Fourth order improved finite difference approach to pure bending analysis of line continuum.
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The document presents a study employing an improved finite difference method (IFDM) for analyzing pure bending in linear continuum, comparing its accuracy to traditional analytical solutions. Results indicate that IFDM provides approximate solutions very close to exact values, with percentage differences ranging from 0 to 0.02%. This numerical method is shown to be effective and practical, especially given the difficulties associated with analytical methods.
![IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
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Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 584
Table -2: Line continua and their boundary conditions.
Linear Continuum Boundary Conditions
Pin – Roller Linear
Continuum ( P – R)
𝑤 0 = 0,
𝑑2
𝑤 0
𝑑𝑥2
= 𝑤 0
𝐼𝐼
= 0;
𝑤 𝑙 = 0,
𝑑𝑤 𝐿
𝑥𝑑2
= 𝑤 𝐿
𝐼𝐼
= 0.
Clamped - Clamped
Linear Continuum (C – C)
𝑤 0 = 0,
𝑑𝑤 0
𝑑𝑤
= 𝑤 0
𝐼
= 0;
𝑤 𝐿 = 0,
𝑑2
𝑤 𝐿
𝑑𝑥2
= 𝑤 𝐿
𝐼
= 0
Clamped – Roller Linear
Continuum (C – R)
𝑤 0 = 0,
𝑑𝑤 0
𝑑𝑤
= 𝑤 0
𝐼
= 0;
𝑤 𝐿 = 0,
𝑑𝑤(𝐿)
𝑑𝑥
= 𝑤(𝐿)
𝐼
= 0.
6. APPLICATION AND RESULT OF IMPROVED
FINITE DIFFERENCE ANALYSIS
The derived coefficients and patterns are applied to the
governing equation at each node using the 3 different cases
with the appropriate boundary conditions being satisfied to
generate the required matrix equations, from which the
deflections at each node were determined. The results of the
pure bending analysis are presented on table 3. Exact result
obtained from previous analysis is also shown in the table
for comparison.
Table -3a: Result data for case 1 and exact values
Linear
continuum
Exact result
Case 1
( n =3)
P-R 5𝑃𝐿4
384𝐸𝐼
0.013048𝑃𝐿4
𝐸𝐼
(0.21% Diff)
C-C 𝑃𝐿4
384𝐸𝐼
0.003348𝑃𝐿4
𝐸𝐼
(28.6% Diff)
C-R 2𝑃𝐿4
384𝐸𝐼
0.005896𝑃𝐿4
𝐸𝐼
(13.2% Diff)
Table -3b: Result data for case 2 and exact values
Linear
continuum
Exact result
Case 2
(n=5)
P-R 5𝑃𝐿4
384𝐸𝐼
0.013026𝑃𝐿4
𝐸𝐼
(0.04% Diff)
C-C 𝑃𝐿4
384𝐸𝐼
0.002919𝑃𝐿4
𝐸𝐼
(12.2% Diff)
C-R 2𝑃𝐿4
384𝐸𝐼
0.005499𝑃𝐿4
𝐸𝐿
(5.6% Diff)
Table -3c: Result data for case 3 and exact values
Linear
continuum
Exact result
Case 3
(n=7)
P-R 5𝑃𝐿4
384𝐸𝐼
0.013021𝑃𝐿4
𝐸𝐼
(0% Diff)
C-C 𝑃𝐿4
384𝐸𝐼
0.002779𝑃𝐿4
𝐸𝐼
(6.7% diff)
C-R 2𝑃𝐿4
384𝐸𝐼
0.005371𝑃𝐿4
𝐸𝐼
(3.13% Diff)
7. CONCLUSIONS
The result for the pure bending analysis of line continua of
three boundary conditions using Improved Finite Difference
are presented on table 3. It can be seen from the analysis that
the results from P-R boundary condition are virtually the
same as exact results. Results emerging from C-C boundary
condition shows a percentage difference ranging from 6.7%
- 28.6% while that of C-R boundary condition ranges from
3.13% - 13.2% from the exact result.
Also, these results suggest that as the number of nodes (or
division) increases, the more accurate the result becomes.
Finally, the results from the pattern developed from this
improved finite difference analysis are effective and is
recommended for use in structural engineering.
REFERENCES
[1]. Alexander Chajes (1974). Structural Stability Theory
New Jersey. Prentice – Hall, Inc.
[2]. Awele, M., Ayodele, J.C., and Osaisai, F.E. (2003).
Numerical Methods for Scientists and Engineers.
Ibadan. Beaver Publications.
[3]. Ibearugbulem, O. M., Ettu, L. O., Ezeh, J. C. and Anya,
U. C. (2013). New Shape function for
Analysis of Line continuum by Direct Variational
Calculus. International Journal of Engineering &
Science, Vol. 2, issue 3; Pp. 66 – 71, ISBN 2319 –
1805.
[4]. Goodwine, B. (2010). Engineering Differential
Equations: Theory and Applications. New York:
Springer Science + Business Media, LLC2011.
[5]. Szilard, R. (2004). Theories and Application of Plate
Analysis. New Jersey: John Wiley and Sons Inc.
