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LECTURE 30 and 31

Deflection of Beams

Introduction:

In all practical engineering applications, when we use the different components, normally we have to operate themwithin the certain limits i.e. the constraints are placed on the performance and behavior of the components. For

instance we say that the particular component is supposed to operate within this value of stress and the deflection ofthe component should not exceed beyond a particular value.

In some problems the maximum stress however, may not be a strict or severe condition but there may be thedeflection which is the more rigid condition under operation. It is obvious therefore to study the methods by which we

can predict the deflection of members under lateral loads or transverse loads, since it is this form of loading whichwill generally produce the greatest deflection of beams.

Assumption: The following assumptions are undertaken in order to derive a differential equation of elastic curve forthe loaded beam

1. tress is proportional to strain i.e. hooks law applies. Thus, the equation is valid only for beams that are notstressed beyond the elastic limit.

!. The curvature is always small.

". #ny deflection resulting from the shear deformation of the material or shear stresses is neglected.

It can be shown that the deflections due to shear deformations are usually small and hence can be ignored.

$onsider a beam #% which is initially straight and hori&ontal when unloaded. If under the action of loads the beamdeflect to a position #'%' under load or infact we say that the axis of the beam bends to a shape #'%'. It is customary

to call #'%' the curved axis of the beam as the elastic line or deflection curve.

In the case of a beam bent by transverse loads acting in a plane of symmetry, the bending moment ( varies alongthe length of the beam and we represent the variation of bending moment in %.( diagram. Futher, it is assumed that

the simple bending theory equation holds good.

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If we look at the elastic line or the deflection curve, this is obvious that the curvature at every point is different) hence

the slope is different at different points.

To express the deflected shape of the beam in rectangular co*ordinates let us take two axes x and y, x*axis coincidewith the original straight axis of the beam and the y + axis shows the deflection.

Futher,let us consider an element ds of the deflected beam. #t the ends of this element let us construct the normalwhich intersect at point denoting the angle between these two normal be di

%ut for the deflected shape of the beam the slope i at any point $ is defined,

This is the differential equation of the elastic line for a beam sub-ected to bending in the plane of symmetry. Itssolution y f/x0 defines the shape of the elastic line or the deflection curve as it is frequently called.

Relationship et!een shear force" endin# moment and deflection:The relationship among shear force,bending

moment and deflection of the beam may be obtained as

ifferentiating the equation as derived

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Therefore, the above expression represents the shear force whereas rate of intensity of loading can also be foundout by differentiating the expression for shear force

$ethods for findin# the deflection: The deflection of the loaded beam can be obtained various methods.The one

of the method for finding the deflection of the beam is the direct integration method, i.e. the method using thedifferential equation which we have derived.

Direct inte#ration method: The governing differential equation is defined as

2here # and % are constants of integration to be evaluated from the known conditions of slope and deflections forthe particular value of x.

Illustrati%e e&amples :let us consider few illustrative examples to have a familiarty with the direct integration

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method

Case 1: $antilever %eam with $oncentrated 3oad at the end4* # cantilever beam is sub-ected to a concentrated load2 at the free end, it is required to determine the deflection of the beam

In order to solve this problem, consider any 5*section 5*5 located at a distance x from the left end or the reference,and write down the expressions for the shear force abd the bending moment

The constants # and % are required to be found out by utili&ing the boundary conditions as defined below

i.e at x 3 ) y 6 ******************** /10

at x 3 ) dy7dx 6 ******************** /!0

8tili&ing the second condition, the value of constant # is obtained as

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Case ': # $antilever with 8niformly distributed 3oads4* In this case the cantilever beam is sub-ected to 8.d.l with rateof intensity varying w 7 length.The same procedure can also be adopted in this case

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%oundary conditions relevant to the problem are as follows4

1. #t x 3) y 6

!. #t x 3) dy7dx 6

The second boundary conditions yields

Case 3: imply upported beam with uniformly distributed 3oads4* In this case a simply supported beam is sub-ectedto a uniformly distributed load whose rate of intensity varies as w 7 length.

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In order to write down the expression for bending moment consider any cross*section at distance of x metre from leftend support.

%oundary conditions which are relevant in this case are that the deflection at each support must be &ero.

i.e. at x 6) y 6 4 at x l) y 6

let us apply these two boundary conditions on equation /10 because the boundary conditions are on y, This yields % 6.

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Futher

In this case the maximum deflection will occur at the centre of the beam where x 37! 9 i.e. at the position where theload is being applied :.o if we substitute the value of x 37!

$onclusions

/i0 The value of the slope at the position where the deflection is maximum would be &ero.

/ii0 Thevalue of maximum deflection would be at the centre i.e. at x 37!.

The final equation which is governs the deflection of the loaded beam in this case is

%y successive differentiation one can find the relations for slope, bending moment, shear force and rate of loading.

Deflection ()*

+lope (d),d&*

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Bendin# $oment

o the bending moment diagram would be

+hear -orce

hear force is obtained by taking

third derivative.

Rate of intensit) of loadin#

Case .: The direct integration method may become more involved if the expression for entire beam is not valid forthe entire beam.3et us consider a deflection of a simply supported beam which is sub-ected to a concentrated load 2

acting at a distance 'a' from the left end.

3et ;1< ;!be the reactions then,

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These two equations can be integrated in the usual way to find =y' but this will result in four constants of integrationtwo for each equation. To evaluate the four constants of integration, four independent boundary conditions will be

needed since the deflection of each support must be &ero, hence the boundary conditions /a0 and /b0 can bereali&ed.

Further, since the deflection curve is smooth, the deflection equations for the same slope and deflection at the pointof application of load i.e. at x a. Therefore four conditions required to evaluate these constants may be defined as

follows4

/a0 at x 6) y 6 in the portion #% i.e. 6 > x > a

/b0 at x l) y 6 in the portion %$ i.e. a > x > l

/c0 at x a) dy7dx, the slope is same for both portion

/d0 at x a) y, the deflection is same for both portion

%y symmetry, the reaction ;1is obtained as

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8sing condition /c0 in equation /"0 and /?0 shows that these constants should be equal, hence letting

@1 @! @

Aence

Bow lastly k"is found out using condition /d0 in equation /C0 and equation /D0, the condition /d0 is that,

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#t x a) y) the deflection is the same for both portion

ALTER/ATE $ETD: There is also an alternative way to attempt this problem in a more simpler way. 3et usconsidering the origin at the point of application of the load,

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%oundary conditions relevant for this case are as follows

/i0 at x 6) dy7dx 6

hence, # 6

/ii0 at x l7!) y 6 /because now l 7 ! is on the left end or right end support since we have taken the origin at thecentre0

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Aence the integration method may be bit cumbersome in some of the case. #nother limitation of the method wouldbe that if the beam is of non uniform cross section,

i.e. it is having different cross*section then this method also fails.

o there are other methods by which we find the deflection like

1. (acaulay's method in which we can write the different equation for bending moment for different sections.

!. #rea moment methods

". Energy principle methods

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