Steel Structures by S. Vinnakota
Chapter 9
page 9-1
CHAPTER 9
P9.1.
Determine the value of S , S x, Z x, "x of the following shapes, using the dimensions given in the LRFDM. (a ) W 36 × 2 3 0 (b ) * W16 × 36 (c ) W T 1 8 × 1 15 (d ) WT8 × 18
Solution a.
1.
Section: W 36 ×230 b f = 16.5 in.; t f = 1.26 in. d = = 35.9 in.; t w = 0.760 in. in. Ela stic sec tio n p rop erti es Flange area area = 16.5 × 1.26 = 20.8 in.2 Web depth, d w = (35.9 - 2 × 1.26) = 33.4 in. Web area area = 0.760 0.760 × 33.4 = 25.4 in.2 Total area, A 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 symm etry axes. above the bottom fiber of the bottom flange
=
14,900 14,900 in. in.4
2.
(Ans.)
Pla stic sec tio n p rop ert ies To satisfy the condition I A Fy dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section section considered , the PNA therefore coincides with the symmetry axis. ÷
= 18.0 in. above the bottom fiber of the bottom flange
Plastic section modulus,
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Steel Structures by S. Vinnakota
Chapter 9
page 9-2
= 934 in.3
(Ans.) (Ans.)
Shape factor,
(Ans.)
The corresponding values given in LRFDM Table 1-1 are: I are: I = 15,000in..4; S = S x = 837 in. in.3 ; and and Z Z = x = 15,000in x = 3 943 in. and take into account the contribution contribution of the web-to-flange web-to-flange fillet fillets. s.
b.
1.
Section: W16×36 b f = 6.99 in.;
t f = 0.430 in.;
d = = 15.9 in.;
t w = 0.295
Ela stic sec tio n p rop erti es Flange area area = 6.99 6.99 × 0.430 = 3.01 in.2 Web depth, d w = (15.9 - 2 × 0.430) 0.430) = 15.0 in. in. 2 Web area = 15.0 × 0.295 = 4.43 in. Total area, A 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 symm etry axes. above the bottom fiber of the bottom flange
=
443 in. in.4
2.
(Ans.)
Pla stic sec tio n p rop ert ies To satisfy the condition I A Fy dA = 0 for a homogeneous section, the PNA divides the section into two equal areas. For the doubly symmetric section section considered , the PNA therefore coincides with the symmetry axis. ÷
= 7.95 in. above the bottom fiber of the bottom flange
Plastic section modulus,
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Steel Structures by S. Vinnakota
Chapter 9
= 63.2 in.3
Shape factor,
page 9-3
(Ans.) (Ans.)
(Ans.)
The corresponding values given in LRFDM Table 1-1 are: I are: I x = 448 in.4; S = S x = 56.5 in.3 ; and Z and Z = x = 3 64.0 in. and take into account the contribution of the web-to-flange web-to-flange fillet fillets. s.
P9.2.
Determine the value of S , S x, Z , Z x, "x of the following built-up sections: (a ) A W1 W1 6 × 3 6 wi wi th th on on e ½ × 12 12 in in . p la late we we ld ld ed ed to to ea eac h fl an an ge ge . (b ) A W16 W16 × 36 wit with h one one ½ × 12 in. in. plat platee wel welde ded d to to the the top top of the the fla flang ngee (c ) A W16 W16 × 36 wit with h a C12 × 20. 20.7 7 wit with h its its web web wel welde ded d to to the the top top fla flang nge. e.
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Steel Structures by S. Vinnakota
P9.3.
Chapter 9
page 9-4
Determine the values of M of M yx and M and M px for the shapes given in Problem P9.1.
Solution
a.
W 3 6 ×230
Yield stress stress,, Fy = 50 ksi Elastic Elastic section section modulus, S x = 828 in.3 (from (from solution solution to P9.1a P9.1a .) Plastic Plastic section section modulus, Z x = 934 in.3 (from (from solution solution to P9.1a P9.1a .) Yield Yield moment, moment, M y = S x F y = 828 × 50.0 ÷ 12 = 3450 ft-ki ft-kips ps Plasti Plasticc moment, M p = Z x F y = 934 × 50.0 ÷ 12 = 3892 ft-ki ft-kips ps b.
P9.4.