[6]. Ugural, A.C (1999). Stress in Plates and Shells
(2nd
Ed.).Singapore: McGraw-Hill.](https://image.slidesharecdn.com/fourth-orderimprovedfinitedifferenceapproachtopurebendinganalysisoflinecontinuum-160812135833/85/Fourth-order-improved-finite-difference-approach-to-pure-bending-analysis-of-line-continuum-6-320.jpg)
![IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
_______________________________________________________________________________________
Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 585
[7]. Yoo, H.C. and Lee, C.S. (2011). Stability of Structures.
Principles and Applications. USA:
Butterworth- Heinemann. Elsevier Inc.](https://image.slidesharecdn.com/fourth-orderimprovedfinitedifferenceapproachtopurebendinganalysisoflinecontinuum-160812135833/85/Fourth-order-improved-finite-difference-approach-to-pure-bending-analysis-of-line-continuum-7-320.jpg)
Fourth order improved finite difference approach to pure bending analysis of line continuum.
- 1. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 579 FOURTH - ORDER IMPROVED FINITE DIFFERENCE APPROACH TO PURE BENDING ANALYSIS OF LINE CONTINUUM. O.M. Ibearugbulem1 , J. C. Ezeh2 , U.O. Okpara3 1, 2, 3 Department of Civil Engineering, Federal University of Technology, Owerri, Nigeria. Abstract Analytical Methods provides exact solutions which have restrictions due to their inherent difficulties. On the other hand, Numerical methods provide approximate solution to governing differential equations. In this research, Improved Finite Difference Method (IFDM) using Polynomial series was used to develop a finite difference pattern. The patterns were applied in analyzing pure bending of line continuum with various boundary conditions. The results obtained from this numerical method were compared with those from exact method to check the accuracy of the solution. The results were found to be identical and very close to the exact results, with percentage difference ranging from 0 – 0.02%. Hence, Improved Finite Difference Method provides simple and approximate solution that are close to the exact value for pure bending analysis of line continuum. Key Words: Improved Finite Difference, Polynomial Series, Linear continuum, Numerical Methods… --------------------------------------------------------------------***--------------------------------------------------------------------- 1. INTRODUCTION The behavior of flexural linear continuum was defined by formulating governing differential