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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 . The column bearing plate is 12 in. measured along the beam, and the bearing plates at the end s 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 (AISC F2.1): Flange and web local buckling slenderness limits, Fy = 50 ksi steel: # " " # # " bf 65 640 h = 5.2 ≤ λp = p = 9.2 ; = 41.9 ≤ λp = p = 90.5 OK 2tf tw Fy Fy 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

(c) Check shear strength (AISC G2.1): q h For rolled I-shapes, when = 41.9 ≤ 2.24 E/Fy = 53.9 , φv = 1.0 and tw Cv = 1.0. φv Vn = φv (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. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(e) Check web crippling strength (AISC J10.3): For interior reactions:  !1.5  s EFyw tf 3N tw  φRn = φ0.80t2w 1 + d tf tw " 1.5 # r 3N 0.515 (29000)(50)(0.875) = (0.75)(0.80)(0.515)2 1 + 24.31 0.875 0.515 Solving for N to give Ru = 580 kips, N = 23.7 in. For exterior reactions, assuming N/d > 0.2:  !1.5  s EFyw tf 4N tw  φRn = φ0.80t2w 1 + − 0.2 d tf tw " 1.5 # r 4N 0.515 (29000)(50)(0.875) = (0.75)(0.40)(0.515)2 1 + − 0.2 24.31 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 web, shall be provided.” The 24.2 in. bearing plate required at the left reaction, the 13.8 in. bearing plate required at the right reaction, and the 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.


7.7. A W16×77 section of A992 steel is to serve on a 10-ft simply ed 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 (b) Check flexural strength assuming adequate lateral (AISC F2.1): Flange and web local buckling slenderness limits, Fy = 50 ksi steel: # " # # " " bf 65 h 640 = 6.8 ≤ λp = p = 9.2 ; = 31.2 ≤ λp = p = 90.5 OK 2tf tw Fy Fy φb Mn = φb Mp = φb Zx Fy = 0.9(150)(50)/12 = 563 ft-kips Mu = 2.5Wu = 563 ft-kips; Wu = 225 kips

(c) Check shear strength (AISC G2.1): q h For rolled I-shapes, when = 31.2 ≤ 2.24 E/Fy = 53.9 , φv = 1.0 and tw Cv = 1.0. φv Vn = φv (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; Wu = 309 kips (e) Check web crippling strength (AISC J10.3): For exterior reactions, for [N/d = 10/16.52 = 0.61] > 0.2:




φRn = φ0.80t2w 1 +

4N − 0.2 d "

= (0.75)(0.40)(0.455)2 1 +

!1.5  s EFyw tf tw  tf tw

4(10) − 0.2 16.52

= 196 kips φRn = 1.0Wu = 196 kips; Wu = 196 kips

#r 0.455 1.5 (29000)(50)(0.76) 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


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 = Mu cos θ = 44.2 cos 30◦ = 38.3 ft-kips Muy = Mu sin θ = 44.2 sin 30◦ = 22.1 ft-kips Mu =

(b) Use AISC H2 with no axial load term and the conservative estimate Mn = SFy : 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 Sx 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


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 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 2(64)(19)2 2Wux x2 = = 1155 ft-kips – Lifted load moment Mux1 = 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 = Mux1 + Mux2 + Mux3 = 1531 ft-kips 2Wuy x2 2(6.4)(19)2 = = 115.5 ft-kips L 40 Tu = Wuy (d/2 + rail height) = 6.4(14/2 + 4.25) = 72.00 in-kips

Muy =

(b) Use the β modified flexure analogy to find the equivalent lateral moment. Use β ≈ 0.5. 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 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(c) Select a beam using AISC H1: Muy + My Sx 1531(12) (115.5 + 92.83)(12) Sx Mux + = Sx ≥ + φb Fy φb Fy Sy 0.9(50) 0.9(50) Sy Sx ≥ 408.2 + 55.56 Sy

For Sx 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. s r GJ (11154)(178) λ= = = 0.02578 ECw (29000)(103000) λL = (0.02578)(40)(12) = 12.38 β = 0.16 for a point load at x and simply ed 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 = 0.16 = 16.19 ft-kips Mf = β L 40 My = 2Mf = 2(16.19) = 32.38 ft-kips Mux Muy My 1531(12) 115.5(12) 32.38(12) + + + + = Sx Sy Sy 558 221 221 = 32.92 + 6.273 + 1.758 = 40.95 ksi ≤ φFy = 0.9(50) = 45 ksi OK

fun =

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.


9.2, Case 1. Determine the maximum concentrated load P that can act at midspan on a simply ed span of 20 ft. Lateral s 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 at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric , 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 = 0.5M 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 AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 6.7 ≤ λp = 0.38 = 46.9 ≤ λp = 3.76 = 90.6 ; = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 6.25 ft and Lr = 18.1 ft. [Lr = 18.1 ft] < [Lb = 20 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). s 2 Cb π 2 E Jc Lb Mn = Fcr Sx = Sx 1 + 0.078 Sx ho rts (Lb /rts )2 s 12(20) 2 (1.32)π 2(29000) (1.83)(1) = (127) 1 + 0.078 (127)(20.4) 2.15 (12(20)/2.15)2 = 415 ft-kips © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Elastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(415) = 374 ft-kips (d) Calculate the maximum service load. Wu L wu L2 + 4 8 (1.46W )(20) (0.0744)(20)2 374 = + = 7.30W + 3.72 4 8 W = 50.7 kips

φb Mn =

Maximum Service Load W = 50.7 kips



(b) Determine the Cb factor, AISC-F1. The beam has lateral at the ends only. The longest unbraced length is Lb = 24 ft. For doubly symmetric , 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 = 0.5M 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 AISC Manual Table 1-1, pp. 1-16 and 1-17. s s # " # " # " bf h E E = 5.86 ≤ λp = 0.38 = 45.9 ≤ λp = 3.76 = 90.6 ; = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 6.89 ft and Lr = 20.3 ft. [Lr = 20.3 ft] < [Lb = 24 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). s 2 Cb π 2 E Jc Lb Mn = Fcr Sx = Sx 1 + 0.078 Sx ho rts (Lb /rts )2 s 12(24) 2 (1.32)π 2(29000) (3.7)(1) = (196) 1 + 0.078 (196)(23.3) 2.37 (12(24)/2.37)2 = 579 ft-kips © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


φb Mn =




(b) Determine the Cb factor, AISC-F1. The beam has lateral at the ends only. The longest unbraced length is Lb = 30 ft. For doubly symmetric , 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 = 0.5M 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 AISC Manual Table 1-1, pp. 1-14 and 1-15. s s # " # " # " bf h E E = 7.8 ≤ λp = 0.38 = 51.9 ≤ λp = 3.76 = 90.6 ; = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 7.42 ft and Lr = 21.4 ft. [Lr = 21.4 ft] < [Lb = 30 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). s 2 Cb π 2 E Jc Lb Mn = Fcr Sx = Sx 1 + 0.078 Sx ho rts (Lb /rts )2 s (1.32)π 2(29000) (3.77)(1) 12(30) 2 = (269) 1 + 0.078 (269)(29) 2.62 (12(30)/2.62)2 = 585 ft-kips © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


φb Mn =



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 , 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 4.28(20)2 wu L2 = = 214 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral , 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. Mu (214)(12) Required Zx = = 57.1 in.3 = φb Fy (0.9)(50) Select: W18×35, Zx = 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 AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.5 ≤ λp = 3.76 = 90.6 ; = 7.06 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(66.5)/12 = 277 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The beam has continuous lateral , 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 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 Zx . 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.


