03 Rectangular

Page 1

RECTANGULAR PLATES

Rect ectang angula ularr Plate Plate - Topi opics cs Page 2

¨ ¨

Fourier Series (5-9) Naviers Solution (10-17) ¤ ¤ ¤ ¤

¨

Uniform Load (18-25) Sine Load (26-35) Uniform Load Over Subregion (36-39) Point Load (40-42)

Levys 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)

Rect ectang angula ularr Plate Plate - Topi opics cs Page 3

Simply Supported Hydrostatic Load (79-80) ¤ Simply Supported Long Rectangular Plates (81-88) ¤

Constant Load (89) n Partial Line Load (90-91) n Concentrated Load (92-93) n

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)

Background: Expressions To Know Page 4

d dx d

(sin u ) = cos u

du

dx du (cos u ) = - sin u dx dx d du (sinh u ) = cosh u dx dx d du (cosh u ) = sinh u dx dx

1

ò (sin ax )dx = - a cos ax 1

ò (cos ax )dx = a sin ax 1

ò (sinh ax )dx = a cosh ax 1

ò (cosh ax )dx = a sinh ax

Fourier Series Page 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: f ( x) =

a0 2

np x np x ö cos sin a b + å æ + ç n ÷ n L L ø n =1 è ¥


¨

Coefficients are found through the use of orthogonality relations: + L

ò cos

- L

mp x L

cos

np x L

dx = 0 ( m ¹ n )

= L ( m = n ) + L

ò

cos

- L + L

ò

sin

- L

mp x L mp x L

sin sin

np x L np x L

dx = 0

all m, n

dx = 0 ( m ¹ n )

= L (m = n)


¨

To determine the coefficients b n of the following Fourier series: f ( x ) =

¥

å b sin n

n =1

¨

Multiply both sides by: sin

¨

f ( x) sin

L

L

mp x L

To obtain: mp x

np x

¥

= å bn sin n =1

np x L

sin

mp x L


¨

Integrate both sides: + L

ò f ( x) sin

mp x L

- L

¨

+ L

dx =

n

- L

L

sin

mp x L

dx

And using the orthogonality relationship, the coefficients are + L found: np x 1 bn

=

f ( x) sin ò L - L

¨

ò b sin

np x

L

dx

n = 1,2,3...

In the same way: an

=

1

+ L

np x

f ( x) cos ò L L - L

dx

n = 0,1,2,3...


¨

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.

Navier Solution for Simply-Supported Rectangular Plates Page 10

¨

a

Rectangular geometry:

x b ¤

Distributed load = p(x,y) y

¤

No closed form solutions like circular plates.


¨

Solution assumes load and deflection can be represented by Fourier series: p ( x, y ) = w( x, y ) =

¥

å å p m =1

n =1

¥

¥

åå m =1

¤

¥

n =1

mn

sin

amn sin

mp x a mp x a

sin sin

np y b np y b

pmn, amn = coefficients to be determined.


¨

Deflections must satisfy the governing differential equation: 4 4 4 w w w p ¶ ¶ ¶ 4 Ñ w= 4 +2 2 2 + 4 = ¶ x ¶ x ¶ y ¶ y D

¨

Boundary conditions (due to sin terms, automatically satisfied): w=0 w=0

¶2w =0 ¶x 2 ¶2w =0 2 ¶y

(x = 0, x = a) (y = 0, y = b)


¨

Physical interpretation: ¤

¤

¤

¤

Consider term m = 1 and n = 2 of w: This term represents a single sin wave deflection in the xdirection, and a double sin wave deflection in the y-deflection, as shown: Note that sin terms ter ms automatically satisfy simply-supported boundary conditions. Increasing the number of terms in the series can improve accuracy.

a12 sin

p x a

sin

2p y

b


¨

The general procedure: ¤

Determine coefficients pmn:

¤

Find pmn from p(x,y) by by multiplying both sides of equation by: m 'p x n 'p y sin

¤

sin

a

b

dxdy

And integrate from 0 to a and 0 to b: b

a

ò ò p( x, y )sin 0

m 'p x a

0

¥

=å

m =1

¥

b

a

sin

n 'p y

å p ò ò sin mn

n =1

0

0

b

dxdy

mp x a

sin

np y b

sin

m 'p x a

sin

n 'p y b

dxdy


¤

Recalling orthogonality relations, in this case: a

ò

sin

0

b

ò

sin

0

¤

mp x a np y b

sin sin

m 'p x a n 'p y b

ì î

dx = í0 if ( m ¹ m ' ),

ì î

dy = í0 if (n ¹ n ' ),

a 2

b 2

ü þ

if (n = n ' )ý

And therefore: pmn

=

4

b

a

ab ò ò

p( x, y ) sin

0

0

mp x a

sin

np y b

ü þ

if ( m = m ' )ý

dxdy


¤

Determine amn by substituting p(x,y) and w(x,y) expressions into governing equation:

ìï éæ mp ö 4 æ mp ö 2 æ np ö 2 æ np ö 4 ù pmn üï mp x np y sin =0 ÷ + 2ç ÷ ç ÷ +ç ÷ úíamn êç ý sin å å a a b b D a b è ø è ø è ø úû m =1 n =1 ï ïþ î êëè ø 2 2 2 é ù p æ m ö æ n ö amnp 4 êç ÷ + ç ÷ ú - mn = 0 êëè a ø è b ø úû D ¥

amn

¥

=

1

pmn

p 4 D éæ m ö 2 æ n ö 2 ù 2 êç ÷ + ç ÷ ú êëè a ø è b ø úû


¤

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

¤

1

¥

p 4 D å m =1

¥

åé n =1

pmn

æ m ö + æ n ö ù êç ÷ ç ÷ ú êëè a ø è b ø úû 2

2

2

sin

mp x a

sin

np y b

The series is convergent  if you use enough terms, you will approach the exact solution.

Simply-Supported Rectangular Plates Under Various Loadings Page 18

¨

Uniformly distributed load: p(x,y) = p 0 pmn

=

4

b

a

ab ò ò

p0 sin

0

pmn pmn

= =

4 p0

ab 4 p0

mp x a

0

a

ò sin 0

a

ab ò

sin

0

mp x é a

sin

b

np y b

dxdy

np y ù

b

êë- np cos b úû dx 0

mp x é b

a êë np

(1 - cos np )ùú û

dx

a b é ù é ù 1 cos p 1 cos p pmn = m n ( ) ( ) úû êë np úû ab êë mp 4 p0


¤

Continue evaluating: pmn pmn

¤

= =

And finally: pmn

=

4 p0

p mn 2

4 p0

p2

( 1 - (- 1) )(1 - (- 1) ) mn

16 p0

p mn 2

(1 - cos mp )(1 - cos np ) m

n

( m, n = 1,3,5,...)


