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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission 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.
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© 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.
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© 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.
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© 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.
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© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
© 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.
7.6. A W24×94 beam on a 6-ft span (see acompanying figure) underpins a column that brings 110 kips dead load and 280 kips live load to its top flange at a location 2.5 ft from the left support. The column bearing plate is 12 in. measured along the beam, and the bearing plates at the end supports are each 8 in. Investigate this beam of A992 steel for (a) flexure, (b) shear, and (c) satisfactory transmission of the reactions and concentrated load (i.e., local web yielding, web crippling, and sidesway web buckling). Specify changes (if any) required to satisfy the AISC Specification . Use LRFD Design Method
(a) Obtain factored loads: W u = 1.2(110) + 1.6(280) = 580 kips W u ab 580(2.5)(3.5) M u = = = 846 ft-kips L 6 W u b 580(3.5) V u = = = 338 kips L 6
(b) Check flexural strength assuming adequate lateral support (AISC F2.1): Flange and web local buckling slenderness limits, F y = 50 ksi steel:
b f = 5.2 2tf
≤
65 λ p = = 9.2 ; F y
h = 41.9 tw
φb M n = φb M p = φb Z x F y = 0.9(254)(50)/12 = 953 ft-kips [φb M n = 953 ft-kips] > [M u = 846 ft-kips] OK
≤
640 λ p = = 90.5 OK F y
(c) Check shear strength (AISC G2.1): h For rolled I-shapes, when = 41.9 ≤ 2.24 E/F y = 53.9 , φv = 1.0 and tw C v = 1.0. φv V n = φ v (0.6F y )Aw C v = φv (0.6F y )dtw C v = 1.0(0.6)(50)(24.31)(0.515)(1.0) = 376 kips [φv V n = 376 kips] > [V u = 338 kips] OK
(d) Check local web yielding strength (AISC J10.2): Rn = (5k + N )F y tw Interior Reaction Rn = (2.5k + N )F y tw Exterior Reaction Rn = (5k + N )F y 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 )F y 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: φRn = φ0.80t2w
3N 1+ d
1.5
tw tf
2
= (0.75)(0.80)(0.515)
EF yw tf tw
3N 1+ 24.31
0.515 1.5 0.875
(29000)(50)(0.875) 0.515
Solving for N to give Ru = 580 kips, N = 23.7 in. For exterior reactions, assuming N/d > 0.2:
φRn = φ0.80t2w 1 +
4N − 0.2 d 2
= (0.75)(0.40)(0.515)
1.5
1+
tw tf
4N 24.31
− 0.2
EF yw tf tw
0.515 1.5 0.875
(29000)(50)(0.875) 0.515
Solving for N to give Ru = 338 kips, N = 24.2 in. Check [N/d = 1.00] > 0.2. Solving for N to give Ru = 242 kips, N = 13.8 in. Check [N/d = 0.57] > 0.2. (f) Check sidesway web buckling strength (AISC J10.4):
When the compression flange is restrained against rotation, for (h/tw )/(Lb/bf ) = ∞ > 2.3, this limit state does not apply. Conclusion:
In accordance with AISC-J10.7, “At unframed ends of beams and girders not otherwise restrained against rotation about their longitudinal axes, a pair of transverse stiffeners , extending the full depth of the 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.
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7.7. A W16×77 section of A992 steel is to serve on a 10-ft simply supported span. The wall bearing length is 10 in. What maximum slowly moving concentrated service load (25% dead load; 75% live load) may be carried? Use LRFD Design Method
(a) Obtain factored loads: W u = 1.2(0.25W ) + 1.6(0.75W ) = 1.5W W u L W u (10) M u = = = 2.5W u with load at midspan 4 4 V u = 1.0W u with load at support (b) Check flexural strength assuming adequate lateral support (AISC F2.1): Flange and web local buckling slenderness limits, F y = 50 ksi steel:
b f = 6.8 2tf
≤
λ p =
65 = 9.2 ; F y
h = 31.2 tw
≤
λ p =
640 = 90.5 OK F y
φb M n = φb M p = φ b Z x F y = 0.9(150)(50)/12 = 563 ft-kips M u = 2.5W u = 563 ft-kips; W u = 225 kips
(c) Check shear strength (AISC G2.1): h For rolled I-shapes, when = 31.2 ≤ 2.24 E/F y = 53.9 , φv = 1.0 and tw C v = 1.0. φv V n = φ v (0.6F y )Aw C v = φv (0.6F y )dtw C v = 1.0(0.6)(50)(16.52)(0.455)(1.0) = 225 kips V u = 1.0W u = 225 kips; W u = 225 kips
(d) Check local web yielding strength (AISC J10.2): Rn = (5k + N )F y tw Interior Reaction Rn = (2.5k + N )F y tw Exterior Reaction Exterior reaction controls because W u is both the interior and exterior load φRn = φ(2.5k + N )F y tw = 1.0 [5(1.4375) + N ] (50)(0.455) = 309 kips φRn = 1.0W u = 309 kips; W u = 309 kips (e) Check web crippling strength (AISC J10.3): For exterior reactions, for [N/d = 10/16.52 = 0.61] > 0.2:
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φRn = φ0.80t2w 1 + =
4N − 0.2 d
(0.75)(0.40)(0.455)2
1.5
1+
tw tf
4(10) 16.52
− 0.2
EF yw tf tw
0.455 1.5 0.76
(29000)(50)(0.76) 0.455
= 196 kips φRn = 1.0W u = 196 kips; W u = 196 kips
(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 W u = 196 kips; Service Load W = W u /1.5 = 131 kips
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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 M u = = = 44.2 ft-kips 8 8 M ux = M u cos θ = 44.2cos30◦ = 38.3 ft-kips M uy = M u sin θ = 44.2sin30◦ = 22.1 ft-kips (b) Use AISC H2 with no axial load term and the conservative estimate M n = S F y : M uy M ux ≤1 + φb M nx φb M ny M uy S x M ux 38.3(12) 22.1(12) S x S x ≥ + = + φb F y φb F y S y 0.9(50) 0.9(50) S y S x ≥ 10.2 + 5.9 S y
For S x on the order of 3 to 4: S x ≈ 27.9 to 33.8 in.3 Assuming Z x ≈ 1.12S x : Z x ≈ 31.2 to 37.8 in.3 Using AISC Table 3-2 Selection by Z x , for W8 beams, find W8×35 with Z x = 34.7 in.3 Check the strength. M uy M ux 38.3(12) 22.1(12) + = + φb F y S x φb F y S y 0.9(50)(31.2) 0.9(50)(10.6) = 0.3272 + 0.5561 = 0.8833 ≤ 1 OK Beam
M nx ft-kips
M ny ft-kips
W8×35 W8×31
117 103
39.8 34.8
Check
0.3272 + 0.5561 = 0.8833 0.3690 + 0.6321 = 1.0011
OK NG
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8.21. Assume a single W section is to serve as a crane runway girder which carries a vertical loading, as shown. In addition, design must include an axial compressive force of 14 kips and a horizontal force of 4 kips on each wheel applied 4 14 in. above the top of the compression flange. Assume torsional simple support at the ends of the beam. Select the lightest W14 section of A992 steel using the β modified flexure analogy approach. Note: All loads except weight of the crane runway girder are live loads.
Use LRFD Design Method
(a) Obtain factored loads: Use an estimated beam weight of 0.342 kips/ft and an estimated beam depth of 14 in. W ux = 1.6(40) = 64 kips – Lifted load W uy = 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 2W ux x2 2(64)(19)2 M ux1 = = = 1155 ft-kips – Lifted load moment L 40 M ux2 = 0.25M ux1 = 0.25(1155) = 288.8 ft-kips – Impact moment wuxx(L − x) (0.4344)(19)(40 − 19) M ux3 = = = 86.7 ft-kips – Dead load moment 2 2 M ux = M ux1 + M ux2 + M ux3 = 1531 ft-kips 2W uy x2 2(6.4)(19)2 M uy = = = 115.5 ft-kips L 40 T u = W uy (d/2 + rail height) = 6.4(14/2 + 4.25) = 72.00 in-kips (b) Use the β modified flexure analogy to find the equivalent lateral moment. Use β ≈ 0.5. T u 72.00 ≈ V f = = 5.143 kips – flange force using h ≈ d h 14 2V f x2 (5.143)(19)2 M f = β = 0.5 = 46.41 kips – flange moment L 40 M y = 2M f = 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: M ux M uy + M y S x 1531(12) (115.5 + 92.83)(12) S x ≥ + = + φb F y φb F y S y 0.9(50) 0.9(50) S x ≥ 408.2 + 55.56 S y
S x S y
For S x on the order of 3 to 5: S x ≈ 574.9 to 686.0 in.3 Assuming Z x ≈ 1.12S x : Z x ≈ 643.8 to 768.3 in.3 Using AISC Table 3-2 Selection by Z x , for W14 beams, find W14×342 with Z x = 672 in.3 (d) Check the beam more accurately using the properties from AISC Table 1-1. λ =
GJ = EC w
(11154)(178) = 0.02578 (29000)(103000)
λL = (0.02578)(40)(12) = 12.38 β = 0.16 for a point load at x and simply supported ends T u = W uy (d/2 + rail height) = 6.4(17.5/2 + 4.25) = 83.20 in-kips T u 83.20 V f = = = 5.536 kips h 15.03 2V f x2 (5.536)(19)2 M f = β = 0.16 = 16.19 ft-kips L 40 M y = 2M f = 2(16.19) = 32.38 ft-kips M ux M uy M y 1531(12) 115.5(12) 32.38(12) + + = + + S x S y S y 558 221 221 = 32.92 + 6.273 + 1.758 = 40.95 ksi ≤ φF y = 0.9(50) = 45 ksi OK
f un =
The beam is sufficient. Check for a lighter beam. Section
M ux ft-kips
β
M y ft-kips
f un ksi
φF y = 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.
