Solucionario vinnakota Capitulo 8.pdf

Steel Structures by S. Vinnakota

Chapter 8

page 8-1

CHAPTER 8: COLUMNS

P8.1.

Plot the LRFDS column curve NFcr / Fy vs. 8c. Show all the salient points.

P8.2.

Use the LRFD Specification and plot the design axial compressive stress NF cr vs. effective slenderness ratio KL/r for steel columns. Include the following steels. (a) A36 (b) A572 Grade 50 (c) A572 Grade 65 (d) A514 Grade 90 (e) A514 Grade 100

P8.3.

Determine the elastic flexural buckling stress and elastic flexural buckling load for the pin-ended columns using the Euler equation. Assume E = 29,000 ksi. A solid square bar 2.0 in. by 2.0 in. (a) L = 6.0 ft (b) L = 12 ft Solution Pin ended column Section: Square b = 2.0 in.; d = 2.0 in. A = bd = 2.0 × 2.0 = 4.00 in.2 I = b d 3/ 12 = 2.0 × 2.0 3 ÷ 12 = 1.33 in.4 a.

b.

Column length, L = 6.0 ft = 72.0 in. Elastic flexural buckling load,

(Ans.)

Elastic flexural buckling stress, fE = PE / A = 73.4 ÷ 4.00 = 18.4 ksi

(Ans.)

Column length, L = 12.0 ft = 144 in. Elastic flexural buckling load,

(Ans.)

Elastic flexural buckling stress, fE = PE / A = 18.4 ÷ 4.00 = 4.60 ksi

(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

P8.4.

Chapter 8

page 8-2

Determine the elastic flexural buckling stress and elastic flexural buckling load for the pin-ended columns using the Euler equation. Assume E = 29,000 ksi. A W 12×96 (a) L = 50 ft (b) L = 25 ft Solution Pin ended column Section: W 12×96 6 A = 28.2 in.2; a.

Ix = 833 in.4;

Iy = 270 in.4

Column length, L = 50 ft = 600 in. From Eq. 8.4.19:

From Eq. 8.4.20, elastic flexural buckling load, (Ans.) Elastic flexural buckling stress, fE = PE / A = 215 ÷ 28.2 = 7.62 ksi

b.

(Ans.)

Column length, L = 25 ft = 300 in. From Eq. 8.4.19:

From Eq. 8.4.20, elastic flexural buckling load, (Ans.) Elastic flexural buckling stress, fE = PE / A = 859 ÷ 28.2 = 30.5 ksi

(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

P8.5.

Chapter 8

page 8-3

Determine the effective lengths of each of the columns of the frame shown in Fig. P8.5. The members are oriented so that the webs are in the plane of the frame. The structure is unbraced in the plane of the frame. All connections are rigid, unless indicated otherwise. In the direction perpendicular to the frame, the frame is braced at the joints. The connections at these points of bracing are simple connections (no rotational restraint). (a) Use the alignment charts. (b) Use the equations given in Section 8.5. See Fig. P8.5 of the text book. Solution Unbraced frame. Column sections: W10×39 6 Ix = 209 in.4; W10×45 6 Ix = 248 in.4 W10×49 6 Ix = 272 in.4 Girder sections: W16×40 6 Ix = 518 in.4; W16×50 6 Ix = 650 in.4 4 W14×30 6 Ix = 291 in. ; W14×34 6 Ix = 340 in.4 Column lengths: c1, c2, c3, c4 = 14 ft; c5, c6 = 12 ft Girder lengths: g1 = 24 ft; g2, g4 = 16 ft; g3 = 32 ft a. Using alignment charts Column c1 "g1 = 0.5 (far end pinned) G A = 10.0 (recommended value for hinged base)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 1.98

(Ans.)

Column c4 "g3 = 0.5 (far end pinned) G A = 10.0 (recommended value for hinged base)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 2.05

(Ans.)

Column c2

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 8

page 8-4

"g1 = 0.0 (near end pinned);

"g2 = 1.0 (both ends rigidly jointed) G A = 1.0 (recommended value for fixed base)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 1.45

(Ans.)

Column c3 "g3 = 0.0 (near end pinned); "g2 = 1.0 (both ends rigidly jointed) G A = 1.0 (recommended value for fixed base)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 1.45

(Ans.)

Column c5 G A = 1.98 (same as G B value calculated for column c2)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 1.30

(Ans.)

Column c6 G A = 1.98 (same as G B value calculated for column c3)

From alignment chart for sidesway uninhited columns (Fig. 8.56b), K x . 1.30

(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

b.

Chapter 8

page 8-5

Using the equations given in Section 8.5 From Eq. 8.5.14, the effective length of a column in an unbraced frame is,

Column c1:

G A = 10.0;

G B = 1.38;

K x . 1.99

(Ans.)

Column c2:

G A = 1.00;

G B = 1.98;

K x . 1.47

(Ans.)

Column c3:

G A = 1.00;

G B = 1.98;

K x . 1.47

(Ans.)

Column c4:

G A = 10.0;

G B = 1.72;

K x . 2.10

(Ans.)

Column c5:

G A = 1.98;

G B = 1.25;

K x . 1.50

(Ans.)

Column c6:

G A = 1.98;

G B = 1.25;

K x . 1.50

(Ans.)

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

Steel Structures by S. Vinnakota

P8.6.

Chapter 8

page 8-6

Repeat Problem P8.5, assuming the structure is braced in the plane of the frame also, by diagonal members in the middle bay. Solution Braced frame. Column sections: W10×39 6 Ix = 209 in.4; W10×45 6 Ix = 248 in.4 W10×49 6 Ix = 272 in.4 Girder sections: W16×40 6 Ix = 518 in.4; W16×50 6 Ix = 650 in.4 4 W14×30 6 Ix = 291 in. ; W14×34 6 Ix = 340 in.4 Column lengths: c1, c2, c3, c4 = 14 ft; c5, c6 = 12 ft Girder lengths: g1 = 24 ft; g2, g4 = 16 ft; g3 = 32 ft a. Using alignment charts Column c1 "g1 = 1.5 (far end pinned) G A = 10.0 (recommended value for hinged base)

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.81

(Ans.)

Column c4 "g3 = 1.5 (far end pinned) G A = 10.0 (recommended value for hinged base)

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.83

(Ans.)

Column c2 "g1 = 0.0 (near end pinned); "g2 = 1.0 (both ends rigidly jointed) G A = 1.0 (recommended value for fixed base)

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Steel Structures by S. Vinnakota

Chapter 8

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.81

page 8-7

(Ans.)

Column c3 "g3 = 0.0 (near end pinned); "g2 = 1.0 (both ends rigidly jointed) G A = 1.0 (recommended value for fixed base)

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.81

(Ans.)

Column c5 G A = 1.98 (same as G B value calculated for column c2)

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.83

(Ans.)

Column c6 G A = 1.98 (same as G B value calculated for column c3)

From alignment chart for sidesway inhited columns (Fig. 8.56a), K x . 0.83

(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

b.

Chapter 8

page 8-8

Using the equations given in Section 8.5 From Eq. 8.5.18, the effective length of a column in an unbraced frame is,

Column c1:

G A = 10.0;

G B = 0.461;

K x . 0.81

(Ans.)

Column c2:

G A = 1.00;

G B = 1.98;

K x . 0.81

(Ans.)

Column c3:

G A = 1.00;

G B = 1.98;

K x . 0.81

(Ans.)

Column c4:

G A = 10.0;

G B = 0.573;

K x . 0.83

(Ans.)

Column c5:

G A = 1.98;

G B = 1.25;

K x . 0.83

(Ans.)

Column c6:

G A = 1.98;

G B = 1.25;

K x . 0.83

(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

P8.7.

Chapter 8

page 8-9

Determine the factored axial compressive strength of an 18 ft long W8×48 column fixed about both axes at its ends. Assume A242 Grade 50 steel. Solve using specification equations only. Check calculations by using various tables in the LRFD Manual.

