Design Of Steel Elements As Per Is 800 3zs38

1. INTRODUCTION Steel has been known since 3000 BC. The Ashoka Pillar made with steel and the iron ts used in Puri temples are more than 1500 years old. The iron pillar made in the 5 th century standing till today in Mehrauli village, Delhi, within a few yards from Kutub Minar evokes the interest and excitement of all the enlightened visitors. Scientists describe this as a “Rustless Wonder”[23]. Another example in south India is the iron post in Kodachadri village in Karnataka, which is 14 meter tall, “Dwaja Stamba” has reported to have remained 1 without rusting for nearly 1 2 millennium. The exciting aspects of these structure is not

Figure1.1 “Rabindra Sethu” Kolkata

merely the obvious fact of technological advances in India at that time, but in the developments of techniques for handling, lifting, erecting and securing such obviously heavy artifacts. India under the British rule experienced growth of iron and steel possibly because of the fallout of technological development of steel in U.K. We can see several steel structures in public buildings, railway stations and bridges, which testifies the growth of steel in the colonial past. The “Rabindra Sethu” Howrah Bridge in Kolkata stands testimony to a marvel in steel. The recent example is the second Hooghly cable stayed bridge at Kolkata called “Vidyasagar Sethu”. Even today the old city area of Kolkata consists of steel structural building built during the British rule. It is not only that the development of steel as structural element is restricted to government sector but has also been in the private sector. Renowned

structural engineer A.B.Karnik has used steel structure for the erection of piles ing a jetty at Dahej in Gujarat. Figure 1.2 “Vidyasagar Sethu” Kolkata

Above statement proves that steel as a structural material was been adopted since the early beginning of its development than how come its use as a structural material is limited in JOURNAL OF INDIAN SOCIETY OF STRUCTURAL ENGINEERS India. Following are the reasons VOLUME 8-4, OCT-NOV-DEC 2006 a) As compared to countries where Figure1.3 Gantry girder for the erection of piles ing a steel is used widely as structural jetty at Dahej designed by Er.A.B.Karnik [11] element for buildings, the cost (of steel) in India is more. b) The old concept of design i.e: working stress method has always proved uneconomical than conventional concrete design. Design of Steel Building Structural Element AsPer IS800:2007: 1 Developing a Computational Tool

c) The design, specially connection is complicated specifically in of selection of type of connection to be adopted, force transfer, check and maintenance of welds. d) Handling and erection of steel structures needs skill along with technology (technology in of machinery). Besides the above disadvantages steel has some inherent advantages too. Steel offers much better compressive and tensile strength, ductile in nature, high quality assurance, long life cycle, light weight construction, recyclable and the most important rapid construction. In metro city like Mumbai structural steel is used as the structural system from the beginning of the decade. It may be possible that the designers may have adopted the foreign design codes as the IS 800 has been just revised in 2007. Hence it can be said the use of structural steel as a structural element in buildings (residential / commercial) is not been arrested totally. 1.1 The revised IS800:2007: The entire design concept of steel structure has been changed to plastic theory of design making the structure more durable by introducing the aspects like corrosion, fatigue, design and detailing for earthquake load, durability, etc. The concept of plastic theory of design is explained later but before that let us have a look on the major difference in IS8001984 and IS800:2007 Table 1.1 Comparison of IS800:2007 and IS800-1984 [8],[9] Section 3

Title Design requirements

4

Method structural analysis Limit State IS800-1984 was based on the working stress method where the Design IS800-2007 is based on the limit state design method. The section defines the partial safety factor for loads, partial safety factor for materials, deflection limits for various member based on the loading acting upon it. Design and This section consist of detailing to meet the ductility requirements Detailing for for a earthquake resisting structure. Unlike concrete structure there Earthquake is no separate IS code for detailing for steel structures, hence it is included in the same code which was not present in IS800-1984 Fatigue Structures like bridges, gantry girders, cranes, slender tower-like (open) structures (subject to wind oscillations), offshore platforms (subject to wave load), and structures ing large rotating equipment are prone to fatigue, under this section the design provision are given for such structures to resist fatigue. Design Assisted This section is special provision where the design is done by testing by Testing the individual member or the prototype of the structure. Durability The most important factor affecting the durability of steel is corrosion. This section provides some surface protection guide for structural steel element.

5

12

13

14 15

Comments The design philosophy has changed for WSM to LSM, which has resulted in a change in the basic design requirement for eg. Section classification to assure that no buckling occurs before the section has reached the plastic stage. of This section describes various analysis methods that could be adopted for the steel structural analysis.

Design of Steel Building Structural Element AsPer IS800:2007: 2 Developing a Computational Tool

16

Fire resistance

Steel is adversely affected with high temperature gradient which generally occurs under the situation of fire. Under this section some guidelines are mention for casing the structural element in concrete to create an insulating effect.

1.2 Need of Computational tool: It is not always possible to have an exact mathematical formulation or modeling for all design problems. There is a considerable scope for exercising engineering judgment. Hence there is no correct solution to a design problem as there could be several so-called correct solutions to the same problem. This is because a) The designs are invariably subject to individual interpretation of standards and codes. b) The solutions are also subject to differing ideas about what is or what in Not required from an engineering and environmental stand point, and c) the individual designers have ingrained ideas from their past experience, which may be valid to-day only to a limited extent, or may not be valid at all. Hence in short it could be said that the designer changes the design changes.[23] Under this circumstances, relying on a particular software which would give such a solution that will be adopted universally is not possible. At the same time neither computer nor the computer program knows about the practical consideration for eg. Talking about Surat, having humid atmosphere and located at the sea cost the, the rebar of concrete building are more prone to corrosion than the building situated in Rajkot or say Bhavnagar. Hence in such a case if a building with similar data in all manner is to be constructed in Bhavnagar and surat the designer will prefer steel in slab at a closer spacing in the latter one then in the former. In practice most of the structural consultant adopts only analysis results from the software and design the structure base on these result, with their experience or using an in house program/software. Also the design procedure as per revised IS800-2007 is much complicated and tedious hence if one could develop such a tool, which on providing the basic design input will give you the design out put will save lots of time of the designer.

