03 Rectangular

8/3/2019 03 Rectangular

1/133

RECTANGULAR PLATES

Page 1


2/133

Rectangular Plate - TopicsPage 2

Fourier Series (5-9)

Naviers Solution (10-17)

Uniform Load (18-25)

Sine Load (26-35) Uniform Load Over Subregion (36-39)

Point Load (40-42)

Levys Solution (43-53)

Uniform Load (54-64)

3 Simply Supported Edges, 1 Clamped Edge (65-69)

Opposite Edges Simply Supported, 1 Free, 1 Clamped (70-72)

Simply Supported with Load Function of x (73-78)


3/133

Rectangular Plate - TopicsPage 3

Simply Supported Hydrostatic Load (79-80)

Simply Supported Long Rectangular Plates (81-88)

n Constant Load (89)

n Partial Line Load (90-91)

n Concentrated Load (92-93)

Symmetric Edge Moments (94-100)

One Edge Moment (101-104)

Superposition (105-112)

Strip Method (113-117) 3 Simply Supported Edges, 1 Fixed Edge (118-121)

Continuous Plates (122-128)


4/133

Background: Expressions To KnowPage 4

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ax

a

dxax

dx

duuu

dx

d

axa

dxaxdx

duuu

dx

d

axa

dxaxdx

duuu

dx

d

axa

dxaxdx

duuu

dx

d

sinh1

coshsinhcosh

cosh1

sinhcoshsinh

sin1

cossincos

cos1

sincossin

==

==

=-=

-==


5/133

Fourier SeriesPage 5

If f(x) is continuous, bounded, periodic function of

period 2L over the range L to +L, i.e. f(x+2L) =

f(x).

Then f(x) may be represented by the Fourier series:

=

++=1

0 sincos2

)(n

nnL

xnb

L

xna

axf

pp


6/133


Coefficients are found through the use of

orthogonality relations:

)(

)(0sinsin

nm,ll0sincos

)(

)(0coscos

nmL

nmdxL

xn

L

xm

adxL

xn

L

xm

nmL

nmdxL

xn

L

xm

L

L

L

L

L

L

==

=

=

==

=

+

-

+

-

+

-

pp

pp

pp


7/133


To determine the coefficients bn of the following Fourier series:

Multiply both sides by:

To obtain:

=

=1

sin)(n

nL

xnbxf

p

L

xmpsin

L

xm

L

xn

bL

xm

xf n n

ppp

sinsinsin)( 1

==


8/133


Integrate both sides:

And using the orthogonality relationship, the coefficients are

found:

In the same way:

dxL

xm

L

xnbdx

L

xmxf

L

L

n

L

L

pppsinsinsin)(

+

-

+

-

=

...3,2,1sin)(1

== +

-

ndxL

xnxf

Lb

L

L

n

p

...3,2,1,0cos)(1

== +

-

ndxL

xnxf

La

L

L

n

p


9/133


If a function is even so that f(x) = f(-x), then

f(x)sin(nx) is odd.

And thus bn = 0 for all n.

If a function is odd so that f(x) = -f(-x), then

f(x)cos(nx) is even.

And thus an = 0 for all n.


10/133

Navier Solution for Simply-Supported

Rectangular PlatesPage 10

Rectangular geometry:

Distributed load = p(x,y)

No closed form solutions like circular plates.

b

a

x

y


11/133



Solution assumes load and deflection can be

represented by Fourier series:

pmn, amn = coefficients to be determined.

( )

( )b

yn

a

xmayxw

byn

axmpyxp

n

mn

m

n

mn

m

pp

pp

sinsin,

sinsin,

11

11

=

=

=

=

=

=


12/133



Deflections must satisfy the governing differential equation:

Boundary conditions (due to sin terms, automatically satisfied):

b)y0,(y0y

w0

a),0(x0

x

w0

2

2

2

2

===

=

===

=

w

xw

D

p

y

w

yx

w

x

ww =

+

+

=

4

4

22

4

4

44 2


13/133



Physical interpretation:

Consider term m = 1 and n = 2of w:

This term represents a single sin

wave deflection in the x-direction, and a double sin wavedeflection in the y-deflection, asshown:

Note that sin terms automaticallysatisfy simply-supported

boundary conditions. Increasing the number of terms in

the series can improve accuracy.

b

y

a

xa

pp 2sinsin12


14/133



The general procedure:

Determine coefficients pmn:

Find pmn from p(x,y) by multiplying both sides of

equation by:

And integrate from 0 to a and 0 to b:

dxdyb

yn

a

xm pp ''sinsin

( )

dxdyb

yn

a

xm

b

yn

a

xmp

dxdyb

yn

a

xm

yxp

n

ab

mn

m

ab

pppp

pp

''

1 001

'

0

'

0

sinsinsinsin

sinsin,

=

=

=


15/133



Recalling orthogonality relations, in this case:

And therefore:

( ) dxdyb

yn

a

xmyxp

abp

ab

mn

ppsinsin,

4

00

=

==

==

)(if2

),(if0sinsin

)(if2

),(if0sinsin

'''

0

'''

0

nnb

nndyb

yn

b

yn

mma

mmdxa

xm

a

xm

b

a

pp

pp


16/133



Determine amn by substituting p(x,y) and w(x,y)

expressions into governing equation:

222

4

222

4

1

4224

1

1

0

0sinsin2

+

=

=-

+

=

-

+

+

=

=

b

n

a

m

p

Da

D

p

b

n

a

ma

b

yn

a

xm

D

p

b

n

b

n

a

m

a

ma

mn

mn

mnmn

n

mnmn

m

p

p

pppppp


17/133



Substituting expression for amn into w(x,y):

The series is convergent if you use enough terms, you

will approach the exact solution.