Solution Given Member: W8 × 48 6 A = 14.1 in.2; rx = 3.61 in.2; ry = 2.08 in. Material: A242 Gr 50 6 Fy = 50 ksi Length, L = 18ft As there are no intermediate supports, L x = L y = 18 ft End conditions: Fixed at both ends about both axes. From Fig. 8.5.4a: K x = K y = 0.65 As the end conditions are the same about both the axes, and as there are no intermediate braces, minor axis buckling controls. So,

a.

Using LRFDS Equations in Section E2 From Eq. E2-4 of the LRFDS, slenderness parameter,

As 8c < 1.5, buckling occurs in the inelastic domain. Use LRFDS Equation E2-2, Fcr =

Fy 6

Fcr = 35.8 ksi

ˆ Pd = Nc Fcr A g = 0.85 × 35.8 × 14.1 = b.

429 kips

(Ans.)

Using LRFDS Table 4 Enter LRFDS Table 4 with 8c = 0.892 and read,

ˆ Pd = Nc Fcr A g = 0.609 × 50 × 14.1 = 429 kips c.

(Ans.)

Using LRFDS Table 3-50 Enter LRFDS Table 3-50, with KL /r = 67.5, and obtain Nc F

ˆ Pd = Nc Fcr A g = 30.5 × 14.1 = 430 kips

cr

= 30.5 ksi by interpolation. (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

P8.8.

Chapter 8

page 8-10

A W12×65 column 30 ft long is fixed at both ends about both axes. In addition it is braced in the weak direction at the one-third points. A242 Grade 42 steel is used. Determine the design axial compressive strength of the column. Solution Member: W12×65 6 A g = 19.1 in.2; rx = 5.28 in.; Material: A242 Gr 42 6 Fy = 42 ksi Column length, L = 30ft For major axis buckling: L x = 30.0 ft; fixed - fixed 6 Kx = For minor axis buckling: L y1= 10.0 ft; fixed - pinned 6 Ky = L y2 = 10.0 ft; pinned - pinned 6 Ky = L y3 = 10.0 ft; pinned - fixed 6 Ky = K x L x = 0.65 × 30.0 = 19.5 ft (K y L y)1 = 0.65 × 10.0 = 6.5 ft = (K y L y)3; K y L y = 10.0 ft

ry = 3.02 in.

0.65 0.80 1.00 0.80

(K y L y)2 = 1.0 × 10.0 = 10.0 ft

;

Slenderness parameter,

Enter LRFDS Table 4 with 8c = 0.537 and obtain,

6

Nc Fcr = 0.753 × 42 = 31.6 ksi

ˆ Pd = Nc Fcr A g = 31.6 × 19.1

= 604 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

P8.9.

Chapter 8

page 8-11

A S10×35 standard I-shape is used as a column in an industrial building. The column considered is 21 ft long. It can be assumed to be hinged about both axes at the top and fixed about both axes at the base. In addition, transverse braces (girts) are connected to the web of the column. What factored axial load is permitted by the LRFDS, if the bracing is positioned at: (a) points 7 ft apart (b) points 5 ft 3 in. apart?

Solution Member: S10 × 35 6 A = 10.3 in.2; rx = 3 .78 in.; ry = 0.898 in. Material: A36 steel (Preferred material specification, for S-shapes) 6 Fy = 36 ksi Length, L = 21 ft L x = 21 ft; fixed-pinned 6 K x = 0.80 6 K x L x = 0.8 × 21 = 16.8 ft

a.

Bracing at 7 ft intervals 3 segments. L y1 = 7.0 ft; fixed-pinned 6 K y1 = 0.8 6 L y2 = 7.0 ft; pinned-pinned 6 K y2 = 1.0 6 L y3 = 7.0 ft; pinned-pinned 6 K y3 = 1.0 6 K y L y = 7.00 ft

(K y L y)1 = 0.8 × 7.0 = 5.60 ft (K y L y)2 = 1.0 × 7.0 = 7.00 ft (K y L y)3 = 1.0 × 7.0 = 7.00 ft

Enter LRFDS Table 3-36 and read,

ˆ Pd = Nc Fcr A g = 19.3 × 10.3 = 199 kips b.

Bracing at 5 ft 3 in. apart 4 segments. L y1 = 5.25 ft; fixed-pinned 6 K y1 = 0.8 6 L y2 = 5.25 ft; pinned-pinned 6 K y2 = 1.0 6 K y L y = 5.25 ft

(Ans.)

(K y L y)1 = 0.8 × 5.25 = 4.20 ft (K y L y)2 = 1.0 × 5.25 = 5.25 ft

Enter LRFDS Table 3-36, and read,

ˆ Pd = Nc Fcr A g = 23.6 × 10.3 = 243 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

P8.10.

Chapter 8

page 8-12

A W 12×72 of A588 Grade 50 steel is used as a column 16 feet long. The column ends are pinned and the weak axis is braced 10 ft from the lower end. The structure is braced in both xx and yy planes of the column. Determine the design axial compressive strength of the column. Solution Section: W 12×72 6 A = 21.1 in.2; rx = 5.31 in.; r y = 3.04 in. Material: A588 Grade 50 steel 6 Fy = 50 ksi Column length, L = 16 ft Structure braced in both xx and yy planes of the column 6 K x # 1.0; K y # 1.0 For buckling about the major axis: 6 pinned at both ends 6 K x = 1.0, L x = 16 ft For buckling about the minor axis: 6 pinned at both ends and brace at 10 ft from base 6 (K y L y)1 = 1.0 × 10.0 = 10.0 ft; (K y L y)2 = 1.0 × 6.0 = 6.00 ft 6 K y L y = max (10.0; 6.00) = 10.0 ft ;

From LRFDM Table 3-50, for K L/r = 39.5, Nc Fcr = 37.9 ksi Design axial compressive strength, Pd = = Nc Fcr A g = 37.9 × 21.1 = 800 kips Alternatively For the W12x72 column given, K x L x = 16.0 ft; From LRFDM Table 4-2, for a W12×72,

(Ans.)

K y L y = 10 ft

From LRFDM Table 4-2, for a W12×72 with KL = 10 ft, Pd = 800 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

P8.11.

Chapter 8

page 8-13

A W12×58 steel shape, strengthened by welding two d× 8 in plates to the outside face of the flanges is used as a column. The member has a length of 15 ft and is assumed to be pin connected at both ends. It is braced at mid-length to prevent movement in the x direction only. Determine the factored axial compressive strength of the column, if A572 Grade 60 steel is used.

Solution Section: W12×58 with two d×8 plates W12×58: A W = 17.0 in.2; IxW = 475 in.4; IyW = 107 in.4 Plates : A pl = 2 × 8 × 3/8 = 6.0 in.2 Material: A572 Gr 60; Fy = 60 ksi Column length, L = 15 ft L x = 15.0 ft; end conditions: pinned-pinned 6 K x = 1.00 L y1 = L y2 = 7.5 ft; end conditions: pinned-pinned 6 K y = 1.00

Properties of built-up section A = 17.0 + 6.0 = 23.0 in.2 Ix = [475 + 0] + 2 [0 + 8.0× 3/8 × (6.29)2] = 712 in.4 Iy = [107 + 0] + 2[1/12 × (3/8) (8.0)3] = 139 in.4

Slenderness parameter,

Enter LRFDS Table 4 with 8c = 0.530 and obtain,

ˆ Pd = Nc Fcr A g = 0.756 × 60 × 23.0 = 1040 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

P8.12.

