Analysis of Rectangular Slabs Supported on Elastic Edge Beams Using Excel Program

(1)

Corner Supports

EDGE BEAMS USING MICROSOFT EXCEL

Indrajit Ghosh1

1_{Adjunct Professor, Indian Institute of Technology, Gandhinagar, India}

ABSTRACT

Since the development of Finite Element Method (FEM) in 60’s and 70’s, its application to structural analysis of nuclear power plants witnessed a tremendous impetus, and gained popularity soon. Now-a-days with the advent of high-speed digital computers its popularity has become even more universal. Even for a minor modification to an existing structure such as change in loading on a concrete slab in a room, engineers take resort to FEM which most of the time becomes very time-consuming and uneconomical. In this paper the author shows how local analysis of slabs be performed using classical structural mechanics principles described in many text books such as Timoshenko and Woinowsky-Krieger (1959). Although some partial derivations are required to solve the governing equations provided in the text books, Microsoft Excel software can be used, and the equations solved. Once the Excel software is developed any one can use it without having to learn the FEM techniques and applications. An example of a slab supported on elastic edge beams under uniformly distributed load has been demonstrated in this paper using Excel. Advantages of using Excel software are many: (a) Excel is available in any office environment, and everyone is familiar with this software. (b) No special training or experience is required to run such an analysis. (c) Maximum deflections are calculated. (d) Unlike FEM there is no need to worry about element size. (e) Results are more accurate than obtained using FEM.

DESCRIPTION

Analysis of a rectangular plate loaded uniformly and supported on four elastic edge beams have been considered in this paper (Fig. 1). The beams are supported rigidly at the corners of the plate. Two beams parallel to each other have the same flexural rigidity. No torsional moment is transferred to the beams.

Figure1. Plate supported on elastic edge beams and on four corners

X

Y

b/2 b/2

Y

(2)

The deflection equation as given by Timoshenko and Woinowsky-Krieger (1959) is as follows:

4 2 2 4 4 2 2 4

n n

4 2 2 4 4 2 2

q

w = (16x - 24a x + 5a ) + (16y - 24b + 5b ) 384D(γ + δ)

n y n x n x n y

A cosh cos + B cosh cos

a a b b

n y n x n x n y

+ C y sinh cos + D x sinh cos

a a b b

Converting 16x - 24a x + 5a and 16y - 24b +

y

2 4

4 2 2 4 4 2 22 44

(16x - 24a x + 5a ) + (16y - 24b4 2 2 4 4 2 22 + 5b )44 (1 4 2 2 44 44 22 22 44 (16x - 24a x + 5a ) + (16y - 24b 22 + 5b )44 (16x - 24a x + 5a ) + (16y - 24b44 22 22 44 44 22 22 + 5b )44

n n

n y

A cosh_n_n hn ycosc + B cosh_n_n cos

a a b b

n n

n y h cos

hn yn yc + + B coshB cosh coscosn y

n n

A c_n _n

n n

n y

+ C y sinh_n_n n ycos + D x sinh_n_n cos

a a b b

n n

n y coss n y n y

s + D x sinh cosn y s + D x sinh cos

s h c

4

5b into cosine series, this equation can be writtem as

a a

b b

where,

q = Unit load on plate

a = Length of short side (X-Direction); b= Length of long side (Y-Direction)

E I = flexural rigity for beam along X-direction

E I = flexural rigity for beam along Y-direction

h =

3 p

p 2

a a b b b b

a a 4

5 5

Thickness of plate;

ν = Poisson's Ratio

E h

D =

where E = Young's Modulus of plate

12(1

ν )

E I

δ

E I

a

γ =

,

δ =

, therefore,

=

aD

bD

γ

E I b

4qa

γ

n

R =

sin(

), p =

, and q =

; Note: sin

2

a

b

n D(

γ + δ)

n

), p =

, and q =

; Note: sin

n

5 5 5 5

n D(

5 5 5 5 5 5

γ +

whe

2

ν )

22

n-1 2

n

(

) = (

1) for n = 1,3,5,7,...

2

) = (

n-1 2

1) for n

2

1)

Boundary conditions

By applying appropriate boundary conditions along the edges y =

b

2

, and x =

a

2

in Equation (2)

below values of

the constants An, Bn, Cn and Dn can be determined for n = 1, 3, 5, 7,….. the following equations are obtained:

n + Bn

n n

n BBn

4 4

5 5

n n

4qa γ n n x δ b n y

w = sin( ) cos + cos

2 a γ a b

n D(γ + δ)

n y n x n x n y

+ A cosh cos + B cosh cos

a a b b

n y n x n x n y

+ C y sinh cos + D x sinh cos (1)

a a b b

γ n

) c

γ n

5 5 5_{n D(}5 5 5 5 5 5 5 _{γ +}

4

γ n

γ δ b n

γ n

γ 44 n y

γ δ b n

γ n

γ 4 n y

γ n

γ n yy

) cos + cc

γ n

γ n y

γ n

γ n yy

γ n

c c c c

γ n

cos c

γ n

cos

γ n

c

γ n

) cos + c ) cos + c

b b

c c cos c

γ a

2 ) c) c) c a γ aγ aγ aγ aγ a bbbbbbbb

n n

n y

+ A cosh_n_n n ycos + B cosh_n_n cos

a a b b

n n

n y coss n y

s + B cosh cosn y s + B cosh cos

s h c

n n

n y

+ C y sinh_n_n n ycos + D x sinh_n_n cos

a a b b

n D x sinhn inh coscos

n n

n y coss n y n y

s + D x sinh cosn y s + D x sinh cos

(3)

2 2

b y=

2

2 2

a x=

2

3 3 4

3 2 4

b b

y= y=

2 2

3 3 4

3 2 4

a a

x= x=

2 2

d w d w

+ = 0 (2a)

dy dx

d w d w

+ = 0 (2b)

dx dy

d w d w EI d w

+(2 ) = (2c)

