Examen

Steel Structures by S. Vinnakota

Chapter 10

page 10-1

CHAPTER 10

P10.1.

Plot to scale the design bending strength M d versus the laterally unsupported length, L b, using C b of 1.0, 1.2, 1.75 and 2.3 for a W14×82 beam. Assume: (a) A992 steel, (b) A572 Grade 42 steel.

P10.2.

Plot to scale the design bending strength M d versus the laterally unsupported length, L b, using C b of 1.0, 1.2, 1.75 and 2.3 for a W24×76 beam. Assume: (a) A992 steel, (b) A572 Grade 65 steel.

P10.3.

Determine the design bending strength of a W18×35 beam segment of A992 steel, if: (a) L b = 4.0 ft (b) L b = 8.0 ft and C b = 1.0 (c) L b = 8.0 ft and C b = 1.1 (d) L b = 8.0 ft and C b = 1.5 (e) L b = 12 ft and C b = 1.0 (f) L b = 12 ft and C b = 1.3 (g) L b = 12 ft and C b = 2.3

Solution From LRFDM Table 5-3, observe that for a W18×35 of Fy = 50 ksi: M px = 249 ft-kips; L p = 4.31 ft; BF = 10.7 kips M rx = 173 ft-kips; L r = 11.5 ft a.

L b = 4 ft As L b = 4 ft < L p = 4.31 ft Md = Mn = M px = 249 ft-kips

(Ans.)

Alternatively, from Beam Selection Plots (LRFDM Table 5-5), for a W18×35 of F y = 50 ksi steel, L b = 4 ft (value to the left of the solid circle), read M d = M od = 249 ft-kips (Ans.) b.

L b = 8 ft and C b = 1.0 As L p = 4.31 ft < L b = 8 ft < L r = 11.5 ft and C b = 1.0: M d = M od = M px - BF (L b - L p) = 249 - 10.7 (8.0 - 4.31) = 210 ft-kips

(Ans.)

Alternatively, from Beam Selection Plots (LRFDM Table 5-5), for a W18×35 of F y = 50 ksi steel, C b = 1.0 and L b = 8 ft, read: M d = M do = 210 ft-kips (Ans.) c.

L b = 8 ft and C b = 1.1 As L p = 4.31 < L b = 8 ft < L r = 11.5 ft and C b > 1.0: M d = min [C b M od; M px] o As M d = 210 ft-kips as calculated in part (b): M d = min [1.1 × 210; 249] = min [231; 249] = 231 ft-kips

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

d.

e.

Chapter 10

L b = 8 ft and C b = 1.5 As L p = 4.31 ft < L b = 8 ft < L r = 11.5 ft and C b > 1.0: M d = min [C b M od; M px] = min [1.50 × 210; 249] = min [315; 249] = 249 ft-kips

page 10-2

(Ans.)

L b = 12 ft and C b = 1.0 As L b = 12 ft > L r = 11.5 ft and C b = 1.0: M d = M od = Mn = M cr The value of M od can conveniently be read from beam selection plots (LRFDM Table 5-5), corresponding to a W18×35 shape of Fy = 50 ksi steel, C b = 1.0 and L b = 12 ft, as 160 ftkips. So, M d = 160 ft-kips (Ans.)

f.

g.

L b = 12 ft and C b = 1.3 As L b = 12 ft > L r = 11.5 ft and C b > 1.0: M d = min [C b M od; M px] o As M d = 160 ft-kips from part (e) above: M d = min [1.3 × 160; 249] = min [208; 249] = 208 ft-kips

(Ans.)

L b = 12 ft and C b = 2.3 As L b = 12 ft > L r = 11.5 ft and C b > 1.0: M d = min [C b M od; M px] = min [2.30 × 160; 249] = min [368; 249] = 249 ft-kips

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

P10.4.

Chapter 10

page 10-3

Determine the design bending strength of a W14×68 beam segment of A992 steel, if: (a) L b = 8.0 ft (b) L b = 16.0 ft and C b = 1.0 (c) L b = 16.0 ft and C b = 1.2 (d) L b = 16.0 ft and C b = 1.67 (e) L b = 30 ft and C b = 1.0 (f) L b = 30 ft and C b = 1.3 (g) L b = 30 ft and C b = 2.27

Solution From LRFDM Table 5-3, observe that for a W14×68 of Fy = 50 ksi steel: M px = 431 ft-kips; L p = 8.69 ft; BF = 6.91 kips M rx = 309 ft-kips; L r = 26.4 ft a.

L b = 8 ft As L b = 8 ft < L p = 8.69 ft Md = Mn = M px = 431 ft-kips

(Ans.)

Alternatively, from beam selection plots (LRFDM Table 5-5), for a W14×68 of Fy = 50 ksi steel, L b = 8 ft (value to the left of the solid circle), read M d = M od = 431 ft-kips (Ans.) b.

L b = 16 ft and C b = 1.0 As L p = 8.69 ft < L b = 16 ft < L r = 26.4 ft and C b = 1.0: M d = M od = M px - BF (L b - L p) = 431 - 6.91 (16.0 - 8.69) = 380 ft-kips

(Ans.)

Alternatively, from beam selection plots (LRFDM Table 5-5), for a W14×68 of Fy = 50 ksi steel, C b = 1.0 and L b = 16 ft, read: M d = M do = 380 ft-kips (Ans.) c.

d.

e.

L b = 16 ft and C b = 1.2 As L p = 8.69 < L b = 16 ft < L r = 26.4 ft and C b > 1.0: M d = min [C b M od; M px] o As M d = 380 ft-kips as calculated in part (b): M d = min [1.2 × 380; 431] = min [456; 431] = 431 ft-kips

(Ans.)

L b = 16 ft and C b = 1.67 As L p = 8.69 ft < L b = 16 ft < L r = 26.4 ft and C b > 1.0: M d = min [C b M od; M px] = min [1.67 × 380; 431] = min [635; 431] = 431 ft-kips

(Ans.)

L b = 30 ft and C b = 1.1 As L b = 30 ft > L r = 26.4 ft and C b = 1.0: M d = M od = Mn = M cr The value of M od can conveniently be read from beam selection plots (LRFDM Table 5-5), corresponding to a W14×68 shape of Fy = 50 ksi steel, C b = 1.0 and L b = 30 ft, as 262 ftkips. So, M d = 262 ft-kips (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

f.

g.