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 at the ends and midspan, 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 4.28(20)2 wu L2 = = 214 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 10 ft. For doubly symmetric , 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 = 0.438M 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 ) (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, Zx = 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 AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.5 ≤ λp = 3.76 = 90.6 ; = 9.15 = 7.06 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 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). Lb − Lp ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(57.6) 10 − 4.31 = (1.30) 277 − 277 − 12 12.4 − 4.31 = 260 ft-kips 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 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 Zx . 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



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 at the ends only, 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 4.28(20)2 wu L2 = = 214 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric , 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 Lb = 20 ft. Mu 214 Required φb Mn = = = 188 ft-kips Cb 1.14 Select: W14×48, φb Mn = 193 ft-kips (d) Correct the moment for the selected beam weight. Mu = 214 +

1.2(48/1000)(20)2 1.2(beam wt)L2 = 214 + = 217 ft-kips 8 8

(e) Compute the design moment strength using the beam properties from the AISC © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 33.6 ≤ λp = 3.76 = 90.6 ; = 9.15 = 6.75 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 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). Lb − Lp ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(70.2) 20 − 6.75 = (1.14) 327 − 327 − 12 21.1 − 6.75 = 243 ft-kips 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 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 Zx . 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: ∗


! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 9.47 − 9.15 0.7(50)(93) = 446 − 446 − 12 24.1 − 9.15 = 442 ft-kips



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 at the ends and midspan, and Fy = 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 3.08(28)2 wu L2 = = 302 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 14 ft. For doubly symmetric , 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 = 0.438M 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 ) (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 Lb = 14 ft. Mu 302 Required φb Mn = = = 232 ft-kips Cb 1.30 Select: W14×48, φb Mn = 239 ft-kips (d) Correct the moment for the selected beam weight. Mu = 302 +

1.2(48/1000)(28)2 1.2(beam wt)L2 = 302 + = 307 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 33.6 ≤ λp = 3.76 = 90.6 ; = 9.15 = 6.75 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 6.75 ft and Lr = 21.1 ft. Lp = 6.75 ft < [Lb = 14 ft] ≤ [Lr = 21.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(70.2) 14 − 6.75 = (1.30) 327 − 327 − 12 21.1 − 6.75 = 327 ft-kips Yielding controls! Calculate the design moment strength. φb Mn = (0.9)(327) = 294 ft-kips The W14×48 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

W14×48 307 294 W14×53 308 327 W21×48 307 366† W21×44 307 244‡ W18×46 307 259‡ W16×45 307 283† W14×48 307 294 † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

6.75 6.11 9.47∗ 7.22 5.01 6.23 6.75

33.6 30.9 53.6 53.6 44.6 41.1 33.6

NG OK 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: ∗


! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 9.47 − 9.15 0.7(50)(93) = 446 − 446 − 12 24.1 − 9.15 = 442 ft-kips




Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(1.4) ≈ 3.08 kips/ft 3.08(28)2 wu L2 = = 302 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 14 ft. For doubly symmetric , 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 = 0.438M 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 ) (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 Lb = 14 ft. Mu 50 ksi 302 50 Required φb Mn = = = 194 ft-kips Cb Fy 1.30 60 Select: W14×43, φb Mn = 208 ft-kips (d) Correct the moment for the selected beam weight. Mu = 302 +

1.2(beam wt)L2 1.2(43/1000)(28)2 = 302 + = 307 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 37.4 ≤ λp = 3.76 = 82.7 ; = 8.35 = 7.54 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (60)(69.6)/12 = 348 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 1.89 29000 = 1.76 Lp = 1.76ry = 6.09 ft Fy 12 60 sp sp Iy Cw (45.2)(1950) = 2.18 in. rts = = Sx 62.6 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 2.18 29000 (1.05)(1) t 0.7(60) (62.6)(13.1) 2 = 1.95 1 + 1 + 6.76 12 0.7(60) (62.6)(13.1) 29000 (1.05)(1) = 17.7 ft 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 14 − 6.09 0.7(60)(62.6) = (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. Section

W14×43 W14×48 W21×48

Mu

φb Mn

ft-kips

ft-kips

307 307 307

304† 350† 404†

bf 2tf

h tw

OKAY?

7.54 6.75 9.47∗

37.4 33.6 53.6

NG OK OK


W21×44 307 244‡ W18×46 307 259‡ W16×45 307 310† W16×40 307 268† W14×43 307 304† W14×38 306 230† W12×45 307 289 † Inelastic lateral torsional buckling controls

7.22 5.01 6.23 6.93 7.54 6.57 7

53.6 44.6 41.1 46.5 37.4 39.6 29.6

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: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 9.47 − 8.35 0.7(60)(93) = 535 − 535 − 12 22.0 − 8.35 = 518 ft-kips ∗



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 , and Fy = 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 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral , 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. Mu (276)(12) Required Zx = = 73.5 in.3 = φb Fy (0.9)(50) Select: W18×40, Zx = 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



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



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 every 7 feet, and Fy = 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 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 7 ft. For doubly symmetric , 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 = 0.990M 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 Lb = 7 ft. Mu 276 Required φb Mn = = = 274 ft-kips Cb 1.00 Select: W21×44, φb Mn = 315 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 +

1.2(44/1000)(35)2 1.2(beam wt)L2 = 276 + = 284 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.6 ≤ λp = 3.76 = 90.6 ; = 9.15 = 7.22 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(95.4)/12 = 398 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, Lp = 4.45 ft and Lr = 13.0 ft. Lp = 4.45 ft < [Lb = 7 ft] ≤ [Lr = 13.0 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(81.6) 7 − 4.45 = (1.00) 398 − 398 − 12 13.0 − 4.45 = 352 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(352) = 316 ft-kips The W21×44 beam is sufficient. To it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 44 lb/ft with a large enough Zx . The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W21×44 284 316† W18×40 283 262† W16×40 283 260† W14×43 284 260† † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

7.22 5.73 6.93 7.54

53.6 50.9 46.5 37.4

OK NG NG NG




Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 17.5 ft. For doubly symmetric , 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 = 0.438M 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 ) (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 Lb = 17.5 ft. Mu 276 Required φb Mn = = = 212 ft-kips Cb 1.30 Select: W21×48, φb Mn = 221 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 +

1.2(48/1000)(35)2 1.2(beam wt)L2 = 276 + = 284 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.6 ≤ λp = 3.76 = 90.6 ; λp = 0.38 = 9.15 < = 9.47 ≤ tw Fy Fy 2tf s # " E = 24.1 λp = 1.0 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: From Table 3-2 in the AISC Manual p. 3-17, Lp = 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). s 2 2 Cb π E Lb Jc Mn = Fcr Sx = Sx 1 + 0.078 2 Sx ho rts (Lb /rts ) s (1.30)π 2(29000) (0.803)(1) 12(17.5) 2 = (93) 1 + 0.078 (93)(20.2) 2.05 (12(17.5)/2.05)2 = 319 ft-kips For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(50)(93) 9.47 − 9.15 = 446 − 446 − 12 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 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 Zx . 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


