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BC-11

20 k. W8  35 A36. 20 k. 20 k. 5 '. 5 '. 10 '. Example on Beam-Columns. Example BC-3. The horizontal beam column shown below is subjected to service live loads as shown. The member is laterally braced at its ends. Check its capacity to LRFD code ?. Solution:-. Column Action:-.

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BC-11

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  1. 20k W8  35 A36 20k 20k 5' 5' 10' Example on Beam-Columns Example BC-3 The horizontal beam column shown below is subjected to service live loads as shown. The member is laterally braced at its ends. Check its capacity to LRFD code? Solution:- Column Action:- Rx = 3.51 in, ry = 2.03 in, Ag = 10.3 in2 BC-11

  2. Example on Beam-Columns (Contd.) c Fcr = 27 ksi (Table 4 - 22 page 4 – 319). c Pn = c Fcr Ag = 27  10.3 = 278.1 k  Use Equation (H1 – 1b) interaction formula. Beam Action:- Lb = 10 ft, Cb = 1.32, Zx = 34.7 in4 Ixx = 127 in3 Sx = 31.2 in3. Mp = Zx Fy = 104.1 k·ft. BC-12

  3. Example on Beam-Columns (Contd.) • Mn = Mp = 104.1k·ft b Mn = 0.9  104.1 = 93.7k·ft BC-13

  4. Example on Beam-Columns (Contd.) Moment Magnification:- Cm = 1.0 (transverse loads and ends unrestrained). Mu = Bi Mnt = 1.013  80.52 = 81.55 k·ft Interaction Formula:- Section is adequate ! BC-14

  5. Laterally Braced Beam-Column Example Example BC – 4 The beam column shown below is part of a braced frame. The factored loads and moments are given as shown. Bending is on strong axis. KxLx = KyLy = 14 ft. The W 14  61 section is made of A572 Gr 50 steel. Is this member safe ? Solution:- • Determine • interaction formula, • and Column Action :- BC-15

  6. Laterally Braced Beam-Column Example (contd.) From the Beam Design Charts with Cb = 1.0, Lb = 14 b Mn = 344 kips·ft. (page 3.121) For actual (Cb)1 see the moment diagram of the column: B) Beam Behavior:- • b Mn = 364.6 kips·ft. b Mp = 383 k·ft (page 3 – 17 – Zx tables). (also confirmed on page 3 – 121). BC-16

  7. Laterally Braced Beam-Column Example (contd.) C) Interaction Equation:- The factored load moments :- Mnt = 82 k·ft , Mlt = 0 Mu = B1Mnt + B2 Mlt = 1.0066  82 + 0 = 82.5 k·ft BC-17

  8. DL = 20k LL = 40k wk/ft P P w = 12  96 20 ft Laterally Braced Beam-Column Example BC – 5:- Determine the service load w(40% DL + 60% LL) that can be applied to the member shown below, Ends are simply supported. Lateral bracing at ends only A 572 Gr 50 steel, strong axis bending. Solution:- Pu = 1.2  20 + 1.6  40 = 88 kips. Let wu = factored ultimate load (k/ft) = ? Mux = B1 Mnt + B2 Mlt = B1 Mnt BC-18

  9. Laterally Braced Beam-Column Column Action:- c Fcr = 28.9 ksi c Pn = c Fcr Ag = 28.9  28.2 = 815.0 k· Beam Action:- For lb = 20 ft & Cb = 1, b Mnx = 498 k·ft (page 3 – 119 LRFD). Actual Cb = 1.14 Actual b Mnx = 1.14  498 =567.7k·ft Beam-Column Action:- BC-19

  10. Laterally Braced Beam-Column Example (contd.) • 1.2 DL + 1.6 LL = wu • 1.2  (0.4 w) + 1.6  (0.6 w) = 10.5 • 1.44 w = 10.5 • w =7.30 k/ft • DL =3.0 k/ft LL = 4.30 k/ft BC-20

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