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9.2, Case 1. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 20 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W21×62 of F y = 50 ksi steel. Use LRFD Design Method
(a) Obtain factored loads: W u = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(62/1000) = 0.0744 kips/ft
(b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.5M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M C b = (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.
b f E = 90.6 ; = 6.7 ≤ λ p = 0.38 F y 2tf The web is compact and the flange is compact so use AISC-F2.
h = 46.9 tw
≤
λ p = 3.76
E = 9.15 F y
For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(144)/12 = 600 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-16, L p = 6.25 ft and Lr = 18.1 ft. [Lr = 18.1 ft] < [Lb = 20 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). M n = F cr S x = S x = (127)
C b π 2 E (Lb /rts )2
(1.32)π 2(29000)
(12(20)/2.15)2 = 415 ft-kips
Jc 1 + 0.078 S x ho
Lb 2 rts
(1.83)(1) 1 + 0.078 (127)(20.4)
12(20) 2 2.15
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Elastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0.9)(415) = 374 ft-kips (d) Calculate the maximum service load. W u L wu L2 φb M n = + 4 8 (1.46W )(20) (0.0744)(20)2 374 = + = 7.30W + 3.72 4 8 W = 50.7 kips Maximum Service Load W = 50.7 kips
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9.2, Case 2. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 24 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W24×84 of F y = 50 ksi steel. Use LRFD Design Method
(a) Obtain factored loads: W u = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(84/1000) = 0.101 kips/ft
(b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 24 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.5M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M C b = (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.
b f E = 90.6 ; = 5.86 ≤ λ p = 0.38 F y 2tf The web is compact and the flange is compact so use AISC-F2.
h = 45.9 tw
≤
λ p = 3.76
E = 9.15 F y
For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(224)/12 = 933 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-16, L p = 6.89 ft and Lr = 20.3 ft. [Lr = 20.3 ft] < [Lb = 24 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). M n = F cr S x = S x = (196)
C b π 2 E (Lb /rts )2
(1.32)π 2(29000)
(12(24)/2.37)2 = 579 ft-kips
Jc 1 + 0.078 S x ho
Lb 2 rts
(3.7)(1) 1 + 0.078 (196)(23.3)
12(24) 2 2.37
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Elastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0.9)(579) = 521 ft-kips (d) Calculate the maximum service load. W u L wu L2 φb M n = + 4 8 (1.46W )(24) (0.101)(24)2 521 = + = 8.76W + 7.26 4 8 W = 58.7 kips Maximum Service Load W = 58.7 kips
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9.2, Case 3. Determine the maximum concentrated load P that can act at midspan on a simply supported span of 30 ft. Lateral supports exist only at the ends of the span. The service load is 65% live load and 35% dead load. The section is W30×99 of F y = 50 ksi steel. Use LRFD Design Method
(a) Obtain factored loads: W u = 1.2(0.35W ) + 1.6(0.65W ) = 1.46W wu = 1.2(99/1000) = 0.119 kips/ft
(b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 30 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.5M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.5M 12.5M C b = (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.
b f E = 90.6 ; = 7.8 ≤ λ p = 0.38 F y 2tf The web is compact and the flange is compact so use AISC-F2.
h = 51.9 tw
≤
λ p = 3.76
E = 9.15 F y
For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(312)/12 = 1300 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-15, L p = 7.42 ft and Lr = 21.4 ft. [Lr = 21.4 ft] < [Lb = 30 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). M n = F cr S x = S x = (269)
C b π 2 E (Lb /rts )2
(1.32)π 2(29000)
(12(30)/2.62)2 = 585 ft-kips
Jc 1 + 0.078 S x ho
Lb 2 rts
(3.77)(1) 1 + 0.078 (269)(29)
12(30) 2 2.62
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Elastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0.9)(585) = 527 ft-kips (d) Calculate the maximum service load. W u L wu L2 φb M n = + 4 8 (1.46W )(30) (0.119)(30)2 527 = + = 11.0W + 13.4 4 8 W = 46.9 kips Maximum Service Load W = 46.9 kips
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9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 1: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has continuous lateral support, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft wu L2 4.28(20)2 M u = = = 214 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. Since the beam has continous lateral support, C b = 1.0. (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Z x , AISC Manual , pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. M u (214)(12) Required Z x = = = 57.1 in.3 φb F y (0.9)(50) Select: W18×35, Z x = 66.5 in.3 (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(35/1000)(20)2 M u = 214 + = 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.
h = 53.5 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 7.06 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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 support, so [Lb = 0] < L p and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb M n = (0.9)(277) = 249 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
W18×35 W16×31 W14×34
M u
φb M n
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 F y = 50 ksi steel.
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9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 2: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft wu L2 4.28(20)2 M u = = = 214 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 10 ft with L b = 10 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.438M M B = moment at 1/2 pt of the unbraced segment = 0.8M M C = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M C b = (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 Z x , AISC Manual , pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. M u (214)(12) Required Z x = = = 57.1 in.3 φb F y (0.9)(50) Select: W18×35, Z x = 66.5 in.3 (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(35/1000)(20)2 M u = 214 + = 214 + = 216 ft-kips 8 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. 1-18 and 1-19.
h = 53.5 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 7.06 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(66.5)/12 = 277 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-18, L p = 4.31 ft and Lr = 12.4 ft.
L p = 4.31 ft < [Lb = 10 ft] ≤ [Lr = 12.4 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (0.9)(260) = 234 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
ft-kips
ft-kips
bf 2tf
W18×35 216 234† 7.06 † W16×31 216 186 6.28 W14×34 216 205 7.41 † Inelastic lateral torsional buckling controls
h tw
OKAY?
53.5 51.6 43.1
OK NG NG
Use W18×35 with F y = 50 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 3: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateral support at the ends only, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.9 + beam wt) + 1.6(2) ≈ 4.28 kips/ft wu L2 4.28(20)2 M u = = = 214 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 20 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.750M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.750M 12.5M C b = (1.0) = 1.14 2.5M + 3(0.750M ) + 4(1.0M ) + 3(0.750M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 20 ft. M u 214 Required φb M n = = = 188 ft-kips C b 1.14 Select: W14×48, φ b M n = 193 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(48/1000)(20)2 M u = 214 + = 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.
h = 33.6 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.75 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(78.4)/12 = 327 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 6.75 ft and Lr = 21.1 ft.
L p = 6.75 ft < [Lb = 20 ft] ≤ [Lr = 21.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (0.9)(243) = 219 ft-kips The W14×48 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 48 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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: © 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.
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0.7(50)(93) = 446 − 446 − 12 = 442 ft-kips
9.47 − 9.15 24.1 − 9.15
Use W14×48 with F y = 50 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 4: dead load is 0.7 kips/ft, live load is 1.4 kips/ft, span is 28 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(1.4) ≈ 3.08 kips/ft wu L2 3.08(28)2 M u = = = 302 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 14 ft with L b = 14 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.438M M B = moment at 1/2 pt of the unbraced segment = 0.8M M C = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M C b = (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 L b = 14 ft. M u 302 Required φb M n = = = 232 ft-kips C b 1.30 Select: W14×48, φ b M n = 239 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(48/1000)(28)2 M u = 302 + = 302 + = 307 ft-kips 8 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. 1-22 and 1-23.
h = 33.6 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.75 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(78.4)/12 = 327 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 6.75 ft and Lr = 21.1 ft.
L p = 6.75 ft < [Lb = 14 ft] ≤ [Lr = 21.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (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
M u
φb M n
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:
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M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0.7(50)(93) = 446 − 446 − 12 = 442 ft-kips
9.47 − 9.15 24.1 − 9.15
Use W21×48 with F y = 50 ksi steel.
© 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.
