[...]... size, in (mm) 8.59 ( 218 .18 6) 9.50 (2 41. 300) 10 .72 (272.228) 9.09 (230.886) 9.94 (252.476) 11 .10 (2 81. 940) 6 × 31/ 2 × 1/ 2 (15 2.4 × 88.9 × 12 .7) 6 × 4 × 7 /16 (15 2.4 × 10 1.6 × 11 .11 ) None Area, in2 (cm2) (cm2) 4.50 (29.034) 4 .18 (26.969) The most economical member is the one with the least area Therefore, use two angles 6 × 4 × 7 /16 in (15 2.4 × 10 1.6 × 11 .11 mm) Downloaded from Digital Engineering Library... CIVIL ENGINEERING 1. 23 For a rectangular section of thickness t, r = 0.289t Then t = 0 .11 5/0.289 = 0.40 in (10 .16 0 mm) Set t = 7 /16 in (11 .11 mm); r = 0 .12 7 in (3.226 mm); L/r = 16 .1/ 0 .12 7 = 12 7; f = 9.59 kips/in2 (66 .12 3 MPa); A = 5.0/9.59 = 0.52 in2 (3.355 cm2) From the Manual, the minimum width required for 1/ 2-in (12 .7 mm) rivets = 11 /2 in (38 .1 mm) Therefore, use a flat bar 11 /2 × 7 /16 in (38 .1 × 11 .11 ... 209 /15 .59 = 13 . 41 kips/in2 (92.4 61 MPa), P(KL)2 = 16 0(240)2 = 9.22 × 10 6 kip·in2 (2.648 × 10 4 kN·m2), and ax/[ax − P(KL)2] = 63.5/(63.5 − 9.22) = 1. 17; then 16 0 + 0.2 21( 31. 5) (12 )(0.793) (13 . 41/ 22) (1. 17) = 207 < 209 kips (929.6 kN) This is acceptable 4 Substitute in the second transformed equation Thus, 16 0 (13 . 41/ 22) + 0.2 21( 31. 5) (12 ) (13 . 41/ 22) = 14 8 < 209 kips (929.6 kN) This is acceptable The W12 × 53 section... helpful in calculating the moment of inertia Thus, Af = 15 in2 (96.8 cm2); Aw = 24.75 in2 (15 9.687 cm2); I = 42,400 in4 (17 6.4 81 dm4); S = 12 56 in3 (20,585.8 cm3) For the T section comprising the flange and one-sixth the web, A = 15 + 4 .13 = 19 .13 in2 (12 3.427 cm2); then I = (1/ 12)(0.75)(20)3 = 500 in4 (20 81. 1 dm4); r = (500 /19 .13 )0.5 = 5 .11 in (12 9.794 mm); L′/r = 18 (12 )/5 .11 = 42.3 2 Ascertain if the member... of steel and thickness of member, the yield-point stress is 50 kips/in2 (344.8 MPa), as given in the Manual Thus, A = 14 .11 in2 ( 910 38 cm2); rx = 3. 61 in ( 91. 694 mm); ry = 2.08 in (52.832 mm) Then KxL/rx = 30 (12 )/3. 61 = 10 0; KyL/ry = 15 (12 )/ 2.08 = 87 5 Determine the allowable stress and member capacity From the Manual, f = 14 . 71 kips/in2 (10 1.425 MPa) with a slenderness ratio of 10 0 Then P = 14 .11 (14 . 71) ... the Manual, A = 3. 61 in2 (23.2 91 cm2); ry = 1. 56 in (39.624 mm); rz = 0.99 in (25 .14 6 mm); rv = (2 × 1. 562 − 0.992)0.5 = 1. 97 in (50.038 mm) 4 Determine the minimum radius of gyration of the built-up section; compute the strut capacity Thus, r = rp = 1. 97 in (50.038 mm); KL/r = 12 (12 ) /1. 97 = 73 From the Manual, f = 16 .12 kips/in2 (766.3 61 MPa) Then P = Af = 2(3. 61) (16 .12 ) = 11 6 kips ( 515 .97 kN) SECTION... 4053 (12 )(35)/[2(22)(34.5)2] − 29.75/6 = 27.54 in2 (17 7.688 cm2) Try 22 × 11 /4 in (558.8 × 31. 75 mm) plates with Af = 27.5 in2 (17 7.43 cm2) The width-thickness ratio of projection = 11 /1. 25 = 8.8 < 16 This is acceptable Thus, the trial section will be one web plate 68 × 7 /16 in (17 27 × 11 .11 mm); two flange plates 22 × 11 /4 in (558.8 × 31. 75 mm) 4 Test the adequacy of the trial section For this test, compute... moments: +10 1.7 + 0 .1( 0.5) (11 5.9 + 13 0.2) = +11 4.0 ft·kips (15 4.58 kN·m); 0.9( 13 0.2) = 11 7.2 ft·kips ( 15 8.92 kN·m); design moment = 11 7.2 ft·kips (15 8.92 kN·m) 3 Select the beam size Thus, S = M/f = 11 7.2 (12 )/24 = 58.6 in3 (960.45 cm3) Use W16 × 40 with S = 64.4 in3 (10 55.52 cm3); Lc = 7.6 ft (2.32 m) SHEARING STRESS IN A BEAM—EXACT METHOD Calculate the maximum shearing stress in a W18 × 55 beam at a section... capacity of 432 kips (19 21. 5 N) Then force per bar = 1/ 2 (0.02)(432) × (16 .1/ 14) = 5.0 kips (22.24 N) Also, L/r ≤ 14 0; therefore, r = 16 .1/ 140 = 0 .11 5 in (2.9 210 mm) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2004 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website CIVIL ENGINEERING. .. = 1. 78 in2 (11 .485 cm2) Try two 4 × 1/ 4 in (10 1.6 × 6.35 mm) plates; Ast = 2.0 in2 (12 .90 cm2); width-thickness ratio = 4/0.25 = 16 This is acceptable Also, (h/50)4 = (68/50)4 = 3.42 in4 (14 2.3 51 cm4); I = (1/ 12)(0.25)(8.44)3 = 12 .52 in4 (5 21. 1 21 cm4) > 3.42 in4 (14 2.3 51 cm4) This is acceptable The stiffeners must be in intimate contact with the compression flange, but they may terminate 13 /4 in (44.45 . Beam 1. 109 Design of Web Reinforcement 1. 111 Determination of Bond Stress 1. 112 Design of Interior Span of a One-Way Slab 1. 113 Analysis of a Two-Way Slab by the Yield-Line Theory 1. 115 DESIGN OF. one-sixth the web, A = 15 + 4 .13 = 19 .13 in 2 (12 3.427 cm 2 ); then I = (1/ 12)(0.75)(20) 3 = 500 in 4 (20 81. 1 dm 4 ); r = (500 /19 .13 ) 0.5 = 5 .11 in (12 9.794 mm); L′/r = 18 (12 )/5 .11 = 42.3. 2. Ascertain. Terms of Use as given at the website. CIVIL ENGINEERING 1. 12 SECTION ONE Modified maximum moments: +10 1.7 + 0 .1( 0.5) (11 5.9 + 13 0.2) = +11 4.0 ft·kips (15 4.58 kN·m); 0.9( 13 0.2) = 11 7.2 ft·kips ( 15 8.92