W16×36 Yield stress stress,, Fy = 50 ksi Elastic Elastic section section modulus, S x = 55.7 in.3 (from (from solution to to P9.1b P9.1b .) 3 Plastic Plastic section section modulus, Z x = 63.2 in. (from (from solution to to P9.1b P9.1b .) Yield Yield moment, moment, M y = S x F y = 55.7 × 50.0 ÷ 12 = 232 ft-ki ft-kips ps Plasti Plasticc moment, M p = Z x F y = 63.2 × 50.0 ÷ 12 = 263 ft-ki ft-kips ps
Determine the values of M of M yx and M and M px for the built-up shapes given in Problem P9.2. Assume A36 steel for all elements.
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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 connection s. Comp ute 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 suppo rts 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 F = 50 ksi y From LRFDM T able 5-3, for a W18×35, Nb M px = 249 ft-kips; Nv V n = 143 kips; Lp = 4.31 ft
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength, M d =
Nb M px = 249 ft-kips
As h /t w < 8pv, the design shear strength is given by Vd =
Nv V n = 143 kips
As per LRFD: Maximum bending moment,
Maximum shear corresponding to this load is, V max = q u L/ 2 = 2.21 × 30 ÷ 2 = 33.2 kips <
V d = 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.)
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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 suppo rted 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: A5 72 Gr 42 6 F = 42 ksi y From LRFDM T able 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)
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Steel Structures by S. Vinnakota
P9.7.
Chapter 9
page 9-7
A W24× 176 o f A572 Grade 60 steel is used for a simple span of 36 ft. If the only dead load presen t is the weight o the b eam, what is the largest service concentrated live load that can be placed at mid -span? Assume continuo us lateral bracing. Deflection need not be checked.
Solution
Simply supported beam: Span, L = 36 ft. Comp ression flange continu ously laterally braced. Section: W24×176 Material: A572 Gr 60 steel 6 F = 60 ksi y 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 Qu = 1.6 QL Maximum bending moment at the center under factored loads is,
From LRFDM Table 1-1
511 in.3; 6 Z = x b f / 2t f
= 4.81;
d = 25 .2 in .; h /t w = 28.7
t w = 0.750 in.
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength is, M d = Nb M px = N b Z x F y = 0.90 × 511 × 60.0 = 27,594 in.-kips = 2300 ft-kips As h /t w < 8pv, the design shear strength is given by Vd = Nv V n = Nv d t w (0.6 F ) y = 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
V max = ½ q u L + ½ Q u = (0.5×0.211×36) + ( 0.5×1.6×157) = 129 kips < V d = 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 be am to be as sh ow n in F ig. P 9.8 . Th e b eam weight is no t in clu ded . Use A588 Gra de 50 stee l. Assum e fu ll lateral suppo rt 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, V max = V A = 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 continu ously laterally braced, Also Vd = 143 kips > 22 kips
O.K.
The W18×35 beam is adequate for the given loads.
No te:
(Ans.)
To be considered ade qu ately late rally su pp orted , br acin g sh ou ld be pro vid ed such tha t L b # 4' 4".
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Steel Structures by S. Vinnakota
P9.9.
Chapter 9
page 9-10
A 34 ft long W27×9 4 shape of A992 steel is used as a simply supported beam. It is subjected to a factored concen trated load of 80 kips at 12 ft from each support. In addition, it is subjected to a factored momen t 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 suppo rt for the comp ression flange.
Solution a.
Da ta 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-mo ment at B = 340 ft-kips (anti-clockwise) F acto re d e nd -m om en t at E = 3 40 ft-k ip s (clo ck wise ) Uniformly distributed load, q u due to self-weight ( = 1.2 × 0.094 = 0.113 klf )
b.
Req uir ed stre ng ths Due to the symmetry of the structure and the loading, the maximum bending moment occurs at the center, C. Consid er 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 M C So: Required bending strength, M u = 620 + 16.3 = 636 ft-kips Required shear strength, V u = 80.0 + 1.92 = 81.9 kips
c.
Des ign stre ng ths Section: W27×94 From LRFDM Table 1-1 6 b f / 2t f = 6.70; h /t w = 49.5 Material: A992 steel 6 F = 50 ksi y From LRFDM Table 5-3, for a W27×94, Nb M px = 1040 ft-kips; Nv V n = 356 kips; Lp = 7.49 ft Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is comp act.
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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 < 8pv, the design shear strength is: As, M u < M d and V u <
page 9-11
Vd =
Nv V n = 356 kips
V d , th e W 27 ×9 4 o f A9 92 s te el c an su pp ort t he g iv en lo ad s.
(A ns.)
