Circular.plates

4

Circular Plates

&

Diaphragms

K.A. | MECH3001Y | Mechanics of Materials & Machines III — Page | 2

LEARNING OBJECTIVES

The object of this chapter is the study of thin plates subjected to

small elastic deformations. “Thin” is a relative term which

indicates that the thickness of the plate is small compared to

the overall geometry.

A ratio of 1:X , where 10X ≥ is the usual criterion.

The deformations are caused by pure bending conditions.


Introduction

Plates are initially flat structural elements (before loading)

having relatively small thicknesses compared with the

remaining dimensions. It is customary in the analysis of plates

to divide the thickness t into equal halves by a plane parallel to

the faces.

The dividing plane is called the midsurface of the plate. The

plate thickness is measured in a direction normal to the

midsurface at each point under consideration.

CLASSIFICATION OF PLATES

Plates may be classified according to:

a) the magnitude of the thickness compared to the

magnitude of the other dimensions

b) the magnitude of the lateral deflection compared to the

thickness

We may thus speak of

1) relatively thick plates with small deflections

2) relatively thin plates with small deflections

3) very thin plates with large deflections


4) extremely thin plates (membranes) that may undergo

either large or small deflections, and so on

PLATES IN ENGINEERING

The purpose of plates in engineering is to cover, generally, a

rectangular or circular area and to support concentrated or

distributed loading normal to the plane of the plate.

A typical example is a pressure diaphragm, as a safety or

control device, supported around its circular periphery, and

subjected to uniform pressure on one face and perhaps a

central point load on the opposite face.


SMALL DEFLECTION OF THIN PLATES

Assumptions

i. No deformation in the middle plane of the plate

(i.e. existence of a neutral surface).

ii. Points in the plate lying initially on a normal to the neutral

surface of the plate remain on the normal during bending.

iii. Normal stresses in the direction transverse to the plate

can be disregarded.

iv. The deflection, w, is a function of the two coordinates in

the plane of the plate, the elastic constants of the material,

and the loading conditions.


MOMENTS AND CURVATURES FOR PURE BENDING General Case in Cartesian Coordinates

We need to relate the bending moments to the curvatures. The

slope and deflection of a plate can then be determined from the

resulting equations.

Bending Moments

Let us consider an element of material (cf. drawing below) cut

from a plate subjected to pure bending (cf. drawing).


The bending moments xM and yM per unit length are positive

as drawn acting on the neutral surface of the plate. The

material above the neutral surface is in a state of compression

while material below it is in a state of tension (biaxial).

Curvatures

Beam bending analogy:

Consider a cantilever beam subjected to a load P acting at the

free end.

Curvature is a measure of how sharply a beam is bent.

The deflections of a beam are usually very small compared to

its length. Small deflections mean that the deflection curve is

nearly flat. .

Under these special conditions of small deflections, the

equation for the curvature becomes


Sign Convention

The sign convention for curvature depends upon the orientation

of the coordinate axes.

Application to plates

Curvatures of the mid-plane in sections parallel to the xz and

yz planes are denoted by 1

xR and 1

yR respectively.


STRESS-STRAIN RELATIONSHIPS

Strains

At a depth z below the middle plane, the strains in the x and y

directions of a lamina such as abcd are

ε = −xx

zR

[1]

ε = −yy

zR

[2]

Stress - Strain relationships

. yx

x E E

ν σσε −= [3]

.y x

y E E

σ ν σε −=

[4]

Combining [1] and [2] with equations [3] and [4] and

rearranging,

2

1

1

νσν

−= +−

xx y

EzR R

[5]

2

1

1

νσν

−= +−

yy x

EzR R

[6]


Equations [5] and [6] show that the bending stresses are

functions of the plate curvatures and are proportional to the

distance from the neutral surface.


BENDING MOMENTS / CURVATURES RELATIONSHIPS

Taking the equilibrium of moments for the infinitesimal plate

element,

2

2

. .

h

x x

h

z dydz M dyσ−

=∫ [7]

2

2

. .

h

y y

h

z dxdz M dxσ−

=∫

[8]

Substituting from equations [5] and [6] for xσ and yσ in

equations [7] and [8], and integrating, we obtain

( )

3

2

1 1

12 1

ν νν

= − + = − +−

xx y x y

EhM DR R R R

[9]

( )

3

2

1 1

12 1

ν νν

= − + = − +−

yy x y x

EhM DR R R R

[10]

( )3

212 1

EhDν

=−

is called the flexural rigidity.


The principal curvatures of the plate are given by

2

2

1 ∂=∂x

wR x

[11]

2

2

1 ∂=∂y

wR y

[12]

where w is the deflection in the z direction.

