a comparative study is800

A Comparative Study of Design Specifications in IS 800-2007 and IS 800-1984 Abstract The introduction of New IS 800-2007 based on Limit State approach as against the previous version based on Working stress approach has generated disarray in the Indian Design industry regarding the basic design approach and the results obtained. This study is a step in bridging the gap and helps throw light on the difference between basic design procedures for Limit State and Working Stress approach. A numerical evaluation of a field example has also been undertaken to depict the differences in the results generated by both the approaches. Introduction Bureau of Indian Standards made a revision of IS 800 – Code of Practice for General Construction in Steel in December 2007. This new revised code is based on the Limit State Approach as against Working Stress Approach in the earlier versions. The two philosophies are explained in brief as follows. Working Stress Method (WSM) This method forms the basis of traditional design of almost all structures irrespective of their material of construction. The method basically assumes that the structural material behaves in a linear elastic manner , and that adequate safety can be ensured by suitably restricting the stresses in the material due to the expected working loads ( service loads) on the structure. He stresses under the working loads are obtained by applying the methods of “Strength of materials”, such as the simple bending theory. The first attainment of yield stress of steel is taken to be the onset of failure. The limitation due to the non linearity (geometric as well as material) and buckling are neglected. The stresses caused by the characteristic loads are checked against the permissible (allowable) stress, which is a fraction of yield stress. Thus the permissible stress may be defined in terms of a factor of Safety, which takes care of the overload or other unknown factors. Thus, Permissible (allowable) stress = Yield Stress__ Factor of Safety Thus in working Stress method Working Stress ≤ Permissible Stress Each member of the structure is checked for a number of different combinations of loadings. Usually a FOS of 1.67 is adopted fro tension members and beams. A value of 1.92 is used for long columns and 1.67 for short columns. A value of 2.5-3 is used for connections (However using the WSM, the ‘real’ safety against ‘failure’ is unknown.). Since dead load, live load and wind load are all unlikely to act on the structure simultaneously the stresses are checked as follows. Stresses due to live load + Dead load < Permissible stress Stresses due to Dead load + Wind Load < Permissible stress Stresses due to Dead Load + Live Load + Wind Load < 1.33 (permissible stress)

Limitations of the Working Stress Method 1) The main assumptions of Linear Elastic behaviour and the implied assumption that the stresses

under Working loads can be kept within the ‘permissible stresses’ are not found to be realistic. 2) It does not consider the consequence of material non linearity and the non-linear behaviour of the

materials in the post buckled State. 3) Steel components have the ability to tolerate high elastic stress by yielding locally and

redistributing the loads, which is not considered in WSM. 4) It fails to discriminate between different types of load that acts simultaneously, but have different

degree of uncertainty, which can lead to nonconservative designs at times when two different loads have counteracting effects.

Inspite of the above shortcomings, the structures designed with WSM have performed well. The size of Tension members is about the same in both LSM and WSM when the LL to DL rate (LDR) is about 3. When DL becomes more predominant, there will be slight economy in using LSM. When LDR greater than 3, WSM will be slightly more economical (3%) However, for other members, WSM results in relatively larger member sizes and hence in less deflection. There are instances where WSM results in considerable over design and where it is not safe (Allen 1972 and Gordon 1978). The WSM is noted for its simplicity in concept as well as application. Limit State Method (LSM) It aims for a comprehensive and rational approach to the design problem, by considering safety at ultimate loads and serviceability at working loads. The selection of various multiple safety factors is supposed to have a sound probabilistic basis, involving the separate considerations of different kinds of failure, types of materials, and types of loads (Kulak and Grondin 2002) Limit State of Strength The objective of limit state design is to ensure that the probability of any limit state being reached acceptably low. This is made possible by specifying appropriate multiple safety factors for each limit state (Level reliability). Of course, in order to b e meaningful, the specified values of safety factors should result in ‘target reliability’ (the reliability that will produce designs which will provide the required amount of safety and at the same time result in economic structures). The multiple safety factor adopted by the Indian code is in the so-called partial safety factor format, which is expressed as Rd ≥ ∑ γif Qid Where Rd is the design strength computed using the reduced material strength Ru/γm , where Ru is the characteristic material strength and γm is the partial safety factor for the material, and allows for uncertainties of element behaviour and possible strength reduction due to manufacturing tolerances and imperfections in the material. The values are given in the Table 4 of IS code. These values are directly taken from Eurocode 3 and have no statistical backing (N Subramanian, 2008)The Partial Safety factors for loads γf make allowances for possible deviation of loads and the reduced possibility of all loads acting together. Even these are based on Eurocode 3 with slight modifications. It is to be noted that the

load factors are reduced when different types of Loads (DL, LL, WL or WL) are acting simultaneously at their peak values. This is because of the reduced probability of all the loads acting concurrently. Structural Stability The code deals with 3 forms of structural Stability. They are 1) General Stability 2) Stability against Overturning 3) Sway Stability

General Stability It should be ensured that the structure as a whole and each of its elements remain stable from the commencement of erection until demolition. When two members are incapable of keeping themselves in equilibrium then sufficient external bracing should be provided for stability. Stability against Overturning The structure as a whole or any part of it should be designed to prevent instability due to overturning or sliding, while designing tall or cantilever structures. While checking for overturning, uplift or sliding, the loads should be multiplied by the relevant γf factors .The code suggests the following a) The loads and forces should be divided into components aiding instability and those resisting

instability. b) The forces and loads causing instability should be combined using the appropriate load facto are

given. c) The permanent loads and effects causing resistance should be multiplied by a partial safety factor

of 0.9 and added together with design resistance( after being multiplied by the appropriate partial safety factor)

d) The resistance effect should be greater than or equal to the destabilising effect. Sway Stability This condition imposes that there must not be excessive lateral deformation under applied loads. a) When the structures are sheltered by an adjoining building, b) When the structure is entirely enclosed, c) When the building is wide in relation to the applied load In order to prevent these situations being critical, the code requires all structures to be checked for a minimum notional horizontal load. The notional loads are specified in Section 4.3.6 of the code as 0.5% of the factored load. The notional horizontal loads should not be 1) applied while considering overturning 2) combined with other horizontal loading such as wind or Earthquake 3) taken to contribute to net shear on the foundation. The notional horizontal load should be applied on the whole structure, in both orthogonal directions, in one direction at a time, and should not be taken to act simultaneously with the factored gravity load. Moreover when the ratio of height to the lateral width of a building is less than unity, such notional loads need not be considered.

Serviceability Limit States In Limit states of serviceability, the variable to be considered is a serviceability parameter (representing deflection, vibration etc.) .A limit state or failure is considered to occur when a specified maximum limit of serviceability, ∆all exceeded (as shown in fig) In the Fig. Pf is the Probability of failure and f∆(∆) is the frequency distribution curve for D. Serviceability limit states relate to satisfactory performance and correspond to excessive deflection, vibration, local deformation, durability and fire resistance. The Load factor γf should be taken as unity for all serviceability limit state calculations, since they relate to the criteria governing normal use.

Pf

∆all Serviceability Variable D (Deflection, Crack width)

Pro

babi

lity

Den

sity

f∆∆∆∆(∆∆∆∆)

Fig. 1 Reliability Model for Serviceability Design

CODAL PROVISON COMPARISON In the newly revised IS 800, stress is laid to make optimum utilisation of the structural member along with provision of making adequate checks for restricting local buckling. Comparison of the critical parameters/clauses of two versions of the code (i.e. IS 800-1984 and IS 800-2007) is as follows SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

1.0 Material (REFER IS:2062) 1.1 Structural

Steels TABLE-1, (Pg.2) • Steel conforming to IS:

2062 (General purpose steel) - Steels designated as Fe410WA, Fe410WB and Fe410WC. This standard covers the requirements of steel plates, strips, sections, flats, bars etc. for use in structural work. The steel is suitable for welded, bolted and riveted structures.

• Steel conforming to IS:8500 - Weldable structural steel (medium and high-strength qualities). Steels designated as Fe-440-HT, Fe-490-HT, Fe-540-HT, Fe-570-HT and Fe-590-HT.

• Steel conforming to

IS:961- Structural steel ( high tensile )-Steels designated as Fe-570-HT and Fe-540 W -HT.

TABLE-1, (Pg.2) • Steel conforming to

IS:2062 (General purpose steel) - Steels designated as Fe410WA, Fe410WB and Fe410WC. This standard covers the requirements of steel plates, strips, sections, flats, bars etc. for use in structural work. The steel is suitable for welded, bolted and riveted structures.

• Structural Steel other than that specified in IS 2062 can be used provided that the permissible stresses and other design provisions are suitably modified and the steel is also suitable for the type of fabrication adopted

• Steel that is not supported by mill test result may be used in structural system after confirming their quality according to IS 1608, or they should be used for unimportant members

1.1.1 Yield stress TABLE-1, (Pg.2) IS 2062 Fe 410 ( WA/WB/WC ) (MPa ) Upto 20mm thk. - 250 20-40 mm - 240 > 40mm - 230

TABLE-1, (Pg.2)IS 2062 Fe 410 ( WA/WB/WC ) (MPa) Upto 20mm thk. - 250 20-40 mm - 240 > 40mm - 230

No change

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

YS - Yield Stress 1.1.2 Ultimate

tensile strength

TABLE-1, (Pg.2) Fe 410 (WA / WB / WC ) - 410 MPa

TABLE-1, (Pg.2) Fe 410 (WA / WB / WC ) - 410 MPa

No Change

1.2 Fasteners / Bolts

Black bolts made from mild steel-IS:2062. High strength friction grip(HSFG) bolts -made from high tensile steel.-IS:961

Black bolts made from mild steel-IS:2062. High strength friction grip(HSFG) bolts -made from high tensile steel.-IS:961

No Change

2.0 GENERAL DESIGN REQUIREMENTS

2.1 Load Combinations

Clause no. 3.4.2 (Pg.-24) 1) DL + IL 2) DL + IL + WL or EL 3) DL + WL or EL

Clause no. 3.5 (Pg.-16) 1) DL + IL 2) DL + IL + WL or EL 3) DL + WL or EL 4) DL + ER (Erection

Load)

More Importance is given to Erection loading

2.2 Section Classification

No such classification has been made

Clause no. 3.7 (Pg. 17) The Sections are classified based on its local buckling strength and the ability to allow rotation before failing. They are

a) Class 1 (Plastic) b) Class 2 (Compact) c) Class 3(Semi-compact)

& d) Class 4 (Slender)

The Class of section governs its design

2.3 Increase of Stresses

Clause no. 3.9 (Pg-31) & IS:816-Cl.7.3(Pg.17) Wind / Earthquake Loads • Strl. Steel members -

33.33 % • Rivets, bolts, Welds etc. -

25 %

Clause no. 5.3.3 (Pg-29) Partial safety factors have to be considered and no increase or decrease of stresses have to be considered for individual loads

2.4 Stability Clause no. 3.12 (Pg-34) Restoring moment > 1.2 x max. Overturning moment (due to DL) + 1.4 x max. Overturning moment (due to IL and WL/EL) IS-800 : In cases where Dead load provides the restoring

Clause no. 5.5.1 (Pg-30) the structure should satisfy 2 limit states

1) Limit state of strength and

2) Limit state of serviceability.

The structure should adhere

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

moment, only 0.9 times DL shall be considered.

to 1) Stability against

Overturning – The loads and effects contributing to the resistance shall be multiplied with 0.9 and added together to get design resistance.

2) Sway Stability 2.5 Limiting

Deflection

Clause no. 3.13 (Pg-34) • Max. Deflection for all

applicable loads (Vertical / Horizontal ) = 1/325 of the Span / height.

Clause no. 5.6.1 (Pg-31) Deflection limits have been provided separately for Industrial buildings and Other Buildings and separate limits have been mentioned for different members.

More importance is given to serviceability requirements for various members in a structure.

3.0 TENSION MEMBERS 3.1 Axial

Stresses Clause no. 4.1 (Pg-37) Stress on the net effective area not to exceed σat = 0.6 fy (in MPa)

Clause no.6 (Pg-32-34) The design strength of a tension member should be lowest of

1) Strength due to yielding of gross c/s

2) Strength due to rupture of critical c/s

3) Strength due to block shear

Additional provision for block shear has been incorporated

3.2 Maximum Slenderness Ratio

Clause no. 3.7 (Pg-30) • Tension member = 400 • Tension Member (

Reversal of stress occurs due to loads other than WL / EL) = 180

• Tension member (subject to possible reverse of stresses due to WL, EL) =350

• No change has been made

3.3 Net Effective Area

Clause no. 4.2 (Pg-37) For area calculation, this clause may be referred.

Clause no. 6.3 &6.4 (Pg 32-34) This clause may be referred.

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

4.0 COMPRESSION MEMBERS 4.1 Axial

stresses Clause no. 5.1 (Pg-38) The direct stress shall not exceed 0.6fy or as calculated by equation given in Cl.-5.1.1. Permissible stress σac shall be taken from Table - 5.1 (Pg-39), for corresponding slenderness ratio.

Clause no. 7.1.2.1 (Pg-34) The allowable axial stress or design compressive stress (fcd) shall be calculated using the formulas given in the clause or can be calculated using tables 9(a),9(b),9(c),9(d) on the basis buckling class of the section

The concept of imperfection factor and buckling class of the section has been introduced

4.2 Effective Length (l), l = KL

Clause no. 5.2 (Pg-38) 1) ‘K’ values shall be taken

appropriately based on degree of end restraint of member as given in Table -5.2 (Pg-41&42) OR follow the procedure given I n Appendix-C.

