6. deflection

INTRODUCTION
1
The cross section of a beam has to be designed in such
a way that it is strong enough to limit the bending moment
and shear force that are developed in the beam. This criterion
is known as the STRENGTH CRITERION of design .
Another criterion for beam design is that the
maximum deflection of the beam must not exceed a given
permissible limit and the beam must be stiff enough to resist
the deflection caused due to loading. This criterion is known
as “STIFFNESS CRITERION of design”

DEFINITIONS:-
(i) DEFLECTION:-
2
The neutral axis in its deflected position after loading of the
beam is known as its elastic curve or deflection curve
(ii) ELASTIC CURVE(OR, DEFLECTION CURVE):-
The vertical distance in transverse direction between
positions of axis before and after loading at the section of
the beam, is defined as the deflection of beam at that
section.

The product of modulus of elasticity and Moment of
Inertia is known as Flexural rigidity.
3
(iii) SLOPE:-
The slope of the beam at any section is defined as the angle
(in radians) of inclination of the tangent drawn at that
section to the axis in its deflected position after loading,
measured w. r. t. The un-deformed axis.
(iv) FLEXURAL RIGIDITY (EI):-

DIFFERENTIAL EQUATION OF ELASTIC CURVE:-
(SLOPE AND DEFLECTION)
Differential equation of elastic curve is,
4
d2y
dx2
= M
E I
ASSUMPTIONS :
(i) Axis of the beam is horizontal before loading.
(ii) Deflection due to S.F. is negligible.

ASSUMPTIONS (CONTD..) :
(d) Plane section remain plane before and after bending.
5
(iii) All the assumptions made in simple bending theory
are valid.
( a) Simple Bending equation M/I=σ/y=E/R is applicable
(b) Material of the beam is homogenous, isotropic and
obey Hooke’s law.
(c) The modulus of elasticity is same in compression as
well as in tension.

Slope ,deflection and radius of curvature
Φ
Φ+dΦ
P
Q
C
dy
dx
A B
y
o x
dΦ
R
6

Therefore, R = ds / dφ
7
Consider a piece of deflected curve of beam PQ of length ds.
Let tangent at P and Q make angles φ and (φ+dφ) with x-axis
at a point A and B respectively.
Point C is the centre of curvature.
The distance CP = CQ = R = radius of curvature
PQ = ds = tangent length
Now ds = R.dφ
P
Q
C
dy
dx
dΦ
R
Φ Φ+dΦ

8
dy
dx
φ dx
ds
φ
ds
dy
If (x,y) is the co-ordinate of the point P
Then
dy
ds
= sinφ = cosφ
dx
ds
= tanφ
dy
dx
R =
ds
dφ
=
ds dx
dφ dx
× =
dφ
dx
ds
dx
=
dφ
dx
secφ
-------(1)
Differentiating tanφ w. r.t. x,
sec2φ =
dφ
dx
d2y
dx2
= sec
2
φ --(2)
Therefore,
dφ
dx
d2y
dx2

In this, (dy/dx) is a small quantity (because curve is almost flat); hence
dy/dx can be ignored. Hence,
9
From equation(1), R =
=
dφ
dx
secφ
=
sec3φ
d2y
dx2
=
sec3φ
d2y
dx2
1
R
=> =
d2y
dx2
dy
dx
2 3/2
1 +
1
R
=
d2y
dx2

(a) Slope ( dy /dx )
(b) Deflection()
(c) The position and magnitude of maximum slope
(d) The posititon and magnitude of maximum deflection.
10
From relation ,
M
I
E
R
=
1
R
=
d2y
dx2
M
EI
=
∴
M = EI
d2y
dx2
=>
SIMPLY SUPPORTED BEAM :-
In case of simply supported beam, terms to be found are:

SIGN CONVENTIONS:
Linear horizontal distance x : positive when measured from left
to right ( )
11
Vertical distance or deflection y is positive when measured above the
axis and is negative when measured below the axis of beam.
Methods for finding slope and deflection of beams:
(i) Double integration method
(ii) Macaulay’s method
(iii) Area moment method
(iv) Conjugate beam method
(v) Unit load method

(i)Simply suported beams:
B
Deflection at support Deflection at
A and B are zero and support A and B
are zero but more at
maximum at the middle of free end and also at the centre
span; slope θ is maximum at of span . Slope is
A and B and zero at middle of maximum at supports
span; θA ( )-ve and θB( )+ve. B and C.
θB
θA
C A B
D
A
12
NOTE : SUPPORT CONDITIONS:

