Break down Lateral Torsional Buckling for me

Lateral Torsional Buckling (LTB for short) for Steel Beams has been explained in depth on this page. In short, LTB is a failure criteria which can limit the flexural capacity of a beam when the beam is spanning long distances with no lateral bracing.

This page is dedicated to showing how to solve for lateral torsional buckling of compact sections using the 13th edition AISC Steel Construction Manual. Even though most beams are compact, this solving for the LTB limits and subsequent Moment Capacities is a very tedious process (which translates in engineering terms to easy to make mistakes).

How to Understand LTB
  1. What assumptions have been made?
  2. How to solve for the LTB limits?
  3. Solve for the flexural capacity of the beam
  4. A Spreadsheet to make everything easier
  5. References

What assumptions have been made?

This is the first section of this page to emphasize that the following assumptions must be met for the following equation to work.

  • For doubly symmetric I-shaped (includes double channel, I-beams, etc.) members bent about their major axis.
  • For compact sections
  • Uses the 13th Edition AISC Steel Construction Manual (aka AISC 360)

How to solve for the LTB limits

Originally Taken From Steel Beam Lateral Torsional Buckling

Lateral Torsional Buckling of Beams
Fig. 1: Moment Capacity during the 3 stages of Lateral Torsional Buckling (L.T.B.)

Three limits exist when solving for lateral torsional buckling (shown in Fig. 1 above):

  • Lb < Lp which defines when a member is not subject to LTB
  • Lp < Lb < Lr which defines when a member is subject to inelastic LTB
  • Lr < Lb which defines when a member is subject to elastic LTB

where:

Lb = Unbraced Length (bracing must resist displacement of the compression flange or twisting of the cross section)

For Compact Cross Sections (most common):

Solving for Lp:

L_p = {1.76{r_y}\sqrt{\frac{E}{F_y}}}

where:

Lp = The limit between no LTB and Inelastic LTB
ry = The radius of gyration? about the weak axis of the cross section
E = The modulus of elasticity of the steel beam (e.g. 29000 ksi)
Fy = The yield strength of the steal beam (e.g. A36 has a yield strength of 36 ksi)

Solving for Lr:

L_r = 1.95r_{ts}\frac{E}{0.7F_y}\sqrt{\frac{Jc}{S_xh_O}}\sqrt{1+\sqrt{1+6.76\left(\frac{0.7F_y}{E}\frac{S_xh_O}{Jc}\right)^2}}

where:

Lr = The limit between Inelastic LTB and Elastic LTB (ft)
rts2 = \frac{\sqrt{I_yC_w}}{S_x} = \frac{I_yh_O}{2S_x}
Iy = The Moment of Inertia about the weak axis of the cross section (in4)
Cw = \frac{I_yh_O^2}{4} (for rectangular flanged doubly symmetric shapes)
Sx = Section Modulus? of the beam about the strong axis of the cross section (in3)
hO = distance between flange centroids = d - tf (in)
E = The modulus of elasticity of the steel beam (e.g. 29000 ksi)
Fy = The yield strength of the steal beam (e.g. A36 has a yield strength of 36 ksi)
J = torsional constant (in4)
c = 1 (for doubly symmetric I-shape) or \frac{h_O}{2}\sqrt{\frac{I_y}{C_w}} (for a channel)
Cw = warping constant (in6)

Note: For standard doubly symmetric I-shapes: rts, ry, J, Sx, hO, Iy, & Cw can all be found in Table 1 of the AISC 360 Steel Construction Manual[1]. This makes these calculations much easier.


Solve for the flexural capacity of the beam

The flexural capacity of the beam will be broken into three sections:

  • Lb < Lp
  • Lp < Lb < Lr
  • Lr < Lb

Lb < Lp

As you can see in Figure 1 above, when Lb < Lp the beam has no LTB instability and thus:

Solving for Lb < Lp:

M_n = M_p = F_yZ_x

where:

Mn = The Nominal Moment (note this is not the design moment)
Fy = The Yield Strength of the Steel (e.g. 36 ksi, 46 ksi, 50 ksi)
Zx = The Plastic Section Modulus in the x or strong axis. Zx is similar to the Section Modulus of a member (it is usually a minimum of 10% greater than the Section Modulus) (in3)

Lp < Lb < Lr

When Lp < Lb < Lr the beam has inelastic LTB and thus:

Solving for Lp < Lb < Lr:

M_n = C_b\left[M_p - (M_p - 0.7F_yS_x)\left(\frac{L_b - L_p}{L_r - L_p}\right)\right] \le M_p

where:

Mn = The Nominal Moment (note this is not the design moment)
Cb = Beam bending Coeeficient
Mp = Mn = Maximum Flexural Capacity
Fy = The yield strength of the steal beam (e.g. A36 has a yield strength of 36 ksi)
Sx = Section Modulus? of the beam about the strong axis of the cross section (in3)
Lb = Unbraced Length (bracing must resist displacement of the compression flange or twisting of the cross section)
Lp = The limit between no LTB and Inelastic LTB for a compact section
Lr = The limit between Inelastic LTB and Elastic LTB (ft)

Lr < Lb

When Lp < Lb < Lr the beam has elastic LTB and thus:

Solving for Lr < Lb:

M_n = F_{cr}S_x \le M_p
F_{cr} = \frac{C_b\pi^2E}{\left(\frac{L_b}{r_{ts}}\right)^2}\sqrt{1+0.078\frac{Jc}{S_xh_O}\left(\frac{L_b}{r_{ts}}\right)^2}

where:

Mn = The Nominal Moment (note this is not the design moment)
Fcr = Elastic Buckling Stress (psi, ksi, etc.)
Cb = Beam bending Coeeficient
E = The modulus of elasticity of the steel beam (e.g. 29000 ksi)
Lb = Unbraced Length (bracing must resist displacement of the compression flange or twisting of the cross section)
rts2 = \frac{\sqrt{I_yC_w}}{S_x} = \frac{I_yh_O}{2S_x}
Iy = The Moment of Inertia about the weak axis of the cross section (in4)
Cw = \frac{I_yh_O^2}{4} (for rectangular flanged doubly symmetric shapes)
J = torsional constant (in4)
c = 1 (for doubly symmetric I-shape)
Sx = Section Modulus? of the beam about the strong axis of the cross section (in3)
Mp = Mn = Maximum Flexural Capacity

Important: Don't forget that Mn is the nominal moment which still needs to be divided by Ωb (for ASD = 1.67) or multiplied by ϕb (for LRFD = 0.9) to find the design flexural strength.



A Spreadsheet to make everything easier

Finally!!! All of these meaningless variables in complex equations broken down into a spreadsheet that is easily used. You can find the spreadsheet and explanations on how to use it on the Steel I-Beam LTB Spreadsheet page. Hopefully this spreadsheet will make quick work of a very complicated process (Any process that requires six equations and more than 20 variables is considered quite complex in my own humble opinion.

References

  1. American Institute of Steel Construction, "Steel Construction Manual 13th edition", 2005
    • This Section is thoroughly covered in Part 3 of the AISC Steel Manual and in Chapter F2 of the Specifications (AISC 360)
    • Table B4.1 in the Specifications (AISC 360) has also been referenced to find λp & λr.
    • Most of the required variables to solve the above complex equations (for standard sized beams) can be found in Table 1 of the manual.

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