Steel Structures by S. Vinnakota
Chapter 9
page 9-1
CHAPTER 9
P9.1.
Determine the value of S x, Z x, "x of the following shapes, using the dimensions given in the LRFDM. (a) W36 × 230 (b)* W16 × 36 (c) WT18 × 115 (d) WT8 × 18
Solution a. Section: W36×230 b f = 16.5 in.; t f = 1.26 in. d = 35.9 in.; tw = 0.760 in. 1. Elastic section properties Flange area = 16.5 × 1.26 = 20.8 in.2 Web depth, d w = (35.9 - 2 × 1.26) = 33.4 in. Web area = 0.760 × 33.4 = 25.4 in.2 Total area, A = 20.8 × 2 + 25.4 = 67.0 in.2 For the doubly symmetric section considered, the center of gravity, G, coincides with the pont of intersection of the two symmetry axes. above the bottom fiber of the bottom flange
=
14,900 in.4
(Ans.)
2.
Plastic section properties To satisfy the condition IA Fy dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section considered, the PNA therefore coincides with the symmetry axis.
÷
= 18.0 in. above the bottom fiber of the bottom flange
Plastic section modulus, 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 9
page 9-2
= 934 in.3
(Ans.)
Shape factor,
(Ans.)
The corresponding values given in LRFDM Table 1-1 are: Ix = 15,000in.4; S x = 837 in.3 ; and Z x = 943 in.3 and take into account the contribution of the web-to-flange fillets.
b.
1.
Section: W16×36 b f = 6.99 in.;
t f = 0.430 in.;
d = 15.9 in.;
tw = 0.295
Elastic section properties Flange area = 6.99 × 0.430 = 3.01 in.2 Web depth, d w = (15.9 - 2 × 0.430) = 15.0 in. Web area = 15.0 × 0.295 = 4.43 in.2 Total area, A = 3.01 × 2 + 4.43 = 10.5 in.2 For the doubly symmetric section considered, the center of gravity, G, coincides with the pont of intersection of the two symmetry axes. above the bottom fiber of the bottom flange
=
443 in.4
(Ans.)
2.
Plastic section properties To satisfy the condition IA Fy dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section considered, the PNA therefore coincides with the symmetry axis.
÷
= 7.95 in. above the bottom fiber of the bottom flange
Plastic section modulus, 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 9
= 63.2 in.3
Shape factor,
page 9-3
(Ans.)
(Ans.)
The corresponding values given in LRFDM Table 1-1 are: Ix = 448 in.4; S x = 56.5 in.3 ; and Z x = 64.0 in.3 and take into account the contribution of the web-to-flange fillets.
P9.2.
Determine the value of S x, Z x, "x of the following built-up sections: (a) A W16 × 36 with one ½ × 12 in. plate welded to each flange. (b) A W16 × 36 with one ½ × 12 in. plate welded to the top of the flange (c) A W16 × 36 with a C12 × 20.7 with its web welded to the top flange.
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Steel Structures by S. Vinnakota
P9.3.
Chapter 9
page 9-4
Determine the values of M yx and M px for the shapes given in Problem P9.1.
Solution
P9.4.
a.
W36×230 Yield stress, F y = 50 ksi Elastic section modulus, S x = 828 in.3 (from solution to P9.1a.) Plastic section modulus, Z x = 934 in.3 (from solution to P9.1a.) Yield moment, M y = S x F y = 828 × 50.0 ÷ 12 = 3450 ft-kips Plastic moment, M p = Z x F y = 934 × 50.0 ÷ 12 = 3892 ft-kips
b.
W16×36 Yield stress, F y = 50 ksi Elastic section modulus, S x = 55.7 in.3 (from solution to P9.1b.) Plastic section modulus, Z x = 63.2 in.3 (from solution to P9.1b.) Yield moment, M y = S x F y = 55.7 × 50.0 ÷ 12 = 232 ft-kips Plastic moment, M p = Z x F y = 63.2 × 50.0 ÷ 12 = 263 ft-kips
Determine the values of M yx and M px for the built-up shapes given in Problem P9.2. Assume A36 steel for all elements.
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
P9.5.
Chapter 9
page 9-5
a) A W18x35 beam of A992 steel spans 30 ft and is connected to columns at either end by means of standard web connections. Compute the uniformly distributed factored load that the member can resist. Assume continuous lateral bracing for the compression flange. b) Find the maximum spacing of lateral supports for the design to still hold good. Solution Simply supported beam: Span, L = 30 ft Loading: Uniformly distributed load, q u Compression flange continuously laterally braced. a. Section: W21×62 From LRFDM Table 1-1 6 b f / 2t f = 7.06; h /t w = 53.5 Material: A992 steel 6 Fy = 50 ksi From LRFDM Table 5-3, for a W18×35, Nb M px = 249 ft-kips; Nv Vn = 143 kips; L p = 4.31 ft
Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength, Md =
Nb M px = 249 ft-kips
As h /tw < 8pv, the design shear strength is given by Vd =
Nv Vn = 143 kips
As per LRFD: Maximum bending moment,
Maximum shear corresponding to this load is, Vmax = q u L/ 2 = 2.21 × 30 ÷ 2 = 33.2 kips <
Vd = 143 kips O.K.
So, the maximum uniformly distributed factored load the W18×35 of A992 steel can support is 2.21 klf. (Ans.) b.
The beam could be considered adequately braced for L b # L p = 4.31 ft say 4' 3"
(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
P9.6.
Chapter 9
page 9-6
Determine the uniformly distributed factored load that can be carried by a W21×62 beam of A572 Grade 42 steel over a simply supported span of 24 ft with lateral supports at 6 ft. Solution Simply supported beam: Span, L = 24 ft Unbraced length, L b = 6 ft Section, W21×62 Material: A572 Gr 42 6 Fy = 42 ksi From LRFDM Table 1-1, for a W21×62, Z x = 144 in.3;
r y = 1.77 in.
From Eq. 9.7.4, limiting unbraced length is,
As L b < L p, the beam is considered adequately laterally braced. From Eq. 9.7.2, the design bending strength is,
Maximum bending moment,
So, the maximum uniformly distributed factored load the W21×62 of A572 Gr 42 can support is 6.3 klf. (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
P9.7.