2) Truss members buckling in the plane of truss, ‘l’ shall be taken as between 0.7 and 1.0 times the distance between the centres of intersections, depending on degree of end restraint. For members buckling in the plane perpendicular to truss, ‘l’ shall be taken as distance between points of restraint.

Clause no. 7.2 (Pg-35-45) 1) ‘K’ values shall be taken

appropriately based on degree of end restraint of member as given in Table-11 (Pg. 45).

2) For Truss & braced frame members buckling in the plane of truss, effective length ‘1’ shall be taken as between 0.7 and 1.0 times the distance between the centres of intersections, depending on degree of end restraint. For members buckling in the plane perpendicular to truss, ‘1’ shall be taken as distance between centres of intersection.

‘K’ values given in both the codes are same.

4.3 Maximum Slenderness Ratio

Clause no. 3.7 (Pg-30) • Comp. Member (DL, IL)

= 180 • Comp. Member (WL, EL)

= 250

Clause no. B7 (Pg-5-37) • Comp. Member (DL, IL)

= 180 • Comp. Member (WL,

EL) = 250

Clause 3.8 (Table 3)(Pg-20) No changes made

4.4 BUILT-UP MEMBERS - Lacing

Clause no. 5.7 (Pg-47) 1) Lacing of comp. member

shall be designed for a transverse shear equal to

Clause no. 7.6 (Pg-48-50) 1) Lacing of comp. member

shall be designed for a transverse shear equal to

No changes have been made in the load

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

at least 2.5 % of the axial force in the member.

2) Slenderness ratio of the lacing bars shall not exceed 145.

3) Angle of Inclination with

the axis of member - (for both single & double lacing) - 40° to 70°.

4) The max. spacing of

lacing shall be such that min. slenderness ratio (l/r) of the components of the member between consecutive connection is not greater than 50 OR 0.7 times the most unfavourable (l/r) of the member as a whole, whichever is less.

at least 2.5 % of the axial force in the member.

2) Slenderness ratio of the lacing bars shall not exceed 145.

3) Angle of Inclination with the axis of member - (for both single & double lacing) - 40° to 70°to the axis of built up section.

4) The max. spacing of lacing shall be such that min. slenderness ratio (l/r) of the components of the member between consecutive connection is not greater than 50 OR 0.7 times the most unfavourable (l/r) of the member as a whole, whichever is less.

calculation and basic design requirements.

4.5 BUILT-UP MEMBERS - Battening / Tie plates

Clause no. 5.8(Pg-51) 1) Battens shall be designed

to carry the bending moment & shears arising from transverse shear force ‘V’ of 2.5 % of the total axial force on the whole comp. member, at any point in the length of the member, divided equally between parallel planes of battens.

2) The spacing of battens

centre to centre of end fastenings shall be such that the slenderness ratio (l/r) of the lesser main component over that distance shall be not greater than 50 OR greater

Clause no. 7.7(Pg-50-52) 1) Battens shall be

designed to carry the bending moment & shears arising from transverse shear force ‘V’ of 2.5 % of the total axial force on the whole comp. member, at any point in the length of the member, divided equally between parallel planes of battens

2) The spacing of battens centre to centre of end fastenings shall be such that the slenderness ratio (l/r)

No changes have been made in the load calculation and basic design requirements.

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

than 0.7 time the slenderness ratio of the member as a whole, about it’s x-x axis. (axis parallel to the battens)

of the lesser main component over that distance shall be not greater than 50 OR greater than 0.7 time the slenderness ratio of the member as a whole, about it’s x-x axis. (axis parallel to the battens)

4.6 Column Base Plate

Clause no. 5.4 (Pg-44) Dimension & thickness of base plate shall be calculated using formulas provided in this clause.

Clause no. 7.4 (Pg-46-48) Dimension & thickness of base plate shall be calculated using formula provided in this clause.

The concept of effective area for load transfer has been introduced.

5.0 MEMBERS SUBJECTED TO BENDING 5.1 Bending

Stress Clause no. 6.2 (Pg-55) 1) Max. permissible, σbt or

σbc = 0.66fy (For Strong & Weak axis bending )

2) Max. permissible σbc for I-beams & Channels (based on section properties and l/ry ) shall be referred from Table-6.1A to 6.1F (Pgs-57 to 62).

3) For beams & plate girders, max. permissible σbc shall be computed as per equation given in Cl. 6.2.3 (Pg-56) OR Table-6.2 (Pg-64) may be referred for σbc calculated from fcb for different values of fy.

( All stresses in MPa)

Clause no 8 (Pg-52-559) 1) For Laterally Supported

Beams: The design strength in bending shall be calculated as per the formulas given on the basis whether the section is susceptible to shear buckling before yielding

2) For Laterally Unsupported Beams, the design strength in bending shall be calculated as per the formulas given and resistance to lateral torsional buckling should not be checked for a) bending is about

minor axis b) Section is hollow or a

solid bar c) In case of major axis

bending, the non dimensional

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

slenderness ratio is less than 0.4.

5.2 Bearing

Stress Clause no. 6.3 (Pg-68) Max. permissible bearing stress, on net area of contact is, σp = 0.75 ƒy

Clause no. 8.7.4 (Pg-67) Should be less than the yield stress of the steel divided by Partial safety factor (ie. fy/1.1)

5.3 Shear stresses

Clause no. 6.4 (Pg-68) 1) Max. permissible Shear

stress is, Tvm = 0.45 ƒy 2) Average shear stress in

member calculated on the cross section of the web shall not exceed the limits as mentioned in Cl. No. 6.4.2 (Pg-69). Also refer Table-6.6A, B, C (Pg-73-75) for stiffened webs.

Clause no. 8.4 (Pg-59-60) The nominal plastic shear resistance under pure shear should be calculated using the formulas =, based on the shear areas specified for various sections. Resistance to shear buckling can be verified based on the value of ratio of depth to web thickness.2 methods have been specified for calculation of nominal shear strength. They are

1) Simple Post Critical Method (can be used for beams with or without intermediate transverse Stiffeners

2) Tension Field Method (can be used for beams with intermediate transverse stiffeners)

5.4 Effective length of Compression Flanges & max. Slenderness Ratio

Clause no. 6.6 (Pg-76) & Clause no. 3.7 (Pg-30) 1) To calculate permissible

bending stress as explained above in 4.0 (I), appropriate effective length shall be considered referring this clause.

2) Max. Slenderness Ratio for Compression flange of beam =300

Clause no. 8.6.1.2 (Pg. 64) No specific criteria are mentioned. But in order to avoid buckling of the compression flange into the web, the web thickness shall satisfy the criteria’s specified

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

3) For cantilever beams of projecting length ‘L’, refer Cl. No. 6.6.3 (Pg.77).

6.0 COMBINED STRESSES 6.1 Axial

Compression & Bending

Clause no. 7.1.1 (Pg-90) The requirements of this clause (Equations) shall be satisfied.

Clause no. 9.3.1 & 9.3.2.2 (Pg. 70-71) The requirements in the above clauses should be satisfied.

Separate governing equations are specified for different types of sections.

6.2 Axial Tension & Bending

Clause no. 7.1.2 (Pg-90) The condition as per this clause shall be satisfied.

Clause no. 9.3.1 & 9.3.2.1(Pg 70-71) The requirements in the above clauses should be satisfied.

Separate governing equations are specified for different types of sections.

6.3 Bending & Shear

Clause no. 7.1.4 (Pg-91) The equivalent stress calculated by the equation given in this clause shall not exceed the value, σe = 0.9Fy.

Clause no 9 (Pg 69-70) The Moment carrying capacity of the section shall be reduced by the amount as specified in the code ( for high shear force). No reduction is required for Shear force value < 60% of allowable shear capacity of the section.

The moment reduction is dictated by the percentage of shear force wrt. allowable shear force in the section.

6.4 Bearing, Bending & Shear

Clause no. 7.1.5 (Pg-92) The equivalent stress calculated by the equation given in this clause shall not exceed the value, σe = 0.9Fy.

No specific criteria are mentioned.

7.0 CONNECTIONS. 7.1 BOLTED :- 7.1.1 Permissible

Stresses for bolts :-

Clause no. 8.9.4 - (Pg-95) IS:2062 (Material) MPa ksi a) Axial tension 120 17.4 b) Shear 80 11.6 c ) Bearing 250 36.25

Clause no. 10 (Pg 73-77) No specific value prescribed. Procedure given for calculation of permissible loads (Axial Tension, Shear, & Bearing).

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

7.1.2 Combined shear and tension in bolts

Clause no. 8.9.4.5 - (Pg-96) The individual stresses should not exceed allowable values and combined stress ratio should not exceed 1.40

Clause no. 10.4.6 (Pg-77) No specific value provided. Procedure given for calculation of permissible loads.

7.1.3 Minimum pitch

Clause no. 8.10.1-(Pg-96) Shall not be less than 2.5 times the nominal diameter of the bolt.

Clause no. 10.2.2(Pg-73) Shall not be less than 2.5 times the nominal diameter of the fastener (Bolt/Rivet).

No change has been made.

7.1.4 Minimum edge distance

Clause no. 8.10.2-(Pg-97) As given in Table 8.2 on page 97

Clause no. 10.2.4.2(Pg-74) Should be >1.7 times hole dia. for sheared or hand-flame cut edges, & >1.5 times hole dia. for rolled, machine-flame cut, sawn and planed edges, from the centre of the hole.

Not much variation is observed in the end results.

7.1.5 Maximum pitch

Clause no. 8.10.1-(Pg-96) Shall not exceed 32t or 300 mm whichever is less, where t is the thickness of the thinner outside plate.

Clause no.10.2.3(Pg-74) Shall not exceed 32t or 300 mm whichever is less, where t is the thickness of the thinner outside plate.

No change has been made.

7.1.6 Maximum edge distance

No specific criteria are mentioned.

Clause no. 10.2.4.3 (Pg-74) Shall not exceed 12tε ,where t is the thickness of the thinner outer plate, and ε = (250/fy)

1/2

Clearance for fastener Holes

Clause no.3.6.1.1 (Pg 28) 1.5 mm for dia. Of bolt <= 25mm, and 2 mm for dia. of bolt > 25mm

Clause no. 10.2.1 (Pg-73) As given in table 19

More practical aspect for clearance has been considered.

7.2 WELDED :- 7.2.1 Fillet welds : Refer IS:816 7.2.1.1 Permissible

stresses. Clause no. 7.1.2-(Pg-17) Shear stress Shall not exceed 110 Mpa

Clause no. 10.5-(Pg-78) Shear stress Shall not exceed 110 Mpa nor as calculated using clause 10.5.7 (Pg. 79-81)

7.2.12 Effective area of weld

Clause no. 6.2.3-(Pg-10) Effective length X Effective throat thickness

Clause no. 6.2.3-(Pg-10) Effective length X effective throat thickness

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

7.2.1.3 Effective

throat thickness

Clause no. 6.2.3-(Pg-10) shall not be less than 3 mm. for calculations it shall be taken as K times the fillet size, where K is a constant. Of a complete penetration weld shall be taken as the thickness of the thinner part joined. And that of an incomplete/partial penetration butt weld shall be taken as minimum thickness of the weld metal common to the parts joined, excluding reinforcement.

Clause no. 10.5.3-(Pg-78) shall not be less than 3 mm and not > 0.7t, where t is the thickness of the thinner plate. for calculations of stresses at faces inclined to each other it shall be taken as K times the fillet size, where K is a constant. Of a complete penetration weld shall be taken as the thickness of the thinner part joined. And that of an incomplete/partial penetration butt weld shall be taken as minimum thickness of the weld metal common to the parts joined, excluding reinforcement.

No Changes have been suggested

7.2.1.4 Effective length

Clause no. 6.2.4-(Pg-11) Shall be the overall length of weld including end returns.

Clause no. 10.5.4-(Pg-78) Shall be the overall length of weld excluding end returns in case of Fillet welds and shall be the overall length of weld including end returns for Butt weld.

7.2.1.5 Minimum length of weld

Clause no. 6.2.4.1-(Pg-11) shall not be less than four times the size of the weld.

Clause no. 10.5.4(Pg-78) shall not be less than four times the size of the weld.


7.2.1.6 Minimum size of the weld

Clause no. 6.2.2-(Pg-10) shall not be less than 3 mm. The minimum size of the first run or the single run weld shall be as given in Table 1 in IS:816

Clause no. 10.5.2-(Pg-78) shall not be less than 3 mm. The minimum size of the first run or the single run weld shall be as given in Table 21 (Pg 78)


7.2.1.8 Field /site weld

Clause no. 7.2-(Pg-17) Permissible stresses in shear and tension reduced to 80%.

Clause no. 10.5.7.2 (Pg. 79) A partial safety factor of 1.5 is to be used for calculation design strength in shear and tension.

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

7.2.2 Butt/Groove Weld :-

7.2.2.1 Permissible stresses.

Clause no. 7.1.1-(Pg-16) Shear stress through throat of butt shall not exceed that of the base material. Tension or compression on section through throat of butt weld shall not exceed the stress in the base metal.

Clause no. 10.5.7(Pg. 79) Permissible stresses I fillet weld shall be calculated as per eqn. in Cl. 10.5.7.1.1 Butt Welds shall be treated a parent metal with a thickness equal to the throat thickness, and shall not exceed those permitted in the parent metal

7.2.2.2 Effective area of weld

Clause no.6.1.61-(Pg-7) Effective length times the effective throat thickness

Clause no. 10.5.4(Pg-78-79) Effective length times the effective throat thickness

8.0 GANTRY GIRDER 8.1 Increase in

stresses Clause no.3.9.3-(Pg-31) While considering the simultaneous effects of vertical & horizontal surge loads of cranes for the combination given in cl. No. 3.4.2.3 & 3.4.2.4, the permissible stresses may be increased by 10 %.

