Designing a Laterally Restrained Steel Beam

steel beam

Understanding the design of steel beams depend to a large extent on whether the compression flange is laterally restrained or not. This is because when a steel beam is subjected to flexure, there is a tendency for it to buckle along its length. This phenomenon is known as lateral-torsional buckling. Its influence reduces the flexural resistance of a steel section and must be given adequate consideration when sizing steel beams.

The subject of this post is the design of laterally restrained steel beams i.e. beams not susceptible to lateral-torsional buckling. How to design laterally unrestrained beams will be covered in a later post.

Whenever one of the following situations occurs in a steel beam, lateral-torsional buckling cannot develop and an assessment of the beam can be carried out based on just the section resistance.

  1. The cross-section of the beam is bent about its minor z-axis
  2. The beam is laterally restrained by means of secondary steel members or by a concrete slab or any other method that prevents lateral displacement of the compressed parts of the cross-section.
  3. Cross-section of the beam has high torsional stiffness and similar flexural stiffness about both principal axes of bending, for example, closed hollow cross-sections.

Member Resistance of Steel Beams

The steps in designing a steel beam basically involve determining the actions which the steel beam is required to support and then selecting a steel section to withstand the actions. However, before selecting a steel section, the beam must be classified because the resistance of the chosen section to a large extent depend on the section category.

Classification of steel section has been covered in a previous post see: Classification of Structural Steel Cross-Section.

Shear Resistance

In almost all steel beams the shear resistance is provided by the webs alone. However, in some steel beams, stiffeners may be provided to strengthen the section and hence increase the shear resistance of the steel section.

Generally, all steel beams must satisfy the following expression in-order to be sufficient in resisting shear.

\frac { { V }_{ Ed } }{ { V }_{ c,Rd } } <1

Where: VEd= Applied Shear Force; Vc,Rd = is the design shear resistance γM0= is the partial factor for the resistance of cross-sections, which is given in the UK National Annex as 1.0; fy= is the yield strength of the steel, based on element thickness

For Class 1 and 2 rolled steel beams, the design shear resistance is defined as Vpl,Rd and is defined in Clause 6.2.6, equation 6.18 of BS EN 1993-1-1 as:

{ V }_{ pl,Rd }=\frac { { A }_{ v }\left( { f }_{ y }/\sqrt { 3 } \right) }{ { \gamma }_{ M0 } }

Where: Av is the cross-section area of the part of the beam that is resisting shear. For ‘I’ and ‘H’ sections this is defined as:

{ A }_{ v }=A-2b{ t }_{ f }+{ t }_{ f }\left( { t }_{ w }+2r \right) \ge { h }{ t }_{ w }

For all other classes of beam sections, you are referred to use Clause 6.2.6 (4) of BS EN 1993-1-1 for determining their design shear resistance.

Flexural Resistance

For laterally restrained beam, the chosen steel section can be verified against flexure by satisfying the expression

\frac { { M }_{ Ed } }{ { M }_{ c,Rd } } \le 1

Where: MEd= design bending moment; Mc,Rd= design bending moment resistance. defined in clause 6.2.5(2) of BS-EN 1993-1-1 as:

{ M }_{ c,Rd }=\frac { { W }_{ y }{ f }_{ y } }{ { \gamma }_{ M0 } }

Where: Wy is the major axis section modulus of the beam based on its classification

  • For class 1 and 2 sections Wy = Wpl,y (Plastic section modulus)
  • Class 3 sections Wy= Wel,y (Elastic section modulus)
  • For class 4 sections Wy = Weff,y ( Minimum effective section modulus).

Where the beam is subjected to a high shear force at the same point as the design maximum moment, such that the shear force is greater than 50% of the shear resistance of the steel section the bending moment resistance must be reduced. Clause 6.2.8 of BS EN 1993-1-1 places a modification factor against the yield strength. This is defined as:

{ f }_{ y,mod }=\left[ 1-{ \left( \frac { 2{ V }_{ Ed } }{ { V }_{ pl,Rd } } -1 \right) }^{ 2 } \right] { f }_{ y }

This modified yield strength is then inserted into the calculation that determines bending moment resistance.