equations. These problems were solved by means of various analytical approaches to obtain exact solution (Chaje, 1974). Due to their inherent difficulties, such analytical and exact solutions have restrictions (Ibearugbulem et al., 20I3; Iyengar, 1988). In areas of considerable practical interest, they are either difficult or impossible to obtain. Hence, numerical methods are engaged to obtain approximate solutions (Ugural, 1999). Thus Finite Difference Method is regarded as a numerical method that has its merit due to its straight forward approach and minimum requirement on hardware. 1.1 FINITE DIFFERENCE METHOD The Finite Difference Method is one of the most general numerical techniques. It refers to the process of replacing the partial derivatives by finite difference quotient and then obtaining solution of resulting system of algebraic equations (Yoo and Lee, 2011). In applying this method, the derivatives in the governing differential equation under consideration are replaced by differences at selected nodes. These nodes make the finite difference mesh. The first order expansion gives the form 𝑓𝑖 𝐼 𝑥 = 𝑓 𝑖+1 −𝑓 𝑖 ∆𝑥 …………….. (1) Equation (1) is known as the First Forward Difference and 𝑓𝑖 𝐼 𝑥 = 𝑓 𝑖− 𝑓 𝑖−1 ∆𝑥 ………… (2) Equation (2) is known as the First Backward Difference Where ∆x means grid spacing/ length of segments along member; i means station number. Hence, the Central Difference is given as 𝑓𝑖 𝐼 𝑥 = 𝑓 𝑖+1− 𝑓 𝑖−1 2∆𝑥 …………. (3) 2. POLYNOMIAL SERIES. Many numerical methods are based on Polynomial series expansion which is given as a function, f(x) 𝑓 𝑥 = 𝑓 𝑥𝑖 + 𝑥−𝑥 𝑖 𝑘 𝑖 ∞ 𝑘=𝑖 𝑓 𝑘 (𝑥𝑖) ..………. (4) Where f (k) = k th derivative of the function. Equation (4) can be re-written as 𝑓 𝑥 = 𝑓 𝑥𝑖 + 𝑥 − 𝑥𝑖 1! 𝑓 𝐼 𝑥𝑖 + 𝑥 − 𝑥𝑖 2 2! 𝑓 𝐼𝐼 𝑥𝑖 + … … . . + 𝑥 – 𝑥𝑖 𝑛 𝑛! 𝑓 𝑛 𝑥𝑖 … … … … … … .. (5) Applying equation (5) to expand the function f(x) at points (x+1) and (x-1) yields
- 2. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 580 𝑓 𝑥+1 = 𝑓 𝑥𝑖 + ∆𝑥 1! 𝑓 𝐼 𝑥𝑖 + ∆𝑥 2 2! 𝑓 𝐼𝐼 𝑥𝑖 + ∆𝑥 3 3! 𝑓 𝐼𝐼𝐼 𝑥𝑖 + ∆𝑥 4 4! 𝑓 𝐼𝑉 𝑥𝑖 + ⋯ + ∆𝑥 𝑛 𝑛! 𝑓 𝑛 𝑥𝑖 … … (6) 𝑓 𝑥−1 = 𝑓 𝑥𝑖 − ∆𝑥 1! 𝑓 𝐼 𝑥𝑖 + ∆𝑥 2 2! 𝑓 𝐼𝐼 𝑥𝑖 − ∆𝑥 3 3! 𝑓 𝐼𝐼𝐼 𝑥𝑖 + ∆𝑥 4 4! 𝑓 𝐼𝑉 𝑥𝑖 − ⋯ + −∆𝑥 𝑛 𝑛! 