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30120140505003

  1. 1. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 28 STATIC ANALYSIS OF COMPLEX STRUCTURE OF BEAMS BY INTERPOLATION METHOD APPROACH TO MATLAB Prabhat Kumar Sinha, Vijay Kumar Yadav*, Saurabha Kumar, Rajneesh Pandey Department of Mechanical Engineering. Shepherd School of Engineering and Technology, Sam Higginbottom Institute of Agriculture, Technology and Sciences, (Formerly Allahabad Agriculture Institute) Allahabad 211007 (INDIA) ABSTRACT In this article a MATLAB programming has shown which is based on the method of super position theory of a beam to investigate the slope and deflection of a complex structure beam. The beam is assumed and it is subjected to several loads in transverse direction. The governing equation of slope and deflection of complex structure beam are obtained by method of super-position theory and Euler-Bernoulli beam theory. Euler-Bernoulli beam theory (also known as Engineer’s beam theory or classical beam theory) is a simplification of the linear theory of elasticity which provides a means of calculating the load carrying and deflection characteristics of beams. It covers the case for Small deflection of a beam which is subjected to lateral loads only for a local point in between the class-interval in ‫-ݔ‬direction by using the interpolation method, to make the table of ‫ݔ‬ and ‫,ݕ‬ then ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ, where, y is a deflection of beam and slope ( ௗ௬ ௗ௫ ሻ at any point in thethin beams, apply the initial and boundary conditions, this can be calculating and plotting thegraph by using the MATLAB is a fast technique method will give results, the result is alsoshown with numerical analytical procedure.Additional analysis tools have been developed such as plate theory and finite elementanalysis, but the simplicity of beam theory makes it an important tool in the science, especially structural and Mechanical Engineering. Keywords: Method of Super Position, Static Analysis, Interpolation Method, Flexural Stiffness, Isotropic Materials, MATLAB. INTRODUCTION When a thin beam bends it takes up various shapes [1]. The shapes may besuperimposed on ‫ݔ‬ െ ‫ݕ‬ graph with the origin at the left or right end of the beam (before itisloaded). At any distance x INTERNATIONAL JOURNAL OF MECHANICAL ENGINEERING AND TECHNOLOGY (IJMET) ISSN 0976 – 6340 (Print) ISSN 0976 – 6359 (Online) Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME: www.iaeme.com/ijmet.asp Journal Impact Factor (2014): 7.5377 (Calculated by GISI) www.jifactor.com IJMET © I A E M E
  2. 2. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 29 meters from the left or right end, the beam will have a deflection‫ݕ‬ and gradient or slope, ௗ௬ ௗ௫ . The statement‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ, ‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡means corresponding toevery value of‫ݔ‬ in the range‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡, there exists one or more values of y. Assumingthat݂ ሺ‫ݔ‬ሻis a single-valued and continuous and that it is known explicitly, then the values of݂ሺ‫ݔ‬ሻcorresponding to certain given values of ‫ݔ‬ , say‫ݔ‬଴ , ‫ݔ‬ଵ,‫ݔ‬ଵ ,…‫ݔ‬௡ can easily be computed andtabulated. The central problem of numerical analysis is the converse one: Given the set of tabular value ሺ‫ݔ‬଴, ‫ݕ‬଴ሻ, ൫‫ݔ‬ଵ ,‫ݕ‬ଵ൯, … … . , ሺ‫ݔ‬௡, ‫ݕ‬௡ሻ satisfying the relation ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ wherethe explicit nature of ݂ ൌ ሺ‫ݔ‬ሻ is not known, it is required to simpler function‫׎‬ ሺ‫ݔ‬ሻsuch that݂ ሺ‫ݔ‬ሻ and‫׎‬ ሺ‫ݔ‬ሻagree at the set of tabulated points. Such a process is interpolation. If ‫׎‬ ሺ‫ݔ‬ሻ is a polynomial, then the process is called polynomial interpolation and ‫׎‬ ሺ‫ݔ‬ሻis called theinterpolating polynomial. As a justification for the approximation of unknown function bymeans of a polynomial, we state that famous theorem due to Weierstrass: If݂ሺ‫ݔ‬ሻiscontinuous in‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡ .Then given any ‫א‬൐ 0, there exists a polynomialܲ ሺ‫ݔ‬ሻ such that ( ) ( )f x p x− ൑ ‫א‬ , for all in ሺ‫ݔ‬଴, ‫ݔ‬ଵሻ, This means that it is possible to find a polynomialܲ ሺ‫ݔ‬ሻwhose graph remains within theregion bounded by ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻെ‫א‬ and‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ+‫א‬ for all ‫ݔ‬ between‫ݔ‬଴ and ‫ݔ‬௡ , however small‫א‬ may be [2].If a beam loaded several point loads, or with some uniformly distributed load and point load the slope and deflection at a point can be found by method of super position in which slope and deflection at any point will be equal to the algebraic some of slope or deflection due to the point loads or uniformly distributed load acting individually. SLOPE, DEFLECTION AND RADIUS OF CURVATURE We have already known the equation relating bending moment and radius of curvature in a beam namely. ெ ௒ = ா ோ Where, M is the bending moment. I is second moment of area about the centroid. E is the Modulus of Elasticity and R is the radius of curvature, Rearranging we have, ଵ ோ = ெ ா Figure-1 illustrates the radius of curvature which is define as the radius of circle that has a tangent the same as the point on x-y graph Y o ݀߰ Q P O Ψ dΨ + Ψ X
  3. 3. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 30 Consider an elemental length PQ = ݀‫ݏ‬ of a curve. Let the tangents at P and Q make anglesΨ and dΨ + Ψ with the axis. Let the normal at P and Q meet at C. Then C is called the center of curvature of the curve at any point between P and Q on the curve. The distance CP = CQ = Ris called the radius of curvature at any point between P and Q on the curve. Obviously, s R d∂ = Ψ or , ܴ ൌ ݀‫ݏ‬ ݀߰ൗ but we know that if (‫,ݔ‬ ‫)ݕ‬ be the coordinate of P. ݀‫ݕ‬ ݀‫ݔ‬ ൌ tan Ψ ௗ௬ ௗ௦ ൌ ௗ௦ ௗ௫⁄ dΨ ௗ௫ൗ = ௦௘௖ట ೏ഗ ೏ೣ …………………………………..……………………………………………(1) Differentiating with respect to, we have ‫ܿ݁ݏ‬ଶ ‫.ݔ‬ ݀߰ ݀‫ݔ‬ ൌ ݀ଶ ‫ݔ݀/ݕ‬ଶ ௗట ௗ௫ ൌ ௗమ௬ ௗ௫మ / ‫ܿ݁ݏ‬ଶ ߰…………………………………………………………………………………………….. (2) Substituting in the equation (1) we have, ܴ ൌ ೞ೐೎ഗ ೏మ೤ ௗ௫మ ‫ܿ݁ݏ‬ଶ‫ݔ‬ ൌ ‫ܿ݁ݏ‬ଷ ߰ ݀ଶ‫ݕ‬ ݀‫ݔ‬ଶ⁄ Therefore, 1 ܴ ൌ ݀ଶ ‫ݕ‬ ݀‫ݔ‬ଶ /‫ܿ݁ݏ‬ଷ ߰ ଵ ோ ൌ ௗమ௬ ௗ௫మ /ሺ‫ܿ݁ݏ‬ଶ ߰ሻ య మor ଵ ோ ൌ ௗమ௬ ௗ௫మ /ሺ1 ൅ ‫݊ܽݐ‬ଶ ߰ሻ య మ For practical member bend due to bending moment the slope ‫߰݊ܽݐ‬ at any point is a small quantity, hence ‫݊ܽݐ‬ଶ ψ can be ignored. Therefore 1 ܴ ൌ ݀ଶ ‫ݕ‬ ݀‫ݔ‬ଶ If M be the bending moment which has produced the radius of curvature R, we have ெ ூ ൌ ா ோ , ଵ ோ ൌ ெ ாூ , ௗమ௬ ௗ௫మ ൌ ெ ாூ ‫ܯ‬ ൌ ‫ܫܧ‬ ௗమ௬ ௗ௫మ………………………………………………………………………………………..(3)
  4. 4. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 31 The product EI is called the flexural stiffness of the beam. In order to solve the slope ௗ௬ ௗ௫ orthedeflection(y) at any point on the beam, an equation for M in terms of position must be substituted into equation (1). We will now examine these cases in the example of cantilever beam [3]. This article presentto developa MATLAB program for a complex structure beam which can workwithout the dependenceupon the beam materials and the aspect ratio. The input shouldbe geometry dimensions of complex beams for example-plate and circular bar such as length, breadth, thickness and diameter, the materials should be isotropic, materials data such as Young’s Modulus and Flexural Stiffness and, to calculate the slope and deflection at any pointin the complex structure by using Super-position method with the help of Interpolation method by analytically as well as MATLAB program and the plot the graph of slope and deflection of thin beams by using MATLAB programming and analyzed the graph and verify for the different values with the results. REVIEW OF LITERATURE AddidsuGezahegnSemie had worked on numerical modeling on thin plates andsolved the problem of plate bending with Finite Element Method and Kirchhoff’s thin platetheory is applied and program is written in FORTRAN and the results were compared with thehelp of ansys and FORTRAN program was given as an open source code. The analysis