No specific criteria are given.

Stresses are to be calculated using adequate Partial Safety factors.

8.2 Limiting deflection

Clause no.3.13.1.3-(Pg-35) This clause shall be referred.

Clause no. 5.6.1 (Pg. 31) Table 6 should be referred.

9.0 DESIGN AND DETAILING FOR EARTHQUAKE LOADS 9.1 Load

Combinations

No such criteria are given

Clause no. 12.2 (Pg. 87) Two more combinations have to be considered

1) 1.2 DL + 0.5 LL ± 2.5 EL

2) 0.9 DL ± 2.5 EL

9.2 Lateral Load Resisting System

No such classification has been made

The Building has been classified as

1) Braced Frame System a) Ordinary

concentrically Braced Frames (OCBF)

b) Special Concentrically Braced Frame(SCBF)

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

c) Eccentrically Braced Frame (EBF)

2) Moment Frame System a) Ordinary Moment

Frame (OMF) b) Special Moment

Frame (SMF) Various criteria for loads on members are specified for different lateral load resisting systems

10.0 FATIGUE 10.1 Reference

design Condition

No such criteria are mentioned.

Clause no. 13.2.1 (Pg. 91) The conditions when Fatigue design becomes necessary are mentioned, along with a plot of standard S-N curve for each category. A capacity reduction factor µr is to be applied when plates greater than 25 mm tk. are joined by transverse fillet or butt welding

10.2 Partial Safety Factors


Clause no. 13.2.3 (Pg. 92) Based on consequences of fatigue failure , component details have been classified and Partial Safety Factors are given for each type.(Refer Table 25 - Pg. 92)

10.3 Detail Category


Clause no. 13.3 (Pg. 92-98) Tables 26 (a) to (d) indicate the classification of different details into various categories for the purpose of assessing fatigue strength.

11 FIRE RESISTANCE 11.1 No such criteria are

mentioned.

Clause no. 13 ( Pg. 105-110) The following points have been discussed and relevant design standards have been mentioned 1) Fire Resistance Level

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

2) Period of Structural Adequacy 3) Variation of mechanical properties of Steel with Temperature 4) Limiting Steel Temperature 5) Thermal Increase with Time in Protected Members 6) Temperature Increase with Time in Unprotected Members 7) Determination of Period of Structural Adequacy from a single test 8) Three-Sided Fire Exposure Condition.

12.0 USEFUL TABLES 1) Maximum Slenderness

ratio - Table 3.1 (Pg-30) 2) Permissible stress in Axial

Compression Table 5.1 (Pg-39)

3) Effective length of compression member of constant Dimension - Table 5.2 (Pg-41-43)

4) Max. permissible bending stress in I-beams or Channels - Table 6.1A/B (Pg-57,58)

5) Permissible avg. shear stress - Table 6.6A (Pg-73)

6) Minimum Thickness of web for Plate Girders with solid web – Table 6.7 (Pg. 83)

7) Max. permissible stress in Rivets and Bolts - Table 8.1 (Pg.-95)

8) Edge Distance of Holes -

1) Tensile Properties of Structural steel products (Pg. 13-14)

2) Limiting Width to Thickness Ratio (Pg. 18)

3) Maximum Values of Effective Slenderness Ratios (Pg.20)

4) Partial Safety Factors for Loads, for Limit States (Pg. 29)

5) Partial Safety Factor for Materials (Pg. 30)

6) Deflection Limits (Pg. 31)

7) Imperfection Factor (Pg. 35)

8) Stress Reduction Factor for columns (Pg. 36-39)

9) Design Compressive Stress for columns (Pg. 40-43)

10) Buckling Class of C/S (Pg. 44)

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

Table 8.2 (Pg. 97) 9) Section Properties - SP-

(6) & IS:808.

11) Effective Length of Prismatic Compression Members (Pg. 45)

12) Constants k1,k2,k3 for calculation of Flexural Torsional buckling in Angle Struts (Pg. 48)

13) Design Bending Compressive Stress Corresponding to Lateral Buckling (Pg. 55-56)

14) Critical Stress wrt elastic Lateral Buckling (Pg. 57)

15) Effective length of Simply Supported Beams (Pg. 58)

16) Effective Length for Cantilever (Pg. 61)

17) Constants α1 & α2 (Pg. 71)

18) Equivalent uniform Moment Factor (Pg. 72)

19) Clearances for Fastener Holes (Pg. 73)

20) Typical Average Values fro coefficient of Friction (Pg. 77)

21) Minimum Size of First run or of a single run Fillet Weld (Pg. 78

22) Values of K for Different Angles Between Fusion Faces (Pg. 78)

23) Response Reduction Factor for a Building System (Pg. 87)

24) Multiplying Factors for Calculated Stress Range (Pg. 92)

25) Partial Safety Factors for Fatigue Strength (Pg. 92)

26) Detail Category Classification for fatigue strength assessment (Pg.

SR NO

Clause Description

IS:800-1984

IS:800-2007

Comments

93-98) 27) Factors to allow for

Variability of Structural Units ( Pg. 103)

28) Environmental Exposure Condition (Pg. 103)

29) Protection Guide for Steel Work Application (Pg. 104-105)

30) Regression Coefficient , k (Pg. 107)

31) Fire Resistance Rating –Encased Steel Columns (Pg. 110)

32) Fire Resistance Rating –Encased Steel Beams (Pg. 111)

33) Normal Tolerance after Erection (Pg. 115)

34) Straightness Tolerance Incorporated in Design Rules (Pg. 115_

Shelter Definition Shelter is a steel structure with roof, and with or without wall cladding provided for equipment to protect from sunshine, rain, show, wind and temperature and noise, which is either fully enclosed or fully opened according to the requirements of equipment.

1) Compressor Shelter For compressor shelter and its related equipment, which would be required in most of the plant facilities. It is usually fully enclosed or partially enclosed with operation platform, hoisting devices such as hoist or crane for maintenance and operation of compressor

2) Ordinary Shelter For equipment other than compressor such as pump, emergency generator, package equipment and etc. include sunshades for truck loading and car parking. It is usually partially enclosed or fully opened and in some cases with operation platform on ground and hoist or crane for maintenance and operation of equipment.

A typical example of shelter for table top type compressor is shown below.

Type of Structure Conventional Steel Frame Building Type A pitched roof of building with rigid forming combined with framing, which are furnished with operating platforms, equipment support, piping support and etc. as required. A typical figure is shown below

Pre Engineered Building (PEB) Type Manufacturer’s standard of Steel framed building with its own materials specification and design code fro steel structure, connection, purlin, girth, cladding material. PEB type, of which the steel weight is less than conventional type, is suitable for simple structure without support for heavy loads nor lots of supports like a warehouse.

Fig. 2 Typical Shelter for Table Type Compressor

Fig. 3 Typical Steel Frame Building Type Compressor Shelter

Comparison of Compressor Shelter members Designed by WSM and LSM Load Combinations The following load combinations have been considered in the analysis of the Portal.

Note: Kindly note that in the LSM , Crane Load is considered as the Leading Live load and the actual Live Load is considered as the accompanying Live Load.

IS 800-2007 IS 800-1984 1) 1.5 (DL+CL)+1.05 (IL) 2) 1.2 (DL+CL)+1.05 IL+0.6WL(-X PREESSURE) 3) 1.2 (DL+CL)+1.05 IL+0.6WL(-X SUCTION) 4) 1.2 (DL+CL)+1.05 IL+0.6WL(X PREESSURE) 5) 1.2 (DL+CL)+1.05 IL+0.6WL(X SUCTION) 6) 1.2 (DL+CL)+1.05 IL+0.6ELX 7) 1.2 (DL+CL)+1.05 IL-0.6ELX 8) 1.2 (DL+CL)+1.05 IL+0.6ELY 9) 1.2 (DL+CL)+1.05 IL-0.6ELY 10) 1.2 (DL+CL+WL(-X PRESSURE))+0.53 IL 11) 1.2 (DL+CL+WL(-X SUCTION))+0.53 IL 12) 1.2 (DL+CL+WL(X PRESSURE))+0.53 IL 13) 1.2 (DL+CL+WL(X SUCTION))+0.53 IL 14) 1.2 (DL+CL+ELX)+0.53 IL 15) 1.2 (DL+CL-ELX)+0.53 IL 16) 1.2 (DL+CL+ELY)+0.53 IL 17) 1.2 (DL+CL-ELY)+0.53 IL 18) 1.5 (DL +WL(-X PRESSURE) 19) 1.5 (DL +WL(-X SUCTION) 20) 1.5 (DL +WL(X PRESSURE) 21) 1.5 (DL +WL(X SUCTION) 22) 1.5 (DL +ELX) 23) 1.5 (DL -ELX) 24) 1.5 (DL +ELY) 25) 1.5 (DL -ELY) 26) 0.9 DL+1.5WL(-X PRESSURE) 27) 0.9 DL+1.5WL(-X SUCTION) 28) 0.9 DL+1.5WL(X PRESSURE) 29) 0.9 DL+1.5WL(X SUCTION) 30) 0.9 DL+1.5ELX 31) 0.9 DL-1.5ELX 32) 0.9 DL+1.5ELY 33) 0.9 DL-1.5ELY 34) 1.2 DL+0.5(IL+CL)+2.5 ELx 35) 1.2 DL+0.5(IL+CL)-2.5 ELx 36) 1.2 DL+0.5(IL+CL)+2.5 ELy 37) 1.2 DL+0.5(IL+CL)-2.5 ELy 38) 0.9 DL+2.5 ELx 39) 0.9 DL-2.5 ELx 40) 0.9 DL+2.5 ELy 41) 0.9 DL-2.5 ELy

1) DL + IL + WL(-X PRESSURE) 2) DL + IL + WL(-X-SUCTION) 3) DL + IL + WL (X-PRESSURE) 4) DL + IL + WL (X-SUCTION) 5) DL + IL 6) DL + IL + CL 7) DL + IL + CL+ WL (-X-PRESSURE) 8) DL + IL + CL+WL (-X-SUCTION) 9) DL + IL + CL+ WL (X-PRESSURE) 10) DL + IL + CL+WL (X-SUCTION) 11) DL + IL + VX + VY 12) DL + IL -VX + VY 13) DL + IL + CL + VX +VY

Member Forces The compressor Shelter is assumed to comprise of the following Different Types of Sections:

1) 2 ISMB 400 Placed at a c/c distance of 1000mm (R1)(Main Columns upto Gantry Girder) 2) ISMB 450 (R2)(Columns supporting Roof Truss) 3) ISA 75 x 75 x 8 (R3)( Truss Top and Bottom Chord Members) 4) ISA 65 x 65 x 6 (R4)( Truss Inclined Members) 5) ISA 50 x 50 x 6 (R5)( Truss Vertical Members) 6) ISMB 150 (R6)(Monitor Portal)

Fig. 4 Staad Model for Portal Frame indicating the locations of Various Members

The member wise STAAD Pro Analysis Output after applying the above mentioned load combinations are as follows: Forces as per LSM Axial Shear Torsion Bending

Section Max Fx kN

Max Fy kN

Max Fz kN

Max Mx kNm

Max My kNm

Max Mz kNm

ISMB400BB100 Max+ve 1067.917 171.777 0 0 0 1366.781 Max -ve -9.306 -222.799 0 0 0 -1505.794 ISMB450 Max+ve 349.129 223.019 0 0 0 280.178 Max -ve -86.659 -154.811 0 0 0 -198.128 ISA75X75X8 Max+ve 376.31 11.804 0 0 0 3.467 Max -ve -387.325 -11.696 0 0 0 -3.016 ISA65X65X6 Max+ve 232.33 0.502 0 0 0 0.587 Max -ve -139.311 -0.528 0 0 0 -0.488 ISA50X50X6 Max+ve 77.453 1.031 0 0 0 0.794 Max -ve -237.676 -1.618 0 0 0 -0.797 ISMB150 Max+ve 17.084 0.647 0 0 0 0.416 Max -ve -17.603 -0.71 0 0 0 -0.566

Forces as per WSM Axial Shear Torsion Bending

Section Max Fx kN

Max Fy kN

Max Fz kN

Max Mx kNm

Max My kNm

Max Mz kNm

ISMB400BB100 Max+ve 779.694 144.088 0 0 0 887.13 Max -ve -119.851 0 0 0 -1072.455 ISMB450 Max+ve 199.834 145.209 0 0 0 178.89 Max -ve -41.206 -106.198 0 0 0 -119.623 ISA75X75X8 Max+ve 166.838 7.847 0 0 0 1.728 Max -ve -181.083 -7.566 0 0 0 -2.067 ISA65X65X6 Max+ve 90.21 0.327 0 0 0 0.446 Max -ve -84.566 -0.421 0 0 0 -0.312 ISA50X50X6 Max+ve 90.303 0.645 0 0 0 0.329 Max -ve -61.541 -0.554 0 0 0 -0.3 ISMB150 Max+ve 11.365 0.58 0 0 0 0.432 Max -ve -11.781 -0.426 0 0 0 -0.283

Saad Pro Sign Convention: + ve Axial Force Denotes Compression - ve Axial Force denotes Tension + ve Shear Force Denotes Shear Force along +ve Y axis - ve Shear Force Denotes Shear Force along -ve Y axis + ve Moment denotes Hogging moment - ve Moment denotes Sagging moment

X

Z

Y

Fig. 5 Staad Global Axis

MATERIAL PROPERTIES :Modulus of elasticity of steel = 200000 MPaYield stress of steel = 250 MPaUltimate Strength of Steel = 410 MPaPartial Safety Factors γm0 = 1.1

