Differentiation-Discrete
Functions
Civil Engineering Majors
Authors: Autar Kaw, Sri Harsha Garapati
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
4/13/2015
http://numericalmethods.eng.usf.edu
1
Differentiation –Discrete
Functions
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Forward Difference
Approximation
lim f x  Δx   f x 
f x  
Δx  0
Δx
For a finite ' Δx'
f  x  x   f  x 
f  x  
x
3
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Graphical Representation Of
Forward Difference
Approximation
f(x)
x
x+Δx
Figure 1 Graphical Representation of forward difference approximation of first derivative.
4
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Example 1
To find the stress concentration around a hole in a plate under a uniform
stress, a finite difference program has been written that calculates the
radial and tangential displacements at different points in the plate. To
find the stresses and hence the stress concentration factor, one needs to
find the derivatives of these displacements. In Table 1 the radial
displacements , u , are given along the y-axes. The radius of the hole is
1.0 cm.
u
a) At x  0 if the radial strain,  r is given by  r 
, find the radial
r
strain at r  1.1cm using forward divided difference method.
b) If the tangential strain at r  1.1cm,  900 is given to you as
0

,
find
the
hoop
stress,
,
at
r

1
.
1
cm
,


90
   0.0029733

E
if   
( r    ), where E  200 GPa and   0.3 .
1  2
5
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Example 1 Cont.
Table 1 Radial displacement as a function of location.
r (cm)
6
u(cm)
1.0
−0.0010000
1.1
−0.0010689
1.2
−0.0011088
1.3
−0.0011326
1.4
−0.0011474
1.5
−0.0011574
1.6
−0.0011650
1.7
−0.0011718
1.8
−0.0011785
1.9
−0.0011857
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Example 1 Cont.
Solution:
u
r
ui 1  ui

r
ri  1.1
a)  r 
ri 1  1.2
r  ri 1  ri
 1.2  1.1
 0.1
ui 1  0.0011088
ui  0.0010689
7
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Example 1 Cont.
ui 1  ui
r
 0.0011088 (0.0010689)

0.1
 0.00039900
r 
b)
 
E
( r    )
2
1 
2  1011

(0.00039900 0.3  0.0029733)
2
1  0.3
 108.35106 Pa
8
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Direct Fit Polynomials
In this method, given ' n  1' data points x0 , y0 , x1 , y1 , x2 , y2 ,, xn , yn 
one can fit a
n th order polynomial given by
Pn x  a0  a1x    an 1xn 1  an xn
To find the first derivative,
Pnx  
dPn ( x )
 a1  2a 2 x    n  1a n 1 x n 2  na n x n 1
dx
Similarly other derivatives can be found.
9
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Example 2-Direct Fit Polynomials
To find the stress concentration around a hole in a plate under a uniform
stress, a finite difference program has been written that calculates the
radial and tangential displacements at different points in the plate. To
find the stresses and hence the stress concentration factor, one needs to
find the derivatives of these displacements. In Table 2 the radial
displacements, u , are given along y-axes. The radius of the hole is
1.0 cm.
u
a) At x  0 if the radial strain,  r is given by  r 
, find the radial
r
strain at r  1.1cm . Use a third order polynomial interpolant for
calculating the radial strain.
b) If the tangential strain at r  1.1cm,   900 is given to you as
0

,
find
the
hoop
stress,
,
at
r

1
.
1
cm,


90
   0.0029733

E
if   
( r    ), where E  200 GPa and   0.3 .
1  2
10
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Example 2-Direct Fit Polynomials cont.
Table 2 Radial displacement as a function of location.
r (cm)
11
u(cm)
1.0
−0.0010000
1.1
−0.0010689
1.2
−0.0011088
1.3
−0.0011326
1.4
−0.0011474
1.5
−0.0011574
1.6
−0.0011650
1.7
−0.0011718
1.8
−0.0011785
1.9
−0.0011857
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Example 2-Direct Fit Polynomials cont.
Solution:
For the third order polynomial (also called cubic interpolation), we choose the
displacement given by
ur   a0  a1r  a2 r 2  a3 r 3
Since we want to find the radial strain at r  1.1 cm , and we are using a
third order polynomial, we need to choose the four points closest to r  1.1
and that also bracket r  1.1 to evaluate it.
The four points are
r0  1.0, r1  1.1, r2  1.2, and r3  1.3
ro  1.0, uro   0.0010000
r1  1.1, ur1   0.0010689
r2  1.2, ur2   0.0011088
r3  1.3, ur3   0.0011326
12
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Example 2-Direct Fit Polynomials cont.
such that
u1.0  0.0010000 a0  a1 1  a2 1  a3 1
2
3
u1.1  0.0010689 a0  a1 1.1  a2 1.1  a3 1.1
2
3
u1.2  0.0011088 a0  a1 1.2  a2 1.2  a3 1.2
2
3
u1.3  0.0011326 a0  a1 1.3  a2 1.3  a3 1.3
2
3
Writing the four equations in matrix form, we have
1
1  a 0   0.0010000
1 1
1 1.1 1.21 1.331  a   0.0010689

