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Steel Structures by S. Vinnakota

Chapter 11

page 11-1

CHAPTER 11 P11.1.

A W12×40 of A992 steel is used as a 13-ft column in a braced frame to carry an axial factored load of 160 kips. Will it be adequate to carry moments about the strong axis of 100 ft-kips at each end, bending the member in single curvature? For buckling about the minor axis consider the column to be pinned at both ends and provided with a brace at mid height. Solution Factored axial load on the column, Pu = 160 kips Length of column, L = 13.0 ft Column part of braced frame. Member pinned at both ends, with a brace at mid height. L x = 13.0 ft; L y = 6.50 ft K x L x = 1.0 × 13.0 = 13.0 ft; K y L y = 1.0 × 6.50 = 6.50 ft W12×40 of A992 steel. From LRFDM Table 4-2, for a W12×40. So, Also from this table, for a W12×40 with KL = 6.50 ft, Pd = 442 kips and Pex (KL)2 = 8790 × 10 4. Axial load ratio,

. Use LRFDS Eq. H1-1a.

Bending is about major axis only 6 M u y* = 0 Structure is braced in yy plane Column is subjected to symmetric single curvature bending moments of 100 ft-kips.

Unbraced length, L b = 6.5 ft. From beam selection tables (LRFDM Table 5-3), corresponding to a W12×40, L p = 6.85 ft > L b , and The LHS of the interaction equation gives

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 11

So, the W12×40 of A992 steel is adequate.

page 11-2

(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

P11.2.

Chapter 11

page 11-3

A W14×61 is used as a 15 ft long column in a braced frame to carry an axial factored load of 200 kips. Determine the maximum moment that may be applied about the strong axis on the upper end when the lower end is hinged.

Solution Factored axial load, Pu = 200 kips. Length of column, L = 15 ft. W14×61 of A992 steel. Column part of braced frame. Lower end hinged. Assume (conservatively), K x = 1.0 and K y = 1.0. K x L x = 1.0 × 15.0 = 15.0 ft; K y L y = 1.0 × 15.0 = 15.0 ft From LRFDM Table 4-2, corresponding to W14×61 and KL = K y L y = 15.0 ft, Pd = 513 kips. Also, for this section, Pex (KL)2 = 18300 × 10 4. Axial load ratio,

. So, use LRFDS Eq. H1.1a.

There are no minor axis moments, and the structure is braced in yy plane. So,

As,

from Table 10.4.1.

From beam selection plots (LRFDM Table 5-5), corresponding to a W14×61 with L b = 15 ft, Also, Nb M px = 383 ft-kips for this shape. So, the design bending strength of the member,

We have

So, the maximum moment that may be applied about the strong axis on the upper end = 263 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

P11.3.

Chapter 11

page 11-4

Figure 11-3 shows a 15-ft long W12×96 column with pinned ends. Two girders and a beam bring in 200 kips of axial load. A column from above delivers 90 kips at an eccentricity of 18-in. through a bracket connected to the web of the column. The structure is braced in both xx and yy planes. The loads are factored loads. Is the column adequate? See Fig. P11.3 of the text book. Solution Length of column, L = 15 ft Column part of braced frame in both xx and yy planes. So, K x # 1.0 and K y # 1.0 Pinned ends. So, K x = 1.0 and K y = 1.0. K x L x = 1.0 × 15.0 = 15.0 ft; K y L y = 1.0 × 15.0 = 15.0 ft From LRFDM Table 4-2, corresponding to W12×96 and KL = K y L y = 15.0 ft, Pd = 935 kips. Also, for this section, Pey (KL)2 = 7730 × 10 4. Factored axial load, Pu = 200 + 90 = 290 kips. Axial load ratio,

. So, use LRFDS Eq. H1.1a.

Structure is braced in xx and yy planes. So M lty and M ltx moments are zero. The 90 kips eccentric load produces moment about minor axis only. So Eccentricity, e = 18 in.

M nty = M 2y = 135 ft-kips From LRFDM Table 5-3 for a W12×96, Nb M py = 250 ft-kips We have

So, W12×96 of A992 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

P11.4.

Chapter 11

page 11-5

An interior column of a building structure has floor girders framing into it at top and bottom with moment resisting connections. It must carry a factored axial load of 800 kips, including the girder and beam reactions, and the self-weight of the column. Live load imbalance in checkerboard loading causes a potential maximum moment at the top and bottom of 180 ft-kips, as shown in Figure P11-4. K for the weak axis is 1.0, and K for the strong axis is estimated as 0.9. The column is 12 ft 6 in. long. Is a W12×106 of A992 steel adequate to carry the load? See Fig. P11.4 of the text book. Solution Factored axial load on the column, Pu = 800 kips Length of column, L = 12.5 ft K x = 0.9; K y = 1.0 K x L x = 0.9 × 12.5 = 11.3 ft; K y L y = 1.0 × 12.5 = 12.5 ft. W12×106 of A992 steel. From LRFDM Table 4-2, corresponding to KL = K y L y = 12.5 ft and a W12×106, Pd = 1115 kips. Also, P ex (KL)2 = 26700 × 10 4 Axial load ratio,

. Use LRFDS Eq. H1-1a.

Structure is braced in xx and yy planes Bending is about major axis only 6 M nty = 0 Column is subjected to symmetric single curvature bending moments of 180 ft-kips.

Unbraced length, L b = 12.5 ft. From beam selection plots (LRFDM Table 5-4), corresponding to a W12×106, L b = 12.5 ft, and C b = 1.0, The LHS of the interaction equation gives

So, the W12×106 of A992 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

P11.5.