γm1 = 1.25

LOADS ACTING ON MEMBER : Axial Force- Tension = 30.764 KN

GEOMETRY :No. of bolts usedfor load Transfer = 6.00 no.Nominal Dia. Of Bolts = 20.00 mmSpacing between bolts = 60.00 mmEdge Distance = 30.00 mmEnd Distance = 30.00 mmEffective Length in "x " dirn. = 892.50 mmEffective Length in "y " dirn. = 892.5 mm

SECTION PROPERTIES :Section details = 2 ISA 50 x 50 x 6, Single Leg connectionGross Area of section (Ag) (one angle) = 568.00 mm2

Width of connected Leg = 50.00 mmWidth of Unconnected Leg (w) = 50.00 mmThickness of Leg (t) = 6.00 mmNet Area of Total Cross Section (An) = 436.00 mm2

Net Area of Connected Leg (Anc) = 150.00 mm2

Gross Area of Ourstanding Leg (Ago) = 282.00 mm3

Clearance for Bolts = 2.00 mm (Table 19, Pg 73)Minimum Gross area in Shear Along Bolt line (Avg) = 1980.00 mmMinimum Net Area in Sheat Along Bolt Line (Avn) = 1254.00 mmMinimum Gross Area in tension (Atg) = 180.00 mmMinimum Net Area in Tension (Atn) = 114.00 mm

w1 = 20.00 mm

SUMMARY OF DESIGN :

30.77 KN SAFE30.77 KN SAFE

Yielding of Gross C/s 85.2 KNFailure of Critical Section 48.51 KN

Design for Tension

TYPE OF STRESSES ACTUAL FORCE PERMISSIBLE FORCE CHECK

DESIGN OF TRUSS MEMBER (2 ISA 50x50x6, Single Leg Connection)IS 800-1984

Design Strength Due to Yielding of Gross Cross SectionDesign Strength of Member under Axial Tension

σper = 150 MPaT per = 85200.00 N > 30764

SAFE

Design Strength due to Rupture of Critical SectionEffective C/s area of Connected Leg (A1)

A1 = 150.00 mm2

Gross C/s area of un connected Leg

A2 = 282.00 mm2

k = 3*A1/(3*A1+A2) = 0.61Net Effective Area (Aeff) = A1 + A2*k = 323.36 mm2

Tensile force resisted by the section = σper * Aeff

= 48504.1 > 30764 N

SAFE

LOADS ACTING ON MEMBER : Axial Force- Compression = 133.729 kN

SECTION PROPERTIES :Section details = 2 ISA 50x50x6, Connected Back to back

Gross Area of section (Ag) = 1136.00 mm2

Width of connected Leg = 50.00 mmWidth of Unconnected Leg (w) = 50.00 mmThickness of Leg (t) = 6.00 mmRadius of Gyratiojn about minor Axis = 15.10 mmRadius of Gyratiojn about major Axis = 23.10 mm

SUMMARY OF DESIGN :

133.729 KN SAFE

Slenderness Check ( Table 3.1, Pg. 30)λ = leff/rmin = 59.11 < 180

SAFE AXIAL FORCE

Allowable Axial stress (Cl. 5.1)σal = 0.6*fcc*fy/(fcc

1.4+fy1.4)1/1.4

fcc = π2*E/(leff/rmin)2

= 564.45 MPaσal = 123.03 Mpa

Allowable Comp Force = 139765.27 > 133729 N SAFE

Axial Force 139.77 KN

TYPE OF CHECKS ACTUAL FORCE PERMISSIBLE FORCE CHECK

Design for Compression

LOADS ACTING ON MEMBER : Factored Loads

Axial Force- Tension = 118.838 kN-m

SUMMARY OF DESIGN :

118.84 KN SAFE118.84 KN SAFE118.84 KN SAFE

Design Strength Due to Yielding of Gross Cross Section

Design Strength iof Member under Axial Tension (Tdg) (Cl. 6.2, Pg. 32)Tdg = Ag*fy/γm0

= 129090.91 N = 129.0909 KN> 118.838 KN

SAFE Design Strength due to Rupture of Critical Section

For Single Angles (Cl. 6.3.3, Pg. 33) β = 1.4 - 0.076*(w/t)*(fy/fu)*(bs/Lc) ≤ {fu*γm0/(fy*γm1)}

≥ 0.7= 1.26 ≤ 1.44

≥ 0.7β = 1.26

Tdn = 0.9*Anc*fu/γm1 + β*Ago*fy/γm0

= 125131.81 > 118838 N

SAFE

Design Strength due to Block Shear (Cl. 6.4,Pg 33)

Tdb1 = [Avg*fy/(30.5*γm0)+0.9*Atnfu/γm1]

= 293460.42Tdb2 = (0.9*Avn*fu/(3

0.5*γm1)+Atg*fy/γm0)283777.26

Tdb = 283777.26 > 118838SAFE

Block Shear 283.78 KN



IS 800-2007Design for Tension

LOADS ACTING ON MEMBER :

Factored Loads

Axial Force- Compression = 133.729 KN

SUMMARY OF DESIGN :


Slenderness Check (Table 3, Pg. 20)

l = leff/rmin = 59.11 < 180

SAFE AXIAL FORCE

Considering that the angles are loaded only through one LegEquivalent Slenderness Ratio

λλλλe = (k1+k2*λλλλvv2+k3*λλλλϕϕϕϕ

2)0.5 (Cl. 7.5.1.2 & Cl. 7.5.2.2)

λvv = (l/rvv)/(ε*(π2E/250)0.5)

= 0.67λϕ = (b1+b2)/(2*t)/(ε*(π2E/250)0.5)

= 0.09

Considering Fixed condition at the Endsk1 = 0.2 (Table 12, Pg. 48)k2 = 0.35 (Table 12, Pg. 48)k3 = 20 (Table 12, Pg. 48)

λe = 0.729

φ = 0.5(1+α*(λe-0.2)+λe2)

= 0.895

X = 1/(φ+(φ2-λe2)0.5)

= 0.707Design Compressive Stress

fcd = X fy/γm0

160.64

Design Comp Force = 182485.16 > 133729

SAFE

Allowable Comp force = 121656.78

Considering Hinged condition at the Endsk1 = 0.7 (Table 12, Pg. 48)k2 = 0.6 (Table 12, Pg. 48)

CHECK

Axial Force (Fixed Ends) 182.49 KN

TYPE OF CHECKS ACTUAL FORCE

Axial Force (Hinged Ends)


138.68 KN

PERMISSIBLE FORCE

k3 = 5 (Table 12, Pg. 48)

λe = 1.005

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.202X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

= 122.07

Design Comp Force = 138674.03 > 133729SAFE


Considering Concentric Axial Force (Angles loaded through both the legs)(Cl. 7.5.1.1, Pg. 47)

λ = 0.67

φ = 0.5(1+α*(λe-0.2)+λe2)

= 0.836X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

169.53


Allowable Comp force = 128387.50 N


γm1 = 1.25

LOADS ACTING ON MEMBER : Axial Force- Tension = 66.624 kN

GEOMETRY :No. of bolts used for load Transfer = 6.00 no.Nominal Dia. Of Bolts = 20.00 mmSpacing between bolts = 60.00 mmEdge Distance = 35.00 mmEnd Distance = 35.00 mmEffective Length in "x " dirn. = 1530.00 mmEffective Length in "y " dirn. = 1400.00 mm

SECTION PROPERTIES :Section details = 2 ISA 65 x 65 x 6, Single Leg connectionGross Area of section (Ag) = 744.00 mm2

Width of connected Leg = 65.00 mmWidth of Unconnected Leg (w) = 65.00 mmThickness of Leg (t) = 6.00 mmNet Area of Total Cross Section (An) = 612.00 mm2




w1 = 30.00 mm

SUMMARY OF DESIGN :


Yielding of Gross C/s 111.6 KNFailure of Critical Section 73 KN

Design for Tension


DESIGN OF TRUSS MEMBER (2 ISA 65x65x6, Single leg connection)IS 800-1984


σper = 150

T per = 111600.00 N > 100000 NSAFE


A1 = 240.00 mm2


A2 = 372.00 mm2

k = 3*A1/(3*A1+A2) = 0.66

Net Effective Area (Aeff) = A1 + A2*k = 485.27 mm2


= 72791.21 > 66624 NSAFE


SECTION PROPERTIES :Section details = 2 ISA 65x65x6, Connected Back to backGross Area of section (Ag) = 1488.00 mm2


SUMMARY OF DESIGN :

90.197 KN SAFE

Slenderness Check( Table 3.1, Pg. 30)λ = leff/rmin = 77.27273 < 180

SAFEAXIAL FORCE

Allowable Axial stress (Cl. 5.1)σal = 0.6*fcc*fy/(fcc

1.4+fy1.4)1/1.4


= 330.25 MPaσal = 103.67 Mpa

Allowable Comp Force = 154265.39 > 90197 SAFE

TYPE OF CHECKS ACTUAL FORCE PERMISSIBLE FORCE


CHECK




SUMMARY OF DESIGN :




= 169090.91 N = 169.09 KN> 150 KN



≥ 0.7= 2.23 ≤ 1.44

≥ 0.7β = 1.44


= 192864.00 > 69655.50 NSAFE



= 306252.90Tdb2 = (0.9*Avn*fu/(3

0.5*γm1)+Atg*fy/γm0)296405.68

Tdb = 296405.68 > 69655.5SAFE





LOADS ACTING ON MEMBER :

Factored LoadsAxial Force- Compression = 223.33 kN

SUMMARY OF DESIGN :

223.33 KN FAILS223.33 KN FAILS

Concentric Axial Force 223.33 KN SAFE


l = leff/rmin = 77.27 < 180SAFE

AXIAL FORCEConsidering that the angles are loaded only through one LegEquivalent Slenderness Ratio


2)0.5 (Cl. 7.5.1.2 & Cl. 7.5.2.2)

λvv = (l/rvv)/(ε*(π2E/250)0.5)

= 0.87λϕ = (b1+b2)/(2*t)/(ε*(π2E/250)0.5)

0.12

Considering Fixed condition at the Endsk1 = 0.2 (Table 12, Pg. 48)k2 = 0.35 (Table 12, Pg. 48)k3 = 20 (Table 12, Pg. 48)λe = 0.873

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.046

X = 1/(φ+(φ2-λe2)0.5)


fcd = X fy/γm0

140.08 MPa

Design Comp Force = 208442.72 < 223330FAILS


Considering Hinged condition at the Endsk1 = 0.7 (Table 12, Pg. 48)k2 = 0.6 (Table 12, Pg. 48)k3 = 5 (Table 12, Pg. 48)

PERMISSIBLE FORCETYPE OF CHECKS ACTUAL FORCE

Axial Force (Hinged Ends)Axial Force (Fixed Ends) 208.45 KN

162.26 KN209.11 KN


CHECK

λe = 1.108

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.337X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

109.04 MPa



Considering Concentric Axial Force (Angles loaded through both the legs)(Cl. 7.5.1.1, Pg. 47)λ = 0.87

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.043X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

= 140.53 MPa




γm1 = 1.25


GEOMETRY :No. of bolts usedfor load Transfer = 11.00 no.Nominal Dia. Of Bolts = 24.00 mmSpacing between bolts = 60.00 mmEdge Distance = 40.00 mmEnd Distance = 40.00 mmEffective Length in "x " dirn. = 1120.00 mmEffective Length in "y " dirn. = 1400 mm

SECTION PROPERTIES :(one angle) = 2 ISA-75x75x8, Single Leg connectionGross Area of section (Ag) = 1138.00 mm2Width of connected Leg = 75.00 mmWidth of Unconnected Leg (w) = 75.00 mmThickness of Leg (t) = 8.00 mmNet Area of Total Cross Section (An) = 930.00 mm2Net Area of Connected Leg (Anc) = 360.00 mm2Gross Area of Ourstanding Leg (Ago) = 568.00 mm3Clearance for Bolts = 2.00 mm (Table 19, Pg 73)Minimum Gross area in Shear Along Bolt line (Avg) = 5120.00 mmMinimum Net Area in Sheat Along Bolt Line (Avn) = 2936.00 mmMinimum Gross Area in tension (Atg) = 320.00 mmMinimum Net Area in Tension (Atn) = 216.00 mmw1 = 35.00 mm

SUMMARY OF DESIGN :




Design for Tension


DESIGN OF TRUSS MEMBER (2 ISA 75x75x8, Single Leg connection) IS 800-1984

Design Strength of Member under Axial Tension

σper = 150.00 Mpa

T per = 170700.00 N > 100000 N


A1 = 360.00 mm2

Gross C/s area of un-connected LegA2 = 568.00 mm2

k = 3*A1/(3*A1+A2)= 0.66

Net Effective Area (Aeff) = A1 + A2*k = 732.23 mm2


= 109835 > 100000 NSAFE

LOADS ACTING ON MEMBER : Axial Force- Compression = 220 kN

SECTION PROPERTIES :Section details = 2 ISA 75 x75 x 8, Back to backGross Area of section (Ag) = 2276.00 mm2


SUMMARY OF DESIGN :

220 KN SAFE

Slenderness Check ( Table 3.1, Pg. 30)λ = leff/rmin = 61.40 < 180

SAFE AXIAL FORCE

Allowable Axial stress (Cl. 5.1)

σal = 0.6*fcc*fy/(fcc1.4+fy

1.4)1/1.4


= 523.00 MPa

σal = 120.69 MpaAllowable Comp Force = 274686.27 > 220000 SAFE

TYPE OF CHECKS ACTUAL FORCE PERMISSIBLE FORCE


CHECK



Axial Force- Tension = 193.66 kN

SUMMARY OF DESIGN :


