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( )[ ]IR
d y d x
d y d x=
+
2 2
23
21
DEFLECTION OF BEAMS
The configuration of the deformed neutral surface is call elastic curve of the beam.
Specifications for design of beams frequently impose limitations on deflections and
stresses.
In many building codes, maximum allowable deflection = 1/300 span components of
aircraft should not exceed certain deformation, else the aero dynamic
characteristics may be altered.
METHODS OF COMPUTING DEFLECTION OF BEAMS
a. Double integration method (Calculus)
b. Moment area method
c. Method of singularity functions (Macaulay’s method)
d. Elastic energy method (Castigliano’s theorem)
(a) Double Integration Method
E Id yd x
M2
2 1= � � � � � ( )
First integration yield slope while second integration yields deflection, yd yd x
Differential equation of the deflection curve of a beam loaded by lateral forces.
It is assumed that deflections caused by shearing action are negligible compared to
these cause by bending action.
where represents Curvature of neutral surfaceME I
IR
= IR
The exact formula for curvature of a curve ( )y f x P=
x
L
AB x
Py
M P L x E Id ydx
= − − =( )2
2
E Id yd x
P L xP x
C= − + +2
12
∴ =C1 0
∴ =1 2
2Rd yd x
where represent slope of curve at any point.dydx
For small deflection, and is small compare to unity d yd x
dydx
2
∴ =E Id yd x
M2
2
(Euler. Bernoulli Equation of bending of a beam loaded by lateral forces).
Exercise
Determine the deflection at every point of the cantilever beam and the maximum
deflection subject to the single concentrated force P.
Condition
Slope at A is zero at x = o
E IyP L x P x
C= − + +2 3
22 6
At x=0, y=0 � C2=0
CW L
1
3
2 4=
−
E IyW L x W x W L x
C= − − +3 4 3
212 24 24
MW L x W x
xE I
d ydx
= − =2 2
2 2
2
xL
=2
Maximum deflection at x = L
(-ve denotes below x – axis)E IyP L
m ax =− 3
3
DP L
E Im ax = 3
3
Exercise
Obtain an expression for deflection curve of the beam subjected to a uniformly
distributed load of a unit length.
If L = 3.5m, P = 60KN, depth = 430mm, I = 2.5 x 108mm2
E = 210GPa, Determine the maximum deflection of beam.
Determine the slope of the right end of the cantilever.
Since
E Idydx
W L x W L xC= − +
2 3
14 6
at centre since beam is symmetical.dydx
= 0
M1
R=M1/L
E Id yd x
ML
x2
21=
E Id yd x
ML
xC= +1
2
12.
E IyM
Lx
C x C= + +13
1 22 3.
xL
=2
DW LE Im ax .=
5384
4
d yd x
xL
= ⇒ =03
Maximum deflection at centre
y = 0 at A, � C2 = 0
Maximum deflection at centre
y = 0 at x = 0, � C2 = 0
y = 0 at x = L, C1 = − M L1
6
Deflection curve is
E IyM x
LM L
x= −13
1
6 6
More deflection when
M1
L/4
wkN/m
A B C DE
L/4
a bP
L
a
P
b c
P
E IyM L
m a x =− 1
2 327
Determine equation of deflection at B and C
6m 3m
W
L
P
( )E Iyw
L xw L
xw L
=−
− − +24 6 24
43 4
Point Load P
E IyP L x P x
= − +2 3
2 6
For uniformly distributed load w
Using method of superpostion
( )E IyP L x P x w
L xw L
xw L
= − + − − − +2 3
43 4
2 6 2 4 6 2 4
(b) Moment – Area Method
1st moment – area theorem
M = EI - (I)
R
Ds = Rdq
R = ds/dce
Sub in (I)
dQ = M ds - (2)
EI
Ds = dx
dQ = Mdx
EI
Angle A and B
Q = dQ = B M dx
A EI
i.e The increase of slope between any two points on a beam is equal to the net area
of the BMD between these points divided by EI.
2nd Moment – Area theorem
vertical contribution of ds = xdQ
substituting (2)
xdQ = MX dx
EI
D = x dQ = A M x dx
B EI
In worlds, this equation states that if A and B are points on the deflection curve of a
beam, the vertical distance of B from the tangent drawn to the curve at A is equal to
the moment write the vertical through B of the area of the BMD between A and B,
divided by EI.