Chapter 10

page 10-4

L b = 30 ft and C b = 1.3 As L b = 30 ft > L r = 26.4 ft and C b > 1.0: M d = min [C b M od; M px] o As M d = 262 ft-kips from part (e) above: M d = min [1.5 × 262; 431] = min [393; 431] = 393 ft-kips

(Ans.)

L b = 30 ft and C b = 2.27 As L b = 30 ft > L r = 26.4 ft and C b > 1.0: M d = min [C b M od; M px] = min [2.27 × 262; 431] = min [595; 431] =

(Ans.)

431 ft-kips

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Steel Structures by S. Vinnakota

P10.5.

Chapter 10

page 10-5

A W24×176 of A992 steel is used for a simple span of 36 ft. If the only dead load present is the weight of the beam, what is the largest service concentrated live load that can be placed at mid-span. Assume: (a) Lateral bracing is provided at the beam ends and at mid-span only; (b) Lateral bracing is provided at the beam ends only; (c) Recalculate, if, in addition, the service live load deflection of the beam is to be limited to L /480. Solution Simply supported beam. Span, L = 36 ft Section: W24×176 Material: A992 steel 6 Fy = 50 ksi Factored bending moment at midspan due to self-weight,

a.

Bracing provided at the beam ends and at mid-span only Unbraced length, L b = 18.0 ft From Table 10.4.3, moment modification factor C b = 1.67 From beam selection plots (LRFDM Table 5-5), for a W24×176 corresponding to L b = 18.0 ft, Also M px = 1920 ft-kips. So, the design bending strength,

So, the largest service concentrated load that can be placed on the beam, with bracing provided at the beam ends and at midspan is 131 kips. (Ans.) b.

Bracing provided at the beam ends only Unbraced length, L b = 36 ft From Table 10.4.3, C b = 1.32 From beam design plots for a W24×176 with L b = 36 ft, M do = 1240 ft-kips. Also M px = 1920 ft-kips. So, the design bending strength,

So, the largest service concentrated load that can be placed on the beam, with bracing provided at the beam end only, is 111 kips (Ans.) c.

Limit state of deflection Span, L = 36.0 ft Allowable deflection,

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Steel Structures by S. Vinnakota

Chapter 10

page 10-6

Section W24×176 6 Ix = 5680 in.4 Maximum service live load deflection,

The serviceability limit state for deflection requires

So, the largest service concentrated live load that can be placed on the beam is only 88.3 kips. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

P10.6.

Chapter 10

page 10-7

Determine the concentrated live load that a W18×50 of A992 steel can carry if the span is 32 ft with the load located at the midspan. Lateral bracing is provided: (a) at the beam ends and at quarter points only, (b) at the beam ends and at the midspan. Shear and deflection need not be checked. Solution Simply supported beam AB. Span, L = 32 ft Section: W18×50 Material: A992 steel 6 Fy = 50 ksi From LRFDM Table 5-3, observe that for a W18×50 of Fy = 50 ksi steel: M px = 379 ft-kips; L p = 5.83 ft; L r = 15.6 ft Loading: Single concentrated live load at center C = Q L Factored bending moment at midspan due to self-weight,

a.

Bracing at the beam ends (A, B) and at quarter points (D, C, and E) Segment DC is more critical than the segment AD. Unbraced length DC = L b = 8 ft > L p = 5.83 ft As the bending moment is linear over AC (assuming an ordinate of 8 units at C and using Eq. 10.4.14):

From beam design plots, for a W18×50 corresponding to L b = 8.0 ft, So, the design bending strength, As per LRFDS,

So, the largest service concentrated load that can be placed on the beam, with bracing provided at the beam ends and at midspan is 24.3 kips. (Ans.) b.

Bracing provided at the beam ends (A, B) and at mid-span (C) only Unbraced length, L b = 16.0 ft > L r = 15.6 ft From Table 10.4.3, moment modification factor C b = 1.67 From beam design plots, for a W18×50 corresponding to L b = 16.0 ft, So, the design bending strength,

So, the largest service concentrated load that can be placed on the beam, with bracing provided at the beam ends and at midspan is 29.0 kips. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

P10.7.

Chapter 10

page 10-8

A W30×108 of A992 steel is used for a simple span of 36 ft. If the only dead load present is the weight of the beam, what is the largest service uniformly distributed live load that can be placed on the beam. Lateral bracing is provided at the beam ends and quarter points only.

Solution Simply supported beam AB. Span, L = 36 ft Section: W30×108 Material: A992 steel 6 Fy = 50 ksi Factored bending moment at midspan due to self-weight,

Bracing provided at the beam ends (A, B) and at quarter points (D, C, E) only. Unbraced length, L b = 9.0 ft From Table 10.4.3d, moment modification factor C b = 1.06 From beam design plots, for a W30×108 corresponding to L b = 9.0 ft, Also M px = 1300 ft-kips. So, the design bending strength, As per LRFDS,

So, the largest service distributed live load that can be placed on the beam, with bracing provided at the beam ends and at the quarter points is 4.93 klf. (Ans.)

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Steel Structures by S. Vinnakota

P10.8.

Chapter 10

page 10-9

Verify the adequacy of the W16×50 beam in Problem 9.11 if continuous lateral bracing is provided over the length AD only. See Fig. P9.11 of the text book. Solution Simply supported beam, AB. Span, L = 40 ft. Material: Fy = 50 ksi steel. Factored loads: Concentrated load at D, 24 ft from A: 10 kips Uniformly distributed load over AD: 1.6 klf Check if the W16×45 section selected in P9.11 is still adequate if the continuous lateral bracing over DB is replaced by point bracing at points D and B only. Unbraced length, L b = 16 ft. Moment at B = M B = 0 Moment at D = M D 280 + 12.0 = 292 ft-kips Moment ratio, r M = 0 From Table 10.4.1, the moment reduction factor, C b = 1.67 corresponding to r M = 0. From LRFDM Table 5-3, for a W16×45, M px = 309 ft-kips ; L p = 5.55 ft; L r = 15.1 ft As L b > L r, design bending strength is obtained using beam selection plots. Corresponding to a W16×45 with C b = 1 and L b = 16 ft, we have

. So,

as the W16×45 section selected earlier is still adequate.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

P10.9.

