Download presentation

Presentation is loading. Please wait.

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.).

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google