Buckling of a Simply Supported Rectangular Plate

8/13/2019 Buckling of a Simply Supported Rectangular Plate

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Buckling of a simply supported Rectangular Plate:

The plate of the structure of a ship is divided into relatively small rectangular panel by means of beams or

longitudinal and girders or transverses. These panels have some constraint at their edges, which is

provided by the torsional rigidity of the stiffening members. Consider a section , the deck of a

transversely framed ship, when buckling of the deck occurs the plating takes up the configuration, thebeam twisting. The twisting of the beams provided an edge constraining moment which is a function of

the torsional properties of the rolled sections. Such sections are known to be very weak in torsion, so that

the constraint moment at the edges is likely to be small and consequently the degree of fixity of the edges

of the panel will be small.

A simply supported panel, length l and breadth b, the load being applied in the direction of the length. The

form of the deflection curve of the panel when buckling occurs can be assumed to be

sin sinm x n y

w Al b

l

b

cr

cr


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This means that the panel will buckle into m half waves in the length and n half wave sin the breadth.

The strain energy due to the bending of the plate is

22 2

2 2

0 0 2

l bD w w

V dxdy

x y

24 2 2 2

2 2

2 2

0 0

sin sin sin sin2

l bDA m n m x n y m x n y

dxdy w Al b l b l b

24 2 2 2

2 28

DblA m n

l b

If the critical stress is crthen the load on a strip length l and width dy is (crx t xdy) and the work done

by this load in buckling is

2

0

1

2

l

cr

wtdy dx

x

So the total work done is obtained by integrating for all such strips across the breadth of the plate.

Hence, Work done2

0 0

1

2

b l

cr

wt dxdy

x

2 2 2

2 22

0 0

1 cos sin2

b l

cr A m m x n yt dxdyl l b

2 2 2

8

crm A b t

l

Now equating the total work done with the strain energy, we get

22 2 2 4 2 2 2

2 28 8

crm A b t DblA m n

l l b

22 2 2 2 2

2 28

crm A b t m n

l l b

22 2

2cr

D b n lm

b t l m b


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The least value of buckling stress is required, and this it will be seen is associated with n = 1, which

means that the panel buckles into one half wave in the breadth, so that

22

2

1cr

D b lm

b t l m b

It remains now to find the value of m which will make the above equation a minimum. This can be done

by differentiating with respect to m and equating to zero.

2

2 2

1 12 0cr

d D b l b l m

dm b t l m b l m b

For maximum and minimum. Multiplying out the brackets

2 2

2 3 2

1 1 10

b lm

l m m m b

4

44,l lm mb b

The wave length is l/m so that this is equal to b. A long panel of plating therefore buckles into a number

of square panels. This will only be true if l is an exact multiple of b and, if this is not so, the panel will

buckle into the nearest whole number of half waves which will make the critical stress a minimum.

Another approach to the problem is to assign various values to m in equation below


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22

2

1cr

D b lm

b t l m b

21b l

k m

l m b

and calculates the value of the quantity multiplying(2D/b

2t). This may be conventionally called k. The

values of k have been worked and shown in table below.

m Vales of K for l/b

1 2 3 4 5 6 7 8

1 4.0 6.25 11.1 18.1 27.4

2 6.25 4.0 4.65 6.25 8.41 11.1

3 11.1 4.65 4.0 4.31 5.16 6.25 7.6 9.2

4 18.1 6.025 4.31 4.0 4.2 4.65 5.4 6.25

These results for k have been plotted in figure below as curves for constant values of m against the ratio

l/b.

1 2 3 4 5 6 7 8

0

5

10

15

20

k

l/b

m = 1

m = 2m = 3

m = 4


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Consider the curves for m = 1, these gives the least value of k at l/b = 1 but at some position between l/b

= 1 and l/b = 2 the curves intersects that for m = 2. At this position the plate would buckle indifferently

into one or two half waves. Similarly between l/b = 2 and 3, the curve for m = 2 intersect that for m =3 in

which case the plate buckles indifferently with two of three half waves and so on. All theses curves are

tangent to a line k = 4 and it will seen that when the length of the plate is an exact multiple of the breadth,

the value for k for minimum buckle stress is 4. Where the length of the plate is not exact multiple ofbreadth , the values of k is a little greater than 4 abut as l/b increases the amount by which k exceeds in

this case diminishes.