Design of Steel Building Structural Element AsPer IS800:2007: 3 Developing a Computational Tool

2. LIMIT STATE DESIGN AND PLASTIC THEORY To safeguard against the risk of failure, safety margins are normally provided in design. The variables such as loads, material strength, and member dimensions are subject to varying degrees of uncertainty and randomness. Hence the structure must be designed to provide for the possibility of overload. Similarly the deviations in the dimensions of or strength of material can result in a member having less than the computed strength. Thus, the design should take into the possibility of under strength also. Further, there are idealization and simplifying assumptions used in the theories of structural analysis and design. There are also several unforeseen factors that influence the prediction of strength and serviceability like construction methods, workmanship and quality control, intended service life of the structure, human errors, possible future change of use, etc. 2.1 Limit states: Limit state is a state of impeding failure, beyond which a structure ceases to perform its intended function satisfactorily [22]. The limit states usually considered for structural steel work are listed in the table.2.1. They are normally grouped into the following two types. Table 2.1: Limit States [22] Limit State of Strength Strength (yield, buckling) Stability against overturning and sway Fracture due to fatigue Plastic collapse Brittle Fracture

Serviceability Limit State Deflection Vibration Fatigue checks (including reparable damage due to fatigue) Corrosion Fire

Ultimate limit states are conditions to be avoided and serviceability limit state are condition that are undesirable. Hence it is clear that any realistic, rational and quantitative representation of safety must be based on statistical and probilistic analysis, which caters for both overload and under strength. Design for the ultimate limit state may be conveniently explained by figure2.1. This figure shows the hypothetical frequency distribution curves for the effect of loads on the structural element and the resistance of the structural element. When the two curves overlap shown by the shaded area, the effect of the loads is greater than the resistance of the element and the element fails.

Design of Steel Building Structural Element AsPer IS800:2007: 4 Developing a Computational Tool

Frequency

Region Corresponding to Failure

Q R

Qm

Rm

Load Effect Q or Resistance of element R

There exist a number of levels of reliability analysis. However for the code the method is simplified by using deterministic data rather than parabolic data, called as level I reliability method.

Figure 2.1 Frequency Distribution Curve [22] 2.2 Limit State Method: It is a comprehensive and rational solution to the design problem, by considering safety at ultimate loads and serviceability at working load. Based on above discussion IS8002007 determines the partial safety factor for load and material as follows Table 2.2: Partial safety factors for loads [8] Limit State of Strength LL’ Combination

DL+LL+CL DL+LL+CL+ WL/EL

DL

Limit state of Serviceability LL’ WL/ WL/ AL DL Leading Accompanying EL Leading Accompanying EL

1.5

1.5

1.05





1.0

1.0

1.0



1.2 1.2

1.2 1.2

1.05 0.53

0.6 1.2 

1.0

0.8

0.8

0.8

DL+WL/EL

1.5  (0.9)* 

1.5



1.0





1.0

DL+ER

1.2 (0.9)

1.2















DL+LL+AL

1.0

0.35

0.35



1.0 







*

This value is to be considered when the dead load contributes to stability against overturning is critical or the dead load causes reduction in stress due to other loads. ‘ When action of different live loads is simultaneously considered, the leading live load is whichever one causes the higher load effects in the member/section. Abbreviations: DL= Dead Load,LL= Imposed Load (Live Loads) WL= Wind Load, CL= Crane Load (Vertical/horizontal), AL=Accidental Load,

Design of Steel Building Structural Element AsPer IS800:2007: 5 Developing a Computational Tool

ER= Erection Load, EL= Earthquake Load.

Table 2.3: Partial safety factors for material [8] Sr.No.

Definition

Partial Safety Factor

1

Resistance, governed by yielding m0

1.10

2

Resistance of member to buckling m0

1.10

3

Resistance, governed by ultimate stress 1.25 m1

4

Resistance of connection m1 Bolts-Friction Type, mf Bolts-Bearing Type, mb (iii)

Rivets, mr

(iv)

Welds, mw

Shop Fabrications

Field Fabrications

1.25 1.25 1.25 1.25

1.25 1.25 1.25 1.50

2.3 Plastic theory: Consider the stress distribution for a typical beam subjected to load as shown in figure 2.2. Point (a) and (b) shows the stress distribution where the stress is within the elastic limit for point (a) and for point (b) the extreme fiber has just reached the elastic limit represented by f=fy. On further increase of load the plastic distribution of stress starts as indicated at point (c), a stage comes when the section completely attains the plastic stage. The point (e) shows the distribution of stress due to strain hardening. However this strength of the section due to strain hardening is not considered for the designed purpose thereby increasing a factor of safety. f < fy

f = fy

f = fy

f = fy PLASTIC STRAIN HARDENING

ELASTIC

ENTIRELY PLASTIC PLASTIC

M < My (a)

M = My (b)

My < M < Mp (c)

M = Mp (d)

(e)

Figure 2.2 Stress distribution for a typical beam[22]

Design of Steel Building Structural Element AsPer IS800:2007: 6 Developing a Computational Tool

