Chapter 5 Analysis and Design of

Beams for Bending 

-- Dealing with beams of different materials:

steel, aluminum, wood, etc.

-- Loading: transverse loads

Concentrated loads

Distributed loads

5.1 Introduction

-- Supports Simply supported

Cantilever Beam

Overhanging

Continuous

Fixed Beam

A. Statically Determinate Beams

-- Problems can be solved using Equations of Equilibrium

B. Statically Indeterminate Beams -- Problems cannot be solved using Eq. of Equilibrium

-- Must rely on additional deformation equations to solve the problems.

FBDs are sometimes necessary:

FBDs are necessary tools to determine the internal

(1) shear force V – create internal shear stress; and

(2) Bending moment M – create normal stress

Where I = moment of inertiay = distance from the N. Surfacec = max distance

m

M cI

xMyI

From Ch 4:

(5.1)

(5.2)

Recalling, elastic section modulus, S = I/c,

m

MS

hence (5.3)

For a rectangular cross-section beam,

216

S bh (5.4)

From Eq. (5.3), max occurs at Mmax

It is necessary to plot the V and M diagrams along the length of a beam.

to know where Vmax or Mmax occurs!

5.2 Shear and Bending-Moment Diagrams•Determining of V and M at selected

points of the beam

2. The bending moment is positive (+) when the external forces acting on the beam tend to bend the beam at the point indicated in fig 5.7c

1. The shear is positive (+) when external forces acting on the beam tend to shear off the beam at the point indicated in fig 5.7b

Moment

Sign Conventions

5.3 Relations among Load, Shear and Bending Moment

0 0: V-(V+ )YF V w x

V w x

dV wdx

(5.5)Hence,

1. Relations between Load and Shear

1

D

C

x

D C xV V wdx

(5.6’)

(5.5’)

Integrating Eq. (5.5) between points C and D

VD – VC = area under load curve between C and D

(5.6)

2. Relations between Shear and Bending Moment

0' :C

M

12

M V xx

02

( ) xM M M V x w x

212

( )M V x w x

dM Vdx

(5.7)

or

0limx

M dMx dx

D

C

x

D C xM M Vdx

dM Vdx

MD – MC = area under shear curve between points C and D

(5.7)

5.4 Design of Prismatic Beams for Bending

max max m m

M c MI S

maxmin

all

MS

(5.1’,5.3’)

(5.9)

-- Design of a beam is controlled by |Mmax|

Hence, the min allowable value of section modulus is:

Question: Where to cut? What are the rules?

Answer: whenever there is a discontinuity in the loading conditions, there must be a cut.

Reminder:

The equations obtained through each cut are only valid to that particular section, not to the entire beam.

5.5 Using Singularity Functions to Determine Shear and Bending

Moment in a Beam

4

4

3

3

2

2

( )

w d yEI dxV d yEI dxM d yEI dx

dydx

y f x

Beam Constitutive Equations

Notes:1.In this set of equations, +y is going upward

and +x is going to the right.2. Everything going downward is “_”, and upward

is +. There is no exception.3. There no necessity of changing sign for an y

integration or derivation. Sign Conventions:

1. Force going in the +y direction is “+”2. Moment CW is “+”

0 1 when {

0 when

x ax a

x a

( ) when {0 when

nn x a x a

x ax a

(5.15)

(5.14)

Rules for Singularity Functions

Rule #1:

Rule #2:

1. Distributed load w(x) is zero order: e.g. wo<x-a>o

2. Pointed load P(x) is (-1) order: e.g. P<x-a>-1

3. Moment M is (-2) order: e.g. Mo<x-a>-2

Rule #3:

2 1

1 0

0 1

1 2

2 3

3 4

x a dx x a

x a dx x a

x a dx x a

1x a dx x a21x a dx x a31x a dx x a4

Rule #4:

1. Set up w = w(x) first, by including all forces, from the left to the right of the beam.

2. Integrating w once to obtain V, w/o adding any constants.3. Integrating V to obtain M, w/o adding any constants.4. Integrating M to obtain EI , adding an integration

constant C1.5. Integrating EI to obtain EIy, adding another constant

C2.6. Using two boundary conditions to solve for C1 and C2.

0 01( )4

V x w a w x a

20 0

1 1( )4 2

M x w ax w x a

0 00 0 0

1( ) (0) ( )4

x x x

M x M V x dx w adx w x a dx

•After integration, and observing that (0) 0M We obtain as before

2

0 01 1( )4 2

M x w ax w x a

00( ) w x w x a

(5.11)

(5.12)

(5.13)

( ) 0 for x< 0w x

0 0( ) 1 x a x a

0( ) for w x w x a

01 for 04

V w a x

0

00 0( ) (0) ( )

x xV x V w x dx w x a dx

1

0 01( )4

V x w a w x a

0 01( )4

V x w a w x a

11 for n 01

n n

x a dx x an

1 for n 1 n nd x a n x a

dx

and

(5.16)

(5.17)

Example 5.05

Sample Problem 5.9

5.6 Nonprismatic Beams

all

MS

01 1( )2 2

V x P P x L

11 1( )2 2

M x Px P x L

0 0

0 0( ) w x w x a w x b

•Load and Resistance Factor Design• D D L L UM M M

Example 5.03