Chapter 5. Beams – Design of Steel Structures

5

Beams

5.1 Introduction

Beams are members that are supported and loaded transversely to the longitudinal axis of the member. The simplest type of beams are known as the simply supported beam (Figure 5.1(a)) in which one of the supports is a hinge which allows only rotation whereas the other support is a roller which allows both translation and rotation. Another type of beam that is commonly encountered is the cantilever beam (Figure 5.1(b)). In this type of beam, one end is rigid or fixed, which does not allow translation and rotation, whereas the other end is free. In a fixed beam (Figure 5.1(c)), both the ends of the beam are fixed or rigid, which do not allow translation and rotation. Other types of beams also encountered in steel structures are continuous and overhang beams. The internal forces developed in beams are usually bending moment (M) and shear force (V).

 

 

Figure 5.1 Types of beams

 

In steel structures, beams are encountered in different applications. Accordingly, they are called as girders, joists, purlins, stringers and girts. The commonly used sections as beams are I, channel, tube, rectangular hallow sections as shown in Figure 5.2(a)–5.2(e). When single sections are not able to resist the loads, built-up sections (Figure 5.2(f), 5.2(g)) are used. In a beam, as some portion is subjected to compression, there is a possibility of buckling. Beams with circular tubular sections and rectangular hallow sections undergo only bending without any buckling. However, beams with I, channel sections undergo lateral buckling in addition to bending when subjected to bending about the major axis since the moment of inertia about the minor axis is much less compared to that about the major axis. The lateral buckling of these sections may be prevented by providing continuous lateral support to the compression flange. Such beams are known as laterally supported beams.

 

 

Figure 5.2 Types of sections

 

The continuous lateral support to the compression flange may be provided by embedding the compression flange in a reinforced concrete (RC) slab (Figure 5.3(a)) or if the steel beam supports RC slab (Figure 5.3(b)) in which case the friction in between the compression flange of the steel beam and the RC slab prevents the lateral movement of the compression flange or by shear connectors in composite beams (Figure 5.3(c)) or by any other means which prevents the lateral movement of the compression flange. Steel beams connected to an RC slab by shear connectors are separately considered in Chapter 13.

 

 

Figure 5.3 Laterally restrained compression flange

5.2 Flexural Behaviour of Beams which do not Undergo Lateral Buckling

Laterally supported beams, in which no local buckling of plate elements occur, fail by flexure/shear or bearing. If a beam is progressively loaded, the beam deflects and the curvature varies along the length of the beam (Figure 5.4). Consider an infinitesimal length of the beam dx where the radius of the curvature is R. Then, the strain in the element at a distance y from the neutral axis is

 

 

Figure 5.4 Deformation of a beam

 

 

The corresponding stress is obtained from the idealized stress-strain diagram in Figure 5.5.

 

 

Figure 5.5 Idealised elasto-plastic stress-strain curve for steel

5.2.1 Elastic Behavior

When the strain in the extreme fibres εmax < εy, i.e., fmax < fy, the behaviour of the beam is elastic. The strain and stress vary linearly across the depth of the beam as shown in Figure 5.6. Therefore, the moment of resistance of the section is

 

 

Figure 5.6 Stress and strain through the depth of the beam for εmax ≤ εy i.e. fmaxfy

 

M = fmax Ze     (5.2)

 

where Ze = the elastic section modulus = I/ymax in which I is the moment of inertia of the cross section about the neutral axis.

5.2.2 Elastic/Plastic Behavior

When εmax = εy i.e., fmax = fy, the yielding of the extreme fibres takes place, whereas the inner fibres of the beam remain elastic. The corresponding moment of resistance is called the yield moment My and is given by

 

My=fy Ze     (5.3)

 

When εmax > εy, there is no increase in the value of fmax which is limited to fy but the inner fibres start yielding as shown in Figure 5.7. In this case, the moment of resistance is obtained by considering both the yielded portion and elastic core.

 

 

Figure 5.7 Stress-strain distribution through the depth of the beam for εmax > εy

 

 

where yc and yt are the distances of the centroids of the areas of the section which are yielded from the neutral axis.

