Steel Structures 5th Edition Solutions Manual

7/21/2019 Steel Structures 5th Edition Solutions Manual

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010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtain

publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopy

ording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ


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7.6. A W2494 beam on a 6-ft span (see acompanying figure) underpins a columnthat brings 110 kips dead load and 280 kips live load to its top flange at a location 2.5 ftfrom the left support. The column bearing plate is 12 in. measured along the beam, andthe bearing plates at the end supports are each 8 in. Investigate this beam of A992 steel for(a) flexure, (b) shear, and (c) satisfactory transmission of the reactions and concentratedload (i.e., local web yielding, web crippling, and sidesway web buckling). Specify changes

(if any) required to satisfy the AISC Specification.Use LRFD Design Method

(a) Obtain factored loads:

Wu = 1.2(110) + 1.6(280) = 580 kips

Mu=Wuab

L =

580(2.5)(3.5)

6 = 846 ft-kips

Vu =Wub

L =

580(3.5)

6 = 338 kips

(b) Check flexural strength assuming adequate lateral support (AISC F2.1):

Flange and web local buckling slenderness limits, Fy = 50 ksi steel:bf2tf

= 5.2

p =

65Fy

= 9.2

;

h

tw= 41.9

p =

640Fy

= 90.5

OK

bMn= bMp= bZxFy= 0.9(254)(50)/12 = 953 ft-kips

[bMn= 953 ft-kips]> [Mu= 846 ft-kips] OK

(c) Check shear strength (AISC G2.1):

For rolled I-shapes, when

h

tw= 41.9

2.24

E/Fy = 53.9

, v = 1.0 and

Cv = 1.0.

vVn= v(0.6Fy)AwCv =v(0.6Fy)dtwCv= 1.0(0.6)(50)(24.31)(0.515)(1.0) = 376 kips

[vVn= 376 kips]> [Vu = 338 kips] OK

(d) Check local web yielding strength (AISC J10.2):

Rn= (5k+ N)Fytw Interior Reaction

Rn= (2.5k+ N)Fytw Exterior ReactionRn = (5k+ N)Fytw= [5(1.625) + N] (50)(0.515)

Solving forRu= 580 kips, N= 14.4 in. at the interior reaction.

Rn = (2.5k+ N)Fytw = [2.5(1.625) + N] (50)(0.515)

Solving forNto giveRu= 338 kips, N= 9.1 in.at the left exterior reaction.

Solving forNto giveRu= 242 kips, N= 5.3 in.at the right exterior reaction.





36/131

(e) Check web crippling strength (AISC J10.3):

For interior reactions:

Rn = 0.80t2w

1 +

3N

d

twtf

1.5

EFywtf

tw

= (0.75)(0.80)(0.515)2

1 + 3N

24.31

0.515

0.875

1.5(29000)(50)(0.875)0.515

Solving forNto giveRu= 580 kips, N= 23.7 in.

For exterior reactions, assuming N/d >0.2:

Rn = 0.80t2w

1 + 4N

d 0.2

twtf

1.5

EFywtftw

= (0.75)(0.40)(0.515)21 + 4N

24.31 0.2

0.515

0.8751.5

(29000)(50)(0.875)

0.515

Solving forNto giveRu= 338 kips, N= 24.2 in. Check [N/d= 1.00]>0.2.

Solving forNto giveRu= 242 kips, N= 13.8 in. Check [N/d= 0.57]>0.2.

(f) Check sidesway web buckling strength (AISC J10.4):

When the compression flangeis restrainedagainst rotation, for

(h/tw)/(Lb/bf) =

>2.3, this limit state does not apply.

Conclusion:

In accordance with AISC-J10.7, At unframed ends of beams and girdersnot otherwiserestrained against rotation about their longitudinal axes, a pair of transverse stiffeners,

extending the full depth of the web, shall be provided. The 24.2 in. bearing plate requiredat the left reaction, the 13.8 in. bearing plate required at the right reaction, and the 23.7 in.bearing plate required at the load are all too long. Bearing stiffeners should be provided.The beam has adequate flexural and shear strength.





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7.7. A W1677 section of A992 steel is to serve on a 10-ft simply supported span.The wall bearing length is 10 in. What maximum slowlymoving concentrated service load(25% dead load; 75% live load) may be carried?