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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.
Da ta 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-mo ment at B = 340 ft-kips (anti-clockwise) Uniformly distributed load, q u due to self-weight ( = 1.2 × 0.094 = 0.113 klf )
b.
Req uir ed stre ng ths Consider the in fluence 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 m oment under applied load occurs at E. Due to self weight: R A = R B = 0.113× 34 ÷ 2 = 1.92 kips M C The maximum bending mo ment now occurs at the center, C. So: Required bending strength, M u = 840 + 16.3 = 856 ft-kips (conservatively) Required shear strength, V u = 90.0 + 1.92 = 91.9 kips
c.
Des ign stre ng ths Section: W27×94 From LRFDM Table 1-1 6 b f / 2t f = 6.70; h /t w = 49.5 Material: A992 steel 6 F = 50 ksi y From LRFDM Table 5-3, for a W27×94, Nb M px = 1040 ft-kips; Nv V n = 356 kips; Lp = 7.49 ft Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is comp act.
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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 < 8pv, the design shear strength is: As, M u < M d and V u <
page 9-13
Vd =
Nv V n = 356 kips
V d , the W27×9 4 of A992 steel can still suppo rt 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 F y = 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: F y = 50 ksi steel Factored loads: Conce ntrated 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, V max = 30.9 kips From LRFDM T able 5-3, for a W16×50 of F y = 50 ksi steel,
As the section is compact and the compression flange continuously laterally supported
So, the W16×50 of Fy = 50 ksi steel is adequate. P9.12.
(Ans.)
A simple beam consists of a W18 ×55 with a PL1×12 co ver 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 continu ously laterally suppo rted See Fig. P2.6 of the text book. Solution
Simply supported beam. Span, L = 36 ft Material: A36 steel. F y = 36 ksi Section: Built-up by welding three plates. a.
Ela stic sec tion p rop erties Total area, A = 8.00 × 2.00 + 1.00×16.0 + 12.0×2.0 = 56.0 in.2 To satisfy the condition I A f dA = 0, the elastic neutral axis must pass through th e center of gravity of the cross section.
=
8.71 in. above the bottom fiber of the bottom flange
=
3,500 in.4
;
b.
Plastic sec tion p rop erties To satisfy the condition I A 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 y p, as measured from the bottom fibe r of the bottom flange, such that the area above the PN A 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 mo ment From LRFDS Eq. F1-1: Plastic mo men t, M p
= m in [ F y Z x ; 1 .5 F y S x min ] = min [ (36.0×408) ÷ 12; (1.5×36×310)÷ 12 ] = min [ 1224 ft-kips; 1395 ft-kips ] = 1224 ft-kips
d.
Des ign stre ng ths Width-to-thickness ratios of the plates: Compression flange: b f / 2t f Web in flexure: h c /t w
= 4.00 / 2.00 = 2.00 = (2×12.0) / 1.00 = 24.0
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength, M d =
Nb M px =
0.90 × 1224 = 1100 ft-kips
As h /t w < 8pv, the design shear strength is given by Vd =
Nv V n = 0.9 × 16.0 × 1.0 × 0.60 × 36.0 = 311 kips
As per LRFD:
Maximum shear corresponding to this load is, V max = q u L/ 2 = 6.79 × 36 ÷ 2 = 122 kips <
V d = 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 continu ous lateral suppo rt, 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 Concen trated load at free end, D: Q D = 16 kips Concen trated load at center of main span, C: Q C = 32 kips Factored distributed load, Factored concentrated load at C, Factored concentrated load at D,
= 2.8 0 klf q u = 1 .2(1.0) + 1.6(1.0) 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, V BC = 63.8 kips. From LRFDM T able 5-3, for a W16×57, As the section is compact and continuously braced,
So, the W16× 57 is adequate.
(Ans.)
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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 (no t including the self weight) is 0.4 klf and the service live load is 0.78 klf. Assume that the beam is continuo usly laterally suppo rted by the floor deck. Check the adequacy of the beam for be nd ing and she ar. Calc ulate the max imu m d ead load an d live load d efle ctio ns of th e b eam . 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, un der 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 So, the W14× 30 is satisfactory for bendin g and shear.
O.K. O.K. (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.)
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Steel Structures by S. Vinnakota
P9.16.
