Chapter 9Deflection of Beams

9.1 Introduction

-- Concerning about the “deflection” of a beam

-- Special interest: the maximum deflection

-- Design: to meet design criteria

1 MEI

1 ( )M xEI

2

2

( )d y M xd x EI

9.1 Introduction

(4.21)

M = bending moment

E = modulus I = moment of inertia

If M is not a constant, i.e. M=M(x)

(9.1)

or

-- will be explained in Sec. 9.3

dydx

2

2

( )d y M xd x EI

y = y(x)

1 PxEI

(9.1)

(9.2)

9.2 Deformation of a Beam under Transverse Loading

1 ( )M xEI

Since M(x) = -Px

Example: a beam subjected to transverse loads

Moment Diagram and the Deformed Configuration:

Mmax occurs at C

In addition to M(x) and 1/, we need further information on:

1. Slope at various locations

2. Max deflection of a beam

3. Elastic curve: y = y(x)

9.3 Equation of the Elastic Curve

10( )

xdyEI M x dx Cdx

2

2

( )d y M xd x EI

(9.4)

(9.5)

2

2

32 2

1

[1 ( ) ]

d yd xdydx

2

2

1 d yd x

2

2

( )d y M xd x EI

9.3 Equation of the Elastic Curve

The curvature of a plane curve at Point Q(x,y) is

For small slope dy/dx 0, hence

2 0( )dydx

Therefore,

Fially,

(9.2)

(9.3)

(9.4)

2

2( )d y M x

d x EI -- the governing diff. equation of a

beam

-- the governing diff. equation of the “elastic curve”

EI = flexural rigidity

tan ( )dy xdx

2

2( )d y M x

d x EI

10( )

xdyEI M x dx Cdx

Integrating Eq. (9.4) once

(9.4)

10( ) ( )

xEI x M x dx C

Since,

It follows,

(9.5’)

(9.5)

1 20 0[ ( ) ]x x

EIy M x C dx C

1 20 0( )

x xEIy dx M x dx C x C (9.6)

10( )

xdyEI M x dx Cdx

(9.5)

Integrating Eq. (9.5) once again,

1 20 0

1( ) ( )x x

y x dx M x dx C x CEI

or

C1 & C2 are determined from the B.C.s

Boundary conditions:

-- the conditions imposed on the beam by its supports

Examples:

Three types of Statically Determinate Beams:

1. Simply supported beams: 2. Overhanging beams:

3. Cantilever beams:

9.4 Direct Determination of the Elastic Curve from the Load Distribution

2

2

( )d y M xd x EI

4

4

( )d y w xdx EI

3

3

1 ( )d y dV w xdx EI dx EI

(9.4)

(9.31)3

31 ( )d y dM V x

dx EI dx EI

(9.32)

3

13

2

1 22

21

2 3

31

22 3 4

12

16

12

( ) ( )

( ) ( )

( ) ( )

( ) ( )

d yEI V x w x dx Cdxd yEI M x dx w x dx C dx CdxdyEI EI x dx dx w x dx C xdx

C x C

EIy x dx dx dx w x dx C x

C x C x C

4

4 ( )d yEI w xdx

(9.33)

9.5 Statically Determinate Beams

0 0 0x y AF F M

Applying equations of equilibrium:

Conclusion:

-- This is a statically indeterminate problem.

-- Because the problem cannot be solved by means of equations of equilibrium

(9.37)

9.5 Statically Determinate Beams

By adding (1) deflection y = y(x) and = (x), the problem can be solved.

i.e. five unknowns with six equations

Statically indeterminate to the 1st degree:

-- one redundant support

Statically indeterminate to the 2nd degree:

-- two redundant supports

9.6 Using Singular functions to Determine the Slope and Deflection of a Beam

03 1( )4 4PV x P x L

3 1( )4 4PM x P x L

2

2

3 14 4

d y PEI x P x Ldx

(9.44)

(9.45)

22

13 1 18 2 4

dyEI EI Px P x L Cdx

33

1 21 1 18 6 4

EIy Px P x L C x C

3

21 10 0 0 06 4P L C

33

11 1 308 6 4PL P L C L

31

124

C wL

(9.46)

(9.47)

9.7 Method of Superposition

9.8 Application of Superposition to Statically Indeterminate Beams

9.9 Moment-Area Theorems

2

2

d d y Mdx dx IE

Md dxEI

D D

C C

x

x

Md dxEI

D

C

x

D C x

M dxEI

(9.54)

(9.55)

area under (M/EI) diagram

between C and D

DC

1dt x d

1Mdt x dxEI

1D

C

x

C xD

Mt x dxEI

1(area between C and D)CD

t x

2(area between C and D)DC

t x

(9.56)

(9.57)

(9.59)

(9.60)

9.10 Application To Cantilever Beams And Beams With Symmetric Loads

9.11 Bending-Moment Diagrams By Parts

9.12 Application Of Moment-area Theorems To Beams With Unsymmetric Loadings

BA

A

t

L

D A DA

or EF= BA

EF HB x tx L L

D D BA A

xy ED EF t tL

(9.61)

(9.62)

(9.63)

9.13 Maximum Deflection

9.14 Use Of Moment-Area Theorems With Statically Indeterminate Beams