978-1337094740 Chapter 10 Part 2

Type Homework Help

Pages 13

Words 926

Textbook Steel Design 6th Edition

Authors William T. Segui

Unlock document.

This document is partially blurred.

Unlock all pages and 1 million more documents.

Get Access

modulus:

Ix1

12 twh32Af

htf

21

12 3/8733257. 573 2. 5

awhctw

bfctfc

733/8

232. 50. 476 1 10

From AISC Equation F5-6,

Rpg 1−aw

−5. 7 E

≤1. 0

0. 7Fy

0. 750577. 9 in. 48.16 ft

[10-21]

48. 16 −14. 11 75. 49 50 ksi ∴use 50 ksi

MnRpgFcrSxc 0. 9796504513/12 1. 842 104ft-kips

bMn0. 90184201. 658 104ft-kips

Mu17000 ft-kips 16600 ft-kips (N.G.)

Try 2 1

cIx

h/2 tf183200

73/2 2. 54697 in.3

To compute the bending strength reduction factor Rpg, the value of awwill be needed:

awhctw

bfctfc

733/8

242. 50. 456 3 10

673

612. 17 in., I1

12 12. 173/831

12 2. 52432880 in.4

[10-22]

A242. 512. 173/864. 56 in.2,rtI

A2880

64. 56 6. 679 in.

Lb25. 0 ft

0. 7Fy

0. 750604. 0 in. 50.33 ft

Since LpLbLr, the girder is subject to inelastic lateral-torsional buckling. From

50. 33 −14. 74 76. 28 50 ksi ∴use 50 ksi

MnRpgFcrSxc 0. 9804504697/12 1. 919 104ft-kips

bMn0. 901. 919 1041. 727 104ft-kips

137. 3 73

137. 3 0. 531 7 in.

For a

≤12. 0 E

12. 0 29000

232. 0 73

232. 0 0. 314 7 in.

Try a 3

8in. 73 in. web.

73

2. 5 23. 52 in.

Try a 2. 5-in. 25-in. flange. Af252. 562. 5 in.2

Check web width-thickness ratio:

h

194. 7, p3. 76 E

3. 76 29, 000

50 90. 55

r5. 70 E

5. 70 29000

cIx

h/2 tf190300

73/2 2. 54879 in.3

From AISC Equation F5-10, the tension flange strength based on yielding is

MnFySxt 5048792. 440 105in.-kips 20330 ft-kips

The compression flange strength is given by AISC Equation F5-7:

For flange local buckling, the relevant slenderness parameters are

bf

2tf

25

22. 55. 0 p0. 38 E

0. 38 29000

50 9. 152

Since p, the flange is compact and Fcr Fy50 ksi

[10-24]

0. 7Fy

0. 750630. 0 in. 52.5 ft

Since LpLbLr, the girder is subject to inelastic lateral-torsional buckling. From

52. 5 −15. 38 77. 01 50 ksi ∴use 50 ksi

MnRpgFcrSxc 0. 9811504879

10.7-3

Assume a girder weight of 160 lb/ft.

wu1. 2wD1. 6wL1. 20. 5 0. 1600. 792 kips/ft

Pu1. 6PL1. 6115184. 0 kips

137. 3 52

137. 3 0. 378 7 in.

From h

≤12 E

12 29000

50 289. 0

and h

≤0. 40E

0. 4029000

Cv1

1. 10 kvE

61. 22

1. 5 6. 419 in.

Try a 1 1

20. 63

13. 5 1. 528 10 (OK)

Girder weight 20. 63 213. 50. 490

144

0. 162 1 kips/ft ≈0.160 kips/ft (Say OK)

htf

21

8.667"

3/8"

(not to scale)

0. 7Fy

0. 750211. 0 in. 17.58 ft

Since LpLbLr,

Lb−Lp

17. 58 −5. 15 41. 13 ksi 50 ksi

13. 5 1. 528 10

Rpg 1−ar

1200 300ar

−5. 7 E

≤1. 0

10.7-4

(a) At the support, VuVLwuL

2Pu248

2120 168 kips

137. 3 46

137. 3 0. 335 in.

[10-28]

232. 0 46

232. 0 0. 198 3 in.

Try a 1

4in. 46 in. web.

46

1/4 184. 0, Aw1/446 2212. 5 in.2

12 twh32Af

12 1/446321846 2

2. 276 104in.4

Sxt Sxc Ix

cIx

h/2 tf2. 276 104

46/2 2910. 4 in.3

0. 7Fy

0. 750223. 4 in. 18.62 ft

[10-29]

18. 62 −5. 453 37. 98 ksi (controls)

(Cb1. 0 is a conservative estimate.)

Compute the plate girder strength reduction factor.

Rpg 1−aw

−5. 7 E

≤1. 0

Cv1

1. 10 kvE

h/tw

2h/tw2Fy

1. 10 kvE

1. 10 9. 59629000

12. 5 12. 8 ksi

h/tw184.

For vVn

12 ksi, a

15 ksi, a

8. 438 1. 896 ksi

For a

1. 10 5. 3429000

50 61. 22

Since h

184. 0 61. 22,

10.7-5

From Problem 10.7-4, use a 1

446 web and 2 9 flanges. Reaction  168 kips

0. 56

E4

0. 56

29000 0. 296 6 in.

