Formulas: general provisions

Equation C-F1-1

Beginning with the 1961 AISC Specification (AISC, 1961) and continuing through the 1986 LRFD Specification (AISC, 1986), the following equation was used to adjust the lateral-torsional buckling equations for variations in the moment diagram within the unbraced length:

C_{b}=1.75+1.05\left(\frac{M_{1}}{M_{2}}\right)+0.3\left(\frac{M_{1}}{M_{2}}\right)^{2} \leq 2.3

Formula 2

$0.7 F_y S_x$ (text_label - position: Point on solid curve at $L_b = L_r$ )

L_b = L_r

Equation C-F1-2a

Kirby and Nethercot (1979) present an equation that is a direct fit to various non-linear moment diagrams within the unbraced segment. Their original equation was slightly adjusted to give Equation C-F1-2a (Equation F1-1 in this Specification):

C_{b}=\frac{12.5 M_{\max }}{2.5 M_{\max }+3 M_{A}+4 M_{B}+3 M_{C}}

Equation C-F1-2b

C_{b}=\frac{4 M_{\max }}{\sqrt{M_{\max }^{2}+4 M_{A}^{2}+7 M_{B}^{2}+4 M_{C}^{2}}}

Equation C-F1-3

The lateral-torsional buckling modification factor given by Equation C-F1-2a is applicable for doubly symmetric sections and singly symmetric sections in single curvature. It should be modified for application with singly symmetric sections in reverse curvature. Previous work considered the behavior of singly symmetric I-shaped beams subjected to gravity loading (Helwig et al., 1997). The study resulted in the following expression:

C_{b}=\left(\frac{12.5 M_{\max }}{2.5 M_{\max }+3 M_{A}+4 M_{B}+3 M_{C}}\right) R_{m} \leq 3.0

Equation C-F1-4

R_{m}=0.5+2\left(\frac{I_{y} T o p}{I_{y}}\right)^{2}

where

$\begin{aligned} I_{y \text { Top }} & =\text { moment of inertia of the top flange about an axis in the plane of the web, } \\ & \text { in. }^{4}\left(\mathrm{~mm}^{4}\right)\end{aligned}$

$\begin{aligned} I_{y} & =\text { moment of inertia of the entire section about an axis in the plane of the } \\ & \text { web, in. }{ }^{4}\left(\mathrm{~mm}^{4}\right)\end{aligned}$

Equation C-F1-5

C_{b}=3.0-\frac{2}{3}\left(\frac{M_{1}}{M_{o}}\right)-\frac{8}{3}\left(\frac{M_{C L}}{\left(M_{o}+M_{1}\right)^{*}}\right)

where

$\begin{aligned} M_{o} & =\text { moment at the end of the unbraced length that gives the largest } \\ & \text { compressive stress in the bottom flange, kip-in. (N-mm) } \\ M_{1} & =\text { moment at other end of the unbraced length, kip-in. (N-mm) } \\ M_{C L} & =\text { moment at the middle of the unbraced length, kip-in. (N-mm) } \\ \left(M_{o}+M_{1}\right)^{*} & =M_{o}, \text { if } M_{1} \text { is positive, causing tension on the bottom flange }\end{aligned}$

Formula 8

C_{b}=3.0-\frac{2}{3}\left(\frac{+200}{-100}\right)-\frac{8}{3}\left(\frac{+50}{-100}\right)=5.67

Formula 9

| Case A | Both end moments are positive or zero | $C_b = 2.0 - \frac{M_o + 0.6M_1}{M_{CL}}$ |

C_b = 2.0 - \frac{M_o + 0.6M_1}{M_{CL}}

Formula 10

| Case B | One end moment is negative ( $M_o$ ) | $C_b = \frac{2M_1 - 2M_{CL} + 0.165M_o}{0.5M_1 - M_{CL}}$ |

C_b = \frac{2M_1 - 2M_{CL} + 0.165M_o}{0.5M_1 - M_{CL}}

Formula 11

| Case C | Both end moments are negative | $C_b = 2.0 - \frac{M_o + M_1}{M_{CL}} \left[ 0.165 + \frac{1}{3} \left( \frac{M_1}{M_o} \right) \right]$ ) |

C_b = 2.0 - \frac{M_o + M_1}{M_{CL}} \left[ 0.165 + \frac{1}{3} \left( \frac{M_1}{M_o} \right) \right]

Found in

GENERAL PROVISIONS