UNRESRAINED BEAM DESIGN II.0 INRODUCION he basic theor of beam buckling was explained in the previous chapter. Doubl smmetric I- section has been used throughout for the development of the theor and later discussion. It was established that practical beams fail b: (i) (ii) (iii) Yielding, if the are short Elastic buckling, if the are long, or Inelastic lateral buckling, if the are of intermediate length. A conservative method of designing beams was also explained and its limitations were outlined. In this chapter a few cases of lateral buckling strength evaluation of beams encountered in practice would be explained. Cantilever beams, continuous beams, beams with continuous and discrete lateral restraints are considered. Cases of monosmmetric beams and non-uniform beams are covered. he buckling strength evaluation of non-smmetric sections is also described..0 CANILEVER BEAMS A cantilever beam is completel fixed at one end and free at the other. In the case of cantilevers, the support conditions in the transverse plane affect the moment pattern. For design purposes, it is convenient to use the concept of notional effective length, k, which would include both loading and support effects. he notional effective length is defined as the length of the notionall simpl supported (in the lateral plane) beam of similar section, which would have an elastic critical moment under uniform moment equal to the elastic critical moment of the actual beam under the actual loading conditions. Recommended values of k for a number of cases are given in able. It can be seen from the values of k that it is more effective to prevent twist at the cantilever edge rather than the lateral deflection. Generall, in framed structures, continuous beams are provided with overhang at their ends. hese overhangs have the characteristics of cantilever beams. In such cases, the tpe of restraint provided at the outermost vertical support is most significant. Effective prevention of twist at this location is of particular importance. Failure to achieve this would result in large reduction of lateral stabilit as reflected in large values of k, in able. Copright reserved Version II -
able Recommended values of k Restraint conditions Loading condition At support At tip Normal Destabilizing Built in laterall and torsionall Free 0.8λ.λ Lateral restraint onl 0.7λ.λ orsional Restraint onl 0.6λ 0.6λ Lateral and orsional Restraint 0.5λ 0.5λ Continuous with lateral and torsional restraint Free Laterall restraint onl.0λ 0.9λ.5λ.5λ orsional Restraint onl 0.8λ.5λ Laterall and orsional Restraint 0.7λ.λ Continuous, with lateral restraint onl Free.0λ 7.5λ Lateral restraint onl.7λ 7.5λ orsional Restraint onl Laterall and orsional Restraint.λ.6λ For continuous cantilevers λ not less than λ where λ is the length of the adjacent span For cantilever beams of projecting length, L, the effective length L L to be used shall be taken as in able (able 8. of New ) for different support conditions..λ.5λ Version II -
able Effective length for Cantilever Beams (able 8. of New, Effective Length, L lt, for Cantilever of Length L) (Section 8..) Restraint condition Loading conditions At Support At tip Normal destabilizing a) Continuous, with lateral restraint to top flange ) Free ) Lateral restraint to top flange ) orsional restraint ) Lateral and torsional restraint.0l.7l.l.l 7.5L 7.5L.5L.6L b) Continuous, with partial torsional restraint ) Free ) Lateral restraint to top flange ) orsional restraint ) Lateral and torsional restraint.0l.8l.6l.l 5.0L 5.0L.0L.L c) Continuous, with lateral and torsional restraint ) Free ) Lateral restraint to top flange ) orsional restraint ) Lateral and torsional restraint.0l 0.9L 0.8L 0.7L.5L.5L.5L.L d) Restrained laterall, torsionall and against rotation on plan ) Free ) Lateral restraint to top flange ) orsional restraint ) Lateral and torsional restraint 0.8L 0.7L 0.6L 0.5L.L.L 0.6L 0.5L op restraint conditions ) Free ) Lateral restraint to top flange ) orsional restraint ) Lateral and torsional restraint Version II -