wascarried out for simple supported plate with distributed load, concentrated load andclamped/fixed edges plates for both distributed and concentrated load. According to the method of super position theory if it is assumed that the beam behaves elastically for a combined loading as well as for the individual loads, the resulting final deflection and slope of the loaded beam is simply the sum of the slope and deflection caused by each of the individual loads. This sum may be algebraic one or it might be vector sum, the type depending on whether or not the individual slope and deflection lie in the same plane. From Euler-Bernoulli beam theory [3] is simplification of the linear theory of elasticity whichprovides a means of calculating the load carrying slope and deflection characteristics of beamindirection. This theory was applicable in Mechanics of Solid [4]. The derivation ofthin beams of slope, deflection and radius of curvature [5] – for example- six cases areoccurred 1- Cantilever thin beam with point load at free end [6], 2- Cantilever thin beam withUniformly Distributed Load (U.D.L.) [7], 3- Cantilever thin Uniformly Varying Load (U.V.L.) [8], 4- Simply supported thin beam point load at mid [9], 5- Simply supported thinbeam with Uniformly Distributed Load (U.D.L.) [10]. 6- Simply supported thin beam withUniformly Varying Load (U.V.L.) [11]. Numerical problem has been taken form Mechanicsof Solids, Derivations or formulations made the table ofand was used ofinterpolation method to found out the unknown value between at any point in between 1- classinterval by using Newton’s forward difference interpolation formula is used from top; Newton’s backward difference interpolation formula is used from bottom starting, Stirlinginterpolation formula is used from the middle to get the results. [12], so it is overcome thisproblem we may use the Interpolation method by using MATLAB programming. There are general assumptions have been made when solving the problems are as follows. 1- Each layer of thin beams undergoes the same transverse deflection. 2- The mass of the point area is not considered as significant in altering the behavior ofthe beams. 3- There is no displacement and rotation of the beam at the fixed end. 4- The material behaves linearly. 5- Materials should be Isotropic. 6- The deflections are small as compared to the beam thickness.[13]
  5. 5. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 32 Case 1- Cantilever thin beam with point and Uniformly Distributed Load at free End ࢞ F ࢞ ࢟ L ࢞ Deflection of thin beam due to point load at the End point A ࡲ ࢞ A B L The bending moment at any position x is simply– Fx. Substituting this into equation (3) we have, ‫ܫܧ‬ ݀ଶ ‫ݕ‬ ݀‫ݔ‬ଶ ൌ െ‫ݔܨ‬ Integrating with respect to ‫ݔ‬ ‫ܫܧ‬ ௗ௬ ௗ௫ ൌ - ி௫మ ଶ + A……………………………………………………………… ………………...(4) Again integrating with respect to ‫ݔ‬ ‫ܫܧ‬ ௗ௬ ௗ௫ ൌ െ ி௫య ଺ + A‫ݔ‬ ൅ ‫……………………………………ܤ‬ …………………………………….(5) A and B are constants of integration and must be found from the boundary conditions. These are at ‫ݔ‬ ൌ ‫,ܮ‬ ‫ݕ‬ ൌ 0 (no deflection) At ‫ݔ‬ ൌ ‫,ܮ‬ ௗ௬ ௗ௫ ൌ 0 (gradient horizontal) Substituting‫ݔ‬ ൌ ‫ܮ‬ & ௗ௬ ௗ௫ ൌ 0 in Equation (4). This gives ‫ܫܧ‬ሺ0ሻ ൌ െ ‫ܮܨ‬ଶ 2 ൅ ‫ܣ‬ ݄݁݊ܿ݁ ‫ܣ‬ ൌ ‫ܮܨ‬ଶ 2 Substitute ‫ܣ‬ ൌ ி௅మ ଶ , ‫ݕ‬ ൌ 0 ܽ݊݀ ‫ݔ‬ ൌ ‫ܮ‬ in to Equation (5) and we get ‫ܫܧ‬ሺ0ሻ ൌ െ ி௅య ଺ ൅ ி௅య ଶ ൅ ‫ܤ‬Hence‫ܤ‬ ൌ െ ி௅య ଷ
  6. 6. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 33 Substitute ‫ܣ‬ ൌ ி௅మ ଶ ܽ݊݀ ‫ܤ‬ ൌ െ ி௅య ଷ in Equation (4)& (5) and Complete Equation are ‫ܫܧ‬ ௗ௬ ௗ௫ ൌ െ ி௫మ ଶ ൅ ி௅మ ଶ ………………………………………………………………………….…....(6) ‫ݕܫܧ‬ ൌ െ ி௫య ଺ ൅ ி௅మ௫ ଶ െ ி௅య ଷ ………………………………………………………………………….(7) The main Point of interest is slope and deflection at free end where x=0 into (6) & (7) gives the standard Equation, Slope at free end ௗ௬ ௗ௫ ൌ ி௅మ ଶாூ ………………………………………………………………………....(8) Deflection at free end ‫ݕ‬ ൌ െ ி௅య ଷாூ …………………………………………..………………………(9) So let consider a cantilever AB of length L loaded with uniformly distributed load w per unit length. x A B L The bending moment at position ‫ݔ‬ is given by ‫ܯ‬ ൌ െ ௪௫మ ଶ substituting this equation (3) We have, ‫ܫܧ‬ ௗమ௬ ௗ௫మ ൌ M= ି௪௫మ ଶ Integrating once we get, ‫ܫܧ‬ ௗ௬ ௗ௫ ൌ െ ௪௫య ଺ + A…………………………………………………………………………….. (10) Again integrating with respect to ‫ݔ‬ ‫=ܫܧ‬ െ ௪௫ర ଶସ +A ‫ݔ‬ + B….…………………………………………………………………………. (11) A and B are the constant of integration and must be found from boundary conditions. These are At ‫ݔ‬ ൌ ‫,ܮ‬ ‫ݕ‬ = 0 (no deflection) At ‫=ݔ‬ L, ௗ௬ ௗ௫ =0 (horizontal)