Chapter 11

page 11-6

A W14×22 tension member of A572 Grade 42 steel is subjected to a factored tensile load T u = 60 kips and factored bending moments M ux of 28 ft-kips and M uy of 6 ft-kips. Is the member satisfactory if L b = 8.0 ft and C b = 1.67? Solution From LRFDM Table A = 6.49 in 2; Z y Z x = 33.2 in.3; S x X 1 = 1600 ksi; X 2

1-1, for a W14×22, we obtain: = 4.39 in 3; S y = 2.80 in.3 = 29.0 in.3; ry = 1.04 in. = 0.0278 (1/ksi) 2

For a W14×22 of A572 Grade 42 steel: Fy = 42 ksi; F L = Fy - Fr = 42 - 10 = 32 ksi

M dy = min [Nb M py ; 1.5 Nb M yy] = min [13.8 ; 1.5 × 8.82] = 13.2 ft-kips

BF = ( Nb M px - Nb M rx ) / (L r - L p) = (105 - 69.6) /(11.0 - 4.01) = 5.06 kips As (L p = 4.01 ft) < (L b = 8 ft) < (L r = 11.0 ft)

Design tensile strength of the member, corresponding to the limit state of yield on gross area T d = T d1 = Nt A g Fy = 0.9 × 6.49 × 42 = 245 kips . So, use Eq. 11.9.11a (LRFDS Eq. H1-1a)

= 0.24 + 0.64 = 0.88 So, the W14×22 of A572 Grade 42 is adequate.


L p, we have : M dx = min [C bM odx; Nb M px] = min [2.17{534}; 551] = 551 ft-kips 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 11

page 11-20

Next B 1x and B 2x are calculated. We have:

M ux* = B 1x M ntx + B 2x M ltx = 1.00 × 205 + 1.20 × 123 = 353 ft-kips

Substituting in the interaction Eq. 11.9.11a: Close. Accept.

Hence the W12×96 of A992 steel is acceptable.

(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 11

page 11-21

P11.13. Check the adequacy of a W14×48 as a 12-ft column in an unbraced frame. The member is subjected to a factored axial load of 200 kips, a no-translation moment M nt = 100 ft-kips, and a translation-permitted moment M lt = 108 ft-kips at the top of the member. One-half of these moments are applied at the other end of the member, bending in single-curvature. All moments are applied about the strong axis. Assume K x = 1.6 and K y = 0.9.

Solution a. Data Length of the column, L = 12 ft

Factored axial load, Pu = 200 kips No bending moments about the y-axis. 6 M uy* = 0.0 Maximum 1 st order moments about the x-axis Top: no-translation moment, M ntx = 100 ft-kips translation moment, M ltx = 108 ft-kips Bottom: no-translation moment = 50.0 ft-kips translation moment = 54.0 ft-kips b.

Check the section Enter column selection table for W14-shapes (LRFDM Table 4-2) with KL = K y L y = 10.8 ft; Fy = 50 ksi, to find that for a W14×48, Pd y = 428 kips, and Ix = 484 in.4.

indicating that minor axis buckling will control. So, P d = P d y = 428 kips So, use interaction Eq. 11.9.11a. The laterally unbraced length L b is 12 ft. From beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W14×48 of Fy = 50 ksi steel Nb M px = 294 ft-kips, L p = 6.75 ft and, for L b = 12 ft and C b = 1.0, M odx = 258 ft-kips. The ratio of (total) end-moments

So, from Table 10.4.1, C b = 1.25. As L b > L p, we have : M dx = min [C bM odx; Nb M px] = min [1.25{258}; 294] = 294 ft-kips Next B 1x and B 2x are calculated. We have:

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 11

page 11-22

M ux* = B 1x M ntx + B 2x M ltx = 1.00 × 100 + 1.08 × 108 = 217 ft-kips

Substituting in the interaction Eq. 11.9.11a:

Hence the W14×48 of A992 steel is not 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 11

page 11-23

P11.15. Select a W12-shape for a pin-ended column with a length of 16 ft to carry a factored load of 360 kips and end moments of 480 ft-kips producing symmetric single curvature bending about the major axis. The member is part of a braced frame in its xx and yy planes.

Solution a. Data Factored axial load, Pu = 360 kips For a column in braced frame, K = 1.0 for design (LFRDS Section C2.1) K x L x = K y L y = 1.0 × 16 ft = 16.0 ft There are no minor axis moments. Because the column is part of a braced frame, M ltx = 0, and Eq. 11.9.3 (LRFDS Eq. C1-2) reduces to Maximum 1 st order major axis moment, M ntx = 480 ft-kips b.

Preliminary selection Assume B 1x = 1.05 6 Method 1: As bending is about major axis only, we select a trail shape using, From Table 11.14.1, for a W12 with KL = 16 ft, m = 1.5, which when substituted in the above relation results in:

Entering the column load table for W12-shapes (LRFDM Table 4-2) with KL = 16 ft and Preq = 1120 kips, we observe that a W12×120 has a design strength Pd = 1140 kips. Also, Ix = 1070 in.4 Check the W12×120.

c.

Column effect For the W12×120 section selected, axial load ratio, So, use the interaction equation 11.9.11a ( LRFDS Eq. H1-1a), which for M uy = 0 reduces to

d.

Beam effect The laterally unbraced length, L b = L y = 16 ft. For a beam segment under uniform moment C b = 1.0 . From LRFDM Table 5-5, for a W12×120, Lp = 11.1 ft. Also, for L b = 16 ft > L p and C b = 1.0, we have,

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 11

page 11-24

The moment magnification factor is calculated next using Eq. 11.9.4. For a pin-ended member under uniform moment C m = 1.0 .

e.

Check the interaction formula

O.K.

So, the W12×120 of A992 steel selected is 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 11

page 11-25

P11.16. Select a W14-shape for a pin-ended column with a length of 24 ft to carry a factored load of 640 kips and end moments of 240 ft-kips producing symmetric single curvature bending about the major axis. The member is part of a braced frame in its xx and yy planes.

Solution a. Data Factored axial load, Pu = 640 kips For a column in braced frame, K = 1.0 for design (LFRDS Section C2.1) K x L x = K y L y = 1.0 × 24 ft = 24.0 ft There are no minor axis moments. Because the column is part of a braced frame, M ltx = 0, and Eq. 11.9.3 (LRFDS Eq. C1-2) reduces to Maximum 1 st order major axis moment, M ntx = 240 ft-kips b.

