SCI ADVISORY DESK

AD 238: Bearing capacity of connected parts with slotted holes

The SCI recently carried out an experimental investigation with the aim of devising design guidance on the bearing capacity of plates with slotted holes. The conclusion was that plates with slotted holes had a lower bearing capacity than standard clearance holes. Therefore, it is recommended that when ordinary bolts are used and the plate is loaded

perpendicular to the direction of the slot the bearing capacity Pbs is reduced by a reduction

factor kbs taken as:

- 0.7 for short slots (d + 6 mm)

- 0.5 for all other lengths up to 3.5d

Bearing capacity covers several different possible failure modes, including possible shear-out or tear-out at the end of the plate as well as actual bearing stresses, but the criterion that is most affected by using slotted holes is the deformation of the connection under

service loads. The reduction factors kbs have been determined so that the deformation of the plate at working load is limited to 1.5 mm, which is in line with the current BS 5950-1 recommendations for clearance holes.

For slotted holes that are designed to act as part of an expansion joint, it is not sufficient to apply these reduction factors. For an expansion joint to slide freely under load, the bearing stresses and deformations permitted by these recommendations (or the existing BS 5950-1 checks for ordinary holes) will be too high to allow efficient operation. A reduced bearing stress based on the Hertz contact pressure would be more appropriate for such joints.

On the other hand, connections with preloaded HSFG bolts are designed not to slip under service loading, so deformation under service loading is not considered in their bearing capacity. Provided that sufficient end distance is available (generally about 3d), much higher bearing capacities are recommended in BS 5950-1 in its check on capacity after slipping into bearing. In such cases the proposed bearing capacity reduction factor for slotted holes need not be applied (though the current reduction factors for HSFG bolts in slotted holes still apply).

Based on the tests carried out, the report recommends that the bearing capacity Pbs is based on:

Pbs = kbs d t pbs

where d is the bolt diameter t is the plate thickness

pbs is the bearing strength of the plate

kbs is a reduction factor

Subject to the limits investigated in the tests:

- 1.4D end distance < 3.0d - edge distance 1.25D - pitch 2.5d

SCIReferences to BS 5950 refer to the 1990 edition. Please note that this has been superseded by the 2000 edition (BS 5950-1: 2000).

SCI ADVISORY DESK

and proposes that the reduction factors be increased to 1.0 when end distance is three times the bolt diameter or greater.

In making proposals to the BSI committee for inclusion in an amendment to BS 5950 : Part 1, the recommendations have been reviewed and refined. The scope has been extended to

include oversize holes, and the kbs factors have been applied to the current BS 5950-1 capacity, including the effect of end distance. This provides a sliding scale and avoids the need for a sudden increase at an end distance of 3d. In addition, it provides a more robust

solution for the small number of cases where the experimentally determined value of kbs would otherwise be less than 0.5. It also provides a simple safe solution to a subsidiary problem not investigated in the tests, the possible effect of slotted holes on shear-out.

In summary, the recommendation is to determine Pbs using:

Pbs = kbs d t pbs but Pbs 0.5 kbs e t pbs

where kbs has the value:

1.0 for normal holes

0.7 for short slotted holes or oversize holes

0.5 for long slotted holes up to 3.5d.

SCI ADVISORY DESK

AD 239: Direct Tension Indicators (load indicating washers)

Readers who look at the small print may have read note 12 under Table 1 from AD 226 and wondered what Direct Tension Indicators are. Direct Tension Indicators are specified in BS 7644: 1993: Part 1 for compressive washers and Part 2 for nut face and bolt face washers.

The note following Clause 1 of Part 1 reads Compressible washer-type direct tension indicators are also know as load indicating washers.

The requirements of Part 1 generally follow the advice given over many years by Cooper and Turner on the use of their Coronet load indicating washers, average indicator gaps being the same but with the possibility of tightening by bolt rotation added. However, this method of tightening is not generally the preferred approach and might require a bolt face washer. Being little used, these washers are not produced as a stock item.

Nut face washers and bolt face washers to Part 2 are very similar, the difference being that the hole in the bolt face washer is slightly bigger. Nut face washers are marked with an indented letter M on one face and bolt face washers with the letters BM.

Annex A of Part 1 describes the use of these indicators and refers to Figures 2 & 3 for the correct assembly for each method of tightening. These figures are reproduced below.

Gap

Indicatorwasher

Beforetightening

a) Under bolt head fitting

Indicatorwasher

Beforetightening

b) Under nut fitting

GapNut facewasher

Through-hardenedwasher

Indicatorwasher

Nut facewasher

Gap

Indicatorwasher

Beforetightening

Beforetightening

a) Under nut fitting

Gap

b) Under bolt head fitting

Bolt facewasher

Through-hardenedwasher

Fig. 2: Bolt tightening by rotation of the nut (normal method of assembly

Fig. 3: Bolt tightening by rotation of the bolt (alternative method of assembly)

It should be noted that average indicator gaps are smaller for the cases shown in Figures 2(b) & 3(b)

SCI ADVISORY DESK

Note: See also AD226

SCI ADVISORY DESK

AD 240: Imposed Load Reduction When Designing Multi-storey Columns

A recent question concerned the application of load reduction when designing multi-storey columns. The query came to light because the engineer was using the multi-storey design example at the end of Chapter 2, Sheet 4, in the Steel Designers Manual (SDM) 5th Edition and had been told (incorrectly) by his checking engineer that this did not properly apply the load reduction given in Table 2 of BS 6399: Part 1: 1996 Specified loading for building: code of practice for dead and imposed loads.

What BS 6399 requires can be illustrated using the simple example in Figure 1.

Fig. 1: Column in a typical multi-storey building

Consider the column between second and third floor levels (marked No 1 in Figure 1). This column carries four floors. According to Table 2, a 30% reduction can be applied to the imposed loadings on all four floors (including the roof) in the design. However, consider also the column at ground level (marked No. 2 in the Figure); it carries six floors and therefore a 40% reduction can be applied to the imposed loads on all six floors carried by this column.

As the design of the various columns throughout a multi-storey building proceeds (downwards, from the roof), the percentage reduction to the imposed loading allowed by Table 2 increases. This means that the design load of a particular column is not simply the load (after reduction) of the column above plus the reduced loading on the floor above; the unreduced load must first be totalled and then the appropriate reduction should be applied. This is the correct way to apply Table 2, and will give a greater reduction than summing reduced loads.

This might appear to require a lot of design work, especially for hand calculations, but the multi-storey design example (Sheet 4) in the SDM shows that it can be carried out easily by a tabular approach. The sheet shows, for example:

First floor

Roof

Fifth floor

Fourth floor

Third floor

Second floor

Ground Level

Column No. 1

Column No. 2

SCI ADVISORY DESK

1. At level 6, for column 1: the column imposed loads (unreduced) for the six floors are given in the third column of the Table as 40.4, 189.2, 152, 152, 152, 152, totalling 837.6 kN. Applying the 40% reduction for 6 floors, gives a value of 502.56 kN, which agrees with the value in the seventh column of the Table.

2. At Level G, for column 2: the column imposed loads (unreduced) for the twelve floors are given in the fourth column of the table as 72.1, 334.9, 269.2 x 10, totally 3099 kN. Applying the 50% reduction for over ten floors, gives a value of 1549.5 kN, which agrees with the value in the eighth column of the Table.

Needless to say, the reductions should never be applied to any dead loading.

SCI ADVISORY DESK

AD 241: Transverse Reinforcement in Composite Beams

Transverse reinforcement in the form of mesh or additional loose bars is required in composite beam design to transfer the longitudinal shear force from the shear connectors into the effective width of the slab. Traditionally, light mesh reinforcement has been used throughout the slab, as a deemed to satisfy approach, but design to BS 5950: Part 3 and Eurocode 4 can lead to the requirement for additional reinforcement, particularly within the high shear regions of primary beams.

The question often posed to SCI by designers is: what is the effect of light mesh reinforcement on the longitudinal shear resistance, and therefore on the bending resistance of a beam, when this reinforcement is less than that required according to code rules for transverse reinforcement? There are two important aspects to this question:

1. Often, composite beams are not designed to their full bending resistance, Mc, because serviceability limits control the design. If the utilisation factor on bending resistance at the ultimate limit state is given by the factor M/Mc, it may be shown that the longitudinal shear force transferred between the beam and the slab reduces conservatively in proportion to M/Mc.

2. Failure by concrete splitting along the line of the shear connectors is a ductile mode of failure, provided the amount of reinforcement is above a certain minimum. In principle, the effective shear resistance of the shear connectors may be down-rated according to the amount of transverse reinforcement provided, rather than having to provide sufficient transverse reinforcement to equate to the design resistance of the shear connectors.