D

dy dx dy dx

d w d w EI d w +(2 ) =

D

dx dy dx dy (2d)

Development of equations using boundary conditions

Equation 2a:

4 2

2 2

2 n n

2

2 2

b _n

y= 2 2

n n

2 2

n n

p A cosh(py)+2pC cosh(py)

d w d w b

+ = q R cos(qy) + cos(px)

+p C ysinh(py) a

dy dx

q B cosh(qx)+D x sinh(qx) cos(qy)

p Rcos(px) p A cosh(py)+C y sinh(py) +

n n

2 2

(qx) s(qy) (qx) s( ) q B_n_n_n _n_n_n

2 2

2 2 _{py)+C y sinh(py)}_{)+C y inh(} ₎ p Rcos(px) p

p Rco

p R s(2222 ) p2222 A cosh(A c h(A c h(h(h(pyh( p Rcos(px) p

p Rcos(px) p A cosh(_n_n_n_n_n_n_n_n_n h(pypy)+C y sinh(py)C yC y_n_n_n_n_n_n_n_n_n h(py)h(py) p R s(px) p _n_n_n_n_n_n_n_n_n_n_n_n C y_n_n_n_n_n_n_n_n_n_n_n_n h(py) p R _n_n_n_n_n_n _n_n_n_n_n_n

2 2

n n n

cos(px)

= 0 + q B cosh(qx)+2qD cosh(qx)+q D x sinh(qx) cos(qy)

n n n

b qb n

Putting y = , and cos(qy) = cos( ) = cos( ) = 0, and eliminating cos(px)

2 2 2

from the equation we get,

pb 2 pb b pb

(1 )A cosh( ) + C cosh( ) + (1 )C sinh( ) = R (3a)

2 p 2 2 2

Similarly,

2 2

4

n n n

d w d w

From + = 0, we get dx dy

qa 2 qa a qa b

(1 )B cosh( ) + D cosh( ) + (1 )D sinh( ) = R (3b)

2 q 2 2 2 a

Development of Eq. 2c:

3 3 4

3 2 4

b b

y= y=

2 2

d w d w EI d w +(2 ) =

D

(4)

Referring to Eq. 2a

2 2 2 2

d w d w d w d w

+ = 0 can be written as =

dy dx dy dx

3 3

3 2

d w d w Differentiating with respect to y, we get =

dy dx dy

3 3 3 3 4

3 2 3 2 4

3 4

a a

2 4

b b

y= y=

2 2

d w d w d w d w EI d w Now, substituting = in +( 2 ) =

D dy dx dy dx dx dy dx

E I

d w d w

Eq. (2c) takes this form: = 2D(1 )

dy dx dx

2

n n n

2

2 2

n n n

4

4 4

b b b b b b

n n

4

Now, d w

= p pA sinh(py)+C sinh(py)+pC y cosh(py) cos(px) dx dy

q q B cosh(qx) + 2qD cosh(qx) + q D x sinh(qx) sin(qy), and

E I d w E I E I

= p Rcos(px) + p A cosh(py) + C 2D(1 ) dx 2D(1 ) 2D(1 )

4 4

4 4_{A c h(} _{) + C}

4 4

n n

p _n_n_n_n_n _n_n_n_n_n p A cosh(py) + C p A cosh(py) + C p A cosh(py) +

4 4

n n

4 3 4

b b

n n n

y sinh(py) cos(px)

E I

+ q B cosh(qx) + 4q D cosh(qx) + q D x sinh(qx) cos(qy) 2D(1 )

Therefore,

2

n n n

2 2

n n n

4 4

b b b b

n n

4 b b

n

p pA sinh(py) + C sinh(py) + pC y cosh(py) cos(px)

q q B cosh(qx) + 2qD cosh(qx) + q D x sinh(qx) sin(qy)

E I E I

= p R cos(px) + p A cosh(py) + C y sinh(py) cos(px)

2D(1 ) 2D(1 )

E I

+ q B cosh

2D(1 )

n n

p A cosh(py) + C ynn nn 2

n n n

p pA s2 _n _n _n

2 2

q q22222 q D22222 sinh(qx) sin( )

q q2222B cosh(qx) + 2qD cosh(qx) + q D x sinh(qx)_n_n_n_n_n_n _n_n_n_n_n_n 2222 _n_n_n_n_n_n sin(qy)

q q B c22 _n_n _n_n 22 _n_n

= p

= b b p b

= p n n

= p R cos(px) + p A cosh(py) +

= b bb b p b bb p nn nn

) 2D(1 )

= p (px) p

b b

)

3 4

n n

(qx) + 4q D cosh(qx) + q D x sinh(qx) cos(qy)

4 3 4

4

q B cosh4 _n_n (qx) 4q D333 _n sh(qx) 444D x inh(_n ) c (qy) q B cosh4 _n

q B co_n_n

q B_n (qx) + 4q D cosh(qx) + q D x sinh(qx) cos(qy)33 _n_n_n 44 _n_n_n

2

n n n

2 2

n n n

4

b b b

qb n

Rearranging and putting cos(qy) = cos( ) = cos( ) = 0,

2 2

we get, p pA sinh(py) + C sinh(py) + pC y cosh(py) cos(px)

q q B cosh(qx) + 2qD cosh(qx) + q D x sinh(qx) sin(qy)

E I E

= p Rcos(px) +

2D(1 )

) = 0,

Rcos

b b b

2

n n n

p pA s2 _n _n _n

2 2

q q22222 22222 sinh(qx) sin( )

q q222B cosh(qx) + 2qD cosh(qx) + q D x sinh(qx)_n_n_n_n_n_n _n_n_n_n_n_n 222 _n_n_n_n_n_n sin(qy)

q q B c22 _n_n _n_n 22 _n_n

= p

= b b p b

= p

= p Rc

= b bb b p bb

)

= p

= p Ib p A cosh(py) + C y sinh(py) cos(px)4 _n _n

2D(1 )

4 4

p A4 _n_n_n sh(py) C_n_n_n sinh(py) s( ) p A cosh(py) + C y sinh(py) cos(px)4 _n_n_n _n_n_n

p A co_n_n _n_n

b

(5)

m

2 2 2 2

m=1,2,3,..

n x

a

Converting cosh(

) and x sinh(

) into Fourier cosine series at interval

< x <

,

b

2

we get,

n x

2b

n

πa

4nab

nπa

2mπx

cosh(

) =>

sinh(

) +

(

1) sinh(

)cos(

)

b

n

πa

2b

π(n a +4m b )

2b

a

n x

) and x sinh(

) into Fourie

n x

r

n x

2

) =>

n x

2

n x

a

< x

a

2

m

1) sinh(

)

mm

(

m=1,2,3,..

