2 = k 21 = 0 and k 31 = k 51 = k 61 = 0
δ1
A displacement at 4 would require a similar set of forces, so that we can also
write;
=
−
AE
L