∗

Flange local buckling limit state must be checked




Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral , 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 Lb = 0 ft. Mu 50 ksi 276 50 Required φb Mn = = = 212 ft-kips Cb Fy 1.00 65 Select: W18×35, φb Mn = 249 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



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 LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 7 ft. For doubly symmetric , 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 = 0.990M 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 Lb = 7 ft. Mu 50 ksi 276 50 Required φb Mn = = = 211 ft-kips Cb Fy 1.00 65 Select: W18×35, φb 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


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.5 ≤ λp = 3.76 = 79.4 ; = 8.03 = 7.06 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (65)(66.5)/12 = 360 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 1.22 29000 = 1.76 Lp = 1.76ry = 3.78 ft Fy 12 65 sp sp Iy Cw (15.3)(1140) = 1.51 in. rts = = Sx 57.6 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 1.51 29000 (0.506)(1) t 0.7(65) (57.6)(17.3) 2 = 1.95 1 + 1 + 6.76 12 0.7(65) (57.6)(17.3) 29000 (0.506)(1) = 10.7 ft 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 7 − 3.78 0.7(65)(57.6) = (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. Section

W18×35 W18×40 W16×40

Mu

φb Mn

ft-kips

ft-kips

282 283 283

266† 321† 324†

bf 2tf

h tw

OKAY?

7.06 5.73 6.93

53.5 50.9 46.5

NG OK OK


W16×36 282 281† 8.12∗ 48.1 NG † W16×31 281 212 6.28 51.6 NG W14×38 283 273† 6.57 39.6 NG W12×40 283 270† 7.77 33.6 NG † W12×35 282 224 6.31 36.2 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: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(65)(56.5) 8.12 − 8.03 = 347 − 347 − 12 21.1 − 8.03 = 346 ft-kips Use W16×40 with Fy = 65 ksi steel.



Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 17.5 ft. For doubly symmetric , 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 = 0.438M 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 ) (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 Lb = 17.5 ft. Mu 50 ksi 276 50 Required φb Mn = = = 163 ft-kips Cb Fy 1.30 65 Select: W14×43, φb Mn = 182 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 +

1.2(beam wt)L2 1.2(43/1000)(35)2 = 276 + = 284 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 37.4 ≤ λp = 3.76 = 79.4 ; = 8.03 = 7.54 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (65)(69.6)/12 = 377 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 1.89 29000 = 1.76 Lp = 1.76ry = 5.86 ft Fy 12 65 sp sp Iy Cw (45.2)(1950) = 2.18 in. rts = = Sx 62.6 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 2.18 29000 (1.05)(1) t 0.7(65) (62.6)(13.1) 2 = 1.95 1 + 1 + 6.76 12 0.7(65) (62.6)(13.1) 29000 (1.05)(1)

= 16.8 ft [Lr = 16.8 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). s 2 2 Jc Lb Cb π E Mn = Fcr Sx = Sx 1 + 0.078 2 Sx ho rts (Lb /rts ) s (1.30)π 2(29000) (1.05)(1) 12(17.5) 2 = (62.6) 1 + 0.078 (62.6)(13.1) 2.18 (12(17.5)/2.18)2 = 290 ft-kips

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. Section

W14×43 W14×48 W21×48

Mu

φb Mn

ft-kips

ft-kips

284 284 284

261‡ 312† 287‡

bf 2tf

h tw

OKAY?

7.54 6.75 9.47∗

37.4 33.6 53.6

NG OK OK


W21×44 284 168‡ W18×46 284 185‡ W16×45 284 227‡ W16×31 281 86.8‡ W14×48 284 312† W14×43 284 261‡ W14×38 283 166‡ W12×45 284 268† W12×35 282 145‡ W10×45 284 252† † Inelastic lateral torsional buckling controls

7.22 5.01 6.23 6.28 6.75 7.54 6.57 7 6.31 6.47

53.6 44.6 41.1 51.6 33.6 37.4 39.6 29.6 36.2 22.5

NG NG NG NG OK NG 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: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 9.47 − 8.03 0.7(65)(93) = 580 − 580 − 12 21.1 − 8.03 = 555 ft-kips ∗




Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. Since the beam has continous lateral , 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. Mu (276)(12) Required Zx = = 36.8 in.3 = φb Fy (0.9)(100) Select: W12×26, Zx = 37.2 in.3 (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

(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-24 and 1-25. s s # " # " # " bf h E E = 47.2 ≤ λp = 3.76 = 64.0 ; λp = 0.38 = 6.47 < = 8.54 ≤ tw Fy Fy 2tf s " # E λp = 1.0 = 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: © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The beam has continuous lateral , 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 = Mp − 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

bf 2tf

W12×26 280 259‡ 8.54∗ W14×26 280 302 5.98 ‡ W16×26 280 313 7.97∗ W14×26 280 302 5.98 W14×22 280 240‡ 7.46∗ ‡ W12×26 280 259 8.54∗ ‡ Elastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked

h tw

OKAY?

47.2 48.1 56.8 48.1 53.3 47.2

NG OK OK OK NG NG





1.2(beam wt)L2 1.2(26/1000)(35)2 = 276 + = 280 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-20 and 1-21. s s # " # " # " bf h E E = 56.8 ≤ λp = 3.76 = 64.0 ; λp = 0.38 = 6.47 < = 7.97 ≤ tw Fy Fy 2tf s # " E = 17.0 λp = 1.0 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: s r 1.12 E 29000 Lp = 1.76ry = 1.76 = 2.80 ft Fy 12 100 sp sp Iy Cw (9.59)(565) rts = = = 1.38 in. Sx 38.4 v s u r u 0.7Fy Sx ho 2 E Jc t 1 + 1 + 6.76 Lr = 1.95rts 0.7Fy Sx ho E Jc v s s u u 0.7(100) (38.4)(15.3) 2 1.38 29000 (0.262)(1) t 1 + 1 + 6.76 = 1.95 12 0.7(100) (38.4)(15.3) 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 Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 0.7(100)(38.4) 7 − 2.80 = (1.00) 368 − 368 − 12 7.65 − 2.80 = 245 ft-kips

For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(100)(38.4) 7.97 − 6.47 = 368 − 368 − 12 17.0 − 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. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Section

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 LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft 1.8(35)2 wu L2 = = 276 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral 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 Lb = 17.5 ft. For doubly symmetric , 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 = 0.438M 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 ) (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 Lb = 17.5 ft. Mu 50 ksi 276 50 Required φb Mn = = = 106 ft-kips Cb Fy 1.30 100 Select: W10×33, φb Mn = 107 ft-kips (d) Correct the moment for the selected beam weight. Mu = 276 +

1.2(beam wt)L2 1.2(33/1000)(35)2 = 276 + = 282 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-24 and 1-25. s s # " # " # " bf h E E = 27.1 ≤ λp = 3.76 = 64.0 ; λp = 0.38 = 6.47 < = 9.15 ≤ tw Fy Fy 2tf s # " E = 17.0 λp = 1.0 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: s r 1.94 E 29000 Lp = 1.76ry = 1.76 = 4.85 ft Fy 12 100 sp sp Iy Cw (36.6)(791) rts = = = 2.20 in. Sx 35 v s u r u 0.7Fy Sx ho 2 E Jc t 1 + 1 + 6.76 Lr = 1.95rts 0.7Fy Sx ho E Jc v s s u u 0.7(100) (35)(9.3) 2 2.20 29000 (0.583)(1) t 1 + 1 + 6.76 = 1.95 12 0.7(100) (35)(9.3) 29000 (0.583)(1)