4nab

2 2 2

4nab

m=1,2,3,.

2 2 2

π(n a +4m

2222 2222 2222

2 2

m

2 2 2 2 2 2 2

m=1,2,3..

2 2 2 2

n x

and, x sinh(

)

b

n

πa

b

nπa

2ab

cosh(

)

sinh(

) +

(

1)

2n

π

2b

n

a

2b

(n a + 4m b )

n

πa

nπa

n a

4m b

nπa

2mπx

cosh(

)

2

sinh(

) cos(

)

b

2b

n a + 4m b

2b

a

See Appendix A for Fourier

n x

)

n x

m=1,2,3..

a

2b

_m=1

(n a + 4m b )

2 2

m

b

22

nπa

2ab

22

sinh(

) +

(

1)

2 2 2

b

nπa

2ab

2 2 2

2 2

sinh(

) +

2

(

2 2 2 2 2 2 2

2 2 2

(

+ 4m b )

2 2 2 22 22 22 2

2 2 2 2 2 2 2

2 2 2 2 2 2

2b

2 2 2

2 2 2 2

2 2 2

n

πa

nπa

n

n a

22222222222222 22222222222222

4

4m b

22222222222222 222

nπa

a

2mπx

) cos(

(

2mπx

2

)

) cos

a

)

aa

4m b

cosh(

)

n

πa

nπa

h(

nπa

)

2

n a

4

sinh(

i h(

n

nπa

) cos

cosh(

)

2

) cos

s(

)

2 2 2 2

sinh(

2 2 2

)

b

(

2b

)

(

2b

)

(

a

)

b

(

)

22

(

2b

)

m b

2 2 2

))

2 2 2 2

2 2 2

(((((

2b

n a + 4m

2222222222222222 2222222222222222 2222222222222222

2 2

m=1,2,3..

(n a +

m=1,2,3..

2ab

2 2

2ab

2 2 2 2 2 2

(n a +

2 2 2 2 2 2

2 2 2

transformations.

b b

3

n n n n 2 n 2 n 4

E I p

qb

b

n

Substituting y =

, x = 0,

= P and cos(qy) = cos(

) = cos(

) = 0, we get

2

2D(1- )

2

pb

1

pb

b

pb

q

2

qb

A sinh(

) + C

sinh(

) + C

cosh(

)

B K +

D K +D K sin(

)

2

p

2

p

q

2

= PR + P A

) = 0, we

= P an

- )

3 3 3

qb

)

qb

b

333

pb

A i h(

pb

q

in(

qb

)

A sinh(

) + C

sinh(

) + C

cosh(

)

A sinh(

pb

b

) + C

sinh(

pb

b

) + C

cosh(

pb

b

)

q

B K

K +

₂₂₂₂₂

+

D K +D K si

_n_n_n_n_n ₂₂ _n_n ₄₄

q

in(

((

2

)

2

p

2

2222

q

nnnn 2222 nnnn 4444

si

2

p

2

n n

A

_n_n _n_n _n_n _n_n

2

p

2

n n n n

n n n n 222222222 nnnnnnnnn

p

n n n n n n n n n n n n n n n n n n n n n n 3

n n 2

n

n 2

pb

b

pb

cosh(

) + C

sinh(

)

2

Rewriting this equation we get

pb

q

qb

A sinh(

)+Pcosh(

) + B K

sin(

)

2

p

2

pb

1

b

+ C

sinh(

) +

cosh(

) + P

sinh(

)

p

2

q

2

+ D

K

q

p

b

A

h(

pb

b

pb

)

A cosh(

) + C

sinh(

)

A cosh(

p

) + C

sinh(

p

)

2

A

_n_n_n

cosh(

) + C

_n_n_n

sinh(

)

2

n n

((

n n

b

33333

b

pb

sinh(

pb

q

n(

q

qb

q

qb

q

qb

q

qb

q

qb

q

qb

b

)

))

)

sinh(

)+Pcosh(

) + B

sinh(

p

)+Pcosh(

p

) + B

K

q

sin(

n(

n((

n(

((

)))))

2

n

sinh(

)+Pcosh(

) + B

n

n n n n

2

n n n n n n n n

n n

K

222

sin((

sin(

((

))

2

(

2 2

p

(

b

pb

1

pb

)

pb

p

1

pb

sinh(

) +

cosh(

) + P

sinh(

)

sinh(

pb

) +

cosh(

pb

) + P

sinh(

pb

p

)

1

pb

p

sinh(

2

) +

2

cosh(

2

) + P

2

sinh(

2

)

p

(

2

)

2

(

2

)

2

(

2

3 3

4

qb

q

qb

sin(

) + K

sin(

) =

PR (3c)

2

p

2

3 3

2

q

3333333333

q

qb

q

qb

b

q

3333333333

qb

q

qb

q

b

((

K

2

K

2

PR

q

qb

q

qb

q

P

qb

q

)

qb

)

qb

i (

qb

q

qb

q

i (

qb

q

qb

)

P

i (

qb

q

b

q

) =

qb

) + K

sin(

qb

q

b

) + K

q

sin(

q

) =

sin(

) + K

sin(

(

n 2

n

K

2

n 2

sin(

((

) + K

)

