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.
ULS-a
ULS-b
Functional & permanent,  F
1.3
1.0
Environmental,  E
0.7
1.3
Table 1: ULS load factors
 V SV   E S E 
R
m
 m  1.15
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 SA in brace A by,
1
qL
2
q   V qF   E qE
2S A sin 45 
(2 braces in the trusswork carry shear force)
We assume that the functional and permanent loads are related to weight of equipment
and weight of deck structure, such that inertia forces caused by the vertical wave-induced
acceleration will be evenly distributed along the main girder with intensity given by,
qE 
2.5
qF
9.81
ULS-a can be found to give the most critical load condition with the following axial force
in brace A,
S A  0.523  qF L
The brace must be checked for beam-column buckling failure, i.e. interaction between
axial compressive force and bending moments should be included in the assessment. The
bending moments cannot be found 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
characteristic column buckling resistance, is outlined in the following. The input
parameter in this procedure is the reduced slenderness ratio,
 Y A 2l 2

,
 2 EI
where A is the cross-sectional area, β is the buckling length factor, l is the brace 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, according to the DNV Classification note 30.1 a design value of β=0.8
should be used instead.
The internal width of the cross-section is denoted bi and the thickness as t. Area and
2.area moment is then expressed as,
A  4bi t
1
4
I   bi  2t   bi4 

12 
The cross-section is assumed to be “hot finished” and therefore assigned buckling curve a,
see DNV Classification Notes 30.1. Then the characteristic buckling stress can be
expressed as,
2


2
2
 4 2 
 1  0.2    0.2     1  0.2    0.2   
 CR   Y 

2 2






A cross-section width, bi, which fulfills the following relationship must be selected,
 A m   CR  bi 
 A  0.131 
qF L
bi t
 m  1.15
The equations above are included in EXCEL to determine the required cross-section
width. Here, it was found that an internal width/height of 450 mm will give sufficient
capacity against column buckling. See Table 2.
bi  450 mm
t [mm]
σY [MPa]
l [mm]
E [MPa]
β [-]
L [m]
bi
[mm]
50
100
150
200
250
300
350
400
450
500
30
355
16971
2.10E05
0.8
72
I
[mm4]
1.17E07
4.63E07
1.2E08
2.47E08
4.44E08
7.25E08
1.1E09
1.6E09
2.22E09
2.99E09
A
[mm2]
6000
12000
18000
24000
30000
36000
42000
48000
54000
60000
λ
[-]
4.03
2.86
2.18
1.75
1.46
1.25
1.10
0.97
0.88
0.80
σcr
σa *γm
[Mpa]
20.8
40.4
67.9
101.5
139.3
178.3
214.6
245.1
268.4
285.5
[Mpa]
2169.4
1084.7
723.1
542.3
433.9
361.6
309.9
271.2
241.0
216.9
σa *γm / σcr
[-]
153.53
36.05
13.66
6.63
3.75
2.35
1.60
1.16
0.89
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,
p   F  gh
N   F NF   E NE
M   FMF   EME
Q   F QF   E QE
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]
2.61E5
2.01E5
ULS-a
ULS-b
N [MN]
79
76
M [MN]
76
127
Q [MN]
26.5
28
Table 3: ULS stress resultant in buckling check
The stresses in the cylindrical shell structure must be determined. According to the DNVRP-C202 code, it is usually permissible to account for the longitudinal stiffeners by an
equivalent shell thickness when axial membrane stresses are calculated,
AS
s
A eq   Dteq
teq  t 
I eq 

8
D 3teq
Here t is the shell thickness, As 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,
N
Aeq
MD
 b max 
2 I eq
a 
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 stress is
given by,
 max 
2Q
 Dt
The hoop (circumferential) stress is calculated as,
h 
pD
2t
According to the DNV-RP-C202 code it is necessary to consider 3 buckling modes for
longitudinally stiffened shells.



Shell buckling
Panel stiffener buckling
Column buckling
Shell buckling
Panel stiffener buckling
Column buckling
Shell buckling can be trigged by three different stress components.



Axial membrane stress caused by axial compressive force and compressive
bending stress in the column.
Shear stresses from torsion or shear force in the column.
Hoop stresses caused by net external pressure.
The three elastic shell buckling stresses are given by formulas valid from curved shell
theory. i.e. we focus on the shell plating enclosed by two longitudinal stiffeners. The
formulas are expressed in the following form,
 2E  t 
  C
12(1  2 )  s 
2
 Ex ,  E ,  Eh 
C  1
  

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 the curved shell solution. Cylindrical shells 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, see 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 buckling stresses in the DNV-RP-C202 code are expressed by formulas
based on orthotropic shell theory where the stiffeners are smeared 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”). The elastic buckling stresses are given by formulas on a
similar form as for the shell buckling problem,
 2E  t 