5.2.3 Plastic Behavior

When εmax >> εy, yielding takes place through the entire depth of the beam as shown in Figure 5.8 and the cross section becomes plastic. At this stage, further deformation is possible at a constant moment and it is known as the formation of plastic hinge.

 

 

Figure 5.8 Stress-strain distribution through the depth of the beam for εmax >> εy

 

The moment capacity of the beam at which the plastic hinge is formed is known as the plastic moment Mp. The value of Mp is obtained from

 

 

where

Ac = Area in compression = A/2

At = Area in tension = A/2

A = Total area of the section

yc, yt = Distances of the centroids of the areas in compression and the tension from the equal area axis, respectively.

Therefore, plastic neutral axis (also known as equal area axis) divides the section into two equal parts. In other words, the plastic neutral axis of a section may be located by equating the area in compression to area in tension (Ac = At).

Mp may be expressed as

 

Mp = fy Zp     (5.6)

 

where is known as the plastic section modulus.

5.3 Flexural Behaviour of Beams which Undergo Lateral Buckling

Beams with I and channel sections which are not provided with continuous lateral restraint to the compression flange undergo lateral torsional buckling in addition to bending since the moment of inertia of these sections about the axis passing through the centroid parallel to flanges (major axis) is much greater than the moment of inertia about the axis passing through the centroid perpendicular to the flanges (minor axis). If a lateral restraint is provided at some spacing by the means of cross beams, then the portion of the main beam in between the cross beams buckles laterally. The lateral torsional buckling of the beam with an I section is shown in Figure 5.9.

 

 

Figure 5.9 Lateral torsional buckling of an I beam

5.3.1 Elastic Critical Moment

Consider the case of a simply supported I beam subjected to pure bending in the plane of the minor axis i.e., the moments are acting about the major axes. The ends of the beam can rotate freely in the planes of the major and minor axes. The critical value of the moment at which the equilibrium of the beam becomes critical is given by

 

 

where

l = Effective span of the beam

E = Young's modulus

G = Shear modulus

Iy = Moment of inertia of the I section about the minor axis

J = Torsion constant

Cw = Warping constant

5.3.2 Shear Centre and its Significance

Shear centre is the point in the plane of the cross section through which the resultant load has to pass so that the cross section is free from twisting. For certain sections, the shear centre coincides with the centroid of the cross section. For example, the hollow rectangular cross section, the hollow circular section, the symmetric I section, the Z section (Figure 5.10(a)). It does not coincide with the centroid of the cross section in certain sections like the channel section, the angle section, the T section, the unequal flange I section (Figure 5.10(b)). It may not matter whether the load is passing through the shear centre or not in the case of laterally supported beams because the compression flange is restrained. However, in the case of laterally unsupported beams, if the load does not pass through the shear centre, it may aggravate (destabilise) or reduce (stabilise) the lateral torsional buckling (Figure 5.11).

 

 

Figure 5.10 Shear centre (SC) of various sections

 

 

Figure 5.11 Effect of the point of application of the load

5.4 Shear Behaviour

In the earlier sections, the effect of bending moment M is considered. The other internal force in a beam, i.e. the shear force produces shear stress τ which is distributed through the depth of the cross section as shown in Figure 5.12 when a beam is in an elastic condition. Mathematically,

 

 

Figure 5.12 Shear stress distribution through the depth of a beam

 

 

where

V = Shear force

Q = Moment of the area above or below the depth where the shear stress is calculated

I = Moment of inertia of the section about axis z

tw = Thickness of the web

Since only the web (the clear depth between flanges) resists shear force, for design purposes, the average shear stress τavg is calculated using the expression

 

 

where dw = the clear depth of web and tw = the thickness of the web.

When the web yields due to shear stress, the nominal shear strength using von Mises criterion is given by

 

 

Then, the shear capacity of a beam when the section becomes plastic may be expressed as

 

Vp = 0.58 fy dw tw          (5.11)

 

A beam may fail due to shear yielding at a support as shown in Figure 5.13.