Use LRFD Design Method


Wu = 1.2(0.25W) + 1.6(0.75W) = 1.5W

Mu=WuL

4 =

Wu(10)

4 = 2.5Wu with load at midspan

Vu = 1.0Wu with load at support

(b) Check flexural strength assuming adequate lateral support (AISC F2.1):

Flange and web local buckling slenderness limits, Fy = 50 ksi steel:

bf

2tf

= 6.8 p = 65Fy

= 9.2 ; htw

= 31.2 p = 640Fy

= 90.5 OKbMn = bMp= bZxFy = 0.9(150)(50)/12 = 563 ft-kips

Mu= 2.5Wu= 563 ft-kips; Wu = 225 kips

(c) Check shear strength (AISC G2.1):

For rolled I-shapes, when

h

tw= 31.2

2.24

E/Fy = 53.9

, v = 1.0 and

Cv = 1.0.

vVn= v(0.6Fy)AwCv =v(0.6Fy)dtwCv= 1.0(0.6)(50)(16.52)(0.455)(1.0) = 225 kips

Vu = 1.0Wu= 225 kips; Wu= 225 kips

(d) Check local web yielding strength (AISC J10.2):

Rn= (5k+ N)Fytw Interior Reaction

Rn= (2.5k+ N)Fytw Exterior Reaction

Exterior reaction controls because Wu is both the interior and exterior load

Rn = (2.5k+ N)Fytw= 1.0 [5(1.4375) + N] (50)(0.455) = 309 kips

Rn = 1.0Wu= 309 kips; Wu= 309 kips

(e) Check web crippling strength (AISC J10.3):For exterior reactions, for [N/d= 10/16.52 = 0.61]>0.2:





38/131

Rn = 0.80t2w

1 + 4N

d 0.2

twtf

1.5

EFywtftw

= (0.75)(0.40)(0.455)21 + 4(10)

16.52 0.2

0.455

0.76 1.5

(29000)(50)(0.76)

0.455

= 196 kips

Rn = 1.0Wu= 196 kips; Wu= 196 kips

(f) Check sidesway web buckling strength (AISC J10.4):

When the compression flangeis restrainedagainst rotation, for

(h/tw)/(Lb/bf) =

>2.3, this limit state does not apply.

Conclusion:

Web crippling controls!

MaxWu= 196 kips; Service Load W =Wu/1.5 = 131 kips





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7.15. Select the lightest W8section of A992 steel to use as a purlin on a roof sloped30 to the horizontal. The span is 21 ft, the load is uniform 0.18 kip/ft dead load plusthe purlin weight and 0.34 kip/ft snow load. Lateral stability is assured by attachment ofthe roofing to the compression flange. Assume the load acts through the beam centroid,there are no sag rods, and biaxial bending must be assumed. Any torsional effect can beresisted by the roofing and therefore it can be neglected.



wu = 1.2(0.18 + 0.035) + 1.6(0.34) = 0.80 kips/ft

Mu =wuL2

8 =

(0.80)(21)2

8 = 44.2 ft-kips

Mux= Mucos = 44.2cos30 = 38.3 ft-kips

Muy = Musin = 44.2sin30 = 22.1 ft-kips

(b) Use AISC H2 with no axial load term and the conservative estimate Mn= S Fy:Mux

bMnx+

MuybMny

1

Sx MuxbFy

+ MuybFy

SxSy

=

38.3(12)

0.9(50) +

22.1(12)

0.9(50)

SxSy

10.2 + 5.9

SxSy

For Sx on the order of 3 to 4: Sx 27.9 to 33.8 in.3

AssumingZx 1.12Sx: Zx 31.2 to 37.8 in.3

Using AISC Table 3-2 Selection byZx, for W8 beams, find W835 withZx= 34.7 in.3

Check the strength.

MuxbFySx

+ MuybFySy

= 38.3(12)

0.9(50)(31.2)+

22.1(12)

0.9(50)(10.6)

= 0.3272 + 0.5561= 0.8833 1 OK

Beam Mnx Mny Checkft-kips ft-kips

W835 117 39.8 0.3272 + 0.5561 = 0.8833 OKW831 103 34.8 0.3690 + 0.6321 = 1.0011 NG





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8.21. Assume a single W section is to serve as a crane runway girder which carries avertical loading, as shown. In addition, design must include an axial compressive force of14 kips and a horizontal force of 4 kips on each wheel applied 4 14 in. above the top of thecompression flange. Assume torsional simple support at the ends of the beam. Select thelightest W14 section of A992 steel using the modified flexure analogy approach. Note:All loads except weight of the crane runway girder are live loads.