  7. 7. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 34 Substituting ‫ݔ‬ ൌ ‫ܮ‬ and ௗ௬ ௗ௫ =0 in equation (10) and we get, ‫ܫܧ‬ሺ0ሻ ൌ െ ௪௅య ଺ +A hence A = ௪௫య ଺ Substituting this in to equation (11) with the known solution y=0 and x=L results in ‫ܫܧ‬ሺ0ሻ ൌ െ ௪௅ర ଶସ ൅ ௪௅య ଺ ൅ ‫ܤ‬ Hence ‫ܤ‬ ൌ ି௪௅ర ଼ Putting the result for A and B into equation in (10) and (11) yield the complete Equations. ‫ܫܧ‬ ௗ௬ ௗ௫ ൌ ି௪௫య ଺ ൅ ‫ݓ‬ ௅య ଺ ………………………………………………………………………..…….(12) ‫ܫܧ‬ሺ‫ݕ‬ሻ ൌ െ ௪௫ర ଶସ ൅ ௪௅య ଺ ‫ݔ‬ െ ௪௅ర ଼ …………………………………………………………………… (13) The main points of intersect is the slope and deflection at free end where slope at free end ௗ௬ ௗ௫ ൌ ௪௅య ଺ாூ …………………………………………………………………………………………..(14) Deflection at free end ‫ݕ‬ ൌ ି௪௅ర ଼ாூ …………………………………………………………………. (15) Numerical Analysis A Cantilever thin beam 6 m long has a point load of 5KN at the free end and uniformly distributed load of 300 N/M. The flexural stiffness is 53.3‫ܰܯ‬ଶ .calculate the slope and deflection. Solution: The solution of the given problem can be obtained by METHOD OF SUPER POSITION. Slope due to point load ࡲ ‫ݕ‬, = ௗ௬ ௗ௫ ൌ ሾെ ி௫మ ଶ ൅ ி௅మ ଶ ] ଵ ாூ …………………………………………………………………………(16) By putting the value of ‫,ܨ‬ variable‫,0=ݔ‬ 2, 4, 6, L=6 and EI in equation (16), we obtain the various value of slope at these points. So ‫ݕ‬ଵ , ൌ ሺ1.6885 ൈ 10ିଷ ሻ (no unit),‫ݕ‬ଶ , = ሺ1.5 ൈ 10ିଷ ሻ (no unit),‫ݕ‬ଷ , =ሺ9.38086 ൈ 10ିସ ሻ(no unit) ‫ݕ‬ସ , = ሺ0ሻ(no unit) Slope due to load U.D.L. ‫ݕ‬,, = ௪ ଺ாூ [െ‫ݔ‬ଷ ൅ ‫ܮ‬ଷ ]………………………………………………………………………………....(17) By putting the value of ‫ݓ‬variable‫,0=ݔ‬ 2, 4, 6, L=6 and EI in equation (17), we obtain the various value of slope at these points. So
  8. 8. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 35 ‫ݕ‬ଵ ,, ൌ ሺ2.0262 ൈ 10ିସ ሻ (no unit), ‫ݕ‬ଶ ,, = ሺ1.9521 ൈ 10ିସ ሻ (no unit), ‫ݕ‬ଷ ,, =ሺ1.4258 ൈ 10ିସ ሻ (no unit), ‫ݕ‬ସ ,, = ሺ0ሻ (no unit) According to the theory of METHOD OF SUPER POSITION. Total slope of the complex beam is y= ‫ݕ‬, ൅ ‫ݕ‬,, Table -1 ‫ݔ‬ 0 2 4 6 Slope y= ‫ݕ‬, ൅ ‫ݕ‬,, 1.8911ൈ 10ିଷ 1.6952ൈ 10ିଷ 1.0806ൈ 10ିଷ 0 %calculate the slope of beam at any point in between 1-class interval %cantilever beam %point load at free end and udl per unit length x=[0 2 4 6]; slope=[1.8911*10.^-3 1.6952*10.^-3 1.0806*10.^-3 0.0]; xi=1; yilin=interp1(x,slope,xi,'linear') yilin = 0.0018 %plote the graph of slope of beam %cantilever thin beam %ponit load at the free end and udl per unit length f=5000; x=[0:1:6]; l=6; EI=53.3*10.^6; w=300; slope=(f/2)*[-x.^2+l.^2]/EI+(w/6)*(-x.^3+l.^3)/EI; plot(x,slope,'--r*','linewidth',2,'markersize',15) xlabel('position along the axies (x)','fontsize',15) ylabel('position along the axis (y)','fontsize',15) title('slope of cantilever beam with point load at free end and udl per unit length','fontsize',15)
  9. 9. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 36 Deflection due to point load equation ‫ݕ‬ ൌ ி ாூ ሾെ ி௫య ଺ ൅ ி௅మ௫ ଶ െ ி௅య ଷ ሿ…………………………………………………………………..…..(18) By putting the value of ‫,ܨ‬ variable‫,0=ݔ‬ 2, 4, 6, L=6 and EI in equation (18), we obtain the various value of deflection at these points. So ‫ݕ‬ଵ , ൌ ሺെ6.7542 ൈ 10ିଷ ሻ݉,‫ݕ‬ଶ , = ሺെ3.5021 ൈ 10ିଷ ሻ݉,‫ݕ‬ଷ , =ሺെ1.0 ൈ 10ିଷ ሻ݉,‫ݕ‬ସ , = ሺ0ሻ݉ Deflection due to uniformly distributed load ሺ‫ݕ‬ሻ ൌ ௪ ாூ ሾെ ௫ర ଶସ ൅ ௅య ଺ ‫ݔ‬ െ ௅ర ଼ ሿ………………………………………………………………………..