= 13.5 ft [Lr = 13.5 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). s 2 2 Jc Lb Cb π E Mn = Fcr Sx = Sx 1 + 0.078 2 Sx ho rts (Lb /rts ) s (1.30)π 2(29000) (0.583)(1) 12(17.5) 2 = (35) 1 + 0.078 (35)(9.3) 2.2 (12(17.5)/2.2)2 = 179 ft-kips

For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(100)(35) 9.15 − 6.47 = 323 − 323 − 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. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Section

Mu

φb Mn

ft-kips

ft-kips

bf 2tf

W10×33 282 161‡ 9.15∗ W10×39 283 219‡ 7.53∗ ‡ W10×45 284 285 6.47 W21×44 284 168‡ 7.22∗ W18×40 283 147‡ 5.73 W16×45 284 227‡ 6.23 W16×31 281 86.8‡ 6.28 ‡ W14×43 284 261 7.54∗ ‡ W14×38 283 166 6.57∗ ‡ W14×26 280 60.6 5.98 ‡ W12×45 284 276 7∗ ‡ W12×35 282 145 6.31 W10×45 284 285‡ 6.47 ‡ W10×39 283 219 7.53∗ W10×30 281 99.2‡ 5.7 W8×40 283 229‡ 7.21∗ ‡ Elastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked

h tw

OKAY?

27.1 25 22.5 53.6 50.9 41.1 51.6 37.4 39.6 48.1 29.6 36.2 22.5 25 29.5 17.6

NG NG OK NG NG NG NG NG NG NG NG NG OK NG NG NG



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 15: dead load is 0 kips/ft, live load is 1 kips/ft, span is 35 ft, the beam has lateral every 5 feet, and Fy = 50 ksi.

Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0 + beam wt) + 1.6(1) ≈ 1.6 kips/ft 1.6(35)2 wu L2 = = 245 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral every 5 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 15 ft to 20 ft with Lb = 5 ft. For doubly symmetric , 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 = 0.995M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.995M 12.5M Cb = (1.0) = 1.00 2.5M + 3(0.995M ) + 4(1.0M ) + 3(0.995M ) (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 Lb = 5 ft. Mu 245 Required φb Mn = = = 244 ft-kips Cb 1.00 Select: W16×40, φb Mn = 274 ft-kips (d) Correct the moment for the selected beam weight. Mu = 245 +

1.2(40/1000)(35)2 1.2(beam wt)L2 = 245 + = 252 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-20 and 1-21. s s # " # " # " bf h E E = 46.5 ≤ λp = 3.76 = 90.6 ; = 9.15 = 6.93 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(73)/12 = 304 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 W Shapes p. 3-17, Lp = 5.55 ft and Lr = 15.9 ft. [Lb = 5 ft] ≤ Lp = 5.55 ft lateral torsional buckling does not apply, AISCF2.1(a). Yielding controls! Calculate the design moment strength. φb Mn = (0.9)(304) = 274 ft-kips The W16×40 beam is sufficient. To 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 Zx . The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W16×40 252 274 W18×40 252 288† W18×35 251 242† W16×40 252 274 W16×36 252 240 W14×38 252 231 † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

6.93 5.73 7.06 6.93 8.12 6.57

46.5 50.9 53.5 46.5 48.1 39.6

OK OK NG OK NG NG



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 at the ends only, and Fy = 50 ksi.

Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0 + beam wt) + 1.6(1) ≈ 1.6 kips/ft 1.6(35)2 wu L2 = = 245 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral at the ends only. The longest unbraced length is Lb = 35 ft. For doubly symmetric , 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 Lb = 35 ft. Mu 245 Required φb Mn = = = 216 ft-kips Cb 1.14 Select: W12×65, φb Mn = 231 ft-kips (d) Correct the moment for the selected beam weight. Mu = 245 +

1.2(65/1000)(35)2 1.2(beam wt)L2 = 245 + = 257 ft-kips 8 8

(e) Compute the design moment strength using the beam properties from the AISC © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 24.9 ≤ λp = 3.76 = 90.6 ; λp = 0.38 = 9.15 < = 9.92 ≤ tw Fy Fy 2tf s " # E = 24.1 λp = 1.0 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: From Table 3-2 in the AISC Manual p. 3-17, Lp = 10.7 ft and Lr = 35.1 ft. Lp = 10.7 ft < [Lb = 35 ft] ≤ [Lr = 35.1 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)(87.9) 35 − 10.7 = (1.14) 403 − 403 − 12 35.1 − 10.7 = 292 ft-kips For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(50)(87.9) 9.92 − 9.15 = 403 − 403 − 12 24.1 − 9.15 = 396 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(292) = 263 ft-kips The W12×65 beam is sufficient. To it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 65 lb/ft with a large enough Zx . The following table shows the moment corrected for the beam weight. Section

W12×65 W24×62 W21×62 W21×57 W18×65 W18×46 W16×57 W14×61

Mu

φb Mn

ft-kips

ft-kips

257 256 256 255 257 253 255 256

263† 104‡ 143‡ 103‡ 155‡ 65.9‡ 122‡ 202‡

bf 2tf

h tw

OKAY?

9.92∗ 5.97 6.7 5.04 5.06 5.01 4.98 7.75

24.9 50.1 46.9 46.3 35.7 44.6 33 30.4

OK NG NG NG NG NG NG NG


W14×53 255 132‡ W14×38 252 58.5‡ W12×65 257 263† W12×58 256 192‡ W12×50 254 121‡ W10×60 256 206† W8×58 256 174† † Inelastic lateral torsional buckling controls ‡ ∗

6.11 6.57 9.92∗ 7.82 6.31 7.41 5.07

30.9 39.6 24.9 27 26.8 18.7 12.4

NG NG OK NG NG NG NG

Elastic lateral torsional buckling controls Flange local buckling limit state must be checked




Use LRFD Design Method (a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(2.8) ≈ 5.32 kips/ft 5.32(48)2 wu L2 = = 1530 ft-kips (without beam) Mu = 8 8 (b) Determine the Cb factor, AISC-F1. The beam has lateral every 16 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 16 ft to 32 ft with Lb = 16 ft. For doubly symmetric , 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 = 0.972M MB = moment at 1/2 pt of the unbraced segment = 1.0M MC = moment at 3/4 pt of the unbraced segment = 0.972M 12.5M Cb = (1.0) = 1.01 2.5M + 3(0.972M ) + 4(1.0M ) + 3(0.972M ) (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 Lb = 16 ft. Mu 1530 Required φb Mn = = = 1510 ft-kips Cb 1.01 Select: W36×135, φb Mn = 1550 ft-kips (d) Correct the moment for the selected beam weight. Mu = 1530 +

