Page 176 - Marine Structural Design

152 Part II Ultimate Strength

kl=nn n=l (8.72)
Eq.(8.72) yields to
Z'EI
P, =- (8.73)
l2
(2) Columns with Fixed Ends
The boundary condition is
dw
w=-=O ut x=O and x=l (8.74)
dx
Applying the boundary conditions to the general solution, we may get:
kl
A=C=O, B=-D, sin-=O (8.75)
2
Hence,

kl= 2nn, n = 1 (8.76)
Eq.(8.76) yields to
4n EI
PE =- (8.77)
I=
(3) Columns One End Fixed and the Other Free
The boundary condition at the fixed end is:
dw
w=-=o atx=O (8.78)
dx
At the free ends, the bending moment and shear force must be equal to zero:
d2w
-- (8.79)
G!K2 -O atx=l
d3w dw
-+k2-=O at x=l (8.80)
dx3 dr
Applying the boundary conditions to the general solution, the elastic buckling force is then
X~EI
P, =- (8.81)
412
(4) Columns with One End Fixed and the Other Pinned
Applying the boundary conditions to the general solution, it may be obtained that:

Z'EI
PE =- (8.82)
(0.71)
The results of this example may be summarized in Figure 8.5, showing the end-fixity
coeficients and effective length for columns with various boundary conditions. A general
buckling strength equation may be obtained as below: