978-1337094740 Chapter 9 Part 2

Type Homework Help

Pages 14

Words 3516

Textbook Steel Design 6th Edition

Authors William T. Segui

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Min. longitudinal spacing 4d40. 52. 0 in.

Min. transverse spacing 4d40. 52. 0 in.

9.5-3

(a) Total load to be supported by the composite section (omit beam weight; check it

later):

Slab: 5

12. 5 85. 36 lb/ft.

For an 18-in. deep beam,

13. 5 79. 04 lb/ft.

Try a W18 86.

0. 85fc´ab AsFy

0. 854a841265, Solution is: a4. 429 in.

yd

2t−a

218. 4

25−4. 429

211. 99 in.

bMnbTy 0. 90126511. 99/12 1138 ft-kips 1067 ft-kips (OK)

Check beam weight:

80. 852230295. 9 ft-kips

bMnbMp698 ft-kips 95.9 ft-kips (OK) UseaW1886

(b) Total load to be supported by the composite section (omit beam weight; check it

later):

Slab: 5

13. 5 70. 45 ft-kips

Try a W18 76.

0. 854a841115, Solution is: a3. 904 in.

yd

2t−a

25

25−3. 904

25. 548 in.

1

Ty 1

0. 85fc´ab AsFy

0. 854a841265, Solution is: a4. 429in.

yd

2t−a

218. 4

25−4. 492

211. 95 in.

b

1

b

Ty 1

1. 67 126511. 95

Asa 5/82

40. 306 8 in.2,Ecwc

1.5 fc´1451.5 43492 ksi

′Ec≤RgRpAsaFu

14. 96 84. 56, round up to 85. total number 285170

Min. longitudinal spacing 4d45/82. 5 in.

Min. transverse spacing 4d45/82. 5 in.

Asa 3/42

40. 441 8 in.2,Ecwc

1.5 fc´1451.5 43492 ksi

′Ec≤RgRpAsaFu

21. 54 58. 73, round up to 59. total number 259118

For one stud at each section, the required spacing will be

9.6-1

(a) Before concrete cures:

Slab: 4

12 1507. 5375. 0 lb/ft

0. 85fc

′b

0. 854901. 06 in.

Area of transformed concrete C

324. 5

50 6. 49 in.2

[9-25]

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

Y2t−a

9.6-2

From Problem 9.2-2,

Beams are W18 97

t5in.

b84

384EIs

3842900017500. 214 4 in.

Δconst 5wconstL4

384EIs

50. 160/1230 124

3842900017505. 746 10−2in.

ΔΔ

DΔ

const 0. 2144 0. 05746 0. 271 9 in. Δ2. 63 in.

0. 85fc

′b

0. 854904. 657 in.

Y2t−a

25−4. 657

22. 672 in., dY218. 6 2. 672 21. 27 in.

Taking moments about the bottom of the steel, we get

9.6-3

(a) From Problem 9.3-1, a W12 16 is used, with t4 in., s7ft,L25 ft,

qconst 20 psf, qpart 15 psf, qL125 psf, A992 steel and 4 ksi concrete.

Before concrete cures:

Slab: 4

∑A

9. 420 10. 76 in., ILB 316. 4 in.4

Δpart 5wpartL4

50. 105/1225 124

9.6-4

(a) From Problem 9.4-1, a W21 57 is used, with t6 in., s9ft,L40 ft,

qconst 20 psf, qL250 psf, A992 steel and 4 ksi concrete.

Before concrete cures:

Slab: 6

0. 85fc

′b

0. 8541082. 274 in.

Y2t−a

26−2. 274

24. 863 in.

Taking moments about the bottom of the steel, we get

[9-29]

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

0. 85fc´ab AsFy

0. 854a108810, Solution is: a2. 206in.

yd

2t−a

223. 6

26−2. 206

216. 70 in.

bMnbTy 0. 9081016. 701. 217 104in.-kips 1014 ft-kips

Loads:

[9-30]

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

wu1. 2wD1. 6wL1. 20. 7301. 60. 1801. 164 k/ft

Mu1

81. 164402233 ft-kips

bMnbMp503 ft-kips 233 ft-kips (OK)

After concrete has cured:

wL25092250 lb/ft

Lower-bound moment of inertia:

Effective flange width (40 12/4 120 in. or 912108 in.