 1   

1 1.2 1.44 1.728 a 2   0.0011088

  

a
1
1
.
3
1
.
69
2
.
197

0
.
0011326

 3  

13
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Example 2-Direct Fit Polynomials cont.
Solving the above four equations gives
a0  0.0041220
a1  -0.0011517
a2  0.0085450
a3  -0.0021500
Hence
ur   a0  a1r  a2 r 2  a3 r 3
 0.0041220 0.0011517r  0.0085450r 2  0.0021500r 3 , 1  r  1.3
14
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Example 2-Direct Fit Polynomials
cont.
Figure 2 Graph of Radial Displacement vs. Location
15
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Example 2-Direct Fit Polynomials cont.
The derivative of radial displacement at
u ' 1.1 
Given that
r  1.1cm is given by
d
u r  r 1.1
dr
ur   0.0041220 0.0011517r  0.0085450r 2  0.0021500r 3 , 1  r  1.3
d
u ' r   u r 
dr
d
 0.0041220 0.0011517r  0.0085450r 2  0.0021500r 3 
dr
 0.0011517 0.017090r  0.0064500r 2 , 1  r  1.3
u ' 1.1  0.0011517 0.017090(1.1) - 0.0064500
(1.1) 2
 0.00052250
cm/cm
 r  0.00052250cm/cm
16
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Example 2-Direct Fit Polynomials cont.
b)   
E
( r    )
2
1 
2  1011

(0.00052250 0.3  0.0029733)
1  0.32
 81.207106 Pa
17
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Lagrange Polynomial
In this method, given x1, y1 ,, xn , yn  , one can fit a n  1th order Lagrangian polynomial
given by
f n ( x) 
where ‘ n ’ in
n
 L ( x) f ( x )
i 0
i
i
f n (x) stands for the n th order polynomial that approximates the function
y  f (x) given at (n  1) data points as x0 , y0 , x1 , y1 ,......,xn1 , yn1 , xn , yn  , and
n
Li ( x)  
j 0
j i
x  xj
xi  x j
Li (x) a weighting function that includes a product of (n  1) terms with terms of
ji
18
omitted.
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Lagrange Polynomial Cont.
Then to find the first derivative, one can differentiate f n x  once, and so on
for other derivatives.
For example, the second order Lagrange polynomial passing through
x0 , y0 , x1, y1 , x2 , y2 
f 2 x  
is
x  x1 x  x2  f x   x  x0 x  x2  f x   x  x0 x  x1  f x 
x0  x1 x0  x2  0 x1  x0 x1  x2  1 x2  x0 x2  x1  2
Differentiating equation (2) gives
19
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Lagrange Polynomial Cont.
2 x  x0  x2 
2 x  x0  x1 
2 x  x1  x2 

f 2 x  
f x0  
f x1  
f x 
x0  x1 x0  x2 
x1  x0 x1  x2 
x2  x0 x2  x1  2
Differentiating again would give the second derivative as
f 2x  
20
2
x0  x1 x0  x2 
f x0  
2
x1  x0 x1  x2 
f x1  
2
x2  x0 x2  x1 
f x2 
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Example 3
To find the stress concentration around a hole in a plate under a uniform
stress, a finite difference program has been written that calculates the
radial and tangential displacements at different points in the plate. To
find the stresses and hence the stress concentration factor, one needs to
find the derivatives of these displacements. In Table 3 the radial
displacements u , are given along the y-axes. The radius of the hole is
1.0 cm
u
a) At x  0, if the radial strain,  r , is given by  r 
, find the radial
r
strain at r  1.1cm . Use a second order Lagrange polynomial
interpolant for calculating the radial strain.
b) If the tangential strain at r  1.1cm,  900 is given to you as
0
,
find
the
hoop
stress,
,
at
r

1
.
1
cm,


90

   0.0029733

E
if   
( r    ) , where E  200 GPa and   0.3 .
1  2
21
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Example 3 Cont.
Table 3 Radial displacement as a function of location.
r (cm)
22
u(cm)
1.0
−0.0010000
1.1
−0.0010689
1.2
−0.0011088
1.3
−0.0011326
1.4
−0.0011474
1.5
−0.0011574
1.6
−0.0011650
1.7
−0.0011718
1.8
−0.0011785
1.9
−0.0011857
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Example 3 Cont.
Solution:
For second order Lagrangian interpolation, we choose the radial
displacement given by
 r  r1  r  r2 
 r  r0  r  r2 
 r  r0  r  r1 

u (r0 )  


u (r1 )  
u (r2 )
u (r )  
r

r
r

r
r

r
r

r
r

r
r

r
 0 1  0 2 
 1 0  1 2 
 2 0  2 1 
Since we want to find the rate of change in the radial displacement at
r  1.1 cm , and we are using second order Lagrangian interpolation, we
need to choose the three points closest to r  1.1 cm that also bracket
r  1.1 cm to evaluate it.
The three points are r0  1.0, r1  1.1, and r2  1.2 .
r0  1.0, ur0   0.0010000
r1  1.1, ur1   0.0010689
r2  1.2, ur2   0.0011088
23
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Example 3 Cont.
The change in the radial displacement at r  1.1 cm is given by
du 1.1 d

u (r ) r 1.1
dr
dr
Hence
u ' r  
2r  r0  r2 
2r  r0  r1 
2r  r1  r2 
ur0  
ur1  
ur2 
r0  r1 r0  r2 
r1  r0 r1  r2 
r2  r0 r2  r1 
u ' 1.1 
21.1  1.1  1.2
 0.0010000  21.1  1.0  1.2  0.0010689
1.0  1.11.0  1.2
1.1  1.01.1  1.2

21.1  1.0  1.1
 0.0011088
1.2  1.01.2  1.1
 5 0.0010000  0 0.0010689  5 0.0011088
  0.00054400 cm/cm
24
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Example 3 Cont.
b)
 
E
( r    )
2
1 
2  1011

(0.00054400 0.3  0.0029733)
2
1  0.3
 76.481106 Pa
25
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit
http://numericalmethods.eng.usf.edu/topics/discrete_02
dif.html
THE END
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