5003 Lectures - Faculty of Engineering and Applied Science

E5003 - Ship Structures I 113
© C.G. Daley
k =
44
AE
L
,
− AE
k 14 = , k 24 = k 34 = k 54 = k 64 = 0
L
This has given us 12 terms, 1/3 of all the terms we need. Next we will find the
terms for the #2 and #5 direction.
Shear Terms
The shear terms are found from the set of forces is required to create a unit
displacement at d.o.f. #2 (and only #2);
For shear of this type, the deflection is;
3
F 2L
F 2 12 EI
δ 2 = = 1 ⇒ = k 22 = 3
12 EI δ
L
Note: to derive this easily, think of the beam as two cantilevers, each L/2
long, with a point load at the end, equal to F2.
The force at d.o.f. #5 is equal and opposite to the force at #2;
F
5
= −F
2
2
2
F 5 ⇒ = k
δ
52
− 12 EI
= 3
L
Following from the double cantilever notion, the end moments (M3, M6) are ;