finite differences and their applications

8/9/2019 Finite Differences and Their Applications

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CHAPTER-3

Finite Differences and their Applications


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.

Most frequently engineering problems are represented by

erent a equat ons, or w c exact so ut ons are genera y not

available.

results, acceptable for most practical purposes.

Numerical methods for the solution of differential e uations

have become popular in the recent years because of the easy

availability of electronic digital computers.

One of these numerical methods is the “Finite Difference

Method”


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1.2 Differentiation Formulas B Inter olatin

ParabolasThe simplest way of obtaining approximate expressions for the

derivatives of a function y(x) at some pivotal point i, consists in

substituting for the function y a parabola passing through a certainnumber of ivotal oints and in takin the derivatives of the

parabola as a approximate values of the derivates of y.

To determine the difference form of the derivative of y, when y is

known at three points i-1,i and i+1 evenly spaced by the interval h

on the x-axis and denoting the corresponding ordinates at these

,− y y y points (Fig.1). Choose origin at point I, we obtain

,

2 c Bx Ax ++= (1)


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y

( ) x y

x x y ++=

1−i y i y 1+i y

x1−i 1+iiFig.1 Interpolating parabola


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−− 2

( ) C y y i

i

==−1

0

i ++== +1

iii y y y

A

+−

= +− 2

2

11

ii

h

y y B

−= −+

2

11

i yC =


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The first derivative of y at i is given by the following relation

(2) 2

1'

h

y y B

dx

dy y iii

i

i−+ −==

=

(3) 22 22

2

''

h y y y A

xd yd y iiiii

i

i +− +−

==

=

Analogous expressions for higher order derivatives may be

obtained by means of higher degree interpolating parabolas.

Pass a cubic parabola through the points and

again choose origin at i.

2,1,,1 ++− iiii

(4) 23 DCx Bx Ax y +++=


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−−==− 23

( ) D y y i

i

==

==

−

0

...

23

( ) DhC h Bh A yh y i +++== +

+

.2.4.82

...

23

2

The third derivative of y at i is given by

3

(5) 63

2

3

'''

h A

dx y iiiiii

i

i++− −−==

=


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r 1+ir

2−nr inr −

n

r 2r 3

r 1−ir

h h h h h h

− + −n −n n x

Fig.2 Evenly spaced pivotal points


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The first backward difference is defined as

(6) 1−−=∇ iii y y y

,

can be written as

(7)2-

-

2-i1-ii

2-i1-i1-ii1iii

y y y

y y y y y y y −

+=

−−=−=

1

1

11

i

n

i

n

i

n

i

n y y y y −−−− ∇∇=∇−∇=∇

two adjoining differences in Table 1 . The nodal points are used in

descending order


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Table 1

x y y∇ y2∇ y3∇ y4∇

i x

1−i x

i y

1−i yi y∇

i y2∇

3

2−i x 2−i y1−i

2−∇ i y

1

2

−∇ i y

2

i

1

3

−∇ i yi y

4∇

−

4−i x

3−i

4−i y3−∇ i y

2−i


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In an analogous manner, forward differences can be derived.

ese are g ven e ow(8)

iii y y y −=∆ +1

Similarly, second and the higher order differences can bewritten as

(9)

.... ....

2- i1i2i2

ii

−−−

++ +=∆∆=∆

The nodal points in forward differences are in the ascending

1 iiii y+ =−=

order.

To relate the derivatives of y with the differences, consider the

+,


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'''3

''2

'= hhh

......

!3!2!1

Using the powers of D to indicate the derivates of y,

( ) ( ) ( ) ( ) ( ) x y Dh

x y D

h

x Dy

h

x yh x y ++++=+ ......!3!2!132

33

22

( )

xe

x y D D D

hD=

++++= ......!3!2!1

1 32

Substituting and indicatingi x x = ( ) r i yh x y =+

i

hD

r ye y = (12)


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In the case of forward differences

−=

−=

e

hD

hD

ii

ir i

y y

y y y(13)

Takin lo on both the sides∆+=

−=

1e

hD

i(14)

( ) ...432

1Log432

+∆

−∆

+∆

−∆=∆+=hD (15)

...6

5

12

11 543222 +∆

−∆

+∆−∆= Dh

(16)

...4

7

2

3

6

54333 +

∆−

∆+∆= Dh

...625444

++∆−∆= Dh


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But in the case of backward differences chan in h into –h in

Eq.(12) we have

i x x = ( ) r i yh x y =+ (17)

i

hD

ye y −

=1

Procee ng n t e same manner, we o ta n

...432

432

+

∇

+

∇

+

∇

+∇=hD

...12

11

54

43222 +

∇+∇+∇= Dh

(18)

17

...42

65444

333

∇

+++∇= Dh

...6


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Table.2

Schematic representation of Backward and Forward Difference Expressions

i1−i2−i3−i4−i

k ir

BACKWARD

TYPE TSCOEFFICIEN

h/1

2/1 h

11−

11 2−

'

1

r

''r 3/1 h

4/1 h 11

11− 3−3

4− 4−6

'''

ir

''''r

h/1

i 1+i 2−i 3−i 4−i

1− 1'1r

FORWARD

2/1 h3/1 h

1

1

1

1−

2−

3−3

''

ir '''

ir

4/1 h 1 14−4− 6''''

ir


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Central Difference

Central differences involving pivotal points symmetrically

oca e w respec o , are more accura e an ac war or

forward differences.,

points forming the differences.

