Properties of the Section
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Transcript of Properties of the Section
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Properties of The Section
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Element Reactions
Previously covered -: reactions at
the supports, SFD, BMD.
VA VB
HB
but what about the reaction of the element itself ?
Previously stated and shown in lectures it doesreact by deflecting and assuming a deformedshape.
Consider different types of loading ?
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Direct Stresses
Strut (shortens) Tie (lengthens)
Compression Tension
Compressive Stress Tension stress
Load W Load W
Area A Area A
= =
W W
= =
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tension
compression
Consider the rubber beam under bending
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If we consider the beam shown as ademonstration, note that :-
The lines at the top have moved closer toeach other
whilst ...Those at the bottom have moved furtherapart.
ieCompression in the top of the section
and
Tension in the bottom of the section
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Along length of beam
Closeness less Closeness great Closeness less
Compressive & Tensile Stresses a min max minIn a simply supported beam the maximumcompressive and tensile stresses occur away from
the supports
Across Section of beamCloseness/Openess of lines greatest at extremeties.
No change at one point. A cross section of beam (simply supported) maxcompressive stress occurs at top, max tensilestrength at bottom and zero stress at a point called‘neutral axis’.
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Neutral
Axis
compression
tensionSection
Max bending ie
max stress
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Look at deflection
The beam used in the demonstration has the
same cross-sectional area.but
fixed in one axis ie
orfixed in the other axis ie
the deflection is different for the same loadWHY ?Depth governs stiffness which dictates deflection(see properties of the section)
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What about the strut ?
Consider the same cross section of beam but with
differing lengths and subjected to the sameloading.
1) 2)w
w
Strut 1) is in direct compressionbut
Strut 2) is being subjected to buckling.
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What is affecting the behaviour of theelement ?
Property of the Section.
Properties of the Section.
What are they and how do they affectbehaviour ?
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Properties of a section
1) Area - tension
2) Second Moment of area -stiffness ie deflection
3) Radius of gyration - buckling
4) Section Modulus - strength
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Area
Used in calculation of :-
Circle = pd
4
b * l
b
2
2
d
bl
1. Direct Stresses(a) Tension(b) Compression
Rectangle =
Square = b
b
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2nd Moment of Area (stiffness)Determined from:1. Centroid / Centre of Area2. Fixed 2nd moment of area.
(a) Rectangle = bd3
12
(b) Circle =p
d4
643. Parallel axis theorem.
y
y
200
x x600
Dimensions in mm
I AA = IXX + Ay2
Ixx = 3.6 * 109 mm4
Iyy = 0.4 * 109 mm4
}or
-then either
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Determine by :-1. Common Sense2. Taking moments of Area about any point.
N.B. Both axes count.
1. Find Centre of Area
X X
Y
Y
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1. Find Centre of Area
Is the point about which first moments of area
balance.A1
A2
LA1
LA2
Centre of area X-X
ie A1 x LA1 = A2 x LA2
where LA1 = lever arm for area A1 & LA2 = lever arm for area A2
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900
900
12
12
First moment ofarea about this axis
Step 1 - Split into rectangles
900
900
12
12
Unsymmetrical section
S 2 S l bi i P P
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Step 2 - Select arbitrary axis P-P 900
900
12
12Step 3 - first moment of area (moa) of each shape
A1
A2
900
900
12
12
A1
A2
894450
Shape A1:
1st moa = (888 x 12) x 894
Shape A2:
1st moa = (900 x 12) x 450
P P
fi f ( ) f h h f
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Step 4 - Equate first moment of area (moa) of each shape to moa of
whole area
900
900
12
12
A1
A2
894450
[(888 x 12) x 894] + [(900 x 12) x 450] = [(888x12) +
(900x12)] x ў
ў = 670.5mm
y
neutral axis (NA) X-XX X
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2nd Moment of Area (I)
This defines, for the same material, the stiffness
of the element.
ie for a simply supported element, thedeflection of the element.
in the 2nd and 3rd years it determines the stressdistribution through the structure
For standard symmetrical elementsie
I is constant but can be Ixx or Iyy
or
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Standard values
b
dX X
Y
Y
Ixx = bd3 /12
Iyy = db3 /12
Ixx = πd4 /64
Iyy = πd4 /64
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If section is not regular then need to use :
Parallel Axis theorem
ie IPP = Ixx + Ay2
b
d NA
PP
IPP = bd3/12 + bd (d/2)2 = bd3/3
X X
R di f G ti G b kli
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Radius of Gyration - Governs buckling
What is radius of gyration and how does it affect
behaviour?