 

 

Figure 5.13 Shear yielding near a support

5.5 Web Buckling and Web Crippling

At the points of application of heavy concentrated loads and at supports, localized compressive stresses of high magnitude act in the web. These localized stresses may cause the web to buckle or cripple as shown in Figure 5.14. Web buckling and web crippling may be prevented by ensuring that the factored support reaction or the factored concentrated load (γf F) is less than or equal to the web buckling strength (Fxd) and the web crippling strength (Fw).

 

 

Figure 5.14 Web buckling and web crippling

5.5.1 Web Buckling

In the case of web buckling, the portion of the web may be considered as a strut with flanges providing restraint. The load spreads out over a finite length of web of known as the dispersion length. The dispersion length is considered as (b1 + n1) at a support and (b1 + 2n1) at the point of application of the concentrated load, where b1 is the stiff bearing length and n1 is the length of dispersion of a 45° line to the level of mid depth of the section as shown in Figure 5.15. Hence, the web buckling strength at a support is

 

 

Figure 5.15 Load dispersion lengths for web buckling

 

Fxd = (b1 + n1)tw fcd     (5.12)

 

where tw is the thickness of the web and fcd is the design compressive stress obtained from Table 4.3 for buckling class ‘c’. Similarly, at the point of application of a concentrated load, the web buckling strength is

 

Fxd = (b1 + 2n1)tw fcd     (5.13)

 

since the load dispersion takes place on either side. The effective length of the web is considered as 0.7 d, where d is the depth of the strut between the flanges.

5.5.2 Web Crippling

In the case of web crippling at a support, the crippling strength of the web is calculated using

 

Fw = (b1 + n2)tw fywm0     (5.14)

 

where

b1 = Stiff bearing length

n2 = Dispersion length through the flange to the web junction at a slope of 1:2.5 to the plane of the flange (Figure 5.16)

 

 

Figure 5.16 Load dispersion lengths for web crippling

 

fyw = Yield strength of the web

At the point of application of a concentrated load,

 

Fw = (b1 + 2n2)tw fywm0     (5.15)

 

since the dispersion of the load takes place on either side as shown in Figure 5.16.

5.6 Design Strength in Bending as per IS 800:2007

5.6.1 Laterally Supported Beams

When the factored design shear force does not exceed 0.6Vd, where Vd is the design shear strength of the cross section, the design bending strength Md is given by

 

Md = βb Zp fym0     (5.16)

 

where

βb = 1.0 for the plastic and compact sections

= Ze /Zp for semi-compact sections in which Ze and Zp are the elastic and plastic section moduli of the cross section, respectively

fy = Yield strength of the steel

γm0 = Partial safety factor = 1.1

To avoid the irreversible deformation under service loads, Md should be less than 1.2 Zp fy/γm0 for simply supported beams and 1.5 Zp fy /γm0 for cantilever beams.

5.6.1.1 Holes in Tension Flange

If there are holes in the tension flange, their effect on the design strength in bending need not be considered if

 

 

where

Anf /Agf = Ratio of the net to gross area of the flange in tension

fy/fu = Ratio of the yield strength to the ultimate strength of the steel

γm1/γm0 = Ratio of the partial safety factors against the ultimate and yield strengths

When Anf /Agf does not satisfy the above requirement, the reduced effective flange area Aef satisfying the above equation may be considered as the effective flange area instead of Agf.

5.6.1.2 Shear-Lag Effect

The shear-lag effect in flanges may be disregarded if

  1. b0 ≤ L0/20 for outstand elements
  2. bi ≤ L0/10 for internal elements

where

L0 = Length between points of zero moment (inflection) in the span

b0 = Width of the outstand

bi = Width of the internal element

5.6.1.3 Biaxial Bending

When a plastic/compact section is subjected to biaxial bending, the section should be designed such that

 

 

where My and Mz are the factored bending moments about the minor and major axes, respectively; Mdy and Mdz are the design bending strengths of the section about the minor and major axes, respectively.

In the above interaction criterion, α1 and α2 are constants. For I and channel sections, α1 = 1.0 and α2 = 2.0; for circular tubes, α1 = 2.0 and α2 = 2.0; for rectangular hollow sections, α1 = α2 = 1.66.

