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7/21/2019 Steel Structures 5th Edition Solutions Manual #$% &'" " (' )) ,-,#- "" . " " " $ ! )"7 1 , - , #-8 &'" " !!)" " " ')" )!!)" " " )> ' " &)") 9 9 9 /0 #1 "$ @ )" " 9 " + 9 6$:6/# )"' !!)" " !" " " 9 " ' 96$32/2? )" 7 ' 9 ' 9 9 ) ) 9# " ) 9 /*0 234 5 26 "$ @ )" " 9 " + 9 6$:6/2 )"' !!)" " " 9 " ' 9 6$32/1 )" 7 ' 9 ' 9 ) 9 ) 9 # " ) 9 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written pe the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic recording, or likewise. 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For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc. http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 15/131 7/21/2019 Steel Structures 5th Edition Solutions Manual #$46$ +* " )+ 4- =& 31 =& > ' ! D )'" 234 5 26 " &"$ /! " D $10 9 4- =&8 9 31 =&8 234 5 26 "> 8 @ )"7 " 9 6$: + 9 6$: /260 9 -2 A )"' !!)" " )"7 " 9 6$32 ' 9 6$32 /120 9 -?$32 )" 7 ' 9 %$4 B %$1 9 %$4/4-0 B %$1/310 9 %26$- =& '9 %$-/4-0 9 ##$1 =& @ )" )" " A D 1 ;' 9 9 9#$#- ; )" " " * " ;' *$ )&'" " * * " " ! " '" *+ ! " / D * 3C%?0$ C # C $ D /-$2 " & )0 /* 9 #$:? ; $0 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written pe the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc. http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 16/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 17/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 22/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 23/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. 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For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 25/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 26/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 27/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 28/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 29/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 30/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 31/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 32/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 33/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 34/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 7.6. A W24×94 beam on a 6-ft span (see acompanying figure) underpins a column that brings 110 kips dead load and 280 kips live load to its top flange at a location 2.5 ft from the left support. The column bearing plate is 12 in. measured along the beam, and the bearing plates at the end supports are each 8 in. Investigate this beam of A992 steel for (a) flexure, (b) shear, and (c) satisfactory transmission of the reactions and concentrated load (i.e., local web yielding, web crippling, and sidesway web buckling). Specify changes (if any) required to satisfy the AISC Specification. Use LRFD Design Method (a) Obtain factored loads: Wu = 1.2(110) + 1.6(280) = 580 kips Wu ab 580(2.5)(3.5) Mu = = = 846 ft-kips L 6 Wu b 580(3.5) Vu = = = 338 kips L 6 (b) Check flexural strength assuming adequate lateral support (AISC F2.1): Flange and web local buckling slenderness limits, F y = 50 ksi steel: bf = 5.2 2tf ≤ λp = 65 = 9.2 ; Fy h = 41.9 tw φb Mn = φb Mp = φb Zx Fy = 0.9(254)(50)/12 = 953 ft-kips [φb Mn = 953 ft-kips] > [Mu = 846 ft-kips] OK ≤ λp = 640 = 90.5 OK Fy (c) Check shear strength (AISC G2.1): h For rolled I-shapes, when = 41.9 ≤ 2.24 E/F y = 53.9 , φv = 1.0 and tw Cv = 1.0. φv Vn = φ v (0.6Fy )Aw Cv = φv (0.6Fy )dtw Cv = 1.0(0.6)(50)(24.31)(0.515)(1.0) = 376 kips [φv Vn = 376 kips] > [Vu = 338 kips] OK (d) Check local web yielding strength (AISC J10.2): Rn = (5k + N )Fy tw Interior Reaction Rn = (2.5k + N )Fy tw Exterior Reaction Rn = (5k + N )Fy tw = [5(1.625) + N ] (50)(0.515) Solving for Ru = 580 kips, N = 14.4 in. at the interior reaction. Rn = (2.5k + N )Fy tw = [2.5(1.625) + N ] (50)(0.515) Solving for N to give Ru = 338 kips, N = 9.1 in. at the left exterior reaction. Solving for N to give Ru = 242 kips, N = 5.3 in. at the right exterior reaction. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 35/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Check web crippling strength (AISC J10.3): For interior reactions: 3N φRn = φ0.80t2w 1 + d tw tf 1.5 = (0.75)(0.80)(0.515)2 1 + EF yw tf tw 3N 24.31 0.515 1.5 0.875 (29000)(50)(0.875) 0.515 Solving for N to give Ru = 580 kips, N = 23.7 in. For exterior reactions, assuming N/d > 0.2: φRn = φ0.80t2w 1+ 4N d − 0.2 = (0.75)(0.40)(0.515)2 1 + tw tf 4N 24.31 1.5 − 0.2 EF yw tf tw 0.515 1.5 0.875 (29000)(50)(0.875) 0.515 Solving for N to give Ru = 338 kips, N = 24.2 in. Check [N/d = 1.00] > 0.2. Solving for N to give Ru = 242 kips, N = 13.8 in. Check [N/d = 0.57] > 0.2. (f) Check sidesway web buckling strength (AISC J10.4): When the compression flange is restrained against rotation, for (h/tw )/(Lb/bf ) = ∞ > 2.3, this limit state does not apply. Conclusion: In accordance with AISC-J10.7, “At unframed ends of beams and girders not otherwise restrained against rotation about their longitudinal axes, a pair of transverse stiffeners, extending the full depth of the shall be provided.” 24.2 in. bearingand plate at the left reaction, the 13.8 in. web, bearing plate required at The the right reaction, therequired 23.7 in. bearing plate required at the load are all too long. Bearing stiffeners should be provided. The beam has adequate flexural and shear strength. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 36/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 7.7. A W16×77 section of A992 steel is to serve on a 10-ft simply supported span. The wall bearing length is 10 in. What maximum slowly moving concentrated service load (25% dead load; 75% live load) may be carried? Use LRFD Design Method (a) Obtain factored loads: Wu = 1.2(0.25W ) + 1.6(0.75W ) = 1.5W Wu L Wu (10) Mu = = = 2.5Wu with load at midspan 4 4 Vu = 1.0Wu with load at support (b) Check flexural strength assuming adequate lateral support (AISC F2.1): Flange and web local buckling slenderness limits, F y = 50 ksi steel: bf = 6.8 ≤ λp = 65 = 9.2 ; h = 31.2 ≤ λp = 2tf Fy tw φb Mn = φb Mp = φ b Zx Fy = 0.9(150)(50)/12 = 563 ft-kips Mu = 2.5Wu = 563 ft-kips; W u = 225 kips 640 = 90.5 OK Fy (c) Check shear strength (AISC G2.1): h For rolled I-shapes, when = 31.2 ≤ 2.24 E/F y = 53.9 , φv = 1.0 and tw Cv = 1.0. φv Vn = φ v (0.6Fy )Aw Cv = φv (0.6Fy )dtw Cv = 1.0(0.6)(50)(16.52)(0.455)(1.0) = 225 kips Vu = 1.0Wu = 225 kips; Wu = 225 kips (d) Check local web yielding strength (AISC J10.2): Rn = (5k + N )Fy tw Interior Reaction Rn = (2.5k + N )Fy tw Exterior Reaction Exterior reaction controls because Wu is both the interior and exterior load φRn = φ(2.5k + N )Fy tw = 1.0 [5(1.4375) + N ] (50)(0.455) = 309 kips φRn = 1.0Wu = 309 kips; W u = 309 kips (e) Check web crippling (AISC J10.3): = 0.61] > 0.2: For exterior reactions,strength for [N/d = 10/16.52 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 37/131 7/21/2019 Steel Structures 5th Edition Solutions Manual φRn = = φ0.80t2w 1+ 4N d − 0.2 (0.75)(0.40)(0.455)2 1+ tw tf 4(10) 16.52 1.5 − 0.2 = 196 kips φRn = 1.0Wu = 196 kips; W u = 196 kips EF yw tf tw 0.455 1.5 0.76 (29000)(50)(0.76) 0.455 (f) Check sidesway web buckling strength (AISC J10.4): When the compression flange is restrained against rotation, for (h/tw )/(Lb/bf ) = ∞ > 2.3, this limit state does not apply. Conclusion: Web crippling controls! Max Wu = 196 kips; Service Load W = Wu /1.5 = 131 kips 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 38/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 7.15. Select the lightest W8section of A992 steel to use as a purlin on a roof sloped 30◦ to the horizontal. The span is 21 ft, the load is uniform 0.18 kip/ft dead load plus the purlin weight and 0.34 kip/ft snow load. Lateral stability is assured by attachment of the roofing to the compression flange. Assume the load acts through the beam centroid, there are no sag rods, and biaxial bending must be assumed. Any torsional effect can be resisted by the roofing and therefore it can be neglected. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.18 + 0.035) + 1.6(0.34) = 0.80 kips/ft wu L2 (0.80)(21)2 = = 44.2 ft-kips 8 8 Mux = M u cos θ = 44.2cos30◦ = 38.3 ft-kips Muy = M u sin θ = 44.2sin30◦ = 22.1 ft-kips Mu = (b) Use AISC H2 with no axial load term and the conservative estimate M n = S Fy : Muy Mux + ≤1 φb Mnx φb Mny Muy Sx Mux 38.3(12) 22.1(12) Sx Sx ≥ + = + φb Fy φb Fy Sy 0.9(50) 0.9(50) Sy Sx ≥ 10.2 + 5.9 Sy For S x on the order of 3 to 4: Sx ≈ 27.9 to 33.8 in.3 Assuming Zx ≈ 1.12Sx : Zx ≈ 31.2 to 37.8 in.3 Using AISC Table 3-2 Selection by Zx , for W8 beams, find W8×35 with Zx = 34.7 in.3 Check the strength. Muy Mux 38.3(12) 22.1(12) + = + φb Fy Sx φb Fy Sy 0.9(50)(31.2) 0.9(50)(10.6) = 0.3272 + 0.5561 = 0.8833 ≤ 1 OK Beam Mnx ft-kips Mny ft-kips Check W8×35 W8×31 117 103 39.8 34.8 0.3272 + 0.5561 = 0.8833 0.3690 + 0.6321 = 1.0011 OK NG 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 39/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 8.21. Assume a single W section is to serve as a crane runway girder which carries a vertical loading, as shown. In addition, design must include an axial compressive force of 14 kips and a horizontal force of 4 kips on each wheel applied 4 14 in. above the top of the compression flange. Assume torsional simple support at the ends of the beam. Select the lightest W14 section of A992 steel using the β modified flexure analogy approach. Note: All loads except weight of the crane runway girder are live loads. Use LRFD Design Method (a) Obtain factored loads: Use an estimated beam weight of 0.342 kips/ft and an estimated beam depth of 14 in. Wux = 1.6(40) = 64 kips – Lifted load Wuy = 1.6(4) = 6.4 kips – Lateral load wux = 1.2(0.020 + 0.342) = 0.4344 kips/ft – Dead load 40 − 4/2 x= = 19 ft – location for maximum moments 2 Mux1 = 2Wux x2 = 2(64)(19)2 = 1155 ft-kips – Lifted load moment L 40 Mux2 = 0.25Mux1 = 0.25(1155) = 288.8 ft-kips – Impact moment wux x(L − x) (0.4344)(19)(40 − 19) Mux3 = = = 86.7 ft-kips – Dead load moment 2 2 Mux = M ux1 + Mux2 + Mux3 = 1531 ft-kips 2Wuy x2 2(6.4)(19)2 Muy = = = 115.5 ft-kips L 40 Tu = Wuy (d/2 + rail height) = 6.4(14/2 + 4.25) = 72.00 in-kips (b) Use the β modified flexure analogy to find the equivalent lateral moment. Use β Tu 72.00 Vf = ≈ = 5.143 kips – flange force using h ≈ d h 14 2Vf x2 (5.143)(19)2 Mf = β = 0.5 = 46.41 kips – flange moment L 40 My = 2Mf = 2(46.41) = 92.83 kips – equivalent lateral moment ≈ 0.5. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 40/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (c) Select a beam using AISC H1: Mux M uy + My Sx 1531(12) (115.5 + 92.83)(12) Sx ≥ + = + φb Fy φb Fy Sy 0.9(50) 0.9(50) Sx ≥ 408.2 + 55.56 S Sx Sy y For S x on the order of 3 to 5: Sx ≈ 574.9 to 686.0 in.3 Assuming Zx ≈ 1.12Sx : Zx ≈ 643.8 to 768.3 in.3 Using AISC Table 3-2 Selection by Zx , for W14 beams, find W14×342 with Zx = 672 in.3 (d) Check the beam more accurately using the properties from AISC Table 1-1. λ= GJ = EC w (11154)(178) = 0.02578 (29000)(103000) λL = (0.02578)(40)(12) = 12.38 β = 0.16 for a point load at x and simply supported ends Tu = Wuy (d/2 + rail height) = 6.4(17.5/2 + 4.25) = 83.20 in-kips Tu 83.20 Vf = = = 5.536 kips h 15.03 2Vf x2 (5.536)(19)2 Mf = β = 0.16 = 16.19 ft-kips L 40 My = 2Mf = 2(16.19) = 32.38 ft-kips fun = Mux + M uy + My = 1531(12) + 115.5(12) + Sx + 6.273 Sy + 1.758 Sy 558 221 = 32.92 = 40.95 ksi ≤ φFy = 0.9(50) = 45 ksi OK 32.38(12) 221 The beam is sufficient. Check for a lighter beam. Section Mux ft-kips β My ft-kips fun ksi φFy = 45 ksi W14×342 W14×311 1531 1523 0.16 0.17 32.38 34.35 32.92 + 6.273 + 1.758 = 40.95 36.12 + 6.966 + 2.072 = 45.16 OK NG Use W14×342, A992 steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 41/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.2, Case 1. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 20 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W21×62 of Fy = 50 ksi steel. Use LRFD Design Method (a) Obtain factored loads: Wu = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(62/1000) = 0.0744 kips/ft (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric members, Rm = 1.0. max Cb = 2.5Mmax +12.5M 3MA + 4MB + 3MC Rm ≤ 3.0 Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.5M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M Cb = (1.0) = 1.32 2.5M + 3(0.5M ) + 4(1.0M ) + 3(0.5M ) (c) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h = 46.9 tw ≤ λp = 3.76 E = 90.6 ; Fy bf = 6.7 2tf ≤ λp = 0.38 AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(144)/12 = 600 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-16, L p = 6.25 ft and Lr = 18.1 ft. [L = 18.1 ft] < [L = 20 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). r b Mn = F cr Sx = Sx = (127) Cb π2E (Lb /rts )2 (1.32)π 2(29000) (12(20)/2.15)2 = 415 ft-kips Jc 1 + 0.078 Sx ho Lb 2 rts (1.83)(1) 1 + 0.078 (127)(20.4) 12(20) 2 2.15 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 42/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(415) = 374 ft-kips (d) Calculate the maximum service load. φM = b Wu L + wu L2 n 4 8 (1.46W )(20) (0.0744)(20)2 374 = + = 7.30W + 3.72 4 8 W = 50.7 kips Maximum Service Load W = 50.7 kips 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 43/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.2, Case 2. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 24 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W24×84 of Fy = 50 ksi steel. Use LRFD Design Method (a) Obtain factored loads: Wu = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(84/1000) = 0.101 kips/ft (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 24 ft. For doubly symmetric members, Rm = 1.0. max Cb = 2.5Mmax +12.5M 3MA + 4MB + 3MC Rm ≤ 3.0 Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.5M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M Cb = (1.0) = 1.32 2.5M + 3(0.5M ) + 4(1.0M ) + 3(0.5M ) (c) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-16 and 1-17. h = 45.9 tw ≤ λp = 3.76 E = 90.6 ; Fy bf = 5.86 2tf ≤ λp = 0.38 AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(224)/12 = 933 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-16, L p = 6.89 ft and Lr = 20.3 ft. [L = 20.3 ft] < [L = 24 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). r b Mn = F cr Sx = Sx = (196) Cb π2E (Lb /rts )2 (1.32)π 2(29000) (12(24)/2.37)2 = 579 ft-kips Jc 1 + 0.078 Sx ho Lb 2 rts (3.7)(1) 1 + 0.078 (196)(23.3) 12(24) 2 2.37 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 44/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(579) = 521 ft-kips (d) Calculate the maximum service load. φM = b Wu L + wu L2 n 4 8 (1.46W )(24) (0.101)(24)2 521 = + = 8.76W + 7.26 4 8 W = 58.7 kips Maximum Service Load W = 58.7 kips 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 45/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.2, Case 3. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 30 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W30×99 of Fy = 50 ksi steel. Use LRFD Design Method (a) Obtain factored loads: Wu = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(99/1000) = 0.119 kips/ft (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 30 ft. For doubly symmetric members, Rm = 1.0. max Cb = 2.5Mmax +12.5M 3MA + 4MB + 3MC Rm ≤ 3.0 Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.5M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M Cb = (1.0) = 1.32 2.5M + 3(0.5M ) + 4(1.0M ) + 3(0.5M ) (c) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-14 and 1-15. h = 51.9 tw ≤ λp = 3.76 E = 90.6 ; Fy bf = 7.8 2tf ≤ λp = 0.38 AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(312)/12 = 1300 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-15, L p = 7.42 ft and Lr = 21.4 ft. [L = 21.4 ft] < [L = 30 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). r b Mn = F cr Sx = Sx = (269) Cb π2E (Lb /rts )2 (1.32)π 2(29000) (12(30)/2.62)2 = 585 ft-kips Jc 1 + 0.078 Sx ho Lb 2 rts (3.77)(1) 1 + 0.078 (269)(29) 12(30) 2 2.62 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 46/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(585) = 527 ft-kips (d) Calculate the maximum service load. φM = b Wu L + wu L2 n 4 8 (1.46W )(30) (0.119)(30)2 527 = + = 11.0W + 13.4 4 8 W = 46.9 kips Maximum Service Load W = 46.9 kips 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 47/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 1: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has continuous lateral support, and Fy = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft Mu = wu L2 4.28(20)2 = = 214 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral support, Cb = 1.0. (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Zx , AISC Manual, pp. 3-11 to 3-19. Assume λ ≤ λp for a compact section. Required Zx = Mu = (214)(12) = 57.1 in.3 φb Fy (0.9)(50) Select: W18×35, Z x = 66.5 in.3 (d) Correct the moment for the selected beam weight. Mu = 214 + 1.2(beam wt)L2 1.2(35/1000)(20)2 = 214 + = 216 ft-kips 8 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. p thw = 53.5 ≤ λ = 3.76 FEy = 90.6 ; b f p 2tf = 7.06 ≤ λ = 0.38 The web is compact and the flange is compact so use AISC-F2. AISC FEy = 9.15 For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(66.5)/12 = 277 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 48/131 7/21/2019 Steel Structures 5th Edition Solutions Manual The beam has continuous lateral support, so [Lb = 0] < Lp and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb Mn = (0.9)(277) = 249 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section W18×35 W16×31 W14×34 Mu φb Mn ft-kips ft-kips 216 216 216 249 203 205 bf 2tf h tw OKAY? 7.06 6.28 7.41 53.5 51.6 43.1 OK NG NG Use W18×35 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 49/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 2: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft Mu = wu L2 4.28(20)2 = = 214 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 10 ft with L b = 10 ft. For doubly symmetric members, Rm = 1.0. 12.5Mmax Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.438M MB = moment at 1/2 pt of the unbraced segment = 0.8M MC = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M Cb = (1.0) = 1.30 2.5M + 3(0.438M ) + 4(0.8M ) + 3(0.938M ) Cb = (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Zx , AISC Manual, pp. 3-11 to 3-19. Assume λ ≤ λp for a compact section. Mu (214)(12) Required Zx = = = 57.1 in.3 φb Fy (0.9)(50) Select: W18×35, Z x = 66.5 in.3 (d) Correct the moment for the selected beam weight. Mu = 214 + 1.2(beam wt)L2 1.2(35/1000)(20)2 = 214 + = 216 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 50/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h = 53.5 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 7.06 ≤ λp = 0.38 2tf AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(66.5)/12 = 277 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-18, L p = 4.31 ft and Lr = 12.4 ft. Lp = 4.31 ft < [Lb = 10 ft] ≤ [Lr = 12.4 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Mn = Cb Mp − Mp − 0.7Fy Sx Lb − Lp Lr − Lp ≤ Mp 0.7(50)(57.6) = (1.30) 277 − 277 − 12 = 260 ft-kips 10 − 4.31 12.4 − 4.31 Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(260) = 234 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section Mu φb Mn ft-kips ft-kips bf 2tf W18×35 216 234† 7.06 † W16×31 216 186 6.28 W14×34 216 205 7.41 † Inelastic lateral torsional buckling controls h tw OKAY? 53.5 51.6 43.1 OK NG NG Use W18×35 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 51/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 3: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateral support at the ends only, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft Mu = wu L2 4.28(20)2 = = 214 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric members, Rm = 1.0. 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.750M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.750M 12.5M Cb = (1.0) = 1.14 2.5M + 3(0.750M ) + 4(1.0M ) + 3(0.750M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 20 ft. Mu 214 Required φ M = = = 188 ft-kips b n b C 1.14 Select: W14×48, φ b Mn = 193 ft-kips (d) Correct the moment for the selected beam weight. Mu = 214 + 1.2(beam wt)L2 1.2(48/1000)(20)2 = 214 + = 217 ft-kips 8 8 (e) Compute the design moment strength using the beam properties from the AISC 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 52/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Manual Table 1-1, pp. 1-22 and 1-23. h = 33.6 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 6.75 ≤ λp = 0.38 2tf E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(78.4)/12 = 327 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 6.75 ft and Lr = 21.1 ft. Lp = 6.75 ft < [Lb = 20 ft] ≤ [Lr = 21.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Mn = Cb Mp − Mp − 0.7Fy Sx Lb − Lp Lr − Lp = (1.14) 327 − 327 − 0.7(50)(70.2) 12 = 243 ft-kips ≤ Mp 20 − 6.75 21.1 − 6.75 Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(243) = 219 ft-kips The W14×48 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 48 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section Mu φb Mn ft-kips ft-kips W14×48 217 219† W21×48 217 200‡ W21×44 217 119‡ W18×46 217 134‡ W16×45 217 163‡ W16×31 216 62.6‡ W14×48 217 219† W14×43 217 187† W14×38 216 120‡ W12×45 217 188† W10×45 217 177† † Inelastic lateral torsional buckling controls bf 2tf h tw OKAY? 6.75 9.47∗ 7.22 5.01 6.23 6.28 6.75 7.54 6.57 7 6.47 33.6 53.6 53.6 44.6 41.1 51.6 33.6 37.4 39.6 29.6 22.5 OK NG NG NG NG NG OK NG NG NG NG ‡ Elastic lateral torsional buckling controls Flange local buckling limit state must be checked (see below) For the limit state of compression flange local buckling, AISC–F3.2, for W21 ×48: ∗ 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 53/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 54/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 4: dead load is 0.7 kips/ft, live load is 1.4 kips/ft, span is 28 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(1.4) ≈ 3.08 kips/ft Mu = wu L2 3.08(28)2 = = 302 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 14 ft with L b = 14 ft. For doubly symmetric members, Rm = 1.0. 