The design of class 4 sections is very complex and it is best to avoid them as far as possible. However, guidance can be found in part 5 of Eurocode 3 on plated structures. Class 4 sections are hardly found in rolled ‘I’ and ‘H steel sections hence it is best to neglect its design in this post.

Serviceability

Deflection

The vertical deflection limits for steel beams can be found in the Clause NA.2.23 of the UK National Annexe to BS EN 1993-1-1. Table 3 is based on these stated limits, which are for the deflection due to unfactored imposed loads/variable actions only

  • L/360 for steel beams supporting brittle finishes
  • L/200 for steel beams not supporting brittle finishes
  • L/180 for cantilevers.

Viberation

In addition to verifying the deflection of steel beams, it is sometimes very necessary to verify the vibration of the beam i.e. its dynamic response. This is easily satisfied by ensuring that the natural frequency of the steel beam is greater than 4Hz. The natural frequency of a simply supported beam can be obtained from:

{ f }_{ 1 }=\frac { 18 }{ { \delta }_{ a } }

Where: f1 =natural frequency of the steel beam; δa = the maximum deflection due to the applied load and self-weight in mm.

Worked Example

Figure 1 shows the part plan of a braced multi-storey steel office building. The steel beams are spaced at 6.0 centres and required to carry a precast concrete slab of 200mm. The slab is considered sufficient to restraint the compression flange against buckling. Select a UKB section in S275 steel for Beam A and verify the adequacy of the chosen section to withstand the design actions. 

Take floor imposed load for offices =3.0kN/m2. The superimposed dead loads on the floor = 2kN/m2 and the actions from demountable partitions =1kN/m2.

Describes the worked example on steel beam
Figure 1: Worked Example on Steel Beam.
Actions

Permanent Actions

i.\quad self\quad weight\quad of\quad slab\quad =0.2\times 25=5kN/{ m }^{ 2 }
ii.\quad superimposed\quad dead\quad load = 2kN/{ m }^{ 2 }
permanent\quad actions,\quad { g }_{ k }=5+2=7kN/{ m }^{ 2 }

Variable Actions

i.\quad imposed\quad load\quad (offices)\quad =3kN/{ m }^{ 2 }
ii.\quad demountable\quad partitions\quad =\quad 1kN/{ m }^{ 2 }
permanent\quad actions,\quad { q }_{ k }=3+1=4kN/{ m }^{ 2 }

By inspection, the permanent actions are less than 4.5 times the variable actions. Hence, we have

{ n }_{ s }=1.35\xi { g }_{ k }+1.5{ q }_{ k }
=\left( 1.35\times 0.925\times 7 \right) +\left( 1.5\times 4 \right) =14.75kN/{ m }^{ 2 }

Load on Beam

load\quad transfered\quad to\quad beam\quad =\quad 14.75\times 6.0\quad =88.5kN/m
assume\quad self\quad weight\quad of\quad beam\quad =2.5kN/m
w=88.5+2.5=91kN/m
Design Actions
{ M }_{ Ed }=\frac { w{ l }^{ 2 } }{ 8 } =\frac { 91\times { 12 }^{ 2 } }{ 8 } =1638kN.m
{ V }_{ Ed }=\frac { wl }{ 2 } =\frac { 91\times 12 }{ 2 } =546kN
Sizing the Steel Beam

Assuming a steel UKB section that is at least compact i.e. a class two section, we have:

{ W }_{ pl,y }\ge \frac { { M }_{ Ed }{ \gamma }_{ M0 } }{ { f }_{ y } } \quad
16\le { t }_{ f }\le 40\quad { R }_{ EH }={ f }_{ y }=265N/{ mm }^{ 2 }
{ W }_{ Pl,y }\ge \frac { 1638\times { 10 }^{ 6 }\times 1.0 }{ 265 } =6181.13{ cm }^{ 3 }
Try\quad a\quad 762\times 267\times \times 173\quad UKB-SECTION
Properties
h=762.2cm\quad b=266.7cm\quad d=686.0cm
{ t }_{ w }=14.3cm\quad { t }_{ f }=21.6cm\quad r=16.5cm
{ W }_{ pl,y }=6200{ cm }^{ 3 }\quad { I }_{ yy }=205000{ cm }^{ 4 }\quad A=220{ cm }^{ 2 }
Section Classification

