NTNU
Faculty of
Engineering Science and Technology
Department of Marine
Technology
SOLUTION
4
& 5
TMR 4195
DESIGN OF OFFSHORE STRUCTURES
Problem 1
In the ULS control
it is necessary to
consider the load cases given
in
Table
1
.
Functional & permanent,
F
Environmental,
E
ULS

a
1.3
0.7
ULS

b
1.0
1.3
Table
1
: ULS load factors
1.15
V V E E m
m
R
S S
a)
The axial force in the brace can be calculated from simple considerations of the
trusswork. We assume that the shear force in the main girder is carried
by
axial forces in
the vertical and inclined braces. Nearby the column (at brace A), some of the shear force
will in reality be carried by the upper and lower chord. We conservatively neglect this
effect and calculate the axial force
S
A
in brace A
by
,
1
2 sin45
2
A
S qL
(2 braces in
the trusswork carry shear force
)
V F E E
q q q
We assume that the
functional and permanent loads are related to weight of equipment
and
weight of deck structure, such that i
nertia forces
caused by the
vertical
wave

induced
acceleration will be evenly distributed along the main girder
with intensity
given by
,
2.5
9.81
E F
q q
ULS

a can be found to give the most critical load condition
with the following axial force
in brace A,
0.523
A F
S q L
The brace must be checked for
beam

column
buckling failure, i.e.
i
nteraction
between
axial compressive force and
bending moments
should be included in the assessment
. The
bending moments cannot be fo
und by simple hand calculations, and for
simplicity we
neglect the effect of bending moments and do a column buckling check instead. This is
non

conservative and should be avoided in a more accurate assessment.
The procedure
in the DNV Classification Notes 30.1, which is
used to calculate the
c
haracteristic column
buckling resistance
,
is outlined in the following
.
The input
parameter in this
procedure is the reduced slenderness ratio,
2 2
2
Y
A l
EI
,
where A is the cross

sectional area, β is the buckling length factor,
l
is the
bra
ce
length
and
I
is the 2.area moment.
Based on the boundary conditions at the ends of brace A
,
it is reasonable to assume that
β=
0.7.
However, a
ccording to
the
DNV Classification note
30.1
a design value of
β=
0.8
should be
used instead
.
The internal widt
h of the cross

section is denoted
b
i
and the thickness as
t
. Area and
2.area moment is then expressed as,
4
i
A bt
4
4
1
2
12
i i
I b t b
The cross

section is
assumed to be “hot finished” and therefore
assi
gned buckling curve a,
see
DNV C
lassification
N
ote
s
30.1. Then the characteristic buckling stress can be
expressed as,
2
2 2 2
2
1 0.2 0.2 1 0.2 0.2 4
2
CR Y
A cross

section width,
b
i
, which fulfills the following
relationship
must be selected,
0.131 1.15
F
A m CR i A m
i
q L
b
bt
The equations above
are included in EXCEL to determine the required cross

section
width. Here, it was found that a
n internal
width/height
of
4
50 mm will give suf
ficient
capacity against column
buckling. See
Table
2
.
450 mm
i
b
b
i
[mm]
I
[mm
4
]
A
[mm
2
]
λ
[

]
σ
cr
[Mpa]
σ
a
*
γ
m
[Mpa]
σ
a
*
γ
m
/
σ
cr
[

]
t
[mm]
30
50
1.17E07
6000
4.03
20.8
2169.4
153.53
σ
Y
[MPa]
355
100
4.63E07
12000
2.86
40.4
1084.7
36.05
l
[mm]
16971
150
1.2E08
18000
2.18
67.9
723.1
13.66
E
[MPa]
2.10E05
200
2.47E08
24000
1.75
101.5
542.3
6.63
β
[

]
0.8
250
4.44E08
30000
1.46
139.3
433.9
3.75
L
[m]
72
300
7.25E08
36000
1.25
178.3
361.6
2.35
350
1.1E09
42000
1.10
214.6
309.9
1.60
400
1.6E09
48000
0.97
245.1
271.2
1.16
450
2.22E09
54000
0.88
268.4
241.0
0.89
500
2.99E09
60000
0.80
285.5
216.9
0.72
Table
2
: Determination of cross

section width
/height
for brace A
b)
It is not obvious which ULS load condition
that will be the critical one. Both ULS

a and
ULS

b must
hence
be checked. The
stress resultants
and external pressure
in the two
ULS
conditions are
calculated as
,
F
p gh
F F E E
N N N
F F E E
M M M
F F E E
Q Q Q
The pressure is calculated at a depth corresponding to 20 m.
With the load factors in
Table
1
the two load conditions are given by,
p
[Pa]
N
[MN]
M
[MN]
Q
[MN]
ULS

a
2.61
E
5
79
76
26.5
ULS

b
2.01
E
5
76
127
28
Table
3
: ULS stress resultant in buckling check
The stresses in the cylindrical shell structure must be
determined. According to the DNV

