I S 8 0 0 - 2 0 0 7

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Horizontal member seen in a structure spanning between columns.
Support loads which are resisted by bending and shear
Supports floors, roof sheeting as purlins, side cladding.

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Floor beams- major beam supporting the secondary beams or joists
Girder- floor beam in buildings
Lintel –beam used to carry wall load over openings, i.e
doors, windows etc
Purlin- roof beam supported by roof trusses
Rafter- roof beam supported by purlins
Spandrel beam- beam at outter most wall of buildings, which carry part of floor load and exterior walls
Stringer beam- longitudinal beam used in bridge floors and supported by floor beams

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Based on how beam is supported
Simply supported beams
Cantilever beams
Fixed beams
Continuous beams

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Universal beams ( rolled sections ): in this material is concentrated in the flanges and very efficient in uniaxial bending
Compound beam : universal beam strengthened by flange plates. Resist bending in vertical as well as horizontal direction.
Composite beam : universal beam with roof slab which gives continuous lateral support. The concrete floor provides the necessary lateral support to the compression flange to prevent lateral buckling.

 Castellated beams: beams made by applying a special technique to wide flange I-beam. This technique consists of making a cut in the web of a wide flange beam in a corrugated pattern. The cut parts are separated and the lower and upper parts are shifted and welded as shown in the next slide

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Light, strong and cheap
Easy to assemble at construction site
Openings simplify the work of installer and electrician, since taking pipes cross the beams do not pose a problem
Constructon elements such as ceiling systems can be installed easily.
Improves aesthetics

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 depth can be determined at will by changing the cutting pattern combining of a lighter upper half with a heavier lower half

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Bending strength of a beam depends upon how well the section performs in bending
Thin projecting flange of an I-beam is likely to buckle prematurely
Web of an I-section can buckle under compressive stress due to bending and shear
In order to prevent such local buckling it is necessary to limit outstand thickness ratios of flanges and depth/thickness ratios or web

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When the design is made using elastic analysis the member should be able to reach the yield stress under compression without buckling.
When design is done by plastic analysis the member should be able to form plastic hinges with sufficient rotation capacity( i.e ductility) without local buckling, so as to permit redistribution of bending moments needed before reaching collapse mechanism

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The transition from elastic to plastic analysis
In elastic design method, the member capacity is based on the attainment of yield stress.
Steel’s unique property of ductility is not utilised
Ductility enables the material to absorb large deformations beyond the elastic limit without fracture, due to which steel possesses reserve strength beyond its yield strength.
The method which utilizes this reserve strength is called Plastic analysis.

 Plastic analysis makes the design more rational, since level of safety is related to collapse load of the structure and not to apparent failure at one point.
Consider I-beam subjected to steadily increasing BM
‘M’ as shown in figure.

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When yield stress reaches the extreme fibre as shown in figure
(b) the nominal moment strength Mn of the beam is referred to as the yield moment My and is given by Mn= My = Ze.fy
Where Ze is the elastic section modulus
Further increase inBM causes the yield to spread inwards from the outter upper surfaces of the beam as shown in figure (C)

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This stage of partial plasticity occurs because of the yielding of the outer fibres without increase of stresses as shown by the horizontal line of the idealised stress strain diagram shown in figure
Upon increasing the BM further, the whole section yields as shown in figure d. When this condition is reached every fibre has a strain equal to or greater than εy = fy/Es.The nominal moment strength Mn at this stage is referred to as the plastic moment Mp and is given by
Mp = fy ∫
A ydA = fy Zp
Where Zp = ∫
A ydA is the plastic modulus.

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Any further increase in BM results only in rotation, since no greater resisting moment than the fully plastic moment can be developed until strain hardening occurs.
The maximum moment Mp is called the Plastic moment of resistance , the portion of the member where Mp occurs is termed as plastic hinge .

