Stiffened Plate Buckling DNV-RP-C201 Rev02-December-2011

 

a. Effective Plate Width (se)

The effective plate width for a continuous stiffener subjected to longitudinal and transverse stress and shear is calculated as follows:

CEK-A

1

Note:

2

Known values Calculated values Final results Known values: l

4 00 000 mm

s

1 00 000 mm

t

20 mm

rw g d fy E

1025 1025.0 .00 0 kg/ kg/m^3 9. 9.81 81 m/s /s^2 ^2 15.00 m 300.00 MPa 200.00 Gpa

3 Where: plate lleength or or st stiffener le len gt gth

 

Loadings:

plate w wiidth, stiffen er er sp spacing.

 

s x,Sd x,Sd

180. 180.00 00 MPa MPa

s y,Sd

50.0 50.00 0 MPa

Where: ax axia iall st stre ress ss in in pl plat atee an and d stif stiffe fene nerr wi with th ccom ompr pres essi sive ve sstr tres esse sess as pos posit itiv ivee

trans ransv ver erse se stre stress ss in p pllate an and d ssttif ifffen ener  er 

plate thickness

pSd 150828. 150828.75 75 Pa=N Pa=N/m^2 /m^2 design design lateral lateral force, force, for this case case is the hydrosta hydrostatic tic pressure pressure at d

wat ater er dens densiity gra gravi vity ty acc ccel eleerat ratio ion n hydrostatic pressure depth

pSd

0.15 0.15 MP MPaa

Reduced plate slenderness:

characteristic yield strength Young modulus of elasticity

Where:

1.017 [] reduced plate slenderness in longitudinal direction 2.130 [] reduced plate slenderness in transverse direction 0.673, then: Cxs= 0.771 [] > where: Cxs is reduction factor due to stresses in the longitudinal direction

Since The reduction factor due to lateral load, k p:

CEK-A

0.24 MPa; it means pSd

Thus k p

1. 1.00 000 0 []

Since, µ (1) (2)

0.405 [], and 1.000 [] 0.198 [] []

(3)

=

s, then k l=

Since k g

>

LG, then k g

axial axial stress stress in pla plate te and stiffe stiffener ner with with com compre pressi ssive ve stress stresses es as pos positi itive ve 5.59 [], where k l is buckling factor for plate between stiffeners. 13.49 [], where k G is buckling factor for plate with the stiffeners removed.

LG

3000 mm

girder length

wfl

180 mm

flange width

tfl

20 mm

hw

400 mm

tw

10 mm

tSd

0.00 MPa

design shear stress

The cross sectional area of stiffener is As

gM

1.15 []

resulting material factor  

and the full plate cross section area s.t =

20000 20000 mm^2 mm^2

Therefore, the equivalent axial force, NSd

4.96 4.968 8 MN

Thus:  

flange thickness web height

 

t crl

40 404. 4.27 27 MP MPaa

t crg

60 60.9 .99 9 MP MPaa

Since tSd