The minimum amount of transverse reinforcement that is provided should be sufficient to achieve the minimum degree of shear connection for the beam span multiplied by the utilisation factor, M/Mc (see below).

These observations are based on full-scale composite beam tests at the University of Cambridge in which only light mesh reinforcement was provided in the primary beam tests. In these 9 m span tests, the secondary beam connections also acted as effective transverse ties to control splitting. In design terms, the minimum longitudinal shear force that should be transferred by the concrete splitting mode of failure is conservatively given by:

Fq,r Fq (M/Mc) Kmin

where Fq,r is the longitudinal shear force per unit length due to the splitting mode of failure, as influenced by the transverse reinforcement that is provided (and any anchorage effect of the steel decking crossing the beams). Fq is the longitudinal shear force per unit length due to the shear connectors, as required for full shear connection, and M/Mc is the utilisation of bending resistance at the ultimate limit state. Kmin is the minimum degree of shear connection required for a symmetrical steel section for a beam span, L, in metres (where Kmin (L-6)/10 but Kmin 0.4).

If this equation is not satisfied, the ratio M/Mc should be reduced so that the magnitude of Fq,r that is determined from the transverse reinforcement provided equates to the right hand side of the equation.

SCI ADVISORY DESK

For example, in the case of 10 m span beam, Kmin = 0.4, and for a typical design M/Mc = 0.8. Therefore, Fq,r should exceed 0.32 Fq. If Fq,r is inadequate, additional reinforcement is required.

Assuming that full shear connection is based on the compressive resistance of the slab and the beam is subject to 2 point loads, it may be shown that the minimum percentage of reinforcement in the shear plane should be at least 0.2% of the cross-sectional area of the slab (in this case the concrete topping). It follows that A193 mesh would be sufficient as transverse reinforcement. Therefore, in practical design cases, additional bar reinforcement would not be necessary.

This simplified approach gives designers a quick and safe way of determining whether a beam designed with a small amount of transverse reinforcement is adequate. If not, a more sophisticated analysis should be performed, based on the partial shear connection principles.

q;bTpW fl W

'0rm03s')IC _______

) cB r B

SCI ADVISORY DESK

AD 242: Varying Depth Beams: BS 5950 : Pt 1 Appendix B3

Recently a question arose concerning the use of Appendix B3 in BS 5950 : Part 1 in checking a beam of varying depth for lateral torsional buckling. On inspection of Appendix B3 this proves more complex than appears at first sight. Consider the following beam subjected to any bending moment arrangement with a maximum value of Ma.

In order to check for lateral torsional buckling, Appendix B3 clearly states that pb (bending strength) must be determined from the cross section properties at the point where maximum moment occurs in the beam. This may be at the maximum or minimum depth, or some intermediate point, and N, u, v, x and ry must be taken from this point in order to determine pb. Once this has been done, the appropriate value of Sx (plastic modulus about the major axis) must be determined for equation 4.3.7.3, Mb=Sx pb.

In fact, Sx must be taken from the critical cross section in the beam, i.e. where the bending

stress

xZM

has its maximum value. This may not be at the point of maximum moment. To

find the critical cross section, the bending stress must be calculated at several locations to find the point of maximum value. When the critical cross section has been located, Sx is determined for this location using the gross section properties at that point. This is what is meant by the sentence in Appendix B3 which states This value of pb applies throughout the length between adjacent restraints.

It is intended to make this issue clearer in the forthcoming amendment to BS 5950 Part 1.

Fig. 1: Beam length L between points of lateral restraint.

m=1.0 (see Note 1,Table 18, BS 5950 : Pt 1) M_

= m Ma=Ma

SCI ADVISORY DESK

AD 243: Splices within unrestrained lengths

Ideally a splice in a compression member, or a laterally unrestrained beam, should be positioned close to a restraint, or, if the member is continuous, at a point of inflexion of the buckled shape. If the splice cannot be located at such a point, it may be possible to achieve the desired effect by adding an additional restraint instead. Nevertheless it is sometimes unavoidable to locate a splice within the unrestrained length of a member, and the purpose here is to explain how to design such splices. The procedures given are based on those expected to be included in the forthcoming amendment to BS 5950: Part 1.

The first thing to note is that these design procedures only affect the splice. The member itself is not affected, provided that it has been correctly designed in the first place and that the recommended procedure is used for the splice.

Design requirements

The basic requirements are that the splice must be both:

stiff enough to avoid reducing the buckling resistance of the member below that required, and also

strong enough to transmit the forces and moments in the member.

Providing a splice that is more flexible than the member itself is considered to be bad practice, because it would be liable to be both uneconomic and potentially unsafe. In any case it would be a complex matter to calculate the buckling resistance of a member in which the splice did not have at least the same stiffness as the member.

It is recommended, if possible, to use a splice with cover plates sized to provide adequate stiffness, by making the moment of inertia of the splice material at least as great as that of the member, considering both axes. This is normally relatively easy to achieve by providing at least as much area in the cover plates as in the relevant element of the member cross-section.

In a splice connection with end plates it is generally more difficult to achieve the same stiffness as the member. An accurate elastic analysis of the connection should be used to verify that it is at least as stiff as the member. It is likely that relatively thick end plates will be needed. Extended end plates may be required if there is a significant moment. Even where a splice connection is entirely in compression, it is advisable to maintain full continuity of stiffness through the connection to safeguard the robustness of the structure.

Whichever type of splice is adopted, preloaded HSFG bolts should be used. In the case of cover-plate splices, they should be designed to prevent slip under ultimate limit state loading, by using a coefficient of 0.9 rather than 1.1 in the formula for slip resistance. An alternative would be to use fitted bolts.

The capacity of the splice connection needs to be checked for all the forces and moments in the member at that point. Besides the internal forces and moments derived from equilibrium with the applied loads, allowance must be made for the following second order effects, as outlined below:

a) moments due to strut action; b) moments due to lateral-torsional buckling, and c) moments due to amplification.

SCI ADVISORY DESK

For an explanation of these effects see AD 244.

Design procedures:

Note: In addition to the procedures that follow, any localised forces and moments resulting from the detailed design of the splice, such as those due to the transfer of shear across the connection, or due to eccentricity between the centroidal axes of the sections either side of the splice, should also be included.

1) Splice for members subjected to compression only:

Design forces and moments for the splice:

Fc,splice = Fc applied force Mx,splice = [Ms(x-x)] (a) strut action major axis My,splice = [Ms(y-y)] (a) strut action minor axis

[Include only one of the terms in square brackets - the one with the most onerous effect.]

where:

Ms is a strut action moment, obtained from: Ms = Mmax sin (180 Lz/LE)

[Ms should be calculated for major (Ms(x-x)) and minor Ms(y-y)) axes, using LEx and LEy with the respective values for Lz];

Lz is the distance along the member to the nearest point of inflexion; LE is the effective length (i.e. the distance between points of inflexion); Mmax is the maximum internal moment due to strut action, given by: Mmax = (py/pc 1) fc S fc is the compressive stress due to axial force; pc is the compressive strength; py is the design strength; S is the plastic modulus of the strut for bending about the axis of buckling.

2) Splice for members subjected to major axis moments only.

Design moments for the splice:

Mx,splice = Mx applied major axis moment at the splice My,splice = Mys (b) lateral-torsional buckling (LTB) where: Mys is the additional minor axis moment at a distance Lz along the member

from the nearest point of inflexion, which is obtained from: Mys = My,max sin (180 Lz/LE)

The additional minor-axis moment (an internal second-order moment equivalent to the strut action moment in a compression member) in a member subject to major axis moment should be taken as having a maximum value My,max midway between points of inflexion of the buckled shape (the points between which the effective length LE is measured), given by:

My,max = (py/pb 1) (Mcy/Mcx) mLT Mx,max

where:

Mcx and Mcy are the major and minor axis moment capacities of the section, for zero shear;

SCI ADVISORY DESK

Mx,max is the maximum major axis moment in the length LLT between lateral restraints; mLT is the equivalent uniform moment factor for lateral-torsional buckling; pb is the bending strength for resistance to lateral-torsional buckling.