12.5Mmax Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.438M MB = moment at 1/2 pt of the unbraced segment = 0.8M MC = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M Cb = (1.0) = 1.30 2.5M + 3(0.438M ) + 4(0.8M ) + 3(0.938M ) Cb = (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L = 14 ft. b Mu 302 Required φb Mn = = = 232 ft-kips Cb 1.30 Select: W14×48, φ b Mn = 239 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(48/1000)(28)2 Mu = 302 + = 302 + = 307 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 55/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 56/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Mn = M p − Mp − 0.7Fy Sx λ − λpf λrf − λpf 0.7(50)(93) = 446 − 446 − 12 9.47 − 9.15 24.1 − 9.15 = 442 ft-kips Use W21×48 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 57/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 58/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-22 and 1-23. h = 37.4 ≤ λp = 3.76 tw E = 82.7 ; Fy bf = 7.54 ≤ λp = 0.38 2tf AISC E = 8.35 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (60)(69.6)/12 = 348 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: Lp = 1.76ry rts = Iy Cw = Sx E Lr = 1.95rts 0.7Fy = 17.7 ft 29000 = 6.09 ft 60 (45.2)(1950) = 2.18 in. 62.6 Jc Sx ho 2.18 = 1.95 12 E 1.89 = 1.76 Fy 12 29000 0.7(60) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (1.05)(1) (62.6)(13.1) 1+ 0.7(60) (62.6)(13.1) 2 1 + 6.76 29000 (1.05)(1) Lp = 6.09 ft < [Lb = 14 ft] ≤ [Lr = 17.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(60)(62.6) 14 − 6.09 = (1.30) 348 − 348 − 12 17.7 − 6.09 = 338 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(338) = 304 ft-kips The W14×43 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. bf h Section Mu φb Mn OKAY? 2tf tw ft-kips ft-kips W14×43 W14×48 W21×48 307 307 307 304† 350† 404† 7.54 6.75 9.47∗ 37.4 33.6 53.6 NG OK OK 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 59/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 60/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 6: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has continuous lateral support, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral support, Cb = 1.0. (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Zx , AISC Manual, pp. 3-11 to 3-19. Assume λ ≤ λp for a compact section. Required Zx = Mu = (276)(12) = 73.5 in.3 φb Fy (0.9)(50) Select: W18×40, Z x = 78.4 in.3 (d) Correct the moment for the selected beam weight. Mu = 276 + 1.2(beam wt)L2 1.2(40/1000)(35)2 = 276 + = 283 ft-kips 8 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. p thw = 50.9 ≤ λ = 3.76 FEy = 90.6 ; b f p 2tf = 5.73 ≤ λ = 0.38 The web is compact and the flange is compact so use AISC-F2. AISC FEy = 9.15 For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(78.4)/12 = 327 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 61/131 7/21/2019 Steel Structures 5th Edition Solutions Manual The beam has continuous lateral support, so [Lb = 0] < Lp and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb Mn = (0.9)(327) = 294 ft-kips The W18×40 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 40 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section W18×40 W16×40 Mu φb Mn ft-kips ft-kips 283 283 294 274 bf 2tf h tw OKAY? 5.73 6.93 50.9 46.5 OK NG Use W18×40 with F = 50 ksi steel. y 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 62/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 7: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. Cb = 2.5M 12.5Mmax Rm ≤ 3.0 max + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.990M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M Cb = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L = 7 ft. b Mu 276 Required φb Mn = = = 274 ft-kips Cb 1.00 Select: W21×44, φ b Mn = 315 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 + 1.2(beam wt)L2 1.2(44/1000)(35)2 = 276 + = 284 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 63/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 64/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 8: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 17.5 ft with L b = 17.5 ft. For doubly symmetric members, Rm = 1.0. 12.5Mmax Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.438M MB = moment at 1/2 pt of the unbraced segment = 0.8M MC = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M Cb = (1.0) = 1.30 2.5M + 3(0.438M ) + 4(0.8M ) + 3(0.938M ) Cb = (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L = 17.5 ft. b Mu 276 Required φb Mn = = = 212 ft-kips Cb 1.30 Select: W21×48, φ b Mn = 221 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(48/1000)(35)2 Mu = 276 + = 276 + = 284 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 65/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h = 53.6 ≤ λp = 3.76 tw λp = 1.0 E = 90.6 ; λp = 0.38 Fy AISC bf E = 9.15 < = 9.47 ≤ Fy 2tf E = 24.1 Fy The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 5.86 ft and Lr = 16.6 ft. [Lr = 16.6 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). Mn = F cr Sx = Sx = (93) Cb π2E Jc 1 + 0.078 Sx ho (Lb /rts )2 (1.30)π 2(29000) (12(17.5)/2.05)2 = 319 ft-kips Lb 2 rts (0.803)(1) 1 + 0.078 (93)(20.2) 12(17.5) 2 2.05 For the limit state of compression flange local buckling, AISC-F3.2: Mn = M p − Mp − 0.7Fy Sx = 446 − 446 − λ − λpf λrf − λpf 0.7(50)(93) 12 9.47 − 9.15 24.1 − 9.15 = 442 ft-kips Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(319) = 287 ft-kips The W21×48 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 48 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section Mu φb Mn ft-kips ft-kips W21×48 284 287‡ W21×44 284 168‡ W18×46 284 185‡ W16×45 284 227‡ W14×48 284 275† † Inelastic lateral torsional buckling controls ‡ bf 2tf h tw OKAY? 9.47∗ 7.22 5.01 6.23 6.75 53.6 53.6 44.6 41.1 33.6 OK NG NG NG NG Elastic lateral torsional buckling controls 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 66/131 7/21/2019 ∗ Steel Structures 5th Edition Solutions Manual Flange local buckling limit state must be checked Use W21×48 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 67/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 9: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has continuous lateral support, and F y = 65 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral support, Cb = 1.0. (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 0 ft. Required φb Mn = Mu 50 ksi = 276 Cb Fy 1.00 Select: W18×35, φ b Mn = 249 ft-kips 50 65 = 212 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 + 1.2(beam wt)L2 1.2(35/1000)(35)2 = 276 + = 282 ft-kips 8 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h tw = 53.5 ≤ λp = 3.76 E Fy = 79.4 ; bf 2tf = 7.06 ≤ λp = 0.38 The web is compact and the flange is compact so use AISC-F2. AISC E Fy = 8.03 For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (65)(66.5)/12 = 360 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 68/131 7/21/2019 Steel Structures 5th Edition Solutions Manual The beam has continuous lateral support, so [Lb = 0] < Lp and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb Mn = (0.9)(360) = 324 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section