¤

Substitute into w:

w=

¤

16 p0

¥

p 6 D å m

¥

å n

1

éæ m ö æ n ö ù mn êç ÷ + ç ÷ ú êëè a ø è b ø úû 2

2

2

sin

mp x a

sin

np y b

( m, n = 1,3,5,...)

Since plate must deflect into a symmetrical shape for a uniform load, knew ahead of time that m and n must be odd.


¤

Maximum deflection occurs at center of plate: wmax wmax

wmax

wmax

a b ö = wæ ç , ÷ è 2 2 ø

=

=

=

16 p0

¥

¥

p D å å 6

16 p0

m

n

¥

¥

p 6 D å m

16 p0

¥

p 6 D å m

å n

¥

å n

1

éæ m ö 2 æ n ö 2 ù mnêç ÷ + ç ÷ ú ëêè a ø è b ø ûú 1

éæ m ö 2 æ n ö 2 ù mnêç ÷ + ç ÷ ú êëè a ø è b ø úû 1

éæ m ö 2 æ n ö 2 ù mnêç ÷ + ç ÷ ú êëè a ø è b ø úû

2

sin

mp 2

sin

m -1

np 2

n -1

(- 1) 2 (- 1) 2 2

é ( m+ n ) -1ù úû 2

(- 1)êë 2

(m, n = 1,3,5,...)


¤

Recall:

¤

Thus:

é ¶ 2w ¶2w ù M x = - D ê 2 +n 2 ú x y ¶ ¶ ë û 2

M x

=

16 p0

p4

¥

å m

2

æ m ö +n æ n ö ç ÷ ç ÷ ¥ mp x np y è a ø ån é 2 è b ø 2 ù 2 sin a sin b æ m ö æ n ö mn êç ÷ + ç ÷ ú êëè a ø è b ø úû

(m, n = 1,3,5,...)


¤

Likewise: 2

2

m ö æ n ö n æ ç ÷ +ç ÷ ¥ ¥ 16 p0 è a ø è b ø sin mp x sin np y (m, n = 1,3,5,...) M y = å å 2 p4 m n a b éæ m ö 2 æ n ö 2 ù mn êç ÷ + ç ÷ ú êëè a ø è b ø úû 16 p0 (1 -n ) ¥ ¥ 1 mp x np y cos cos (m, n = 1,3,5,...) M xy = å å 2 4 2 2 a b p ab m n éæ m ö æ n ö ù mn êç ÷ + ç ÷ ú êëè a ø è b ø úû

¤

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

¤

However, Mxy does not vanish at edges and corners.


¤

Assume square plate: a = b

¤

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

wmax

wmax

=

=

16 p0

¥

¥

p 6 D å m

å

16 p0

1

n

1

éæ m ö 2 æ n ö 2 ù mnêç ÷ + ç ÷ ú êëè a ø è b ø úû

p 6 D é æ 1 ö 2 ù 2 ê2ç ÷ ú êë è a ø úû

= 0.00416

p0 a 4 D

=

4 p0 a 4

p 6 D

é ( m+ n ) -1ù úû 2

(- 1)êë 2


¤

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

¤

¤

= 0.00406

p0 a 4 D

(exact solution )

Also: M x , max

= 0.0534 p0 a 2

( first term)

M x , max

= 0.0469 p0 a 2

( four terms )

Moment does not converge as rapidly as deflection.


¨

Loading: p ( x, y ) = p0 sin

p x a

sin

p y b


¤

Calculate pmn: pmn

=

4

b

ab ò ò

p0 sin

0

pmn ¤

=

a

0

4 p0 ab

ab

4

p x a

sin

= p0

p y b

sin

mp x a

sin

np y b

( m = n = 1)

Calculate w: w=

p0

¥

p D å å 4

m

w=

¥

p0

p 4 D é 1 êë a 2

1

éæ m ö 2 æ n ö 2 ù mn êç ÷ + ç ÷ ú ëêè a ø è b ø ûú 1 p x p y sin sin 2 a b 1ù + 2ú b û n

2

sin

mp x a

sin

np y b

dxdy


¤

Calculate moments:

M x

¤

=

é ¶ 2w ¶2w ù M x = - D ê 2 + n ¶y 2 úû ë ¶x

é 1 + n ù sin p x sin p y ê 2 b2 úû a b 1 ù ëa 2é 1 p ê 2 + 2ú ëa b û p0

2

And likewise:

M y

=

M y

é ¶2 w ¶ 2w ù = - D ê 2 + n 2 ú ¶x û ë ¶y

é n + 1 ù sin p x sin p y ê 2 b 2 úû a b 1 ù ëa 2é 1 p ê 2 + 2ú ëa b û p0

2


¤

Calculate moments:

M xy

=-

¶2w M xy = - D(1 +n ) ¶x¶y

p0 (1 -n ) 1 1 p éê 2 + 2 ùú ëa b û 2

1 2

ab

cos

p x a

cos

p y b


¤

Calculate shears: Q x

¤

=

p0 p x p y cos sin 1 1 a b pa éê 2 + 2 ùú ëa b û

Q y

And likewise: Q y

¶ æ ¶ 2 w ¶ 2 w ö + 2 ÷÷ Q x = - D çç 2 ¶ x è ¶x ¶y ø

=

¶ æ ¶ 2 w ¶ 2 w ö = - D çç 2 + 2 ÷÷ ¶ y è ¶x ¶y ø

p0 p x p y sin cos 1 1ù a b é pb ê 2 + 2 ú ëa b û


¤

Calculate total applied load: b

ptotal =

a

ò ò

p0 sin

0

0

a

Note :

