solucionario vinnakota capitulo 8.pdf
DESCRIPTION
solucionario del libro Vinnakota capitulo 8 completoTRANSCRIPT
Steel Structures by S. Vinnakota Chapter 8 page 8-1
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
CHAPTER 8: COLUMNS
cr y cP8.1. Plot the LRFDS column curve NF / F vs. 8 . Show all the salient points.
crP8.2. Use the LRFD Specification and plot the design axial compressive stress NF 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 / 12 = 2.0 × 2.0 ÷ 12 = 1.33 in. 3 3 4
a. Column length, L = 6.0 ft = 72.0 in.
Elastic flexural buckling load, (Ans.)
E E Elastic flexural buckling stress, f = P / A = 73.4 ÷ 4.00 = 18.4 ksi (Ans.)
b. Column length, L = 12.0 ft = 144 in.
Elastic flexural buckling load, (Ans.)
E E Elastic flexural buckling stress, f = P / A = 18.4 ÷ 4.00 = 4.60 ksi (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-2
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.4. 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
x ySection: W 12×96 6 A = 28.2 in. ; I = 833 in. ; I = 270 in.2 4 4
a. Column length, L = 50 ft = 600 in.
From Eq. 8.4.19:
From Eq. 8.4.20, elastic flexural buckling load,
(Ans.)
E E Elastic flexural buckling stress, f = P / A = 215 ÷ 28.2 = 7.62 ksi (Ans.)
b. Column length, L = 25 ft = 300 in.
From Eq. 8.4.19:
From Eq. 8.4.20, elastic flexural buckling load,
(Ans.)
E E Elastic flexural buckling stress, f = P / A = 859 ÷ 28.2 = 30.5 ksi (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-3
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.5. 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.
x xColumn sections: W10×39 6 I = 209 in. ; W10×45 6 I = 248 in.4 4
xW10×49 6 I = 272 in. 4
x xGirder sections: W16×40 6 I = 518 in. ; W16×50 6 I = 650 in.4 4
x xW14×30 6 I = 291 in. ; W14×34 6 I = 340 in. 4 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)
AG = 10.0 (recommended value for hinged base)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.98 (Ans.)
Column c4
g3" = 0.5 (far end pinned)
AG = 10.0 (recommended value for hinged base)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 2.05 (Ans.)
Column c2
Steel Structures by S. Vinnakota Chapter 8 page 8-4
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
g1 g2" = 0.0 (near end pinned); " = 1.0 (both ends rigidly jointed)
AG = 1.0 (recommended value for fixed base)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.45 (Ans.)
Column c3
g3 g2" = 0.0 (near end pinned); " = 1.0 (both ends rigidly jointed)
AG = 1.0 (recommended value for fixed base)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.45 (Ans.)
Column c5
A BG = 1.98 (same as G value calculated for column c2)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.30 (Ans.)
Column c6
A BG = 1.98 (same as G value calculated for column c3)
xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.30 (Ans)
Steel Structures by S. Vinnakota Chapter 8 page 8-5
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
b. Using the equations given in Section 8.5
From Eq. 8.5.14, the effective length of a column in an unbraced frame is,
A B xColumn c1: G = 10.0; G = 1.38; K . 1.99 (Ans.)
A B xColumn c2: G = 1.00; G = 1.98; K . 1.47 (Ans.)
A B xColumn c3: G = 1.00; G = 1.98; K . 1.47 (Ans.)
A B xColumn c4: G = 10.0; G = 1.72; K . 2.10 (Ans.)
A B xColumn c5: G = 1.98; G = 1.25; K . 1.50 (Ans.)
A B xColumn c6: G = 1.98; G = 1.25; K . 1.50 (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-6
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.6. 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.
x xColumn sections: W10×39 6 I = 209 in. ; W10×45 6 I = 248 in.4 4
xW10×49 6 I = 272 in. 4
x xGirder sections: W16×40 6 I = 518 in. ; W16×50 6 I = 650 in.4 4
x xW14×30 6 I = 291 in. ; W14×34 6 I = 340 in. 4 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)
AG = 10.0 (recommended value for hinged base)
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.81 (Ans.)
Column c4
g3" = 1.5 (far end pinned)
AG = 10.0 (recommended value for hinged base)
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.83 (Ans.)
Column c2
g1 g2" = 0.0 (near end pinned); " = 1.0 (both ends rigidly jointed)
AG = 1.0 (recommended value for fixed base)
Steel Structures by S. Vinnakota Chapter 8 page 8-7
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.81 (Ans.)
Column c3
g3 g2" = 0.0 (near end pinned); " = 1.0 (both ends rigidly jointed)
AG = 1.0 (recommended value for fixed base)
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.81 (Ans.)
Column c5
A BG = 1.98 (same as G value calculated for column c2)
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.83 (Ans.)
Column c6
A BG = 1.98 (same as G value calculated for column c3)
xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.83 (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-8
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
b. Using the equations given in Section 8.5
From Eq. 8.5.18, the effective length of a column in an unbraced frame is,
A B xColumn c1: G = 10.0; G = 0.461; K . 0.81 (Ans.)
A B xColumn c2: G = 1.00; G = 1.98; K . 0.81 (Ans.)
A B xColumn c3: G = 1.00; G = 1.98; K . 0.81 (Ans.)
A B xColumn c4: G = 10.0; G = 0.573; K . 0.83 (Ans.)
A B xColumn c5: G = 1.98; G = 1.25; K . 0.83 (Ans.)
A B xColumn c6: G = 1.98; G = 1.25; K . 0.83 (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-9
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.7. 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
x yMember: W8 × 48 6 A = 14.1 in. ; r = 3.61 in. ; r = 2.08 in.2 2
yMaterial: A242 Gr 50 6 F = 50 ksi
Length, L = 18ft
x yAs there are no intermediate supports, L = L = 18 ft
x yEnd conditions: Fixed at both ends about both axes. From Fig. 8.5.4a: K = K = 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,
cAs 8 < 1.5, buckling occurs in the inelastic domain. Use LRFDS Equation E2-2,
cr y crF = F 6 F = 35.8 ksi
d c cr g� P = N F A = 0.85 × 35.8 × 14.1 = 429 kips (Ans.)
b. Using LRFDS Table 4
cEnter LRFDS Table 4 with 8 = 0.892 and read,
d c cr g� P = N F A = 0.609 × 50 × 14.1 = 429 kips (Ans.)
c. Using LRFDS Table 3-50
c crEnter LRFDS Table 3-50, with KL /r = 67.5, and obtain N F = 30.5 ksi by interpolation.
d c cr g� P = N F A = 30.5 × 14.1 = 430 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-10
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.8. 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
g x yMember: W12×65 6 A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2
yMaterial: A242 Gr 42 6 F = 42 ksi
Column length, L = 30ft
For major axis buckling:
x xL = 30.0 ft; fixed - fixed 6 K = 0.65
For minor axis buckling:
y1 yL = 10.0 ft; fixed - pinned 6 K = 0.80
y2 y L = 10.0 ft; pinned - pinned 6 K = 1.00
y3 y L = 10.0 ft; pinned - fixed 6 K = 0.80
x xK L = 0.65 × 30.0 = 19.5 ft
y y 1 y y 3 y y 2(K L ) = 0.65 × 10.0 = 6.5 ft = (K L ) ; (K L ) = 1.0 × 10.0 = 10.0 ft
y yK L = 10.0 ft
;
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.537 and obtain,
c cr 6 N F = 0.753 × 42 = 31.6 ksi
d c cr g� P = N F A = 31.6 × 19.1 = 604 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-11
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.9. 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
x yMember: S10 × 35 6 A = 10.3 in. ; r = 3 .78 in.; r = 0.898 in.2
yMaterial: A36 steel (Preferred material specification, for S-shapes) 6 F = 36 ksi
Length, L = 21 ft
x x x xL = 21 ft; fixed-pinned 6 K = 0.80 6 K L = 0.8 × 21 = 16.8 ft
a. Bracing at 7 ft intervals
3 segments.
y1 y1 y y 1L = 7.0 ft; fixed-pinned 6 K = 0.8 6 (K L ) = 0.8 × 7.0 = 5.60 ft
y2 y2 y y 2L = 7.0 ft; pinned-pinned 6 K = 1.0 6 (K L ) = 1.0 × 7.0 = 7.00 ft
y3 y3 y y 3L = 7.0 ft; pinned-pinned 6 K = 1.0 6 (K L ) = 1.0 × 7.0 = 7.00 ft
y yK L = 7.00 ft
Enter LRFDS Table 3-36 and read,
d c cr g� P = N F A = 19.3 × 10.3 = 199 kips (Ans.)
b. Bracing at 5 ft 3 in. apart
4 segments.
y1 y1 y y 1L = 5.25 ft; fixed-pinned 6 K = 0.8 6 (K L ) = 0.8 × 5.25 = 4.20 ft
y2 y2 y y 2L = 5.25 ft; pinned-pinned 6 K = 1.0 6 (K L ) = 1.0 × 5.25 = 5.25 ft
y yK L = 5.25 ft
Enter LRFDS Table 3-36, and read,
d c cr g� P = N F A = 23.6 × 10.3 = 243 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-12
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.10. 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
x ySection: W 12×72 6 A = 21.1 in. ; r = 5.31 in.; r = 3.04 in.2
yMaterial: A588 Grade 50 steel 6 F = 50 ksi
Column length, L = 16 ft
x yStructure braced in both xx and yy planes of the column 6 K # 1.0; K # 1.0
x xFor buckling about the major axis: 6 pinned at both ends 6 K = 1.0, L = 16 ft
For buckling about the minor axis: 6 pinned at both ends and brace at 10 ft from base
y y 1 y y 26 (K L ) = 1.0 × 10.0 = 10.0 ft; (K L ) = 1.0 × 6.0 = 6.00 ft
y y6 K L = max (10.0; 6.00) = 10.0 ft
;
c crFrom LRFDM Table 3-50, for K L/r = 39.5, N F = 37.9 ksi
Design axial compressive strength,
d c cr gP = = N F A = 37.9 × 21.1 = 800 kips (Ans.)
Alternatively
x x y yFor the W12x72 column given, K L = 16.0 ft; K L = 10 ft
From LRFDM Table 4-2, for a W12×72,
dFrom LRFDM Table 4-2, for a W12×72 with KL = 10 ft, P = 800 kips. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-13
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.11. 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
W xW yW W12×58: A = 17.0 in. ; I = 475 in. ; I = 107 in.2 4 4
pl Plates : A = 2 × 8 × 3/8 = 6.0 in.2
yMaterial: A572 Gr 60; F = 60 ksi
Column length, L = 15 ft
x xL = 15.0 ft; end conditions: pinned-pinned 6 K = 1.00
y1 y2 yL = L = 7.5 ft; end conditions: pinned-pinned 6 K = 1.00
Properties of built-up section
A = 17.0 + 6.0 = 23.0 in. 2
x I = [475 + 0] + 2 [0 + 8.0× 3/8 × (6.29) ] = 712 in.2 4
y I = [107 + 0] + 2[1/12 × (3/8) (8.0) ] = 139 in.3 4
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.530 and obtain,
d c cr g� P = N F A = 0.756 × 60 × 23.0 = 1040 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-14
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.12. 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
x ystrength of the column as per the LRFDS. Assume K = 1.0 and K = 1.4.