W18×35 W16×31 W14×34 W12×35 Mu φb Mn ft-kips ft-kips 282 281 282 282 324 263 266 250 bf 2tf h tw OKAY? 7.06 6.28 7.41 6.31 53.5 51.6 43.1 36.2 OK NG NG NG Use W18×35 with Fy = 65 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 69/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 10: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 65 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. Cb = 2.5M 12.5Mmax Rm ≤ 3.0 max + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.990M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M Cb = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L = 7 ft. b Mu 50 ksi 276 50 Required φb Mn = = = 211 ft-kips Cb Fy 1.00 65 Select: W18×35, φ b Mn = 217 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 + 1.2(beam wt)L2 1.2(35/1000)(35)2 = 276 + = 282 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 70/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h = 53.5 ≤ λp = 3.76 tw E = 79.4 ; Fy bf = 7.06 ≤ λp = 0.38 2tf AISC E = 8.03 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (65)(66.5)/12 = 360 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: Lp = 1.76ry rts = Iy Cw = Sx E Lr = 1.95rts 0.7Fy = 10.7 ft 29000 = 3.78 ft 65 (15.3)(1140) = 1.51 in. 57.6 Jc Sx ho 1.51 = 1.95 12 E 1.22 = 1.76 Fy 12 29000 0.7(65) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (0.506)(1) (57.6)(17.3) 1+ 0.7(65) (57.6)(17.3) 2 1 + 6.76 29000 (0.506)(1) Lp = 3.78 ft < [Lb = 7 ft] ≤ [Lr = 10.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(65)(57.6) 7 − 3.78 = (1.00) 360 − 360 − 12 10.7 − 3.78 = 295 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(295) = 266 ft-kips The W18×35 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. bf h Section Mu φb Mn OKAY? 2tf tw ft-kips ft-kips W18×35 W18×40 W16×40 282 283 283 266† 321† 324† 7.06 5.73 6.93 53.5 50.9 46.5 NG OK OK 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 71/131 7/21/2019 Steel Structures 5th Edition Solutions Manual W16×36 W16×31 W14×38 W12×40 W12×35 282 281 283 283 282 281† 212† 273† 270† 224† 8.12∗ 6.28 6.57 7.77 6.31 48.1 51.6 39.6 33.6 36.2 NG NG NG NG NG † Inelastic lateral torsional buckling controls Flange local buckling limit state must be checked (see below) For the limit state of compression flange local buckling, AISC–F3.2, for W16 ×36: ∗ Mn = M p − Mp − 0.7Fy Sx λ − λpf λrf − λpf 0.7(65)(56.5) = 347 − 347 − 12 = 346 ft-kips 8.12 − 8.03 21.1 − 8.03 Use W16×40 with Fy = 65 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 72/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 73/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-22 and 1-23. h = 37.4 ≤ λp = 3.76 tw E = 79.4 ; Fy bf = 7.54 ≤ λp = 0.38 2tf AISC E = 8.03 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (65)(69.6)/12 = 377 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: Lp = 1.76ry rts = E 1.89 = 1.76 Fy 12 Iy Cw = Sx E Lr = 1.95rts 0.7Fy 2.18 = 1.95 12 29000 = 5.86 ft 65 (45.2)(1950) = 2.18 in. 62.6 Jc Sx ho 29000 0.7(65) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (1.05)(1) (62.6)(13.1) 1+ 0.7(65) (62.6)(13.1) 2 1 + 6.76 29000 (1.05)(1) = 16.8 ft [Lr = 16.8 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). 2 2 Mn = F cr Sx = Sx Cb π E 2 (Lb /rts ) = (62.6) (1.30)π 2(29000) (12(17.5)/2.18)2 = 290 ft-kips 1 + 0.078 Jc Sx ho Lb rts (1.05)(1) 1 + 0.078 (62.6)(13.1) 12(17.5) 2 2.18 Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(290) = 261 ft-kips The W14×43 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. bf h Section Mu φb Mn OKAY? 2tf tw ft-kips ft-kips W14×43 W14×48 W21×48 284 284 284 261‡ 312† 287‡ 7.54 6.75 9.47∗ 37.4 33.6 53.6 NG OK OK 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 74/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 75/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 76/131 7/21/2019 Steel Structures 5th Edition Solutions Manual The beam has continuous lateral support, so [Lb = 0] < Lp and lateral torsional buckling does not apply, AISC-F2.1(a). For the limit state of compression flange local buckling, AISC-F3.2: λ − λpf Mn = M p − Mp − 0.7Fy Sx λrf − λpf 0.7(100)(33.4) 8.54 − 6.47 = 310 − 310 − 12 17.0 − 6.47 = 287 ft-kips Flange local buckling controls! Calculate the design moment strength. φb Mn = (0.9)(287) = 259 ft-kips The W12×26 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. Section Mu φb Mn ft-kips ft-kips W12×26 280 259‡ W14×26 280 302 W16×26 280 313‡ W14×26 280 302 W14×22 280 240‡ W12×26 280 259‡ ‡ Elastic lateral torsional buckling controls bf 2tf h tw OKAY? 8.54∗ 5.98 7.97∗ 5.98 7.46∗ 8.54∗ 47.2 48.1 56.8 48.1 53.3 47.2 NG OK OK OK NG NG ∗ Flange local buckling limit state must be checked Use W14×26 with Fy = 100 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 77/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 13: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 100 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft Mu = wu L2 1.8(35)2 = = 276 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. Cb = 2.5M 12.5Mmax Rm ≤ 3.0 max + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.990M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M Cb = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L = 7 ft. b Mu 50 ksi 276 50 Required φb Mn = = = 137 ft-kips Cb Fy 1.00 100 Select: W16×26, φ b Mn = 138 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 + 1.2(beam wt)L2 1.2(26/1000)(35)2 = 276 + = 280 ft-kips 8 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 78/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-20 and 1-21. h = 56.8 ≤ λp = 3.76 tw λp = 1.0 E = 64.0 ; λp = 0.38 Fy AISC bf E = 6.47 < = 7.97 ≤ Fy 2tf E = 17.0 Fy The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: Lp = 1.76ry rts = E 1.12 = 1.76 Fy 12 29000 = 2.80 ft 100 Iy Cw = Sx (9.59)(565) = 1.38 in. 38.4 E Lr = 1.95rts 0.7Fy Jc Sx ho 1.38 = 1.95 12 29000 0.7(100) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (0.262)(1) (38.4)(15.3) 1+ 0.7(100) (38.4)(15.3) 2 1 + 6.76 29000 (0.262)(1) = 7.65 ft Lp = 2.80 ft < [Lb = 7 ft] ≤ [Lr = 7.65 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Lr − Lp 0.7(100)(38.4) = (1.00) 368 − 368 − 12 = 245 ft-kips Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp 7 − 2.80 7.65 − 2.80 For the limit state of compression flange local buckling, AISC-F3.2: Mn = M p − Mp − 0.7Fy Sx λ − λpf λrf − λpf = 368 − 368 − 0.7(100)(38.4) 12 7.97 17.0 − − 6.47 6.47 = 348 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(245) = 220 ft-kips The W16×26 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 79/131 7/21/2019 Section Steel Structures 5th Edition Solutions Manual Mu φb Mn ft-kips ft-kips bf 2tf W16×26 280 220† 7.97∗ W16×31 281 280† 6.28 † W16×36 282 391 8.12∗ † W18×35 282 356 7.06∗ † W16×31 281 280 6.28 W14×34 282 335† 7.41∗ † W14×30 281 286 5.74 W14×26 280 196† 5.98 † W12×30 281 264 7.41∗ W10×30 281 218† 5.7 † Inelastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked h tw OKAY? 56.8 51.6 48.1 53.5 51.6 43.1 45.4 48.1 41.8 29.5 NG NG OK OK NG OK OK NG NG NG Use W14×30 with Fy = 100 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 80/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 81/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-24 and 1-25. h = 27.1 ≤ λp = 3.76 tw λp = 1.0 E = 64.0 ; λp = 0.38 Fy AISC bf E = 6.47 < = 9.15 ≤ Fy 2tf E = 17.0 Fy