Use b108 in.

0. 85fc

′b

0. 8541082. 206 in.

Y2t−a

26−2. 206

24. 897 in.

Taking moments about the bottom of the steel, we get

[9-31]

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

2. 0 −1. 239 0. 761 in.

Required Is50. 730/1240 124

9.6-5

(a) From Problem 9.4-2, a W14 22 is used, with t4 in., s8ft,L27 ft,

qconst 20 psf, qpart 20 psf, qL120 psf, A992 steel and 4 ksi concrete.

Before concrete cures:

Slab: 4

0. 85fc

′b

0. 854811. 178 in.

Y2t−a

∑A

12. 98 11. 98 in., ILB 540. 7 in.4

Δpart 5wpartL4

50. 160/1227 124

9.7-1

(a) Lower-bound moment of inertia:

0. 85fc

Use CV′515 kips.

From CT,0.85fc

0. 854a72515, Solution is: a2. 104 in.

Y2t−a

24. 5 −2. 104

23. 448 in.

∑A

20. 60 15. 0 in. ILB 1289 in.4

9.7-2

Steel headed stud anchors:

Asa 3/42

40. 441 8 in.2,Ecwc

1.5 fc´1451.5 43492 ksi

′Ec≤RgRpAsaFu

0.85 fc´bt 0. 854904. 5 −2765. 0 kips

Since AsFyis the smallest of the three possibilities, C735 kips, and there is full

0. 85fc

′b

0. 854902. 402 in.

Moment arm for concrete compressive force is

9.7-3

Asa 3/42

40. 441 8 in.2,Ecwc

1.5 fc´1451.5 43492 ksi

[9-35]

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

′Ec≤RgRpAsaFu

0.85 fc´bt 0. 854664. 5 −1. 5673. 2 kips

Since ∑Qnis the smallest of the three possibilities, C173.2 kips, there is partial

173. 2 Fybft′−FyAs−bft′0

173. 2 505. 03t′−507. 69 −5. 03t′0, Solution is: t′0. 420 1

Since tf0. 420 in., the PNA is at the bottom of the flange.

∑A

5. 577 9. 505 in.

0. 85fc

′b

0. 854660. 771 8 in.

Moment arm for concrete compressive force is

̄t−a

9.7-4

ThebeamisaW1840.

Asa 3/42

40. 441 8 in.2,Ecwc

1.5 fc´1451.5 43492 ksi

′Ec≤RgRpAsaFu

0.85 fc´bt 0. 8541204. 5 −1. 51224 kips

Since ∑Qnis the smallest of the three possibilities, C292.9 kips, there is partial

[9-37]

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

composite action, and the PNA is in the steel section. Determine whether the PNA is in

∑A

8. 829 11. 88 in.

0. 85fc

′b

0. 8541200. 717 9 in.

Moment arm for concrete compressive force is

̄t−a

211. 88 4. 5 −0. 7179

216. 02 in.

Moment arm for compressive force in the steel is

[9-38]

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

Before the concrete cures,

wu1. 20. 47131. 60. 2000. 885 6 kips/ft

Mu1

80. 8856402177 ft-kips

Ma1

80. 6713402134 ft-kips

Mnx

b

Mpx

b

196 ft-kips 134 ft-kips (OK)

After the concrete cures,

9.8-1

From the solution to Problem 9.7-3, for ¾-in. studs and fc´ 4ksi,Qn 17.23 kips

0.85 fc´bt 0. 854664. 5 −1. 5673. 2 kips

0. 85fc´b258. 5

0. 854661. 152 in.

Y2t−a

24. 5 −1. 152

23. 924 in.

9.8-2

0.85 fc´bt 0. 854905−2918. 0 kips

0. 85fc´b378. 4

0. 854901. 237 in.

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978-1337094740 Chapter 9 Part 2

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978-1337094740 Chapter 9 Part 2

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