Use an estimated beam weight of 0.342 kips/ft and an estimated beam depth of 14 in.

Wux = 1.6(40) = 64 kips Lifted load

Wuy = 1.6(4) = 6.4 kips Lateral load

wux = 1.2(0.020 + 0.342) = 0.4344 kips/ft Dead load

x=40 4/2

2 = 19 ft location for maximum moments

Mux1= 2Wuxx2L

=2(64)(19)240

= 1155 ft-kips Lifted load moment

Mux2= 0.25Mux1= 0.25(1155) = 288.8 ft-kips Impact moment

Mux3=wuxx(L x)

2 =

(0.4344)(19)(40 19)

2 = 86.7 ft-kips Dead load moment

Mux = Mux1+ Mux2+ Mux3= 1531 ft-kips

Muy =2Wuyx2

L =

2(6.4)(19)2

40 = 115.5 ft-kips

Tu= Wuy(d/2 + rail height) = 6.4(14/2 + 4.25) = 72.00 in-kips

(b) Use themodified flexure analogy to find the equivalent lateral moment. Use 0.5.

Vf =Tu

h

72.00

14 = 5.143 kips flange force using h d

Mf =2Vfx

2

L = 0.5

(5.143)(19)2

40 = 46.41 kips flange moment

My = 2Mf= 2(46.41) = 92.83 kips equivalent lateral moment





41/131

(c) Select a beam using AISC H1:

Sx MuxbFy

+Muy+ My

bFy

SxSy

=

1531(12)

0.9(50) +

(115.5 + 92.83)(12)

0.9(50)

SxSy

408.2 + 55.56SxS

y

For Sx on the order of 3 to 5: Sx 574.9 to 686.0 in.3

AssumingZx 1.12Sx: Zx 643.8 to 768.3 in.3

Using AISC Table 3-2 Selection by Zx, for W14 beams, find W14342 withZx = 672 in.3

(d) Check the beam more accurately using the properties from AISC Table 1-1.

=

GJ

ECw=

(11154)(178)

(29000)(103000)= 0.02578

L= (0.02578)(40)(12) = 12.38

= 0.16 for a point load at xand simply supported endsTu=Wuy(d/2 + rail height) = 6.4(17.5/2 + 4.25) = 83.20 in-kips

Vf =Tu

h =

83.20

15.03= 5.536 kips

Mf =2Vfx

2

L = 0.16

(5.536)(19)2

40 = 16.19 ft-kips

My = 2Mf= 2(16.19) = 32.38 ft-kips

fun =Mux

Sx

+Muy

Sy

+My

Sy

=1531(12)

558

+115.5(12)

221

+32.38(12)

221= 32.92 + 6.273 + 1.758

= 40.95 ksi

Fy = 0.9(50) = 45 ksi

OK

The beam is sufficient. Check for a lighter beam.

Section Mux My fun Fy = 45 ksift-kips ft-kips ksi

W14342 1531 0.16 32.38 32.92 + 6.273 + 1.758 = 40.95 OKW14311 1523 0.17 34.35 36.12 + 6.966 + 2.072 = 45.16 NG

Use W14342, A992 steel.





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9.2, Case 1. Determine the maximum concentrated load P that can act at midspan ona simply supported span of 20 ft. Lateral supports exist only at the ends of the span. Theservice load is 65% live load and 35% dead load. The section is W2162 ofFy = 50 ksisteel.



Wu= 1.2(0.35W) + 1.6(0.65W) = 1.46W

wu= 1.2(62/1000) = 0.0744 kips/ft

(b) Determine theCb factor, AISC-F1.

The beam has lateral support at the ends only. The longest unbraced length isLb= 20 ft. For doubly symmetric members, Rm= 1.0.

Cb= 12.5Mmax2.5Mmax+ 3MA+ 4MB+ 3MCRm 3.0

Mmax= max moment in the unbraced segment = M

MA= moment at 1/4 pt of the unbraced segment = 0.5M

MB = moment at 1/2 pt of the unbraced segment = 1.0M

MC= moment at 3/4 pt of the unbraced segment = 0.5M

Cb= 12.5M

2.5M+ 3(0.5M) + 4(1.0M) + 3(0.5M)(1.0) = 1.32

(c) Compute the design moment strength using the beam properties from the AISCManualTable 1-1, pp. 1-18 and 1-19.h

tw= 46.9

p = 3.76

E

Fy= 90.6

;

bf2tf

= 6.7

p= 0.38

E

Fy= 9.15

The web is compact and the flange is compact so use AISC-F2.