(19) By putting the value of‫,ݓ‬ variable ‫,0=ݔ‬ 2, 4, 6, L=6 and EI in equation, we obtain the various value of deflection at these points. So ‫ݕ‬ଵ ,, ൌ െሺ9.1182 ൈ 10ିସ ሻm,‫ݕ‬ଶ ,, =െሺ5.1028 ൈ 10ିସ ሻm,‫ݕ‬ଷ ,, =െሺ0.03027 ൈ 10ିସ ሻm,‫ݕ‬ସ ,, = ሺ0ሻm According to the theory of METHOD OF SUPER POSITION Total deflection of the complex beam is y= ‫ݕ‬, ൅ ‫ݕ‬,, Table-2 ‫ݔ‬ 0 2 4 6 y= ‫ݕ‬, ൅ ‫ݕ‬,, െ7.7650 ൈ 10ିଷ െ4.0123 ൈ 10ିଷ െ1.0032 ൈ 10ିଷ 0 %calculate the deflection of beam at any point in between 1-class interval %cantilever beam %point load at free end and udl per unit length x=[0 2 4 6]; y=[-7.7650*10.^(-3) -4.0123*10.^(-3) -1.0032*10.^3 0.0]; xi=1; yilin=interp1(x,y,xi,'linear') yilin = 0.0018 %plote the graph of slope of beam %cantilever thin beam %ponit load at the free end and udl per unit length f=5000; x=[0:1:6]; l=6; EI=53.3*10.^6; w=300; y=(f/EI)*[(-x.^3/6)+(l.^2*x/2)-(l.^3)/3]+(w/EI)*[(-x.^4/24)+(l.^3 *x/6)-(l.^4/8)]; plot(x,y,'--r*','linewidth',2,'markersize',15) xlabel('position along the axies (x)','fontsize',15) ylabel('position along the axis (y)','fontsize',15) title('deflection of cantilever beam with point load at free end and udl per unit length','fontsize',15)
  10. 10. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 37 Case: 2 A simply supported beam subjected a point load at mid-point and uniformly distributed load per unit length F By the method of super position first find the slope and deflection due to each load F and The resulting slope and deflection is the sum of individual slope and deflection. 1:-slope and deflection cause by point load for simply supported beam F x The beam is symmetrical so the reactions are , the bending moment will be change at the center but because the bending will be symmetrical each side of the center we need only solve for the left hand. The bending moment at position x up to the middle is given by So
  11. 11. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 38 Integrating w.r.t. ‫ݔ‬ once ‫ܫܧ‬ ௗ௬ ௗ௫ = ி௫మ ସ + A………………………………………………………………………………….…(20) Integrating w.r.t. ‫ݔ‬ again, ‫ݕܫܧ‬= ி௫య ଵଶ +A‫ݔ‬ + B……………………………………………………………………………….(21) A and b are constant of integration and must be founded from the boundary condition .These are At ‫ݔ‬=0, ‫ݕ‬=0 (no deflection at the ends) At ‫ݔ‬= ௅ ଶ , ௗ௬ ௗ௫ =0 (horizontal at the middle) Putting ‫ݔ‬= ௅ ଶ and ௗ௬ ௗ௫ =0 in question (20) ‫ܫܧ‬ሺ0ሻ= ி௅మ ଵ଺ +A hence A= െ ி௅మ ଵ଺ Substituting A= െ ி௅మ ଵ଺ , ‫ݕ‬ ൌ 0 and ‫ݔ‬ ൌ 0 in equation (21) and we get, ‫ܫܧ‬ሺ0ሻ, hence B=0 Substituting A= െ ி௅మ ଵ଺ and B=0 in equation (20) and (21), so ‫ܫܧ‬ ௗ௬ ௗ௫ = ி௫మ ସ െ ி௅మ ଵ଺ ` ………………………………………………………………………………..(22) ‫ݕܫܧ‬= ி௫య ଵଶ െ ி௅మ ଵ଺ ‫ݔ‬……………………………………………………………………………..….. (23) The main point of interest is slope at the end and the deflection at the middle. Substituting ‫ݔ‬=0 into (22) gives the standard equation for the slope at the left end. The slope at the right end is equal but opposite sign. Slope at the ends ௗ௬ ௗ௫ = ± ி௅మ ଵ଺ாூ …………………………………………………………………...…..(24) The slope is negative on the left end and positive at right end. Substituting ‫ݔ‬= ௅ ଶ in to the equation (23) gives the standard equation for the deflection at the middle. Deflection at the middle ‫ݕ‬ ൌ െ ி௅య ସ଼ாூ …………………………………………………………...…(25) 2:- slope and deflection cause by the U.D.L. for simply supported beam ‫ݔ‬
  12. 12. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 39 The beam is symmetrical so the reaction are ௪௅ ଶ . The bending moment at point ‫ݔ‬ is ‫ܯ‬ ൌ ௪௅௫ ଶ - ௪௫మ ଶ Substituting value of ‫ܯ‬ in equation (3) we have ‫ܫܧ‬ ௗమ௬ ௗ௫మ = ௪௅௫ ଶ - ௪௫మ ଶ Integrating w.r.t. ‫ݔ‬ ‫ܫܧ‬ ௗ௬ ௗ௫ = ௪௅௫మ ସ - ௪௫య ଺ + A …………………………………………………………………………………………………....(26) Integrating w.r.t. ‫ݔ‬ again ‫ݕܫܧ‬= ௪௅௫య ଵଶ - ௪௫ర ଶସ + A‫ݔ‬ + B………………………………………………………………………...(27) A and B are constant of integration and must be find from the boundary condition .these are at ‫ݔ‬ ൌ 0, ‫ݕ‬ ൌ 0 (no deflection at the ends) at ‫ݔ‬ ൌ ௅ ଶ , ( ௗ௬ ௗ௫ )=0 (horizontal at the middle) putting‫ݔ‬ ൌ ௅ ଶ , ( ௗ௬ ௗ௫ )=0 in equation (26) we get ‫ܫܧ‬ሺ0ሻ = ௪௅య ଵ଺ - ௪௅య ସ଼ + A, hence A= - ௪௅య ଶସ Substituting‫ݔ‬ ൌ 0, ‫ݕ‬ ൌ 0 and A= - ௪௅య ଶସ in to equation (27), the complete equation are ‫ܫܧ‬ ௗ௬ ௗ௫ = ௪௅௫మ ସ - ௪௫య ଺ - ௪௅య ଶସ …………………………………………………………………………… .(28) ‫ݕܫܧ‬= ௪௅௫య ଵଶ - ௪௫ర ଶସ - ௪௫௅య ଶସ ……………………………………………………………………………...(29) The main point ofinterest to find out the slope at the ends and deflectionin the middle. Substituting ‫ݔ‬ ൌ 0 into (28) gives the standard equation for the slope at the left end. The slope at the right end will be equal and opposite sign. Slope at the free end ௗ௬ ௗ௫ = ± ௪௅య ଶସாூ ………………………………………………………………….(30) The slope at the left end is negative and positive at right end. Substituting ‫ݔ‬ ൌ ௅ ଶ in equation (29) gives the standard equation for the deflection at the middle. Deflection at the middle ‫ݕ‬ ൌ- 5 ௪௅ర ଷ଼ସாூ ………………………………………………………….….(31)
