solucionario capitulo 9

8/10/2019 solucionario capitulo 9

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Steel Structures by S. Vinnakota Chapter 9 page 9-319-31

PROPRIETARY MATER IAL. 2006 T he McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may bedisplayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limiteddistribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, youare using it without permission.

P9.26. Select the most economical wide flange section for the 30-ft-long overhang beam, with supports at the leftend and 10 ft from the right end. It is subjected to the follow ing service loads: 1.5 klf dead load and 2 klf liveload 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 iscontinuously laterally braced. Assume A572 Grade 42 steel.

Solution

Section: W-shape. Comp ression flange continuo usly laterally braced.

yMaterial: A572 G rade 42 steel. 6 F = 42 ksiOverhanging beam ACBD.

Main span AB: L = 20 ftOverhang BD: a = 10 ft

Service loads:Uniformly distributed over AD: Dead load = 1.5 klf

LLive load, q = 2.0 klf LD Concen trated live load at free end, D: Q = 10 kips

LC Concen trated live load at center of main span, C: Q = 30 kipsAssume self weight of the beam = 60 plf = 0.06 klf

uFactored distributed load, q = 1.2(1.5 + 0.06) + 1.6(2.0) = 5.07 klf

uCFactored concentrated load at C, Q = 1.6(30.0) = 48.0 kips

uDFactored concentrated load at D, Q = 1.6(10.0) = 16.0 kips

Maximum positive moment, occurs at C.

Maximum negative moment occurs at support B.

So,

max BCMaximum shear, V = V = 162 - 16.0 - 5.07 10.0 = 95.3 kips.

From Eq. 9.7.18:

reqEntering LRFDM Table 5-3 with Z = 131 in. , we observe that a W2455 (in bold face) is the3

lightest section that provides adequ ate flexural strength. For this shape, we have:

x w f f w Z = 135 in. ; d = 23.6 in.; t = 0.395 in.; b /2 t = 6.94; h / t = 54.13

Limiting b/t ratios for plate buckling:


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Comp ression flange:

Web in flexure:

Web in shear:Alternatively, these values can be read from T able 9.5.1.

f f pf w pwAs b / 2t < 8 and h / t < 8 , the section is compact. So, the design bending strength is,

d b px b x y M = N M = N Z F = 0.90 135 42.0 = 5103 in.-kips = 425 ft-kips

w pvAs h / t < 8 , the design shear strength is given by

d v n v w yV = N V = N d t (0.6 F ) = 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 W2455. (Ans.)


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P9.27. Select the lightest W-shape for a 18 -ft-long cantilever beam of A572 Grade 42 steel. It is subjected toconcen trated, service live loads of 16 kips and 10 kips acting at 12 ft and 18 ft, respectively, from the fixedend. 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.

P9.28. In a building, roof beams are spaced at 10-ft intervals and are connected to girder we bs by simpleconn ections. The span of the beams is 24 ft. The beam is designed for the following service loads: deadload (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

beam is co ntinuou sly laterally sup ported by the ro of d eck ing . Specify any camber 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 isto carry a concentrated live load (movable) of 6 kip. The load consists of precision machinery whichrequires that the deflection be limited to a maximum of L/800 . The beam h as an effective span of 40 ft and iscontinu 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 thatthe beam be no more than 18 in. in depth. Assume full lateral support to the compression flange and select asuitable A 992 steel beam. Is the deflection OK, if the beam carries a plastered ceiling? Live load is b rd thetotal load on the beam.

P9.31. A simply supported beam with continuous lateral support for the entire span of 30 ft is subjected to afactored, 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.


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P9.32. To save construction depth with a precast plank system, a structural tee is used for the overhang beam ABCwith 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 liveload 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

yYield stress o f material, F = 50 ksiBeam with overhang, ABC.

1Main span (AB) = L = 20 ft

2Overhang (BC) = L = 6 ftLoading:

D Dead load, q = 0.6 klf

LLive load, q = 0.3 klf

uFactored load, q = 1.2 0.6 + 1.6 0.3 = 1.20 klf

Total load = 1.2026.0 = 31.2 kips

;

Location of point of zero shear, from support A 6

Maximum positive moment,

Maximum negative moment occurs at suppo rt B,

reqRequired bending strength, M = 49.5 ft-kips

As the beam is continuously laterally supported

xSelect a WT668 with S = 9.46 in. (Ans.)3


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P9.33. 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 weightconcrete (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 oncomposite 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 load deflection to L /36 0. Use A99 2 steel .

See Fig. 3.2.2 of the text book for guidance.

Solution Simply supported beam.

bSpan, L = 30 ft. Spacing, S = 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 W1835 (in bold face):

Design bending strength,

Allowable deflection,

Maximum deflection, under nominal live load:

Required

req xEnter LRFDM Table 5-2 with I = 602 in. and observe that a W1840 with I = 612 in. is the4 4

lightest section that satisfies this cond ition.

Use a W18 40 of A992 steel. (Ans.)