1.2(135/1000)(48)2 1.2(beam wt)L2 = 1530 + = 1580 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-12 and 1-13. s s # " # " # " bf h E E = 54.1 ≤ λp = 3.76 = 90.6 ; = 9.15 = 7.56 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 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 ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(439) 16 − 8.41 = (1.01) 2120 − 2120 − 12 24.2 − 8.41 = 1740 ft-kips 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 W36×150 W40×149 W36×135 W33×141 W33×130 W30×132 W27×129 W24×131 † Inelastic lateral

1580 1570† 1580 1830† 1580 1810† 1580 1570† 1580 1610† 1580 1440† 1580 1330† 1580 1210† 1580 1270† 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





1.2(beam wt)L2 1.2(130/1000)(48)2 = 1530 + = 1580 ft-kips 8 8


(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-12 and 1-13. s s # " # " # " bf h E E = 51.7 ≤ λp = 3.76 = 82.7 ; = 8.35 = 6.73 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (60)(467)/12 = 2340 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 2.39 29000 = 1.76 Lp = 1.76ry = 7.71 ft Fy 12 60 sp sp Iy Cw (218)(56600) = 2.94 in. rts = = Sx 406 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 2.94 29000 (7.37)(1) t 0.7(60) (406)(32.2) 2 = 1.95 1 + 1 + 6.76 12 0.7(60) (406)(32.2) 29000 (7.37)(1) = 21.8 ft Lp = 7.71 ft < [Lb = 16 ft] ≤ [Lr = 21.8 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 16 − 7.71 0.7(60)(406) = (1.01) 2340 − 2340 − 12 21.8 − 7.71 = 1820 ft-kips

Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.9)(1820) = 1640 ft-kips The W33×130 beam is sufficient. To it is the lightest beam use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 130 lb/ft with a large enough Zx . The following table shows the moment corrected for the beam weight. Section

W33×130 W30×124

Mu

φb Mn

ft-kips

ft-kips

1580 1580

1640† 1390†

bf 2tf

h tw

OKAY?

6.73 5.65

51.7 46.2

OK NG


W27×129 W24×117 W21×122 † Inelastic lateral

1580 1370† 1570 1280† 1570 1220† torsional buckling controls

4.55 7.53 6.45

39.7 39.2 31.3

NG NG NG



9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (a) A992 steel; continuous lateral

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 , 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 ≤ λp for a compact section. Mu (1, 172)(12) = 313 in.3 Required Zx = = 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 moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-14 and 1-15. s s # " # " # " b h E E f = 49.6 ≤ λp = 3.76 = 90.6 ; = 6.89 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mn = Mp = Fy Zx = (50)(346)/12 = 1, 440 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: The beam has continuous lateral , 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 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 Zx . The following table shows the moment corrected for the beam weight. Section

W30×108 W27×102 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



9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (b) A992 steel; lateral 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 at the ends. The unbraced length is Lb = 30 ft. For doubly symmetric , Rm = 1.0. Use statics to find the moments in the beam: 12.5Mmax 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 Lb = 30 ft. Mu 1, 172 Required φb Mn = = = 1, 030 ft-kips Cb 1.14 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 AISC Manual Table 1-1, pp. 1-16 and 1-17. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

s s # " # " # " bf h E E = 33.2 ≤ λp = 3.76 = 90.6 ; = 9.15 = 5.92 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2.

For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(418)/12 = 1, 740 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-14, Lp = 10.6 ft and Lr = 33.7 ft. Lp = 10.6 ft < [Lb = 30 ft] ≤ [Lr = 33.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 30 − 10.6 0.7(50)(371) = (1.14) 1, 740 − 1, 740 − 12 33.7 − 10.6 = 1, 350 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(1, 350) = 1, 210 ft-kips The W24×146 beam is sufficient. To it is the lightest beam, use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 146 lb/ft with a large enough Zx . The following table shows the moment corrected for the beam weight. Section

W24×146 W36×135 W33×141 W30×132 W27×146 W27×129 W24×146 W24×131 W21×132 W18×143

Mu

φb Mn

ft-kips

ft-kips

1,190 1,190 1,190 1,190 1,190 1,190 1,190 1,190 1,190 1,190

1,210† 910‡ 994‡ 796‡ 1,350† 760‡ 1,210† 1,030† 973† 1,010†

†

Inelastic lateral torsional buckling controls

‡


bf 2tf

h tw

OKAY?

5.92 7.56 6.01 5.27 7.16 4.55 5.92 6.7 6.01 4.25

33.2 54.1 49.6 43.9 39.4 39.7 33.2 35.6 28.9 22

OK NG NG NG OK NG OK NG NG NG



9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (c) A992 steel; lateral 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 at the ends and at 10 ft. The worst loading and longest unbraced length occur on the segment from 10 ft to 30 ft with Lb = 20 ft. For doubly symmetric , Rm = 1.0. Use statics to find the moments in the beam: 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 Lb = 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 AISC Manual Table 1-1, pp. 1-12 and 1-13. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

s s # " # " # " bf h E E = 54.5 ≤ λp = 3.76 = 90.6 ; = 9.15 = 7.76 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2.

For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 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 20 − 8.19 0.7(50)(359) = (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 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 Zx . 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 2tf

h tw

OKAY?

7.76 6.17 5.41 7.53

54.5 47.8 42.5 39.2

OK NG NG NG




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 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 at the ends and at 10 ft. The worst loading and longest unbraced length occur on the segment from 10 ft to 30 ft with Lb = 20 ft. For doubly symmetric , Rm = 1.0. Use statics to find the moments in the beam: 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 Lb = 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 AISC Manual Table 1-1, pp. 1-16 and 1-17. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

s s # " # " # " bf h E E = 43.1 ≤ λp = 3.76 = 82.7 ; λp = 0.38 = 8.35 < = 8.5 ≤ tw Fy Fy 2tf s " # E λp = 1.0 = 22.0 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: s r E 2.91 29, 000 = 1.76 Lp = 1.76ry = 9.38 ft Fy 12 60 sp sp Iy Cw (259)(35200) = rts = = 3.42 in. Sx 258 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 3.42 29, 000 (4.72)(1) t 0.7(60) (258)(23.3) 2 = 1.95 1 + 1 + 6.76 12 0.7(60) (258)(23.3) 29, 000 (4.72)(1) = 26.1 ft Lp = 9.38 ft < [Lb = 20 ft] ≤ [Lr = 26.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 20 − 9.38 0.7(60)(258) = (1.15) 1, 440 − 1, 440 − 12 26.1 − 9.38 = 1, 270 ft-kips

For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(60)(258) 8.5 − 8.35 = 1, 440 − 1, 440 − 12 22.0 − 8.35 = 1, 440 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(1, 270) = 1, 140 ft-kips The W24×104 beam is not sufficient. Try heavier sections at the same depth. The following table shows the moment corrected for the beam weight.


Section

W24×104 W24×117 W30×116 W30×108 W27×114 W24×104 W24×103 W21×111

Mu

φb Mn

ft-kips

ft-kips

1,190 1,190 1,190 1,190 1,190 1,190 1,190 1,190

1,140† 1,320† 1,210† 1,060‡ 1,120† 1,140† 849‡ 1,130†

bf 2tf

h tw

OKAY?