Thus the first order central difference of (x) at I is defined by

11

11

−+ +−+

=−= iiiii

y y y y

y y yδ (19)

( )11

22

1 −+

−

−= ii y y


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differences.

2/12

12/1

2

12/1

2

12/1

2

1−

−−

++

−+

+

−=

−−

−==iiii

ii y y y y y y

(20)

The nth central difference is defined by

11 −+ iii

i

n

i

n y y 1−= δ δ δ


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r r rr r

11r 1r ir

2/h 2/h

x

2/h 2/h 2/h 2/h 2/h

− − r

Fig.3 Pivotal points for central differences

+ +r rr


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Table 3

x y yδ y2δ y3δ y4δ

2−i x

1−i x

2−i y

1−i y2/3−i yδ

1

2

−i yδ

i x i y2/1−i

2/1+i yδ

i y2δ

2

2/1−i

2/1

3

+i yδ

i y4δ

+

2+i x

1+i

2+i y2/3+i yδ

1+i


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a e

Schematic representation of Central Difference Expressions

2+i1+ii1−i2−i

k

ir TSCOEFFICIEN

h2/1

2/1 h

1

11 2−

'

1r

''ir

1−

32/1 h

4

/1 h11

12 2−

4− 4−6

'''

ir

''''

ir

1−


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1.4 Derivation of Differential formulas using

Taylors seriesIt is possible to get derivatives of y in terms of the difference

x=a ( ) ( ) ( ) ( ).. ''2' −− a ya xa ya x (21)

To obtain evaluate the series for the oints

...!2!1

y '' 11, +− == ii x x x x

...

!3

.

!2

.

!1

. '''3''2'

1 ++++=+iii

ii

yh yh yh y y (22)

...!3

.

!2

.

!1

. '''3''2'

11 +−+−=−iii

i

yh yh yh y y

(23)


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.

Solution of the governing equilibrium equations for beams,

plates or any other engineering problem by finite difference

method requires proper finite difference representation of the boundar conditions.

Consequently the derivatives at the boundaries are to be

replaced by finite difference expressions which require then ro uc on o c ous po n s ou s e e oun ar es.

The boundaries may be fixed, simply supported, completely

free or mixed t e.The finite difference expressions for different boundary

conditions or the relations of fictitious points to inner points are

er ve .


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Fixed boundar

A fixed end in beam or plate has deflection and the slope zero

0=i y

( )11 021

−+ =−=

ii

i y yhdx

dy

(24)

Sim l su orted

11 −+ = ii

The boundary condition representing a simple support has

deflection and bending moment is directly proportional to the

2

11

2

2

02 −+ =

+−=

iii

h

y y y

dx

yd (25)

11 +− −= ii y y


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The free end has always bending moment, i.e., the secondderivative and the shear force, i.e., the third derivative zero. The

or na es o e ec on a c ous po n s can e expresse n

terms of inner points of the beam.2 −

2

0

11

2

11

2

−=

==

−+

−+

y y y

hdx

iii

iii

i

(26)

02

223

2112

3

3

=+−+−

=

++−−

h

y y y y

dx

yd iiii

i


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1.6 Beam deflection

To illustrate the application of finite difference method to

,

flexural rigidity EI simply supported over the span L.

e eam s oa e w t a un orm y str ute oa o

intensity q per unit length.

The transverse deflection y is given by the following differential

equation.

qdx

yd EI =

4

4

(27)


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The differential equation (27) can be written in the difference

(28)

EI hq y y y y y iiiii

4

2112.464 =+−+− −−++

Since, both ends of the beam are simply supported, we have

0 and 00 == == L x x y y

(29)0 02

0

2 =

=

== L x x

dx

y

dx

y

Considering the beam to be divided into 4 equal parts with

interval h=L/4 and numberin the ivotal oints as shown in fi .3,the differential equation (b) can be written at each pivotal point

.


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1− 5

h h h h h h

4

Point 1: (30) EI

q y y y y y 32101

.464 =+−+−−

4

Point 2: (31) EI

q y y y y y 43210

.464 =+−+−

Point 3: (32) EI

q y y y y y 54321

.464 =+−+−


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From the end conditions in equation (29), we obtain

(33)351140 0 0 y y y y y y −=−=== −

o v ng t e a ove equat ons we get

qL yqL yqL y4

3

4

2

4

1

5 7 5 ===

The exact solution at the mid-span is whereas the EI

qL y

4

2 .512

5=

finite difference solution by considering beam into 4 parts, is

4

an suc a percen age error s . y ncreas ng e

number of subintervals, the error can be reduced to a large extent.

EI y2 .