See Figure P8.12 of the text book.
Solution
Column length = 20ft
x yGiven: K = 1.00, K = 1.4
Dubly symmetric section built-up from:
C xC yCMC 13×50: A = 14.7 in. ; I = 314 in. ; I = 16.4 in.2 4 4
wCt = 0.787 in.; x = 0.974 in.
plPlate ½×12: A = 6.0 in.2
For the built-up section:
A = 14.7×2 + 12× ½ = 35.4 in.2
xI = = 628 + 0 = 628 in.4
1d = 6 + 0.787 - 0.974 = 5.813 in.
yI = = 1100 in.4
;
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.730 and obtain,
d c cr g� P = N F A = 0.680 × 42 × 35.4 = 1010 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-15
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.13. 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
x yNo intermediate braces 6 L = L = 20.0 ft
x yEnd conditions: fixed at the base and pinned at the top 6 K = 0.80; K = 0.80
yMaterial: A572 Gr 65 6 F = 65 ksi
Mono-symmetric section built-up from two W-shapes:
1 x1 y1W8×48: A = 14.1 in. ; I = 184 in. ; I = 60.9 in.2 4 4
1 d = 8.50 in.
2 x2 y2W12×26: A = 7.65 in. ; I = 204 in. ; I = 17.3 in.2 4 4
w2 t = 0.230 in.
For the built-up section:
A = 14.1 + 7.65 = 21.8 in.2
from bottom fibre
1 2d = 5.77 - 4.25 = 1.52 in.; d = 8.615 - 5.77 = 2.845 in.
xI = [184 + 14.1×1.52 ] + [17.3 + 7.65×2.845 ] = 296 in.2 2 4
yI = [ 60.9 + 0] + [204 + 0] = 265 in.4
x yr = 3.68 in.; r = 3.49 in.
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.829 and obtain,
d c cr g� P = N F A = 0.637 × 65 × 21.8 = 903 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-16
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.14. 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
x yNo intermediate braces 6 L = L = 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.
x y Hence, K = 2.10; K = 2.10.
yMaterial: A572 Gr 60 6 F = 60 ksi
Doubly symmetric section built-up from:
W wW xW yW W24×76 6 A = 22.4 in. ; t = 0.44 in.; I = 2100 in. ; I = 82.5 in.2 4 4
T T xT yT WT 9×53 6 A = 15.6 in. ; d = 9.365 in.; I = 104 in. , I = 110 in. 2 4 4
T y = 1.97 in.
For the built-up section:
A = 22.4 + 2 × 15.6 = 53.6 in.2
xI = [2100 + 0] + [2×110 + 0] = 2320 in.4
yI = [82.5 + 0] + 2 [104 + 15.6 (0.22 + 9.365 - 1.97) ] = 2100 in.2 4
;
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 1.17 and obtain,
So, the design strength of the column is
d c cr gP = N F A = 0.479 × 60 × 53.6 = 1540 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-17
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.15. 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
x x y yaxial compressive strength of the column, if K L = 32 ft and K L = 28 ft. Use the LRFD Specification.
See Figure P8.15 of the text book.
Solution
x x y yGiven: K L = 32 ft; K L = 28 ft
yMaterial: A572 Gr 42 6 F = 42 ksi
Doubly symmetric section built-up from:
W fW xW yW W14×730: A = 215 in. ; b = 17.9 in.; I = 14,300 in. ; I = 4,720 in.2 4 4
pl PL 24 × 4 : A = 24 × 4 = 96 in. ; 2
For the built-up section:
A = 215 + 2 (96) = 407 in.2
xI = [14300 + 0] + 2 × [4608 + 0] = 23,520 in.4
yI = [4720 + 0] + 2 × [128 + 96 (8.95 + 2.0) ] = 28,000 in.2 4
;
For the column:
controls;
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.612 and obtain,
So, the design strength of the column is
d c cr gP = N F A = 0.727 × 60 × 407 = 12,430 kips (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-18
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
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.
Steel Structures by S. Vinnakota Chapter 8 page 8-19
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.18. 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.
x y x yAssume that K = K = 1.0 and L = L = 14 ft.
Solution
x x y yGiven: K L = 14.0 ft; K L = 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
rRoof live load, L = 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
r1.2D + 1.6L + 0.5L
Contribution to axial load, from roof,
Contribution to axial load, from the seven floors,
So, factored load on the column,
uP = 153 + 1770 = 1920 kips
y y reqEnter LRFDM Table 4-2 for W14 shapes with KL = K L = 14 ft, P = 1920 kips and select a
dW14×176 with P = 1940 kips.
Provide a W14×176 of A992 steel. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-20
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.19. Select the lightest W section for a column supporting a factored axial load of 620 kips. Support conditions,
x yapplicable for both principal axes, are full fixity at the bottom and a pinned connection at the top. L = L = 17
ft 6 in. Assume A992 steel is used.
Solution
yMaterial: A 992 steel 6 F = 50 ksi
uFactored load, P = 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:
x x y yK L = K L = 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 :
dW10 series 6 W10 × 68 6 P = 625 kips > 620 kips O.K.
dW12 series 6 W12 × 65 6 P = 647 kips > 620 kips O.K.
dW14 series 6 W14 × 74 6 P = 662 kips > 620 kips O.K.
Use W12 × 65, the lightest section.
Steel Structures by S. Vinnakota Chapter 8 page 8-21
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.20. 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 ft . The structure is braced in both the xx and yy planes of the column. Select a suitable W shape of2
A588 Grade 50 steel.
See Figure P8.20 of the text book.