.0 CONINUOUS BEAMS Beams, extending over a number of spans, are normall continuous in vertical, lateral or in both planes. In the cases, where such continuit is not provided lateral deflection and twisting ma occur. Such a situation is tpicall experienced in roof purlins before sheeting is provided on top of them and in beams of temporar nature. For these cases, it is alwas safe to make no assumption about possible restraints and to design them for maximum effective length. Another case of interest with regard to lateral buckling is a beam that is continuous in the lateral plane i.e. the beam is divided into several segments in the lateral plane b means of full effective braces. he buckled shapes for such continuous beams include deformation of all the segments irrespective of their loading. Effective length of the segments will be equal to the spacing of the braces if the spacing and moment patterns are similar. Otherwise, the effective length of each segment will have to be determined separatel. o illustrate the behaviour of continuous beams, a single-span beam provided with equall loaded cross beams is considered (see Fig. ). W W A B C λ λ λ D λ b Fig. Single - span beam wo equall spaced, equall loaded cross beams divide the beam into three segments laterall. In this case, true M cr of the beam and its buckling mode would depend upon the spacing of the cross beams. he critical moment M crb for an ratio of λ / λ b would lie in between the critical moment values of the individual segments. he critical moments for the two segments are obtained using the basic equation given in the earlier chapter. π.75 λ ( E I GJ) π λ M cr E Γ GJ (In the outer segment, m 0.57. Using / m and the basic moment, the critical moment is determined) π λ ( EI GJ) π λ M cr E Γ GJ (his segment is loaded b uniform moments at its ends basic case) () () Version II -
M cr and M cr values are plotted against λ / λ b and shown in Fig. for the particular case considered with equal loading and a constant cross section throughout. Dimensionless Critical Moment Mcr / 8 6 McruB 0 M crc M cr M cr Unbraced beams M cr M crub 0. 0. 0. 0.7 0. 0.5 Zero interaction point Dimensionless Load Position λ /λ b Fig. Interaction between M cr and M cr It is seen that for λ / λ b 0.7, M cr and M cr are equal and the two segments are simultaneousl critical. he beam will buckle with no interaction between the two segments. For an other value of λ / λ b there will be interaction between the segments and the critical load would be greater than the individual values, as shown in the figure. For values of λ / λ b < 0.7, outer segments will restrain the central segment and viceversa when λ / λ b > 0.7. he safe load for a laterall continuous beam ma be obtained b calculating all segmental critical loads individuall and choosing the lowest value assuming each segment as simpl supported at its ends. w (udl) w (udl) w (udl) λ λ λ λ Continuous beam and loading λ Deflected shape λ λ λ Buckled shape Fig. Continuous beam deflected shape and buckled shape It is of interest to know the behaviour of beams, which are continuous in both transverse and lateral planes. hough the behaviour is similar to the laterall unrestrained beams, Version II -5
their moment patterns would be more complicated. he beam would buckle in the lateral plane and deflect in the vertical plane. here is a distinct difference between the points of contraflexure in the buckled shape and points of contraflexure in the deflected shape. hese points will not normall occur at the same location within a span, as shown in Fig.. herefore, it is wrong to use the distance between the points of contraflexure of the deflected shape as the effective length for checking buckling strength.. PROVISIONS OF NEW (LSM VERSION) For beams, which are provided with members giving effective lateral restraint to the compression flange at intervals along the span, in addition to the end torsional restraint required, the effective length for lateral torsional buckling shall be taken as the distance, centre-to-centre, of the restraint members in the relevant segment under normal loading condition and. times this distance, where the load is not acting on the beam at the shear and is acting towards the shear centre so as to have destabilizing effect during lateral torsional buckling deformation. Where a member is provided as intermediate lateral supports to improve the lateral buckling strength, these restraints should have sufficient strength and stiffness to prevent lateral movement of the compression flange at that point, relative to the end supports. he intermediate lateral restraints