Chapter 5 Approximate.doc

7/28/2019 Chapter 5 Approximate.doc

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Chapter 5 Approximate Methods of Analysis of Plates

1. Validity of Classical Plate Theory

As in the case of the classical beam theory, there are obvious discrepancies in the classical theory

of plates that has been presented with regard to transforming a three-dimensional object into two-

dimensional entity. Most notably, we have employed a plane stress assumption in which

0zz xz yz = = = . From equilibrium considerations of an isolated plate free-body, we expressed

the vertical flexural shear stresses as functions of vertical bending shearing forces, andx yQ Q . It

is, therefore useful to assure that our approach will yield meaningful results despite some

obvious discrepancies.

Recall an equilibrium equation of a 3D element given in terms of stresses

0x

xyxx xz

y z

+ + =

(2-12a)

where

3 3,

/12 /12xyx

xx xy M zM zh h

= =

Hence,

3 30

/12 /12

xyx xzMMz z

h x h y z

+ + =

3 3/12 /12

xyxz xx

MMz zQ

z h x y h

= + =

Now integrating with respect to z, gives

( )2

3

/ 2,

/12xz x

zQ f x y

h = +

Know 0xz = at / 2z h= , which leads

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2

3

/ 80

/12x

hQ f

h= +

3

2xf hQ=

Therefore,

2

2

31 4

2

xxz

Q z

h h

=

Similarly,

2

2

31 4

2

y

yz

Q z

h h

=

From Eqs. (2-8) and (2-12c), we have

2 2

2 2

3 31 4 1 4

2 2

xy yxz xzzQQz q z

z x y h h x y h h

= = + =

Integrating with respect to z, we have

( )3

2

3 4,

2 3zz

q zz g x y

h h

= +

The boundary conditions for zz are

(1) at / 2 and (2) 0 at / 2zz zzq z h z h = = = =

From condition (1), we have

( )3

2

3 4,

2 2 24

q h hq g x y

h h

= + +

( )

1,

2g x y q =

Hence,

3

3

1 3 2

2 2zz

z zq

h h

= +

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3/21

We thus obtained a set of stresses , , andxz yz zz from rigorous equilibrium considerations

using results from a simplified theory that neglected these very stresses. If it turns out that the

above stresses which are assumed to be zero in the simplified theory are indeed small, we have

inner consistency of our theory. In order to accomplish this, we shall make an order-of-

magnitude study (adapted from Shames and Dym, Energy and Finite Element Methods in

Structural Mechanics, Hemisphere, 1985) of all six stresses for comparison.

From Fig. 5-1 showing inner portion of the plate, it becomes obvious

Fig. 5-1

which implies that the order of magnitude of the total force is equal to the order of magnitude of

the average applied load intensity, q, times the order of magnitude of the area of the inner

portion. If we denote an average vertical shearing force around the perimeter of the inner

portion, Qv, we may write the following order-of-magnitude equation:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 v vO Q O L O q O L O Q O q O L O qL= = =

Next, consider an order-of-magnitude study of moment acting on the inner portion shown in the

sketch.

Equating moments about an axis C-C going through the line of action of the resultant of the force

distribution q, gives

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Fig. 5-2

( ) ( ) ( ) ( ) ( ) ( ) ( )0 v v vsO Q O L O L O M O L O M O L= + +

where Mv is vertical bending moment and Mvs is in-plane twisting moment. As the vertical

moments are proportional to qL2, we have

( ) ( ) ( ) ( )2 2andv vsO M O qL O M O qL= =

Since stresses are computed from the moment divided by the section modulus of unit width as

2/ 6

xxx

M

h =

Hence,

( ) ( ) ( )2 2 2

2 2 2, , andxx yy xy

qL qL qLO O O O O O

h h h

= = =

Recall the vertical stresses are computed from the vertical shearing force, Qv, divided by the area

of unit width. Hence,

( ) ( ) ( ) ( )1 1

= andxz yzL L

O O qL O O q O O qL O O q

h h h h

= = =

Recall also that

( ) ( )zzO O q =

We can now make comparisons.