It follows that the relationships between the bending moments

and the curvatures become,

2 2

2 2xw wM D

x yν

∂ ∂= − +∂ ∂

[13]

2 2

2 2ν

∂ ∂= − +∂ ∂

yw wM D

y x [14]


BENDING MOMENTS / BENDING STRESSES RELATIONSHIPS

The bending stresses can be expressed as functions of the

bending moments by eliminating the curvatures between

equations [5], [6], [9] and [10].

3

12 σ =x xz M

h [15]

3

12 σ =y yz M

h [16]


SYMMETRICAL BENDING OF CIRCULAR PLATES Polar Coordinates

The deflection w of a plate of circular shape depends on its

radial position alone if the applied load and conditions of end

support are independent of the angle θ .

Consider a circular plate, subjected to the symmetrical loading

condition (cf. figure below). Any diametral section may be used

to indicate the deflection curve, the associated slope ψ and

deflection w at any radius r .

The curvature of the plate in the diametral plane rz is

2

2

1 =r

d wR dr

[17]


and for small values of w (+ve downwards) the slope at any

point is given by

ψ = dwdr

[18]

The second principal radius of curvature Rθ

is in a plane

perpendicular to rz and is represented by lines such as PQ.

1

rRθ

ψ= [19]


Let us now consider an element of the circular plate subjected

to bending moments along the edges. Let rM and Mθ

be the

bending moment per unit length acting on the plate element

(cf. drawing below). The circular plate element can be analysed

in the same way as in the case of the plate in Cartesian

coordinates.

Thus, the equations giving the bending moments in terms of the

slope ψ are as follows:

( )3

212 1

ψ ψνν

= − +−

rdEhM

rdr [20]

( )3

212 1θ

ψ ψνν

+

= −−

dEhMr dr

[21]

These equations can also be written in terms of curvatures or

deflections.


2

2rd w dwM D

r drdr

ν

= − + [22]

2

2

1 dw d wM Dr dr drθ

ν +

= − [23]


LOAD, SHEAR FORCE AND BENDING MOMENT

Consider an infinitesimal element taken from a circular plate.

The plate element must be in equilibrium under the action of

the uniformly distributed loading p per unit area, the resulting

shear forces Q per unit length and the bending moments per

unit length.

For vertical equilibrium,

. . . 0dQ

Q r d p r d dr Q dr r dr ddr

θ θ θ

× + × − + + = [24]

⇒ dQ Q

prdr

+ = [25]


For equilibrium of moments in the radial plane, taking moments

about the outer edge of the plate element, we obtain

...

2 .sin . . 02

rr r

dMM dr r dr d M rd

dr

dM dr Q r d drθ

θ θ

θ θ

+ + − × −

× + × =

[26]

Equation [26] may be simplified to give

0rr

dMr M M Qr

dr θ+ − + = [27]


RELATIONSHIPS BETWEEN DEFLECTION, SLOPE AND

LOADING

Substituting equations [20] and [21] in equation [27], and after

simplifying, we obtain

2

2 2

1 Qd dr Ddrdr r

ψ ψ ψ+ − = − [28]

Substituting equations [22] and [23] in equation [27], we obtain

the variation of the deflection with the radius;

3 2

3 2 2

1 1 Qd w d w dwr Ddrdr dr r

+ − = [29]

Equations [28] and [29] can be expressed as follows:

( )1.

Qd d rr Ddr dr

ψ

= − [30]

1 Qd d dwrr Ddr dr dr

= [31]

A solution for ψ and w is readily obtainable from equations

[30] and [31] by integration.


The shear force Q is a function of the applied loading p and

may be obtained by integration of equation [25].

. . .r dQ Qdr p r dr+ = or ( ) . .d Qr p r dr=

Therefore,

0

r

Qr pr dr∫=

Substituting in equation [31],

0

1 1r

d d dwr r pr drr Ddr dr dr

= ∫ [32]

If we know ( )p f r= or if p happens to be constant, then

equation [32] can be integrated to find the deflection of the

circular plate at any radius.

(integration constants are evaluated from boundary conditions)


CIRCULAR PLATE SUBJECTED TO UNIFORM PRESSURE

Starting with equation [32], obtained in the previous section,

and through successive integrations, we obtain the equation

giving the deflection of the plate.