True stress-strain curve Upper yield stress Engg. stress-strain curve fu

Stress, MPa

fy

Yield Plateau

ey

Strain-hardenign region

esh

Post-ultimate stress region

Strain, mm/mm

eu = 0.23

Figure 2.3 shows the stress-strain relationship for the typical test specimen of mild steel. The figure represent the point of significance of the test where up till εy the specimen remains in the elastic range where the stress-strain relationship is linear. Then the upper yield stress is reached, after which there is a significant increase in strain with constant value of stress. This is because the steel flows plastically or plastic deformation starts. Figure 2.3 Stress-Strain Relationship The increase in the strain with constant for Mild Steel [22] stress continues until steel reaches strain hardening stage, where there is a increase in the stress with strain. This continues until the ultimate stress is reached after which due to a large local reduction in the cross section (neck formation) occur and the load capacity decreases until tensile failure takes place.

3. DESIGN OF CONNECTION AS PER IS800:2007 Determining the exact design procedure for connection is not possible as there is a wide classification of connection based on the material used for connection, its behavior to the applied action (loads). Hence in these section only the required check to be taken or how one can calculate the capacity of connection under shear, tension, bearing, block shear, etc is shown. 3.1 Design Provision for Bolted Connection: Table no 3.1 IS800:2007 Provision for Pitch, Gauge, Edge Distance [8] 1) Maximum Pitch a) In General 32t or 300mm (smaller to the two) Where t is thickness of smaller plate b) For tension member 16t or 200mm (smaller to the two) c) For compression member 12t or 200mm (smaller to the two) Design of Steel Building Structural Element AsPer IS800:2007: 7 Developing a Computational Tool

Clause 10.2.3 10.2.3. 2

d) For any two consecutive fasteners in a line adjacent and parallel to an edge of an outside plate 10.2.3. 3 2) Minimum Pitch Clause 2.5 times nominal diameter of the bolt 10.2.2 3) Edge distance Minimum Edge distance = 1.7 times hole diameter for shear or hand flame cut edges = 1.5 times hole diameter for hot rolled, machine flame cut edges Maximum Edge distance = 12tε 250 ⁄ fy (¿) Where ε = √¿

Clause 10.2.4. 2

10.2.4. 3

t is the thickness of thinner plate Table no 3.2 IS800:2007 Provision for Strength of Bolted Connection 1) Bearing Strength in Shear Vsb < Vnsb / γmb

Clause 10.3.3

where Vnsb = nominal shear capacity of a bolt, calculated as follows:

Vnsb 

fu  nn Anb  ns Asb  lj   lg  pkg  3

where fu = ultimate tensile strength of a bolt

nn = number of shear planes with threads intercepting the shear plane ns = number of shear planes without threads intercepting the shear plane Asb = nominal plain shank area of the bolt

Clause 10.3.3. 1

Anb = net tensile area at threads, may be taken as the area corresponding to root diameter at the thread Reduction factor for long t for distance between first and last row of bolt more than 15d lj = 1.075 – lj / (200 d)

but 0.75 < lj < 1.0

Where, d= nominal diameter of the fastener Reduction factor for large grip length is to be considered when total thickness of the connected plates exceeds 5 times the diameter, d, of the bolts.

lg = 8 d /(3 d+lg)

10.3.3. 2

10.3.3. 3

lg < lj

Design of Steel Building Structural Element AsPer IS800:2007: 8 Developing a Computational Tool

Reduction factor for bolts carrying calculate shear through a packing plate in excess of 6 mm

pk = (1 - 0.0125 tpk) where tpk = thickness of packing plate

2) Strength of bolt in tension Tb < Tnb / γmb

Clause 10.3.5

where, Tnb = nominal tensile capacity of the bolt, calculated as follows: Tnb = 0.90 fub An

< fyb Asb (γm1 / γm0)

where, fub = ultimate tensile stress of the bolt fyb = yield stress of the bolt An = net tensile stress area as specified in the appropriate Indian Standard. For bolts where the tensile stress area is not defined, An shall be taken as the area at the

3) Strength of bolt in bearing

Clause 10.3.4

Vdpb = Vnpb / mb Where Vnpb = 2.5 kb d tp fu where, kb is smaller of [ e/(3do) , (p/3do) - 0.25 , fub/fu , 1.0 ] e ,p = end and pitch distances of fastener along bearing direction, do = diameter of the bolt hole , d = nominal diameter of the bolt , fub = ultimate tensile stress of bolt , fu = ultimate tensile stress of plate , tp = thickness of connected plates experiencing bearing stress in the same direction For countersunk bolt t = thickness of plate - 1/2 x depth of countersinking

4) Strength of plate in tension

Tb < Tnb / γmb Tdn where ,

=

Clause 10.3.5

0.9 fu An / m1 m1 An An An dh g

= partial safety factor for failure at ultimate stress = net effective area of plate = ( b - n dh ) x t for chain bolting = [b - n dh + (S p2/4 g )] x t for zig - zag bolting b,t = width and thickness of plate = diameter of bolt hole ( 2 mm in addition to the diameter of bolt for directly punched holes ) = gauge distance i.e. the distance between the two consecutive rivets / bolts in a chain measured at right angles to the direction of tension (stress) in the member

Design of Steel Building Structural Element AsPer IS800:2007: 9 Developing a Computational Tool

5) Check for block shear The design strength for block shear shall be 3

Tdb = ( Avg fy /( or

m0) + 0.9Atn fu /m1 ) 3

Tdb = (0.9Avn fu /(

Clause 6.4

m1) + Atg fy /m0 )

where Avg and Avn = minimum gross and net area in shear along a line of transmitted force respectively.