Chapter 9
page 9-19
A W30×148 of F = 50 ksi steel has been selected for the simply supported beam shown in Fig. P9.16. y Loads Q consist of 50 kips dead load and 80 kips live load. Check the adequ acy of the beam for bending and shear. Calculate the maximum dead load and live load deflections. Assume continuo us lateral suppo rt 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 o f LRFDM T able 5-17), the maximum deflection which occurs at C:
(Ans.)
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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 concen trated 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 = 1.6QL = 1.6×80 = 128 kips
From LRFDM T able 5-3, for a W12×30 Shape is compact and laterally supported, therefore V d = Nv Vn = 86.3
<
So, the W12×30 is not satisfactory.
V max = 128 kips
N.G. (Ans.)
Entering LRFDM Table 5-4 with V req = 128 kips and limiting the selection to W12 shapes, observe that a W12×72 with
So, select a W12×72 of A992 steel.
(Ans.)
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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 suppo rted beam. Continu ously laterally suppo rted. Material: A992 steel. 6 F = 50 ksi y 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 (9 0 plf) is less than the value assumed (10 0 plf) the design is O.K. Select a W30×90.
(Ans.)
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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 suppo rted span of 34 ft. The beam is continuously laterally supported. Assume no deflection limitations. Use A572 G rade 42 steel. Solution
Simply suppo rted beam. Continu ously laterally suppo rted. Material: A572 Grade 42 steel. 6 F = 42 ksi y Span, L = 34 ft Assume a self-weight of 40 plf or 0.04 0 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 adequ ate flexural strength. For this shape, we have: Z = 66.5 in.3; x
d = 17.7 in.; t w = 0.300 in.; b f /2t = 7.06; f
h /t w = 53.5
As the section is compact for F = 50 ksi steel (no ind ication to the con trary in this table), it is y compact for Grade 42 steel considered. So, the design bending strength is, M d = Nb M px = Nb Zx 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 /t w < 8pv, the design shear strength is given by Vd = Nv V n = Nv d t w (0.6 F ) y = 0.90 × 17.7 × 0.300 × 0.60 × 42.0 = 120 kips As, As the weight of the beam (3 5 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 suppo rt for the compression flange. Select the most econo mical 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 Neg lect in flue nce of s elf-w eight to s tart with. To tal fac tore d lo ad on the beam = 30 0 k ips Due to symmetry of the structure and the loading, reaction, R A = 150 kips Maximum bending mom ent occurs at midspan.
Maximum shear, V max = 150 kips Entering the LRFDM Table 5-3 with M req = 2400 ft-kips, we observe that a W30×19 1(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 mome nt, 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 po ints D and E, an d m ust car ry a s ervice floo r lo ad o f 10 0 p sf (5 0% dea d load and 50 % live loa d) and a wa ll load of 0.6 kips per linear foot of wall (100% dead load). Assume adequate lateral suppo rt, A992 steel, and select the lightest W-12 shape for flexure and shear. The maximu m 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), L2 = 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 N M 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.)
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Steel Structures by S. Vinnakota
P9.22.
Chapter 9
page 9-25
A simply supported b eam 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 suppo rt the given factored loads. Assume full lateral suppo rt.
Solution
Simply suppo rted beam. Span AB: L = 24 ft Section: W-shape. Comp ression flange continuo usly laterally braced. Material: A572 G rade 60 steel. 6 F = 60 ksi y Factored loads: Uniformly distributed over AB: = 0.6 klf Concen trated load at point D (with AD = 8 ft): QuD = 40.0 kips Concen trated 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, V max = 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 adequ ate flexural strength. For this shape, we have: Z = 78.4 in.3; x
d = 17.9 in.; t w = 0.315 in.; b f /2t = 5.73; f
h /t w = 50.9
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, 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 Zx Fy = 0.90 × 78.4 × 60.0 = 4234 in.-kips = 353 ft-kips As h /t w < 8pv, the design shear strength is given by Vd = Nv V n = Nv d t w (0.6 F ) y = 0.90 × 17.9 × 0.315 × 0.60 × 60.0 = 183 kips As, As the weight of the beam (4 0 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 suppo rted beam AB, with a span o f 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 A D . Select a Wshape of A572 Grade 5 0 steel that can safely support the given factored loads. Assume full lateral suppo rt to the compression flange. Solution
Simply supported beam, AB. Span, L = 28 ft Material: F y = 50 ksi steel Factored loads: Conce ntrated 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 m oment 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 du e to self weight, Required bending strength, M u = 362 + 5.88 = 368 ft-kips (conservatively) Required shear strength, V u = 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 (5 0 plf) is the same as the value assumed (50 plf) the design is O.K. So, select a W21×50. No te:
(Ans.)
