Solver Vinnakota Cap 11

S te e l S tr u c tu r es b y S . V in n a k o t a

C h ap t e r 1 1

p ag e 1 1 - 1

CHAPTER 11

P11.1.

A W12 ×40 o f 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 mome nts about the strong axis of 100 ft-kips at each end, bendin g the memb er 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, P u = 160 kips Length of column, L column, L = 13.0 ft Column p art of braced frame. Member pinn ed at both ends, with a brace at mid height. L x = 13.0 ft; L y = 6.50 ft 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 with KL = 6.50 ft, Pd = 442 kips and P and P ex ( KL KL )2 = 8 79 79 0 × 1 04.

Axial load ratio, Bending is about major axis only Structure is braced in yy in yy plane plane

. Use LRFDS Eq. H1-1 a . 6

M u y* = 0

Column is subjected to symmetric single curvature bending moments of 100 ft-kips.

Unbraced length, L length, L b = 6.5 ft. From beam selection tables tables (LRFDM T able 5-3), corresponding corresponding to a W 12×40, L p = 6.85 ft > L b , and The LHS of the interaction equation gives

PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


C h ap t e r 1 1

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

p ag e 1 1 - 2

(Ans.)



C h ap t e r 1 1


p ag e 1 1 - 2

(Ans.)



P11.2.

C h ap t e r 1 1

p ag e 1 1 - 3

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

Solution

Factored axial load, P load, P u = 200 kips. Length of column, L column, L = 15 ft. W14×61 of A992 steel. steel. Colum n part of braced frame. Lower end hinged. Assume (conservatively), K (conservatively), K x = 1.0 and K 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 and KL = K y L y = 15.0 ft, P ft, P d = 513 kips. Also, for this section, P ex ( KL )2 = 1 83 83 00 00 × 1 04 . A x ial lo ad r ati o ,

. S o , u s e L R F D S E q . H 1 .1 a .

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

As,

from Table 10.4.1.

From beam selection plots (LRFDM Table 5-5), corresponding to a W1 4×61 with L with L b = 15 ft, Also, Nb M px = 383 ft-kips for this shape. 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.)



P11.3.

C h ap t e r 1 1

p ag e 1 1 - 4

Figure 11-3 shows a 15-ft 15-ft long W12×96 column with pinned end s. Two girders and and a beam bring in 200 kips of axial load. load. A column from above d elivers 90 kips at an eccentricity of 18-in. through a bracket connected to the web of the column. The structure is is braced in in both xx and yy and yy pla planes. nes. The loads are are factored loads. loads. Is the column adequate? See Fig. P11.3 of the text book. Solution

Length of column, L column, L = 15 ft Column part of braced frame in both xx and xx and yy yy planes. So, K So, K x # 1.0 and K and K y # 1. 0 Pinned ends. So, K So, K x = 1.0 and K 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 and KL = K y L y = 15.0 ft, P ft, P d = 935 kips. Also, for this section, P ey ( KL )2 = 7 73 73 0 × 1 04. Factored axial load, P load, P u = 200 + 90 = 290 kips.

A x ial lo ad r ati o ,

. S o , u s e L R F D S E q . H 1 .1 a .

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

Eccentricity, e = 18 in.

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

S o , W 1 2 × 9 6 o f A9 9 2 st e e l i s ad e q u ate .

(A n s.)



P11.4.

C h ap t e r 1 1

p ag e 1 1 - 5

An interior column of a building structure has floor girders framing into it at top and bottom with moment resisting connection s. It must carry a factored factored axial load of 800 kips, including the girder and beam reactions, and the self-weight self-weight of the column. Live load imbalance in checkerboard loading causes a potential maximum moment at the top and bottom of 18 0 ft-kips, ft-kips, as shown in Figure P11-4. K for for the weak axis is 1.0, and K and K for 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, P u = 800 kips kips Length of column, L 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 T able 4-2, corresponding corresponding to KL to KL = K y L y = 12.5 ft ft and and a W12×106, P W12×106, P d = 1115 kips. 2 4 Also, P Also, P ex ( KL ) = 2 67 67 00 00 × 1 0

Axial load ratio,

. Use LRFDS Eq. H1-1 a .

Structure is braced in xx in xx and and yy yy planes 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 length, L b = 12.5 ft. From beam selection plots (LRFDM Table 5-4), corresponding to a W12 ×106, L ×106, L b = 12.5 ft, and C b = 1.0,

The LHS of the interaction equation gives


(Ans.)


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 T able 1-1, for a W14×22, we obtain: A = 6.49 in2; Z y = 4.39 in3; S y = 2.80 in.3 Z x = 33.2 in.3; S x = 29.0 in.3; r y = 1.04 in. X 1 = 1600 ksi; X 2 = 0.0278 (1/ksi) 2 For a W14×2 2 of A572 Grade 42 steel: F = 42 ksi; F L = Fy - F r = 42 - 10 = 32 ksi y

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 d 1 = Nt A g F y = 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.

<

1.0 safe (Ans.)

PROPRIETARY MATER IAL. © 2006 T he 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 11

page 11-7

No te that, for convenien ce, we hav e ass um ed tha t ten sio n yi eld ing (T d 1) controls over tension rupture (T d 2). If this is not the case, the design tensile strength should b e based upon the tension rupture design strength.



P11.6.