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For equilibrium of normal forces, the tensile and compressive forces should be equal. In elastic stage, when bending varies from zero at neutral axis to a max at the extreme fibres, this condition is achieved when the neutral axis passes through the centroid of the section.
In fully plastic stage, because the stress is uniformly equal to the yield stress, equilibrium is achieved when the neutral axis divides the section into two equal areas.

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 Considering the general cross section in fig. and equating the compressive and tensile forces we get.
Fy .A1 = fy. A2
Since A1= A2= A/2
A= A1+A2
Plastic moment of resistance
Mp = fy.A1. y1 + fy. A2.y2
=fy. A/2 (y1+y2)
Thus Mp = fy. Zp
Where Zp = A/2( y1+y2) is the plastic modulus of section

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The ratio Mp/My is a property of the cross sectional shape and is independent of the material properties. This ratio is known as the shape facot v and is given by v = Mp/My = Zp/ Ze
For wide flange I – sections in flexure about the strong axis (Z-Z) the shape factor ranges from1.09 to about 1.18 with average value being about 1.14.
One may conservatively take the plastic moment strength of I-sections bent about their strong axis to be atleast 15% greater than the strength
My when the extreme fibre reaches the yield stress fy.
On the other hand the shape factor for I bent about their minor axis is same as for rectangle i.e about 1.5
When material at the centre of the section is increased, the value of v increases.

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As external load increases, thus BM increases, rotation at a section increases proportionally up to the yield point.
Further increase in moment will generate a non linear relation between the stress and strain and the curvature increases rapidly to reach an unbounded value as moment tends to reach the value of Mp.
The part on the moment rotation curve where there is plasticity may be projected onto the moment diagram which can then be projected to the section and the area that’s plasticized can be determined.
At the centre of the area there will be full plasticity over the full depth of the section and section behaves as a hinge called plastic hinge.

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 Thus the plastic hinge can be defined as a yielded zone. In this zone the bending of a structural member can cause an infinite rotation to take place at constant plastic moment Mp of the section.
Plastic hinges in a member are formed at:-
Max moment locations,
At intersections of two members where BM is same; in weaker section
Restrained ends,
Below point loads
Hinges may not form simultaneously as the loading increases

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Considering the figure again
Let Wu be the load on a simply supported beam as shown
Mp= Wu.L/4 also Mp= fy. Zp= fy.bh²/4---(1)
And My = fy.Ze = fy. bh²/6---(2)
We can write Eq. 2 as
My = fy. (bh²/4)(2/3)
= Mp . (2/3)-------(3)
Let x be the length of plasticity zone. From similar triangles
Mp/(L/2)= My /(L/2 – x/2)
Mp/L = My / (L-x)
My/Mp = (L-x)/L
Substituting My =( 2/3)Mp
------->x= L/3
Hence for SS beam the plastic hinge length is equal to 1/3 rd of the span.

 For uniformly loaded fixed-end beam, Bowles( 1980) showed that the hinge length at the ends is given by x



L
2 .
83


1

1 v
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For an I-section, the length of the plastic hinge at the centre of the beam is ( taking v=1.12)
L hinge = 2L/8.645 = 0.23L i.e 23 % of the span length

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Class 1(Plastic): cross section which can develop plastic hinges and have rotation capacity required for failure of the structure by formation of plastic mechanism. The section having width to thickness ratio of plate element less than that specified under class 1 as shown in table 2 (page 18) ..\is.800.2007- code of practice for gener steel.pdf
Class 2 (compact section): cross section which can develop plastic moment of resistance, but have inadequate plastic hinge rotation capacity for formation of plastic mechanism, due to local buckling. The section having width to thickness ratio of plate elements between those specified for class 2 and class 1 shown in table 2.