Chapter 9
page 9-7
A W24×176 of A572 Grade 60 steel is used for a simple span of 36 ft. If the only dead load present is the weight o the beam, what is the largest service concentrated live load that can be placed at mid-span? Assume continuous lateral bracing. Deflection need not be checked.
Solution Simply supported beam: Span, L = 36 ft. Compression flange continuously laterally braced. Section: W24×176 Material: A572 Gr 60 steel 6 Fy = 60 ksi Loading: Dead load: Self weight of the beam 6 q D = 176 plf = 0.176 klf Live load: A central concentrated load 6 QL Factored loads: Uniformly distributed load 6 q u = 1.2× 0.176 = 0.211 klf Central concentrated load 6 Q u = 1.6 Q L Maximum bending moment at the center under factored loads is,
From LRFDM Table 1-1 6 Z x = 511 in.3; b f / 2t f = 4.81;
d = 25.2 in.; h /tw = 28.7
t w = 0.750 in.
Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength is, M d = Nb M px = Nb Z x Fy = 0.90 × 511 × 60.0 = 27,594 in.-kips = 2300 ft-kips As h /tw < 8pv, the design shear strength is given by Vd = Nv V n = Nv d tw (0.6 Fy) = 0.90 × 25.2 × 0.750 × 0.60 × 60.0 = 612 kips As per LRFD:
Maximum shear corresponding to this load is,
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Steel Structures by S. Vinnakota
Chapter 9
page 9-8
Vmax = ½ q u L + ½ Q u = (0.5×0.211×36) + ( 0.5×1.6×157) = 129 kips < Vd = 612 kips O.K. So, the maximum concentrated service live load the W24×176 of A572 Grade 60 steel can support is 157 kips. (Ans.)
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Steel Structures by S. Vinnakota
P9.8.
Chapter 9
page 9-9
A triangular opening in the floor of an industrial building results in the factored loads on a W18×35 simple beam to be as shown in Fig. P9.8. The beam weight is not included. Use A588 Grade 50 steel. Assume full lateral support for the compression flange. Check the adequacy of the beam. See Fig. P9.8 of the text book. Solution Simply supported beam, AB. Span, L = 32 ft Factored loads: Concentrated load at center, C = 20 kips Two triangular loads, with maximum intensity (at supports) of 1.5 klf. Total factored load on the beam = 20.0 + 2×½×1.5×16 = 44.0 kips Reaction, Due to symmetry of the structure and the loading the maximum bending moment occurs at the center, C.
Maximum shear, Vmax = VA = 22 kips Additional bending moment at midspan due to self weight of the beam,
Required bending strength, M req = 224 + 5.38 = 229 ft-kips From LRFDM Table 5-3 for a W18×35,
As the beam is continuously laterally braced, Also Vd = 143 kips > 22 kips
O.K.
The W18×35 beam is adequate for the given loads.
Note:
(Ans.)
To be considered adequately laterally supported, bracing should be provided such that L b # 4' 4".
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Steel Structures by S. Vinnakota
P9.9.
Chapter 9
page 9-10
A 34 ft long W27×94 shape of A992 steel is used as a simply supported beam. It is subjected to a factored concentrated load of 80 kips at 12 ft from each support. In addition, it is subjected to a factored moment of 340 ft-kips, one at its left-end (anti-clockwise) and one at its right-end (clockwise). Neglect self weight of the beam in the calculation and check if the beam is safe. Assume full lateral support for the compression flange.
Solution a. Data Simply supported beam, AB: Span, L = 34 ft Compression flange continuously laterally braced. Loading: Factored concentrated load at D and E (with AD = BE = a = 12 ft) = 80 kips Factored end-moment at B = 340 ft-kips (anti-clockwise) Factored end-moment at E = 340 ft-kips (clockwise) Uniformly distributed load, q u due to self-weight ( = 1.2 × 0.094 = 0.113 klf ) b.
Required strengths Due to the symmetry of the structure and the loading, the maximum bending moment occurs at the center, C. Consider the influence of self weight separately. From the applied loads: R A = R B = 80.0 kips M A = - 340 ft-kips M C = 80.0×17.0 - 80.0×5.0 - 340 = + 620 ft-kips Due to self weight: R A = R B = 0.113× 34 ÷ 2 = 1.92 kips MC So: Required bending strength, M u = 620 + 16.3 = 636 ft-kips Required shear strength, Vu = 80.0 + 1.92 = 81.9 kips
c.
Design strengths Section: W27×94 From LRFDM Table 1-1 6 b f / 2t f = 6.70; h /t w = 49.5 Material: A992 steel 6 Fy = 50 ksi From LRFDM Table 5-3, for a W27×94, Nb M px = 1040 ft-kips; Nv Vn = 356 kips; L p = 7.49 ft Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact.
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Steel Structures by S. Vinnakota
Chapter 9
So, the design bending strength is: M d =
Nb M px = 1040 ft-kips
Also, as h /t w < 8 pv, the design shear strength is: As, M u <
M d and Vu <
page 9-11
Vd =
Nv V n = 356 kips
Vd, the W27×94 of A992 steel can support the given loads.
(Ans.)
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Steel Structures by S. Vinnakota
P9.10.
Chapter 9
page 9-12
Repeat Problem P9.9, with the end moment acting at the left end only.
Solution a. Data Simply supported beam, AB: Span, L = 34 ft Compression flange continuously laterally braced. Loading: Factored concentrated load at D and E (with AD = BE = a = 12 ft) = 80 kips Factored end-moment at B = 340 ft-kips (anti-clockwise) Uniformly distributed load, q u due to self-weight ( = 1.2 × 0.094 = 0.113 klf ) b.
Required strengths Consider the influence of self weight separately. From the applied loads: R A = 90.0 kips; R B = 70.0 kips M A = - 340 ft-kips M D = 90.0 × 12.0 - 340 = + 740 ft-kips M E = 90.0×22.0 - 80.0×10.0 - 340 = + 840 ft-kips M B = 0 ft-kips So, the maximum bending moment under applied load occurs at E. Due to self weight: R A = R B = 0.113× 34 ÷ 2 = 1.92 kips MC The maximum bending moment now occurs at the center, C. So: Required bending strength, M u = 840 + 16.3 = 856 ft-kips (conservatively) Required shear strength, Vu = 90.0 + 1.92 = 91.9 kips
c.