For a semi-compact section,

 

5.6.2 Laterally Un-supported Beams

Lateral torsional buckling need not be considered if

  1. the bending is about the minor axis (y) of the section
  2. the ratio between the moments of inertia about the major and minor axes is not high
  3. the non-dimensional slenderness ratio λLT is less than 0.4 in the case of major axis bending

5.6.2.1 Design Bending Strength

The design bending strength of laterally un-supported beams is governed by lateral torsional buckling and is given by

 

Md = βb Zp fbd     (5.20)

 

where fbd = the design bending compressive stress.

The design bending compressive stress fbd is expressed as

 

fbd = χLT fy/γm0     (5.21)

 

where

χLT = bending stress reduction factor to account for lateral torsional buckling

 

 

in which

 

 

αLT being imperfection parameter.

αLT = 0.21 for the rolled steel section and

αLT = 0.49 for the welded steel section

The non-dimensional slenderness ratio λLT is given by

 

 

where Mcr = elastic critical moment

fcr, b = extreme fibre bending compressive stress corresponding to the elastic lateral buckling moment

5.6.2.2 Elastic Critical Moment/Elastic Critical Stress

The elastic lateral buckling moment Mcr of simply supported prismatic members with a symmetric cross section is given by

 

 

The extreme fibre elastic critical stress fcr, b may be expressed as

or

 

 

where It = torsional constant

     Iw = warping constant

     Iy = moment of inertia about the minor axis

     ry = radius of gyration about the minor axis

     LLT = effective length for lateral torsional buckling

     E = Young's modulus

     hf = centre to centre distance between flanges

     tf = thickness of flange

5.6.2.3 Effective Length for Lateral Torsional Buckling

For simply supported beams and girders of the span L where no lateral restraint to the compression flange is provided but each end of the beam is restrained against torsion, the effective LLT for the lateral torsional buckling is given in Table 5.1. Restraint against torsion at the ends may be provided by

 

Table 5.1 The effective length LLT for lateral torsional buckling of simply supported beams

Note L = span of the beam D = overall depth of the beam

 

  1. web or flange cleats (Figure 5.17(a)), or
  2. bearing stiffeners acting in conjunction with the bearing of the beam (Figure 5.17(b)), or
  3. lateral end frames or external supports providing lateral restraint to the compression flanges at the ends (Figure 5.17(c)), or
  4. the ends of the beam built into walls (Figure 5.17(d)).

 

 

Figure 5.17 Types of torsion restraint

 

It is usually assumed that plane cross sections remain plane during twisting of circular bars subjected to torsion. However, when members with cross sections like solid rectangular or I or channel are subjected to torsion, the warping of cross sections takes place as shown in Figure 5.18. Warping means the out of plane deformation of a cross section when a member is subjected to torsion. If warping is allowed at both the ends, no longitudinal stresses (tension/compression) are developed in the members. Warping may be prevented (restrained) at one or both the ends by making the ends built-in or rigidly connected to a supporting member like a column. This induces additional longitudinal tensile or compressive stresses in the member.

 

 

Figure 5.18 Warping of an I beam

 

In simply supported beams with intermediate lateral restraints against lateral torsional buckling, the effective length for lateral torsional buckling LLT should be taken as the distance centre to centre of the restraint members in the relevant segment under a normal loading condition and 1.2 times this distance where the load is not acting on the beam at the shear centre and is acting towards the shear centre so as to have destabilizing effect during lateral torsional buckling.

5.7 Design Strength in Shear as per IS 800:2007

 

The factored shear force in a beam, VVd          (5.27)

 

where Vd = the design shear strength of a beam

 

= Vn /γm0     (5.28)

 

in which Vn is the nominal shear strength of the section and γm0 is the partial safety factor.

The nominal plastic shear resistance under pure shear is given by

 

 

where Av = shear area

fyw = yield strength of the web

The shear area Av for different sections is given below.

