Static analysis of thin beams by interpolation method approach

International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 2, March - April (2013) © IAEME
254
STATIC ANALYSIS OF THIN BEAMS BY INTERPOLATION
METHOD APPROACH TO MATLAB
Prabhat Kumar Sinha Vijay Kumar*, Piyush Pandey Manas Tiwari
Mechanical Engineering Department
Sam Higginbottom Institute of Agriculture Technology and sciences, Allahabad
ABSTRACT
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 ሺ
ௗ௬
ௗ௫
ሻ at any point in the
thin beams, apply the initial and boundary conditions, this can be calculating and plotting the
graph by using the MATLAB is a fast technique method will give results, the result is also
shown with numerical analytically procedure. The successful demonstrated it quickly because
engineering and an enabler of the Industrial Revolution.
Additional analysis tools have been developed such as plate theory and finite element
analysis, but the simplicity of beam theory makes it an important tool in the science,
especially structural and Mechanical Engineering.
Keywords: Static Analysis, Interpolation Method, Flexural Stiffness, Isotropic Materials,
MATLAB.
INTRODUCTION
When a thin beam bends it takes up various shapes [1]. The shapes may be
superimposed on ‫ݔ‬ െ ‫ݕ‬ graph with the origin at the left or right end of the beam (before it is
loaded). At any distance x meters from the left or right end, the beam will have a deflection ‫ݕ‬
and gradient or slopeሺ
ௗ௬
ௗ௫
ሻ. The statement ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ, ‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡ means: corresponding to
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ISSN 0976 – 6340 (Print)
ISSN 0976 – 6359 (Online)
Volume 4, Issue 2, March - April (2013), pp. 254-271
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255
every value of ‫ݔ‬ in the range ‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡, there exists one or more values of y. Assuming
that ݂ሺ‫ݔ‬ሻ is a single-valued and continuous and that it is known explicitly, then the values of
݂ሺ‫ݔ‬ሻ corresponding to certain given values of ‫,ݔ‬ say ‫ݔ‬଴, ‫ݔ‬ଵ,…, ‫ݔ‬௡, can easily be computed and
tabulated. The central problem of numerical analysis is the converse one: Given the set of
tabular values ሺ‫ݔ‬଴, ‫ݕ‬଴ሻ, ሺ‫ݔ‬ଵ , ‫ݕ‬ଵሻ, ሺ‫ݔ‬ଶ, ‫ݕ‬ଶሻ, … , ሺ‫ݔ‬௡, ‫ݕ‬௡ሻ satisfying the relation ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ where
the explicit nature of ݂ሺ‫ݔ‬ሻ is not known, it is required to simpler function, ‫׎‬ሺ‫ݔ‬ሻ such that
݂ሺ‫ݔ‬ሻ and ‫׎‬ሺ‫ݔ‬ሻ agree at the set of tabulated points. Such a process is interpolation. If ‫׎‬ሺ‫ݔ‬ሻ is
a polynomial, then the process is called polynomial interpolation and ‫׎‬ሺ‫ݔ‬ሻ is called the
interpolating polynomial. As a justification for the approximation of unknown function by
means of a polynomial, we state that famous theorem due to Weierstrass: If ݂ሺ‫ݔ‬ሻ is
continuous in ‫ݔ‬଴ ൑ ‫ݔ‬ ൑ ‫ݔ‬௡, then given any ߳ ൐0, there exists a polynomial ܲሺ‫ݔ‬ሻ such that
( ) ( )f x P x− <∈, for all in ሺ 0x , nx ).
This means that it is possible to find a polynomial ܲሺ‫ݔ‬ሻ whose graph remains within the
region bounded by ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ-߳ and ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ+߳ for all ‫ݔ‬ between ‫ݔ‬଴ and ‫ݔ‬௡, however small ߳
may be [2].
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,
1/ܴ ൌ ‫ܧ/ܯ‬
Figure-1 illustrates the radius of curvature which is defined as the radius of circle that has a
tangent the same as the point on x-y graph.
Figure-1
Consider an elemental length ܲܳ ൌ ݀‫ݏ‬ of a curve. Let the tangents at P and Q make angles ߰
and ߰+݀߰ with the axis. Let the normal at P and Q meet at C. Then C is called the centre of
curvature of the curve at any point between P and Q on the curve. The distance CP = CQ = R
is called the radius of curvature at any point between P and Q on the curve.
Obviously, ݀‫ݏ‬ ൌ ܴ݀߰