Preliminary selection Assume B 1x = 1.15 6 Method 1: As bending is about major axis only, we select a trail shape using, From Table 11.14.1, for a W12 with KL = 24 ft, m = 1.2, which when substituted in the above relation results in:

Entering the column load table for W12-shapes (LRFDM Table 4-2) with KL = 24 ft and Preq = 971 kips, we observe that a W14×120 has a design strength P d = 972 kips. Also, I x = 1380 in.4 and L p = 13.2 ft. Check the W14×120.

c.

Column effect For the W14×120 section selected, axial load ratio, So, use the interaction equation 11.9.11a ( LRFDS Eq. H1-1a), which for M uy = 0 reduces to

d.

Beam effect The laterally unbraced length, L b = L y = 16 ft. For a beam segment under uniform moment C b = 1.0 . From LRFDM Table 5-5, for a W14×120, for L b = 24 ft > L p and C b = 1.0, we have,

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 11

page 11-26

The moment magnification factor is calculated next using Eq. 11.9.4. For a pin-ended member under uniform moment C m = 1.0 .

e.

Check the interaction formula .

So, the W14×120 of A992 steel selected is 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 11

page 11-27

P11.17. Repeat Problem P11.16, if the member is provided with a brace at mid-length. [P11.16. Select a W14-shape for a pin-ended column with a length of 24 ft to carry a factored load of 640 kips and end moments of 240 ft-kips producing symmetric single curvature bending about the major axis. The member is part of a braced frame in its xx and yy planes. ] Solution Column length, L = 24 ft Pinned at both ends; brace at mid-height. Column part of braced frame in both planes. K x L x = 24.0 ft; K y L y = 12.0 ft; L b = 12.0 ft Factored axial load, Pu = 640 kips End moments produce, symmetric single curvature bending about x-axis. M ux = 240 ft-kips a.

Select a preliminary section Assume r x /r y . 1.66 6 (K x L x)y = 24.0 ÷ 1.66 . 14 ft. From Table 11.14.1, read m = 1.4. Assume B 1 x = 1.2 6 M ux* = B 1 x M ux = 1.2 × 240 = 288 ft-kips From Eq. 11.14.5a, Pu eq = Pu + m M ux* = 640 + 1.4 × 288 = 1040 kips Enter LRFDM Table 4-2 for W14-shapes, with KL = 14 ft and P req = 1040 kips, and select a W14×99 with Pd = 1060 kips.

b.

Check the selected section From LRFDM Table 4-2, for a W14×99: rx /ry = 1.66 6 (K x L x)y = 24.0 ÷ 1.66 = 14.5 ft > K y L y = 12.0 ft For KL = 14.5 ft, Pd = 1050 kips 6 Pu /Pd = 0.610 > 0.2 6 Use Eq. 11.9.11a For symmetric, single curvature no translation moments, C mx = 1.0 From LRFDM Table 4-2, for a W14×99: Pex (KL)2 = 31,800 × 10 - 4. As, (KL)ntx = 24.0 ft, ;

We obtain, M ux* = B 1x M ntx = 1.20 × 240 = 288 ft-kips From LRFDM Table 5-3, for a W14×99 and Fy = 50 ksi: Nb M px = 646 ft-kips; L p = 13.5 ft As L b = 12.0 ft


Preq = 1,120 kips So, the W14×109 is a potential trial shape. From Table 11.14.2, it is seen that this shape has a u value of 1.97 resulting in a revised value:

Reentering the column selection tables, it is seen that the W14×109 with an axial load capacity of 1270 kips appears to be an acceptable shape. So, let us try W14×109-shape. c.

Design strengths From Table 1-1 of the LRFDM, for a W14×109-shape:

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 11

page 11-35

A = 32.0 in.2; Ix = 1240 in.4; Iy = 447 in.4 Also, from column load tables, P d = 1,270 kips. Axial load ratio,

. Therefore, use Eq. 11.9.11a.

Unbraced length, L b = L = 14 ft From beam selection plots for W-shapes (LRFDM Table 5-5), for a W14×109-shape of Fy = 50 ksi steel, L b = 14 ft and C b = 1.0, M odx = 715 ft-kips As L p < L b < L r we have M dx = min [ C bM odx; NbM px] = min [l.25{715}; 720 ] = 720 ft-kips M dy = NbM py = 344 ft-kips d.

Second-order moments The magnification factors B 1x and B 2x are calculated from Eq. 11.9.4.

;

The second-order moments are therefore = B 1xM ntx = 1.00 × 240 = 240 ft-kips = B 1yM nty = 1.15 × 60.0 = 69.0 ft-kips The LHS of the interaction equation 11.9.11a ( LRFDS Eq. H1-1a) is:

So, select a W14×109 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 11

page 11-36

P11.23. Solve Problem P11.18, if the solution is limited to W12-shapes. [P11.18. Select a W14 section of A992 steel for a 14-ft-long beam-column in a framed structure braced in both directions. Factored axial load, Pu = 840 kips. The first order, symmetric, single curvature endmoments are: M ntx = 280 ft-kips and M nty = 40 ft-kips. Assume K x = K y = 1.0.] See Fig. 11.18 of the text book. Solution a. Data L = 14 ft; K x = K y = 1.0; K x L x = K y L y = 1.0 × 14.0 = 14.0 ft Pu = 840 kips; M ntx = 280 ft-kips; M nty = 40.0 ft-kips Braced frame, so M ltx = M lty = 0 and M ux* = B 1x M ntx; M uy* = B 1yM nty b.