Design strengths Section: W27×94 From LRFDM Table 1-1 6 b f / 2t f = 6.70; h /t w = 49.5 Material: A992 steel 6 Fy = 50 ksi From LRFDM Table 5-3, for a W27×94, Nb M px = 1040 ft-kips; Nv Vn = 356 kips; L p = 7.49 ft Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact.
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Steel Structures by S. Vinnakota
Chapter 9
So, the design bending strength is: M d =
Nb M px = 1040 ft-kips
Also, as h /t w < 8 pv, the design shear strength is: As, M u <
M d and Vu <
page 9-13
Vd =
Nv V n = 356 kips
Vd, the W27×94 of A992 steel can still support the given loads. (Ans.)
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Steel Structures by S. Vinnakota
P9.11.
Chapter 9
page 9-14
A wide flange beam AB frames across an open well in a building and is subjected to the factored loads shown in Fig. P9.11. Determine if a W16×50 of Fy = 50 ksi steel is sufficient under these conditions. Assume continuous lateral support to the compression flange. See Fig. P9.11 of the text book. Solution Simply supported beam, AB. Span, L = 40 ft Material: Fy = 50 ksi steel Factored loads: Concentrated load at D, 24 ft from A: 10 kips Uniformly distributed load over AD: 1.6 klf
Location of point of zero shear ( z o from A): Maximum moment, Additional moment due to self-weight, Required bending strength, M u = 298 + 12.0 = 310 ft-kips Maximum shear, Vmax = 30.9 kips From LRFDM Table 5-3, for a W16×50 of Fy = 50 ksi steel,
As the section is compact and the compression flange continuously laterally supported
So, the W16×50 of F y = 50 ksi steel is adequate. P9.12.
(Ans.)
A simple beam consists of a W18×55 with a PL1×12 cover plate welded to each flange. Determine the factored uniform load the beam can support in addition to its own weight for a 28-ft simple span. Assume A572 Grade 50 steel and full lateral support.
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Steel Structures by S. Vinnakota
P9.13.
Chapter 9
page 9-15
A 36-ft long simple beam consists of a built-up section obtained by welding three A36 steel plates as shown in Fig. P2.6. Determine the factored uniformly distributed load the beam can support, in addition to its selfweight. The beam is continuously laterally supported See Fig. P2.6 of the text book. Solution Simply supported beam. Span, L = 36 ft Material: A36 steel. Fy = 36 ksi Section: Built-up by welding three plates. a.
Elastic section properties Total area, A = 8.00 × 2.00 + 1.00×16.0 + 12.0×2.0 = 56.0 in.2 To satisfy the condition IA f dA = 0, the elastic neutral axis must pass through the center of gravity of the cross section.
=
8.71 in. above the bottom fiber of the bottom flange
=
3,500 in.4
;
b.
Plastic section properties To satisfy the condition IA Fy dA = 0, for the unequal flanged I-section, the PNA divides the section into two equal areas. As, A /2 = 56/2 = 28.0 in.2 is greater than the area of the bottom flange, it follows that the PNA will be in the web. To locate the PNA, find the distance yp, as measured from the bottom fiber of the bottom flange, such that the area above the PNA is equal to the area below. (12.0× 2.00) + l.00×(
- 2.00) =
1.00× (18.0 -
) + 8.00× 2.00 ÷
= 6.0
Plastic section modulus,
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Steel Structures by S. Vinnakota
Chapter 9
page 9-16
= 408 in.3 c.
Plastic moment From LRFDS Eq. F1-1: Plastic moment, M p
= min [ F y Z x ; 1.5 F y S x min ] = min [ (36.0×408) ÷ 12; (1.5×36×310)÷ 12 ]
d.
= min [ 1224 ft-kips; 1395 ft-kips ] = 1224 ft-kips Design strengths Width-to-thickness ratios of the plates: Compression flange: b f / 2t f = 4.00 / 2.00 = 2.00 Web in flexure: h c /t w = (2×12.0) / 1.00 = 24.0 Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength, Md =
Nb M px =
0.90 × 1224 = 1100 ft-kips
As h /tw < 8pv, the design shear strength is given by Vd =
Nv Vn = 0.9 × 16.0 × 1.0 × 0.60 × 36.0 = 311 kips
As per LRFD:
Maximum shear corresponding to this load is, Vmax = q u L/ 2 = 6.79 × 36 ÷ 2 = 122 kips <
Vd = 311 kips O.K.
Self-weight of the beam = 56 × 3.4 = 190 plf = 0.190 klf Factored value of the self-weight = 1.2 × 0.190 = 0.228 klf So, the maximum uniformly distributed factored load the built-up section can support, in addition to its self-weight is: 6.56 klf. (Ans.)
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Steel Structures by S. Vinnakota
P9.14.
Chapter 9
page 9-17
A W16×57 of A992 steel is used for the over hanging beam ABCD (Fig. P9.14) under the service loads (50% dead and 50% live ) shown. Assume continuous lateral support, and check the adequacy of the beam for bending and shear. See Fig. P9.14 of the text book. Solution Overhanging beam ACBD. Main span AB: L = 20 ft Overhang BD: a = 8 ft Service loads: Uniformly distributed load over AD: q = 2 klf Concentrated load at free end, D: Q D = 16 kips Concentrated load at center of main span, C: Q C = 32 kips Factored distributed load, Factored concentrated load at C, Factored concentrated load at D,
q u = 1.2(1.0) + 1.6(1.0) = 2.80 klf Q uC = 1.2(16.0) + 1.6(16.0) = 44.8 kips Q uD = 1.2(8.0) + 1.6(8.0) = 22.4 kips
Maximum positive moment, occurs at C. Maximum negative moment occurs at support B.
So, Maximum shear, VBC = 63.8 kips. From LRFDM Table 5-3, for a W16×57, As the section is compact and continuously braced,
So, the W16×57 is adequate.
(Ans.)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Steel Structures by S. Vinnakota
P9.15.