256
Or, ܴ ൌ ݀‫߰݀/ݏ‬
But we know that if ሺ‫,ݔ‬ ‫ݕ‬ሻ be the coordinate of P,
ௗ௬
ௗ௫
ൌ ‫߰݊ܽݐ‬
ௗ௬
ௗ௦
ൌ
ௗ௦/ௗ௫
ௗట/ௗ௫
ൌ
௦௘௖ట
ௗట/ௗ௫
………………………………………………………(1)
‫߰݊ܽݐ‬ ൌ
݀‫ݕ‬
݀‫ݔ‬
Differentiating with respect to ‫,ݔ‬ we have
‫ܿ݁ݏ‬ଶ
‫.ݔ‬
݀߰
݀‫ݔ‬
ൌ ݀ଶ
‫ݔ݀/ݕ‬ଶ
ௗట
ௗ௫
ൌ
ௗమ௬
ௗ௫మ
/‫ܿ݁ݏ‬ଶ
߰……………………………………………………………(2)
Substituting in equation (1) we have,
ܴ ൌ
௦௘௖ట
೏మ೤
೏ೣమ
‫ܿ݁ݏ‬ଶ‫ݔ‬
ൌ
‫ܿ݁ݏ‬ଷ
߰
݀ଶ‫ݔ݀/ݕ‬ଶ
Therefore,
1
ܴ
ൌ
݀ଶ
‫ݕ‬
݀‫ݔ‬ଶ
/‫ܿ݁ݏ‬ଷ
߰
ଵ
ோ
ൌ
ௗమ௬
ௗ௫మ / 2 3/2
(sec )Ψ or
ଵ
ோ
ൌ
ௗమ௬
ௗ௫మ / 2 3/2
(1 tan )+ Ψ
For practical member bent due to the bending moment the slope ‫߰݊ܽݐ‬ at any point is a small
quantity, hence ‫݊ܽݐ‬ଶ
߰ can be ignored.
Therefore,
1
ܴ
ൌ ݀ଶ
‫ݔ݀/ݕ‬ଶ
If M be the bending moment which has produced the radius of curvature R, we have,
‫ܯ‬
‫ܫ‬
ൌ
‫ܧ‬
ܴ
1
ܴ
ൌ
‫ܯ‬
‫ܫܧ‬
݀ଶ
‫ݕ‬
݀‫ݔ‬ଶ
ൌ
‫ܯ‬
‫ܫܧ‬
‫ܯ‬ ൌ ‫ܫܧ‬
ௗమ௬
ௗ௫మ…………………………………………………………..(3)
The product EI is called the flexural stiffness of the beam. In order to solve the slope ሺ
ௗ௬
ௗ௫
ሻ or
the deflection ሺ‫ݕ‬ሻ at any point on the beam, an equation for M in terms of position ‫ݔ‬ must be
substituted into equation (1). We will now examine these cases in the example of cantilever
beam [3].
OBJECTIVE OF THE PRESENT WORK
The objective of the present work is to develop a MATLAB program which can work
without the dependence upon the thin beam materials and the aspect ratio. The input should
be geometry dimensions of thin 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 point
in between 1-class interval by using interpolation method by analytically as well as

257
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 for
results.
LITERATURE REVIEW
Addidsu Gezahegn Semie had worked on numerical modelling on thin plates and
solved the problem of plate bending with Finite Element Method and Kirchoff’s thin plate
theory is applied and program is written in Fortran and the results were compared with the
help of ansys and Fortran program was given as an open source code. The analysis was
carried out for simple supported plate with distributed load, concentrated load and
clamped/fixed edges plates for both distributed and concentrated load.
From Euler-Bernoulli beam theory [3] is simplification of the linear theory of elasticity which
provides a means of calculating the load carrying slope and deflection characteristics of beam
in direction. This theory was applicable in Mechanics of Solid [4]. The derivation of
thin beams of slope, deflection and radius of curvature [5] – for example- six cases are
occurred 1- Cantilever thin beam with point load at free end [6], 2- Cantilever thin beam with
Uniformly 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 thin
beam with Uniformly Distributed Load (U.D.L.)[10]. 6- Simply supported thin beam with
Uniformly Varying Load (U.V.L.) [11]. Numerical problem has been taken form Mechanics
of Solids, Derivations or formulations made the table of and was used of
interpolation method to found out the unknown value between at any point in between 1-class
interval by using Newton’s forward difference interpolation formula is used from top;
Newton’s backward difference interpolation formula is used from bottom starting, Stirling
interpolation formula is used from the middle to get the results. [12], so it is overcome this
problem 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 behaviour of
the 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]
Case 1- Cantilever Thin Beam with Point Load at Free End-
Figure-2
The bending moment at any position x is simply– Fx. Substituting this into equation (3) we
have, [14]