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 5: dead load is 0.7 kips/ft, live load is 1.4 kips/ft, span is 28 ft, the beam has lateral support at the ends and midspan, and F y = 60 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(1.4) ≈ 3.08 kips/ft wu L2 3.08(28)2 M u = = = 302 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 14 ft with L b = 14 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.438M M B = moment at 1/2 pt of the unbraced segment = 0.8M M C = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M C b = (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 L b = 14 ft. M u 50 ksi 302 50 Required φb M n = = = 194 ft-kips C b F y 1.30 60 Select: W14×43, φ b M n = 208 ft-kips
(d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(43/1000)(28)2 M u = 302 + = 302 + = 307 ft-kips 8 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. 1-22 and 1-23.
h = 37.4 ≤ λ p = 3.76 tw
b f E = 82.7 ; = 7.54 ≤ λ p = 0.38 F y 2tf
E = 8.35 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (60)(69.6)/12 = 348 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 1.89 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29000 = 6.09 ft 60
(45.2)(1950) = 2.18 in. 62.6
Jc S x ho
2.18 29000 = 1.95 12 0.7(60)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(1.05)(1) (62.6)(13.1)
= 17.7 ft
1+
0.7(60) (62.6)(13.1) 2 1 + 6.76 29000 (1.05)(1)
L p = 6.09 ft < [Lb = 14 ft] ≤ [Lr = 17.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(60)(62.6) 14 − 6.09 = (1.30) 348 − 348 − 12 17.7 − 6.09 = 338 ft-kips
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (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
M u
φb M n
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
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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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0.7(60)(93) = 535 − 535 − 12 = 518 ft-kips
9.47 − 8.35 22.0 − 8.35
Use W16×45 with F y = 60 ksi steel.
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9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 6: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has continuous lateral support, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. Since the beam has continous lateral support, C b = 1.0. (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Z x , AISC Manual , pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. M u (276)(12) Required Z x = = = 73.5 in.3 φb F y (0.9)(50) Select: W18×40, Z x = 78.4 in.3 (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(40/1000)(35)2 M u = 276 + = 276 + = 283 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.
h = 50.9 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 5.73 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(78.4)/12 = 327 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 support, so [Lb = 0] < L p and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb M n = (0.9)(327) = 294 ft-kips The W18×40 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 40 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
W18×40 W16×40
M u
φb M n
ft-kips
ft-kips
283 283
294 274
bf 2tf
h tw
OKAY?
5.73 6.93
50.9 46.5
OK NG
Use W18×40 with F y = 50 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 7: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.990M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M C b = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 7 ft. M u 276 Required φb M n = = = 274 ft-kips C b 1.00 Select: W21×44, φ b M n = 315 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(44/1000)(35)2 M u = 276 + = 276 + = 284 ft-kips 8 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. 1-18 and 1-19.
h = 53.6 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 7.22 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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, L p = 4.45 ft and Lr = 13.0 ft.
L p = 4.45 ft < [Lb = 7 ft] ≤ [Lr = 13.0 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (0.9)(352) = 316 ft-kips The W21×44 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 44 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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 W21×44 with F y = 50 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 8: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support at the ends and midspan, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 17.5 ft with L b = 17.5 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.438M M B = moment at 1/2 pt of the unbraced segment = 0.8M M C = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M C b = (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 L b = 17.5 ft. M u 276 Required φb M n = = = 212 ft-kips C b 1.30 Select: W21×48, φ b M n = 221 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(48/1000)(35)2 M u = 276 + = 276 + = 284 ft-kips 8 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. 1-18 and 1-19.
h = 53.6 ≤ λ p = 3.76 tw
λ p = 1.0
E = 90.6 ; λ p = 0.38 F y
b f E = 9.15 < = 9.47 ≤ F y 2tf
E = 24.1 F y
The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 5.86 ft and Lr = 16.6 ft. [Lr = 16.6 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c).
C b π 2 E M n = F cr S x = S x (Lb /rts )2 = (93)
(1.30)π 2(29000)
(12(17.5)/2.05)2 = 319 ft-kips
Jc 1 + 0.078 S x ho
Lb 2 rts
(0.803)(1) 1 + 0.078 (93)(20.2)
12(17.5) 2 2.05
For the limit state of compression flange local buckling, AISC-F3.2:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (0.9)(319) = 287 ft-kips The W21×48 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 48 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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
© 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.
∗
Flange local buckling limit state must be checked
Use W21×48 with F y = 50 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 9: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has continuous lateral support, and F y = 65 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. Since the beam has continous lateral support, C b = 1.0. (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 0 ft. M u 50 ksi 276 50 Required φb M n = = = 212 ft-kips C b F y 1.00 65 Select: W18×35, φ b M n = 249 ft-kips
(d) Correct the moment for the selected beam weight. M u = 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.
h = 53.5 ≤ λ p = 3.76 tw
b f E = 79.4 ; = 7.06 ≤ λ p = 0.38 F y 2tf
E = 8.03 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (65)(66.5)/12 = 360 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 support, so [Lb = 0] < L p and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb M n = (0.9)(360) = 324 ft-kips The W18×35 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 35 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
W18×35 W16×31 W14×34 W12×35
M u
φb M n
ft-kips
ft-kips
282 281 282 282
324 263 266 250
bf 2tf
h tw
OKAY?
7.06 6.28 7.41 6.31
53.5 51.6 43.1 36.2
OK NG NG NG
Use W18×35 with F y = 65 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 10: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 65 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.990M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M C b = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 7 ft. M u 50 ksi 276 50 Required φb M n = = = 211 ft-kips C b F y 1.00 65 Select: W18×35, φ b M n = 217 ft-kips
(d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(35/1000)(35)2 M u = 276 + = 276 + = 282 ft-kips 8 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. 1-18 and 1-19.
h = 53.5 ≤ λ p = 3.76 tw
b f E = 79.4 ; = 7.06 ≤ λ p = 0.38 F y 2tf
E = 8.03 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (65)(66.5)/12 = 360 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 1.22 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29000 = 3.78 ft 65
(15.3)(1140) = 1.51 in. 57.6
Jc S x ho
1.51 29000 = 1.95 12 0.7(65)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(0.506)(1) (57.6)(17.3)
= 10.7 ft
1+
0.7(65) (57.6)(17.3) 2 1 + 6.76 29000 (0.506)(1)
L p = 3.78 ft < [Lb = 7 ft] ≤ [Lr = 10.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(65)(57.6) 7 − 3.78 = (1.00) 360 − 360 − 12 10.7 − 3.78 = 295 ft-kips
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (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
M u
φb M n
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
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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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0.7(65)(56.5) = 347 − 347 − 12 = 346 ft-kips
8.12 − 8.03 21.1 − 8.03
Use W16×40 with F y = 65 ksi steel.
© 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.
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 11: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support at the ends and midspan, and F y = 65 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends and midspan. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 17.5 ft with L b = 17.5 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.438M M B = moment at 1/2 pt of the unbraced segment = 0.8M M C = moment at 3/4 pt of the unbraced segment = 0.938M 12.5M C b = (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 L b = 17.5 ft. M u 50 ksi 276 50 Required φb M n = = = 163 ft-kips C b F y 1.30 65 Select: W14×43, φ b M n = 182 ft-kips
(d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(43/1000)(35)2 M u = 276 + = 276 + = 284 ft-kips 8 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. 1-22 and 1-23.
h = 37.4 ≤ λ p = 3.76 tw
b f E = 79.4 ; = 7.54 ≤ λ p = 0.38 F y 2tf
E = 8.03 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (65)(69.6)/12 = 377 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 1.89 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29000 = 5.86 ft 65
(45.2)(1950) = 2.18 in. 62.6
Jc S x ho
2.18 29000 = 1.95 12 0.7(65)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(1.05)(1) (62.6)(13.1)
1+
0.7(65) (62.6)(13.1) 2 1 + 6.76 29000 (1.05)(1)
= 16.8 ft [Lr = 16.8 ft] < [Lb = 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c). C b π 2 E M n = F cr S x = S x (Lb /rts )2 = (62.6)
(1.30)π 2(29000)
(12(17.5)/2.18)2 = 290 ft-kips
Jc 1 + 0.078 S x ho
Lb 2 rts
(1.05)(1) 1 + 0.078 (62.6)(13.1)
12(17.5) 2 2.18
Elastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (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
M u
φb M n
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
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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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0.7(65)(93) = 580 − 580 − 12 = 555 ft-kips
9.47 − 8.03 21.1 − 8.03
Use W14×48 with F y = 65 ksi steel.
© 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.
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 12: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has continuous lateral support, and F y = 100 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. Since the beam has continous lateral support, C b = 1.0. (c) Since the unbraced length is relatively short, select a beam using Table 3-2 Selection by Z x , AISC Manual , pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. M u (276)(12) Required Z x = = = 36.8 in.3 φb F y (0.9)(100) Select: W12×26, Z x = 37.2 in.3 (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(26/1000)(35)2 M u = 276 + = 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.
h = 47.2 ≤ λ p = 3.76 tw
λ p = 1.0
E = 64.0 ; λ p = 0.38 F y
b f E = 6.47 < = 8.54 ≤ F y 2tf
E = 17.0 F y
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 support, so [Lb = 0] < L p and lateral torsional buckling does not apply, AISC-F2.1(a). For the limit state of compression flange local buckling, AISC-F3.2:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (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
M u
φb M n
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
Use W14×26 with F y = 100 ksi steel.
© 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.