Chapter 8

page 8-14

A 20 ft long, A572 Grade 42 steel column is obtained by welding a ½ × 12 in. plate to two MC 13 × 50 channel shapes, to form a doubly symmetric section as shown in Fig. P8.12. Determine the factored axial compressive strength of the column as per the LRFDS. Assume K x = 1.0 and K y = 1.4. See Figure P8.12 of the text book. Solution Column length = 20ft Given: K x = 1.00, K y = 1.4 Dubly symmetric section built-up from: MC 13×50: A C = 14.7 in.2; IxC = 314 in.4; IyC = 16.4 in.4 twC = 0.787 in.; x = 0.974 in. Plate ½×12: A pl = 6.0 in.2 For the built-up section: A = 14.7×2 + 12× ½ = 35.4 in.2 Ix = = 628 + 0 = 628 in.4 d 1 = 6 + 0.787 - 0.974 = 5.813 in. = 1100 in.4

Iy =

;

Slenderness parameter,

Enter LRFDS Table 4 with 8c = 0.730 and obtain,

ˆ Pd = Nc Fcr A g = 0.680 × 42 × 35.4 = 1010 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

P8.13.

Chapter 8

page 8-15

The cross-section of a structural steel column is a built-up section obtained by welding the web of a W12×26 shape to the flange of a W8×48 to form a mono-symmetric section shown in Fig. P8.13. The 20-ft long column is fixed at the base and pinned at the top about both axes. The column is part of a braced frame in both the xx and yy planes. Assume A572 Grade 65 steel. Determine the design axial compressive strength of the column. See Figure P8.13 of the text book. Solution Column length, L = 20 ft No intermediate braces 6 L x = L y = 20.0 ft End conditions: fixed at the base and pinned at the top 6 K x = 0.80; K y = 0.80 Material: A572 Gr 65 6 Fy = 65 ksi Mono-symmetric section built-up from two W-shapes: W8×48: A 1 = 14.1 in.2; Ix1 = 184 in.4; Iy1 = 60.9 in.4 d 1 = 8.50 in. W12×26: A 2 = 7.65 in.2; Ix2 = 204 in.4; Iy2 = 17.3 in.4 tw2 = 0.230 in. For the built-up section: A = 14.1 + 7.65 = 21.8 in.2 from bottom fibre

d 1 = 5.77 - 4.25 = 1.52 in.;

d 2 = 8.615 - 5.77 = 2.845 in.

Ix = [184 + 14.1×1.52 2] + [17.3 + 7.65×2.845 2 ] = 296 in.4 Iy = [ 60.9 + 0] + [204 + 0] = 265 in.4 rx = 3.68 in.;

ry = 3.49 in.

Slenderness parameter, Enter LRFDS Table 4 with 8c = 0.829 and obtain,

ˆ Pd = Nc Fcr A g = 0.637 × 65 × 21.8 = 903 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

P8.14.

Chapter 8

page 8-16

A cantilever column (fixed at the base and free at the top, about both axes) is obtained by welding two WT 9 × 53's to the web of a W24 × 76 section, to form a doubly symmetric compound section shown in Fig. P8.14. The column is 20 ft long. Assume A572 Grade 60 steel and determine the design axial compressive strength of the column as per the LRFD Specification. See Figure. P8.14 of the text book. Solution Column length, L = 20 ft No intermediate braces 6 L x = L y = 20.0 ft End conditions: fixed at the base and free at the top The recommended value for the effective length factor of a cantilever column is 2.1, from Fig. 8.5.4. Hence, K x = 2.10; K y = 2.10. Material: A572 Gr 60 6 Fy = 60 ksi Doubly symmetric section built-up from: W24×76 6 A W = 22.4 in.2; twW = 0.44 in.; IxW = 2100 in.4; IyW = 82.5 in.4 WT 9×53 6 A T = 15.6 in.2; d T = 9.365 in.; I xT = 104 in.4, I yT = 110 in.4 y T = 1.97 in. For the built-up section: A = 22.4 + 2 × 15.6 = 53.6 in.2 Ix = [2100 + 0] + [2×110 + 0] = 2320 in.4 Iy = [82.5 + 0] + 2 [104 + 15.6 (0.22 + 9.365 - 1.97)2] = 2100 in.4

;

Slenderness parameter, Enter LRFDS Table 4 with 8c = 1.17 and obtain, So, the design strength of the column is Pd = Nc Fcr A g = 0.479 × 60 × 53.6 = 1540 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

P8.15.

Chapter 8

page 8-17

A heavy column used in the ground floor of a high-rise building is formed by welding two 4×24 in. plates to the flanges of a W14×730 section as shown in Fig. P8.15. Assume A572 Grade 42 steel and determine the design axial compressive strength of the column, if K x L x = 32 ft and K y L y = 28 ft. Use the LRFD Specification. See Figure P8.15 of the text book. Solution Given: K x L x = 32 ft; K y L y = 28 ft Material: A572 Gr 42 6 Fy = 42 ksi Doubly symmetric section built-up from: W14×730: A W = 215 in.2; b fW = 17.9 in.; IxW = 14,300 in.4; IyW = 4,720 in.4 PL 24 × 4 : A pl = 24 × 4 = 96 in.2;

For the built-up section: A = 215 + 2 (96) = 407 in.2 Ix = [14300 + 0] + 2 × [4608 + 0] = 23,520 in.4 Iy = [4720 + 0] + 2 × [128 + 96 (8.95 + 2.0)2] = 28,000 in.4 ;

For the column: controls;

Slenderness parameter, Enter LRFDS Table 4 with 8c = 0.612 and obtain, So, the design strength of the column is Pd = Nc Fcr A g = 0.727 × 60 × 407 = 12,430 kips

(Ans.)

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

Steel Structures by S. Vinnakota

Chapter 8

page 8-18

P8.16.

Determine the design axial compressive strength of the columns considered in the Problem P8.5, assuming A992 steel is used.

P8.17.

Determine the design axial compressive strength of the columns considered in the Problem P8.5, assuming A992 steel is used.

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

P8.18.

Chapter 8

page 8-19

An eight-story office building has a bay size of 40 feet by 36 feet. The structure is designed for a roof dead load of 80 psf, a roof live load of 20 psf, a floor dead load of 80 psf, and a floor live load of 50 psf. Make a preliminary design, and select a suitable W14-shape of A992 steel for an interior column at ground level. Assume that K x = K y = 1.0 and L x = L y = 14 ft. Solution Given: K x L x = 14.0 ft; K y L y = 14.0 ft Tributary area at each level = 40 × 36 = 1440 ft2 The column under consideration receives loads from roof and seven floors. Roof dead load = 80 psf Roof live load, L r = 20 psf Floor dead load = 80 psf Floor live load, L = 50 psf With only gravity loads acting on the column, load combination LC-4 controls the required axial strength of the column 1.2D + 1.6L + 0.5L r Contribution to axial load, from roof,

Contribution to axial load, from the seven floors,

So, factored load on the column, Pu = 153 + 1770 = 1920 kips Enter LRFDM Table 4-2 for W14 shapes with KL = K y L y = 14 ft, Preq = 1920 kips and select a W14×176 with Pd = 1940 kips. Provide a W14×176 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

P8.19.

Chapter 8

page 8-20

Select the lightest W section for a column supporting a factored axial load of 620 kips. Support conditions, applicable for both principal axes, are full fixity at the bottom and a pinned connection at the top. L x = L y = 17 ft 6 in. Assume A992 steel is used. Solution Material: A 992 steel 6 Fy = 50 ksi Factored load, Pu = 620 kips From Fig. 8.5.4c, for a column fixed at one end and pinned at the other end, the recommended value of K = 0.80. Hence, we have: K x L x = K y L y = 0.80 × 17.5 = 14.0 ft As KL is same about both axes, buckling in the weak direction will control the strength of the column. So, we enter the column selection tables in Part 4 of the LRFDM with a value of KL = 14 ft : W10 series 6 W10 × 68 6 Pd = 625 kips > 620 kips O.K. W12 series 6 W12 × 65 6 Pd = 647 kips > 620 kips O.K. W14 series 6 W14 × 74 6 Pd = 662 kips > 620 kips O.K. Use W12 × 65, the lightest section.