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P9.34. Design a spandrel beam of the shopping center given in Problem P9.33. The spandrel beam suppo rts, inaddition to floor loads, a facia panel weighing 0.6 klf. There are no clearance limitations, but the totalservice load deflection is to be limited to 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

bSpacing of beams, S = 8 ft

bTributary width for the spandrel beam = S = 4.00 ftWeight of slab = (5/12) 100 4.00 = 167 plf Weight of flooring and ceiling = 10.04.00 = 40.0 plf Facia panel weight = 600 plf Self-weight of spandrel beam (assumed) = 60.0 plf

DTotal dead load, q = 0.867 klf

LTotal live load, q = 120 4= 48 0 = 0.480 klf sTotal service load, q = 0.867 + 0.480 = 1.35 klf

uTotal factored load, q = 1.2 0.867 + 1.6 0.480 = 1.81 klf

allAllowable deflection under total service load, * = 0.5 in. As this deflection limitation is quitestringent, we will select a section to satisfy the deflection criteria and check for stren gth.

xSo, I $ 1700 in. 4

reqEnter LRFDM Table 5-2 with I = 1700 in. and select a W2468 for which:4

x I = 1830 in. > 1700 in. O.K.4 4

b px v nN M = 664 ft-kips; N V = 266 kips


Maximum shear,

So, select a W24 68 of A992 steel. (Ans.)

P9.35. Design an interior girder of the shopping center described in Problem 9.33. Due to clearance limitations thenomin al depth of the girder is limited to 24 inches. Limit the live load deflection to L/360. Determine thecamber to be specified. Assume continuou s lateral suppo rt.


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P9.36. The floor framing system of an office building has bay sizes of 27 ft by 34 ft. The floor beams are at 9 ftcenters 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

beams , and carri es a live load of 1 00 psf . Design the beam with ou t co un ting on composite action betweenthe 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.

bSpan, L = 34 ft. Spacing, S = 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,


From LRFDM T able 5-3 select a W2144 (in bold face):

Design bending strength,

Allowable deflection,

Maximum deflection, under nominal live load:

Deflection under nominal dead load,

Provide a camber of in.

Use a W21 44 of A992 steel. As the self-weight assumed is close to the actual value, no revision isnecessary. (Ans.)


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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, inaddition to beam reactions, a facia panel weighing 0.8 klf. There are no clearance limitations. Assumecontinuous lateral support.


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P9.39. 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 ftLoading:

A concentrated load at D (with AD = a = 2 ft; BD = b = 4 ft).

DDead load, Q = 40 kips

LLive load, Q = 30 kips

uFactored loads, Q = 1.2 40 + 1.630 = 96 kips

reqEntering LRFDM Table 5-3 with M = 128 ft-kips, we observe that a W1226 is the lightestW12 -shape that provides adequate flexural strength. For this shape,

f f wFrom LRFDM Table 1-1, for a W1226 6 b / 2t = 8.54; h / t = 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.

f f pf w pwAs b / 2t < 8 and h / t < 8 , the section is compact. So, the design bending strength usedabove is O.K.

w pvAs h / t < 8 , the design shear strength used above is also O.K.

So, select a W1226 of A992 steel. (Ans.)


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P9.40. Select the lightest beam section to carry a uniformly distributed, factored load of 20 klf on a simple span o f 8ft. 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 doesnot include self weight of the beam.


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P9.42. Determine the size o f the bearing plate required for an end reaction, under a factored load of 90 kips, for aW16 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 taFrom Table 1-1 of the LRFDM for a W1645 :

f d = 16.1 in.; b = 7.04 in.

w f t = 0.345 in.; t = 0.565 in.

1k = 0.967 in.; k = 13 16 = 0.813 in.

b. Len gth o f bearing 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:

v nN V = 150 kips1 2N R = 41.7 kips; N R = 17.3 kli

r 5 r 6N R = 49.8 kips; N R = 6.52 kliMinimum 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 , BAssume conse rvatively that the bearing plate covers full area of the support. Hence, theconfinement factor $ = 1 and the required plate width can be found from:

6 B $ 7.35 say 8 in. roundin g to the nearest inch.The flange width of a W2462 is 7.04 in., making the bearing plate slightly wider than the beamflange.

d. Thickness of the bearing plate , t

The span o f the cantilever strip is,


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The required thickness is

Use a PL 18 0'-8 of A36 steel. (Ans.)


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P9.43. A W144 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

uFactored axial load, P = 5310 kipsW14426 column

f b = 16.7 in.; d = 18.7 in.

f f b d = 312 in. ; b + d = 35.4 in.2

Concrete: = 5.0 ksi

yBase plate: A36 steel F = 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 ensions of base p lateOptimize plan dimensions.

f > b = 16.7 in. O.K.

N $ > d = 18.7 in. O.K.

So, the plate dimensions are 44 48 in.

O.K.

3. Plate th ickn ess


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n = (conservatively)*

Provide t = 2 in. > 2.28 in.

Use a PL 2 44 4'-0 of A36 steel and concrete strength = 5 ksi. (Ans.)


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P9.44. A W104 5 column of A992 steel suppo rts a factored axial load of 565 kips. Design a base plate for thecolumn 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

uFactored axial load, P = 565 kipsW1045 column

f b = 8.02 in.; d = 10.1 in.

f f b d = 81.0 in. ; b + d = 18.1 in.2

Concrete: = 3.0 ksi

yBase plate: A36 steel F = 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 ensions of base p lateOptimize plan dimensions.

f > b = 8.02 in. O.K.

N $ > d = 10.1 in. O.K.

So, the plate dimensions are 18 21 in.O.K.

3. Plate th ickn ess

dp dp 1 P = F A = 1.53 378 = 578 > 565 kips O.K.

solucionario capitulo 9

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