  1. The major axis bending of I and channel sections

    Av = htw (Hot rolled)

    Av = dtw (Welded)

  2. The minor axis bending of I and channel sections

    Av = 2btf (Hot rolled or welded)

  3. Rectangular hollow tubes of uniform thickness

    Av = Ah/(b + h) (Loaded parallel to depth h)

    Av = Ab/(b + h) (Loaded parallel to width b)

  4. Circular hollow tubes of uniform thickness, Av = 2A/π
  5. Plates and solid bars, Av = A

    A = cross sectional area

    b = overall breadth of the tubular section, the breadth of I section flanges

    d = clear depth of the web between the flanges

    h = overall depth of the section

    tf = thickness of the flange

    tw = thickness of the web

Fastener holes need not be accounted for the design shear strength calculation, if

 

 

If Avn does not satisfy this condition, the effective shear area may be considered satisfying this limit.

5.8 Limit State Serviceability – Deflection

A beam designed for limit state of collapse in bending and shear has to be checked for the limit state of serviceability. Deflection is one of the important parameter of the limit state of serviceability that has to be kept within acceptable limits. The excessive deflection of a beam not only causes a sense of insecurity to the occupants of the structure but also a damage to non-structural components like cladding. Hence, IS 800: 2007 limits the maximum deflection of a flexural member as given in Table 1.6.

In general, the maximum deflection of a beam may be expressed in the form

 

 

where W = total load on the span

L = span

E = Young's modulus

I = moment of inertia about neutral axis

K = coefficient depending on the type of beam and loading

= 5/384 for a simply supported beam with a uniformly distributed load

= 1/48 for a simply supported beam with a central concentrated load

= 1/8 for a cantilever beam with a uniformly distributed load

= 1/3 for a cantilever beam with a concentrated load at the free end

Example 5.1

A simply supported beam MB400 @ 61.6 kg/m has an effective span of 5 m. Find (i) the design bending strength of the beam (ii) the design shear strength of the beam (iii) the intensity of udl that the beam may carry under service condition (iv) the maximum deflection. Assume that the beam is laterally supported. The grade of the steel is E250.

From Appendix A, Iz = 20,458.4 cm4; Zez = 1,022.9 cm3; Zpz = 1,176.18 cm3

 

 

∴ From Table 1.7, the section is plastic. Hence, βb = 1.0

  1. Md = βb Zpz fy/γm0

      = 1.0 × 1,176.18 × 103 × 250/1.1 = 267.3 kNm

    This should not be greater than 1.2 Zez fy/γm0 = 1.2 × 1,022.9 × 103 × 250/1.1

                                                                         =279 kNm

    Md = 267.3 kNm

  2.  

     

    Figure 5.19

     

    The intensity of the load in working condition, w = wu /γf = 85.5/1.5 = 57 kN/m

Example 5.2

Design a simply supported I section to support the slab of a hall 9 m × 24 m with beams spaced at 3 m c/c. The thickness of the slab is 100 mm. Consider a floor finish load of 0.5 kN/m2 and a live load of 3 kN/m2. The grade of the steel is E250. Assume that an adequate lateral support is provided to the compression flange.

 

Weight of the slab = 0.1 × 3 × 25 = 7.5 kN/m

Weight of the floor finish = 3 × 0.5 = 1.5 kN/m

Live load = 3.0 × 3.0 = 9 kN/m

Total load = 7.5 + 1.5 + 9 = 18 kN/m

Factored load = 1.5 × 18 = 27 kN/m

Maximum bending moment, M = 27 × 92/8 = 273.4 kNm

Maximum shear force, V = 27 × 9/2 = 121.5 kN

Assuming that βb = 1.0, the plastic section modulus to be provided

From Appendix B, NPB 400 × 180 × 66.3 may be tried.

 

 

∴ From Table 1.7, the section is plastic. Hence, βb = 1.0

Md = βb Zpz fy/γm0
                                                      = 1.0 × 1,307.26 × 103 × 250/1.1 = 297 kNm

 

This should not be greater than 1.2 Ze fy /γm0 = 1.2 × 1,156.4 × 103 × 250/1.1 = 315 kNm

Md = 297 kNm > 273.4 kNm

 

 

 

Figure 5.20

 

 

Figure 5.21

 

Check for deflection

 

 

Not satisfactory.

Hence, NPB 450 × 190 × 77.57 may be tried.

