Lecture Bending of Plates

8/13/2019 Lecture Bending of Plates

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Assumptions of Thin-plate theory

1. Before loading, the midsurface is flat.

2. Plate thickness t is small in comparison with spanwise dimensions of

the plate.

3. The plate material is homogeneous, isotropic, and linearly elastic.4. A line normal to the midsurface before loading remains normal to the

midsurface after loading.

5. Stress normal to the midsurface szis negligible compared to sxand

sy.

6. Lateral deflections are small in comparison with the plate thickness,

and the slope of the midsurface is small.

7. The midsurface is a neutral surface, i.e., the strains ex, ey, gxyare zero.


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0 Integrating, we get ( , )zw

w w x y

z

e

00 ( , )zxw u w

u z u x yx z x

g

00 ( , )yzv w w

v z v x yz y y

g

0Here ( , ) and ( , ) represent, respectively, the values

of and on the midplane.

ou x y v x y

u v

According to the last assumption, 0 0 0u v


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andw w

u z v z x y

Therefore,

2 2 2

2 22x y xy

w w wz z z

x yx ye e g

Before loading After loading


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Generalized Hookes law

x x y z Ee s s s

y y z x Ee s s s

z z x y Ee s s s

xy xy Gg

yz yz Gg

zx zx Gg

where

2 (1 )

EG


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For 0, 0, 0, we getz yz zxe g g

2 2

2 2 2 21 1x x y

E E w wz

x ys e e

2 2

2 2 2 21 1y y x

E E w wz

y xs e e

2

1xy xy

E wG z

x y g


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Stressresultants

/ 2

/ 2

x x

y yt

xy xyt

x zx

y yz

M z

M z

M z dz

Q

Q

s

s


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Substituting the expressions for , and , we getx y xys s

2 2

2 2xw wM D

x y

2 2

2 2y

w wM D

y x

2

1xy

wM D

x y

where3

212 (1 )

E tD

is the flexural rigidity of the plate


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The stresses could be written as

3

12 xx

M z

ts

3

12 yy

M z

ts

3

12 xyxy

M z

t

The maximum stresses occur on the bottom and top

surfaces (atz = t/2) of the plate.


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Load is distributed over the upper surface of the plate.

The intensity of the load is q, so that the load acting onthe element is q dx dy.


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Summation of all the forces in thez-direction

0yx

QQdx dy dy dx q dx dy

x y


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0yx

QQq

x y

Taking moments of all the forces about x-axis, we get

0xy y

y

M Mdx dy dy dx Q dx dy

x y

The moment of the load qand the moment due to change in

the force Qyare neglected, since they are small quantities of

a higher order than those retained. After simplification,

0xy y

y

M MQ

x y


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Similarly, by taking moments with respect to they-axis, we get

0

yx x

x

M M

Qy x

Substituting the expressions of Qx, Qyand noting that

Mxy=Myx, we obtain

2 22

2 22

y xyx M MM

qx yx y

Or,

4 4 4

4 2 2 42 qw w w

Dx x y y

Symbolically, 4 q

wD

2 is the Laplacian operator


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4 qwD

To determine w, it is required to integrate this equation

with the constants of integration dependent upon the

appropriate boundary conditions.

Also we can write

2 2

2

2 21 1x y

w wM M D D w

x y

Or, the moment function 2

1

x yM MM D w

Shear forcesx y

M MQ Q

x y

2 2

2 2

M Mp

x y

2 2

2 2

w w M

Dx y


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Boundary conditions

Built-in edge 0 and 0x a

x a

wwx

Simply-supported edge

2 2

2 20 and 0

x a

x a

w ww

x y

The second condition is analogous to

2

2

20 or also 0

x ax a

ww

x

which do not involve Poissons ratio .


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Boundary conditions

Free edge 0 0 0x xy xx a x ax aM M Q

Kirchhoff showed that these three boundary conditions proposed

by Poisson are too many and that two conditions are sufficient forthe complete determination of the delfection w.


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Boundary conditions

Free edge

xy

x

x a

MQ

y

0xy

x x

x a

MV Q

y


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Boundary conditions

Therefore for a free edge at x= a

3 3

3 22 1 0

x a

w w

x x y

For zero bending moments along the free edge requires

2 2

2 20

x a

w wx y


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Example

Determine the deflection and stress in a very long and narrowrectangular plate if it is simply supported at edgesy = 0 andy =b.

(a) The plate carries a nonuniform loading expressed by

0

( ) sin y

p y pb

(b) The plate is under a uniform loadp0. Letp0= 10 kPa, b = 0.4 m,

t = 10 mm,= 1/3, and E = 200 GPa.

x

y

b


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2 2

2 2x

w wM D

x y

2 2

2 2y

w wM D

y x

2

1xyw

M D

x y

where3

212 (1 )

E tD



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4 4 4

4 2 2 42

qw w w

Dx x y y

4

4

d w p

dx D


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The stresses could be written as

3

12x

x

M z

ts 312

yy M z

ts 312

xyxy M z

t

The maximum stresses occur on the bottom and top

surfaces (atz = t/2) of the plate.