𝑓 𝑛 𝑥𝑖 … … . . … (7) 3. IMPROVED FINITE DIFFERENCE FORMULATION. For the derivation of improved derivatives, the polynomial series is applied once the grid parts are evenly spaced. Figure 1 schematically indicates the grid parts for central differences. Figure 1: Computational grid for finite difference approximation h is the grid spacing / segment length. Taking y3 as the pivotal point and expanding the function y3 using the series and truncating at the fourth-order derivative yIV gave; 𝑦4 = 𝑦3 + ℎ𝑦3 𝐼 + ℎ2 2 𝑦3 𝐼𝐼 + ℎ3 6 𝑦3 𝐼𝐼𝐼 + ℎ4 24 𝑦3 𝐼𝑉 … … (8) 𝑦2 = 𝑦3 − ℎ𝑦3 𝐼 + ℎ2 2 𝑦3 𝐼𝐼 − ℎ3 6 𝑦3 𝐼𝐼𝐼 + ℎ4 24 𝑦3 𝐼𝑉 … … (9) Subtracting equations (9) from equation (8) gave: 𝑦4 − 𝑦2 = 2ℎ𝑦3 𝐼 + ℎ3 3 𝑦3 𝐼𝐼𝐼 . That is 𝑦3 𝐼 = 𝑦4 2ℎ − 𝑦2 2ℎ − ℎ2 6 𝑦3 𝐼𝐼𝐼 … … … … (10) Adding equations (8) and (9) gave: 𝑦4 + 𝑦2 = 2𝑦3 + ℎ2 𝑦3 𝐼𝐼 + ℎ4 12 𝑦3 𝐼𝑉 . That is 𝑦3 𝐼𝐼 = 1 ℎ2 𝑦2 − 2𝑦3 + 𝑦4 − ℎ2 12 𝑦3 𝐼𝑉 … … … … (11) Taking differences between nodes y3 and y5 and y3 and y1 equations (8) and (9) shall give: 𝑦5 = 𝑦3 + 2ℎ𝑦3 𝐼 + 4ℎ2 2 𝑦3 𝐼𝐼 + 8ℎ3 6 𝑦3 𝐼𝐼𝐼 + 16ℎ4 24 𝑦3 𝐼𝑉 … (12) 𝑦1 = 𝑦3 – 2ℎ𝑦3 𝐼 + 4ℎ2 2 𝑦3 𝐼𝐼 − 8ℎ3 6 𝑦3 𝐼𝐼𝐼 + 16ℎ4 24 𝑦3 𝐼𝑉 … (13) Subtracting equation (13) from equation (12) gives: 𝑦5 − 𝑦1 = 4ℎ𝑦3 𝐼 + 8ℎ3 3 𝑦3 𝐼𝐼𝐼 … … … (14) Adding equations (12) and (13) gives: 𝑦3 + 𝑦1 = 2𝑦3 + 4ℎ2 𝑦3 𝐼𝐼 + 16ℎ4 12 𝑦3 𝐼𝑉 … … (15) From equation (14), we have: 𝑦3 𝐼 = 1 4ℎ 𝑦5 + 𝑦1 − 2ℎ2 3 𝑦3 𝐼𝐼𝐼 … … … (16) From equation (15), we have: 𝑦3 𝐼𝐼 = 1 4ℎ2 𝑦5 + 𝑦1 − 2𝑦3 − 4ℎ2 12 𝑦3 𝐼𝑉 … … (17) Subtracting equation (16) from equation (10) gave: 0 = 𝑦4 2ℎ − 𝑦5 4ℎ − 𝑦2 2ℎ + 𝑦1 4ℎ − ℎ2 6 𝑦3 𝐼𝐼𝐼 + 2ℎ2 3 𝑦3 𝐼𝐼𝐼 That is; 0 = 3 𝑦4 ℎ − 3 𝑦5 2ℎ − 3 𝑦2 ℎ + 3 𝑦1 2ℎ − ℎ2 𝑦3 𝐼𝐼𝐼 + 4ℎ2 𝑦3 𝐼𝐼𝐼 4 5 6 y h h h h h h0 1 2 3 4 5 6 y0 h h h h h h yy1 y2 yy3 yy4 yy5 6 0 1 2 3
- 3. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 581 . This implies; 3ℎ2 𝑦3 𝐼𝐼𝐼 = −3𝑦1 2ℎ + 3𝑦2 ℎ − 3𝑦4 ℎ + 3𝑦5 2ℎ ∴ 𝑦3 𝐼𝐼𝐼 = 1 2ℎ3 −𝑦1 + 2𝑦2 − 2𝑦4 + 𝑦5 … … … (18) Subtracting equation (17) from equation (11) gives: 0 = 𝑦2 ℎ2 − 2𝑦3 ℎ2 + 𝑦4 ℎ2 − ℎ2 𝑦3 𝐼𝑉 12 − 𝑦5 4ℎ2 + 2𝑦3 4ℎ2 + 4ℎ2 12 𝑦3 𝐼𝑉 That is; 0 = 12𝑦2 − 24𝑦3 + 12𝑦4 − ℎ4 𝑦3 𝐼𝑉 − 3𝑦5 − 3𝑦1 + 6𝑦3 + 4ℎ4 𝑦3 𝐼𝑉 This implies; 3ℎ4 𝑦3 𝐼𝑉 = 3𝑦1 − 12𝑦2 + 18𝑦3 − 12𝑦4 + 3𝑦5 ∴ 𝑦3 𝐼𝑉 = 1 ℎ4 𝑦1 − 4𝑦2 + 6𝑦3 − 4𝑦4 + 𝑦5 … … … (19) Substituting equations (18) into equation (10) gives: 𝑦3 𝐼 = 𝑦4 2ℎ − 𝑦2 2ℎ − ℎ2 6 1 2ℎ3 −𝑦1 + 2𝑦2 − 2𝑦4 + 𝑦5 That is; 𝑦3 𝐼 = 𝑦4 2ℎ − 𝑦2 2ℎ + 𝑦1 12ℎ − 𝑦2 6ℎ + 𝑦4 6ℎ − 𝑦5 12ℎ ∴ 𝒚 𝟑 𝑰 = 𝟏 𝟏𝟐𝒉 𝒚 𝟏 − 𝟖𝒚 𝟐 + 𝟖𝒚 𝟒 − 𝒚 𝟓 … … … … (20) Equation (20) shows the first derivative of the improved finite difference. Substituting equation (19) into equation (11) gives: 𝑦3 𝐼𝐼 = 𝑦2 ℎ2 − 2𝑦3 ℎ2 + 𝑦4 ℎ2 − ℎ2 12 1 ℎ4 𝑦1 − 4𝑦2 + 6𝑦3 − 4𝑦4 + 𝑦5 That is; 𝑦3 𝐼𝐼 = 𝑦2 ℎ2 − 2𝑦3 ℎ2 + 𝑦4 ℎ2 − 𝑦1 12ℎ2 + 𝑦2 3ℎ2 − 𝑦3 2ℎ2 + 𝑦4 3ℎ2 − 𝑦5 12ℎ2 ∴ 𝒚 𝟑 𝑰𝑰 = 𝟏 𝟏𝟐𝒉 𝟐 −𝒚 𝟏 + 𝟏𝟔𝒚 𝟐 − 𝟑𝟎𝒚 𝟑 + 𝟏𝟔𝒚 𝟒 − 𝒚 𝟓 … … (21) Equation (21) shows