0

1 1r

d d dwr r pr drr Ddr dr dr

= ∫

We assume in this configuration that the pressure p is

constant. Therefore,

2

0

1 1

2

r pd d dwr r pr dr rr D Ddr dr dr

= =∫

Eliminating r from both sides and integrating:

2

1

1

2

1

4

pd d dwr rr Ddr dr dr

pd dwr r Cr Ddr dr

=

= +


Multiplying both sides by r and integrating again:

3

1

4 212

4

16 2

pd dwr r C rDdr dr

Cpdwr r r CDdr

= +

= + +

Dividing both sides by r and integrating:

3 1 2

16 2

C Cpdw r rrDdr

= + +

4 21

2 3ln

64 4

Cpw r r C r C

D= + + + [33]

For a circular plate subjected to a uniform pressure p ,

equation [33] gives the deflection of the plate in terms of the

radial position r . Integration constants 1

C , 2

C and 3

C are

evaluated based on the boundary conditions and slope of the

plate at preset positions.


Problem 4.1 – Uniformly loaded plate (Clamped edges)

Consider a circular plate of radius a (cf. drawing below).

Find

a) the equation of the deflection curve

b) the maximum deflection of the plate

c) the equations of bending moments

d) the maximum stresses in the plate


Equation of deflection curve

• Since the periphery of the plate is clamped, the slope of

the plate is 0 at r a= 0dwdr

=

• Due to the obvious symmetry of the problem, the slope of

the plate is also 0 at 0r = 0dwdr

=

• The deflection of the plate is 0 at the clamped boundary.

4 21

2 3ln

64 4

Cpw r r C r C

D= + + +

3 1 2

16 2

C Cpdw r rrDdr

= + +

Since the slope cannot be infinite at 0r = , 2

0C =

3 1

16 2

Cpdw r rDdr

= +

⇒ 3

1016 2

C apaD

= + 2

1 8

paC

D=−


⇒ 4 4

30

64 32

pa paC

D D= − +

4

3 64

paC

D=

Therefore the deflection curve is given by

22 2

64

pw a r

D

= −

Maximum deflection

The maximum deflection occurs at the centre of the plate.

Therefore,

( )0

4

max64

rpa

w wD

== =

Equations of bending moments

The bending moment equations are determined from the

moment/curvature relationships (equations [22] & [23]):

2

2rd w dwM D

r drdr

ν

= − +

2

2

1 dw d wM Dr dr drθ

ν

= − +


23

16 16

p padw r rD Ddr

= − 22

2

2

3

16 16

p pad w rD Ddr

= −

⇒

2 22 2

2 2 2 2

2 2

3

16 16 16 16

316

1 316

rp pa p pa

M D r rD D D D

pr a r a

pa r

ν

ν

ν ν

= − − + −

= − − + −

= + − +

( )

2 22 2

2 2 2 2

2 2

3

16 16 16 16

316

1 1 316

p pa p paM D r r

D D D D

pr a r a

pa r

θν

ν

ν ν

= − − + −

= − + − −

= + − +

Maximum stresses in the plate

At the periphery, r a=

2

8r

paM = −

2

8

paM

θν= −


At the centre 0r = , and

2

116

rpa

M Mθ

ν

= = +

Similar to the case of Cartesian coordinates, where

3

12. xx

M z

hσ = and

3

12. yy

M z

hσ =

the bending stresses for the circular plate are as follows:

3

12. rr

M z

hσ = [34]

3

12.M z

hθ

θσ = [35]

On inspecting equations [34] and [35], we observe that the

radial and the circumferential stresses will have maximum

values when the bending moments are maximum and for

maximum value of z (all values here refer to absolute values).


The table below summarises these information.

0r = r a=

rM 2

116

pa ν

+ 2

8

pa−

Mθ

2

116

pa ν

+ 2

8

paν−

max2

hz = ±

Radial stress is maximum at the periphery while the

circumferential stress is maximum at the centre.

Therefore,

( )( )max

max

2

2 2

3.6.

4

rr

paM

h hσ = ± = ±

( )( )max

max

2

2 2

6. 31

8

M pa

h hθ

θσ ν

=± =± +


Problem 4.2 – Plate simply supported at periphery

A circular plate subjected to a uniform load pressure is simply

supported at its edge (cf. drawing below).

Find

a) the equation of the deflection curve

b) the maximum deflection of the plate

c) the equations of bending moments in the plate

d) the maximum stresses in the plate


Equation of deflection curve

Since the periphery of the plate is simply supported, 0rM = at

r a= .

Due to the symmetry of the problem, the slope of the plate is 0

at 0r = 0dwdr

=

The deflection of the plate is 0 at the simply supported

periphery.