(ii)Cantilever Beam:
Deflection and slope both are zero
at fixed support.
Slope and deflections are
maximum at free end
θ increases from
point A towards B.
B
θmax
ymax
A
13

We have from differential equation of flexure,
EId2
y/dx2
=M
Integrating w. r. t.. x both sides, we get
EI (dy /dx) =∫M dx +C1
Integrating again w .r .t. x both sides ,we get
EI (y)=∫∫M dx. dx +C1(x)+C2
where C1 and C2 are constant of integration
Thus, integrating the differential equation w .r .t. x, we get the
equation for slope (dy /dx) ,and integrating it twice w. r .t. x, we get
the equation for deflection ( y).
14
DOUBLE INTEGRATION METHODS :

The method of double integration is convenient only in
few cases of loadings. In general case of loading, where
several loads are acting the B.M. expression for each of
the different regions of loading is different .Then the
method of double integration will become extremely
lengthy and laborious. Therefore ,it is not generally used.
15
The two constants of integration can be determined using
the known conditions of deflection and slope at the
supports

Case--1(i): Determine the slope and deflection equation for
the beam loaded as shown in fig.
(ii) Find the maximum deflection and maximum slope.
Solution:
B.M. at section 1-1
M=- P( x)
EI d
2
y/dx
2
=M=- P (x)
EI (dy /dx ) =-P(x
2
/2) +C1
EIy=-P(x3
/2*3) + C1x+ C2
At fixed end, when x= L (x=0, at free
end)
16
P
1
1
x
A B
θA l
yA

(dy /dx) =0
Therefore,
-(PL
2
/2)+C1=0
C1=PL
2
/2
At x=L,y=0,
-PL
3
/6+(PL
2
/2)l+C2=0
or,C2=PL
3
/6-PL
3
/2
=PL3
/6[1-3]=-PL3
/3
Therefore ,C2=-PL
3
/3
17

Equation of slope; EI (dy/ dx) =-Px
2
/2+PL
2
/2-----(1)
Equation of deflection ,EI (y)=-Px3
/6+PL2
x/2-PL3
/3-----(2)
Maximum deflection :
When x=0 (at free end) ,then from equation (2),
EI(y)=-0+0-PL3
/3
ymax= -PL
3
/3EI
Maximum Slope:
Slope is maximum at free end (at x=0).hence from equation (1),
EI (dy/ dx)=-0+PL2
/2
(dy /dx) max=PL
2
/2EI
18

θA=8.036*10
-5
radian =0.46
0
Deflection at free end (i.e; at A):= yA=PL
3
/3EI
=(6*3
3
) /3 (210*10
6
*16*10
-4
) => yA=0.161mm
Solution: I =16x10
4
cm
4
=16x10
4
x(10
-2
)
4
m
4
=16x10
-4
m
4
.
θA=PL
2
/2EI
slope θA=(dy/dx) at A=[1/(210x106x16x10-4)]x[6x(3)2/2]
19
Exercise Problem-(1) What will be the deflection and slope
at free end when P=6kN, L=3m, E=210GPa, I=16x104cm4.

Case (2) : When cantilever is subjected to an u .d. L. of
intensity w unit/m run over entire span
Here A is the origin. Take a
section X-X at a distance x from
A.
B.M at distance x from A= Mx =EId
2
y/dx
2
=-w.x.x/2=-wx
2
/2
Integrating once,
EI (dy/dx) =-wx
3
/6+C1 ------------------------(1)
where C1 is constant of integration
B
A
L
x
w unit /m
X
X
20

Applying boundary conditions:-
at x=L, dy/dx=0 from equation(1)
0=-wL
3
/6+C1 C1=wL
3
/6
therefore,
EI dy/dx = -wx3
/6+wL3
/6---------(2)
Integrating once again,
EI y=-wx
4
/24 +wL
3
.x/6 +C2 ---------------- (3)
where C2 is 2nd constant of integration
Applying boundary condition;
at x=L, y=0
21