2

1

x yM MM D w

Shear forcesx y

M MQ Q

x y


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23 2

1

2

y

yz

Q z

t t

23 2

1

2

xzx

Q z

t t

3

3 2 2 1 2

4 3 3z

p z z

t ts

The maximum shear stress, as in the case of a rectangular beam,

occurs atz= 0, and can be expressed as

,max

3

2

y

yz

Q

t ,max

3

2

xzx

Q

t


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Naviers Solution for Simply Supported

Rectangular Plates

In general, the solution of the

bending problem involves the

following Fourier Series for

load and deflection.

1 1

( , ) sin sinmnm n

m x n yp x y p

a b

1 1

( , ) sin sinmnm n

m x n yw x y a

a b

wherepmnand amnrepresent undetermined coeffiecients.


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This approach was introduced by Navier in 1820. The deflection

must satisfy the following boundary conditions.

2

20 0 ( 0, )ww x x a

x

2

20 0 ( 0, )

ww y y b

y

12

1, 2

2sin sin

m n

x yaa b


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Determination ofpmn

1 1

( , ) sin sinmnm n

m x n yp x y p

a b

Multiply each side by sin sin and integrate

between limits (0, ) and (0, ).

m x n ya b

a b

0 0

0 01 1

( , )sin sin

sin sin sin sin

a b

a b

mn

m n

m x n yp x y dx dya b

m x n y m x n yp dx dy

a b a b


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It can be shown that

0

0 ( )

sin sin / 2 ( )

a m mm x m x

dx a m ma a

0

0 ( )sin sin

/ 2 ( )

b n nn y n ydy

b n nb b

Therefore, after simplification, we can wrtite

0 0

4( , )sin sin

a b

mn

m x n yp p x y dx dy

ab a b


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Evaluation of amn

1 1

( , ) sin sinmnm n

m x n yw x y a

a b

Substitute the expression of deflection into the plate bending equation.

4 4 4

4 2 2 42

qw w w

Dx x y y

where q is the load andD is the flexural rigidity of the plate.


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After simplification,

4 2 2 4

1 1

2 sin sin 0mnmnm n

pm m n n m x n ya

a a b b D a b

This equation must apply for allxandy. Therefore,

4 2 2 4

2 0mnmnpm m n n

a

a a b b D

Or,

242 2

1

( / ) ( / )

mnmn

pa

D m a n b

And

242 2

1 1

1sin sin

( / ) ( / )

mn

m n

p m x n yw

D a bm a n b


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Special Cases

1. Uniformly distributed load 0( , )p x y p

0 0

4( , )sin sin

a b

mn


ab a b

0

2

0

2

41 cos 1 cos

4

1 1 1 1

mn

m n

pp m n

mn

p

mn

Or,0

2

16( , 1,3, )mn

pp m n

mn


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242 2

1 1

1sin sin

( / ) ( / )

mn

m n

p m x n yw

D a bm a n b

Or,

0

262 2

16 sin( / ) sin( / )

( / ) ( / )m n

p m x a n y bw

D mn m a n b

( , 1,3, )m n

Maximum deflection occur at the centre of the plate (x=a/2, y=b/2).

( ) / 2 10

max 262 2

116

( / ) ( / )

m n

m n

pw

D mn m a n b

1 / 2 1 / 2

Note that sin( / 2) 1 and sin( / 2) 1m n

m n

( , 1,3, )m n


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2 2

2 2x

w wM D

x y

2 2

2 2y

w wM D

y x

2

1xyw

M D

x y

where3

212 (1 )

E tD



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2 2

0

242 2

/ ( / )16sin sin

( / ) ( / )

x

m n

m a n bp m x n yM

a bmn m a n b

2 2

0

24

2 2

/ ( / )16sin sin

( / ) ( / )y

m n

m a n bp m x n yM

a bmn m a n b

0

24 2 2

16 (1 ) 1cos cos

( / ) ( / )xy m n

p m x n yM

ab a bm a n b


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2. Patch load

( , ) / 4p x y P cd

1 1

1 1

sin siny d x c

mny d x c

P m x n yp dx dy

abcd a b

Or,

1 1

2

4sin sin sin sinmn

m x n yP m c n dp

mncd a b a b

0 0

4( , )sin sin

a b

mn


ab a b


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3. Point load

1 1

2

4sin sin sin sinmn

m x n yP m c n dp

mncd a b a b

For patch load we wrote

The above expression could be used for point loadby putting the limiting values of c and d to zero.

1 14 sin sinmnm x n yP

p

ab a b

Deflection

1 1

242 2

sin( / ) sin( / )4sin sin

( / ) ( / )m n

m x a n y bP m x n yw

Dab a bm a n b


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When the loadP is applied at the centre of the plate (x1=a/2, y1=b/2),

the previous equation reduces to

( ) / 2 1 1 1

242 2

sin( / ) sin( / )41

( / ) ( / )

m n

m n

m x a n y bPw

Dab m a n b

( , 1,3, )m n

Furthermore, if the plate is square (a=b), the maximum deflection,

which occurs at the centre can be written as

2

max 242 2

4 1

m n

P aw

D m n

( , 1,3, )m n

2 2

maxRetaining the first nine terms 0 01142 (Exact value 0 01159 )

Pa Pa

w D D

Lecture Bending of Plates

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