b

yn

a

xm

bn

am

p

Dw

n

mn

m

ppp

sinsin1

1

222

1

4

=

=

+

=


18/133

Simply-Supported Rectangular Plates

Under Various LoadingsPage 18

Uniformly distributed load: p(x,y) = p0

( )

( ) ( )

-

-=

-=

-=

=

pp

pp

pp

p

pp

p

pp

nn

bm

m

a

ab

pp

dxnn

b

a

xm

ab

pp

dxb

yn

n

b

a

xm

ab

pp

dxdyb

yn

a

xmp

abp

mn

a

mn

a

mn

ab

mn

cos1cos14

cos1sin4

cossin4

sinsin4

0

0

0

b

00

0

0

0

0


19/133



Continue evaluating:

And finally:

( )( )

( )( ) ( )( )

,...)5,3,1,(16

11114

cos1cos14

2

0

2

0

2

0

==

----=

--=

nmmn

pp

mn

pp

nmmn

pp

mn

nm

mn

mn

p

p

ppp


20/133



Substitute into w:

Since plate must deflect into a symmetrical shape for a

uniform load, knew ahead of time that m and n must be

odd.

,...)5,3,1,(sinsin116

222

6

0 =

+

=

nmb

yn

a

xm

b

n

a

m

mn

D

pw

nm

ppp


21/133



Maximum deflection occurs at center of plate:

( ) ( )

( )( )

-

+

--

-

+

=

--

+

=

=

+

=

=

12

222

6

0max

2

1

2

1

222

6

0max

2226

0

max

max

1116

11116

,...)5,3,1,(2sin2sin

116

2,

2

nm

nm

nm

nm

nm

b

n

a

mmn

D

pw

b

n

a

mmn

D

pw

nm

nm

b

n

a

mmn

D

p

w

baww

p

p

pp

p


22/133



Recall:

Thus:

,...)5,3,1,(sinsin16

222

22

4

0 =

+

+

=

nmb

yn

a

xm

b

n

a

mmn

b

n

a

m

pM

nm

x

ppn

p

y

w

x

w

2

2

2

2

+

-= nDMx


23/133



Likewise:

Mx and My are zero at x = 0, x = a, y = 0, and y = b.

However, Mxy does not vanish at edges and corners.

( ),...)5,3,1,(coscos

1116

,...)5,3,1,(sinsin16

222

4

0

222

22

4

0

=

+

--=

=

+

+

=

nmb

yn

a

xm

b

n

a

mmn

ab

pM

nmb

yn

a

xm

b

n

a

m

mn

b

n

a

m

pM

nm

xy

nm

y

ppp

n

ppn

p


24/133



Assume square plate: a = b

Use first term of Fourier series: m = n = 1

( )

( )

D

apw

D

ap

a

D

pw

b

n

a

mmn

D

p

w

nm

nm

4

0max

6

4

0

22

6

0max

12

2226

0

max

00416.0

4

12

116

1

116

=

=

=

-

+

=

-+

pp

p


25/133



Using first four terms: m,n = 1,3

Also:

Moment does not converge as rapidly as deflection.

)(00406.04

0max solutionexact

D

apw =

)(0469.0

)(0534.0

2

0max,

2

0max,

termsfourapM

termfirstapM

x

x

=

=


26/133



Loading:( )

b

y

a

xpyxp

ppsinsin, 0=


27/133



Calculate pmn:

Calculate w:

)1(4

4

sinsinsinsin4

00

0

0

0

====

=

nmpabp

abp

dxdyb

yn

a

xm

b

y

a

xp

abp

mn

ab

mn

pppp

b

y

a

x

ba

D

pw

b

yn

a

xm

b

n

a

mmn

D

pw

nm

ppp

ppp

sinsin11

1

sinsin1

2

22

4

0

222

4

0

+

=

+

=


28/133



Calculate moments:

And likewise:

y

w

x

w

2

2

2

2

+

-= nDMx

b

y

a

x

ba

ba

pM

b

y

a

x

baba

pM

y

x

ppn

p

ppn

p

sinsin1

11

sinsin1

11

222

22

2

0

222

222

0

+

+

=

+

+

=

x

w

y

w

2

2

2

2

+

-= nDMy


29/133



Calculate moments:

( )yx

w1

2

+-= nDMxy

( ) by

ax

ab

ba

pMxy ppp

n coscos111

1 2

22

2

0

+

--=


30/133



Calculate shears:

And likewise:

+

-=2

2

2

2

y

w

x

w

xDQx

by

ax

bab

pQ

b

y

a

x

baa

pQ

y

x

ppp

pp

p

cossin11

sincos

11

22

0

22

0

+

=

+

=

+

-=2

2

2

2

y

w

x

w

yDQy


31/133



Calculate total applied load:

[ ]

2

0

00

0

0

0

4

211cossin:

sinsin

p

ppp

pp

pp

abpp

aa

a

xadx

a

xNote

dxdyb

y

a

xpp

total

aa

ab

total

=

=---=

-=

=


32/133



Find reactions at edges:

Setting x = a:

( )

( )

b

y

ab

ba

p

b

y

baa

pV

b

y

a

x

ab

ba

p

b

y

a

x

baa

p

V

y

MQV

x

x

xy

xx

p

p

np

p

pp

p

npp

p

sin1

11

1sin

11

sincos

1

11

1

sincos11

22

22

0

22

0

22

22

0

22

0

+

--

+-=

+

-

+

+=

+=


33/133



Simplifying:

Likewise for Vy

at edge y = b:

( )

( )

( )a

x

ab

bab

pV

b

y

ba

baa

pV

b

y

bba

baa

pV

y

x

x

pn

p

pn

p

pn

p

sin21

11

sin21

11

sin111

11

222

22

0

222

22

0

2222

22

0

-+

+

-=

-+

+

-=

-++

+

-=


34/133



Sum of all reactions:

By simple manipulation:

( ) ( )

( )2

22

2

0

2

0

222

22

0

222

22

0

00

11

184

21

11

421

11

4

22

+

---=

-+

+

-

-+

+

-=

+

baab

pabp

ba

baa

pb

ab

bab

pa

dyVdxV

b

x

a

y

p

np

n

pp

n

pp


35/133



Note that first term equals total applied load, soreactions are larger.