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

P8.20.

Chapter 8

page 8-21

The interior column of an industrial building supports a roof dead load of 55 psf and a snow load of 35 psf. The column is 25 feet long. The bottom end is fixed, and the top end pinned. In addition, its weak axis is braced by struts at a point 6 feet down from the column top as shown in Figure P8.20. The column has a tributary area of 1800 ft2. The structure is braced in both the xx and yy planes of the column. Select a suitable W shape of A588 Grade 50 steel. See Figure P8.20 of the text book. Solution Tributary area Dead load, D Snow load, S

= 1800 ft2 = 55 psf = 35 psf

Assuming 50 plf for self weight of column, dead load of column =

say 1.5

Factored axial load on the column,

As the structure is braced in both the xx and yy planes of the column, K x L x # 25 ft, and K y L y # 25 ft For buckling about major axis 6 Fixed at base and pinned at top 6 K x L x = 0.80 × 25 = 20 ft For buckling about the minor axis 6 Fixed at base, pinned at top and braced 19 ft from the base: 6 (K y L y)1 = 0.8 × 19 = 15.2 ft (K y L y)2 = 1.0 × 6 = 6.00 ft 6 K y L y = max [15.2; 6.00] = 15.2 ft Enter LRFDM Table 4-2 for W10 shapes with KL = K y L y = 15.2 and observe that a W10×39 has a Pdy = 263 kips > 221 kips. Also, resulting in:

So, major axis buckling will not control the design strength. So, select a W10×39 of A588 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

P8.21.

Chapter 8

page 8-22

Select the lightest W12 of A588 Grade 50 steel to carry an axial compressive load of 500 kips dead load and 800 kips live load. The 30 ft long truss member is assumed to be pinned at both ends about both axes. Make all checks. Repeat the problem if, in addition, the column is provided with weak direction support at mid-height.

Solution Service loads: D = 500 kips, L = 800 kips Use load combination, LC-2 6 P u = 1.2×500 + 1.6×800 = 1880 kips Column length, L = 30 ft End conditions: Pinned at both ends about both axes. Material: A588 Grade 50 6 Fy = 50 ksi a.

No intermediate bracing K xL x = K yL y = 1.00 × 30 ft = 30.0 ft. So, buckling about the minor axis will control the design. Enter Column Tables for W12 shapes, in LRFDM Table 4-2, with KL = K y L y = 30 ft and select a W12×336 that provides P d = P dy = 1910 kips > P u = 1880 kips. Check: For a W12×336: A = 98.8 in.2;

rx = 6.41 in.;

ry = 3.47 in.

;

Enter LRFDS Table 3-50 and read Nc F cr = 19.3 ksi

ˆ Pd = Nc Fcr A g = 19.3 × 98.8

= 1907 kips > 1880 kips

So, select a W12×336 of A588 Grade 50 steel.

b.

O.K. (Ans.)

Intermediate bracing provided at midheight K x L x = 1.00 × 30 ft = 30.0 ft; K y L y = 1.00 × 15 ft = 15.0 ft Enter Column tables for W12 shapes in LRFDM Part 4, with KL = K y L y = 15.0 ft and tentatively select a W12×190 with Pdy = 1900 kips > P u = 1880 kips. Also r x /r y = 1.79 for this section.

Reenter Column tables for W12 shapes, with KL = (K x L x)y = 16.8 ft and select a W12×210 with Pdx = 1992 kips > Pu = 1880 kips. For this section : A = 61.8 in.2 ;

r x = 5.89 in.;

r y = 3.28 in.

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

Steel Structures by S. Vinnakota

Chapter 8

page 8-23

;

From LRFDS Table 3-50, Nc Fcr = 32.4 ksi Pd = Nc Fcr A g = 32.4 × 61.8 = 2,000 kips So, select a W12×210 of A588 Grade 50 steel.

> 1,880 kips

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

P8.22.

Chapter 8

page 8-24

Select the lightest W14 of A572 Grade 50 steel to carry an axial compression load of 100 kips dead load and 360 kips live load. The 28 ft long column is pinned at both ends about both axes and in addition has weak direction support at mid height. Make all checks.

Solution Column length, L = 28 ft Material: A572 Grade 50 steel 6 Fy = 50 ksi Service loads: D = 100 kips; L = 360 kips Factored axial load on the column (from LC 4-2), Pu = 1.2×100 + 1.6×360 = 696 kips Column pinned at both ends about both axes, and has support at midheight for buckling about the minor axis, resulting in: K x L x = 1.0×28.0 = 28.0 ft; K y L y = 1.0×14.0 = 14.0 ft Enter Column Load Tables for W14 shapes, in LRFDM Part 4, with KL = K y L y = 14.0 ft and observe that for a W14×82, Pdy = 729 kips > Pu = 696 kips. Also, for this section r x /r y = 2.44

ˆ W14×82 with Pd = 729 kips is the lightest W14 shape. Checks 1. Strength For a W14 × 82, LRFDM Table 1-1 gives: A = 24.0 in.2; r x = 6.05 in;

r y = 2.48 in.

; From LRFDS Table 3-50, for

ˆ Pd = Nc Fcr A g = 30.4 × 24.0 = 730 kips

> 696 kips

O.K.

2. Local buckling of plates With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain for Flanges:

Web:

ˆ O.K.

ˆ O.K.

So, plate local buckling will not reduce the member design strength calculated above. So, select W14×82 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

P8.23.

Chapter 8

page 8-25

Design the lightest W-shape of A992 steel to support a factored axial load of 960 kips. The effective length with respect to its minor axis is 14 ft and the effective length with respect to its major axis is 30 ft.

Solution Column length, L = 14 ft Material: A572 Grade 50 6 Fy = 50 ksi Factored axial compressive load, Pu = 960 kips Given: K x L x = 30ft; K y L y = l4 ft a.

Member selection Enter LRFDM Table 4-2 for W10 shapes, with KL = K yL y = 14.0 ft, P req = 960 kips and observe that a W10×112 gives P dy = 1050 kips. Also, for this section, r x /r y = 1.74 resulting in (K x /L x)y = 17.2 > 14.0 ft Reentering the table with KL = (K x L x)y = 17.2 ft, we observe that the heaviest W10 section (W10 ×112) does not provide adequate axial strength. Next, enter LRFDM Table 4-2 for W12 shapes, with KL = K y L y = 14.0 ft, Preq = 960 kips and note that a W12×96 gives Pdy = 966 kips and has r x /r y = 1.76. ˆ (K xL x)y = 17.0 > 14.0 ft Reentering Table 4-2 with KL = (K x /L x)y = 17.0 ft, we observe that a W12×106 provides, Pd = 968 kips > P u = 960 kips. Proceeding in a similar manner, we observe that for a W14×90: Pdy = 969 kips; r x /r y = 1.66. ˆ (K x/L x)y = 18.1 > 14.0 ft Reentering Table 4-2 with KL = (K x /L x)y = 18.1 ft, we find that a W14×99 provides, P d = 962 kips > Pu = 960 kips. From W12×106 and W14×99, W14×99 is the lighter, so we select W14×99.

b.

Checks 1. Strength W14 × 99 : A = 29.1 in.2,

r x = 6.17 in.,

r y = 3.71 in. ;

From LRFDS Table 3-50, for , NcF cr = 33.1 ksi ˆ Pd = Nc Fcr A g = 963 kips > 960 kips 2. Local buckling of plates With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain 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 8

Flanges:

Web:

page 8-26

ˆ O.K.

ˆ O.K.

So, plate local buckling will not reduce the member design strength calculated above.

So, select a W14×99 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

P8.24.