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( ) ( ) ( ) ( )2

2/ and /xz xy zz xx

h hO O O O O O

L L

= =

which implies that the vertical shear stresses are smaller by an order of magnitude h/L. Likewise

the transverse normal stress, zz , is smaller by an order of magnitude2 2

/h L . Thus for a very

thin plate whereL/h >>10 we have inner consistency and the classical theory may be expected to

give good results.

(b) Strain Energy of Plate bending

Strain energy per unit volume of the stressed body is

/2

/2

12

h

ij ijh

U dzdxdy

=

/2

/2

1( )

2

h

xx xx yy yy zz zz xy xy xz xz yz yzh

dzdxdy

= + + + + + (5-1)

In the case of a thin plate, we have

0 (plane stress)zz xz yz = = =

Recalling and substituting the following stress strain relationship into Eq. (5-1), gives

( )

( )

( )

1

1

2 1

xx xx yy

yy yy xx

xy xy

E

E

E

= = +

=

( )

( )

( )

2

2

1

1

2 1

xx xx yy

yy yy xx

xy xy

E

E

E

= +

= +

= +

( )/2

2 2 2/2

12 2 1

2

h

xx yy xx yy xyhU dzdxdyE

= + + +

Recall

( ), , ,w w

w w x y u z v z x y

= = =

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2 2 2

2 2, ,xx yy xy

u w w w v u wz z z

x x y x x y x y

= = = = = + =

( )

2 2 2 2 2

2 2 2 2 2 2, ,

1 1 1

xx yy xy

Ez w w Ez w w Ez w

x y y x x y

= + = + =

+

Substituting these relationships into the strain energy expression, yields

( ) ( ) ( )2

2, , 2 1 , , ,2

xx yy xy xx yy

DU w w w w w dxdy = + + (5-2)

where

( )

3

2

12 1

EhD

=

plate flexural rigidity per unit width (k-in2/in)

The loss of potential energy of applied loads is

( ),V p x y wdxdy= (5-3)

U V = + (5-4)

The problem of the bending of plates reduces in each particular case to that of finding a function,

w, that satisfies the given boundary conditions and makes the functional, Eq. (5-4), a

minimum.

(c) Ritz Method

An energy method developed by Walter Ritz (1908) applies the principle of minimum potential

energy. According to this theory, of all displacements that satisfy the boundary conditions (at

least the GBC), those making the total potential energy functional of the structure, Eq. (5-4) a

minimum are the displacement sought pertinent to the stable equilibrium. Let

( ) ( ) ( ) ( ) ( )1 1 2 21

, , , . . . , ,n

n n i i

i

w x y c f x y c f x y c f x y c f x y=

= + + + = (5-5)

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7/21

where ( ),if x y are continuous functions that satisfy individually at least the GBCs and are

capable of representing the deflected shape of the structure. The unknown coefficients, c1, . . . ,

cn are determined from

1 2

0, 0, . . . . , 0nc c c

= = =

(5-6)

Eq. (5-6) gives n simultaneous equations for c1, . . . , cn. c1, . . . , cn are called the natural

coordinates. The accuracy of the Ritz method relies primarily on the good selection of the trial

displacement function, Eq. (5-5)

Example #1

Hydrostatically Loaded Rectangular Plate

Assume the Navier solution as the trial function.

( ), sin sinmnm x n y

w x y aa b

=

22

2

2, sin sinxx mn

w m m x n yw ax a a b

= =

22

2

2, sin sinyy mn

w n m x n yw a

y b a b

= =

2

2, cos cosxy mn

w mn m x n yw a

x y ab a b

= =

Substituting these partial derivatives into Eq. (5-2), gives

( ) ( ) ( )2

2

0 0, , 2 1 , , ,

2

a b

xx yy xy xx yy

DU w w w w w dxdy = + +

Let

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sin , sin , cos , and cosm n m nm x n y m x n y