8.5∗ 7.53 6.17 6.89 5.41 8.5∗ 4.59 7.05

43.1 39.2 47.8 49.6 42.5 43.1 39.2 34.1

NG OK OK NG NG NG NG NG

†


‡


∗



9.6. Select the lightest W sections for the conditions shown in the accompanying figure. Assume there is no deflection limitation. Use (a) A992 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 at the ends, at 15 ft, and at 27 ft. For doubly symmetric , Rm = 1.0. 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 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 Lb = 12 ft. Mu 1, 124 Required φb Mn = = = 1, 120 ft-kips Cb 1.00 Select: W30×108, φb Mn = 1, 140 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 1, 124 + (108/1000)(42)2 = 1, 150 ft-kips 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-14 and 1-15. s s # " # " # " bf h E E = 49.6 ≤ λp = 3.76 = 90.6 ; = 9.15 = 6.89 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy 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, Lp = 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 ≤ Mp Mn = Cb Mp − Mp − 0.7Fy Sx Lr − Lp 0.7(50)(299) 12 − 7.59 = (1.00) 1, 440 − 1, 440 − 12 22.0 − 7.59 = 1, 270 ft-kips 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


Check segment A Section

W30×116

Mu

φb Mn

ft-kips

ft-kips

1,110

1,420




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 at the ends, at 15 ft, and at 27 ft. For doubly symmetric , Rm = 1.0. 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 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 Lb = 12 ft. Mu 50 ksi 1, 124 50 Required φb Mn = = = 933 ft-kips Cb Fy 1.00 60 Select: W30×99, φb 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 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-14 and 1-15. s s # " # " # " bf h E E = 51.9 ≤ λp = 3.76 = 82.7 ; = 8.35 = 7.8 ≤ λp = 0.38 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (60)(312)/12 = 1, 560 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 2.1 29, 000 = 1.76 Lp = 1.76ry = 6.77 ft Fy 12 60 sp sp Iy Cw (128)(26800) = 2.62 in. rts = = Sx 269 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u u 2.62 29, 000 (3.77)(1) t 0.7(60) (269)(29) 2 = 1.95 1 + 1 + 6.76 12 0.7(60) (269)(29) 29, 000 (3.77)(1)

= 19.3 ft 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 12 − 6.77 0.7(60)(269) = (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 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 Zx . The following table shows the moment corrected for the beam weight.


Section

Mu

φb Mn

ft-kips

ft-kips

1,150 1,150 1,150

1,180† 1,060† 948†

W30×99 W27×94 W24×94 †

bf 2tf

h tw

OKAY?

7.8 6.7 5.18

51.9 49.5 41.9

OK NG NG

bf 2tf

h tw

OKAY?

7.8

51.9

OK


Check segment A Section

W30×99

Mu

φb Mn

ft-kips

ft-kips

1,110

1,400



9.7. A floor beam, laterally ed at the ends only and ing vibration inducing heavy machinery, is subject to the loads shown in the accompanying figure. Select the lightest W section of A992 steel. Compare the result when there is no deflection limit with that when L/d is limited to a minimum of 20 under full load, a traditional limit to minimize perceptible vibration due to pedestrian traffic.

Use LRFD Design Method (a) Obtain the factored loads (neglecting the beam weight): Wu = 1.2(0) + 1.6(14)(1.5) = 33.6 kips (includes impact) Mu = 7(33.6) = 235.2 ft-kips (b) Determine the Cb factor, AISC-F1. The beam has lateral at the ends. For doubly symmetric , Rm = 1.0. 12.5Mmax Rm ≤ 3.0 Cb = 2.5Mmax + 3MA + 4MB + 3MC 12.5(235.2) = (1.0) = 1.05 2.5(235.2) + 3(210.0) + 4(235.2) + 3(210.0) (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 Lb = 25 ft. Mu 235.2 Required φb Mn = = = 223 ft-kips Cb 1.05 Select: W12×58, φb Mn = 233 ft-kips (d) Correct the moment for the beam weight. 1 Mu = 235.2 + (58/1000)(25)2 = 241 ft-kips 8 (e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-24 and 1-25. s s # " # " # " bf h E E = 27 ≤ λp = 3.76 = 90.6 ; = 7.82 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(86.4)/12 = 360 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-17, Lp = 8.87 ft and Lr = 29.9 ft. Lp = 8.87 ft < [Lb = 25 ft] ≤ [Lr = 29.9 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)(78) 25 − 8.87 = (1.05) 360 − 360 − 12 29.9 − 8.87 = 272 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(272) = 245 ft-kips The W12×58 beam is sufficient. To it is the lightest beam, use Table 1-1 of the AISC Manual, and examine the first beam in each group lighter than 58 lb/ft with a large enough Zx . The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W12×58 241 245† W24×55 240 121‡ W21×55 240 169‡ W21×57 241 148‡ W18×55 240 162‡ W18×46 240 92.4‡ W16×57 241 169‡ W14×53 240 185‡ W14×38 239 82.3‡ W12×50 240 167‡ W10×54 240 200† W8×58 241 189† † Inelastic lateral torsional buckling controls ‡

bf 2tf

h tw

OKAY?

7.82 6.94 7.87 5.04 5.98 5.01 4.98 6.11 6.57 6.31 8.15 5.07

27 54.6 50 46.3 41.1 44.6 33 30.9 39.6 26.8 21.2 12.4

OK NG NG NG NG NG NG NG NG NG NG NG


With no deflection, use W12×58 with Fy = 50 ksi steel. Considering the L/d deflection limit the minimum depth is d = following table shows the moment corrected for the beam weight.

25(12) = 15 in. The 20


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


With deflection use W16×67 with Fy = 50 ksi steel.


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 ed beam. Lateral occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume Cb = 1.0. Case 1: MD = 49 ft-kips, ML = 98 ft-kips, L = 28 ft, Fy = 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 5ML L2 (5)(98 × 12)(28 × 12)2 = = 511 in.4 Min Ix = 48E∆lim (48)(29, 000)(0.933) (c) The problem statement says to use Cb = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual, pp. 3-96 to 3-131 with Lb = 7 ft. Mu 215.6 Required φb Mn = = = 216 ft-kips Cb 1.00 Select: W18×35, φb Mn = 217 ft-kips

Unbraced Length, AISC

(e) Correct the moment for the beam weight. 1 Mu = 215.6 + 1.2 (35/1000)(28)2 = 220 ft-kips 8 (f) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.5 ≤ λp = 3.76 = 90.6 ; = 7.06 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(66.5)/12 = 277 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-18, Lp = 4.31 ft and Lr = 12.4 ft. Lp = 4.31 ft < [Lb = 7 ft] ≤ [Lr = 12.4 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)(57.6) 7 − 4.31 = (1.00) 277 − 277 − 12 12.4 − 4.31 = 241 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(241) = 217 ft-kips The W18×35 beam does not have sufficient strength. The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W18×35 220 217† W18×40 220 261† W16×36 220 225† W14×38 220 218† † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

7.06 5.73 8.12 6.57

53.5 50.9 48.1 39.6

NG OK OK NG

(g) Check the beams for deflection. Section

Ix in.4

LL Defl. in.

LL Defl. Limit in.

OKAY?