From concentrated reaction at each corner:

4 reactions (one at each corner). Equilibrium is therefore satisfied.

This is physical corners tend to rise.

( ) ( )2

22

2

0

2

22

2

0

11

12

11

12,2

+

-=-=

+

--=

===

baab

pFR

baab

pF

byaxatMF

ccc

xyc

p

n

p

n


36/133



Total load P distributed uniformly over a sub-region

4cd:

( )cd

Pyxp

4, =


37/133



Determine pmn:

How? See next slide.

( )

b

dn

a

cm

b

yn

a

xm

mncd

Pp

dxdyb

yn

a

xm

abcd

Pp

dxdyb

yn

a

xmyxp

abp

mn

cx

cx

dy

dy

mn

ab

mn

ppppp

pp

pp

sinsinsinsin4

sinsin

sinsin,4

11

2

00

1

1

1

1

=

=

=

+

-

+

-


38/133



Note:

Using identities:

Obtain:

( ) ( )

( )( )

( ) ( )

a

cm

a

xm

m

a

a

cxm

a

cxm

m

a

a

xm

m

adx

a

xmcx

cx

cx

cx

ppp

bababa

bababa

bababa

pp

p

pp

p

sinsin2

sinsin2coscos

sinsincoscoscos

sinsincoscoscos

coscos

cossin

1

11

1

1

1

1

=-=--+

+=-

-=+

--+

-=

-=+

-

+

-


39/133



If expand load to cover entire plate surface by

substituting the following:

Obtain the previous solution:

2

2

2

2

11

bd

ac

by

ax ====

mn

P

pmn 216

p=


40/133



Point load at x = x1, y = y1:


41/133



Use previous derivation, with c,d 0.

b

yn

a

xm

ab

Pp

b

n

db

dn

d

a

ma

cm

a

m

c

ca

cm

c

b

dn

a

cm

b

yn

a

xm

mncd

Pp

mn

mn

11

11

2

sinsin4

sin)0lim(similarly,

1

cos)0lim(

:RulesHopital'L'use

00sin)0lim(

sinsinsinsin4

pp

pp

ppp

p

ppppp

=

=

=

=

=


42/133



Leads to:

If P is applied to the center of a square plate (a=b):

Using first nine terms (m,n = 1, 3, 5):

sinsinsinsin14 11

222

4 b

yn

a

xm

b

yn

a

xm

b

n

a

mDab

Pw

nm

ppppp

+

=

( )...),,(m,n

nmD

Paw

by

ax

nm

53114

2

2

2224

2

11 =+

===

p

==

D

Paexact

D

Paw

22

max 01159.001142.0


43/133

Levys Solution for Rectangular PlatesPage 43

Navier solution slow convergence of series forbending moments.

Levys solution:

Overcomes slow convergence. Uses single series (Navier uses double series)

Allows for more general boundary conditions (Navieronly valid for simply-supported conditions on all sides).n

Particular BC on two opposite sides (x = 0 & x = a)n Arbitrary BC on remaining edges (y = -b/2 & y = +b/2)


44/133


Procedure:

As before, result is sum of homogeneous and particular

solution:

Particular solution is obtained for each specific loading.

Homogeneous solution is independent of loading:

phwww +=

04 = hw


45/133


Homogeneous solution assumed to be:

( ) ( )

( )2

atconditionsboundaryfulfill

cossin 11

byyf

a

xmyfwor

a

xmyfw

m

mmh

mmh

=

=

=

=

=

pp


46/133


For example, if plate is simply-supported at x = 0

and x = a:

Use the sin term, which will automatically satisfy the

boundary conditions at these edges.

( )

=

= a

xmyfw

m

mh

psin

1

a),0(x0x

w0 2

2

===

= xw


47/133


Substitute into governing equation:

Term inside brackets must be zero linear

differential equation with constant coefficients.

0sin2

0

1

4

2

22

4

4

4

=

+

-

=

= a

xmf

a

m

dy

fd

a

m

dy

fd

w

m

mmm

h

ppp

024

2

22

4

4

=

+

- m

mm fa

m

dy

fd

a

m

dy

fd pp


48/133


Solution (from differential equations):

Can be written as:

( )AAm

yeCyeCeCeCmAm

AmAm

fa

m

dy

fd

a

m

dy

fd

AyAyAyAy

mmm

=

+++==-

=+-

=

+

-

--

,

0

02

02

4321

222

4224

4

2

22

4

4 pp


49/133


And thus the general solution is:

Can also be written as (usually easier to work with):

Note:

a

ym

ma

ym

ma

ym

ma

ym

mm yeDyeCeBeAfpppp

--+++= ''''

a

ymyD

a

ymyC

a

ymB

a

ymAf mmmmm

ppppcoshsinhcoshsinh +++=

( ) ( )uuuu

eeueeu--

+=-= 21

cosh2

1sinh


50/133


Therefore:

casesspecificfordeterminedconstants,,,

sincoshsinhcoshsinh1

+++=

=

mmmm

m

mmmmh

DCBA

a

xm

a

ymyD

a

ymyC

a

ymB

a

ymAw

ppppp


51/133


Particular solution:

Last equation found by multiplying both sides by the sin term

below and integrating:

( )

( ) ( )

( ) ( )

=

=

=

=

=

a

m

m

m

m

mp

dxa

xmyxp

ayp

a

xmypyxp

a

xmykw

0

1

1

sin,2

sin,

sin

p

p

p

a

xm p'sin


52/133


Plugging into differential equation:

=

+-

=

=

= axm

Dp

axmk

am

dykd

am

dykd

D

pw

m

m

m

mmm

p

pppp sinsin211

4

2

22

4

4

4

D

pk

a

m

dy

kd

a

m

dy

kd mm

mm =

+

-4

2

22

4

4

2pp


53/133


The general procedure thus becomes:

From loading p(x,y), find pm.