should be either connected to an appropriate bracing sstem capable of transferring the restraint force to the effective lateral support at the ends of the member, or should be connected to an independent robust part of the structure capable of transferring the restraint force. wo or more parallel member requiring such lateral restraint shall not be simpl connected together assuming mutual dependence for the lateral restraint. he intermediate lateral restraints should be connected to the member as close to the compression flange as practicable. Such restraints should be closer to the shear centre of the compression flange than to the shear centre of the section. However, if torsional restraint preventing relative rotation between the two flanges is provided, the intermediate lateral restraint ma be connected at an appropriate level. For beams which are provided with members giving effective lateral restraint at intervals along the span, the effective lateral restraint shall be capable of resisting a force of.5 percent of the maximum force in the compression flange taken as divided equall between the points at which the restraint members are provided. Further, each restraint point should be capable of resisting percent of the maximum force in the compression flange. For Purlins which are adequatel restrained b sheeting need not be normall checked for the restraining forces required b rafters, roof trusses or portal frames that carr predominatel roof loads provided there is bracing of adequate stiffness in the plane of rafters or roof sheeting which is capable of acting as a stressed skin diaphragm. Version II -6
.0 EFFECIVE LAERAL RESRAIN Providing proper lateral bracing ma increase the lateral stabilit of a beam. Lateral bracing ma be either discrete (e.g. cross beams) or continuous (e.g. beam encased in concrete floors). he lateral buckling capacit of the beams with discrete bracing ma be determined b using the methods described in a later Section. For the continuousl restrained beams, assuming lateral deflection is completel prevented, design can be based on in-plane behaviour. It is important to note that in the hogging moment region of a continuous beam, if the compression flange (bottom flange) is not properl restrained, a form of lateral deflection with cross sectional distortion would occur.. Discrete bracing In order to determine the behaviour of discrete braces, consider a simpl supported beam provided with a single lateral support of stiffness K b at the centroid, as shown in Fig.. M K b M Fig. Beam with single lateral support he relationship between K b and M cr is shown in Fig. 5..5 K bλ Mcr with bracing Mc r without bracing.0.5.0.5 K bλ K bλ 00 5 K bλ λ G J / E Γ 0.0 0 6 8 0 6 Non dimensional bracing stiffness K b λ / 8 E I Fig. 5 Relationship between K b and M cr Version II -7
It is seen that M cr value increases with K b, until K b is equal to a limiting value of K bλ. he corresponding M cr value is equal to the value of buckling for the two segments of the beam. M cr value does not increase further as the buckling is now governed b the individual M cr values of the two segments. Generall, even a light bracing has the abilit to provide substantial increase in stabilit. here are several was of arranging lateral bracing to improve stabilit. he limiting value of the lateral bracing stiffness, K bλ, is influenced b the following parameters. Level of attachment of the brace to the beam i.e. top or bottom flange. he tpe of loading on the beam, notabl the level of application of the transverse load pe of connection, whether capable of resisting lateral and torsional deformation he proportion of the beam. Provision of bracing to tension flanges is not so effective as compression flange bracing. Bracing provided below the point of application of the transverse load would not be able to resist twisting and hence full capacit of the beam is not achieved. For the design of effective lateral bracing sstems, the following two requirements are essential. Bracing should be of sufficient stiffness so that buckling occurs between the braces Lateral bracing should have sufficient strength to withstand the force transferred b the beam. A general rule is that lateral bracing can be considered as full effective if the stiffness of the bracing sstem is at least 5 times the