Preliminary selection Assume B 1x = 1.05; B 1y = 1.20 M ux* = 1.05 × 280 = 294 ft-kips; M uy* = 1.20 × 40.0 = 48.0 ft-kips For a W12 column (Fy = 50 ksi steel) KL = 14 ft, m = 1.5 from Table 11.14.1. Assume u = 2.0. So:

From column selection tables (LRFDM Table 4-2), for KL = K y L y = 14 ft select: W12×152 with Pd = 1,550 kips > Preq = 1,425 kips So, the W12×152 is a potential trial shape. From Table 11.14.2, it is seen that this shape has a u value of 2.11 resulting in a revised value:

So, the W12×152 with an axial load capacity of 1550 kips appears to be an acceptable shape. For the W12×152-shape: Ix = 1430 in.4; Iy = 454 in.4 c.

Check W12×152 for strength As

use Eq. 11.9.11a (LRFDS Eq. H1-1a).

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 11

page 11-37

For symmetric, single curvature moments given, C mx = C my = 1.0.

M ux* = 1.06 × 280 = 297 ft-kips M uy* = 1.22 × 40 = 48.8 ft-kips The laterally unbraced length, L b = L y = 14 ft W12×152 of A992 steel. From LRFDM Table 5-3: Lp = 11.3 ft; L r = 62.1 ft Nb M px = 911 ft-kips; BF = 5.59 kips With uniform moment over the unbraced length, C b = 1.0 As L p = 11.3 ft < L b = 14.0 ft < L r = 62.1 ft M dx = M odx = Nb M px - BF (L b - L p)

Also, from LRFDM Table 5-3, NbM py = 410 ft-kips, for this shape. So, M dy = NbM py = 410 ft-kips Using Eq. 11.9.11a (or, LRFDS Eq. H1-1a):

So, select a W12×152 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 11

page 11-38

P11.24. Solve Example 11.15.4, if the solution is limited to W12-shapes. [EXAMPLE 11.15.4: A 30 ft long chord member of a truss is subjected, under factored loads, to an axial compressive force of 690 kips and a uniformly distributed factored load of 0.3 klf causing bending about its weak axis. Use A992 steel and select the appropriate W14 section. Include the influence of the self weight of the member which causes bending about the major axis.] See Fig. X11.15.4 of the text book. Solution a. Data Pu = 690 kips Assume self-weight of the beam = 190 plf The factored distributed load due to self weight,

. The

maximum major-axis bending moment at the center of the beam, Factored uniformly distributed transverse load, q ux = 0.30 klf The maximum, minor-axis bending moment at the center of the beam is:

b.

Preliminary selection As the member is quite flexible, we will assume B 1x = 1.2 and B 1y = 2.5, resulting in M *ux = B lx M ntx = 1.20 × 25.7 = 30.8 ft-kips M *uy = B ly M nty = 2.50 × 33.8 = 84.5 ft-kips For the biaxially bent beam-column:

For a W12 with KL = 30 ft and Fy = 50 ksi, m = 1.2 from Table 11.14.1. Also, for W12-sections heavier than a W12×152, u . 2.11 from Table 11.14.2. Hence, we have

In the column load tables for W-Shapes (LRFDM Table 4-2) for Fy = 50 ksi and KL = 30 ft, P d = 967 kips ( > = 941 kips) for a W12×190 selected. So, try a W12×190. c.

Check the section Axial strength of the column, P d = 967 kips The axial load ratio,

. So, use Eq. 11.9.11a.

Major axis bending From LRFDM Table 5-3, for a W12×190 and Fy = 50 ksi: Nb M px = 1170 ft-kips; Ix = 1890 in.4; Iy = 589 in.4 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

Nb M py = 523 ft-kips

Chapter 11

L p = 11.5 ft;

page 11-39

L r = 76.6 ft;

BF = 5.79

As L p = 11.5 ft < L b = 14.0 ft < L r = 76.6 ft M odx = Nb M px - BF (L b - L p) A simply supported beam under uniformly distributed load and laterally braced at the ends only, has a C b value of 1.14 from Table 5-1 of the LRFDM. As the unbraced length, L b, is between L p and L r, the design major axis bending strength of the member is: M dx = min[ C b M odx ; Nb M px ] = min[ 1.14 × 1063 ; 1170] = 1170 ft-kips For a beam-column simply supported at the ends and subjected to uniformly distributed lateral load, C m = 1.0. Also,

M *ux = B lx M ntx = 1.20 × 25.7 = 30.8 ft-kips Minor axis bending M dy = Nb M py = 523 ft-kips For a beam-column with uniformly distributed transverse load, C m = 1.0. We have:

d.

Check for limit state of strength

The W12×190 of A992 steel is therefore acceptable.

(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 11

page 11-40

P11.25. Solve Example 11.15.5, if the solution is limited to W12-shapes. [EXAMPLE 11.15.5: A W10-shape. is used as a continuous top chord member of the 128 ft long roof truss of an industrial building. The eight-panel Pratt truss is 12 ft deep at the center and 8 ft deep at the supporting columns. The top chord supports purlins at the panel points and midway between the panel points. Under factored gravity loads, each purlin transmits a load of 35 kips (end purlin 17.5 kips). These loads include provision for the dead weight of the truss too. Use A992 steel and select the lightest section, in the center panel.] See Fig. X11.15.5 of the text book. Solution a. General From Example 11.15.5, member U 4U 5 is the critical element of the truss top-chord. It acts as a beam-column and is to be designed for:

As there is no relative translation of the ends of a truss member, K = 1.0. We assume that lateral support is provided for the top-chord member at its ends (panel points) and at the center by the purlins. Thus, b.

Trial section We will assume a KL value of 10 ft (as major axis buckling may control). For a W12-shape of A992 steel, KL = 10 ft, we read a m value of 1.7 from Table 11.14.1. Assume B 1x = 1.1. Entering column selection table for W12 series (LRFDM Table 4-2) with KL = K y L y = 8 ft,

6 W12×79 has P dy = 917 kips > 878 kips required;

r x /r y = 1.75

As the design strength about both axes is greater than the required value of 878 kips, we will consider the W12×79-shape. c.