the second derivatives of the improved finite difference. From equation (8), we can obtain: 𝑦4 𝐼 = 𝑦3 𝐼 + ℎ𝑦3 𝐼𝐼 + ℎ2 2 𝑦3 𝐼𝐼𝐼 + ℎ3 6 𝑦3 𝐼𝑉 + ℎ4 24 𝑦3 𝑉 … … … (22) From equation (9), we can obtain: 𝑦2 𝐼 = 𝑦3 𝐼 − ℎ𝑦3 𝐼𝐼 + ℎ2 2 𝑦3 𝐼𝐼𝐼 − ℎ3 2 𝑦3 𝐼𝑉 + ℎ4 24 𝑦3 𝐼𝑉 … … … (23) Subtracting equation (23) from equation (22) gives: 𝑦4 1 − 𝑦2 1 = 2ℎ𝑦3 𝐼𝐼 + 𝑦 𝐼𝐼 3 𝑦3 𝐼𝑉 ∴ 𝑦3 𝐼𝐼 = 1 2ℎ 𝑦4 𝐼 − 𝑦2 𝐼 − ℎ2 6 𝑦3 𝐼𝑉 … … … … . (24) Adding equation (22) and equation (23) gives: 𝑦4 𝐼 + 𝑦2 𝐼 = 2𝑦3 𝐼 + ℎ2 𝑦3 𝐼𝐼𝐼 + ℎ4 12 𝑦3 𝑉 ∴ 𝑦3 𝐼𝐼𝐼 = 1 ℎ2 𝑦2 𝐼 − 2𝑦3 𝐼 + 𝑦4 𝐼 − ℎ2 12 𝑦3 𝑉 … … … (25) From equation (12) and equation (13), we have: 𝑦5 𝐼 = 𝑦3 𝐼 + 2ℎ𝑦3 𝐼𝐼 + 4ℎ2 2 𝑦3 𝐼𝐼𝐼 + 8ℎ3 6 𝑦3 𝐼𝑉 + 16ℎ4 24 𝑦3 𝑉 … … (26) 𝑦1 𝐼 = 𝑦3 𝐼 − 2ℎ𝑦3 𝐼𝐼 + 4ℎ2 2 𝑦3 𝐼𝐼𝐼 − 8ℎ3 6 𝑦3 𝐼𝑉 + 16ℎ4 24 𝑦3 𝑉 … … … (27) Subtracting equation (27) from equation (26) gives: 𝑦5 𝐼 − 𝑦1 𝐼 = 4ℎ𝑦3 𝐼𝐼 + 8ℎ3 3 𝑦3 𝐼𝑉 … … … . (28) Adding equation (26) and equation (27) gives: 𝑦5 𝐼 + 𝑦1 𝐼 = 2𝑦3 𝐼 + 4ℎ𝑦2 𝑦3 𝐼𝐼𝐼 + 16ℎ4 12 𝑦3 𝑉 … … … … (29) From equation (28), we can obtain: 𝑦3 𝐼𝐼 = 1 4ℎ 𝑦5 𝐼 − 𝑦1 𝐼 − 2ℎ2 3 𝑦3 𝐼𝑉 … … … . . (30) From equation (29), we can obtain: 𝑦3 𝐼𝐼𝐼 = 1 4ℎ2 𝑦1 𝐼 − 2𝑦3 𝐼 + 𝑦5 𝐼 − 4ℎ2 12 𝑦3 𝑉 … … (31) Subtracting equation (30) from equation (24) gives: 0 = 𝑦4 𝐼 2ℎ − 𝑦2 𝐼 2ℎ − ℎ2 6 𝑦3 𝐼𝑉 − 𝑦5 𝐼 4ℎ + 𝑦1 𝐼 4ℎ + 2ℎ2 3 𝑦3 𝐼𝑉 That is; 0 = 6𝑦4 𝐼 ℎ3 − 6𝑦2 𝐼 ℎ3 − 2𝑦3 𝐼𝑉 − 3𝑦5 𝐼 ℎ3 + 3𝑦1 𝐼 ℎ3 + 8𝑦3 𝐼𝑉 This implies that 6𝑦3 𝐼𝑉 = 1 ℎ3 −3𝑦1 𝐼 + 6𝑦2 𝐼 − 6𝑦4 𝐼 + 3𝑦5 𝐼 ∴ 𝑦3 𝐼𝑉 = 1 2ℎ3 −𝑦1 𝐼 + 2𝑦2 𝐼 − 2𝑦4 𝐼 + 𝑦5 𝐼 … … … (32) Subtracting equation (31) from equation (25) gives: 0 = 𝑦2 𝐼 ℎ2 − 2𝑦3 𝐼 ℎ2 + 𝑦4 𝐼 ℎ2 − ℎ4 12 𝑦3 𝑉 − 𝑦1 𝐼 4ℎ2 + 2𝑦3 𝐼 4ℎ2 − 𝑦5 𝐼 4ℎ2 + 4ℎ2 12 𝑦3 𝑉 That is ℎ2 𝑦3 𝑉 = 𝑦1 𝐼 ℎ2 − 4𝑦2 𝐼 ℎ2 + 6𝑦5 𝐼 ℎ2 − 4𝑦4 𝐼 ℎ2 + 𝑦5 𝐼 ℎ2 ∴ 𝑦3 𝐼 = 1 ℎ4 𝑦1 𝐼 − 4𝑦2 𝐼 + 6𝑦3 𝐼 − 4𝑦4 𝐼 + 𝑦5 𝐼 … … … (33) Substituting equation (32) into equation (24), we have: 𝑦3 𝐼𝐼 = 1 2ℎ 𝑦4 𝐼 − 𝑦2 𝐼 − ℎ2 6 1 2ℎ3 − 𝑦1 𝐼 + 2𝑦2 𝐼 − 2𝑦4 𝐼 + 𝑦5 𝐼
- 4. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 582 That is; = 𝑦4 𝐼 2ℎ − 𝑦2 𝐼 2ℎ + 1 12ℎ 𝑦1 𝐼 − 1 6ℎ 𝑦2 𝐼 + 1 6ℎ 𝑦4 𝐼 − 1 12ℎ 𝑦5 𝐼 ∴ 𝑦3 𝐼𝐼 = 1 12ℎ 𝑦1 𝐼 − 8𝑦1 𝐼 + 8𝑦4 𝐼 − 𝑦5 𝐼 … … … (34) Substituting equation (33) into equation (25), we have: 𝑦3 𝐼𝐼𝐼 = 1 ℎ2 𝑦2 𝐼 − 2𝑦3 𝐼 + 𝑦4 𝐼 − ℎ2 12 1 ℎ4 𝑦1 𝐼 − 4𝑦2 𝐼 + 6𝑦3 𝐼 − 4𝑦4 𝐼 + 𝑦5 𝐼 That is; = 1 ℎ2 𝑦2 𝐼 − 2𝑦3 𝐼 + 𝑦4 𝐼 − 1 12ℎ2 𝑦1 𝐼 − 4𝑦2 𝐼 + 6𝑦3 𝐼 − 4𝑦4 𝐼 + 𝑦5 𝐼 ∴ 𝑦3 𝐼𝐼𝐼 = 1 12ℎ2 𝑦1 𝐼 − 16𝑦2 𝐼 − 30𝑦3 𝐼 + 16𝑦4 𝐼 − 𝑦5 𝐼 … … … (35) Ignoring higher order term in equation (10) gives: 𝑦3 𝐼 = 1 2ℎ 𝑦4 − 𝑦2 … … … … … (36) Similarly; 𝑦2 𝐼 = 1 2ℎ 𝑦3 − 𝑦1 … … … … … (37) 𝑦1 