3) Splice for members subjected to combined axial compression and bending:

Design forces and moments for the splice:

Fc,splice = Fc applied Mx,splice = Mx + [Ms(x-x)] + (Madd,xs) applied + (a) strut + (c) amplification My,splice = My + [Ms(y-y)] + (Mys) + Madd,ys applied + (a) strut + (b) LTB + (c) amplification

[Include only the more onerous of the strut action terms in square brackets and the more onerous of the two terms in round brackets.]

where:

Ms(x-x) and Ms(y-y) are strut action moments, see (1); My is the minor axis moment at the splice; Mys is the additional minor-axis moment due to lateral-torsional buckling, see (2); Madd,xs and Madd,ys are additional moments due to amplification of Mx and My by Fc each at

its appropriate distance Lz along the member from the relevant point of inflexion, obtained from:

Madd,xs = Madd,x,max sin (180 Lz/LEx) Madd,ys = Madd,y,max sin (180 Lz/LEy). LEx is the effective length for flexural buckling about the major axis; LEy is the effective length for flexural buckling about the minor axis;

Madd,x,max = ( )1/ -cExxx

fpMm

Madd,y,max = ( )1/ -cEyyy

fp

Mm

mx is the equivalent uniform moment factor for flexural buckling about the major axis; my is the equivalent uniform moment factor for flexural buckling about the minor axis;

pEx = 2

2

x

E

l

p

pEy = 2

2

y

E

l

p

Note: See also AD244

SCI ADVISORY DESK

AD 244: Second order moments

The aim in this item is to give a simple explanation of the three types of additional second-order moment encountered in AD 243, namely:

moments due to strut action;

moments due to lateral-torsional buckling;

moments due to amplification.

Moments due to strut action:

It is well known that compression members need to be designed using a compressive strength pc, which is less than the design strength py of the steel. The value of pc depends on the design strength py and the slenderness l. The formula from which the tables giving pc are derived are based on the fact that the actual stress in a slender compression member is higher than the apparent stress fc, and at failure the actual stress at mid-length reaches the design strength py . The strut action moment accounts for the difference between the compression strength pc and the design strength py .

The strut action moment arises from the eccentricity of the axial force relative to the deformed shape of the member as it buckles. This process is initiated by the naturally occurring imperfections in a real member. The moment varies along the length of a member. It is zero at pinned ends. Its effect on an un-spliced member is covered by using pc, but in a spliced member it needs to be accounted for explicitly at the splice, unless this is at a pinned end or a point of contraflexure.

Moments due to lateral-torsional buckling:

By analogy with a compression member, the effects on a beam subject to lateral-torsional buckling resulting in a bending strength pb less than py can be simply represented by an additional moment similar to a strut action moment. The analogy is convenient, but only approximate. This type of additional moment arises from a moment applied about the major axis of the section, but needs to be considered as acting about its minor axis. Like a strut action moment, it varies along the length of the member and is zero at points of contraflexure.

Moments due to amplification:

In a similar way to strut action, the axial force in a member acts at an eccentricity from the deformed shape caused by an applied moment, producing an extra moment. The difference between the applied moment and the final amplified moment is an additional moment due to flexure, which needs to be accounted for in designing a splice within the length of the member. Unlike one due to lateral-torsional buckling, an additional moment due to flexure acts in the same plane as the applied moment causing it. As in the other cases, it varies along the length of the member and is zero at points of contraflexure.

Note:

In line with the treatment of member buckling adopted in BS 5950 Part1, it is logical to assume that applying a major axis moment to an axially loaded member leads to only one

SCI ADVISORY DESK

additional moment, either Mys about the minor axis due to lateral-torsional buckling, or else Madd,xs about the major axis due to amplification, but not both (see (3) in AD 243).

Note: See also AD243

SCI ADVISORY DESK

AD 245: Allowance for Bolt Holes

The effect of bolt holes on moment capacity and shear capacity is not as clear in BS 5950: Part 1: 1990 as is the effect on tension capacity or compression resistance, although the basic rules on net and gross areas are intended to apply in all cases. The question has also arisen of the effect of bolt holes at plastic hinge locations. This answer is based on clarifications expected to be included in the forthcoming amendment to the code.

Tension: A bar in tension will fail either at its yield strength Ys on its gross cross-sectional area Ag or in rupture at its ultimate tensile strength Us on its net area An at bolt holes. It should also not suffer permanent deformation under service loading. The resulting design criteria are:

yielding: Pt Ag Ys / gm1

rupture: Pt An Us / gm2

deformation: Pt gf Pt,ser = gf An Ys / gser

and, allowing for the values of the various g factors, this leads to the code rule:

Pt = Ae py

in which:

Ae = Ke An but Ae Ag Ke = (Us / py) / 1.2 but Ke 1.2 py = Ys but py Us / 1.2

Accordingly, no allowance need be made for bolt holes in a tension member unless An < Ag / Ke .

Compression: Resistance to buckling of a compression member is based on the gross area Ag . The capacity of a cross-section can also be based on Ag because reaching the yield strength in compression on the net area cannot lead to necking, as in tension, and strain hardening takes over.

Bending moment: The tension flange of a beam is similar to a bar in tension, whereas the compression flange is similar to a compression member. Thus, bolt holes in the tension flange may need to be allowed for. However, the code does not currently describe how this should be done. It is recommended that, as in Eurocode 3, the same rules as for a tension member should be applied to the tension flange itself. Holes in the tension zone of the web need not be allowed for unless there are also holes in the tension flange. If there are bolt holes in the web and the flange at the same point in the length of the member, the rules for a tension member should be applied to the complete tension zone comprising the tension flange plus the tension zone of the web.

Shear: Nothing is currently said in the code about the effect of bolt holes on shear, but current good practice is to follow the procedure given in the Green book, which covers this issue along with the block shear effect. The forthcoming amendment to the code is expected to cover these two items separately. The resistance of a web to shear rupture is about 70% of its resistance to rupture in tension, whereas its resistance to yielding in shear is about 60% of its resistance to yielding in tension. Accordingly, it is recommended that no allowance be made for the effect of bolt holes on shear capacity unless 70% of Av,net is less than 60% of Av / Ke , that is, unless Av,net < 0.85 Av / Ke .

SCI ADVISORY DESK

Tension flanges:

From the above it can be seen that no allowance need be made for bolt holes in a tension flange provided that An Ag / Ke . If this is not satisfied, either the bolt holes can be staggered or a reduced moment capacity can be calculated by taking an effective area for the tension flange of Ke An . For a simple but conservative check, the moment capacity can be reduced by multiplying it by the ratio Ke An / Ag .

Plastic hinges:

In the case of plastic analysis, in the region of a plastic hinge location it is necessary for the tension flange to be able to stretch plastically without any other damage.

Punching of bolt holes hardens the steel locally around the holes, making it liable to crack in such circumstances. Accordingly, at plastic hinge locations, and within a length each side equal to the depth of the member:

full size punching of holes is not acceptable.

For a plastic hinge to rotate, it is necessary for the flange to yield uniformly over a similar length, rather than just deforming locally at bolt holes. Accordingly at plastic hinge locations:

drilled holes are acceptable - but only if, for the tension flange: An Ag / Ke

This is not currently stated in the code (although it is implied in Eurocode 3) but it is expected to be included in the forthcoming amendment.

SCI ADVISORY DESK

AD 246: Standards Organisations

BSI (British Standards Institution)

BSI is an independent body incorporated by Royal Charter. Representative BSI committees, each with an independent chairman and secretariat, are responsible for issuing and maintaining British Standards.

BSI is the UK member body of CEN and ISO. Appropriate BSI committees "mirror" the relevant CEN and ISO committees and nominate the UK delegates to them. BSI shares in providing the chairmen and technical secretariats of CEN and ISO committees. All CEN Standards and appropriate ISO Standards are also issued as BSI Standards.

The contents of all BSI Standards, including those produced by CEN or ISO, are copyright.

The documents of all BSI committees are marked "private circulation" and not to be used for any purpose other than the work of the committee. However, for briefing purposes, committee members may circulate these documents within the organization they represent.

Contact information

BSI Standards, British Standards House, 389 Chiswick High Road, London W4 4AL. Tel: 0181 996 9000 Fax: 0181 996 7400 E-mail: info@bsi.org.uk Web-site:www.bsi.org.uk

CEN (European Committee for Standardization)

CEN is an international association, established under Belgian law by an agreement between 15 national standards bodies. Its aim is to implement standardization on a Europe-wide basis. It now has 19 members, the national standards bodies of Austria, Belgium, Czech Republic, Denmark, Finland, France, Germany, Greece, Iceland, Ireland, Italy, Luxembourg, Netherlands, Norway, Portugal, Spain, Sweden, Switzerland and United Kingdom.

CEN Standards exist in three official versions (English, French and German). Versions in any other language made by translation under the responsibility of a CEN member into its own language and notified to the Central Secretariat has the same status as the official versions.

CEN Standards are issued in member countries by the national standards body - BSI in the UK.

Delegates to CEN meetings are experts chosen to represent their national standards organization, which is the committee member. National delegations must always speak as one, but positions must be flexible for the work to result in European standards.