Chapter 9
page 9-18
A W14×30 of A992 steel is used as a simply supported, uniformly loaded beam with a span length of 28 ft. The service dead load (not including the self weight) is 0.4 klf and the service live load is 0.78 klf. Assume that the beam is continuously laterally supported by the floor deck. Check the adequacy of the beam for bending and shear. Calculate the maximum dead load and live load deflections of the beam. Solution Simply supported beam. Span, L = 28 ft Self weight of the beam = 30 plf Service dead load, q D = 0.40 + 0.03 = 0.43 klf Service live load, q L = 0.78 klf Since the dead load is less than 8 times the live load, load combination LC-2 controls:
For a simply supported beam, under uniformly distributed load, the maximum bending moment occurs at midspan, and is equal to:
Maximum shear, which occurs at support, is equal to
From LRFDM Table 5-3 for a W14×30 As the beam is continuously braced, We have O.K. O.K. So, the W14×30 is satisfactory for bending and shear.
(Ans.)
For a simply supported beam, under uniformly distributed load, the maximum deflection occurs at midspan (Case 1, Table 5-17 of the LRFDM). The central deflection under service live load is, (Ans.)
The central deflection under service dead load is
(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
P9.16.
Chapter 9
page 9-19
A W30×148 of Fy = 50 ksi steel has been selected for the simply supported beam shown in Fig. P9.16. Loads Q consist of 50 kips dead load and 80 kips live load. Check the adequacy of the beam for bending and shear. Calculate the maximum dead load and live load deflections. Assume continuous lateral support for the compression flange. See Fig. P9.16 of the text book. Solution Simply supported beam AB. Span, L = 26 ft Concentrated loads, at center C, and at points D and E (CD = CE = 10 ft) Service load, Q: 50 kips dead load, and 80 kips live load Factored load, Q u = 1.2 × 50.0 + 1.6 × 80.0 = 188 kips Reaction at A, Bending moment at D, M D = 282×3.0 = 846 ft-kips Bending moment at C, Bending moment at C due to self weight, So, the required bending strength, Required shear strength,
From LRFDM Table 5-3, for a W30×148: We have
From Beam Diagrams and Formulas (Case 7 and Case 9 of LRFDM Table 5-17), the maximum deflection which occurs at C:
(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
P9.17.
a)
b)
Chapter 9
page 9-20
A W12×30 of A992 steel is used as a simply supported beam with a span length of 10 ft. It is subjected to two concentrated loads of 80 kips each, located 1 ft from each support. The loads given are service live loads. Neglect self weight of the beam in the calculations. Check the adequacy of the beam for bending and shear, assuming that the beam is continuously laterally supported. Redesign, if necessary (limit the selection to W12s only).
Solution Simply supported beam AB. Span, L = 10 ft Loading: Two concentrated loads at C and D (AC = BD = a = 1 ft) Service live load, Q L = 80 kips Factored loads, Q u = 1.6Q L = 1.6×80 = 128 kips
From LRFDM Table 5-3, for a W12×30 Shape is compact and laterally supported, therefore Vd = Nv Vn = 86.3
<
So, the W12×30 is not satisfactory.
Vmax = 128 kips
N.G. (Ans.)
Entering LRFDM Table 5-4 with Vreq = 128 kips and limiting the selection to W12 shapes, observe that a W12×72 with
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
P9.18.
Chapter 9
page 9-21
A large steel beam simply spans 60 feet and carries an applied load of 1.5 klf ( D = 0.5 klf and L = 1.0 klf). Select the lightest adequate W-shape of A992 steel. Include the effect of member self weight. Assume that the beam is continuously laterally supported. Solution Simply supported beam. Continuously laterally supported. Material: A992 steel. 6 Fy = 50 ksi Span, L = 60 ft As the span is large, assume a self-weight of 100 plf or 0.100 klf. Dead load, q D = 0.500 + 0.100 = 0.600 klf Live load, q L = 1.00 klf Factored load, q u = 1.2(0.600) + 1.6(1.00) = 2.32 klf Maximum factored moment,
Maximum shear, Entering LRFDM Table 5-3 with M req = 1044 ft-kips, we observe that a W30×90 is the lightest section that provides adequate flexural strength. For this shape,
As the weight of the beam (90 plf) is less than the value assumed (100 plf) the design is O.K. Select a W30×90.
(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
P9.19.
Chapter 9
page 9-22
Select the lightest W section to carry a uniform dead load of 0.3 klf and a live load of 0.6 klf on a simply supported span of 34 ft. The beam is continuously laterally supported. Assume no deflection limitations. Use A572 Grade 42 steel. Solution Simply supported beam. Continuously laterally supported. Material: A572 Grade 42 steel. 6 Fy = 42 ksi Span, L = 34 ft Assume a self-weight of 40 plf or 0.040 klf. Dead load, q D = 0.300 + 0.040 = 0.340 klf Live load, q L = 0.600 klf Factored load, q u = 1.2(0.340) + 1.6(0.600) = 1.37 klf Maximum factored moment,
Maximum shear, From Eq. 9.7.18:
Entering LRFDM Table 5-3 with Z req = 62.9 in.3, we observe that a W18×35 (in bold face) is the lightest section that provides adequate flexural strength. For this shape, we have: Z x = 66.5 in.3;
d = 17.7 in.; t w = 0.300 in.; b f /2t f = 7.06;
h /t w = 53.5
As the section is compact for Fy = 50 ksi steel (no indication to the contrary in this table), it is compact for Grade 42 steel considered. So, the design bending strength is, M d = Nb M px = Nb Z x Fy = 0.90 × 66.5 × 42.0 = 2514 in.-kips = 210 ft-kips Limiting b/t ratio for plate (web) buckling in shear: As h /tw < 8pv, the design shear strength is given by Vd = Nv V n = Nv d tw (0.6 Fy) = 0.90 × 17.7 × 0.300 × 0.60 × 42.0 = 120 kips As, As the weight of the beam (35 plf) is less than the value assumed (40 plf) the design is O.K. Select a W18×35 of A572 Gr 42 steel.
(Ans.)
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Steel Structures by S. Vinnakota
P9.20.