258
EI
ௗమ௬
ௗ௫మ
ൌ െ‫ݔܨ‬
Integrate with respect to ‫ݔ‬, we get
EI
ௗ௬
ௗ௫
ൌ
ିி௫మ
ଶ
൅ ‫ܣ‬………………………………………………………(4)
Integrate again and we get
EIy=െ
ி௫య
଺
൅ ‫ݔܣ‬ ൅ ‫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)
Substitute ‫ݔ‬ ൌ ‫ܮ‬ ܽ݊݀
ௗ௬
ௗ௫
ൌ 0, in equation (4). This gives
‫ܫܧ‬ሺ0ሻ ൌ െ
‫ܮܨ‬ଶ
2
൅ ‫ܣ‬ ݄݁݊ܿ݁ ‫ܣ‬ ൌ ‫ܮܨ‬ଶ
/2
Substitute ‫ܣ‬ ൌ
ி௅మ
ଶ
, ‫ݕ‬ ൌ 0 ܽ݊݀ ‫ݔ‬ ൌ ‫ܮ‬ ݅݊‫݋ݐ‬ ݁‫݊݋݅ݐܽݑݍ‬ ሺ5ሻ ܽ݊݀ ‫݁ݓ‬ ݃݁‫ݐ‬
‫ܫܧ‬ሺ0ሻ ൌ െ
‫ܮܨ‬ଷ
6
൅
‫ܮܨ‬ଷ
2
൅ ‫ܤ‬ ݄݁݊ܿ݁ ‫ܤ‬ ൌ െ
‫ܮܨ‬ଷ
3
Substitute ‫ܣ‬ ൌ
ி௅మ
ଶ
ܽ݊݀ ‫ܤ‬ ൌ െ‫ܮܨ‬ଷ
/3 into equations (4) and (5) and the complete equations
are
‫ܫܧ‬
ௗ௬
ௗ௫
ൌ െ
ி௫మ
ଶ
൅
ி௅మ
ଶ
…………………………………………………(6)
‫ܫܧ‬ ൌ
ିி௫య
଺
൅
ி௅మ௫
ଶ
െ
ி௅య
ଷ
……………………………………………….(7)
The main points of interest is slope and deflection at free end where ‫ݔ‬ ൌ 0.
Substituting ‫ݔ‬ ൌ 0 into (6) and (7) gives the standard equations,
Slope at free end
ௗ௬
ௗ௫
ൌ
ி௅మ
ଶாூ
……………………………….(8)
Deflection at free end
‫ݕ‬ ൌ
ିி௅య
ଷாூ
…………………………………………….(9)
Numerical Analysis-1
A cantilever thin beam is 4m long and has a point load of 5KN at the free end. The
flexural stiffness is 53.3MN2
. Calculate the slope and deflection at the free end.
Solution:-
Slope equation ‫ݕ‬′
ൌ ݂′ሺ‫ݔ‬ሻ
݀‫ݕ‬
݀‫ݔ‬
ൌ ‫ݕ‬′
ൌ ሾെ
‫ݔܨ‬ଶ
2
൅
‫ܮܨ‬ଶ
2
ሿ
1
‫ܫܧ‬
‫ݕ‬′
ൌ
‫ܨ‬
2‫ܫܧ‬
ሾെ‫ݔ‬ଶ
൅ ‫ܮ‬ଶሿ
‫ݐݑ݌‬ ‫ݔ‬ ൌ 0݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬′
ൌ
5000
2 ‫כ‬ 53.3 ‫כ‬ 10଺
ሾെ0 ൅ 6ଶሿ
‫ݎ݋‬ ‫ݕ‬ଵ
′
ൌ ሺ1.6885 ‫כ‬ 10ିଷሻ° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ

259
Similarly, ‫ݐݑ݌‬ ‫ݔ‬ ൌ 2݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଶ
′
ൌ
5000
2 ‫כ‬ 53.3 ‫כ‬ 10଺
ሾെ2ଶ
൅ 6ଶሿ
‫ݎ݋‬ ‫ݕ‬ଶ
′
ൌ ሺ1.5 ‫כ‬ 10ିଷሻ°ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
Also, ‫ݐݑ݌‬ ‫ݔ‬ ൌ 4݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଶ
′
ൌ
5000
2 ‫כ‬ 53.3 ‫כ‬ 10଺
ሾെ4ଶ
൅ 6ଶሿ
‫ݕ‬ଷ
′
ൌ ሺ9.38086 ‫כ‬ 10ିସሻ° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
And, ‫ݐݑ݌‬ ‫ݔ‬ ൌ 6݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ସ
′
ൌ
5000
2 ‫כ‬ 53.3 ‫כ‬ 10଺
ሾെ6ଶ
൅ 6ଶሿ
‫ݕ‬ସ
′
ൌ ሺ0ሻ° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
Table-1
‫ݔ‬ሺ݉ሻ 0 2 4 6
‫ݕ‬′ሺ°ሻ ‫݁݌݋݈ݏ‬ 1.6885*10-3
1.5*10-3
9.38086*10-4
0.000
MATLAB PROGRAM-USING INTERPOLATION METHOD
% calculate the slope of beam at any point in between 1-class interval
% cantilever beam
% point load at free end
x=[0 2 4 6];
slope=[1.66885*10.^-3 1.5*10.^-3 9.38086*10.^-4 0.0];
xi=1;
yilin=interp1(x,slope,xi,'linear')
yilin = 0.0016° (Answer)
Plot the graph of slope of beam
% plot the graph of slope of beam
% cantilever thin beam
F=5000;
x=[0:1:6];
L=6;
EI=53.3*10.^6;
slope=(F/2)*[-x.^2+L.^2]/(EI);
plot(x,slope,'--r*','linewidth',2,'markersize',12)
xlabel('position along the axis (x)','Fontsize',12)
ylabel('position along the axis (y)','Fontsize',12)
title('slope of cantilever beam with point load at free end','Fontsize',12)