Chapter 11

page 11-8

A W10×68 of A992 steel is used as an exterior column of a building structure braced in both directions (see Fig. 5.2.5c for a plan view). It must carry 310 kips from the girders con nected to the web, plus a 100 -kip reaction from the floor beam that frames into the colum n flange with an eccentricity of 8 in. The colum n is 15-ft long and is assumed to be p inned about both axes as both ends. The structure is braced in both xx and yy planes of the column. Is the member adequate? See Fig. 5.2.5c of the text book . Solution

Length of column, L = 15 ft Column part of a braced frame in both xx and yy planes

6

In the absence of rational analysis assume (LRFDS Section C2.1), Unbraced length, L b = 15 ft. Factored axial load, P u = 310 + 100 = 410 kips. Eccentricity of 100 kip load, ex = 8 in. Factored major axis moment, As the structure is braced in both xx and yy planes, there are no M lt moments. Also, as there are no be nd ing mo ments abo ut min or axis,

From LRFDM T able 4-2, for a W10×68 with KL = K y L y = 15.0 ft, P d = 597 kips, and P ex ( KL )2 = 11300 × 104 Axial load ratio,

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

6

From Table 10.4.1, corresponding to a linear variation of moments over the unbraced length and r M = 0, C b = 1.67. From beam selection plots (LRFDM Table 5-5), corresponding to a W10×68 with L b = 15 ft,

Also, Nb M px = 320 ft-kips for this shape. So, the design

be nd ing stre ngth of t he mem ber,



Chapter 11

So, the W10×68 of A992 steel is adequate.

page 11-9

(Ans.)



P11.7.

Chapter 11

page 11-10

Figure P11.7 sho ws a 15-ft-long hinged b ase column pinne d at both ends. It must carry a factored axial load of 400 kips and a uniformly applied wind load of 1.5 klf that causes bending abou t the major axis. The structure is braced in both xx and yy planes. Check if a W10× 60 of A992 steel will be adequate. See Fig. P11.7 of the text book. Solution

Length of column, L = 15 ft Structure braced in both xx and yy planes of the column 6 K x # 1.0 ; K y # 1.0 No inte rme diate b raci ng. Co lum n p inn ed at b oth ends. So, K x L x = 15 ft, K y L y = 15 ft. Section W10×60 of A992 steel From LFRDM Table 4-2, for a W10×60 with KL = 15 ft, P d = 523 kips Also, P ex ( KL )2 = 9 76 0 × 1 04.

Axial load ratio,

. So, Eq. 11.9.11 a (LRFDS Eq. H1-1a) governs.

Uniformly distributed load, q u = 1.5 klf Maximum first-order moment, Unbraced length, L b = L y = 15.0 ft. From beam selection plots (LRFDM Table 5-5), for a W10×60 with L b = 15 ft,

From Fig. 10.4.3a, for a simple beam with uniformly distributed transverse load and braced at the ends only, C b = 1.14. We have

From LRFDC Table C.C2.1, for a beam-column with pinned ends and uniformly distributed transverse load, C m = 1.0.

;

Substituting in the interaction equation, we get:

The W10×60 of A992 steel selected is adequate.

(Ans.)



P11.8.

Chapter 11

page 11-11

A 18-ft-long hinged base column carries a factored axial load of 160 kips as shown in Fig. P11.8. The column h as its weak axis braced at third-points by channel girts. The wind load on the wall is picked up by the girts and transferred to the column as point loads (8 kips each, under factored wind). The structure is br aced in b oth xx and yy planes. Check if a W10× 33 of A992 is adequate. See Fig. P11.8 of the text book. Solution

Length of column, L = 18 ft L x = 18 ft; L y1 = L y2 = L y3 = L y = 6 ft Structure braced in both xx and yy planes of the column section. Member pinned at both ends. So, K x L x = 18 ft; K y L y = 6 ft. Unbraced length, L b = 6 ft. Column section: W10×33 o f A992 steel. From LRFDM Table 4-2, for a W10×33. So,

Also from LRFDM Table 4-2, for a W10×33 with KL = 8.33 ft, Pd = 340 kips, and P ex ( KL )2 = 4890 × 104 Axial load ratio, The pin ended member un der two equal concentrated loads at third points is subjected to uniform moment over the middle bay. From LRFDM T able 5-3, for a W10×33, L p = 6.85 ft and Nb M px = 146 ft-kips As L b < Lp, M dx = Nb M px = 146 ft-kips The column is part of a braced structure in both xx and yy planes We have

6

M ltx = 0.0, M lty = 0.0

As per LRFDS Section C1.2, for compression members subject to transverse loading between their supports, C m = 1.00 when member ends are unrestrained.



Chapter 11

page 11-12

The interaction equation gives

So, the W10×33 of A992 steel provided is adequate.

(Ans.)



P11.9.

Chapter 11

page 11-13

Investigate the adequacy of a W8×24 beam-column used with its web vertical, as the top chord member of a truss. Under factored loads of a load combination , the member is subjected to an axial force of 110 kips and a uniformly distributed lateral load of 0.32 klf. The memb er is 12 ft long and could be assumed to be pinned about both axes at both ends. There is no intermediate bracing present. Assume A242 steel. Neglect selfweight. See Fig. P11.9 of the text book . Solution a.

Da ta From LRFDM T able 1-1, we have for a W8×24: 3 A = 7.08 in.2; S y = 5.63in. Z y = 8.57 in.3; I y = 18.3 in.4; r y = 1.61 in. From LRFDM Tables 2-4 and 2-1, we observe that the W8×24, a Group 1 shape, is available in A242 steel in Grade 50 only. That is, F = 50 ksi. y Required axial compressive strength, P u = 110 kips Column length, L = 12 ft; K x = K y = 1.0

b.

Des ign axi al stre ng th

From LRFDS Table 3-50, for KL /r = 89.4, Nc Fcr = 23.7 ksi, resulting in: P d = 23.7 × 7.08 = 168 kips Therefore, Eq. 11.9.11a (LRFDS Eq. H1-1a) governs.

c.

Des ign ben ding stre ng th As the bending is about the minor axis, there is no concern for lateral-torsional buckling. We have:

d.

Req uir ed ben din g stren gth There is no bendin g about the major axis. Hence, M *ux = 0 As the member is part of a truss, there is no relative translation of the ends; i.e., M lt momen ts are zero. Hence,



Chapter 11

page 11-14

Maximum factored, first-order, no-translation moment,

From Table 11 .13.1, for

As the beam-column is pinned at the ends, and subjected to a uniformly distributed lateral load, C m = 1.0 from LRFDC Table C-C1.1. So,

The factored, second-order bending moment,

e.