3.2 Design Provision for Welded Connection Table No 3.3 IS800:2007 Provision for Size, Throat Thickness, Effective Length of weld [8] 1) Minimum size of weld Clause 10.5.2.3 Thickness of thicker part Minimum size of weld - to 10 mm 3mm 10 to 20mm 5mm 20 to 32mm 6mm 32 to 50mm 8mm for first run, 10mm for minimum size of weld 2) Throat of fillet weld Angle between fusion face in degree Constant K

60-90

91-100

101-106

107-113

114-120

0.70

0.65

0.60

0.55

0.5

3) Effective length Effective length shall not be less than 4times the size of weld. Table no 3.4 IS800:2007 Provision for Strength of Welded Connection 1) Design stress in weld a) Shop weld Fwd = fwn / γmw = fu/(√3γmw)

Clause 10.5.3.2

Clause 10.5.4.1

Clause 10.5.7.1. 1

Where, γmw = partial safety factor fwn = fu / √3 fu = smaller of ultimate stress of weld or of parent material

b) Site weld Fwd = fwn / 1.5γmw 2) Reduction in design capacity a) for long ts βLw = 1.2 - 0.2 Lj / (150 x tt ) <= 1.0 Where,

tt Lj

= thickness of throat = Length of t in the direction of force transfer

3) Design capacity of weld in either tensile or compressive or shear force fa or q = P / ( Lw tt ) Where, fa q Lw tt

= = = =

10.5.7.1. Clause 10.5.7.2

Clause 10.5.9

calculated normal stress due to axial force in MPa shear stress in MPa effective length of weld in mm effective throat thickness in mm

Design of Steel Building Structural Element AsPer IS800:2007: 10 Developing a Computational Tool

P

=

force being transmitted (axial force or shear force)

4. DESIGN OF FLEXURAL MEMBER AS PER IS800:2007 As far as steel frame structures are concerned the flexural member are classified to laterally ed beam, laterally uned beam and beam-column. In this section we will discus the design procedure for laterally ed and laterally uned beams. 4.1 Design of Laterally ed Beam: Table No 4.1Detail Procedure for Design of Laterally ed Beam [8] Step 1: Trial section: M γ m0 Zp = Where M = The design moment fy Fy = Yield stress γm0 = Partial safety factor for material

Step 2: Classification of section: Once the section is selected it is to be classified asper table 2 page 18 IS800-2007 b d For that one has to calculate t f , t w

Clause 3.7.2 3.7.4

Where b = Clear outstanding length of the flange from the web to the end of flange tf = Thickness of flange d = Clear depth of web tw = thickness of web

Step 3: Check for shear buckling: d 250/f y t w < 67 ε Where ε = √ If

d tw

Clause 8.2.1.1

> 67 ε the flange of the beam must be design to resist bending

and web must be designed to resist shear. In this case the moment capacity of the section shall be calculated as shown in step 7 Step 4: Calculation for moment capacity of the section: βb Z p f y 1.2 Ze f y Md = < γ m0 γ m0

Clause 8.2.1.2

Where Fy = Yield stress γm0 = Partial safety factor for material βb = 1 for plastic and compact section =

z e / z p for semi compact section

Zp = Plastic section modulus

Design of Steel Building Structural Element AsPer IS800:2007: 11 Developing a Computational Tool

Step 5: Calculation for shear capacity of the section: A v f yw Vd = √3

Clause 8.4.1

Where Av = Area of element resisting shear = d tw

Step 6: Check for high shear V< 0.6Vd

Clause 8.2.1.2

Where V = Factored Design shear force Vd = Shear capacity of the section

If V> 0.6Vd than the moment capacity of the section shall be calculated as shown in step 7

Step 7: Calculation for moment capacity of section when V> 0.6Vd or 67 ε Mdv = Md-β(Md-Mfd) < Where β =

1.2 Ze f y γ m0

(2 V /V d−1)

d tw

>

Clause 9.2.2

for plastic and compact section

2

Mfd = Plastic design strength fo the area of the cross section excluding the shear area, considering partial safety factor γm0

=

Mdv = Ze fy/γm0 <

Z ¿ 2 p−¿ d t w ¿ ¿ f yw ¿ ¿ ¿

1.2 Ze f y γ m0

for semi compact section

Design of Steel Building Structural Element AsPer IS800:2007: 12 Developing a Computational Tool

Step 8: Check for web buckling The concentrated load or reaction at shall not exceed the capacity of the web as a column Strength of web = fcd Awb

Clause 8.7.3.1

Where Awb = (b1+ n1)tw b1 = stiff bearing on the flange n1 = (d/2)-tf tw = thickness of web tf = thickness of flange fcd can be directly calculated from table 9c page no 42 Figure no 4.1 Web Buckling IS800-2007 if the KL/r ratio is known. Where L = d = clear depth of web, K = 0.7 Table 11 page no 45 IS800-2007 give the value of KL based on various Condition and r=

√ I / A wb1

Step 9: Check for web crippling The concentrated load or reaction at shall not be more than Fw = (b1+n2)fyw tw/γm0 Where b1 = stiff bearing on the flange n2 = 2.5(r1+tf) tw = thickness of web fyw = Yield stress of web γm0 = Partial safety factor

=

√

b 13 t w /12 b1 t w

=

7.1.2.1

tw 2 √3 Clause 8.7.4

Figure no 4.2 Web Crippling

Step 10 Check for holes in tension zone The effect of tension need not be considered if Anf/Agf > (fy/fu) (γm1/γm0)/0.9 Step 11 Effect of shear lag The shear lag effect in the flange may be disregarded provided a) for outstanding elements (ed along one edge) b0< L0/20 b) for internal elements (ed along two edges) bi
Clause 8.2.1.4 Clause 8.2.1.5

