New Steel Construction, March/April 2001 Page 25

It is clearly not possible to give meaningful details of all the changes, soa few fairly major ones have been selected for illustration, mainly byexamples. As many of these indicate an advantage in using the newedition, it is important to recall that the prime purpose of theamendments was safety, thus there are some cases where the newversion is more conservative.

1. Moment capacity. The moment capacity of class 3 semi-compactand class 4 slender sections is increased by using two new parametersSeff and Zeff in 4.2.5. The effective plastic modulus Seff is found in 3.5.6and answers the question of what happens if a section is only justoutside the limits for class 2 compact. The effective section modulus Zeffis found in 3.6.2 and covers class 4 slender sections.

Figure 1 shows how the moment capacity of a welded sectionvaries with the d/t ratio of the web. The solid curve between d/t valuesof 100 to 120 represents Seff /Z and the solid curve beyond d/t = 120represents Zeff /Z. The dotted lines show the equivalent momentcapacity according to the 1990 edition. The new method for slendersections is a variant of that given in EC3.

For rolled sections, the flange is more likely to be slender, not theweb. The value of Seff will then be based on the flange b/T ratio (or b/tfor an RHS). In 3.5.6.2 and 3.5.6.3 two values are given for Sx.eff .The first applies when the web is slender and the second applies whenthe flange is slender.

Figure 1: Variation of moment capacity with web slenderness ratio d/t

2. Compression member with a slender section. The newmethod for class 4 slender sections under axial compression, see4.7.4(b), uses the effective area Aeff . For example, the web of a45719167 UB with d/t = 48.0 may be slender, depending on the load.For any value of axial compression, the basic limit in Table 11 is 40.For S275 steel, using 3.6.2.2, Aeff = A - (d - 40t) t = A - (d/t - 40) t 2 ,see figure 2.

So Aeff = 85.5 - (48 - 40)(8.5/10)2 = 85.5 - 5.8 = 79.7 cm2. If theslenderness = L/ry = 100 then using 4.7.4(b) the reduced slendernessis [Aeff /A]0.5 = 100(79.7/85.5)0.5 = 96.5. From Table 23, curve (b) applies

and from Table 24 sheet (3) the value of pcs is 147 N/mm2. So thecompression resistance Pcs = Aeff pcs = 79.7147/10 = 1172 kN. The newmethod is as given by EC3.

Figure 2: Effective area

3. Classification. Classification now depends on the axial load insome cases. Due to this, although the compression resistance in (2) isabout 20% more than given by the 1990 edition, it is still an under-estimate. Even under a load of 1205 kN, from 3.5.5(a) the ratio r2 =120510/85.5/275 = 0.5125 so the other expression in Table 11 gives anincreased d/t limit of 120/(1 + 20.5125) = 59.26 which is greater than48.0 and thus the section is not slender after all. If the memberslenderness is 100 then pc is 141 N/mm2 and the compressionresistance Pc = Apc= 85.5141/10 = 1205.5 kN.

If the member slenderness is 60 and it is assumed that thesection is not slender, then pc = 221N/mm2 and the compressionresistance Pc would be 1890 kN. However a check under this load willshow that the section is now slender. Provided that the memberslenderness is low enough, the section will become slender when120/(1 + 2r2 ) exceeds 48.0, which happens for S275 steel when the loadexceeds 1763 kN.

For simplicity, to avoid doubt whether the section is slender or not,the limit of 40 is useful. On the other hand, for maximum economy theexpression using r2 is more appropriate.

Most rolled sections will not become slender except when themember slenderness is low. Thus a practical design strategy is toinitially assume that sections are not slender, but include a check thatthis is the case.