Chapter 10

page 10-10

Check the adequacy of the beam in Problem P9.16, if lateral bracing is provided at the supports and at the concentrated load points only. See Fig. P9.16 of the text book. Solution Simply supported beam AB. Span, L = 26 ft. Concentrated loads, at center C, and at points D and E (CD = CE = 10 ft) Lateral bracing at points A, D, C, E, and B. Unbraced lengths: Segment, AD : L b1 = 3 ft Segment, DC : L b2 = 10 ft Bending moment at D = M D = 846 ft-kips Bending moment at C = M C = 1786 ft-kips Section: W30×148 of F y = 50 ksi steel For a W30×148, from LRFDM Table 5.3 BF = 38.5 kips; L r = 22.8 ft Factored bending moment at midspan due to self-weight,

Segment DC is critical

From Table 10.4.1, corresponding to r M = - 0.474, the moment reduction factor C b = 1.27 As L p < L b2 < L r,

Maximum moment at C including the contribution from self-weight is 1786 + 15 = 1800 ft-kips < M d = 1880 ft-kips. The W30×148 of F y = 50 ksi steel is still O.K.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-11

P10.10. A 34-ft-long W27×94 shape of A572 Grade 50 steel is used as a simply supported beam. It is subjected to a factored concentrated load of 90 kips at 12 ft from each support. In addition, the beam is subjected to a factored moment of 340 ft-kips at its left end (anticlockwise). Neglect self-weight of the beam in the calculations, and check if the beam is safe as per the LRFDS. Lateral bracing is provided at the end points and load points only. Solution a. Data Simply supported beam, AB. Span, L = 34 ft. Factored loads: Concentrated loads at D and E (with AD = BE = 12 ft) = 90 kips End moment, M A = - 340 ft-kips

M D = 80.0×12.0 = 960 ft-kips M C = 80.0×22.0 - 90.0×10.0 = 860 ft-kips M A = - 340 ft-kips From LRFDM Table 5-3, for a W27×94, M px = 1040 ft-kips; L p = 7.49 ft M rx = 729 ft-kips; L r = 19.9 ft Vn = 356 kips; BF = 25.2 kips b.

Segment DE Unbraced length, L b = 10 ft M D = 860 ft-kips; M E = 960 ft-kips; Moment ratio, From Table 10.4.1, moment reduction factor C b = 1.04 for r M = 0.896 Maximum moment, M max = 960 ft-kips As L p < L b < L r

As, c.

, segment 2 is O.K.

Segment EB Unbraced length, L b = 12 ft M E = 960 ft-kips; M B = 0 ft-kips; Moment ratio, r M = 0 Maximum moment, M max = 960 ft-kips

C b = 1.67

As M d = 1040 > M max = 960 ft-kips, segment 3 is O.K. too. Segment AD has the same unbraced length, smaller maximum moment, and more favorable moment distribution compared to Segment EB. So, Segment AD is adequate too. So, the W27×94 of A572 Grade 50 steel is adequate.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-12

P10.11. A W18×86 is used for a partially restrained beam, supporting the factored loads shown in Fig. P10.11. Lateral supports are provided at the ends and at load points only. Using A992 steel and neglecting the dead weight of the beam, find if the beam is overloaded. Shear and deflection need not be checked. See Fig. P10.11 of the text book. Solution a. Data Simply supported beam AB. Span, L = 34 ft. Factored loads: Concentrated loads at C and D (with AC = BD = 12 ft) = 70 kips End moments M A = M B = - 200 ft-kips Reaction, R A = 70 kips Moment (uniform over segment CD) = M C = 70 ×12 - 200 = 640 ft-kips Segment AC is subjected to a linearly varying moment having a maximum negative value of 200 ftkips at the left-end and a maximum positive value of 640 ft-kips at the right-end. Segment CD is subjected to a uniform moment of 640 ft-kips while Segment DB is similar to Segment AC. Section W18×86. Material: A992 steel 6 Fy = 50 ksi. From the LRFDM Table 5-3, for a W18×86 of A 992 steel: M px = 698 ft-kips; Lp = 9.29 ft; M rx = 498 ft-kips; Lr = 26.0 ft BF = 11.9 kips; Ix = 1530 in.4 Member braced at points A, C, D, and B. b.

Vn

= 238 kips

Limit state of flexural strength Segment AC L p = 9.29 ft < L b = 12.0 ft < L r = 26.0 ft As the segment is bent in reverse curvature under end moments only, we have

From Table 10.4.1 we obtain, corresponding to rM = 0.313, C b = 2.09. M d1 = min [C b { M px - BF (L b - L p)}; M px] = min [ 2.08{698 - 11.9 (12.0 - 9.29)}; 698 ] = 698 ft-kips M max = 640 < M d = 698 ft-kips O.K. Segment CD L p = 9.29 ft < L b = 10.0 ft < L r = 26.0 ft C b = 1.0, as the segment is subjected to uniform moment. M d = 1.0 [698.0 - 11.9 (10.0 - 9.29)] = 690 ft-kips M max = 640 < M d = 690 ft-kips O.K. Self-weight of the beam = 86 lb/ft say 0.09 klf Additional bending moment due to self-weight, if it were to be included:

M max = 640 + 12.2 =

652 < M d = 690 ft-kips

So, the W18×86 of A992 steel has adequate strength.

O.K. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-13

P10.12. Determine the concentrated live load that a W18×50 of A572 Grade 60 steel can carry if the span is 40 ft with the load located at the midspan. Lateral bracing is provided: (a) at the beam ends and at quarter points only, (b) at the beam ends and at the midspan only, and (c) at the beam ends only. Solution Simply supported beam, AB. Span, L = 40 ft Loading: Single concentrated live load, Q L at center C. Material: A572 Grade 60 steel. 6 Fy = 60 ksi Section W18×50; FL = F y - 10 = 50 ksi From LRFDM Table 1-1, Z x = 101 in.3; r y = 1.65 in.; S x = 88.9 in.3 X 1 = 1920 ksi; X 2 = 12400 × 10 -6 (1/ksi)2 b f /2tf = 6.57; h /t w = 45.2 Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2t f <

and h /tw <

, the section is compact.

Bending moment at C due to self-weight

a.