Where L0 = Length between point of zero moment in span bi =Width of flange as an internal element b0 = Width of flange with outstand

4.2 Design of Laterally uned Beam: Table No 4.2Detail Procedure for Design of Laterally uned Beam [8] The design procedure of laterally uned beam is same as that of laterally ed beam the only difference is the calculation of the design bending strength or the moment capacity, as it is governed by laterally torsional buckling and is given by Md = βb Zp fbd 1) Calculate Mcr

Clause

8.2.2.1 Design of Steel Building Structural Element AsPer IS800:2007: 13 Developing a Computational Tool

Mcr =

2 0.5

[ ( )]

π 2 EI y hf 1 L¿ / r y 1+ 2 20 hf /t f 2L ¿

(note the formula is for the simplified case of I

section) Where Iy,ry = moment of inertia and radius of gyration respectively about the weaker axis. LLT = effective length for lateral torsional buckling as per clause 8.3 hf = center to center distance between the flange tf = thickness of flange

2) Calculate λLT λLT = √ βb Z p f y / M cr <

8.2.2

√ 1.2 Z e f y /M cr

where βb = 1 for plastic and compact section = Ze/Zp for semi-compact section Zp,Ze = plastic and elastic section modulus respectively fy = yield stress

3) Calculate fbd If λLT < 0.4 the design moment shall be calculate as done in laterally ed beams Or else following steps shall be followed Calculate effective length KL from table 15 page 58 IS800-2007 Based on this value determine slenderness ratio KL/r Calculate hf/tf, based on the value of Kl/r and hf/tf one can determine the value of fcr,b from table 14 page no 57 IS800:2007

8.2.2.1

8.2.2

From the value of fcr,b and the value of imperfection parameter αLT the value of fbd can be determined using table 13(a)/(b) page 55/56 IS800:2007 Other steps are similar to that of laterally restrained beam

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Laterally ed beam

Design of Steel Building Structural Element AsPer IS800:2007: 14 Developing a Computational Tool

Done By :- Darshit Vejani

Sheet no:-1 of 3

Date:- 15 December, 2009

Statement Design a simply ed beam laterally ed to carry a uniform distributed load of 44KN/m. The effective span of beam is 8m.[16] 1)

Trial section: Factored moment = 528 KNm Factored shear force = 176 KN M γ m0 Zp = = 232320 mm3 fy 2

Fy = Yield stress= 250N/mm γm0 = Partial safety factor for material =1.1 Select ISWB500 @ 95.2 Kg/m Dept of Section H 500 mm Width of Flange bf 250 mm Thickness of flange tf 14.7 mm Thickness of web tw 9.9 mm Plastic section modulus Zp 2351350 mm3 Elastic section modulus ZE 2091600 mm3

2)

3)

Classification of Section b d t f = 8.5 , t w = 47.535 Clear outstanding length of the flange from the web to the end of flange b = 120.05 mm Clear depth of web d = 470.6 mm The section is plastic Check for Shear Buckling d tw

= 47.535 < 67 ε Ok, No shear buckling

IS 800:2007 Clause 10.3.3 Clause 5.4.1 Table 5

IS 800:2007 Clause 3.7.2 Clause 3.7.4 Table 2 Page no 18

IS 800:2007 Clause 8.2.1.2

Design of Steel Building Structural Element AsPer IS800:2007: 15 Developing a Computational Tool

4.3 Illustration of Design of Laterally ed beam:

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Laterally ed beam Done By :- Darshit Vejani 4)

Sheet no:-2 of 3

Date:- 15 December, 2009

Calculation for Bending Moment capacity βb Z p f y 1.2 Ze f y Md = < γm0 γ m0

IS 800:2007 Clause 8.2.1.2

= 534.39 < 570.436 KNm Ok, No plastic deformation occurs at 5)

Calculation of Shear Capacity A v f yw Vd = √3 Av = d tw = 4658.94 mm2 fyw/ √ 3 = 250/ √ 3

6)

Check for High Shear V< 0.6Vd 176< 0.6 x 383,284 176 < 230.27 KN Hence No need to reduce the design moment capacity

7)

Check for Web Buckling Radius of gyration r = tw/2

√ 3 = 2.85 mm

K = 0.7, L = 470.6 ie d KL/r = 115.58 From fig and assuming a seat angle ISA 120x120x6 used for connection b1= 120 mm n1= 235.3 mm tw = 9.9 mm Awb = 3517.47 mm2 fcd = 89.15 N/mm2 base on value of KL/r from table 9c Strength of Web = Awbfcd = 313.582 KN > 176 KN

IS 800:2007 Clause 8.4.1

IS 800:2007 Clause 8.2.1.2

IS 800:2007 Clause 8.7.1.5

IS 800:2007 Clause 7.1.2.1 Clause 8.7.3.1

Design of Steel Building Structural Element AsPer IS800:2007: 16 Developing a Computational Tool

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Laterally ed beam Done By :- Darshit Vejani

8)

Sheet no:-3 of 3

Check for Web Crippling b1 =120 mm r1 = 15 mm tw = 9.9 mm n2 = 74.25 mm fyw = 250 N/mm2 Fw = (b1+n2)fyw tw/γm0 = 437.06 KN > 176 KN Hence Safe against web crippling.