  13. 13. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 40 NUMERICAL ANALYSIS-2 A simply supported thin beam 0.5m long subjected a point load at the middle of the beam of 1N and a uniformly distributed load 1N/݉݉ has a section of uniform circular 5݉݉diameter the modulus of elasticity (E) is 2 ൈ 10ହ N/݉݉ଶ .calculate the slope and deflection at any point in between any 1-class interval and plot the graph. By the method of super position the total slope and deflection cause by point load and u.d.l. is equal to the algebraic summation of slope and deflection of beam cause by individual load. so slope cause by point load is ‫ܫܧ‬ ݀‫ݕ‬ ݀‫ݔ‬ ൌ ‫ݔܨ‬ଶ 4 െ ‫ܮܨ‬ଶ 16 By putting the value of variable ‫ݔ‬=0.00, 50.0݉݉,100 ݉݉,200 ݉݉…….500 ݉݉, ‫ܮ‬ ൌ 500݉݉, EI = 6136000N-݉݉ଶ and F = 1N in above equation we get the various value of slope at that point, these are ‫ݕ‬ଵ , ൌ െሺ0.0025ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଶ , =െሺ0.0024ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଷ , =െሺ0.0021ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ସ , =െ(0.0016)‫݊ܽ݅݀ܽݎ‬, ‫ݕ‬ହ , =െሺ0.0091ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬଺ , =െሺ0.0000ሻ ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬଻ , =ሺ0.0011ሻ‫ݕ݊ܽ݅݀ܽݎ‬଼ , =ሺ0.0024ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଽ , = ሺ0.0039ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଵ଴ , =ሺ0.00570ሻ ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଵଵ , =ሺ0.007639ሻ ‫݊ܽ݅݀ܽݎ‬ Slope due to U.D.L is ‫ݕ‬,, ൌ ௗ௬ ௗ௫ = ௪ ாூ [ ௅௫మ ସ - ௫య ଺ - ௅య ଶସ ] By putting the value of variable ‫ݔ‬ = 0.00, 50.0݉݉, 100 ݉݉, 200 ݉݉……500 ݉݉, ‫ܮ‬ ൌ 500݉݉, EI=6136000N-݉݉ଶ and ‫ݓ‬ ൌ 1ܰ/݉݉ଶ in above equation we get the various value of slope at that point, these are ‫ݕ‬ଵ ,, ൌ െሺ0.8488ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଶ ,, = െሺ0.8012ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଷ ,, =െሺ0.6722ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ସ ,, = െ(0.3910) ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ହ ,, =െሺ0.25124ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬଺ , =െሺ0.0000ሻ ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬଻ ,, =ሺ0.2512ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬଼ ,, =ሺ0. .4821ሻ‫݊ܽ݅݀ܽݎ‬, ‫ݕ‬ଽ ,, =ሺ0.6722ሻ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଵ଴ ,, =ሺ0.80128ሻ ‫݊ܽ݅݀ܽݎ‬,‫ݕ‬ଵଵ ,, =ሺ0.848815ሻ ‫݊ܽ݅݀ܽݎ‬ Resultant slope due to combined load is (by method of super position)‫ݕ‬= ‫ݕ‬, +‫ݕ‬,, TABLE-3 ‫ݔ‬ሺ݉݉ሻ 0 50 100 150 200 250 ‫ݕ‬ሺslpoe ) -0.8473 -0.8036 -0.6743 -0.3894 -0.2521 0.0000 ‫ݔ‬ሺ݉݉ሻ 300 350 400 450 500 ‫ݕ‬ሺslpoe ) 0.2523 0.4845 0.6722 0.6779 0.8558 %calculate the slope of simply supported thin beam at any point in between any 1-class interval %simplysupported thin beam with point load and udl per unit length x=[0 50 100 150 200 250 300 350 400 450 500]; y=[-0.8473 -0.8036 -0.6743 -0.3894 -0.2521 0.0000 0.2523 0.4845 0.6722 0.6779 0.8558]; xi=125; yilin=interp1(x,y,xi,'linear') yilin = -0.5319 %plote graph of slope of simply supported beam %simply supported beam with point load at middle and udl per unit length
