Buckling of Stiffened Panels 1
overall buckling vs plate buckling
PCCB 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.
2
Timoshenko does so by calculating k in σ cr := k⋅
π ⋅D
and observing the
2
b ⋅t
value of γ which results in a critical stress above that which will cause plate
buckling alone. where;
γ :=
flexural_rigidity_of_combined_section
γ :=
flexural_rigidity_of_plate
E⋅ Ix
where;
D⋅ b
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.
α :=
for the following:
α := 0.1 , 0.2 .. 10
a
δ :=
B
Ax
γ :=
B⋅ t
E⋅ Ix
D⋅ B
κ 1( α , n) := n⋅
δ := 0.1
3
π
α
⋅
4⋅ α
n
κ 2( α , n) := n⋅
+1
π
α
⋅
4⋅ α
n
−1
α
2 n
2
π
α

(
)
γ o α , δ , n :=
+ 16⋅ 
⋅δ
n
 κ 1( α , n) 
 κ 2( α , n) 
1
1
⋅ tanh
−
⋅ tan
κ 1( α , n)
 2  κ 2( α , n)  2 
16
⋅
  γ o( α , δ , 1)  

γ omax( α , δ ) := max  γ o( α , δ , 2)  
  γ ( α , δ , 3)
 o

40
40
γ o ( α , δ , 1)
γ o ( α , δ , 2)
20
γ omax( α , δ )
γ o ( α , δ , 3)
0
0
5
20
0
10
α
0
5
α
1
10
curve fit:
one stiffener, combining Hughes and Bleich terms, N=1 stiffener
 2.5 + 16⋅ δ ⋅ α 2 − 10.8 ⋅ α

2
2
 2

  γ bx1( α , δ )  
0.5
48.8
γ bx2( α , δ ) :=
+ 112⋅ δ ⋅  1 +
⋅ 2⋅ δ 
γ bx( α , δ ) := min 
2
2


  γ bx2( α , δ )  
γ bx1( α , δ ) :=
22.8
⋅α +
Hughes: 13.1.4
slightly
modified in
terms
40
30
γ omax( α , δ )
20
γ bx( α , δ )
Fig. 180, Bleich
10
0
0
5
10
α
to move to Bleich relationships in Hughes terms (ratios) to match text:
where:
Π :=
a
δ x:=
B
Ax
γ :=
b⋅ t
E⋅ Ix
D⋅ b
one stiffener
(
)
(
)
two stiffeners
(
)
(
2
γ bx11 Π , δ x := 22.8⋅ Π + 2.5 + 16⋅ δ x ⋅ Π − 10.8⋅ Π
(
)
3
2
γ bx21 Π , δ x := 43.5⋅ Π + 36⋅ Π ⋅ δ x
)
(
γ bx12 Π , δ x := 48.8 + 112⋅ δ x⋅ 1 + 0.5⋅ δ x
)
2
γ bx22 Π , δ x := 288 + 610⋅ δ x + 325⋅ δ x
Π := 1 , 1.1 .. 5
800
300
( )
γ bx12( Π , δ x)
γ bx11 Π , δ x
( )
γ bx22( Π , δ x)
200
γ bx21 Π , δ x
100
0
600
400
200
0
2
4
0
6
Π
1
2
3
Π
2
4
5
combination
(
)

  γ bx11( Π , δ x)  

  γ bx12( Π , δ x)  
γ b Π , δ x , N := if  N = 1 , min 
  γ bx21( Π , δ x)   

  γ bx22( Π , δ x)   
, min 
minimum ratio of EI/Db (A/bt = 0.1)
500
gamma = EI/Db
400
300
200
100
0
1
2
3
4
panel aspect ratio = a/B
5
2 stiffeners
1 stiffener
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 b
the plate is a width b some typical scantlings:
stiffener web
stiffener flange
Aw := 0.8
Af := 0.85
stiffener area
As := Aw + Af
stiffener depth panel length
d := 5
3
L := 96
breadth between set above
stiffeners b
b := 30
t = 0.5
calculate in terms of be , initially be = b
moment of inertia using 8.3.6 to calculate radius of gyration:
Ae( be) := As + be⋅ t
C1( be) :=
 Ae( be) Aw 
−
+ Af⋅ be⋅ t
4 
 3
Aw⋅ 
(Ae( be))2
Ie( be) := Ae( be) ⋅ ( d) ⋅ C1( be)
2
Ie( be)
ρ e( be) :=
Ae( be)
column critical stress:
σ e_cr( be) :=
2
π ⋅E
 L 
 ρ (b )
 e e 
plate critical stress: σ a_cr( be) := 3.615⋅ E⋅ 
2

 be 
t
; initial value σ e_cr( b) = 49341
2
initial value σ a_cr( b) = 30125
things are ok as column > plate => plate "fails" first. now consider increasing stress
beyond σ a_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
σ a⋅ b
now consider what happens to the values of critical stress as effective
σ e( be) :=
be
breadth is reduced. the definitions above are still active; repeated here for info:
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.
be := 15 .. b
5
1 .10
4
critical stress
8 .10
column:
 L 
 ρ (b )
 e e 
4
6 .10
plate: σ a_cr( be ) := 3.615⋅ E⋅ 

b
 e
4
4 .10
4
2 .10
σ e_cr( be ) :=
2
π ⋅E
15
20
25
effective breadth
30
column (stiffener with plate)
plate alone
4
t
2
2
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:
Cπ :=
B
a
⋅
γ x( be)
(
2⋅ 1 +
1 + γ x( be)
= Cπ ⋅
and L_over_ρ
eq
L
ρ ( be)
)
or
then
Cπ := 1 whichever is less. γ x( be) :=
E⋅ Ix( be)
D⋅ be
2
π ⋅E
σ e_cr( be) :=
C ⋅ L 
 π ρ (b )
e e 