s s c ca b a b

= = = =

( )2

2 22

4

10 0 0 0

, ,a b a b

xx yy mn m n mn m n

m nU w w dxdy a s s a s s dxdy

a b

= + = +

22 2

4

2 20 0

a b

mn m n

m na s s dxdy

a b

= +

22 2

2 2 2

2 2

4

0 0 2 2 2 2

' ' ' '2 2 2 2

' '

mn m na b

mn m n m n m n

m na s s

a bdxdy

m n m na a s s s s

a b a b

+

= + + +

The second term in the square bracket is for m m and n n and vanishes. Hence,

2 22 2 4 2 2

4 2 2 2 2

1 2 2 2 20 0 4

a b

mn m n mn

m n ab m nU a s s dxdy a

a b a b

= + = +

2 24

2 4 2 2 2 2,4

xy mn m n mn

mn ab mnw dxdy a c c dxdy a

ab ab

= =

2 24

4 2 2 2 2, ,

4xx yy mn m n mn

mn ab mnw w dxdy a s s dxdy a

ab ab

= =

Hence,

( )22 , , , 0xy xx yyU w w w dxdy= =

( )

24 2 2

2

1 2 2 22 8

mn

D abD m nU U U a

a b

= + = +

(5-7)

The loss of potential energy of the applied load is then

( ) ( ), ,V f x y w x y dA= (5-8)

where

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( ) ( )0, and , sin sinmnq m x n y

f x y x w x y aa a b

= =

2

0 0

0 0 0 0

a b bmn

mn m n m m n

n

aq x q a a axV a s s dxdy s c s dya a m m

= =

( ) ( )1 12 2

0 0

0

1 2 1 2cos

m mb

mn mnn

m n m n

ba aq a q a b n ys dy

oa m a m n b

+ + = =

( ) ( )1 120 0

2

1,2, 1,3,

1 2 142 1m m

mn mn

m n m n

aq a q a abb

a m n m n

+ +

= =

= =

The total energy functional is minimized with respect to the arbitrarily chosen natural

coordinates

( )2 1

4 2 2

2 0

2 2 2

14

8

m

mnmn

q a ababD m nU V a

a b mn

+ = + = +

0 gives specific values of and .mn mn mn

U Vm n

a a a

= + =

24 2 2

2 24mn

mn

U abD m na

a a b

= +

and ( )1

0

2

14m

mn

q abV

a mn

+=

24 2 2

2 24 mn

abD m na

a b

+

( )

1

0

2

14

0

m

q ab

mn

+

=Hence,

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10/21

( )1

0

262 2

2 2

116m

mn

qa

D m nmn

a b

+=

+

and

( )( )

1

0

262 2

1,2, 1,3,

2 2

116, sin sin

m

m n

q m x n yw x y

D a bm nmn

a b

+

= =

=

+

For a square plate (a=b),

( )( )

( )

14

0

262 2

1,2, 1,3,

116, sin sin

m

m n

q a m x n yw x y

D a bmn m n

+

= =

=

+

4 4

0 0

/ 2, 1 6

16 10.00416

4x y a m n

q a q aw

D D= = = = = = checks with Navier solution

The accuracy of the Ritz method relies primarily on the good selection of the trial function, Eq.

(5-5). A double trigonometric series is one of the best trial functions to be used. When the trial

function satisfies NBCs as well as GBCs, the trial function renders an extremely fast

converging solution. Recall the suggested form of trigonometric functions representing various

combinations of edge conditions.

Example #2

A simply supported square plate subjected to a triangular prism as shown in Fig. 5-3 is

examined.

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11/21

Fig. 5-3

Assume ( )x

, sin sina

mn

m n

m n yw x y w

a

= . As the load and the deflection shape are

symmetrical with respect to the centerline axes of the plate, the summation of sine series is

executed on odd (1,3,5, ) indexes.