Bracing at the beam ends (A, B) and at quarter points (D, C, and E) Unbraced length DC = L b = 10 ft As the bending moment is linear over AC (assuming an ordinate of 8 units at C and using Eq. 10.4.14):

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

As L p = 5.32 ft

Chapter 10

<

L b = 10.0 ft <

page 10-14

L r = 13.6 ft,

If Q u is the factored load at C, we have

So, the maximum service live load at C is 27.7 kips. b.

(Ans.)

Lateral bracing at beam ends (A, B) and at midspan (C) As L b = 20 ft > L r = 13.6 ft

From Fig. 10.4.3f, C b = 1.67 M d = min [1.67 × 180; 455] = 301 ft-kips Also, as per LRFDS, So, the maximum service live load is 18.1 kips c.

(Ans.)

Lateral bracing at beam ends (A, B) only As L b = 40 ft > L r = 13.6 ft

From Fig. 10.4.3e, C b = 1.32 M d = min [1.32 × 70.1; 455] = 92.5 ft-kips Also, as per LRFDS, The largest service concentrated live load that can be placed on the beam is only 5.03 kips. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-15

P10.13. Repeat Problem P10.5 if the shape is of A572 Grade 60 steel. Solution Simply supported beam AB. Span, L = 36 ft. Section W24×176 Material: A572 Grade 60 steel. From LRFDM Table 1-1, for a W24×176, Zx = 511 in.3; Sx = X1 = 3130 ksi; X2 = Fy = 60 ksi; FL = b f /2tf = 4.81; h /t w =

450 in.3; ry 1610 × 10 -6 (1/ksi)2 60 - 10 = 50 ksi 28.7

=

3.04 in.

Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2t f < and h /tw < , the section is compact.

Factored bending moment at midspan due to self-weight,

a.

Bracing at the beam ends and at mid-span Unbraced length, L b = 18 ft From Figure 10.4.3f, C b = 1.67 As L p < L b < L r

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-16

and

So, the largest Q L is 157 kips b.

(Ans.)

Bracing at the beam ends only Unbraced length, L b = 36 ft From Fig. 10.4.3e, C b = 1.32 As L b > L r ,

Observe that this value is the same as that obtained from the beam selection plots for the section with F y = 50 ksi. It is so, because the nominal bending strength is the elastic lateral-torsional buckling strength of the beam, which is independent of the yield stress of the material.

Q L = 111 kips as in Problem P10.1b. c.

(Ans.)

Limit state of deflection This is independent of the yield stress of the material. So, the calculations are the same as those given in the solution to Problem P10.5c. The largest service concentrated live load that can be placed on the beam is 88.3 kips. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-17

P10.14. Select the lightest available W-section of A992 steel for a factored moment of 530 ft-kips, if (a) L b = 6 ft; (b) L b = 10 ft; and (c) L b = 19 ft. Assume C b = 1.0.

Solution a. L b = 6 ft; C b = 1.0 Moment under factored loads = M u = 530 ft-kips Enter beam selection plots (LRFDM Table 5-5) with M req = 530 ft-kips, C b = 1, L b = 6 ft and observe that a W21×62 with M od = 540 ft-kips is the lightest section (in continuous line) that provides the required bending strength. Check W21×62. From LRFDM Table 5-3: As L b < L p, So, select a W21×62 of A992 steel. b.

(Ans.)

L b = 10 ft; C b = 1.0 Moment under factored loads = M u = 530 ft-kips Enter beam selection plots (LRFDM Table 5-5) with M req = 530 ft-kips, C b = 1.0, L b = 10 ft and observe that a W21×68 with M d = 540 ft-kips is the lightest section (in continuous line) that provides the required bending strength. Check W21×68. From LRFDM Table 5-3: BF = 21.5 kips As L p < L b < L r and C b = 1.0, > M u = 530 ft-kips So, select a W21×68 of A992 steel.

c.

O.K. (Ans.)

L b = 19 ft; C b = 1.0 Moment under factored loads = M u = 530 ft-kips Enter beam selection plots (LRFDM Table 5-5) with M req = 530 ft-kips, C b = 1.0, L b = 19 ft and observe that a W24×84 with M d = 556 ft-kips is the lightest section (in continuous line) that provides the required bending strength. Check W24×84 From LRFDM Table 5-3:

From LRFDM Table 5-1, for a W24×84, Zx = 224 in.3; Sx = X1 = 1950 ksi; X2 = Fy = 50 ksi; FL =

196 in.3; ry 12200 × 10 -6 (1/ksi)2 50 - 10 = 40 ksi

=

1.95 in.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-18

As L b > L r and C b = 1.0,

> M u = 530 ft-kips So, select a W24×84 of A992 steel.

O.K. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-19

P10.15 (a)A 30-ft-long simply supported beam is loaded at the third-points of the span with concentrated loads of 10 kips (5 kips dead load and 5 kips live). Lateral bracing is provided at the supports and load points. The self-weight of the beam may be neglected. Select the lightest W-shape of A572 Grade 50 steel. (b) Redesign if the deflections are to be limited to L/360, under combined service dead and live loads. Solution Simply supported beam AB. Span, L = 30 ft Concentrated loads at points D and E (with AD = BE = 10.0 ft). Factored load, Q u = 1.2 × 5.00 + 1.6 × 5.00 = 14.0 kips M D = M E = 14.0 × 10.0 = 140 ft-kips A572 Grade 50 steel. Fy = 50 ksi Lateral bracing at A, D, E, and B. a.

Design for strength only Unbraced lengths: L b = 10 ft Segment DE with uniform moment over L b is more critical. Maximum moment, M max = 140 ft-kips Enter beam selection plots (LRFDM Table 5-5) with M req = 140 ft-kips, C b = 1, L b = 10 ft and observe that a W14×30 with M d = 149 ft-kips is the lightest section that provides the required bending strength. Check W14×30 BF = 6.05 kips; Vn = 101 kips As L p < L b < L r and C b = 1.0, > 140 ft-kips Vd = 101 kips > Vmax = 14 kips So, select a W14×30 of A527 Gr 50.

b.

O.K.

O.K. (Ans.)