260
Figure-3
Deflection equation ‫ݕ‬ ൌ ݂ሺ‫ݔ‬ሻ[15]
‫ݕ‬ ൌ
‫ܨ‬
‫ܫܧ‬
ሾ
െ‫ݔ‬ଷ
6
൅
‫ܮ‬ଶ
‫ݔ‬
2
െ
‫ܮ‬ଷ
3
ሿ
‫ݐݑ݌‬ ‫ݔ‬ ൌ 0݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଵ ൌ
5000
53.3 ‫כ‬ 10଺
ቈ0 ൅ 0
െ6ଷ
3
቉ ൌ െ6.7542 ‫כ‬ 10ିଷ
݉
Similarly, ‫ݐݑ݌‬ ‫ݔ‬ ൌ 2݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଶ ൌ
5000
53.3 ‫כ‬ 10଺
ቈ
െ2ଷ
6
൅
6ଶ
‫כ‬ 2
2
െ6ଷ
3
቉ ൌ െ3.5021 ‫כ‬ 10ିଷ
݉
‫ݐݑ݌‬ ‫ݔ‬ ൌ 4݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଷ ൌ
5000
53.3 ‫כ‬ 10଺
ቈ
െ4ଷ
6
൅
6ଶ
‫כ‬ 4
2
െ6ଷ
3
቉ ൌ െ1.0 ‫כ‬ 10ିଷ
݉
‫ݐݑ݌‬ ‫ݔ‬ ൌ 6݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ସ ൌ
5000
53.3 ‫כ‬ 10଺
ቈ
െ6ଷ
6
൅
6ଶ
‫כ‬ 6
2
െ6ଷ
3
቉ ൌ െ0.0݉
Table-2
x(m) 0 2 4 6
y(m) -6.7542*10-3
-3.5021*10-3
-1.0*10-3
0.0
MATLAB PROGRAM- USING INTERPOLATION METHOD
% calculate the deflection of cantilever thin beam at any point in between
% any 1-class interval
x=[0 2 4 6];
y=[-6.7542e-3 -3.5021e-3 -1.0e-3 0.0];
xi=1;
yilin=interp1(x,y,xi,'linear')
yilin =-0.0051(Answer)
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
x 10
-3
position along the axis (x)
positionalongtheaxis(y)
slope of cantilever beam with point load at free end

261
Plot the graph of deflection
% plot the graph of deflection of beam
% cantilever beam
F=5000;
x=[0:1:6];
EI=53.3*10.^6;
L=6;
y=(F/EI)*[((-x.^3)/6)+((L.^2*x)/2)-((L.^3)/3)];
plot(x,y,'--r*','linewidth',2,'Markersize',12)
xlabel('position along the axis (x)','Fontsize',12)
ylabel('position along the axis (y)','Fontsize',12)
title('deflection of cantilever beam with point load at free
end','fontsize',12)
Figure-4
Case 2- Cantilever Thin Beam with Uniformly Distributed Load (U.D.L.)-
Figure-5
The bending moment at position ‫ݔ‬ is given by ‫ܯ‬ ൌ
ି௪௫మ
ଶ
. Substituting this equation (3) we
have,
‫ܫܧ‬
݀ଶ
‫ݕ‬
݀‫ݔ‬ଶ
ൌ
െ‫ݔݓ‬ଶ
2
Integrate wrt ‫ݔ‬ and we get,
0 1 2 3 4 5 6
-7
-6
-5
-4
-3
-2
-1
0
x 10
-3
deflection of cantilever beam with point load at free end