The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: Lp = 1.76ry rts = E 1.94 = 1.76 Fy 12 29000 = 4.85 ft 100 Iy Cw = Sx (36.6)(791) = 2.20 in. 35 E Lr = 1.95rts 0.7Fy Jc Sx ho 2.20 = 1.95 12 29000 0.7(100) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (0.583)(1) (35)(9.3) 1+ 0.7(100) (35)(9.3) 2 1 + 6.76 29000 (0.583)(1) = 13.5 ft [Lr = 13.5 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). C π2E Mn = F cr Sx = Sx (L b/r )2 b ts = (35) (1.30)π 2(29000) (12(17.5)/2.2)2 = 179 ft-kips Jc 1 + 0.078 Sx ho L 2 rtsb (0.583)(1) 1 + 0.078 (35)(9.3) 12(17.5) 2 2.2 For the limit state of compression flange local buckling, AISC-F3.2: Mn = M p − Mp − 0.7Fy Sx λ − λpf λrf − λpf = 323 − 323 − 0.7(100)(35) 9.15 − 6.47 12 17.0 − 6.47 = 293 ft-kips Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(179) = 161 ft-kips The W10×33 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 82/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 83/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 84/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 85/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 16: dead load is 0 kips/ft, live load is 1 kips/ft, span is 35 ft, the beam has lateral support at the ends only, and F y = 50 ksi. Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0 + beam wt) + 1.6(1) ≈ 1.6 kips/ft Mu = wu L2 1.6(35)2 = = 245 ft-kips (without beam) 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 35 ft. For doubly symmetric members, Rm = 1.0. 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC Mmax = max moment in the unbraced segment = M MA = moment at 1/4 pt of the unbraced segment = 0.750M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.750M 12.5M Cb = (1.0) = 1.14 2.5M + 3(0.750M ) + 4(1.0M ) + 3(0.750M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 35 ft. Mu 245 Required φ M = = = 216 ft-kips b n b C 1.14 Select: W12×65, φ b Mn = 231 ft-kips (d) Correct the moment for the selected beam weight. Mu = 245 + 1.2(beam wt)L2 1.2(65/1000)(35)2 = 245 + = 257 ft-kips 8 8 (e) Compute the design moment strength using the beam properties from the AISC 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 86/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 87/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 88/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 89/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-12 and 1-13. h = 54.1 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 7.56 ≤ λp = 0.38 2tf AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(509)/12 = 2120 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-14, L p = 8.41 ft and Lr = 24.2 ft. Lp = 8.41 ft < [Lb = 16 ft] ≤ [Lr = 24.2 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(439) = (1.01) 2120 − 2120 − 12 = 1740 ft-kips ≤ Mp 16 − 8.41 24.2 − 8.41 Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(1740) = 1570 ft-kips The W36×135 beam is not sufficient. Check heavier sections at the same depth. The following table shows the moment corrected for the beam weight. Section Mu φb Mn ft-kips ft-kips W36×135 1580 1570† W36×150 1580 1830† W40×149 1580 1810† W36×135 1580 1570† W33×141 1580 1610† W33×130 1580 1440† W30×132 1580 1330† W27×129 1580 1210† W24×131 1580 1270† † Inelastic lateral torsional buckling controls bf 2tf h tw OKAY? 7.56 6.37 7.11 7.56 6.01 6.73 5.27 4.55 6.7 54.1 51.9 54.3 54.1 49.6 51.7 43.9 39.7 35.6 NG OK OK NG OK NG NG NG NG Use W33×141 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 90/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 91/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 92/131 7/21/2019 Steel Structures 5th Edition Solutions Manual W27×129 1580 1370† W24×117 1570 1280† W21×122 1570 1220† † Inelastic lateral torsional buckling controls 4.55 7.53 6.45 39.7 39.2 31.3 NG NG NG Use W33×130 with Fy = 60 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 93/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (a) A992 steel; continuous lateral support Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft Wu = 1.2(28) + 1.6(7) = 44.8 kips 1 Mu = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral support, Cb = 1.0. (c) Since the unbraced length is zero, select a beam using Table 3-2 Selection by Zx , AISC Manual, pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. Mu (1, 172)(12) Required Zx = = = 313 in.3 φb Fy (0.90)(50) Select: W30×108, Zx = 346 in.3 (d) Correct the moment for the beam weight. 1 Mu = 1, 172 + (108/1000)(30)2 = 1, 190 ft-kips 8 (e) Compute the design Manual Table 1-1, pp.moment 1-14 andstrength 1-15. using the beam properties from the h = 49.6 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 6.89 ≤ λp = 0.38 2tf AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 94/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Mn = M p = F y Zx = (50)(346)/12 = 1, 440 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: The beam has continuous lateral support, so [Lb = 0] < Lp and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb Mn = (0.90)(1, 440) = 1, 300 ft-kips The W30×108 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 108 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section W30 W27× ×108 102 W24×104 Mu φb Mn ft-kips ft-kips 1,190 1,190 1,190 1,300 1,140 1,080 bf 2tf h tw OKAY? 6.89 6.03 8.5 49.6 47.1 43.1 OK NG NG Use W30×108 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 95/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (b) A992 steel; lateral support at the ends Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft Wu = 1.2(28) + 1.6(7) = 44.8 kips 1 Mu = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends. The unbraced length is Lb = 30 ft. For doubly symmetric12.5M members, R m = 1.0. Use statics to find the moments in the beam: max Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC 12.5(1, 172) = (1.0) = 1.14 2.5(1, 172) + 3(879.4) + 4(1, 172) + 3(879.4) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 30 ft. Mu 1, 172 Required φb Mn = = = 1, 030 ft-kips 1.14 Cb Select: W24×146, φb Mn = 1, 070 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 1, 172 + (146/1000)(30)2 = 1, 190 ft-kips 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-16 and 1-17. AISC 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 96/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 97/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (c) A992 steel; lateral support at the ends and at 10 ft Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft Wu = 1.2(28) + 1.6(7) = 44.8 kips 1 Mu = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends and at 10 ft. The worst loading and longest unbraced occurRon the segment from to 10 find ft tothe 30 ft with L b in = the 20 ft. For doubly symmetriclength members, 1.0. Use statics moments beam: m= 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC 12.5(1, 172) = (1.0) = 1.15 2.5(1, 172) + 3(1, 172) + 4(1, 092) + 3(626.5) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 20 ft. Mu 1, 172 Required φb Mn = = = 1, 020 ft-kips Cb 1.15 Select: W33×118, φb Mn = 1, 080 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 1, 172 + (118/1000)(30)2 = 1, 190 ft-kips 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-12 and 1-13. AISC 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 98/131 7/21/2019 Steel Structures 5th Edition Solutions Manual h = 54.5 