Design of Beams (Unit - 8)

Design of Beams (Unit - 8) Contents Introduction Beam types Lateral stability of beams Factors affecting lateral stability Behaviour of simple and built - up beams in bending (Without vertical stiffeners) Design strength of laterally supported beams Design strength of laterally unsupported beams Shear strength of beams Maximum deflection Design of Purlins

Introduction Beams are structural elements subjected to transverse loads in the plane of bending causing BMs and SFs. Symmetrical sections about z-z axis are economical and geometrical properties of such sections are available in SP (6) The compression flange of the beams can be laterally supported (restrained) or laterally unsupported (unrestrained) depending upon whether restraints are provided are not. The beams are designed for maximum BM and checked for maximum SF, local effects such as vertical buckling and crippling of webs and deflection. Beams can be fabricated to form different types of c/s for the specific requirements of spans and loadings. Section 8 shall be followed in the design of such bending members. Types of beam cross sections Beams can be of different cross sections depending on the span and loadings and are shown below -

C / S of Plate Girders

Simple I sections are used for normal spans and loadings with all the geometrical properties available in IS 800 : 2007. All the other sections indicated in the figure are built up sections. These sections are used when the normal I sections become inadequate due to large spans and loadings. These sections are also used due to other functional requirements. I section with cover plates are used when the loads are heavy and the spans are large. If the depth of the beam is restricted due to functional reasons, smaller depth I sections with cover plates can be used Additional cover plates increases the lateral load resistance with increase in I YY. The properties of ISMB and ISWB sections with cover plates are available in SP (6). (Z PZ has to be obtained from calculations)

Two I sections with cover plates can be used when very heavy loads and spans act on the beam. The properties of these sections are not available in SP (6) and have to be calculated. Two I sections placed one above the other are used when the loads are light with large spans, where deflection is the main criteria. The properties of these sections are not available in SP (6) and have to be calculated Gantry girders are used in industrial buildings to lift loads and typical sections used are indicated in the figure. The properties of these sections are available in SP (6), (Z PZ has to be obtained from calculations) Plate girders are used where the spans exceed 20m and the loads are heavy. The properties of these sections are available in SP (6), (Z PZ has to be obtained from calculations) Box sections have large torsional rigidity and can be used as single cell, twin cell or multi - cell sections. The openings are advantageously used for service lines. Castellated beams are special sections fabricated from I sections and are used for light loads and large spans. The openings are advantageously used for service lines. In all built up beams, the fabrication cost is higher due to the provision of connections between the elements. Section Classification There are four classes of section namely Plastic, Compact, Semi - Compact and Slender sections as given in IS 800 : 2007. [cl. 3.7.2 pp - 17] For design of beams, only Plastic and Compact sections are used. Lateral Stability of Beams A beam transversely loaded in its own plane can attain its full capacity (Plastic moment) only if local and lateral instabilities are prevented. Local buckling of beams can be due to web crippling and web buckling. They are avoided by proper dimensioning of the bearing plate and through secondary design checks. Flanges shall always satisfy the outstand to thickness ratio as per IS 800 : 2007 so that local failures of flanges are avoided. Plastic and Compact sections are used. Lateral buckling of beams is the out of plane bending and is due to compressive force in the flange and is controlled by providing sufficient lateral restraint to the compressive flange.

Lateral stability of beams is affected by span of the beam, moment of inertia and the support conditions. Local failures of flanges (Secondary design checks) The local failure of flanges (plates) reduces the plastic moment capacity of the section due to buckling and is avoided by limiting the outstand to thickness ratios as given in IS 800: 2007. Local failures of web (Secondary design checks) The web of a beam is thin and can fail locally at supports or where concentrated loads are acting. There are two types of web failure - Web Crippling (or Crimpling) Web crippling causes local crushing failure of web due to large bearing stresses under reactions at supports or concentrated loads. This occurs due to stress concentration because of the bottle neck condition at the junction between flanges and web. It is due to the large localized bearing stress caused by the transfer of compression from relatively wide flange to narrow and thin web. Web crippling is the crushing failure of the metal at the junction of flange and web. Web crippling causes local buckling of web at the junction of web and flange.

For safety against web crippling, the resisting force shall be greater than the reaction or the concentrated load. It will be assumed that the reaction or concentrated load is dispersed into the web with a slope of 1 in 2.5 as shown in the figure Let Resisting force = F wc Thickness of web = t w Yield stress in web = f yw Width of bearing plate = b 1 Width of dispersion = n 2 = 2.5 h 2 Depth of fillet = h 2 (from SP [6]) F wc = [(b 1 + n 2 ) t w f yw ] / γ mo Reaction, R U For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as F wc = [(b 1 + 2 n 2 ) t w f yw ] / γ mo Concentrated load, W U

Web Buckling The web of the beam is thin and can buckle under reactions and concentrated loads with the web behaving like a short column fixed at the flanges. The unsupported length between the fillet lines for I sections and the vertical distance between the flanges or flange angles in built up sections can buckle due to reactions or concentrated loads. This is called web buckling.

For safety against web buckling, the resisting force shall be greater than the reaction or the concentrated load. It will be assumed that the reaction or concentrated load is dispersed into the web at 45 as shown in the figure. Let Resisting force = F wb Thickness of web = t w Design compressive stress in web = f cd Width of bearing plate = b 1 Width of dispersion = n 1 F wb = (b 1 + n 1 ) t w f cd Reaction, R U For concentrated loads, the dispersion is on both sides and the resisting force can be expressed as F wc = [(b 1 + 2 n 1 ) t w f cd ] Concentrated load, W U The design compressive stress f cd is calculated based on a effective slenderness ratio of 0.7 d / r y, where d = clear depth of web between the flanges. r y = radius of gyration about y-y axis and is expressed as = (I yy / area) = [(t w ) 3 / 12] / t = (t w ) 2 / 12] kl / r y = (0.7 d) / (t w ) 2 / 12] = 2.425 * d / t w Design compressive stress in web, f cd for the above slenderness ratio is obtained from curve, C (Buckling class C) (Table 9c, pp 42) Shear lag effects In simple theory of bending, plane sections remain plane before and after bending require that no SF is present in the beam. But in practice, SF influences the bending stress in the flanges and causes the section to warp. This results in non uniform distribution of flexural stresses in the flanges with stress being greater at the junction of flange and web. This is known as shear lag effect. In built up beams with wide flanges, this can be considerable, while in normal simple I sections, it is negligible. Shear lag depends on the width to span ratio and is specified in cl. 8.2.1.5 (pp - 53 and 54)

Distribution of compressive stress in the flange Laterally Supported (Restrained) beams Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (I YY ) and its presence reduces the plastic moment capacity of the section. Beams where lateral buckling of the compression flange are prevented are called laterally restrained beams. Such continuous lateral supports are provided in two ways - i) The compression flange is connected to an RC slab throughout by shear connectors. ii) External lateral supports are provided at closer intervals to the compression flange so that it is as good continuous lateral support. Cl 8.2.1 (pp - 52 and 53) gives the specifications in this regard. Typical lateral supports are shown in the figure.

Typical lateral supports Design of such laterally supported beams are carried out using Clauses 8.2.1.2, 8.2.1.3, 8.2..1.5, 8.4, 8.4.1, 8.4.1.1, 8.4.2.1 and 5.6.1 (Deflection) In addition, the beams shall be checked for vertical buckling of web and web crippling. The design is simple, but lengthy and does not involve trial and error procedure. Design steps for laterally supported beams The design of laterally supported beams consists of selecting a section based on the plastic section modulus and checking for its shear capacity, deflection, web buckling and web crippling. Most of the equations are available in IS 800 : 2007. The steps are - i) The maximum BM and SF at collapse is calculated based on the service loads (characteristic loads) and the span of the beam. Factored load at collapse = 1.5 * characteristic loads.