Chapter 9
page 9-23
A simply supported beam AB with a span of 48 feet must carry three column loads of 100 kips each (Fig. P9.20). The loads are factored loads. Assume full lateral support for the compression flange. Select the most economical W-section assuming A992 steel, if the nominal depth is limited to 30 in. Include the effect of member self-weight. See Fig. P9.20 of the text book. Solution Simply supported beam, AB. Span, L = 48 ft Factored, concentrated load at center C = 100 kips Factored, concentrated loads at points D and E (AD = BE = a = 12 ft) = 100 kips Neglect influence of self-weight to start with. Total factored load on the beam = 300 kips Due to symmetry of the structure and the loading, reaction, R A = 150 kips Maximum bending moment occurs at midspan.
Maximum shear, Vmax = 150 kips Entering the LRFDM Table 5-3 with M req = 2400 ft-kips, we observe that a W30×191(in bold face) is the lightest W30 shape that provides the required flexure strength. For this section:
Additional bending moment at C due to factored self-weight is,
As the revised maximum moment, 2400 + 66.0 = 2470 ft-kips is less than 2530 ft-kips, the W30×191 selected is adequate.
(Ans.)
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Steel Structures by S. Vinnakota
P9.21.
Chapter 9
page 9-24
The 32 foot long steel beams AB of Figure P9.21 are spaced 8 feet on center. They are simply supported at points D and E, and must carry a service floor load of 100 psf (50% dead load and 50% live load) and a wall load of 0.6 kips per linear foot of wall (100% dead load). Assume adequate lateral support, A992 steel, and select the lightest W-12 shape for flexure and shear. The maximum service load deflection of the beam should be within ± ½ in. See Fig. P9.21 of the text book. Solution Beam with overhangs, ADCEB. Main span (DE), L 1 = 24 ft; overhangs (DA, EB), L 2 = 4 ft Spacing of beams, S b = 8 ft Service distributed dead load, q D = 50.0 × 8.0 = 400 plf Service distributed live load, q L = 50.0 × 8.0 = 400 plf Factored distributed load, Wall load = 0.6 klf Factored concentrated load at A and B, Symmetric and symmetrically loaded structure. Reactions, R D = R E = 5.76 + 1.12×16.0 = 23.7 kips The maximum positive bending moment occurs at midspan (point C). We have:
Maximum negative bending moment occurs at the supports B and D:
Maximum moment, M max = 48.6 ft-kips From LRFDM Table 5-3, a W12×14 with NM px = 65.2 ft-kips > 48.9 ft-kips is the lightest section that has adequate bending strength. Select a W12×14 section 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
P9.22.
Chapter 9
page 9-25
A simply supported beam with a span of 24 ft has a uniform load of 0.6 klf over the entire span. In addition, it has a 40 kip load 8 ft from left end and a 30 kip load 8 ft from the other end of the beam. Select a Wshape of A 572 Grade 60 steel that can safely support the given factored loads. Assume full lateral support.
Solution Simply supported beam. Span AB: L = 24 ft Section: W-shape. Compression flange continuously laterally braced. Material: A572 Grade 60 steel. 6 Fy = 60 ksi Factored loads: Uniformly distributed over AB: = 0.6 klf Concentrated load at point D (with AD = 8 ft): Q uD = 40.0 kips Concentrated load at point E (with BE = 8 ft): Q uE = 30.0 kips Assume self weight of the beam = 50 plf = 0.05 klf Factored distributed load, q u = 1.2( 0.05) + 0.6 = 0.660 klf
Maximum moment occurs at D. Maximum shear, Vmax = R A = 44.5 kips. From Eq. 9.7.18:
Entering LRFDM Table 5-3 with Z req = 74.4 in.3, we observe that a W18×40 (in bold face) is the lightest section that provides adequate flexural strength. For this shape, we have: Z x = 78.4 in.3;
d = 17.9 in.; t w = 0.315 in.; b f /2t f = 5.73;
h /t w = 50.9
Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength is,
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Steel Structures by S. Vinnakota
Chapter 9
page 9-26
M d = Nb M px = Nb Z x Fy = 0.90 × 78.4 × 60.0 = 4234 in.-kips = 353 ft-kips As h /tw < 8pv, the design shear strength is given by Vd = Nv V n = Nv d tw (0.6 Fy) = 0.90 × 17.9 × 0.315 × 0.60 × 60.0 = 183 kips As, As the weight of the beam (40 plf) is less than the value assumed (50 plf) the design is O.K. Select a W18×40 of A572 Gr 60 steel.
(Ans.)
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Steel Structures by S. Vinnakota
P9.23.
Chapter 9
page 9-27
A simply supported beam AB, with a span of 28 ft, has a 40 kip load at a point D, 16 ft from end A. In addition, the beam is subjected to a uniformly distributed load of 1.6 klf over the portion AD. Select a Wshape of A572 Grade 50 steel that can safely support the given factored loads. Assume full lateral support to the compression flange. Solution Simply supported beam, AB. Span, L = 28 ft Material: Fy = 50 ksi steel Factored loads: Concentrated load at D, 16 ft from A: Q u D = 40 kips Uniformly distributed load over AD: 1.6 klf
As the shear changes sign at D, the maximum bending moment under these loads occurs at D.
Assume self-weight of the beam = 50 plf = 0.05 klf Additional moment due to self-weight, Additional shear due to self weight, Required bending strength, M u = 362 + 5.88 = 368 ft-kips (conservatively) Required shear strength, Vu = 35.4 + 0.84 = 36.2 kips
Entering LRFDM Table 5-3 with M req = 368 ft-kips, we observe that a W21×50 is the lightest compact section that provides adequate flexural strength. For this shape,
As the weight of the beam (50 plf) is the same as the value assumed (50 plf) the design is O.K. So, select a W21×50. Note:
(Ans.)
From LRFDM Table 5-3, we observe that a W21×48 of Grade 50 steel has adequate bending and shear strengths; and could be selected. However, the section is non-compact. Design of noncompact beams is considered in Section 10.5.2.
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Steel Structures by S. Vinnakota
P9.24.
Chapter 9
page 9-28
A 20 ft long cantilever beam AB is fixed at end A, and free at end B. It is subjected to a concentrated factored load of 24 kips at point C, where AC = 12 ft. In addition, the beam is subjected to a uniformly distributed factored load of 2.5 klf over its entire length. Select a W-shape of A572 Grade 60 steel that can support the loads. Assume adequate lateral support to the beam flanges.