From pm, find km and thus wp.

Using boundary conditions, find Am, Bm, Cm, Dm and thuswh.

Result is sum of wp and wh.


54/133


Simply-supported rectangular plate under uniform

loading.

Find pm.

( )

( )

( ) ( )...5,3,14

22cos

2sin

2

,

0

0

0

0

0

0

0

==

=

-=

=

=

mm

pyp

m

ap

aa

xm

m

ap

adx

a

xmp

ayp

pyxp

m

aa

m

p

pp

pp


55/133


Find km.

n Since the right hand side is not a function of y, then kmcannot be a function of y, so derivatives of km are zero:

Dm

pk

a

m

dy

kd

a

m

dy

kdm

mm

ppp 0

4

2

22

4

4 42 =

+

-

Dmapk

Dmpk

am mm 55

4

00

4

44ppp ==


56/133


Thus wp becomes:

This particular solution represents

the deflection of a uniformly

loaded, simply-supported strip

parallel to the y-axis (very longin the y-direction).

a

xm

mD

apw

m

p

pp

sin14

...3,1

55

4

0

=

=


57/133


Now must evaluate constants in homogeneous solution.

Observing that the deflection must be symmetrical with

respect to the x-axis: Am = Dm = 0.

n sinh function is not symmetric.n cosh function is symmetric.

n y*sinh is symmetric

n y*cosh is not symmetric.

sinh

cosh


58/133

Levys Solution for Rectangular Plates

Page 58

Expression for deflection is now:

Satisfies governing equations and simply-supported

boundary conditions at x = 0 and x = a.

Remaining boundary conditions:

a

xm

Dm

ap

a

ymyC

a

ymBwww

m

mmhp

pp

ppsin

4sinhcosh

...3,155

4

0

=

++=+=

==

=

2y0

y

w0

2

2 bw


59/133


Page 59

Substituting:

Gives:

02sinh22cosh22

04

2sinh

22cosh

55

4

0

=+

+

=++

a

bm

a

bmCa

bmCa

mB

Dm

ap

a

bmbC

a

bmB

mmm

mm

ppppp

pp

a

bmDm

apC

a

bmDm

a

bmbapmap

B

m

m

2cosh

22

cosh

2tanh4

44

3

0

55

3

0

4

0

pp

pp

pp

=

+

-=


60/133


Page 60

Setting:

Simplifies constants:

m

m

m

mm

Dm

apC

Dm

bapmapB

ap

apap

cosh

2

cosh

tanh4

44

3

0

55

30

40

=

+-=

a

bmm

2

pa =


61/133


Page 61

Deflection:

Maximum deflection is at x = a/2, y = 0:

axm

by

aym

b

y

mD

apw

m

m

m

m

m

mm

papa

aa

aap

sin2sinhcosh2

1

2cosh

cosh2

2tanh1

14

...3,155

4

0

+

+-=

=

2sincosh2

2tanh1

14

...3,155

4

0max

pa

aap

m

mD

apw

m

mm

m

+-=

=


62/133


Page 62

Maximum deflection can be simplified:

Can manipulate to the following form:

Note that first term is the previous solution for a

uniformly loaded simply-supported strip.

( )( )

+-

-=

=

-

m

mm

m

m

mD

apw

aaa

p cosh22tanh

114

...3,15

2

1

5

4

0max

( )( )

+--=

=

-

m

mm

m

m

mD

ap

D

apw

aaa

p cosh22tanh14

384

5

...3,15

2

1

5

4

0

4

0max


63/133


Page 63

For a square plate (a = b):

n Note that the series converges rapidly.

Can rewrite equation as:

( )D

ap

D

ap

D

apw

4

0

5

4

0

4

0max 00406.0...00025.068562.0

4

384

5=+--=

p

( )( )

+--==

=

-

m

mm

m

m

mD

apw

a

aa

pdd

cosh2

2tanh14

384

5

...3,15

2

1

51

4

01max


64/133


Page 64

Can also write similar expressions for moments:

Parameters d1, d2, d3 can be determined for variouscases.

Note that as plate becomes long (b/a tends to infinity),

results equal that of simply-supported strip.

If ratio of sides is large (b/a > 4), effect of short sidesis negligible and plate can be considered as infinite

strip.

2

03max,

2

02max, apMapM yx dd ==


65/133


Page 65

3 edges simply-supported, 1 edge clamped,

uniform pressure.

Deflection is symmetrical about x = a/2.

Thus, w uses only odd m.


66/133


Page 66

Deflection:

Note that the particular solution is the same as before, since it is for a

uniform load with simply-supported boundary conditions at x = 0 and x

= a, and has nothing to do with the boundary conditions on the y sides.