4. Equivalent uniform moment for lateral-torsional buckling.In the new edition, the shape of the bending moment diagram is alwaysallowed for by new equivalent uniform moment factors mLT instead ofusing either m or n factors, depending on the type of loading, as in the1990 edition. Table 18 gives values of mLT for common cases. Inaddition, it gives a simple formula to allow the appropriate value of mLTto be estimated for any shape of moment diagram. It should be notedthat, in each case, the relevant pattern of moments is that over the

BS 5950-1: 2000Technical detailsColin Taylor

Page 26 New Steel Construction, March/April 2001

STEEL DESIGN

length LLT of a segment of the member between adjacent lateralrestraints.

For example, consider the beam shown in figure 3. All loads arefactored.

Figure 3: Main beam supporting secondary beams at 3m centres.

The secondary beams provide lateral restraint to the main beam at B, Cand D. The segments from B to C and from C to D have the largestmoments and are therefore critical. The moments in segment BC areshown in figure 4.

Figure 4: Moment diagram for segment BC.

From Table 18, the equivalent uniform moment factor mLT is given by:

mLT = 0.2 + 0.15M2 + 0.5M3 + 0.15M4 but mLT 0.44

Mmax

mLT = 0.2 + 0.151348 + 0.51453 + 0.151554 = 0.904

1650

Thus the equivalent uniform moment mLT Mmax is 0.9041650 = 1492kNm. So the beam needs to have a buckling resistance moment Mb of atleast 1492 kNm and a moment capacity Mc of at least 1650 kNm.

5. Buckling resistance moment. The buckling resistance momentMb of a class 3 semi-compact or class 4 slender section is reduced, see4.3.6.4, by basing it on the effective plastic modulus or effective elasticmodulus, as for the moment capacity, see (1), whereas the 1990 editionuses the plastic modulus. This is off-set to some extent by using areduced slenderness, as for axial compression of a slender section, see(2).

For example, a 15215223 UC has a b/T ratio of 11.2 compared tothe limits of 10 for class 2 compact and 15 for class 3 semi-compact,i.e. 10 and 15 for S275 steel. From 3.5.6.2 the effective plasticmodulus Sx.eff = Zx + (Sx - Zx )(3f /(b/T ) - 1)/(3f /2f - 1)= 164 + (182 - 164)(15/11.2 - 1)/(15/10 - 1) = 164 + 180.339/0.5 = 176.If the equivalent slenderness LT = uv = 100, the reduced equivalentslenderness = uv[w ]0.5 = uv[Sx.eff/Sx]0.5 = 100(176/182)0.5 = 98.3, see4.3.6.7.

6. Plate girders. The beam shown in figure 5 is a plate girder inS275 steel, with a web 1200mm 10mm and flanges 400mm 40mm.Secondary beams at B, C, D and E apply loads to the web of the mainbeam through the top flange. They also provide the main beam withlateral restraint at those points. All loads are factored.

Figure 5: Plate girder

The maximum bending moment is 5550 kNm (at the centre of the span)and the moments at B and C are 3696 kNm and 5544 kNm respectively.The maximum shear is 1240 kN and the shears at B and C are 1224 kNand 608 kN respectively.

The shear capacity Pv = 0.6275120010/1000 kN = 1980 kN.Thus, given sufficient stiffeners, the web is adequate. Assume initiallythat load carrying or bearing stiffeners will be required at A and F toresist the end reactions and at B, C, D and E to resist the loads appliedthrough the top flange, then check the shear buckling resistance of theweb to see whether any intermediate web stiffeners are required.

For the panel A-B, the ratio a/d = 3000/1200 = 2.5 and d/t = 120, sofrom Table 21(1) qw = 102 N/mm2 thus from 4.4.5.2 the shear bucklingresistance of the web Vw = dtqw 120010102/1000 = 1224 kN which isless than the applied shear Fv of 1240 kN. Therefore an intermediatestiffener is needed between A and B.

