PRESENTATION ON DESIGN OF STEEL STRUCTURE

Subject code- 1615502UNIT-01

DOS INTRODUCTION

What are steel structures

• In steel structures, structural steel is the main load carrying material to transfer the load within them and to transfer load to the ground

• Ex: - I-Beam, Tee section, [ - Channel section, Steel plate etc..,

• Steel concrete composite structures are also used in high-rise buildings but we are only going to study about steel structures in this paper

Common Steel structures

1. Roof truss in factories, cinema halls, railways etc.,

2. Crane girders, columns, beams

3. Plate girders, bridges

4. Transmission towers, water tank, chimney etc.,

Framed Building

Framed Building

Industrial Building

Truss Bridge

Advantages• High comp. & tensile strength per unit weight hence low

construction weight, saves space• Good aesthetic view• Good quality and durability• Very high speed of construction• Reusability and scrap value – env. Friendly• Better solution to cover large span and tall structures

Disadvantages• Highcost – Initial• Corrosion• Low fire resistance

Adv. & Disadv.

Steel

Steel making

• First iron is extracted from iron ores like haematite, limestone, magnetite in furnace

• Oxygen is passed through molten iron to remove carbon and impurities to make steel.

• Magnese is added to strengthen the steel

• Adding chrome, nickel, phosphorous can impart special properties in steel

Cont….

• Semi finished products from the machine is hot rolled to different sections like bars, plates, angles, sections etc..,

• Adding carbon increases the tensile strength and hardness but lowers ductility and toughness

• In building we use structural steel which has low carbon of upto 0.1% to have ductility and yield.

Properties of steel

• Physical properties (IS800:2.2.4)

1. r = 7850 kg/m3 = 78.5kN/m3

2. E = 2x105 N/mm2

3. Poison ratio µ = 0.3

Type l Design-

ation

UTS

(MPa)Yield strength (Mpa)

Thickness (mm)

<20 20-40 >40

Standard

structural

steel

Fe 410A 410 250 240 230

Fe 410B 410 250 240 230

Fe 410C 410 250 240 230

High tensile

structural

steel

St58HT 580 360 0.05 1.00

ST55-HTW 550 360 .05 1.00

Mechanical properties

FE 410 A W

IRON

GRADE

Tensile test

specimen before

ruptureF

Area=S-

F

F

Standard tensile

test specimen

F

rt

d

L

Area=S0-

L

Ductility• Ability of material to change its shape without fracture

Mild steel – high ductility

High carbon steel – low ductility

Toughness & brittle fracture

• Ability of material to resist (absorb) impact load like earthquake load, machine load etc..,

• Requires both strength and ductility

• At low temp. steel fails on impact loading due to reduction in ductilityand toughness called brittle fracture

TempAt high temp strength reduces

Corrosion

Steel corrodes in moist air, sea water and acid. Adopt Painting, metallic coating, plastic coating, using corrosion resistant steel to resist corrosion

Hardness

• Resistance of the material to intentions and scratching

• Brinell harness, rockwell hardness number are used to measure hardness

Fatigue

• Damage of material to cyclic loading

• Occurs due to moving loads, vibration in bridge

Residual stress

• Latent stress present in the steel sections due to uneven heating and cooling during steel making

Stress concentration

• Under loading, stress is concentrated at places at abrubtchange in geomentry like holes bolts

Common Steel members

Beam – Column construction

Rolled steel Channel - section

• ISLC, ISMC, ISLC, ISSC(Indian standard special section)

• Example ISMC 300 & 0.351 kN/m

300 mm

Rolled Steel Angle section

• ISA Equal angle – ISA 150 x 150 x 12

• ISA unequal angle – ISA 150 x 115 x 12 Thickness

Prakash Kumar Sekar from Civilrnd.com

THANK YOU

PRESENTATION ON DESIGN OF STEEL STRUCTURE

Subject code- 1615502UNIT-02

CONNECTIONS

CONNECTION

Connections are the devises which are used to join elementsof a structure together at a point such that forces can betransferred between them safely.

