978-1337094740 Chapter 5 Part 4

Type Homework Help

Pages 12

Words 1069

Textbook Steel Design 6th Edition

Authors William T. Segui

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5.11-4

Dead loads: Slab weight 4. 5

12 15056. 25 psf

wD56. 25 55. 5336. 9 lb/ft (neglect beam wt. and check

it later.)

Ix510 in.4588 in.4required (N.G.)

Try a W18 40, Ix612 in.4,MnbMp294 ft-kips

Check beam weight:

From the Zxtable,

[5-61]

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

Try a W18 40, Mn

b

Mp

b

196 ft-kips 143 ft-kips (OK)

Ix612 in.4588 in.4required (OK)

5.11-5

Dead loads: Slab and deck weight 43 psf

wD43 5212600. 0 lb/ft (neglect beam wt. and check it

later.)

Live load: wL160 20122160 lb/ft

Try a W21 44, bMnbMp358 ft-kips 326 ft-kips (OK)

Check beam weight:

Mu326 1

81. 2 0. 044252

330 ft-kips 358 ft-kips (OK)

Try a W21 44, Mn

b

Mp

b

238 ft-kips 216 ft-kips (OK)

Ix843 in.4786 in.4required (OK)

Check beam weight:

Ma216 1

5.11-6

Live load: wL80 20101000 lb/ft

From the Zxtable,

Try a W14 22, bMnbMp125 ft-kips 117 ft-kips (OK)

Try a W14 22, Mn

b

Mp

b

82. 8 ft-kips 80.5 ft-kips (OK)

Ix199 in.4186 in.4required (OK)

Check beam weight:

Ma80. 5 1

5.11-7

Dead loads: Slab weight 5

12 15062. 5 psf

wD62. 55312. 5 lb/ft (neglect beam wt. and check it later.)

Live load: wL80 205500 lb/ft

From the Zxtable,

Try a W16 31, bMnbMp203 ft-kips 132 ft-kips (OK)

Ix375 in.4340 in.4required (OK)

Check beam weight:

Mu132 1

Try a W16 31, Mn

b

Mp

b

135 ft-kips 91.4 ft-kips (OK)

[5-65]

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

Ix375 in.4340 in.4(OK)

Check moment for beam weight:

5.11-8

Slab weight 5

Girder reaction 35. 253

252. 88 kips

Mu52. 8810−35. 255353 ft-kips

UseaW2144

(b) wawDwL312. 5 500 812. 5 lb/ft

5' 5' 5' 5'

36.57 k 36.57 k

5.11-9

Slab weight 5

12 15062. 5 psf. For beams, use 35 lb/ft 30 ft 

1050 lb/beam

Expressed as a distributed load, the weight of the beams is

Mu362 1

81. 2 0. 048202

Vu≈7. 24020

5.12-1

(a) For a W16 31, bf5. 53 in., tf0. 440 in., and Sx47. 2 in.3

Afg

2. 433 47. 21957 in.-kips 163.1 ft-kips.

Mn163 ft-kips

5.12-2

Afg

3. 500 93. 04746 in.-kips 395. 5 ft-kips.

Mn396 ft-kips

(b) This is a noncompact shape (see footnote in Dimensions and

Properties tables). Since there is continuous lateral support, the

24. 08 −9. 152

442. 1 100 −10. 54%

5.12-3

(a) For a W18 35, bf6. 00 in., tf0. 425 in., and Sx57. 6 in.3

Check to see if the flange holes need to be accounted for. The gross area

of one flange is

Afg

2. 55 57. 62652 in.-kips 221 ft-kips.

Mn221 ft-kips

5.13-1

Dead loads: Slab weight 4

12 15050. 0 psf

5.13-2

Assuming a joist weight 4psf, wD432 54164 lb/ft

Use a 16K4

5.14-1

(a) Factored load Ru1. 685136. 0 kips

1. 0500. 37551. 24ℓ

b≥136 ℓb≥1. 05 in.

213ℓb

1.5 EFytf

0. 750. 800. 375213ℓb

13. 9

0. 645

1.5 29000500. 645

0. 375 ≥136

ℓb≥0. 2158 in.

whose dimensions conform to the details of the applied load.

1. 50 ≥85 ℓb≥0. 6 in.

0. 80tw

≥Ra

13. 9

0. 375

0. 645

0. 375

2. 00 ≥85

5.14-2

(a) The factored load reaction is

Ru1. 2D1. 6L1. 2281. 656123. 2 kips

1. 0500. 5202. 51. 32ℓ

b≥123. 2 ℓb≥1. 438 in.

0. 750. 400. 520214N

29. 7 −0. 2 0. 520

0. 670

0. 520 ≥123. 2

ℓb≥2. 692 in.

29. 7

0. 670

0. 520 ≥123. 2

ℓb≥1. 609 in., and ℓb

d1. 609

29. 7 5. 418 10−20. 2 (OK)

Try ℓb6 in. Determine B.

0. 600. 853A11. 234≥123. 2 A1≥65. 25 in.2

Try B65. 25/6 10. 88 in. Try a 6 11 plate. Support width is

0. 600. 853A11. 255≥123. 2, Solution is: 64. 16 ≤A1

1. 7fc´A11. 7366337 kips

0. 85fc´A1A2

0. 853661. 255211 kips 337 kips

(OK)

Plate thickness: nB−2k

4. 18 in.

1. 50 2. 51. 32ℓ

Use AISC Equation J10-5 to determine the value of ℓbrequired to

2. 00 0. 400. 520213ℓb

29. 7

0. 520

0. 670

0. 520 ≥84

ℓb≥1. 975 in., and ℓb

d1. 975

29. 7 6. 650 10−20. 2 (OK)

Try ℓb6 in. Determine B.

2. 50 ≥84 A1≥66. 74 in.2

Try B66. 74/6 11. 12 in. Try a 6 12 plate. Support width is

0. 853A11. 247

2. 50 ≥84 A1≥66. 04 in.2

0. 85fc´A1A2

0. 853721. 247229 kips 367 kips

(OK)

5.14-3

(a) The factored load is

0. 65 0. 853. 5A1

≥390, Solution is: 158. 9 in.2

Check upper limit:

c1. 7fc´A10. 601. 73. 5158. 9567. 3 kips 390

kips (OK)

Conservatively, let 1. 0, resulting in

[5-77]

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

ℓmaxm,n,n´max1. 063, 2. 16, 3. 0753. 075 in.

1. 08

Required t ℓ0. 90

2Pu

BNFy

3. 075 0. 90

390



in.; use 1 in.

2. 31 0. 853. 5A1

The upper limit is

c

1. 7fc´A11

2. 31 1. 73. 5187

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978-1337094740 Chapter 5 Part 4

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978-1337094740 Chapter 5 Part 4

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