  C
12(1  2 )  l 
2
 Ex ,  E ,  Eh
C  1
  

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, Ieff, which is defined
by the effective shell flange, se. See Figure 1.
Figure 1: Effective shell flange
In this problem the effective shell flange was set equal to 275 mm. The distance ez,
stiffener area, As and stiffener 2.area moment, Is, are found in Figure A in the exercise
appendix. Ieff can then be calculated by,
set
set  AS
1
2
 set 3   ez  0.5t  d z  set  I s  d z2 As
12
dz 
I eff
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 formula for the equivalent reduced slenderness,
eq2 
Y
j
 a b
 



 h ,

  Ex  Ex  E  Eh 
where σY is the yield strength, σa is axial compressive stress from axial force, σb is
compressive bending stress, τ is shear stress and σh 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.
Plasticity is accounted for by the equivalent stress, σj,
j 
  b   a     h     b   a    h   3 2
2
2
2
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,
 ksd 
Y
 m 1  eq4
Where the material factor γm depends on the equivalent reduced slenderness,
1.15

 m  0.85  0.60eq

1.45
for eq  0.5
for 0.5 < eq  1.0
for eq  1.0
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. Two locations of the column are checked in the buckling assessment.


The “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 minimum shell thickness
which can be used.
If the stiffener scantlings are increased, a pronounced increase of equivalent 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 L200stiffeners and a shell thickness of 21 mm. See results in the next pages.
Stress analysis based on equivalent shell thickness
Stiffener
L200
L300
L400
AS
teq
Ieq
τmax
σa
σbmax
σh
[mm2]
[mm] [mm4]
[Mpa] [Mpa] [Mpa] [Mpa]
3.51E+03
26.4 1.04E+13
95.3
36.7
80.3
62.1
4.90E+03
28.5 1.12E+13
88.1
33.9
80.3
62.1
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
L200
L300
L400
AS
teq
Ieq
τmax
σa
σbmax
σh
[mm2]
[mm] [mm4]
[Mpa] [Mpa] [Mpa] [Mpa]
3.51E+03
26.4 1.04E+13
91.6
61.3
84.9
47.9
4.90E+03
28.5 1.12E+13
84.8
56.7
84.9
47.9
7.75E+03
32.9 1.29E+13
73.5
49.1
84.9
47.9
Table 5: Stress analysis ULS-b
failure by
axial stress
Shear stress
Hoop stress
ρ [-] Zs [-]
ξ [-]
C [-]
σE [Mpa]
4 0.50
3.84
2.69 4.22
836.2
5.53 0.60
3.84
1.09 5.57
1102.8
1.10 0.60
3.84
0.44 1.13
223.4
ψ[-]
Table 6: Shell elastic buckling stress
stiffener
L200
L300
L400
σj [Mpa]
162.4
159.7
155.9
 [-] γm [-] σksd [Mpa]
1.01
1.00
0.98
1.45
1.45
1.44
172.4
173.3
175.8
Table 7: Buckling control at neutral axis of column, ULS-a
stiffener
L200
L300
L400
σj [Mpa]
114.3
105.7
92.1
 [-] γm [-] σksd [Mpa]
1.16
1.19
1.25
1.45
1.45
1.45
159.6
157.2
153.0
Table 8: Buckling control at outer fibre of column, ULS-a
stiffener
L200
L300
L400
σj [Mpa]
167.1
164.4
160.6
 [-] γm [-] σksd [Mpa]
0.85
0.83
0.81
1.36
1.35
1.33
198.6
201.9
207.4
Table 9: Buckling control at neutral axis of column, ULS-b
stiffener
L200
L300
L400
σj [Mpa]
135.5
124.6
107.0
 [-] γm [-] σksd [Mpa]
1.04
1.09
1.20
1.45
1.45
1.45
169.6
165.3
157.0
Table 10: Buckling control at outer fibre of column, ULS-b
Panel stiffener buckling control
failure by
axial stress
Shear stress
Lateral pressure
αC [-] Zt [-] ψ [-] ξ [-] ρ[-]
106.9 81.8 67.1 57.4
0.5
106.9 81.8 71.7 23.3
0.6
106.9 81.8 22.8 9.4
0.6
C [-] σksd [Mpa]
73.0
679.0
73.1
679.6
23.5
218.2
Table 11: Elastic panel L200-stiffener buckling stresses
failure by
axial stress
Shear stress
Lateral pressure
αC [-] Zt [-] ψ [-] ξ [-] ρ[-]
C [-]
σksd [Mpa]
281.6 81.8 152.9 57.4
0.5 155.5
1446.5
281.6 81.8 97.0 23.3
0.6 98.0
911.5
281.6 81.8 35.6 9.4
0.6 36.1
335.4
Table 12: Elastic panel L300-stiffener buckling stresses
failure by
axial stress
Shear stress
Lateral pressure
αC [-] Zt [-] ψ [-] ξ [-] ρ[-]
C [-]
σksd [Mpa]
692.8 81.8 296.2 57.4
0.5 297.6
2768.0
692.8 81.8 129.1 23.3
0.6 129.8
1207.6
692.8 81.8 54.7 9.4
0.6 55.0
511.2
Table 13: Elastic panel L400-stiffener buckling stresses
stiffener
L200
L300
L400
σj [Mpa]
162.4
159.7
155.9
 [-] γm [-] σksd [Mpa]
1.09
0.86
0.70
1.45
1.37
1.27
165.5
196.7
228.8
Table 14: Buckling control at neutral axis of column, ULS-a
stiffener
L200
L300
L400
σj [Mpa]
114.3
105.7
92.1
 [-] γm [-] σksd [Mpa]
1.22
0.95
0.78
1.45
1.42
1.32
155.2
181.0
211.4
Table 15: Buckling control at outer fibre of column, ULS-a
σj [Mpa]
stiffener
L200
L300
L400
 [-] γm [-] σksd [Mpa]
167.1
164.4
160.6
1.01
0.80
0.65
1.45
1.33
1.24
172.3
209.0
240.3
Table 16: Buckling control at neutral axis of column, ULS-b
σj [Mpa]
stiffener
L200
L300
L400
 [-] γm [-] σksd [Mpa]
135.5
124.6
107.0
1.08
0.83
0.68
1.45
1.35
1.26
166.4
203.1
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
 A 
E
kL 
 2.5