Solution
Tributary area = 1800 ft2
Dead load, D = 55 psf
Snow load, S = 35 psf
Assuming 50 plf for self weight of column, dead load of column = say 1.5
Factored axial load on the column,
x x y yAs the structure is braced in both the xx and yy planes of the column, K L # 25 ft, and K L # 25 ft
x xFor buckling about major axis 6 Fixed at base and pinned at top 6 K L = 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:
y y 16 (K L ) = 0.8 × 19 = 15.2 ft
y y 2(K L ) = 1.0 × 6 = 6.00 ft
y y6 K L = max [15.2; 6.00] = 15.2 ft
y yEnter LRFDM Table 4-2 for W10 shapes with KL = K L = 15.2 and observe that a W10×39 has a
dyP = 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-22
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.21. 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
uUse load combination, LC-2 6 P = 1.2×500 + 1.6×800 = 1880 kips
Column length, L = 30 ft
End conditions: Pinned at both ends about both axes.
yMaterial: A588 Grade 50 6 F = 50 ksi
a. No intermediate bracing
x x y yK L = K L = 1.00 × 30 ft = 30.0 ft. So, buckling about the minor axis will control the design.
y yEnter Column Tables for W12 shapes, in LRFDM Table 4-2, with KL = K L = 30 ft and select a
d dy uW12×336 that provides P = P = 1910 kips > P = 1880 kips.
Check:
x yFor a W12×336: A = 98.8 in. ; r = 6.41 in.; r = 3.47 in.2
;
c crEnter LRFDS Table 3-50 and read N F = 19.3 ksi
d c cr g� P = N F A = 19.3 × 98.8 = 1907 kips > 1880 kips O.K.
So, select a W12×336 of A588 Grade 50 steel. (Ans.)
b. Intermediate bracing provided at midheight
x x y yK L = 1.00 × 30 ft = 30.0 ft; K L = 1.00 × 15 ft = 15.0 ft
y yEnter Column tables for W12 shapes in LRFDM Part 4, with KL = K L = 15.0 ft and tentatively
dy u x yselect a W12×190 with P = 1900 kips > P = 1880 kips. Also r /r = 1.79 for this section.
x x y dxReenter Column tables for W12 shapes, with KL = (K L ) = 16.8 ft and select a W12×210 with P
u= 1992 kips > P = 1880 kips. For this section :
x yA = 61.8 in. ; r = 5.89 in.; r = 3.28 in.2
Steel Structures by S. Vinnakota Chapter 8 page 8-23
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
;
c crFrom LRFDS Table 3-50, N F = 32.4 ksi
d c cr g P = N F A = 32.4 × 61.8 = 2,000 kips > 1,880 kips O.K.
So, select a W12×210 of A588 Grade 50 steel. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-24
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.22. 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
yMaterial: A572 Grade 50 steel 6 F = 50 ksi
Service loads: D = 100 kips; L = 360 kips
Factored axial load on the column (from LC 4-2),
uP = 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:
x x y yK L = 1.0×28.0 = 28.0 ft; K L = 1.0×14.0 = 14.0 ft
y yEnter Column Load Tables for W14 shapes, in LRFDM Part 4, with KL = K L = 14.0 ft and observe
dy u x ythat for a W14×82, P = 729 kips > P = 696 kips. Also, for this section r /r = 2.44
d� W14×82 with P = 729 kips is the lightest W14 shape.
Checks
1. Strength
x y For a W14 × 82, LRFDM Table 1-1 gives: A = 24.0 in. ; r = 6.05 in; r = 2.48 in.2
;
From LRFDS Table 3-50, for
d c cr g� P = N F A = 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: � O.K.
Web: � 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-25
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.23. 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
yMaterial: A572 Grade 50 6 F = 50 ksi
uFactored axial compressive load, P = 960 kips
x x y yGiven: K L = 30ft; K L = l4 ft
a. Member selection
y y reqEnter LRFDM Table 4-2 for W10 shapes, with KL = K L = 14.0 ft, P = 960 kips and observe that
dy x ya W10×112 gives P = 1050 kips. Also, for this section, r /r = 1.74 resulting in
x x y (K /L ) = 17.2 > 14.0 ft
x x yReentering the table with KL = (K L ) = 17.2 ft, we observe that the heaviest W10 section (W10
×112) does not provide adequate axial strength.
y y reqNext, enter LRFDM Table 4-2 for W12 shapes, with KL = K L = 14.0 ft, P = 960 kips and note that
dy x ya W12×96 gives P = 966 kips and has r /r = 1.76.
x x y� (K L ) = 17.0 > 14.0 ft
x x y dReentering Table 4-2 with KL = (K /L ) = 17.0 ft, we observe that a W12×106 provides, P = 968
ukips > P = 960 kips.
dy x yProceeding in a similar manner, we observe that for a W14×90: P = 969 kips; r /r = 1.66.
x x y� (K /L ) = 18.1 > 14.0 ft
x x y dReentering Table 4-2 with KL = (K /L ) = 18.1 ft, we find that a W14×99 provides, P = 962 kips
u> P = 960 kips.
From W12×106 and W14×99, W14×99 is the lighter, so we select W14×99.
b. Checks
1. Strength
x y W14 × 99 : A = 29.1 in. , r = 6.17 in., r = 3.71 in.2
;
c cr From LRFDS Table 3-50, for , N F = 33.1 ksi
d c cr g � P = N F A = 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
Steel Structures by S. Vinnakota Chapter 8 page 8-26
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
Flanges: � O.K.