  14. 14. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 41 F=1; w=1; x=[0:50:250:50:500]; EI=6136000; L=500; y=(1/4)*(F/EI)*[(x.^2)-(L.^2/4)]+(1/2)*(w/EI)*[L*x.^2/2-x.^3/3-L.^3/12]; plot(x,y,'--r*','linewidth',2,'markersize',15) xlabel('position along the axies (x)','fontsize',15) ylabel('position along the axis (y)','fontsize',15) title('slope of simply supported beam with point load at middle and udl per unit length','fontsize',15) Deflection cause by point load Deflection equation ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ ‫ݕܫܧ‬= ி௫య ଵଶ െ ி௅మ ଵ଺ ‫ݔ‬ ‫ݕ‬ ൌ ி ସாூ [ ௫య ଷ െ ௅మ ସ ‫ݔ‬] By putting the value of variable ‫ݔ‬=0.00, 50.0݉݉, 100 ݉݉, 200 ݉݉……500 ݉݉, ‫ܮ‬ ൌ 500݉݉, EI=6136000N-݉݉ଶ and F=1N in above equation we get the various value of deflection at that point, these are ‫ݕ‬ଵ , ൌ ሺ0.0000ሻ݉݉,‫ݕ‬ଶ , = െሺ0.1256ሻ݉݉,‫ݕ‬ଷ , =െሺ0.2410ሻ݉݉,‫ݕ‬ସ , = െ(0.3361)݉݉, ‫ݕ‬ହ , =െሺ0.4006ሻ݉݉,‫ݕ‬଺ , =െሺ0.4244ሻ݉݉,‫ݕ‬଻ , =െሺ0.3972ሻ݉݉,‫ݕ‬଼ , =െሺ0.3089ሻ݉݉ ‫ݕ‬ଽ , =െሺ0.1494ሻ݉݉,‫ݕ‬ଵ଴ , =ሺ0.0916ሻ ݉݉,‫ݕ‬ଵଵ , =ሺ0.4244ሻ ݉݉ Deflection cause by U.D.L. ‫ݕ‬,, ൌ ‫ݓ‬ ‫ܫܧ‬ ሾ ‫ݔܮ‬ଷ 12 െ ‫ݔ‬ସ 24 െ ‫ܮݔ‬ଷ 24 െ ‫ܮݔ‬ଷ 24 ሿ ‫ݕ‬ଵ ,, ൌ െሺ0.0000ሻ݉݉,‫ݕ‬ଶ ,, =െሺ41.63ሻmm,‫ݕ‬ଷ ,, =െሺ78.77ሻ݉݉,‫ݕ‬ସ ,, = െ(107.84) ݉݉ ‫ݕ‬ହ ,, =െሺ126.30ሻ݉݉,‫ݕ‬଺ , =െሺ132.62ሻ݉݉,‫ݕ‬଻ ,, =െሺ126.30ሻ݉݉,‫ݕ‬଼ ,, =െሺ108.47ሻ݉݉ ‫ݕ‬ଽ ,, =െሺ78.77ሻ݉݉,‫ݕ‬ଵ଴ ,, =െሺ41.63ሻ ݉݉,‫ݕ‬ଵଵ ,, =െሺ0.000ሻ݉݉ ‫ݕ‬= ‫ݕ‬, +‫ݕ‬,,
  15. 15. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 42 TABLE-4 ‫ݔ‬ሺ݉݉ሻ 0 50 100 150 200 250 ‫ݕ‬ሺDeflection) 0.0000 -41.7500 -79.0110 -108.17 -126.70 - 133.04 ‫ݔ‬ሺ݉݉ሻ 300 350 400 450 500 ‫ݕ‬ሺDeflection ) -126.69 -108.77 -78.91 -41.72 -0.4244 %calculate the deflection of simply supported beam at any point in between 1-class interval %simply supported thin beam with point load at middle and udl per unit length x=[0 50 100 150 200 250 300 350 400 450 500]; y=[0.0000 -41.7500 -79.0110 -108.17 -126.70 -133.04 -126.69 -108.77 -78.91 -41.72 -0.4244]; xi=90; yilin=interp1(x,y,xi,'linera') yilin =-71.5588 %plot the graph of deflection of beam %simply supported beam with point load at middle and udl per unit length F=1; w=1; x=[0:50:500]; L=500; EI=6136000; y=(1/4)*(F/EI)*[((x.^3)/3)-((L.^2*x)/4)]+(1/12)*(w/EI)*[(L*x.^3)-(x.^4/2)-(L.^3*x/2)]; plot(x,y,'--r*','linewidth',2,'markersize',15) xlabel('position along the axies (x)','fontsize',15) ylabel('position along the axis (y)','fontsize',15) title('deflection of simply supported beam with point load at middle and udl per unit length','fontsize',15)
  16. 16. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 43 DISCUSSION AND CONCLUSION It was observed that in case of Cantilever complex beams (point load at free and u.d.l.) and simply supported complex beam (point load and u.d.l.) are carried out by the numerical analysis and MATLAB programming, made the table of x verse slope and x verse deflection after the taken at any one point in between any 1-class- interval in complex beam and then calculated value at same point by using interpolation method through MATLAB programming to analysed the value at that point is slope and deflection ,we have analysed by plotted the graph of static slope and deflection of complex beam through the MATLAB programming . In future work it can be applied for composite materials of beam which are non- isotropic,it can be extending that is used in trusses like perfect frame, deficient and redundant.it can alsoused in tapered and rectangular beam. REFERENCES [1]- Zhang, G.Y., 2010, “A Thin Beam Formulation Based on Interpolation Method “International Journal of numerical methods in engineering, volume 85, pp. 7-35. [2]- Wang, Hu, Guang, Li Yao, 2007, “Successively Point Interpolation for OneDimensional Formulation”, Engineering Analysis with Boundary Elements, volume31, pp. 122-143. [3]- Ballarini, Roberto, S, 2003 “Euler-Bernoulli Beam Theory”, Mechanical EngineeringMagazine Online. [4]- Liu, Wing Kam, 2010, “Meshless Method for Linear One-Dimensional Interpolation Method”, International Journal of Computer Methods in Applied Mechanics andEngineering, volume 152, pp. 55-71. [5]- Park, S.K. and Gao, X.L., 2007, “Bernoulli-Euler Beam Theory Model Based on aModified Coupled Stress Theory”, International of Journal of Micro-mechanics and Micro- engineering, volume 19, pp. 12-67. [6]- Paul, Bourke, 2010, “Interpolation