Solution Cantilever beam. Span AB: L = 20 ft Section: W-shape. Flanges adequately laterally braced. Material: A572 Grade 60 steel. 6 Fy = 60 ksi Factored loads: Uniformly distributed over AB: = 2.50 klf Concentrated load at point C (with AC = 12 ft): Q uC =
24.0 kips
Assume self weight of the beam = 100 plf = 0.10 klf Factored distributed load, q u = 1.2( 0.10) + 2.5 = 2.62 klf
From Eq. 9.7.18:
Entering LRFDM Table 5-3 with Z req = 176 in.3, we observe that a W24×68 (in bold face) is the lightest section that provides adequate flexural strength. For this shape, we have: Z x = 177 in.3;
d = 20.1 in.; t w = 0.415 in.; b f /2tf = 7.66;
h /t w = 52.0
Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength is, M d = Nb M px = Nb Z x Fy = 0.90 × 177 × 60.0 = 9558 in.-kips = 797 ft-kips As h /tw < 8pv, the design shear strength is given by Vd = Nv V n = Nv d tw (0.6 Fy) = 0.90 × 20.1 × 0.415 × 0.60 × 60.0 = 270 kips PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Steel Structures by S. Vinnakota
Chapter 9
page 9-29
As, As the weight of the beam (68 plf) is less than the value assumed (100 plf) the design is O.K. Select a W24×68 of A572 Gr 60 steel.
(Ans.)
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Steel Structures by S. Vinnakota
P9.25.
Chapter 9
page 9-30
A simply supported beam ABCD with an overhanging end is supported at points B and D such that AB = 8 ft and BC = CD = 12 ft. The beam is subjected to a uniformly distributed, factored load of 3 klf over the entire length. In addition, it is subjected to concentrated, factored loads of 10 kips and 40 kips, at points A and C, respectively. Select a W-shape of A572 Grade 50 steel that can support the factored loads. Assume adequate lateral support for the compression flange. Solution Section: W-shape. Compression flange adequately laterally braced. Material: A572 Grade 50 steel. 6 Fy = 50 ksi Overhanging beam ABCD. Overhang AB: a = 8 ft Main span BD: L = 24 ft Factored loads: Uniformly distributed over AD: = 3.0 klf Concentrated load at free end, A: Q uA = 10 kips Concentrated load at center of main span, C: Q uC = 40 kips Assume self weight of the beam = 50 plf = 0.05 klf Factored distributed load, q u = 1.2( 0.05) + 3.0 = 3.06 klf
Maximum positive moment, occurs at C. Maximum negative moment occurs at support B.
So, Maximum shear, Vmax = VBC = 98.6 - 10.0 - 3.06 × 8.0 = 64.1 kips. Entering LRFDM Table 5-3 with M req = 371 ft-kips, we observe that a W21×50 is the lightest compact section that provides adequate flexural strength. For this shape,
As the weight of the beam (50 plf) is the same as the value assumed (50 plf) the design is O.K. So, select a W21×50. Note:
(Ans.)
From LRFDM Table 5-3, we observe that a W21×48 of Grade 50 steel has adequate bending and shear strengths; and could be selected. However, the section is non-compact. Design of noncompact beams is considered in Section 10.5.2.
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Steel Structures by S. Vinnakota
P9.26.
Chapter 9
page 9-31
Select the most economical wide flange section for the 30-ft-long overhang beam, with supports at the left end and 10 ft from the right end. It is subjected to the following service loads: 1.5 klf dead load and 2 klf live load uniformly distributed over the entire length of the beam. In addition, the beam is subjected to concentrated live loads of 30 kips and 10 kips at 10 ft and 30 ft respectively from the left end. The beam is continuously laterally braced. Assume A572 Grade 42 steel.
Solution Section: W-shape. Compression flange continuously laterally braced. Material: A572 Grade 42 steel. 6 Fy = 42 ksi Overhanging beam ACBD. Main span AB: L = 20 ft Overhang BD: a = 10 ft Service loads: Uniformly distributed over AD: Dead load = 1.5 klf Live load, q L = 2.0 klf Concentrated live load at free end, D: Q LD = 10 kips Concentrated live load at center of main span, C: Q LC = 30 kips Assume self weight of the beam = 60 plf = 0.06 klf Factored distributed load, Factored concentrated load at C, Factored concentrated load at D,
q u = 1.2(1.5 + 0.06) + 1.6(2.0) Q uC = 1.6(30.0) = 48.0 kips Q uD = 1.6(10.0) = 16.0 kips
= 5.07 klf
Maximum positive moment, occurs at C. Maximum negative moment occurs at support B.
So, Maximum shear, Vmax = VBC = 162 - 16.0 - 5.07 × 10.0 = 95.3 kips. From Eq. 9.7.18:
Entering LRFDM Table 5-3 with Z req = 131 in.3, we observe that a W24×55 (in bold face) is the lightest section that provides adequate flexural strength. For this shape, we have: Z x = 135 in.3; d = 23.6 in.; tw = 0.395 in.; b f /2tf = 6.94; h /tw = 54.1 Limiting b/t ratios for plate buckling:
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Steel Structures by S. Vinnakota
Chapter 9
page 9-32
Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength is, M d = Nb M px = Nb Z x Fy = 0.90 × 135 × 42.0 = 5103 in.-kips = 425 ft-kips As h /tw < 8pv, the design shear strength is given by Vd = Nv V n = Nv d tw (0.6 Fy) = 0.90 × 23.6 × 0.395 × 0.60 × 42.0 = 211 kips As, As the weight of the beam (55 plf) is less than the value assumed (60 plf) the design is O.K. Select a W24×55.
(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 9
page 9-33
P9.27.
Select the lightest W-shape for a 18-ft-long cantilever beam of A572 Grade 42 steel. It is subjected to concentrated, service live loads of 16 kips and 10 kips acting at 12 ft and 18 ft, respectively, from the fixed end. The beam is continuously laterally braced. The maximum service load deflection is limited to 1/600 of its span length. Neglect beam weight in all the calculations.
P9.28.
