Lateral-torsional buckling of beams

A deep, narrow beam bent about its strong axis is like a column lying on its side: its compression flange wants to buckle, and the easiest way is sideways. Because the tension flange resists moving sideways, the section twists as it deflects laterally. This combined sideways bending and twisting is lateral-torsional buckling (LTB). It governs beams whose compression flange is not braced, such as girders between their connected beams, beams during erection before the deck is fixed, crane runways and monorails. This lesson explains the behaviour, gives the design rules, and checks the mezzanine edge girder.

How a beam buckles laterally

Picture a long scale ruler loaded on edge between two supports that stop it twisting at the ends. Under a small load it bends downward in its stiff plane. At a critical load it suddenly kicks sideways and twists, and it can carry little more. The critical moment depends on two kinds of resistance: the lateral bending stiffness of the section, \(EI_y\), and its torsional stiffness. Torsional stiffness itself has two parts. St. Venant torsion, \(GJ\), is the resistance of the plates to shear as they twist. Warping torsion, \(EC_w\), is the resistance of the flanges to bending in opposite directions in their own planes as an I-section twists.

For a doubly symmetric beam with simple end supports that prevent twist but allow warping, under uniform moment, the elastic critical moment is

where \(L\) is the unbraced length. S16 uses the symbol \(M_u\) for this elastic critical moment. The first term under the root dominates for long beams and for compact sections such as HSS; the warping term dominates for short, deep I-beams.

Torsional stiffness falls quickly with length, so the critical moment falls even faster than a column's buckling load: halving the unbraced length can more than double the critical moment.

Real beams: three regions

Like columns, real beams have initial crookedness and residual stresses. The design curves therefore have three regions:

• Short unbraced lengths: the beam reaches \(M_p\) (for compact or class 1 and 2 sections) before buckling.
• Intermediate lengths: part of the section yields before the beam buckles, and the strength falls between \(M_p\) and a value near \(0.7M_y\).
• Long lengths: the beam buckles elastically at the critical moment.

S16 gives a single expression for the inelastic region and uses the elastic moment beyond it:

For class 3 sections, \(M_y\) replaces \(M_p\) in these expressions. The transition at \(0.67M_p\) reflects the onset of yielding with residual stresses.

Moment gradient: the factor ω2

The expressions above assume a uniform moment over the unbraced length, which is the worst case: the whole compression flange is equally stressed. When the moment varies along the segment, only part of the flange is highly stressed and the beam is stronger. The equivalent moment factor accounts for this, using the maximum moment and the moments at the quarter points of the unbraced segment:

It is 1.0 for uniform moment, 1.13 for a simply supported span under uniform load, about 1.26 for a central point load, and 1.75 for a moment varying linearly from zero to a maximum. It multiplies \(M_u\), so it raises the resistance in the inelastic and elastic regions but never above \(\phi M_p\).

Bracing and load height

A brace for lateral-torsional buckling must prevent either lateral movement of the compression flange or twist of the section. Beams framing in at mid-depth with simple connections, cross frames, and a deck fastened to the compression flange all do this. A tension flange brace does little. Braces need both strength and stiffness; lesson 11 gives the requirements.

The expressions assume the load is applied at the shear centre. Load on the top flange, free to move with the beam, tips the section further as it twists and lowers the critical moment. S16 handles this by increasing the effective unbraced length, for example to 1.2 times the actual length for a simply supported beam with an unrestrained top-flange load, and more for cantilevers. Load hanging from the bottom flange, as from a monorail trolley, is stabilizing.

The mezzanine edge girder

The edge girders span 9.0 m between columns and receive the mezzanine beams at their third points. The beams frame in with double-angle connections at mid-depth and brace the girder there, so the girder has three unbraced segments of 3.0 m. The deck spans parallel to the girder and does not brace it. The factored moment, including self-weight, is \(M_f = 454 + 1.25(0.73)(9.0)^2/8 = 464\) kN·m, nearly uniform over the middle segment.

If the same girder had no braces between columns, its resistance over 9 m with \(\omega_2 = 1.13\) would fall to about 118 kN·m, a quarter of what it needs. Bracing, not the size of the beam, is what makes it work.

Sofia's program computes the resistance of each segment, tabulates the resistance against unbraced length, which is how the curve in the figure was drawn, and checks the girder during erection, when it is in place but the beams are not yet connected.