The members of CEN are also members of ISO. In recent years the so-called "Vienna agreement" has helped avoid duplication of work between CEN and ISO. Many draft standards are now presented for parallel approval by both organizations.

Contact information

CEN Central Secretariat, Rue de Stassart 36, B-1050 Brussels, Belgium Tel: 00 32 2 519 6811

SCI ADVISORY DESK

Fax: 00 32 2 519 68 19 E-mail: infodesk@cenorm.be Web-site: www.cenorm.be

ISO (International Standards Organisation)

ISO and IEC (International Electrotechnical Commission) together form a system for international standardization as a whole. They develop and maintain international standards through technical committees. ISO working languages are English, French and Russian. All ISO Standards are published in English and French (and sometimes in multi-lingual editions including Russian and other languages). These versions are equivalent, and each is regarded as an original-language version.

The character of ISO Standards varies. Some are intended for direct adoption as (or in place of) national standards, whilst others are intended as a basis for drafting national standards. Some ISO Standards become British Standards through being adopted as CEN Standards. Others are implemented directly as British Standards, but some are not adopted by BSI for various reasons. This can be due to their character, or because the relevant BSI committee does not consider their content to be appropriate, or regards them as irrelevant to the UK. Alternatively there may be an established BSI or CEN Standard, or work in progress in CEN, that is regarded as more relevant.

In recent years the so-called "Vienna agreement" has helped avoid duplication of work, and many drafts are now presented for parallel approval by both CEN and ISO.

Contact information

ISO Central Secretariat, 1 Rue de Varamb, Case postale 56, CH-1211 Genve 20, Switzerland.

Tel: 00 41 22 749 01 11 Fax: 00 41 22 733 34 30 E-mail: central@isocs.iso.ch Web-site: www.iso.ch

SCI ADVISORY DESK

AD 247: Use of Composite Construction in an Aggressive Environment

Composite construction, using hot-rolled steel beams and composite steel deck slabs, is widely used for medium and high rise buildings in the UK. In internal mild exposure conditions the G275 galvanising layer on the decking ensures the desired design life can be achieved. Leaving the upper surfaces of the beam top flanges unpainted, so that the shear studs can be through-deck welded after laying out the decking, does not present a corrosion risk in these conditions.

In more aggressive (moderate or severe) environments, protection of the steel decking and steel beams requires further consideration. The following recommendations identify ways in which the steel components may be detailed/specified in order to overcome problems of increased exposure to the underside of the slabs. In some situations, such as car parks, it will also be necessary to take measures to prevent corrosion of the steel components due to water percolating through cracks in the slab. This may be achieved by controlling cracking, or by applying protection to the top surface of the slab.

Decking

G275 galvanising (275 g of zinc coating per square metre) is the standard for decking. Although thicker coatings (up to G600) can be applied, decking with this degree of galvanising is difficult to obtain and likely to require a large minimum order. A far better solution than trying to obtain non-standard galvanising may be to specify coated steel or additional paint protection. Such an additional layer of protection has the advantage that it can be regularly inspected and remedial work undertaken. Clearly, manufacturers/specialist advice should be sought before specifying a particular product.

Beams

Through-deck welding of the shear studs is beneficial because it enables continuous sheets of decking to be laid on the steel beams prior to fixing the studs. It also enhances the way in which the decking behaves as transverse reinforcement of the slab. However, in a potentially severe corrosive environment the need to keep the upper surfaces of the beams free from paint (to avoid contamination of the stud welds) is not acceptable. This leaves the designer who wishes to use composite beams and slabs with two options:

Use shear connectors that are attached to the beams without the need for welding. A number of connectors that use shot-fired pins are available.

Weld the studs to the beams in the fabrication shop, prior to applying the corrosion protection.

An implication of adopting the latter approach is that the decking can no longer be laid in sheets that are continuous over two spans. The sheets must be laid in single spans and butted up to the studs, or have pre-punched holes to allow the decking to be dropped over the studs. Both these approaches have disadvantages:

Single span decking is less structurally efficient than when it can be made continuous.

Trying to align holes and studs is easier said than done, and the resistance of the shear studs may be reduced.

SCI ADVISORY DESK

The presence of paint on the shear studs should not compromise their performance, which does not rely on surface bond.

SCI ADVISORY DESK

AD 248: Use of High Strength Friction Grip Connections

Recently we have received several questions concerning where and when to use HSFG connections. Many designers are familiar with the use of M20 and M24 8.8 bolts in clearance holes, but do not regularly specify HSFG connections. An ordinary bolt assembly, comprising, for example an M20 bolt in an M22 clearance hole, transmits applied forces by shear in the bolt and bearing between the shank of the bolt and the connected ply. This means that a small amount of slip takes place as the connection is loaded and the clearances are taken up. Some small deformation as the bolt and plies reach equilibrium in bearing will also occur.

HSFG connections transfer force quite differently. Although HSFG connections still use bolts in clearance holes, load is transferred by frictional resistance between the contact surfaces of the plies. The frictional resistance is achieved by preloading the bolt in considerable tension, which clamps the plies together. The preload in the bolt is achieved by applying a controlled torque, or by the so-called part-turn method (where the nut is rotated further, following tightening to snug tight). Alternatively, torque may be applied and measured by load-indicating washers, which deform under a pre-determined load. Other fasteners may be used where a sacrificial length of the shank breaks off at a predetermined load. The frictional resistance depends on the applied load, and the condition of the contact surfaces. Most commonly, the contact surfaces are shot blasted and left unpainted, although other surface finishes may be accommodated. The contact surfaces are frequently called the faying surfaces.

Compared with the use of bolts in clearance holes, friction grip assemblies are generally considered to be more expensive, due to the protection of the contact surfaces (usually masked during painting), the equipment and time taken during installation, and the subsequent application of protection systems to any unpainted steelwork which remains exposed. The primary advantage of friction grip assemblies is that slip is prevented. Secondary advantages include the pre-tension in the bolts, which prevents the bolt loosening under vibration. Clause 6.1.7 of BS5950-1:2000 reminds designers that special consideration is required when connections are subject to vibration, load reversal, or fatigue.

There are generally four reasons for using HSFG connections:

1. Fatigue

In structures where fatigue is a design consideration, the use of preloaded assemblies is recommended. The high level of preload means that the tension in the bolt does not vary significantly, provided the plies are stiff and truly in contact at the bolt.

2. Vibration

Severe, continued vibration has been known to lead to ordinary (non-preloaded) bolt assemblies becoming loose. Preloaded HSFG bolts may be used when vibration is a concern. Other effective solutions exist to prevent the nuts from coming loose, including the use of lock nuts, nuts with nylon inserts, or split pins. Bolts may be avoided altogether by designing a welded detail. It should be noted that nuts working loose due to vibration is not normally a consideration in orthodox building structures, and in most cases no special measures are required.

SCI ADVISORY DESK

3. Load Reversal

When load reversal occurs (excluding reversal solely from wind loading) the possibility of slip in the connections (as the clearance holes allow movement to and fro) should be prevented. Significant reversal may take place in structures supporting moving loads such as cranes. Preloaded HSFG bolts are recommended in these situations. Alternative solutions exist to prevent slip in the connections, including close tolerance bolts and injection bolts. Close tolerance bolts (also called fitted bolts) are used in connections where the normal clearance is reduced to an absolute minimum, so the bolts are in bearing immediately load is applied. This generally involves drilling the holes undersize and reaming through both components simultaneously to ensure an exact match of holes in each ply. Injection bolts permit the injection of resin into the cavity left by bolts in clearance holes. At present, injection bolts are not commonly used in the UK.

It may be possible to re-arrange the connection, such that the bolts are used in tension, or to substitute a welded detail

4. Dimensional Stability

Even if load reversal is not possible, it may be important to eliminate slip in connections. Examples include splices in moment-resisting members, where slip in a cover plate type splice may lead to additional deflections in the member and an unsightly kink. In some situations, accumulated slip in a number of connections may lead to additional deflection of the bolted assembly. Unexpected additional deflection due to accumulated connection slip may lead, for example, to loss of fall on roof trusses with shallow slopes. Whereas shallow trusses and similar assemblies of numerous bolted components can be subject to this effect, in most sloping roof trusses, the pitch is sufficient for the slip in the connections to be neglected. To prevent the additional deflections from accumulated connection slippage, shallow slope trusses and the vertical bracing connections in tall, multi-storey buildings are often made with preloaded HSFG bolts.

These four situations demonstrate when and why a designer would specify a preloaded (HSFG) bolt. However, for most building structures the use of ordinary bolts (non-preloaded) in clearance holes is perfectly acceptable, even in moment connections, and the small amount of slip associated with clearance holes has no practical consequences for the structure.