Check interaction

So, the W8×2 4 of A242 Grade 50 steel is O.K.

(Ans.)



P11.10.

Chapter 11

page 11-15

A W12×50 of A992 steel is used as the bottom chord member of a welded roof truss. The 12-ft long member, between panel points, is subjected to an axial tensile force of 200 kips under factored loads. Determine the maximum factored concentrated load that could be suspend ed at midspan. Assume that the load causes bending about the major axis and that lateral supports are provided at the panel points only. Solution

Truss bottom chord member. Section: W12×50 of A992 steel Member length, L = 12 ft; unbraced length, L b = 12 ft Axial tension under factored loads, T u = 200 kips Factored concentrated load acting at midspan, Q u Transverse load causes bending about major axis. Assume pinned end s (neglect restraint provided by t he end c on nectio ns) . Max imu m b ending mo ment occur s at mid span. We have: M ux = ( Q u L) ÷ 4 = 3Qu ft-kips From LRFDM T able 1-1, for a W12×50, we obtain, A = 14.6 in. 2 From beam selection tables for W-Shapes (LRFDM Table 5-3, for example), for a W12×50 shape of F = 50 ksi steel, we have: y Nb M px = 270 ft-kips L p = 6.92 ft; L r = 21.5 ft; BF = 5.30 kips From Fig. 10.4.3e, a simply supported beam under a concentrated load at midspan and laterally bra ced at th e en ds on ly h as a C b vallue of 1.32. As, ( L p = 6.92 ft) < ( L b = 12 ft) < ( Lr = 21.5 ft),

Alternatively, the value of 243 ft-kips for M dxo could be read from LRFDM Table 5-5. Design tensile strength of the member, corresponding to the limit state of yield on gross area T d = T d 1 = Nt A g F y = 0.90 × 14.6 × 50.0 = 657 kips

Axial load ratio, So, use Eq. 11.9.11a (LRFDS Eq. H1-1a)



Chapter 11

page 11-16

So, the maximum factored concentrated load that could be suspended at midspan is 70.5 kips.

(Ans.) No te: For lack of information about the end connection details, we have assumed that tension yielding (T d 1) controls over tension rupture (T d 2). If this is not the case, the design tensile strength should be bas ed up on the ten sion r up tur e design stre ngth.



P11.11.

Chapter 11

page 11-17

A W14×8 2 is used as a column in an unbraced portal frame. It is 13 ft long and pinned at the base. Under factored loads of a load comb ination, it is subjected to an axial load of 191 kip s, M nt = 156 ft-kips and M lt = 104 ft-kips at the top. Assume K = x 1.7 and K y = 1.0. Check the adequacy of the member.

Solution a.

Da ta Column length, L = 13 ft K x = 1.7; K y = 1.0 6 K x L x = 22.1 ft; K y L y = 13.0 ft Factored axial load, P u = 191 kips Major axis moments at the top: M nt = 156 ft-kips; M lt = 104 ft-kips Major axis momen ts at the base: 0.0 (pinned base) There are no minor axis moments. 6 M uy* = 0.0 A 992 steel: 50 ksi F = y W14×82 section: I x = 881 in.4 ; A = 24.0 in.2 r r x = 6.05 in.; y = 2.48 in.

b.

Des ign axi al stre ng th ;

From LRFDS Table 3-50, design axial compressive stress, Nc F cr = 31.8 ksi Axial strength of the column, P d = Nc Fcr A = 31.8 × 24.0 = 763 kips Axial load ratio, So, we have to use Eq. 11.9.11a (or, LRFDS Eq. H1-1a) for verification of memb er strength. c.

Des ign flex ura l str ength For major axis bending, the laterally unbraced length, L b is 13.0 ft From LRFDM Table 5-3, for a W14×82 section of F = 50 ksi steel, read y

From beam selection plots for W-Shapes (LRFDM Table 5-5), for a W14×82 of F y = 50 ksi steel, C b = 1.0, and Lb = 13 ft, M odx = 490 ft-kips. From Table 10.4.1, a beam segment that has a linear distribution of bending moment over its unbraced length, with a zero value at one end, has a C b value of 1.67. As the unbraced length, L b, is greater than L p, the design major axis bending strength of the member is: M dx = min[ C b M odx ; Nb M px ] = min[ 1.67 × 490 ; 521] = 521 ft-kips



d.

Chapter 11

page 11-18

Second-order moments For bending about the major axis, the maximum 1 st order moment,

The linear variation of major axis mome nt over the heigh t of the column, w ith a zero value at the bas e, re sults in C mx = 0.60 (see Table 10.4.1).

6

M ux* = B 1 x M ntx + B 2 x M ltx = 1.00 × 156 + 1.06 × 104 = 266 ft-kips

e.

Check interaction formula

Thus, the W14×82 of A992 steel is acceptable as a beam-column, to carry the given factored loads, according to LRFDS . (Ans.)



P11.12.

Chapter 11

page 11-19

An unbraced frame contains a 14-ft column which is subjected to an axial load of 428 kips, a no-translation moment M nt = 205 ft-kips, and a translation-permitted moment M lt = 123 ft-kips, at the top-end with half of these momen ts at the bottom-end cau sing reverse curvature. The forces given are under factored loads of a load combin ation, and the bending is about the major axis. Check if a W12 ×96 will be able to carry this loading. Assume that the effective length factor in the plane is K = 1.8 and K y = 1.0. x

Solution a.