) K

4444

sin(

sin((

(

)

) =

P

PR

q

(

n 2

q

p

(

2

)

44

p

(

2

)

m

2 2 _{2 2} _{2 2}

m=1,2,3,.. 2

4 4 2 2

2

2 2 2 m=1,2,3..

where,

2b n a 4n n a

K [K corresponds to cosh(qx)] = sinh( ) + ( 1) sinh( ) n a 2b (n a + 4m b ) 2b

b n a b n a

K [K corresponds to x sinh(qx)] = cosh( ) sinh( ) +

2n 2b n a 2b

2ab

(n a

ab

2b 2b

2b nn aa 2b

2 2 2 2

m

2 2 2 2 2 2

n a n a n a 4m b n a ( 1) cosh( ) 2 sinh( )

b 2b 2b

(6)

Similarly, (2d) can be written as 3

a a

n 1 n

3 3

n 1 3

n

E I q

p np qa qa

A K sin( ) +B sinh( )+ cosh( )

q 2 2 2D(1 ) 2

2 p np p np

+C K sin( )+K sin( )

p q 2 q 2

1 qa a qa a qa

+D sinh( )+ cosh( )+Q sinh( ) =

q 2 2 2 2 2

4

b

QR (3d)

a

m

1 1 _{2 2} _{2 2}

m=1,2,3,.. where,

2a n b 4nab n b

K [K corresponds to cosh(py)] = sinh( ) + ( 1) sinh( )

n b 2a (n b + 4m a ) 2a

2

3 3 2 2

2 2 2 2 2

m

2 2 2 2 2 2 2 2 2

m=1,2,3..

a n b a n b

K [K corresponds to y sinh(py)] = cosh( ) sinh( ) +

2n 2a n b 2a

2a b n b n b n b 4m a n b ( 1) cosh( ) 2 sinh( )

a 2a 2a

(n b + 4m a ) n b + 4m a

Let us make the following substitutions:

4

b b a a

5 5

1 1 2 2

1 2

3 4

5

E I p

E I q

4qa

n

P =

;

Q =

; R =

sin(

)

2D(1

)

2D(1

)

n D(

)

2

n

2m

n

2m

p =

, q =

, p =

and q =

a

b

a

pb

qa

T = (1

)cosh(

);

T = (1

)cosh(

)

2

pb

qa

2

b

2

a

T =

cosh(

) + (1

)

sinh(

);

T =

cosh(

) + (1

)

sinh(

)

p

2

q

2

T

sin(

)

=

)

2

sin(

)

(

n

2m

n

2m

, q =

, p =

and q =

1 1 2 2

n

2m

n

2

q

2

E I p

_{b b}_{b b}

E I q

_{a a}_{a a}

;

Q =

; R =

b b a a

E I p

_{b b}_{b b}

E I q

_{a a}_{a a}

2D(1

; R =

a a

=

)

2D(1

)

555555 555555 b b

;

Q =

a a

; R =

)

;

Q

2D(1

)

;

)

sin(

(

2

1 2

pb

qa

)cosh(

);

T = (1

)cosh(

)

1 2

pb

b

qa

2

1 2

(

)

(

1 2

3 4

qa

pb

qa

pb

qa

)

sinh(

);

T =

cosh(

) + (1

)

sinh(

3 4

pb

qa

pb

qa

pb

qa

pb

qa

pb

qa

)

2

q

2

3 4

(

)

(

)

(

3 4

6

pb

qa

= sinh(

) + Pcosh(

);

T = sinh(

) + Qcosh(

)

(7)

3 3

7 2 8 1

9

10

11 2 4

q qb p pa pa n

T = K sin( ); T = K sin( ) NOTE: sin( ) = sin( )

p 2 q 2 2 2

pb pb pb

1 b b

T = sinh( ) + cosh( ) + P sinh( );

p 2 2 2 2 2

qa qa qa

1 a a

T = sinh( )+ cosh( ) + Q sinh( )

q 2 2 2 2 2

q 2

T = ( K +K )

q p

( )

3 3

q 3 q 3

q 33 qb p 33 pa

q qb p pa

q q

q qb p pa

q q

q qb p pa

E: s

q qb p pa

q q

OTE:

sin( ); T = K sin( ) NOT

q qb p pa

p 2

7 2 8

p 2

7 2 8

p 2 q 2

7 2 8 1

7 2 8 (( ))

7 2 8 1

7 2 8

7 2 8 (( ))

7 2 8 1

7 2 8

7 2 sin( ); T = K8 K sin(s ( ) NOT)

7 2 8 1

7 2 8 1 (( ))

7 2 8 1

7 2 8

7 2 sin( ); T = K8 sin( ) NOT

q q q q p q p 3 3

12 1 3

4 4

1 2 3 4

qb 2 p pa

sin( ); T = ( K +K ) sin( )

2 p q 2

b b

P = R; P = R ; P = PR; P = QR

a a

1 R; P = R2 3

1 2 3

3 3 3 3 3 3 p p 3 3 3 3 p p 3 3 a p pa p p p p p pa

( a)

p pa p pa p p p p p p p p p p p p ( )

); T = ( K +K ) s

); sin( sin(n(n( sin(

sin(qbqbqbqbqb ₁₂₁₂ ( K₁₁ ₃₃

2 p

2 ););); 121212 (pK111 33) s ((( ))) a ( ) p pa p p p p p p

) p s p

) pp sin(pp )

q 2

q ( 2 )

q ((((((( 2 )))))))

q 2 q 2 ) s ) s ) s ) s ) s q 2 q 2 q 2

) s ( )

) sin(( ))

) s ( )

) sin( )