SCI ADVISORY DESK

AD 249: Design of Members Subjected to Torsion

We have recently received a number of questions in relation to structural steel members subjected to torsion. From these calls it is evident that there is some confusion over the difference in behaviour of open and closed cross-sections. In response to these queries, it therefore seems timely to present a qualitative background to the theory of torsion.

Torsional loading can arise within members in two ways: an externally applied torque; or when the applied load acts eccentrically to the shear centre of the cross-section. In both cases, the member will twist about its longitudinal axis, which passes through the shear centre of the cross-section.

Categories of cross-sections

Torsional loading has a significant influence on the initial choice of section for maximum structural efficiency. I-shaped sections are particularly poor in resisting torsion while hollow sections can be very effective. A distinction is normally made between these two types of sections, by calling I- and channel sections (which are poor in resisting torsion), Open Sections (Figure 1); while rectangular and circular hollow sections (which are more effective in resisting torsion), are referred to as Closed Sections (Figure 1).

Closed sectionsOpen sections

Fig. 1: Open and closed structural sections

Location of shear centre and its significance

The position of the shear centre is of particular importance in design, since when a load is applied to a member, a torque will develop if the applied load does not act through the shear centre of the cross-section. In these circumstances, the torque is simply equal to the applied load multiplied by its eccentricity from the shear centre. A cross-section having two axes of symmetry has its shear centre located at the centre of gravity of the cross-section (see Figure 2(a)). With one axis of symmetry, the shear centre lies on that axis, but will in general not coincide with the centre of gravity (see Figure 2(b)). However, for a section having skew-symmetry, the shear centre and the centre of gravity do coincide (see Figure 2(c)). For angle sections, the shear centre is located at the intersection of the two legs (see Figure 2(d)).

SCI ADVISORY DESK

cs

(a) (b) (c) (d)

cs cs

sc

Fig. 2: Location of shear centre s and centre of gravity c

Reference will be made below to warping, which is best described by considering the rectangular hollow section shown in Figure 3. In the initial conditions, the ends of the hollow section are rectangular and plane. Suppose that a slit is made along one side of this section (thereby transforming it from a closed section to an open section), and one end is twisted relative to the other. As can be seen in Figure 3, in response to this applied torque, the ends of the hollow section remain rectangular, but are no longer plane. This distortion of the cross-section is called warping, and is particularly pronounced in I-beams.

O

E sc

s

c

yy

zx

E

O

xd f

(a) Cross-section of splitrectangular hollow section

(b) Warping displacements

Warping displacement, w

Fig. 3: Warping of a split (open) rectangular hollow section

The total resistance of a structural member to torsional loading may be considered to be the sum of two components namely, uniform torsion and warping torsion. In some cases, only uniform torsion occurs. Whereas, when both uniform torsion and warping torsion are included in the torsional resistance, the member is in a state of non-uniform torsion. A diagrammatic representation of uniform and non-uniform torsion, on a member composed of an I-section, is shown in Figure 4.

SCI ADVISORY DESK

Uniform torsion

When a member is subjected to uniform torsion (sometimes referred to as pure or St Venant torsion), the rate of change of the angle of twist is constant along the member, and the longitudinal warping deflexions are also constant along the member (see Figure 4(a)). In this case, the torque acting at any cross-section is resisted by a single set of shear stresses distributed around the cross-section (see Figure 5). The ratio of the torque T to the twist rotation f, per unit length, is defined as the torsional rigidity GJ of the member; where G is the shear modulus and J is the torsion constant (sometimes called the St Venant torsion constant).

(a)

t

t

t

t

t

t

(b)

Constant shearflow tt

Fig. 5: Shear stresses due to uniform torsion of (a) closed sections; and (b) open sections

Non-uniform torsion

When a member is subjected to non-uniform torsion, the rate of change of the angle of twist varies along the member (see Figure 4(b) and (c)). In this case, the warping deflexions vary along the member and, to resist the applied torque, an additional set of shear stresses act in conjunction with those due to uniform torsion. The stiffness of the member associated with

Constant torqueends free to warp

Varying torque End warping prevented

(b) Non-uniform torsion (c) Non-uniform torsion(a) Uniform torsion

Fig. 4: Uniform and non-uniform torsion of a member composed of an I-section (viewed on plan)

SCI ADVISORY DESK

these additional shear stresses is proportional to the warping rigidity EH; where E is the modulus of elasticity and H is the warping constant.

For a member composed of an I-section, the action of warping resistance can be visualised as follows: the torque T is resisted by a moment comprising of forces equal to the shear forces in each flange, which are separated by the lever arm, df equal to the depth between the centroids of the two flanges. If each flange is treated as a beam, the bending moments produced by these forces lead to the warping normal stresses, as shown in Figure 6(b).

For an I- or H-section, this approach provides a reasonable approximation, but will generally over-estimate the warping normal stress whilst under-estimating the warping shear stress (since the approach ignores the shear stresses from uniform torsion). However, it cannot be readily applied to channel sections; in such circumstances, more rigorous methods of analysis need to be adopted.

x z

y

f

f

f

Warping shear stresses

Bimoment

Flange shears

wt

Warping normal stresses

fwd M

wf

fV

M

V

V

fM

fd

(b) Bimoment and warping stresses(a) Rotation of cross-section

y

xfd

u

fc,s

Fig. 6: Warping stresses in an open section member composed of an I-beam

Effect of cross-section on torsional behaviour

Before examining how different types of sections perform in resisting torsion, it is useful to first introduce the non-dimensional torsion parameter for the member:

( )22 / GJLEHK p= . For sections that have a very high torsional rigidity GJ compared to their warping rigidity EH, K becomes small; in these circumstances the member will effectively be in a state of uniform torsion (as indicated in Figure 7). Closed sections, whose torsional rigidities are very large, behave in this way, as do sections whose warping rigidities are negligible, such as angle and T-sections.

Conversely, for sections whose warping rigidity EH is very high compared to their torsional rigidity GJ, K becomes very large, and the member is in the limiting state of warping torsion (as indicated in Figure 7). Very thin-walled open sections, such as light gauge cold-formed

SCI ADVISORY DESK

sections, whose torsional rigidities are very small, behave in this way. Between these extremes, the applied torque is resisted by a combination of the uniform and warping torsion components, and the member is in the general state of non-uniform torsion. This occurs for intermediate values of the parameter K, as shown in Figure 7, which are appropriate for most open sections such as hot-rolled I- or channel sections.

00.1 1 10

0.2

0.4

0.6

0.8

1.0Uniformtorsion

Non-uniformtorsion

Warpingtorsion

Closedsections

Anglesections

Hot-rolledsections

Very thin-walledchannel sections

(War

ping

tor

que/

tota

l tor

que)

at

mid

-spa

n

Torsion parameter, K

Fig. 7: Effect of cross-section on torsional behaviour

Whether a member is in a state of uniform or non-uniform torsion also depends on the loading arrangement and the warping restraints. If the torsion resisted is constant along the member and warping is unrestrained (as shown in Figure 4(a)), then the member will be in uniform torsion, even if the torsional rigidity is very small. If, however, the torsion resisted varies along the length of the member (Figure 4(b)), or if the warping displacements are restrained in any way (Figure 4(c)), then the rate of change of the angle of twist rotation will vary, and the member will be in a state of non-uniform torsion.

As a result of applying a torque to a member, the torsional stresses induced within the section, which should be considered in design, are:

a) Shear stresses due to uniform torsion.

b) Shear stresses due to warping torsion.

c) Bending stresses due to warping.

Each of the above stresses is associated with the angle of twist f, or its derivatives. Hence, if f is determined for different positions along the member length, the corresponding stresses can be evaluated at each position.

Torsion of closed sections

As discussed above, the torsional rigidity GJ of a closed section is very large compared with its warping rigidity EH (see Figure 7). Therefore, a member composed of a closed section may be considered to be subject only to uniform torsion. For a rectangular or circular hollow section, the uniform torsion will result in a uniform shear stress developing within the walls of the cross-section (see Figure 5(a)). In these circumstances, the problem is statically

SCI ADVISORY DESK

determinate, and the shear stress t, as well as the angle of twist f, may be determined from simple statics.

Torsion of open sections

As discussed above, for members composed of open sections such as hot-rolled I- or channel sections, the section may be considered to be subject to the general state of non-uniform torsion (see Figure 7). In these circumstances, the applied torque is resisted by a combination of uniform torsion and warping torsion components.