In a building, roof beams are spaced at 10-ft intervals and are connected to girder webs by simple connections. The span of the beams is 24 ft. The beam is designed for the following service loads: dead load (not including the self weight), 65 psf; roof live load, 20 psf; snow load, 40 psf; and rain load, 25 psf. Deflection of the beam is limited to L/240. Select a suitable W-shape of A242 Grade 42 steel. Assume the beam is continuously laterally supported by the roof decking. Specify any camber provided.
P9.29.
A simple beam in a research lab is to support a dead load of 1 klf and a live load of 1.5 klf. In addition, it is to carry a concentrated live load (movable) of 6 kip. The load consists of precision machinery which requires that the deflection be limited to a maximum of L/800. The beam has an effective span of 40 ft and is continuously laterally supported. Use A992 steel and select a suitable section.
P9.30.
A W-shape is used to support a factored load of 12 klf on a 20 ft simple span. The architect specifies that the beam be no more than 18 in. in depth. Assume full lateral support to the compression flange and select a suitable A 992 steel beam. Is the deflection OK, if the beam carries a plastered ceiling? Live load is b rd the total load on the beam.
P9.31.
A simply supported beam with continuous lateral support for the entire span of 30 ft is subjected to a factored, uniformly distributed load of 2.4 klf. The beam depth is limited to not more than 14 in. (nominal), and the deflection is limited to L/360. Select a suitable A992 shape.
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Steel Structures by S. Vinnakota
P9.32.
Chapter 9
page 9-34
To save construction depth with a precast plank system, a structural tee is used for the overhang beam ABC with supports at A and B, as shown in Fig. P9.32. The beam is subjected to a dead load of 0.6 klf and a live load of 0.3 klf. The dead load includes provision for self-weight of the beam. Assume continuous lateral support, and select a suitable WT6 of A992 steel. See Fig. P9.32 of the text book. Solution Yield stress of material, Fy = 50 ksi Beam with overhang, ABC. Main span (AB) = L 1 = 20 ft Overhang (BC) = L 2 = 6 ft Loading: Dead load, q D = 0.6 klf Live load, q L = 0.3 klf Factored load, q u = 1.2 × 0.6 + 1.6 × 0.3 = 1.20 klf Total load = 1.20×26.0 = 31.2 kips ; Location of point of zero shear, from support A 6 Maximum positive moment,
Maximum negative moment occurs at support B,
Required bending strength, M req = 49.5 ft-kips As the beam is continuously laterally supported
Select a WT6×68 with S x = 9.46 in.3
(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
P9.33.
Chapter 9
page 9-35
The bay size for a shopping center is 32'×30'. The beams are spaced at 8 ft centers and have a span of 30 ft. All the beams and girders are simply supported. The floor deck consists of 5-in.-thick slab of light weight concrete (100 pcf) directly supported by the beams. Dead load from flooring and ceiling is 10 psf. The floor live load, including provision for partitions, is 120 psf. Design an interior beam without counting on composite action between the slab and the steel beam, but assuming that lateral buckling of the beam is prevented. Limit the live load deflection to L/360. Use A992 steel. See Fig. 3.2.2 of the text book for guidance. Solution Simply supported beam. Span, L = 30 ft. Spacing, S b = 8 ft. Weight of slab = Weight of flooring and ceiling = 10 psf Assume the self-weight of the beam to be 40 plf. Distributed dead load, Distributed live load, Distributed factored load on the beam,
Maximum bending moment,
From LRFDM Table 5-3 select a W18×35 (in bold face):
Design bending strength,
Allowable deflection, Maximum deflection, under nominal live load:
Required
Enter LRFDM Table 5-2 with Ireq = 602 in.4 and observe that a W18×40 with Ix = 612 in.4 is the lightest section that satisfies this condition. Use a W18×40 of A992 steel.
(Ans.)
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Steel Structures by S. Vinnakota
P9.34.
Chapter 9
page 9-36
Design a spandrel beam of the shopping center given in Problem P9.33. The spandrel beam supports, in addition to floor loads, a facia panel weighing 0.6 klf. There are no clearance limitations, but the total service load deflection is to be limited to ½ in. Assume continuous lateral support. See Fig. 3.2.2 of the text book for guidance. Solution Simply supported beam. Span, L = 30 ft Spacing of beams, S b = 8 ft Tributary width for the spandrel beam = ½ S b = 4.00 ft Weight of slab = (5/12) ×100 × 4.00 = 167 plf Weight of flooring and ceiling = 10.0×4.00 = 40.0 plf Facia panel weight = 600 plf Self-weight of spandrel beam (assumed) = 60.0 plf Total dead load, q D = 0.867 klf Total live load, q L = 120 × 4= 480 = 0.480 klf Total service load, q s = 0.867 + 0.480 = 1.35 klf Total factored load, q u = 1.2 × 0.867 + 1.6 × 0.480 = 1.81 klf Allowable deflection under total service load, *all = 0.5 in. As this deflection limitation is quite stringent, we will select a section to satisfy the deflection criteria and check for strength.
So, Ix
$ 1700 in.4
Enter LRFDM Table 5-2 with Ireq = 1700 in.4 and select a W24×68 for which: Ix = 1830 in.4 > 1700 in.4 O.K. Nb M px = 664 ft-kips; Nv Vn = 266 kips Maximum bending moment,
Maximum shear,
So, select a W24×68 of A992 steel. P9.35.
(Ans.)
Design an interior girder of the shopping center described in Problem 9.33. Due to clearance limitations the nominal depth of the girder is limited to 24 inches. Limit the live load deflection to L/360. Determine the camber to be specified. Assume continuous lateral support.
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
P9.36.
Chapter 9
page 9-37
The floor framing system of an office building has bay sizes of 27 ft by 34 ft. The floor beams are at 9 ft centers and have a span of 34 ft. Assume that the beams and girders are simply supported. The floor deck consists of 5-in.-thick reinforced concrete “one way” slab of ordinary concrete, directly supported by the beams, and carries a live load of 100 psf. Design the beam without counting on composite action between the slab and the steel beam, but assuming that lateral buckling of the beams is prevented. Check deflection, and specify camber. Use A992 steel.
