wk3 sections
TRANSCRIPT
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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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Universi t of S dne Structures SECTIONS
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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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Universi t of S dne Structures SECTIONS
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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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Universi t of S dne Structures SECTIONS
Peter Smith & Mike Rosenman
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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Universi t of S dne 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
Z bd2
6mm3
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Universi t of S dne Structures SECTIONS
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The bigger the Section Modulus, the
strongerthe section
Contains d2, so depth is important
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Peter Smith & Mike Rosenman
Strength --> Failure of Element
Stiffness -->Amount of Deflection
depth is important
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Peter Smith & Mike Rosenman
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
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Can be calculated, with a little extra work Manufacturers publish tables of properties
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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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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=
Peter Smith & Mike Rosenman
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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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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 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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Peter Smith & Mike Rosenman
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
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W
L
Central point load
WL3
8EI
48d
WL3
3EI
128d
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whereWis theTOTALload
Uniformly Distributed Load(w per metre length)
L
Total load = W
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The deflection is only one-fifth of a
simply supported beam
Continuous beams are generally stif fer than
simply supported beam
whereWis theTOTALload
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
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Check the deflection of the steel channel
previously designed for strength
The maximum deflection
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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
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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