9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 13: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support every 7 feet, and F y = 100 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.3 + beam wt) + 1.6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support every 7 feet. The loading is uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 14 ft to 21 ft with L b = 7 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.990M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.990M 12.5M C b = (1.0) = 1.00 2.5M + 3(0.990M ) + 4(1.0M ) + 3(0.990M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 7 ft. M u 50 ksi 276 50 Required φb M n = = = 137 ft-kips C b F y 1.00 100 Select: W16×26, φ b M n = 138 ft-kips
(d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(26/1000)(35)2 M u = 276 + = 276 + = 280 ft-kips 8 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. 1-20 and 1-21.
h = 56.8 ≤ λ p = 3.76 tw
λ p = 1.0
E = 64.0 ; λ p = 0.38 F y
b f E = 6.47 < = 7.97 ≤ F y 2tf
E = 17.0 F y
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:
L p = 1.76ry
rts =
E 1.12 = 1.76 F y 12
I y C w = S x
29000 = 2.80 ft 100
(9.59)(565) = 1.38 in. 38.4
E Lr = 1.95rts 0.7F y
Jc S x ho
1.38 = 1.95 12
29000 0.7(100)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(0.262)(1) (38.4)(15.3)
= 7.65 ft
1+
0.7(100) (38.4)(15.3) 2 1 + 6.76 29000 (0.262)(1)
L p = 2.80 ft < [Lb = 7 ft] ≤ [Lr = 7.65 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(100)(38.4) = (1.00) 368 − 368 − 12 = 245 ft-kips
≤ M p
7 − 2.80 7.65 − 2.80
For the limit state of compression flange local buckling, AISC-F3.2:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (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
M u
φb M n
ft-kips
ft-kips
bf 2tf
W16×26 2 80 220† 7.97∗ W16×31 28 1 280† 6.28 † W16×36 28 2 39 1 8.12∗ W18×35 28 2 356† 7.06∗ W16×31 28 1 280† 6.28 † W14×34 28 2 33 5 7.41∗ W14×30 28 1 286† 5.74 † W14×26 28 0 19 6 5.98 † W12×30 28 1 26 4 7.41∗ W10×30 28 1 218† 5.7 † Inelastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked
h tw
OKAY?
56.8 51.6 48.1 53.5 51.6 43.1 45.4 48.1 41.8 29.5
NG NG OK OK NG OK OK NG NG NG
Use W14×30 with F y = 100 ksi steel.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission 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.
9.3. Select Select the lightest lightest W section section as a beam. Assume Assume only flexure must must be considered considered;; i.e., omit treating treating shear and deflection. deflection. The dead load given is in addition addition to the weight weight of the beam. Case 14: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateral support at the ends and midspan, and F and F y = 100 ksi.
Use LRFD Design Method
(a) Obtain Obtain factored factored loads: loads: wu = 1.2(0. 2(0.3 + beam wt) + 1.6(0. 6(0.9) ≈ 1.8 kips/ft wu L2 1.8(35)2 M u = = = 276 ft-kips ft-kips (without (without beam) 8 8 (b) Determ Determine ine the the C b factor, AISC-F1. The beam has lateral lateral support at the ends and midspan. midspan. The loading is uniform uniform and symmetric, so the worst loading will occur on a segment containing the midpoint of the beam. Use the segment from 0 ft to 17.5 ft with L b = 17. 17.5 ft. For doubly symmetric members, Rm = 1.0. 12. 12.5M max max C b = Rm ≤ 3.0 2.5M max 3 M A + 4M 4 M B + 3M 3 M C max + 3M M max max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0. 0.438M 0.8M M B = moment at 1/2 pt of the unbraced segment = 0. M C = moment at 3/4 pt of the unbraced segment = 0. 0.938M 12. 12.5M C b = (1. (1.0) = 1. 1.30 2.5M + + 3(0. 3(0.438 438M M )) + 4(0. 4(0.8M ) M ) + 3(0. 3(0.938 938M M )) (c) Since Since the unbrac unbraced ed length length is fairly fairly long, select select a beam using Table Table 3-10 Av Avail ailabl ablee Manual , pp. 3-96 to 3-131 with L b = 17. Moment vs. Unbraced Length, AISC Manual 17.5 ft. M u 50 ksi 276 50 Required φb M n = = = 106 ft-kips C b F y 1.30 100 Select: W10×33, 33, φ φ b M n = 107 ft-kips
(d) Correct Correct the moment moment for the selected selected beam weight. weight. 1. 1 .2(beam wt)L wt)L2 1. 1 .2(33/ 2(33/1000)(35)2 M u = 276 + = 276 + = 282 ft-kips ft-kips 8 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission 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) Comput Computee the design design mom momen entt streng strength th using using the beam propert properties ies from the AISC Manual Table Table 1-1, pp. 1-24 and 1-25.
h = 27. 27.1 ≤ λ p = 3.76 tw
λ p = 1.0
b f E = 6.47 < = 9.15 ≤ F y 2tf
E = 64. 64.0 ; λ p = 0.38 F y
E = 17. 17.0 F y
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:
L p = 1.76r 76ry rts =
E 1.94 = 1.76 F y 12
I y C w = S x
29000 = 4.85 ft 100
(36. (36.6)(791) = 2. 2 .20 in. 35
E Lr = 1.95r 95rts 0.7F y
Jc S x ho
2.20 = 1.95 12
29000 0.7(100)
1+
0.7F y S x ho 2 1 + 6. 6.76 E J c
(0. (0.583)(1) (35)(9. (35)(9.3)
1+
0.7(100) (35)(9. (35)(9.3) 2 1 + 6. 6 .76 29000 (0. (0.583)(1)
= 13. 13.5 ft [Lr = 13 1 3.5 ft] < ft] < [ [L Lb = 17. 17.5 ft] elastic lateral torsional buckling applies, AISC-F2.1(c).
C b π 2 E M n = F = F cr = S x cr S x = S (Lb /rts )2 = (35)
(1. (1.30)π 30)π 2(29000)
(12(17. (12(17.5)/ 5)/2.2)2 = 179 ft-kips
Jc 1 + 0. 0 .078 S x ho
Lb 2 rts
(0. (0.583)(1) 1 + 0. 0 .078 (35)(9. (35)(9.3)
12(17. 12(17.5) 2 2.2
For the limit state of compression flange local buckling, AISC-F3.2:
M n = M = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0. 0 .7(100)(35) 9.15 − 6.47 = 323 − 323 − 12 17. 17.0 − 6.47 = 293 ft-kips Elastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0. (0.9)(179) = 161 ft-kips The W10×33 beam is not not suffic sufficie ien nt. Chec Check k hea heavier vier secti section onss at the same depth depth.. The 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 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
M u
φb M n
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
Use W10×45 with F y = 100 ksi steel.
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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 support every 5 feet, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0 + beam wt) + 1.6(1) ≈ 1.6 kips/ft wu L2 1.6(35)2 M u = = = 245 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support 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 L b = 5 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.995M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.995M 12.5M C b = (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 L b = 5 ft. M u 245 Required φb M n = = = 244 ft-kips C b 1.00 Select: W16×40, φ b M n = 274 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(40/1000)(35)2 M u = 245 + = 245 + = 252 ft-kips 8 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. 1-20 and 1-21.
h = 46.5 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.93 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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, L p = 5.55 ft and Lr = 15.9 ft. [Lb = 5 ft] ≤ L p = 5.55 ft lateral torsional buckling does not apply, AISCF2.1(a).
Yielding controls! Calculate the design moment strength. φb M n = (0.9)(304) = 274 ft-kips The W16×40 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 40 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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
Use W16×40 with F y = 50 ksi steel.
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9.3. Select the lightest W section as a beam. Assume only flexure must be considered; i.e., omit treating shear and deflection. The dead load given is in addition to the weight of the beam. Case 16: dead load is 0 kips/ft, live load is 1 kips/ft, span is 35 ft, the beam has lateral support at the ends only, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0 + beam wt) + 1.6(1) ≈ 1.6 kips/ft wu L2 1.6(35)2 M u = = = 245 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends only. The longest unbraced length is Lb = 35 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.750M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.750M 12.5M C b = (1.0) = 1.14 2.5M + 3(0.750M ) + 4(1.0M ) + 3(0.750M ) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 35 ft. M u 245 Required φb M n = = = 216 ft-kips C b 1.14 Select: W12×65, φ b M n = 231 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(65/1000)(35)2 M u = 245 + = 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.
h = 24.9 ≤ λ p = 3.76 tw
λ p = 1.0
E = 90.6 ; λ p = 0.38 F y
b f E = 9.15 < = 9.92 ≤ F y 2tf
E = 24.1 F y
The web is compact and the flange is noncompact so use AISC-F3. For the limit state of lateral torsional buckling, AISC-F3.1 uses AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-17, L p = 10.7 ft and Lr = 35.1 ft.
L p = 10.7 ft < [Lb = 35 ft] ≤ [Lr = 35.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (0.9)(292) = 263 ft-kips The W12×65 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 65 lb/ft with a large enough Z x . 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
M u
φb M n
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
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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 W12×65 with F y = 50 ksi steel.
