978-1337094740 Chapter 6 Part 1

Type Homework Help

Pages 14

Words 3384

Textbook Steel Design 6th Edition

Authors William T. Segui

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CHAPTER 6 -BEAM-COLUMNS

6.2-1

The total axial compressive load is Pa 250 kips

6.2-2

Compute the compressive strength.

14. 11 1. 20. 097  1. 6wLwL7. 52 kips/ft

wL7. 52 kips/ft

[6-2]

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

(b) For Lc20 ft, Pn

c

468 kips

6.6-1

In the plane of bending,

Pe12EI∗

6.6-2

In the plane of bending,

Pe12EI

6.6-3

Lcx KxL0. 91412. 6 ft, Lcy KyL1. 01414 ft.

0. 4443 8

378. 1

646 00. 965 1. 0 (OK)

This member satisfies the AISC Specification

(b) ASD solution:

8

Max

May

0. 456 8

258. 5

6.6-4

(a) LRFD solution:

The factored-load axial force is

[6-6]

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

2. 5199. 83125. 1450. 32324. 422. 173

For Cb2. 173, bMn2. 173283615. 0 ft-kips

199. 8 0. 401 5

Pe12EI

cPn

8

Mux

bMnx

Muy

bMny

0. 3559 8

199. 8

324 0

0. 904 1. 0 (OK)

2. 5Mmax 3MA4MB3MC

12. 5135

2. 5135384. 5434316. 52. 173

For Cb2. 173,

b

2. 173189411 ft-kips Mp

b

∴use Mn

b

Mp

b

216 ft-kips

Pa120 kips, Mnt 135 ft-kips

6.6-5

(a) LRFD solution:

84. 0 1. 0

Pe12EI

Lc122EIx

KxL2229000171

10 1223399 kips

B1Cm

Member is satisfactory.

(b) ASD solution: Pa32 kips, Mnt 60 ft-kips

[6-9]

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

For the axis of bending,

Cm0. 6 −0. 4 M1

0. 6 −0. 4 −60

b

For Lc10 ft, Pn

c

220 kips.

Pn/c

32

220 0. 145 5 0. 2 ∴use Equation 6.6 (AISC Equation H1-1b)

Max

May

0. 1455 60. 9

6.6-6

The factored-load axial force is Pu285 kips

The factored-load end moments are

2. 5120362443542. 259

For Cb2. 259, bMn2. 259257580. 6 ft-kips

Since 580.6 ft-kips bMp,usebMnbMp280 ft-kips

8

Mux

Muy

0. 5126 8

120. 0

6.6-7

(a) LRFD solution:

Qu1. 271. 61837. 2 kips

cPn

8

Mux

bMnx

Muy

bMny

0. 3069 8

385. 6

488 01. 009

1. 01 1. 0 (N.G.)

Member is unsatisfactory.

8

Max

May

0. 3105 8

282. 4

6.6-8

2. 5Mmax 3MA4MB3MC

2. 5306. 03275. 44285. 63295. 81. 056

306. 0 0. 946 7

Pe12EI

K1L22EIx

KxL22290005900

11 1229. 692 104kips

B1Cm

2. 5Mmax 3MA4MB3MC

12. 5225

2. 52253202. 542103217. 51. 056

For Cb1. 056,

b

1. 0569601014 ft-kips Mp

b

For the axis of bending,

6.6-9

(a) LRFD Solution:

1. 2MD1. 6ML1. 249. 501. 6100. 5220. 2 ft-kips

For the axis of bending,

220. 2 0. 2

B1Cm

1−Pr/Pe1Cm

1−1. 0Pu/Pe10. 2

1−Pu/7359

Assume B11. 0 and check it later.

cPn

bMnx

bMny

Let Pu

990 8

220. 2

551 01. 0, Solution is: 638. 3 kips

1. 20. 33P1. 60. 67P638. 3, Solution is: 434. 8 kips

P435 kips

(b) ASD solution:

For the axis of bending,

b

Since 796.8 ft-kips Mp

b

,use Mn

b

Mp

b

367 ft-kips

From Table 6-2 with Lc  15 ft, Pn/c  659 kips

Assume that Pa

0.2 and use Equation 6.5 (AISC Eq. H1-1a):

6.6-10

(a) LRFD solution:

cPn

8

Mux

bMnx

Muy

bMny

0. 4466 8

22. 5

86. 6 0

8

Max

May

0. 4492 8

16. 2

6.6-11

Mnty Muy150 ft-kips

1. 13 (N.G.)

The W14 145 is adequate.

6.6-12

(a) LRFD solution:

Pu1. 220/21. 620/228. 0 kips

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978-1337094740 Chapter 6 Part 1

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978-1337094740 Chapter 6 Part 1

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