ii) The plastic section modulus, Z PZ is calculated using Z PZ = M U / (f Y / 1.1) M U = Maximum BM f Y = Yield stress of the given grade of steel 1.1 = Partial safety factor in yielding iii) A trial section having the appropriate plastic section modulus is adopted using IS 800 or SP (6) depending upon the type of section required. The section shall be plastic or compact section. iv) The beam shall be checked for shear lag and design bending strength as given in cl. 8.2.1.5 and 8.2.1.2 (pp -53) v) The beam is checked for deflection using appropriate formula depending on the type of loadings. vi) The section is checked for shear as given in cl. 8.4, 8.4.1 and 8.4.1.1. If V U 0.6 V d it is a case of high shear or otherwise low shear. For high shear, the design bending strength is calculated from cl. 9.2 vii) The section is checked for web buckling and crippling using appropriate formula. Laterally unsupported beam Beams subjected to BM develop compressive and tensile forces and the flange subjected to compressive forces has the tendency to deflect laterally. This out of plane bending is called lateral bending or buckling of beams. Lateral buckling of beams involves three kinds of deformations namely lateral bending, twisting and warping. The lateral bending of beams depends on the effective span between the restraints, minimum moment of inertia (I YY ) and can reduce the plastic moment capacity of the section.

Factors affecting lateral stability Type of C/S - The lateral buckling strength can be improved by choosing an appropriate c/s where I YY is large. Box sections satisfies this and also has large torsional rigidity as it is a closed section. Open sections like I sections have low torsional rigidity and are more susceptible to lateral instability. Cl 8.2.2 pp - 54 mentions that hollow sections need not be checked for lateral buckling strength. Support conditions - The lateral restraint provided depends on the restraint provided by the supports. The effect of various support conditions is taken into account using the concept of effective length as given in Table 15 - pp 58 for simply supported beams and Table 16 - pp 61 for cantilever beams. Effective length - This concept incorporates the various types of restraints to the flanges and for simply supported beams Table 15 - pp 58 can be used. The same information for cantilever beams is given in Table 16 - pp 61. Beams without proper restraint of the compression flange undergoes lateral buckling resulting in lesser load carrying capacity. Behavior of beams The actual behavior of beams depends on whether the beam is allowed to reach its plastic moment capacity. If the beam is prevented from local and lateral buckling, the beam reaches its full plastic capacity with plastic hinges formed at critical points of maximum BM which has been described in plastic analysis. If the beam is not restrained laterally, the beam can undergo elastic lateral torsional buckling and can fail due to instability with large lateral deflections, rotations and warping. If the web is too thin, the beam can fail in shear due to diagonal compression. The beam can also fail due to local effects such as web buckling, web crippling or distortion of flanges if these effects are not considered in the design. Lateral Torsional Buckling of Beams The presence of compression in the flanges causes lateral deflection (side sway) along with rotation known as Lateral Torsional Buckling of Beams. The assumptions made in the analysis are - i) The beam is initially undistorted without residual stresses. ii) The beam behaves elastically upto failure. iii) The beam is subjected to pure bending in the plane of web

If lateral restraint is not provided to the compression flange, buckling of a beam takes place about its minor axis, accompanied by twisting moment and warping. The load at which the beam buckles will be much less than the load causing full plastic moment. The design bending compressive stress is dependent on a factor called non - dimensional slenderness ratio, λ LT, which in turn is dependent on the elastic lateral torsional buckling moment, M cr. The value of M cr can be obtained by solving a fourth order differential equation. The value of M cr can b calculated using the equations given in cl. 8.2.2.1 pp -54 for doubly symmetric c/s and annex E (pp 128-129) for c/s symmetrical about the minor axis. The design bending compressive strength can be calculated using a set of equations as specified in cl.8.2.2 (Table 13a and 13b) pp -54 to 57. Design steps for laterally unsupported beams The design of laterally unsupported beams consists of selecting a section based on the plastic section modulus and checking for its shear capacity, deflection, web buckling and web crippling. Most of the equations are available in IS 800 : 2007. The steps are - All the steps given in design of laterally supported beams shall be used here. Also, The plastic section modulus required is increased by 25-50%. The design bending strength is calculated using the appropriate provisions in the code for lateral supported beams. Other checks like deflection, shear and local criteria will be same.