Chapter 8

page 8-27

A column is built into a wall so that it may be considered as continuously braced in the weak direction. The column is 18 ft long and may be considered hinged at both ends with respect to strong axis buckling. The factored axial load to be carried is 700 kips. Use A992 steel. Select the lightest W-shape used as columns. See Fig. X8.5.1 f of the text book for guidance. Solution Column length, L = 18 ft Factored load, Pu = 700 kips Material: A992 steel Because column is continuously braced in the weak direction, K y L y = 0. The calculation is shown in a tabular form below. The loads and other information are from LRFDM Table 4-2. The first trial shape is obtained by entering the table with KL = 0. Section Pdy

W10 × 60

W12 × 58

W14 × 61

748

723

761

1.71

2.10

2.44

10.5

8.57

7.38

(kips)

r x /r y (K x L x)y

(ft)

Revised section Pdx

W10×68

(kips)

715

W12×65

W14×68

745

773

The lightest section is a W12 × 65. 1. Strength W12×65: A = 19.1 in.2;

r x = 5.28 in.; r y = 3.02 in.

From LRFDS Table 3-50, Nc F cr = 37.6 ksi Pd = Nc Fcr A g = 37.6 × 19.1 = 718 > 700 kips

ˆ O.K.

2. Local buckling of plates With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain for Flanges:

Web:

ˆ O.K. ˆ O.K.

So, plate local buckling will not reduce the member design strength calculated above. So, select a W12×65 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 8

page 8-28

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

P8.25.

Chapter 8

page 8-29

Repeat Problem P8.24, if the selection is extended to W-shapes up to 18-in. nominal depth. Solution Given: L = 18ft; Pu = 700 kips; Fy = 50 ksi K y Ly = 0 Aim is to see, if a lighter section than W12x65 (obtained in Problem 8.24) could be found by selecting a deeper section. The design calculation is given below in a tabular form. Section rx

(in)

W16 × 67

W16 × 57

W18 × 60

W18 × 65

6.96

6.72

7.47

7.49

31

32

28.9

28.8

39.62

39.43

39.99

40.00

(K x L x)/r x

Nc Fcr (ksi)1 A

(in.2)

19.7

16.8

17.6

19.1

Pd

(kips)

781

662 (NG)

704

764

7.7

5.0

5.4

5.1

13.5

13.5

13.5

13.5

35.9 (NG)

33.0

38.7 (NG)

35.7

35.8

35.8

35.8

35.8

b f /2t f

h /tw

Note 1: Values are from LRFDS Table 3-50. The lightest acceptable W-shape is a W18×65; as the web of the W18×60 shape does not satisfy the web local buckling criteria. Checks 1. Strength W18×65: A = 19.1 in.2;

r x = 7.49 in.; r y = 1.69 in.

From LRFDS Table 3-50, Nc F cr = 40.0 ksi Pd = Nc Fcr A g = 40.0 × 19.1 = 764 > 700 kips

ˆ O.K.

2. Local buckling of flange plates and local buckling of web plate are O.K. as shown in the table. So, select a W18×65 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

P8.26.

Chapter 8

page 8-30

A steel column is built into a wall so that it may be considered as continuously braced for buckling about its yaxis. If the factored load on the column is 1000 kips and K x L x = 22 ft, select the lightest W section for the column. The depth of the section provided is to be limited to 18 in. (nominal). See Fig. X8.5.1f of the text book. Solution Given: K y L y = 0; K x L x = 22 ft; Pu = 1000 kips Material: A572 Grade 50 6 Fy = 50 ksi a.

Member selection Enter Column Tables for W10 shapes in LRFDM Table 4-2, with K y L y = 0 and Preq = 1000 kips, and tentatively select a W10×88 with Pdy = 1100 kips and rx /ry = 1.73. (K x L x)y = 22/1.73 = 12.71 Re-enter the table and select a W10×112 with Pdx = 1102 kips. O.K. Use similar procedure to select a W12×96 and a W14×99, as the lightest W12 and W14 shapes. Deeper shapes are not tabulated in column tables. But, select sections weighing close to 99 plf and lighter, and verify if they provide adequate strength. Observe that: Lighest W16 ÷ W16×89 Lighest W18 ÷ W18×97 The lightest of all is a W16×89

b.

Checks 1. Strength W16×89: A = 26.4 in.2;

rx = 7.05 in.

From LRFDS Table 3-50, Nc F cr = 38.4 ksi Pd = Nc Fcr A g = 38.4 × 26.4 = 1014 > 1000 kips

ˆ O.K.

2. Local buckling of plates With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain for Flanges:

Web:

ˆ O.K.

ˆ O.K.

So, plate local buckling will not reduce the member design strength calculated above. Select a W16×89 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 8

page 8-31

See the design calculation shown in the following table

From LRFDM Table 4-2 Section

From LRFDM Table 1-1

W10×112

W12×96

W14×99

Section

W16×89

W18×97

Pdy (kips)

1,400

1,200

1,240

A (in.2)

26.4

28.5

r x /r y

1.74

1.76

1.66

rx

7.05

7.82

(in.) (K xL x)y (ft)

12.6

12.5

13.3

37.4

33.5

Pdx (kips)

1,106

1,008

1,081

1,014

1,114

b f /2t f

4.17

6.76

9.34

5.92

6.41

13.5

13.5

13.5

13.5

13.5

10.4

17.7

23.5

25.9

30.0

35.8

35.8

35.8

35.8

35.8

h/tw

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

P8.27.

Chapter 8

page 8-32

Select the lightest W12 shape of A588 Grade 50 steel to carry an axial compressive force of 600 kips under factored loads. Assume the member to be part of a braced frame in both planes. The idealized support conditions are that the member is hinged in both principal directions at the top of a 30 ft height; supported in the weak direction at 12 ft and 22 ft from the bottom; and fixed in both directions at the base.

Solution Factored axial compressive load, Pu = 600 kips Material: A588 Grade 50 6 Fy = 50 ksi Column length, L = 30 ft Column part of a braced frame in both planes 6 No relative translation of the top end with respect to the bottom end. For buckling about x-axis: fixed at base - hinged at top K xL x = 0.8 × 30 = 24.0 ft For buckling about y-axis: fixed at base -hinged at top and intermediate braces at 12 ft and 22 ft from base (K y L y)1 = 0.8 × 12 = 9.60 ft; (K y L y)2 = 1.0 × 10 = 10.0 ft; (K y L y)3 = 1.0 × 8 = 8.00 ft K y L y = 10.0 ft a.

Member selection Enter LRFDM Table 4-2 with KL = K y L y = 10.0 ft and tentatively select a W12 × 58 with Pdy = 611 kips and r x /r y = 2.10. So, (K x L x)y = 24.0 ÷ 2.10 = 11.4 > 10.0 ft Re-entering Table 4-2 with KL = (K x L x)y = 11.4 ft, we observe that the W12×58 gives Pdx = 581 kips < 600. We also observe that for sections heavier than W12×58, r x /r y . 1.75. So, (K x L x) y = 13.7 > 10.0 ft Re-entering LRFDM Table 4-2 with KL = (K x L x)y = 13.7 ft, we select a W12×65 with Pdx = 653 kips.

b.

Checks 1. Strength W12 × 65:

A = 19.1 in.2;

rx = 5.28 in.;

ry = 3.02 in.

;

Enter LRFDS Table 3-50 and read, NcF cr = 34.2 ksi ˆ Pd = Nc Fcr A g = 653 kips > 600 kips O.K. 2. Local buckling of plates Verify and show O.K. So, select a W12×65 of A588 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 8

page 8-33

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

P8.28.

Chapter 8

page 8-34

The roof structure of an industrial building is supported by A992 steel columns, located between the roof beam and floor girder as shown in Fig. P8.28. The columns are connected in such a manner that the ends may be assumed pinned against rotation about the strong (xx) axis, and fixed with respect to the weak (yy) axis. If the length of the column is 24 ft and the factored axial load is 242 kips, find the lightest W10 shape. See Fig. P8.28 of the text book. Solution Column length, L = 24 ft Material: A572 Grade 50 Factored Load: 242 kips K x = 1.00; K x L x = 1.00 × 24 = K y = 0.65; K y L y = 0.65 × 24 =

24.0 ft 15.6 ft

a.

