978-1337094740 Chapter 6 Part 2

Type Homework Help

Pages 14

Words 4047

Textbook Steel Design 6th Edition

Authors William T. Segui

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Pe1y2EIy

K1L22EIy

KyL2229000116

12 1221601 kips

B1yCmy

6.6-13

(a) LRFD solution:

2. 5Mmax 3MA4MB3MC

12. 54Q

2. 54Q3044Q301. 923

For Cb1. 923,

bMn1. 923331636. 5 ft-kips bMp600 ft-kips

∴use bMnbMp600 ft-kips

8

Mux

Muy

0. 7769 8

4. 096Q

Let 0. 7769 8

600 01, Solution is: 36. 77 kips

Maximum Q36. 8 kips

2. 5Mmax 3MA4MB3MC

12. 52. 5Q

2. 52. 5Q3042. 5Q301. 923

For Cb1. 923,

b

1. 923220423. 1 ft-kips Mp

b

399 ft-kips

8

Max

May

0. 7803 8

2. 563Q

Let 0. 7803 8

2. 563Q

6.7-1

Determine the axial compressive design strength. Use LcKxLfor the unbraced

2. 5Mmax 3MA4MB3MC

12. 5135. 7

2. 5135. 7335. 11421. 83378. 772. 208

For Cb2. 208, bMn2. 2084791058 ft-kips

8

Mux

Muy

0. 5182 8

135. 7

6.7-2

(a) LRFD solution

98. 4 0. 266 7

For the braced case,

Pe12EIx

2. 598. 4336. 948. 20353. 32. 239

For Cb2. 239, bMn2. 239415929. 2 ft-kips

Since 929.2 ft-kips bMp,usebMnbMp473 ft-kips

528

cPn

8

Mux

bMnx

Muy

bMny

0. 8341 8

98. 4

473 0

b

Since 618.0 ft-kips Mp

b

,use Mn

b

Mp

b

314 ft-kips

360

6.8-1

(a) LRFD solution

The factored axial load is

Pu1. 20. 30 2361. 60. 70 236349. 3 kips

cPn

8

Mux

bMnx

Muy

bMny

0. 7308 8

248. 6

258 0

cPn

8

Mux

bMnx

Muy

bMny

0. 4318 8

248. 6

422 00. 955 4

0. 955 1. 0 (OK)

Calculate B1:

is not needed.)

UseaW1279

selection.

Mnx

281 ft-kips and Pn

538 kips

Pn/c

538 0. 438 7 0. 2 ∴use Equation 6.5 (AISC Eq. H1-1a):

Pn/c

8

Max

Mnx/b

May

Mny/b

0. 4387 8

168

281 0

0. 970 1. 0 (OK)

6.8-2

Mux (top) 182 ft-kips

400

6.8-3

(a) LRFD Solution

Mntx 1

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

Mux

Muy

0. 1918 135

Cb

bMn1. 14157179. 0 ft-kips bMp176 ft-kip

∴use bMn176 ft-kips

2cPn

Mux

bMnx

Muy

bMny

0. 1918 138. 2

176 0

Pn/c234 kips Mnx/b104 ft-kips

As a preliminary check (remember that B1has not yet been computed and Cbhas not

been accounted for),

Pn/c

45

234 0. 192 3 0. 2 ∴use Equation 6.6 (AISC Equation H1-1b)

Max

May

45

Pe1x

2885

Max B1xMntx 1. 0269092. 34 ft-kips

From Figure 5.15 in the textbook, Cb1. 14. Modify Mnx/bto account for Cb.

6.8-4

Pu140 kips

The factored moments at the top are

Mntx Mnty 150 ft-kips

The factored moments at the bottom are

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

2cPn

Mux

bMnx

Muy

bMny

140

21020150

525 150

260 0. 931

0. 931 1. 0 (OK)

[6-33]

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

bMnx 568 ft-kips, cPn979 kips, and bMny 273 ft-kips

Determine which interaction equation to use:

cPn

140

979 0. 143 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

As a preliminary check,

Mux

Muy

140

B1xCmx

1−Pu

Pe1x

0. 4

1−140

7756

0. 407 1.0 ∴B1x1. 0 as assumed

Pe1y2EIy

6.8-5

(a) LRFD Solution

Pu1. 21351. 6415826. 0 kips

The factored moments are

Mntx (top) 1. 2901. 6270540. 0 ft-kips

cPn

8

Mux

bMnx

Muy

bMny

826

1620 8

540

1030 0

0. 976 1/0 (OK)

Calculate B1for the axis of bending:

[6-35]

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

(Since this shape is adequate for Cb1.0, computation of the actual value of Cbis not

necessary.)

UseaW14159

(b) ASD Solution

been accounted for),

Pn/c

550

1070 0. 514 0. 2 ∴use Equation 6.5 (AISC Equation H1-1a)

8

Max

May

0. 514 8

360

Pn/c

Mnx/b

Mny/b

618 01. 08

Calculate B1:

[6-36]

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

Cmx 0. 6 −0. 4 M1

0. 6 −0. 4 120

360 0. 466 7

2. 536030412032402. 143

CbMnx/b2. 1436181324 ft-kips Mpx/b649 ft-kips

∴use Mnx/b649 ft-kips

6.8-6

(a) LRFD Solution

Mntx 1. 2160/21. 6160/2224. 0 ft-kips

bMnx 461 ft-kips cPn898 kips bMny 229 ft-kips

Determine which interaction equation to use:

cPn

128. 8

898 0. 143 4 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

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

Mux

Muy

128. 8

Check the W14s. Try a W14 82. From Table 6-2,

bMnx 462 ft-kips cPn697 kips bMny 168 ft-kips

cPn

128. 8

697 0. 184 8 0. 2 ∴use Equation 6.4 (AISC Eq. H1-1b):

Mux

Muy

128. 8

299. 6 0. 300 9

Cmy 0. 6 −0. 4 M1

0. 6 −0. 4 33. 6

43. 4 0. 290 3

Pe1x2EIx

Lcx22EIx

KxL2229000662

0. 8 16 1228031 kips

2. 5299. 6393. 1437. 83168. 72. 222

∴use bMn446 ft-kips

128. 8

The service-load moments at the bottom are

Mntx 214 ft-kips, Mnty 31 ft-kips

selection. For the two axes,

Max B1xMntx 1. 0214214 ft-kips

Check the W12s. Try a W12 79. From Table 6-2,

Mnx/Ωb277 ft-kips Pn/Ωc520 kips Mny/Ωb135 ft-kips

As a preliminary check (remember that B1has not yet been computed and Cbhas not

been accounted for),

92

Cmx 0. 6 −0. 4 M1

0. 6 −0. 4 160

214 0. 300 9

Cmy 0. 6 −0. 4 M1

0. 6 −0. 4 24

31 0. 290 3

Pe1x2EIx

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978-1337094740 Chapter 6 Part 2

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978-1337094740 Chapter 6 Part 2

Unlock document.

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Trusted by Thousands of
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