Flange in compression

outstand\quad of\quad compression\quad flange\quad c=\frac { b-{ t }_{ w }-2r }{ 2 }
\frac { 266.7-14.3-\left( 2\times 16.5 \right) }{ 2 } =109.7mm
\frac { c }{ { t }_{ f } } =\frac { 109.7 }{ 21.6 } =5.08<8\quad (flange\quad is\quad class\quad 1)

Web under pure bending

c=d=686.0mm
\frac { c }{ { t }_{ w } } =\frac { 686 }{ 14.3 } =47.97<66\quad (web\quad is\quad class\quad 1)

Therefore since the flanges and web is class one therefore the section can be taken as class one.

Shear Resistance
verify\quad that\quad \frac { { V }_{ Ed } }{ { V }_{ c,Rd } } \le 1.0
{ V }_{ c,Rd }={ V }_{ pl,Rd }=\frac { { A }_{ v }\left( { f }_{ y }/\sqrt { 3 } \right) }{ { \gamma }_{ M0 } }
{ A }_{ v }=A-2b{ t }_{ f }+{ t }_{ f }\left( { t }_{ w }+2r \right) \ge { h }_{ w }{ t }_{ w }
=220\times { 10 }^{ 2 }-2\times 266.7\times 21.6+21.6\left( 14.3+2\times 16.5 \right) =115.0{ cm }^{ 2 }
{ V }_{ c,Rd }=\frac { 11500\times \left( 265/\sqrt { 3 } \right) }{ 1.0 } =1759.5kN
\frac { { V }_{ Ed } }{ { V }_{ c,Rd } } =\frac { 546 }{ 1759.5 } =0.31<1\quad o.k
Bending Resistance
verify\quad that\quad \frac { { M }_{ Ed } }{ { M }_{ c,Rd } } <1.0
{ M }_{ c,Rd }=\frac { { W }_{ pl,y }{ f }_{ y } }{ { \gamma }_{ M0 } } =\frac { 6200\times { 10 }^{ 3 }\times 265 }{ 1.0 } =1643kN.m
\frac { { M }_{ Ed } }{ { M }_{ c,Rd } } =\frac { 1638 }{ 1643 } =0.997<1\quad (o.k)
Deflection

Deflection of steel beams is carried out under the characteristic imposed loading only. This is because deflection is a serviceability limit state.

w=4.0\times 6=24.0kN/m
{ \delta }_{ a }=\frac { 5w{ l }^{ 4 } }{ 384EI } =\frac { 5\times 24\times { \left( 12000 \right) }^{ 4 } }{ 384\times 205000\times { 10 }^{ 4 }\times 210000 } =15.05mm
{ \delta }_{ lim }=\frac { span }{ 360 } =\frac { 12000 }{ 360 } =33.33>15.05mm\quad (o.k)
Viberation
{ f }_{ 1 }=\frac { 18 }{ \sqrt { { \delta }_{ a } } } =\frac { 18 }{ \sqrt { 15.05 } } =4.64>4Hz\quad (o.k)
Adopt\quad 762\times 267\times 173-UKB\quad Section

Thankyou! leave a comment if you find this post helpful

Omotoriogun Victor
About Omotoriogun Victor 66 Articles
A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

3 Trackbacks / Pingbacks

  1. Designing a Laterally Unrestrained Steel Beam - STRUCTURES CENTRE
  2. Design of Steel Elements in Tension to Eurocode - STRUCTURES CENTRE
  3. Design and Detailing of Windposts to Masonry - STRUCTURES CENTRE

Leave a Reply

Your email address will not be published.


*