1. B-14 1 – Full section Yielding (L b ≤ L p ): M n = M p = FyZ x (AISC F 2.1) 2 – Inelastic Lateral Torsional Buckling ( L p < L b ≤ L r ): 3 – Elastic Lateral Torsional Buckling (L r < L b ): M n = F cr S x ≤ M p (AISC F-2.3) where: J = Torsional constant (in 4 ). ho = Distance between flanges centroids (in). C = 1.0 for w shapes. r ts = radius of gyration of the compression flange plus one-sixth of the web.

2. B-15 (C b ) Equations (F 2.2) & (F2.4) for compact beams affected by lateral torsional buckling, require the introduction of the “Moment Gradient Factor” (C b ) for non-uniform bending moment values between the lateral bracing points for (L b ). AISC provides value for C b as: The effect of C b on Nominal Strength is shown below:

3. B-16 (C b ) Example B - 5 Determine (C b ) for a uniformly loaded, simply supported beam with lateral supports at its ends only. Solution

4. B-17 (C b ) For unbraced cantilever beams, AISC recommends the value of C b = 1.0. A value of C b = 1.0 is always conservative and represent uniform banding throughout the unbraced length (L b ), (See Table 3-1) AISC.

5. B-18 Example B - 6 Solution Determine the design strength (  b M n ) for W14  68 made of A-572-Gr50 steel and: A)Continuous lateral support. B)Unbraced length = 20 ft, C b = 1.0 C)Unbraced length = 20 ft, C b = 1.75 A) Check compactness: web is always compact !  M n = M p = FyZ = 20  115 = 5750 in·k = 479 ft·k.  b M n = 0.9  479 = 431 ft·k. B)

6. B-19 Continued: SinceL p < (L b = 20 ft) < L r  Equation F – 2.2 controls:  b M n = 0.9  316.25 = 284.6 ft·kip. C) For C b = 1.75, other conditions unchanged:  M n = 1.75  316.25 = 553.4 ft·k. Since M n ≤ M p, thenM n = M p = 479 ft·k  b M n = 0.9  479 =431 ft·k.

7. Example B - 7 Solution A simply supported beam of span = 20ft is to carry static dead load of (1.0 k/ft) and a live load of (2.0 k/ft) in addition to its own dead weight. The flange is laterally supported at support points only. Select the most economical W shape using A572-Gr50 steel. B-20 w L = 20 ft Estimate self weight = 0.06 k/ft. W u = 1.2 x 1.06 + 1.6 x 2 = 4.47 k/ft Determine (C b ) for UDL = 1.14 (see B-16)  b M n  M u   b M n  224 k.ft Enter Beam Design Moments Chart at AISC for L b = 20 ft, and  b M n = 234, select: W12 x 53 (page 3.126 but for C b = 1.0) Check your selection: From Load Factor Design Selection Table 3.2 in AISC (page 3-17) : Zx = 77.9 in 3, L p = 8.76 ft, L r = 28.2 ft.

8. B-21 SinceL p <(L b =20 ft) < L r then :equation ( F-2 – 2b AISC) where C b = 1.14 M p = F y Z x = 50x77.9 = 3895 k. in = 324.6 k·ft. S x = 70.6 in 3  b M n = 0.9 x 292 = 262.7  (M u = 223.6 k·ft) OK

9. B-22 As noted earlier, most W,M & S shapes are compact for Fy = 36 ksi and Fy = 50 ksi, very few sections are non-compact because of their flanges, but non are slender. The effect of non-compact flange is recognized in the AISC as the smaller value of LTB (AISC F 2.2) and where

10. B-23 A simple supported beam with span = 45 ft is laterally supported at ends only, and is subject to the following service loading: D.L. = 0.4 k/ft ( including self wt.) L.L = 0.7 k/ft Is W 14 x 90 made of A572-Gr50 steel adequate ? Example B - 8 Solution Wu = 1.2 x 0.4 = 1.6 k·ft

11. B-24  p < < r The shape is non-compact. Section properties: Z x = 157 in 3, S x = 143 in 3 (properties 1) L p = 15.2 ft, L r = 42.6 ft (Table 3.2 p. 3.16)  b M n = 0.9 x 638.2 = 574.4  M u OK  Now we check the capacity due to LTB: L r < ( L b = 45 ft)  Elsatic LTB controls

12. B-25 M n = F cr S x  M p  b M n = 398.7 k. ft < 405 k·ft (probably this beam is O.K.).