Design strength of laterally supported and unsupported beams These are analysis problems where the strength of the beam is required. The design strength will be based on flexural or bending strength and shear strength. Bending strength of laterally supported beams are calculated using the provisions given 8.2.1.2 (pp -53) by knowing the plastic section modulus Z pz Bending strength of laterally unsupported beams are calculated using the provisions given 8.2.2 and 8.2.2.1 (pp -54) by knowing the plastic section modulus Z pz The shear strength of the c/s is obtained from cl. 8.4(pp -59). Shear strength of Beams Shear forces always exists with BMs and the maximum shear stress has to be checked with the shear yield stress. Shear stresses can become important if the depth of the beams are restricted and when beams are subjected to large concentrated loads near the supports. The distribution of shear stress at limit state (plastic) is shown below - The nominal shear yielding strength is based on the Von Mises yield criteria which assumes wide and thin webs without any local failures. The shear strength is expressed as τ y = f yw / 3 f yw = yield strength of the web The design strength V d = [A v f yw / 3] / 1.1 A v = Shear area as specified in cl. 8.4.1.1 pp - 59

The web can buckle elastically or inelastically depending on the ratio of d / t w. if this ratio exceeds 67 ε, where ε = (250 /f y ) and d = clear depth of the web between flanges, resistance to shear buckling has to be verified. Shear failure can occur due to excessive yielding of the web area if the shear capacity is exceeded. The beam will be a high shear condition if V U > 0.6 V d and the moment capacity of the section decreases and has to calculated using the provisions given in cl. 9.2.2 pp - 70 Maximum Deflection A beam may have adequate strength in flexure and shear and can be unsuitable if it deflects excessively under the service loads. Excessive deflection causes problems in the functioning of the structure. It can harm floor finishes, cause cracks in partitions and excessive vibrations in industrial buildings and ponding of water in roofs. Cl.5.6.1, 5.6.1.1 and Table 6 gives relevant specifications with respect to deflection. The beam size may have to be taken based on deflection, if the spans and loadings are large. Typical maximum deflection formulae for simple loadings are given below -

Example 1 Design a simply supported beam of span 8 m. The spacing of the beams are 4m with thickness of RC slab =150 mm, floor finishes = 1.4 kn / m 2 and light partitions = 1kN / m 2. The beam also carries a central concentrated load of 250 kn with all the loads being characteristic loads. The beam is laterally restrained. with grade of steel being Fe 490. Check the beam for deflection, shear, web buckling and crippling. Load on the RC slab = 0.15 * 25 + 1.4 + 1.0 + 4.0 = = 10.15 kn/m 2 To calculate the self weight of the beam - Total load on the beam = 10.15 * 4 * 8 + 250 = 574.8 kn Self weight of the beam = (Total load on the beam) / 350 = 574.8 / 350 = 1.64 kn/m Total UDL on the beam, w = 10.15 * 4 + 1.64 = 42.24 kn/m The beam loaded is shown in the figure. Design for flexure - Maximum BM at centre, M U = (42.24 * 8 2 / 8 + 250 * 8 / 4) * 1.5 = 1256.88 knm Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl. 8.2.1.2 pp 53) = (1256.88 x 10 6 * 1.1) / (1.0 * 350) = 3950.2 x 10 3 mm 3 Adopt ISWB 600 @ 145.1 kg/m (1.45 kn / m < 1.64 kn / m)

Thickness of flange, t f = 23.6 mm > 20mm, f Y = 330 N / mm 2, (Table 1, pp 14) Z P ) REQD = 4189.6 x10 3 mm 3 Z P ) PRO = 4341.63 x10 3 mm 3 (OK) Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * 4341.63 x10 3 * 330) / 1.1 = 1302.49 x 10 6 N mm = 1302.49 knm > M U (OK) To check the type of section - Outstand = b / 2 = 125 mm (fig 2, pp 19) Outstand / t f = 125 / 23.6 = 5.3 < 9.4 ε < 9.4 * (250 / 330) < 8.18 Hence the section is plastic (OK) Check for deflection - Maximum deflection = (5wL 4 ) / (384EI) + (WL 3 ) / (48EI) = (5 * 42.05 * 8000 4 ) / (384 * 2 x 10 5 * 115626.6 x 10 4 ) + (250 x 10 3 * 8000 3 ) / (48 * 2 x 10 5 * 115626.6 x 10 4 ) = 21.23 mm < L / 360 < 8000 / 360 < 22.22 mm (OK) 42.05 kn / m = 42.05 N / mm (with actual self weight of the beam) E = 2 x 10 5 N / mm 2 and I = I ZZ = I XX (from SP - 6, Table I) Check for shear lag effect - width of flange = 250 mm L / 20 800 / 20 = 400 mm (cl. 8.2.1.5, pp - 53) (OK)