512.=


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1.7 Solution of characteristic value roblems

Consider the Euler buckling problem of a simply supported

beam as shown in Fig.4, acted upon by compressive axial force P.

The beam is of uniform cross section throughout.The deflection of the axis of the beam are overned b the

following characteristic problem.

340'' =+ y

P y iv


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The differential e uation 34 can be written in the difference

form as:

02

.464 112112 =

+−+

+−+− −+−−++ y y y P y y y y y iiiiiiii

Let n=L/n and simplifying Eq.(35)

(36)02..

464 112

2

2112 =+−++−+− −+−−++ iiiiiiii y y y

EI n

L P y y y y y

Put K EI

=

.

(37)042

64 2122122 =+

−+

−+

−+ −−++ iiiii y y K

y K

y K

y


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The boundary conditions for the problem shall be:

( ) ( ) ( ) ( ) 000 '''' ==== L y L y y y

0, −−== L

(38)

(39)

Approximation n=2

−

1−

0 1 2

3 2/h =

h h h h

upon solving Eq. (d) and using the boundary conditions we get

the value of K

2


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The smallest root is K 4=9.367

2367.9 L

EI P cr =(43)

The exact value of and this gives an error of28775.9 L EI P cr =


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1.8 Richardson’s Extrapolation

The approximate values of the functions obtained by using two

different sets of subintervals can be improved by Richardson’s

extrapolation.

and An2, the approximate values obtained by using n1 and n2

su n erva s respec ve y.

The error e(x) can be written in series form as

( ) ( ) ( ) ( ) ....... 634

2

2

1 +++= h x f h x f h x f xe (44)


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of x, the error series has constant coefficients and may be writtenas

Let and

....634

2

2

1

abh

abh

hchchce

−=−=

+++=

46

Taking first term of the series only we have

21 nn

2

2

1

111

n

ac A Ae n

−=−=

2

Where (b-a) is the total interval or the limits

2

2

122

n A Ae n −=−=


37/46


38/46

1.9 Use of Unevenly Spaced Pivotal Pointsons er a s mp y suppor e co umn as s own n g. , ac e

upon by compressive axial force P. The governing differential

equation is given by

00

0.''

==

=+

L y y

y EI

P y (48)

(49)

04.0 I 04.0 I P P

0 I

L6.0 L2.0 L2.0

1− 0 1 2 2 1 0 1−

. . . ..


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The beam is unevenl divided into arts as shown in Fi .4

Since the spacing between pivotal points varies from point to point the second derivative of y at I can be written as

( ) ( )[ ]112''

11

2

.

1

+− ++−+= iiii y y yh y α α α α (50)

For the iven roblem1

11

−

+

−

−=−=

ii

iii

x x

x x x xh α

2.0

15.0

2

1

=

=

Lh

Lh(51)

3.0

3

4

15.0

2.01 ==

L

Lα

.2.02 == Lα


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difference form as:

2

( )0.1

111 =+++−

+ +− iiii y

EI y y y α α

α α (52)

Substituting Eq. (c) in the Eq.(d) and solving by using the

boundary conditions that at both the ends of the column0= .

The value of as against the exact value of 20855.7

L P cr =

0 EI

= 2cr


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1.10 Inte ration Formulas b Inter olatin

Parabolas In many engineering problems, evaluation of integrals is too

cumbersome if the integration is not possible within the finite

terms.,

methods of integration are employed.

If a parabola is assumed to pass through a number of pivotal points, then the area under the curve between the limits will be

approximately equal to the area under the integrand.

,.......y,y,y,y,y ....., 2i1ii1-i2-i ++


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Evenly spaced by h and taking the origin at x=xi, the straight

line

Will ass throu h the oints 0 and h rovided

( ) B x A x y += .

h y y A ii −= +

1

The approximate first degree parabola thus

ii y −+1

And the area under x curve between 0 and h is a roximated

ih

= . (53)

by

2

1+ − iih

hh y y (54)1

0 2

.

2

.. +iiih


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2−h hh1

0 2.

2.. ++=+== iii y y y

h x x y

This is known as the “Trapezoidal Rule Formula”, since it

approximates the area under one strip by area of a trapezoid.

0γ 1γ 2γ 3γ 4γ 5γ 6γ 7γ 8γ

x

Fig.4

a x = b x =


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If the second de ree arabola is assed( ) C x B x A x y ++= ..2

through points (-h, yi-1), (0,yi), (h, yi+1)

C h Bh +−=− ..2

C h Bh A y

C y

i

i

++=

=

+ ..2

1

(56)

Solving for A,B and C, we have

iii

h A

−

+−= −−

2 211

i yC

h

=

= −2


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The area under this arabola between –h and +h ives

( ) ( )112

43

..2.3

.2. −+

+

−++=+== ∫ iii

h

h y y y

hC h A

hdx x y B (57)

This is known as “Simpsons‘ 1/3 Rule Formula” for the area

under the two strips of width h.

m ar y, e area un er our s r ps e ween - an

becomes2 4+ −==

h h 582 3

−+− h


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finite differences and their applications

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