Axial and bending strengths W12×79: A = 23.2 in.2 ;

r x = 5.34 in. ;

r y = 3.05 in.

From LRFDS Table 3-50, for

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 11

page 11-41

So, use Eq. 11.9.11a. From beam selection tables for W-Shapes (LRFDM Table 5-3, for example), for a W12×79, As L b = 8 ft < d.

L p = 10.8 ft,

Second-order moment, For bending about the x-axis, we have from Table 11.9.1, for:

From Table C-C1.1 of the LRFDC, we have for a fixed ended beam with a central transverse load:

e.

Check for limit state of strength For the beam-column under major axis bending, we have: O.K.

So, select a W12×79 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 11

page 11-42

P11.26. Solve Example 11.15.6, if the solution is limited to W12-shapes. [EXAMPLE 11.15.6: A W14 column, 14 ft long, is part of an unbraced frame in the plane of the web and part of a braced frame in the perpendicular direction such that K x = 1.5 and K y = 1.0. First-order factored load analysis gives an axial load of 650 kips under (1.2D + 0.5L + 1.6W), the non-sway gravity moments M nt under (1.2D + 0.5L) and the sway moments M Rt under 1.6W, as shown in Fig. X11.15.6. For the story under consideration, 3Pu = 20,000 kips. The allowable story drift index is 0.002 due to total horizontal (unfactored) wind forces 3H = 500 kips. Select the lightest section using A992 steel.] See Fig. X11.15.6 of the text book. Solution a. Data Length of the column, L = 14 ft

Maximum 1 st order no-translation moment, M ntx = 100 ft-kips Maximum 1 st order translation moment, M ltx = 200 ft-kips Story drift index, )o /h = 0.002 b.

Preliminary selection Assume B 1x = 1.0, and B 2x = 1.1, giving

For a W12 section of Grade 50 steel and KL = K y L y = 14 ft, value of m is 1.5 from Table 11.14.1. Hence Enter column selection table for W12-shapes (LRFDM Table 4-2) with Preq = 1130 kips; KL = 14 ft; Fy = 50 ksi, to find that for a W12×120, Pd y = 1220 kips

> Preq = 1130 kips

and

indicating that major axis buckling will not control, and Pd = Pdy = 1,220 kips c.

Check the section For a W12×120 section, A = 35.3 in.2 ; Ix = 1070 in.4 So, use interaction Eq. 11.9.11a.

The laterally unbraced length L b is 14 ft. From beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W12×120 of Fy = 50 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 11

page 11-43

ksi steel, Nb M px = 698 ft-kips and, for L b = 14 ft and C b = 1.0, M odx = 692 ft-kips. The ratio of (total) end-moments

So, from Table 10.4.1, C b = 2.16. As L b > L p, we have : M dx = min [C bM odx; Nb M px] = min [2.16{692}; 698 ] = 698 ft-kips Next B 1x is calculated from Eq. 11.9.4. We have:

Factored, second-order moment is

Substituting in the interaction Eq. 11.9.11a : OK

Hence adopt a W12×120 of A572 Grade 50 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 11

page 11-44

P11.27. Solve Example 11.15.7, if the solution is limited to W12-shapes. [EXAMPLE 11.15.7: Select a W14 section of A992 steel for a 14 ft long beam-column, part of an unbraced frame in y-plane and part of a braced frame in the x-plane. K x = 1.4 and K y = 1.0. Factored axial load on column Pu is 840 kips. First order, single curvature moments under gravity loads are: M ntx = 280 ft-kips, M nty = 40 ft-kips. The first order reverse-curvature moments under factored wind loads are M Rtx1 = 200 ft-kips and M Rtx2 = 50 ft-kips. The drift index is 1/400 under EH = 100 kips. Total factored gravity load above this story is 6,800 kips.] See Fig. X11.15.7 of the text book. Solution a. Data W12-shape. of A992 steel L = 14.0 ft; K x = 1.4; K y = 1.0 K x L x = 1.4×14 = 19.6 ft; K y L y = 1.0×14.0 = 14.0 ft = L b Pu = 840 kips; M ntx = 280 ft-kips; M nty = 40.0 ft-kips M ltx1 = 200 ft-kips; M ltx2 = 50.0 ft-kips; M ltx = 200 ft-kips M lty = 0 (column part of braced frame in xx plane) )o /h = 1/400; EH = 100 kips; EPu = 6,800 kips b.

Preliminary selection Assume B 1x = 1.05; B 1y = B 2x = 1.2 M ux* = B 1x M ntx + B 2x M ltx = 1.05 × 280 + 1.2 × 200 = 534 ft-kips M *uy = B 1y M nty = 1.2 × 40.0 = 48.0 ft-kips From Table 11.14.1, for a W12-shape with KL = 14 ft, get m = 1.. Assume u = 2.0. = 840 + 1.6 × 534 + 1.6 × 2.0 × 48.0 = 1850 kips From LRFDM Table 4-2, for a W12×190 and KL = K y L y = 14 ft, Pdy = 1950 kips > 1850 kips. Also rx /ry = 1.79 for this section, resulting in (K x L x)y = 19.6 /1.79 = 11.0 < K y L y = 14 ft. So, K y L y controls the design. For the W12×190 section, u = 2.11 from Table 11.14.2, resulting in a revised value, Pu eq = 840 + 1.6 × 534.0 + 1.6 × 2.11 × 48.0 = 1856 kips The W12×190 is still valid.

c.