RP

C202 code, it is usu
ally permissible to account for the longitudinal stiffeners
by an
equivalent
shell
thickness when
axial
membrane
stresses are calculated,
S
eq
A
t t
s
eq eq
A Dt
3
8
eq eq
I D t
Here
t
is the shell thickness,
A
s
is the cross

sectional area of the stiffener
without
shell
plating
,
D
is column diameter
and
s
is the stiffener spacing.
The axial
stress
and bending stress
at “outer fiber”
can then be calculated as,
a
eq
N
A
max
2
b
eq
MD
I
The shear stress will be zero at the location where maximum bending stress occurs. At the
“neutral axis” of the column
, where the bending stress is zero
, maximum shear stre
ss
is
given by,
max
2
Q
Dt
The hoop
(circumferential)
stress is calculated a
s
,
2
h
pD
t
According to the DNV

RP

C202 code it is necessary to consider 3 buckling modes for
longitudinally stiffened shells.
Shell buck
ling
Panel stiffener buckling
C
olumn buckling
Shell buckling
Panel stiffener buckling
Column buckling
S
hell
buckling
can
be trigged by th
ree different stress components.
A
xial membrane stress
caused by
axial
compressive force
and
compressive
bending stress in the column.
S
hear stresses from
torsion or
shear force
in the column.
H
oop stresses c
aused by net external pressure.
The
three
elastic shell buckling stress
es
are given by formulas valid from curved she
ll
theory. i.e.
we focus on
the shell plating
enclosed by two
longitud
inal stiffeners
.
The
formulas are expressed in the following form,
2
2
2
,,
12(1 )
Ex E Eh
E t
C
s
1
C
Here
E
is Young’s modulus,
ν
is Poisson’s ratio,
s
is the stiffener spacing and
t
is the shell
thickness.
The formula for C is obtained by an elliptic interpolation of the asymptotic
solutions for a flat plate and a curved shell. The parameter ψ accounts for the flat plate
solution
, while ξ accounts for t
he curved shell
solution
.
Cylindrical s
hells are sensitive for
initial
imperfections and will never reach the elastic buckling stress
for an ideal cylinder
.
Therefore a knock

down factor
of typically 0.5

0.6
, expressed by ρ, is applied to the shell
part of
the solution.
The parameters ψ, ξ and ρ depend on geometrical quantities
of the
shell, s
ee the DNV

RP

C202 code
Sec 3.3
included in the exercise appendix for further
details
.
Panel stiffener buckling is caused by the following effects
Compressive
membrane axial stress due to bending and axial force in the column
Shear stresses due to torsion or shear force in the column
Lateral load due to net external pressure
Panel stiffener
b
uckling stress
es
in the DNV

RP

C202 code
are expressed
by formulas
bas
ed on ort
h
otr
op
ic shell theory where the stiffener
s
are
s
meared over the shell
thickness.
Orthotropic shell theory is in principle only valid if the stiffeners are “light”
and if the spacing is small (Compendium in “TMR4205

Buckling and Ultimate Strength
Analysis of Marine Structures”).
T
he elastic buckling stresses are given by formulas on
a
similar
form as for
the
shell buckling
problem
,
2
2
2
,,
12(1 )
Ex E Eh
E t
C
l
1
C
Here
l
is the length of the stiffener, and
C
accounts for the same effects as described for
shell buckling. Since the stiffeners are smeared over
the shell thickness,
the parameters
involved in
C depends
also
on the stiffener effective 2.area moment,
I
eff
, which is
defined
by the
effective shell flange,
s
e
.
See
Figure
1
.
Figure
1
: Effective shell flange
In this problem the effective shell flange was set equal to 275 mm.
The distance
e
z
,
stiffener area,
A
s
and stiffener 2.area mom
ent,
I
s
, are
found
in
Fi
gure A in the exercise
appendix.
I
eff
can then be calculated by,
e
z
e S
s t
d
s t A
2
3 2
1
0.5
12
eff e z z e s z s
I s t e t d s t I d A
In the following the procedure
recommended
in the DNV