Also, the simply-supported boundary conditions at x = 0 and x = a are

automatically satisfied due to the sin term in x.

a

xm

Dm

ap

a

ym

yD

a

ymyC

a

ymB

a

ymAw

m

m

mmm

p

p

p

ppp

sin

4

cosh

sinhcoshsinh

55

4

0

...3,1

++

++=

=


67/133


Page 67


The boundary conditions provide 4 equations to solve

the 4 unknown constants, shown on next page.

b)(y0y

w0

)0(y0y

w0

2

2

==

=

==

=

w

w


68/133


Page 68

After some manipulation, the constants are:

a

bm

ba

m

a

m

a

m

C

DmapB

Dm

a

Dm

apA

m

mmm

mmmm

mm

m

m

mmm

mmmmm

pb

bbb

bp

bp

bbp

b

p

pbbbbbbb

p

=

-

-

-

-=

-=

-=---

=

sinhcosh

coshsinhcoshsinh2

2

1

4

sinhcosh

sinhcosh2cosh22

2

55

40

2

55

4

0


69/133


Page 69

For a square plate (a=b):

===

===

0,2

084.0

2,

20028.0

20max,

4

0

yaxapM

ay

ax

D

apw

y

center


70/133


Page 70

Opposite edges simply-supported, 3rd edge free,

4th edge clamped, uniform pressure.


71/133


Page 71

Deflection:

Note that the particular solution is the same as before, since it is for a

uniform load with simply-supported boundary conditions at x = 0 and x

= a, and has nothing to do with the boundary conditions on the y sides.

Also, the simply-supported boundary conditions at x = 0 and x = a are

automatically satisfied due to the sin term in x.

a

xm

Dm

ap

a

ym

yD

a

ymyC

a

ymB

a

ymAw

m

m

mmm

pp

p

ppp

sin

4

cosh

sinhcoshsinh

55

4

0

...3,1

++

++=

=


72/133


Page 72


The boundary conditions provide 4 equations to solve the 4

unknown constants. Not shown here, but it is straightforwardif not tedious.

Same procedure for any similar uniformly loaded plate.

( )

shearmoment

b)(y0yx

w2

y

w0

)0(y0y

w0

2

3

3

3

2

2

2

2

==

-+

=

+

==

=

nnx

w

y

w

w


73/133

Levys Method Applied to Non-

uniformly Loaded Rectangular PlatesPage 73

Assume rectangular plate simply-supported at x = 0

and x = a.

Assume load is a function of x only.

( )

( )

=

=

=

a

m

m

m

dxa

xmxp

ap

a

xmpxp

0

1

sin2

sin

p

p


74/133



As before, assume for the particular solution:

Plugging into governing equation:

( )

=

= a

xmykw

m

mp

psin

1

0sin21

4

2

22

4

4

4

=

-

+

-

=

= a

xmDpk

am

dykd

am

dykd

D

pw

m

mm

mm

p

ppp


75/133



Since pm is only a function of x, km cannot be a function

of y, and the derivatives with respect to y are zero.

Particular solution, which represents deflection of a strip

under load p(x), becomes:

D

pk

a

m

dy

kd

a

m

dy

kd mm

mm =

+

-4

2

22

4

4

2pp

Dm

apk

D

pk

a

m mm

mm 44

44

p

p==

a

xm

m

p

D

aw

m

mp

pp

sin1

44

4

=

=


76/133



Assume the two arbitrary edges, y = -b/2 and y =

b/2, are also simply-supported:

n Note that, as before, Am and Dm are eliminated since they

are associated with terms that are not symmetric in y.

Bm and Cm are determined from boundary conditions at

y edges:

a

xm

Dm

ap

a

ymyC

a

ymBwww

m

mmmhp

pp

ppsinsinhcosh

144

4

=

++=+=

==

=2

y0y

w0

2

2 bw

M A N


77/133



After evaluation of boundary conditions, the deflection

is:

Given p(x) find pm find w find M find .

a

bm

a

xm

a

ym

a

ym

a

ym

m

p

D

aw

m

m

m m

mmm

2

sinsinhcosh2

1

coshcosh2

2tanh1

144

4

pa

pppa

pa

aap

=

+

+-=

=

L M h d A l d N


78/133



Values of pm for

various loadings.

L M h d A li d N


79/133



Example: Hydrostatically loaded plate.

( )

( )

( ) ( )...2,11-2

00cos0

2

cossin2

sin

2

sin

2

1m0

2

2

0

0

2

2

0

0

2

0

0

0

0

==

+-

-=

-

=

=

=

=

+

mm

pp

mm

a

a

p

p

a

xm

m

xa

a

xm

m

a

a

pp

dxa

xm

xa

p

dxa

xm

a

x

pap

a

xpxp

m

m

a

m

aa

m

p

pp

pp

pp

pp

L M h d A li d N


80/133



Can now find w by substituting expression for pm.

If plate is square (a = b), for example, the maximum

deflection is:

=== 0,2

00203.04

0 ya

xD

apwcenter

L M h d A li d N


81/133



Example: Line loads on long rectangular plates.

L M h d A li d N


82/133



Only need to consider one half of plate (positive y)

due to symmetry:

Each half takes half the load:

)0(y0

y

w==

( ) )0(yy2

)( 2 =

-=-= wDxP

Qy

L M th d A li d t N


83/133



Use alternate form of governing equation:

There is no surface pressure, so the particular solution is

zero and:

Also, w and its derivatives should vanish at y = :

0'' == mm CA

a

xmyeDyeCeBeAw

m

a

ym

ma

ym

ma

ym

ma

ym

mh

pppppsin''''

1

=

--

+++=

hww =

L M th d A li d t N


84/133



Equations for w:

Evaluating symmetry boundary condition:

a

xmyeDeBw

m

a

ym

ma

ym

m

pppsin''

1

=

--

+=

==+

-

=

-+

-

==

---

a

mBDD

a

mB

ye

a

meDe

a

mB

mmmm

a

ym

a

ym

ma

ym

m

pp

pp ppp

''0''

0''

)0(y0y

w

L M th d A li d t N


85/133



Evaluating remaining boundary condition:

Need to evaluate derivatives (next slide).