For the limit state of yielding, AISC-F2.1:

Mn= Mp= FyZx= (50)(144)/12 = 600 ft-kips

For the limit state of lateral torsional buckling, AISC-F2.2:

From Table 3-2 in the AISC Manualp. 3-16, Lp= 6.25 ft andLr = 18.1 ft.

[Lr

= 18.1 ft]< [Lb

= 20 ft] elastic lateral torsional buckling applies, AISC-F2.1(c).

Mn= FcrSx= SxCb

2E

(Lb/rts)2

1 + 0.078

Jc

Sxho

Lbrts

2

= (127)(1.32)2(29000)

(12(20)/2.15)2

1 + 0.078

(1.83)(1)

(127)(20.4)

12(20)

2.15

2= 415 ft-kips





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Elastic lateral torsional buckling controls! Calculate the design moment strength.

bMn = (0.9)(415) = 374 ft-kips

(d) Calculate the maximum service load.

bMn

=WuL

4 +

wuL2

8

374 =(1.46W)(20)

4 +

(0.0744)(20)2

8 = 7.30W+ 3.72

W = 50.7 kips

Maximum Service Load W = 50.7 kips





44/131




Wu= 1.2(0.35W) + 1.6(0.65W) = 1.46W

wu= 1.2(84/1000) = 0.101 kips/ft








Cb= 12.5M

2.5M+ 3(0.5M) + 4(1.0M) + 3(0.5M)(1.0) = 1.32


tw= 45.9

p = 3.76

E

Fy= 90.6

;

bf2tf

= 5.86

p = 0.38

E

Fy= 9.15



Mn= Mp= FyZx= (50)(224)/12 = 933 ft-kips



[Lr

= 20.3 ft]< [Lb


Mn= FcrSx= SxCb

2E

(Lb/rts)2

1 + 0.078

Jc

Sxho

Lbrts

2

= (196)(1.32)2(29000)

(12(24)/2.37)2

1 + 0.078

(3.7)(1)

(196)(23.3)

12(24)

2.37

2= 579 ft-kips





45/131


bMn = (0.9)(579) = 521 ft-kips


bMn

=WuL

4 +

wuL2

8

521 =(1.46W)(24)

4 +

(0.101)(24)2

8 = 8.76W+ 7.26

W = 58.7 kips






46/131




Wu= 1.2(0.35W) + 1.6(0.65W) = 1.46W

wu= 1.2(99/1000) = 0.119 kips/ft








Cb= 12.5M

2.5M+ 3(0.5M) + 4(1.0M) + 3(0.5M)(1.0) = 1.32


tw= 51.9

p = 3.76

E

Fy= 90.6

;

bf2tf

= 7.8

p= 0.38

E

Fy= 9.15



Mn= Mp= FyZx= (50)(312)/12 = 1300 ft-kips



[Lr

= 21.4 ft]< [Lb


Mn= FcrSx= SxCb

2E

(Lb/rts)2

1 + 0.078

Jc

Sxho

Lbrts

2

= (269)(1.32)2(29000)

(12(30)/2.62)2

1 + 0.078

(3.77)(1)

(269)(29)

12(30)

2.62

2= 585 ft-kips





47/131


bMn = (0.9)(585) = 527 ft-kips


bMn

=WuL

4 +

wuL2

8

527 =(1.46W)(30)

4 +

(0.119)(30)2

8 = 11.0W+ 13.4

W = 46.9 kips






48/131

9.3. Select the lightest W section as a beam. Assume only flexure must be considered;i.e., omit treating shear and deflection. The dead load given is in addition to the weightof the beam.

Case 1: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has continuouslateral support, and Fy = 50 ksi.



wu = 1.2(0.9 + beam wt) + 1.6(2) 4.28 kips/ft

Mu=wuL

2

8 =

4.28(20)2

8 = 214 ft-kips (without beam)


Since the beam has continous lateral support, Cb= 1.0.

(c) Since the unbraced length is relatively short, select a beam using Table 3-2 SelectionbyZx, AISC Manual, pp. 3-11 to 3-19. Assume p for a compact section.