TYPES OF CONNECTIONS

BASED ON MEANS OF CONNECTION

WELDED CONNECTIONS

RIVITED CONNECTIONS

BOLTED CONNECTIONS

BASED ON FORCES TO BE TRANSFERED

TRUSS CONNECTIONS

FULLY RESTRAINED CONNECTONS

PARTIALLY RESTRAINED CONNECTIONS

SPLICES

BARCKETS

BASED ON PLACEMENT OF PARTS TO BE JOINED

LAP JOINTS

BUTT JOINTS

TRUSS CONNECTIONS

In truss connections only the axial forces are to be transferred. These are simplest in all types of connections which may either be welded

or bolted.

FULLY RESTRAINED/MOMENT CONNECTIONS

These connections are particularly used when continuitybetween the members of the building frame is required toprovide more flexural resistance and reduce lateral deflectiondue to wind loads. In this type of connection both the websand flanges are connected. In this connection greater than 90%moment can be transferred with full transfer of shear andother forces

PARTIALLY RESTRAINED CONNECTIONS

These connections have rigidity less than 90%. The originalangles between the connected members may change up to acertain limit after the application of loads. These connectionscan transfer some percentage of moment along with full shearforce.

SIMPLE\SHEAR CONNECTIONS

These connections have less than 20% rigidity. These areconsidered flexible and beams become simply supported. Inthis case only the web is connected with the other memberbecause most of shear stresses are concentrated in the web.

SEMI-RIGID CONNECTIONS

These types of connections provide rigidity in between fullyrestrained and simple connections and approximately 20% to90% moment compared with ideal rigid joint may betransferred. These type of connections are mostly used inpractice because their performance is exceptionally well under

cyclic loads and earthquakes

SPLICES

These are used to extend the length of a particular member. These connections may be bolted or welded.

BRACKETS

These are the connections which used to transfer moment

besides other type of forces. The term bracket is generally

used for an extra plate projecting out of column and acting like a seat for the beam.

COLUMN BASE CONNECTIONSThese connections can either be pined or fixed depending upon the type

of forces to be transferred.

THANK YOU

PRESENTATION ON DESIGN OF STEEL STRUCTURE

Subject code- 1615502UNIT-03

Design of tension members

Modes of failure

Gross section yielding

Net section yielding

Block shear failure

Design strength of member is least of:-

• Strength due to yielding of gross section

• Rupture of critical section

• Block shear

ti

b ndh

i 4gi

p2

•Block shear failure is

also seen in welded

connections.

•A typical failure of a

gusset in the welded

connection is shown in

the figure.

•The planes of failure are

chosen around the weld

•Here plane B-C is under

tension and planes A-B

and C-D are in shear

Lug angles

Short angles used to

connect the gusset

and outstanding leg

of the main member

as shown in figure

Refer Clause

10.12.2 of IS 800-

2007

Problem• Determine the tensile strength of the plate 120 mm x 8 mm

connected to a 12 mm thick gusset plate with bolt holes as shown in the figure. The yield strength and ult. Strngth of the steel used are 250 MPa and 400 MPa. The diameter of bolt used is 16 mm.

Solution• The design strength Td of the plate is

calculated based on following criteria

• A) Gross section yielding:

The design strength Tdg of the plate limited to the yielding of gross cross section Ag is given by

fy = 250 M pa

Ag = 120 x 8 = 960 mm2 ;

ti

b ndh

i 4gi

p2

PRESENTATION ON DESIGN OF STEEL STRUCTURE

Subject code- 1615502UNIT-04

DESIGN OF COMPRESSION MEMBERS

INTRODUCTION• The strength of steel compression members is usually limited by their

tendency to buckle.

• The load at which a compression member becomes unstable is the

buckling load.

• The buckling load depends on the length, cross-section, and end

conditions of the column and the stiffness of the material

• Steel Compression members are

a) Building columns & Beams

b) Frame Bracing

c) Truss members

• Useful in pure compression as well as in beam columns

• Design Clauses from IS codes

COLUMNS & BEAMS

• Columns form the main component of a structure which serves the basic purpose of supporting and transmitting the entire loads both vertical and horizontal for which the overall structure is intended to the foundation system.

• Beams are generally subjected only to flexure about the horizontal axis whereas columns are subjected to axial load along with bending moment about the major axis.

• The minor axis moment in columns are generally nil or very nominal since in standard structural system, the columns are so oriented that the frames along the major axis of the columns are moment resistant frames, and column bracings are provided in the frames along the other perpendicular direction.