( )

2)(

yxy

)0(yy2

)(

2

2

2

2

2

xPwwD

wDxP

Qy

=

+

=

-=-=

L M th d A li d t N


86/133



( )a

xmeD

a

mw

a

xmeD

a

mw

a

xmyeD

a

meD

a

meD

a

meB

a

mw

a

xmyeD

a

meDeB

a

mw

a

xmyeDeB

a

mw

a

ym

m

a

ym

m

a

ym

ma

ym

ma

ym

ma

ym

m

a

ym

m

a

ym

m

a

ym

m

a

ym

ma

ym

m

pp

pp

ppppp

ppp

pp

p

p

pppp

ppp

pp

sin'2y

sin'2

sin''''y

sin'''yy

sin''x

2

2

2

22

2

2

2

2

2

2

2

=

-=

+

-

-

=

-+

-

=

+

-=

-

-

----

---

--

L M th d A li d t N


87/133



Substituting into boundary condition:

And from previous expression:

Dm

apD

D

pD

a

my

DpeD

am

xPwwD

mm

mm

ma

ym

m

22

22

2

2

2

2

2

4'

2'20

cancel)(sin terms2

'2

2

)(

yxy

pp

pp

==

=

=

+

-

Dm

apB mm 33

3

4'

p=

L M th d A li d t N


88/133



Thus deflection becomes:

Equations for P(x):

a

xme

a

ym

m

p

D

aw

a

xme

Dm

yape

Dm

apw

aym

m

m

m

a

ym

ma

ym

m

ppp

ppp

p

pp

sin14

sin44

133

3

122

2

33

3

-

=

=

--

+=

+=

( )

=

= a

xmpxP

m

m

psin

1

Levys Method Applied to Non


89/133



If p(x) = p0:

( )D

ap

D

ap

mD

apw

m

mD

apaww

a

xme

a

ym

mD

apw

m

m

m

a

ym

m

1536

5

1536

51

2sin

10,

2

sin11

3

0

5

4

3

0

...3,14

2

1

4

3

0max

...3,1 44

3

0

max

...3,144

3

0

pppp

p

p

ppp

p

=

=

-=

=

=

+=

=

-

=

-

=

[ ]

( )...5,3,14

cos12

cos2

sin2

0

0

0

0

0

0

==

-=

-=

=

mm

pp

mm

p

a

xm

m

pdx

a

xmp

ap

m

aa

m

p

pp

pp

p



90/133

Levy s Method Applied to Non-


If p(x) is a constant load p0 but only partially along

line:



91/133



And plugging into expression for w:

n For x1 = a/2 and c = a/2, get previous result.

( ) ( )

=

-+

+-=

=

+-=

+

-

a

cm

a

xm

m

pp

a

m

a

m

m

pdx

a

xmp

ap

pxp

m

m

ppp

ppp

p

sinsin4

cxcos

cxcos

2sin

2

cxtocxfrom)(

10

110

cx

cx

0

110

1

1

a

xm

a

cm

a

xm

ea

ym

mD

ap

wa

ym

m

pppp

p

p

sinsinsin1

1 1

144

3

0-

=

+=



92/133



For case of concentrated load:



93/133



Substitute the following expressions into w:

a

cm

a

cm

c

PpcpP

pp

== sin2

or200

a

xm

a

xme

a

ym

mD

Paw

a

xm

a

cm

a

xme

a

ym

mD

a

c

Pw

a

ym

m

a

ym

m

ppp

p

ppppp

p

p

sinsin11

2

sinsin11

2

1

1

33

2

1

144

3

-

=

-

=

+=

+

=

Rectangular Plates Under Distributed


94/133


Edge MomentsPage 94

Simply-supported rectangular plate, symmetrically

distributed edge moments at y = +b/2 and y = -b/2:



95/133


Edge MomentsPage 95

Assume a Fourier sin series for the moment distribution:

Boundary conditions:

( )

( )

=

=

=

=

a

m

m

m

dxa

xm

xfaM

by

a

xmMxf

0

1

sin

2

2sin

p

p

( )

( )

===

===

=

2by

ywD-0

,0x0x

w0

2

2

2

2

xfw

axw



96/133


Edge MomentsPage 96

Since p0 = 0 and plate deflection is symmetric about x-

axis:

n Above expression satisfies governing equation and first set

of boundary conditions.

n Use other two boundary conditions to determine constants.

a

xm

a

ymyC

a

ymBw

m

mm

pppsinsinhcosh

1

=

+=



97/133


Edge MomentsPage 97

First boundary condition:

a

xm

a

ymb

a

ymyCw

bCB

a

bmb

CB

w

m

mm

mmm

mmmmm

ppa

p

a

p

aaa

sincoshtanh2

sinh

tanh2

20sinh2cosh

2

by0

1

=

-=

-=

==+

==



98/133


Edge MomentsPage 98

Second boundary condition:

Using basic derivation:

( )

a

xm

a

ymy

a

ymb

m

M

D

aw

Dm

aMC

a

xmM

a

xm

a

mCD

xf

m

m

m

m

m

mm

m

m

m

mm

pppa

ap

ap

ppa

p

sinsinhcoshtanh2cosh2

cosh2

sinsincosh2

2

by

y

wD-

1

11

2

2

=

=

=

-=

-=

=

-

==



99/133


Edge MomentsPage 99

If f(x) = M0 (uniformly distributed moments):

For square plate (a = b), results at center:

a

xm

a

ymy

a

ymb

mD

aMw

mm

M

a

xm

m

Mdx

a

xmM

aM

m

m

m

ax

x

a

m

pppa

ap

pp

pp

sinsinhcoshtanh2cosh

12

...3,14

cos2

sin2

...3,122

0

00

0

0

0

=

==

-=

==-==

00

20 256.0,394.0,0368.0 MMMMD

aMw yx ===



100/133


Edge MomentsPage 100

Displacement along axis of symmetry (y = 0):

For a very long strip (a >> b), i.e. center deflection ofa strip of length b subjected to two equal and opposite

bending moments at ends:

a

xm

mD

abMw

m m

m pa

ap

sincosh

tanh

...3,122

0

=

=

D

bM

D

bMw

a

xm

mD

bM

a

xm

mD

abMw

a

bm

mm

m

mmmm

842

sin1

2sin

21coshtanh

2

0

2

0

...3,1

20

...3,122

0

=

=

==

=

=

=

pp

pp

pap

paaaa



101/133



Plate with one edge moment:

n Note: This problem switches previous x and y directions.