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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 17: dead load is 0.7 kips/ft, live load is 2.8 kips/ft, span is 48 ft, the beam has lateral support every 16 feet, and F y = 50 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(2.8) ≈ 5.32 kips/ft wu L2 5.32(48)2 M u = = = 1530 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support 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 L b = 16 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.972M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.972M 12.5M C b = (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 L b = 16 ft. M u 1530 Required φb M n = = = 1510 ft-kips C b 1.01 Select: W36×135, φb M n = 1550 ft-kips (d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(135/1000)(48)2 M u = 1530 + = 1530 + = 1580 ft-kips 8 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. 1-12 and 1-13.
h = 54.1 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 7.56 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(509)/12 = 2120 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From Table 3-2 in the AISC Manual p. 3-14, L p = 8.41 ft and Lr = 24.2 ft.
L p = 8.41 ft < [Lb = 16 ft] ≤ [Lr = 24.2 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(50)(439) = (1.01) 2120 − 2120 − 12 = 1740 ft-kips
≤ M p
16 − 8.41 24.2 − 8.41
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (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
M u
φb M n
ft-kips
ft-kips
W36×135 1580 1570† W36×150 1580 1830† W40×149 1580 1810† W36×135 1580 1570† W33×141 1580 1610† W33×130 1580 1440† W30×132 1580 1330† W27×129 1580 1210† W24×131 1580 1270† † Inelastic lateral torsional buckling controls
bf 2tf
h tw
OKAY?
7.56 6.37 7.11 7.56 6.01 6.73 5.27 4.55 6.7
54.1 51.9 54.3 54.1 49.6 51.7 43.9 39.7 35.6
NG OK OK NG OK NG NG NG NG
Use W33×141 with F y = 50 ksi steel.
© 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.
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 18: dead load is 0.7 kips/ft, live load is 2.8 kips/ft, span is 48 ft, the beam has lateral support every 16 feet, and F y = 60 ksi.
Use LRFD Design Method
(a) Obtain factored loads: wu = 1.2(0.7 + beam wt) + 1.6(2.8) ≈ 5.32 kips/ft wu L2 5.32(48)2 M u = = = 1530 ft-kips (without beam) 8 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support 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 L b = 16 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C M max = max moment in the unbraced segment = M M A = moment at 1/4 pt of the unbraced segment = 0.972M M B = moment at 1/2 pt of the unbraced segment = 1.0M M C = moment at 3/4 pt of the unbraced segment = 0.972M 12.5M C b = (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 L b = 16 ft. M u 50 ksi 1530 50 Required φb M n = = = 1260 ft-kips C b F y 1.01 60 Select: W33×130, φb M n = 1420 ft-kips
(d) Correct the moment for the selected beam weight. 1.2(beam wt)L2 1.2(130/1000)(48)2 M u = 1530 + = 1530 + = 1580 ft-kips 8 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. 1-12 and 1-13.
h = 51.7 ≤ λ p = 3.76 tw
b f E = 82.7 ; = 6.73 ≤ λ p = 0.38 F y 2tf
E = 8.35 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (60)(467)/12 = 2340 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 2.39 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29000 = 7.71 ft 60
(218)(56600) = 2.94 in. 406
Jc S x ho
2.94 29000 = 1.95 12 0.7(60)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(7.37)(1) (406)(32.2)
= 21.8 ft
1+
0.7(60) (406)(32.2) 2 1 + 6.76 29000 (7.37)(1)
L p = 7.71 ft < [Lb = 16 ft] ≤ [Lr = 21.8 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(60)(406) = (1.01) 2340 − 2340 − 12 = 1820 ft-kips
≤ M p
16 − 7.71 21.8 − 7.71
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0.9)(1820) = 1640 ft-kips The W33×130 beam is sufficient. To verify it is the lightest beam use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 130 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
W33×130 W30×124
M u
φb M n
ft-kips
ft-kips
1580 1580
1640† 1390†
bf 2tf
h tw
OKAY?
6.73 5.65
51.7 46.2
OK NG
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W27×129 1580 1370† W24×117 1570 1280† W21×122 1570 1220† † Inelastic lateral torsional buckling controls
4.55 7.53 6.45
39.7 39.2 31.3
NG NG NG
Use W33×130 with F y = 60 ksi steel.
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9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (a) A992 steel; continuous lateral support
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft W u = 1.2(28) + 1.6(7) = 44.8 kips 1 M u = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the C b factor, AISC-F1. Since the beam has continous lateral support, C b = 1.0. (c) Since the unbraced length is zero, select a beam using Table 3-2 Selection by Z x , AISC Manual , pp. 3-11 to 3-19. Assume λ ≤ λ p for a compact section. M u (1, 172)(12) Required Z x = = = 313 in.3 φb F y (0.90)(50) Select: W30×108, Z x = 346 in.3 (d) Correct the moment for the beam weight. 1 M u = 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.
h = 49.6 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.89 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
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.
M n = M p = F y Z x = (50)(346)/12 = 1, 440 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: The beam has continuous lateral support, so [Lb = 0] < L p and lateral torsional buckling does not apply, AISC-F2.1(a). Yielding controls! Calculate the design moment strength. φb M n = (0.90)(1, 440) = 1, 300 ft-kips The W30×108 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 108 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
W30×108 W27×102 W24×104
M u
φb M n
ft-kips
ft-kips
1,190 1,190 1,190
1,300 1,140 1,080
bf 2tf
h tw
OKAY?
6.89 6.03 8.5
49.6 47.1 43.1
OK NG NG
Use W30×108 with F y = 50 ksi steel.
© 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.
9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (b) A992 steel; lateral support at the ends
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): wu = 1.2(0.7) + 1.6(3.5) = 6.44 kips/ft W u = 1.2(28) + 1.6(7) = 44.8 kips 1 M u = (6.44)(30)2 + 10(44.8) = 1, 172 ft-kips 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends. The unbraced length is Lb = 30 ft. For doubly symmetric members, R m = 1.0. Use statics to find the moments in the beam: 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C 12.5(1, 172) = (1.0) = 1.14 2.5(1, 172) + 3(879.4) + 4(1, 172) + 3(879.4) (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 30 ft. M u 1, 172 Required φb M n = = = 1, 030 ft-kips 1.14 C b Select: W24×146, φb M n = 1, 070 ft-kips (d) Correct the moment for the beam weight. 1 M u = 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.
h = 33.2 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 5.92 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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, L p = 10.6 ft and Lr = 33.7 ft.
L p = 10.6 ft < [Lb = 30 ft] ≤ [Lr = 33.7 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(50)(371) 30 − 10.6 = (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 M n = (0.90)(1, 350) = 1, 210 ft-kips The W24×146 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 146 lb/ft with a large enough Z x . 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
M u
φb M n
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
‡
Elastic 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
Use W24×146 with F y = 50 ksi steel.
© 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.
9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (c) A992 steel; lateral support at the ends and at 10 ft
Use LRFD Design Method
(a) Obtain Obtain the factored loads (neglectin (neglecting g the beam weight weight): ): wu = 1.2(0. 2(0.7) + 1. 1.6(3. 6(3.5) = 6. 6.44 kips/ft W u = 1.2(28) + 1. 1.6(7) = 44. 44.8 kips 1 M u = (6. (6.44)(30)2 + 10(44. 10(44.8) = 1, 1, 172 ft-kips 8 (b) Determ Determine ine the the C b factor, AISC-F1. The beam has lateral support 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 L with L b = 20 ft. For doubly symmetric members, Rm = 1.0. Use statics to find the moments in the beam: 12. 12.5M max max C b = Rm ≤ 3. 3 .0 2.5M max 3 M A + 4M 4 M B + 3M 3 M C max + 3M 12. 12.5(1, 5(1, 172) = (1. (1.0) = 1. 1.15 2.5(1, 5(1, 172 172)) + 3(1, 3(1, 172) + 4(1, 4(1, 092) + 3(626. 3(626.5) (c) Since Since the unbrac unbraced ed length length is fairly fairly long, select select a beam using Table Table 3-10 Av Avail ailabl ablee Manual , pp. 3-96 to 3-131 with L b = 20 ft. Moment vs. Unbraced Length, AISC Manual M u 1, 172 Required φb M n = = = 1, 020 ft-kips C b 1.15 Select: W33×118, φb M n = 1, 080 ft-kips (d) Correct Correct the moment moment for the beam weight. weight. 1 (118/1000)(30)2 = 1, 190 ft-kips M u = 1, 172 + (118/ 8 (e) Comput Computee the design design mom momen entt streng strength th using using the beam propert properties ies from the AISC Manual Table 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 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.
h = 54. 54.5 ≤ λ p = 3.76 tw
b f E = 90. 90.6 ; = 7.76 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M = M p = F = F y Z x = (50)(415)/ (50)(415)/12 = 1, 1, 730 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: Manual Table From the AISC Manual Table 3-2 p. 3-14, L 3-14, L p = 8.19 ft and Lr = 23. 23.5 ft.