ò sin 0

ptotal =

4 abp0

p2

p x

p x a

sin

p y b

dxdy a

a 2a é p x ù dx = - êcos ú = - [- 1 - 1] = pë p p a a û0 a


¤

¤

Find reactions at edges: V x

= Q x +

V x

=

¶M xy ¶ y

p x p y p0 cos sin 1 1 a b pa éê 2 + 2 ùú ëa b û

+

p0 (1 - n )

p éê 2 + 2 ùú ëa b û 1

1

1 2

ab2

cos

Setting x = a: V x

=-

p0 p y sin 1 1 b pa éê 2 + 2 ùú ëa b û

-

p0 (1 - n ) 1 1 p éê 2 + 2 ùú ëa b û

1 2

ab2

sin

p y b

p x a

sin

p y b


¤

Simplifying: V x

V x

¤

é 1 + 1 + (1 -n )ù sin p y 2 ê 2 b2 b 2 úû b 1 1 ù ëa é pa ê 2 + 2 ú ëa b û p0 é 1 + (2 -n )ù sin p y =2 ê 2 b 2 úû b 1 1 ù ëa é a p ê 2 + 2ú ëa b û =-

p0

Likewise for Vy at edge y = b: V y

=-

é 1 + (2 -n ) ù sin p x 2 ê 2 a 2 úû a 1 ù ëb + 2ú b û

p0 1 pb éê 2 ëa


¤

Sum of all reactions: a

b

ò

ò

2 V y dx + 2 V x dy 0

=-

¤

0

p0 é 1 + (2 - n )ù - 4b é 1 + (2 - n ) ù 2 ê 2 a 2 úû p b2 úû p é 1 1 ù 2 êë b2 ëa 1 1 é ù pb ê 2 + 2 ú pa ê 2 + 2 ú ëa b û ëa b û

4a

p0

By simple manipulation: 4 p0 ab 8 p0 (1 - n ) =-

p

2

-

p 2ab éê 2 + 2 ùú ëa b û 1

1

2


Note that first term equals total applied load, so reactions are larger. ¤ From concentrated reaction at each corner: ¤

= 2M xy at x = a, y = b 2 p0 (1 -n ) F c = 2 1 1 é ù p 2 ab ê 2 + 2 ú ëa b û F c

Rc

= - F c =

4 reactions (one at each corner). ¤ Equilibrium is therefore satisfied. ¤ This is physical  corners tend to rise. ¤

2 p0 (1 -n ) 1 1 p 2 ab éê 2 + 2 ùú ëa b û

2


¨

Total load P distributed uniformly over a sub-region 4cd: p ( x, y ) =

P 4cd

Simply-Suppor ted Rectangular Simply-Supported Rectangular Plates Under Various Loadings Page 37

¤

Determine pmn: pmn

=

4

b

ab ò ò

p ( x, y ) sin

0

¤

pmn

=

pmn

=

a

P

mp x a

0

y1 + d x1 + c

ò ò

abcd y1 - d x1 -c 4 P

p 2 mncd

sin

sin

mp x

mp x1 a

How? See next slide .

a sin

sin

sin

b

np y

np y1 b

np y

b sin

dxdy

dxdy mpc a

sin

npd b


¤

Note:

x1 + c

mp x

x1 + c

mp x ù

écos ò êë úû a m p a x - c x - c a é mp ( x1 + c ) mp ( x1 - c )ù cos =- cos ê úû mp ë a a sin

dx = -

a

1

1

¤

Using identities:

¤

Obtain:

=

cos(a

2a

mp

+ b ) = cos a cos b - sin a sin b cos(a - b ) = cos a cos b + sin a sin b cos(a + b ) - cos(a - b ) = -2 sin a sin b mp x1 mpc

sin

a

sin

a


¤

If expand expand load to cover entire plate surface by substituting the following: following: x1

¤

=

a 2

y1

=

b 2

c=

Obtain the previous solution: pmn

=

16 P

p 2 mn

a 2

d =

b 2


¨

Point load at x = x 1, y = y1:


¤

Use previous derivation, with c,d pmn

à 0. mp x1 np y1 mpc npd 4 P sin sin sin sin = 2 a b a b p mncd mpc sin a =0 lim(c ® 0) 0 c

use L' Hopital' s Rule :

mp lim(c ® 0) a

cos

similarly, lim(d ® 0)

pmn

=

4 P

ab

sin

mp x1 a

sin

mpc a

1 sin

npd

b d np y1 b

= =

mp a np b


¤

Leads to:

w=

¤

¥

p 4 Dab å m

¥

1

åé n

æ m ö + æ n ö ù êç ÷ ç ÷ ú êëè a ø è b ø úû 2

2

2

sin

mp x1 a

sin

np y1 b

sin

mp x a

sin

np y b

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

x1

¤

4 P

=

a 2

y1

=

b 2

®

w=

4 Pa 2

¥

¥

p D å å (m 4

m

n

1 2

+n

)

2 2

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

= 0.01142

Pa 2 D

æ Pa 2 ö çç exact = 0.01159 ÷÷ D è ø

(m,n = 1 ,3 ,5...)

Levys Solution for Rectangular Plates Page 43

¨

¨

Navier solution à slow convergence of series for bending moments. Levys solution: Overcomes slow convergence. ¤ Uses single series (Navier uses double series) ¤ Allows for more general boundary conditions (Navier only valid for simply-supported conditions on all sides). ¤

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


¨

Procedure: ¤

As before, result is sum of homogeneous and particular solution: w = wh + w p

¤

Particular solution is obtained for each specific loading.

¤

Homogeneous solution is independent of loading:

Ñ 4 wh = 0


¨

Homogeneous solution assumed to be:

¥ mp x ö æ æ mp x ö wh = å f m ( y ) sin ç ÷ or wh = å f m ( y ) cosç ÷ a a è ø è ø m =1 m =1 ¥

f m ( y ) Þ fulfill boundary conditions at y = ±

b 2


¨

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. mp x ö = å f m ( y ) sinæ ç ÷ è a ø m =1 ¥

wh

w=0

¶2w =0 2 ¶x

(x = 0, x = a)


¨

Substitute into governing equation: Ñ 4 wh = 0 2 ¥ é 4 d f m mp ö d 2 f m æ ÷ ê 4 - 2ç å 2 dy a dy è ø m =1 ê ë

¨

4 ù æ mp x ö mp ö æ +ç f ÷ m ú sin ç ÷=0 è a ø úû è a ø

Term inside brackets must be zero à linear differential equation with constant coefficients. d 4 f m dy

4

mp ö - 2æ ç ÷ è a ø

2

d 2 f m dy

2

mp ö + æ ç ÷ è a ø

4

f m

=0


¨

Solution (from differential equations): 2

4

mp ö d 2 f m æ mp ö æ - 2ç +ç ÷ ÷ f m = 0 4 2 dy è a ø dy è a ø

d 4 f m

¨

Can be written as: m 4 - 2 A2 m 2 + A 4

(m 2 - A2 )