Check for shear - V U = (42.05 * 8 / 2 + 250 / 2) * 1.5 = 439.8 kn V d = (A V * f yw ) / ( 3 * 1.1) (cl. 8.4, 8.4.1 and 8.4.1.1, pp - 59) = (600 * 11.8 * 330) / ( 3 * 1.1) = 1226.29 x 10 3 N = 1226.29 kn V U < 0.6 * V d < 735.77 kn (Low shear) (OK) d / t w = (600-23.6 * 2) / 11.8 = 46.85 < 67ε < 58.32 (cl 8.4.2.1, pp -59) (OK) Check for web Buckling - kl/r y = 2.425 * d / t w = 2.425 * (600-23.6 * 2) / 11.8 = 113.6 From Table 9(c), pp - 42, f cd = 97.82 N / mm 2 (f cd can also be calculated using the equations given in cl.7.1.2.1 as in compression members) F wb = (b 1 + n 1 ) t w f cd = (b 1 + 600 / 200) * 11.8 * 97.82 To get b 1, F wb = V U = 439.8 x 10 3 b 1 = 81.02 mm, say 90 mm 90 mm wide bearing plate is provided. (minimum of 75 mm shall be provided) Check for web crippling - F wc = [(b 1 + n 2 ) t w f yw ] / γ mo = [(90 + 2.5 * 46.05) * 11.8 * 330] / 1.1 = 726.14 x 10 3 N = 726.14 kn > V U (OK) Hence ISWB 600 @ 145.1 kg / m satisfies all the specifications and can be used for the given problem. Example 2 A floor plan has a series of secondary beams spaced at 2 m c/c supported on main beams spaced at 12 m c/c. The main beams are supported on columns spaced at 12 m c/c. The floor is used for commercial purpose. Design the main beam by assuming suitable loads. The beam is laterally restrained. with grade of steel being Fe 490. Check the beam for deflection, shear, web buckling and crippling. The layout for the given problem is down in the figure. SB = Secondary Beams MB = Main Beams C = Columns Design of Secondary Beam - Load on the RC slab = 0.1 * 25 + 1.4 + 1.0 + 4.0 = 8.9 kn/m 2

To calculate the self weight of the beam - Total load on the beam = 8.9 * 2 * 12 = 213.6 kn Self weight of the beam = (Total load on the beam) / 350 = 213.6 / 350 = 0.61 kn/m Total UDL on the beam, w = 8.9 * 2 + 0.61 = 18.41 kn / m Design for flexure - Maximum BM at centre, M U = (18.41 * 12 2 / 8) * 1.5 = 497.07 knm Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl. 8.2.1.2 pp 53) = (497.07 x 10 6 * 1.1) / (1.0 * 350) = 1562.22 x 10 3 mm 3 Adopt ISLB 500 @ 75 kg/m (0.75 kn / m >0.61 kn / m) Thickness of flange, t f = 15 mm < 20mm, f Y = 350 N / mm 2, (Table 1, pp 14) Z P ) REQD = 1574.1 x10 3 mm 3 ( with new self weight)