L p = 8.19 ft < [Lb = 20 ft] ≤ [Lr = 23. 23.5 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C = C b M p − M p − 0. 0.7F y S x Lr − L p 0. 0 .7(50)(359) 20 − 8. 8.19 = (1. (1.15) 1, 730 − 1, 730 − 12 23. 23.5 − 8. 8.19 = 1, 390 ft-kips
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0. (0.90)(1, 90)(1, 390) = 1, 1, 250 ft-kips The W33×118 beam is sufficien sufficient. t. To verif verify y it is the lightes lightestt beam, beam, use Tabl Tablee 1-1 of the AISC Manual , and examine the first beam in each group lighter than 118 lb/ft with a large enough Z enough Z x . The following table shows the moment corrected for the beam weight. Section
W33×118 W30×116 W27×114 W24×117 †
M u
φb M n
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
Inelastic lateral torsional buckling controls
Use W33×118 with F y = 50 5 0 ksi steel.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission 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.
9.4. Select the lightest W sections for the stituation shown in the accompanying figure, under the following conditions: (d) A572 Grade 60 steel; lateral support at the ends and at 10 ft
Use LRFD Design Method
(a) Obtain Obtain the factored loads (neglectin (neglecting g the beam weight weight): ): wu = 1.2(0. 2(0.7) + 1. 1.6(3. 6(3.5) = 6. 6.44 kips/ft W u = 1.2(28) + 1. 1.6(7) = 44. 44.8 kips 1 M u = (6. (6.44)(30)2 + 10(44. 10(44.8) = 1, 1, 172 ft-kips 8 (b) Determ Determine ine the the C b factor, AISC-F1. The beam has lateral support 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 L with L b = 20 ft. For doubly symmetric members, Rm = 1.0. Use statics to find the moments in the beam: 12. 12.5M max max C b = Rm ≤ 3. 3 .0 2.5M max 3 M A + 4M 4 M B + 3M 3 M C max + 3M 12. 12.5(1, 5(1, 172) = (1. (1.0) = 1. 1.15 2.5(1, 5(1, 172 172)) + 3(1, 3(1, 172) + 4(1, 4(1, 092) + 3(626. 3(626.5) (c) Since Since the unbrac unbraced ed length length is fairly fairly long, select select a beam using Table Table 3-10 Av Avail ailabl ablee Manual , pp. 3-96 to 3-131 with L b = 20 ft. Moment vs. Unbraced Length, AISC Manual M u 50 ksi 1, 172 50 Required φb M n = = = 846 ft-kips C b F y 1.15 60 Select: W24×104, φb M n = 875 ft-kips
(d) Correct Correct the moment moment for the beam weight. weight. 1 M u = 1, 172 + (104/ (104/1000)(30)2 = 1, 190 ft-kips 8 (e) Comput Computee the design design mom momen entt streng strength th using using the beam propert properties ies from the AISC Manual Table 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 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.
h = 43.1 ≤ λ p = 3.76 tw
λ p = 1.0
E = 82.7 ; λ p = 0.38 F y
b f E = 8.35 < = 8.5 ≤ F y 2tf
E = 22.0 F y
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:
L p = 1.76ry
rts =
E 2.91 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29, 000 = 9.38 ft 60
(259)(35200) = 3.42 in. 258
Jc S x ho
3.42 29, 000 = 1.95 12 0.7(60)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(4.72)(1) (258)(23.3)
= 26.1 ft
1+
0.7(60) (258)(23.3) 2 1 + 6.76 29, 000 (4.72)(1)
L p = 9.38 ft < [Lb = 20 ft] ≤ [Lr = 26.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(60)(258) 20 − 9.38 = (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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (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.
© 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
W24×104 W24×117 W30×116 W30×108 W27×114 W24×104 W24×103 W21×111
M u
φb M n
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
†
Inelastic lateral torsional buckling controls
‡
Elastic lateral torsional buckling controls Flange local buckling limit state must be checked
∗
Use W30×116 with F y = 60 ksi steel.
© 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.
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 W u = 1.2(15) + 1.6(15) = 42.0 kips 1 M u = (2.24)(42)2 + 10(42.0) = 1, 124 ft-kips 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends, at 15 ft, and at 27 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C 12.5(1, 084) = (1.0) = 1.56 segment A 2.5(1, 084) + 3(318.2) + 4(604.8) + 3(860.0) 12.5(1, 124) = (1.0) = 1.00 segment B 2.5(1, 124) + 3(1, 114) + 4(1, 124) + 3(1, 114) Assume segment B controls. (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 12 ft. M u 1, 124 Required φb M n = = = 1, 120 ft-kips C b 1.00 Select: W30×108, φb M n = 1, 140 ft-kips (d) Correct the moment for the beam weight. 1 M u = 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.
h = 49.6 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.89 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (50)(346)/12 = 1, 440 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2: From the AISC Manual Table 3-2 p. 3-15, L p = 7.59 ft and Lr = 22.0 ft.
L p = 7.59 ft < [Lb = 12 ft] ≤ [Lr = 22.0 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (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 †
M u
φb M n
ft-kips
ft-kips
1,150 1,150 1,150 1,150 1,150 1,150
1,150† 1,260† 1,150† 1,050† 918† 1,020†
bf 2tf
h tw
OKAY?
6.89 6.17 5.41 8.5 4.59 7.05
49.6 47.8 42.5 43.1 39.2 34.1
NG OK NG NG NG NG
bf 2tf
h tw
OKAY?
6.17
47.8
OK
Inelastic lateral torsional buckling controls
Check segment A Section
W30×116
M u
φb M n
ft-kips
ft-kips
1,110
1,420
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Use W30×116 with F y = 50 ksi steel.
© 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.
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 W u = 1.2(15) + 1.6(15) = 42.0 kips 1 M u = (2.24)(42)2 + 10(42.0) = 1, 124 ft-kips 8 (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends, at 15 ft, and at 27 ft. For doubly symmetric members, Rm = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C 12.5(1, 084) = (1.0) = 1.56 segment A 2.5(1, 084) + 3(318.2) + 4(604.8) + 3(860.0) 12.5(1, 124) = (1.0) = 1.00 segment B 2.5(1, 124) + 3(1, 114) + 4(1, 124) + 3(1, 114) Assume segment B controls. (c) Since the unbraced length is fairly long, select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 12 ft. M u 50 ksi 1, 124 50 Required φb M n = = = 933 ft-kips C b F y 1.00 60 Select: W30×99, φ b M n = 1, 020 ft-kips
(d) Correct the moment for the beam weight. 1 M u = 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.
h = 51.9 ≤ λ p = 3.76 tw
b f E = 82.7 ; = 7.8 ≤ λ p = 0.38 F y 2tf
E = 8.35 F y
The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (60)(312)/12 = 1, 560 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 2.1 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29, 000 = 6.77 ft 60
(128)(26800) = 2.62 in. 269
Jc S x ho
2.62 29, 000 = 1.95 12 0.7(60)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(3.77)(1) (269)(29)
1+
0.7(60) (269)(29) 2 1 + 6.76 29, 000 (3.77)(1)
= 19.3 ft
L p = 6.77 ft < [Lb = 12 ft] ≤ [Lr = 19.3 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p Lr − L p 0.7(60)(269) 12 − 6.77 = (1.00) 1, 560 − 1, 560 − 12 19.3 − 6.77 = 1, 310 ft-kips
M n = C b M p − M p − 0.7F y S x
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0.90)(1, 310) = 1, 180 ft-kips The W30×99 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 99 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight.
© 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
M u
φb M n
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
Inelastic lateral torsional buckling controls
Check segment A Section
W30×99
M u
φb M n
ft-kips
ft-kips
1,110
1,400
Use W30×99 with F y = 60 ksi steel.
© 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.
9.7. A floor beam, laterally supported at the ends only and supporting 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): W u = 1.2(0) + 1.6(14)(1.5) = 33.6 kips (includes impact) M u = 7(33.6) = 235.2 ft-kips (b) Determine the C b factor, AISC-F1. The beam has lateral support at the ends. For doubly symmetric members, R m = 1.0. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C 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 L b = 25 ft. M u 235.2 Required φb M n = = = 223 ft-kips C b 1.05 Select: W12×58, φ b M n = 233 ft-kips (d) Correct the moment for the beam weight. 1 M u = 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.
h = 27 ≤ λ p = 3.76 tw
E = 90.6 ; F y
b f = 7.82 ≤ λ p = 0.38 2tf
E = 9.15 F y
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The web is compact and the flange is compact so use AISC-F2. For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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, L p = 8.87 ft and Lr = 29.9 ft.
L p = 8.87 ft < [Lb = 25 ft] ≤ [Lr = 29.9 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (0.90)(272) = 245 ft-kips The W12×58 beam is sufficient. To verify it is the lightest beam, use Table 1-1 of the AISC Manual , and examine the first beam in each group lighter than 58 lb/ft with a large enough Z x . The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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
Elastic lateral torsional buckling controls
With no deflection, use W12×58 with F y = 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) 20
= 15 in. The
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Section
M u
φb M n
ft-kips
ft-kips
W16×67 241 336† W24×62 241 153‡ W21×62 241 213‡ W21×57 241 148‡ W18×65 241 219‡ W18×46 240 92.4‡ W16×57 241 169‡ † Inelastic lateral torsional buckling controls ‡
bf 2tf
h tw
OKAY?