2

=0

m = ± A,± A

=0 m = C 1e Ay

+ C 2e - Ay + C 3 ye Ay + C 4 ye - Ay


¨

And thus the general solution is: mp y

f m ¨

a

+ B'm e

mp y a

mp y

+ C 'm ye

a

+ D'm ye

-

mp y a

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

¨

= A'm e

-

= Am sinh

mp y a

Note: sinh u =

+ Bm cosh

1 2

(e

u

- e -u )

mp y a

+ C m y sinh

cosh u =

mp y a 1 2

(e

u

+ Dm y cosh + e -u )

mp y a


¨

Therefore: ¥

wh

mp y mp y mp y mp y ö mp x A B C y D y = å æ + + + sinh cosh sinh cosh sin ç m ÷ m m m a a a a a è ø m =1

Am , Bm , C m , Dm

® constants determined for specific cases


¨

¨

Particular solution:

mp x ö = å k m ( y )sinæ ç ÷ è a ø m =1 ¥ æ mp x ö p( x, y ) = å pm ( y )sinç ÷ è a ø m =1 a 2 æ mp x ödx pm ( y ) = ò p ( x, y )sinç ÷ a0 a è ø ¥

w p

Last equation found by multiplying both sides by the sin term below and integrating:

æ m'p x ö ÷ è a ø

sin ç


¨

Plugging into differential equation:

Ñ 4 w p =

p D

é d 4 k m æ mp ö 2 d 2 k m æ mp ö 4 ù æ mp x ö ¥ é pm ù æ mp x ö +ç ÷ ÷ k m ú sin ç ÷ = å ê ú sinç ÷ ê 4 - 2ç å 2 dy a dy a a D a è ø è ø úû è ø m=1 ë û è ø m =1 ê ë ¥

d 4 k m dy 4

mp ö - 2æ ç ÷ a è ø

2

d 2 k m dy 2

mp ö + æ ç ÷ a è ø

4

k m

=

pm D


¨

The general procedure thus becomes: ¤

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

¤

From pm, find km and thus wp.

¤

Using boundary conditions, find A m, Bm, Cm, Dm and thus wh.

¤

Result is sum of wp and wh.


¨

Simply-supported rectangular plate under uniform loading. ¤

Find pm.

p( x, y ) = p0

æ mp x ödx = 2 p é- a pm ( y ) = p0 ò sin ç ÷ 0ê a 0 a a è ø ë mp 2

pm ( y ) =

4 p0

mp

a

(m = 1,3,5...)

a

æ mp x öù = 2 p é 2a ù cosç ÷ú 0ê a a è øû 0 ë mp úû


¤

Find km. d 4 k m dy 4

n

mp ö - 2æ ç ÷ a è ø

2

d 2 k m dy 2

mp ö + æ ç ÷ a è ø

4

k m

=

4 p0

mp D

Since the right hand side is not a function of y, then k m cannot be a function of y, so derivatives of k m are zero:

æ mp ö ç è

a

÷ ø

4

k m

=

4 p0

mp D

k m

=

4 p0 a 4

m 5p 5 D


¤

Thus wp becomes: w p

¤

=

4 p0 a 4

p 5 D

¥

å

m =1, 3...

1

m5

sin

mp x a

This particular solution represents the deflection of a uniformly loaded, simply-supported strip parallel to the y-axis (very long in the y-direction).


¤

Now must evaluate constants in homogeneous solution.

¤

Observing that the deflection must be symmetrical with respect to the x-axis: Am = Dm = 0. sinh function is not symmetric. n cosh function is symmetric. n y*sinh is symmetric n y*cosh is not symmetric. n

cosh

sinh


¤

Expression for deflection is now:

æ mp y mp y 4 p0 a 4 ö mp x w = w p + wh = å çç Bm cosh + C m y sinh + 5 5 ÷÷ sin a a m p D ø a m =1, 3... è ¥

¤

Satisfies governing equations and simply-supported boundary conditions at x = 0 and x = a.

¤

Remaining boundary conditions: w=0

¶2w =0 2 ¶y

æ y = ± b ö ç ÷ 2 ø è


¤

Substituting: Bm cosh

mpb

æ Bm mp è 2a

¤

Gives:

sinh

mpb

+

4 p0a 4

2

2a

5

5

æ mpb ö ÷ a 2 è ø

4 p0 a 4 + mp p0 a 3b tanhç

Bm

C m

=0

m p D mpb mpb mpb + C m ö÷ cosh + C m sinh 2a 2a 2a ø

2a

2ç

+ C m

b

=m5p 5 D cosh

=

2 p0 a 3

m 4p 4 D cosh

mpb 2a

mpb 2a

=0


am =

¤

Setting:

¤

Simplifies constants: Bm C m

==

mpb 2a

4 p0 a 4 + mp p0 a 3b tanh a m

m5p 5 D cosh a m 2 p0 a 3

m 4p 4 D cosh a m


¤

Deflection: w=

¤

4 p0 a 4

p 5 D

æ a m tanh a m + 2 2a m y ç 1cosh å 5 ç m b 2 cosh a m m =1, 3... è 1 2a m y ö mp y m x ÷÷ sin p sinh + 2 cosh a m a b ø a ¥

1

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

=

4 p0 a 4

p 5 D

æ a m tanh a m + 2 ö mp ç1 ÷÷ sin å 5ç 2 cosh a m ø 2 m =1,3... m è ¥

1


¤

Maximum deflection can be simplified: wmax

¤

p 5 D

¥

å

(- 1)

( m -1) 2

m5

m =1, 3...

æ a m tanh a m + 2 ö çç1 ÷÷ 2 cosh a m ø è

Can manipulate to the following following form: wmax

¤

=

4 p0 a

4

=

5 p0 a

4

384 D

-

4 p0 a

4

p 5 D

¥

å

m =1, 3...

( m -1)

(- 1)

m5

2

æ a m tanh a m + 2 ö çç ÷÷ è 2 cosh a m ø

Note that first term is the previous solution for a uniformly loaded simply-supported strip.


¤

For a square plate (a = b):

wmax n

¤

=

5 p0 a

4

384 D

-

4 p0 a

4

p 5 D

(0.68562 - 0.00025 + ...) = 0.00406

p0 a

4

D

Note that the series converges rapidly.

Can rewrite equation as:

wmax

= d1

p0 a D

4

d1 =

5 384

-

4

¥

p 5 må =1, 3...