Design Strength Due to Yielding of Gross Cross SectionDesign Strength iof Member under Axial Tension (Tdg) (Cl. 6.2, Pg. 32)

Tdg = Ag*fy/γm0

= 258636.36 N = 258.64 KN> 150 KN



≥ 0.7= 1.12 ≤ 1.44

≥ 0.7β = 1.12


= 250220.17 > 193662.5 NSAFE



= 735588.97Tdb2 = (0.9*Avn*fu/(3

0.5*γm1)+Atg*fy/γm0)

= 641356.40

Tdb = 641356.40 > 193662.5SAFE

MATERIAL PROPERTIES :Modulus of elasticity of steel = 200000 MPaYield stress of steel = 250 MPaUltimate Strength of Steel = 410 MPa






Partial Safety Factors γm0 = 1.1

γm1 = 1.25


Axial Force- Compression = 376.31 kN

SECTION PROPERTIES :Section details = 2 ISA-75x75x8, Back to backGross Area of section (Ag) = 2276.00 mm2

Width of connected Leg = 75.00 mmWidth of Unconnected Leg (w) = 75.00 mm

Thickness of Leg (t) = 8.00 mmRadius of Gyratiojn about minor Axis = 22.80 mmRadius of Gyratiojn about major Axis = 34.90 mm

SUMMARY OF DESIGN :

376.31 KN FAILS376.31 KN FAILS

Concentric Axial Force 376.31 KN SAFE

Slenderness Check (Table 3, Pg. 20)l = leff/rmin = 49.12 < 180.0

SAFE AXIAL FORCE

Considering that the angles are loaded only through one LegEquivalent Slenderness Ratio


2)0.5 (Cl. 7.5.1.2 & Cl. 7.5.2.2)

λvv = (l/rvv)/(ε*(π2E/250)0.5)

= 0.55λϕ = (b1+b2)/(2*t)/(ε*(π2E/250)0.5)

0.11

Considering Fixed condition at the Endsk1 = 0.2 (Table 12, Pg. 48)k2 = 0.35 (Table 12, Pg. 48)k3 = 20 (Table 12, Pg. 48)λe = 0.728

φ = 0.5(1+α*(λe-0.2)+λe2)

= 0.894

X = 1/(φ+(φ2-λe2)0.5)

= 0.707

Design Compressive Stress

PERMISSIBLE FORCETYPE OF CHECKS ACTUAL FORCE

Axial Force (Hinged Ends)Axial Force (Fixed Ends) 365.88 KN

288.66 KN420.45 KN

CHECK

fcd = X fy/γm0

160.75 Mpa



Considering Hinged condition at the Endsk1 = 0.7 (Table 12, Pg. 48)k2 = 0.6 (Table 12, Pg. 48)k3 = 5 (Table 12, Pg. 48)λe = 0.969

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.158X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

126.83



Considering Concentric Axial Force (Angles loaded through both the legs)(Cl. 7.5.1.1, Pg. 47)

λ = 0.55φ = 0.5(1+α*(λe-0.2)+λe

2)

= 0.739X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

= 184.73




γm1 = 1.25


GEOMETRY :No. of bolts usedfor load Transfer = 6.00 no.Nominal Dia. Of Bolts = 20.00 mmSpacing between bolts = 60.00 mmEdge Distance = 35.00 mmEnd Distance = 35.00 mmDistance between bolt lines 60.00 mmEffective Length in "x " dirn. = 2000.00 mmEffective Length in "y " dirn. = 2000 mm

SECTION PROPERTIES :Section details = ISMB 150 (connected via 2 bolt lines)


Width of connected Leg = 150.00 mmWidth of Unconnected Leg (w) = 40.00 mmThickness of Web (tw) = 4.80 mmThickness of flange (t) = 7.60 mmNet Area of Total Cross Section (An) = 1688.80 mm2




w1 = 45.00 mm

SUMMARY OF DESIGN :


DESIGN OF TRUSS MEMBER (ISMB 150)IS 800-1984


Design for Tension

Yielding of Gross C/s 285 KNFailure of Critical Section 168.96 KN

x

y


σper = 150 Mpa

T per = 285000.00 N > 100000 NSAFE


A1 = 485.76 mm2


A2 = 1143.04 mm2

k = 3*A1/(3*A1+A2) = 0.56Net Effective Area (Aeff) = A1 + A2*k = 1126.35 mm2


= 168951.9 > 11781 NSAFE


SECTION PROPERTIES :Section details = ISMB 150


Width of connected Leg = 150.00 mmWidth of Unconnected Leg (w) = 40.00 mmThickness of Web (tw) = 4.80 mmThickness of flange (tf) = 7.60 mmRadius of Gyration about minor Axis = 16.60 mmRadius of Gyration about major Axis = 61.80 mm

SUMMARY OF DESIGN :

11.368 KN SAFE

Slenderness Check ( Table 3.1, Pg. 30)λ = leff/rmin = 120.48 < 180 SAFE

AXIAL FORCEAllowable Axial stress (Cl. 5.1)σal = 0.6*fcc*fy/(fcc

1.4+fy1.4)1/1.4


= 135.85 MPaσal = 63.27 MpaAllowable Comp Force = 120204.53 > 11368 N SAFE






SUMMARY OF DESIGN :




= 431818.18 N = 431.82 KN> 150 KN

SAFEDesign Strength due to Rupture of Critical Section


≥ 0.7= 1.53 ≤ 1.44

≥ 0.7β = 1.44


= 518313.47 > 17603 NSAFE

Design Strength due to Block Shear


= 249093.84

Tdb2 = (0.9*Avn*fu/(30.5*γm1)+Atg*fy/γm0)

241156.30

Tdb = 241156.30 > 17603

SAFE



Yielding of Gross C/s 431.82 KNFailure of Critical Section 518.32 KNBlock Shear 241.16 KN


Axial Force- Compression = 17.084 kN

SUMMARY OF DESIGN :

17.09 KN SAFE


λ = leff/rmin = 120.48 < 180SAFE

AXIAL FORCE (Cl. 7.1.2.1)h/b = 1.875 > 1.2

tf = 7.6 < 40 mm

(buckling about y-y axis)α = 0.34 (class b)λ = 1.36φ = 0.5(1+α*(λe-0.2)+λe

2)

= 1.617X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

91.04



(buckling about x-x axis)α = 0.21 (class a)λ = 1.36

φ = 0.5(1+α*(λe-0.2)+λe2)

= 1.542X = 1/(φ+(φ2-λe

2)0.5)


fcd = X fy/γm0

= 799.60Design Comp Force = 1519234.09 > 16600

SAFEAllowable Comp force = 1012822.73 N




MATERIAL PROPERTIES :Modulus of elasticity of steel = 200000 MPaYield stress of steel = 250 MPaPartial Safety Factors γm0 = 1.1

γm1 = 1.25

LOADS ACTING ON COLUMNS : Axial force (P) = 199.818 kNBending moment @ major axis (Mx) = 179 kN-mBending moment @ minor axis (My) = 0 kN-m

GEOMETRY :Eff. Length @ x-axis (leffx) = 3.13 mEff. Length @ y-axis (leffy) = 2.61 m

SECTION PROPERTIES :Section details = ISMB 450

Area of Combined section (Ax) = 9227.00 mm2

Moment of inertia @ major axis (Ixx) = 303908000 mm4

Moment of inertia @ minor axis (Iyy) = 8340000 mm4

Elastic Moduli of section @ major axis (Zexx) = 1350700.00 mm3

Elastic Moduli of section @ minor axis (Zeyy) = 111200.00 mm3

Plastic Modulus of section @ major axis (Zpxx) = 1533360 mm3

Radius of gyration @ major axis (rxx) = 181.50 mmRadius of gyration @ minor axis (ryy) = 30.10 mmWeight of section = 72.40 Kg / mThickness of flange (Tf) of I Section = 17.40 mmThickness of web (tw) = 9.40 mmOverall depth (D) = 450.00 mmClear depth of web (d1) = 415.20 mmEffective Web Depth (h1) = 379.20 mmCentre to centre distance between flanges = 432.60 mmWidth of Flange of Individual I section = 150.00 mm

SUMMARY OF DESIGN :

21.66 MPa SAFE132.53 MPa SAFE0.00 MPa SAFE

0.84 SAFE

CALCULATION OF ACTUAL STRESSES :Actual compressive stress (σac,cal) = P / Ax

= 199818 / 9227

DESIGN OF COLUMN ( ISMB 450)IS 800-1984

TYPE OF STRESSES ACTUAL STRESS PERMISSIBLE STRESS CHECK

Axial compression stress 93.61 MPaBending stress @ major axis 133.13 MPaBending stress @ minor axis 150.00 MPaRatio of combined stresses 1.00

x

y

= 21.65579278 MPa

Actual bending stress @ major axis (σbcx,cal) = Mx / Zxx

= 179000000 / 1350700= 132.52 MPa

Actual bending stress @ minor axis (σbcy,cal) = My / Zyy

= 0 / 111200= 0.00 MPa

CALCULATION OF PERMISSIBLE STRESSES : AXIAL STRESSES :

Slenderness ratio in major direction (λx) = lx / rxx

= 3132 / 181.5= 17.26

Slenderness ratio in minor direction (λy) = lY / ryy

= 2610 / 30.1= 86.71

Maximum slenderness ratio ( λmax ) = 86.71 < 180 = λperm.

SAFEElastic critical stress in major dir. (fccx ) = π2 ∗ Ε / λx

2

= π2 ∗ 200000 / 297.78= 6622.15 MPa

Elastic critical stress in minor dir. (fccy ) = π2 ∗ Ε / λy2

= π2 ∗ 200000 / 7518.8= 262.27 MPa

Minimum elastic critical stress ( fcc ) = 262.27 MPa

Permissible axial stress (σac ) 0.6 * fcc * fy

[(fcc)1.4 + (fy)

1.4 ] 1/1.4

0.6 * 262.27 * 250[(262.27)^1.4 + (250)^1.4 ]^1/1.4

= 93.60 MPa (as per clause 5.1.1)Ratio of axial compression = σac,cal

σac

= 21.6693.60

= 0.231

BENDING STRESS :Y = 26.5 * 105 / ( l / ryy)

2 (as per clause 6.2.4)= 26.5 * 10^5 / ( 86.72 )^2= 352.45 MPa

X = Y * [1+ 1/20 * { (l/ryy)*(T/D) }2 ]0.5 (as per clause 6.2.4)= 352.46 * [1+ 1/20 * { ( 86.72)*(17.4 / 450) }^2 ]^0.5= 440.50 MPa

C1 = Distance Between NA and Top Extreme Fiber

C2 = Distance Between NA and Bottom Extreme Fiber

=

=

C1 = C2

K1 = 1.0 For ψ = 1.0 (as per table 6.3)

K2 = 0.0 For ω = 0.5 (as per table 6.4)

fcbx = K1 * ( X + K2 * Y ) * C1/C2 (as per clause 6.2.4)= 1 * ( 440.51 + 0 * 352.46 ) * 1.0= 440.50

( If T/t < 2.0 and d1/t < 1344 / sqrt(fy), fcbx = 1.2 * fcbx &

If T/t > 2.0 or d1/t > 1344 / sqrt(fy) , fcbx = fcbx ) (as per clause 6.2.4.1)

fcbx = 1.2 * 440.51

= 528.60 MPa

0.66 x fcbx x fy

[(fcbx)1.4 + (fy)

1.4 ] 1/1.4

0.66 x 528.61 x 250[(528.61)^1.4 + (250)^1.4 ]^1/1.4

= 133.13 MPaσbcy = 150.00 MPa (as per clause 6.2.5)

CHECK FOR COMBINED STRESSES :Cmx = 0.6 (as per 7.1.3)

Cmy = 0.6 (as per 7.1.3)

σac,cal σbcx,cal x Cmx σbcy,cal x Cmy

σac 1 - σac,cal x σbcx 1 - σac,cal x σbcy

0.6 x fccx 0.6 x fccy

21.66 132.53 x 0.6 0 x 0.693.61 1 - 21.66 x134 1 - 21.66 x 150

0.6 x 6623 0.6 x 262.3= 0.23 0.60 0.00= 0.83 < 1.00

SAFE

σbcx =

=

=

=

LOADS ACTING ON COLUMNS : Factored LoadsAxial force (P) = 349.129 kNBending moment @ major axis (Mx) = 280.178 kN-mBending moment @ minor axis (My) = 0 kN-m

Shear Force = 223.019 KN

SUMMARY OF DESIGN :

37.84 MPa SAFE207.44 MPa FAILS

0 MPa SAFE

1.197 FAILS


= 349129/9227= 37.84 MPa


= 280178000/1350700= 207.43 MPa


= 0/111200= 0.00 MPa

SECTION CLASSIFICATION (Table 2)For CompressionFlange b/Tf = 4.31 < 9.4 ε PlasticWeb d/Tw = 40.34 < 42 ε Semi-Compact

Semi CompactFor FlexureFlange b/Tf = 4.31 < 9.4 ε Plastic

d/Tw = 47.87 < 84 ε Plasticε = (250/fy)0.5 Plastic

CALCULATION OF PERMISSIBLE STRESSES :

AXIAL STRESSES :Slenderness ratio in major direction (λx) = lx / rxx

= 3132/181.5= 17.26


= 2610/30.1= 86.71

IS 800-2007


Axial compression stress 139.21 MPaBending stress @ major axis 201.28 MPaBending stress @ minor axis 0 MPa

Ratio of combined stresses 1.00

Maximum slenderness ratio ( λmax ) = 86.71 < 180 = λperm.