Check the selected section For the W12×190 of A992 steel column with (K x L x)y = 11.0 ft and K y L y = 14 ft, Pd = 1,950 kips. so use Eq. 11.9.11a. For symmetric, single curvature no translation moments, C mx = C my = 1.0. From LRFDM Table 4-2, for a W12×190: Pex (KL)2 = 54,100 × 10 - 4; Pey (KL)2 = 16,900 × 10 - 4

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 11

page 11-45

Conservatively take (KL)ntx = (KL)nty = 1.0 × 14 = 14.0 ft. Then,

;

Also,

We obtain M ux* = B 1x M ntx + B 2x M ltx = 1.05 × 280 + 1.20 × 200 = 534 ft-kips M uy* = B 1y M nty = 1.16 × 40.0 = 46.4 ft-kips From LRFDM Table 5-3, for a W12×190 and Fy = 50 ksi: Nb M px = 1170 ft-kips; L p = 11.5 ft Nb M py = 523 ft-kips L r = 76.6 ft; BF = 5.79 As L p = 11.5 ft < L b = 14.0 ft < L r = 76.6 ft M odx = Nb M px - BF (L b - L p) The ratio of (total) end-moments

So, from Table 10.4.1, C b = 1.27. As L b > L p, the design major axis bending strength of the member is: M dx = min[ C b M odx ; Nb M px ] = min[ 1.27 × 1063 ; 1170] = 1170 ft-kips Also, We therefore have:

So, the W12×190 of A992 steel selected is 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 11

page 11-46

P11.28. Solve Example 11.15.8, if the solution is limited to W14-shapes. [EXAMPLE 11.15.8: Select a W12 section of A992 steel for a 16 ft long beam-column. The member is part of a symmetric unbraced frame about both axes. Under factored loads, P u = 600 kips, M ntx = M nty = 0, M Rtx = 240 ft-kips and M Rty = 100 ft-kips. Also K x = 1.5 and K y = 1.3. For all columns in the story under consideration, EPui = 10,800 kips, EPe2x = 120,000 kips and EPe2y = 82,000 kips.] See Fig. X11.15.8 of the text book. Solution a. Data L = 16.0 ft; Kx = 1.5; Ky = 1.3; Pu = 600 kips; M ltx = 240 ft-kips; EPui = 10,800 kips; From Eq. 11.9.9 ( LRFDS

= Lb K x Lx = K y Ly = M ntx = M lty = EPe2x = Eq. C1-5) :

16.0 ft 1.5 × 16.0 = 24.0 ft 1.3 × 16.0 = 20.8 ft M nty = 0 100 ft-kips 120,000 kips;

EP e2y

= 82,000 kips

M ux* = B 1x M ntx + B 2x M ltx = 0 + 1.10 × 240 = 264 ft-kips M uy* = B 1y M nty + B 2y M lty = 0 + 1.15 × 100 = 115 ft-kips b.

Preliminary selection Entering Table 11.14.1 with KL = K y L y = 20.8 ft, Fy = 50 ksi and W14-shapes we obtain m = 1.2. Assume u = 2.0. We obtain = 600 + 1.2 × 264 + 1.2 × 2.0 × 115 = 1193 kips From LRFDM Table 4-2 for W-Shapes, for KL = K yL y = 20.8 ft and Fy = 50 ksi, a W14×132 has Pdy = 1192 kips . 1193 kips. Also, u = 1.99 from Table 11.14.2, resulting in: = 600 + 1.2 × 264 + 1.2 × 1.99 × 115 = 1191 kips

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 11

page 11-47

So, try a W14×132. c.

Check the selected section From LRFDM Table 5-3, for a W14×132 of A992 steel: Lp = 13.3 ft; L r = 49.7 ft Nb M px = 878 ft-kips; BF = 6.88 kips;

Nb M py = 419 ft-kips

With equal, reverse-curvature moments at the ends C b = 2.27 As L p = 13.3 ft < L b = 16.0 ft < L r = 49.7 ft M odx = Nb M px - BF (L b - L p) M dx = min [

] = min [2.27 × 859; 878] = 878 ft-kips

The axial load ratio,

so use Eq. 11.9.11a (LRFDS Eq.H1-1a).

LHS

So, the W14×132 of A992 steel selected is acceptable.

(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 11

page 11-48

P11.29. Solve Example 11.15.9, if the solution is limited to W12-shapes. [EXAMPLE 11.15.9: Select a W14 section of A992 steel for a beam-column in an unbraced frame with factored loads: Pu = 400 kips; symmetric single curvature moments, M ntx = 150 ft-kips, M nty = 50 ft-kips, reverse curvature moments M Rtx = 116 ft-kips and M Rty = 72 ft-kips. There are no transverse loads along the span. Story height is 15 ft. K x = 1.4 and K y = 1.2. The allowable story drift index is 1/500, due to unfactored horizontal wind forces in the yy plane of 120 kips (causing bending about xx axis) and 82 kips in x-x direction (causing bending about yy axis). Total factored gravity load above this level is EP u = 7,200 kips.] See Fig. X11.15.9 of the text book. Solution a. Data L Kx Ky Pu M ntx M ltx EP u

= 15 ft; = 1.4; = 1.2; = 400 kips = 150 ft-kips; = 116 ft-kips; = 7,200 kips

Drift index, – –

b.

Lb K x Lx K y Ly M nty M lty

= L = 15.0 ft = 1.4 × 15 = 21.0 ft = 1.2 × 15 = 18.0 ft = 50.0 ft-kips; symmetric single curvature = 72.0 ft-kips; reverse curvature

, corresponding to:

a horizontal force H = 120 kips for bending about x-axis of the column, and a horizontal force H = 82 kips for bending about y-axis.

Preliminary selection Assume B 1x = 1.05; B 1y = 1.10. From LRFDS Eq. C1-1: M *ux = B 1x M ntx + B 2x M ltx = 1.05 × 150 + 1.14 × 116 = 290 ft-kips M *uy = B 1y M nty + B 2y M lty = 1.10 × 50.0 + 1.21 × 72.0 = 142 ft-kips From Table 11.14.1, for a W12 column of Fy = 50 ksi steel and KL = K y L y = 18 ft, the coefficient m = 1.4. Assume u = 2.0. Pueq = 400 + 1.4 × 290 + 1.4 × 2.0 × 142 = 1200 kips

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Steel Structures by S. Vinnakota

Chapter 11

page 11-49

By interpolation in the KRFDM Table 4-2, corresponding to KL = K y L y = 18 ft and Fy = 50 ksi, we observe that a W12×136 has Pd = 1210 kips ( > Preq = 1200 kips). This selection has r x /r y = 1.77, resulting in:

indicating that x-axis buckling will not control the design. Also, u = 2.09 for a W12×136, from Table 11.14.2. The revised value of Pueq = 400 + 1.4×290 + 1.4×2.09×142 = 1220 kips Still try a W12×136. c.