RequiredZx = MubFy

=(214)(12)(0.9)(50)

= 57.1 in.3

Select: W1835, Zx = 66.5 in.3

(d) Correct the moment for the selected beam weight.

Mu= 214 +1.2(beam wt)L2

8 = 214 +

1.2(35/1000)(20)2

8 = 216 ft-kips

(e) Compute the design moment strength using the beam properties from the AISCManualTable 1-1, pp. 1-18 and 1-19.

htw

= 53.5

p = 3.76

EFy= 90.6

;b

f2tf

= 7.06

p = 0.38

EFy= 9.15



Mn= Mp= FyZx= (50)(66.5)/12 = 277 ft-kips






49/131

The beam has continuous lateral support, so [Lb= 0]< Lp and lateral torsionalbuckling does not apply, AISC-F2.1(a).

Yielding controls! Calculate the design moment strength.

bMn = (0.9)(277) = 249 ft-kips

The W1835 beam is sufficient. To verify it is the lightest beam use Table 1-1 of theAISC Manual, and examine the first beam in each group lighter than 35 lb/ft with a largeenough Zx. The following table shows the moment corrected for the beam weight.

Section Mu bMnbf

2tf

h

twOKAY?

ft-kips ft-kips

W1835 216 249 7.06 53.5 OKW1631 216 203 6.28 51.6 NGW1434 216 205 7.41 43.1 NG

Use W1835 with Fy = 50 ksi steel.





50/131


Case 2: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateralsupport at the ends and midspan, and Fy = 50 ksi.



wu = 1.2(0.9 + beam wt) + 1.6(2) 4.28 kips/ft

Mu=wuL

2

8 =

4.28(20)2



The beam has lateral support at the ends and midspan. The loading is uniform andsymmetric, so the worst loading will occur on a segment containing the midpoint ofthe beam. Use the segment from 0 ft to 10 ft with Lb= 10 ft. For doubly symmetricmembers,Rm= 1.0.






Cb= 12.5M

2.5M+ 3(0.438M) + 4(0.8M) + 3(0.938M)(1.0) = 1.30


RequiredZx = MubFy

=(214)(12)

(0.9)(50) = 57.1 in.3

Select: W1835, Zx = 66.5 in.3


Mu= 214 +1.2(beam wt)L2

8 = 214 +

1.2(35/1000)(20)2

8 = 216 ft-kips





51/131


h

tw= 53.5

p = 3.76

E

Fy= 90.6

;

bf2tf

= 7.06

p = 0.38

E

Fy= 9.15

The web is compact and the flange is compact so use AISC-F2.For the limit state of yielding, AISC-F2.1:

Mn= Mp= FyZx= (50)(66.5)/12 = 277 ft-kips


From Table 3-2 in the AISC Manualp. 3-18, Lp= 4.31 ft andLr = 12.4 ft.Lp= 4.31 ft

< [Lb= 10 ft] [Lr = 12.4 ft] inelastic lateral torsional buckling

applies, AISC-F2.1(b).

Mn= Cb Mp Mp 0.7FySxLb LpLr Lp

Mp= (1.30)

277

277

0.7(50)(57.6)

12

10 4.31

12.4 4.31

= 260 ft-kips

Inelastic lateral torsional buckling controls! Calculate the design moment strength.

bMn = (0.9)(260) = 234 ft-kips


Section Mu bMnbf

2tf

h

twOKAY?

ft-kips ft-kips

W1835 216 234 7.06 53.5 OKW1631 216 186 6.28 51.6 NGW1434 216 205 7.41 43.1 NG Inelastic lateral torsional buckling controls






52/131


Case 3: dead load is 0.9 kips/ft, live load is 2 kips/ft, span is 20 ft, the beam has lateralsupport at the ends only, andFy = 50 ksi.



wu = 1.2(0.9 + beam wt) + 1.6(2) 4.28 kips/ft

Mu=wuL

2

8 =

4.28(20)2



The beam has lateral support at the ends only. The longest unbraced length is Lb=20 ft. For doubly symmetric members, Rm= 1.0.