SAFE

Column Buckling Curve = a (As per Table 10)Imperfection Factor (α1) = 0.21 (As per Table 7)Imperfection Factor (α2) = 0.34Moodification factor for effective length (MF)MF = 1 (not required as the column is not built up (Cl. 7.7.1.4))

Leffx = 3.132 mLeffy = 2.61 m

Modified Slenderness ratio in major direction (λx) = lx / rxx

= 3132/181.5= 17.26

Mopdified Slenderness ratio in minor direction (λy) = lY / ryy

= 2610/30.1= 86.71= 2543.77 MPa

Euler Buckling stress in major dir. (fccx ) = π2 ∗ Ε / λy2 (Cl. 7.1.2.1)

= π2∗200000/297.78= 6630.59 MPa

Euler Buckling stress in minor dir. (fccy ) = π2 ∗ Ε / λy2

= π2∗200000/7518.8= 262.60 MPa

λxx =(fy/fccx)0.5= (250/6630.59)^0.5 = 0.19 (Cl. 7.1.2.1)

λyy =(fy/fccy)0.5 = (250/262.6)^0.5 = 0.98φχ =0.5∗[1+α (λ−0.2)+λ2] = 0.518φγ =0.5∗[1+α (λ−0.2)+λ2] = 1.108

fcdx = (fy/γm0)/(φx +[φx 2-λxx 2]0.5)=Xfy/γm0≤fy/γm0 = 227.56 Mpa

fcdy = (fy/γm0)/(φy +[φy 2-λyy 2]0.5)=Xfy/γm0≤fy/γm0 = 139.20 Mpa

Ndx = fcdx * Ax = 2099714.57 NNdy = fcdy * Ax = 1284443.59 N

Permissible compressive Stress fcd = 139.20 MpaDesign Compressive Strength.= 139.21x9227/1000 = 1284.44 KN

BENDING STRESS :(for Laterally Unrestrained Case: )(Cl. - 8.2.2)fcr,b = 365.85Mcr = 662216997.19λLT 0.76 ≤ 0.78223981

= 0.83

λLT = 0.83 ≥ 0.4

therefore the member is to be considered as laterally unsupported

Check whether webs are susceptible to Shear Buckling before Yielding

Shear force = 223019.00 NPermissible Shear Force = 1506884.20 NRatio V/Vd = 0.15 < 0.6Therefore the webs are not susceptible to shear buckling before yielding

βb = Zp/Ze = 1 (for Plastic Section)

αLT = 0.21 (for rolled steel section - Cl. 8.2.2)

φLT = 0.5[1+αLT(λLT-0.2)+λLT2] = 0.91

X LT = 1/(φLT + [ φLT2 - λLT

2}0.5) = 0.78

fbd = X LT.fy/γm0 = 177.30

Design Bending Moment (Cl. 8.2.2)Mdx = βb*Zp*fbd = 271866985.46 N-mm = 271.87 KNmDesign Bending Stress = 201.28 MPa

Check of Resistance of the C/s to the combined action of axial compressive force & BM (Cl. 9.3.1.3)

Nd = Ag*fy/γmo = 2097045.45 N = 2097.05 KN

Basic Governing Equation (Cl. 9.3.1.3) -------- N/Nd + My/Mdy +Mz/Mdz <= 1.0

= (349.129 / 2099.72) + (280.178 / 271.87) = 1.197 ≥ 1 FAILS

Check for Overall Member Strength

ny = N/Ndy = 0.272nx = N/Nx = 0.166 ψ = = 0.00

Cmx = 0.6 (As per Table 18, Pg. 72)Cml = 0.4

λy = (fy/fccy)^0.5 = 0.976λx = (fx/fccx)^0.5 = 0.194

Ky = 1+(λy-0.2)*ny < 1+0.8*ny= 1.211 < 1.217

Kx = 1+(λx-0.2)*nx < 1+ 0.8*nx= 0.999 < 1.133

KLT = 1- 0.1*λLTny/(CmLT - 0.25) ≥ 1 - 0.1ny/(CmLT - 0.25)

= 0.850 ≤ 0.922KLT = 0.922

P/Pdy + KLT.Mx/Mdx ≤ 1.0 (Cl. 9.3.2.2)

1.222 ≥ 1.000 FAILS

P/Pdx + Kz.Cmx.Mx/Mdx ≤ 1.0 (Cl. 9.3.2.2)

0.784 ≤ 1.000 SAFE

MATERIAL PROPERTIES :Modulus of elasticity of steel = 200000 MPaYield stress of steel = 250 MPa

LOADS ACTING ON COLUMNS : Axial force (P) = 780 kNBending moment @ major axis (Mx) = 1073 kN-mBending moment @ minor axis (My) = 0 kN-mShear Force = 150 KN

GEOMETRY :Eff. Length @ x-axis (leffx) = 8.50 m

Eff. Length @ y-axis (leffy) = 3.50 m

SECTION PROPERTIES :Section details = Double ISMB400 BB 1000Area of Single Section = 7846.00 mm2

Moment of Inertia @ major axis of single section = 204584000.00 mm4

Moment of Inertia @ minor axis of single section = 6221000.00 mm4

Plastic Section modulus of section @ major axis = 1176180.00 mm3

Centre to centre distance between 2 ISMB (b1) = 1000.00 mmArea of Combined section (Ax) = 15692.00 mm2

Moment of inertia @ major axis (Ixx) = 3935442000.00 mm4 ########Moment of inertia @ minor axis (Iyy) = 409168000.00 mm4 ########Elastic Moduli of section @ major axis (Zexx) = 6904284.21 mm3


Plastic Modulus of section @ major axis (Zpxx) = 7846000.00 mm3

Plastic Modulus of section @ minor axis (Zpyy) = 2352360.00 mm3

Radius of gyration @ major axis (rxx) = 500.79 mmRadius of gyration @ minor axis (ryy) = 161.48 mm

Weight of section = 123.20 Kg / mThickness of flange (Tf) of I Section = 16.00 mmThickness of web (tw) = 8.90 mm

Overall depth (D) = 400.00 mmClear depth of web (d1) = 368.00 mm

DESIGN OF MAIN COLUMN (ISMB 400 BB 1000)IS 800-1984

x

y

b

b1

Effective Web Depth (h1) = 334.40 mm

Width of Flange of Individual I section (b) = 140.00 mm

SUMMARY OF DESIGN :

49.71 MPa SAFE155.42 MPa SAFE0.00 MPa SAFE

0.92 SAFE


= 780000 / 15692= 49.706857 MPa


= 1073000000 / 6904285= 155.41 MPa


= 0 / 511460= 0.00 MPa

CALCULATION OF PERMISSIBLE STRESSES : AXIAL STRESSES :


= 8500 / 500.8= 16.97


= 3500 / 161.48= 21.67

Maximum slenderness ratio ( λmax ) = 21.67

< 180 = λperm. SAFE

Elastic critical stress in major dir. (fccx ) = π2 ∗ Ε / λx2

= π2 ∗ 200000 / 288.09= 6844.89 MPa

Elastic critical stress in minor dir. (fccy ) = π2 ∗ Ε / λy2

= π2 ∗ 200000 / 469.8= 4197.36 MPa

Minimum elastic critical stress ( fcc ) = 4197.36 MPa

Permissible axial stress (σac ) 0.6 * fcc * fy

[(fcc)1.4 + (fy)

1.4 ] 1/1.4

0.6 * 4197.37 * 250 (as per clause 5.1.1)[(4197.37)^1.4 + (250)^1.4 ]^1/1.4

= 147.97 MPa Ratio of axial compression = σac,cal

σac

=

Bending stress @ minor axis 165.00 MPa

=

Bending stress @ major axis 163.88 MPa

Ratio of combined stresses 1.00

Axial compression stress 147.97 MPa


= 49.71147.97

= 0.336(as per clause 6.2.4)

BENDING STRESS :

Y = 26.5 * 105 / ( l / ryy)2

= 26.5 * 10^5 / ( 21.68 )^2 (as per clause 6.2.4)= 5640.70 MPa

X = Y * [1+ 1/20 * { (l/ryy)*(T/D) }2 ]0.5

= 5640.71 * [1+ 1/20 * { ( 21.68)*(16 / 400) }^2 ]^0.5= 5745.72 MPa


C2 = Distance Between NA and Bottom Extreme Fiber (as per table 6.3)

C1 = C2 (as per table 6.4)

K1 = 1.0 For ψ = 1.0 (as per clause 6.2.4)K2 = 0.0 For ω = 0.5

fcbx = K1 * ( X + K2 * Y ) * C1/C2

= 1 * ( 5745.73 + 0 * 5640.71 ) * 1.0= 5745.72 (as per clause 6.2.4.1)


If T/t > 2.0 or d1/t > 1344 / sqrt(fy) , fcbx = fcbx )

fcbx = 1.2 * 5745.73= 6894.87 MPa

0.66 x fcbx x fy

[(fcbx)1.4 + (fy)

1.4 ] 1/1.4

0.66 x 6894.87 x 250[(6894.87)^1.4 + (250)^1.4 ]^1/1.4 (as per clause 6.2.5)

= 163.88 MPaσbcy = 165.00 MPa

(as per 7.1.3) CHECK FOR COMBINED STRESSES : (as per 7.1.3)

Cmx = 0.6

Cmy = 0.6

σac,cal + σbcx,cal x Cmx + σbcy,cal x Cmy

σac 1 - σac,cal x σbcx 1 - σac,cal x σbcy

0.6 x fccx 0.6 x fccy

49.71 + 155.42 x 0.6 + 0 x 0.6147.97 1 - 49.71 x 164 1 - 49.71 x 165

0.6 x 6845 0.6 x 4197.4= 0.34 + 0.58 + 0.00= 0.91 < 1.00

SAFE

=

=

σbcx =

=

LOADS ACTING ON COLUMNS : Factored LoadsAxial force (P) 1067.917 kNBending moment @ major axis (Mx) 1505.79 kN-mBending moment @ minor axis (My) 0 kN-mShear Force 222.799 KN

SUMMARY OF DESIGN :

68.06 MPa SAFE

218.1 MPa SAFE

0 MPa SAFE

1.259 FAILS


= 1067.917/15692= 68.05 MPa


= 1505790000/6904284.22= 218.10 MPa


= 0/511460= 0.00 MPa

SECTION CLASSIFICATION (Table 2)For CompressionFlange b/Tf = 4.38 < 9.4 ε PlasticWeb d/Tw = 37.57 < 42 ε Semi-Compact

Semi CompactFor FlexureFlange b/Tf = 37.57 < 42 ε Semi Compact

d/Tw = 4.38 < 84 ε PlasticSemi Compact

ε = (250/fy)0.5


AXIAL STRESSES :


= 4500 / 161.48

= 27.87


= 11500 / 400.99

1.00

Bending stress @ major axis 227.28 MPa

CHECK

Axial compression stress 219.39 MPa

Bending stress @ minor axis 0 MPa

Ratio of combined stresses

IS 800-2007

TYPE OF STRESSES ACTUAL STRESS PERMISSIBLE STRESS

= 28.68

Maximum slenderness ratio ( λmax ) = 28.68< 180.00 = λperm. SAFE

Column Buckling Curve = c (As per Table 10)Imperfection Factor (α) = 0.49 (As per Table 7)

Moodification factor for effective length (MF)MF - 1.1 (for battened Column) (Cl. 7.7.1.4)

Leffx = 9.35 mLeffy = 3.85 m

Modified Slenderness ratio in major direction (λx) = lx / rxx

= 9350/500.8= 18.67

Mopdified Slenderness ratio in minor direction (λy) = lY / ryy

= 3850/161.48= 23.84= 2543.77 MPa

Euler Buckling stress in major dir. (fccx ) = π2 ∗ Ε / λy2 (Cl. 7.1.2.1)

= π2∗200000/348.59= 5664.14 MPa

Euler Buckling stress in minor dir. (fccy ) = π2 ∗ Ε / λy2

= π2∗200000/568.46= 3473.32 MPa

λxx =(fy/fccx)0.5 = (250/5664.15)^0.5 = 0.21 (Cl. 7.1.2.1)λyy =(fy/fccy)0.5 = (250/3473.32)^0.5 = 0.27φχ =0.5∗[1+α (λ−0.2)+λ2] = 0.525φγ =0.5∗[1+α (λ−0.2)+λ2] = 0.553fcdx = (fy/γm0)/(φx +[φx 2-λxx 2]0.5)=Xfy/γm0≤fy/γm0 = 226.10

fcdy = (fy/γm0)/(φy +[φy 2-λyy 2]0.5)=Xfy/γm0≤fy/γm0 = 219.38

Ndx = fcdx * Ax = 3548018.48 N

Ndy = fcdy * Ax = 3442573.99 N

Permissible compressive Stress, (fcd) = 219.38 MpaDesign Compressive Strength. = 3129.61 KN

BENDING STRESS :(for Laterally Unrestrained Case: ) (Cl. - 8.2.2)

fcr,b = 146503.79 N

Mcr = 27247573270.14 N-mm

λLT = 0.04 < 0.4therefore the member is to be considered as laterally supported

Check whether webs are susceptible to Shear Buckling before Yielding

Shear force = 222799 N

Permissible Shear Force = 1293264.60 NRatio V/Vd = 0.17 < 0.6Therefore the webs are not susceptible to shear buckling before yielding

βb = Zp/Ze

Design Bending Moment (Cl. 8.2.1.2)Mdx = βb*Zp*fy/γmo = 1569155502.39 N-mm = 1569.16 KNm

