Peter Smith & Mike Rosenman

The size and shape of the crosssection of the piece of material
used

For timber, usually a rectangle

For steel, various formed
sections are more efficient

For concrete, either rectangular,
or often a Tee
A timber and plywood I-beam
University of Sydney – Structures
SECTIONS
1/28
Peter Smith & Mike Rosenman

What shapes are
possible in the
material?

What shapes are
efficient for the
purpose?

Obviously, bigger is
stronger, but less
economical
University of Sydney – Structures
Some hot-rolled steel sections
SECTIONS
2/28
Peter Smith & Mike Rosenman

Beams are oriented one way

Depth around the X-axis is the strong way

Some lateral stiffness is also needed

Columns need to be stiff both ways (X and Y)
Y
X
Timber
beam
X
Cold-formed Timber
steel
post
University of Sydney – Structures
Hot-rolled
steel
Y
Steel
tube
SECTIONS
3/28
Peter Smith & Mike Rosenman
 ‘Stress is proportional to strain’

Parts further from the centre strain more

The outer layers receive greatest stress
Unchanged length
Most shortened
Most lengthened
University of Sydney – Structures
SECTIONS
4/28
Peter Smith & Mike Rosenman

The stresses developed resist bending

Equilibrium happens when the resistance equals
the applied bending moment
All the compressive
stresses add up to form a
compressive force C
C
a
T
MR = Ca
= Ta
Internal
Moment of
Resistance
All the tensile stresses add
up to form a tensile force T
University of Sydney – Structures
SECTIONS
5/28
Peter Smith & Mike Rosenman


Simple solutions for rectangular sections
b
Doing the maths (in the Notes)
gives the Moment of Inertia
d
For a rectangular section
3
bd
I=
12
University of Sydney – Structures
mm4
SECTIONS
6/28
Peter Smith & Mike Rosenman

The bigger the Moment of Inertia, the stiffer the
section

It is also called Second Moment of Area

Contains d3, so depth is important

The bigger the Modulus of Elasticity of the
material, the stiffer the section

A stiffer section develops its Moment of
Resistance with less curvature
University of Sydney – Structures
SECTIONS
7/28
Peter Smith & Mike Rosenman


Simple solutions for rectangular sections
b
Doing the maths (in the Notes)
gives the Section Modulus
d
For a rectangular section
bd
Z=
6
University of Sydney – Structures
2
mm3
SECTIONS
8/28
Peter Smith & Mike Rosenman

The bigger the Section Modulus, the
stronger the section

Contains d2, so depth is important
University of Sydney – Structures
SECTIONS
9/28
Peter Smith & Mike Rosenman

Strength --> Failure of Element

Stiffness --> Amount of Deflection
depth is important
University of Sydney – Structures
SECTIONS
10/28
Peter Smith & Mike Rosenman

The area tells how much stuff there is
● used for columns and ties
● directly affects weight and
cost

A = bd
The radius of gyration is a derivative of I
Y
b
● used in slenderness ratio
rx = d/√12
ry = b/√12
X
X
d
Y
University of Sydney – Structures
SECTIONS
11/28
Peter Smith & Mike Rosenman

Can be calculated, with a little extra work
 Manufacturers publish tables of properties
University of Sydney – Structures
SECTIONS 12/28
Peter Smith & Mike Rosenman
University of Sydney – Structures
SECTIONS 12/28
Peter Smith & Mike Rosenman

Checking Beams
● given the beam section
● check that the stresses & deflection
are within the allowable limits

Designing Beams
● find the Bending Moment and Shear
Force
● select a suitable section
University of Sydney – Structures
SECTIONS 13/28
Peter Smith & Mike Rosenman

Go back to the bending moment diagrams

Maximum stress occurs where bending moment is
a maximum
M is maximum here
Bending Moment
Stress =
Section Modulus
University of Sydney – Structures
M
f=
Z
SECTIONS 14/28
Peter Smith & Mike Rosenman
b
d

Given the beam size and material

Find the maximum Bending Moment

Use Stress = Moment/Section Modulus
Compare this stress to the Code allowable stress

M = max BM
Z = bd2 / 6
Actual Stress = M / Z
Allowable Stress (from Code)
Actual < Allowable?
University of Sydney – Structures
SECTIONS 15/28
Peter Smith & Mike Rosenman

Given a softwood timber beam 250 x 50mm

Given maximum Bending Moment = 4kNm

Given Code allowable stress = 8MPa
250
50
4 kNm
Section Modulus Z = bd2 / 6
= 50 x 2502 / 6
= 0.52 x 106 mm3
Actual Stress
f=M/Z
= 4 x 103 x 103 / 0.52 x 106
= 7.69 MPa < 8MPa
Actual Stress < Allowable Stress
University of Sydney – Structures
SECTIONS 16/28
Peter Smith & Mike Rosenman