262
‫ܫܧ‬
݀‫ݕ‬
݀‫ݔ‬
ൌ
െ‫ݔݓ‬ଷ
6
൅ ‫ܣ‬ … … … … … … … … … … … … … … … … … … … … ሺ10ሻ
Integrate again we get,
‫ݕܫܧ‬ ൌ
െ‫ݔݓ‬ସ
24
൅ ‫ݔܣ‬ ൅ ‫ܤ‬ … … … … … … … … … … … … … … … … … … ሺ11ሻ
A and B are constants of integration and must be found from the boundary conditions. These
are,
ܽ‫ݐ‬ ‫ݔ‬ ൌ ‫,ܮ‬ ‫ݕ‬ ൌ 0 ሺ݊‫݋‬ ݂݈݀݁݁ܿ‫݊݋݅ݐ‬ሻ
ܽ‫ݐ‬ ‫ݔ‬ ൌ ‫,ܮ‬
݀‫ݕ‬
݀‫ݔ‬
ൌ 0 ሺ݄‫݈ܽݐ݊݋ݖ݅ݎ݋‬ሻ
Substitute ‫ݔ‬ ൌ ‫ܮ‬ ܽ݊݀
ௗ௬
ௗ௫
ൌ 0 ݅݊ ݁‫݊݋݅ݐܽݑݍ‬ ሺ10ሻ and we get,
‫ܫܧ‬ሺ0ሻ ൌ
െ‫ܮݓ‬ଷ
6
൅ ‫ܣ‬ ݄݁݊ܿ݁ ‫ܣ‬ ൌ
‫ܮݓ‬ଷ
6
Substitute this into (11) with the known solution
‫ݕ‬ ൌ 0 ܽ݊݀ ‫ݔ‬ ൌ ‫ܮ‬ ‫ݏݐ݈ݑݏ݁ݎ‬ ݅݊
‫ܫܧ‬ሺ0ሻ ൌ
െ‫ܮݓ‬ସ
24
൅
‫ܮݓ‬ଷ
6
൅ ‫ܤ‬ ݄݁݊ܿ݁ ‫ܤ‬ ൌ
െ‫ܮݓ‬ସ
8
Putting the results for A and B into equations in (10) and (11) yields the complete equations.
‫ܫܧ‬
݀‫ݕ‬
݀‫ݔ‬
ൌ
െ‫ݔݓ‬ଷ
6
൅ ‫ݓ‬
‫ܮ‬ଷ
6
… … … … … … … … … … … … … … … … … … … … … ሺ12ሻ
‫ݕܫܧ‬ ൌ
െ‫ݔݓ‬ସ
24
൅
‫ܮݓ‬ଷ
‫ݔ‬
6
െ
‫ܮݓ‬ସ
8
… … … … … … … … … … … … … … … … … … … … ሺ13ሻ
The main points of interest is the slope and deflection at free end where
‫ݔ‬ ൌ 0, ‫݃݊݅ݐݑݐ݅ݐݏܾݑݏ‬ ‫ݔ‬ ൌ 0 ݅݊‫݋ݐ‬ ሺ12ሻ ܽ݊݀ ሺ13ሻ gives the standard equations.
Slope at free end
ௗ௬
ௗ௫
ൌ
௪௅య
଺ாூ
… … … … … … … … … … … … … … … … … … … … … ሺ14ሻ
Deflection at free end ‫ݕ‬ ൌ
ି௪௅ర
଼ாூ
… … … … … … … … … … … … … … … … … … ሺ15ሻ
A cantilever thin beam is 6m long and has a U.D.L. of 300N/m. The flexural stiffness
is 60MN2
. Calculate the slope and deflection at the free end. [16]
Solution:-
Given data:- ‫ݓ‬ ൌ
ଷ଴଴ே
௠
, ‫ܫܧ‬ ൌ 60‫ܰܯ‬ଶ
, ‫ܮ‬ ൌ 6݉
Slope equation ‫ݕ‬′
ൌ ݉ ൌ
ௗ௬
ௗ௫
ൌ
௪
଺ாூ
ሾെ‫ݔ‬ଷ
൅ ‫ܮ‬ଷ
ሿ
‫ݐݑ݌‬ ‫ݔ‬ ൌ 0݉, ‫ܮ‬ ൌ 6݉
݉ଵ ൌ
ଷ଴଴
଺‫଺כ‬଴‫כ‬ଵ଴ల
ሾെ0ଷ
൅ 6ଷሿ ൌ 1.7999 ‫כ‬ 10ିସ
ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
‫ݐݑ݌‬ ‫ݔ‬ ൌ 2݉, ‫ܮ‬ ൌ 6݉
݉ଶ ൌ
ଷ଴଴
଺‫଺כ‬଴‫כ‬ଵ଴ల
ሾെ2ଷ
൅ 6ଷሿ ൌ 1.7332 ‫כ‬ 10ିସ
ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ,
‫ݐݑ݌‬ ‫ݔ‬ ൌ 4݉, ‫ܮ‬ ൌ 6݉
݉ଷ ൌ
ଷ଴଴
଺‫଺כ‬଴‫כ‬ଵ଴ల
ሾെ4ଷ
൅ 6ଷሿ ൌ 1.2667 ‫כ‬ 10ିସ
ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ and
‫ݐݑ݌‬ ‫ݔ‬ ൌ 6݉, ‫ܮ‬ ൌ 6݉
݉ସ ൌ
300
6 ‫כ‬ 60 ‫כ‬ 10଺
ሾെ6ଷ
൅ 6ଷሿ ൌ 0.00 ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ

263
Table-3
0 2 4 6
Slope(m) 1.7999*10.^-
4
1.7332*10.^-
4
1.2667*10.^-
4
0.00
% calculate the slope of thin beam at any point in between 1-class interval
% cantilever UDL
x=[0 2 4 6];
m=[1.7999e-4 1.7332e-4 1.2666e-4 0.00];
xi=1;
yilin=interp1(x,m,xi,'linear')
yilin =(1.7666e-004) (Answer)
Plot the graph of slope
% plot the graph of slope of beam
% cantilever udl thin beam
w=300;
x=[0:1:6];
EI=60*10.^6;
L=6;
m=(w/6)*(-x.^3+L.^3)/(EI);
plot(x,m,'--r*','linewidth',2,'markersize',12)
xlabel('position along the axis (x)','fontsize',12)
ylabel('position along the axis (y)','fontsize',12)
title('slope of cantilever udl thin beam','fontsize',12)
Figure-6
0 1 2 3 4 5 6
0
1
x 10
-4
slope of cantilever udl thin beam

264
Deflection equation:-
‫ݕ‬ ൌ
‫ݓ‬
‫ܫܧ‬
ቈ
െ‫ݔ‬ସ
24
൅
‫ܮ‬ଷ
‫ݔ‬
6
െ
‫ܮ‬ସ
8
቉
‫ݐݑ݌‬ ‫ݔ‬ ൌ 0݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଵ ൌ
300
60 ‫כ‬ 10଺
ቈെ0 ൅ 0 െ
6ସ
8
቉ ൌ െ8.1 ‫כ‬ 10ିସ
݉,
‫ݐݑ݌‬ ‫ݔ‬ ൌ 2݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଶ ൌ
300
60 ‫כ‬ 10଺
ቈെ
2ସ
24
൅
6ଷ
‫כ‬ 2
6
െ
6ସ
8
቉ ൌ െ4.533 ‫כ‬ 10ିସ
݉,
‫ݐݑ݌‬ ‫ݔ‬ ൌ 4݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ଷ ൌ
300
60 ‫כ‬ 10଺
ቈെ
4ସ
24
൅
6ଷ
‫כ‬ 4
6
െ
6ସ
8
቉ ൌ െ1.433 ‫כ‬ 10ିସ
݉
And ‫ݐݑ݌‬ ‫ݔ‬ ൌ 6݉, ‫ܮ‬ ൌ 6݉
‫ݕ‬ସ ൌ
300
60 ‫כ‬ 10଺
ቈെ
6ସ
24
൅
6ଷ
‫כ‬ 6
6
െ
6ସ
8
቉ ൌ 0.00݉
Table-4
‫ݔ‬ሺ݉ሻ 0 2 4 6
‫ݕ‬ሺ݉ሻ -8.1*10.^-4 -4.533*10.^-4 -1.433*10.^-4 0.00
% calculate the deflection of thin beam at any point in between 1-class
% interval
% cantilever udl thin beam
x=[0 2 4 6];
y=[-8.1*10.^-4 -4.533*10.^-4 -1.433*10.^-4 0.00];
xi=1;
yilin = -6.3165e-004 (Answer)
Plot the graph of deflection
% plot the graph of deflection of beam
% cantilever thin beam
% cantilever udl
w=300;
x=[0:1:6];
EI=60*10.^6;
L=6;
y=(w/EI)*[(-x.^4/24)+(L.^3*x/6)-(L.^4/8)];
plot(x,y,'--r*','linewidth',2,'markersize',12)
title('deflection of cantilever udl thin beam','fontsize',12)

265
Figure-7
Case 3- Cantilever Thin Beam with Uniformly Varying Load (U.V.L)-
Figure-8
Consider a section ‫ݔ‬ െ ‫ݔ‬ at a distance ‫ݔ‬ the fixed end AB intensity of loading at ‫ݔ‬ െ ‫ݔ‬.[17]
ൌ
ሺ‫ܮ‬ െ ‫ݔ‬ሻ‫ݓ‬
‫ܮ‬
‫ݎ݁݌‬ ‫ݐ݅݊ݑ‬ ‫݊ݑݎ‬
The bending moment at section ‫ݔ‬ െ ‫ݔ‬ is given by
‫ܫܧ‬
݀ଶ
‫ݕ‬
݀‫ݔ‬ଶ
ൌ െ
1
2
ሺ‫ܮ‬ െ ‫ݔ‬ሻ
‫ݓ‬
‫ܮ‬
ሺ‫ܮ‬ െ ‫ݔ‬ሻ.
ሺ‫ܮ‬ െ ‫ݔ‬ሻ
3
ൌ
െ‫ݓ‬ሺ‫ܮ‬ െ ‫ݔ‬ሻଷ
6‫ܮ‬
Integrating we get,
‫ܫܧ‬
݀‫ݕ‬
݀‫ݔ‬
ൌ
‫ݓ‬ሺ‫ܮ‬ െ ‫ݔ‬ሻସ
24‫ܮ‬
൅ ‫ܥ‬ଵ
At B the slope is 0,
0 1 2 3 4 5 6
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
x10
-4
deflectionof cantilever udlthinbeam