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 7.76 ≤ λp = 0.38 2tf E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(415)/12 = 1, 730 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-14, L p = 8.19 ft and Lr = 23.5 ft. Lp = 8.19 ft < [Lb = 20 ft] ≤ [Lr = 23.5 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(50)(359) 20 − 8.19 = (1.15) 1, 730 − 1, 730 − 12 23.5 − 8.19 = 1, 390 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(1, 390) = 1, 250 ft-kips The W33×118 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 118 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section W33×118 W30×116 W27×114 W24×117 † Mu φb Mn ft-kips ft-kips 1,190 1,190 1,190 1,190 1,250† 1,110† 1,020† 1,160† bf h 2tf tw 7.76 6.17 5.41 7.53 54.5 47.8 42.5 39.2 OKAY? OK NG NG NG Inelastic lateral torsional buckling controls Use W33×118 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 99/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (d) A572 Grade 60 steel; lateral support at the ends and at 10 ft Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft Wu = 1.2(28) + 1.6(7) = 44.8 kips 1 Mu = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends and at 10 ft. The worst loading and longest unbraced occurRon the segment from to 10 find ft tothe 30 ft with L b in = the 20 ft. For doubly symmetriclength members, 1.0. Use statics moments beam: m= 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC 12.5(1, 172) = (1.0) = 1.15 2.5(1, 172) + 3(1, 172) + 4(1, 092) + 3(626.5) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 20 ft. Mu 50 ksi 1, 172 50 Required φb Mn = = = 846 ft-kips Cb Fy 1.15 60 Select: W24×104, φb Mn = 875 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 1, 172 + (104/1000)(30)2 = 1, 190 ft-kips 8 (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-16 and 1-17. AISC 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 100/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 101/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 102/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 103/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-14 and 1-15. h = 49.6 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 6.89 ≤ λp = 0.38 2tf AISC E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(346)/12 = 1, 440 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-15, L p = 7.59 ft and Lr = 22.0 ft. Lp = 7.59 ft < [Lb = 12 ft] ≤ [Lr = 22.0 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp ≤ Mp 0.7(50)(299) = (1.00) 1, 440 − 1, 440 − 12 = 1, 270 ft-kips 12 − 7.59 22.0 − 7.59 Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(1, 270) = 1, 150 ft-kips The W30×108 beam is not sufficient. Try heavier sections at the same depth. The following table shows the moment corrected for the beam weight. Section W30×108 W30×116 W27×114 W24×104 W24×103 W21×111 † Mu φb Mn ft-kips ft-kips 1,150 1,150 1,150 1,150 1,150 1,150 1,150† 1,260† 1,150† 1,050† 918† 1,020† bf 2tf h tw OKAY? 6.89 6.17 5.41 8.5 4.59 7.05 49.6 47.8 42.5 43.1 39.2 34.1 NG OK NG NG NG NG bf 2tf h tw OKAY? 6.17 47.8 OK Inelastic lateral torsional buckling controls Check segment A Section W30×116 Mu φb Mn ft-kips ft-kips 1,110 1,420 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 104/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Use W30×116 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 105/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.6. Select the lightest W sections for the conditions shown in the accompanying figure. Assume there is no deflection limitation. Use (b) A572 Grade 60 steel. Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.4) + 1.6(1.1) = 2.24 kips/ft Wu = 1.2(15) + 1.6(15) = 42.0 kips 1 Mu = (2.24)(42)2 + 10(42.0) = 1, 124 ft-kips 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral support at the ends, at 15 ft, and at 27 ft. For doubly symmetric members, Rm = 1.0. Cb = 2.5M 12.5Mmax Rm ≤ 3.0 max + 3MA + 4MB + 3MC 12.5(1, 084) = (1.0) = 1.56 segment A 2.5(1, 084) + 3(318.2) + 4(604.8) + 3(860.0) 12.5(1, 124) = (1.0) = 1.00 segment B 2.5(1, 124) + 3(1, 114) + 4(1, 124) + 3(1, 114) Assume segment B controls. (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L b = 12 ft. Mu 50 ksi 1, 124 50 Required φb Mn = = = 933 ft-kips Cb Fy 1.00 60 Select: W30×99, φ b Mn = 1, 020 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 1, 124 + (99/1000)(42)2 = 1, 150 ft-kips 8 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 106/131 7/21/2019 Steel Structures 5th Edition Solutions Manual (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-14 and 1-15. h = 51.9 ≤ λp = 3.76 tw E = 82.7 ; Fy bf = 7.8 ≤ λp = 0.38 2tf AISC E = 8.35 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (60)(312)/12 = 1, 560 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: Lp = 1.76ry rts = Iy Cw = Sx E Lr = 1.95rts 0.7Fy = 19.3 ft 29, 000 = 6.77 ft 60 (128)(26800) = 2.62 in. 269 Jc Sx ho 2.62 = 1.95 12 E 2.1 = 1.76 Fy 12 29, 000 0.7(60) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (3.77)(1) (269)(29) 1+ 0.7(60) (269)(29) 2 1 + 6.76 29, 000 (3.77)(1) Lp = 6.77 ft < [Lb = 12 ft] ≤ [Lr = 19.3 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(60)(269) 12 − 6.77 = (1.00) 1, 560 − 1, 560 − 12 19.3 − 6.77 = 1, 310 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(1, 310) = 1, 180 ft-kips The W30×99 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 99 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 107/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 108/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 109/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 110/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Section Mu φb Mn ft-kips ft-kips W16×67 241 336† W24×62 241 153‡ W21×62 241 213‡ W21×57 241 148‡ W18×65 241 219‡ W18×46 240 92.4‡ W16×57 241 169‡ † Inelastic lateral torsional buckling controls ‡ bf 2tf h tw OKAY? 7.7 5.97 6.7 5.04 5.06 5.01 4.98 35.9 50.1 46.9 46.3 35.7 44.6 33 OK NG NG NG NG NG NG Elastic lateral torsional buckling controls With deflection use W16×67 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 111/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 1: MD = 49 ft-kips, M L = 98 ft-kips, L = 28 ft, F y = 50 ksi, and the deflection limit is L/360. Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): Mu = 1.2MD + 1.6ML = 1.2(49) + 1.6(98) = 216 ft-kips (b) Obtain the minimum moment of inertia Ix required using the service live load moment. L 28(12) ∆lim = = = 0.933 in. 360 360 5MLL2 (5)(98 × 12)(28 × 12)2 Min I x = = = 511 in.4 48E ∆lim (48)(29, 000)(0.933) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual, pp. 3-96 to 3-131 with L b = 7 ft. Mu 215.6 Required φb Mn = = = 216 ft-kips Cb 1.00 Select: W18×35, φ b Mn = 217 ft-kips Unbraced Length, AISC (f) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. AISC (e) Correct the moment for the beam weight. 1 Mu = 215.6 + 1.2 (35/1000)(28)2 = 220 ft-kips 8 h = 53.5 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 7.06 ≤ λp = 0.38 2tf E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 112/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 113/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 114/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 115/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Section Ix in.4 LL Defl. in. LL Defl. Limit in. OKAY? W14×34 W14×38 W16×40 340 385 518 1.40 1.24 0.921 0.933 0.933 0.933 NG NG OK W16 W18× ×36 35 448 510 1.06 0.935 0.933 0.933 NG NG Use W16×40 with Fy = 60 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 116/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 3: MD = 0 ft-kips, M L = 240 ft-kips, L = 48 ft, F y = 50 ksi, and the deflection limit is L/300. Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): Mu = 1.2MD + 1.6ML = 1.2(0) + 1.6(240) = 384 ft-kips (b) Obtain the minimum moment of inertia Ix required using the service live load moment. L 48(12) ∆lim = = = 1.92 in. 300 300 5MLL2 (5)(240 × 12)(48 × 12)2 Min I x = = = 1, 790 in.4 48E ∆lim (48)(29, 000)(1.92) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual, pp. 3-96 to 3-131 with L b = 12 ft. Mu 384.0 Required φb Mn = = = 384 ft-kips Cb 1.00 Select: W21×62, φ b Mn = 440 ft-kips Unbraced Length, AISC (f) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. AISC (e) Correct the moment for the beam weight. 1 Mu = 384.0 + 1.2 (62/1000)(48)2 = 405 ft-kips 8 h = 46.9 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 6.7 ≤ λp = 0.38 2tf E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 117/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 118/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 119/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 4: MD = 0 ft-kips, M L = 240 ft-kips, L = 48 ft, F y = 65 ksi, and the deflection limit is L/300. Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): Mu = 1.2MD + 1.6ML = 1.2(0) + 1.6(240) = 384 ft-kips (b) Obtain the minimum moment of inertia Ix required using the service live load moment. L 48(12) ∆lim = = = 1.92 in. 300 300 5MLL2 (5)(240 × 12)(48 × 12)2 Min I x = = = 1, 790 in.4 48E ∆lim (48)(29, 000)(1.92) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, Manual, pp. 3-96 to 3-131 with L b = 12 ft. Mu 50 ksi 384.0 50 Required φb Mn = = = 295 ft-kips Cb Fy 1.00 65 Select: W21×48, φ b Mn = 313 ft-kips AISC (e) Correct the moment for the beam weight. 1 M = 384.0 + 1.2 (48/1000)(48)2 = 401 ft-kips u 8 (f) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-18 and 1-19. h = 53.6 ≤ λp = 3.76 tw E = 79.4 ; λp = 0.38 Fy AISC bf E = 8.03 < = 9.47 ≤ Fy 2tf 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 120/131 7/21/2019 Steel Structures 5th Edition Solutions Manual λp = 1.0 E = 21.1 Fy The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: Lp = 1.76ry rts = Iy Cw = Sx E Lr = 1.95rts 0.7Fy = 14.3 ft 29, 000 = 5.14 ft 65 (38.7)(3950) = 2.05 in. 93 Jc Sx ho = 1.95 2.05 12 E 1.66 = 1.76 Fy 12 29, 000 0.7(65) 1+ 0.7Fy Sx ho 2 1 + 6.76 E Jc (0.803)(1) (93)(20.2) 1+ 2 1 + 6.76 0.7(65) 29, 000 (93)(20.2) (0.803)(1) Lp = 5.14 ft < [Lb = 12 ft] ≤ [Lr = 14.3 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(65)(93) 12 − 5.14 = (1.00) 580 − 580 − 12 14.3 − 5.14 = 409 ft-kips For the limit state of compression flange local buckling, AISC-F3.2: Mn = M p − Mp − 0.7Fy Sx λ − λpf λrf − λpf 0.7(65)(93) 9.47 − 8.03 = 580 − 580 − 12 21.1 − 8.03 = 555 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φ M = (0.90)(409) = 369 ft-kips b n The W21×48 beam does not have sufficient strength. The following table shows the moment corrected for the beam weight. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 121/131 7/21/2019 Section Steel Structures 5th Edition Solutions Manual Mu φb Mn ft-kips ft-kips bf 2tf h tw W21×48 401 369† 9.47∗ 53.6 † W21×83 413 748 5 36.4 † W24×76 410 764 6.61 49 W24×68 408 662† 7.66 52 † W24×62 405 461 5.97 50.1 W18×65 406 488† 5.06 35.7 † Inelastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked (see below) OKAY? NG OK OK OK OK OK (g) Check the beams for deflection. Section Ix in.4 LL Defl. in. LL Defl. Limit in. OKAY? W21×48 W21×83 W24×76 W24×68 W24×62 W18×65 959 1830 2100 1830 1550 1070 3.58 1.88 1.63 1.88 2.21 3.21 1.92 1.92 1.92 1.92 1.92 1.92 NG OK OK OK NG NG Use W24×68 with Fy = 65 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 122/131 7/21/2019 Steel Structures 5th Edition Solutions Manual 9.13, Part a. Select the lightest W section for the situation shown in the accompanying figure. The concentrated load W is 5 kips dead load and 15 kips live load. Assume lateral support is provided at the reactions and the concentrated loads. Use A992 steel. Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): The maximum moment is at the support. Wu = 1.2(5) + 1.6(15) = 30.0 kips 1 Mu = (beam wt.)(10)2 + 10Wu = 10(30.0) = 300.0 ft-kips w/o beam 2 (b) Determine the Cb factor, AISC-F1. The moment is uniform over the 30 ft span, so C b = 1.0. (c) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC (e) Compute the design moment strength using the beam properties from the Manual Table 1-1, pp. 1-22 and 1-23. AISC Manual , pp. 3-96 toM3-131300.0 with L b = 30 ft. u Required φb Mn = = = 300 ft-kips Cb 1.00 Select: W14×74, φ b Mn = 302 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 300.0 + 1.2 (74/1000)(10)2 = 304 ft-kips 2 h = 25.4 ≤ λp = 3.76 tw E = 90.6 ; Fy bf = 6.41 ≤ λp = 0.38 2tf E = 9.15 Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = M p = F y Zx = (50)(126)/12 = 525 ft-kips 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 123/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 124/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 125/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 126/131 7/21/2019 Steel Structures 5th Edition Solutions Manual Section Mu φb Mn ft-kips ft-kips W21×48 303 305† W21×44 303 190‡ W18×46 303 204‡ W16×45 303 235† W14×43 303 228† † Inelastic lateral torsional buckling controls ‡ ∗ bf 2tf h tw OKAY? 9.47∗ 7.22 5.01 6.23 7.54 53.6 53.6 44.6 41.1 37.4 OK NG NG NG NG Elastic lateral torsional buckling controls Flange local buckling limit state must be checked (f) Check Case 1. The following table shows the moment corrected for the beam weight. Section Mu φb Mn bf 2t h t ft-kips ft-kips f w W21×48 394 398‡ 9.47∗ ‡ W21×44 393 311 7.22 ‡ W18×46 393 333 5.01 W16×45 393 309 6.23 W14×43 393 261 7.54 ‡ Elastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked 53.6 53.6 44.6 41.1 37.4 OKAY? OK NG NG NG NG Use W21×48 with Fy = 50 ksi steel. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 127/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 128/131 7/21/2019 Steel Structures 5th Edition Solutions Manual F2.1(a). Yielding controls! Calculate the design flexural strength. φb Mn = (0.90)(892) = 802 ft-kips (e) Moment magnification. Obtain the slenderness ratio for the axis of bending. KL Kx Lx Axis of bending = = 47.3 (from above) r rx π 2EI π 2 (29, 000)(1240) Pe1 = = = 5, 090 kips (KL)2 [(1.0)(22)(12)]2 According to AISC-C1.1b, for transverse loading, C m = 1.0. Cm 1.0 B1 = = = 1.076 1 − Pu /Pe1 1 − 360/5, 090 (f) Use AISC-H1 to check the beam-column. Pu 360 ≥ φc Pn = 1, 520 = 0.236 0.2 so use AISC Formula (H1-1a) omitting the y-axis bending term. Pu 8 Mux 8 (1.076)(5.5)Wu + = 0.236 + ≤ 1.0 φc Pn 9 φb Mnx 9 802 1.0 − 0.2362 Wu = = 116.5 kips 0.006556 Calculate the service load. 1.2(0.2W ) + 1.6(0.8W ) = 116.5 kips W = 76.7 kips Service concentrated load is W = 76.7 kips. 010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 129/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 130/131 7/21/2019 Steel Structures 5th Edition Solutions Manual http://slidepdf.com/reader/full/steel-structures-5th-edition-solutions-manual 131/131