Different types of Cross sections

Compression members in trusses

DESIGN METHODS

ALLOWABLE STRESS DESIGN

• With the development of linear elastic theories, the stress-strain behavior of new

materials like wrought iron & mild steel could be accurately represented.

• The allowable stress is defined in terms of a “factor of safety” which represented a

margin for overload and other unknown factors which could be tolerated by thestructure.

Allowable stress =

LIMIT STATE DESIGN

• An improved design philosophy to make allowances for the shortcomings in the

“allowable stress design” was developed in formulating design standards and codes.

Although there are many variations the basic concept is broadly similar.

• The probability of operating conditions not reaching failure conditions formsthe

basis of “Limit States Design” adopted.

• In order to reduce the probability of its occurrence to a very low level.”

Serviceability limit state” refers to the limits on acceptable performance of the

structure.

Yield Stress

Factor of Safety

DESIGN PROCEDURE

DESIGN A MEMBER SUBJECTED HAVING A SPAN OF 3M

WHICH IS FIXED @ BOTH ENDS LSM (As per IS: 800-2007)

Solution:

Let us take ISMB 200 @ 254 N/m

Area = 3233mm2

Depth (d) = 200mm

Width of flange (b) = 100mm

Thickness of the flange (tf) = 10.8mm,

Thickness of the web (tw) = 5.7mm

Buckling Load

2EIPcr (KL)2

• Pcr is the load at which the compression member

becomes unstable

• E is modulus of elasticity of steel

• I is moment of inertia of the cross section

• L is the length of the compression member

• K is the effective length factor

Step 1: Type of the Section Step 4: Determination of Non

DimensionalPd Axfc d 428.82kN

1 7 8

3 1 . 2 25 . 7

d

t w

The section is Compact

Step 2: Determination of Effective

Length

Leff = 0.65 x 3000 = 1950mm

Step 3: Calculate the Slenderness

Ratio

KL

lx 1950 23.43

r x rx 83.2

Step 5: Calculation of ø

0.510.22

0.5 1.341.02 0.2 1.022 1.16

ry ry 21.5

1950 90.69

K L

l y

250x23.432

2x2x1052E

KL2

0.2638

fy r

x

fyfcc

2

1.020y

cc

fy 250 x90.692r

f 2E 2 x2x105

fKL

y

Step 6: Calculation of Stress

Reduction Factor

Step 7: Determination of design

Compressive Stress fcd

Step 8: Determination of

Compressive stress Pd1 1

0.58[1.16 (1.162 1.022 )]220.5

f y / mo

2 2 0.5cd

f

132.64 fy

fy

mo mo

Pd A fcd 428.82kN

CONCLUSION

• The load carrying capacity of the compression

members as per IS 800-2007 is controlled by

‘stress reduction factor, inclination of tension

field stress in web and effective slenderness ratio.

• The slenderness ratio is inversely proportional to

the stress reduction factor. The design

compressive stress is directly proportional to

‘stress reduction factor’.

THANK YOU

PRESENTATION ON DESIGN OF STEEL STRUCTURE

Subject code- 1615502UNIT-05

BEAM

Beams

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.

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 isconcentrated in the flanges and very efficient in uni-axial bending

Compound beam: universal beam strengthened byflange plates. Resist bending in vertical as well ashorizontal direction.

Composite beam: universal beam with roof slabwhich gives continuous lateral support. The concretefloor provides the necessary lateral support to thecompression 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 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

The sectional dimensions should be such that the following conditions are satisfied:-

When the design is made using elastic analysis themember should be able to reach the yield stressunder compression without buckling.

When design is done by plastic analysis the membershould be able to form plastic hinges with sufficientrotation capacity( i.e ductility) without localbuckling, so as to permit redistribution of bendingmoments needed before reaching collapsemechanism

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 largedeformations beyond the elastic limit withoutfracture, due to which steel possesses reservestrength beyond its yield strength.

The method which utilizes this reserve strength iscalled 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 toas 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 fromthe 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 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.

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-sectionsbent about their strong axis to be atleast 15% greater than the strengthMy when the extreme fibre reaches the yield stress fy.

