Section Properties 2

7/31/2019 Section Properties 2

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AS4100 Standard Grades and

Sections

Asst. Prof. Hang Thu Vu

[email protected]


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Lecture outline Industry uses AS4100 for general purpose steel structure analysis

and design. Within CIVL3111, we will refer to this standard as themain design code.

General material properties to use for design to AS4100

Discuss available standard grades and sections for design toAS4100

Grades: overview of availability. Yield stress and tensile strength Sections: overview of availability. Effects of shapes on section

capacity against loading actions

Study the meanings, usages and how to compute sectionparameters I, Z, S

Bending of steel members in elastic and plastic ranges Second moment of area I

Elastic section modulus Z

Plastic section modulus S


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Material properties to use for design to

AS4100


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Basic parameter values and adjustment for

elevated temperatures


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Standard Grades


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Hot-rolled and cold-form grades Hot-rolled products (plates and sections) are in

grades 200, 250, 300, 350, 400, 450, 500.These may be obtained with notch ductilequalities (with suffix L0 or L15) and/or weather-resistant qualities (with prefix WR). Weatheringsteels are in Grade 350 only

Hot-rolled welded sections are produced fromAS/NZS 3678 plates; popularly in Grades 300,400, WR350

Cold-formed hollow sections are with prefixes C.

They are produced in Grades C250, C350, andC450. They may come with notch ductile L0quality


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Hot-rolled and cold-form grades

Data for design to AS4100 forthe most commonly used

sections and plates in Grades250, 300, 350


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Values of yield stress fy


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Value of tensile strength fu


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Existence of residual stress

The material usually does not cool downuniformly after rolling for the hot-rolledsections. A similar effect occurs when plates

are welded at the junctions This induces initial residual stress in the

members. Residual stresses may be as highas fy/2. They greatly affect the behaviour of

steel members. Allowance must always be made for residual

stress effect on structure behaviours


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Existence of residual stress

Tips of flanges and middle of web cool and harden, morequickly than the rest of the cross-section.

The harder parts are in a state of compression. Thejunctions, being held by the harder parts and unable tocontract as far as they would otherwise, are placed intension.


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Standard sections


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Standard sections and plates

Steel products are provided in standard sizesand shapes. For material availability and costaspect, it is recommended to use standardsections in your design. Below are fundamental

sections


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Standard sections and plates


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Effects of shapes on member strength

It is observed that steel is produced in various sections.

These include "plate", "rounds, bars and rods", "angles","channels" and "I-sections".

The I-sections are produced as "beam" sections with Iyymuch less than Ixx and "column" sections with Iyy of closervalue to I

xx

.

Most of these sections are produced by rolling red hotsteel. Most sections have parallel flanges (the "universal"beam and column sections, and the channels). Some"taper flange" Is and channels are also produced.

It is possible to make very large sections (e.g. for bridgebeams) by welding plate into the form of an I or box.Lengths are available from a minimum of 6 metres to amaximum of 30 metres. Refer to Table 2, OneSteel "HotRolled and Structural Products", 5th ed.


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The main reason for existence of various section

shapes is to maximize member strength whileminimize the amount of used material for costpurpose

Consider shape effects on member strength

subjected to different types of loading Tension: stress depends on cross sectional area. It

is independent of section shape.

Compression: if the member is bulky, stress in thesquashed member is independent of section shape.It is similar to tension case

A

P


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However, if buckling occurs, the buckling loadthat the member can withstand depends onsecond moment of area Ixx and Iyy.

As the column buckle about the axis which is ofweaker I, it is important to have Ixx and Iyy ofsimilar magnitude (Universal columns) when

there is no lateral bracing for weak axis

2

2

)(kL

EIPcr


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Bending: stress in a beam depends on the second

moment of area I

The bending moment which a beam can carry before itsflange starts to yield is M=Zfy, where Z is the elasticmodulus.

If the moment is increased further, yielding spreadsthroughout the cross section. Total collapse occurs at amoment M=Sfy where S is the plastic section modulus(study later).

I

My

maxy

IZ


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To maximize the section capacity for bending isto maximize Z and S. For a given amount ofmaterial, it is done by spreading the materialaway from the neutral axis xx associated with

the bending. The I section is ideal for thispurpose. A weak point is for the same amount of material

when Ixx increases to support bending about xxaxis, I

yy

is reduced. It makes the beam flexible inthe lateral direction, and prone to "lateralbuckling (study later


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Compare section properties for sections of

same amount of material

Area A = 16000 mm2

Spread material further away from neutral axis tomake rectangle, I shape (610 UB 125), truss


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Compare section properties for sections of

same amount of material

Section I (mm4) Z (mm3) S (mm3) Ratio S/Z

Square 21.3 x 106 337.4 x 103 506.1 x 103 1.5

Rectangle 42.7 x 106 477.4 x 103 716.1 x 103 1.5

I beam 986 x 106

3230 x 103

3680 x 103

1.14Truss (Ad2)/4 (Ad)/2 (Ad)/2 1

The I shape is ideal for achieving high bending stiffness for thesame amount of cross sectional area

For some sections, S is computed by factoring Z with a factor k. Square and rectangle: k = 1.5 exactly

I beam: k ~ 1.15

You will learn how to compute S directly later. It is expected that youdo not use the factoring approach in your calculation within this unit.