266
Therefore,
At ‫ݔ‬ ൌ 0
݀‫ݕ‬
݀‫ݔ‬
ൌ 0
0 ൌ
‫ܮݓ‬ଷ
24
൅ ‫ܥ‬ଵ
Then ܿଵ ൌ െ
௪௅య
ଶସ
‫ܫܧ‬
ௗ௬
ௗ௫
ൌ
௪ሺ௅ି௫ሻర
ଶସ௅
െ
௪௅య
ଶସ
… … … … … … … … … … … … … … . . ሺ16ሻ Slope equation
Integrating again we get,
‫ݕܫܧ‬ ൌ െ
‫ݓ‬ሺ‫ܮ‬ െ ‫ݔ‬ሻହ
120‫ܮ‬
െ
‫ܮݓ‬ଷ
24
‫ݔ‬ ൅ ܿଶ
The deflection at B is 0
‫ݔ‬ ൌ 0, ‫ݕ‬ ൌ 0
0 ൌ െ
‫ܮݓ‬ସ
120
൅ ܿଶ
ܿଶ ൌ
‫ܮݓ‬ସ
120
Therefore,
௪ሺ௅ି௫ሻఱ
ଵଶ଴௅
െ
௪௅య௫
ଶସ
൅
௪௅ర
ଵଶ଴
… … … … … … … … … … ሺ17ሻ Deflection equation
To find the slope at C at the free end putting the value ‫ݔ‬ ൌ ‫ܮ‬ in the slope equation we
get,[18]
‫ܫܧ‬
݀‫ݕ‬
݀‫ݔ‬
ൌ െ
‫ܮݓ‬ଷ
24
Therefore,
݀‫ݕ‬
݀‫ݔ‬
ൌ െ
‫ܮݓ‬ଷ
24‫ܫܧ‬
To find the deflection at C putting ‫ݔ‬ ൌ ‫ܮ‬ in the deflection equation we get,
‫ܮݓ‬ସ
24
൅
‫ܮݓ‬ସ
120
ൌ െ
‫ܮݓ‬ସ
120
ሺ5 െ 1ሻ ൌ െ
‫ܮݓ‬ସ
30
Therefore,
‫ݕ‬ ൌ െ
‫ܮݓ‬ସ
30‫ܫܧ‬
Download deflection of ܿ ൌ
௪௅ర
ଷ଴ாூ
Cantilever of length ‫ܮ‬ ൌ 6݉ carrying a distributed load whose intensity varies
uniformly from zero at the free end to 800N/m at fixed end, flexural stiffness ሺ‫ܫܧ‬ ൌ
643300ܰ݉ଶ
ሻ.[19]
Solution:-
Given data:-
Span lengthሺ‫ܮ‬ሻ ൌ 6݉, ‫ݓ‬ ൌ
଼଴଴ே
௠
, ‫ܫܧ‬ ൌ 643300ܰ݉ଶ
,
Slope equation
ௗ௬
ௗ௫
ൌ ‫ݕ‬′
ൌ ݉ ൌ
௪
ଶସாூ
ቂ
ሺ௅ି௫ሻర
௅
െ ‫ܮ‬ଷ
ቃ