Check selected section From LRFDM Table 4-2, for a W12×136 column with Fy = 50 ksi, KL = K y L y = 18 ft, Pd = 1210 kips; Pex (KL)2 /10 4 = 35,500; P ey (KL)2 /10 4 = 11,400 The magnification factors B 1x and B 1y are determined, using conservatively, (KL)ntx = L = 15 ft and (KL)nty = L = 15 ft. Thus

For the symmetric, single-curvature, first-order, no-translational moments given, C mx = 1.0 and C my = 1.0. Thus

The second-order factored moments are M *ux = 1.04 × 150 + 1.14 × 116 = 288 kips M *uy = 1.13 × 50.0 + 1.21 × 72.0 = 144 kips From LRFDM Table 5-3, for a W12×136 of A992 steel: Lp = 11.2 ft; L r = 55.7 ft Nb M px = 803 ft-kips; BF = 5.49 kips; We will conservatively assume that C b = 1.0. As L p = 11.2 ft < L b = 15.0 ft < L r = 55.7 ft and C b = 1.0

Nb M py = 361 ft-kips

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Steel Structures by S. Vinnakota

Chapter 11

page 11-50

M dx = Nb M px - BF (L b - L p) M dy = Nb M py = 361 ft-kips As the axial load ratio,

, we use Eq. 11.9.11a.

Substituting the various terms in the interaction formula , we obtain:

Select a W12×136 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 11

page 11-51

P11.30. Solve Example 11.15.10, if the solution is limited to W14-shapes. [EXAMPLE 11.15.10: It is required to design an interior crane column for a mill building. The column of A992 steel and 28 ft high carries two crane girders in addition to the roof load. The crane runway on each side of the column is to be supported by means of a bracket welded to the flange of the column at a distance 12 ft below the top. The roof covering including decking, insulation, joists, truss, and piping is 30 psf of the roof surface and the snow load is assumed to be 30 psf. The contributory area for the column under consideration is 1800 sq. ft. Maximum vertical load from wheels, including impact of wheels is given as 40 kips. The horizontal load from the crane wheels is 6 kips. The weight of the crane girder and rails is 3.2 kips and the weight of each bracket is estimated as 1.8 kips. The eccentricity of the crane loads is 16 in. Select a suitable W12-shape.. Assume the column is pinned at both ends about both axes and laterally supported at bracket level by the crane girders.] See Fig. X11.15.10 of the text book. Solution a. Factored loads The column will be designed for an axial load Pu of 228 kips and a bending moment, M ltx = 114.0 ft-kips (for details see Example 11.15.10). Also, we have

b.

Preliminary member selection The equivalent axial load on the beam-column is: Assume minor axis buckling controls the design axial strength. For a W14 column with KL = L y = 16 ft, m = 1.3 from Table 11.14.1. Assume B 1x = 1.2. So

Ky

Using the column load table for W-Shapes, select a W14×613, which for KL = 16 ft has P d = 486 kips ( > = 406 kips) and . We have:

as assumed. So, try W14×61 c.

Column action Axial load ratio, So, the interaction formula to be checked is Eq. 11.9.11a.

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Steel Structures by S. Vinnakota

d.

Chapter 11

page 11-52

Beam action From LRFDM Table 5-3, for a W14×61 beam of A992 steel L p = 8.65 ft; Nb M px = 383 ft-kips; BF = 6.50 kips L r = 25.0 ft; Ix = 640 in.4 The lower segment which is longer, subjected to heavier axial load, and higher maximum bending moment, is more critical. As L p < L b = 16 ft < L r

The variation of bending moment over the the segment is linear, and as from Table 10.4.1, and

M *ux = B 1x M ntx with

The moment reduction factor C mx is conservatively obtained from LRFDC Table C-C1.1 corresponding to a pin ended column with a central concentrated load. As the bending is about the major axis

e.

Limit state of strength

So, select a W14×61 of A992 steel.

(Ans.)

Note: The next lighter W14 (a W14×53) will not work (LHS = 1.04).

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 11

page 11-53

P11.31. Solve Example 11.15.11, if the portal frame is fixed at the base. Limit the selection of columns to W14shapes. [EXAMPLE 11.15.11: Design the members of the hinged-base, rigid jointed portal frame ABCD shown in Fig. 5.1.2, studied in Example 5.3.1. The beam BC is subjected to a uniformly distributed service load of 2.5 k Rf (1 k Rf D + 1.5 k Rf S). In addition, the frame is subjected to a concentrated dead load of 30 kips at the column tops B and C. The wind load on the frame consists of a 15 kip horizontal force acting at the joint B. The dead loads given include provision for self weight of members. Use A992 W-shapes with their webs in the plane of the frame. The columns are braced at the top and bottom against y-axis displacement and at mid-height against y-axis buckling. Lateral bracing for the beam is provided at the ends B and C, and at the quarter points F, E, and G. Limit the drift to h/250 and the live load deflection to L/360.]

P11.32. A 24-ft long W21×93 of A992 steel is used as a simply supported beam with regard to both principal axes. Lateral braces are provided only at the supports. The beam is subjected to a single concentrated load Q at midspan. The load passes through the shear center of the cross section, and is inclined at an angle of 15 o with the vertical axis (web axis). Neglect the self-weight of the bam and determine the maximum factored load Q u as per LRFDS.