Given the maximum Bending Moment
Given the Code allowable stress for the material
Use Section Modulus = Moment / Stress
Look up a table to find a suitable section
M = max BM
Allowable Stress (from Code)
b?
required Z = M / Allowable Stress
a) choose b and d to give
Z >= than required Z or
d?
b) look up Tables of Properties
University of Sydney – Structures
SECTIONS 17/28
Peter Smith & Mike Rosenman


Given the maximum Bending Moment = 4 kNm
b?
Given the Code allowable stress for
structural steel = 165 MPa
d?
required Z = 4 x 106 / 165 = 24 x 103 mm3
(steel handbooks give Z values in 103 mm3)
looking up a catalogue of steel purlins we find
C15020 - C-section 150 deep, 2.0mm thickness has a
Z = 27.89 x 103 mm3
University of Sydney – Structures
(smallest section Z >= reqd Z)
SECTIONS 18/28
Peter Smith & Mike Rosenman

Both E and I come into the deflection formula
(Material and Section properties)

The load, W, and span, L3

Note that I has a d3 factor

Span-to-depth ratios (L/d) are often used
as a guide
W
Depth, d
Span, L
University of Sydney – Structures
SECTIONS 19/28
Peter Smith & Mike Rosenman
W
Central point load
3
L
8d
Total load = W
(w per metre length)
L
5d
where W is the TOTAL load
University of Sydney – Structures
WL
=
48EI
Uniformly Distributed Load
3
5WL
=
384EI
SECTIONS 20/28
Peter Smith & Mike Rosenman
Central point load
W
L
128d
Total load = W
(w per metre length)
WL
=
3EI
Uniformly Distributed Load
48d
L
where W is the TOTAL load
University of Sydney – Structures
3
3
WL
=
8EI
SECTIONS 21/28
Peter Smith & Mike Rosenman
Uniformly Distributed Load
Total load = W
(w per metre length)
L
d
3
WL
=
384EI
where W is the TOTAL load


The deflection is only one-fifth of a
simply supported beam
Continuous beams are generally stiffer than
simply supported beam
University of Sydney – Structures
SECTIONS 22/28
Peter Smith & Mike Rosenman

Given the beam size and material

Given the loading conditions

Use formula for maximum deflection

Compare this deflection to the Code allowable
deflection
Given load, W, and span, L
Given Modulus of Elasticity, E, and Moment of Inertia, I
Use deflection formula to find deflection
Be careful with units (work in N and mm)
Compare to Code limit (usually given as L/500, L/250 etc)
University of Sydney – Structures
SECTIONS 23/28
Peter Smith & Mike Rosenman
W = 8kN


Check the deflection of the steel channel
previously designed for strength
The maximum deflection <= L / 500
Section = C15020
E = 200 000 MPa
d = (5/384) x WL3/EI mm
L = 4m
Loading Diagram
I = 2.119 x 106 mm4
( Let us work in N and mm )
d = (5/384) x 8000 x 40003 / (200000 x 2.119 x 106)
= 16 mm
Maximum allowable deflection = 4000 / 500
= 8 mm
deflects too much - need to chose stiffer section
University of Sydney – Structures
SECTIONS 24/28
Peter Smith & Mike Rosenman
65


Need twice as much I
150
Could use same section back to back
100% more material
75

A channel C20020 (200 deep 2mm thick)
has twice the I but only 27% more material
200
design for strength
check for deflection
strategy for heavily loaded beams
University of Sydney – Structures
SECTIONS 25/28
Peter Smith & Mike Rosenman

Given the loading conditions

Given the Code allowable deflection

Use deflection formula to find I

Look up a table to find a suitable section
Given load, W, span, L, and Modulus of Elasticity, E
Use the Code limit — e.g., turn L/500 into millimetres
Use deflection formula to find minimum value of I
Look up tables or use I = bd3/12 and choose b and d
University of Sydney – Structures
SECTIONS 26/28
Peter Smith & Mike Rosenman
= better sections for beams


Beams need large I and Z in direction of bending
Need stiffness in other direction to resist
lateral buckling
Columns usually need large value of r
in both directions
 Some sections useful for both

University of Sydney – Structures
SECTIONS 27/28
Peter Smith & Mike Rosenman

Deep beams are economical but subject to
lateral buckling
University of Sydney – Structures
SECTIONS 28/28