( )[ ]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