Da ta Length of the column, L = 14 ft

Factored axial load, Pu = 428 kips No bend ing mo ments abo ut the y-axis. 6 M uy* = 0.0 Maximum 1 st order moments about the x-axis Top: no-translation moment, M ntx = 205 ft-kips translation moment, M ltx = 123 ft-kips B otto m:

b.

n o-tran slatio n m om en t translation mome nt

= - 1 03 ft-kip s = - 61.5 ft-kips

Check the section Enter column selection table for W12-shapes (LRFDM Table 4-2) with KL = K y L y = 14 ft; F = y 4 50 ksi, to find that for a W12×96, Pd y = 966 kips, and I = 833 in. x

indicating that major axis buckling will control. Reenter the table with KL = 14.3 ft and get Pd = P d x = 957 kips. So, use interaction Eq. 11.9.11 a. The laterally unbraced length L b is 14 ft. From beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W12×9 6 of F y = 5 0 ksi steel Nb M px = 551 ft-kips, L p = 10.9 ft and, for L b = 14 ft and C b = 1.0, M odx = 534 ft-kips. The ratio of (total) end-moments

So, from Table 10.4.1, C b = 2.17. As L b > L p, we have : M dx = min [C bM odx; Nb M px] = min [2.17{534}; 551] = 551 ft-kips PROPRIETARY MATER IAL. © 2006 T he 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 11

page 11-20

Nex t B 1 x and B 2 x are calculated. We have:


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

Hence the W12 ×96 of A992 steel is acceptable.

(Ans.)



P11.13.

Chapter 11

page 11-21

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 momen ts are applied at the other end of the member, bending in single-curvature. All momen ts are applied about the strong axis. Assume K x = 1.6 and K = 0.9. y

Solution a.


Factored axial load, Pu = 200 kips No bend ing mo ments abo ut 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 B otto m: n o-tran slatio n m om en t = 5 0.0 ft-k ip s translation mome nt = 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; F = y 50 ksi, to find that for a W14×48, Pd y = 428 kips, and I = 484 in.4. x

indicating that minor axis buckling will control. So, P d = P d y = 428 kips So, use interaction Eq. 11.9.11 a. The laterally unbraced length L b is 12 ft. From beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W14×4 8 of F y = 5 0 ksi steel Nb M px = 294 ft-kips, Lp = 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 Nex t B 1 x and B 2 x are calculated. We have:



Chapter 11

page 11-22


Substituting in the interaction Eq. 11.9.11 a:

Hence the W14 ×48 of A992 steel is not adequate.

(Ans.)



P11.15.

Chapter 11

page 11-23

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 mom ents of 480 ft-kips producing symmetric single curvature bending about the major axis. The memb er is part of a braced frame in its xx and yy planes.

Solution a.

Da ta Factored axial load, P u = 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 colu mn is part of a braced frame, M ltx = 0, and Eq. 11.9.3 (LRFDS Eq . C1-2) reduces to Maximum 1st order major axis moment, M ntx = 480 ft-kips

b.

Pre limi na ry s electio n Assume B 1 x = 1.05

6

Method 1 :

As bending is abo ut 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 P req = 1120 kips, we observe that a W12×120 h as a design strength P d = 1140 kips. Also, I = 1070 in.4 x 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.

Bea m e ffec t The laterally unbraced length, L b = L y = 16 ft. For a beam segment under u niform moment C b = 1.0 . From LRFDM Table 5-5, for a W12×120 , L p = 11.1 ft. Also, for L b = 16 ft > Lp and C b = 1.0, we have,



Chapter 11

page 11-24

The mom ent magnification factor is calculated next using Eq. 11.9.4. For a pin-ended m ember 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.)



P11.16.

Chapter 11

page 11-25

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 mom ents of 240 ft-kips producing symmetric single curvature bending about the major axis. The memb er is part of a braced frame in its xx and yy planes.

Solution a.

Da ta Factored axial load, P u = 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 colu mn is part of a braced frame, M ltx = 0, and Eq. 11.9.3 (LRFDS Eq . C1-2) reduces to Maximum 1st order major axis moment, M ntx = 240 ft-kips

b.


6

Method 1 :

As bending is abo ut 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 P req = 971 kips, we observe that a W14×120 has a design strength P d = 972 kips. Also, I = 1380 in.4 and L p x = 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.

Bea m e ffec t The laterally unbraced length, L b = L y = 16 ft. For a beam segment under u niform moment C b = 1.0 . From LRFDM Table 5-5, for a W14×120 , for L b = 24 ft > Lp and C b = 1.0, we have,



Chapter 11

page 11-26


e.

Check the interaction formula .


(Ans.)



P11.17.

Chapter 11

page 11-27

Repeat Problem P11.16, if the member is provided with a brace at mid-length. [ P11 .16 . Sel ect a W 14 -sh ap e fo r a pin -en ded colum n wi th a len gth of 24 ft to carry a fa cto red loa d o f 64 0 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. Colum n 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, P u = 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 = P u + m M ux* = 640 + 1.4 × 288 = 1040 kips Enter LRFDM T able 4-2 for W14-shapes, with KL = 14 ft and P req = 1040 kips, and select a W14×99 with P d = 1060 kips.

b.

Check the selected section From LRFDM Table 4-2, for a W14×99: r x /r y = 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, P d = 1050 kips

6

P u / P d = 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: P ex ( KL )2 = 3 1,8 00 × 1 0 - 4. As, ( KL ) ntx = 2 4.0 ft,

;

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

< L p = 13.5 ft, M dx = Nb M px = 646 ft-kips

We therefore have:

So, select a W14 ×99 of A992 steel.

(Ans.)



P11.18.

Chapter 11

page 11-28

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, P u = 840 kips. The first order, symmetric, single curvature end-moments 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.

Da ta A572 Gr 50 steel L = 14 ft; K x = K y = 1.0; K x L x = K y L y = 1.0 × 14.0 = 14.0 ft P u = 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 1 x M ntx; M uy* = B 1 yM nty

b.

Pre limi na ry s electio n Assume B 1 x = 1.05; B 1 y = 1.20 M ux* = 1.05 × 280.0 = 294.0 ft-kips;

M uy* = 1.20 × 40.0 = 48.0 ft-kips

Method 2 :

Since the axial load is relatively large, we get the med ian value of m from LRFDM Table 6-1. For W14-shapes with L b = 14 ft, we obtain m = 0.826×10-3. Also, the median value of n for W14shapes is 1.64×10-3 . Equation 11.14.10 a therefore yields:

From LRFDM Table 6-2, a W14×120 with KL = 14 ft has b = 0.773 × 10-3 < 0.808 × 10-3 . Also, m = 1.13×10 -3 for a member with L b = 14 ft. Addition ally, n = 2.34×10 -3. Substituting these values of b, m, and n in Eq. 11.14.10a, we obtain:

By repeating the calculations for the next heavier section, namely, W14×132, we obtain LHS = 0.991 < 1.0. O.K. Select W14×132. c.