0=-wL
4
/24+wL
4
/6+C2
Therefore,
C2=wL4
/24-wL4
/6
C2=-wL
4
/8.
Therefore, equation (3) becomes,
EI(y)=-wx4
/24 + wL3
.x/6 –wL4
/8--------(4)
Maximum deflection
It occurs at free end where x=0
From (4),EIy=-0+0-wL4
/8
22

ymax=-wL4
/8EI
similarly maximum slope
It occurs at free end where x=0
from (2),
EI (dy/dx) =-0+wL
3
/6
(dy/dx )max=wL
3
/6EI
23

Case 3:-When simply supported beam is
subjected to a single concentrated
load at mid-span.
[Ans;
Downward deflection at
centre ;at point
C=PL
3
/48EI
Slope max. at the
ends=-PL
2
/16EI ]
L/2 L/2
P
P
C
A B
RA=P/2
RB=P/2
x X
X
24

Case 4:-Simply supported beam of span l carrying a
uniformly distributed load of intensity w per unit run over
the whole span.
[Ans: ymax=-5wL
4
/(384EI)
(dy/dx)A =-wL
3
/24EI]
A
x
W unit / run
B
RA=WL/2 RB=WL/2
25
X
X

MACAULAY’S METHOD
For a general case of loading on the beam ,though the
expression for B.M. varies from region to region ,the
constants of integration remain the same for all regions.
Macaulay recognized this fact and proposed an
ARTIFICE (a method )which is known as the Macaulay’s
method to obtain the slope and deflection equations.
It is essentially modified method of double
integration of the B.M. expression but following a set of
rules given below:-
26

(1)Assuming origin of the beam at extreme left end, take a
section in the last region of the beam at a distance x from the
origin and write the expression for B.M. at that section
considering all the force on the left of section.
(2)Integrate the term of the form (x-a)
n
using the formula
∫(x-a)
n
dx=(x-a)
n+1
/n+1
where a=distance of load from origin.
(3)While finding slope and deflection in the form (x-a)
n
,if
(x-a) becomes negative on substituting the value of x,
neglect the term containing the factor (x – a)
n
.
27

(4)If a couple (moment) of magnitude ‘c’ is acting at a
distance ‘a’ from the origin of the beam, then write the BM
due to couple in the form c (x-a).
(5)If the beam carries a U.D.L, extend it up to the extreme
right end and superimpose an UDL equal and opposite to
that which has been added while extending the given UDL
on the beam.
28

Practice problems:-
(Q-1) A cantilever beam of span L carries a udl of intensity
w/unit length for half of its span as shown in figure. If E is the
modulus of elasticity and I is moment of inertia, determine the
following in terms of w, L, E and I.
(i) A expression for slope (dy/dx) at free end
(ii) An expression for deflection( y ) at free end
(iii) The magnitude of upward vertical force to be applied at free
end in order to resume this end to the same horizontal level as built
in end.
45
[Ans (i)θA=WL
3
/48EI (ii)A =7wL
4
/384EI( ) (iii)P=7wL/128]
L/2 L/2
w/m
A B

Q- (2 ) Determine the values of deflections at points C,D and E in
the beam as shown in figure. Take E=2*10
5
MPa ; I= 60 *10
8
mm
4
1m 2m
10kN/m
1m 1m
20kN
30kN
A C D E B
[C=0.0603mm(downward), D=0.0953mm(downward)
E=0.0606mm(downward)]
47

Q-(3) Find the position and magnitude of maximum deflection for
the beam loaded as shown in fig. Take E=200GPa ,I=7500cm
4
.
3KN/m
20 KN
A
D B
X
4m 4m 4m
C
X
X
[Ans: ymax at 3.7 m from A=-118/EI=7.99mm
yc=-32/EI=-2.13mm]
48

Q-(4) Determine the magnitude and position of maximum deflection
for the beam loaded as shown in fig. Take EI=800Nm
2
[Ans:ymax =80 mm at 1.59m from A , yE =73mm]
E
B
D
A
C
1m
1m 1m 1m
120kNm
80kN 20kN
49

Q-(5) Find the deflection and slope at free end for loaded
beam shown in fig.
4kN/m
10 KN
1m 2m 1m
D
A
B
C
[Ans:θD=62.33/EI, y=-191/EI ]
Q-(6) Find the deflection at C and magnitude of maximum
deflection. Take EI=40MN-m
2
4m 2m
200KN
1m
B
A
C
[Ans: ymax=-13.45mm, yC=-13.33mm ]
50

6. deflection

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6. deflection