Member selection Enter LRFDM Table 4-2 with KL = K y L y = 15.6 ft, and note that for a W10 × 39, P dy = 254 > P req and rx /ry = 2.16. So, (K x L x)y = 24.0 ÷ 2.16 = 11.1 < 15.6 ˆ W10 × 39 selected is O.K.

b.

Checks 1. Strength W10 × 39:

A = 11.5 in.2;

rx = 4.27 in.;

ry = 1.98 in.

;

From LRFDS Table 3-50, Nc F cr = 22.1 ksi ˆ Pd = Nc Fcr A g = 254 kips > 242 kips

ˆ O.K.

2. Local buckling of plates Show O.K. Select a W10×39 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

P8.29.

(a)

(b)

Chapter 8

page 8-35

Find the most economical W12 section to carry an 800 kip factored axial compressive load. The 22-ft long column is pinned at both ends about both axes. In addition, it is braced at mid-height. Use A572 Grade 50 steel and the LRFD Specifications. After the column sections were ordered, the architect of the project decided not to provide lateral bracing at mid-height of the column as assumed in (a). The fabricator strengthened the shapes received, by welding two d ×14 in. plates of A572 Gr 50 steel in stock. What arrangement did he use? Substantiate your results. See Fig. P8.29 of the text book.

Solution Length, L = 22 ft Material: A572 Grade 50 Factored load: 800 kips a.

With bracing at mid-height K x L x = 1.0 × 22 = 22.0 ft K y L y = 1.0 × 11 = 11.0 ft Enter LRFDM Table 4-2 with KL = 11 ft and tentatively select a W12×79 which provides a Pdy of 860 kips. Also, r x / r y = 1.75. ˆ (K x L x)y = 22/1.75 = 12.6 ft > 11.0 ft Re-enter the table and observe that for KL = (K x L x)y = 12.6 ft, W12×79 provides P dx = 824 kips. So, the W12×79 section is still O.K.

b.

Checks 1. Strength check W12×79:

A = 23.2 in.2;

rx = 5.34 in.;

ry = 3.05 in.

;

From LRFDS Table 3-50, Nc F cr = 35.6 ksi Pd = Nc Fcr A g = 826 kips > 800 kips ˆ O.K. 2. Local buckling of plates With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain for Flanges:

Web:

ˆ O.K.

ˆ O.K.

So, plate local buckling will not reduce the member design strength calculated above.

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

b.

Chapter 8

page 8-36

With no intermediate bracing K x L x = 1.0 × 22 = 22.0 ft;

K y L y = 1.0 × 22 = 22.0 ft

Alternative 1: Plates parallel to the xx axis, welded to the flanges 1.

Strength of revised member W12×79: A W = 23.2 in.; IxW = 662 in.4; IyW = 216 in.4 d W = 12.38 in.; b fW = 12.1 in. For the built-up section: A = 23.2 + (14 × 3/8 × 2) = 33.7 in.2

From LRFDS Table 3-50, Nc Fcr = 27.2 ksi ˆ Pd = Nc Fcr A g = 27.2 × 33.7 = 917 kips 2.

> 800 kips

ˆ O.K.

Local buckling of cover plates. Plate between welds is to be checked as a stiffened element. Distance between welds equals the width of the flange.

Alternative 2: The plates are welded to the flanges, parallel to the web plane. 1.

Strength of revised member b f W = 12.1 in.; d 2 = 6.05 + 0.19 = 6.24 in

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

Steel Structures by S. Vinnakota

Chapter 8

page 8-37

;

;

From LRFDS Table 3-50, Nc Fcr = 32.3 ksi Pd = Nc Fcr A g = 32.3 × 33.7 = 1089 kips > 800 kips 2.

Local buckling of plates. Distance between welds equals the depth of the section.

So second arrangement, with the plates welded to the flanges, parallel to the web, is better!

(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

P8.30.

Chapter 8

page 8-38

An interior column of a building is to support a factored axial load of 640 kips. It is 22 ft long and is to be of A992 steel. The column base is rigidly fixed to the footing, and the top of the column is rigidly connected to stiff trusses. Assume that bracing is provided to the structure to prevent side sway in the xx (strong) plane of the column, but the side sway in the yy (weak) plane of the column is not prevented. Select the lightest column shape.

Solution Column length = 22 ft Material: A992 Factored load, Pu = 640 kips For buckling about: x-axis: Column fixed at base and top, and part of an unbraced structure K x = 1.2 y-axis: Column fixed at base and top, and part of a braced frame, K y = 0.65 K x L x = 1.2 × 22 = 26.4 ft K y L y = 0.65 × 22 = 14.3 ft a.

Member selection Enter LRFDM Table 4-2 for W10 shapes, with KL = K y L y = 14.3 ft, P u = 640 kips and observe that a W10×77 gives Pdy = 699 kips and has r x /r y = 1.73. ˆ (K x L x)y = 15.3 > 14.3 ft Reentering Table 4-2 with KL = (K x L x)y = 15.3 ft, we observe that the W10×77 provides, P d = 667 kips > P req = 640 kips. Similarly, for a W12 × 72, Pdy = 710 kips, rx / r y = 1.75, (K x L x)y = 15.1 ft, and Pdx = 692 kips. ˆ W12 x 72 is O.K. Finally, for a W14 × 74, Pdy = 652 kips, rx / r y = 2.44, (K x L x)y = 10.8 ft, and Pd = Pdy = 652 kips. ˆ W14 × 74 is O.K. Of these three sections, W12 × 72 is the lightest.

b.

Checks 1. Strength check W12×72:

A = 21.1 in.2;

ˆ Select it

rx = 5.31 in.;

ry = 3.04 in.

;

From LRFDS Table 3-50, NcF cr = 32.8 ksi Pd = Nc Fcr A g = 692 kips > Pu = 640 kips 2) Local buckling of plates Flange and web plates satisfy the requirements (show). So, select a W12×72 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 8

page 8-39

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

P8.31.

Chapter 8

page 8-40

Select the lightest W shape of A588 Grade 50 steel for a 18 ft long column to support an axial dead load of 290 kips and a live load of 280 kips. The column is assumed to be hinged at the base, while the top of the column is rigidly framed to very stiff girders. Assume that bracing is provided to the structure to prevent side sway in the xx (strong) plane of the column, but the side sway in the yy (weak) plane is not prevented. Select the lightest column shape. Solution Material: A588 Grade 50 Loads: Dead load = 290 kips; Live load = 280 kips Factored load, Pu = 1.2D + 1.6L = 1.2 × 290 + 1.6 × 280 = 796 kips Column length, L = 18 ft Buckling about xx axis: Sway permitted. Hinged base. Fixed top. 6 K x = 2.0 from Fig. 8.5.4 f Buckling about yy axis: Sway prevented. Hinged base. Fixed top. 6 K y = 0.8 from Fig. 8.5.4 b K x L x = 2.0 × 18 = 36.0 ft; K y L y = 0.80 × 18 = 14.4 ft a.