Design for strength and serviceability For the W14×30 section selected, Ix = 288 in.4 For a simply supported beam under two symmetric concentrated loads, the maximum deflection occurs at midspan. From Case 9 of Beam Diagrams in LRFDM Table 5-17, we have:

Serviceability limit state of deflection 6

Entering LRFDM Table 5-2 with Ireq = 571 in.4, we observe that a W18×40 with Ix = 612 in.4 is the lightest section that satisfies this criteria. Bending and shear are adequate. So, select a W18×40 of A992 steel. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-20

P10.16. Select the lightest W-section to carry a uniform dead load of 0.3 klf and a live load of 0.6 klf on a simply supported span of 34 ft. Assume no deflection limitations. Use A572 Grade 50 steel. The beam is laterally supported at: (a) the beam ends and midspan only, (b) the beam ends only. Solution Simply supported beam AB. Span, L = 34 ft Applied, uniformly distributed loads: Dead load, q D = 0.30 klf; Live load, q L = 0.60 klf Factored load = 1.2 × 0.30 + 1.6 × 0.60 = 1.32 klf Bending moment at the center, under applied loads = (1.32 × 34.0 2 ) ÷ 8 = 191 ft-kips a.

Bracing at the ends and midspan only Unbraced length, L b = 17 ft From Fig. 10.4.3, C b = 1.30 Assume self-weight of the beam = 40 plf Additional bending moment at the center, due to the assumed self-weight = (1.2 ×0.04 × 34.0 2 ) ÷ 8 = 6.94 ft-kips So, required bending strength, M u = 191 + 6.94 = 198 ft-kips The limit state of flexure requires,

;

and

Enter beam selection plots with L b = 17 ft, M req = 152 ft-kips and select a W16×40, which has

So, select a W16×40 for the beam provided with bracing at the ends, and at midspan only. b.

(Ans.)

Bracing at the ends only Unbraced length, L b = 34 ft From Fig. 10.4.3, C b = 1.14 Assume self-weight of the beam = 60 plf Additional bending moment at the center, due to the assumed self-weight = (1.2 ×0.06 × 34.0 2 ) ÷ 8 = 10.4 ft-kips So, required bending strength, M u = 191 + 10.4 = 201 ft-kips The limit state of flexure requires,

;

and

Enter beam selection plots with L b = 34 ft, M req = 176 ft-kips and select a W12×58, which has

So, select a W12×58 for the beam provided with bracing at the ends only.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-21

P10.17. A 24-ft-long simply supported beam is loaded with a uniform dead load of 1.6 klf, including the beam weight and a uniform snow load of 2.5 klf. Lateral bracing is provided at beam ends and at 8 ft intervals. Select the lightest W-shape of A992 steel.

Solution Simply supported beam AB. Span, L = 24 ft Uniformly distributed loads: Dead load, q D = 1.60 klf (includes beam weight) Snow load, q S = 2.50 klf Factored load = 1.2 × 1.60 + 1.6 × 2.50 = 5.92 klf Bending moment at the center, under factored loads, M u = (5.92 × 24.0 2 ) ÷ 8 = 426 ft-kips Shear at the supports, Vu = (5.92 × 24.0 ) ÷ 2 = 71.0 kips Bracing provided at the ends and third-points only Unbraced length, L b = 8 ft From Fig. 10.4.3c, C b = 1.01 The limit state of flexure requires,

;

and

Enter beam selection plots with L b = 8 ft, M req = 422 ft-kips and select a W21×57, which has

O.K. V d = 231 kips > 71.0 kips O.K. So, select a W21×57 for the given beam, provided with bracing at the ends, and at third-points only. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-22

P10.18. Repeat Problem P9.23 if lateral bracing is provided at beam ends and point D only.

Solution Simply supported beam, AB. Span, L = 28 ft Material: Fy = 50 ksi steel Factored loads: Concentrated load at D, 16 ft from A: Q uD = 40 kips Uniformly distributed load over AD: 1.6 klf Bending moment at a point z from A, for 0 ft

z

16 ft is:

The bending moments at quarter points 1, 2, 3 of the segment AD and the maximum moment (at D) are: M1 = 129 ft-kips; M2 = 232 ft-kips M3 = 310 ft-kips; MD = 362 ft-kips

Assume self-weight of the beam = 50 plf = 0.05 klf Additional moment due to self-weight, Additional shear due to self weight, Required bending strength, M u = 362 + 5.88 = 368 ft-kips (conservatively) Required shear strength, Vu = 35.4 + 0.84 = 36.2 kips So, Entering beam selection plots (LRFDM Table 5-5) with L b = 16 ft and M oueq = 256 ft-kips, we observe that a W18×50 with is a possible solution. Check: M d = min [1.44 × 256; 379 ] = 369 ft-kips > M max = 368 ft-kips

O.K.

Segment DB (with L b = 12 ft; C b = 1.67) is less critical than AD. As the self-weight 50 plf is the same as the assumed value of 50 plf, no redesign is necessary. So, select a W18×50.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-23

P10.19. A simple beam supporting elevator machinery carries a concentrated load of 6 kips at mid-span. It is laterally unsupported and has an effective span of 14 ft. The beam deflection is to be limited to L /1000. Allowance is to be made for a 10 % impact, which is to be included in computing the deflection. Select a suitable A992 steel beam. Solution Simply supported beam AB. Span, L = 14 ft Loading: Concentrated live load at center C = Q L = 6 kips Allowance for impact: 10 % Service live load, Q L+I = 6.60 kips Allowable deflection, As the deflection limitation is quite stringent, we will select a section to satisfy the deflection criteria first, and then check the section selected for strength.

So, Ix

134 in.4

Enter LRFDM Table 5-2 with Ireq = 134 in.4 and select a W12×22 for which: Ix = 156 in.4 > 134 in.4 O.K. M px = 110 ft-kips; L p = 3.00 ft; L r = 8.42 ft; Vn = 86.3 kips From LRFDM Table 1-1, S x = 25.4 in.3; ry = 0.848 in. X 1 = 2170 ksi; X 2 = 8460 × 10 -6 (1/ksi)2 As L b = 14 ft > L r = 8.42 ft

From Fig. 10.4.3e, C b = 1.32 M d = min [1.32 × 36.3; 110] = 47.9 ft-kips Maximum bending moment,

Maximum shear,

So, select a W14×22 of A992 steel.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-24

P10.20. Redesign the beam in Problem P9.18, if lateral bracing is provided at: (a) intervals of 12 feet, (b) the ends and mid-span only. Solution Simply supported beam AB. Span, L = 60 ft Applied, uniformly distributed loads: Dead load, q D = 0.50 klf; Live load, q L = 1.0 klf Factored load = 1.2 × 0.50 + 1.6 × 1.0 = 2.20 klf Bending moment at the center, under applied loads = (2.20 × 60.0 2 ) ÷ 8 = 990 ft-kips a.