Fro m L RF DM Table 5-3 , we ob serv e th at a W 21 ×48 o f Gra de 50 steel h as ad eq uate be nd ing and shear strengths; and could be selected. Howeve r, the section is non-comp act. 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 concen trated 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 G rade 60 steel. 6 F = 60 ksi y Factored loads: Uniformly distributed over AB: = 2.50 klf Concen trated 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 adequ ate flexural strength. For this shape, we have: Z = 177 in.3; x
d = 20.1 in.; t w = 0.415 in.; b f /2t = 7.66; f
h /t w = 52.0
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength is, M d = Nb M px = Nb Zx Fy = 0.90 × 177 × 60.0 = 9558 in.-kips = 797 ft-kips As h /t w < 8pv, the design shear strength is given by Vd = Nv V n = Nv d t w (0.6 F ) y = 0.90 × 20.1 × 0.415 × 0.60 × 60.0 = 270 kips PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Steel Structures by S. Vinnakota
Chapter 9
page 9-29
As, As the weight of the beam (6 8 plf) is less than the value assumed (10 0 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 b eam is subjected to a un iformly distributed, factored load of 3 klf over the entire length. In addition, it is subjected to con centrated, 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 suppo rt the factored loads. Assume adequate lateral support for the compression flange. Solution
Section: W-shape. Comp ression flange adequately laterally braced. Material: A572 G rade 50 steel. 6 F = 50 ksi y Overhanging beam ABCD. Overhang AB: a = 8 ft Main span BD: L = 24 ft Factored loads: Uniformly distributed over AD: = 3.0 klf Concen trated load at free end, A: Q uA = 10 kips Concen trated 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, V max = V BC = 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 (5 0 plf) is the same as the value assumed (50 plf) the design is O.K. So, select a W21×50. No te:
(Ans.)
Fro m L RF DM Table 5-3 , we ob serv e th at a W 21 ×48 o f Gra de 50 steel h as ad eq uate be nd ing and shear strengths; and could be selected. Howeve r, the section is non-comp act. 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 follow ing 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 concen trated 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. Comp ression flange continuo usly laterally braced. Material: A572 G rade 42 steel. 6 F = 42 ksi y 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 Concen trated live load at free end, D: Q LD = 10 kips Concen trated 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, V max = V BC = 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 adequ ate flexural strength. For this shape, we have: Z = 135 in.3; d = 23.6 in.; t w = 0.395 in.; b f /2t x f = 6.94; h /t w = 54.1 Limiting b/t ratios for plate buckling:
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Steel Structures by S. Vinnakota
Chapter 9
page 9-32
Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength is, M d = Nb M px = Nb Zx Fy = 0.90 × 135 × 42.0 = 5103 in.-kips = 425 ft-kips As h /t w < 8pv, the design shear strength is given by Vd = Nv V n = Nv d t w (0.6 F ) y = 0.90 × 23.6 × 0.395 × 0.60 × 42.0 = 211 kips As, As the weight of the beam (5 5 plf) is less than the value assumed (60 plf) the design is O.K. Select a W24×55.
(Ans.)
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Steel Structures by S. Vinnakota
P9.27.
P9.28.
Chapter 9
page 9-33
Select the lightest W-shape for a 18 -ft-long cantilever beam of A572 Grade 42 steel. It is subjected to concen trated, 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 w eight in all the calculations. In a building, roof beams are spaced at 10-ft intervals and are connected to girder we bs by simple conn ections. The span of the beams is 24 ft. The beam is designed for the following service loads: dead load (not inclu ding the self weight), 65 p sf; roof live load, 20 p sf; snow load, 40 psf; and rain load, 25 p sf. Deflection of the beam is limited to L/240. Select a suitable W-shape of A242 Grade 42 steel. Assume the be am is co ntinu ou sly l ater ally s up po rted b y th e ro of d eck ing . Specify an y cam be r prov ide d.
P9.29.
A simple beam in a research lab is to sup port a dead load o f 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 h as an effective span of 40 ft and is continu ously laterally suppo rted. Use A992 steel and select a suitable section.
P9.30.
A W-shape is used to supp ort 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. (nomin al), and the deflection is limited to L/360. Select a suitable A992 shape.
PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 supp orts 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 continuo us lateral support, and select a suitable WT6 of A992 steel. See Fig. P9.32 of the text book. Solution
Yield stress o f material, F y = 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 suppo rt 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.)