102/133



Assume same Fourier series expression:

Deflection is in the following form:

( ) ( )

( )

=

=

=

=

b

xn

n

nx

dyb

yn

MbM

xb

ynMM

0

0

10

sin

2

0sin

p

p

b

yn

b

xn

xDb

xn

xCb

xn

Bb

xn

Aw nnnnn

pppppsincoshsinhcoshsinh 1

=

+++=



103/133




Last two boundary conditions are automatically

satisfied, as always. First boundary condition results in: Bn = 0.

( ) ( )

( )

( )byw

aw

Mw x

===

=

==

=

==

=

,0y0y

w0

x0x

w

0

0xx

wD-0

2

2

2

2

02

2



104/133



Remaining 3 boundary conditions can be used to solve

for remaining 3 constants, resulting in:

Based on moment, can find Mn and then substitute into

w expression above.

( )b

yn

b

ynb

xna

b

xan

x

b

ann

M

D

b

wn

n p

p

pp

pp sinsinh

sinh

coshsinh

2 1

=

-

-

=

Method of Superposition Applied to


105/133


Bending of Rectangular PlatesPage 105

Complex problem replaced by several simplersituations. Simpler solutions: Navier or Levy approach.

Superposed so that the overall governing equation and

boundary conditions are fulfilled. For example, consider a rectangular plate under any

lateral load, one edge (y=0) clamped and otherssimply-supported. Start with plate with all edges simply-supported.

Add the solution of plate with bending moment appliedalong y=0, with magnitude to eliminate rotations alongclamped edge.



106/133



Example: Two edges simply-supported, two edges

clamped, uniform pressure.



107/133



Plate 1 (from previous Levy approach):

For plate 2, need to determine the moment along the edges

so that the sum of the rotations of plate 1 and plate 2 alongthese edges equals zero (clamped boundary condition).

Thus, need to find the rotation along these edges in plate 1.

a

xm

b

y

a

ym

b

y

mD

apw

m

m

m

m

m

mm

papa

aa

aap

sin2

sinhcosh2

1

2cosh

cosh2

2tanh1

14

...3,155

4

01

+

+-=

=

a

bmm

2:where

pa =



108/133



Rotation along edge found by taking derivative of w

with respect to y:

Substitute y = b/2:

a

xm

b

y

a

m

b

y

ba

ym

b

y

bmD

ap

y

w

m

m

mm

m

m

mm

m

mm

papa

aapa

aaa

aap

sin2

sinhcosh2

12cosh

2

cosh2

1

2sinh

2

cosh2

2tanh14

...3,155

4

01

+

+

+-=

=

a

xm

a

m

b

bmD

ap

y

w

m

m

mm

m

m

m

m

m

mm

pa

pa

aa

a

aa

a

aa

p

sinsinhcosh2

1cosh

2

cosh2

1

sinh2

cosh2

2tanh14

2

...3,1

55

4

01

+

+

+-=

=



109/133



Further manipulation:

( )

( ( ))

( ( ))

abm

a

xm

mD

ap

y

w

a

xm

bmD

ap

y

w

a

xm

bbbmD

ap

y

w

m

mmm

m

m

mmm

m

mm

mmm

mmm

m

m

2:where

sin1tanhtanh12

sin1tanhtanh4

sintanh2tanhtanh14

...3,144

3

01

...3,155

4

01

2

...3,155

4

01

pa

paaaa

p

paaaa

ap

pa

aaaaa

ap

=

+-=

+-=

+++

-=

=

=

=



110/133

Me od o Supe pos o pp ed o


Plate 2, from previous solution:

Edge rotation for this case:

=

=

= 2sin

1

by

a

xmMM

m

my

p

a

xm

a

ym

ya

ymb

m

M

D

a

wm

mm

m ppp

aap sinsinhcoshtanh2cosh2 12

=

-=

a

xm

a

ym

a

my

a

yma

ym

a

mb

m

M

D

a

y

w

m

m

m

m

pppp

ppa

apsincoshsinh

sinhtanh

2cosh2 1

2

--

=

=



111/133

M p p pp


Further manipulation, substituting y = b/2:

(

)

( )

( )[ ] axm

m

M

D

a

y

w

a

xm

m

M

D

a

y

w

a

xm

m

M

D

a

y

w

mmmmm

m

mm

m

mmmm

mmm

m

mmm

m

m

p

aaaap

paaaaa

p

paaa

aaaap

sin1tanhtanh2

sintanhtanhtanh2

sincoshsinh

sinhtanhcosh2

1

2

1

2

1

2

--=

--=

--

=

=

=

=



112/133

p p pp


Slopes must be equal and opposite:

Final result is then found by:

n Substituting Mm into w2.

n Add solutions of w1 and w2.

( )( )1tanhtanhtanh1tanh4

:gives

2

33

2

0

21

--+-

=

=

-=

mmmm

mmmmmm

apM

by

y

w

y

w

aaaaaaaa

p

S i M h d


113/133

Strip MethodPage 113

Simple, approximate approach for computing

deflection and moment in rectangular plate with

arbitrary boundary condition.

Plate is assumed to be divided into two systems ofstrips at right angles to one another.

Each strip is regarded as functioning as a beam.

Not accurate, but this method gives conservative values

for deflection and moment. Employed in practice.