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Class 3(Semi compact section):cross section in which extreme fibre in compression can reach yield stress but cannot develop plastic moment of resistance, due to local buckling. The width to thickness ratio of plate shall be less than that specified under class 3,but greater than that specified under class 2 as shown in the table 2 is.800.2007- code of practice for gener steel.pdf
Class 4 (slender): cross-section in which the elements buckle locally even befor reaching yield stress. The width to thickness ratio of plate shall be greater than that specified under class 3. IS code 800 considers the design of members belonging to class 4 beyond its scope.

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Design bending strength (bs) of beam, supported against lateral torsional buckling( laterally supported beam) is governed by the yield stress.
The factored design moment, M at any section, in a beam due to external actions, shall satisfy the relationship M<=Md where Md is the design bending strength of the section.

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A beam may be assumed to be adequately supported at the supports provided the compression flange has full lateral restraint and nominal torsional restraint at support supplied by web cleats, partial depth of plates etc.
Full lateral restraint to cmpression flange may be assumed to exist if the frictional or other positive restraint of a floor connection to the compression flange of the member is capable of resisting a lateral force not less than 2.5 percent of the max force in the compression flange of the member. This may be considered to be uniformly distributed along the flange.

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Laterally supported beams of plastic,compact or semicompact sections are classified into the following cases;
Case i: Web of section susceptible to shear buckling before yielding
Case ii : Web of section not susceptible to shear buckling before yielding

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Web of section susceptible to shear buckling before yielding
When the flanges are plastic, compact, semi-compact but the web is susceptible to shear buckling before yielding (d/tw <= 67 ε) the design yielding stress may be calc using one of the following methods: i) the BM and axial force acting on the section may be assumed to be resisted by flanges only and web is designed only to resist shear.
ii) the whole section resist the BM and axial force acting on the section and therefore the web has to be designed for combined shear and its share of normal stresses. This is done by using simple elastic theory in case of semi-compact webs and simple plastic theory in case of compact and plastic webs.

 Web of section not susceptible to buckling under shear before yielding (page 59)
 d/tw >= 67 ε
Beams in this case are stoky beams where ε is given by ( see right)
For these beams the factored SF V does not exceed 0.6Vd, where Vd is the design shear strength given by
Vd

Av .
fy
3 .
 mo
 γm0 = 1.1

 when factored design SF does not exceed 0.6 Vd , the design bending strength Md shall be taken as
Md

 b .
zp .
 m 0 fy

1 .
2 .
ze .
 m 0 fy in .
case .
of .
simply sported .
beam
Md

 b .
zp .
 m 0 fy

1 .
5 ze .
 m 0 fy for .
.
beam
Where βb = 1.0 for plastic and compact section
βb = ze/zp for semi-compact sections
Ze,zp = plastic and elastic section moduli of the cross sections

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 When factored SF V exceeds 0.6 Vd, the design strength Md will be taken as Md = Mdv
Where Mdv = design bending strength under high shear
As per IS code this is calculate as follows:-
Mdv

Md
 
( Md

Mfd )

1 .
2 Ze .
 mo fy
( forplastic sec tion )
 Where  

 2
V
Vd


1

2
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Md = plastic design moment of the whole section considering web buckling effect
Mfd= pdm of the area of c/s excludign the shear area considering psf given as

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.
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
.
2
4

.

.

.
sec
Ze = elastic section modulus of the whole section.

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Simple theory of bending is based on the assumption that plane sections remain plane after bending. But presence of shear strain causes section to warp. Its effect is to modify the bending stresses obtained by simple theory, producing higher stresses near junction of a web and lower stresses at points far away from it as shown in the figure. This effect is called shear lag.
the effect is minimal in rolled sections, which have narrow and thick flanges and more pronounced in plate girder sections, having wide thin flanges when they are subjected to high shear forces esp in the region of concentrated loads.

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Under increasing tranverse loads, a beam shoul attain its full plastic moment capacity.
This type of behaviour in laterally supported beams have been already covered
Two imp assump made to achieve the ideal behaviour are i) the compression flange is restrained from moving laterally
Any form of local buckling is prevented.
A beam experiencing bending about major axis and its compression flange not restrained against buckling may not attain its material capacity. If the laterally unrestrained length of a beam is relatively long then a phenomenon known as lateral buckling or lateral torsional bucking of the beam may take place and the beam would fail well before it can attain its full moment capacity. (similar to eulers buckling of columns).