Area of shaded and A = 1/2 (WC/y) (
1/2)
= WL2/16
slope at support = - A/EI
= WL2/16 EI
deflection of support reduction to centre
= Ax/EI
= WL2 L/3
16 EI
= WL3
48EI
exa. Store and deflections of a S. S beam of span A will n.d.l
Shaded area A = 2/3 (WL2/8) (1/2)
= WL 3/24
Slope at support = - A/EI
= - WL3/24 EI
Deflection of support relation to certain = A x/EI
= (WL3/24) (5/16l)
EI
= 5WL 4/384 EI
Moment Area Method (Mohr and Greene)
(q) First Area Theorem
Bo
x
A
x
A
y
d yd x
=ME I
2
2
Integrating between A and B
d yd x
d yd x
M dxE I
AE I
i e QA
E I
B A A
B
−
= ∫ =
=. .
The first moment � Area theorem (for slope) states that: The increase of slope (angle
between the tangents) between two points A and B is equal to the area of the B.M.D
between these two points divided by EI
i e QA
E I. =
(b) Second Moment � Area Theorem
If �
is the deflection of point B on the deflection curve to the tangent to this curve drawn
at point A.
The second moment � Area Theorem (for deflection) states that: The vertical deflection of
point B on a deflection curve from the tangent drawn to the curve at A is equal to the
moment of the area of the bending moment diagram between A and B form B divided by
EI.
i eM xd x
E IA xE IA
B
.
_
∆ = ∫ =
Hence, deflection at nay point can be found by chosen a point A where the slope is zero,
and taking moments about the point where deflection is required.
e.g.
To find the deflection under the point load to the end of a cantilever beam
BL
P
Tangent at B
2L/3
PLB.M.D
w kN/m
L
a b
1.0m 2.0m
L
BA
A B
A = Area of bending moment diagram = − − =−
P LL P L2 2
2
∆ =−
=−P L
xL
xE I
P LE I
2 2
223
13
-ve indicates final psition of B
Determine the slope at the end B of the cantilever beam from first moment � area theorem
QA
E IP LE I
= =− 2
2
-ve indicates clockwise angle at B
∆ =− w L
E Ia t end
4
8
deflection at centre of span
dcentre = 6.18mm (Steel designer)
M = P ab
m ax L
Dmax always occurs within 0.07744 of the centre after beam, when
b a≥
Dcentre= −
P LE I
aL
aL
3 3
4 83
4
also within 2.5% of dmax
L/2 L/2
P
P/2 P/2
Dmax =P L
E Ia t cen tre
3
4 8
the max slope and deflection
WL/8
5/8 .L/2
L/2
2
A BC
max
RA=wL/2 RB=wL/2
WL/2
L/2
B.M.D
Solution
By symmetry, the slope is zero at the centre, hence, max, slope and deflection
can be found from the area of the B.M.D over half the beam i.e. between A
and C.
Shaded Area (Parabola)
Aw l L
w L
=
=
23 8 2
2 4
2
3
Slope between support A and point C = slope at support A ≡
maximum slope = =AE I
w LE I
' 3
24
Deflection of support A relative to centre C where slope is zero (note is−x
taken moment of B.M.D about point A)
A xE I
w LL
E Iw L
E I
−
=
=
3
4
2 45
1 6
53 84
Macaulay’s method
One continous expression for bending moment
Example
10m
BA
15m20m
5kN 8kN
x
Question :
(a) Calculate deflection at C and D
(b) Maximum deflection
E = 2 X 105 N/mm2, I = 1 X 109 mm4
Solution :
20RA = 5 X 10 + 8 X 5
RA = 4.5kN
RB = 8.5kN
( ) ( )
( ) ( )
( ) ( )
E Id yd x
x x x
E Id yd x
xx x A
E Iyx
x x A x B
2
2
22 2
33 3
4 5 5 1 0 8 1 5
4 52
52
1 082
1 5
4 56
56
1 086
1 5
= − − − −
= − − − − +
= − − − − + +
.
.
.
when x = 0, y = 0, � B = 0
x = 20, y = 0, A = -250
( ) ( )E Iy x x x x= − − − − −0 75 0 83 10 1 33 15 2503 3 3. . .