St i M th d


114/133


Based on deflections and moments of beams with

various end conditions derived from mechanics of

materials.

8

3

8

5 pLR

pLR BA ==

St i M th d


115/133


Consider the rectangular plate:

Strip of span a carries uniform load pa. Strip of span b carries uniform load pb.

St i M th d


116/133


Two conditions that must be satisfied:

For simply-supported edges (same for clamped plate):

( )00 ==+== yxpppww baba

44

4

044

4

00

44

44

384

5

384

5

baapp

babppppp

bpapww

EI

bpw

EI

apw

baba

baba

bb

aa

+=

+=+=

==

==

St i M th d


117/133


And thus:

Using the expressions for beams in the given tables,

once the center deflection is determined, can find the

deflection and moment anywhere along either strip.

( )

384

5

384

5

384

544

44

0

44

maxbaEI

bap

EI

bp

EI

apw ba

+===

St i M th d


118/133


Example: 3 edges simply-supported, 1 edge clamped,uniform p0.

Use expressions for displacement of a beam simply-

supported on each end along y, and a beam clamped onone end and simply-supported on another along x.

Determine mid-span deflection and maximum moments

St i M th d


119/133


Setting mid-span deflections equal (note: this is not

always the location of maximum deflection):

( )EIba bapEIbpw

ba

app

ba

bppppp

bpapwwEI

bp

EI

apww

bcenter

baba

baba

baba

44

44

0

4

44

4

044

4

00

44

44

521925

3845

52

2

52

5

52384

5

192

+==

+=

+=+=

==

==

Strip Method


120/133


To find maximum moments, must first find where

maximum occurs for each beam, then substitute for

appropriate load and span.

For example, for the y-span (simply-supported to

simply-supported), it is known that the maximum

moment occurs at midspan:

( )

( )( )0

524

084222

44

24

0

2

=+

=

==

-

=

yba

bapM

ybpbb

pbb

pM

y

bbby

Strip Method


121/133


In the x-direction span, must first determine where the

maximum moment occurs:

n Reaction force at simply-supported end = 3paa/8.

( )

( )

=+

=

-=

-

=

==-=

-=

axba

bapM

apaapapaM

axapxpdx

dM

axpxpxM

x

aaax

aax

aax

8

3

52128

45

128

9

8

3

8

3

8

3

2

1

8

3

8

30

8

3

8

3

2

1

44

42

0

2

2

2

Simply-Supported Continuous


122/133


Continuous plate:

A uniform plate that extends over a support and has more

than one span that may be of varying length.

Intermediate supports provided by beams or columns.

Analysis is based on equilibrium of individual panels

and compatibility of displacements or force at

adjoining edges.

Assume only rigid intermediate beams: Plate has zero deflection along axis of supporting beam.

Beam does not prevent rotation, i.e. simply-supported.



123/133


Consider two-paneled continuous plate, as shown:

( ) y

sin...3,1 b

mMyf

m

m

p

=

=



124/133


Procedure:

Consider each panel separately.

On connecting edges, assume a moment is applied

(value unknown). Use regular plate boundary conditions to solve for

constants.

Need one additional equation since there is an

additional unknown (moment). This equation is basedon continuity at the connecting edges, i.e. the slopes

must be equal.



125/133


Expressions for deflections of each plate:

(

() yxxH

xxGxFxEw

b

m

yDmbpxxD

xxCxBxAw

mmm

mmmmmmm

m

mmm

m

mmmmmm

lllll

pl

lp

l

lll

sincosh

sinhcoshsinh

sin4cosh

sinhcoshsinh

22

...3,122222

55

4

011

...3,1

11111

+++=

=

++

++=

=

=



126/133



( )

( )

( )

( )0ysinD-0

00

ysinD-0

000

2

...3,12

2

2

2

2

22

2

2

2

2

1

...3,12

1

1

2

1

12

1

1

2

1

==

=

==

=

===

==

=

=

=

xMx

ww

axx

ww

axMxww

xx

ww

m

m

m

m

m

m

l

l



127/133


Additional boundary condition continuity:

Find constants Am Hm and Mm (9 unknown) using

given 9 equations. Straightforward, results on next page.

02

2

1

1

21 ==

=

xax x

w

x

w



128/133


aD

M

D

MG

aD

aME

ab

pM

D

aD

Db

p

Db

pB

aa

ab

pa

b

pM

a

aa

DA

m

m

m

m

mm

m

m

mm

m

m

o

m

mmm

mm

om

mm

m

m

om

m

o

m

m

m

mm

lll

ll

lll

lll

ll

ll

llll

l

coth2H2

0F)coth1(2

)cosh1(2

2

csch

2C

4

csch2

coth4

)cosh1(2

2sinh

coth1

m

m

2

4

4

0m5

54

=-=

=-=

+-+-=

=-=

--+

+-+=

Homework Problem 6


129/133

Homework Problem 6Page 129

A simply supported rectangular plate is under the action of

hydrostatic pressure expressed by p = p0x/a, where constant

p0 represents the load intensity along the edge x = a.

b

Homework Problem 6


130/133


Determine, employing Naviers approach and retaining the

first term of the series solution, the equation for the resulting

solution:

by

ax

ba

Dpw pp

psinsin

11

182

22

6

0

+

=

Homework Problem 7


131/133


A water level control structure consists of a vertically

positioned simply supported plate. The structure is filled with

water up to the upper edge level at x = 0.

Homework Problem 7


132/133


Show, using Levys method, that by taking only the first term

of the series solution, the value for the deflection at the center

for a = b/3 is:

Dapw

4000614.0=

Homework Problem 8


133/133


A rectangular plate is subjected to a uniform loading p0 and

supported as shown.

Determine the maximum stresses using the strip method.

03 Rectangular

Documents

Transcript of 03 Rectangular