Z P ) PRO = 1773.67 x10 3 mm 3 (OK) Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * 1773.67 x10 3 * 350) / 1.1 = 564.35 x 10 6 N mm = 564.35 knm > M U (OK) Check for deflection - Maximum deflection = (5wL 4 ) / (384EI) = (5 * 18.55 * 12000 4 ) / (384 * 2 x 10 5 * 38579 x 10 4 ) = 64.91 mm > L / 360 < 12000 / 360 < 33.33 mm (UNSAFE) E = 2 x 10 5 N / mm 2 and I = I ZZ = I XX (from SP - 6, Table I) I ZZ = I XX ) REQD = (5wL 4 ) / (384E * 33.33) = (5 * 18.55 * 12000 4 ) / (384 * 2 x 10 5 * 33.33) = 75135 x 10 4 mm 4 Adopt ISMB 600 @ 122.6 kg / m (1.23 kn/m) Maximum deflection = 27.98 mm < 33.33 mm (OK) NOTE - The chosen section is heavier than ISLB 500 @ 75 kg/m and hence the design bending strength will be satisfactory. Other checks like shear lag effect, shear, web buckling and web crippling can be satisfied on similar lines. Design of Main Beam - Reaction from each secondary beam = (8.9 * 2 + 1.23) * 12 = 228.36 kn say, 230 kn Self weight of main beam = (Total load on the beam) / 350 = (230 * 5) / 300 = 3.8 kn/m The beam is shown in the figure below - Maximum BM at centre, M U = (597.8 * 6-230 * 4-230 * 2 - [3.8 * 6 * 6] / 2) * 1.5 = 3207.6 kn m Z P ) REQD = (M U * 1.1) / (β b * f Y ) ( from cl. 8.2.1.2 pp 53) = (3207.6 x 10 6 * 1.1) / (1.0 * 320) = 11026 x 10 3 mm 3 Single I section is not possible and it is proposed to provide I section with cover plates on either side. Properties are available in Table XIV for ISMB and ISWB sections in SP - 6

Z EZ = Z XX ) REQD = Z P ) REQD / 1.14 = 9672 x 10 3 mm 3 The section shall also be worked using deflection condition. δ MAX = (WL 3 / 24EI) * β * (3-4 * β 2 ) for each pair of concentrated loads + (WL 3 ) / (48EI) + (5wL 4 ) / (384EI) ; β = a / L Equating this to the maximum deflection, L / 360 = 33.33 mm, we get 33.33 = (230 * 1000 * 12000 3 ) / (24 * 2 x 10 5 * I ZZ ) * [ (2 / 12) * {3-4 * (2 / 12) 2 } + (4 / 12) * {3-4 * (4 / 12) 2 }] + (230 x 10 3 * 12000 3 ) / (48 * 2 x 10 5 * I ZZ ) + (5 * 3.8 * 12000 4 ) / (384 * 2 x 10 5 * I ZZ ) Solving for I ZZ, we get 33.33 = (1.57 x10 11 ) / I ZZ I ZZ ) REQD = 4.71 x 10 9 mm 4 = 471000 cm 4 No section in Table XIV gives this value of I ZZ. Adopt ISMB 600 @ 122.6 kg/m with 40mm thick plates. To calculate the width of the plate - 4.71 x 10 9 = 91813 x 10 4 + [(b * 40 3 ) / 12 + b * 40 * 320 2 ] * 2 b = 462.27 mm say, 475 mm I ZZ ) PRO = 4.81 x 10 9 mm 4 > I ZZ ) REQD Defection is within limits.