Web: � 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-27
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.24. 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
uFactored load, P = 700 kips
Material: A992 steel
y yBecause column is continuously braced in the weak direction, K L = 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 W10 × 60 W12 × 58 W14 × 61
dyP (kips) 748 723 761
x yr /r 1.71 2.10 2.44
x x y(K L ) (ft) 10.5 8.57 7.38
Revised
section
W10×68 W12×65 W14×68
dxP (kips) 715 745 773
The lightest section is a W12 × 65.
1. Strength
x yW12×65: A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2
c crFrom LRFDS Table 3-50, N F = 37.6 ksi
d c cr gP = N F A = 37.6 × 19.1 = 718 > 700 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: � O.K.
Web: � O.K.
So, plate local buckling will not reduce the member design strength calculated above.
So, select a W12×65 of A992 steel. (Ans.)
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 McGraw-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 8-29
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.25. Repeat Problem P8.24, if the selection is extended to W-shapes up to 18-in. nominal depth.
Solution
u yGiven: L = 18ft; P = 700 kips; F = 50 ksi
y y K L = 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 W16 × 67 W16 × 57 W18 × 60 W18 × 65
xr (in) 6.96 6.72 7.47 7.49
x x x(K L )/r 31 32 28.9 28.8
c crN F (ksi) 39.62 39.43 39.99 40.001
A (in. ) 19.7 16.8 17.6 19.12
dP (kips) 781 662 (NG) 704 764
f fb /2t 7.7 5.0 5.4 5.1
13.5 13.5 13.5 13.5
wh /t 35.9 (NG) 33.0 38.7 (NG) 35.7
35.8 35.8 35.8 35.8
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
x yW18×65: A = 19.1 in. ; r = 7.49 in.; r = 1.69 in.2
c crFrom LRFDS Table 3-50, N F = 40.0 ksi
d c cr gP = N F A = 40.0 × 19.1 = 764 > 700 kips � 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-30
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.26. A steel column is built into a wall so that it may be considered as continuously braced for buckling about its y-
x xaxis. If the factored load on the column is 1000 kips and K L = 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
y y x x uGiven: K L = 0; K L = 22 ft; P = 1000 kips
yMaterial: A572 Grade 50 6 F = 50 ksi
a. Member selection
y y reqEnter Column Tables for W10 shapes in LRFDM Table 4-2, with K L = 0 and P = 1000 kips, and
dy x ytentatively select a W10×88 with P = 1100 kips and r /r = 1.73.
x x y(K L ) = 22/1.73 = 12.71
dxRe-enter the table and select a W10×112 with P = 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
xW16×89: A = 26.4 in. ; r = 7.05 in.2
c crFrom LRFDS Table 3-50, N F = 38.4 ksi
d c cr gP = N F A = 38.4 × 26.4 = 1014 > 1000 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: � O.K.
Web: � 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-31
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
See the design calculation shown in the following table
From LRFDM Table 4-2 From LRFDM Table 1-1
Section W10×112 W12×96 W14×99 Section W16×89 W18×97
dyP
(kips)
1,400 1,200 1,240 A
(in. )2
26.4 28.5
x y xr /r 1.74 1.76 1.66 r
(in.)
7.05 7.82
x x y(K L )
(ft)
12.6 12.5 13.3 37.4 33.5
dxP
(kips)
1,106 1,008 1,081 1,014 1,114
f fb /2t 4.17 6.76 9.34 5.92 6.41
13.5 13.5 13.5 13.5 13.5
wh/t 10.4 17.7 23.5 25.9 30.0
35.8 35.8 35.8 35.8 35.8
Steel Structures by S. Vinnakota Chapter 8 page 8-32
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.27. 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
uFactored axial compressive load, P = 600 kips
yMaterial: A588 Grade 50 6 F = 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
x xK L = 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
y y 1 y y 2 y y 3(K L ) = 0.8 × 12 = 9.60 ft; (K L ) = 1.0 × 10 = 10.0 ft; (K L ) = 1.0 × 8 = 8.00 ft
y yK L = 10.0 ft
a. Member selection
y y dyEnter LRFDM Table 4-2 with KL = K L = 10.0 ft and tentatively select a W12 × 58 with P = 611
x ykips and r /r = 2.10. So,
x x y(K L ) = 24.0 ÷ 2.10 = 11.4 > 10.0 ft
x x y dxRe-entering Table 4-2 with KL = (K L ) = 11.4 ft, we observe that the W12×58 gives P = 581 kips
x y < 600. We also observe that for sections heavier than W12×58, r /r . 1.75. So,
x x y(K L ) = 13.7 > 10.0 ft
x x y dxRe-entering LRFDM Table 4-2 with KL = (K L ) = 13.7 ft, we select a W12×65 with P = 653
kips.
b. Checks
1. Strength
x yW12 × 65: A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2
;
c crEnter LRFDS Table 3-50 and read, N F = 34.2 ksi
d c cr g� P = N F A = 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.)
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 McGraw-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 8-34
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.28. 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
x x xK = 1.00; K L = 1.00 × 24 = 24.0 ft
y y yK = 0.65; K L = 0.65 × 24 = 15.6 ft
a. Member selection
y y dy reqEnter LRFDM Table 4-2 with KL = K L = 15.6 ft, and note that for a W10 × 39, P = 254 > P
x yand r /r = 2.16. So,
x x y (K L ) = 24.0 ÷ 2.16 = 11.1 < 15.6
� W10 × 39 selected is O.K.
b. Checks
1. Strength
x yW10 × 39: A = 11.5 in. ; r = 4.27 in.; r = 1.98 in.2
;
c crFrom LRFDS Table 3-50, N F = 22.1 ksi
d c cr g� P = N F A = 254 kips > 242 kips � O.K.
2. Local buckling of plates
Show O.K.
Select a W10×39 of A992 steel. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-35
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.29. (a) 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.