Try a stiffener at a point A* located 1200 mm from A, see figure 6.For the end panel A-A*, the ratio a/d = 1200/1200 = 1.0, so qw= 128 N/mm2 and Vw = 120010128/1000 = 1536 kN, which is morethan the applied shear Fv . As an end panel, it will need an end post,unless the critical shear resistance Vcr is also more than Fv , see 4.4.5.4.The ratio Vw/Pv = 1536/1980 = 0.78 which is more than 0.72, so from4.4.5.4 Vcr = (9Vw - 2Pv)/7 = (91536 - 21980)/7 = 1409 kN > 1240 kN.The 1200 mm end panel is therefore satisfactory, but note in passingthat for a 1500 mm panel due to putting the stiffener A* mid-way fromA to B, the value of Vcr would be less than the applied shear Fv .

The maximum shear Fv in the 1800 mm panel A*-B is 1234 kN, theratio a/d = 1.5 and qw = 111 N/mm2 , hence Vw = 1332 kN > 1234 kN, sothis panel is also satisfactory. In the panel B-C, Fv = 624 kN, a/d = 2.5

New Steel Construction, March/April 2001 Page 27

and Vw = 1224 kN, so no intermediate stiffener is needed. The panel C-Ddoes not need checking.

With a web d/t ratio of 120, the section is just class 3 semi-compactand so the section modulus Zx should be used. For 40 mm flanges, py= 265 N/mm2 , so the moment capacity Mc = py Zx = 26521477/1000= 5691 kNm, which is more than the maximum applied moment of5550 kNm.

Checks are also needed on the moments at C and B, combined withthe maximum shears in the adjacent web panels. At C, panel B-Capplies, so the ratio Fv /Vw = 624/1224 = 0.51 < 0.6. Thus the momentcapacity at C is not reduced by shear buckling, see 4.4.4.2(a). At B,the moment capacity of the flanges alone is 400401240265/106

= 5258 kNm which is more than the applied moment of 3696 kN, soagain no reduction is needed for shear.

A check is also needed on lateral-torsional buckling between thelateral restraints, but this is omitted here for brevity. In addition theweb stiffeners have to be sized, see the next example.

NOTE: This example has been devised to illustrate variousprovisions of the code. However, if the web plate is made 12 mm thickinstead of 10 mm, the required shear buckling resistance can be achievedwithout introducing an additional stiffener at A*, nor needing to anchorany tension fields. Also because the web d/t ratio is increased to 100, thesection becomes class 2 compact and smaller flange plates can be used toobtain the same moment capacity, leading to a reduction in the weight ofsteel used.

7. Web bearing and buckling. New rules are given for the webbearing capacity and buckling resistance under point loads and endreactions applied to the web through a flange, see 4.5.2 and 4.5.3.That for buckling is simpler than before, but both are liable to be moreconservative. Bearing now comes first, because the bearing capacity isnow used in determining the buckling resistance.

The plate girder used in (6) will be used again here. Figure 6shows one end of this girder, including the intermediate stiffener addedbetween A and B. The ends of the girder have end plates, but thestiffeners at B and C need to be designed to resist a local load of 600 kNapplied through the flange. The stiff bearing length b1 , see 4.5.1.3 andFigure 13 of BS 5950-1: 2000, depends on the cross-section geometry ofthe member that applies the load. In this case b1 = 60 mm.

Figure 6: Web stiffeners of welded plate girder

The bearing capacity Pbw of the unstiffened web of a welded Isection, except at the end of the member, see 4.5.2.1, is given by(b1 + 5T)tpyw . Thus Pbw = (60 + 540)10275/1000 kN = 715 kN whichis more than 600 kN, so no stiffener is needed to prevent crushing of theweb. The buckling resistance Px of the unstiffened web, in this case, see4.5.3.1 , is given by 25tPbw/[(b1 + 5T ) d ]0.5. So Px= 2510715/[(60 + 540)1200]0.5 = 320 kN which is less than 600 kNand thus a stiffener is required to prevent web buckling.