Design Bending Stress = 227.27 MPa

Check for overall buckling from bending consideration (Considering Push Pull effect)(As shown in Subramanian - Pg 1114)

Axial force due to BM in One I section = 1505790 N

Maxm. Compressive force in One I section = 2573707 N = 2573.71 KN

Compression resistance of the section = 3129612.72 N = 3129.61 KN

> 2573.71 KN

SAFE

Check of Resistance of the C/s to the combined action of axial compressive force & BM (Cl. 9.3.1.3)

Nd = Ag*fy/γmo = 3566363.64 N = 3566.36 KN

Basic Governing Equation (Cl. 9.3.1.3) -------- N/Nd + My/Mdy +Mz/Mdz <= 1.01067.917 / 3566.37 + 1505.79 / 1569.16 = 1.259 > 1 FAILS

Check for Overall Member Strengthny = N/Ndy = 0.310nx = N/Nx = 0.301

ψ = 0.0Cmx = 0.6 (As per Table 18, Pg. 72)

CmLT = 0.4λy = (fy/fccy)0.5 = 0.268λx = (fx/fccx)0.5

= 0.210

Ky = 1+(λy-0.2)*ny < 1+0.8*ny

= 1.021 < 1.248

Kx = 1+(λx-0.2)*nx < 1+ 0.8*nx

= 1.003 < 1.241

P/Pdy + Ky.Cmy.My/Mdy + KLT.Mx/Mdx <= 1.0 (Cl. 9.3.2.2)0.70 < 1.000

SAFE

P/Pdx + 0.6.Ky.Cmy.My/Mdy +Kx.Cmx.Mx/Mdx <= 1.0 (Cl. 9.3.2.2)0.879 < 1.000

SAFE

MATERIAL PROPERTIES :Modulus of elasticity of steel = 200000 MPaYield stress of steel = 250 MPaPartial Safety Factors γm0 = 1.1

γm1 = 1.25

LOADS ACTING ON COLUMNS : Bending moment @ major axis (Mx) = 145 kN-mBending moment @ minor axis (My) = 0 kN-m

GEOMETRY :Eff. Length @ x-axis (leffx) = 6.50 m

Eff. Length @ y-axis (leffy) = 1.10 m

SECTION PROPERTIES :Section details = ISMB 300+160x10 PLT(T/B)Area of section (Ax) = 8826.00 mm2

Moment of inertia @ major axis (Ixx) = 162942666.67 mm4

Moment of inertia @ minor axis (Iyy) = 11365666.67 mm4

Elastic Moduli of section @ major axis (Zexx) = 1018391.67 mm3


Plastic Moduli of section @ major axis (Zpxx) = 1147740.00 mm3

Plastic Moduli of section @ minor axis (Zpyy) = 126693.59 mm3

Radius of gyration @ major axis (rxx) = 135.87 mmRadius of gyration @ minor axis (ryy) = 35.89 mmWeight of section = 69.32 Kg / mThickness of flange (T) = 20.85 mmThickness of web (t) = 7.50 mmOverall depth (D) = 320.00 mmClear depth of web (d1) = 275.20 mmCentre to Centre distance of flanges = 298.34 mmShear Resisting Area (Av) = 2400.00 mm2

SUMMARY OF DESIGN :

142.39 MPa SAFE0.00 MPa SAFE

CALCULATION OF ACTUAL STRESSES :Actual bending stress @ major axis (σbcx,cal) = Mx / Zxx

= 145000000 / 1018392= 142.38 MPa


162.35 MPaBending stress @ minor axis 165.00 MPa

DESIGN OF BEAM IS 800-1984


Bending stress @ major axis

x

y

= 0 / 142071= 0.00 MPa



= 6500 / 135.88= 47.84


= 1100 / 35.89= 30.65



BENDING STRESS :Y = 26.5 * 105 / ( l / ryy)

2 (as per clause 6.2.4)

= 26.5 * 10^5 / ( 30.66 )^2= 2820.28 MPa

X = Y * [1+ 1/20 * { (l/ryy)*(T/D) }2 ] (as per clause 6.2.4)

= 2820.28 * [1+ 1/20 * { ( 30.66)*(20.85 / 320) }^2 ]

= 3088.75 MPa


C2 = Distance Between NA and Bottom Extreme Fiber

C1 = C2

K1 = 1.0 For ψ = 1.0 (as per table 6.3)

K2 = 0.0 For ω = 0.5 (as per table 6.4)

fcbx = K1 * ( X + K2 * Y ) * C1/C2 (as per clause 6.2.4)= 1 * ( 3088.76 + 0 * 2820.28 ) * 1.0= 3088.75


If T/t > 2.0 or d1/t > 1344 / sqrt(fy) , fcbx = fcbx ) (as per clause 6.2.4.1)

fcbx = 1.2 * 3088.76= 3706.50 MPa

0.66 x fcbx x fy

[(fcbx)1.4 + (fy)

1.4 ] 1/1.4

0.66 x 3706.51 x 250[(3706.51)^1.4 + (250)^1.4 ]^1/1.4

= 162.35 MPa

σbcy = 165.00 MPa (as per clause 6.2.5)

=

σbcx =

LOADS ACTING ON BEAMS : Factored Loads

Bending moment @ major axis (Mx) = 217.5 kN-mBending moment @ minor axis (My) = 0 kN-m

SUMMARY OF DESIGN :

213.58 MPa SAFE0.00 MPa SAFE

SECTION CLASSIFICATION (Cl. 3.7.4, Pg. 18-20)

Flange b/(Tf+Tp) = 3.56 < 8.4ε Plasticbe/Tp = 1.00 < 8.4ε Plasticbi/Tf = 14.00 < 29.3ε Plastic

Web d/Tw = 32.20 < 84ε Plastic

ε = (250/fy)0.5 Plastic

CALCULATION OF ACTUAL STRESSES :Actual bending stress @ major axis (σbcx,cal) = Mx / Zxx

= 217500000 / 1018392= 213.57 MPa


= 0 / 142071= 0.00 MPa



= 6500 / 135.88= 47.84


= 1100 / 35.89= 30.65



CHECK

Bending stress @ major axis 227.27

IS 800-2007

TYPE OF STRESSES ACTUAL STRESS PERMISSIBLE STRESS

Bending stress @ minor axis 227.27

BENDING STRESS : (considering beam as laterally Unsupported) : Cl. 8.2.2 - Pg. No. 54

fcr-bxx = 1.1*π2*E/(LLT/ry)2*[1+1/20*((LLT/ry)/(hf/tf))

2]0.5 = 2559.68 Mpa

fcr-byy 1.1*π2*E/(LLT/rx)2*[1+1/20*((LLT/rx)/(hf/tf))

2]0.5 = 10183.07

Mcr xx = π2*E*Iy*hf/(2LLT)*[1+1/20*((LLT/ry)/(hf/tf))2]0.5 = 3063642872.39 N-mm

= 3063.64 KN-m= βb*Zp*fcr-b = 2937.85 KN-m

Mcr xx = min. of above two values = 2937.85 KN-m

Mcr yy = π2*E*Ix*hf/(2LLT)*[1+1/20*((LLT/rx)/(hf/tf))2]0.5 = 3268208535.03 N-mm

= 3268.21 KN-m

= βb*Zp*fcr-b 1290.13 KN-m

Mcr yy = min. of above two values = 1290.13 KN-m

λLT xx = (βb*Zp*fy/Mcr)0.5 <= (1.2*Ze*fy/Mcr)0.5

= 0.313 <= 0.322λLT xx = (fy/fcr-b)0.5

= 0.313

λLT xx = 0.313

As λLT is < 0.4 , the beam can be considered as laterally supported.

λLT yy = (βb*Zp*fy/Mcr)0.5 <= (1.2*Ze*fy/Mcr)0.5

= 0.157 <= 0.182

λLT yy = (fy/fcr-b)0.5= 0.157

λλλλLT yy = 0.157As λLT is < 0.4 , the beam can be considered as laterally supported.

Check whether Webs are Susceptible to Shear Buckling before yielding (Cl. 8.2.1.1)Shear Force Acting = Mx*8/Leff2

= 41.18 KN/m

Shear force at ends (V) = 133846.15 N

Design Shear Strength = 314918.3286 NRatio = V/Vd = 0.43 < 0.6Therefore the webs are not susceptible to shear buckling before yielding

Design Moment

Mdxx = βb*Zp*fy/γm0 = 260850000.00 N-mm

= 260.85 KN-m > 217.5 KN-mSAFEPermissible Bending Stress = 227.27 Mpa

Mdyy = βb*Zp*fy/γm0 = 28793998.64 N-mm28.79 KN-m

Permissible Bending Stress 227.27 Mpa > 0 SAFE

Base Plate Design and Anchor Bolt Design

Input DataGrade of Concrete = 30 N/mm2Yield Stress of Steel 250 N/mm2Axial Force (P) = 266.67 KN

Base Plate dimensionLength (L) = 450 mmBreadth (B) = 450 mm

Pedestal DimensionsLength (Lp) = 500 mmBreadth (Bp) = 500 mm

Dimensions of the rectangle circumscribing the columnLength = 300 mmBreadth = 250 mm

Projection of slab base beyond the rectangle circumscribing the columna = 100 mmb = 75 mm

CalculationAllowable stress in steel ( σbs) = 185 (Cl. 5.4.3, pg.45, IS 800-1984) Stress under Base platew = Px 103/(L x B)

= 1.32 N-mmSupporting Area A1 = 250000 mm2

Loaded Area A2 = 202500 mm2

Allowable stress in conc. = sqrt(A1/A2)x0.25xfck (Cl.34.4,Pg.65, IS 456)8.33 > 1.32 SAFE

Thickness of base plate (Cl. 5.4.3, Pg. 45)

t = (3 x w (a2 - b2/4) / σbs)0.5 = 13.55 mm

LoadingAxial Force (P) = 400 KN

CalculationStress under Base plate w = Px 103/(L x B)

= 1.98 N-mmAllowable stress in conc. = 0.45 x fck (Cl.34.4,Pg.65, IS 456)

13.5 > 1.98 SAFE

Thickness of base plate (Cl. 7.4.3.1, Pg 47)t = (2.5 x w (a2 - 0.3b2) x γm0 / fy)0.5 = 13.44 mm

IS 800-2007

The following examples depict the general procedure for the design of base plates and anchor bolts (if required)as per WSM and LSM for various loading conditions at column end.

Base Plate ( Without Moment)IS 800-1984

w

B

L

Input DataGrade of Concrete = 25 MPa Yield Stress of Steel 250 MPa Column section ISHB 250Axial Force (P) = 333.33 KNMoment (M) = 30 KN-mShear Force (Fx) = 0 KNShear Force (Fy) = 0 KN




Projection of slab base beyond the rectangle circumscribing the columna = 145 mmb = 75 mm

Calculationeccentricity = 90 mmSection Modulus = 19440000 mm4

Maxm pressure at the base = P/A + M/Z= 3.09 MPa

Min Pressure at the Base = P/A - M/Z= 0.00 MPa

Pressure at the critical section of the column = 2.22Distance of Critical Section from the face pof the Plate = 151.25 mm

Supporting Area A1 = 360000Loaded Area A2 = 216000Allowable stress in conc. = sqrt(A1/A2)x0.25xfck (Cl.34.4,Pg.65, IS 456)

8.07 > 3.09 SAFE

= 32007.27 N-mm

Allowable stress in steel ( σbs) = 185 MPa (Cl. 5.4.3, pg.45, IS 800-1984)

Thickness = sqrt(6 M / σbs)

= 32.22 mm

Moment at Critical Section of Column considering Cantilever effect and Trapezoidal Loading

Base Plate ( With Moment but no tension at base)IS 800-1984

w

B

L

Input DataColumn section ISHB 250Axial Force (P) = 500 KNMoment (M) = 45 KN-mShear Force (Fx) = 0 KNShear Force (Fy) = 0 KN

Calculationeccentricity = 90 mmSection Modulus = 19440000 mm4

Maxm pressure at the base = P/A + M/Z= 4.63 N/mm2

Min Pressure at the Base = P/A - M/Z= 0.00 N/mm2

Pressure at the face of the column = 3.39

Allowable stress in conc. = 0.45 x fck (Cl.34.4,Pg.65, IS 456)11.25 > 4.63

SAFE

= 44312.8072 N-mm

Allowable BM = Zp x fy / γm0

Thickness = sqrt(M x γm0 / (Zp x fy)) < sqrt(M x γm0 / (1.5/6 x fy))

27.93 mm

IS 800-2007

Moment at Face of Column considering Cantilever effect and Trapezoidal Loading

Input DataGrade of Concrete = 30 N/mm2

Yield Stress of Steel 250 N/mm2

Column section W310x310x143Axial Force (P) = 475.00 KNMoment (M) = 333.33 KN-mShear Force (Fx) = 0 KN

Shear Force (Fy) = 0 KN




Projection of slab base beyond the rectangle circumscribing the columna = 212.5 mmb = 162.5 mm

Bolt Diameter (φ) = 30 mmNo. Of bolts = 3 no.Edge Distance = 50 mm

= 325 mmDistance of Tension Bolt from Opp Edge (d) = 700 mmNet Area of Bolt = 561 mm2

(from IS 1367-1967 & IS 1364-1967)

Permissible Bond Stress (τb) = 0.9 N/mm2(Table 21, IS 456-2000)

Calculationeccentricity = 701.75 mmSection Modulus = 60937500 mm4


Min Pressure at the Base = P/A - M/Z= -4.50 N/mm2

= 0.00Distance of Critical Section from the face of the Plate = 220.63 mm

Supporting Area A1 = 1100000 mm2

Loaded Area A2 = 487500 mm2

Base Plate ( With Moment and tension at base)

Pressure at the critical section of the column

IS 800-1984

c/c Distance of Bolt hole and CG of

T

k=n

w

d

g g

B

L

Allowable stress in conc., (w ) = sqrt(A1/A2)x0.25xfck (Cl.34.4,Pg.65, IS 456)11.27 > 6.44 N/mm2 SAFE

Taking Moments @ centre of Tension Bolt0.5nd2pL(1-n/3) = P.d + M

n = (3 - sqrt(9 -24*((Pg+M)/(d2wB)))/2 = 0.30Length of loaded area , k = nd = 211.61 mm

Balancing Vertical ForcesForce in Bolt , T = 0.5 kw B - P = 299800.20 N

M = 182792.55 N-mm per mm width of the plate

Allowable stress in steel ( σbs) = 185 N/mm2(Cl. 5.4.3, pg.45, IS 800-1984)

Thickness = sqrt(6 M / σbs)

= 77.00 mm

Anchor Bolt DesignAllowable Tensile Steress in Bolts = 120 N/mm2 (Table 8.1, IS 800 1984)Tension Capacity of 1 Bolt = 561 x 120 = 67320 N/mm2No. of Bolts Reqd. = 299800.21 / 67320 = 5 no.