Its Iz = 33,743 cm4

 

 

∴ NPB 450 × 190 × 77.57 may be provided.

Example 5.3

Re-design Example 5.2 using the MB 300 section and rectangular flats. M16 bolts of property class 4.6 and product Grade C may be used for connection.

The plastic section modulus to be provided = 1,202.8 cm3 (from Example 5.2). A section consisting of MB 300 with two plates 300 mm × 16 mm on the flanges may be tried.

 

 

 

Figure 5.22

 

 

∴ From Table 1.7, the section is plastic. Hence, βb = 1.0

 

                       Agf = 140 × 13.1 + 300 × 16 = 6,634 mm2
                                    Anf = 6,634 − 2 × 18 × 13.1 − 2 × 18 × 16 = 5,586 mm2

 

∴ The effect of holes need not be considered in the tension flange.

Zp = 2[300 × 16 × 158 + 140 × 13.1 × 143.8 + 136.9 × 7.5 × 136.9/2] = 2,183 cm3 > 1,202.8 cm3

Md = 1.0 × 2,183 × 103 × 250/1.1 = 496 kNM > 273.4 kNm                OK

 

Check for deflection

Iz of MB 300 = 8,980 cm4 (from Appendix-A)

Iz of entire section = 8,990 × 104 + 2 × (300 × 16) × (150 + 8)2 = 32,955 × 104 mm4

 

 

Design of connection

The horizontal shear force per unit length between the flange of I section and the flange plate

 

 

Assuming that bolt threads intervene in the shear plane,

 

(From Table 2.2,Anb = 157 mm2)

 

 

Assuming an end distance e = 40 mm,

 

 

∴ 2 × 29,731 = 288p

or          p = 206 mm

          pmax = 16t = 16 × 12.4 = 198.4 mm or 200 mm for tension flange

          pmax = 12t = 12 × 12.4 = 148.8 mm or 200 mm for compression flange

A pitch of 180 mm in the tension flange and 120 mm in the compression flange may be provided.

Example 5.4

Design an I section purlin of the span of 4m subjected to an udl of 1.5 kN/m in the plane of the minor axis and 0.5 kN/m in the plane of the major axis under service condition. Assume that the purlin is continuous over the supports and no lateral buckling occurs. The grade of the steel is E250.

The factored bending moments about the major and minor axes

 

 

Considering bending only about the major axis, the plastic section modulus required

 

 

From Appendix A, LB75 @ 6.1 kg/m (Figure 5.3) may be tried.

 

 

∴ The section is plastic and βb = 1.0

 

Mdz = 1.0 × 22.35 × 103 × 250/1.1 = 5.1 kNm
Mdy = 1.0 × 12.5 × 103 × 250/1.1 = 2.8 kNm
α1 = 1.0 and α2 = 2.0                                

 

The following interaction condition should be satisfied.

 

Example 5.5

Re-do Example 5.4 with a channel section.

From Appendix A, JC 100 @5.8 kg/m (Figure 5.24) may be tried. Its Zpz = 28.38 cm3

 

 

Figure 5.24

 

 

Figure 5.23

 

 

∴ The section is plastic and βb = 1.0

The equal area axis parallel to the flanges coincides with the major axis passing through the centroid whereas the equal area axis parallel to the web may be determined by equating the areas on either side of this axis.

 

          (100 − 2 × 5.1) × 3.0 + 2 × z × 5.1 = 2 × (45 − z) × 5.1

or        z = 9.3 mm

          Zpz = 2[45 × 5.1 − (50 − 5.1/2) + (50 − 5.1) × 3.0 × (50 − 5.1)/2] = 27.8 cm3

          Zpy = (100 − 2 × 5.1) × 3.0 × (9.3 − 3.0/2) + 2 × 9.3 × 5.1

            × 9.3/2 + 2 × (45 − 9.3) × 5.1 × (45 − 9.3)/2 = 9.0 cm3

          Mdz = 1.0 × 27.8 × 103 × 250/1.1 = 6.3 kNm

          Mdy = 1.0 × 9.0 × 103 × 250/1.1 = 2.0 kNm

          α1 = 1.0 and α2 = 2.0

Example 5.6

Re-do Example 5.4 using a square hollow section.