As a general working rule, it can be assumed that k is 4 and the critical stress becomes2

2

4cr

D

b t

Case Study:

(a) When the length (l ) of the panel is less than the breadth (b), and here it wi ll be noted that m must be

equal to uni ty

22

2cr

D b l

b t l b

After re arrange

22 2

2 21cr

D l

l t b

Now it is possible to examine the relative merits of stiffening a large sheet of plating in the longitudinal

direction or transvers direction.

Largest sheet of plating, length, L and breadth B is stiffened by

(i) Longitudinal, spaced b apart. The buckling of the plating between the stiffeners is then2

2

4cr

D

b t

(ii) The stiffeners is turned round, so that there are now transverse stiffeners spaced b apart, itwill seen that the buckling strength of the plating between stiffeners is obtained from

By putting l = b and b = B

22 2

2 21cr

D l

l t b

22 2

2 21cr

D b

b t B


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In ship structures the ratio b/B would be about 1/6, so that the term in the brackets is not greatly different

from unity, and from this it can be concluded that the buckling strength of the plating with longitudinal

stiffening is nearly four times as great when transvers stiffening is employed. This shows the great

advantages of longitudinal over transvers stiffening in ship structure and the trend in the modern ships is

to use the former type of stiffening in nearly all classes of ships. Apart from the intrinsic advantages of

using this type of stiffening to increase the buckling strength of the plating, it also has the advantages of

preventing the development of permanent set in the plating in welded ships where there is a greater

tendency to have initial unfairness. Classification Societies almost compel the structural designer to use

longitudinal stiffening in welded ships by imposing quite severe penalties on the scantlings of bottom

plating, for example, when transverse framing is employed.

Problem : Breadth of Panel = 15 Ft.

Stiffener Spacing = 30 inch

P late thickness = 0.6 inch

E =13500 Tons/inch^23 3

2 2

2 2

13500 0.6267 / 13500 / , 0.3

12(1 ) 12(1 0.3 )

EtD tons in in E tons in

(a) Longitudinal Stiffeners:

2 22

2 2

4 4 267

19.5 /30 0.6cr long

D

tons inb t

(b) Transverse Stiffeners:

2 2

2 2 2 2

2 2 2 2

267 2.51 1 5.15

30 0.6 15cr transvers

D b

b t B

(c) Ratio of buckling Stress in two cases:

19.5

3.78 4.05.15

cr long

cr transvers


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Influence of longitudinal stiffeners on the buckling strength of

plating:

In order to develop the full buckling strength of a panel of plating between stiffeners, it is important that

the stiffeners themselves should remain rigid so as to provided support to the small panels. If a large sheet

of plating stiffened by longitudinal members is considered, there is the possibility that the sheet as whole

may buckle, and this possibility will now be examined. Consider a sheet of plating simply supported

round it edges, length l and breadth b, as shown in Fig below. If this sheet buckles, then the equation for

the whole surface can be written

sin sinx y

w Al b

To some extent this assumption for the deflection of the stiffened panel is an approximation, since if a

section through the plating is considered the deflection curve would probably be more accurately

represented by the Figure below, because the influence of the stiffeners is to restraint the plating at the

attachment to the stiffeners.

The strain energy of the bending of the stiffened panel consists of that of the plating, plus in this case the

additional strain energy of the stiffeners. The rigidity of the stiffeners are represented by the moment of

inertia I may be difficult to define, but it will be assumed that it is possible to put a value to I which

should really include the influence of a certain breadth of plating to which the stiffeners is attached.