𝐼 = 1 2ℎ 𝑦2 − 𝑦2 … … … … … (38) 𝑦4 𝐼 = 1 2ℎ 𝑦5 − 𝑦3 … … … … … (39) 𝑦5 𝐼 = 1 2ℎ 𝑦6 − 𝑦4 … … … … … (40) Substituting equations (36) to (34) into equation (35) gives: 𝑦3 𝐼𝐼𝐼 = 1 24ℎ3 (– 𝑦2 + 𝑦0 + 16𝑦3 − 16𝑦1 − 30𝑦4 + 30𝑦2 + 16𝑦5 − 16𝑦3 − 𝑦6 + 𝑦4) That is; = 1 24ℎ3 𝑦0 − 16𝑦1 + 29𝑦2 − 29𝑦4 + 16𝑦5 − 𝑦6 ∴ 𝒚 𝟑 𝑰𝑰𝑰 = 𝟏 𝟐𝟒𝒉 𝟑 (𝒚 𝟎 − 𝟏𝟔𝒚 𝟏 + 𝟐𝟗𝒚 𝟐 + 𝟎𝒚 𝟑 − 𝟐𝟗𝒚 𝟒 +𝟏𝟔𝒚 𝟓 − 𝒚 𝟔) …. (41) Equation (41) shows the third derivatives of the improved finite difference. From equation (22) and (23), we can obtain: 𝑦4 𝐼𝐼 = 𝑦3 𝐼𝐼 + ℎ𝑦3 𝐼𝐼𝐼 + ℎ2 2 𝑦3 𝐼𝑉 + ℎ3 6 𝑦3 𝑉 + ℎ4 24 𝑦3 𝑉𝐼 … … (42) 𝑦2 𝐼𝐼 = 𝑦3 𝐼𝐼 − ℎ𝑦3 𝐼𝐼𝐼 + ℎ2 2 𝑦3 𝐼𝑉 − ℎ3 6 𝑦3 𝑉 + ℎ4 24 𝑦3 𝑉𝐼 … … (43) Adding equations (42) and (43) gives: 𝑦4 𝐼𝐼 + 𝑦2 𝐼𝐼 = 2𝑦3 𝐼𝐼 + ℎ2 𝑦3 𝐼𝑉 + ℎ4 12 𝑦3 𝑉𝐼 This implies that 𝑦3 𝐼𝑉 = 1 ℎ2 𝑦2 𝐼𝐼 − 2𝑦3 𝐼𝐼 + 𝑦4 𝐼𝐼 − ℎ2 12 𝑦3 𝑉𝐼 … … … (44) From equations (26) and (27), we can obtain; 𝑦5 𝐼𝐼 = 𝑦3 𝐼𝐼 + 2ℎ𝑦3 𝐼𝐼𝐼 + 4ℎ2 2 𝑦3 𝐼𝑉 + 8ℎ3 6 𝑦3 𝑉 + 16ℎ2 24 𝑦3 𝑉𝐼 … … . (45) 𝑦1 𝐼𝐼 = 𝑦3 𝐼𝐼 − 2ℎ𝑦3 𝐼𝐼𝐼 + 4ℎ2 2 𝑦3 𝐼𝑉 − 8ℎ3 6 𝑦3 𝑉 + 16ℎ4 24 𝑦3 𝑉𝐼 … … . (46) Adding equations (45) and (46) gives: 𝑦5 𝐼𝐼 + 𝑦1 𝐼𝐼 = 2𝑦3 𝐼𝐼 + 4ℎ2 𝑦3 𝐼𝑉 + 16ℎ4 12 𝑦3 𝑉𝐼 This implies that: 𝑦3 𝐼𝑉 = 1 4ℎ2 𝑦1 𝐼𝐼 − 2𝑦3 𝐼𝐼 + 𝑦5 𝐼𝐼 − 4ℎ2 12 𝑦3 𝑉𝐼 … … . . (47) Subtracting equation (47) from equation (44) gives: 𝑦2 𝐼𝐼 ℎ2 − 2𝑦3 𝐼𝐼 ℎ2 + 𝑦4 𝐼𝐼 ℎ2 − ℎ2 12 𝑦3 𝑉𝐼 − 𝑦1 𝐼𝐼 4ℎ2 + 2𝑦3 𝐼𝐼 4ℎ2 − 𝑦5 𝐼𝐼 4ℎ2 + 4ℎ2 12 𝑦3 𝑉𝐼 = 0 This implies that: 𝑦3 𝑉𝐼 = − 4𝑦2 𝐼𝐼 ℎ4 + 6𝑦3 𝐼𝐼 ℎ4 − 4𝑦4 𝐼𝐼 ℎ4 + 𝑦1 𝐼𝐼 ℎ4 + 𝑦5 𝐼𝐼 ℎ4 ∴ 𝑦3 𝑉𝐼 = 1 ℎ4 𝑦1 𝐼𝐼 − 4𝑦2 𝐼𝐼 + 6𝑦3 𝐼𝐼 − 4𝑦4 𝐼𝐼 + 𝑦5 𝐼𝐼 … (48) Substituting equation (48) into equation (47) gives: 𝑦3 𝐼𝑉 = 1 4ℎ2 𝑦1 𝐼𝐼 − 2𝑦3 𝐼𝐼 + 𝑦5 𝐼𝐼 − 4ℎ2 12 1 ℎ4 𝑦1 𝐼𝐼 − 4𝑦2 𝐼𝐼 + 6𝑦3 𝐼𝐼 − 4𝑦4 𝐼𝐼 + 𝑦5 𝐼𝐼 That is; = 𝑦1 𝐼𝐼 4ℎ2 − 2𝑦3 𝐼𝐼 4ℎ2 + 𝑦5 𝐼𝐼 4ℎ2 − 𝑦1 𝐼𝐼 3ℎ2 + 4𝑦2 𝐼𝐼 3ℎ2 − 6𝑦3 𝐼𝐼 3ℎ2 + 4𝑦4 𝐼𝐼 3ℎ2 − 𝑦5 𝐼𝐼 3ℎ2 That is; = −𝑦1 𝐼𝐼 12ℎ2 + 4𝑦2 𝐼𝐼 3ℎ2 − 5𝑦3 𝐼𝐼 2ℎ2 + 4𝑦4 𝐼𝐼 3ℎ2 − 𝑦5 𝐼𝐼 12ℎ2
- 5. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 583 ∴ 𝑦3 𝐼𝑉 = 1 12ℎ2 −𝑦1 𝐼𝐼 + 16𝑦2 𝐼𝐼 − 30𝑦3 𝐼𝐼 + 16𝑦4 𝐼𝐼 − 𝑦5 𝐼𝐼 … . . (49) Ignoring higher order term in equation (11) gives: 𝑦3 𝐼𝐼 = 1 ℎ2 𝑦2 − 2𝑦3 + 𝑦4 … … … (50) Similarly; 𝑦2 𝐼𝐼 = 1 ℎ2 𝑦1 − 2𝑦2 + 𝑦3 … … … (51) 𝑦1 𝐼𝐼 = 1 ℎ2 𝑦0 − 2𝑦1 + 𝑦2 … … … (52) 𝑦4 𝐼𝐼 = 1 ℎ2 𝑦3 − 2𝑦4 + 𝑦5 … … … (53) 𝑦5 𝐼𝐼 = 1 ℎ2 𝑦4 − 2𝑦5 + 𝑦6 … … … (54) Substituting equations (50) to (54) into equation (49) gives: 𝑦3 𝐼𝑉 = 1 12ℎ4 (– 𝑦0 + 2𝑦1 − 𝑦2 + 16𝑦1 − 32𝑦2 +16𝑦3 − 30𝑦2 + 60𝑦3 − 30𝑦4 + 16𝑦3 −32𝑦4 + 16𝑦5 − 𝑦4 + 2𝑦5 − 𝑦6) That is; = 1 12ℎ4 −𝑦0 + 18𝑦1 − 63𝑦2 + 92𝑦3 − 63𝑦4 + 18𝑦5 − 𝑦6 ∴ 𝒚 𝟑 𝑰𝑽 = 𝟏 𝟏𝟐𝒉 𝟒 −𝒚 𝟎 + 𝟏𝟖𝒚 𝟏 − 𝟔𝟑𝒚 𝟐 + 𝟗𝟐𝒚 𝟑 − 𝟔𝟑𝒚 𝟒 + 𝟏𝟖𝒚 𝟓 − 𝒚 𝟔 … (𝟓𝟓) Equation (55) shows the fourth derivative of the improved finite difference. The coefficients of the corresponding function values in the improved finite difference expressions of the first-order, second-order, third-order and fourth-order derivatives as given in equations (20), (21), (41) and (55) are used to develop finite difference patterns and are schematically given on table 1. Table -1: Schematic representation of Higher order Finite Difference Expression. 4. BASIC EQUATION FOR PURE BENDING. The fourth – order differential equation of a bent line continuum was given by Ugural (Ugural, 1999) as: 𝐸𝐼 𝑑4 𝑤 𝑑𝑥4 = 𝑃 Where w is the deflection and P is the applied uniformly distributed load per meter length of the continuum. 5. LINEAR CONTINUUM WITH VARIOUS BOUNDARY CONDITIONS. Three line continua with different boundary conditions were studied in this work namely: Pin – Roller supports (P – R), Clamped – Clamped supports (C – C) and Clamped – Roller supports (C – R). Their boundary conditions are shown in table 2. Also, the line continua were studied using 3 different cases namely: Case 1: Dividing the linear continuum into 4 equal lengths Case 2: Dividing the linear continuum into 6 equal lengths Case 3: Dividing the linear continuum into 8 equal lengths 𝒚 𝟑 𝒏 COEFFICIENTS 𝟏𝟐𝒉 𝟐 𝒚 𝟑 𝑰𝑰 𝟏𝟐𝒉𝒚 𝟑 𝑰 𝟐𝟒𝒉 𝟑 𝒚 𝟑 𝑰𝑰𝑰 𝟏𝟐𝒉 𝟒 𝒚 𝟑 𝑰𝑽 1 -8 0 8 -1 -1 16 -1-30 16 029-1 16 -1-29 16 92-63-1 18 -1-63 18
- 6. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 584 Table -2: Line continua and their boundary conditions. Linear Continuum Boundary Conditions Pin – Roller Linear Continuum ( P – R) 𝑤 0 = 0, 𝑑2 𝑤 0 𝑑𝑥2 = 𝑤 0 𝐼𝐼 = 0; 𝑤 𝑙 = 0, 𝑑𝑤 𝐿 𝑥𝑑2 = 𝑤 𝐿 𝐼𝐼 = 0. Clamped - Clamped Linear Continuum (C – C) 𝑤 0 = 0, 𝑑𝑤 0 𝑑𝑤 = 𝑤 0 𝐼 = 0; 𝑤 𝐿 = 0, 𝑑2 𝑤 𝐿 𝑑𝑥2 = 𝑤 𝐿 𝐼 = 0 Clamped – Roller Linear Continuum (C – R) 𝑤 0 = 0, 𝑑𝑤 0 𝑑𝑤 = 𝑤 0 𝐼 = 0; 𝑤 𝐿 = 0, 𝑑𝑤(𝐿) 𝑑𝑥 = 𝑤(𝐿) 𝐼 = 0. 6. APPLICATION AND RESULT OF IMPROVED FINITE DIFFERENCE ANALYSIS The derived coefficients and patterns are applied to the governing equation at each node using the 3 different cases with the appropriate boundary conditions being satisfied to generate the required matrix equations, from which the deflections at each node were determined. The results of the pure bending analysis are presented on table 3. Exact result obtained from previous analysis is also shown in the table for comparison. Table -3a: Result data for case 1 and exact values Linear continuum Exact result Case 1 ( n =3) P-R 