lateral stiffness of the member to be braced. Provisions in BS 5950 stipulate that adequate lateral and torsional restraints are provided if the are capable of resisting ) a lateral force of not less than % of the maximum factored force in the compression flange for lateral restraints, ) and a couple with lever arm equal to the depth between centroid of flanges and a force not less than % of the maximum factored compression flange force.. Continuous restraint In a framed building construction, the concrete floor provides an effective continuous lateral restraint to the beam. As a result, the beam ma be designed using in-plane strength. A few examples of full restrained beams are shown in Fig. 6. he lateral restraint to the beam is effective onl after the construction of the floor is completed. he beam will have to be temporaril braced after its erection till concreting is done and it has hardened. For the case shown in Fig.6 (a), the beam is full encased in concrete, and hence there will be no lateral buckling. In the arrangement shown in Fig.6 (b), the slab rests directl upon the beam, which is left unpainted. Full restraint is generall developed if the load Version II -8
transmitted and the area of contact between the slab and the beam are adequate to develop the needed restraint b friction and bond. Friction connection Concrete topping (a) Beam full encased in concrete (b) Beam in friction connection with concrete Light weight Steel Decking Fig. 6 Beams with continuous lateral restraint (c) Beam with metal decking Shear studs For the case shown in Fig.6(c), the metal decking along with the concrete provides adequate bracing to the beam. However, the beam is susceptible to buckling before the placement of concrete due to the low shear stiffness of the sheeting. Shear studs are provided at the steel-concrete interface to enhance the shear resistance. he codal provisions require that for obtaining full effective continuous lateral bracing, it must withstand not less than % of the maximum force in the compression flange. 5.0 BUCKLING OF MONOSYMMERIC BEAMS For beams smmetrical about the major axis onl e.g. unequal flanged I- sections, the non-coincidence of the shear centre and the centroid complicates the torsional behaviour of the beam. he monosmmetric I-sections are generall more efficient in resisting loads provided the compressive flange stresses are taken b the larger flange. When a monosmmetric beam is bent in its plane of smmetr and twisted, the longitudinal bending stresses exert a torque, which is similar to torsional buckling of short concentricall loaded compression members. he longitudinal stresses exert a torque, M given b M M x β x dφ / dz () where β x / I x ( x ) da 0 () A is the monosmmetr propert of the cross section. Explicit expression for β x for a monosmmetric I-section is given in Fig. 7. he torque developed, m, changes the effective torsional rigidit of the section from GJ to (GJM x β x ). In doubl smmetric beams the torque exerted b the compressive bending stresses is completel balanced b the restoring torque due to the tensile stresses and therefore β x is zero. In monosmmetric beams, there is an imbalance of torque due to larger stresses in the smaller flange, which is farther from the shear centre. Version II -9
( ) / D - h B D S C t 0 h B ( )( ) ( )t D B B t D D h B / h α 0 x ) / ( ) / ( B B α Γ αb h / ( ) ( ) ( ) ( ) / / / t h B B h B B h x x I β Fig. 7 Properties of monosmmetric I-sections Hence, when the smaller flange is in compression there is a reduction in the effective torsional rigidit; M x β x is negative and when the smaller flange is in tension M x β x is positive. hus, the principal effect of monosmmetr is that the buckling resistance is increased when the larger flange is in compression and decreased when the smaller flange is in compression. his effect is similar to the Wagner effect in columns. he value of critical moment for unequal flange I beam is given b. M c G J E GJ E I m m πρ ρ π Γ π π λ λ (5) Where ρ m c I I (6) I c is the section minor axis second moment of area of the compression flange. he monosmmetr propert is approximated to β x 0.9h ( ρ m -) ( I / I x ) (7) and the warping constant Γ b Γ ρ m (- ρ m ) I h (8) Version II -0