Member selection Enter LRFDM Table 4-2 for W10 shapes, with KL = K y L y = 14.4 ft, Preq = 796 kips and observe that a W10×88 provides P dy = 803 kips. Also, for this section, r x / r y = 1.73 resulting in : (K x L x)y = 20.8 > 14.4 ft Reentering the table with KL = (K x L x)y = 20.8 ft, we observe that the heaviest W10 section (W10 ×112) does not provide adequate axial strength. Next, enter LRFDM Table 4-2 for W12 shapes, with KL = K y L y = 14.4 ft, P req = 796 kips and observe that a W12×87 gives Pdy = 863 kips and has r x /r y = 1.75. ˆ (K x L x)y = 20.6 > 14.4 ft Reentering Table 4-2 with KL = (K x L x)y = 20.6 ft, we observe that a W12×106 provides, Pd = 835 kips > P req = 796 kips. Proceeding in a similar manner, we observe that for a W14×90: Pdy = 960 kips; rx/ry = 1.66. ˆ (K x L x)y = 21.7 > 14.4 ft Reentering Table 4-2 with KL = (K x L x)y = 21.7 ft, we find that a W14×99 provides, Pd = 862 kips > Preq = 796 kips. From W12×106 and W14×99, W14×99 is the lighter, so choose W14×99.

b.

Checks W14×99: A = 29.1 in.2;

r x = 6.17 in;

From LRFDS Table 3-50, NcF cr = 29.7 ksi ˆ Pd = Nc Fcr A g = 864 kips > 796 kips So, select a W14×99 of A588 Grade 50 steel.

r y = 3.71 in.

ˆ 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

P8.32.

Chapter 8

page 8-41

Design a 12 ft long W12 interior column of a building structure shown in Fig. P8.32, to support a factored axial load of 820 kips. The column is part of an unbraced, rigid jointed frame in the plane of its web. W27×94 girders, 28 ft long, are rigidly connected to each flange, at the top and bottom of the column under consideration. Assume same sections are used for columns just above and below. The column considered is braced normal to its web at the top and bottom so that side sway is prevented in that plane. Assume A992 steel. See Fig. P8.32 of the text book. Solution Factored load, Pu = 820 kips Material: A992 steel L c = 12 ft; L g = 28 ft Buckling about the xx axis: Column part of an unbraced frame. 6 K x $ 1.0 Buckling about the yy axis: Column part of a braced frame. 6 K y # 1.0. Assume K y = 1.0. Girders: W27×94 a.

Member selection Assume Ix for the column as 600 in.4 From Eq. 8.5.7:

Enter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read, K x = 1.15. K x L x = 1.15 × 12 = 13.8 ft K y L y = 1.0 × 12 = 12.0 ft Enter LRFDM Table 4-2 with Preq = 820 kips, KL = K y L y = 12.0 ft and tentatively select a W12 × 79 with Pdy = 838 kips. For this section, r x / r y = 1.75 6 (K x L x)y = 7.89 ft < 12.0 ft. ˆ W12 × 79 is O.K. b.

Strength check W12×79: A = 23.2 in.2; Ix = 662 in.4; r x = 5.34 in.; r y = 3.05 in. Use the revised value of Ix and find G B = G A = 0.472, K x = 1.16

ˆ

;

From LRFDS Table 3-50, NcF cr = 36.1 ksi ˆ Pd = Nc Fcr A g = 36.1 × 23.2 = 838 kips So, select a W12×79

>

820 kips

ˆ 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

P8.33.

Chapter 8

page 8-42

Select a suitable W12 shape for the columns of a fixed-base, rigid-jointed, unbraced, portal frame shown in Fig. P8.33, using A992 steel and LRFD Specification. The horizontal girder is a W18×76. The webs of these rolled sections lie in the plane of the frame. In the plane perpendicular to the frame, bracing (girts) are provided at top, and mid height of columns using simple flexible beam-to-column connections. The factored axial load on the columns is 400 kips. (Hint: As a starting point assume that Ix of column section lies between 300 and 500 in.4). See Figure P8.33 of the textbook. Solution Factored load, Pu = 400 kips Material: A992 steel Length, L = 16 ft; L x = 16 ft; L y1 = 8 ft; L y2 = 8 ft Buckling about the x-axis: Column part of an unbraced frame. 6 K x $ 1.0 Buckling about the y-axis: Column part of a braced frame. Fixed base. Girts at midheight. K y1 = 0.8; K y2 = 1.0 (K y L y) 1 = 0.8 × 8.0 = 6.40 ft; (K y L y)2 = 1.0 × 8.0 = 8.00 ft; 6 K y L y = 8.00 ft L g = 32 ft; L c = 16 ft W18×76 girder: Ix = 1330 in.4 Assume Ix of column = 400 in.4 G A = 1 (Recommended value for a fixed base)

From Eq. 8.5.7:

Enter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain, K x = 1.21 ˆ K x L x = 1.21 × 16 = 19.4 ft Enter LRFDM Table 4-2 with P u = 400 kips, KL = K y L y = 8.00 ft and tentatively select a W12× 40 with Pdy = 416 kips. For this section, rx / ry = 2.64 6 (K x L x)y = 7.5 ft < 8.00 ft. ˆ W12×40 is O.K. b.

Checks W12×40: A = 11.7 in.2; Ix = 307 in.4; rx = 5.13 in.; ry = 1.94 in. Revised value of G B = 0.47. So, K x changes to 1.10, and K x L x = 1.10 × 16 = 17.6 ft ;

From LRFDS Table 3-50, NcF cr = 35.6 ksi, and Pd = Nc Fcr A g = 417 kips > 400 kips So, select a W12×40 of A992 steel.

O.K. (Ans.)

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

Steel Structures by S. Vinnakota

P8.34.

Chapter 8

page 8-43

The column AB of a two bay, single story, rigid-jointed steel frame is subjected to a factored axial compressive load of 1930 kips (Fig. P8.34). The W30×90 girders are 30 ft long, and the columns are 16 ft long. The bases are hinged. Side sway is possible in the plane of the frame. In the perpendicular direction K y = 0.8. Assume A242 Grade 46 steel and check the adequacy of the W14×193. Include inelastic behavior of the column. See Figure P8.34 of the text book. Solution Factored load, Pu = 1930 kips Material: A242 Grade 46 6 Fy = 46 ksi L x = 16 ft; L y = 16 ft; K y = 0.8 L g = 30 ft; L c = 16 ft Column: W14×193 6 Ix = 2400 in.4; A = 56.8 in.2; 4 Girders: W30×90 6 Ix = 3610 in. G A = 10 (Recommended value for a pinned base)

r x = 6.50 in.;

r y = 4.05 in.

From Eq. 8.5.7:

Enter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain K x = 1.80. ˆ K x L x = 1.80 × 16 = 28.8 ft Also, K y L y = 0.80 × 16.0 = 12.8 ft ;

Slenderness parameter,

Enter LRFDS Table 4 with 8c = 0.674 and obtain,

ˆ Pd = Pd e = Nc Fcr A g = 0.702 × 46 ×56.8 = 1835 kips < 1930 kips

N.G.

Axial stress,

From LRFDS Table 4-1, corresponding to Fy = 46 ksi, we obtain SRF = 0.331 ˆ G Bi = 0.623 × 0.331 = 0.205; G A = 10 Re-enter nomograph and obtain,

K x = 1.75

6 8c = 0.655

ˆ Pd i = Nc Fcr A g = 0.710 × 46 × 56.8 = 1860 kips <

6

1930 kips

N.G.

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 8

The W14×193 is therefore inadequate.

page 8-44

(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

P8.35.

Chapter 8

page 8-45

Check the adequacy of a W12×152 for column AB of the rigid-jointed steel frame shown in Fig. P8.35. The columns are 12 ft long, and the columns above and below AB may be assumed to be approximately of the same size as AB. The adjacent girders are all of W18×50 and 32 ft long. The webs of all columns and girders shown lie in the plane of the paper. For buckling in the perpendicular plane, take K = 0.9. All members are of A992 steel. Take inelastic behavior of columns into account. The factored axial load in the member is 1540 kips. See Fig. P8.35 of the text book. Solution Material: A992 steel Column length, L = 12 ft K y = 0.9; K y L y = 0.9 × 12.0 = 10.8 ft L c = 12 ft; L g = 32 ft Columns: W12×152 6 Ix = 1430 in.4; Girders: W18×50 6 Ix = 800 in.4

A = 44.7 in.2;

rx = 5.66 in.;

ry = 3.19 in.

