978-1337094740 Chapter 6 Part 3

Type Homework Help

Pages 9

Words 1085

Textbook Steel Design 6th Edition

Authors William T. Segui

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2Pn/c

Max

Mnx/b

May

Mny/b

92

2520214

297 31

135

1−1. 60Pa

Pe1x

1−1. 6092

10690

∴B1x1. 0 as assumed

Pe1y2EIy

KyL2229000148

16 1221149 kips

Max

May

92

6.8-7

(a) LRFD Solution

Determine which interaction equation to use:

cPn

120. 0

686 0. 174 9 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

As a preliminary check (remember that Cbhas not been accounted for)

Mux

Muy

120. 0

120. 0

2cPn

bMnx

bMny

2709199. 5

376 64. 5

185 0. 964

Check the W14s. Try a W14 68. From Table 6-2,

bMnx 374 ft-kips cPn577 kips bMny 138 ft-kips

Determine which interaction equation to use:

120. 0

64. 5 0. 683 7

Pe1x2EIx

Lcx2229000597

16 1224635 kips

B1xCmx

0. 6812

0. 699 3 1.0 ∴B1x1. 0 as assumed

1−Pu

Pe1y

1−120

1514

Since the amplification factors are as assumed, and this shape is adequate for Cb1.0,

computation of the actual value of Cbis not necessary. The preliminary evaluation is

sufficient.

UseaW1272

(b) ASD Solution (abbreviated version; rejected trials not shown.)

Try a W12 72. From Table 6-2, with LcLb16 ft,

Mnx/Ωb250 ft-kips Pn/Ωc472 kips Mny/Ωb123 ft-kips

Pn/c

80

472 0. 169 5 0. 2 ∴use Equation 6.4 (AISC Equation H1-1b)

Max

May

80

1−1. 60Pa

Pe1x

1−1. 6080

4635

∴B1x1. 0 as assumed

6.8-8

Assume that B1B21.0 for purposes of making a trial selection.

MuB1Mnt B2Mℓt1. 02701. 030300. 0 ft-kips

Lcx

KxL

2. 016

cPn

75

584 0. 128 4 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

2cPn

Mux

bMnx

Muy

bMny

75

2584300. 0

356 00. 907

Consider the braced condition first.

[6-45]

or posted to a publicly accessible website, in whole or in part.

For the axis of bending,

Pe2x2EIx

K2xL2229000533

2. 0 16 1221035 kips

6.8-9

(a) LRFD Solution

Load combinations involving dead load (D), roof live load (Lr),andwindload(W)are

Mℓt0. 53216. 0 ft-kips

Load Combination 4: 1. 2D1. 0W0. 5Lr

[6-46]

or posted to a publicly accessible website, in whole or in part.

Pnt 1. 2191. 0−110. 53329. 7 kips

Pℓt1. 01. 41. 4 kips

Pestory

1224

MuB1Mnt B2Mℓt1. 0279. 81. 06116. 0296. 8 ft-kips

PuPnt B2Pℓt70. 1 1. 0610. 770. 84 kips

Cm0. 6 −0. 4 M1

0. 6 −0. 4 0

0. 6

B1Cm

1−Pr/Pe1Cm

1−1. 0Pnt Pℓt/Pe1

Determine which interaction equation to use:

cPn

70. 84

456 0. 155 4 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

2cPn

Mux

bMnx

Muy

bMny

70. 84

2456296. 8

383 0

(Since the frame and loading are symmetrical, there are no sidesway moments for this

load combination.)

Pℓt1. 4 kips

MaB1Mnt B2Mℓt1. 02090209 ft-kips

PaPnt B2Pℓt52 052 kips

Try a W14 61. From Table 6-2, using LcLb18 ft,

Mnx/Ωb208 ft-kips Pn/Ωc304 kips

Determine the critical effective length for compression. Use the sway condition.

1−Pr/Pe1Cm

1−1. 60Pa/Pe10. 6

1−1. 6052/3926

0. 613 0 1. 0 ∴B11. 0 as assumed

Modify Mnx/bto account for Cb.

CbMnx/b1. 67208347 ft-kips Mpx/b254 ft-kips

∴use Mnx/b254 ft-kips

52

0. 609 8 1. 0 ∴B11. 0 as assumed

the maximum drift caused by service loads.)

6.8-10

(a) LRFD Solution

Determine the total vertical load to be stabilized by the bracing.

0. 9Fy

0. 9366. 948 10−2in.2

From AISC Equation A-6-2a, the required lateral stiffness is

br 1

2Pr

Lbr

1

0. 75

2418. 1

18 125. 162 kips/in.

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978-1337094740 Chapter 6 Part 3

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978-1337094740 Chapter 6 Part 3

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