import math
# Mezzanine edge girder W530x74: lateral-torsional buckling, CSA S16 clause 13.6
E, G, Fy, phi = 200000.0, 77000.0, 345.0, 0.90
Zx, Iy, J, Cw = 1800e3, 10.4e6, 475e3, 690e9     # mm3, mm4, mm4, mm6
Mp = Zx * Fy / 1e6
P, L = 151.5, 9.0                                 # beam reactions (kN) at third points
Mf = P * L / 3 + 1.25 * 0.73 * L**2 / 8
print(f"Mf = {Mf:.1f} kN.m, Mp = {Mp:.0f} kN.m, phiMp = {phi*Mp:.0f} kN.m")

def omega2(Mmax, Ma, Mb, Mc):
    return min(4 * Mmax / math.sqrt(Mmax**2 + 4 * Ma**2 + 7 * Mb**2 + 4 * Mc**2), 2.5)

def Mr(Lb, w2):
    Mu = w2 * math.pi / Lb * math.sqrt(E * Iy * G * J + (math.pi * E / Lb)**2 * Iy * Cw) / 1e6
    if Mu > 0.67 * Mp:
        return min(1.15 * phi * Mp * (1 - 0.28 * Mp / Mu), phi * Mp), Mu
    return phi * Mu, Mu

w_mid = omega2(1, 1, 1, 1)
w_end = omega2(1, 0.25, 0.5, 0.75)
print(f"omega2: middle segment {w_mid:.2f}, end segments {w_end:.2f}")
for name, Lb, w2, M in (("middle 3 m", 3000, w_mid, Mf), ("end 3 m", 3000, w_end, Mf)):
    r, Mu = Mr(Lb, w2)
    print(f"Segment {name}: Mu = {Mu:.0f} kN.m, Mr = {r:.0f} kN.m, Mf = {M:.0f} kN.m, ratio {M/r:.2f}")

print("Mr (omega2 = 1.0) against unbraced length:")
for Lb in (1.5, 2, 3, 4, 5, 6, 8, 9):
    r, Mu = Mr(Lb * 1000, 1.0)
    print(f"  L = {Lb:3.1f} m: Mr = {r:4.0f} kN.m")
# Erection: girder in place before the beams, braced only at the columns, self-weight plus a worker
Me = 1.25 * 0.73 * 81 / 8 + 1.5 * 1.0 * 9 / 4
r9, _ = Mr(9000, 1.13)
print(f"Erection, L = 9 m, omega2 = 1.13: Mr = {r9:.0f} kN.m against {Me:.1f} kN.m")

Mf = 463.7 kN.m, Mp = 621 kN.m, phiMp = 559 kN.m
omega2: middle segment 1.00, end segments 1.75
Segment middle 3 m: Mu = 655 kN.m, Mr = 472 kN.m, Mf = 464 kN.m, ratio 0.98
Segment end 3 m: Mu = 1143 kN.m, Mr = 545 kN.m, Mf = 464 kN.m, ratio 0.85
Mr (omega2 = 1.0) against unbraced length:
  L = 1.5 m: Mr =  559 kN.m
  L = 2.0 m: Mr =  559 kN.m
  L = 3.0 m: Mr =  472 kN.m
  L = 4.0 m: Mr =  356 kN.m
  L = 5.0 m: Mr =  246 kN.m
  L = 6.0 m: Mr =  185 kN.m
  L = 8.0 m: Mr =  123 kN.m
  L = 9.0 m: Mr =  105 kN.m
Erection, L = 9 m, omega2 = 1.13: Mr = 118 kN.m against 12.6 kN.m

During erection the girder carries only its own weight and a worker, so even with a 9 m unbraced length it is far from buckling. That is not true of every member: long beams lifted by a crane at two points, or set on their supports with nothing yet attached, can buckle under their own weight. Erectors brace long members during lifting and connect the bracing members early.

Using the CISC beam tables

The CISC Handbook has beam selection tables that list \(M_r\) at a series of unbraced lengths for \(\omega_2 = 1.0\), and charts that plot the same information for many shapes on one sheet. To use them with a moment gradient, divide the factored moment by \(\omega_2\) in the elastic and inelastic ranges, or compute the resistance directly as above. The tables also give \(L_u\), the longest unbraced length at which the beam still develops \(\phi M_p\) under uniform moment.

A good habit is to look up the critical length for a beam before choosing its bracing. If the unbraced length is well below it, the beam develops its full plastic moment and LTB can be ignored. If it is well above, consider adding braces before choosing a heavier beam.

Does a stronger steel help?

Owen asked whether the girder could be lighter in a higher-strength steel. For short unbraced lengths, yes: \(M_p\) grows with \(F_y\). For long ones, no: the elastic critical moment depends on \(E\), \(G\) and the section constants, none of which change with the grade, so a long unbraced beam in 450 MPa (65 ksi) steel buckles at the same moment as one in 345 MPa (50 ksi) steel. In the inelastic range the benefit is partial, and the critical length for full plastic moment actually shrinks as \(F_y\) rises, because the beam has further to go before it yields. The lesson for design is general: when stability governs, add bracing or choose a stiffer section; do not expect strength alone to solve the problem. The same reasoning applies to slender columns in lesson 4 and to the brace angles in lesson 11.

Practical cases

• Cantilevers: the free end must be braced against twist, or the effective length is much longer than the cantilever. Canopy beams and overhangs at roof edges are common problem areas.
• Continuous beams: in negative moment regions the bottom flange is in compression and is not braced by the deck. Bottom flange braces (kickers to the deck or to joists) are added near supports when needed.
• Channels and angles bent about their strong axis twist because their shear centre is outside the section, and need their own checks.
• HSS are torsionally stiff and almost never fail by LTB.