Mean Thickness of flange, t f = (475 * 40 + 210 * 20.8) / 475 =49.2 mm > 40 mm, f Y = 320 N / mm 2, (Table 1, pp 14) Z PZ = 3510.63 x 10 3 + 475 * 40 * 320 * 2 = 15670 x 10 3 mm 3 Design bending strength, M d = (β b * Z P * f Y) / 1.1 = (1.0 * 15670 x10 3 * 320) / 1.1 = 4558 x 10 6 N mm = 4558 knm > M U (OK) Check for shear - Self weight = 1.23. + [(475 * 40 * 2) / 1000 2 ] * 78.50 = 4.21 kn/m V U = (4.21 * 12 / 2 + 230 * 5) / 2) * 1.5 = 900.4 kn V d = (A V * f yw ) / ( 3 * 1.1) (cl. 8.4, 8.4.1 and 8.4.1.1, pp - 59) = (680 * 12 * 320) / ( 3 * 1.1) = 1370.5 x 10 3 N = 1370.5 kn > V U (OK) 0.6 * V d = 822.3 kn V U > 822.3 kn (High shear) Design bending strength has to be modified. d / t w = (600-20.8 * 2) / 12 = 46.53 < 67ε < 59.22 (cl 8.4.2.1, pp -59) (OK) M dv = M d - β * (M d - M fd ) (cl. 9.2.2, pp - 70) β = [(2V U ) / V d - 1] 2 = 0.0985 M fd = Z P ) fd * f y / 1.1 Z P ) fd = [(475-12) * 40 * 320 + (210-12) * 20.8 * 289.6] * 2 = 14.24 x 10 6 mm 3 M fd = 4142.55 knm M dv = 4558-0.0985 * (4558-4142.55) = 4517 knm > M U (OK) The other checks for web buckling, crippling and shear lag can be calculated as earlier. Example 3 A simply supported beam has an effective span of 8 m and the beam has a c/s ISWB 600 @ 145.1 kg/m. Calculate the design BM and the safe UDL, the beam can support. The beam is laterally restrained against torsion but partially restrained against warping. The grade of the structural steel is Fe 490. From Table 15, L LT = 0.85 * 8000 = 6800 mm The value of the Elastic Lateral Torsional Buckling moment, M cr is calculated using all the equations given in IS 800 : 2007

i) M cr = { (π 2 EI Y ) / (L LT ) 2 }* [GI t + (π 2 EI w ) / (L LT ) 2 ]} (cl. 8.2.2.1, pp 54) E = 2 x 10 5 N/mm 2, I Y = 5298.3 x 10 4 mm 4,G = 0.769 x 10 5 N/mm 2 ( cl. 2.2.4.1 pp - 12), The equations given below are from Annex E - 1.2 pp - 129 I t = Torsional constant = Σ b i (t i ) 3 / 3 = 2 * 250 * 23.6 3 / 3 + (600-23.6 * 2) * 11.8 3 / 3 = 2.493 x 10 6 mm 4 I w = warping constant = (1- β f ) β f I Y (h y ) 2 β f = I fc / (I fc + I ft ) = 0.5 ( for symmetrical sections about both the axis) h Y = c/c distance between the flanges = 600-23.6 = 576.4 mm I w = 4.4 x 10 12 mm 6 Substituting, M cr = 926.52 x10 6 N mm = 926.52 kn m ii) M cr = (π 2 EI Y h f ) / [2 * (L LT ) 2 ] }* [1+ 1 / 20 { (L LT / r y ) / (h f / t f )} 2 ] 0.5 h f = c/c distance between the flanges = 600-23.6 = 576.4 mm r y = 53.5 mm and t f = 23.6 mm Substituting, M cr = 1000.13 x 10 6 N mm = 1000.13 kn m iii) f cr,b = (1.1π 2 E) / (L LT / r y ) 2 * [1+ 1 / 20 { (L LT / r y ) / (h f / t f )} 2 ] 0.5 = 206.22 N/mm 2 Alternately, f cr,b can also be obtained from Table 14 - pp 57 M cr = β b * Z P * f cr,b = 1.0 * 4341.63 x10 3 * 206.22 = 895.33 x10 6 N mm = 895.33 kn m Each equation gives different values of M cr and the procedure given in Annex E is more accurate but lengthy and cumbersome. Consider the third approach as it gives the least values required and simple. If more accuracy is required, averaging can be done. M cr = 895.33 x10 6 N mm and f cr,b = 206.22 N/mm 2 will be used in further calculations.

