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. 0.7 ULS-b 1.0 1. Table 1: ULS load factors R V SV ES E m 1.15 m 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 SA sin 45 ql ( braces in the trusswork carry shear force) q q q V F E E We assume that the functional and permanent loads are related to weight of uipment 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, q E.5 q 9.81 F ULS-a can be found to give the most critical load condition with the following axial force in brace A, S A 0.5 q L F 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 0.1, 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, A l Y EI, where A is the cross-sectional area, β is the buckling length factor, l is the brace length and I is the.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 0.1 a design value of β=0.8 should be used instead. The internal width of the cross-section is denoted b i and the thickness as t. Area and.area moment is then expressed as, A 4bt i 1 4 4 I 1 b i t b i The cross-section is assumed to be hot finished and therefore assigned buckling curve a, see DNV Classification Notes 0.1. Then the characteristic buckling stress can be expressed as, 1 0. 0. 1 0. 0. 4 CR Y A cross-section width, b i, which fulfills the following relationship must be selected, ql F b 0.11 1.15 A m CR i A m bt i The uations above are included in EXCEL to determine the ruired cross-section width. Here, it was found that an internal width/height of 450 mm will give sufficient capacity against column buckling. See Table. bi 450 mm
b i [mm] I [mm 4 ] A [mm ] λ [-] σ cr σ a *γ m σ a *γ m / σ cr [-] t [mm] 0 50 1.17E07 6000 4.0 0.8 169.4 15.5 σ Y [MPa] 55 100 4.6E07 1000.86 40.4 1084.7 6.05 l [mm] 16971 150 1.E08 18000.18 67.9 7.1 1.66 E [MPa].10E05 00.47E08 4000 1.75 101.5 54. 6.6 β [-] 0.8 50 4.44E08 0000 1.46 19. 4.9.75 L [m] 7 00 7.5E08 6000 1.5 178. 61.6.5 50 1.1E09 4000 1.10 14.6 09.9 1.60 400 1.6E09 48000 0.97 45.1 71. 1.16 450.E09 54000 0.88 68.4 41.0 0.89 500.99E09 60000 0.80 85.5 16.9 0.7 Table : 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 gh F N N N F F E E M M M F F E E Q Q Q F F E E The pressure is calculated at a depth corresponding to 0 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.61E5 79 76 6.5 ULS-b.01E5 76 17 8 Table : ULS stress resultant in buckling check The stresses in the cylindrical shell structure must be determined. According to the DNV- RP-C0 code, it is usually permissible to account for the longitudinal stiffeners by an uivalent shell thickness when axial membrane stresses are calculated, AS t t s A Dt I D t 8
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, N a A MD bmax 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 stress is given by, max Q Dt The hoop (circumferential) stress is calculated as, h pd t According to the DNV-RP-C0 code it is necessary to consider 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,,, Ex E Eh E t C 1(1 ) s 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-C0 code Sec. 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-C0 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 TMR405 - 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,,, Ex E Eh E 1(1 ) t C l 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.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 ual to 75 mm. The distance e z, stiffener area, A s and stiffener.area moment, I s, are found in Figure A in the exercise appendix. I eff can then be calculated by, st e dz set AS 1 I 0.5 eff set ez t dz set Is dz As 1 In the following the procedure recommended in the DNV-RP-C0 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 uivalent reduced slenderness, Y a b h, j Ex Ex E Eh where 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 ual to zero in the uation for the uivalent reduced slenderness.
Plasticity is accounted for by the uivalent stress, σ j, j b a h b a h Note that the normal stress components are not set ual to zero if they are tensile in the uivalent stress uation. Thereafter, the DNV-RP-C0 code calculate the design buckling strength as, Y ksd 4 1 m Where the material factor γ m depends on the uivalent reduced slenderness, 1.15 for 0.5 m 0.85 0.60 for 0.5 < 1.0 1.45 for 1.0 Finally, the buckling capacity is found acceptable if the uivalent 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 uivalent 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 ual. Based on this, the design which gives optimal material costs must be based on the L00- stiffeners and a shell thickness of 1 mm. See results in the next pages.