Check W14×132 for stre ng th As

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

A = 38.8 in.2; I x = 1530 in.4; I y = 548 in.4



Chapter 11

page 11-29

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

M ux* = 1.06 × 280 = 297 ft-kips M uy* = 1.18 × 40 = 47.2 ft-kips From beam selection plots for W-Shapes (LRFDM Table 5-5), for a W14×132 of F y = 50 ksi steel, L b = 14 ft, and uniform bending (C b = 1.0), M dx = 872 ft-kips. Also, from LRFDM Table 5-3, Nb M py = 419 ft-kips, for this shape. M dy = Nb M py = 419 ft-kips Using Eq. 11.13.11a (or, LRFDS Eq. H1-1a):

6


P11.19.

0.990 < 1.0

O.K.

(Ans.)

A pin-ende d colum n in a braced frame must carry a factored axial load of 222 kip s along with a factored uniformly distributed transverse load of 1.2 klf. The column is 14-ft long and is braced at mid length. The transverse load is applied to put bendin g about the strong axis. Select the lightest W10 -shape of A992 steel.



P11.20.

Chapter 11

page 11-30

Solve Example 11.15.1, if the solution is limited to W12-shapes. [EXAMPLE 11.15.1: Select a W14-shape. for a column to support, under factored loads, an axial load of 1600 kips and a major axis moment at the top and bottom of 210 ft-kips as determined by a first-order analy sis of the structure. The moments are in the opposite direction resulting in single curvature. The interior column considered is part of a braced frame in b oth directions. Use A992 steel.] See Fig. X11.15.1 of the text book . Solution a.

Da ta Factored axial load, P u = 1600 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 × 11 ft = 11 ft There are no minor axis moments. 6 M uy* = 0 Colum n is part of a braced frame 6 M ltx = 0

b.

6

Maximum 1st order major axis moment, M ntx = 210 ft-kips Preliminary selection Assume B 1 x = 1.1

6

Method 1 :

As bendin g is about major axis only, we select a trail shape using

For a W12 with KL = 11 ft, m = 1.65 from Table 11.14.1. So, Entering the column load table for W12-shapes (LRFDM Table 4-2) with KL = 11 ft and P req = 1980 kips, we observe that a W12×190 h as a design strength P d = 2100 kips. Check the W12×190. c.

Column effect For the W12×1 90 section selected, with P d = 2100 kips, Axial load ratio, So, use the interaction equation 11.9.11 a ( LRFDS Eq. H1-1a), which for M uy = 0 reduces to

d.

Bea m e ffec t The laterally unbraced length, L b = L y = 11 ft From LRFDM T able 5-3, L p = 11.5 ft, Nb M px = 1170 ft-kips and I = 1890 in.4 , for a W12×190. x



Chapter 11

page 11-31

As L b < L p, we have, The mom ent magnification factor is calculated next using Eq. 11.9.4. For a pin-ended m ember under uniform moment C m = 1.0 .

e.

Check the interaction formula O.K.


(Ans.)



P11.21.

Chapter 11

page 11-32

Solve Example 11.15.1, if the member is 16-ft long. [EXAMPLE 11.15.1: Select a W14-shape. for a column to support, under factored loads, an axial load of 1600 kips and a major axis moment at the top and bottom of 210 ft-kips as determined by a first-order analy sis of the structure. The moments are in the opposite direction resulting in single curvature. The interior column considered is part of a braced frame in b oth directions. Use A992 steel.] See Fig. X11.15.1 of the text book. Solution a.

Da ta Factored axial load, P u = 1600 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 × 11 ft = 11 ft There are no minor axis moments. 6 M uy * = 0 Colum n is part of a braced frame 6 M ltx = 0

6

Maximum 1st order major axis moment, M ntx = 210 ft-kips b.


6

Method 1 :

As bendin g is about major axis only, we select a trail shape using

For a W14 with KL = 11 ft, m = 1.3 from Table 11.14.1. So, Entering the column load table for W14-shapes (LRFDM Table 4-2) with KL = 16 ft and P req = 1915 kips, we observe that a W14×193 h as a design strength P d = 2050 kips. Check the W14×193. c.

Column effect For the W14×193 section selected, with P d = 2050 kips, Axial load ratio, So, use the interaction equation 11.9.11 a ( LRFDS Eq. H1-1a), which for M uy = 0 reduces to

d.

Bea m e ffec t The laterally unbraced length, L b = L y = 16 ft W14×193 of A992 steel. From LRFDM Table 5-3:



Chapter 11

L r = 70.1 ft B F = 7 .1 9 kip s With uniform moment over the unbraced length, C b = 1.0 As L p = 1 4.3 ft < L b = 16.0 ft < L r = 70.1 ft M dx = M odx = Nb M px - BF ( Lb - Lp) L p

=

page 11-33

14.3 ft;

Nb M px = 1 33 0 ft-kip s;


e.

Check the interaction formula O.K.


(Ans.)



P11.22.

Chapter 11

page 11-34

Solve Example 11.15.2, if the solution is limited to W14-shapes. [EXAMPLE 11.15.2: Select the lightest W12 section of A992 steel for an interior column in a framed stru ctu re b raced in bo th xx a nd yy d irectio ns. The c olu mn is 1 4 ft lon g. Un der fa cto red gra vity loa ds the column is subjected to an axial load of 600 kips and the major and minor axis moments shown in Fig. X1 1.1 5.2 . Ass um e K x = K y = 1.0 an d n o l ateral brac ing b etwe en the flo or leve ls.] See Fig. X11.15.2 of the text book . Solution a.