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Resistance to lateral buckling need not be checked separately for the following cases:i) bending is about minor axis of the section ii) section is hollow( rect/tubular) or solid bars iii) in case of major axis bending, λLT <= 0.4 where λLT is the non dimensional slenderness ratio for torsional buckling.
The design bending strength of laterally unsupported beam is given in is.800.2007- code of practice for gener steel.pdf
(page
54)
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The values of fcr,b can also be determined from the table 14 IS-800 page 57 is.800.2007- code of practice for gener steel.pdf

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Consider an I-beam subjected to max SF (at supp of SSB). The external shear ‘V’ varies along the longitudinal axis ‘x’ of the beam with BM as
V= dM/dx, while beam is in the elastic stage, the internal stresses
τ,which resist external shear V1’ and can be as :
 
V Q
Iz .
t
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V=SF under consideration
Q= Ay= static moment of the cross section about N.A
Iz= MI of entire cross sectio @ Z-Z axis(NA)
T= the thickness of the portion at which τ is calculated

Pattern of shear stress distribution

LATERAL TORSIONAL LOADING
When a beam fails by lateral torsional buckling, it buckles about it weak axis, even though it is loaded in the strong plane. The beam bends about its strong axis up to the critical load at which it buckles laterally, refer figure.

The lateral torsional bucking of an I-section is considered with the following assumptions
• The beam is initially undistorted.
• Its behaviour is elastic.
• It is loaded by equal and opposite end moment in the plane of the web.
• The load acts in the plane of web only.

 For SS beams and girders for span length L, where no lateral restraint to the compressive flanges are provided, but where each end of beam is restrained agains torsion, the effective length L
LT of the lateral buckling can be taken as given in the table as per IS
800 (page 58)

In SS beams with intermediate lateral restraints against torsional buckling the effective length for lateral torsional buckling should be equal to 1.2 times the length of the relevant segment in between the lateral restraints.
Restraint against torsional rotation at supports in these beams can be provided by web or flange cleats or bearing stiffeners acting in conjunction with the bearing of the beam or lateral end frames or external supports providng lateral restraint to the compression flanges at the ends, or they can be built into walls

For cantilever beams for projecting length L, the effective length LLt to be used shall be given in table cl 8.3.3 page 61 of is.800.2007- code of practice for gener steel.pdf

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Web crippling Web buckling
A heavy load or reaction conc. on a short length produces a region of high compressive stresses in the vertical elements of the web either under the load of at the support. The web under a load or above buckling as shown in figure
(right) and a web reaction point, may cause web failures such as web crippling or crushing as shown in figure(left) above.
Web buckling occurs when the intensity of vertical compressive stress near the centre of section becomes greater than the critical buckling stress for the web acting as a column.
Tests indicate that for rolled S B the initial failure is by web crippling rather than by buckling.

Dispersion of concentrated load for evaluation of web buckling
But for built up beams having greater rations of depth to thickness of web, failure by vertical buckling may be more probable than by failure by web crippling. This can be avoided by spreading the load over a large portion of the flange or by providing stiffeners in the web at points of load and reactions by thickening the web plate.
The above figure shows a plate girder SS at ends. The max diagonal compression occurs at the NA and will be inclined at 45 to it.

The web buckling strength at support will be
Fwb = (b1+n1).tw. fc
Where (b1+n1) is the length of the stiff portion of the bearing plus the additional length given by the dispersion at 45 to the level of NA, fc is the allowable compressive stress corresponding to the assumed web strut according to buckling curve ‘c’ , tw is the thickness of web plate.
Effective length = (d1√2)/2
Mini radius of gyration = t/√12
Slenderness ratio = le/ry
= [(d1√2)/2]/[t/√12] or d1 √6 / t
λ
= 2.45 d1/t
Hence the slenderness ratio of the idealised compression strut is taken as
λ
= 2.45
d1/t.