Uniform torsion

If a torque is applied to the ends of a member, in such a way that the ends are free to warp, then the member will only develop uniform torsion (Figure 4(a)). The resulting shear stresses will vary linearly across the thickness of the element (see Figure 5(b)): they are maximum at the element surfaces, with two equal values, but opposite in direction. These stresses are a function of the rate of change of the angle of twist, and are greatest in the thickest element of the cross-section i.e., typically the flanges in an I-beam. (At junctions between the web and the flanges, the local shear stresses may exceed the stresses in the thickest element of the cross-section; for rolled sections, this effect may be neglected by the designer, as allowance for the root fillet radii are made in determining the torsional constant J in section property tables).

Warping torsion

When a uniform torque is applied to a member restrained against warping, the section itself will be subject to non-uniform torsion with the rate of change of the angle of twist varying along the length of the member (see Figure 4(c)). The rotation of the section with respect to a restrained end will be accompanied by bending of the flanges in their own plane (sometimes referred to as the Bimoment). The warping normal and warping shear stresses developed by this condition are shown in Figure 6. Warping stresses are also generated in members of open section when the applied torque varies along the length, even if the ends are free to warp (Figure 4(b)).

End conditions

As discussed above, the end conditions will also greatly influence the torsional stresses along the member. Note that end conditions for torsion calculations may be quite different from those for bending e.g., a beam may be supported at both ends, but torsionally restrained at only one end: the torsional equivalent of a cantilever. Torsional fixity must be provided by at least one point along the length of the member, otherwise it will simply physically twist when the torque is applied. Also, warping fixity cannot be provided without also providing torsional fixity. As a result, there are three possible boundary conditions which may sensibly considered in torsion calculations:

i. Torsion fixed, warping fixed This condition is satisfied when, at the ends of the member, both twisting about the longitudinal axis and warping of the cross-section are prevented (sometimes referred to as a Fixed torsional end condition). Effective warping fixity is practically almost impossible to achieve in most structures. A connexion providing fixity in both directions is not sufficient, it is also necessary to restrain the two flanges either side of the web. Details such as those shown in Figure 8(a) need to be provided to achieve this type of boundary condition. It should be noted, however, that the provision of warping fixity does not produce such a large reduction in torsion stresses as is obtained from bending

SCI ADVISORY DESK

fixity. Therefore, it is more practical to assume warping free connexions (see (ii) below), even when fixity is provided in terms of bending.

ii. Torsion fixed, warping free This is satisfied when the cross-section at the ends of the member is prevented from twisting, but is allowed to warp freely (sometimes referred to as a Pinned torsional end condition). Such a condition may be readily achieved by providing the relatively simple standard connexions, such as shown in Figure 8(b).

iii. Torsion free, warping free This condition is achieved when the end is free to warp and twist (sometimes referred to as a Free torsional end condition): the unsupported end of a cantilever illustrates this case.

Plate Channel

(a) Torsion fixed, warping fixed (b) Torsion fixed, warping free

Fig. 8: Practical end conditions

The above gives a basic overview of the considerations that should be made in the design process for structural steel sections, which are to be subjected to torsion. Design equations for estimating the stresses due to torsion, in combination with bending, may be found in SCI publication 057 entitled Design of Members Subject to Combined Bending and Torsion.

SCI ADVISORY DESK

AD 250: Choice of Steel Sub-Grade

During a recent seminar on the Amendments to BS 5950-1: 1990, a straw poll of the (large) audience indicated that only 5% of the structural engineers present specified the steel sub-grade. A further 2% of the audience indicated that they checked that the steelwork contractor chose the correct sub-grade. This left over 90% of the audience who neither specified the sub-grade themselves, nor ensured that others were specifying the correct sub-grade.

The implication of this poll is that there is potential for an inappropriate sub-grade of material used in structures, which could lead to an unacceptably high risk of brittle fracture.

This advisory note serves as a reminder that an appropriate sub-grade must be specified. It should be noted that the requirement to specify an appropriate sub-grade is not new, being covered in BS 5950-1:1990.

Brittle fracture becomes increasingly likely:

At lower temperatures.

As the stress level increases.

As the strain rate increases.

The more stress raisers that are present (which disrupt the uniform state of stress).

The thicker the element.

If the circumstances make brittle fracture possible, a tougher steel must be specified. Tougher steels are generally more expensive.

BS 5950-1:2000 requires the choice of a K factor from Table 3, which covers the circumstances in which the steel is being used. Table 3 addresses stress level, type of detail and strain conditions which combine to indicate a K factor between 0.25 and 4. The lower the K factor, the tougher the steel will need to be.

In most circumstances, the K factor will be at least as low as 1, since this is the K factor for welded generally, with a stress higher than 30% of the normal yield strength.

The requirement in Clause 2.4.4 is that:

t < Kt1 where t is the thickness of each element (commonly the flange)

t1 is the limiting thickness at the minimum service temperature from Table 4 or 5 of BS 5950-1:2000

Knowing the element thickness and knowing the K factor, Table 4 (plates, flats and rolled sections), or Table 5 (hollow sections), may be used to choose an appropriate sub-grade, such that t1 > t/K.

Who should specify the Steel Sub-grade?

The designer of the structure knows the information needed to specify the steel sub-grade and should take responsibility for ensuring an appropriate steel is used. Unless steelwork contractors are designing the structure, they will need the sub-grade at tender stage to prepare a quotation. To leave the contractor to choose the sub-grade is not satisfactory, as

SCI ADVISORY DESK

it is not necessarily clear from arrangement drawings if steel is exposed, or if it is highly stressed.

Note that changes in the construction details could affect the choice of sub-grade. For example, a welded beam to column connection might be stiffened or unstiffened. The K factor is higher if the column flange is stiffened. Thus if a heavier, unstiffened column section is used to replace a stiffened column, the K factor would reduce, and a tougher steel may be required because of the revised detail and the increased flange thickness.

Steel Used in External Conditions

Specifiers should note that, generally, it will be necessary to specify at least a J0 sub-grade for steel used in external conditions. This represents a change from BS 5950-1:1990 (as amended in 1992), which allowed less tough steels to be used in exposed conditions.

Why no history of failures to date?

If, according to the straw poll, there is every possibility of a less capable steel being used than the design standard would indicate, why is there no extensive catalogue of failure in service? Clearly, even though the risk is increased, this does not mean that failure occurs. One further contribution is the steel itself which, when of UK origin, has typically been tougher than specified. This, however, does not represent an excuse to avoid proper specification, particularly as more steel is sourced from foreign manufacturers.

SCI ADVISORY DESK

AD 251: Equivalent Uniform Moment Factor, m

BS5950-1:2000 sees the demise of the n factor method when calculating buckling resistance Mb. In Section 4, the m factor method is the only approach available, and will become a familiar friend.

The equivalent uniform moment factor, m, allows for the fact that, for two identical beams with the same maximum applied bending moment, the buckling resistance is greater for a varying bending moment diagram than for a uniform bending moment diagram.

In the two examples shown below, with the same maximum applied moment, the lateral displacement of the compression flange is more pronounced when the member is subject to a uniform bending moment. Thus, the member will buckle at a lower maximum moment if the moment is uniform. Alternatively, the buckling resistance of a member subject to a varying bending moment is larger than when the same member is subject to a uniform bending moment.

In BS5950-1:1990, the m factor method could only be used if there were no intermediate loads. In BS5950-1:2000 the method has wide application. Table 18 contains members subject to end moments only, some standard cases of intermediate loads and a formula to establish mLT for a general loadcase.

BS5950-1:2000 distinguishes between the equivalent uniform moment factor for lateral torsional buckling, mLT, and the equivalent uniform moment factors for flexural buckling, mx, my, and myx. Lateral torsional and flexural buckling are different (one does not have the torsional component) and thus different Tables for m are required. Table 26 covers m values for flexural buckling, presenting cases for end moments only, standard cases of intermediate load, and a general case.

Fig. 1: Member with a uniform applied moment

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Fig. 2: Member with a varying applied moment

m Factor for General Case

To use the general case formula, one needs to know the bending moment at quarter points, and the maximum moment on the segment. Software may already provide this information (or may be configured to provide this information). No doubt new software to BS5950-1: 2000 will automatically determine this information to calculate m.

Flexural buckling m factors when used with Sway-Sensitive frames

Clause 4.8.3.3.4, in the section covering member buckling checks, contains important information about the use of flexural buckling m factors mx, my, and myx, when designing sway-sensitive frames.

If using Annex E (effective length factors greater than 1.0), the minimum value of the factors mx, my, and myx is 0.85. If amplifying the sway moments, the factors mx, my, and myx can only be applied to the non-sway component of the total bending moment. Neither of these restrictions applies to mLT.

Designers should note the potential for misapplication of the m factors, particularly if applying them to sway sensitive frames. The following table summarises the position.