W18×35 W18×40 W16×36 W14×38

510 612 448 385

0.935 0.779 1.06 1.24

0.933 0.933 0.933 0.933

NG OK NG NG




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 5ML L2 (5)(98 × 12)(28 × 12)2 = = 511 in.4 Min Ix = 48E∆lim (48)(29, 000)(0.933) (c) The problem statement says to use Cb = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with Lb = 7 ft. Mu 50 ksi 215.6 50 Required φb Mn = = = 180 ft-kips Cb Fy 1.00 60 Select: W14×34, φb Mn = 193 ft-kips (e) Correct the moment for the beam weight. 1 Mu = 215.6 + 1.2 (34/1000)(28)2 = 220 ft-kips 8 (f) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " bf h E E = 43.1 ≤ λp = 3.76 = 82.7 ; = 7.41 ≤ λp = 0.38 = 8.35 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (60)(54.6)/12 = 273 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: s r E 1.53 29, 000 = 1.76 Lp = 1.76ry = 4.93 ft Fy 12 60 sp sp Iy Cw (23.3)(1070) = rts = = 1.80 in. Sx 48.6 v s u r u 0.7Fy Sx ho 2 E Jc t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u 1.80 29, 000 (0.569)(1) u 0.7(60) (48.6)(13.5) 2 t = 1.95 1 + 1 + 6.76 12 0.7(60) (48.6)(13.5) 29, 000 (0.569)(1) = 13.9 ft Lp = 4.93 ft < [Lb = 7 ft] ≤ [Lr = 13.9 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − Lp Mn = Cb Mp − Mp − 0.7Fy Sx ≤ Mp Lr − Lp 7 − 4.93 0.7(60)(48.6) = (1.00) 273 − 273 − 12 13.9 − 4.93 = 249 ft-kips

Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(249) = 224 ft-kips The W14×34 beam has sufficient strength. The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W14×34 220 224† W14×38 220 255† W16×40 220 302† W16×36 220 262† W18×35 220 249† † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

7.41 6.57 6.93 8.12 7.06

43.1 39.6 46.5 48.1 53.5

OK OK OK OK OK

(g) Check the beams for deflection.


Section

Ix in.4

LL Defl. in.

LL Defl. Limit in.

OKAY?

W14×34 W14×38 W16×40 W16×36 W18×35

340 385 518 448 510

1.40 1.24 0.921 1.06 0.935

0.933 0.933 0.933 0.933 0.933

NG NG OK NG NG




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 5ML L2 (5)(240 × 12)(48 × 12)2 = = 1, 790 in.4 Min Ix = 48E∆lim (48)(29, 000)(1.92) (c) The problem statement says to use Cb = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual, pp. 3-96 to 3-131 with Lb = 12 ft. Mu 384.0 Required φb Mn = = = 384 ft-kips Cb 1.00 Select: W21×62, φb Mn = 440 ft-kips


(e) Correct the moment for the beam weight. 1 Mu = 384.0 + 1.2 (62/1000)(48)2 = 405 ft-kips 8 (f) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 46.9 ≤ λp = 3.76 = 90.6 ; = 6.7 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(144)/12 = 600 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-16, Lp = 6.25 ft and Lr = 18.1 ft. Lp = 6.25 ft < [Lb = 12 ft] ≤ [Lr = 18.1 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)(127) 12 − 6.25 = (1.00) 600 − 600 − 12 18.1 − 6.25 = 489 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(489) = 440 ft-kips The W21×62 beam has sufficient strength. The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W21×62 405 440† W21×68 408 494† W21×73 409 536† W21×83 413 620† W24×76 410 633† W24×68 408 549† W24×62 405 402† W18×65 406 409† † Inelastic lateral torsional buckling controls

bf 2tf

h tw

OKAY?

6.7 6.04 5.6 5 6.61 7.66 5.97 5.06

46.9 43.6 41.2 36.4 49 52 50.1 35.7

OK OK OK OK OK OK NG OK


Ix in.4

LL Defl. in.

LL Defl. Limit in.

OKAY?

W21×62 W21×68 W21×73 W21×83 W24×76 W24×68 W24×62 W18×65

1330 1480 1600 1830 2100 1830 1550 1070

2.58 2.32 2.15 1.88 1.63 1.88 2.21 3.21

1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92

NG NG NG OK OK OK NG NG





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 5ML L2 (5)(240 × 12)(48 × 12)2 = = 1, 790 in.4 Min Ix = 48E∆lim (48)(29, 000)(1.92) (c) The problem statement says to use Cb = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with Lb = 12 ft. Mu 50 ksi 384.0 50 Required φb Mn = = = 295 ft-kips Cb Fy 1.00 65 Select: W21×48, φb Mn = 313 ft-kips (e) Correct the moment for the beam weight. 1 Mu = 384.0 + 1.2 (48/1000)(48)2 = 401 ft-kips 8 (f) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.6 ≤ λp = 3.76 = 79.4 ; λp = 0.38 = 8.03 < = 9.47 ≤ tw Fy Fy 2tf © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

"

λp = 1.0

s

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: s r E 1.66 29, 000 = 1.76 Lp = 1.76ry = 5.14 ft Fy 12 65 sp sp Iy Cw (38.7)(3950) = rts = = 2.05 in. Sx 93 v s u r 0.7Fy Sx ho 2 E Jc u t Lr = 1.95rts 1 + 1 + 6.76 0.7Fy Sx ho E Jc v s s u 0.7(65) (93)(20.2) 2 2.05 29, 000 (0.803)(1) u t = 1.95 1 + 1 + 6.76 12 0.7(65) (93)(20.2) 29, 000 (0.803)(1)

= 14.3 ft 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: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 9.47 − 8.03 0.7(65)(93) = 580 − 580 − 12 21.1 − 8.03 = 555 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(409) = 369 ft-kips The W21×48 beam does not have sufficient strength. The following table shows the moment corrected for the beam weight.


Section

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


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



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 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 . 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 Cb = 1.0. (c) Select a beam using Table 3-10 Available Moment vs. Manual, pp. 3-96 to 3-131 with Lb = 30 ft. 300.0 Mu = = 300 ft-kips Required φb Mn = Cb 1.00 Select: W14×74, φb 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 (e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-22 and 1-23. s s # " # " # " b h E E f = 25.4 ≤ λp = 3.76 = 90.6 ; = 6.41 ≤ λp = 0.38 = 9.15 tw Fy 2tf Fy The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(126)/12 = 525 ft-kips © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-17, Lp = 8.76 ft and Lr = 31.0 ft. Lp = 8.76 ft < [Lb = 30 ft] ≤ [Lr = 31.0 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)(112) 30 − 8.76 = (1.00) 525 − 525 − 12 31.0 − 8.76 = 336 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(336) = 302 ft-kips The W14×74 beam does not have sufficient strength. The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

W14×74 304 302† W14×82 305 348† W12×79 305 337† W12×72 304 297† W16×77 305 317‡ W16×67 304 251‡ W18×76 305 329‡ W18×71 304 192‡ † Inelastic lateral torsional buckling controls ‡

bf 2tf

h tw

OKAY?