Bracing at 12 ft intervals Unbraced length, L b = 12 ft As the segments are of equal length, the middle segment, where the maximum bending moment occurs, is the critical segment. Over this segment the moment is essentially uniform. So, we will conservatively assume C b = 1.0. Assume self-weight of the beam = 100 plf Additional bending moment at the center, due to the assumed self-weight = (1.2 ×0.100 × 60.0 2 ) ÷ 8 = 54.0 ft-kips So, required bending strength, M u = 990 + 54.0 = 1040 ft-kips Enter the beam selection plots with L b = 12 ft, and M req = 1040 ft-kips and select a W24×104 shape (in continuous line) for which From LRFDM Table 5-3, for a W24×104 M px = 1080 ft-kips; L p = 10.3 ft BF = 18.6 kips; L r = 26.9 ft As L p < L b < L r and C b = 1.0,

Select a W24×104 for the beam provided with bracing at 12 ft intervals.

(Ans.)

b. Bracing at the ends and midspan only Unbraced length, L b = 30 ft From Fig. 10.4.3, C b = 1.30 Assume self-weight of the beam = 130 plf Additional bending moment at the center, due to the assumed self-weight = (1.2 ×0.130 × 60.0 2 ) ÷ 8 = 70.2 ft-kips So, required bending strength, M u = 990 + 70.2 = 1060 ft-kips The limit state of flexure requires,

,

and

Enter beam selection plots with L b = 30 ft, M req = 815 ft-kips and select a W24×131, which has

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-25

So, select a W24×131 for the beam provided with bracing at the ends, and at midspan only.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-26

P10.21. Redesign the beam in Problem P9.20, if lateral bracing is provided at the support points A and B and at the load point C only. See Fig. P9.20 of the text book. Solution Simply supported beam AB. Span, L = 48 ft. Factored loads: At center C = 100 kips At points D and E (with CD = CE = 12 ft) = 100 kips Bracing at A, B, and C. Reaction, R A = 150 kips The maximum bending moment occurs at midspan and M C = 150×24 - 100×12 = 2400 ft-kips Unbraced length, L b = 24 ft The moments at quarter points 1, 2, 3 of the segment AC are calculated next. We have: M 1 = 150 × 6 = 900 ft-kips ; M 2 = 150 × 12 = 1800 ft-kips M 3 = 150 × 18 - 100 × 6 = 2100 ft-kips ; M max = 2400 ft-kips

Assume self weight of the beam = 200 plf Additional bending moment at the center, due to the assumed self-weight = (1.2 × 0.200 × 48.0 2 ) ÷ 8 = 69.1 ft-kips So, required bending strength, M u = 2400 + 69.1 = 2470 ft-kips The limit state of flexure requires,

Entering the beam selection plots (LRFDM Table 5-5) with L b = 24 ft; C b = 1, and we observe that the W30×191 with

is still a possible

solution. The design bending strength, is:

O.K. So, the W30×191 selected earlier (for continuous bracing) is adequate for the present case where bracing is provided at the supports and at the midspan only. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-27

P10.22. A simple beam carries a uniformly distributed dead load of 2.1 klf plus its self-weight, and three 12 kip concentrated live loads at the quarter points of 40-ft span. Determine the lightest W-section to carry the loads, if lateral bracing is provided at the supports and the load points only. Use A992 steel. Solution Span, L = 40 ft Bracing at supports A, B, and quarter points D, C, and E. Unbraced lengths: AD = DC = L b = 10 ft Assume the self-weight of the beam to be 100 plf. The factored distributed load, q u = 1.2(2.10 + 0.100) = 2.64 klf The factored concentrated load, at quarter points D, C, E ; Q u = 1.6×12.0 = 19.2 kips Total load on the beam = 2.64 × 40.0 + 3×19.2 = 163 kips Bending moment at a point z from A, for 10 ft z 20 ft is:

Segment DC is more critical than Segment AD. The bending moments at quarter points 1, 2, 3 of the segment DC and the maximum moment (at C) are: M1 = M3 =

768 ft-kips; 880 ft-kips;

M2 = MC =

831 ft-kips 912 ft-kips

So

Entering beam selection plots (LRFDM Table 5-5) with L b = 10 ft and M oueq = 844 ft-kips, we observe that a W27×84 with is a possible solution. Check: M d = min [1.08 × 856; 915 ] = 915 ft-kips > M max = 912 ft-kips

O.K.

As the self-weight 84 plf is less than the assumed value of 100 plf, no redesign is necessary. So, select a W27×84.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-28

P10.23. Select the lightest W-shape of A572 Grade 50 steel for a simple beam of span 36 ft. The beam is subjected to a uniformly distributed dead load of 1 klf acting over the entire length. In addition, the beam receives concentrated loads (6 kips dead and 14 kips live, respectively) from purlins located at quarter points of the beam. Lateral bracing is provided at beam ends and quarter points only. The loads given are service loads. Solution Span, L = 36 ft Assume the self-weight of the beam = 80 plf Uniformly distributed dead load, q D = 1.08 klf Concentrated loads at quarter points D, C, and E: Dead load, Q D = 6 kips Live load, Q L = 14 kips Factored uniformly distributed load, q u = 1.2×1.08 = 1.30 klf Factored concentrated load, Q u = 1.2×6 + 1.6×14 = 29.6 kips Reaction, Maximum moment, Bracing provided at supports A, B, and quarter points D, C, and E. Unbraced length, L b = 9 ft Segment DC is more critical than the segment AD. For uniformly distributed load acting alone (Case d in Fig. 10.4.3) C b = 1.30. For the three equal concentrated loads (Case j in Fig. 10.4.3) C b = 1.11. In the present problem where both these two sets of loads are acting simultaneously, we assume C b = 1.08. So,

Enter beam selection plots with L b = 9 ft, M req = 688 ft-kips and select a W24×76 with M do = 706 ft-kips, M px = 750 ft-kips and L p = 6.8 ft.

Self-weight = 76 plf So, provide a W24×76.

<

80 plf assumed.