Stress analysis based on uivalent shell thickness Stiffener A S [mm] t [mm] I [mm4] σ a σ bmax τ max σ h L00.51E+0 6.4 1.04E+1 95. 6.7 80. 6.1 L00 4.90E+0 8.5 1.1E+1 88.1.9 80. 6.1 L400 7.75E+0.9 1.9E+1 76.4 9.4 80. 6.1 Table 4: Stress analysis ULS-a Shell buckling control Stiffener A S [mm] t [mm] I [mm4] σ a σ bmax τ max σ h L00.51E+0 6.4 1.04E+1 91.6 61. 84.9 47.9 L00 4.90E+0 8.5 1.1E+1 84.8 56.7 84.9 47.9 L400 7.75E+0.9 1.9E+1 7.5 49.1 84.9 47.9 Table 5: Stress analysis ULS-b failure by ψ[-] ρ [-] Z s [-] ξ [-] C [-] σ E axial stress 4 0.50.84.69 4. 86. Shear stress 5.5 0.60.84 1.09 5.57 110.8 Hoop stress 1.10 0.60.84 0.44 1.1.4 Table 6: Shell elastic buckling stress stiffener σ j [-] γ m [-] σ ksd L00 16.4 1.01 1.45 17.4 L00 159.7 1.00 1.45 17. L400 155.9 0.98 1.44 175.8 Table 7: Buckling control at neutral axis of column, ULS-a stiffener σ j [-] γ m [-] σ ksd L00 114. 1.16 1.45 159.6 L00 105.7 1.19 1.45 157. L400 9.1 1.5 1.45 15.0 Table 8: Buckling control at outer fibre of column, ULS-a stiffener σ j [-] γ m [-] σ ksd L00 167.1 0.85 1.6 198.6 L00 164.4 0.8 1.5 01.9 L400 160.6 0.81 1. 07.4 Table 9: Buckling control at neutral axis of column, ULS-b
stiffener σ j [-] γ m [-] σ ksd L00 15.5 1.04 1.45 169.6 L00 14.6 1.09 1.45 165. L400 107.0 1.0 1.45 157.0 Table 10: Buckling control at outer fibre of column, ULS-b Panel stiffener buckling control failure by α C [-] Z t [-] ψ [-] ξ [-] ρ[-] C [-] σ ksd axial stress 106.9 81.8 67.1 57.4 0.5 7.0 679.0 Shear stress 106.9 81.8 71.7. 0.6 7.1 679.6 Lateral pressure 106.9 81.8.8 9.4 0.6.5 18. Table 11: Elastic panel L00-stiffener buckling stresses failure by α C [-] Z t [-] ψ [-] ξ [-] ρ[-] C [-] σ ksd axial stress 81.6 81.8 15.9 57.4 0.5 155.5 1446.5 Shear stress 81.6 81.8 97.0. 0.6 98.0 911.5 Lateral pressure 81.6 81.8 5.6 9.4 0.6 6.1 5.4 Table 1: Elastic panel L00-stiffener buckling stresses failure by α C [-] Z t [-] ψ [-] ξ [-] ρ[-] C [-] σ ksd axial stress 69.8 81.8 96. 57.4 0.5 97.6 768.0 Shear stress 69.8 81.8 19.1. 0.6 19.8 107.6 Lateral pressure 69.8 81.8 54.7 9.4 0.6 55.0 511. Table 1: Elastic panel L400-stiffener buckling stresses stiffener σ j [-] γ m [-] σ ksd L00 16.4 1.09 1.45 165.5 L00 159.7 0.86 1.7 196.7 L400 155.9 0.70 1.7 8.8 Table 14: Buckling control at neutral axis of column, ULS-a stiffener σ j [-] γ m [-] σ ksd L00 114. 1. 1.45 155. L00 105.7 0.95 1.4 181.0 L400 9.1 0.78 1. 11.4 Table 15: Buckling control at outer fibre of column, ULS-a
stiffener σ j [-] γ m [-] σ ksd L00 167.1 1.01 1.45 17. L00 164.4 0.80 1. 09.0 L400 160.6 0.65 1.4 40. Table 16: Buckling control at neutral axis of column, ULS-b stiffener σ j [-] γ m [-] σ ksd L00 15.5 1.08 1.45 166.4 L00 14.6 0.8 1.5 0.1 L400 107.0 0.68 1.6 4. Table 17: Buckling control at outer fibre of column, ULS-b Column buckling should according to the DNV-RP-C0 code also be checked if, A kl I column E.5 Y Here k is the buckling length factor, A is the cross-sectional area,.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 ual to, and with the L00-stiffeners and a shell thickness of 1 mm this gives, A 6.4mm 1010 mm kl 5 10 mm 9 1 4 I 1.0410 mm column E 1010 MPa.5.5 1479 55MPa Y Therefore it is not necessary to assess the column buckling mode.
c) Brace A has a plate thickness of 0 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, D and D in DNV-OS-C101 Sec. 4 be assigned to brace A. The calculated column shell has thickness 1 mm and is a primary member. Steel grade B can therefore be used here. d) According to Sec...1 in DNV-RP-C0 the stress concentration factor, SCF, for the joint can be set ual to.9 since we have a gusset plate with favourable geometry. The brace cross-sectional area was found to be 54 000 mm in problem 1a). The largest stress range in the joint during 0 years is then ual to, SCF Nbrace.9100 10 N 0 yr 19MPa A 54000mm brace 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=.0, 0.5 0mm log N log11.699 log 5mm The parameters needed in the fatigue calculations are then, log N log11.640 log log a 11.640, m.0 The fatigue damage is given by, m TD q m D 1 T a h Z Here T D [s] is the design life and T Z [s] is the zero up-crossing period. As described in DNV-RP-C0 the q-parameter is calculated by, 19 q 600 465 0 ln 6. 1/1.1 9.16 The accumulated fatigue damage can then be calculated, 600 4655 9.16 1.0 D 1 0.94< 11.64 6. 10 1.1 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 ual 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 5 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, ma 0.89 cr By inserting a yield stress of 5.5 MPa in the Excel workbook, and thereafter adjusting the thickness, it can be found that a plate thickness of 5 mm gives the same utilization ratio with respect to buckling as in problem 1a). b i [mm] I [mm 4 ] A [mm ] λ [-] σ cr σ a *γ m σ a *γ m / σ cr [-] t [mm] 5 50 7.81E+06 5000 5.51 17.0 60. 15.5 σ Y [MPa] 5.5 100.9E+07 10000.74 6.1 101.6 6.05 l [mm] 16971 150 911458 15000.79 6.5 867.7 1.66 E [MPa].10E05 00 1.9E+08 0000. 98. 650.8 6.6 β [-] 0.8 50.49E+08 5000 1.84 19.0 50.6.75 L [m] 7 00 5.76E+08 0000 1.57 184.4 4.9.5 50 8.8E+08 5000 1.7. 71.9 1.60 400 1.8E+09 40000 1.1 79.8 5.4 1.16 450 1.79E+09 45000 1.09.8 89. 0.89 500.4E+09 50000 0.99 61.7 60. 0.7 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 0 years is hence ual to, SCF Nbrace.9100 10 N 0 yr 155MPa A 45000mm brace The new q-parameter and the fatigue damage is, 155 q 10.965 1/1.1 600 465 0 ln 6. 600 465 5 10.965 D 1 1.4 11.699 6. 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.