4 4

b b

b 4 b 4

b 44 b 44

b b b b b b b b b b b b b b b b b b

1 2 3

b b b b b b b b b b b b b b b b b b b b b b b b b b

; P = P

; P

; P = P

b b

; P

b b

= PR; P = QR

b b

= PR; P = QR

b b

1 2 ; P =3 P

1 2 3

1 2 ; P3

1 2 ;;;; P =PP =P =3== PR;PR;PR;PP P =P =P =44 QRQRQR

1 2 3

1 2 3 P =P =444 QRQR

1 2 3

a a

1 2 3

a a

1 2 3

a 4444 QRQR a

Then, re-writing (3a) through (3d), we get

n n

n 1 n 3 1

pb 2 pb b pb

(1 )A cosh( ) + C cosh( ) + (1 ) sinh( ) = RL

2 p 2 2 2

Or, A T + C T = P (4a)

2 p pb pb pb pb n n pb )A cosh(_n_n (pbpb) + C) _n_n

2 p n n n n n n n n n n pb pb pb

( ) RL

pb pbpb

pb pb pb

cosh(( ) + (1) ) sinh(( ) = RL) cosh(pbpbpbpbpb) + (1 ) sinh(pbpb) = RL

2 p 2 2 2

2 pcosh( 2 ) + (1 )2sinh( 2 ) = RL 2 p (( 2 ) (1) (1 ))2 (( 2 ))

4

n n

n 2 n 4 2

qa 2 qa a qa b

(1 )B cosh( ) + D cosh( ) + (1 ) sinh( ) = R

2 q 2 2 2 a

Or, B T + D T = P (4b)

2 q

4 4 4 4

2 a bb 4

n n

qa )B cosh(_n_n (qaqa) + D) _n_n

2 q n n n n n n n n n n

2 qaqaqaqa a qaqaqa bbbbbbb cosh(( ) + (1) ) sinh(( ) = R) cosh(qaqaqa) + (1 ) sinh(qaqaqa) = R bb

2 q 2 2 2 a

2 qcosh( 2 ) + (1 )2sinh( 2 ) = R a 2 q (( 2 ) (1) (1 ))2 (( 222222222222222222222 )) aaaaaaaaaaaaaaaaaaaaa

3

n n 2

n

3 3

n 2 4

pb pb q qb

A sinh( ) + Pcosh( ) + B K sin( )

2 2 p 2

pb pb pb

1 b b

+ C sinh( ) + cosh( ) + P sinh( )

p 2 2 2 2 2

q qb q qb

2

+ D K sin( ) + K sin( ) =

q p 2 p 2

pb pb 33333 bb

pb pb

A sinh(pbpb pbpb qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq ((((((((qqqqqqqqbqqqqqqqqqqbqbqqqqqqbqqqqqbqqqqqqqbqbqbbbbb)))))))) A sinh( ) + Pcosh( ) + B

A sinh(pbpb) + Pcosh(pbpb) + B KKKKKKKKKKKKKKKK qqq ss (ssin(sin(ssssin(ssin(sssssin((((((((((((((((((((((((((qqq )))))))))))))))))))))))))))))))

2 2

n n 2

A sinh(_n_n ) + Pcosh( ) + B_n_n ₂₂

2 2

n n 2

n n K2 s

n n 2

n n K2 s

n n 2

n n KKKK2 sin(sin(s (s (((( )))))))

p 2

p ( 2 ) p (( 2 ))

p 2

p ( 2 )

n n 2

p 2 p 2 p 2 p 2 p 2 p 2 p 2 p 2 p 2

p ( 2 )

pb pb pb

1 pb b pbpb

C 11sinh(pbpb) + bbcosh(pbpbpb) + P sinh(pbpb) C sinh( ) + cosh( ) + P sinh( )

C s ( ) ( )

C 11sinh(pbpbpbpb) + bbcosh(pbpbpbpbpb) + P sinh(pbpb)

p 2 2 2 2 2

C sinh( ) + cosh( ) + P sinh( ) p h(h( 2 ) ) 2 (( 2 )) 2 (( 2 ))

3 3

2 qqqqq 33333 qbqbqqqb qqq 33333 qbqbqb

D K

D 22K

D_n_n 2K₂₂ qqqqqqqqqqqqqqqq sin(sin(sin(qbqqqbqbqbqqbqbqqqbqbqbqbqb) + K) + K) + KK₄₄ qqqqqqqqqqq s (sin(sin(sin((((qbqbqbqbqbqbqbqbqbqbqb) =)))) =) =) =

D K

D_n_n_n_n_n_n_n_n_n_n_n_n_n_n_n_n_n_n_n K₂₂₂₂₂₂₂₂₂₂₂₂₂₂₂₂₂₂₂ sin( ) + KK₄₄₄₄₄₄₄₄₄₄₄₄₄₄₄₄₄₄₄ s (sin(((((((((( )))) =))))))) =

q p

n 2 4

q p

n 2 4

q p

n 2 4

2 p 2

2 p ( 2 )

2 p 2

n 2 4

n 2 4 (( ))

n 2 4

n 5 n 7 n 9 n 11 3

PR

Or, A T + B T + C T + D T = P (4c)

PR PR P P P PR P

3 3 3

a a

n 1 n n 1 3

n

E I q

p np qa qa 2 p np p np

A K sin( ) + B sinh( )+ cosh( ) + C K sin( ) + K sin( )

q 2 2 2D(1 ) 2 p q 2 q 2

qa qa qa

1 a a

+ D sinh( ) + cosh( ) + Q sinh( )

q 2 2 2 2 2

3 3 3

E I

3 3 3

2

E I q 2

p n np p np

p npp qqa E I qE I q qaaa 2 ppp np A K

A_n K₁ pppppppppppppppp sin(sin(sin(npnnnnpnpnpnpnpnpnnnpnpnpnpppppppppppppp) + B s) + B s) + B sB sinh(B sinh(B sinh(_n qqaqaqqaqqaqaqaqaqqaqqqqqaqaqaqqaqqqa)+)+)+ E I qE I qE I qE I qa aa aa aa aa aa aa aa aa aa aa aqqqq cosh(cosh(cosh(cosh(cosh(((((((qaqaqaqaqaqaqaqaqaqaqaqaqaaaaaaaaaaaaa) + C)))))))) + C) + C) KKKK ppppppppppppppppppppppppp sin(sin(s (sin(((((npnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnpnp)) + K)))) + K)) + K pppppppppppppppppppppp sisin(sin(n(nnnnpnpnnpnnpnnnpnnpnnnpnnnnn ))))))