9

Check for Effect of Shear Lag b0< L0/20 b0 = 120 mm L0/20 = 400 mm Effect of shear lag may be disregarded

Design As per IS800:2007

Design As per IS800:1984[16]

[email protected]

Date:- 15 December, 2009 IS 800:2007 Clause 8.7.4

IS 800:2007 Clause 8.2.1.5

% Difference in weight/meter 40.44 %

Design of Steel Building Structural Element AsPer IS800:2007: 17 Developing a Computational Tool

5. DESIGN OF MEMBER SUBJECTED TO AXIAL AND BENDING FORCES AS PER IS800:2007 Column in a multistory, multibay structure are generally referred as “Beam-Column”. Beam column are aptly named, as they behave essentially like restrained beams, forming plastic hinges, and under other conditions fail by buckling in a similar way to axially loaded columns or by lateral torsional buckling similar to unrestrained beams. Columns of multistory, multibay structure are subjected to axial forces along with biaxial moments, hence these columns are to be designed as member subjected to combined axial and bending forces. 5.1 Design of Beam-Column: Table No 5.1Detail Procedure for Design of Beam-Column [8] Step1: Trial section Selection of the trial section is a trial and error method. However after gaining experience or by referring some design aids one can judge the required section. Step2: Check for axial load capacity The selected section shall be check for axial load capacity. Calculate effective length based on condition from table no 11 page no 45 IS800-2007. Calculate the slenderness ratio KL/r, where r = √ I / A . Using the h/bf ratio and tf classify the buckling class of cross-section using table 10 page no 44 IS800:2007 where h = depth of I-section

Clause 7.2.2

7.1.2.2

bf = width of flange tf = thickness of flange

Now determine the design compressive stress fcd form table no 9 a/b/c/d page no 40/41/42/43. Thus the design compressive strength Pd = Aefcd Where Ae = effective sectional area as defined in 7.3.2 Pd > P, where P is the factored axial load Step3: Classification of section Once the section is selected it is to be classified asper table 2 page 18 IS800-2007 b d For that one has to calculate t f , t w

7.1.2.1 7.1.2

Clause 3.7.2 3.7.4

Where b = Clear outstanding length of the flange from the web to the end of flange tf = Thickness of flange d = Clear depth of web tw = thickness of web

Step4: Check for section strength The section shall be checked for following equation. For plastic and compact section. α α My Mz + <1 M ndy M ndz

( ) ( ) 1

Clause 9.3.1.1.

2

Design of Steel Building Structural Element AsPer IS800:2007: 18 Developing a Computational Tool

And N M y Mz + + N d M dy M dz

<1 Clause 9.3.1.3

For semi compact section fx < fy/γm0 N M y Mz + + N d M dy M dz < 1 Where My,Mz = factored applied moment about major and minor axis respectively. Mndy, Mndz = design reduced flexural strength under combined axial force and the respective uniaxial moment acting alone. N = factored applied axial force (tension or compression) Nd = design strength in tension, Td as obtained by = Agfy/γm0 Mdy,Mdz = design strength under corresponding moment acting alone (as per 8.2) Ag = Gross area of the cross- section α1,α2 = constants as given in table 17 page no71 IS800-2007 γm0 = partial safety factor For I-section Mndy = Mdy for n < 0.2 Mndy = 1.56 Mdy (1-n) (n+0.6) for n > 0.2 where n = N/Nd Mndz = 1.11 Mdz (1-n) < Mdz

Step5: Check for Overall Member Strength The following equation shall be checked C M M P + K y my y + K ¿ z < 1 Pd y M dy M dz C M C M P +0.6 K y my y + K z mz z P dz M dy M dz

Clause 9.3.2.2

<1

Where P = Applied axial load Cmy,cmz CmLT = Equivalent uniform moment factor obtained from table 18 page no72 IS 800:2007, which depends on the shape of the bending moment diagram between lateral bracing points in the appropriate plane of bending. Pdy, Pdz = design strength under axial compression as governed by buckling about minor(y) and major (z) axis respectively. Mdy, Mdz = design bending strength about y(minor) and z (major axis respectively considering laterally uned length of cross section Caluse 8.2.2 λy = λLT = Non- dimensional slenderness ratio for lateral buckling about minor axis λz = Non- dimensional slenderness ration for lateral buckling about major axis Ky = 1 + (λy-0.2)ny < 1+0.8ny Kz = 1 + (λz-0.2)nz < 1+0.8nz

KLT =

1−

0.1 λ ¿ n y CmLT −0.25

0.1 n y > 1− CmLT −0.25

ny,nz = ratio of actual applied axial force to the design axial strength for buckling about the y and z axis respectively.

Design of Steel Building Structural Element AsPer IS800:2007: 19 Developing a Computational Tool

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Beam-Column Done By :- Darshit Vejani

Sheet no:-1 of 3

Date:- 15 December, 2009

Statement A column is subjected to an axial load of 500 KN. The beam connected to the flange of a column has an eccentric load of 100KN and the beam connected to web of column has an eccentric load 50KN. If the effective length of column is 4 meters design the column for both axial force and biaxial moments.[16] 1)

Trial section: Factored axial force = 975 KN Assuming fcd = 120 N/mm2 Area of section required = 8125 mm2 Select ISHB400 @ 77.4 Kg/m Depth of Section h 400 mm Width of Flange bf 250 mm Thickness of flange tf 12.7 mm Thickness of web tw 9.1 mm Plastic section modulus Zpz 2351350 mm3 Elastic section modulus ZEz 2091600 mm3 Plastic section modulus Zpy 202315.078 mm3 Elastic section modulus ZEy 218300 mm3 Radius of gyration about zz rz = 168.7mm Radius of gyration about yy ry = 52.6

