30120140505003

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
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IJMET
© I A E M E

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

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)

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]

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
ி௅య
଺
൅
ி௅య
ଶ
൅ ‫ܤ‬Hence‫ܤ‬ ൌ െ
ி௅య
ଷ

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)

34
Substituting ‫ݔ‬ ൌ ‫ܮ‬ and
ௗ௬
ௗ௫
=0 in equation (10) and we get,
௪௅య
଺
+A hence A =
௪௫య
଺
Substituting this in to equation (11) with the known solution
y=0 and x=L results in
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

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)

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)
title('deflection of cantilever beam with point load at free end and udl per unit length','fontsize',15)

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

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

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)

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

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];
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,,

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)];
title('deflection of simply supported beam with point load at middle and udl per unit
length','fontsize',15)

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-classinterval 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.
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