4 212 3ln

64 4

Cpw r r C r C

D= + + +

3 1 2

16 2

C Cpdw r rrDdr

= + +

Since the slope cannot be infinite at 0r = , 2

0C =

3 1

16 2

Cpdw r rDdr

= +

22 1

2

3

16 2

Cpd w rDdr

= +

2

2rd w dwM D

r drdr

ν

= − +


2 21 13

16 2 16 2r

C Cp pM D r r

D Dν

= − + + +

⇒ 2 2

1 130

16 2 16 2

C Cpa paD D

ν

= + + +

2

1

3

8 1

paC

D

ν

ν

+= −

+

⇒ 4 2 2

3

30

64 8 41

pa pa a CD D

ν

ν

+= − +

+

4

3

5

64 1

paC

D

ν

ν

+=

+

Therefore the deflection curve is given by

2 44 2

3 5

64 32 641 1

p pa paw r r

D D D

ν ν

ν ν

+ += − +

+ +


Maximum deflection

The maximum deflection occurs at the centre of the plate.

Therefore,

( )0

4

max

5

64 1r

paw w

D

ν

ν

=

+= =

+

Equations of bending moments

2 22 2

2 2

3 33

16 16 16 161 1

3

16

rp pa p pa

M D r rD D D D

pa r

ν νν

ν ν

ν

−

+ += − − + −

+ +

+=

2 22 2

2 2

3 33

16 16 16 161 1

3 1 316

p pa p paM D r r

D D D D

pa r

θ

ν νν

ν ν

ν ν

+ += − − + −

+ +

= + − +

Maximum bending moments occur for 0r = .

2 3

16r

paM M

θ

ν

+= =


Maximum stresses in the plate

3

12. rr

M z

hσ =

3

12.M z

hθ

θσ =

Maximum stresses are obtained at the centre of the plate;

Therefore,

( )( )max

max

2

2 2

3 36.

8

rr

paM

h h

νσ

+= ± = ±


Problem 4.3 – Cylinder Head Valve

A cylinder head valve of diameter 38mm is subjected to a gas

pressure of 1.4MPa. It may be regarded as a uniform thin

circular plate simply supported around the periphery by the

seat, as shown below. Assuming that the valve stem applies a

concentrated force at the centre of the plate, calculate the

movement of the stem necessary to lift the valve from its seat.

The flexural rigidity of the valve is 260N.m, and Poisson’s ratio

for the material is 0.3.


Solution: Problem 4.3:

The above problem can be solved as the superposition of two

simpler problems.

1. a circular flat plate subjected to a uniform pressure and

having its edges freely supported, and

2. a circular flat plate with a concentrated load at its centre

and with its edges simply supported too.

However, care must be taken regarding the orientation of the

uniform pressure and the concentrated load, and subsequently,

the sign of the deflection.

We found at a previous section that the deflection curve for a

circular flat plate subjected to a uniform pressure and with

freely supported edges is given by:

2 44 2

3 5

64 32 641 1

p pa paw r r

D D D

ν ν

ν ν

+ += − +

+ +

We now need to find the deflection curve for sub problem #2.

Plate with Concentrated Load and Freely Supported Edges

Starting with equation [31], i.e.

1 Qd d dwrr Ddr dr dr

=


we first need to find the expression for Q (shear stress per unit

length).


Problem 4.4 – Thin Steel Diaphragm

A circular thin steel diaphragm having an effective diameter of

200mm is clamped around its periphery and is subjected to a

uniform gas pressure of 180kPa. Calculate a minimum

thickness for the diaphragm if the deflection at the centre is not

to exceed 0.5mm. ( 208E GPa= , 0.287ν = )

ANS: 3.1mm


Problem 4.4 – Circular Plate

A circular plate 500mm diameter and 2.5mm thick is clamped

around its edge and is subjected to a concentrated load of

900N at its centre.

Calculate the radial and tangential bending stresses at the

fixed edge. ( 208E GPa= , 0.29ν = )

ANS: 68.75

19.94


Bibliography

Advanced Strength and Applied Elasticity [Book] / auth. Ugural Ansel C. and Fenster Saul K.. - New

Jersey : Prentice Hall, 2003. - ISBN 0-13-047392-8.

Mechanics of Engineering Materials [Book] / auth. Benham P. P., Crawford R. J. and Armstrong C.

G.. - London : Longman Group Limited, 1996. - 2nd Edition. - ISBN 978-0-582-25164-9.

Mechanics of Materials 2 - An Introduction to the Mechanics of Elastic and Plastic Deformation of

Solids and Structural Materials [Book] / auth. Hearn E. J.. - Oxford : Butterworth-Heinemann, 1997. -

3rd Edition : pp. 193-218. - ISBN 0 7506 3266 6.

Mechanics of Solids and Structures [Book] / auth. Rees David W. A.. - London : Imperial College

Press, 2000. - pp. 137-183. - ISBN 1-86094-217-2.

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