A_n_n_n_n_n_n_n K₁₁₁₁₁₁₁ _n_n_n_n_n_n_n )

q 2

n 1 3

n 1 3 n(n( )

q 2

n 1 3

n 1 3 sin(

n 1 3

n 1 3 n( )

q 2

n 1 3

n 1 (( ) + K) 3 sin(n(

n 1 3

n 1 3 n( )

n 1 n

n 1 sin( ) + B sn

n 1 n

n 1 B sinh(n )+ cosh((((( ) + C)))) nnnnnnnnnnnnnnn KK111111111111111 sin(s ( )) 333333333333333

n 1 n

q 2 q 2

q 2 q 2 )

q 2 q 2

n 1 3

n 1 3 n(n(

q 2

n 1 n

q 2

n 1 n

q 2

n 1 n

2 p q 2

2 p

2 2D(1 ) 22 pp q 2

2 2D(1 ) 222222 nnnnn pppppp 11111 qqq 222 33333 2 2D(1 )

2 2D(1 ) 2 2D(1111111111111 )))))))))))))

qa

1 qa a qaqa a

) + cosh( ) + Q sinh( )

qa qa qa

qa qaqa

qa qa

1 qa a

D 11s qa aa D sinh(

D s

D 11sinh(qaqa aa

q 2

D sinh( q h(h( 2

4

n 8 n 6 n 12 n 10 4

b = QR

a

Or, A T +B T +C T +D T = P (4d)

4 4 4 b 4 Q

QR bbb = Q = QQRQR = QQR

aa

n n n n

(8)

3 1

n n

1 1

2 4

n n

2 2

T P

A = C from (4a) (5a)

T T

P T

B = D from (4b) (5b)

T T

3

T₃

C f

C 3 f

n n

T₃

T T

n n

2 4

P₂₂ T₄₄

D f

2 4

n n

2 4

T T

n n

n n n n

Substituting A and B in (4c) and (4d), and we get C and D as follows:

6 7 3 7 10 6

4 1 2 11 4 1 2 4

8 1 2 8 5 2 5 5 1 2 5 8 2 8 C C

n

3 7 9 3 10 6

12 11 4 4 C C

8 1 5 2 5 5 1 8 2 8

6 7

4 1 2 11 4

C

8 1 2 8 5 2 5

T T P T T T

P P P T T P P T

( )( ) ( )( )

T T T T T T T T T T T T T T W -X

C = = (5c)

T T T T T T

T T T T Y -Z

( )( ) ( )( )

T T T T T T T T T T

T T

P P P T T

where, W = ( )(

T T T T T T T

6

T T₆

P₄₄ P₁₁ P₂₂ TT₆₆₆₆₆₆₆ T TT₇₇₇₇₇₇₇ PP₃₃₃₃₃₃₃ TT₇₇₇₇₇₇ TT₁₀₁₀₁₀₁₀₁₀₁₀ TT

( )( ) ( )( )

( )

( 444 111 222 66666666666666 )((( 77777777777777 ))) (((( 33333333333333 77777777777 )))()(( 1010101010101010101010 666 )))

T T T T T T T T T T T T T T W

-( )( ) ( )( )

( )( ) ( )( )

6 T T₆₆ T₁₂₁₂₁₂ TT₃₃₃₃₃₃₃ T₁₁₁ TT₇₇₇₇₇₇₇ TT₉₉₉₉₉₉₉ TT₃₃₃₃₃₃ TT₁₀₁₀₁₀₁₀₁₀₁₀ TT

( )( ) ( )( )

( 12121212 3333333333333 )((( 1111 7777777777777 ))) (( 9999999999999 33333333333 )()(( 1010101010101010101010 666 )))

T T T T T T T T T T

( )( ) ( )( )

6 11 4 7

6 7

T₆₆₆₆₆₆₆ T₇₇₇₇₇₇₇ P₄₄ P₁₁ P₂₂ TT₆₆₆₆₆ T TT₇₇₇₇₇

)(

4 1 2 66 77

4 1 2

4 1 2 666 777 T T T T )( T T T

3 1 2 7 10 4 6

C

5 1 2 5 8 2 8

3 7 9 3 10 6

12 11 4 4

C C

8 1 5 2 5 5 1 8 2 8

P P P T T T T

); X = ( )( );

T T T T T T T

T T T T T T

T T T T

Y = ( )( ) and Z = ( )( )

T T T T T T T T T T

0 6

P₃ T₇ T₁₀₀₀₀ T₆₆₆₆

P T T T

P₃₃₃₃₃₃₃ T₇₇₇₇₇₇₇ T₁₀₁₀₁₀₁₀₁₁₁₀ T₆₆₆₆₆

( )( );

( )

( 33333333333333333 11 22 77777777777777777 )((( 11110110101101010101110110000 44 666666666666 )))

T T T T T T T

( )( )

6

T T₆

T₁₂₁₂₁₂ TT₃₃₃₃₃₃₃ T₁₁₁ TT₇₇₇₇₇₇₇ TT₉₉₉₉₉₉₉ TT₃₃₃₃₃₃ TT₁₀₁₀₁₀₁₀₁₀₁₀ TT

( )( ) and Z = ( )( )

( 1212 333)((( 1111 44 777 ))) (( 999 333 ))((( 101010 44 6 )))

C C

6 6 12 3333 1 7777 9999 333 101010

12 1

12 33333 1 77777 99999 3333 10101010

T T T T T T T T T T

C C

C C (( )()( ))