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Bending of steel member in elastic and

plastic ranges


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Revision: Structural analysis


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We want to calculate the maximum values ofactions (moment, shear force, axial force ..)for member design

Free body diagram

Vertical reaction Ay, equivalent load P. At thecut of distance X from left end: shear force V,bending moment M


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The reaction at A is

The equivalent point load at a distance x/2 from A is

Take moment about the cut,

Force equilibrium for y direction,

Maximum shear force happens at end

Maximum bending moment happens at middle

wLAy2

1

wxP

22

1

022

1

2wx

wLxM

xwxwLxM

wxwLV

wxwLV

2

1

02

1

wLV2

1max

8

2

max

wLM


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Bending of a beam member

Bending stress xx of the cross-section of a beam varies alongthe beam height

The beam is under maximum stress when M=Mmax andy=ymax=d/2

The extreme fibre of the cross-section starts to yield when xx =fy. Hence, the moment capacity that a cross-section can take is

Z

M

I

My

ZfM yy


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When the whole cross section yields the momentcapacity increases to Ms.

For design purpose (lower bound of plasticity), strainhardening is ignored. The material is termed as pureplasticity (see below figure).

It is assumed that the maximum stress that anywhere inthe cross section can reach is fy (see below figure)


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Z is termed elastic section modulus

S is termed plastic section modulus


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Second moment of area


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Formulae

Second Moment of Area: also known with other names

Second Moment Of Inertia, Area Moment of Inertia The mathematical equations to calculate the Second

Moment of Area :

y is the distance from the neutral axis xx to aninfinitesimal area dA

x is the distance from the neutral axis yy to aninfinitesimal area dA

A

yy

A

xx

dAxI

dAyI

2

2


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Example 1: Rectangular section

12

8833

3

332

2

32

2

22

bdI

ddbybbdyydAyI

xx

d

d

d

dA

xx


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Example 2: Hollow sections

Rectangular hollow sections

I sections

1212

3

22

3

11dbdb

Ixx

122

12

33chbd

Ixx


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Example 3: Circular sections and Circular

hollow sections

4

4r

Ixx

4

2

4

14 rrI

xx


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Neutral axis (NA) of non-symmetric

sections. When the section is non-symmetric, we need to find the

location yc of the neutral axis with respect to a datumwhich is usually chosen at the base of the section

The sign shows the contributions from all nelements of the cross section.

n

i

i

n

i

ii

c

A

hAy

1

1


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Neutral axis of non-symmetric sections.

Ai is the area of the element ith,

hi is distance between centroid (NA) of theelement ith and the chosen datum


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Parallel axis theorem

The second moment of area for the whole section

with respect to the located neutral axis is calculatedfrom the Parallel Axis Theorem

I: the second moment of area

Ii: the second moment of area of element ith

Ai: area of element ith di: distance between the neutral axis of element i

thand the neutral axis of the whole section

n

i

iii dAII1

2


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Example: Calculate Ixx for T section

(200x20 + 500x10) yc= 200x20x510 +

500x10x250

Hence

yc = 365.5 mm

n

i

i

n

iii

c

A

hA

y

1

1


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Example: Calculate Ixx for T section

d1 = 510 - 365.5 = 144.5 mm

d2 = 365.5 - 250 = 115.5 mm

Ixx = 200x203/12 +

200x20x(144.5)2

+ 10x5003/12 + 10x500x(115.5)2

Hence

Ixx = 254,522,250 mm4

n

i

iii dAII1

2


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Elastic section modulus


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Formulae

Elasticsection modulus Z of a beam is the ratio ofa cross section's second moment of area I to thedistance of the extreme compressive fibrefrom theneutral axis

The elastic section modulus marks the yield point ofthe material when the most outer fibre starts to yielddue to bending moment My=Zfy

maxy

IZ


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Plastic section modulus


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Meaning and usage

It is also known by other name First Momentof Area

Plastic section modulusS associates withthe full plasticity of the whole section. It is the

state where no strain hardening occurred isassumed and the stress anywhere of the crosssection equals to yield stress fy

It is used to compute the plastic moment (fullcapacity of a cross-section) Mp = Sfy


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Define location of the PNA

The plastic neutral axis PNA, which is also referred to asthe equal area axis, is the axis that splits the crosssection into two equal areas. These areas refer to theequal amount of fibres yielded under compression and

tension respectively. For symmetric section, the plastic and elastic neutral

axis coincide. They are the axis through the centroid ofthe section.

For non-symmetric section, location y the PNA withrespect to a datum which is usually chosen at the baseof the section is defined


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Define location of the PNA

To locate the PNA,compute distance ybetween the PNA andthe bottom base

Total area = (200 x 40) + (400 x 40) = 24000 mm2

Check: 200x40 < 24000/2

Hence we have: 40 x y = 24000/2

Hence y=300


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Calculate S

S is computed as the sum of moment of elementareas about the PNA

S: the plastic section modulus

Ai: area of element ith

ei: distance between the neutral axis of element ith

and the plastic neutral axis of the whole section.

n

iiieAS 1


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Example: T section

y=300

S = (200 x 40 x 120) + (100 x 40 x 50) + (300x 40 x 150)

Hence, S = 2960 x 103 mm3


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Next lecture

Investigate the loads acting on the structurein terms of

Permanent load (dead load)

Imposed load (live load) Load factors to comply with limit state design and

member design to AS4100

ReadAS/NZS 1170.1:2002

Documents

Section Properties 2