Y
 I column
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 conservatively assume that all 4 columns will
buckle simultaneously in a side-sway mechanism. The buckling length factor is therefore
set equal to 2, and with the L200-stiffeners and a shell thickness of 21 mm this gives,
2
   26.4mm 10 103mm

 A 
3
kL


2

35

10
mm

  392




1.04 1013mm4
 I column 

2
2.5
E
Y
 2.5 
210 103 MPa
 1479
355MPa
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 a primary member since it is part of the global force transfer of waveinduced loads, deck loads etc. Thus, steel grade B can according to Table D1, D2 and D3
in DNV-OS-C101 Sec. 4 be assigned to brace A.
The calculated column shell has thickness 21 mm and is a primary member. Steel grade B
can therefore be used here.
d) According to Sec. 3.3.12 in DNV-RP-C203 the stress concentration factor, 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 mm2 in problem 1a). The largest stress
range in the joint during 20 years is then equal to,
SCF  Nbrace
2.9 1200 103 N
 20 yr  2 
 2
 129MPa
Abrace
54000mm2
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

 30mm  
log N  log11.699  3  log   
 

25
mm

 

The parameters needed in the fatigue calculations are then,
log N  log11.640  3log   
 log a  11.640 , m  3.0
The fatigue damage is given by,
T qm  m 
D D
 1  
TZ a 
h
Here TD [s] is the design life and TZ [s] is the zero up-crossing period. As described in
DNV-RP-C203 the q-parameter is calculated by,
q
129
1/1.1
  3600  24  365  20  

 ln 
6.3

 
 9.126
The accumulated fatigue damage can then be calculated,
3600  24  365  25 9.1263 
3 
1.0
D
 11.64  1    0.94<
6.3
10
DFF
 1.1 
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,
 
  m a  0.89
 cr
By inserting a yield stress of 532.5 MPa in the Excel workbook, 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).
t [mm]
σY [MPa]
l [mm]
E [MPa]
β [-]
L [m]
25
532.5
16971
2.10E05
0.8
72
bi
[mm]
50
100
150
200
250
300
350
400
450
500
I
[mm4]
7.81E+06
3.39E+07
91145833
1.92E+08
3.49E+08
5.76E+08
8.83E+08
1.28E+09
1.79E+09
2.42E+09
A
[mm2]
5000
10000
15000
20000
25000
30000
35000
40000
45000
50000
λ
[-]
5.51
3.74
2.79
2.22
1.84
1.57
1.37
1.21
1.09
0.99
σcr
σa *γm
[Mpa]
17.0
36.1
63.5
98.2
139.0
184.4
232.3
279.8
323.8
361.7
[Mpa]
2603.2
1301.6
867.7
650.8
520.6
433.9
371.9
325.4
289.2
260.3
σa *γm /
σcr
[-]
153.53
36.05
13.66
6.63
3.75
2.35
1.60
1.16
0.89
0.72
Table 18: Modified plate thickness and yields stress of brace A
It is assumed that the change in stiffness of members in the deck structure does not
change the global force transfer of the wave-induced loads. The largest stress range in the
joint during 20 years is hence equal to,
SCF  Nbrace
2.9 1200 103 N
 20 yr  2 
 2
 155MPa
Abrace
45000mm2
The new q-parameter and the fatigue damage is,
155
q
 10.965
1/1.1
  3600  24  365  20  

 ln 
6.3

 
3600  24  365  25 10.9653 
3 
D
 11.699  1    1.42
6.3
10
 1.1 
The accumulated fatigue damage is larger than 1.0, and the fatigue life of the joint must
be improved. This can be done by applying weld improvement methods such as grinding,
heat relief methods, peening, redressing of welds etc. These methods are expensive, 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.