( )

( )

0

0

0

2, for 0

2

2, 2 for

2

q x aq x y x

a

q x aq x y q x a

a

=

= <

Potential energy of applied load, Eq. (5-8)

( ) ( )/ 2

0

0 0

2, , 2 sin sin

a a

mn

A

q x m x m yV q x y w x y dA w dxdy

a a a

= =

/ 2 / 20

0 0 0 0

4sin sin , let sin sin

a a a a

mn

m n

q m x n y m x n yw x dx dy I x dx dy

a a a a a

= =

( )

( )

2 21

2

31

22 3

/ 2 2sin cos cos 1

0 0

21

m

m

a aa m x ax m x a n y a aI

m a m a n a m n

am n

= =

=

2

0

3 2

1,3, 1,3,

8 1sin

2mn

m n

q a mV w

m n

= =

=

The strain energy (for the square plate), U, Eq. (5-7), is

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12/21

( )2

4 2 2 2 42

2 2 2 2

2 2 28 8

mn mn

m n

a D m n DU w m n w

a a a

= + = +

0mn mn mn

U V

w w w

= + =

gives specific values of mnw .

( )4 2

22 2 0

2 3 2

8 1sin 0

4 2

mnD w q a mm na m n

+ =

Hence,

( )

4

0

27 2 2 2

32sin

2mn

q a mw

Dm n m n

=+

and

( )( )

4

0

272 2

1,3, 1,3,

32 1, sin sin sin

2 2 2m n

q a m m x n yw x y

D m n

= =

=+

4

0

max / 2, 10.00265x y a m n

q aw

D= = = = = vs.

4

0

max classical0.00263

q aw

D= 0.76% difference

Example #3

A rectangular plate with a=1.5b is fixed at all four edges as shown in Fig. 5-4. Compute the

maximum deflection at the center of the plate subjected to a uniformly distributed load of 0q .

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13/21

Fig. 5-4

Assume for simplicity ( ) ( ) ( ), with 1mnm n

w x y w f x g y m n= = = .

( ) 11, 1 cos 1 cosx y

w x y wa b

= + +

(note the coordinate translation)

The expression in the parenthesis of the second term in Eq. (5-2) is called the Gaussian curvature

and the first variation of the integral of the Gauss curvature vanishes if simply w=0 on the

boundary. Hence, the second term in Eq. (5-2) is not evaluated in the expression of the strain

energy.

( ) ( ) ( )2

2, , 2 1 , , ,

2xx yy xy xx yy

DU w w w w w dxdy = + + (5-2)

( )4 2

22 11

1 2 1 3 3

3 3 2

2 2

a b

a b

D wD b aU U U U w dxdy

a b ab

= + = = = + +

Since

11sin 1 cos

w x yw

x a a b

= + and

2 2

112 2cos 1 cos

w x yw

x a a b

= +

yaa

b

b

0q

x

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14/21

11sin 1 cos

w y xw

y b b a

= + and

2 2

112 2cos 1 cos

w y xw

y b b a

= +

( )