Try two plates, 5

16 in. 4 in., with 1-in. cutouts.

1. 8Fy

1. 8502. 489 in.2

From Apb 2b−1t,tApb/2b−12. 489

24−10. 414 8 in.

Try 7/16 4 plates.

2. 195 15. 72

Since Lc/r25, use Fcr Fy.

10.7-6

(a) wu1. 2wD1. 6wL1. 21. 01. 624. 4 kips/ft

Pu1. 6PL1. 6475760. 0 kips

Left reaction VLwuLPu

24. 480760

2556. 0 kips

htf

21

For a 1.5-in. length, Rn1. 513. 9220. 88 kips

Required spacing:

Rn

20. 88

sVuQ

13. 5 Vu2214

198000 Vu138. 3 kips

Shear at mid-span, left of load, 556 −4. 440380. 0 kips, so maximum spacing

will never be used.

Spacing required at mid-span RnIx

VuQ20. 88198000

38022144. 914 in.

Ix1

12 twh32Af

htf

12 0. 578322. 5 2278 2. 5

1. 980 105in.4

Use E70 electrodes, Rn/0. 9279Dkips/in., where Dweld size in sixteenths.

Try 5

16 -in. 11

2-in. intermittent fillet welds. For two welds,

Weld strength 20. 927959. 279 kips/in.

Base metal shear yield strength 0. 4tFy0. 40. 55010. 0 kips/in.

For a 1.5-in. length, Rn/1. 59. 27913. 92 kips

Required spacing:

Rn/

13. 92

sVaQ

13. 5 Va2214

198000 Va92. 21 kips

Shear at mid-span, left of load, 357. 5 −340237. 5 kips, so maximum spacing

will never be used.

10.7-7

(a) Assume a girder weight of 200 lb/ft.

wu1. 2wD1. 6wL1. 21. 3 0. 2001. 62. 35. 48 kips/ft

137. 3 77

137. 3 0. 560 8 in.

From h

≤12. 0 E

12. 0 29000

232. 0 77. 0

232. 0 0. 331 9 in.

0. 5 154. 0, Aw0. 577 21. 540. 0 in.2

0. 9hFy

6544712

0. 97750−40. 0

612. 20 in. 2

bf≥12. 20

1. 5 8. 133 in.

Try a 1 1

2-in. 81

4-in. flange, Af1. 58. 2512. 38 in.2

Girder weight 40. 0 212. 380. 490

1. 5 8. 207 in.

Try a 1 1

2-in. 81

4-in. flange

Weight is the same as before: 0.2204 kips/ft 0.25 kips/ft assumed (OK)

Af1. 58. 2512. 38 in.2

htf

21

Compression flange: Compute the plate girder strength reduction factor.

Rpg 1−aw

1200 300aw

−5. 7 E

≤1. 0

awhctw

bfctfc

770. 5

8. 251. 53. 111 10

0. 7Fy

0. 750175. 0 in. 14.58 ft

Since LbLr,

Try a 0.5 77 web and a 1.5 14 flange. (Preliminary trials not shown.)

Af1. 51421. 0 in.2

htf

21

Rpg 1−1. 833

1200 3001. 833154 −5. 7 29000

50 0. 982 5 1. 0

Check FLB:

bf

2tf

14

21. 54. 667, p0. 38 E

0. 38 29, 000

50 9. 152

0. 7Fy

0. 750319. 9 in. 26.66 ft

Since LpLbLr,

Lb−Lp

26. 66 −7. 808 38. 71 ksi (controls)

(Cb1. 0 is a conservative estimate.)

MnRpgFcrSxc 0. 982538. 7120937. 96 104in.-kips

bMn0. 907. 96 104/12 5970 ft-kips

Trusted by Thousands of
Students

Here are what students say about us.

Albert

University of Michigan

“I found almost every finance case study paper for my MBA courses.”.

Anna

University of Massachutsetts

“Wow! Solution manual for 3 out of 4 courses.”.

Collins

Jacksonville State University

“One-stop shop for college students. I passed all my exams thanks to Coursepaper”.

Jill

Boston University

“A helpful studying resources, a combination of all studying material in one place”.

Drake

Clark Atlanta University

“I graduated thanks to Coursepaper”.

Karen

College of Charleston

“I invested in Coursepaper, and it is paid off after the first semester. I got straight A”.

Hill

Concordia University Irvine

“Awesome awesome awesome site”.

Rachel

Coppin State University

“The one website that I recommend to every college students”.

978-1337094740 Chapter 10 Part 2

Unlock document.

Trusted by Thousands of
Students

Albert

Anna

Collins

Jill

Drake

Karen

Hill

Rachel

Kristopher

Resources

Company

Legal

978-1337094740 Chapter 10 Part 2

Unlock document.

Trusted by Thousands ofStudents

Albert

Anna

Collins

Jill

Drake

Karen

Hill

Rachel

Kristopher

Resources

Company

Legal

Trusted by Thousands of
Students