Factor

mLT mx, my, myx

Association

Lateral-torsional buckling Flexural buckling

Use with

MLT Mx, My

Determined from:

Table 18 Table 26

Restrictions

In sway-sensitive frames: i) mimimum 0.85 if using Annex E ii) only applied to non-sway moments if amplifying sway effects

SCI ADVISORY DESK

AD 252: Axially Loaded Columns - Base Plate Design - Effective Area Method BS 5950-1: 2000

The empirical method for determining the size of base plates in BS 5950-1:1990 has been replaced by the effective area method in BS 5950-1:2000. Doubt has been raised as to the theoretical validity of the empirical method when used with deep UBs or with bases that have very small outstand dimensions a and b (see Figure 1). The effective area method offers more economy than the empirical method, while still producing safe designs when compared to test results.

The effective area method for base plate design may initially seem to be more complex than the empirical method given in BS 5950-1:1990. However, the approach is much more reliable and can be used for all column sections. The forthcoming revision of the SCI and BCSA green book Joints in Steel Construction: Simple Connections will include design procedures for base plates using the effective area method and will include standard base plates and their capacities for I, H and hollow section columns.

We have received several questions concerning the effective area method for base plate design, and in particular what to do when the effective area does not fit on the base plate selected.

The basic design procedure is set out below.

1. Calculate required area = axial load / 0.6 fcu. Where: fcu is the cube strength of either the concrete or the grout, which ever is weaker.

2. Calculate outstand c (see Figure 1) by equating required area to actual area expressed as a function of c. The expression for the actual effective area of an I or H section may be approximated to 4c2 + (column perimeter) c + column area.

Fig. 1 Fig. 2 Fig. 3

3. Check that there is no overlap of effective area between flanges (see Figure 2). This will occur if 2c > the distance between the inner faces of the flanges. If an overlap exists, modify the expression for effective area and recalculate c.

4. Check the effective area fits on the size of base plate selected (see Figure 3). If effective area does not fit on the base plate, modify the expression for effective area to allow for the limitations of the plate size and recalculate c, or select a larger base

c

c c

c

Effective area

a

b

Overlap area

c

c

c

c c

c

Effective area

a B

SCI ADVISORY DESK

plate. For the case shown in Figure 3, the modified expression for the effective area will be,

4c2 + (column perimeter) c + column area 2 (B+2c) (c a).

5. If c has been recalculated, step 3 will need to be repeated.

6. Calculate required plate thickness tp using expression below (given in Clause 4.13.2.2):

5.03

=

yppw

cpt

Where:

w = 0.6fcu

pyp is the plate design strength

The expression for the plate thickness can be derived from equating the moment produced by the uniform load w to the elastic moment capacity of the base plate (both per unit length).

Moment from uniform load on cantilever = Elastic moment capacity of plate w c2 / 2 = py Z (per unit length) w c2 / 2 = py tp2 / 6

Rearranging gives tp = c(3 w /py)0.5

When the outstand of the effective area is equal either side of the flange (as in Figure 1 and Figure 4), the cantilevers are balanced and there is no resultant moment induced in the flange. However, if the cantilevers do not balance either side of the flange, as would be the case in Figure 3, then theoretically to satisfy equilibrium there is a resultant moment induced in the flange (see Figure 5 and Figure 6).

Fig. 4 Fig. 5 Fig. 6

It is important to remember that the method given in BS 5950-1:2000 is only a design model and does not exactly represent what will happen in practice. The remainder of the plate (not only the effective area) does exist and does carry load. With this in mind, the moment induced in the column flange due to unbalanced cantilevers does not need to be explicitly considered in the design of either the column or the base plate.

M= wc2/2

c c

M= wc2/2

c a

M= wa2/2

M=(c2-a2)/2

M= a2/2 M= c2/2

SCI ADVISORY DESK

AD 253: Design Considerations for the Vibration of Floors Part 1 (Amended)

Introduction

Designers are more frequently facing the need to consider, in detail, the vibration response of floors. This Advisory Desk article has been written as a summary of the key issues.

A structure that is composed of any material will vibrate if subjected to cyclic or sudden loading. In most cases, the vibrations are imperceptible, and so can be neglected in building design. However, in some circumstances, the response of the structure to a common cyclic load (e.g. walking activities, etc.) is sufficient to produce a response that is perceptible to the occupants of the building. The most common example of such small, but perceptible, vibrations occurs in floor structures. This is not a new phenomenon, but it is more noticeable within the working environment of modern offices.

What disturbs building occupants?

For structures that are subjected to static loading, it is normal for the engineer to calculate the vertical deflection of the floor. This is to avoid undesirable deflections and to limit the possibility of damage to brittle finishes that will occur in serviceability conditions. However, for floors that are subject to cyclic or sudden loading, human perception of motion is usually related to acceleration levels rather than displacement. In most practical building structures, if the magnitude of these acceleration levels is not limited, the typical reaction of the building occupants varies between irritation and a feeling of insecurity. This is based on the instinctive human perception that motion in a solid building structure indicates structural inadequacy or failure.

The working environment also affects the perception of motion. For busy environments, where the occupant is surrounded by the activity that is producing the vibrations, the perception of motion is reduced. In contrast, for quiet environments (such as laboratories and residential dwellings), where the source of the vibration is unseen, the perception of motion is significantly heightened.

A simple model of a vibrating system

A simple model that illustrates the vibration behaviour of a structure is shown in Fig. 1 (a). The bending stiffness is modelled as a spring of stiffness k, and the floor mass is modelled by a point of mass m. All practical structures will have some damping, conveniently modelled as a viscous (or oil-pot) damper. Damping refers to the loss in mechanical energy within a mechanical system. Practical floor structures possess a low level of natural damping (normally in the order of 1%), which does not affect the natural frequency of the system. However, the magnitude of the damping is very important, as this determines the maximum possible magnification to the acceleration response of the floor (see below).

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Mass,

Viscousdamper,

k c

m

Externalforce, p(t)

Period, T

Dis

plac

emen

t,

v

tTime,

v(t)

Stiffness,

Displacement,

(a) (b)

Fig. 1(a): Idealised single-degree-of-freedom system, basic components

Fig. 1(b): Idealised single-degree-of-freedom system, undampened free vibration response

Higher damping depends on the energy dissipation through non-structural components such as partitions, which are largely dependant on frictional forces. Intuitively, it might be expected that the presence of the false flooring would have an effect on the damping. However, experimental studies have shown that this is not the case. The reason is that deflections of modern false floors do not produce sufficient movement to develop significant friction forces.

A disturbance to the floor by a suddenly applied load will cause the system to vibrate as shown in Fig. 1 (b). If no other external forces are applied to the system, the damping will cause the displacements to die away with time.

Walking activities

Walking produces a cyclic loading that is repeated at regular intervals called periods, which are inversely proportional to the pace frequency. A typical example of a load-time function that is produced by walking activities is shown in Fig. 2 (a).

Fig. 2(a): Loading caused by an individual walking at 2.0Hz, load-time function

In general, a repeated force can be represented by a combination of sinusoidal forces, whose frequencies are multiples (or harmonics) of the pace frequency, as shown in Fig. 2 (b). The magnitude of the force for each of these harmonic components is taken as a proportion of the static weight of the person multiplied by a Fourier coefficient na . The values of the Fourier coefficients have been established experimentally for different activities and different activity frequencies. For walking activities, Fig. 3 shows how the magnitude of the first three harmonic components vary with frequency.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 0.25 0.5 0.75 1 1.25 1.5

Time (s)

Forc

e / st

atic

wei

ght

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Fig. 2(b): Loading caused by an individual walking at 2.0 Hz, decomposition of the load-time function into the first three harmonic load components

The trend exhibited by Fig. 3 is that the magnitude of the Fourier coefficient is lower for higher harmonics. For example, the average Fourier coefficient for the first three harmonics of walking is typically taken as 1a = 0.4, 2a = 0.1 and 3a = 0.1. In buildings, it is considered that a walking frequency of between 1.7 and 2.4 Hz can realistically occur. (However, a higher frequency range may be found for other activities e.g., dancing, aerobics, etc.).

Fig. 3: Fourier coefficients for walking activities

Floor response

When cyclic forces (such as those from walking activities) are applied to a structure, it will begin to vibrate. If the cyclic forces are applied continuously to the simple model in Figure 1(a), the motion of the structure will reach a steady-state i.e., vibration of a constant amplitude and frequency will be achieved (see Fig. 1 (b)). The magnitude of the peak acceleration response for a long continued excitation is given by:

factor ionmagnificat Dynamicmass

force applieda peak

=

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 0.25 0.5 0.75 1 1.25 1.5

Time (s)

Forc

e / st

atic

wei

ght

1st Harmonic component

2nd Harmonic component

3rd Harmonic component

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7 8 9 10

Harmonic component frequency (Hz)

Four

ier

coef

fici

ent,

an

Harmonic, n1 r2 3

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As can be seen from the above expression, the peak acceleration response decreases as the mass of the floor increases and increases as the dynamic magnification factor increases.