RP

C202
for buckling strength
assessment
is outlined.
This calculation procedure must be performed separately for both
the shell buckling check and the panel stiffener buckling check.
There will be interaction between the stress components which cause buckling. This
interaction is
taken into account
in
a
for
mula for the equivalent reduced slenderness,
2
a b h
Y
eq
j Ex Ex E Eh
,
w
here
σ
Y
is the yield strength,
σ
a
is axial compressive stress from axial force, σ
b
is
compressive bending stress, τ is shear stress and σ
h
is hoop stress from net external
pressure. If any of the
normal
stress components are tensile, they must be set equal to zero
in the equation for the
equivalent
reduced slenderness.
P
lasticity is accounted for by the equivalent stress, σ
j
,
2 2 2
2
3
j b a h b a h
Note that
the normal stress components are not set equal to zero if they are tensile
in the
equivalent stress equation
.
Thereafter,
the DNV

RP

C202 code
calculate
the design buckling strength
as
,
4
1
Y
ksd
m eq
Where the mater
ial factor
γ
m
depends on the
equivalent
reduced slenderness,
1.15 for 0.5
0.85 0.60 for 0.5 < 1.0
1.45 for 1.0
eq
m eq eq
eq
Finally, the buckling capacity is found acceptable if the equivalent stress is less than the
design buckling strength,
j ksd
The shell
buckling check and panel stiffener buckling check must be performed for both
ULS

a and ULS

b.
T
wo locations of the column are
checked in the buckling assessment
.
T
he
“outer column fibre”
where
maximum compressive
bending stress
occurs.
At the neutral axis
where
maximum shear stress is found.
In the following pages
results from the buckling control
are
presented.
The key
observation
from the analysis
can be summarized as,
Shell buckling at neutral axis comes out to be the critical point
, and
here
scantlings of
the L

stiffeners
give
minor effect on
the
shell buckling strength.
The
buckling control at this location therefore determines the m
inimum shell thickness
which can be used.
If the stiffener scantlings are increased, a
pronounced
increase of e
quivalent shell
thickness occurs. This is negative for material costs and structural weight.
The utilization
of buckling capacities
in ULS

a and ULS

b
are
almost equal.
Based on this, the design which gives optimal material costs must be based on the L200

stiffeners and a shell thickness of 21 mm.
See results in
the
next
pages.
Stress analysis
based on
equivalent shell thickness
Stiffener
A
S
[mm2]
t
eq
[mm]
I
eq
[mm4]
σ
a
[Mpa]
σ
bmax
[Mpa]
τ
max
[Mpa]
σ
h
[Mpa]
L200
3.51E+03
26.4
1.04E+13
95.3
36.7
80.3
62.1
L300
4.90E+03
28.5
1.12E+13
88.1
33.9
80.3
62.1
L400
7.75E+03
32.9
1.29E+13
76.4
29.4
80.3
62.1
Table
4
:
Stress analysis ULS

a
Shell buckling control
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
162.4
1.01
1.45
172.4
L300
159.7
1.00
1.45
173.3
L400
155.9
0.98
1.44
175.8
Table
7
: Buckling control at neutral axis of column, ULS

a
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
114.3
1.16
1.45
159.6
L300
105.7
1.19
1.45
157.2
L400
92.1
1.25
1.45
153.0
Table
8
:
Buckling control at
outer fibre of column
, ULS

a
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
167.1
0.85
1.36
198.6
L300
164.4
0.83
1.35
201.9
L400
160.6
0.81
1.33
207.4
Table
9
:
Buckling control
at neutral axis of column, ULS

b
Stiffener
A
S
[mm2]
t
eq
[mm]
I
eq
[mm4]
σ
a
[Mpa]
σ
bmax
[Mpa]
τ
max
[Mpa]
σ
h
[Mpa]
L200
3.51E+03
26.4
1.04E+13
91.6
61.3
84.9
47.9
L300
4.90E+03
28.5
1.12E+13
84.8
56.7
84.9
47.9
L400
7.75E+03
32.9
1.29E+13
73.5
49.1
84.9
47.9
Table
5
: Stress analysis ULS