22

22

2211

2

cos 1 cos

, ,

cos 1 cos

a b a b

xx yya b a b

x y

a a b dxdyw w dxdy

wy x

b b a

+ + = + +

2 22 22 2

2 2

22 2

11

2 2

cos 1 cos cos 1 cos

2 cos 1 cos cos 1 cos

a b

a b

x y y x

a a b b b a dxdy

wx y y x

a a b b b a

+ + + = + + +

I I I

w-

+ +=

1 2 3

2

11

Then

2 2 2 42

1 2

2cos 1 cos 1 cos sin

2 4

a b b

a b b

ax y y x a xI dxdy dy

aa a b b a a

= + = + +

24 4

2

3 31 cos 1 2cos cos

b b

b b

y y ydy dy

a b a b b

= + = + +

( )4 4 4

3 3 3

2 2 3sin sin 2 0 0

2 4

bb y y b y by b b

ba b b a a

= + + + = + + + =

Likewise,

4

2 3

3aI

b

=

2 2

2 2

4

2 2

cos 1 cos cos 1 cos

cos 1 cos cos 1 cos

x y y x

a a b b b a

x x y y

a b a a b b

+ +

= + +

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15/21

4

3

2 2cos 1 cos cos 1 cos

2

b a

b a

I y y x xdy dx

a b b b a a

= + +

2 2cos 1 cos cos cos sin sin

2 4

a a

a a

ax x x x a x x a xdx dx a

aa a a a a a

+ = + = + + =

Likewise,

2 2cos 1 cos cos cos sin sin2 4

b b

b b

by y y y b y y b ydy dy b

bb b b b b b

+ = + = + + =

4

3

2I

ab

=

4 2

113 33 3 2

2D w b aU

a b ab = + +

( )0 0 11, 1 cos 1 cosa b a b

a b a b

x yV q w x y dxdy q w dxdy

a b

= = + +

0 11 1 cos sinb

b

ay a xq w dy x

ab a

= + + 0 114abq w=

U V = +

4

11 03 3

11 11 11

3 3 24 0

U V b aD w abq

w w w a b ab

= + = + + =

4

0 0

11 4 24 4

3 3

4 41 1

3 3 23 3 2

abq q aw

b aD D a a

a b ab b b

= =

+ + + +

Fora/b=1.04 4 42

0 0 0

max 0 4

4 10.02053 . 0.02016

3 3 2

x y

q a q a q aw vs

D D D

= = = =

+ +

1

1.6% difference

1 Table 35 (0.00126qoa4 /D), Timoshenko and Woinowsky-Krieger (1950). Since the dimension is twice the one

used in Table 35, adjustment is needed; 0.00126x16=0.02016.

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16/21

Example #4

A simply supported rectangular plate with a=1.5b as shown in Fig. 5-5 is subjected to a

uniformly distributed load of 0q . Compute the deflection at the center.

Fig. 5-5

Assume a deflection function

( ) ( ) ( )4 3 4 3

11, 2 2

x x x y y yw x y f x g y w

a a a b b b

= = + +

which satisfies GBCs and NBCs. Hence, it is expected to converge fast.