Radius of gyration is that distance between a singlepoint of concentrated area and a given axis such as
I = Ar 2
where r = radius of gyration
r =I
A
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Radius of Gyration
r xx
= IXX
A
or
r yy = Iyy
A b
dX X
Y
Y
In the figure shown,
by inspectionr xx > r yy
Therefore section will
buckle about y-y axis
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Section Modulus
Zxx = Ixx y
Zyy = Iyy
y
This property governs strength of the sectionunder elastic behaviour, only if buckling isprevented
b
dX X
Y
Y
y
Zxx here based
on y shown
y1
Zxx here based
on y1 shown
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Only if buckling is prevented?
Lateral torsional buckling
It is the buckling of the compression flange
which governs this behaviour. Therefore need to provide restraints to the
compression flange.
It is the length of any unrestrained length ofmember in compression which governs behaviour.
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Consider diagonal bracing in a
frame (struts)Smaller diagonal bracing membersin frame 1 than frame 2
Frame 1 Frame 2
Struts,pinned ateach end
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Restraining a beam from buckling
Wall loading beam
Laterally unrestrained
Ie buckling occurs
RC slab loading beam
Laterally restrained
Ie buckling cannot occur
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Plastic Section Modulus
This property govern strength of the sectionunder plastic behaviour.
Neutral
Axis
compression
tension Section
Elastic Behaviour
compression
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NeutralAxis
compression
tension
σy
σy
d/2
d/2
C
Th
b
T = C = σy * (b* d/2/2)
Moment Capacity = T*h + C *h
= [σy * (b* d/2 *1/2) * d/2 *2/3] *2
= σy* ( b * d2 /6) = σy* z
(NB h = 2/3 * d/2)
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Neutral
Axis
compression
tension Section
Plastic Behaviour
Whole section yields
ie goes plastic
compression
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NeutralAxis
compression
tension
σy
σy
d/2
d/2
C
Th
b
T = C = σy * (b* d/2)
Moment Capacity = T*h + C *h
= [σy * (b* d/2 ) * d/2 *1/2] *2
= σy* ( b * d2 /4) = σy* s
(NB h = 1/2 * d/2)
Y
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24mm
48mm
Section 6mm thick
X X
Y
Y
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If asked which is weaker axis ie ifconsidering best shape to use as
a strut??
By inspection it is Y-Y axis
Where Iyy = (6x243/12) + (42x63/12)
= 7668 mm4
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But if Ixx also requ’d
Step 1 – Find position of X-X axis using
first moment of area
Arbitrary axis P-PP P
X Xŷ
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[(24x6)+(42x6) x ŷ ] = (24x6x3) + (42x6x27)
ŷ = 18.27mm
Now need to apply parallel axis theorem
ie Ixx = ∑( Ixx self +Ah2 )
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X X
X-X self
X-X self
h
h
Ixx = [(24 x 63 / 12) + (24 x 6 x (18.27-3) 2)]+ [(6 x 423 /12) + (6 x 42 x (27-18.27))2]
= 90258.5 mm4
18.27
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Consider the following shapes24
24
18
24
24
18
Both 6mm thick
24
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Ixx = [(24 x 63 / 12) + (24 x 6 x 122)] x 2+ [(6 x 183 / 12) + (6 x 18 x 02)]
= 45252 mm4
Iyy = [(6 x 243/12) + (6x24x92)] x 2
+ (18x6
3
/12) = 37476 mm
4
24
24
18 XX
Since the axes of the
side area is not the same
as the whole then cannot
use subtraction. Need to
use parallel axis
theorem. UGGHH!!!
24
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Ixx = (24 x 303/12) - 2 x( 9 x 183/12)= 45252 mm4
Iyy = 2 x(6 x 243 / 12) + (18 x 63 /12)=14148 mm4
Notice anything strange about these
values?
24
24
18
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Consider the shapes24
24
18
24
24
18
What have I done???
I h d th h ith
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I have moved the shape withrespect to Y-Y when
determining Ixx24
24
18
XX
Y
Y
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Simpler Iyy calc??24
24
18
24
24
18
Y
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Iyy = (6 x 423 /12) + 2 x (12x63 /12)
=37476 mm4
We have therefore moved shape relativeto X-X axis only in order to find Iyy
42
12
Y
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Other properties from Ixx and Iyy
rxx = = (45252/396)0.5 = 10.7mm
ryy = = (14148/396)0.5 = 5.98mm
zxx = = (45252/15) = 3016.8mm3
Zyy= = (14148/15) = 943.2 mm3
Ixx A
Iyy A
Ixx
yIyyy
Consider the channel section
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Consider the channel section24
30
Ixx = 24 x 303 /12 - 18 x 183 /12
45252 mm4
X X
Consider the channel section
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Consider the channel section24
30
Y
Y
=
24
Y
Y