Ver little is known of the effects of variations in the loading and the support conditions on the lateral stabilit of monosmmtric beams. However, from the available results, it is established that for top flange loading higher critical loads are alwas obtained when the larger flange is used as the compression flange. Similarl for bottom flange loading higher critical loads can be obtained when this is the larger flange. For ee- sections β x can be obtained b substituting the flange thickness or equal to zero; also for ee sections the warping constant, Γ is zero. 6.0 BUCKLING OF NON-UNIFORM BEAMS Non-uniform beams are often used in situations, where the strong axis bending moment varies along the length of the beam. he are found to be more efficient than beams of uniform sections in such situations. he non-uniformit in beams ma be obtained in several was. Rectangular sections generall have taper in their depths. I-beams ma be tapered in their depths or flange widths; flange thickness is generall kept constant. However, steps in flange width or thickness are also common. apering of narrow rectangular beams will produce considerable reduction in minor axis flexural rigidit, EI, and torsional rigidit, GJ; consequentl, the have low resistance to lateral torsional buckling. Reduction of depth in I-beams does not affect EI, and has onl marginal effect on GJ. But warping rigidit, EΓ, is considerabl reduced. Since the contribution of warping rigidit to buckling resistance is marginal, depth reduction does not influence significantl the lateral buckling resistance of I beams. However, reduction in flange width causes large reduction in GJ, EI and EΓ. Similarl, reduction in flange thickness will also produce large reduction in EI, EΓ, and GJ in that order. For small degrees of taper there is little difference between width-tapered beams and thickness tapered beams. But for highl tapered beams, the critical loads of thickness tapered ones are higher. hus, the buckling resistance varies considerabl with change in the flange geometr. Based on the analsis of a number of beams of different cross sections with a variet of loading and support conditions, the elastic critical load for a tapered beam ma be determined approximatel b appling a reduction factor r to the elastic critical load for an equivalent uniform beam possessing the properties of the cross section at the point of maximum moment r 7 γ 5 γ (9) S x0 D B γ S x D0 B S x section modulus. 0 0 (0) flange thickness. Version II -
D depth of the section. B flange width. Subscripts 0 and relate to the points of maximum and minimum moment respectivel. For the design of non-uniform sections, BS 5950 provides a simple method, in which the properties where the moment is maximum ma be used and the value of n is suitabl adjusted. he value of n is given b n.5 0.5 A sm / A lm.0 () Where A sm and A lm are flange areas at the points of the smallest and largest moment, and m.0 7.0 BEAMS OF UNSYMMERICAL SECIONS he theor of lateral buckling of beams developed so far is applicable onl to doubl smmetrical cross sections having uniform properties throughout its length. Man lateral buckling problems encountered in design practice belong to this categor. However, cases ma arise where the smmetr propert of the section ma not be available. Such cases are described briefl in this Section. he basic theor can also be applied to sections smmetrical about minor axis onl e.g. Channels and Z-sections. In this section, the shear centre is situated in the axis of smmetr although not at the same point as the centroid. In the case of channel and Z sections, instabilit occurs onl if the loading produces pure major axis bending. he criterion is satisfied for the two sections if: () for the channel section, the load must act through the shear centre Fig.8 (a) and, () for Z-section in a direction normal to the horizontal principal plane [Fig.8 (b)]. v x W Centroid (c) u x x W u x Shear center (s) v Centroid and shear center (a) Channel section (b) Z-section Fig. 8 Loading through shear centre Version II -
