thin plate theory

7/28/2019 Thin Plate Theory

1/8

Thin Plate Theory

Bending of a thin elastic plate: Dd

4w

dx 4+ P

d2w

dx 2= q(x)

where D

EH

= ( )

3

212 1 is the flexural rigidity,H= thickness of the plate, E & are

Young's Modulus and Poisson's Ratio respectively. Also, w=w(x) is the vertical

displacement that varies with horizontal location x , P is the Horizontal Force of

compression and q(x) is the distributed load on the surface of the plate.

For the earth, any displacement of the lithospheric plate (density o) into the fluid mantleresult in an upward buoyancy force( )gwm 0 . On the surface of the lithosphere, atrough is associated with such displacement. In the ocean, water is drawn in to fill the

trough and on the continents, sediments will fill the basin. This infill material acts as an

extra load (downward force)that further deform the lithosphere. Let the density of this

infill be f then the downward load is

( )gwfo . Combining the upward force at the

base and the downward force on the surface of the lithosphere,

gwxqxq fmapplied )()()( = where )(xqapplied is the applied load that cause thedeformation in the first place.

In the following, we will take P=0, i.e. no horizontal compression force.

The bending/fiber stress at some height z above the neutral plane is given by:

xx x zEz d w

dx,( ) =

( )1 22

2, so that it becomes maximum at the surfaces z H= / 2 with

xx xEH d w

dx

D

H

d w

dx

max( ) =( )

=2 1

62

2

2 2

2

2or

rr r

D

H

w

r r

w

r

max( ) = +

6

2

2

2(in cylindrical coordinates)

Flexure of a thin elastic plate due to a line load V0 applied at x=0:

Dd w

dxgw V xm f

4

4 0+ = ( ) ( )

The solution of the above can be expressed in terms of the flexural parameter

(wavelength)

( )

4/1

4

=g

D

fm

and is: w xV

D

ex x x( ) cos sin/= +[ ]0

3

8

The maximum deflection is at x=0 with amplitude wV

D0

03

8=

.

The deflection of the lithosphere under a line load is characterized by a forebulge.

The zero crossing of w(x) is at x01 3

41= ( ) = tan

The peak of the bulge is at xb = ( ) = sin 1 0

The height of the bulge is w w e wb = =

0 00 0432 .


2/8

Flexure of a thin elastic plate due to a box-car load with half-width L:

{0

)(4

4 ghgw

dx

wdD

L

fm

=+ for

Lx

Lx

where the box-car load has density f and height h.The solution of the above can be expressed in terms of the flexural parameter

(wavelength)

=( )

41 4

D

gm f

/

and the isostatic displacement wh

isosL

m f

=( )

.

w xw C C

w e F F

isosx x x x

isosx x x

( )cos cosh sin sinh

cos sin/= {

+ { }+{ }

1 2

1 2

1

forLx

Lx

where C eL L

1 = /

cos

, C e

L L2 =

/sin

, F L L1 = cos sinh , FL L

2 = sin cosh .This solution show a depression under the load and a peripheral bulge (peak) outside.

At x=0, the amplitude of the depression is ( )11 CwisosThe zero crossing of w(x) is at ( )2110 tan FFx =

The peak of the bulge is at ( )4

tan 211 += FFxb

The height of the bulge is w w eb isosx= 0 4

4/ /

sin

Flexure of a thin elastic plate over a liquid-filled shell of thickness h and radius R:

D wEH

Rw gw qm f applied + + =

42

( )

where the 2nd term represents the effect of shell action (i.e. in plane forces and curvature

of the shell). The solution can be expressed in terms of flexural parameter

l=+ ( )

( )

D

g

D

gEHR m f m f

2

1 4 1 4

/ /

and the Bessel-Kelvin functions of zero

o r d e r ber, bei, ker & k e i where ber x i bei x J xei( ) + ( ) = ( )0 3 4/ and

ker/x i kei x K xei( ) + ( ) = ( )0 4 and the primes indicate their derivatives.

For a concentrated (point) load of magnitude P:

w rP

Dkei

r P

gkei

r

m f

( ) =

( )

l

l l l

3

2 2

For a disc load of radius A and wq

D

q

g

q

gisos

applied applied

EH

R m f

applied

m f

= =+ ( )

( )

l4

2 :

w rw ber kei bei

w ber bei kei

isosA A r A A r

isosA A r A A r

( )ker' '

' ker '= {

( ) ( ) ( ) ( ) +{ }( ) ( ) ( ) ( ){ }

l l l l l l

l l l l l l

1for

r A

r A

Thin Plate Theory is valid only ifA/l is small (or A H 100 3 4/ )


3/8

-30

-20

-10

0

10

VerticalDef

lection

0.0E+0 2.0E+5 4.0E+5 6.0E+5 8.0E+5 1.0E+6

Distance from center of the Disc Load (M)