Tension in each bolt = 299800.21 / 5 = 59960.04 N

Bond force produced = πφlτb = T

Length of Anchor Bolt Required = T/(πφτb) = 707.24 mm

Provide 10 no. of 750 mm long, 30 mm dia. bolts.

Moment at Section of Column considering Cantilever effect and Trapezoidal Loading

Input DataColumn section W310x310x143Axial Force (P) = 712.5 KNMoment (M) = 500 KN-mShear Force (Fx) = 0 KNShear Force (Fy) = 0 KN

Design Bond Stress of concrete (τbd) = 1.5 N/mm2(Cl. 26.2.1.1, IS 456-2000)

Ultimate tensile stress of bolt = 400 N/mm2 (as per IS 1367-Part 3)Yield Tress of the bolt = 240 N/mm2 (as per IS 1367-Part 3)





Allowable stress in conc. = 0.45 x fck (Cl.34.4,Pg.65, IS 456)13.5 > 9.67 N/mm2 SAFE

Length of Loaded Area considering Plastic Deformation

y = L/2+a-[(L/2+g)2-2x(M+Pg)/(0.45xfckxB)]0.5

131.44 < 212.5< 750 SAFE

= 260450.73 N-mm per mm width of the plate

Allowable BM = Zp x fy / γm0

Thickness = sqrt(M x γm0 / (1/4 x fy)) < sqrt(M x γm0 / (1.5/6 x fy))

67.70 mm (Cl.

Anchor Bolt DesignForce in Anchor Bolt = 0.45 x fck x y x B - P

= 440873.39 NTensile Capacity of 1 30 mm dia bolt = 0.9 x fub x An / gmb < fyb x Asb / gm0 (Cl. 10.3.5)

0.9*400 x 561/1.25 < 240x561/1.1 = 122400 N

No. of bolts required , (nb) = 4 no.

Length of Anchor Bolt Required(Concrete break out failure in Tension Criteria) = ((Nu/nb)/(k x sqrt(fck)))1/1.5 = 119.01 mm

(Pull out criteria) = φ x σs /(4 τbd) = 982.34 mmProvide 8 no of 1000 mm long, 30 mm dia bolts.

Moment at Face of Column considering Cantilever effect and Rectangular Loading

IS 800-2007

w

y

T

P

M

d

Input DataGrade of Concrete = 30 N/mm2Yield Stress of Steel 250 N/mm2Column section W310x310x143Axial Force (P) = 300.00 KNMoment (M) = 166.67 KN-mShearForce (Fx) = 30 KNShearForce (Fy) = 33.33 KN




Projection of slab base beyond the rectangle circumscribing the columna = 97.5 mmb = 97.5 mm

Bolt Diameter = 30 mmEdge Distance = 50 mmc/c Diatance of Bolt hole and CG of Column (g) = 210 mmDistance of Tension Bolt from Opp Edge (d) = 470 mmNet Area of Bolt (from IS 1367-1967 & IS 1364-1967) = 561.00 mm2

Permissible Bond Stress (τb) (Table 21, IS 456-2000) = 0.9 N/mm2




= 5.25= 105.63 mm

Supporting Area A1 = 360000Loaded Area A2 = 270400Allowable stress in conc. = sqrt(A1/A2)x0.25xfck (Cl.34.4,Pg.65, IS 456)

8.65 > 8.22 SAFE

Gusseted Base Plate ( With Moment and tension at base)

Pressure at the critical section of the columnDistance of Critical Section from the face of

IS 800-1984

k = nd

T

P

M

d

w

g g

h

a

a

b

Taking Moments @ centre of Tension Bolt0.5nd2pL(1-n/3) = P.d + M

n = (3 - sqrt(9 -24*((Pg+M)/(d2wB)))/2 = 0.57Length of loaded area = k = nd = 268.19 mm

Moment at Section of Column considering Cantilever effect and Trapesoidal Loading= (5.25+2*8.65)/(5.25+8.65)*(105.63/3)*(5.25+8.65)/2*105.63 = 41936.467 N-mm

= 20968.23357 N-mmShear Force = 176175.282 N

Allowable stress in steel ( σbs) = 185 MPa (Cl. 5.4.3, pg.45, IS 800-1984)

Assume: Gusset Plate Thickness = 16 mm

Height of Gusset = 6.52 mm200 mm SAY

Let us adopt the gusset plate of size 200 x 97.5 x 15 mm

Moment of resistance = (1/6xtxd2xσbs)

= 19733333.33 > 20968.234N-mm N-mm

SAFE

Design of Thickness of Base plate(Considering the base plate to act as Simply Supported Beam with overhangs on 2 ends)Moment at the face of gusset = 28740.50481 N-mmMoment at the centre of the SS span = 85517.30769 N-mm

Plate thickness required = sqrt(M/(1/6* σbs))

52.66 mm

Anchor Bolt DesignBalancing Vertical ForcesForce in Bolt , T = 0.5 kwB - P = 303428.01 N

Allowable Tensile Stress in Bolt = 120 N/mm2 (Table 8.1, Pg. 95, IS 800 1984)Tension Capacity of 1 Bolt = 561 x 120 = 67320 NNo. of Bolts Reqd. on Tension Face = 303428.02 / 67320 = 5 no.

Tension in each bolt = 303428.02 / 5 = 60685.602 N

Bond force produced = πφlτb

Length of Anchor Bolt Required = T/(πφτb) = 715.80 mm

Provide 10 no. of 750 mm long, 30 mm dia. bolts.

Moment about column face for 1 gusset

Input DataColumn section W310x310x143Axial Force (P) = 450 KNMoment (M) = 250 KN-mShearForce (Fx) = 45 KNShearForce (Fy) = 50 KN

Design Bond Stress of concrete = 1.5 N/mm2 (Cl. 26.2.1.1, IS 456-2000)Ultimate tensile stress of bolt = 400 N/mm2 (as per IS 1367-Part 3)Yield stress of the bolt = 240 N/mm2 (as per IS 1367-Part 3)

Calculationeccentricity = 555.56 mmSection Modulus = 23434666.67 mm4




Allowable stress in conc. = 0.45 x fck (Cl.34.4,Pg.65, IS 456)13.5 > 12.33

SAFE

Length of Loaded Area considering Plastic Deformationy = L/2+d-[(L/2+d)2-2x(M+Pd)/(0.45xfckxB)]0.5

119.64 > 97.5< 520

SAFEMoment at Face of Column considering Cantilever effect and Rectangular Loading

= 33366937.50 N-mm

Assume: Gusset Plate Thickness = 20 mm(Considering The Gusset Plate as the outstanding element of compression Flangeb/tf = 4.875 < 13.6h = 272 mm

300 mm - SAY

Let us adopt the gusset plate of size 200 x 97.5 x 15 mm

= 16683468.75 N-mmShear Force = 419937.7913 NCheck Vp = 0.6xAvxfyw/(3)0.5/γm0

= 472377.5 > 419937.79(Cl.8.2.1, Pg. 52-53) SAFETherefore the gusset plate is not succeptible to shear buckling

Moment of resistance = (1/6xtd2fy/γm0) (Considering Elastic limit is not exceeded)= 68181818 > 16683469

N-mm N-mm

IS 800-2007

(again considering it as an outstanding element of

Moment about column face for 1 gusset

w

y

T

P

M

d

SAFE

Design of Thickness of Base plate Moment at the face of gusset = 40542.19 N-mmMoment at the centre of the SS span = 137700 N-mm

Plate thickness required = sqrt(M x γm0 /(1/4* fy))

49.23 mm

Horizontal Shear CheckResultant Horizontal Shear = (Fx2+Fy2)0.5 = 67.27Permissible horizontal Shear = 0.45*P = 202.5 > 67.27

KN SAFE KN

Anchor Bolt Design

force in Anchor Bolt, (Nu) = 0.45 x fck x y x B - P389875.58 N

Tensile Capacity of 1-30 mm dia bolt. = 0.9 x fub x An / γmb < fyb x Asb / γm0 (Cl. 10.3.5)(Cl. 10.3.5) = 0.9*400 x 561/1.25 < 240x561/1.1 = 122400 N

No. of Bolts Reqd. on Tension Face = 389875.59 / 122400 = 4 no.

Length of Anchor Bolt Required(Concrete break out failure in Tension Criteria) = ((Nu/nb)/(k x sqrt(fck)))1/1.5 = 109.64 mm

(Pull out criteria) = φ x σs /(4 τbd) = 868.71 mm

Provide 8 no of 900 mm long, 30 mm dia bolts.

Observation: as the Concrete strength goes on reducing the thickness of the base plate goes on reducing

Conclusion: The above examples shows a general approach towards analysing/designing of members through IS 800-2007 based on Limit state design. It can be concluded here that the Limit state approach tends to reduce tension member size in general and increase the size of compression member accompanied with bending, while designing. Though a generalised concept guides us that Limit state theory leads to smaller member size, the same does not hold the ground for steel design according to IS 800-2007.The main contributing factor for the same seems to be a large Factor of Safety and Various Load combinations as mentioned above. Also in the present code the Live load has been bifurcated into two parts namely, Leading and Accompanying Live Load. The example taken here being that of an industrial structure that has sufficiently large amount of Crane load for it to be considered as Leading Live Load. This consideration also amplifies the Loading on the structures as the FOS for the Leading Live load is more than that of Accompanying Live load. The following table shows that Stress utilisation ratio which is the ratio of Actual Stress (Force) to Permissible Stress (Force) which signifies percentage of material utilised for resisting external forces. Out of a total of 15 Cases, 10 Cases indicated that Limit state Method based IS 800-2007 yields results on a higher side as compared to working stress approach based IS 800-1984

Stress Utilisation Ratio

Section Action Working Stress Method

Limit State Method

Remarks

Tension 0.63 0.95 LSM is Conservative 2 ISA 50x50x6 Compression 0.96 0.73 Tension 0.91 0.41 2 ISA 65x65x6 Compression 0.58 1.07 LSM is Conservative Tension 0.97 0.77 2 ISA 75x75x8 Compression 0.80 1.03 LSM is Conservative Tension 0.07 0.07 LSM is Conservative ISMB 150 Compression 0.09 0.10 LSM is Conservative Compression 0.23 0.31 LSM is Conservative Bending 1.00 0.96 ISMB 450 Combined Forces 0.84 1.26 LSM is Conservative Compression 0.34 0.31 Bending 0.95 0.96 LSM is Conservative A ISMB 400 BB 1000 Combined Forces 0.92 1.26 LSM is Conservative

ISMB300+160x10 PLT(T/B) Beam Bending 0.88 0.94 LSM is Conservative The same does not hold true in case of Base Plate and Anchor bolt design, where the results of Limit State Design, which yield less thickness of the base plate as compared to Working Stress approach.

Thickness Type

Working Stress Method Limit State Method Base Plate Without Moment 13.55 13.44 Base Plate With Moment but no Tension 32.22 27.93 Base Plate With Moment & Tension 77.00 67.70 Gusseted Base plate- With Moment & Tension 52.66 49.23

While carrying out the above exercise it was realised that though the newly developed IS code is quite elaborate, there are some areas that still needs some light to be thrown on, which are as mentioned below 1) Thickness of Flange and centre to centre distance between flanges in Built up members:

In the Clause 8.2.2.1, for finding out the value of Mcr, thickness of flange and the centre to centre distance between flanges are required, but the code is yet to provide any guidelines.

2) For Primary compression members , accompanied with bending moment, the code recommends considering Semi Compact and Slender sections only, but for calculation of Section strength of such section , guidelines are given for Plastic and Compact sections. The code is yet to throw any light as to which sections shall be considered in this category.

3) Elastic Lateral Torsional Buckling capacity has to be calculated for beams and columns not restrained laterally, but the code remains silent on the method of calculating the same for unsymmetrical sections and sections symmetric about major axis.

4) For calculation of bearing pressure for size of gusset plate larger than required, a parameter “c” has been introduced, but the code is yet to come up with the guidelines for calculating the same.

Inspite of a few shortcomings (some of which have been mentioned above) in the new code which would be taken care of in the future amendments, the code has explained and given guidelines for almost each and every thing that are required in day to day’s work. Though the results obtained by the new code seems to be conservative as compared to the previous version of the code (which may be because of more importance given to Local Buckling Criteria), the Limit State Method ( as adopted by the new version of the code ) ensures that local failures in the member section are avoided and the structure lives upto its designed life.

a comparative study is800

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