(b) 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
x xK L = 1.0 × 22 = 22.0 ft
y yK L = 1.0 × 11 = 11.0 ft
dyEnter LRFDM Table 4-2 with KL = 11 ft and tentatively select a W12×79 which provides a P of 860
x ykips. Also, r / r = 1.75.
x x y� (K L ) = 22/1.75 = 12.6 ft > 11.0 ft
x x y dxRe-enter the table and observe that for KL = (K L ) = 12.6 ft, W12×79 provides P = 824 kips. So,
the W12×79 section is still O.K.
b. Checks
1. Strength check
x yW12×79: A = 23.2 in. ; r = 5.34 in.; r = 3.05 in.2
;
c crFrom LRFDS Table 3-50, N F = 35.6 ksi
d c cr g P = N F A = 826 kips > 800 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: � O.K.
Web: � O.K.
So, plate local buckling will not reduce the member design strength calculated above.
Steel Structures by S. Vinnakota Chapter 8 page 8-36
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
b. With no intermediate bracing
x x y yK L = 1.0 × 22 = 22.0 ft; K L = 1.0 × 22 = 22.0 ft
Alternative 1: Plates parallel to the xx axis, welded to the flanges
1. Strength of revised member
W xW yWW12×79: A = 23.2 in.; I = 662 in. ; I = 216 in. 4 4
W fWd = 12.38 in.; b = 12.1 in.
For the built-up section:
A = 23.2 + (14 × 3/8 × 2) = 33.7 in.2
c crFrom LRFDS Table 3-50, N F = 27.2 ksi
d c cr g� P = N F A = 27.2 × 33.7 = 917 kips > 800 kips � O.K.
2. 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
f W 2b = 12.1 in.; d = 6.05 + 0.19 = 6.24 in
Steel Structures by S. Vinnakota Chapter 8 page 8-37
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
;
;
c crFrom LRFDS Table 3-50, N F = 32.3 ksi
d c cr gP = N F A = 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.)
Steel Structures by S. Vinnakota Chapter 8 page 8-38
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.30. 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
uFactored load, P = 640 kips
For buckling about:
xx-axis: Column fixed at base and top, and part of an unbraced structure K = 1.2
y y-axis: Column fixed at base and top, and part of a braced frame, K = 0.65
x xK L = 1.2 × 22 = 26.4 ft
y yK L = 0.65 × 22 = 14.3 ft
a. Member selection
y y uEnter LRFDM Table 4-2 for W10 shapes, with KL = K L = 14.3 ft, P = 640 kips and observe that
dy x ya W10×77 gives P = 699 kips and has r /r = 1.73.
x x y� (K L ) = 15.3 > 14.3 ft
x x y dReentering Table 4-2 with KL = (K L ) = 15.3 ft, we observe that the W10×77 provides, P = 667
reqkips > P = 640 kips.
dy x y x x y dxSimilarly, for a W12 × 72, P = 710 kips, r / r = 1.75, (K L ) = 15.1 ft, and P = 692 kips.
� W12 x 72 is O.K.
dy x y x x y d dyFinally, for a W14 × 74, P = 652 kips, r / r = 2.44, (K L ) = 10.8 ft, and P = P = 652 kips.
� W14 × 74 is O.K.
Of these three sections, W12 × 72 is the lightest. � Select it
b. Checks
1. Strength check
x yW12×72: A = 21.1 in. ; r = 5.31 in.; r = 3.04 in.2
;
c crFrom LRFDS Table 3-50, N F = 32.8 ksi
d c cr g u P = N F A = 692 kips > P = 640 kips
2) Local buckling of plates
Flange and web plates satisfy the requirements (show).
So, select a W12×72 of A992 steel. (Ans.)
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 McGraw-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 8-40
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.31. 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
uFactored load, P = 1.2D + 1.6L = 1.2 × 290 + 1.6 × 280 = 796 kips
Column length, L = 18 ft
Buckling about xx axis:
xSway permitted. Hinged base. Fixed top. 6 K = 2.0 from Fig. 8.5.4 f
Buckling about yy axis:
ySway prevented. Hinged base. Fixed top. 6 K = 0.8 from Fig. 8.5.4 b
x x y yK L = 2.0 × 18 = 36.0 ft; K L = 0.80 × 18 = 14.4 ft
a. Member selection
y y reqEnter LRFDM Table 4-2 for W10 shapes, with KL = K L = 14.4 ft, P = 796 kips and observe that
dy x ya W10×88 provides P = 803 kips. Also, for this section, r / r = 1.73 resulting in :
x x y (K L ) = 20.8 > 14.4 ft
x x yReentering the table with KL = (K L ) = 20.8 ft, we observe that the heaviest W10 section (W10
×112) does not provide adequate axial strength.
y y reqNext, enter LRFDM Table 4-2 for W12 shapes, with KL = K L = 14.4 ft, P = 796 kips and observe
dy x ythat a W12×87 gives P = 863 kips and has r /r = 1.75.
x x y� (K L ) = 20.6 > 14.4 ft
x x y dReentering Table 4-2 with KL = (K L ) = 20.6 ft, we observe that a W12×106 provides, P = 835
reqkips > P = 796 kips.
dy x yProceeding in a similar manner, we observe that for a W14×90: P = 960 kips; r /r = 1.66.
x x y� (K L ) = 21.7 > 14.4 ft
x x y dReentering Table 4-2 with KL = (K L ) = 21.7 ft, we find that a W14×99 provides, P = 862 kips
req > P = 796 kips.
From W12×106 and W14×99, W14×99 is the lighter, so choose W14×99.
b. Checks
x yW14×99: A = 29.1 in. ; r = 6.17 in; r = 3.71 in.2
c crFrom LRFDS Table 3-50, N F = 29.7 ksi
d c cr g� P = N F A = 864 kips > 796 kips � O.K.