b
failure
by
ψ
[

]
ρ
[

]
Z
s
[

]
ξ
[

]
C
[

]
σ
E
[Mpa]
axial stress
4
0.50
3.84
2.69
4.22
836.2
Shear stress
5.53
0.60
3.84
1.09
5.57
1102.8
Hoop stress
1.10
0.60
3.84
0.44
1.13
223.4
Table
6
: Shell elastic buckling stress
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
135.5
1.04
1.45
169.6
L300
124.6
1.09
1.45
165.3
L400
107.0
1.20
1.45
157.0
Table
10
:
Buckling control
at outer fibre of column, ULS

b
Panel stiffener buckling control
Table
12
: Elastic
panel L3
00

stiffener buckling stresses
Table
13
: Elastic panel L4
00

stiffener buckling stresses
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
162.4
1.09
1.45
165.5
L300
159.7
0.86
1.37
196.7
L400
155.9
0.70
1.27
228.8
Table
14
:
Buckling control
at neutral axis of column, ULS

a
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
114.3
1.22
1.45
155.2
L300
105.7
0.95
1.42
181.0
L400
92.1
0.78
1.32
211.4
Table
15
:
Buckling control
at outer fibre of column, ULS

a
failure
by
α
C
[

]
Z
t
[

]
ψ
[

]
ξ
[

]
ρ
[

]
C [

]
σ
ksd
[Mpa]
axial stress
106.9
81.8
67.1
57.4
0.5
73.0
679.0
Shear stress
106.9
81.8
71.7
23.3
0.6
73.1
679.6
Lateral pressure
106.9
81.8
22.8
9.4
0.6
23.5
218.2
Table
11
: Elastic panel L200

stiffener buckling stresses
failure
by
α
C
[

]
Z
t
[

]
ψ
[

]
ξ
[

]
ρ
[

]
C [

]
σ
ksd
[Mpa]
axial stress
281.6
81.8
152.9
57.4
0.5
155.5
1446.5
Shear stress
281.6
81.8
97.0
23.3
0.6
98.0
911.5
Lateral pressure
281.6
81.8
35.6
9.4
0.6
36.1
335.4
failure
by
α
C
[

]
Z
t
[

]
ψ
[

]
ξ
[

]
ρ
[

]
C [

]
σ
ksd
[Mpa]
axial stress
692.8
81.8
296.2
57.4
0.5
297.6
2768.0
Shear stress
692.8
81.8
129.1
23.3
0.6
129.8
1207.6
Lateral pressure
692.8
81.8
54.7
9.4
0.6
55.0
511.2
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
167.1
1.01
1.45
172.3
L300
164.4
0.80
1.33
209.0
L400
160.6
0.65
1.24
240.3
Table
16
:
Buckling control at neutral axis of column,
ULS

b
stiffener
σ
j
[Mpa]
[

]
γ
m
[

]
σ
ksd
[Mpa]
L200
135.5
1.08
1.45
166.4
L300
124.6
0.83
1.35
203.1
L400
107.0
0.68
1.26
234.2
Table
17
:
Buckling control at outer fibre of column, ULS

b
Column buckling should according to
the
DNV

RP

C202
code
also be checked if,
2
2.5
Y
column
A E
kL
I
Here
k
is the buckling length factor,
A
is the cross

sectional area, 2.area moment
is
denoted
I
,
and
L
is
the column length
. Since we have no information of the structural
design of the hull and the platform deck, we c
onservatively assume that all 4 columns will
buckle simultaneously in a side

sway
mechanism.
The buckling length factor is
therefore
set equa
l to 2
,
and
w
ith the L200

stiffeners
and a shell thickness of 21 mm
this gives,
2
2
3
3
13 4
26.4 10 10
2 35 10 392
1.04 10
column
A mm mm
kL mm
I mm
3
210 10
2.5 2.5 1479
355
Y
E MPa
MPa
Therefore it is
not necessary to assess the column
buckling mode.
c)
Brace A has a plate thickness of 30 mm and the service temperature is 0 °C.
The brace
is
considered to be
part of the primary structure
since it transfer
global forces from the deck
loads, and is t
herefore
assigned steel grade
B
a
ccor
ding to
Table
D1, D2 and
D3 in DNV

