Chapter Objectives

� Design a beam to resist both bending and shear

loads

A Bridge Deck under Bending Action

Castellated Beams

Post-tensioned Concrete Beam

Lateral Distortion of a Beam Due to Lateral Load

Action

T-Beam Roof on a Office Building

Modern Steel Frame Building

Joint in steel frame building

Beams with Different Boundary/Support

Conditions

Beams with Different Vertical Load Conditions

Triangular distributed load and concentrated load on an over-hang beam.

Beams with Different Vertical Load Conditions

Uniform distributed load a simply supported beam.

Internal Actions of a simply supported beam

loaded with uniform load

Internal Actions of a cantilever beam loaded

with triangular distributed load

Mmax

Vmax

MmaxVmax

Internal Actions of a simply-supported beam

loaded with an arbitrary loading

Deformations and Stresses of a Beam Under

Bending Effect

Bending Moment

Undeformed state – No bending

Deformed state – Negative

Bending

Deformation shape of a Cantilever Beam under

Bending Action

Deformed state – Positive Bending

“Almost” Undeformed state – No

bending

External Moment

Acting

Stress and Strain Profile of a Section under

Bending

Observations:

• Maximum strain and stress occur at the outer

most layers/fibers,

• Strain and stress profiles vary linearly with the

distance y from the centroid,

• There is no strain and stress at the neutral axis

(neutral axis is the centroid of the section),

• The value c defines the distance from the

centroid to the outer most layer.

Stress and Strain Profile of a Section under

Bending

Bending Formula

(General)

yI

M

z

x =σ

Bending Formula

(Maximum Stress at

a arbitrary section)

cI

M

z

=maxσ

Bending Formula (Absolute

Maximum Stress at a section

where max. mom. is observed)

cI

M

z

maxmax =σ

Cross Section

PRISMATIC BEAM DESIGN

• Basis of beam design

• Strength concern (i.e., provide safety margin to

normal/shear stress limit)

• Serviceability concern (i.e., deflection limit)

• Section strength requirement

allow

dreq

MS

σmax

' =

c

IS =

S: Section modulus

I: Moment of inertia about the

centroidal axis

c: half-height of the section

Ch

b

c

Cross Section View

PRISMATIC BEAM DESIGN (cont)

• Choices of section (refer to standard tables):

• Section modulus are given for example for

standard steel sections in codes such as AISC

standard

W 460 X 68

Height = 459 ≈ 460 mm

Weight = 0.68 kN/m

PRISMATIC BEAM DESIGN (cont)

• For other sections made of other materials such as

wood, you need to calculate it yourself,

Nominal dimensions (in multiple of 25mm) e.g. 50 (mm)

x 100 (mm) actual or “dressed” dimensions are smaller,

e.g. 50 x 100 is 38 x 89.

• Built-up sections

PRISMATIC BEAM DESIGN (cont)

Procedure:

• Shear and Moment Diagram

• Determine the maximum shear and moment in

the beam. Often this is done by constructing the

beam’s shear and moment diagrams.

• For built-up beams, shear and moment diagrams

are useful for identifying regions where the shear

and moment are excessively large and may

require additional structural reinforcement or

fasteners.

PRISMATIC BEAM DESIGN (cont)

• Normal Stress

– If the beam is relatively long, it is designed by

finding its section modulus and using the flexure

formula, Sreq’d = Mmax/σallow.

– Once Sreq’d is determined, the cross-sectional

dimensions for simple shapes can then be

computed, using Sreq’d = I/c.

– If standard steel sections are to be used, several

possible values of S may be selected from the

tables. Of these, choose the one having the smallest

cross-sectional area, since this beam has the least

weight and is therefore the most economical.

PRISMATIC BEAM DESIGN (cont)

• Normal Stress (cont.)

– Make sure that the selected section modulus, S, is

slightly greater than Sreq’d, so that the additional

moment created by the beam’s weight is

considered.

• Shear Stress

– Normally beams that are short and carry large

loads, especially those made of wood, are first

designed to resist shear and then later checked

against the allowable-bending-stress requirements.

– Using the shear formula, check to see that the

allowable shear stress is not exceeded; that is, use

τallow ≥ (Vmax /As), As is the shear area and is a

function of the cross sectional shape.

PRISMATIC BEAM DESIGN (cont)

• Shear Stress (cont.)

– If the beam has a solid rectangular cross section,

the shear formula becomes τallow ≥ 1.5(Vmax/A), and

if the cross section is a wide flange, it is generally

appropriate to assume that the shear stress is

constant over the cross-sectional area of the beam’s

web so that τallow ≥ Vmax/Aweb, where Aweb is

determined from the product of the beam’s depth

and the web’s thickness.

Web Area

Flange Area

Flange Area

PRISMATIC BEAM DESIGN (cont)

• Adequacy of Fasteners

– The adequacy of fasteners used on built-up beams

depends upon the shear stress the fasteners can

resist. Specifically, the required spacing of nails or

bolts of a particular size is determined from the

allowable shear flow, qallow = VQ/I, calculated at

points on the cross section where the fasteners are

located.

EXAMPLE 1

A beam is to be made of steel that has an allowable bending

stress of σallow = 170 MPa and an allowable shear stress of τallow

= 100 MPa. Select an appropriate W shape that will carry the

loading shown in Fig. a.

EXAMPLE 1 (cont)

( )( )( )( )( )( ) 33

33

33

33

33

33

mm 10987 100200

mm 10984 80250

mm 101060 74310

mm 101030 64360

mm 101200 67410

mm 101120 60460

=×

=×

=×

=×

=×

=×

SW

SW

SW

SW

SW

SW

m3m

= 706x103 mm3

EXAMPLE 1 (cont)

• The beam having the least weight per metre is chosen, W460 x 60

• The beam’s weight is

• From Appendix B (tables), for a W460 x 60, d = 455 mm and tw = 8 mm.

• Thus

• The beam is designed to be a W460 x 60.

Solutions

( )( ) kN 60.36/60.0 == mmkNW

( )( )( )

(OK) MPa 100 MPa 7.248455

10900 3

max <===web

avgA

Vτ

EXAMPLE 2

The laminated wooden beam shown in Fig. 11–8a supports a

uniform distributed loading of 12 kN/m. If the beam is to have a

height-to-width ratio of 1.5, determine its smallest width.

The allowable bending stress is 9 MPa and the allowable shear

stress is 0.6 MPa. Neglect the weight of the beam.

EXAMPLE 2 (cont)

• Applying the flexure formula,

• Assuming that the width is a,

the height is 1.5a.

Solutions

( )( )

3

6

3

max m 00119.0109

1067.10===

allow

req

MS

σ

( )( )( )

m 147.0

m 003160.0

75.0

5.100119.0

33

3

121

=

=

===

a

a

a

aa

c

ISreq

Mmax=

= Vmax

EXAMPLE 2 (cont)

• Applying the shear formula for rectangular sections,

• Since the design fails the shear criterion, the beam must be redesigned

on the basis of shear.

• This larger section will also adequately resist the normal stress.

Solutions

( )( )

( )( )( ) MPa6.0929.0

147.0147.05.1

10205.15.1

3

maxmax >===

A

Vτ

( )( )

(Ans) mm 183 m 183.0

5.1

10205.1600

5.1

3

max

==

=

=

a

aa

A

Vallowτ