Solution Simply supported beam. Span, L = 34 ft. Spacing, S b = 9 ft. Slab thickness, t = 5 in. Weight of slab = Assume the self-weight of the beam to be 50 plf. Distributed dead load, Distributed live load, Distributed factored load on the beam,
Maximum bending moment,
From LRFDM Table 5-3 select a W21×44 (in bold face):
Design bending strength,
Allowable deflection, Maximum deflection, under nominal live load:
Deflection under nominal dead load, Provide a camber of ¾ in. Use a W21×44 of A992 steel. As the self-weight assumed is close to the actual value, no revision is necessary. (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 9
page 9-38
P9.37.
Design an interior girder of the floor framing system of Problem 9.36. Due to clearance limitations, depth of girder cannot exceed 27 in. (nominal). Assume continuous lateral support.
P9.38.
Design a spandrel girder of the floor framing system of Problem 9.36. The spandrel girder supports, in addition to beam reactions, a facia panel weighing 0.8 klf. There are no clearance limitations. Assume continuous lateral support.
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
P9.39.
Chapter 9
page 9-39
Select the most economical W12 section that can carry a concentrated dead load of 40 kips and a live load of 30 kips at the third point of a 6 ft simple span. Check for bending, shear and local buckling. Neglect self weight of beam. Assume adequate lateral bracing to the compression flange.
Solution Simply supported beam AB. Span, L = 6 ft Loading: A concentrated load at D (with AD = a = 2 ft; BD = b = 4 ft). Dead load, Q D = 40 kips Live load, Q L = 30 kips Factored loads, Q u = 1.2 × 40 + 1.6×30 = 96 kips
Entering LRFDM Table 5-3 with M req = 128 ft-kips, we observe that a W12×26 is the lightest W12-shape that provides adequate flexural strength. For this shape,
From LRFDM Table 1-1, for a W12×26 6 b f / 2t f
= 8.54;
h /tw = 47.2
Limiting b/t ratios for plate buckling: Compression flange: Web in flexure: Web in shear: Alternatively, these values can be read from Table 9.5.1. As b f / 2tf < 8pf and h /tw < 8pw, the section is compact. So, the design bending strength used above is O.K. As h /tw < 8pv, the design shear strength used above is also O.K. So, select a W12×26 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 9
page 9-40
P9.40.
Select the lightest beam section to carry a uniformly distributed, factored load of 20 klf on a simple span of 8 ft. Compression flange is adequately supported. Use A992 steel.
P9.41.
Select the lightest W section to support a factored uniformly distributed load of 2.6 klf over a simple span of 25 ft. Assume A572 Grade 60 steel and continuous lateral support. Make all checks. The given load does not include self weight of the beam.
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
P9.42.
Chapter 9
page 9-41
Determine the size of the bearing plate required for an end reaction, under a factored load of 90 kips, for a W16×45 A992 steel beam. The beam rests on a concrete wall with a 28-day compressive strength of 3.0 ksi.
Solution a. Data From Table 1-1 of the LRFDM for a W16×45 : d = 16.1 in.; b f = 7.04 in. tw = 0.345 in.; tf = 0.565 in. k = 0.967 in.; k 1 = 13 ÷ 16 = 0.813 in. b.
Length of bearing plate, N Assume
, which requires N > 0.2 × 16.1 = 3.22 in.
From LRFDM Table 9-5: Beam End Bearing Constants, for a W16×45 of A992 steel: NvVn = 150 kips NR 1 = 41.7 kips; NR 2 = 17.3 kli NrR 5 = 49.8 kips; NrR 6 = 6.52 kli Minimum length of bearing plate to prevent local web yielding,
Minimum length of bearing plate to prevent web crippling,
Try N = 8 in. Check: N /d = 8.00 /16.1 = 0.497 > 0.2, as assumed. O.K. c.
Width of bearing plate, B Assume conservatively that the bearing plate covers full area of the support. Hence, the confinement factor $ = 1 and the required plate width can be found from:
6 B $ 7.35 say 8 in. rounding to the nearest inch. The flange width of a W24×62 is 7.04 in., making the bearing plate slightly wider than the beam flange. d.
Thickness of the bearing plate, t The span of the cantilever strip is,
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 9
page 9-42
The required thickness is
Use a PL 1×8×0'-8 of A36 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
P9.43.
Chapter 9
page 9-43
A W14×426 column of A992 steel supports a factored axial load of 5310 kips. Design a base plate for the column if the supporting concrete has a cylinder strength
= 5 ksi. Assume that (a) The full area of the
concrete support is covered by the base plate, (b) The support will be a 4 ft by 4 ft concrete pier. Solution a. 1. Data Factored axial load, P u = 5310 kips W14×426 column bf = 16.7 in.; bf d = 312 in.2; Concrete: Base plate: A36 steel
d = 18.7 in. b f + d = 35.4 in.
= 5.0 ksi Fy = 36.0 ksi
As the full area of the support is covered by the base plate, $ = 1.0
O.K. 2.
Plan dimensions of base plate Optimize plan dimensions.
> b f = 16.7 in. N $
>
d = 18.7 in.
O.K. O.K.
So, the plate dimensions are 44 × 48 in. O.K. 3.
Plate thickness
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 9
n* =
Provide t = 2½ in.
page 9-44
(conservatively)
>
2.28 in.
Use a PL 2½ ×44×4'-0 of A36 steel and concrete strength
= 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
P9.44.
Chapter 9
page 9-45
A W10×45 column of A992 steel supports a factored axial load of 565 kips. Design a base plate for the column if the supporting concrete has a cylinder strength . Assume that (a) The full area of the concrete support is covered by the base plate, (b) The footing size is 20 in. by 20 in.
Solution a. 1. Data Factored axial load, P u = 565 kips W10×45 column bf = 8.02 in.; bf d = 81.0 in.2; Concrete: Base plate: A36 steel
d = 10.1 in. b f + d = 18.1 in.
= 3.0 ksi Fy = 36.0 ksi
As the full area of the support is covered by the base plate, $ = 1.0
O.K. 2.
Plan dimensions of base plate Optimize plan dimensions.
> b f = 8.02 in. N $
>
d = 10.1 in.
O.K. O.K.
So, the plate dimensions are 18 × 21 in. O.K. 3.
Plate thickness
Pdp = Fdp A 1 = 1.53 × 378 = 578 > 565 kips
O.K.
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 9
page 9-46
n* =
Provide t = 1¾ in. . 1.76 in. Use a PL 1¾×18×1'-9 of A36 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.