Cb= 12.5Mmax

2.5Mmax+ 3MA+ 4MB+ 3MCRm 3.0

Mmax= max moment in the unbraced segment = MMA= moment at 1/4 pt of the unbraced segment = 0.750M



Cb= 12.5M

2.5M+ 3(0.750M) + 4(1.0M) + 3(0.750M)(1.0) = 1.14

(c) Since the unbraced length is fairly long, select a beam using Table 3-10 AvailableMoment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with Lb= 20 ft.

RequiredbM

n=

Mu

Cb=

214

1.14= 188 ft-kips

Select: W1448, bMn= 193 ft-kips


Mu= 214 +1.2(beam wt)L2

8 = 214 +

1.2(48/1000)(20)2

8 = 217 ft-kips

(e) Compute the design moment strength using the beam properties from the AISC





53/131

ManualTable 1-1, pp. 1-22 and 1-23.h

tw= 33.6

p = 3.76

E

Fy= 90.6

;

bf2tf

= 6.75

p = 0.38

E

Fy= 9.15



Mn= Mp= FyZx= (50)(78.4)/12 = 327 ft-kips


From Table 3-2 in the AISC Manualp. 3-17, Lp= 6.75 ft andLr = 21.1 ft.Lp= 6.75 ft

< [Lb= 20 ft] [Lr = 21.1 ft] inelastic lateral torsional buckling

applies, AISC-F2.1(b).

Mn= Cb

Mp

Mp 0.7FySx

Lb LpLr Lp

Mp

= (1.14)

327

327 0.7(50)(70.2)12

20 6.7521.1 6.75

= 243 ft-kips


bMn = (0.9)(243) = 219 ft-kips


Section Mu bMn bf2tfh

twOKAY?

ft-kips ft-kips

W1448 217 219 6.75 33.6 OKW2148 217 200 9.47 53.6 NGW2144 217 119 7.22 53.6 NGW1846 217 134 5.01 44.6 NGW1645 217 163 6.23 41.1 NGW1631 216 62.6 6.28 51.6 NGW1448 217 219 6.75 33.6 OKW1443 217 187 7.54 37.4 NGW1438 216 120 6.57 39.6 NGW1245 217 188 7 29.6 NGW1045 217 177 6.47 22.5 NG Inelastic lateral torsional buckling controls Elastic lateral torsional buckling controls Flange local buckling limit state must be checked (see below)

For the limit state of compression flange local buckling, AISCF3.2, for W2148:





54/131


55/131


Case 4: dead load is 0.7 kips/ft, live load is 1.4 kips/ft, span is 28 ft, the beam has lateralsupport at the ends and midspan, and Fy = 50 ksi.



wu = 1.2(0.7 + beam wt) + 1.6(1.4) 3.08 kips/ft

Mu=wuL

2

8 =

3.08(28)2



The beam has lateral support at the ends and midspan. The loading is uniform andsymmetric, so the worst loading will occur on a segment containing the midpoint ofthe beam. Use the segment from 0 ft to 14 ft with Lb= 14 ft. For doubly symmetricmembers,Rm= 1.0.






Cb= 12.5M

2.5M+ 3(0.438M) + 4(0.8M) + 3(0.938M)(1.0) = 1.30

(c) Since the unbraced length is fairly long, select a beam using Table 3-10 AvailableMoment vs. Unbraced Length, AISC Manual, pp. 3-96 to 3-131 with L

b= 14 ft.

RequiredbMn =Mu

Cb=

302

1.30= 232 ft-kips



Mu= 302 +1.2(beam wt)L2

8 = 302 +

1.2(48/1000)(28)2

8 = 307 ft-kips





56/131


57/131

Mn= Mp

Mp 0.7FySx pf

rf pf

= 446

446

0.7(50)(93)

12

9.47 9.15

24.1 9.15

= 442 ft-kips






58/131


59/131


h

tw= 37.4

p = 3.76

E

Fy= 82.7

;

bf2tf

= 7.54

p = 0.38

E

Fy= 8.35

The web is compact and the flange is compact so use AISC-F2.For the limit state of yielding, AISC-F2.1:

Mn= Mp= FyZx= (60)(69.6)/12 = 348 ft-kips


Lp= 1.76ry

E

Fy= 1.76

1.89

12

29000

60 = 6.09 ft

rts=

IyCw

Sx=

(45.2)(1950)

62.6 = 2.18 in.