Data Column length, L = 14 ft; K x L x = 14 ft; K y L y = 14 ft Braced frame: M ltx = 0.0; M lty = 0.0 M x2 = 240 ft-kips; M x1 = 120 ft-kips; From Table 10 .4.1, corresponding to r M = - 0.5, C b = 1.25; M y2 =

b.

C m = 0.8 6

C my =

1.0

Selection of trial section Assume: B 1 x = 1.0; B 1 y = 1.1 So ,

For a W14 column ( F y = 50 ksi steel) KL = 14 ft, m = 1.4 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: W14×109 with P d = 1,270 kips > P req = 1,120 kips So, the W14×1 09 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.

Des ign stre ng ths From Table 1-1 of the LRFDM, for a W14×109 -shape:



Chapter 11

page 11-35

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

. Therefore, use Eq. 11.9.11 a.

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

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

;

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


(Ans.)



P11.23.

Chapter 11

page 11-36

Solve Problem P11.18, if the solution is limited to W12-shapes. [ P11 .18 . Sel ect a W 14 sec tion o f A9 92 stee l fo r a 14 -ft-lon g b eam-c olumn i n a framed s tructu re b raced in both directions. Factored axial load, Pu = 840 kips. The first order, symmetric, single curvature endmoments are: M ntx = 280 ft-kips and Mnty = 40 ft-kips. Assume K x = K y = 1.0.] See Fig. 11.18 of the text book . Solution a.

Da ta L = 14 ft; K x = K y = 1.0; K x L x = K y L y = 1.0 × 14.0 = 14.0 ft P u = 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 1 x M ntx; M uy* = B 1 yM nty

b.

Pre limi na ry s electio n Assume B 1 x = 1.05; B 1 y = 1.20 * M ux = 1.05 × 280 = 294 ft-kips; M uy* = 1.20 × 40.0 = 48.0 ft-kips For a W12 column ( F y = 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 P d = 1,550 kips > P req = 1,425 kips So, the W12×1 52 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 W12×1 52 with an axial load capacity of 1550 kip s appears to be an acceptable shape. For the W12 ×152-shape: I x = 1430 in.4; I = 454 in.4 y c.

Check W12×152 for stre ng th As

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



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: L p = 11.3 ft; L r = 62.1 ft Nb M px = 9 11 ft-kips; BF = 5.5 9 kips With uniform moment over the unbraced length, C b = 1.0 As L p = 1 1.3 ft < L b = 14.0 ft < L r = 62.1 ft M dx = M odx = Nb M px - BF ( Lb - Lp)

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


(Ans.)



P11.24.

Chapter 11

page 11-38

Solve Example 11.15.4, if the solution is limited to W12-shapes. [EXAMPLE 11.15.4: A 3 0 ft lo ng chord mem ber of a tr uss is sub jected , un der fa cto red load s, to an axial compressive force of 690 kips and a uniformly distributed factored load of 0.3 klf causing bending abou t its weak axis. Use A992 steel and select the appro priate W14 section. Include the influence of the self weight of the member which causes bending about the major a xis.] See Fig. X11.15.4 of the text book . Solution a.

Da ta 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.

Pre limi na ry s electio n As the membe r is quite flexible, we will assume B 1 x = 1.2 and B 1 y = 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 F y = 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 F = 50 ksi and KL = 30 ft, P d = y 967 kips ( > c.

= 941 kips) for a W12×190 selected. So, try a W12×190.

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

. So, use Eq. 11.9.11 a.

Ma jor axis b endin g From LRFDM Table 5-3, for a W12×190 and F = 50 ksi: y Nb M px = 1170 ft-kips; I x = 1890 in.4; I y = 589 in.4 PROPRIETARY MATER IAL. © 2006 T he 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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


N b 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 = 1 1.5 ft < L b = 14.0 ft < L r = 76.6 ft M odx = Nb M px - BF ( Lb - Lp) 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 Lr , 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 Mino r a xis bend ing 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.)



P11.25.

Chapter 11

page 11-40

Solve Example 11.15.5, if the solution is limited to W12-shapes. [EXAMPLE 11.15.5: A W 10 -sh ap e. is use d a s a conti nuou s to p c ho rd mem ber of the 12 8 ft lo ng roof truss of an industrial building. The eight-panel Pratt truss is 12 ft deep at the center and 8 ft deep at the sup po rtin g c olu mn s. The to p c ho rd sup po rts purlins at the pa nel poin ts and mid way betwee n th e pan el po ints. Under facto red gra vity loa ds, ea ch pu rlin transmit s a loa d o f 35 k ips (en d p urlin 17 .5 kip s). These loads include pro vision 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 U4 U5 is the critical element of the truss top-chord. It acts as a be am-colum n a nd is to be designed fo r:

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 pu rlin s. Th us, 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 1 x = 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 W12×7 9-shape. c.

Ax ial an d b endin g s treng ths W12×79: A = 23.2 in.2 ;

r = 5.34 in. ; x

r = 3.05 in. y

From LRFDS Table 3-50, for



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 < L p = 10.8 ft, d.

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.


(Ans.)



P11.26.

Chapter 11

page 11-42

Solve Example 11.15.6, if the solution is limited to W12-shapes. [EXAMPLE 11.15.6: A W 14 colum n, 14 ft lo ng , is pa rt o f an un braced fra me in the pla ne of t he web an d pa rt o f a braced frame in t he perpe nd icu lar dir ection suc h tha t K x = 1.5 an d K y = 1.0 . First-order fac tored loa d a na lysis g ives an axi al loa d o f 650 kip s und er ( 1.2 D + 0.5 L + 1.6 W), the no n-s way gravi ty moments Mnt under (1.2D + 0.5L) and the sway moments MRt under 1.6W, as shown in Fig. X11.15.6. For the story und er consideration, 3 P u = 20 ,00 0 k ips . The a llowab le stor y d rift ind ex is 0 .00 2 d ue to total horizontal (unfactored) wind forces 3 H = 500 k ips . Select the lightest sect ion usi ng A9 92 stee l.] See Fig. X11.15.6 of the text book . Solution a.