( m -1)

(- 1)

m5

2

æ a m tanh a m + 2 ö çç ÷÷ è 2 cosh a m ø


¤

Can also write similar expressions for moments: M x ,max

= d 2 p0 a 2

M y ,max

= d 3 p0 a 2

¤

Parameters d1, d2, d3 can be determined for various cases.

¤

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 sides is negligible and plate can be considered as infinite strip.


¨

3 edges simply-supported, 1 edge clamped, uniform pressure. ¤

Deflection is symmetrical about x = a/2.

¤

Thus, w uses only odd m.


¤

Deflection:

w=

æ A sinh mp y + B cosh mp y + C y sinh mp y ç m å m m a a a m =1, 3... è mp y 4 p0 a 4 ö mp x + Dm y cosh + 5 5 ÷÷ sin a m p D ø a ¥

¤

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.


¤

Remaining boundary conditions: w=0 w=0

¤

¶w =0 ¶y ¶2w =0 2 ¶y

(y = 0) (y = b)

The boundary conditions provide 4 equations to solve the 4 unknown constants, shown on next page.


¤

Am Bm

After some manipulation, the constants are:

=

2 p0a 4 2 cosh 2 b m

m 5p 5 D

=-

a - 2 cosh b m - b m sinh b m = - Dm cosh b m sinh b m - b m mp

4 p0a 4

m 5p 5 D 2

æ mp ö sinh b cosh b - æ mp ö sinh b - æ mp ö b cosh b 2ç ÷ ç ÷ ç ÷ m m m m 1 a a a è ø è ø è ø C m = - b m 2 cosh b m sinh b m - b m mpb bm = a


¤

For a square plate (a=b): wcenter = 0.0028 M y ,max

p0a 4 D

= 0.084 p0a 2

æ x = a , y = a ö ç ÷ 2 2 è ø æ x = a , y = 0 ö ç ÷ 2 è ø


¨

Opposite edges simply-supported, 3 rd edge free, 4th edge clamped, uniform pressure.


¤

Deflection:

w=

æ A sinh mp y + B cosh mp y + C y sinh mp y ç m å m m a a a è m =1, 3... mp y 4 p0 a 4 ö mp x + Dm y cosh + 5 5 ÷÷ sin a m p D ø a ¥

¤

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.


¤

Remaining boundary conditions: ¶w (y = 0) w=0 =0 ¶y ¶2 w ¶2w ¶ 3w ¶ 3w +n 2 = 0 + (2 - n ) 2 = 0 2 3 ¶ y ¶ x ¶y ¶x ¶y moment

(y = b)

shear

¤

The boundary conditions provide 4 equations to solve the 4 unknown constants. Not shown here, but it is straightforward if not tedious.

¤

Same procedure for any similar uniformly loaded plate.

Levys Method Applied to Nonuniformly Loaded Rectangular Plates Page 73

¤

Assume rectangular plate simply-supported at x = 0 and x = a.

¤

Assume load is a function of x only. æ mp x ö ÷ å è a ø m =1 a 2 æ mp x ö dx pm = ò p ( x )sin ç ÷ a0 è a ø p ( x ) =

¥

pm sin ç


¤

As before, assume for the particular solution: mp x ö = å k m ( y )sin æ ç ÷ è a ø m =1 ¥

w p

¤

Plugging into governing equation: Ñ 4 w p =

p D

é d 4 k m æ mp ö 2 d 2 k m ÷ ê 4 - 2ç å 2 dy a dy è ø m =1 ê ë ¥

æ mp ö

4

pm ù

æ mp x ö = 0 k +ç sin ÷ m ÷ ú ç a D a è ø úû è ø


¤

Since pm is only a function of x, k m cannot be a function of y, and the derivatives with respect to y are zero. 2 4 d 4 k m mp ö d 2 k m æ mp ö pm æ 2 k + = ç ÷ ç ÷ m 2 dy 4 a dy a D è ø è ø

æ mp ö ç ÷ è a ø ¤

4

k m

=

pm

k m

D

=

pm a 4 m 4p 4 D

Particular solution, which represents deflection of a strip under load p(x), becomes: w p

=

a4

¥

pm

p D å m 4

m =1

4

sin

mp x a


Assume the two arbitrary edges, y = -b/2 and y = b/2, are also simply-supported: ¥ æ mp y mp y pm a 4 ö mp x w = w p + wh = å çç Bm cosh + C m y sinh + 4 4 ÷÷ sin a a m p D ø a m =1 è

¤

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: w=0

¶2w =0 2 ¶y

æ y = ± b ö ç ÷ 2 è ø


¤

After evaluation of boundary conditions, the deflection is: a 4 ¥ pm æ a m tanh a m + 2 mp y cosh w = 4 å 4 çç1 a 2 cosh a m p D m=1 m è 1 mp y mp y ö mp x ÷÷ sin sinh + 2 cosh a m a a ø a mpb am = 2a

¤

Given p(x)

à

find pm à find w à find M à find .


¤

Values of pm for various loadings.


¤

Example: Hydrostatically loaded plate. p ( x ) = p0

x a

æ mp x ödx = 2 p0 x sinæ mp x ödx pm = ò p0 sin ç ÷ ç ÷ 2 a0 a a ò a è a ø è ø 0 2

a

a

x

a

pm

=

2 p0

pm

=

2 p0

pm

=

2 p0

a2

a

2

mp

éæ a ö 2 æ mp x ö æ xa ö æ mp x öù ÷ sinç ÷-ç ÷ cosç ÷ú êç m a m a p p è ø è ø è øúû 0 êëè ø é æ a 2 ö ù ÷÷ cos(mp ) - 0 + 0ú ê0 - çç m p ë è ø û

(- 1)m +1

(m = 1,2...)


¤

Can now find w by substituting expression for p m.

¤

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

wcenter = 0.00203

p0 a 4 D

æ x = a , y = 0 ö ç ÷ è 2 ø


¤

Example: Line loads on long rectangular plates.