7.7 5.97 6.7 5.04 5.06 5.01 4.98
35.9 50.1 46.9 46.3 35.7 44.6 33
OK NG NG NG NG NG NG
Elastic lateral torsional buckling controls
With deflection use W16×67 with F y = 50 ksi steel.
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9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 1: M D = 49 ft-kips, M L = 98 ft-kips, L = 28 ft, F y = 50 ksi, and the deflection limit is L/360.
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): M u = 1.2M D + 1.6M L = 1.2(49) + 1.6(98) = 216 ft-kips (b) Obtain the minimum moment of inertia I x required using the service live load moment. L 28(12) ∆lim = = = 0.933 in. 360 360 5M LL2 (5)(98 × 12)(28 × 12)2 Min I x = = = 511 in.4 48E ∆lim (48)(29, 000)(0.933) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual , pp. 3-96 to 3-131 with L b = 7 ft. M u 215.6 Required φb M n = = = 216 ft-kips C b 1.00 Select: W18×35, φ b M n = 217 ft-kips
Unbraced Length, AISC
(e) Correct the moment for the beam weight. 1 M u = 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.
h = 53.5 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 7.06 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
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: M n = M p = F y Z x = (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, L p = 4.31 ft and Lr = 12.4 ft.
L p = 4.31 ft < [Lb = 7 ft] ≤ [Lr = 12.4 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (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
M u
φb M n
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
I x 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 W18×40 with F y = 50 ksi steel.
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9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 2: M D = 49 ft-kips, M L = 98 ft-kips, L = 28 ft, F y = 60 ksi, and the deflection limit is L/360.
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): M u = 1.2M D + 1.6M L = 1.2(49) + 1.6(98) = 216 ft-kips (b) Obtain the minimum moment of inertia I x required using the service live load moment. L 28(12) ∆lim = = = 0.933 in. 360 360 5M LL2 (5)(98 × 12)(28 × 12)2 Min I x = = = 511 in.4 48E ∆lim (48)(29, 000)(0.933) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Unbraced Length, AISC Manual , pp. 3-96 to 3-131 with L b = 7 ft. M u 50 ksi 215.6 50 Required φb M n = = = 180 ft-kips C b F y 1.00 60 Select: W14×34, φ b M n = 193 ft-kips
(e) Correct the moment for the beam weight. 1 M u = 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.
h = 43.1 ≤ λ p = 3.76 tw
b f E = 82.7 ; = 7.41 ≤ λ p = 0.38 F y 2tf
E = 8.35 F y
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: M n = M p = F y Z x = (60)(54.6)/12 = 273 ft-kips For the limit state of lateral torsional buckling, AISC-F2.2:
L p = 1.76ry
rts =
E 1.53 = 1.76 F y 12
I y C w = S x
E Lr = 1.95rts 0.7F y
29, 000 = 4.93 ft 60
(23.3)(1070) = 1.80 in. 48.6
Jc S x ho
1.80 29, 000 = 1.95 12 0.7(60)
1+
0.7F y S x ho 2 1 + 6.76 E Jc
(0.569)(1) (48.6)(13.5)
= 13.9 ft
1+
0.7(60) (48.6)(13.5) 2 1 + 6.76 29, 000 (0.569)(1)
L p = 4.93 ft < [Lb = 7 ft] ≤ [Lr = 13.9 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 0.7(60)(48.6) 7 − 4.93 = (1.00) 273 − 273 − 12 13.9 − 4.93 = 249 ft-kips
Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (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
M u
φb M n
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.
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Section
I x 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 W16×40 with F y = 60 ksi steel.
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9.8. For the case assigned by the instructor, select the lightest W section to serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. Assume C b = 1.0. Case 3: M D = 0 ft-kips, M L = 240 ft-kips, L = 48 ft, F y = 50 ksi, and the deflection limit is L/300.
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): M u = 1.2M D + 1.6M L = 1.2(0) + 1.6(240) = 384 ft-kips (b) Obtain the minimum moment of inertia I x required using the service live load moment. L 48(12) ∆lim = = = 1.92 in. 300 300 5M LL2 (5)(240 × 12)(48 × 12)2 Min I x = = = 1, 790 in.4 48E ∆lim (48)(29, 000)(1.92) (c) The problem statement says to use C b = 1.0. (d) Select a beam using Table 3-10 Available Moment vs. Manual , pp. 3-96 to 3-131 with L b = 12 ft. M u 384.0 Required φb M n = = = 384 ft-kips C b 1.00 Select: W21×62, φ b M n = 440 ft-kips
Unbraced Length, AISC
(e) Correct the moment for the beam weight. 1 M u = 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.
h = 46.9 ≤ λ p = 3.76 tw
b f E = 90.6 ; = 6.7 ≤ λ p = 0.38 F y 2tf
E = 9.15 F y
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: M n = M p = F y Z x = (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, L p = 6.25 ft and Lr = 18.1 ft.
L p = 6.25 ft < [Lb = 12 ft] ≤ [Lr = 18.1 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (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
M u
φb M n
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
(g) Check the beams for deflection. Section
I x 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
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Use W24×68 with F y = 50 ksi steel.
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9.8. For the case assigned assigned by the instructo instructor, r, select the lightest lightest W section section to serve serve as a uniformly loaded library floor beam on a simply supported beam. Lateral support occurs at the ends and at L/4, L/2, and 3L/4. Given dead load moment does not include beam weight. weight. Assume C Assume C b = 1.0. Case 4: M D = 0 ft-kips, M ft-kips, M L = 240 ft-kips, L ft-kips, L = = 48 ft, F ft, F y = 65 ksi, and the deflection limit is L/300. L/300.
Use LRFD Design Method
(a) Obtain Obtain the factored loads (neglectin (neglecting g the beam weight weight): ): M u = 1.2M D + 1. 1 .6M L = 1.2(0) + 1. 1.6(240) = 384 ft-kips ft-kips (b) Obtain the minimum moment of inertia I inertia I x required using the service live load moment. L 48(12) ∆lim = = = 1.92 in. 300 300 5M LL2 (5)(240 × 12)(48 × 12)2 Min I Min I x = = = 1, 790 in.4 48E 48E ∆lim (48)(29, (48)(29, 000)(1. 000)(1.92) (c) The problem problem statemen statementt says says to use C use C b = 1.0. (d) (d) Sele Select ct a beam beam usin usingg Table able 33-10 10 Av Avai aila labl blee Mo Mome men nt vs. vs. Unbr Unbrac aced ed Leng Length th,, AISC Manual , pp. 3-96 to 3-131 with L b = 12 ft. M u 50 ksi 384..0 50 384 Required φb M n = = = 295 ft-kips C b F y 1.00 65 Select: W21×48, 48, φ φ b M n = 313 ft-kips
(e) Correct Correct the momen momentt for the beam weigh weight. t. 1 M u = 384. 384.0 + 1. 1 .2 (48/ (48/1000)(48)2 = 401 ft-kips 8
(f) Comp Comput utee the the desi design gn mo mome men nt stre streng ngth th usin usingg the the beam beam prope propert rtie iess from from the the AISC Manual Table Table 1-1, pp. 1-18 and 1-19.
h = 53. 53.6 ≤ λ p = 3.76 tw
E = 79. 79.4 ; λ p = 0.38 F y
b f E = 8.03 < = 9.47 ≤ F y 2tf
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λ p = 1.0
E = 21. 21.1 F y
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:
L p = 1.76r 76ry
rts =
E 1.66 = 1.76 F y 12
I y C w = S x
E Lr = 1.95r 95rts 0.7F y
29, 29, 000 = 5. 5 .14 ft 65
(38. (38.7)(3950) = 2.05 in. 93
Jc S x ho
2.05 29 29,, 000 = 1.95 12 0.7(65)
1+
0.7F y S x ho 2 1 + 6. 6.76 E J c
(0. (0.803)(1) (93)(20. (93)(20.2)
1+
0.7(65) (93)(20. (93)(20.2) 2 1 + 6. 6 .76 29, 29, 000 (0. (0.803)(1)
= 14. 14.3 ft
L p = 5.14 ft < [Lb = 12 ft] ≤ [Lr = 14. 14.3 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p Lr − L p 0. 0 .7(65)(93) 12 − 5.14 = (1. (1.00) 580 − 580 − 12 14. 14.3 − 5.14 = 409 ft-kips
M n = C = C b M p − M p − 0.7F y S x
For the limit state of compression flange local buckling, AISC-F3.2:
M n = M = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
0. 0 .7(65)(93) 9.47 − 8.03 = 580 − 580 − 12 21. 21.1 − 8.03 = 555 ft-kips Inelastic lateral torsional buckling controls! Calculate the design moment strength. φb M n = (0. (0.90)(409) = 369 ft-kips The W21×48 beam beam does does not not hav have suffic sufficie ien nt stre streng ngth th.. The The foll folloowing wing tabl tablee show showss the the moment corrected for the beam weight.
