978-1337094740 Chapter 10 Part 1

Type Homework Help

Pages 14

Words 2689

Textbook Steel Design 6th Edition

Authors William T. Segui

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CHAPTER 10 -PLATE GIRDERS

10.4-1

5. 70 29, 000

50 137. 3

Since 3. 76 E

h

5. 70 E

, the web is noncompact, and the provisions of

AISC F4 apply. However, the slender-web provisions of AISC F5 may be used (see

User Note in AISC F4).

Because the girder has continuous lateral support, lateral-torsional buckling does not

apply.

10.4-2

0. 5 140. 0 5. 70 E

50 137. 3

Since h

5. 70 E

, this is a slender-web shape (plate girder).

htf

21

10.4-3

60

3/8 160, 5. 70 E

5. 70 29, 000

50 137. 3

Since h

5. 70 E

, the web is slender.

htf

21

2tf

27/86. 857 p0. 38 E

50 9. 152

∴Fcr Fy50 ksi

Rpg 1−aw

1200 300aw

−5. 7 E

≤1. 0

awAw

603/8

(not to scale)

660

610 in., I1

12 103/831

12 7/8123126. 0 in.4

A103/8127/814. 25 in.2,rtI

A126

14. 25 2. 974 in.

Lb40/2 20 ft

0. 7Fy

0. 750268. 9 in. 22.41 ft.

Since LpLbLr,

Lb−Lp

22. 41 −6. 566 48. 47 ksi 50 ksi

where Cb1. 30 is from Figure 5.15 in the textbook.

10.4-4

52

1/4 208, 5. 70 E

5. 70 29, 000

50 137. 3

[10-4]

Since h

5. 70 E

, the web is slender.

htf

21

35. 0 16. 18

Since pr,

−p

16. 18 −9. 152 43. 92 50 ksi

8.667"

1/4"

(not to scale)

0. 7Fy

0. 750436. 1 in. 36.34 ft.

Since LpLbLr,

Lb−Lp

36. 34 −10. 65 54. 11 ksi 50 ksi

where Cb1. 30 is from Figure 5.15 in the textbook. Since Fcr Fy,use

Fcr Fy50 ksi

FLB controls and Fcr 42. 94 ksi. Compute the plate girder strength reduction factor.

10.4-5

Since h

5. 70 E

, the web is slender.

htf

21

2tf

22. 54. 4 p0. 38 E

50 9. 152

∴Fcr Fy50 ksi

Rpg 1−ar

1200 300ar

−5. 70 E

≤1. 0

awAw

780. 5

8wuL2PuL

41

84. 480276080

41. 872 104ft-kips

Since 18,700 ft-kips 17,700 ft-kips, flexural strength is not adequate

(b) ASD solution

b

1. 969 104

1. 67 1. 179 104ft-kips

10.5-1

(a) h

70

1. 10 kvE

1. 10 1029000

Cv1

1. 10 kvE

h/tw

83. 77

140 0. 598 4

1. 10 kvE

1. 10 5. 61329000

50 62. 76

Since h

62. 76, use AISC Equation G2-7 or G2-8

1. 15 1 a/h2

Compute Cv2.1.37

kvE

1. 37 5. 61329000

1. 15 1 2. 8572

Check the no-tension field case:

1. 10 kvE

62. 76

Since h/tw62. 76,

Cv1

1. 10 kvE

h/tw

62. 76

140 0. 448 3

Since h

61. 22, Cv1

1. 10 kvE

61. 22

10.5-2

(a) h

90

0. 90 533. 3 kips

Aw90 119/1651. 75 in.2

Vn0. 6AwFyCv10. 651. 7550Cv1533. 3, Solution is: Cv10. 343 5

1. 10 kvE

1. 10 5. 3429000

50 61. 22

Since h

61. 22, Cv1

1. 10 kvE

h/tw

61. 22

160 0. 382 6

Vn0. 6FyAwCv10. 65051. 750. 3826594. 0 kips 533.3 kips

No intermediate stifferers are required.