3 2 4 3

11 4 6 1 2ww x x y y y

x a a a b b b

= + +

2 4 32

11

2 212 12 2

ww x x y y y

x a a a b b b

= +

4 3 3 2

11 2 4 6 1ww x x x y y

y b a a a b b

= + +

4 3 22

11

2 22 12 12

ww x x x y y

y b a a a b b

= +

x

a

b0q

y

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17/21

3 2 3 22

11 4 6 1 4 6 1ww x x y y

x y ab a a b b

= + +

( ) ( )2 2

2

0 0 0 0, ,

2 2

a b a b

xx yy

D DU w dxdy w w dxdy= = +

2

2 4 3

2

2

114 3 20 0

2

12

721

2

a b

x x y y y

a a a b b bDw dxdy

x x x y y

b a a a b b

+ = + +

2 22 4 3

4

2 24 3 2

4

2

112 4 3

2 2

1

2

12

72

22

x x y y y

a a a b b b

x x x y y

b a a a b bDw

x x y y y

a b a a b b b

x

a

+

+ + =

+ +

0 0

4 3 2

2

a b

dxdy

x x y y

a a b b

+

( )211 1 2 372Dw I I I= + +

17


18/21

2 22 4 3

4

10 0

24 3 4 2 3

0 0

2

2 2

a b

b a

x x y y ya I dxdy

a a b b b

y y y x x xdy dx

b b b a a a

= +

= + +

2 24 3 4 3

0 0= 2 2

5 3 2 30

b by y y a a a a y y ydy dy

b b b b b b

+ + = +

24 3 8 2 6 5 7 4

0 02 4 2 4 4

30 30

b ba y y y a y y y y y ydy dy

b b b b b b b b b

= + = + + +

4 4 70 210 360 210 315 504 31

30 9 3 7 3 2 5 30 630 30 630

a b b b b b b ab ab+ + + = + + + = =

1 3

31

18900

bI

a=

Likewise

2 3

31

18900

aI

b=

From Maple

2 4 3 4 3 2

3 2 2 0

12 2 2

a x x y y y x x x y yI dxdy

a b a a b b b a a a b b

= + +

4 3 2 2 4 3

2 2 0 0

22 2

b ay y y y y x x x x xdy dx

a b b b b b b a a a a a

= + +

2 2

2 17 17 289

420 420 88200

b a

a b ab

= =

2 2

11 113 3 3 3

31 31 289 62 62 28972

18900 18900 88200 525 525 1225

b a b aU Dw Dw

a b ab a b ab

= + + = + +

18


19/21

4 3 4 3

0 11 0 110 0

2 225

a b x x x y y y abV q w dxdy q w

a a a b b b

= + + =

2

11 0 113 3

62 62 289/ 25

525 525 1225

b aU V Dw q abw

a b ab

= + = + +

11 03 3

11 11 11

62 62 2892 / 25 0

525 525 1225

U V b aDw q ab

w w w a b ab

= + = + + =

( ) ( )

0 0

11

3 3 3 3

62 62 289 124 124 5782 25

525 525 1225 21 21 49

q ab q abw

b a b aD D

a b ab a b ab

= = + + + +

( )

4 4

0 0

4 4 2 2 4 4 2 2

3 3

1029

6076 6076 12138 6076 12138

1029

q ab q a b

b a a b a b a b DD

a b

= = + + + +

Fora=b

4 4

0 0

max / 2, / 2

1029 1 1 1 1 1 10.004137

12152 12138 16 4 2 16 4 2x a y b

q a q aw

D D= =

= + + = +

Comparing the coefficient of the classical solution (Lvy solution) 0,00406, this is 1.9% higher.

There are other classical approximate methods such as Galerkin and Kantorovich and numerical

methods such as finite difference and finite element. Other than finite element method, these

approximate methods are now historical interest as they do not necessarily contribute to a better

understanding of the subject matter or provide computational efficiency.

Suggested form of trig series to be used in approximate solutions

1. fixed - free

1 cos2

xy a

= l

, sin2 2

xy a

= l l

,

2

cos2 2

xy a

= l l

,

3

sin2 2

xy a

= l l

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20/21

0 00 and 0x xy y= == = meet GBC

0 and 0x xy y= = l l one-half of NBC violated may be acceptable

2. simple - simple

sinx

y a

=l

,

2

cos , sinx x

y a y a = =

l l l l

00 and 0x xy y= == =l meet GBC

0 0 and 0x xy y= = = =l meet NBC fast converging solution

3. fixed - fixed

2 2 21 cos , sin

x xy a y a

= = l l l

0 x=00 and y 0xy = = = meet GBC atx=0

0 and 0x xy y= == =l l meet GBC atx= l fast converging solution

4. fixed - simple

22 2 2 2 2

1 cos , sin , cosx x x

y a y a y a = = =

l l l l l

00 and 0x xy y= == =l meet GBC

00 and 0x xy y= = = l meet all GBC but one NBC violated may be acceptable

20


21/21

5. simple - free

2 3

sin , cos , sin , cos2 2 2 2 2 2 2

x x x xy a y a y a y a

= = = = l l l l l l l

0 x=00 and y 0xy = = = meet one GBC and one NBC

0 and 0x xy y= = =l l meet one NBC but violated one NBC may be acceptable

Assumed displacement functions

1. must satisfy the GBC - if violated, divergence may occur.

2. satisfy the NBC as well as GBC - an extremely fast converging solution is assured.

3. satisfy the GBC but violate the NBC - may be acceptable but convergence may be slow.

HW#8

A rectangular plate of dimension a and b is clamped on all four edges. The plate is subjected to

a uniformly distributed load of 0q . Using the deflection equation of both ends clamped beam for

the assumed plate deflection equation, ( ) ( ) ( ) ( ) ( )2 22 2

11,w x y f x g y w x x a y y b = = ,

develop the plate deflection equation. Ifa=b, find the maximum deflection at the center of the

plate and compare with other classical solution.

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