For the required plastic modulus of 15.9 cm3, 63.5 × 63.5 × 4.5 HF SHS (from Appendix E) may be tried.

 

Zpy = Zpz = 21.93 cm3                  

 

 

∴           The section is plastic and hence βb = 1.0.

 

Mdz = Mdy = 1.0 × 21.93 × 103 × 250/1.1 = 4.98 kNm α1 = α2 = 1.66

 

63.5 × 63.5 × 3.6 HF SHS may be tried.

 

 

Figure 5.25

 

 

∴       The section is plastic and hence βb = 1.0.

 

Mdz = Mdy = 1.0 × 18.36 × 103 × 250/1.1 = 4.2 kNm
α1 = α2 = 1.66

Example 5.7

Re-do Example 5.4 using a tubular section.

The resultant intensity of the uniformly distributed load

The maximum factored bending moment,

The maximum factored shear force,

The plastic section modulus to be provided

 

From Appendix D, a tube with nominal bore = 65 mm, outside diameter = 76.1 mm, class: medium may be tried. Its thickness t = 3.6 mm, the area of the cross section A = 820 mm2 and the mass = 6.42 kg/m.

The plastic section modulus is given by

d/t = 76.1/3.65 = 20.9 < 42 ε2 where ε = 1.0

∴       From Table 1.7, the section is plastic and hence βb = 1.0

Md = 1.0 × 19.13 × 103 × 250/1.1 = 4.4 kNm < 3.8 kNm

OK

 

 

Figure 5.26

Example 5.8

Find the design bending strength of MB 400 in Example 5.1 if the compression flange is laterally unsupported. Assume that full torsional and warping restraints are provided at the supports. Also, assume that the load acts on the upper flange which will have a destabilizing effect.

From Appendix A, Iy = 622.1 × 104 mm4, tf = 16 mm, tw = 8.9 mm, ry = 28.2 mm hf = 400 − 16 = 384 mm

From Table 5.1, the length for lateral torsional buckling, LLT = 0.85 L = 0.85 × 5.0 = 4.25 m

 

 

fbd = 0.51 × 250/1.1 = 115.9 MPa

Md = bb Zp fbd = 1.0 × 1176.18 × 103 × 115.9 = 136.3 kNm

 

Compared to the value of Md in Example 5.1, this is about 51%. This example shows that if the compression flange of a beam is not laterally supported, there will be a reduction in the design bending strength.

 

 

Figure 5.27

Example 5.9

Re-do Example 5.8, if a lateral support to the compression flange is provided at mid-span.

From Section 5.6.2.3,

LLT = 1.2 × distance between the lateral supports

   = 1.2 × 2.5 = 3 m = 3,000 mm

 

 

By providing a lateral support in the middle of the span, there is about 36% increase in the bending strength compared to that when there is no lateral support in the middle.

Example 5.10

Design a laterally unsupported I beam with simply supported ends of effective span 6 m subjected to a working load of 35 kN/m. Assume that full torsional and warping restraints are provided at the supports and the load acts on the upper flange which will have destabilizing effect.

The factored load = 1.5 × 35 = 52.5 kN/m

The maximum bending moment,

The maximum shear force,

The design of a laterally unsupported beam involves trail and error procedure.

Let fbd = 120 N/mm2

The plastic section modulus to be provided

 

From Appendix A, HB450 @ 92.5 kg/m may be tried. Its properties are

Zpz = 2,030.95 cm3, bf = 250 mm, tf = 13.7 mm, tw = 11.3 mm, ry = 50.8 mm

 The length for the lateral torsional buckling, LLT = 0.85 × 6,000 = 5,100 mm

                                           hf = 450 − 13.7 = 436.3 mm

 

 

Figure 5.28

 

 

Hence the section is plastic and βb = 1.0.

Md = 1.0 × 2,030.95 × 103 × 156.5 = 317.9 kNm >> 236.25 kNm Not satisfactory

HB400 @82.2 kg/m may be tried. Its properties are

Zpz = 1,626.36 cm3,b = 250 mm, tf = 12.7 mm, tw = 10.6 mm, ry = 51.6 mm

hf = 400 − 12.7 = 387.3 mm

 

 

 

The section is compact and hence βb = 1.0.