Lake Bonneville

DISC

H=200km

H=120km

H=75km

Disc Radius = 117km

Ave.Height of Ice Load=305m

-800

-600

-400

-200

0

200

V

erticalDeflection(M)

0E+0 5E+5 1E+6 2E+6 2E+6

Fennoscandia Ice

DISC

H=200km

H=120km

H=75km

-1000

-800

-600

-400

-200

0

200

VerticalDeflection(M)

0E+0 5E+5 1E+6 2E+6 2E+6 2E+6 3E+6

Prediction from Thin Plate Theory

DISC

H=200km

H=120km

H=75km

Disc Radius = 1300km

Ave.Height of Ice Load=3kmLaurentian Ice

Disc Radius = 675km

Ave.Height of Ice Load=2.2km


4/8

-4

-3

-2

-1

0

1

Vertic

alDisplacement(km)

0 200 400 600 800 1000 1200

X (km)

Boxcar Load:

1/ 2width=200km, height=3km, density=2800kg/ m^3

Lithosphere: E=1E11 Pa, Poisson =0.25

Mantle: density=3300 kg/ m^3

150 km Lith

100 km Lith

50 km Lith

25 km Lith

-400

-200

0

200

400

Max.FibreStress(MPa)

0 200 400 600 800 1000 1200

X (km)

Maximum Bending Stress (MPa)


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-4000

-3000

-2000

-1000

0

1000

W(r)(M)

0 1000 2000 3000 4000

Load Radius = 2000 km

A / l=26.14

Thin Plate Theory

FE results

-2000

-1500

-1000

-500

0

W(r)(M)

0 100 200 300 400

R (km)

Load Radius = 100 km, A / l=1.30

-4000

-3000

-2000

-1000

0

1000

W(r)(M)

0 250 500 750

Load Radius = 250 km

A / l=3.27

-600

-500

-400

-300

-200

-100

0

W

(M)

0 50 100 150 200

R (km)

Load Radius = 50 km,

A / l=0.65

Thin Plate Theory breaks down when A / l< 1.5

l= Flexural parameter

A= Load Radius

H = 50 km lithosphere


6/8

Degree of Compensation:

Let the load be a harmonic load of wavelength : q x ghx

applied c o( ) sin=

2, then

Dd w

dxgw ghm f c o

x4

42+ = ( )( ) sin

Let the solution be: w x w xo( ) sin=

2

, by substituting into the equation above, we get

wh

oo

Dg

m

c c

= + ( ) 1

24

.

Since the isostatic displacement (due to buoyancy force alone) is wh

isosc o

m c

=( )

then

w w Co isos= where C is the degree of compensation: Cm c

m cDg

=( )

+ ( )

24

If the wavelength of the load is long, >> ( )

2

1 4

Dgm c

/

, then C=1 and

w wo isos= .

If the wavelength of the load is short,


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Let us now calculate the gravity of such harmonic mountain chain:

There are 2 contributions to the surface Free-Air gravity anomaly. The first is the

contribution due to the topography: g x Ghx

topo c o( ) sin=

2

2

The second is due to the deflection at the base of the lithosphere. The anomalous mass

associated with the deflection is:

= ( ) = ( ) + ( )

c mm c o

Dg

w h xm

c c1

22

4sin , but this

mass distribution is buried at depth H, so this mass contribution to surface gravity is:

gG h e x

mm c o

H

Dg

m

c c

= ( )

+ ( )

2

1

22

24

/

sin

Thus the surface Free-Air gravity anomaly is: g x g x g xFA topo m( ) ( ) ( )= +

g G he x

FA c o

H

Dgm c

= + ( )

( )

2 1

1

22

2 4

/

sin

Since Bouguer gravity anomaly is g x g x G hB FA c( ) ( )= 2

therefore, gG h e x

Bc o

H

D

gm c

= + ( )

( )

2

1

22

24

/

sin

If the wavelength of the load is long,

>>( )

2

1 4D

gm c

/

, and >>H

then g xFA( ) = 0 and g G hx

G hB c o c= = 22

2

sin , thus the surface is totally

compensated.

If the wavelength of the load is short,


8/8

-500

-400

-300

-200

-100

0

100

Vertic

alDisplacement(M)

0 500 1000 1500 2000 2500

Distance from the Load Center (km)

Effect of Lithospheric Thickness

Boxcar load magnitude = 15 MPa

w(z=0) 25km

w(z=0) 50km

w(z=0) 75km

w(z=0)100km

w(0)150km

w(0)200km

0.0

0.2

0.4

0.6

0.8

1.0

Degreeof

Compensation

0 500 1000 1500 2000

Wavelength (km)

Harmonic Load on top of Floating Beam

thin plate theory

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