Method”, International Journal of NumericalMethods in Engineering, volume 88, pp. 45-78. [7]- Launder, B.E. and Spading D.B., 2010, “The Numerical Computation of Thin Beams”, International Journal of Computer Methods in Applied Mechanics andEngineering, volume 3, pp. 296-289. [8]- Ballarini and Roberto, 2009, “Euler-Bernoulli Beam Theory Numerical Study of ThinBeams”, International of Computer in Applied Mechanics and Engineering, volume 178, pp. 323-341. [9]- Thomson, J.F., Warsi Z.U.A. and Mastin C.W., 1982 “Boundary Fitted Co-ordinatesystem for Numerical Solution of Partial Differential Equations”, Journal ofComputational Physics, volume 47, pp. 1-108. [10]- Gilat, Amos, January 2003, “MATLAB An Introduction with Application, Publication- John Wiley and Sons. [11]- Hashin, Z and Shtrikman, S., 1963 “A Variation Approach to the Theory of ElasticBehaviour of Multiphase Materials”, Journal of Mechanics and Physics of Solids,volume 11, pp. 127-140. [12]- Liu, G.R. and Gu, Y.T., 2001, “A Point Interpolation Method for One-DimensionalSolids”, International Journal of Numerical Methods Engineering, pp. 1081-1106. [13]- Katsikade, J.T. and Tsiatas, G.C., 2001, “Large Deflection Analysis of Beams with Variable Stiffness”, International Journal of Numerical Methods Engineering, volume 33, pp. 172-177. [14]- Atluri, S.N. and Zhu, T., 1988, “A new meshless local Petrov-Galerkin (MLPG) approach in computational mechanics”, Computational Mechanics volume 22, pp.117-127.
  17. 17. International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 – 6340(Print), ISSN 0976 – 6359(Online), Volume 5, Issue 5, May (2014), pp. 28-44 © IAEME 44 [15]- Atluri, S. N. and Zhu, T., 2000, “New concepts in meshless methods”, International Journal of Numerical Methods Engineering, volume 47, pp. 537-556. [16]- Newmark, N.M., 2009, “A Method of Computation for Structural Statics”, Journal ofEngineering Mechanics Division, ASCE, volume 85, pp. 67-94. [17]- Bickley, W.G., 1968 “Piecewise Cubic Interpolation and Two-point Boundary ValueProblem”, Computer Journal, volume 11, pp. 200-206. [18]- Sastry, S.S., 1976, “Finite Difference Approximations to One Dimensional ParabolicEquation” Journal Computer and Applied Maths., volume 2, pp. 20- 23. [19]- Liu, G. R. and Gu,Y. T., 2001, “A Point Interpolation Method for two-dimensionalsolids”, International Journal for Numerical Methods in Engineering, volume 50, pp. 55-60. [20]- Timoshenko, Stephen P. and Gere, James M., 1962, “Theory of Elastic Stability”, International Student Edition by McGraw-Hill Book Company, New York. [21]- Prabhat Kumar Sinha and Rohit, “Analysis of Complex Composite Beam by usingTimoshenko Beam Theory & Finite Element Method”, International Journal of Design and Manufacturing Technology (IJDMT), Volume 4, Issue 1, 2013, pp. 43 - 50, ISSN Print: 0976 – 6995, ISSN Online: 0976 – 7002. [22]- Dr. Ibrahim A. Assakkaf, “Beams: Deformation by superposition spring 2003 ENES 220 Mechanics of materials” department of civil and environmental engineering university of MARYLAND, COLLAGE PARK. [23]- Prabhat Kumar Sinha, Ishan Om Bhargava and Saifuldeen Abed Jebur, “Non Linear Dynamic and Stability Analysis of Beam using Finite Element in Time”, International Journal of Mechanical Engineering & Technology (IJMET), Volume 5, Issue 3, 2014, pp. 10 - 19, ISSN Print: 0976 – 6340, ISSN Online: 0976 – 6359. [24]- Jn Mahto, Sc Roy, J Kushwaha and Rs Prasad, “Displacement Analysis of Cantilever Beam Using FEM Package”, International Journal of Mechanical Engineering & Technology (IJMET), Volume 4, Issue 3, 2013, pp. 75 - 78, ISSN Print: 0976 – 6340, ISSN Online: 0976 – 6359. [25]- Prabhat Kumar Sinha, Vijay Kumar, Piyush Pandey and Manas Tiwari, “Static Analysis of Thin Beams by Interpolation Method Approach To MATLAB”, International Journal of Mechanical Engineering & Technology (IJMET), Volume 4, Issue 2, 2013, pp. 254 - 271, ISSN Print: 0976 – 6340, ISSN Online: 0976 – 6359.

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