If these conditions are satisfied, M cr of these sections can be obtained using their properties and the theoretical equation. he warping constant Γ for the sections are: B h B ht Γ 6 B ht for a channel () Γ B h ( Bt h) [ t ( B Bh h ) t B h] for a Zsection () where, h distance between flange centroids t thickness of web B total width of flange flange thickness. It is ver difficult to obtain such loading arrangements so as to satisf the restrictions mentioned above. In such cases, the behaviour ma not be one of lateral stabilit; instead a combination of bending and twisting or bi-axial bending. For sections, which have smmetr about minor axis onl, their shear centre does not coincide with the centroid. his results in complicated torsional behaviour and theoretical predictions are not applicable. Fig. 9 shows the instabilit behaviour of sections with flanges of varing sizes and positions (top or bottom). It can be seen from the figure that sections with flange in the compression region are more advantageous. Non dimensional critical moment Mcr /(E I G J) / 0 8 6 0 0 00 000 Non dimensional torsional parameter λ G J / E I h Fig. 9 Effect of flange position and proportion on lateral stabilit While considering the case of tapered beams, it has been established, based on lateral stabilit studies, that variation of the flange properties can cause large changes in the lateral buckling capacit of the beam, whereas tapering of depth has insignificant influence on the buckling capacit. Version II -
8.0 SUMMARY In this chapter, lateral torsional buckling of some practical cases of beams has been explained. It is pointed out that for cantilever beams the tpe of restraint provided at fixed -end plas a significant role in their buckling capacities. orsional restraint of the cantilever beam has been found to be more beneficial than lateral restraint. In the case of beams with equall spaced and loaded cross beams the critical moment of the main beam and the associated buckling mode will depend on the spacing of the cross beams. Requirements for effective lateral restraint have been presented. Continuous restraint provided b concrete floors to beams in composite constructions of buildings is discussed. As discussed in an earlier chapter, the local buckling effects should be taken into account b satisfing the minimum requirements of the member cross-section. Cases of monosmmetric beams and non- uniform beams are also briefl explained. Finall cases of beams with un-smmetric sections are discussed and concluded that beams with flanges in the compression zone are more advantageous from the point of view of lateral torsional buckling. 9.0 REFERENCES. rahair N.S., he behaviour and design of steel structures, Chapman and Hall London, 977. Kirb P.A. and Nethercot D.A., Design for structural stabilit, Granada Publishing, London, 979 Version II -
Structural Steel Design Project Calculation sheet Problem I : Job No. Sheet of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB Date.8/0/07 A propped cantilever has a span of 9.8 m. it is loaded b cross beams at. m and 6.6 m from its left hand end. he ends of the beam and the loaded points are assumed to be full braced laterall and torsionall. 90 kn 65 kn. m. m. m A B he loads as given are factored. Design a suitable section for the beam: C D he bending moment diagram of the beam is as shown below. 60 kn m 08 kn m -0 KN m Section classification of ISMB 50: he properties of the section are: B Depth, h 50 mm h t wtf Width, B 50 mm Web thickness, t w 9. mm Flange thickness, t f 7. mm Version II -5
Structural Steel Design Project Calculation sheet Depth between fillets, d 79. mm. Radius of gration about minor axis, r 0. mm. Job No. Sheet of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB Date.8/0/07 Plastic modulus about major axis, Z p 5.6 x 0 mm Assume f 50 N / mm, E 00000 N / mm, γ m.0 Appendix I of () pe of section (i) flange criterion: b b b B 50 75 mm 75 7.. < 9.ε, where ε 50 f (ii) Web criterion: Hence o.k. d 79. 0. t 9. d < 8 ε t Hence o.k. b d Since < 9. ε and < 8 ε, the section is classified as plastic t t f w Now the moment gradients are different for different segments of the entire span viz. for AB, BC and CD. able. (Section.7.) of So based on the moment gradients, each segment will be checked against Lateral orsional Buckling and the corresponding safe bending stress will also be checked for each segment. Version II -6