C C

3 1 2 7 12 3 4 1 2 6 9 3

5 1 2 5 8 1 8 1 2 8 5 1 D D

n

7 3 10 6 9 3

11 4 12 4 D D

5 2 5 8 1 8 2 8 5 1

P P P T T T P P P T T T

( )( ) ( )( )

T T T T T T T T T T T T W X

D = = (5d)

T T T T T T

T T T T Y Z

( )( ) ( )( )

T T T T T T T T T T

6 9 3

P₃ T₇ T₃ T₆₆ T₉₉ T₃₃

P T T T

P₃₃₃₃₃₃₃ T₇₇₇₇₇₇₇ T₃₃₃₃₃₃₃ T

( )( ) ( )( )

( )

( 3333333333333333333 7777777777777777777 )(( 3333333333333333333 )) (((( 66666 ))())( 99999 33333 ))

T T T T T T T T T T T T

( )( ) ( )( )

WD XD WDD XDD

(

D D

( (

D D

Y_D_D Z_D_D ((((

D D

3

( T₀₀₀₀ T₆₆₆₆ T₉₉₉₉ T₃₃

T₁₁₁₁ T₄₄ TT₇₇₇₇₇₇₇ TT₃₃₃₃₃₃₃ TT₁₀₁₀₁₁₀₁₁₀₁₀ TT₆₆₆₆₆ ₉₉₉₉₉

( )( ) ( )( )

( )

( 111111 444 777777777777 )(( 1212 333333333333 )) ((( 11010101101011101010000000 44 66666666666666 )))(( 99999999999999 333 ))

T T T T T T T T T T

( )( ) ( )( )

3 1 2 7 12 3 4 1 2 6 9 3

D D

5 1 2 5 8 1 8 1 2 8 5 1

7 3 10 6 9 3

11 4 12 4

D D

5 2 5 8 1 8 2 8 5 1

P P P T T T P P P T T T

where, W = ( )( ); X = ( )( );

T T T T T T

T T T T

Y = ( )( ) and Z = ( )( )

T T T T T T T T T T

6 9 3

P₃ T₇ T₃ T₆₆ T₉₉ T₃₃

P T T T

P₃₃₃₃₃₃₃ T₇₇₇₇₇₇₇ T₃₃₃₃₃₃₃ T

)( ); X = ( )( );

3 1 2 7 12 3 4 1 2 66 99 33

3 7 3

D D

3 7 3 666 999 333

3 7 3

D D

D D ( )(

D D

3 T₀₀₀₀ T₆₆₆₆ T₉₉₉₉ T₃₃ T₁₁₁₁ T₄₄ TT₇₇₇₇₇₇₇ TT₃₃₃₃₃₃₃ TT₁₀₁₀₁₁₀₁₁₀₁₀ TT₆₆₆₆₆ ₉₉₉₉₉

( )( ) and Z = ( )( )

( )

( 11 4 777 )(( 1212 333 )) ((( 1010100 44 6666 )())( 9999 3 ))

D D

0 6 9 3

0 6 9

11 4

11 4 777777777 333333333 10110101011110 66666 99999

T T T T T T T T T T

D D

D D (( )()( ))

D D

Substituting values of Cn and Dn from (5c) and (5d) in (5a) and (5b), values of An and Bn can be found.

Given the values of dimensions a and b, loading q, plate thickness, h, Young’s Moduli of plate, Ep, Beams, Ea and Eb, Moments of Inertia of beams, Ia and Ib, Poisson’s ratio, , numerical values of An, Bn, Cn and Dn are found out using Microsoft Excel for n = 1, 3, 5, 7…..

Verification of Excel software

Let us consider an example: A rectangular plate simply-supported on four rigid edge beams with the input data (lengths of sides, load on plate, plate thickness, Young’s moduli, moments of inertia of the beams and Poisson’s ratio) shown in Table 1. Maximum deflection is calculated at the end of Table 1.

Table 1: Input data and maximum deflection.

Length in X-Direction, a (in.) 60

(9)

Uniform Load, q (psi) 0.694

Plate thickness, h (in.) 1

Young's Modulus for Plate, Ep (psi) 30,000,000 Young's Modulus for Beam on side a, Ea (psi) 30,000,000 Young's Modulus for Beam on side b, Eb (psi) 30,000,000 Moment of Inertia for beam on side a, Ia (in.4) 1,000,000,000* Moment of Inertia for beam on side b, Ib (in.4) 1,000,000,000*

Poisson’s ratio, n 0.3

3 p

2

E h

D =

12(1 - )

2,666,667

Deflection Coefficient at the center of the plate, a 0.008308 Deflection at the center of the plate, w (in.) [w = a_qa4_/D] _0.028020

*High values of Ia and Ib signify that the beams are robust and rigid, and thus the plate is effectively simply supported on its edges.

Table 2 provides a comparison of maximum deflections of plates with various a and b dimensions with the same boundary conditions (i.e., rectangular plates simply-supported on four rigid edge beams) as above. Results are almost the same as given in Table 8 of

Timoshenko and

Woinowsky-Krieger (1959).

Table 2: Comparison of maximum deflections - EXCEL vs. Timoshenko and Woinowsky (1959)* Short Side, a (in.)

(X-Direction)

Long Side, b (in.) (Y-Direction)

Deflections (in.) (Using Excel)

Deflections

(in.)* % Difference

60 60 0.013700 0.013694 -0.05%

60 72 0.019058 0.019023 -0.18%

60 84 0.023895 0.023779 -0.49%

60 96 0.028020 0.027995 -0.09%

60 108 0.031419 0.031401 -0.06%

60 120 0.034160 0.034167 0.02%

* Table 8, Timoshenko and Woinowsky-Krieger (1959)

CONCLUSIONS

1. Maximum deflections are calculated using classical plate theory formulation as given in Timoshenko and Woinowsky-Krieger (1959).