IS 800:2007 Clause 7.2.2

Design of Steel Building Structural Element AsPer IS800:2007: 20 Developing a Computational Tool

2)

Check for axial load capacity KL = 4000 mm KL/ry = 76.04 h/bf = 1.6 , tf = 12.7 mm there for buckling class (b) fcd = 156.4 N/mm2 Pdy = 1543 KN > 975 KN ok

IS 800:2007 Clause 7.1.2.2 Table 10 Page no 44 Clause 7.1.2.1 Table 9b page no 41

5.2 Illustration of Design of Beam-Column:

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Beam-Column Date:- 15 December, 2009 Classification of Section IS 800:2007 b d Clause 3.7.2 t f = 9.48 , t w = 41.16 Clause 3.7.4 Table 2 Clear outstanding length of the flange from the web to the end of Page no 18 flange b = 195.45 mm Clear depth of web d = 374.6 mm The section is compact Done By :- Darshit Vejani

3)

Sheet no:-2 of 3

Design of Steel Building Structural Element AsPer IS800:2007: 21 Developing a Computational Tool

4)

Check for section strength 9.1 75 + 3 My = 50 x 10 x 1.5 = 3.153KNm (Assuming the 2 2

(

)

beam connected with a ISA75x75x6) Similarly Mz = 35.625 KNm N = 975 KN Nd = Agfy/γm0 = 9866 x 250/1.1 = 2242.272 KN N/Nd = n = 0.43 Mndy = 1.56 Mdy (1-n) (n+0.6) = 42.171 Mndz = 1.11 Mdz (1-n) < Mdz = 223.7929 α1 = 1 α2 = 2 My M ndy

α1

Mz + M ndz

5)

IS 800:2007 Clause 9.3.1.1 Table no 17 Page 71

α2

( ) ( )

IS 800:2007 Clause 9.3.1.1

= 0.1138< 1

Check for over all member strength KL/rz = 4000/168.7 = 23.71 fcd = 223 N/mm (Table 9b Page no 41 recall as done in step 2) Pdz = 2200.118 Pdy = 1543.0 (from step 2)

IS 800:2007 Clause 7.1.2.2 Table 10 Page no 44 Clause 7.1.2.1 Table 9b page no 41

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Beam-Column Done By :- Darshit Vejani

Sheet no:-3 of 3

Date:- 15 December, 2009

Design of Steel Building Structural Element AsPer IS800:2007: 22 Developing a Computational Tool

From KL/ry =76.04

KL/rz = 23.71 And h/tf = 30.49

fcr,b y = 437 N/mm2

fcr,bz = 4373.44 N/mm2

IS 800:2007 Clause 8.2.2.1 Table 14 Page no 57

αLT = 0.21 fbdy = 185.802 N/mm2 Mdy = βZpfcdy = 37.59 KNm λ = λ = √ f y / f cr ,by y

fbdz = 227.3 N/mm2 Mdy = βZpfcdy = 353.75KN/m λ = √ f y / f c r ,bz

LT

z

= 0.756 Ky = 1 + (λy-0.2)ny < 1+0.8ny = 1.504 Kz = 1 + (λz-0.2)nz < 1+0.8nz = 1.017 0.1 λ ¿ n y 1− KLT = CmLT −0.25

Clause 8.2.2. Table 13a page no 55

>

1−

= 0.239

0.1 n y CmLT −0.25

= 0.908

Cmy = Cmz = 1 C M M P + K y my y + K ¿ z P dy M dy M dz

= 0.853 < 1

C M C M P +0.6 K y my y + K z mz z P dz M dy M dz

= 0.621 < 1 IS 800:2007 Clause

Design As per IS800:2007 [email protected]

Design As per IS800:1984 [16] % Difference in weight/meter [email protected] 6.2 %

6. DESIGN OF SIMPLE COLUMN BASE PLATE AS PER IS800:2007 There is no doubt in the fact that when one is dealing with the design of a multistoried steel frame structure, he/she will have to go for gusseted base plates as the column base are Design of Steel Building Structural Element AsPer IS800:2007: 23 Developing a Computational Tool

subjected to high bending moments. However in this section we will discus the design procedure for the simple column base only and the gusseted plate will be considered in the future scope of the study. 6.1 Design of Simple Column Base Plate: Table No 6.1Detail Procedure for Design of Simple Column Base Plate[8] Step1: Calculation bearing strength of concrete Bearing strength of concrete = 0.45fck as per clause 34.4 IS 456-2000 Step2: Calculation of area of base plate Area of base plate A = Pu/ bearing strength of concrete. Try to have a equal projection of base plate. Thus A = (D+2b) (bf+2a) Where D = depth of column section bf = with of flange of column section a,b = larger and smaller projection of the base plate beyond column

Step3: Calculation of intensity of pressure w acting below the base plate w = Pu/A1 where A1 is the provided area of base plate. Step4: Calculate the minimum thickness of the base plate 2.5 w ( a2 −0.3 b2 ) γ m 0 /f y > tf ts =

√

Clause 7.4.3.1

Where ts = Thickness of slab base w = uniform pressure from below on the slab base under the factored axial load tf = thickness of flange of column section a,b = larger and smaller projection of the base plate beyond column

Design of Steel Structural Building Asper IS800:2007: Developing a Computational Tool Job Title :- Design of Column Base Plate

Design of Steel Building Structural Element AsPer IS800:2007: 24 Developing a Computational Tool