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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 supp orted. The floor deck consists of 5-in.-thick slab of light weight concrete (100 pcf) directly suppo rted 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 pre ven ted . Lim it th e live l oad deflec tion to L /36 0. Use A99 2 st eel. 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 T able 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 I req = 602 in.4 and observe that a W18×40 with I x = 612 in.4 is the lightest section that satisfies this cond ition. Use a W18× 40 of A992 steel.
(Ans.)
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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 suppo rts, 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 continu ous lateral suppo rt. 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= 48 0 = 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 stren gth.
So, I x
$ 1700 in.4
Enter LRFDM Table 5-2 with I req = 1700 in.4 and select a W24×68 for which: I x = 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 nomin al depth of the girder is limited to 24 inches. Limit the live load deflection to L/360. Determine the camber to be specified. Assume continuou s lateral suppo rt.
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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 suppo rted. The floor deck consists of 5-in.-thick reinforced concrete “one way” slab of ordinary concrete, directly supported by the be ams , and carri es a live load of 1 00 psf. Design the beam with ou t co un ting o n c om po site action betwe en 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 T able 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.)
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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 spand rel girder supports, in addition to beam reactions, a facia panel weighing 0.8 klf. There are no clearance limitations. Assume continuous lateral support.
PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
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 bend ing, 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, QD = 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 /t w = 47.2
Limiting b/t ratios for plate buckling: Comp ression flange: Web in flexure: Web in shear: Alternatively, these values can be read from T able 9.5.1. As b f / 2t f < 8 pf and h /t w < 8 pw, the section is compact. So, the design bending strength used above is O.K. As h /t w < 8pv, the design shear strength used above is also O.K. So, select a W12×26 of A992 steel.
(Ans.)
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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 o f 8 ft. Comp ression 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 continuou s lateral suppo rt. Make all checks. The given load does not include self weight of the beam.
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Steel Structures by S. Vinnakota
P9.42.
Chapter 9
page 9-41
Determine the size o f the bearing plate required for an end reaction, under a factored load of 90 kips, for a W16 ×45 A99 2 steel beam. The beam rests on a concrete wall with a 28-day compressive strength of 3.0 ksi.
Solution a.
Da ta From Table 1-1 of the LRFDM for a W16×45 : d = 16.1 in.; b f = 7.04 in. t w = 0.345 in.; t f = 0.565 in. k = 0.967 in.; k 1 = 13 ÷ 16 = 0.813 in.
b.
Len gth o f be aring pla te, 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: NvV n = 150 kips N R 1 = 41.7 kips; N R 2 = 17.3 kli Nr R 5 = 49.8 kips; Nr R 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 conse rvatively that the bearing plate covers full area of the support. Hence, the confinement factor $ = 1 and the required plate width can be found from:
say 8 in. roundin g to the nearest inch. 6 B $ 7.35 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 o f the cantilever strip is,
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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.)
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Steel Structures by S. Vinnakota
P9.43.
Chapter 9
page 9-43
A W14×4 26 column o f A992 steel supports a factored axial load of 5310 kips. Design a base plate for the column if the suppo rting 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 sup port will be a 4 ft by 4 ft concrete pier. Solution a. 1.
Da ta Factored axial load, P u = 5310 kips W14×426 column b f = 16.7 in.; b f d = 312 in.2; Concrete: Base plate: A36 steel
=
d = 18.7 in. b f + d = 35.4 in.
5.0 ksi
F y = 36.0 ksi
As the full area of the supp ort is covered by the base p late, $ = 1.0
O.K.
2.
Plan dim ensio ns of ba se p late 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.
Plat e th ickn ess
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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.)
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Steel Structures by S. Vinnakota
P9.44.
Chapter 9
page 9-45
A W10×4 5 column of A992 steel suppo rts a factored axial load of 565 kips. Design a base plate for the column if the suppo rting 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.
Da ta Factored axial load, P u = 565 kips W10×45 column b f = 8.02 in.; = 81.0 in.2; b f d Concrete: Base plate: A36 steel
d = 10.1 in. b f + d = 18.1 in.
= 3.0 ksi F y = 36.0 ksi
As the full area of the supp ort is covered by the base p late, $ = 1.0
O.K.
2.
Plan dim ensio ns of ba se p late 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.
Plat e th ickn ess
P dp = F dp A 1 = 1.53 × 378 = 578 > 565 kips
O.K.
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