¤

Only need to consider one half of plate (positive y) due to symmetry: ¶w =0 ¶y

¤

(y = 0)

Each half takes half the load: Q y

=-

P ( x) 2

= - D

¶ 2 (Ñ w) ¶y

(y = 0)


¤

Use alternate form of governing equation: mp y mp y mp y mp y æ ö mp x a a a a = å çç A'm e + B'm e + C 'm ye + D'm ye ÷÷ sin a m =1 è ø

¥

wh

¤

There is no surface pressure, so the particular solution is zero and:

w = wh

¤

Also, w and its derivatives should vanish at y =

A'm = C 'm = 0

¥:


¤

Equations for w: mp y mp y ¥ æ ö mp x w = å çç B'm e a + D 'm ye a ÷÷ sin a m =1 è ø

¤

Evaluating symmetry boundary condition: ¶w =0 ¶y

(y = 0)

mp y é - map y æ mp ö - map y ù mp ö - a æ + D'm êe -ç B'm ç ÷e ÷ ye ú = 0 è a ø è a ø ë û æ mp ö + D' = 0 æ mp ö B'm ç D B = ' ' ÷ ÷ m m mç è a ø è a ø


¤

Evaluating remaining boundary condition: ¶ 2 (Ñ w) ¶y 2 ¶ æ ¶ 2 w ¶ 2 w ö P ( x) çç 2 + 2 ÷÷ = D 2 ¶y è ¶x ¶y ø Q y

¤

=-

P ( x)

= - D

(y = 0)

Need to evaluate derivatives (next slide).


mp y mp y æ ö mp x a çç B'm e + D'm ye a ÷÷ sin a è ø mp y mp y mp y ù mp x ¶ 2 w ¶ é æ mp ö mp ö æ a a a ' ' ' B e D e D ye = + ç ÷ ç ÷ m ú sin m ¶y 2 ¶y êë è a ø m a a è ø û 2 2 mp y mp y mp y mp y ù mp x m m m ¶ 2 w éæ mp ö p p p æ ö æ ö æ ö a a a a ' ' ' ' B e D e D e D ye = + ç ÷ ç ÷ ç ÷ ç ÷ ê ú sin m m m a a a a ¶y 2 êëè a ø m è ø è ø è ø úû mp y é æ mp ö ù mp x 2 Ñ w = ê- 2ç ÷ D'm e a ú sin a ë è a ø û mp y é æ mp ö2 ù ¶ 2 (Ñ w) = ê2ç ÷ D'm e a ú sin mp x a ¶y êë è a ø úû

¶ 2 w æ mp ö = -ç ÷ ¶x 2 è a ø

2


¤

Substituting into boundary condition: ¶ æ ¶ 2 w ¶ 2 w ö P ( x) çç 2 + 2 ÷÷ = D ¶y è ¶x ¶y ø 2 2

mp ö æ 2ç ÷ D'm e a è ø

mp y a

=

pm 2 D

(sin terms cancel)

2

y ® 0

¤

æ mp ö D' = pm 2ç ÷ m 2 D a è ø

And from previous expression:

D'm =

pm a 2 4m 2p 2 D

B'm =

pm a 3 4m 3p 3 D


¤

Thus deflection becomes:

æ pm a 3 - map y pm a 2 y - map y ö mp x w = å çç e + 2 2 e ÷÷ sin 3 3 a 4m p D m =1 è 4m p D ø mp y a 3 ¥ pm æ mp y ö - a mp x w= e 1 sin + ÷ å 3 3 ç 4p D m =1 m è a ø a ¥

¤

Equations for P(x): mp x ö æ ( ) P x = å pm sin ç ÷ a è ø m =1 ¥


¤

If p(x) = p0: a

é æ mp x öù = 2 p0 [1 - cos mp ] = pm = p0 ò sin ç dx cos ÷ ç ÷ú ê p a 0 a m a è ø è øû 0 mp ë 2

pm

=

æ mp x ö

a

4 p0

2 p0

(m = 1,3,5...)

mp

mp y mp y ö - a mp x æ w= 4 e + 1 sin ÷ 4 ç m a a p D må è ø =1,3... a ö p0 a 3 ¥ 1 mp æ wmax = wç ,0 ÷ = 4 sin å 2 è 2 ø p D m =1,3... m 4

p0 a 3

wmax

=

p0 a

¥

3

1

¥

(- 1)

m -1

m4 p 4 D må =1,3...

2

é 5p 5 ù 5p p0 a 3 = 4 ê = ú p D ë1536 û 1536 D p0 a 3


¤

If p(x) is a constant load p 0 but only partially along line:


p( x) = p0 from x1 - c to x1 + c x1 + c

mp x ö mp (x1 + c ) ö mp (x 1 - c ) öù 2 p0 é æ æ æ pm = p0 sin ç ÷dx = ÷ + cosç ÷ú ê- cosç p a x ò a m a a è ø è ø è øû ë -c 2

1

pm = ¤

é æ mp x1 ö æ mpc öù sin ç ÷ sin ç ÷ú ê mp ë è a ø è a øû

4 p0

And plugging into expression for w:

mp y ö æ w = 4 å 4 ç1 + ÷e p D m =1 m è a ø p0 a 3

n

¥

1

mp y a

sin

mp x1 a

sin

mpc a

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

sin

mp x a


¤

For case of concentrated load:


¤

Substitute the following expressions into w:

P = 2cp0 or p0

=

P 2c

æ mpc ö » mpc ÷ a è ø a

sinç

mp y P ü a 3 ¥ 1 æ mp y ö - a mp x1 ì mpc ü mp x ì w = í ý 4 å 4 ç1 + e sin sin ÷ í ý p 2 c D m a a a a î þ î þ è ø m =1 mp y Pa 2 ¥ 1 æ mp y ö - a mp x1 mp x w= + e 1 sin sin å ç a ø÷ a a 2p 3 D m =1 m3 è

Rectangular Plates Under Distributed Edge Moments Page 94

¤

Simply-supported rectangular plate, symmetrically distributed edge moments at y = +b/2 and y = -b/2:


Assume a Fourier sin series for the moment distribution: ¥ mp x ö b ö æ æ f ( x ) = å M m sin ç ÷ ç y = ± ÷ 2 ø è a ø è m =1 a 2 æ mp x ödx M m = ò f ( x )sin ç ÷ a0 è a ø ¤ Boundary conditions: ¤

w=0 w=0

¶2w =0 2 ¶x ¶2w = f ( x ) -D ¶y 2

(x = 0, x = a ) æ y = ± b ö ç ÷ 2 è ø


¤

Since p0 = 0 and plate deflection is symmetric about xaxis: ¥ mp y mp y ö mp x æ w = å ç Bm cosh + C m y sinh ÷ sin a a ø a m =1 è Above expression satisfies governing equation and first set of boundary conditions. n Use other two boundary conditions to determine constants. n