267
‫ݐݑ݌‬ ‫ݔ‬ ൌ 0, ‫ܮ‬ ൌ 6݉
‫′ݕ‬ଵ
ൌ ݉ଵ ൌ
଼଴଴
ଶସ‫଺כ‬ସଷଷ଴଴
ቂ
ሺ଺ି଴ሻర
଺
െ 6ଷ
ቃ ൌ 0.00° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ,
‫ݐݑ݌‬ ‫ݔ‬ ൌ 2, ‫ܮ‬ ൌ 6݉
‫′ݕ‬ଶ
ൌ ݉ଶ ൌ
଼଴଴
ଶସ‫଺כ‬ସଷଷ଴଴
ቂ
ሺ଺ିଶሻర
଺
െ 6ଷ
ቃ ൌ െ0.0090° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ,
‫ݐݑ݌‬ ‫ݔ‬ ൌ 4, ‫ܮ‬ ൌ 6݉
‫ݕ‬Ԣଷ ൌ ݉ଷ ൌ
800
24 ‫כ‬ 643300
ቈ
ሺ6 െ 4ሻସ
6
െ 6ଷ
቉ ൌ െ0.0111° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
‫ݐݑ݌‬ ‫ݔ‬ൌ 6, ‫ܮ‬ൌ 6݉
‫′ݕ‬4
ൌ ݉4 ൌ
800
24 ‫כ‬ 643300
ቈ
ሺ6 െ 6ሻ4
6
െ 63
቉ ൌ െ0.0112° ሺ݊‫݋‬ ‫ݏݐ݅݊ݑ‬ሻ
Table-4
‫ݔ‬ሺ݉݁‫݁ݎݐ‬ሻ 0 2 4 6
‫݁݌݋݈ݏ‬ሺ݉ሻ° 0 -0.0090 -0.0111 -0.0112
% calculate the slope of thin beam at any point in between 1-class interval
% cantilever U.V.L.
x=[0 2 4 6];
slope=[0 -0.0090 -0.0111 -0.0112];
xi=1;
yilin=interp1(x,slope,xi,'linear')
yilin = -0.0045°
Plot the graph of slope of cantilever U.V.L.
MATLAB PROGRAM-
% plot the graph of slope of thin beam
% cantilever U.V.L.
w=800;
x=[0:1:6];
EI=643300;
L=6;
m=(1/24)*(w/EI)*[((L-x).^4/L)-L.^3];
plot(x,m,'--r*','linewidth',2,'markersize',12)
title('slope of cantilever U.V.L. thin beam','fontsize',12)

268
Figure-9
Deflection equation W - ሺL-xሻ5 L4
y ൌ - L3 x ൅ ሾ20ሿ
24 EI 5L 4
Put x = 0m, L = 6m
800 - (6-0)5
64
y 1 = - 63
* 0 + = 0.00m
24*70*109
*9.19*10- 6
5*6 5
x = 2m, L = 6m
800 - (6-2)5
64
y 2 = - 63
* 2 + = -0.011m,
24*70*109
*9.19*10- 6
5*6 5
x = 4m, L = 6m
800 - (6-4)5
64
y 3 = - 63
* 4 + = -0.031m,
24*70*109
*9.19*10- 6
5*6 5
x = 4m, L = 6m
800 - (6-6)5
64
y 4 = - 63
* 6 + = -0.054m,
24*70*109
*9.19*10- 6
5*6 5
Table-5
x (metre) 0 2 4 6
y(metre) 0.0000 -0.0110 -0.0310 -0.0540
0 1 2 3 4 5 6
-0.012
-0.01
-0.008
-0.006
-0.004
-0.002
0
slope of cantilever U.V.L. thin beam

269
% calculate the deflection of beam at any point in between 1-class interval
% cantilever U.V.L. thin beam
x=[0 2 4 6];
y=[0.0000 -0.0110 -0.0310 -0.0540];
xi=1;
yilin =-0.0055
Answer is -0.0055m
Plot the graph of deflection of cantilever U.V.L.-
MATLAB PROGRAM-
% plot the graph of deflection of thin beam
% cantilever U.V.L.
w=800;
x=[0 2 4 6];
EI=643300;
L=6;
y=(1/24)*(w/EI)*[(-(L-x).^5/(5*L))-L.^3*x+L.^4/5];
plot(x,y,'--r*','linewidth',2,'markersize',12)
title('deflection of cantilver U.V.L.thin beam','fontsize',12)
Figure-10
FUTURE SCOPE
1- It can be extending as apply in simply supported thin beam (point load at mid, uniformly
distributed load and uniformly varying load).
2- It can be extending in case of composite materials of beam which are non-isotropic.
3- It can be extending that is used in trusses like perfect, deficient and redundant.
4- It can be extending that is used in tapered and triangular beam.
5- It can be extending in Aeronautics, Aerodynamics and Space Engineering which is consisting
of fixed vanes and crossed moving fixed vanes in rotor.
6- It can extending in Orthopaedics in Medical Sciences which is applicable in replace or
support to the bones.
0 1 2 3 4 5 6
-0.06
-0.05
-0.04
-0.03
-0.02
-0.01
0
deflection of cantilver U.V.L. thin beam

270
DISCUSSION AND CONCLUSION
It was observed that in case of Cantilever thin beams (point load at free end, u.d.l. and
u.v.l.) are carried out by the numerical analysis and MATLAB programming, made the table
of ‫ݔ‬ verse slope and ‫ݔ‬ verse deflection after that taken at any one point in between any 1-
class-interval in thin beam and then calculated value at same point by using Interpolation
Method through the MATLAB programming to analysed the value at that point is slope and
deflection, we have analyzed by plotted the graph of static Slopes and Deflections of thin
beams through the MATLAB programming.
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Static analysis of thin beams by interpolation method approach

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