On the other hand the shape factor for I bent about their minor axis issame as for rectangle i.e about 1.5

When material at the centre of the section is increased, the value of vincreases.

Plastic hinge concept

As external load increases, thus BMincreases, rotation at a sectionincreases proportionally up to the yieldpoint.

Further increase in moment willgenerate a non linear relation betweenthe stress and strain and the curvatureincreases rapidly to reach anunbounded value as moment tends toreach the value of Mp.

The part on the moment rotation curvewhere there is plasticity may beprojected onto the moment diagramwhich can then be projected to thesection and the area that’s plasticizedcan be determined.

At the centre of the area there will befull plasticity over the full depth of thesection and section behaves as a hingecalled plastic hinge.

Thus the plastic hinge can be defined as a yielded zone. In thiszone the bending of a structural member can cause an infiniterotation to take place at constant plastic moment Mp of thesection.

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

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 plasticityzone. 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

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

L 1

1

2.83 v

x

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 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.

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 tobuckling under shear beforeyielding (page 59)

d/tw >= 67 ε

Beams in this case are stoky beamswhere ε is given by ( see right)

For these beams the factored SF Vdoes not exceed 0.6Vd, where Vdis the design shear strength givenby

3.mo

γm0 = 1.1

Av. fyVd

when factored design SF does not exceed 0.6 Vd, the design bendingstrength Md shall be taken as

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

M d b . z p . f y

1 . 2 . z e . f y

i n . c a s e . o f . s i m p l y . s p o r t e d . b e a m m 0 m 0

f o r . c a n t i l e v e r . b e a mM d b . z p . f y

1 . 5 z e . f y

m 0 m 0

When factored SF V exceeds 0.6 Vd, the design strength Md will betaken as Md = Mdv

Where Mdv = design bending strength under high shear

As per IS code this is calculate as follows:-

m oM d v M d ( M d Mfd )

1.2Ze . fy( forplastiacndcompactsec tion)

2

Md = plastic design moment of the whole section considering webbuckling effect

Mfd= pdm of the area of c/s excludign the shear area considering psf given as

V Where 2

V d 1

4

tw.h2

Mdv Ze. fy

for.semi compact.section

where Zfd Zp

Mfd Zfd. fy

mo

m

Ze = elastic section modulus of the whole section.

Effect of holes in the tension zone

Shear lag effects

Simple theory of bending is based onthe assumption that plane sectionsremain plane after bending. Butpresence of shear strain causessection to warp. Its effect is to modifythe bending stresses obtained bysimple theory, producing higherstresses near junction of a web andlower stresses at points far awayfrom it as shown in the figure. Thiseffect is called shear lag.

the effect is minimal in rolledsections, which have narrow andthick flanges and more pronouncedin plate girder sections, having widethin flanges when they are subjectedto high shear forces esp in the regionof concentrated loads.

Design of laterally unsupported beams

Under increasing tranverse loads, a beam shoul attain its full plasticmoment 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).

The values of fcr,b can also be determined from the table 14 IS-800 page 57.

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

I z . t

V Q

Pattern of shear stress distribution

LATERAL TORSIONAL LOADINGWhen 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.

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

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

Deflection

Table (Page 31 )IS 800 gives recommended limits for deflection for certain structural members

reasons for limiting deflections:i. 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 byi. Increasing depth of the beam sectionii. Reducing the spaniii. Providing greater end restraints

Purlins

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 = 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 figure

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

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 bedetermined from the following expression given byGaylord

A trial section is selected from IS handbook or steel table and properties b and d are noted

fyp

fy

Zpz Mz.mo

2.5d

My.mo

5. The deign capacities of the section Mdz and Mdy are given by:

And,

For safety Mdz >= Mz and Mdy >= My

f y

m o

f y

m o

M d y Zpy .f y

f .Zey .f y

m o m o

1.2Zez .M d z Zpz .

6. The load capacity of the section is checked using the following interaction equation:-

My1

Mdz Mdy

Mz

7. 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)

A v y

A v zf y

3 . m o

Where Avz = h.tw and Avy = 2.bf. tfThe deflection of the purlin calculated should be less than l/180

8. 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

3 . m o

f yV d y

V d z

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

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

MZ

1.33 066 fy

THANK YOU

UNIT - 6

STEEL ROOF TRUSS