¤

First boundary condition: w=0

æ y = ± b ö ç ÷ 2 è ø

Bm cosh a m Bm

= -C m

w=

¥

å m =1

+ C m

b 2

æ è

b 2

sinh a m

=0

am =

mpb 2a

tanh a m

C m ç y sinh

mp y a

-

b 2

tanh a m cosh

mp y ö a

÷ sin

ø

mp x a


¤

Second boundary condition: ¶2w æ y = ± b ö ( ) -D f x = ç ÷ 2 ¶y

¤

è

2 ø

Using basic derivation: ¥ mp mp x mp x æ ö cosh a m ÷ sin - 2 D å C m ç = å M m sin a a è a ø m =1 m =1 ¥

C m

=-

w=

aM m 2mp D cosh a m

æ b tanh a cosh mp y - y sinh mp y ö sin mp x ç ÷ å m 2p D m =1 m cosh a m è 2 a a ø a a

¥

M m


¤

M m

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

2

a

a ò

M 0 sin

0

w=

¤

mp x a

¥

dx = -

2 M 0

mp

cos

mp x a

x = a x = 0

=

4 M 0

mp

m = 1,3...

æ b tanh a cosh mp y - y sinh mp y ö sin mp x ç ÷ m 2 cosh 2 m a a a p 2 D må a è ø =1, 3... m

2 M 0a

1

For square plate (a = b), results at center: w = 0.0368

M 0 a 2 D

, M x

= 0.394 M 0 , M y = 0.256M 0


¤

Displacement along axis of symmetry (y = 0): w=

¤

tanh a m

¥

M 0 ab

å

p 2 D

m =1, 3...

m 2 cosh a m

sin

mp x a

For a very long strip (a >> b), i.e. center deflection of a strip of length b subjected to two equal and opposite bending moments at ends: tanh a m

w=

» am

M 0 ab

p D 2

¥

å

m =1, 3...

cosh a m

am m

2

sin

p ù M 0b 2 é = w= ê ú 2p D ë 4 û 8 D M 0b 2

»1

am =

mpb 2a

é ¥ 1 mp x ù sin = êå ú 2p D ëm =1,3... m a a û

mp x

M 0b 2


¤

Plate with one edge moment:

n

Note: This problem switches previous x and y directions.


¤

Assume same Fourier series expression: ¥ np y ö (M x )0 = å M n sinæ ( x = 0 ) ç ÷ è b ø n =1 b 2 æ np y ödy M n = ò (M x )0 sinç ÷ b0 b è ø

¤

Deflection is in the following form:

w=

æ A sinh np x + B cosh np x + C x sinh np x + D x cosh np x ö ç n ÷ å n n n b b b b ø n =1 è ¥

sin

np y b


¤

Boundary conditions: w=0 w=0 w=0

¶2w -D = (M x )0 ¶x 2 ¶2w =0 2 ¶x ¶2w =0 ¶y 2

( x = 0) (x = a ) (y = 0, y = b )

¤

Last two boundary conditions are automatically satisfied, as always.

¤

First boundary condition results in: B n = 0.


¤

Remaining 3 boundary conditions can be used to solve for remaining 3 constants, resulting in:

np x ö æ a sinh ç ÷ np y M n b ¥ np (a - x ) b ÷ sin ç x cosh w= å npa ç np y ÷ 2p D n=1 b b sinh n sinh ç ÷ b è b ø

¤

Based on moment, can find M n and then substitute into w expression above.

Method of Superposition Applied to Bending of Rectangular Plates Page 105

¨

Complex problem replaced by several simpler situations. 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 others simply-supported. Start with plate with all edges simply-supported. ¤ Add the solution of plate with bending moment applied along y=0, with magnitude to eliminate rotations along clamped edge. ¤


¨

Example: Two edges simply-supported, two edges clamped, uniform pressure.


¤

Plate 1 (from previous Levy approach): æ a m tanh a m + 2 2a m y ç 1 cosh w1 = 5 5 ç b p D må 2 cosh a m =1, 3... m è 1 2a m y ö mp y ÷÷ sin mp x + sinh b ø a 2 cosh a m a mpb where : a m = 4 p0 a 4

¥

1

2a

For plate 2, need to determine the moment along the edges so that the sum of the rotations of plate 1 and plate 2 along these edges equals zero (clamped boundary condition). ¤ Thus, need to find the rotation along these edges in plate 1. ¤


¤

Rotation along edge found by taking derivative of w with respect to y:

¶w1 4 p0 a 4 ¥ 1 æ a m tanh a m + 2 æ 2a m ö 2a m y ç = 5 sinh ç ÷ 5 ç 2 cosh a m è b ø b ¶ y p D må =1,3... m è + ¤

1 2 cosh a m

mp y æ 2a m ö a

ç è

b

÷ cosh

ø

2a m y

b

+

1

mp

2 cosh a m a

sinh

2a m y ö

b

÷÷ sin ø

Substitute y = b/2: a m tanh a m + 2 æ 2a m ö ¶w1 4 p0 a 4 ¥ 1 æ ç = 5 ç ÷ sinha m 5 ç 2 cosh m b ¶ y p D må a è ø =1, 3... m è ö mp x æ 2a m 2 ö 1 mp ç ÷ cosh a m + + sinh a m ÷ sin ÷ 2 cosh a m çè b ø÷ 2 cosh a m a a ø 1

mp x a


¤

Further manipulation: ö mp x am2 a m tanha m (a m tanh a m + 2)+ + tanh a m ÷÷ sin b b a ø mp x am tanh tanh 1 sin ( ( )) a a a a + m m m m 5

¶w1 4 p0 a 4 ¥ 1 æ a m = 5 ç m ¶ y p 5 D må è b =1,3...