(10)

3. Fourier transformations (see Appendix A) have been used for hyperbolic functions in the equations.

4. No high speed computer is required to solve this problem.

5. Excel is commercial Microsoft software. No special skill is required to use this software.

6.

Commercial Finite Element software (such as SAP, STAAD, ANSYS, etc.) can be used to solve such problems. Unfortunately procurement of such software, and special training, skill and experience may not be always available to the advantage of the user.

REFERENCE

Timoshenko, S.P. and Woinowsky-Krieger, S. (1959), “Theory of Plates and Shells,” International Student Edition,McGraw-Hill Book Company, Inc., Second Edition, New York.

APPENDIX A: Fourier Transformations

The following are Fourier transformations of cosh(n y) a

n y

( )

(n y), cosh(nπx) b ,

nπy y sinh( )

a and

n x x sinh( )

b

n x ) n x

in cosine series:

m

2 2 2 2

m=1,2,3,..

n y

b

(i) cosh(

), at interval range

< y <

a

2

2m y

2a

n b

4nab

n b

Fourier =

sinh(

) +

(

1) sinh(

)cos(

)

n b

2a

(n b + 4m a )

2a

b

n y

) at int

n y

2m

n b

4nab

n b

) +

(

1) sinh(

)cos(

2m

)

n b

4nab

n b

4nab

n

m y

)

m y

sinh(

n b

(

2

b

< y

b

2

y

1) sinh(

)

m

(

1) sinh(

mm m=1,2,3,..

2 2 2

m=1,2,3,.

2 2

4m

2

2 2 2

4nab

2 2 2

(n b + 4

2222 2222 2222

m

2 2 2 2

m=1,2,3,..

n

πx

a

(ii) cosh(

), at interval range

< x <

b

2

2b

n a

4nab

n a

2m

πx

Fourier =

sinh(

) +

(

1) sinh(

)cos(

)

n a

2b

π(n a + 4m b )

2b

a

n a

4nab

n a

2m

m

πx

) +

(

1) sinh(

)cos(

)

n a

4nab

n a

2m

n a

4nab

n a

2

m

πx

sinh(

n a

s

(

2

a

< x

a

2

1) sinh(

) s

m

(

1) sinh(

mm m=1,2,3,..

2 2 2

m=1,2,3,.

2 2 2

π(n a + 4m

2222 2222 2222

4nab

2 2

m

2 2 2 2 2 2 2

m=1,2,3..

2 2 2 2

n

πy

b

(iii) y sinh(

) at interval range

< y <

a

2

a

n

πb

a

nπb

2a b

Fourier =

cosh(

)

sinh(

) +

(

1)

2n

π

2a

n

π b

2a

(n b + 4m a )

n

πb

nπb

n b

4m a

nπb

cosh(

)

2

sinh(

)

a

2a

n b + 4m a

2a

+

b

< y

b

2

y

m

a

22

nπb

2a b

22

sinh(

) +

(

1)

2 2 2 2 2 2 2

2 2 2

a

nπb

b

2 2 2

(n b + 4m a )

2 2 2 22 22 22 22

2 2 2 2 2 2 2

2 2

2a

2 2 2 2

2 2 2

2 2 2 222 222

+

222

b

)

b

2 2 2 2

4m a

2 2 2 2

4m a

b

2 2 2

n

πb

nπb

2

n

2 2 2

4

2 2 2

n b

22 22 22

nπb

i h(

nπb

n

cosh(

n

πb

cosh(

n b

₂₂ ₂₂

4

₂₂ ₂₂₂

sinh(

)

m a

2 2 2

sinh(

((

2a

n b + 4m

22222222222222 22222222222222 22222222222222

cosh(

a

cosh((

(

2 2

m=1,2,3..

2 2

m=1,2,3..

(n b +

2a

2 2

2 2 2

2a

2 2 2

(n b +

2 2 2

2 2 2 2 2 2

2m

πy

cos(

)

b

2m

m

πy

cos(

cos

s(

2m

m

πy

cos

s(

2m

m

πy

coss(

b

s(

cos

2 2

m

2 2 2 2 2 2 2

m=1,2,3..

2 2 2 2

n x

a

(iv) x sinh(

) at interval range

< x <

b

2

b

n

πa

b

nπa

2ab

Fourier =

cosh(

)

sinh(

) +

(

1)

2n

π

2b

n

a

2b

(n a + 4m b )

n

πa

nπa

n a

4m b

nπa

cosh(

)

2

sinh(

)

b

2b

n a + 4m b

2b

n x

) at inte

n x

m=1,2,3..

a

2b

_m=1

(n a + 4m b )

a

< x

a

2

2 2

m

b

22

nπa

2ab

22

sinh(

) +

(

1)

2 2 2

b

nπa

n

2 b

2 2 2

2 2

sinh(

) +

2

(

2 2 2 2 2 2 2

2 2 2

+ 4m b )

2 2 2 22 22 22 22

2 2 2 2 2 2 2

2 2

2b

2 2 2 2

2 2 2

nπa

(

nπa

)

a

)

2 2 2 2

4m b

2 2 2 2

2 2 2

4m b

2 2 2

n

πa

nπa

2

n

2 2 2

4

2 2 2

n a

22 22 22

nπa

i h(

nπa

n

cosh(

n

πa

cosh(

n a

₂₂₂ ₂₂₂

4

₂₂₂ ₂

sinh(

sinh

sinh(

)

cosh((

)

2b

(

)

(

2

₂₂ ₂₂ ₂₂ ₂₂₂

sinh(

)

m b

2 2 2

sinh(

((

2b

n a

2222222222222 2222222222222

4m

2222222222222

b

cosh(

(

b

(

2 2

m=1,2,3..

(n a +

m=1,2,3..

2ab

2 2

2

2ab

2

2 2 2

(n a +

2 2 2

2 2 2 2 2 2

2 2 2

2m

πx

cos(

)

a

2m

m

π

2m

m

π

cos(

s(

cos

s(

2m

m

π