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

Pre limi na ry s electio n Assume B 1 x = 1.0, and B 2 x = 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 P req = 1130 kips; KL = 14 ft; F = 50 ksi, to find that for a W12 ×120 , y Pd y = 1220 kips

> P req = 1130 kips

and

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

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

The laterally unbraced length L b is 14 ft. From beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W12×1 20 of F y = 5 0



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 Nex t B 1 x 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.)



P11.27.

Chapter 11

page 11-44

Solve Example 11.15.7, if the solution is limited to W12-shapes. [ EXAM PLE 11 .15 .7: Sel ect a W 14 sec tion o f A9 92 steel fo r a 14 ft lo ng beam -co lum n, p art of an unbraced frame in y-plane and part of a braced frame in the x-plane. K x = 1.4 and Ky = 1.0. Factored axial load on column Pu is 840 kips. First order, single curvature moments under gravity loads are: Mntx = 280 ft-kips, M nty = 40 ft-kips. The first order reverse-curvature moments under factored wind loads are M Rtx1 = 20 0 ft-kips an d M Rtx2 = 50 ft-k ips. The d rift ind ex i s 1/40 0 u nd er E H = 10 0 k ips . Total fac tor ed gravi ty loa d a bo ve t his story i s 6,80 0 k ips .] See Fig. X11.15.7 of the text book . Solution a.

Da ta 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 P u = 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; E H = 100 kips; E P u = 6,800 kips

b.

Pre limi na ry s electio n Assume B 1 x = 1.05; B 1 y = B 2 x = 1.2 M ux* = B 1 x M ntx + B 2 x M ltx = 1.05 × 280 + 1.2 × 200 = 534 ft-kips M *uy = B 1 y 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.. Assum e 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, P dy = 1950 kips > 1850 kips. Also r x /r y = 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, P u eq = 840 + 1.6 × 534.0 + 1.6 × 2.11 × 48.0 = 1856 kips The W 12×190 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, P d = 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: P ex ( KL )2 = 54,100 × 10 - 4; P ey ( KL )2 = 16,900 × 10 - 4



Chapter 11

page 11-45

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

;

Also,

We obtain M ux* = B 1 x M ntx + B 2 x M ltx = 1.05 × 280 + 1.20 × 200 = 534 ft-kips M uy* = B 1 y M nty = 1.16 × 40.0 = 46.4 ft-kips From LRFDM Table 5-3, for a W12×190 and F = 50 ksi: y Nb M px = 1170 ft-kips; L p = 11.5 ft N b M py = 523 ft-kips Lr = 76.6 ft; BF = 5.79 As L p = 1 1.5 ft < L b = 14.0 ft < L r = 76.6 ft M odx = Nb M px - BF ( Lb - Lp) 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:


(Ans.)



P11.28.

Chapter 11

page 11-46

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 pa rt o f a sym metric un bra ced frame ab ou t both axe s. Un der fa cto red loa ds, Pu = 60 0 k ips, 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, E P ui = 10 ,80 0 k ips, E P e2x = 12 0,0 00 kip s a nd E Pe2y = 82 ,00 0 k ips .] See Fig. X11.15.8 of the text book . Solution a.

Da ta L = 16.0 ft; K x = 1.5; Ky = 1.3; = 600 kips; P u M ltx = 240 ft-kips; E P ui = 10,800 kips; From Eq. 11.9.9 ( LRFDS

= L b K x L x = K y L y = M ntx = M lty = E P e2 x = 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;

EPe2y

= 8 2,000 kips

M ux* = B 1 x M ntx + B 2 x M ltx = 0 + 1 .10 × 240 = 264 ft-kips M uy* = B 1 y M nty + B 2 y M lty = 0 + 1.15 × 100 = 115 ft-kips b.

Pre limi na ry s electio n Entering Table 11.14.1 with KL = K y L y = 20.8 ft, F = 50 ksi and W14-shapes we obtain m = y 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 . 1 193 kips. Also, u = 1.99 from Table 11.14.2, resulting in: = 600 + 1.2 × 264 + 1.2 × 1.99 × 115 = 1191 kips



Chapter 11

page 11-47

So, try a W14×132. c.

Check the selected section From LRFDM T able 5-3, for a W14×132 of A992 steel: L p = 13.3 ft; L r = 49.7 ft Nb M px = 87 8 ft-kip s; B F = 6 .8 8 k ip s;

Nb M py = 419 ft-kips

With equal, reverse-curvature moments at the ends C b = 2.27 As L p = 1 3.3 ft < L b = 16.0 ft < L r = 49.7 ft M odx = Nb M px - BF ( Lb - Lp) M dx = min [

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

The axial load ratio,

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

LHS

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

(Ans.)



P11.29.

Chapter 11

page 11-48

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 fac tor ed loa ds: P u = 40 0 k ips ; sy mm etri c si ng le cu rva tur e mom ents, M ntx = 15 0 ft-kips, M nty = 50 ft-ki ps, 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 direc tion ( causing ben din g a bo ut yy a xis) . Total factor ed gravi ty lo ad ab ove this lev el is E P u = 7,2 00 kips.] See Fig. X11.15.9 of the text book . Solution a.

Da ta L K x K y Pu M ntx M ltx E P u

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

Drift index, – –

b.

L b K x L x K y L y 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

, correspon ding 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.