O.K. (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-29

P10.24. A W27×146 of A992 steel has four bolt holes in each flange for f-in.-dia. bolts. (a) Determine the reduced elastic section modulus S x. (b) Determine the reduced plastic section modulus Z x. What is the percentage reduction in Z x? Solution A992 steel: From LRFDM bf = d = =

Fy = 50 ksi; Fu = 65 ksi Table 1-1 for a W27×146: 14.0 in.; t f = 0.975 in.; Ix = 5660 in.4; 27.4 in.; c = d /2 = 13.7 in.; Z x = 464 in.3 (d - tf )/2 = (27.4 - 0.975) /2 = 13.21 in.

S x = 414 in.3

Bolts: Diameter, d = f in. Assume standard punched holes; d e = d + c = 1.00 in. A fg = b f tf = 14.0 × 0.975 = 13.65 in.2 A fn = A fg - n d e t f = 13.65 - 4 × 1.00 × 0.975

=

9.75 in.2

Design yield strength of the tension flange, T dfg = 0.9 A fg Fy = 0.9 × 13.65 × 50 = 614 kips Design fracture strength of the tension flange, T dfn = 0.75 A fn Fu = 0.75 × 9.75 × 65 = 475 kips As the fracture strength is less than the yield strength, the flexural properties must be based on effective tension flange area given by Eq. 10.5.1,

Effective moment of inertia,

Effective section modulus,

(Ans.)

Effective plastic section modulus, (Ans.) Reduction in plastic section modulus, due to the presence of holes in the tension flange

=

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-30

P10.25. A W24×104 of A572 Grade 60 steel has two bolt holes in each flange for 1 c-in.-dia. bolts. (a) Determine the reduced elastic section modulus S x. (b) Determine the reduced plastic section modulus Z x. What is the percentage reduction in Z x? Solution A572 Grade 60 steel: Fy = 60 ksi; F u = 75 ksi From LRFDM Table 1-1 for a W24×104: b f = 12.8 in.; t f = 0.750 in.; Ix = 3100 in.4; d = 24.1 in.; c = d /2 = 12.05 in.; Z x = 289 in.3 = (d - tf )/2 = (24.1 - 0.750) /2 = 11.68 in.

S x = 258 in.3

Bolts: Diameter, d = 1 c in. Assume standard punched holes; d e = d + c = 1.25 in. A fg = b f tf = 12.8 × 0.750 = 9.60 in.2 A fn = A fg - n d e t f = 9.60 - 2 × 1.25 × 0.750

=

7.725 in.2

Design yield strength of the tension flange, T dfg = 0.9 A fg Fy = 0.9 × 9.60 × 60 = 518 kips Design fracture strength of the tension flange, T dfn = 0.75 A fn Fu = 0.75 × 7.725 × 75 = 435 kips As the fracture strength is less than the yield strength, the flexural properties must be based on effective tension flange area given by Eq. 10.5.1,

Effective moment of inertia,

Effective section modulus,

(Ans.)

Effective plastic section modulus, (Ans.) Reduction in plastic section modulus, due to the presence of holes in the tension flange

=

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-31

P10.26. Select the lightest W-section to support a factored uniformly distributed load of 2.6 klf over a simple span of 25 ft. Assume A572 Grade 60 steel and continuous lateral support. Make all checks. The given load does not include self-weight of the beam. Solution Simply supported beam AB. Span, L = 25 ft Assume self-weight = 40 plf Factored load, q u = 2.6 + 0.04 = 2.64 klf Required strengths:

Yield stress, F y = 60 ksi From LRFDM Table 5-3, select W14×30 in bold face (lightest). From LRFDM Table 1-1, we have

d = 13.8 in.;

tw = 0.270 in.

For the 60 ksi rolled steel section (from Table 9.5.1):

So, the beam bending strength is to be reduced for flange local buckling.

Design shear strength, = 121 kips > 33.0 kips O.K. So, select a W14×30 of A572 Grade 60 steel

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-32

P10.27. A W8×10 beam with a yield stress of 50 ksi has an unbraced segment length of 6 ft and is subjected to a uniform bending moment. Calculate the values of L p, M p, L pN, and M pN. Also determine the maximum uniform factored bending moment that may be applied to the beam segment. Solution Length of unbraced segment, L b = 6.00 ft. Loading: Uniform moment over unbraced length. C b = 1.0 Section W8×10. Material: A992 steel. Fy = 50 ksi From LRFDM Table 1-1, for a W8×10, Zx = 8.87 in.3; Sx = 7.81 in.3; ry X1 = 1820 ksi; X2 = 15800 (1/ksi)2 Fy = 50 ksi; FL = 50 - 10 = 40 ksi b f /2tf = 9.61; h /t w = 40.5

=

0.841 in.

Limiting b/t ratios for plate buckling (see Table 9.5.1): Compression flange:

Web in flexure: Web in shear: Also, (Ans.)

As L p < L b < L r and C b = 1.0,

As < b f / 2tf < the section is non-compact, and inelastic local buckling of flange plate may limit the design bending strength.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-33

Design bending strength, (Ans.) (Ans.) (Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-34

P10.28. A M10×8 beam is simply supported over a span of 12 ft. Adequate lateral support is provided to the beam, which supports a uniformly distributed factored load, including its own weight, of 0.5 klf and a factored live load consisting of two factored concentrated loads of 5 kips each acting at the third points of the span. Assume A572 Grade 50 steel, and check the adequacy of the beam for bending and shear. Solution Simply supported beam AB. Span, L = 12 ft. Factored distributed load, q u = 0.5 klf Factored concentrated loads, Q u = 5 kips acting at the third-points (D and E). Maximum bending moment occurs at the center, C. Maximum shear, Section M10×8 Material: A572 Grade 50 steel. From LRFDM Table 1-2, for a M10×8, Zx = 8.20 in.3; Sx = X1 = 1330 ksi; X2 = Fy = 50 ksi; FL = b f /2tf = 7.39; h /t w =

6.95 in.3; ry 73200 (1/ksi)2 50 - 10 = 40 ksi 65.0

=

0.500 in.

Limiting b/t ratios for plate buckling (see Table 9.5.1): Compression flange: Web in flexure: Web in shear: As b f / 2t f <

and h /tw <

, the section is compact.

As the member is provided with adequate lateral bracing, the design bending strength, O.K. As h /tw >

, shear buckling of web occurs.

Web in shear: As h /tw < , shear buckling of web occurs in the inelastic domain and the shear strength is given by LRFDS Eq. F2-2.

