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Design of beam
IS 800- 2007
 Horizontal member seen in a structure spanning
between columns.
 Support loads which are resisted by bending and
 Supports floors, roof sheeting as purlins, side
 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
Stringer beam under stairs
Stringer beam in bridge
 Based on how beam is supported
 Simply supported beams
 Cantilever beams
 Fixed beams
 Continuous beams
Commonly used type beam sections
 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
What is advantage of castellation
 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
 depth can be determined
at will by changing the
cutting pattern
 combining of a lighter
upper half with a heavier
lower half
Classification of beam section
 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
 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
The sectional dimensions should be such that the
following conditions are satisfied: 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
Plastic analysis
 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.
Concepts of plastic analysi
 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.
 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)
 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
 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
 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.
 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.
 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
Shape factor
 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
Plastic hinge concept
 As external load increases, thus BM
increases, rotation at a section
increases proportionally up to the yield
 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.
 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
Plastic hinges in a member are formed at:Max moment locations,
At intersections of two members where BM is same; in weaker
Restrained ends,
Below point loads
Hinges may not form simultaneously as the loading increases
 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 1/3rd of
the span.
 For uniformly loaded fixed-end beam, Bowles( 1980)
showed that the hinge length at the ends is given by
 L 
 1
 2.83 
 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
Based on above the beam sections are classified
as follows as per IS 800-2007
 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.
 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 2is.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.
Failure modes of beams
Design strength in bending (flexure)
 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.
Laterally supported beam
 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
 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.
Classification of laterally supported beams
 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
Case I
 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.
Case II
 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
Vd 
 γm0 = 1.1
Av. fy
3 .mo
 when factored design SF does not exceed 0.6 Vd, the design bending
strength Md shall be taken as
Md 
b.zp. fy 1.2.ze. fy
in.case.of .simply.sported.beam
Md 
b.zp. fy
1.5 ze. fy
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
 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) 
 Where
 V
  2
 1
 Vd
1.2Ze. fy
( forplastic
andcompactsec tion)
 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
Zfd. fy
Mfd 
tw.h 2
where   Zfd  Zp 
Ze. fy
Mdv 
for.semi  compact. sec tion
Ze = elastic section modulus of the whole section.
Effect of holes in the tension zone(page 53)
Shear lag effects
 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.
Design of laterally unsupported beams
 Under increasing tranverse loads, a beam shoul attain its full plastic
moment capacity.
This type of behaviour in laterally supported beams have been already
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
 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
The values of fcr,b can also be determined from the table 14 IS-800
page 57is.800.2007- code of practice for gener steel.pdf
Shear strength of beams
 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=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
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
Lateral buckling
Local buckling
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.
Effective length lateral torsional buckling
 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 LLT 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
Web buckling and web crippling
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
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:
Excessive deflection creates problem for floors or roof
drainage called ponding which leads to corrosion of
steel reinforcement inside floor
ii. In case of beams framed together are of different
sections must deflect in the same way
Designer can reduce deflections by
Increasing depth of the beam section
ii. Reducing the span
iii. Providing greater end restraints
 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
Simple beams yields largest moments and deflections(
BM max = WL²/8)
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
Design procedure of Channel/I-section purlins
 Design of purlin is by trial and error procedure and
various steps involved in design are as follows
1. The span of purlin is taken as c/c distance between
adjacent trusses.
2. Gravity loads P, due to sheeting and LL, and the
horizontal load H due to wind are computed.
3. 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
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
mo 
Zpz  Mz.
 2.5 My.
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.
 1.2 Zez.
Mdy  Zpy.
 f .Zey.
For safety Mdz >= Mz and Mdy >= My
6. The load capacity of the section is checked using
the following interaction equation:-
Mdz Mdy
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)
Vdz 
Vdy 
3 .mo
3 .mo
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
Design of angle purlins
 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
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
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