λ LT = (f y / f cr,b ) = (330 / 206.22) = 1.265 Also, λ LT [(1.2 * Z E * f Y ) / M cr ] [(1.2 * 3854.2 x 10 3 * 330) / 895.33 x10 6 1.306 Adopt λ LT = 1.265 ( > 0.4 Hence lateral buckling analysis required) φ LT = 0.5 * [1+ α LT * (λ LT - 0.2) + (λ LT ) 2 ] = 0.5 * [1 + 0,21 * (1.265-0.2) + 1.265 2 ] = 1.412 χ LT = 1/ { φ LT + [ (φ LT ) 2 - (λ LT ) 2 ] 0.5 } Substituting, χ LT = 0.49 < 1.0 f bd = χ LT * f Y / 1.1 = 147 N/mm 2 M d = β b * Z P * f bd = 1.0 * 4341.63 x10 3 * 147 = 638.22 x 10 6 Nmm = 638.22 knm M SAFE = 638.22 / 1.5 = 425.48 knm w SAFE * L 2 / 8 = M SAFE w SAFE = 53.185 kn/m (including self weight) Other calculations can be carried out using an effective span of 8m as in laterally supported beams. Design of laterally unsupported beam The section can be chosen based on two conditions - i) Z P ) REQD = (M U * 1.1) / (β b * f Y ) * 1.25 to 1.5 ii) δ) PER = L / 360 for Simply supported beams = L / 180 for cantilever beams Relevant expressions for deflection are used based on the loadings. The moment of Inertia required is calculated..

A suitable section based on the above requirement is chosen and the design bending strength, M d is calculated as in Example 3. This shall be greater than M U. If required the section has to be modified for economy. Once the section is chosen, other checks shall be using the effective span as in laterally supported beams. Example1 can be treated as laterally unrestrained beam and worked. Design of Purlins Purlins are flexural members used in trusses to support the roof covering and spans between the trusses. Purlins are provided on the top rafter (top chord) at all the joints. The spacing of the purlins depends on the type of the roofing material and for normal materials, it ranges from 1.4 to 1.8 m. The sections used for purlins are usually angles (equal or unequal) as they are economical and variety of sections is available. A typical view of purlin is shown in the figure. The new code do not provide the design specifications. Therefore the specifications as per the old code IS:800 1984 is followed. Cl. 8.9 pp - 69 shall also be followed. Based on IS : 875 Part 2, LL on inclined roofs shall be taken as - LL = 0.75-0.02 / of the slope for slopes > 10 subjected to a minimum of 0.4 kn / m 2 For slopes 10, LL = 0.75 kn / m 2 DL of AC sheets = 0.17 kn / m 2 and GI sheets = 0.13 kn / m 2

Example 4 Design a suitable single angle purlin having AC sheets as covering with spacing of trusses = 4.5m. Pitch of the truss is 1 in 5 with spacing of the purlins = 1.6m. Wind pressure normal to the roof is 1,3 kn / m 2. Span of the truss = 18m. Rise of the truss = 1/5 * 18 = 3.6m Slope of the truss, tan θ = 3.6 / 9.0 = 0.4 θ = 21.8 For the given problem, LL = 0.75-0.02 * (21.8-10) = 0.514 kn / m 2 (DL + LL) on plan area = 0.17 + 0.514 = 0.684 kn / m 2 Vertical load on each purlin = 0.684 * 1.6 = 1.094 kn/m Self weight of the purlin = 0.1 kn/m Total vertical load = 1.194 kn/m Load acting normal to the purlin = 1.194 * cos θ = 1.12 kn/m (DL + LL) DL acting normal to the purlin = (0.17 * 1.6 + 0.1) = 0.37 kn/m (Downwards) WL acting normal to the purlin = 1.3 * 1.6 = 2.08 kn/m (Upwards) (DL + WL) = 1.71 kn/m (Upwards) (DL + WL) is governing for the design. Maximum BM, M = wl 2 / 10 = (1.71 * 4.5 2 ) / 10 = 3.46 knm (assumed as continuous spanned purlins) Z E ) REQD = M / 0.66 f Y = 3,46 x 10 6 / (0.66 * 250) assuming Fe410 grade steel = 21 x 10 3 mm 3 (21 cm 3 ) Minimum depth of angle = L / 45 =100 mm Minimum width of angle = L / 60 =75 mm From Table IV, SP - 6, choosing an angle ISA 125 x 75 x 6 (9.2 Kg/m) Z E ) PRO = 22.2 x 10 3 mm 3 > 21 x 10 3 mm 3 (OK)

Maximum deflection, δ = (w * L 4 ) / (384EI) (Continuous span) = (1.71 * 4500 4 ) / (384 * 2 x 10 5 * 187.8 x 10 4 ) = 4.86 mm < L / 360 = 12.5 mm (OK) The chosen section ISA 125 x 75 x 6 is OK