Lr = 1.95rtsE

0.7Fy

Jc

Sxho

1 +

1 + 6.76

0.7Fy

E

SxhoJc

2

= 1.95

2.18

12

29000

0.7(60)

(1.05)(1)

(62.6)(13.1)

1 +

1 + 6.76

0.7(60)

29000

(62.6)(13.1)

(1.05)(1)

2= 17.7 ft

Lp= 6.09 ft

< [Lb= 14 ft] [Lr = 17.7 ft] inelastic lateral torsional bucklingapplies, AISC-F2.1(b).

Mn= Cb

Mp

Mp 0.7FySx

Lb LpLr Lp

Mp

= (1.30)

348

348

0.7(60)(62.6)

12

14 6.09

17.7 6.09

= 338 ft-kips


bMn = (0.9)(338) = 304 ft-kips

The W1443 beam is not sufficient. Check heavier sections at the same depth. The

following table shows the moment corrected for the beam weight.

Section Mu bMnbf

2tf

h

twOKAY?

ft-kips ft-kips

W1443 307 304 7.54 37.4 NGW1448 307 350 6.75 33.6 OKW2148 307 404 9.47 53.6 OK





60/131


61/131


Case 6: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam hascontinuous lateral support, and Fy = 50 ksi.



wu = 1.2(0.3 + beam wt) + 1.6(0.9) 1.8 kips/ft

Mu=wuL

2

8 =

1.8(35)2



Since the beam has continous lateral support, Cb= 1.0.


RequiredZx = MubFy

=(276)(12)(0.9)(50)

= 73.5 in.3

Select: W1840, Zx = 78.4 in.3


Mu= 276 +1.2(beam wt)L2

8 = 276 +

1.2(40/1000)(35)2

8 = 283 ft-kips


htw

= 50.9

p = 3.76

EFy= 90.6

;b

f2tf

= 5.73

p = 0.38

EFy= 9.15



Mn= Mp= FyZx= (50)(78.4)/12 = 327 ft-kips






62/131

The beam has continuous lateral support, so [Lb= 0]< Lp and lateral torsionalbuckling does not apply, AISC-F2.1(a).

Yielding controls! Calculate the design moment strength.

bMn = (0.9)(327) = 294 ft-kips


Section Mu bMnbf

2tf

h

twOKAY?

ft-kips ft-kips

W1840 283 294 5.73 50.9 OKW1640 283 274 6.93 46.5 NG

Use W1840 with Fy

= 50 ksi steel.





63/131


Case 7: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateralsupport every 7 feet, and Fy = 50 ksi.



wu = 1.2(0.3 + beam wt) + 1.6(0.9) 1.8 kips/ft

Mu=wuL

2

8 =

1.8(35)2



The beam has lateral support every 7 feet. The loading is uniform and symmetric,so the worst loading will occur on a segment containing the midpoint of the beam.Use the segment from 14 ft to 21 ft with Lb = 7 ft. For doubly symmetric members,Rm= 1.0.






Cb= 12.5M

2.5M+ 3(0.990M) + 4(1.0M) + 3(0.990M)(1.0) = 1.00


b= 7 ft.

RequiredbMn =Mu

Cb=

276

1.00= 274 ft-kips



Mu= 276 +1.2(beam wt)L2

8 = 276 +

1.2(44/1000)(35)2

8 = 284 ft-kips





64/131


65/131


Case 8: dead load is 0.3 kips/ft, live load is 0.9 kips/ft, span is 35 ft, the beam has lateralsupport at the ends and midspan, and Fy = 50 ksi.



wu = 1.2(0.3 + beam wt) + 1.6(0.9) 1.8 kips/ft

Mu=wuL

2

8 =

1.8(35)2



The beam has lateral support at the ends and midspan. The loading is uniform andsymmetric, so the worst loading will occur on a segment containing the midpoint of thebeam. Use the segment from 0 ft to 17.5 ft with Lb = 17.5 ft. For doubly symmetricmembers,Rm= 1.0.






Cb= 12.5M

2.5M+ 3(0.438M) + 4(0.8M) + 3(0.938M)(1.0) = 1.30


b= 17.5 ft.

RequiredbMn =Mu

Cb=

276

1.30= 212 ft-kips



Mu= 276 +1.2(beam wt)L2

8 = 276 +

1.2(48/1000)(35)2

8 = 284 ft-kips





66/131


h

tw= 53.6

p = 3.76

E

Fy= 90.6

;

p = 0.38

E

Fy= 9.15

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Steel Structures 5th Edition Solutions Manual