Similarly in case of web crippling the crippling strength can also be calculated assuming an empirical dispersion length = b1 + n2
The dispersion length is b1= b +n2
Where n2 is the length obtained by dispersion through flange , to the flange to web connection (web toes of fillets), at slope of 1:2.5 to the plane of the flange (i.e. n2 =
2.5d1) as shown in figure above.the crippling strength of the web (also called as the web bearing cpacity) at supports is calculated as
Fcrip = (b1 + n2 ) t fyw where fyw is the design yield strength of the web
If the above bearing capacity or crippling strength is exceeded stiffeners must be provided to carry the load

Table (Page 31 ) IS 800 gives recommended limits for deflection for certain structural members
 reasons for limiting deflections: i.
ii.
Excessive deflection creates problem for floors or roof drainage called ponding which leads to corrosion of steel reinforcement inside floor
In case of beams framed together are of different sections must deflect in the same way
Designer can reduce deflections by i.
Increasing depth of the beam section ii.
iii.
Reducing the span
Providing greater end restraints

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Beams used on trusses to support sloping roof systems bet adjacent trusses
Channels, angle sections, cold formed C or Z sections
Placed in inclined position over the main rafters
Purlins may be designed as simple,continuous,cantilever beams
Simple beams yields largest moments and deflections(
BM max = WL
²
/8)
Continuous (moment = WL²/10)
When erecting purlin it is desirable that they are erected over the rafter with flange facing up slope as shown in figure

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Design of purlin is by trial and error procedure and various steps involved in design are as follows
The span of purlin is taken as c/c distance between adjacent trusses.
Gravity loads P, due to sheeting and LL, and the horizontal load H due to wind are computed.
The components of these loads in directions perp. and parallel to sheeting are determined. These loads are multiplied by p.s.f
P=γf .P1
H= γf .H1

4.
The BM max Muu and Mvv are calculated by
Muu = Pl/10 and Mvv = Hl/10
Purlins are subjected to biaxial bending and require trial and error method for design.
The required value of section modulus may be determined from the following expression given by
Gaylord
Zpz

Mz .
 mo fy

2 .
5 d p

 My .
 mo fy


A trial section is selected from IS handbook or steel table and properties b and d are noted

5.
The deign capacities of the section Mdz and Mdy are given by:
Mdz

Zpz .
fy
 mo

1 .
2 Zez .
fy
 mo
And,
Mdy

Zpy .
fy
 mo
  f .
Zey .
fy
 mo
For safety Mdz >= Mz and Mdy >= My

6.
The load capacity of the section is checked using the following interaction equation:-
Mz
Mdz

My
Mdy

1

7.
8.
Check whether the shear capacity of the section for both the z and y axes, ( for purlins shear capacity will always be high and may not govern the design)
V dz

V dy
 fy
3 .
 m o
A vz fy
3 .
 m o
A vy
Where Avz = h.tw and Avy = 2.bf. tf
The deflection of the purlin calculated should be less than l/180
Under wind (combined with DL) the bottom flange of purlins, which is laterally unsupported will be under compression. Hence, under this load case, the lateral torsional buckling capacity of the section has to be calculated

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An angle is unsymmetrical @ both axes.
May be used when slope of the roof is < than 30
Vertical loads and horizontal loads acting on the purlins are determined and the max BM is calculated as PL/10 and HL/10, where P and H are the vertical and horizontal loads respectively.
The section modulus is calculated by
Z

M
1 .
33

066
 fy
 The trial section is the selected assuming the depth of angle section as 1/45 of th span and width of the angle section as 1/60 of the span.
The depth and width must not be less than the specified value to ensure that the deflection is within limits