6.41 5.92 8.22 8.99 6.77 7.7 8.11 4.71

25.4 22.4 20.7 22.6 31.2 35.9 37.8 32.4

NG OK OK NG OK NG OK NG




9.13, Part b. 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 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): Wu = 1.2(5) = 6.00 kips (dead load only) Wu = 1.2(5) + 1.6(15) = 60.0 kips There are two loading cases: For Loading Case 1, live load is on the span but not on the cantilevers. Maximum moment at midspan. 1 (60.0)(30) = 390 ft-kips Mu = (beam wt.)(302 − 102 ) − (6.00)(10) + 8 4 For Loading Case 2, live load is on the cantilevers but not on the 30 ft span. Maximum moment at the s. 1 Mu = (beam wt.)(10)2 + (30.0)(10) = 300 ft-kips 2 (b) Determine the Cb factor, AISC-F1. For the 30 ft span in Case 1: 12.5Mmax Cb = 2.5Mmax + 3MA + 4MB + 3MC 12.5(390) = = 1.86 2.5(390) + 3(52.5) + 4(165) + 3(278) For the 30 ft span in Case 2: 12.5(300) Cb = = 1.14 2.5(300) + 3(278) + 4(255) + 3(232) Assume Case 2 controls. (c) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with Lb = 15 ft. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Mu 300.0 = = 264 ft-kips Cb 1.14 Select: W21×48, φb Mn = 268 ft-kips Required φb Mn =

(d) Correct the moment for the beam weight. 1 (48/1000)(10)2 = 303 ft-kips Mu = 300.0 + 1.2 2 (e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19. s s # " # " # " bf h E E = 53.6 ≤ λp = 3.76 = 90.6 ; λp = 0.38 = 9.15 < = 9.47 ≤ tw Fy Fy 2tf s " # E λp = 1.0 = 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 the AISC Manual Table 3-2 p. 3-17, Lp = 6.09 ft and Lr = 16.6 ft. Lp = 6.09 ft < [Lb = 15 ft] ≤ [Lr = 16.6 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)(93) 15 − 6.09 = (1.14) 446 − 446 − 12 16.6 − 6.09 = 338 ft-kips For the limit state of compression flange local buckling, AISC-F3.2: ! λ − λpf Mn = Mp − Mp − 0.7Fy Sx λrf − λpf 0.7(50)(93) 9.47 − 9.15 = 446 − 446 − 12 24.1 − 9.15 = 442 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb Mn = (0.90)(338) = 305 ft-kips The W21×48 beam has sufficient strength. The following table shows the moment corrected for the beam weight.


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


(f) Check Case 1. The following table shows the moment corrected for the beam weight. Section

Mu

φb Mn

ft-kips

ft-kips

bf 2tf

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

h tw

OKAY?

53.6 53.6 44.6 41.1 37.4

OK NG NG NG NG



12.3. Determine the maximum service load W (kips) at the mid-height of the beamcolumn shown in the accompanying figure. Assume the member is hinged with respect to bending in both x and y directions at the top and bottom. Additionally, lateral occurs in the weak direction at mid-height. Use LRFD Design Method (a) Obtain factored loads. Pu = 1.2(100) + 1.6(150) = 360 kips 1 Mnt = Wu (22) = 5.5Wu 4 (b) Check the section type. h i h i q h = 12.3 ≤ λ = 3.76 E = 90.6 p Fy htw i i h q bf E ; 2tf = 4.96 ≤ λp = 0.38 F = 9.15 y

The web is compact and the flange is compact. (c) Column action with a compact web and flange, use AISC-E3 with Kx = Ky = 1.0; Lx = 22 ft; Ly = 11 ft. Find the maximum KL r K y Ly K x Lx 1.0(22)(12) 1.0(11)(12) = 47.3 ; = 41.8 = = rx 5.58 ry 3.16 h i h i q KL E For r = 47.3 ≤ 4.71 F = 113 y

Fe =

π2E

π 2 (29, 000)

= = 128 ksi (KL/r)2 (47.3)2 h i h i Fcr = 0.658Fy /Fe Fy = 0.65850/128 (50) = 42.5 ksi

φc Pn = φc Fcr Ag = 0.90(42.5)(39.9) = 1, 520 kips

(d) Beam action with a compact web and flange, use AISC-F2 with Lb = 11 ft. Use AISC-F1 to find Cb with Rm = 1.0 for doubly symmetric . 12.5Mmax Cb = Rm ≤ 3.0 2.5Mmax + 3MA + 4MB + 3MC 12.5M = (1.0) = 1.67 2.5M + 3(0.25M ) + 4(0.50M ) + 3(0.75M ) For the limit state of yielding, AISC-F2.1: Mn = Mp = Fy Zx = (50)(214)/12 = 892 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-16, Lp = 11.2 ft and Lr = 63.3 ft. [Lb = 11 ft] ≤ Lp = 11.2 ft lateral torsional buckling does not apply, AISC-


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 K x Lx = 47.3 (from above) Axis of bending = r rx π 2 EI π 2 (29, 000)(1240) Pe1 = = = 5, 090 kips (KL)2 [(1.0)(22)(12)]2 According to AISC-C1.1b, for transverse loading, Cm = 1.0. Cm 1.0 B1 = = = 1.076 1 − Pu /Pe1 1 − 360/5, 090 (f) Use AISC-H1 to check the beam-column. 360 Pu = 0.236 ≥ 0.2 so use AISC Formula (H1-1a) omitting the y-axis = φc Pn 1, 520 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.


12.4. Investigate the adequacy of the given section of the accompanying figure. No t translation can occur and external lateral is provided at the ends only. Use LRFD Design Method (a) Obtain the factored loads. MaDL MaLL MbDL MbLL Mnta Mntb Pu

= McDL = 2.90 ft-kips = McLL = 3.80 ft-kips = 2.00 ft-kips = 2.50 ft-kips = 1.2(2.9) + 1.6(3.8) = 9.56 ft-kips = 1.2(2) + 1.6(2.5) = 6.4 ft-kips = 1.2(14) + 1.6(50) = 96.8 kips

(b) Column action: KL r φc Fcr φc Pn Pu φc Pn

Largest

(KL)y 1.0(15)12 = = 111 ry 1.62 = 18.2 ksi = φc Fcr Ag = 18.2(8.25) = 151 kips 96.8 = = 0.643 > 0.2 Use AISC Formula (H1-1a) 151 =

(c) Beam Action. Neither lateral torsional buckling nor web local buckling are limit states for weak axis bending. Check flange local buckling. "

s # " # bf E = 7.0 < λp = 0.38 = 9.15 ; section is compact. 2tf Fy

1 φb Mn = φb Mp = φb Fy Zy = 0.9(50)(10.1) = 37.9 ft-kips 12 (d) Moment magnification. Obtain the slenderness ratio for the axis of bending. The beam is bending about the weak axis. KL KL Axis of bending = = 111 r r y π 2 (29, 000)(21.7) π 2 EI Pe1 = = = 191 kips (KL)2 [(1.0)(15)(12)]2 Pu 96.8 Cm = 1 − 0.4 = 1 − 0.4 = 0.798 Pe1 191 Cm 0.798 B1 = = = 1.615 1 − Pu /Pe1 1 − 96.8/191

(e) Check AISC Formula (H1-1a), omitting the x-axis bending term. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Pu 8 + φc Pn 9

Muy φb Mny

96.8 8 1.615(9.56) = + 191 9 37.9 = 0.643 + 0.362 = 1.005 > 1.0 No good!

Section has insufficient strength for the given loading.


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