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Section
M u
φb M n
ft-kips
ft-kips
bf 2tf
h tw
W21×48 401 369† 9.47∗ 53.6 † W21×83 413 748 5 36.4 † W24×76 410 764 6.61 49 † W24×68 408 662 7.66 52 † W24×62 405 461 5.97 50.1 † W18×65 406 488 5.06 35.7 † Inelastic lateral torsional buckling controls ∗ Flange local buckling limit state must be checked (see below)
OKAY?
NG OK OK OK OK OK
(g) Check the beams for deflection. Section
I x in.4
LL Defl. in.
LL Defl. Limit in.
OKAY?
W21×48 W21×83 W24×76 W24×68 W24×62 W18×65
959 1830 2100 1830 1550 1070
3.58 1.88 1.63 1.88 2.21 3.21
1.92 1.92 1.92 1.92 1.92 1.92
NG OK OK OK NG NG
Use W24×68 with F y = 65 ksi steel.
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9.13, Part a. Select the lightest W section for the situation shown in the accompanying figure. The concentrated load W is 5 kips dead load and 15 kips live load. Assume lateral support is provided at the reactions and the concentrated loads. Use A992 steel.
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): The maximum moment is at the support. W u = 1.2(5) + 1.6(15) = 30.0 kips 1 M u = (beam wt.)(10)2 + 10W u = 10(30.0) = 300.0 ft-kips w/o beam 2 (b) Determine the C b factor, AISC-F1. The moment is uniform over the 30 ft span, so C b = 1.0. (c) Select a beam using Table 3-10 Available Moment vs. Manual , pp. 3-96 to 3-131 with L b = 30 ft. M u 300.0 Required φb M n = = = 300 ft-kips C b 1.00 Select: W14×74, φ b M n = 302 ft-kips
Unbraced Length, AISC
(d) Correct the moment for the beam weight. 1 M u = 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.
b f E = 90.6 ; = 6.41 ≤ λ p = 0.38 F y 2tf The web is compact and the flange is compact so use AISC-F2.
h = 25.4 ≤ λ p = 3.76 tw
E = 9.15 F y
For the limit state of yielding, AISC-F2.1: M n = M p = F y Z x = (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, L p = 8.76 ft and Lr = 31.0 ft.
L p = 8.76 ft < [Lb = 30 ft] ≤ [Lr = 31.0 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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 M n = (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
M u
φb M n
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
Elastic lateral torsional buckling controls
Use W18×76 with F y = 50 ksi steel.
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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 support is provided at the reactions and the concentrated loads. Use A992 steel.
Use LRFD Design Method
(a) Obtain the factored loads (neglecting the beam weight): W u = 1.2(5) = 6.00 kips (dead load only) W u = 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) M u = (beam wt.)(302 − 102 ) − (6.00)(10) + = 390 ft-kips 8 4 For Loading Case 2, live load is on the cantilevers but not on the 30 ft span. Maximum moment at the supports. 1 M u = (beam wt.)(10)2 + (30.0)(10) = 300 ft-kips 2 (b) Determine the C b factor, AISC-F1. For the 30 ft span in Case 1: 12.5M max C b = 2.5M max + 3M A + 4M B + 3M C 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) C b = = 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 L b = 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.
M u 300.0 = = 264 ft-kips C b 1.14 Select: W21×48, φ b M n = 268 ft-kips Required φb M n =
(d) Correct the moment for the beam weight. 1 M u = 300.0 + 1.2 (48/1000)(10)2 = 303 ft-kips 2
(e) Compute the design moment strength using the beam properties from the AISC Manual Table 1-1, pp. 1-18 and 1-19.
h = 53.6 ≤ λ p = 3.76 tw
λ p = 1.0
E = 90.6 ; λ p = 0.38 F y
b f E = 9.15 < = 9.47 ≤ F y 2tf
E = 24.1 F y
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, L p = 6.09 ft and Lr = 16.6 ft.
L p = 6.09 ft < [Lb = 15 ft] ≤ [Lr = 16.6 ft] inelastic lateral torsional buckling applies, AISC-F2.1(b). Lb − L p ≤ M p M n = C b M p − M p − 0.7F y S x Lr − L p 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:
M n = M p − M p − 0.7F y S x
λ − λ pf λrf − λ pf
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 M n = (0.90)(338) = 305 ft-kips The W21×48 beam has sufficient strength. The following table shows the moment corrected for the beam weight.
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Section
M u
φb M n
ft-kips
ft-kips
W21×48 303 305† W21×44 303 190‡ W18×46 303 204‡ W16×45 303 235† W14×43 303 228† † Inelastic lateral torsional buckling controls ‡ ∗
bf 2tf
h tw
OKAY?
9.47∗ 7.22 5.01 6.23 7.54
53.6 53.6 44.6 41.1 37.4
OK NG NG NG NG
Elastic lateral torsional buckling controls Flange local buckling limit state must be checked
(f) Check Case 1. The following table shows the moment corrected for the beam weight. Section
M u
φb M n
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
Use W21×48 with F y = 50 ksi steel.
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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 support occurs in the weak direction at mid-height. Use LRFD Design Method
(a) Obtain factored loads. P u = 1.2(100) + 1.6(150) = 360 kips 1 M nt = W u (22) = 5.5W u 4 (b) Check the section type.
;
h = 12.3 ≤ λ = 3.76 E = 90.6 p tw F y b f E 2tf = 4.96 ≤ λ p = 0.38 F y = 9.15
The web is compact and the flange is compact. (c) Column action with a compact web and flange, use AISC-E3 with K x = K y = 1.0; Lx = 22 ft; L y = 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 For
KL r
= 47.3
4.71
≤
E F y
= 113
π 2 E π 2 (29, 000) F e = = = 128 ksi (KL/r)2 (47.3)2 F cr =
0.658F y /F e
F y =
0.65850/128
(50) = 42.5 ksi
φc P n = φc F cr Ag = 0.90(42.5)(39.9) = 1, 520 kips (d) Beam action with a compact web and flange, use AISC-F2 with L b = 11 ft. Use AISC-F1 to find C b with R m = 1.0 for doubly symmetric members. 12.5M max C b = Rm ≤ 3.0 2.5M max + 3M A + 4M B + 3M C 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: M n = M p = F y Z x = (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, L p = 11.2 ft and L r = 63.3 ft.
[Lb = 11 ft] ≤ L p = 11.2 ft lateral torsional buckling does not apply, 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.
F2.1(a). Yielding controls! Calculate the design flexural strength. φb M n = (0.90)(892) = 802 ft-kips (e) Moment magnification. Obtain the slenderness ratio for the axis of bending. KL K x Lx Axis of bending = = 47.3 (from above) r rx π 2EI π 2 (29, 000)(1240) P e1 = = = 5, 090 kips 2 (KL)2 [(1.0)(22)(12)] According to AISC-C1.1b, for transverse loading, C m = 1.0. C m 1.0 B1 = = = 1.076 1 − P u /P e1 1 − 360/5, 090 (f) Use AISC-H1 to check the beam-column. P u 360 = = 0.236 ≥ 0.2 so use AISC Formula (H1-1a) omitting the y-axis φc P n 1, 520 bending term. P u 8 M ux 8 (1.076)(5.5)W u ≤ 1.0 + = 0.236 + φc P n 9 φb M nx 9 802 1.0 − 0.2362 W u = = 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.
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12.4. Investigate the adequacy of the given section of the accompanying figure. No joint translation can occur and external lateral support is provided at the ends only. Use LRFD Design Method
(a) Obtain the factored loads. M aDL = M cDL = 2.90 ft-kips M aLL = M cLL = 3.80 ft-kips M bDL = 2.00 ft-kips M bLL = 2.50 ft-kips M nta = 1.2(2.9) + 1.6(3.8) = 9.56 ft-kips M ntb = 1.2(2) + 1.6(2.5) = 6.4 ft-kips P u = 1.2(14) + 1.6(50) = 96.8 kips (b) Column action: (KL)y KL 1.0(15)12 = = = 111 r ry 1.62 φc F cr = 18.2 ksi φc P n = φc F cr Ag = 18.2(8.25) = 151 kips P u 96.8 = = 0.643 > 0.2 Use AISC Formula (H1-1a) φc P n 151
Largest
(c) Beam Action. Neither lateral torsional buckling nor web local buckling are limit states for weak axis bending. Check flange local buckling.
b f = 7.0 < λ p = 0.38 2tf
E = 9.15 ; section is compact. F y
φb M n = φ b M p = φb F y Z y = 0.9(50)(10.1)
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 Axis of bending = r
KL = 111 r y π 2 EI π 2 (29, 000)(21.7) P e1 = = = 191 kips (KL)2 [(1.0)(15)(12)]2 P u 96.8 C m = 1 − 0.4 = 1 − 0.4 = 0.798 P e1 191 C m 0.798 B1 = = = 1.615 1 − P u /P e1 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.