 

Md = 1.0 × 1,626.36 × 103 × 154.3 = 250.9 kNm > 236.25 kNm

OK

Example 5.11

A concentrated load of 400 kN acts on MB 400 through a stiff bearing plate of dimension 150 mm along the length of the beam. Check the web for buckling and crippling.

Check for web buckling

b1 = 150 mm, n1 = 200 mm

b1 + 2n1 = 550 mm

The effective length of the strut, KL = 0.7 (400 − 2 × 16) = 257.6 mm

 

 

KL/r = 257.6/2.57 = 100

From Table 4.4, for buckling class c, fcd = 107 MPa

 

 

Figure 5.29

 

Fxd = 550 × 8.9 × 107 = 523.8 kN < 1.5 × 400 kN Not satisfactory
So, the bearing length may have to be increased.

 

Check for web crippling

From Appendix A, for MB 400, R1 = 14 mm and tf = 16 mm

R1 + tf = 30 mm

n2 = 30 × 2.5 = 75 mm

b1 + 2n2 = 150 + 2 × 75 = 300 mm

Fw = 300 × 8.9 × 250/1.1 = 607 kN > 1.5 × 400 kN

OK

Problems

For the following problems, consider the grade of steel as E250

  1. WPB 300 × 300 × 88.33 is used to span a gap of 4 m with ends simply supported. Find (a) the intensity of the uniformly distributed load it can carry over the entire span in the service condition (b) the concentrated load it can carry at mid-span in the service condition. Calculate the maximum deflection in each case. Assume that the beam is laterally supported throughout.
  2. The roof of a hall of 12 m × 8 m consists of a RC slab 100 mm thick and a 50 mm floor finish. The slab is supported on steel beams spaced at 3 m centre to centre. The live load on the slab is 2 kN/m2. Design an intermediate steel beam using I section. Assume that the slab provides adequate lateral restraint to the compression flange of the steel beam.
  3. A beam of span of 5 m carrying a load of 20 kN/m on the left half and a concentrated load of 50 kN at the quarter span on the right half. Design the beam assuming that it is laterally supported along its length.
  4. A beam of 6 m of an effective span carries a uniformly distributed load of 30 kN/m with concentrated loads of 15 kN each at one-third points of the span. The depth of the beam is limited to 300 mm. Design the beam with additional plates to the flanges. Assume that the beam is laterally supported throughout.
  5. A beam with a section of WB 500 has an effective span of 6 m. If the beam is connected to columns by framed connections and is otherwise laterally unsupported, determine the intensity of the uniformly distributed load it can carry in addition to its own weight in a service condition.
  6. Redesign the beam in problem 5.3 if the beam is laterally unsupported. Assume that full torsional and warping restraints are provided at the ends of the beam.
  7. A RC slab 100 mm thick is supported on cantilever beams at a spacing of 4m. The span of the cantilever beam is 2 m. Consider a floor finish load of 1 kN/m2 and a live load of 3 kN/m2. Assume that the built-in end provides full torsional and warping rigidity.
  8. Redesign the beam in Problem 5.6 if a lateral support is provided at mid-span.
  9. Design the continuous beam shown in Figure 5.30.
  10. Design a channel purlin of a roof truss subjected to a maximum bending moment 8 kNm about the major axis and 5 kNm about minor axis. Assume that lateral support is provided by the sheeting.
  11. Redesign Problem 5.10 using an I section
  12. Redesign Problem 5.10 using tubular section.
  13. Redesign Problem 5.10 using a rectangular hollow section (RHS).
  14. Design the beam AB of the portal frame shown in Figure 5.31. Assume that lateral restraints are provided at A and B, otherwise, the beam is laterally unsupported. Assume EI is the same for the entire frame.
  15. Design the beam shown in Figure 5.32. Assume that the beam is laterally supported.

 

 

Figure 5.30

 

 

Figure 5.31

 

 

Figure 5.32