OS

C101 Sec. 4.
The calculated column shell has thickness 21 mm
and is a primary member.
S
teel grade
B
can
therefore
be used here.
d)
Accordin
g to Sec. 3.3.12 in DNV

RP

C203
the stress concentration facto
r,
SCF
,
for the
joint can be set equal to 2.9 since we have a gusset plate with favourable geometry
.
The
brace cross

sectional area was found to be 54
000 mm
2
in problem 1a).
The largest
stress
range in the joint during 20 years is then equal to,
3
20
2
2.9 1200 10
2 2 129
54000
brace
yr
brace
SCF N
N
MPa
A mm
A type F1
SN

curve should be used
. There is no fatigue limit since the environment is
corrosive, and the SN

curve is therefore one

sloped with m=3.0,
0.25
30
log log11.699 3 log
25
mm
N
mm
The parameters needed in the fatigue calculations
are
then
,
log log11.640 3log log 11.640 , 3.
0
N a m
The fatigue damage is given by,
1
m
D
Z
T
q m
D
T a h
Here
T
D
[s]
is the design life and
T
Z
[s]
is the zero up

crossing period.
As described in
DNV

RP

C203 the q

parameter is calculated by,
1/1.1
129
9.126
3600 24 365 20
ln
6.3
q
The
accumulated
fatigue damage can then be calculated,
3
11.64
3600 24 365 25 9.126 3 1.0
1 0.94<
6.3 10 1.1
D
DFF
The accumulated fatigue damage
,
D
,
multiplied with a design fatigue factor,
DFF
, must
be less than 1.0. According to DNV

OS

C101 Sec. 6, the
DFF
is equal to 1.0 for an
external structural member that is accessible for inspection and repair in dry and clean
conditions. Therefore, the fatigue life is sufficient for a service life of 25 years.
e)
The yield stress
in brace A
is increased by
50
% and the utilization with respect to
buckling shall
remain unchanged. Utilization with respect to
buckling
calculated
in
problem
1a) is found in
Table
2
,
0.89
m a
cr
By inserting a yield stress of 532.5 MPa in the
Excel
work
book, and thereafter adjusting
the thickness, it
can be
found that a plate thickness of 25 mm gives the same utilization
ratio with respect to buckling
as in problem 1a).
b
i
[mm]
I
[mm
4
]
A
[mm
2
]
λ
[

]
σ
cr
[Mpa]
σ
a
*
γ
m
[Mpa]
σ
a
*
γ
m
/
σ
cr
[

]
t
[mm]
25
50
7.81E+06
5000
5.51
17.0
2603.2
153.53
σ
Y
[MPa]
532.5
100
3.39E+07
10000
3.74
36.1
1301.6
36.05
l
[mm]
16971
150
91145833
15000
2.79
63.5
867.7
13.66
E
[MPa]
2.10E05
200
1.92E+08
20000
2.22
98.2
650.8
6.63
β
[

]
0.8
250
3.49E+08
25000
1.84
139.0
520.6
3.75
L
[m]
72
300
5.76E+08
30000
1.57
184.4
433.9
2.35
350
8.83E+08
35000
1.37
232.3
371.9
1.60
400
1.28E+09
40000
1.21
279.8
325.4
1.16
450
1.79E+09
45000
1.09
323.8
289.2
0.89
500
2.42E+09
50000
0.99
361.7
260.3
0.72
Table
18
: Modified plate thickness of brace A
It is assumed that
the
change in
stiffness of
members in
the deck structure members
does
not change
the
global
force transfer
of the wave loads
. The largest stress range in the joint
during 20 years is then equal to,
3
20
2
2.9 1200 10
2 2 155
45000
brace
yr
brace
SCF N
N
MPa
A mm
The
new
q

parameter
and the fatigue damage is equal to
,
1/1.1
155
10.965
3600 24 365 20
ln
6.3
q
3
11.699
3600 24 365 25 10.965 3
1 1.42
6.3 10 1.1
D
The accumulated fatigue damage is larger than 1.0, and the fatigue life of the joint must
thus
be improved. This can be done by applying weld improvement methods such as
grinding, heat relief
methods
,
peening, re
dressing of welds etc. These methods are
exp
ensive, and the best option is
usually
to change the joint design
.
In general, a large
increase
of
yield stress
should
never be accepted
without reassessment
of the
fatigue life.
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