Wk3 Sections

7/21/2019 Wk3 Sections

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Universi t of S dne Structures SECTIONS

Peter Smith & Mike Rosenman

The size and shape of the cross-sectionof 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

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What shapes arepossible in the

material?

What shapes areefficient for the

purpose?

Obviously, biggeris stronger, but less

economical

Some hot-rolled steel sections

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Beams are oriented one way

Depth around the X-axis is the strong way

Some lateral stif fness is also needed

Columns need to be stif f both ways (X and Y)

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Timber

post

Hot-rolled

steel

Steel

tube

Y

Y

Cold-formed

steel

Timber

beam

X X


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Stress is proportional to strain

Parts further from the centre strain more

The outer layers receive greatest stress

Most shortened

Most lengthened

Unchanged length

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The stresses developed resist bending

Equil ibrium 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

Internal

Moment ofResistance

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The bigger the Moment of Inertia, the stifferthesection

It is also called Second Moment of Area

Contains d3, so depth is important

The bigger the Modulus of Elasticityof the

material, the stiffer the section

Astiffer sectiondevelops itsMoment ofResistance withless curvature

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Simple solutions for rectangular sections

b

d

Doing the maths (in the Notes)

gives the Section Modulus

For a rectangular section

Z bd2

6mm3

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The bigger the Section Modulus, the

strongerthe section

Contains d2, so depth is important

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10/29Universi t of S dne Structures SECTIONS


Strength --> Failure of Element

Stiffness -->Amount of Deflection

depth is important

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The areatells how much stuff there is used for columns and ties

directly affects weight and

cost

rx= d/12

ry= b/12

A = bd

The radius of gyrationis a derivative of I

used in slenderness ratio

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b

dX X

Y




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

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P t S ith & Mik R




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P t S ith & Mik R




Checking Beams

Designing Beams

given the beam section

check that thestresses & deflectionare within the allowable limits

find the Bending Moment and Shear

Force

select a suitable section

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P t S ith & Mik R




Go back to the bending moment diagrams

Maximum stress occurs where bending moment is

a maximum

f=M

Z

M is maximum here

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Bending Moment

Section ModulusStress=





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?= than required Zor

b) look up Tables of Properties

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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 103mm3

looking up a catalogue of steel purlins we findC15020- C-section 150 deep, 2.0mm thickness has a

Z = 27.89 x 103mm3

(steel handbooks give Z values in 103mm3)

(smallest section Z >= reqd Z)

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Both Eand Icome into the deflection formula(Material and Section properties)

Depth, d

Span, L

W

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The load, W, and span, L3

Note that Ihas a d3

factor

Span-to-depth ratios (L/d) are often used

as a guide





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WL3

48EI

8d

Central point loadW

L

5WL3

384EI5d

Uniformly Distributed Load

whereWis theTOTALload

(w per metre length)

Total load = W

L





W

L

Central point load

WL3

8EI

48d

WL3

3EI

128d

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Uniformly Distributed Load(w per metre length)

L

Total load = W





The deflection is only one-fifth of a

simply supported beam

Continuous beams are generally stif fer than

simply supported beam


WL3

384EId

(w per metre length)

L

Total load = W

Uniformly Distributed Load

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Given load,W, and span,L

Given Modulus of Elasticity,E,and Moment of Inertia,IUse 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)

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Given the beam size and material

Given the loading conditions

Use formula for maximum deflection

Compare this deflection to the Code allowabledeflection





Check the deflection of the steel channel

previously designed for strength

The maximum deflection



Need twice as muchI

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 theIbut only 27% more material

strategyfor heavily loaded beams25/28




Given the loading conditions

Given the Code allowable deflection

Use deflection formula to findI

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/500into millimetresUse deflection formula to find minimum value of I

Look up tables or use I= bd3/12and choose band d

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Beams need largeI andZ in direction of bending

Need sti ffness in other direction to resist

lateral buckling

Some sections useful for both

Columns usually need large value of rin both directions

= better sections for beams

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Deep beams are economical but subject to

lateral buckling

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