University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

The size and shape of the cross-section 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

1/28

University of Sydney – Structures SECTIONS

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

Some hot-rolled steel sections

2/28

University of Sydney – Structures SECTIONS

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)

3/28

Timberpost

Hot-rolledsteel

Steeltube

Y

Y

Cold-formedsteel

Timberbeam

X X

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

‘Stress is proportional to strain’

Parts further from the centre strain more

The outer layers receive greatest stress

Most shortened

Most lengthened

Unchanged length

4/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

The stresses developed resist bending

Equilibrium happens when the resistance equals the applied bending moment

C

T

All the tensile stresses add up to form a tensile force T

All the compressive stresses add up to form a

compressive force C

a

MR = Ca = Ta

InternalMoment ofResistance

5/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Simple solutions for rectangular sections

For a rectangular section d

b Doing the maths (in the Notes)

gives the Moment of Inertia

I =bd3

12mm4

6/28

University of Sydney – Structures SECTIONS

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

7/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Simple solutions for rectangular sections

b

d

Doing the maths (in the Notes)

gives the Section Modulus

For a rectangular section

Zbd2

6mm3

8/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

The bigger the Section Modulus, the stronger the section

Contains d2, so depth is important

9/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Strength --> Failure of Element

Stiffness --> Amount of Deflection

depth is important

10/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

The area tells how much stuff there is● used for columns and ties● directly affects weight and

cost

rx = d/√12ry = b/√12

A = bd The radius of gyration is a derivative of I

● used in slenderness ratio

11/28

Y

b

dX X

Y

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Can be calculated, with a little extra work Manufacturers publish tables of properties

12/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

12/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Checking Beams

Designing Beams

● given the beam section● check that the stresses & deflection

are within the allowable limits

● find the Bending Moment and Shear Force● select a suitable section

13/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Go back to the bending moment diagrams

Maximum stress occurs where bending moment is a maximum

f =M

Z

M is maximum here

14/28

Bending Moment

Section ModulusStress =

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Given the beam size and material

Z = bd2 / 6M = max BM

Actual Stress = M / Z

Allowable Stress (from Code)

b

d

Find the maximum Bending Moment

Use Stress = Moment/Section Modulus Compare this stress to the Code allowable stress

Actual Allowable?<

15/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Given a softwood timber beam 250 x 50mm

Section Modulus Z = bd2 / 6

Actual Stress f = M / Z

50

250

Given maximum Bending Moment = 4kNm

Given Code allowable stress = 8MPa 4 kNm

= 4 x 103 x 103 / 0.52 x 106

= 50 x 2502 / 6 = 0.52 x 106 mm3

= 7.69 MPa < 8MPa

Actual Stress < Allowable Stress

16/28

University of Sydney – Structures SECTIONS

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

b?

d?

M = max BMAllowable Stress (from Code)

required Z = M / Allowable Stress

a) choose b and d to give Z >= than required Z or

b) look up Tables of Properties

17/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Given the maximum Bending Moment = 4 kNm Given the Code allowable stress for

structural steel = 165 MPa

b?

d?

required Z = 4 x 106 / 165 = 24 x 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

(steel handbooks give Z values in 103 mm3)

(smallest section Z >= reqd Z)

18/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Both E and I come into the deflection formula (Material and Section properties)

Depth, d

Span, L

W

19/28

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

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

20/28

WL3

48EI8d

Central point loadW

L

5WL3

384EI5d

Uniformly Distributed Load

where W is the TOTAL load

(w per metre length)

Total load = W

L

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

W

L

Central point load

WL3

8EI

48d

WL3

3EI

128d

21/28

where W is the TOTAL load

Uniformly Distributed Load(w per metre length)

L

Total load = W

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

The deflection is only one-fifth of a

simply supported beam Continuous beams are generally stiffer than

simply supported beam

where W is the TOTAL load

WL3

384EId

(w per metre length)

L

Total load = W

Uniformly Distributed Load

22/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Given load, W, and span, LGiven Modulus of Elasticity, E, and Moment of Inertia, IUse deflection formula to find deflectionBe careful with units (work in N and mm)Compare to Code limit (usually given as L/500, L/250 etc)

23/28

Given the beam size and material

Given the loading conditions

Use formula for maximum deflection

Compare this deflection to the Code allowable deflection

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Check the deflection of the steel channel

previously designed for strength The maximum deflection <= L / 500

W = 8kN

L = 4m

Loading Diagram

Section = C15020 E = 200 000 MPa I = 2.119 x 106 mm4

= (5/384) x 8000 x 40003 / (200000 x 2.119 x 106)

= (5/384) x WL3/EI mm ( Let us work in N and mm )

Maximum allowable deflection = 4000 / 500

= 16 mm

= 8 mm

deflects too much - need to chose stiffer section

24/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Need twice as much I

design for strength check for deflection

65

150

75

200

Could use same section back to back

100% more material

A channel C20020 (200 deep 2mm thick)

has twice the I but only 27% more material

strategy for heavily loaded beams25/28

University of Sydney – Structures SECTIONS

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, EUse the Code limit — e.g., turn L/500 into millimetres

Use deflection formula to find minimum value of ILook up tables or use I = bd3/12 and choose b and d

26/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Beams need large I and Z in direction of bending Need stiffness in other direction to resist lateral buckling

Some sections useful for both

Columns usually need large value of r in both directions

= better sections for beams

27/28

University of Sydney – Structures SECTIONS

Peter Smith & Mike Rosenman

Deep beams are economical but subject to lateral buckling

28/28