So, select a W14×99 of A588 Grade 50 steel. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-41
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.32. 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
uFactored load, P = 820 kips
Material: A992 steel
c gL = 12 ft; L = 28 ft
xBuckling about the xx axis: Column part of an unbraced frame. 6 K $ 1.0
y yBuckling about the yy axis: Column part of a braced frame. 6 K # 1.0. Assume K = 1.0.
Girders: W27×94
a. Member selection
xAssume I for the column as 600 in. From Eq. 8.5.7: 4
xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read, K = 1.15.
x xK L = 1.15 × 12 = 13.8 ft
y yK L = 1.0 × 12 = 12.0 ft
req y yEnter LRFDM Table 4-2 with P = 820 kips, KL = K L = 12.0 ft and tentatively select a W12 × 79
dy x y x x ywith P = 838 kips. For this section, r / r = 1.75 6 (K L ) = 7.89 ft < 12.0 ft.
� W12 × 79 is O.K.
b. Strength check
x x yW12×79: A = 23.2 in. ; I = 662 in. ; r = 5.34 in.; r = 3.05 in.2 4
x B A xUse the revised value of I and find G = G = 0.472, K = 1.16
� ;
c crFrom LRFDS Table 3-50, N F = 36.1 ksi
d c cr g� P = N F A = 36.1 × 23.2 = 838 kips > 820 kips � O.K.
So, select a W12×79 (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-42
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.33. 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
xcolumns is 400 kips. (Hint: As a starting point assume that I of column section lies between 300 and 500 in. ).4
See Figure P8.33 of the textbook.
Solution
uFactored load, P = 400 kips
Material: A992 steel
x y1 y2Length, L = 16 ft; L = 16 ft; L = 8 ft; L = 8 ft
xBuckling about the x-axis: Column part of an unbraced frame. 6 K $ 1.0
Buckling about the y-axis: Column part of a braced frame. Fixed base. Girts at midheight.
y1 y2K = 0.8; K = 1.0
y y 1 y y 2 y y(K L ) = 0.8 × 8.0 = 6.40 ft; (K L ) = 1.0 × 8.0 = 8.00 ft; 6 K L = 8.00 ft
g cL = 32 ft; L = 16 ft
xW18×76 girder: I = 1330 in.4
xAssume I of column = 400 in.4
AG = 1 (Recommended value for a fixed base)
From Eq. 8.5.7:
xEnter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain, K = 1.21
x x� K L = 1.21 × 16 = 19.4 ft
u y yEnter LRFDM Table 4-2 with P = 400 kips, KL = K L = 8.00 ft and tentatively select a W12× 40
dy x y x x ywith P = 416 kips. For this section, r / r = 2.64 6 (K L ) = 7.5 ft < 8.00 ft.
� W12×40 is O.K.
b. Checks
x x yW12×40: A = 11.7 in. ; I = 307 in. ; r = 5.13 in.; r = 1.94 in.2 4
B x x xRevised value of G = 0.47. So, K changes to 1.10, and K L = 1.10 × 16 = 17.6 ft
;
c crFrom LRFDS Table 3-50, N F = 35.6 ksi, and
d c cr g P = N F A = 417 kips > 400 kips O.K.
So, select a W12×40 of A992 steel. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-43
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.34. 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
yare hinged. Side sway is possible in the plane of the frame. In the perpendicular direction K = 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
uFactored load, P = 1930 kips
yMaterial: A242 Grade 46 6 F = 46 ksi
x y yL = 16 ft; L = 16 ft; K = 0.8
g cL = 30 ft; L = 16 ft
x x yColumn: W14×193 6 I = 2400 in. ; A = 56.8 in. ; r = 6.50 in.; r = 4.05 in.4 2
xGirders: W30×90 6 I = 3610 in.4
AG = 10 (Recommended value for a pinned base)
From Eq. 8.5.7:
xEnter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain K = 1.80.
x x� K L = 1.80 × 16 = 28.8 ft
y yAlso, K L = 0.80 × 16.0 = 12.8 ft
;
Slenderness parameter,
cEnter LRFDS Table 4 with 8 = 0.674 and obtain,
d d e c cr g� P = P = N F A = 0.702 × 46 ×56.8 = 1835 kips < 1930 kips N.G.
Axial stress,
yFrom LRFDS Table 4-1, corresponding to F = 46 ksi, we obtain SRF = 0.331
Bi A� G = 0.623 × 0.331 = 0.205; G = 10
x cRe-enter nomograph and obtain, K = 1.75 6 8 = 0.655 6
d i c cr g� P = N F A = 0.710 × 46 × 56.8 = 1860 kips < 1930 kips N.G.
Steel Structures by S. Vinnakota Chapter 8 page 8-44
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
The W14×193 is therefore inadequate. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-45
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.35. 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
y y yK = 0.9; K L = 0.9 × 12.0 = 10.8 ft
c gL = 12 ft; L = 32 ft
x x yColumns: W12×152 6 I = 1430 in. ; A = 44.7 in. ; r = 5.66 in.; r = 3.19 in.4 2
xGirders: W18×50 6 I = 800 in. 4
From Eq. 8.5.7:
xEnter nomograph for sway uninhibited columns (Fig. 8.5.6 b) and get K = 2.23
;
c crFrom LRFDS Table 3-50, N F = 33.6 ksi
d c cr g�P = N F A = 33.6 × 44.7 = 1500 kips < 1540 kips N.G.