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
σ a⋅ ( b⋅ t + Ax)
=
σ e⋅ ( be⋅ t + Ax) and
σ a( be) :=
( be⋅ t + Ax)
⋅σ
(b )
( b⋅ t + Ax) e_cr e
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
2.5
( )
ρ e be
5
2 .10
( )
σ e_cr be
2
 be 
 be 
ρ e
2
σ e_cr
 2
1.5
1
5
1.5 .10
5
1 .10
4
5 .10
0
0
5
10
0
2
4
6
be⋅ t
be⋅ t
As
As
now for our design rules for PCCB Panel Collapse Combined Buckling:
5
8
10
- PCCB - Panel Collapse Combined Buckling
function of be , graphical approach
b
input := 1
HSF
tp
b
input data
ignore numerical
values shown
here
stiffener
BSF := 3.94
TSF := .215
TSW := .17
try this with SDEPTH = 6, to see yield effect
plate
a
SDEPTH := 5.
a := 8⋅ 12
b := 25
t := .5
N := 2
L = 96
L := a
material
3
σ Y := 80⋅ 10
E⋅ t
6
υ := 0.3 E := 29.6⋅ 10 D :=
3
(
12⋅ 1 − υ
general parameters:
Aw := ( SDEPTH − TSF) ⋅ TSW
HSW := SDEPTH − TSF
Aw = 0.81
HSW = 4.79
d := SDEPTH −
TSF
2
+
t
B := ( N + 1) ⋅ b
2
d = 5.14
As := Aw + Af
Af = 0.85
As = 1.66
b⋅ t
Ap := b⋅ t
B = 75
Af := BSF⋅ TSF
As
= 7.53
δ x :=
Ap = 12.5
As
b⋅ t
δ x = 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 and
compare with br( be). If br( be)/ b < start use result until converges.
Ae( be) := As + be⋅ t
C1( be) :=
Ae( be) = 11.23
(Ae( be))
2
C1( be) = 0.09
Ie( be)
ρ e( be) :=
 Ae( be) Aw 
−
+ Af⋅ be⋅ t
4 
 3
Aw⋅ 
γ x( be) :=
Ae( be)
2
Ie( be) = 25.87
(
)
2
12⋅ 1 − υ ⋅ Ie( be)
be⋅ ( t)
ρ e( be) = 1.52
Ie( be) := Ae( be) ⋅ ( d) ⋅ C1( be)
3
γ x( be) = 118.07
 B
γ x( be)

 ⋅

Cπ := min  a 2⋅ ( 1 + 1 + γ x( be) )  


1


for iteration
br( be) :=
Cπ ⋅ a⋅ t
(
ρ e( be) ⋅ 3⋅ 1 − υ
)
2
br( be)
b
= 0.7657
Checks??
br( be) = 19.14
6
brat :=
br( be)
b
Cπ = 1
brat = 0.766
)
2
critical stress relationships:
stiffener & plate as a column effective portion of plate
2
−π ⋅ E
σ ecr( be) :=
 Cπ ⋅ L 

 ρ e( be) 
σ ecr_pl( be) := −4⋅
2
2
π ⋅D
2
2
2
be ⋅ t
σ ecr( be) = −73015
original plate
σ ecr_pl( be) = −73008
π ⋅D
t
σ o := 3.62⋅ E⋅   OR....
 b
σ o := −4⋅
σ o = −42804
σ o = −42804
2
b ⋅t
for plotting to determine intersection;
be :=
1.8 .10
5
1.6 .10
5
1.4 .10
5
b b
, + 0.1 .. b
2 2
( )
5
1.2 .10
− σ ecr_pl( be)
− σ ecr be
σY
− σo
5
1 .10
4
8 .10
4
6 .10
4
4 .10
12
14
16
18
20
be
now obseving intersection:
be := brat⋅ b
be = 19.14
assuming iteration complete and matched above.
7
22
24
26
stiffener & plate as a column
σ ecr( be) :=
2
−π ⋅ E
 Cπ ⋅ L 

 ρ e( be) 
2
σ ecr( be) = −73018
effective portion of plate
σ ecr_pl( be) := −4⋅
translation back to applied stress
2
π ⋅D
σ axcr :=
2
be ⋅ t
σ ecr_pl( be) = −73015
 be⋅ t + As 
 b⋅ t + A ⋅ σ ecr( be)
s 

σ axcr = −57913
partial safety factor
γ C := 1.5
γRPCCB := γ C⋅
σ C := −20000
σC
input
is it slender?? λ>1 ??
λ :=
σ axcr
γRPCCB = 0.51801
8
a
π ⋅ ρ e( be)
⋅
σY
E
λ = 1.05