# Designing a Laterally Unrestrained Steel Beam

Laterally unrestrained steel beams are beams in which the compression flange do not have sufficient restraint and as a result highly susceptible to lateral-torsional buckling. A classical example of laterally unrestrained beams is a steel beam supporting light roof cladding. As a general rule, a restraint to the top flange of a laterally restrained beam must be capable of resisting a force that is equivalent to 2.5% of the ultimate compression load in the top flange of the beam element it is restraining. If this isn’t the case, then the steel beam is unrestrained and the effect of lateral-torsional buckling on the flexural resistance of the section must be fully taken into account.

There are, however, exceptions, beams with certain types of cross-sections are not susceptible to lateral-torsional buckling even though they are laterally unrestrained. For example, Closed hollow sections with a height/depth ratio of less than or equal to 2, are not susceptible to lateral-torsional buckling and can be designed using the exact procedures given in the last post on Designing a Laterally Restrained Steel Beam.

The design of laterally unrestrained beams follows the same procedure as laterally restrained beams. The added level of complexity is from the need to consider the effect of lateral-torsional buckling.

#### Lateral Torsional Buckling

To address the effect of lateral torsional buckling on laterally unrestrained steel beams BS-EN 1993-1-1 applies a reduction factor to the flexural resistance of a steel section. This can be obtained from equation 6.55 of Clause 6.3.2.1 (3) of BS-EN 1993-1-1 as.

{ M }_{ b,Rd }={ \chi }_{ LT }\frac { { { W }_{ y }f }_{ y } }{ { \gamma }_{ M1 } }

Where: Wy is the major axis section modulus of the beam based on its classification and is the same for restrained beams; fy is the yield strength of the steel, based on element thickness γM1; is the partial factor for the resistance of members subject to instability, which in the UK N.A is set at 1.0; χLT is the reduction factor that takes into account lateral-torsional buckling.

The value of the reduction factor is related to the slenderness of the beam which in turn relative to the span of the unrestrained portion of the beam subjected to compression. This is defined as the non-dimensional slenderness (λLT )

{ \lambda }_{ LT }=\sqrt { \frac { { W }_{ y }{ f }_{ y } }{ { M }_{ cr } } }

Where: Wy and fy are as defined previously; Mcr is the elastic critical moment for lateral-torsional buckling, which is based on the slenderness of the beam.

BS-EN 1993-1-1 is silent on how the elastic critical moment should be determined. However, several methods have been developed on calculating the elastic critical moment. The simplest of them all will be presented in this post.

##### Slenderness

The most fastest and conservative method of calculating the relative slenderness of a steel beam can be obtained from Table 1.1 of NCCI: Determination of non-dimensional slenderness of I and H sections SN002aEN-EU. It is based on applying the following equations that vary depending on the grade of steel being used.

{ \lambda }_{ LT }=\left[ \frac { L/{ i }_{ zz } }{ 96 } \right] \quad for\quad S275
{ \lambda }_{ LT }=\left[ \frac { L/{ i }_{ zz } }{ 85 } \right] \quad for\quad S355

Where: L = distance between restraints to the compression flange of the beam izz = is the radius of gyration about the minor axis of the beam

This method of estimating the non-dimensional slenderness of steel beams is very conservative and could possibly result in oversized members because it neglects the bending moment diagram. There are, however, more accurate but complex methods available in the NCCI: Determination of non-dimensional slenderness of I and H sections SN002aEN-EU.

Haven determined the non-dimensional slenderness, the reduction factor can be calculated based on equation 6.57 of Clause 6.3.2.3 of BS EN 1993-1-1 as:

{ \chi }_{ LT }=\frac { 1 }{ { \Phi }_{ LT }+\sqrt { { \Phi }_{ LT }^{ 2 }-\beta { \lambda }_{ LT }^{ 2 } } } \quad \le 1.0\le \frac { 1 }{ { \lambda }_{ LT }^{ 2 } }
{ \Phi }_{ LT }=0.5\left[ 1+{ \alpha }_{ LT }\left( { \lambda }_{ LT }-{ \lambda }_{ LT,0 } \right) +\beta { \lambda }_{ LT }^{ 2 } \right]

Where: The values of β and (λLT,0) are defined as 0.75 and 0.40 respectively in Clause NA.2.17 of U.K- NA to BS EN 1993-1-1; αLT is the imperfection factor and is found in Table 6.3 of BS EN 1993-1-1, which reads against the steel beam’s buckling curve. The buckling curves are labelled ‘a’ to ‘d’ and can be found in Clause NA.2.17 of NA to BS EN 1993-1-1. The buckling curve is dependent upon the h/b ratio of the beam section.

Finally, the member resistance to lateral-torsional buckling is carried out according to clause 6.3.2.1(1) by satisfying the following condition.

\frac { { M }_{ Ed } }{ { M }_{ b,Rd } } \le 1.0

#### Worked Example

A 10m span laterally unrestrained steel beam is subjected to point load from supporting a secondary steel beam at mid-span. The ultimate load from the secondary beam is 650kN. Assuming the secondary beam provides sufficient restraint to the primary beam, determine the adequacy of an 838 × 292 × 226 UKB section in S275 steel against lateral torsional buckling only if the section is class one.

Properties of Section: Wpl,y= 9160cm3; h= 850.9; b=293.8 izz= 6.27cm

##### Design Moment
{ M }_{ Ed }=\frac { Pl }{ 4 } +\frac { w{ l }^{ 2 } }{ 8 } =\frac { 650\times 10 }{ 4 } +\frac { 2.5\times { 10 }^{ 2 } }{ 8 } =1656.3kN.m
##### Flexural Buckling Resistance

Since the secondary beam restraints the primary beam at mid-span, the unrestrained length can be taken as half the beam span.

l=\frac { 10 }{ 2 } =5m
{ \lambda }_{ LT }=\left( \frac { l/{ i }_{ zz } }{ 96 } \right) =\frac { 5000/62.7 }{ 96 } =0.83
\frac { h }{ b } =\frac { 850.9 }{ 293.8 } =2.89
{ \Phi }_{ LT }=0.5\left[ 1+{ \alpha }_{ LT }\left( { \lambda }_{ LT }-{ \lambda }_{ LT,0 } \right) +\beta { \lambda }_{ LT }^{ 2 } \right]
=0.5\left[ 1+0.49\left( 0.83-0.4 \right) +0.75\times { 0.83 }^{ 2 } \right] =0.86
{ \chi }_{ LT }=\frac { 1 }{ { \Phi }_{ LT }+\sqrt { { \Phi }_{ LT }^{ 2 }-\beta { \lambda }_{ LT }^{ 2 } } } \quad \le 1.0\le \frac { 1 }{ { \lambda }_{ LT }^{ 2 } }
{ \chi }_{ LT }=\frac { 1 }{ 0.86+\sqrt { { 0.86 }^{ 2 }-0.75\left( { 0.83 }^{ 2 } \right) } } =0.75<1<1.45
{ M }_{ b,Rd }={ \chi }_{ LT }\frac { { W }_{ y }{ f }_{ y } }{ { \gamma }_{ M1 } } =0.75\times \frac { 9160\times 265 }{ 1.0 } =1820.55kN.m
\frac { { M }_{ Ed } }{ { M }_{ b,Rd } } =\frac { 1656.3 }{ 1820.55 } =0.91<1.0\quad o.k

This worked example assumes that the section has been classified, the flexural and shear resistance checked and other serviceability limit state criteria satisfied. For practical purposes, all of these must be verified, the same way as for a laterally restrained beam in addition to the lateral-torsional buckling check carried out in this post.

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