From Eq. 8.5.7:

Enter nomograph for sway uninhibited columns (Fig. 8.5.6 b) and get K x = 2.23 ; From LRFDS Table 3-50, Nc Fcr = 33.6 ksi ˆPd = Nc Fcr A g = 33.6 × 44.7 = 1500 kips < 1540 kips

N.G.

Axial stress, Enter LRFDM Table 4-1 for F y = 50 ksi and obtain, SRF = 0.461

ˆ G iA = G iB = SRF G e = 0.461 × 4.76 = 2.20

ˆ K x = 1.65

Enter LRFDS Table 3-50, and obtain Nc F cr = 37.4 ksi Pd = Nc Fcr A g = 37.4 × 44.7 = 1670 kips > 1540 kips

ˆ The W12×152 of A992 steel is adequate

ˆ 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

P8.36.

Chapter 8

page 8-46

A W14×193 column is part of a rigid-jointed, unbraced frame in both xx and yy planes. Two W24×84 girders, 36 ft long are rigidly connected to the flanges, while two W18 × 60 girders, 18 ft long, are rigidly connected to the web (Fig. P8.36). A992 steel is used for all members. The column in the tier above is also a W14×193, while the column in the tier below is a W14×211. The story height is 12 ft 6 in. A particular load combination results in a factored axial compressive load of 2000 kips in the column under consideration. Is the section safe according to the LRFDS? See Fig. P8.36 of the text book. Solution Column length, L = 12.5 ft; Factored load, Pu = 2000 kips A992 steel

L x = L y = 12.5 ft

Buckling about the x- axis: Column part of an unbraced frame. 6 K x $ 1.0 L b = 36 ft; L c = 12.5 ft W14 × 193: Ix = 2400 in.4; r x = 6.50 in.; A = 56.8 in.2 W14 × 211: Ix = 2660 in.4 W24 × 84: Ix = 2370 in.4 Buckling about the y-axis: Column part of an unbraced frame. 6 K y $ 1.0 L b = 18 ft; L c = 12.5 ft W14 × 193: Iy = 931 in.4; r y = 4.05 in. W14 × 211: Iy = 1030 in.4 W18 × 60: Ix = 984 in.4 For buckling about the major axis of the column section (Eq. 8.5.7):

;

Enter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K x = 1.80 For buckling about the minor axis of the column section (Eq. 8.5.7):

;

Again, enter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K y = 1.40

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 8

page 8-47

;

From LRFDS Tables 3-50, Nc Fcr = 34.9 ksi Pd = Nc Fcr A g = 34.9 × 56.8 = 1980 kips

ˆ The column is inadequate according to the LRFDS.

< 2000 kips

N.G. (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

P8.37.

Chapter 8

page 8-48

Select suitable W12 shapes for columns AB and CD of the axially loaded, unbraced frame with leaning column shown in Fig. P8.37. The girder BC is rigidly connected to the left column but has only a simple connection to the right column. The columns are braced top and bottom against sidesway out of the plane of the frame. In addition, the columns are braced at midheight by girts. Assume A992 steel. The loads given are factored loads. See Fig. P8.37 of the text book. Solution Factored load, Pu = 800 kips Column height, L = 15 ft Girder: W18×60 6 Ix = 984 in.4 a.

Design of the leaning column CD K x = 1.0; K y = 0.50; Required strength = 800 kips

K y L y = 0.5×15 = 7.50 ft

From LRFDM, Table 4-2, for F y = 50 ksi and KL = K y L y = 7.50 ft, tentatively select a W14×74 for which, Pdy = 841 kips > 800 kips O.K. rx /ry = 2.44

6

(K x L x)y =

So, Pd = Pdy = 841 kips > 800 kips Use a W14×74 for column CD. b.

Design of Column AB (Yura’s Method ) Required axial strength (out of plane), Puy Required axial strength (in-plane), Pux L c = 15 ft; L b = 32 ft Try W14×233: A = 68.5 in.2;

< 7.70 ft

OK (Ans.)

= 800 kips = 800 + 800 = 1,600 kips

Ix = 3,010 in.4;

r x = 6.63 in.

Enter nomograph for sway uninhibited columns (Fig. 8.5.6b), and get K x = 3.20

Pdx = 24.4×68.5 = 1,671 kips > Pux = 1,600 kips K y L y = 0.5 × 15 = 7.50 ft Use W14×233

O.K.

Y Pdy = 2,810 kips > Puy = 800 kips

OK

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

P8.38.

Chapter 8

page 8-49

Design the columns AB and CD of the axially loaded, unbraced frame with leaning columns shown in Fig. P8.38. The girders are rigidly connected to the interior column while the other connections are simple. The columns are braced top and bottom against sidesway out of the plane of the frame. In addition, the exterior columns are braced at midheight by girts. Assume A992 steel. Use W12 shapes. The loads given are factored loads. See Fig. P8.38 of the text book. Solution W30×124 girders 6 a.

Ix = 5,360 in.4

Design of the leaning column CD Factored load on the column, Pu = 2200 kips Column length, L = 16 ft L x = L y = 16.0 ft K x = K y = 1.0 6 K y L y = 1.0×16.0 = 16.0 ft From LRFDM, Table 4-2, try W14×211 having P dy = 2,240 kips

> 2,200 kips

Use W14×211 for column CD. b.

O.K. (Ans.)

Design of columns AB and EF (Yura’s Method) Column length, L = 16 ft = L c = L x Girder length, L g = 36 ft Required axial strength (out of plane), Puy =

1,000 kips

Required axial strength (in-plane),

Pux

Try W14×233: A = 68.5 in.2;

Ix = 3010 in.4;

=

due to symmetry r x = 6.63 in.

Enter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K x = 2.20

Pdx = 31.6×68.5 • 2170 kips > Pux = 2100 kips K y L y = 16 ft

Y

Pdy = 2760 kips > Puy = 1000 kips

Use W14×233 for columns AB and EF.

O.K. 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

P8.39.

Chapter 8

page 8-50

Design the columns AB and CD of the axially loaded, unbraced frame with leaning columns shown in Fig. P8.39. The girders are rigidly connected to the interior columns while the connections to the exterior columns are simple. The columns are braced top and bottom against sidesway out of the plane of the frame. In addition, the exterior columns are braced at midheight by girts. Assume A992 steel. Use W12-shapes. The loads given are factored loads. See Fig. P8.39 of the text book. Solution W30×124 girders 6 a.

Ix = 5,360 in.4

Design of the leaning columns AB and EF Factored load on the column, Pu = 1000 kips Column length, L = 16 ft L x = 16.0 ft; L y1 = L y2 = 8.0 ft K x L y = 16.0; K y L y = 1.0×8.0 = 8.00 ft Enter LRFDM Table 4-2 for W14 shapes, with KL = K y L y = 8.0 ft, P req = 1000 kips and observe that a W14×90 gives P dy = 1070 kips and has r x /r y = 1.66. ˆ (K x L x)y = 9.64 > 8.0 ft Reentering Table 4-2 with KL = (K x L x) y = 9.64 ft, we observe that a W14×90 provides, P d = 1047 kips > P req = 1000 kips. Use W14×90 for columns AB and EF.

b.

(Ans.)

Design of column CD (Yura’s Method) Column length, L = 16 ft = L c = L x Girder length, L g = 36 ft Required axial strength (out of plane), Puy = 2200 kips Required axial strength (in-plane), Pux = 2200 + 1000 = 3200 kips due to symmetry Try W14×342: A = 101 in.2; Ix = 4900 in.4; r x = 6.98 in.

Enter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K x = 2.14

Pdx = 33.0× 101 • 3330 kips > Pux = 3200 kips K y L y = 16 ft Use W14×342

Y

Pdy = 3690 kips > Puy = 2200 kips

O.K. O.K. (Ans.)

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