Structural Steel eaching Project (iii) Job No. Sheet of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB Calculation sheet Lateral torsional buckling for segment AB: he beam length AB. m Check for Slenderness Ratio: Effective length criteria: Date.8/0/07 With ends of compression flanges full restrained for torsion at support but both the flanges are not restrained against Warping, effective length of simpl supported beam L L.0 L, where L is the span of the beam. able 8. of Hence, L L.0 x. M 00 mm, L L /r 00/0..86 Since the moment is varing from -0 k-nm to 60 k-nm, there will be moment gradient. So for calculation of f bd, Critical Moment, M cr is to be calculated. Now, Critical Moment, M c c c c c ( KL) 0.5 π EI K I GIt ( KL) w cr ( g j) ( g j Kw I π EI Where, c, c, c factors depending upon the loading and end restraint conditions (able F.) K, K w effective length factors of the unsupported length accounting for boundar conditions at the end lateral supports, Here, both K and K w can be taken as.0. and g distance between the point of application of the load and the shear centre of the cross section and is positive when the load is acting towards the shear centre from the point of application ) Clause F.. of Appendix F of j s 0.5 A (z - ) da /I z Version II -7
Structural Steel Design Project Calculation sheet Job No. Sheet of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB s coordinate of the shear centre with respect to centroid, positive when the shear centre is on the compression side of the centroid Here, for plane and equal flange I section, g 0.5 x h 0.5 x 0.5 0.5 M 5 mm j.0 ( β f ) h /.0 (when β f 0.5) h distance between shear centre of the two flanges of the cross section h - t f here, β f 0.5, and h h - t f 50 7..6 mm hence, j.0 x ( x 0.5 ) x.6/.0 0 and s 0 I b t /, for open section t i i x 50 x 7. (50 x 7.) x 9. 9.57 x 0 mm he warping constant, I w, is given b I w (-β f ) β f I h for I sections mono-smmetric about weak axis (-0.5) x 0.5 x 8 x 0 x.6 90965.6 x 0 mm 6 Modulus of Rigidit, G 0.769 x 0 5 N/mm Date.8/0/07 Here, ψ -0/60-0.5 and K.0, for which c.70, c 0 and c 0.676 Hence, Critical Moment, able F. of Appendix F of M c c c c c ( KL) 0.5 π EI K I GI w t ( KL) cr g j g j Kw I π EI π x00000.70 ( ) ( ) 0.5 5 x8x0 90965x0 0.769x0 x9.57x0 x00 (.0x00) 8x0 π x00000x8x0 Version II -8
Structural Steel Design Project Calculation sheet 9606.60 N-mm Job No. Sheet 5 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB Date.8/0/07 Calculation of f bd : Now, λ L β Z f M b p cr.0x 5.6x0 x50 96.06x 0.587 for which, φ L 0.5x[ α L ( λl 0.) λl ] 0.5x 0.( 0.587 0.) 0.587 0.7 [ ] 0 Clause 8.. of for which, χ L 5 { φl 0. [ φ L λ L ] } 5 { 0.7 0.895 f bd χ L f / γ m0 0.895 x 50 /.0 0. N/mm 0. [ 0.7 0.587 ] } Hence, M d β b Z p f bd.0 x 5.6 x 0./000 900.76/000 Max. Bending Moment, M max 60 kn-m.90 kn-m Hence, M d > M max (.90 > 60) ISMB 50 is adequate against lateral torsional buckling for the applied bending moments for the segment AB of the beam. (iv) Lateral torsional buckling for segment BC: he beam length BC. m Check for Slenderness Ratio: Effective length criteria: With ends of compression flanges full restrained for torsion at support but both the flanges are not restrained against Warping, effective length of simpl supported beam L L.0 L, where L is the span of the beam. able 8. of Version II -9
Structural Steel Design Project Job No. Sheet 6 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Calculation sheet Checked b. KB Date.8/0/07 Hence, L L.0 x. M 00 mm, L L /r 00/0. 76. Since the moment is varing from 60 k-nm to 08 k-nm, there will be moment gradient. So for calculation of f bd, Critical Moment, M cr is to be calculated. Now, Critical Moment, M c c c c c ( KL) 0.5 π EI K I GIt ( KL) w cr ( g j) ( g j Kw I π EI Where, c, c, c factors depending upon the loading and end restraint conditions (able F.) K, K w effective length factors of the unsupported length accounting for boundar conditions at the end lateral supports, Here, both K and K w can be taken as.0. and g distance between the point of application of the load and the shear centre of the cross section and is positive when the load is acting towards the shear centre from the point of application ) Clause F.. of Appendix F of j s 0.5 A (z - ) da /I z s coordinate of the shear centre with respect to centroid, positive when the shear centre is on the compression side of the centroid Here, for plane and equal flange I section, g 0.5 x h 0.5 x 0.5 0.5 M 5 mm j.0 ( β f ) h /.0 (when β f 0.5) h distance between shear centre of the two flanges of the cross section h - t f here, β f 0.5, and h h - t f 50 7..6 mm Version II -0