5𝑃𝐿4 384𝐸𝐼 0.013048𝑃𝐿4 𝐸𝐼 (0.21% Diff) C-C 𝑃𝐿4 384𝐸𝐼 0.003348𝑃𝐿4 𝐸𝐼 (28.6% Diff) C-R 2𝑃𝐿4 384𝐸𝐼 0.005896𝑃𝐿4 𝐸𝐼 (13.2% Diff) Table -3b: Result data for case 2 and exact values Linear continuum Exact result Case 2 (n=5) P-R 5𝑃𝐿4 384𝐸𝐼 0.013026𝑃𝐿4 𝐸𝐼 (0.04% Diff) C-C 𝑃𝐿4 384𝐸𝐼 0.002919𝑃𝐿4 𝐸𝐼 (12.2% Diff) C-R 2𝑃𝐿4 384𝐸𝐼 0.005499𝑃𝐿4 𝐸𝐿 (5.6% Diff) Table -3c: Result data for case 3 and exact values Linear continuum Exact result Case 3 (n=7) P-R 5𝑃𝐿4 384𝐸𝐼 0.013021𝑃𝐿4 𝐸𝐼 (0% Diff) C-C 𝑃𝐿4 384𝐸𝐼 0.002779𝑃𝐿4 𝐸𝐼 (6.7% diff) C-R 2𝑃𝐿4 384𝐸𝐼 0.005371𝑃𝐿4 𝐸𝐼 (3.13% Diff) 7. CONCLUSIONS The result for the pure bending analysis of line continua of three boundary conditions using Improved Finite Difference are presented on table 3. It can be seen from the analysis that the results from P-R boundary condition are virtually the same as exact results. Results emerging from C-C boundary condition shows a percentage difference ranging from 6.7% - 28.6% while that of C-R boundary condition ranges from 3.13% - 13.2% from the exact result. Also, these results suggest that as the number of nodes (or division) increases, the more accurate the result becomes. Finally, the results from the pattern developed from this improved finite difference analysis are effective and is recommended for use in structural engineering. REFERENCES [1]. Alexander Chajes (1974). Structural Stability Theory New Jersey. Prentice – Hall, Inc. [2]. Awele, M., Ayodele, J.C., and Osaisai, F.E. (2003). Numerical Methods for Scientists and Engineers. Ibadan. Beaver Publications. [3]. Ibearugbulem, O. M., Ettu, L. O., Ezeh, J. C. and Anya, U. C. (2013). New Shape function for Analysis of Line continuum by Direct Variational Calculus. International Journal of Engineering & Science, Vol. 2, issue 3; Pp. 66 – 71, ISBN 2319 – 1805. [4]. Goodwine, B. (2010). Engineering Differential Equations: Theory and Applications. New York: Springer Science + Business Media, LLC2011. [5]. Szilard, R. (2004). Theories and Application of Plate Analysis. New Jersey: John Wiley and Sons Inc. [6]. Ugural, A.C (1999). Stress in Plates and Shells (2nd Ed.).Singapore: McGraw-Hill.
- 7. IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308 _______________________________________________________________________________________ Volume: 03 Issue: 02 | Feb-2014, Available @ http://www.ijret.org 585 [7]. Yoo, H.C. and Lee, C.S. (2011). Stability of Structures. Principles and Applications. USA: Butterworth- Heinemann. Elsevier Inc.