Done By :- Darshit Vejani

Sheet no:-1 of 1

Date:- 15 December, 2009

Statement A column section ISHB 250@51 Kg/m carries an axial load of 600KN. Design a slab base for the column allowable bearing pressure of concrete is 4 N/mm2[16] 1)

Calculation for Factored Load Pu = 600x1.5 = 900 KN

2)

Calculation of Area of Slab Base A = Pu / bearing strength of concrete = 225000 mm2 A = (D+2b) (bf+2a) =(250+2a) (125+2a) for a=b  4 a2 +750 a−193750=0

3)

4)

a = 150 Area provided A1 LxB = 550 x 425 = 233750 Calculation of Intensity of Pressure w acting below the base plate w= Pu/A1 w = 3.85 n/mm2 Calculate the minimum Thickness of the base Plate ts =

√ 2.5 w ( a −0.3 b ) γ 2

2

m0

/f y

> tf = 25.82mm

IS800:2007 Clause 7.4.3.1

Provide 32mm thick 550 x 425 Plate or provide two 16mm 550 x 425 and connect the plates by fillet weld at the perifery 6.2 Illustration of Design of Column Base Plate:

7. COMMENTS AND SCOPE OF STUDY Design of Steel Building Structural Element AsPer IS800:2007: 25 Developing a Computational Tool

In order to develop a computational tool (MS Excel program with Visual Basic Application on the front hand, as proposed) for design of steel building structural element as per the revise IS800-2007 the first and most important requirement was to understand and get familiar with the design concepts and procedure of IS800:2007. In this report the design clauses for connection, design procedure for flexure member, beam-column and simple column base plate are discussed. The future scope of the dissertation is listed in a tabular form as follows. Table 7.1 List of future scope of the dissertation Sr.no Task 1 Understanding the design concept and procedure for gusseted base plate asper IS800:2007. 2 Understanding the design concept and procedure for bracing systems asper IS800:2007. 3 Learning Visual Basic programming language. 4 Developing a design tool in form of spreadsheet for design of individual steel building structural element.

Design of Steel Building Structural Element AsPer IS800:2007: 26 Developing a Computational Tool

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Awari, U. R., Design of Compression member (Part II-Built up column, lacing and Battening), In proceeding of National Workshop on Design of Steel Structures using

[2]

Revised IS800:2007, Pune, July 2009, 54-73. Bhirud, Y. L., Design of Compression member (Part I-Angle Struts), In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune,

[3]

July 2009, 43-53. Bhusari, J. P., Introduction to Plastic Analysis, In proceeding of National Workshop

[4]

on Design of Steel Structures using Revised IS800:2007, Pune, July 2009, 105-124. Dugal, S. K., Design of Steel Structures, 2nd Edition, Tata McGraw-Hill Publication

[5]

company Limited, 2003. Ghosh, M. M., Philosophy of Limit State Design and Classification of Sections, In proceeding of National Workshop on Design of Steel Structures using Revised

[6]

IS800:2007, Pune, July 2009, 11-22. Ghosh, M. M., Design of Tension Member, In proceeding of National Workshop on

[7]

Design of Steel Structures using Revised IS800:2007, Pune, July 2009, 23-42. Guha, A., and Bandyopadhyay, T. K., Innovation Structural Design of Tension , Compression Member, Subjected to Flexure And Combined Bending and Axial Force Based on IS:800:2007, Insdag’s Steel in Construction,

[8] [9] [10]

Special Issue 2008, 11-20. IS800:2007, Indian Standard Code of Practice for General Construction in Steel. IS800:1984, Indian Standard Code of Practice for General Construction in Steel. Karkare, B., Design of column Base: Slab base and Gusseted base, Moment Resisting Column base, Anchor bolt, In proceeding of National Workshop on Design of Steel

[11]

Structures using Revised IS800:2007, Pune, July 2009, 223-239. Karnik, A. B., Innovation in off-Shore Pilling for Jetty Construction, Journal of

[12]

Indian Society of Structural Engineers, Volume 8-4, Oct-Nov-Dec-2003, 17-20. Kodag, P. B., Design of Flexural Member Laterally Uned, In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune,

[13]

July 2009, 132-139. Kulkarni, P., Combined Stresses: Moment with High Shear, In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune, July 2009,

[14]

182-187. Kulkarni, P., Combined Stresses: Beam Column axial with Biaxial Bending, In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune, July 2009,187-200. Design of Steel Building Structural Element AsPer IS800:2007: 27 Developing a Computational Tool

[15]

Negi, L. S., Design of Steel Structures, 2nd Edition, Tata McGraw-Hill Publication

[16] [17]

company Limited, 2006. Ram Chandra, Design of Steel Structures, 10th Edition, Standard Book House, 2002. Santhakumar, A. R., Earthquake Consideration in Design of Frames, In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune,

[18]

July 2009,251-274. Shah, V. L., Limit State Design of Steel Structures IS800:2007, Structures

[19]

Publications, 2009. Shah, V. L., Design of Bolted Connections, In proceeding of National Workshop on

[20]

Design of Steel Structures using Revised IS800:2007, Pune, July 2009,74-93. Shah, V. L., Design of Welded Connections, In proceeding of National Workshop on

[21]

Design of Steel Structures using Revised IS800:2007, Pune, July 2009, 94-104. Shiyekar, M. R., Design of Flexural Member, Laterally ed, In proceeding of National Workshop on Design of Steel Structures using Revised IS800:2007, Pune,

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July 2009, 125-131. Subramanian, N., Design of Steel Structures, Oxford University Press, 2008. Teaching Resource Material, Institute of Steel Development and Growth.

Design of Steel Building Structural Element AsPer IS800:2007: 28 Developing a Computational Tool