¶w1 4 p0 a 4 ¥ = 5 a ¶ y p D må =1,3... m b mp x ¶w1 2 p0 a 3 ¥ 1 ( ( )) = 4 a tanh a a tanh a + 1 sin m m m m 4 a ¶ y p D må =1,3... m mpb where : a m = 2a


¤

Plate 2, from previous solution: ¥ æ mp x ö M y = å M m sin ç ÷ a è ø m =1

æ y = ± b ö ç ÷ 2 è ø a ¥ M m æ b tanh a cosh mp y - y sinh mp y ö sin mp x w2 = ç ÷ å m 2p D m=1 m cosh a m è 2 a a ø a ¤

Edge rotation for this case: æ b a ¥ M m ¶w2 æ mp ö sinh mp y = ç a tanh ÷ mç ç2 m a a ¶ y 2p D å a cosh è ø m =1 m è mp y mp ö mp y ö mp x - sinh - yæ ÷ cosh sin ç ÷ a a a ø÷ a è ø


¤

Further manipulation, substituting y = b/2:

a ¥ M m ¶w2 (a m tanh a m sinh a m = å ¶ y 2p D m=1 m cosh a m mp x - sinh a m - a m cosh a m )sin a

a ¥ M m mp x ¶w2 ( ) = a tanh a tanh a tanh a a sin m m m m m m a ¶ y 2p D å m =1 a ¥ M m mp x ¶w2 ( ) tanh tanh 1 sin [ ] = a a a a m m m m m a ¶ y 2p D å m =1


¤

Slopes must be equal and opposite: ¶w1 ¶w =- 2 ¶ y ¶ y gives :

¤

æ y = ± b ö ç ÷ 2 ø è 4 p0 a 2 a m - tanh a m (1 + a m tanh a m ) M m = 3 3 m p a m - tanh a m (a m tanh a m - 1)

Final result is then found by: Substituting Mm into w2. n Add solutions of w1 and w2. n

Strip Method Page 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 of strips 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.


¨

Based on deflections and moments of beams with various end conditions derived from mechanics of materials.

R A

=

5 pL 8

R B

=

3 pL 8


¨

Consider the rectangular plate:

¤

Strip of span a carries uniform load p a.

¤

Strip of span b carries uniform load p b.


¤

Two conditions that must be satisfied:

wa ¤

= wb

p0

= pa + pb

( x = y = 0)

For simply-supported edges (same for clamped plate): 5 pa a 4

wa

=

wa

= wb

p0

384 EI

wb

=

5 pb b 4 384 EI

Þ pa a 4 = pbb 4

= pa + pb Þ pa = p0

b4 a

4

+b

4

pb

= p0

a4 a4 + b4


¤

And thus: wmax

¤

=

5 pa a 4 384 EI

=

5 pbb 4 384 EI

=

5 p0a 4b 4

384 EI (a

4

+ b4 )

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.


¨

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

Use expressions for displacement of a beam simplysupported on each end along y, and a beam clamped on one end and simply-supported on another along x. ¤ Determine mid-span deflection and maximum moments ¤


¤

Setting mid-span deflections equal (note: this is not always the location of maximum deflection): wa

= wb

Þ

wa

= wb

Þ

p0

pa a 4 192 EI 2 pa a 4

=

wcenter =

384 EI

=

384 EI

= 5 pbb 4

= pa + pb Þ pa = p0 5 pbb 4

5 pbb 4

5b 4 2a

4

5 p0 a 4b 4

+ 5b

192(2a 4 + 5b 4 ) EI

4

pb

= p0

2a 4 2a

4

+ 5b 4


¤

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: b öæ b ö b öæ b ö pb b 2 æ æ M y = pb ç ÷ç ÷ - pb ç ÷ç ÷ = 8 è 2 øè 2 ø è 2 øè 4 ø M y

=

p0 a 4b 2

4(2a

4

+ 5b

4

)

( y = 0)

( y = 0)


¤

In the x-direction span, must first determine where the maximum moment occurs: n

Reaction force at simply-supported end = 3p aa/8. M x ( x ) = dM x dx

1 2

pa x

2

3

- pa ax 8

3

3

8

8

= pa x - pa a = 0 Þ x = a 2

9 æ 3 ö 1 æ 3 ö 3 æ 3 ö M x ç a ÷ = pa ç a ÷ - pa aç a ÷ = pa a 2 è 8 ø 2 è 8 ø 8 è 8 ø 128 2 4 45 p0 a b æ x = 3 a ö M x = ç ÷ 128(2a 4 + 5b 4 ) 8 è ø

Simply-Supported Continuous Rectangular Plates Page 122

¨

¨

¨

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.


¨

Consider two-paneled continuous plate, as shown:

f ( y ) =

¥

å

M m sin

m =1,3...

mpy b


¨

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 based on continuity at the connecting edges, i.e. the slopes must be equal.


¨

Expressions for deflections of each plate: w1

=

¥

å ( A

m

sinh lm x1 + Bm cosh lm x1 + C m x1 sinh lm x1

m =1, 3...

+ Dm x1 cosh lm x1 + lm = w2

=

4 p0b 4 ö

÷÷ sin lm y m p D ø 5

5

mp b ¥

å ( E sinh l x m

m

2

+ F m cosh lm x2 + Gm x2 sinh lm x2

m =1, 3...

+ H m x2 cosh lm x2 )sin lm y


¨

Boundary conditions: w1

=0

w1

=0

w2

=0

w2

=0

¶ 2 w1 ( x1 = 0 ) =0 2 ¶ x1 ¥ ¶ 2 w1 -D = å M m sin lm y 2 ¶ x1 m =1,3... ¶ 2 w2 ( x2 = a ) =0 2 ¶ x2 ¥ ¶ 2 w2 -D = å M m sin lm y 2 ¶ x2 m =1,3...

( x1 = a )

( x2 = 0)


¨

Additional boundary condition

à

continuity:

¶w1 ¶w2 = ¶ x1 x = a ¶x2 x =0 1

¨

2

Find constants Am ® Hm and Mm (9 unknown) using given 9 equations. ¤

Straightforward, results on next page.


ì coth lma é M m 2 po ü ù 4 po æ lma ö ( -1 + cosh lma )ú + 5 ç coth lma Am = ía + - cschlm a ÷ ý ê 2 D î sinh lm a ë 2lm l4mb l b ø þ û m è 1

Bm

=-

Dm

=-

E m

=

Gm

=-

4 po

Cm

l5m Db cschlma é M m

D

M m a 2lm D

ê 2l ë m

+

=

2 po

l4mb

2 p0

l4m Db ù û

( -1 + cosh lma )ú

(1 - coth 2 lm a )

M m 2lm D

Hm

Fm

=

=0

M m 2lm D

coth lm a

Homework Problem 6 Page 129

¨

A simply supported rectangular plate is under the action of hydrostatic pressure expressed by p = p 0x/a, where constant p0 represents the load intensity along the edge x = a.

b


¨

Determine, employing Navier s approach and retaining the first term of the series solution, the equation for the resulting solution: w=

8 p0

1

p 6 D é 1 1 ù 2 êë a 2 + b 2 úû

sin

p x a

sin

p y b


¨

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.


¨

Show, using Levys 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:

w = 0.00614

p0 a 4 D

03 Rectangular

Recommend Documents