Buckling of Stiffened Panels 1overall buckling vs plate bucklingPCCB Panel Collapse Combined Buckling

Various estimates have been developed to determine the minimum size stiffener to insure the plate buckles while the stiffener remains straight. this is equivalent to insuring that plate buckling occurs before overall buckling.

Timoshenko does so by calculating k in σcr kπ

2D⋅

b2 t⋅⋅:= and observing the

value of γ which results in a critical stress above that which will cause plate buckling alone. where;

γflexural_rigidity_of_combined_section

flexural_rigidity_of_plate:= γ

E Ix⋅

D b⋅:= where;

Ix is the inertia of the plate with the attached plate associated with individual stiffener.

Bleich pg 365, 367, for plates with longitudinal stiffness determines minimum γ to insure the plate buckles before the stiffener (overall buckling). unfortunately Bleich uses different ratios than Hughes. Bleich uses B where Hughes uses b.

for the following: αaB

:= δAx

B t⋅:= γ

E Ix⋅

D B⋅:=

α 0.1 0.2, 10..:= δ 0.1:= κ1 α n,( ) nπ

α⋅

4 α⋅

n1+⋅:= κ2 α n,( ) n

π

α⋅

4 α⋅

n1−⋅:=

γo α δ, n,( )

16

π2

α

n

3⋅

1κ1 α n,( ) tanh

κ1 α n,( )2

⋅1

κ2 α n,( ) tanκ2 α n,( )

2

⋅−

16α

n

2⋅ δ⋅+:=

γomax α δ,( ) max

γo α δ, 1,( )

γo α δ, 2,( )

γo α δ, 3,( )

:=

0 5 100

20

40

γo α δ, 1,( )

γo α δ, 2,( )

γo α δ, 3,( )

α

0 5 100

20

40

γomax α δ,( )

α

1

1 2 3 4 50

200

400

600

800

γbx21 Π δx,( )γbx22 Π δx,( )

Π

0 2 4 60

100

200

300

γbx11 Π δx,( )γbx12 Π δx,( )

Π

Π 1 1.1, 5..:=

γbx22 Π δx,( ) 288 610 δx⋅+ 325 δx2

⋅+:=γbx12 Π δx,( ) 48.8 112 δx⋅ 1 0.5 δx⋅+( )⋅+:=

γbx21 Π δx,( ) 43.5 Π3

⋅ 36 Π2

⋅ δx⋅+:=γbx11 Π δx,( ) 22.8 Π⋅ 2.5 16 δx⋅+( ) Π2

⋅+ 10.8 Π⋅−:=

two stiffenersone stiffener

to move to Bleich relationships in Hughes terms (ratios) to match text:

where: ΠaB

:= δxAx

b t⋅:= γ

E Ix⋅

D b⋅:=

Fig. 180, Bleich

0 5 100

10

20

30

40

γomax α δ,( )

γbx α δ,( )

α

γbx α δ,( ) minγbx1 α δ,( )

γbx2 α δ,( )

:=γbx2 α δ,( ) 48.8

2112 δ⋅ 1

0.52

2⋅ δ⋅+

⋅+:=Hughes: 13.1.4slightly modified in terms

γbx1 α δ,( ) 22.82

α⋅2.52

16 δ⋅+

α2

⋅+10.8

2α⋅−:=

one stiffener, combining Hughes and Bleich terms, N=1 stiffenercurve fit:

2

t 0.5=b 30:=L 96:=d 5:=As Aw Af+:=Af 0.85:=Aw 0.8:=

set abovebreadth betweenstiffeners b

panel lengthstiffener depth stiffener areastiffener flangestiffener web

A more direct approach is to calculate the overall buckling stress and insure it is larger than the plate critical stress.

The overall buckling stress is the value at which the stiffeners reach critical stress, modeling each stiffener as a column of stiffener with attached (portion) of plate with some equivalent slenderness ratio.