Axial stress,
yEnter LRFDM Table 4-1 for F = 50 ksi and obtain, SRF = 0.461
iA iB e x� G = G = SRF G = 0.461 × 4.76 = 2.20 � K = 1.65
c crEnter LRFDS Table 3-50, and obtain N F = 37.4 ksi
d c cr gP = N F A = 37.4 × 44.7 = 1670 kips > 1540 kips � O.K.
� The W12×152 of A992 steel is adequate (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-46
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.36. 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
x yColumn length, L = 12.5 ft; L = L = 12.5 ft
uFactored load, P = 2000 kips
A992 steel
xBuckling about the x- axis: Column part of an unbraced frame. 6 K $ 1.0
b cL = 36 ft; L = 12.5 ft
x xW14 × 193: I = 2400 in. ; r = 6.50 in.; A = 56.8 in.4 2
xW14 × 211: I = 2660 in. 4
xW24 × 84: I = 2370 in. 4
yBuckling about the y-axis: Column part of an unbraced frame. 6 K $ 1.0
b cL = 18 ft; L = 12.5 ft
y yW14 × 193: I = 931 in. ; r = 4.05 in.4
yW14 × 211: I = 1030 in.4
xW18 × 60: I = 984 in.4
For buckling about the major axis of the column section (Eq. 8.5.7):
;
xEnter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K = 1.80
For buckling about the minor axis of the column section (Eq. 8.5.7):
;
yAgain, enter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K = 1.40
Steel Structures by S. Vinnakota Chapter 8 page 8-47
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
;
c crFrom LRFDS Tables 3-50, N F = 34.9 ksi
d c cr gP = N F A = 34.9 × 56.8 = 1980 kips < 2000 kips N.G.
� The column is inadequate according to the LRFDS. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-48
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.37. 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
uFactored load, P = 800 kips
Column height, L = 15 ft
xGirder: W18×60 6 I = 984 in.4
a. Design of the leaning column CD
x y y yK = 1.0; K = 0.50; K L = 0.5×15 = 7.50 ft
Required strength = 800 kips
y y yFrom LRFDM, Table 4-2, for F = 50 ksi and KL = K L = 7.50 ft, tentatively select a W14×74 for
which,
dyP = 841 kips > 800 kips O.K.
x y x x yr /r = 2.44 6 (K L ) = < 7.70 ft
d dySo, P = P = 841 kips > 800 kips OK
Use a W14×74 for column CD. (Ans.)
b. Design of Column AB (Yura’s Method )
uyRequired axial strength (out of plane), P = 800 kips
uxRequired axial strength (in-plane), P = 800 + 800 = 1,600 kips
c bL = 15 ft; L = 32 ft
x xTry W14×233: A = 68.5 in. ; I = 3,010 in. ; r = 6.63 in.2 4
xEnter nomograph for sway uninhibited columns (Fig. 8.5.6b), and get K = 3.20
dx uxP = 24.4×68.5 = 1,671 kips > P = 1,600 kips O.K.
y y dy uyK L = 0.5 × 15 = 7.50 ft Y P = 2,810 kips > P = 800 kips OK
Use W14×233
Steel Structures by S. Vinnakota Chapter 8 page 8-49
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 McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.38. 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
xW30×124 girders 6 I = 5,360 in.4
a. Design of the leaning column CD
uFactored load on the column, P = 2200 kips
Column length, L = 16 ft
x yL = L = 16.0 ft
x y y yK = K = 1.0 6 K L = 1.0×16.0 = 16.0 ft
dyFrom LRFDM, Table 4-2, try W14×211 having P = 2,240 kips > 2,200 kips O.K.
Use W14×211 for column CD. (Ans.)
b. Design of columns AB and EF (Yura’s Method)
c xColumn length, L = 16 ft = L = L
gGirder length, L = 36 ft
uyRequired axial strength (out of plane), P = 1,000 kips
uxRequired axial strength (in-plane), P = due to symmetry
x xTry W14×233: A = 68.5 in. ; I = 3010 in. ; r = 6.63 in.2 4
xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K = 2.20
dx uxP = 31.6×68.5 � 2170 kips > P = 2100 kips O.K.
y y dy uyK L = 16 ft Y P = 2760 kips > P = 1000 kips O.K.
Use W14×233 for columns AB and EF. (Ans.)
Steel Structures by S. Vinnakota Chapter 8 page 8-50
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be
displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you
are using it without permission.
P8.39. 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
xW30×124 girders 6 I = 5,360 in.4
a. Design of the leaning columns AB and EF
uFactored load on the column, P = 1000 kips
Column length, L = 16 ft
x y1 y2L = 16.0 ft; L = L = 8.0 ft
x y y yK L = 16.0; K L = 1.0×8.0 = 8.00 ft
y y reqEnter LRFDM Table 4-2 for W14 shapes, with KL = K L = 8.0 ft, P = 1000 kips and observe that
dy x ya W14×90 gives P = 1070 kips and has r /r = 1.66.
x x y� (K L ) = 9.64 > 8.0 ft
x x y dReentering Table 4-2 with KL = (K L ) = 9.64 ft, we observe that a W14×90 provides, P = 1047
reqkips > P = 1000 kips.
Use W14×90 for columns AB and EF. (Ans.)
b. Design of column CD (Yura’s Method)
c xColumn length, L = 16 ft = L = L
gGirder length, L = 36 ft
uyRequired axial strength (out of plane), P = 2200 kips
uxRequired axial strength (in-plane), P = 2200 + 1000 = 3200 kips due to symmetry
x xTry W14×342: A = 101 in. ; I = 4900 in. ; r = 6.98 in.2 4
xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K = 2.14
dx uxP = 33.0× 101 � 3330 kips > P = 3200 kips O.K.
y y dy uyK L = 16 ft Y P = 3690 kips > P = 2200 kips O.K.
Use W14×342 (Ans.)