At C , x = 10m
EIy = 750 2500 = -1750kNm3
yx
x x−
=1 7 5 0 1 0
2 1 0 1 0
2
5 9
At D, x = 15m
y =
( c) Maximum deflection can be judged to be between the loads and writing
the term in (x-15)3, for = 0, we have
2.25x2 2.5(x-10)2 250 = 0
0.25x2 50x + 500 = 0
x2 200x + 2000 = 0
x =± −
= ± =2 0 0 2 0 0 8 0 0 0
21 0 0 5 3 2 0 1
2
Maximum deflection,
ymax. = − − − − x
x x1 0
2 1 0 1 0
2
5 9
Exercise
BA
10m
x/3
x/2xRA RB
Determine the deflection of the centroid of loading
Solution :
�MA : 10RB = (1 x 10) x 5 + (1/2 x 4 x10) x 10/3
RB = 11.7kN
( )E Id yd x
xx
x xx
x x x
2
2
2
2 3
1 1 72
12
0 43
1 1 7 0 5 0 0 6 7
= − −
= − −
. .
. . .
E Id yd x
x x xA= − − +
1 1 72
0 53
0 0 6 74
2 3 4. . .
At x = 0, y = 0 , B = 0
At x = 10, y = 0 , A =
Centroid of loading is at x = (10)m = 6.67m from B23
Assignment
A horizontal beam simply supported at its ends carries a load which varies
uniformly from 2kN/m from one end to 6kN/m at the other over a span of
8m.
E = 2.05 x 105N/mm2, I = 8.05 x 108mm4
(a) Find the deflection at the centre of the beam
(b) Position magnitude of the maximum deflection.
A C10m 10m
2kN/m2kN/m
B
BAC
2m
20kN
2m 2m2m8m
20kN 15kN
30kN/m
Examples
(1) Draw Shear force diagram and Bending moment diagram and determine
the moment at C.
(2)
CA
B
30kN
1m1.5m
3m
40kN
D
50kN/m
(3)
(4)30kN25kN
C5m5m
10m10m2m
B
A
15m
8m
Hine at A , B and C. Determine the reactions at A and B and draw the
Bending moment diagram.
Solutions
(1)
RA
CL/2 L/2
2kN/m2kN/m
RB
x
w xx
wL
wL
w xw xL
x
w w x
= =
= =
−
2
2
20 2.
( )m w wx
w x xw L
x
ww xL
x w xL
w Lx
w x w xL
w xL
w Lx x x
x
x x x= − + −
+
= −
+
+
= + − + = − +
2
2 3
2 3 32
3
212
23 4
22
13
24
22
3 410
30
( )
m xL
m w w
w L w L w LK N m
c
c
= =
= − +
= + = =
213
1 2 8 2 43 3 3
2 2 2
.
( )
mm
x L
M a xim um w hendmdx
x w wx w
Lw xL
w L
w Lw x
w xL
w xL
w xL
Lx
xL
x xLL
xL L L L
m
A
B
x
==
=
=
− + −
+
+
= − − +
= −
− + =
=+ ± −
= =
00
0
22 2
4
42 2 2
4
40
2 210
2 2
2 2 2
2
22
2 2
,
,
CB.M.D
Cubic
xL
w Lw x
w xL
w L w L w L
V
V
x
L
> ≤
= − +
= − + =
= − + =
= − + =
02
4
4 2 40
1 0 1 02 51 0
2 5
1 0 5 0 6 25 5 6 25
2
5
2 5
2
,
.
. ..
γ
γ
Shear Force,
w Lw x
xL
w xL
w Lw x
w xL
w xL
41
2
42
2
2 2
− −
−
− + −
S.F.D
L/2 L/2
CA
B
30kN
1m1.5m
3m
40kN
D
50kN/m
Draw the Shear Force and Bending Moment Diagram for the beam shown.
Solution:
�MB: -30 x 4 40 x 1.5 50 x 1.5 x1.5/2 + 3RA = 0
RA = 60 + 18.75 = 78.75kN
RB = 30 + 40 + 50(1.5) 78.75 = 66.25kN
x < 1:
Vx = -30kN
V < x < 2.5:
-30-66.25
AB
D
8.75
48.75
C
S.F.D
Vx = -30 = 78.75 = 48.75
From B:
0 < x < 1.5
Vx = -66.25 + 50x
V1.5 = -66.25 + 75 = 8.75kN
0 < x < 1
Mx = -30x, M1 = -30
1 < x < 2.5
Mx = -30x + 78.75(x � 1)
M2.5 = -30 x 2.5 + 78.75(1.5)
= -75 + 118.12
= 43.12kNm
From B,
0 < x < 1.5
Mx = -66.25x + 50x2/2
Mmax. occurs when Vx = 0 i.e x = = 1.325m from B6 6 .25
5 0
Mmax. = -66.25(1.325) + 25(1.325)2 = 43.89kNm
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