λ L_over_ρeq:=

We will continue to model the plate failure as a gradual failure i.e the center of the plate "fails" in buckling while the outer section remains effective at an effective breadth be paradoxically, the column is "stiffer" when the plate flange (be) is reduced for ratios typical of ship structure: let's first evaluate the plate and column critical stresses for a short panel. As we assumed in plate buckling (and bending) the width is such that we can model a slice independently:

the column is a stiffener with an attached plate of width bthe plate is a width b some typical scantlings:

1 2 3 4 50

100

200

300

400

500

2 stiffeners1 stiffener

minimum ratio of EI/Db (A/bt = 0.1)

panel aspect ratio = a/B

gam

ma

= EI

/Db

γb Π δx, N,( ) if N 1= minγbx11 Π δx,( )γbx12 Π δx,( )

, min

γbx21 Π δx,( )γbx22 Π δx,( )

,

:=

combination

3

calculate in terms of be, initially be = bmoment of inertia using 8.3.6 to calculate radius of gyration:

Ae be( ) As be t⋅+:= C1 be( )Aw

Ae be( )3

Aw

4−

⋅ Af be⋅ t⋅+

Ae be( )( )2:= Ie be( ) Ae be( ) d( )2

⋅ C1 be( )⋅:=

ρe be( )Ie be( )Ae be( ):=

column critical stress: σe_cr be( ) π2

E⋅

Lρe be( )

2:= ; initial value σe_cr b( ) 49341=

plate critical stress: σa_cr be( ) 3.615 E⋅tbe

2⋅:= initial value σa_cr b( ) 30125=

things are ok as column > plate => plate "fails" first. now consider increasing stress beyond σa_cr b( ) and plate gradually fails reducing effective breadth. Note that we are using an assumption due to von Karman, that the "failed" center region has no compressive stress while the outer regions are fully effective at σe defined from force equilibrium as

σe be( )σa b⋅

be:= now consider what happens to the values of critical stress as effective

breadth is reduced. the definitions above are still active; repeated here for info:

be 15 b..:= these values are not "short" but with appropriate scantlings the result is the same. this set of values assumes yield stress is > 3x10^4 or so.

15 20 25 302 .104

4 .104

6 .104

8 .104

1 .105

column (stiffener with plate)plate alone

effective breadth

criti

cal s

tress

column: σe_cr be( ) π2

E⋅

Lρe be( )

2:=

plate: σa_cr be( ) 3.615 E⋅t

be

2⋅:=

4

even though the numbers above represent a long column (plate), the analysis is only appropriate for short panels. to analyze long panels a correction is needed to λ = L_over_ρ to account for the tendency of the plate to deflect with more than one half sine wave making the column slightly stiffer (smaller slenderness ratio λ). Τhe correction is as follows:

CπBa

γx be( )2 1 1 γx be( )++( )⋅

⋅:=be or Cπ 1:= whichever is less. γx be( )

E Ix be( )⋅

D be⋅:=

and L_over_ρeq

= CπL

ρ be( )⋅ then σe_cr be( ) π

2E⋅

CπL

ρe be( )⋅

2:=

this adds a non linearity to the problem so an iterative method is used to make the initial comparison of whether the column critical stress is greater than the plate critical stress or to solve for the common critical stress. The text proposes an iterative method similar to that below for PCCB to determine the common value of critical stress. After doing so it is necessary to bring this stress (based on effective breadth) back to applied (average) stress; again using statics

σa b t⋅ Ax+( )⋅ = σe be t⋅ Ax+( )⋅ and σa be( )be t⋅ Ax+( )b t⋅ Ax+( ) σe_cr be( )⋅:= at the effective

breadth that the iteration converges.