Pre limi na ry s electio n Assume B 1 x = 1.05; B 1 y = 1.10. From LRFDS Eq. C1-1: M *ux = B 1 x M ntx + B 2 x M ltx = 1.05 × 150 + 1.14 × 116 = 290 ft-kips * M uy = B 1 y M nty + B 2 y M lty = 1.10 × 50.0 + 1.21 × 72.0 = 142 ft-kips From Table 11.14.1, for a W12 column of F y = 50 ksi steel and KL = K y L y = 18 ft, the coefficient m = 1 .4. Assum e u = 2.0. P ueq = 400 + 1.4 × 290 + 1.4 × 2.0 × 142 = 1200 kips



Chapter 11

page 11-49

By interpolation in the KRFDM Table 4-2, corresponding to KL = K y L y = 18 ft and F = 50 ksi, y we observe that a W12×136 has Pd = 1210 kips ( > P req = 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 W12×1 36, from Table 11.14.2. The revised value of P ueq = 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 F y = 50 ksi, KL = K y L y = 18 ft, P d = 1210 kips; P ex ( KL )2 /104 = 35,500; P ey ( KL )2 /1 04 = 1 1,4 00 The magnification factors B 1 x and B 1 y are determined, using co nservatively, ( 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 T able 5-3, for a W12×136 of A992 steel: L p = 11.2 ft; L r = 55.7 ft B F = 5 .4 9 k ip s; Nb M px = 80 3 ft-kip s; We will conservatively assume that C b = 1.0. As L p = 11.2 ft < L b = 1 5.0 ft < L r = 55.7 ft and C b = 1.0

Nb M py = 361 ft-kips



Chapter 11

page 11-50

M dx = Nb M px - BF ( Lb - Lp) M dy = Nb M py = 361 ft-kips

As the axial load ratio,

, we use Eq. 11.9.11 a.

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

Select a W12 ×136 of A992 steel.

(Ans.)



P11.30.

Chapter 11

page 11-51

Solve Example 11.15.10, if the solution is limited to W14-shapes. [EXAMPLE 11 .15.10: It is req uir ed to desig n a n inte rior cr an e c olu mn for a m ill b uil din g. The c olu mn of A9 92 stee l an d 2 8 ft high carrie s two cra ne girders in a dd itio n to the roo f lo ad . The c ran e ru nwa y o n e ach sid e o f th e c olu mn is to b e sup po rted by mean s of a bracket weld ed to the flang e o f th e c olu mn at a d istan ce 12 ft below the top. The roof covering including dec king, 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 conside ration is 1800 sq. ft. Maximu m vertical load from wheels, including impact of wheels is given as 40 kips. The horizonta l 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 supp orted at bra cke t le vel by the cr an e g ird ers. ] See Fig. X11.15.10 of the text book. Solution a.

Fac tored loa ds The column will be designed for an axial load Pu of 228 kips and a bending mom ent, M ltx = 114.0 ft-kips (for details see Example 11.15.10). Also, we have

b.

Pre limi na ry m ember se lect ion The equivalent axial load on the beam-column is: Assume minor axis buckling controls the design axial strength. For a W14 colu mn with KL = K y L y = 16 ft, m = 1.3 from Table 11.14.1. Assume B 1 x = 1.2. So

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. 11.9.11 a.



d.

Chapter 11

page 11-52

Be am ac tion From LRFDM T able 5-3, for a W14×61 beam of A992 steel L p = 8.65 ft; Nb M px = 3 83 ft-kips; B F = 6 .5 0 kip s 4 L r = 25.0 ft; Ix = 640 in. The lower segment which is longer, subjected to heavier axial load, and higher maximum bending momen t, 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 1 x 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.

Lim it st ate of stre ng th


(Ans.)

No te: T he next li gh ter W1 4 ( a W1 4× 53 ) will n ot wo rk ( LHS = 1.0 4).



P11.31.

Chapter 11

page 11-53

Solve Example 11.1 5.11, if the portal frame is fixed at the base. Limit the selection of columns to W14shapes. [EXAMPLE 11 .15.11: Des ign the memb ers of the hing ed-ba se, r igid join ted po rtal fram e ABC D sho wn i n Fig . 5. 1.2 , stu die d in Exa mp le 5 .3.1 . The b eam B C is sub jected to a u niformly dis trib ute d se rvic e lo ad of 2.5 k R f (1 k R f D + 1 .5 k R f S). In ad dition , th e fr am e is subje cted to a c on cen tra ted de ad loa d o f 30 k ips at the column tops B and C. The wind load on the frame consists of a 15 kip horizontal force acting at the joi nt B. The d ead loa ds given inc lude provi sio n for self weig ht of m em ber s. Use A9 92 W-sha pes with the ir 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 q uarter 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 suppo rts. 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.



P11.33.

Chapter 11

page 11-54

Redesign th e crane runway beam o f 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 perpend icular to the beam, at each wheel, 4¼ in. above the top flange. [ EXAM PLE 10 .4.1 0: A 1 8 ft lo ng , cra ne runwa y beam c arries the two end whe els of a cra ne. Th e wh eels are spaced 12 ft on centers and each transfers a maximum vertical live load of 24 kips to the beam. Ne glect the lat era l fo rce an d lon git ud ina l fo rce on the memb er. The cr an e b eam is wit hout latera l su pp ort 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.

Re qu ired st ren gth s 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 d ue to live load,

Required bending strength about major axis, M ux = 1.2 M D + 1.6 M 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 on ly).



Chapter 11

page 11-55

Entering the beam selection plots for W-Shapes (LRFDM Table 5-5) , with C b = 1, F = 50 ksi and y 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.

P11.34.

(Ans.)

A crane runway girder 24 ft in length is to be designed to carry the two end wh eels of a 5-ton crane. Lateral suppo rts are provided at the ends only. The whee ls of the crane are 8 ft on centers and each wheel transfers a maximum load of 12.5 kips to the top o f 60-lb rail. Assume 10 p ercent 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.



P11.35.

Chapter 11

page 11-56

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 roo f 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 = 1.2 D + 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 22.6o = 0.403 klf o q ux = q uV sin 2 = 0.437 sin 22.6 = 0.168 klf No te that q uy pro du ces major axis mome nt, wh ile q ux pro du ces min or axis mo men t. Th e p url in a cts as a simple beam of span L (= 18 ft) for major axis bendin g. It is not laterally suppo rted. 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 bend ing strengths are:

Or,

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


Solver Vinnakota Cap 11

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