Since h

61. 22, Cv1

1. 10 kvE

61. 22

10.5-3

Before developing the LRFD and ASD solutions, compute the nominal shear strength

of each panel.

66

1. 10 kvE

1. 10 8. 97929000

50 79. 38

Since h

79. 38, Cv1

1. 10 kvE

h/tw

79. 38

211. 2 0. 375 9

1. 10 kvE

1. 10 8. 49029000

50 77. 19

[10-11]

1. 15 1 a/h2

Compute Cv2.1.37

kvE

1. 37 8. 49029000

1. 10 kvE

1. 10 5. 3429000

Cv1

1. 10 kvE

h/tw

61. 22

211. 2 0. 289 9

Vn0. 6FyAwCv10. 65021. 560. 2899187. 5 kips

(a) LRFD solution

51. 2 kips 169 kips (OK)

Girder has enough shear strength

(b) ASD solution

End panel: Allowable strength Vn

v

243. 1

1. 67 146 kips

10.5-4

1. 10 kvE

1. 10 18. 229, 000

Cv1

1. 10 kvE

h/tw

113

156 0. 724 4

1. 10 kvE

Fyw

1. 10 6. 46729000

50 67. 37

Since h

67. 37, use AISC Equation G2-7 or G2-8

1. 15 1 a/h2

Compute Cv2.1.37

kvE

1. 37 6. 46729000

1. 15 1 1. 8462

(a) Compute the factored-load shear at the beginning of each panel (this will be the

maximum shear in the panel).

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

The shear strength of all of the 12-ft panels will be the same, and the shear is less than

617 kips in each 12-ft panel. Therefore, there will be enough shear strength in all of the

12-ft panels.

The girder has enough shear strength.

(b) Compute the required strength at the beginning of each panel (this will be the

v

1. 67 540 kips 358 kips (OK)

First interior panel:

Allowable strength Vn

v

685. 4

1. 67 410 kips 346 kips (OK)

[10-15]

10.6-1

Bearing strength: Apb 6−0. 51/225. 5 in.2

3. 036 13. 83

From AISC J4.4, for compression elements with Lc/r25, the nominal strength is

RnFyAg368. 442303. 9 kips

10.6-2

Bearing strength: Apb 6−0. 59/16412. 38 in.2

Compressive strength: The maximum permissible length of web is

3. 492 7. 732

From AISC J4.4, for compression elements with Lc/r25, the nominal strength is

RnFyAg5013. 92696. 0 kips

10.7-1

Try tf¾ in., h73 −20. 7571. 5 in.

≥5. 70 E

5. 70 29, 000

∴tw≤h

137. 3 71. 5

137. 3 0. 520 8 in.

From h

≤11. 7 E

11. 7 29, 000

241. 4 71. 5

241. 4 0. 296 2 in.

Try a 5

0. 9hFy

6360012

0. 971. 550−22. 34

69. 703 in. 2

bf≥9. 703

0. 75 12. 94 in.

Try a 3

4-in. 14-in. flange, Af0. 751410. 5 in.2

htf

21

19. 24 −9. 152 49. 55 ksi

10. 5 2. 128 10

RPG 1−ar

1200 300ar

−5. 70 E

Fcr

≤1. 0

1−2. 128

1200 3002. 128228. 8 −5. 70 29, 000

49. 55 0. 894 8 1. 0

SxIx

cIx

h/2 tf43, 790

71. 5/2 0. 751200 in.3

Tension flange: MnSxReFyf 12001. 0506. 0 104in.-kips

Compression flange: Check FLB:

r−p

1. 0501−1

11. 67 −9. 152

19. 24 −9. 152 43. 76 ksi

13. 13 1. 701 10

RPG 1−ar

1200 300ar

−5. 70 E

Fcr

≤1. 0

10.7-2

(a) Try tf2.5 in., h78 −22. 573. 0 in.

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

Try a 2. 5-in. 23-in. flange. Af232. 557. 5 in.2

Check web width-thickness ratio:

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978-1337094740 Chapter 10 Part 1

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978-1337094740 Chapter 10 Part 1

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