Basic Lateral Torsional Buckling Steel Beam Example

In our previous lesson (here) we considered basic Lateral Torsional Buckling and `used it' on a wood beam. Now let's look at a steel beam.

Another Example

Consider a steel beam W 12 x 40 of F_y 36 ksi steel simply spanning 30 ft.

Determine the allowable uniform load based on a) extreme fiber in (strong axis) bending yielding and FOS 1.6, and b) lateral torsional buckling and FOS of 1.6.

Solution

Part a)

M = F_y S

S = strong axis section mod = 51.9 in.³ from Manual of Steel Construction, ...

F_y = 36 ksi

M = 36 ksi (51.9 in.³) = 1,868 k-in. = 156 k-ft.

M _allow = M/FOS = 156 k-ft / 1.6 = 97 k-ft

And,

M = ω L² / 8

ω _allow = 8 M / L² = 8 (97 k-ft)/(30 ft)² = 0.865 k/ft = 865 plf

Part b)

From Manual of Steel Const., ASD, 9^th ed. ...

E = 29,000,000 psi

I_y = 44.1 in.⁴

G = 11,200,000 psi

J = 0.95 in.⁴

M _critical = (π/L_e)√(E I_y G J)

= (π/360 in.)√(29,000,000*44.1*11,200,000*0.95 in.⁴)

= 1,018,000 lb-in. = 84.8 k-ft

Dividing by FOS of 1.6 ...

M _allow = 53 k-ft

ω _allow = 8 (53) / 30² = .471 k/ft = 471 plf ...

So, we are limited by stability at 471 plf ... about half the value based on extreme fiber in flexure.

Note that if we used a higher strength steel, we'd still be limited at 471 plf.

In many real life applications whatever is delivering the distributed load (a floor system, roof, etc.) will also be counted upon to provide lateral bracing. Then we can use the full flexural capacity (865 plf in this example).

NOTE: THESE CALCULATIONS ARE FOR A MECHANICS OF MATERIALS CLASS AND NOT FOR THE DESIGN OF STRUCTURAL STEEL MEMBERS. FOR THE DESIGN OF STEEL MEMBERS IN BUILDINGS AND OTHER STRUCTURES USE THE PROVISIONS OF THE APPLICABLE BUILDING CODES FOR YOUR PARTICULAR SITUATION.

Bracing Length

Now let us consider the maximum unbraced length for which we will have no reduction in bending capacity due to Lateral Torsional Buckling. Using the above example we will set the flexural and Lateral Torsional Buckling capacities equal, and solve for L. If the factors of safety for each are different we would have to decide on which side of the `FOS' we will do the calculation, but since they are equal we would get the same answer either way.

So,

M _{capacity, flex} = M _{capacity, LTB}

S Fy = (π/L)√(E I_y G J)

Plugging in the numbers,

(51.9 in.³)(36,000 lb/in.²) = (π/L)√(29,000,000*44.1*11,200,000*0.95 in.⁴)

L = (π)√(29,000,000*44.1*11,200,000*0.95 in.⁴) / [(51.9 in.³)(36,000 lb/in.²)]

L = 196 in. = 16.3 ft.

So, what this means is that if we provide bracing against LTB at intervals not exceeding 16.3 ft, we may use the full flexural tension capacity of the beam. In much of our construction we are able to get such bracing by `connecting' the top of the beam to joists or trusses or perhaps a deck on top the beam.

Note: we have considered beams in positive bending. If we have a beam in negative bending (whole or part) then the bracing against LTB needs to be on the bottom of the beam (the zone in compression).

Check:

M _{allowable, flexure} = S Fy / 1.6 = 51.9 (36,000) / 1.6 = 1,168,000 lb-in. = 97 k-ft.

M _{allow, LTB} = (π/196 in.)√(29,000,000*44.1*11,200,000*0.95 in.⁴) = 1,169,000 lb-in. = 97 k-ft.

Sweet.

Again, note: in the design of steel buildings and other structures, use `the Code'. It will be more complicated, and more precise, but the results should not be hugely different.

References

Stability of Beams, Jeff Filler, Associated Content

Manual of Steel Construction, ASD, 9^thEdition, American Institute of Steel Construction, Inc., 1 East Wacker Drive, Suite 3100, Chicago, Illinois, 60601.