properties of the section

7/27/2019 Properties of the Section

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Properties of The Section



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 ?



Direct Stresses

Strut (shortens) Tie (lengthens)

Compression Tension

Compressive Stress Tension stress

Load W Load W

Area A Area A

= =

W W

= =



tension

compression

Consider the rubber beam under bending



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



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’.



Neutral

Axis

compression

tensionSection

Max bending ie

max stress



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)



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.



What is affecting the behaviour of theelement ?

Property of the Section.

Properties of the Section.

What are they and how do they affectbehaviour ?



Properties of a section

1) Area - tension

2) Second Moment of area -stiffness ie deflection

3) Radius of gyration - buckling

4) Section Modulus - strength



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



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



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



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



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



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



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



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



Standard values

b

dX X

Y

Y

Ixx = bd3 /12

Iyy = db3 /12

Ixx = πd4 /64

Iyy = πd4 /64



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



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



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



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



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.



Consider diagonal bracing in a

frame (struts)Smaller diagonal bracing membersin frame 1 than frame 2

Frame 1 Frame 2

Struts,pinned ateach end



Restraining a beam from buckling

Wall loading beam

Laterally unrestrained

Ie buckling occurs

RC slab loading beam

Laterally restrained

Ie buckling cannot occur



Plastic Section Modulus

This property govern strength of the sectionunder plastic behaviour.

Neutral

Axis

compression

tension Section

Elastic Behaviour

compression



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)



Neutral

Axis

compression

tension Section

Plastic Behaviour

Whole section yields

ie goes plastic

compression



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



24mm

48mm

Section 6mm thick

X X

Y

Y



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



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



[(24x6)+(42x6) x ŷ ] = (24x6x3) + (42x6x27)

ŷ = 18.27mm

Now need to apply parallel axis theorem

ie Ixx = ∑( Ixx self +Ah2 )



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



Consider the following shapes24

24

18

24

24

18

Both 6mm thick

24



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



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



Consider the shapes24

24

18

24

24

18

What have I done???

I h d th h ith



I have moved the shape withrespect to Y-Y when

determining Ixx24

24

18

XX

Y

Y



Simpler Iyy calc??24

24

18

24

24

18

Y



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



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



Consider the channel section24

30

Ixx = 24 x 303 /12 - 18 x 183 /12

45252 mm4

X X

Consider the channel section



Consider the channel section24

30

Y

Y

=

24

Y

Y

properties of the section

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