Mathematically, the dynamic magnification factor is given by the following identity:

( ) ( )222 211

zbb +-=D

where b is the ratio of the activity frequency to the natural frequency of the structure and z is the damping ratio.

As can be seen from the second equation, the dynamic magnification factor is related to the level of damping that the floor possesses, and the ratio b (e.g., from walking, etc.) This variation is shown graphically in Fig. 4 below.

Fig. 4: Variation of dynamic magnification factor with damping and frequency.

Fig. 4 shows different types of response depending on the frequency ratio b. When the cyclic force from the activity is applied at a frequency much lower than the natural frequency of the system (b > 1), the steady-state response is governed by the mass (inertia) of the structure. In this case, the amplitude is less than the static deflection. Neither of these ranges of b are of great practical importance.

The range of b of greatest importance is where the dynamic force is applied at a frequency close to the natural frequency of a structure. In any structure which is lightly damped (as is found in most practical floor systems), very large responses will occur. The condition when the frequency ratio is unity is called resonance (i.e., the frequency of the applied load equals the natural frequency of the structure; b = 1). At resonance, very large dynamic magnification factors are possible and, for undamped systems (i.e. z = 0), the steady-state

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3

Frequency ratio b (activity frequency / frequency of structure)

Dyn

amic

mag

nifica

tion

fac

tor

D

z = 0.015

z = 0.1

z = 0.5

z = 1.0

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response tends towards infinity. A more general result may be obtained from the second equation, which shows that for resonance (b = 1) the dynamic magnification factor is inversely proportional to the damping ratio, and: (Fig. 4).

zb 21

1 ==D

Since in many practical structural systems the natural damping z is of the order of 1%, if precautions against resonance are not made, magnification factors of up to 50 may result. Given that the force in the structure is proportional to the displacement, the dynamic magnification factor also applies to structural forces. In spite of this, for office floors, the magnifying effects to this load are normally neglected when assessing ultimate limit state criteria because the static weight of an individual walker is so small. This is certainly not the case when large groups of people (crowds) take part in synchronised activities (i.e. dancing, aerobics, etc.). In these circumstances, the magnification on the static load of the crowd may cause extreme loading to the floor, and should be considered as an additional imposed loading case for design at the ultimate limit state. In the UK, guidance is given in BS 6399-1:1996 for structures that fall within this special category.

The above gives a basic overview of dynamics, and what considerations should be made in the design process for floors, which are to be subjected to occupant-induced vibrations. Design equations for estimating the fundamental frequency, and response of a floor, may be found in SCI publication 076. In addition, further design recommendations can be found in a recently completed SCI technical report RT 852 entitled Design Guide for Vibrations of Long Span Composite Floors, which also compares the performance of present design guide predictions with measured results. As well as acting in an advisory rle, for the past three years, the SCI has also offered a consultancy service for designers. Our experts have produced all the necessary calculations for clients, on a variety of different types of floor. In addition to the initial design stage, we have evaluated existing floors and, where necessary, offered advice on remedial measures. We can also offer an in-situ testing service on floors where it is perceived that there may be a potential vibration problem.

Note: See also AD254

Note: This Note has been amended following the correction to the formula for D given in New Steel Construction, issue January/February 2002.

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AD 254: Design Considerations for the Vibration of Floors Part 2

Introduction

AD 253 Design considerations for the vibration of floors- Part 1 warned that where vibrations disturb the occupants of a building it is usually excessive accelerations that cause the disturbance. It also warned that having the frequency of the floor above a certain threshold frequency, such as 3Hz, does not necessarily guarantee that the floor will prove satisfactory in service. In addition, AD 253 described the forcing function produced by walking activities, and summarised the dynamic behaviour of floor structures. This Advisory Desk article follows on from AD 253 by giving information on acceptable levels of acceleration and explaining how the expected acceleration may be calculated.

Acceptability criteria

The evaluation of the exposure of humans to vibrations within buildings is covered by BS 6472: 1992 Guide to evaluation of human exposure to vibration in buildings (1Hz to 80Hz), which is written to cover many vibration environments in buildings. This Standard presents acceleration limits for vibrations as a function of the exposure time and the frequency. It contains graphs of acceleration against frequency for both lateral and vertical vibration. These graphs are called base curves. The base curve for vertical vibration is in Figure 1. Note that the acceleration is given as the rms (root mean square) value.

1.000

0.010

0.0011001 10

0.100

rms

acce

lera

tion

(m

/s)

Frequency (Hz)

Fig. 1: Building vibration z-axis base curve for root-mean-square (rms) acceleration according to BS 6472:1992

The accelerations acceptable for different uses of buildings are described using the base curve. Multiplying factors are used to increase the acceleration level of the base curve

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according to the intended use of the building. These multiplying factors are referred to as response factors within the SCI guide P 076, Design Guide on the Vibration of Floors. Suitable design values for different office accommodation will also be given in a future Advisory Desk article.

The maximum calculated value of acceleration is found as follows. The base curve acceleration, basea , is found from Figure 1 for the fundamental frequency of the floor. This acceleration, basea , is multiplied by the response factor, R, appropriate to the use of the floor to give a limiting acceleration Rabase .

For example, for a floor with the fundamental frequency equal to 5Hz, the base curve acceleration is 0.005 m/sec2 rms. If the R value appropriate for the use of the building is 8, then acceptable acceleration is 0.005 8 = 0.04 m/sec2 rms.

Calculation of the accelerations

Resonance can occur even if the fundamental frequency of the floor is above a given minimum design value. This is because vibrations arise from components of the walking activity. These components of walking activity occur because the force versus time graph of walking is made up of many different sine curves, as shown in AD 253. The largest response will generally occur when the lowest whole number multiple (harmonic) of the activity frequency is equal to the fundamental frequency of the floor (i.e., resonance). For example, consider a floor with a fundamental frequency 0f = 4.6 Hz. The frequency range for walking activities is 1.7 Hz to 2.4 Hz, so the pace rate that would cause the highest floor response would be from the second harmonic of the walking activity (2 2.3 Hz = 4.6 Hz). If the floor design was changed so that the fundamental frequency was raised to 6Hz, the highest floor response would be from the third harmonic of the walking activity (3 2.0 Hz = 6.0 Hz.

The peak acceleration response is calculated from the following equation:

za

210

MP

a npeak = (1)

where 0P is the weight (=mass gravity) of the person in Newtons (N),

na is the Fourier coefficient of the nth harmonic component of the walking activity,

M is the effective vibrating mass (kg) and z is the damping ratio.

(The above equation can be derived from AD 253 from the equation for peaka , using 1=bD as

the value of the Dynamic magnification factor.)

The response assumed in developing the above equation is purely sinusoidal, whereas the limits to vibrations in BS 6472 are given in terms of the root-mean-square (rms) acceleration

rmsa . The rms acceleration is found from this sinusoidal response by dividing peaka by 2 .

The value of 2

peaka should be compared with acceleration from the curve in Figure 1, basea ,

(for the fundamental frequency of the floor) multiplied by the response factor, R, appropriate

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to the use of the floor. If 2

peaka is greater than Rabase , the floor may be unsuitable for

the intended use.

The mass of a person is commonly taken as 76 kg.

The appropriate Fourier coefficient of the walking activity, na , is the value that coincides with the fundamental frequency of the floor, which may be calculated by the methods given below. Simple recommendations for design values of na will be given in a future Advisory Desk article. The highest values of na occur at the first harmonic of walking activity, as shown in AD 253 Figure 3. Therefore, minimum frequency limits are normally given in design guides and codes of practice to ensure that the fundamental frequency of the structure does not coincide with the first harmonic of walking activities. By ensuring that the fundamental frequency of the floor is above a the first harmonic of walking activities, the magnitude of the floor acceleration is reduced because the higher harmonic components of the walking activity have lower values of na . However, as can be seen from Equation (1), the acceleration response from the floor may still be very high if the mass is too low. A practical example of a floor with a low mass would be a steel framed structure with a timber floor.

The response calculation in Equation (1) uses only one harmonic of the walking activity, which is the harmonic that coincides with the fundamental frequency of the floor. This is normally sufficient for checking the response to walking activity. However, for more vigorous activities such as dancing, aerobics etc, it is recommended that more harmonics are considered.

Floor mass for vibration and the effective vibrating mass, M

For offices in the UK, 10% of the design imposed loading is nor