Assuming that I represents the increased moment of inertia provided by one stiffener, then

22

2

02

lEI w

Stain energy of stiffener dxx

2 42 2

4

0

sin sin2

lEI A x y

dxl l b

4 22

3 sin

4

EIA y

l b

The value of y in this expression should be that appropriate to the particular stiffener. If there are n evenly

spaced stiffeners, then the value of y for the pthstiffeners is yp = pb/(n+1), so that

4 22

3 sin

4 1

EIA pStrain energy of stiffeners

l n

To obtain the total strain energy this requires summing for all the stiffeners. Thus


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4 22

31

sin4 1

p n

p

EIA pStrain energy of all stiffeners

l n

It can be shown that2

1

1sin

1 2

p n

p

p n

n

4 2

3

( 1)

8

EIA nStrain energy of stiffeners

l

The strain energy of bending of plate is

24 2

2 2

1 1

8

DblAStrain energy of bending of plate

l b

The work done by the critical load on the plate will be given by the following equation with m = 1,

2 2 2

8

crm A b t

l

but to this must be added the work done by the load on the stiffeners. If a is the additional area provided

by one stiffeners, then

2

02

l

cra wwork done on one stiffener dxx

2 22 2

2

0

cos sin2

l

cra A x y

dxl l b

2 22sin

4

craA y

l b

2 22

1

sin4 1

p n

cr

p

aA pFor all stiffeners work done

l n

2 2

( 1)8

craA nl

Now equating total work done to the total strain energy

22 2 2 2 4 2 4 2

2 2 3

( 1) ( 1) 1 1 ( 1)

8 8 8 8

cr cr btA n aA n DblA EIA n

l l l b l


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From this it will be seen that

2

2 2

2 2 2

1 1 ( 1)

( 1)

cr

EI nDbl

l b l

bt a n

For economical design from a buckling point of view, the critical stress determined from above equation

should be equal to the critical stress for a panel of platting between two stiffeners. The breadth of small

panel will be b/(n+1) and the corresponding buckling stress will be

2 2

2

4 ( 1)cr

D n

b t

And if the buckling stress for the two modes of failure is to be same,

2

2 2

2 2 22 2

2

1 1 ( 1)

4 ( 1)

( 1)

EI nDbl

l b lD n

b t bt a n

Having decided upon the required buckling stress for a particular problem for a plate of given principal

dimensions and thickness, it is possible to determine the value of n. An equation involving two unknowns

a and I would be obtained from the above equation. Unless a relation existed between I and a it would not

be possible to solve for I completely. The best away to deal with this would be to assume a as some

percentage of cross sectional area btof the plate.

Problem:

Consider a sheet of plating 15 ft long, 20ft breadth and 0.6 inch thick, with stiffener spacing 30 inch apart.

L = 15 ft

B = 20 ft

T = 0.6 inch

Stiffener spacing l = 30 inch

3 32 2

2 2

13500 0.6267 / 13500 / , 0.3

12(1 ) 12(1 0.3 )

EtD tons in in E tons in


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If the stiffeners are spaced longitudinally, then the buckling stress of the plating between stiffeners is

2 22

2 2

4 4 26719.5 /

30 0.6cr

Dtons in

b t

Let the area of longitudinal is 20% , the area of plating

2( 1) 0.2 240 0.6 28.8

2401 7

30

a n in

n

2

2 2

2 2 2

1 1 ( 1)

( 1)cr

EI nDbl

l b l

bt a n

22 2

2 2 2

1 1 13500 (7 1)267 240 180

240 180 180

( 1) 240 0.6 1.2cr

I

bt a n

47.52 32.86

172.8cr

I

Equating with the buckling strength of the stiffened panel to that for the panel between stiffeners, the

value of I

4172.8 19.5 47.52 101.132.86

I in

A section which would satisfy the requirement is a 7 in. x 3.5 in. x 0.44 in. toe welded angle bar and this

has area of 4.43 in2. The total area is (4.43 x 7) = 31.01 which is slightly higher than 28.8 in2.

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Buckling of a Simply Supported Rectangular Plate