Some calculations to validate the column becoming "stiffer" with less effective plate. Think of the location of the neutral axis or the radius of gyration as the plate is reduced. With a wide plate flange the neutral axis is close to the plate. The radius of gyration which plays in the critical stress varies as follows: (recall larger ρ => larger critical stress:

be 0 b..:=

0 5 101

1.5

2

2.5

ρ e be( )

ρ ebe

2

be t⋅

As

0 2 4 6 8 100

5 .104

1 .105

1.5 .105

2 .105

σe_cr be( )

σe_crbe

2

be t⋅

As

now for our design rules for PCCB Panel Collapse Combined Buckling:

5

d SDEPTHTSF

2−

t2

+:= B N 1+( ) b⋅:= Ap b t⋅:=b t⋅As

7.53= δxAs

b t⋅:=

d 5.14= B 75= Ap 12.5= δx 0.13=

parameters that are functions of be

let: be 0.7657 b⋅:= as a place to start and for printing purposes. Can also use be directly andcompare with br be( ). If br be( )/ b < start use result until converges.

Ae be( ) As be t⋅+:=

C1 be( )Aw

Ae be( )3

Aw

4−

⋅ Af be⋅ t⋅+

Ae be( )( )2:= Ie be( ) Ae be( ) d( )2

⋅ C1 be( )⋅:=

Ae be( ) 11.23= C1 be( ) 0.09= Ie be( ) 25.87=

ρe be( )Ie be( )Ae be( ):=

γx be( )12 1 υ

2−( )⋅ Ie be( )⋅

be t( )3⋅

:=Cπ min

Ba

γx be( )2 1 1 γx be( )++( )⋅

⋅

1

:=

ρe be( ) 1.52=γx be( ) 118.07=

for iterationCπ 1=

br be( )Cπ a⋅ t⋅

ρe be( ) 3 1 υ2

−( )⋅⋅

:= br be( )b

0.7657= Checks?? bratbr be( )

b:= brat 0.766=

br be( ) 19.14=

- PCCB - Panel Collapse Combined Bucklingfunction of be, graphical approach

btp

HSF

a

b input dataignore numerical values shown here

input 1:=

stiffener

BSF 3.94:= SDEPTH 5.:= TSF .215:= TSW .17:=

try this with SDEPTH = 6, to see yield effect plate

a 8 12⋅:= b 25:= t .5:=

As 1.66=Af 0.85=Aw 0.81=HSW 4.79=

As Aw Af+:=Af BSF TSF⋅:=Aw SDEPTH TSF−( ) TSW⋅:=HSW SDEPTH TSF−:=

general parameters:

DE t3⋅

12 1 υ2

−( )⋅

:=E 29.6 106⋅:=υ 0.3:=σY 80 103

⋅:=

material

L 96=L a:=

N 2:=

6

assuming iteration complete and matched above.

be 19.14=be brat b⋅:=now obseving intersection:

12 14 16 18 20 22 24 264 .104

6 .104

8 .104

1 .105

1.2 .105

1.4 .105

1.6 .105

1.8 .105

σecr be( )−

σecr_pl be( )−

σY

σo−

be

beb2

b2

0.1+, b..:=

for plotting to determine intersection;

σecr_pl be( ) 73008−=σecr be( ) 73015−=

σo 42804−=σo 42804−=

σo 4−π

2D⋅

b2 t⋅⋅:=OR....σo 3.62 E⋅

tb

2⋅:=

σecr_pl be( ) 4−π

2D⋅

be2 t⋅

⋅:=σecr be( ) π2

− E⋅

Cπ L⋅

ρe be( )

2:=

original plateeffective portion of platestiffener & plate as a column

critical stress relationships:

7

γRPCCB 0.51801=

λ 1.05=γRPCCB γC

σC

σaxcr⋅:=

λa

π ρe be( )⋅

σY

E⋅:=

is it slender?? λ>1 ??inputσC 20000−:=γC 1.5:=

partial safety factor

σaxcr 57913−=σecr_pl be( ) 73015−=σecr be( ) 73018−=

σaxcrbe t⋅ As+

b t⋅ As+

σecr be( )⋅:=σecr_pl be( ) 4−

π2

D⋅

be2 t⋅

⋅:=σecr be( ) π2

− E⋅

Cπ L⋅

ρe be( )

2:=

translation back to applied stresseffective portion of platestiffener & plate as a column

8