So, the M10x8 is adequate for the given loading.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-35

P10.29. A propped cantilever, with a span L = 30 ft, is subjected to a single concentrated load, Q, at mid span. The load consists of 50 kips dead load and 40 kips snow load. The self-weight of the beam may be neglected. Assume full lateral support, and select the lightest W24-shape. Use elastic analysis with redistribution of moments taken into account.

Solution Beam AB, simply supported at A and fixed at B. Span AB = L = 30 ft. Full lateral support is provided. Loading: Single concentrated load, Q, at midspan, C. Dead load, Q D = 50 kips; Snow load, Q S = 40 kips Factored load, Q u = 1.2×50 + 1.6×40 = 124 kips From Case 13 in LRFDM Table 5-17: Shears, Moments and Deflections, the internal moments from an elastic analysis of the propped cantilever are obtained as: MA =

0.0 ft-kips

LRFDS Section A5.1 permits the beam to be designed for a reduced moment. Thus,

Required bending strength, M req = 628 ft-kips Enter beam selection tables (LRFDM Table 5-3) with M req = 628 ft-kips and observe that a W24×68 shape is the lightest section (in bold face) with:

So, select a W24×68 of A992 steel.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-36

P10.30. Determine if the section selected in Problem 10.29 is satisfactory, if the beam is laterally supported at the supports and at the point of application of the concentrated load.

Solution The adjusted moment diagram consists of M A = 0.0; M C = + 616 ft-kips; and M B = - 628 ft-kips. Section selected in P10.29 is a W24×68 of A992 steel. From LRFDM Table 5-3:

Check if the section is adequate, if lateral bracing is provided at points A, C, and B only. For segment CB, check if L b < L pd, where the unbraced length is given by LRFDS Eq. F1-17 as:

As L b = 15 ft < L pd = 17.6 ft and M max = 628 < 664 ft-kips Segment CB is O.K. As the last plastic hinge forms at C, flexural design strength of Segment AC shall be determined according to Section F1.2. For segment AC, M max = 616 ft-kips; L b = 15.0 ft; rM = 0.0; C b = 1.67 From beam design plots, for a W24×68 corresponding to L b = 15.0 ft, Also

M px = 664 ft-kips.

So, the design bending strength,

So, the W24×68 of A992 steel selected earlier is still adequate.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-37

P10.31. A fixed ended beam, with a span L = 30 ft, is subjected to a single concentrated load, Q, at a distance a = 12 ft from the left support. The load Q consists of 40 kip dead load and 80 kip live load. The self-weight of the section may be neglected. Assume full lateral support and select the lightest W24-shape. Use elastic analysis with redistribution of moments taken into account.

Solution Beam AB, fixed at A and B. Span AB = L = 30 ft. Full lateral support is provided. Loading: Single concentrated load, Q, at point D (with AD = a = 12 ft, and DB = b = 18 ft). Dead load, Q D = 40 kips; Snow load, Q L = 80 kips Factored load, Q u = 1.2×40 + 1.6×80 = 176 kips From Case 13 in LRFDM Table 5-17: Shears, Moments and Deflections, the internal moments from an elastic analysis of the propped cantilever are obtained as:

LRFDS Section A5.1 permits the beam to be designed for a reduced moment. Thus,

Required bending strength, M req = 684 ft-kips Enter beam selection tables (LRFDM Table 5-3) with M req = 684 ft-kips and observe that a W24×76 shape is the lightest W24-section (in bold face) with:

So, select a W24×76 of A992 steel.

(Ans.)

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-38

P10.32. Determine if the section selected in Problem 10.31 is satisfactory, if the beam is laterally supported at the supports and at the point of application of the concentrated load.

Solution The adjusted moment diagram consists of M A = - 684 ft-kips; M D = + 671 ft-kips; and M B = - 456 ft-kips. Section selected in P10.29 is a W24×76 of A992 steel. From LRFDM Table 5-3:

Check if the section is adequate, if lateral bracing is provided at points A, D, and B only. For segment AD, check if L b < L pd, where the unbraced length is given by LRFDS Eq. F1-17 as:

As L b = 12 ft < L pd = 18.1 ft and M max = 684 < 750 ft-kips Segment AD is O.K. As the last plastic hinge forms at B, flexural design strength of Segment DB shall be determined according to Section F1.2. For segment DB, M max = 671 ft-kips; L b = 18.0 ft; rM = + 0.680; C b = 2.21 From beam design plots, for a W24×76 corresponding to L b = 18.0 ft, Also

M px = 750 ft-kips.

So, the design bending strength,

So, the W24×76 of A992 steel selected earlier is still adequate.

(Ans.)

P10.33. Select the lightest W-section required for a two span continuous beam ABC with L 1 = L 2 = L = 32 ft. The beam is hinged at A and roller supported at B and C. It supports a factored distributed load, including its self-weight, of 12 klf. Assume full lateral support to the beam. Use elastic analysis with redistribution of moments taken into account. Check for shear.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Steel Structures by S. Vinnakota

Chapter 10

page 10-39

P10.34. Select the lightest W-section required for a three span continuous beam ABCD with L 1 = L 2 = L 3 = L = 24 ft. The beam is hinged at A and roller supported at B, C and D. It supports a factored distributed load, including its self-weight, of 10 klf. Assume full lateral support to the beam. Use elastic analysis with redistribution of moments taken into account.. Check for shear.

Solution Beam ABCD. Hinged at A and roller supported at B, C, and D. Spans AB = BC = CD = L = 24 ft. Full lateral support is provided. Loading: Factored uniformly distributed load, q u = 10 klf. All spans loaded. From Case 39 in LRFDM Table 5-17: Shears, Moments and Deflections, the internal moments from an elastic analysis of the three span continuous beam are obtained as:

Also,

The point E, where the positive moment in the span AB is a maximum, is located at a distance, z o = 0.400 L = 9.6 ft from A. LRFDS Section A5.1 permits the beam to be designed for a reduced moment. Thus,

Required bending strength, M req = 518 ft-kips Enter beam selection tables (LRFDM Table 5-3) with M req = 518 ft-kips and observe that a W21×62 shape is the lightest W-section (in bold face) with:

So, select a W21×62 of A992 steel.

(Ans.)

Note: Other load combinations (partial loading) have to be checked before the final selection of the member.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.