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 11

page 11-54

P11.33. Redesign the crane runway beam of Example 10.4.10, if it has to carry a lateral force of 3 kips in addition to the loads given there. The lateral force acts perpendicular to the beam, at each wheel, 4¼ in. above the top flange. [EXAMPLE 10.4.10: A 18 ft long, crane runway beam carries the two end wheels of a crane. The wheels are spaced 12 ft on centers and each transfers a maximum vertical live load of 24 kips to the beam. Neglect the lateral force and longitudinal force on the member. The crane beam is without lateral support except at the columns. The specified impact is 25% of the live load. Design the beam, using A242 Grade 50 steel. The maximum service live load deflection is to be limited to L/500.] See Fig. X10.4.10 of the text book. Solution Span, L = 18 ft; Unbraced length, L b = 18 ft Wheel load, Q = 24 kips; Wheel spacing, a = 12 ft Impact factor, I = 25% a.

Required strengths Equivalent statically applied concentrated wheel load, Q yL = 1.25×24.0 = 30.0 kips The maximum live load moment will occur at midspan, with one of the crane wheels at the center of the span (Fig. X10.4.11b). Assume a 90-lb rail, weighing 90 lb/yard or 30 lb/ft. Also, assume the weight of the crane beam to be 90 lb/ft.

Bending moment due to dead load,

Bending moment due to live load,

Required bending strength about major axis, M ux = 1.2M D + 1.6M L = 1.2(4.86) + 1.6(135) = 222 ft-kips Horizontal load on the beam = Lateral thrust from the moving crane = Q xL = 3.0 kips Assume a W18×76 shape: d = 18.2 in.; b f = 11.0 in.; t f = 0.680 in. Height of rail = 4 ¼ in. From Eq. 11.16.4, equivalent flange force, Q f = 1.6 Q xL (½ d + 4¼ ) ÷ (½ d - ½ t f) = 7.32 kips Factored bending moment about the y-axis of the flange = M u f = 7.32 × 18.0 ÷ 4 = 32.9 ft-kips b.

Strength limit states As the bending moment is essentially due to the concentrated load at midspan, use C b = 1.32 (value from Fig. 10.4.1e, given for a simple beam with lateral bracing at the supports only).

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 11

page 11-55

Entering the beam selection plots for W-Shapes (LRFDM Table 5-5) , with C b = 1, F y = 50 ksi and L b = 18 ft, observe that a W18×76 has a bending strength M do = 514 ft-kips and Nb M px equals 612 ft-kips, resulting in: M d = min [C b M do ; Nb M px]= min [1.32×514 ; 612] = 612 ft-kips Also from LRFDM Table 5-3, Nb M py = 155 ft-kips = M dy Using Eq. 11.16.6:

As the weight of the beam selected (76 lbs) is less than the assumed value of 90 lbs used in the dead load calculations, no revision is necessary. The section satisfies the requirements for compactness as there is no footnote in the beam selection tables stating otherwise. Therefore, use a W18×76 of A242 Grade 50 steel.

(Ans.)

P11.34. A crane runway girder 24 ft in length is to be designed to carry the two end wheels of a 5-ton crane. Lateral supports are provided at the ends only. The wheels of the crane are 8 ft on centers and each wheel transfers a maximum load of 12.5 kips to the top of 60-lb rail. Assume 10 percent of the wheel load acting as a lateral load applied at 4¼ in. above the top of the compression flange. Select the lightest W14-shape of A992 steel.

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 11

page 11-56

P11.35. Select the lightest MC-shape of A36 steel to be used as purlins in an industrial building. The roof pitch is 5 on 12; the rafters are 18 ft on centers; and the purlins are spaced at 8-ft intervals along the roof. The roof covering, including purlins, weighs 16 psf of the roof surface and the snow load is 24 psf of the horizontal surface. The purlins should be designed without sag rods.

Solution Inclination of roof, 2 = tan -1 (5 ÷ 12) = 22.6 o Spacing of trusses, L = 18 ft Spacing of purlins along roof = 8 ft Horizontal Spacing of purlins = Uniform service loads per foot of purlin: Dead load, D = Snow load, S = With only dead and snow loads acting on the purlin, load combination LC - 3, namely, [1.2D + 1.6S] controls the design. Both these loads are vertical loads. q uV = 1.2D + 1.6S = 1.2 × 0.128 + 1.6×0 .177 = 0.437 klf The components of the factored loads acting in y- and x- directions at the mid-width of the purlin are: q uy = q uV cos 2 = 0.437 cos 22.6 o = 0.403 klf q ux = q uV sin 2 = 0.437 sin 22.6 o = 0.168 klf Note that q uy produces major axis moment, while q ux produces minor axis moment. The purlin acts as a simple beam of span L (= 18 ft) for major axis bending. It is not laterally supported. So, L b = 18 ft. With no sag rods placed, the purlin acts as a simple beam with respect to weak axis bending, with a span, L = 20 ft. The required bending strengths are:

Or,

To select a trial shape, use the beam selection plots for C-shapes (LRFDM Table 5-11) and choose

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 11

page 11-57

a shape with a relatively large margin of strength with respect to major axis bending. For an unbraced length, L b = 18 ft, Fy = 36 ksi and C b = 1 an MC10×33.6 provides

Also, from LRFDM Table 5-10, for an MC10×33.6,

From Table 10.4.1, C b = 1.14 for a simply supported beam under uniformly distributed load, with lateral supports at the ends only. Design flexural strength for major axis bending,

For C-shapes bent about their minor axis, the shape factor is greater than 1.5. So, the design flexural strength for minor axis bending,

Substituting in the interaction Eq. 11.16.5 results in:

So, select a MC10×33.6

(Ans.)

P11.36. Solve Problem P11.36, if one line of sag rods is used. P11.37. Solve Problem P11.36, if W-shapes of A992 steel are used. P11.38. Select the lightest W section of A992 steel to carry 0.4 klf dead load, in addition to the weight of the beam, and live load of 1.5 klf. The superimposed load is applied eccentrically 4 in. from the center of the web. The beam is simply supported and has a span of 28 ft. Assume that lateral bracing is provided at the end only.

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.