finite differences and their applications
TRANSCRIPT
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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 −=−=
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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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