Structural Steel Design Project Calculation sheet hence, j.0 x ( x 0.5 ) x.6/.0 0 and s 0 I b t /, for open section t i i Job No. Sheet 7 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB x 50 x 7. (50 x 7.) x 9. 9.57 x 0 mm he warping constant, I w, is given b I w (-β f ) β f I h for I sections mono-smmetric about weak axis (-0.5) x 0.5 x 8 x 0 x.6 90965.6 x 0 mm 6 Modulus of Rigidit, G 0.769 x 0 5 N/mm Date.8/0/07 Here, ψ 08/60-0.5 and K.0, for which c., c 0 and c 0.998 Hence, Critical Moment, able F. of Appendix F of M c c c c c ( KL) 0.5 π EI K I GI w t ( KL) cr g j g j Kw I π EI π x00000. ( ) ( ) 0.5 5 x8x0 90965x0 0.769x0 x9.57x0 x00 (.0x00) 8x0 π x00000x8x0 0697985.9 N-mm Calculation of f bd : Now, λ L β Z f M b p cr.0x 5.6x0 x50 0697.98x 0.600 for which, φ L 0.5x[ α L ( λl 0.) λl ] 0.5x[ 0.( 0.600 0.) 0.600 ] 0.7 for which, χ L 0. 5 { φl [ φ L λ L ] } 0.7 0.7 0.600 0.89 { 5 0. [ ] } 0 Clause 8.. of Version II -
Structural Steel Design Project Job No. Sheet 8 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Calculation sheet Checked b. KB Date.8/0/07 f bd χ L f / γ m0 0.89 x 50 /.0 0.7 N/mm Hence, M d β b Z p f bd.0 x 5.6 x 0.7/000 05.7/000 Max. Bending Moment, M max 60 kn-m 0.5 kn-m Hence, M d > M max (0.5 > 60) ISMB 50 is adequate against lateral torsional buckling for the applied bending moments for the segment BC of the beam. (v) Lateral torsional buckling for segment CD: he beam length CD. m Check for Slenderness Ratio: Effective length criteria: With ends of compression flanges full restrained for torsion at support but both the flanges are not restrained against Warping, effective length of simpl supported beam L L.0 L, where L is the span of the beam. able 8. of Hence, L L.0 x. M 00 mm, L L /r 00/0. 06. Since the moment is varing from 08 k-nm to 0 k-nm, there will be moment gradient. So for calculation of f bd, Critical Moment, M cr is to be calculated. Now, Critical Moment, M c c c c c ( KL) 0.5 π EI K I GIt ( KL) w cr ( g j) ( g j Kw I π EI Where, c, c, c factors depending upon the loading and end restraint conditions (able F.) ) Clause F.. of Appendix F of Version II -
Structural Steel Design Project Job No. Sheet 9 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Calculation sheet Checked b. KB Date.8/0/07 K, K w effective length factors of the unsupported length accounting for boundar conditions at the end lateral supports, Here, both K and K w can be taken as.0. and g distance between the point of application of the load and the shear centre of the cross section and is positive when the load is acting towards the shear centre from the point of application j s 0.5 A (z - ) da /I z s coordinate of the shear centre with respect to centroid, positive when the shear centre is on the compression side of the centroid Here, for plane and equal flange I section, g 0.5 x h 0.5 x 0.5 0.5 M 5 mm j.0 ( β f ) h /.0 (when β f 0.5) h distance between shear centre of the two flanges of the cross section h - t f here, β f 0.5, and h h - t f 50 7..6 mm hence, j.0 x ( x 0.5 ) x.6/.0 0 and s 0 I b t /, for open section t i i x 50 x 7. (50 x 7.) x 9. 9.57 x 0 mm he warping constant, I w, is given b I w (-β f ) β f I h for I sections mono-smmetric about weak axis (-0.5) x 0.5 x 8 x 0 x.6 90965.6 x 0 mm 6 Modulus of Rigidit, G 0.769 x 0 5 N/mm Version II -
Structural Steel Design Project Calculation sheet Here, ψ 0/08 0 and K.0, for which c.879, c 0 and c 0.99 Hence, Critical Moment, Job No. Sheet 0 of 0 Rev. Job title: UNRESRAINED BEAM DESIGN Worked example: Made b. GC Date.6/0/07 Checked b. KB 0.5 π EI K I GIt ( KL) w Mcr c ( c g c j) ( c g c j) ( KL) Kw I π EI 0.5 5 π x00000x8x0 90965x0 0.769x0 x9.57x0 x00.879 (.0x00) 8x0 π x00000x8x0 5775.56 N-mm Date.8/0/07 able F. of Appendix F of Calculation of f bd : Now, λ L β Z f M b p cr 0.58.5x α λ 0. λ for which, φ L ( ).0x 5.6x0 x50 57.8x [ L L ] [ 0.( 0.58 0.) 0.58 ] 0 L 0.5x 0.7 for which, χ L 0. φ φ L λ L 0.7 0.7 0.58 5 { L [ ] } { 5 0. [ ] } 0 Clause 8.. of 0.898 f bd χ L f / γ m0 0.898 x 50 /.0 0.09 N/mm Hence, M β Z f d b p bd.0 x 5.6 x 0.09/000 9./000 Max. Bending Moment, M max 08 kn-m Hence, M d > M max (.9 > 08).9 kn-m ISMB 50 is adequate against lateral torsional buckling for the applied bending moments for the segment CD of the beam. Version II -