The design of the load carrying stiffener is given in 4.5.3.3. Try apair of stiffeners each 100 mm 10 mm (b/(ts) < 13, see 4.5.1.2).Including 300 mm of web, the complete cruciform area As of the stiffeneris 5000 mm2 , Is about the centreline of the web is 10(2100 + 10)3/12+ (300 - 10)103/12 = 7741667 mm4 and hence rs = (7741667/5000)0.5

= 39.3 mm. The top flange is restrained in position and direction in theplane of the stiffener by the secondary beam, thus = 0.71200/39.3= 21.4 and using curve (c) of Table 24, pc = 268 N/mm2. So the bucklingresistance is 5000268/1000 = 1340 kN. This is much more than 600 kN,but a smaller stiffener would seem disproportionate.

These stiffeners also serve as intermediate stiffeners. The rule in4.4.6.4 is unchanged, giving a required Is for the stiffener (using theactual web thickness t for simplicity) of 900000 mm4, which is less thanthe 7717500mm4 contributed by the stiffeners alone. A pair of100 mm10 mm stiffeners could also be used for the intermediatestiffener in the end panel.

The tension field in the 1800mm panel A*-B needs to be anchoredby the end panel A-A*, and this imposes extra forces on the stiffenersbounding it. For the panel A*-B, the ratio Vw /Pv = 1332/1980 = 0.67 sothe value of Vcr is given by (Vw /0.9)2 /Pv = (1332/0.9)2/1980 = 1106 kN.Then from H.4.1, using the option for Fv < Vw , the longitudinal anchorforce Hq = 0.5dtpy ((Fv - V cr )/(Vw - V cr ))[1 - V cr /Pv ] 0.5

= 0.5120010275/1000((1234 - 1106)/(1332-1106))[1 - 1106/1980] 0.5

= 621 kN. From H.4.4 the additional compression in the end stiffenerFtf = 0.15Hqd/ae = 0.156211200/1200 = 93 kN.

If the end of the girder is supported from beneath, this 93 kNshould be added to the end reaction of 1240 kN. However, if the girder issupported by bolting its end plate to the supporting member, this doesnot apply. Assuming this applies here, a 250 mm15 mm end plate isprobably suitable, subject to checking the details of the boltedconnection. There is also a balancing tension of 93 kN in the next-to-endstiffener at A*, producing a tensile stress of 931000/10/200 = 47 N/mm2,which is less than 275 N/mm2 and is therefore satisfactory.

From H.4.4, the anchor force produces a longitudinal shear forceRtf in the end panel of 0.75Hq = 466 kN. As the end panel is square, itsresistance Vcr,ep = Vcr = 1409 kN, which is sufficient.

8. Combined axial load and bending. The most important changein 4.8 is that the use of equivalent uniform moment factors m has beenclarified. The m factor to be used in each case is related to the patternof moments about the axis being considered, taken over the lengthbetween lateral restraints against the form of buckling beingconsidered. Thus a member subject to biaxial bending, with 2 axes ofbending and two possible planes of buckling, will have 2 2 = 4different m factors (m LT, m x, m y and myx), see 4.8.3.3.4.

As for lateral-torsional buckling, the m factor for in-plane bucklingcan now be estimated approximately for any pattern of moments byusing the general method given in Table 26 of the code.

An innovation is the recognition that a sufficiently large moment inone plane can, in some cases, reduce the resistance of a member tobuckling in the perpendicular plane due to axial compression, see4.8.3.3.2 and 4.8.3.3.3. However in other cases, the method forinteraction between buckling about two axes can be more severe andhas therefore been retained alongside the new method.

It should be noted that the use of the more exact method for themember buckling resistance of compression members with moments isnow limited to I or H sections with equal flanges, and to CHS, RHS andbox sections. For other sections, such as channels, the simplifiedmethod given in 4.8.3.3.1 should be used.

It should also be noted that, in the more exact method, the 0.5factor in the expression for in-plane buckling about the minor axis hasbeen increased to 1.0 for I and H sections only.

9. Connections. There are a number of important changesconcerning the design of bolted and welded connections, which it isintended to cover in detail in a subsequent article.