978-1337094740 Chapter 5 Part 2

Type Homework Help

Pages 14

Words 2959

Textbook Steel Design 6th Edition

Authors William T. Segui

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Cb1.67

5.5-11

=15'

26.41 k 29.29 k

- 29.29 k

- 8.59 k

- 15.29 k

Mmax 26. 4118. 86−1. 418. 862/2 249. 1 ft-kips

2. 5249. 13230. 84247. 83245. 11. 024

Cb1.02

=15'

38.57 k

- 42.87 k

Mmax 38. 5718. 91−2. 0418. 912/2 364. 6 ft-kips

2. 5Mmax 3MA4MB3MC

12. 5364. 6

5.5-12

First, determine the nominal flexural strength. Verify that this shape is

compact. For the flange,

p0. 38 29, 000

6. 04 p∴flange

0. 750

14020. 42. 451. 0

14020. 4

29000

18. 73 −6. 358 .

7078 in.-kips 589.8 ft-kips

1. 2wD1. 6wL1. 20. 0681. 6wL1. 669, Solution is:

wL0. 992 1

[5-23]

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

1. 130 −0. 068 1. 06 kips/ft

wL1. 06 kips/ft

5.5-13

Verify that this shape is compact. For the flange,

p0. 38 29, 000

50 9. 15, bf

2tf

6. 57 p∴flange

54. 6 3. 319 in.2

rts 3. 319 1. 822 in.

(rts can also be found in the dimensions and properties tables.

Lr1. 95rts E

0. 7Fy

Sxho

Jc

Sxho

1. 951. 82229000

0. 750

0. 7981. 0

54. 813. 590. 7981. 0

54. 813. 59

6. 76 0. 750

29000

195. 0 in. 16. 26 ft

The unbraced length is14 ft. For Lb 14 ft, Lp Lb Lr,so

16. 26 −5. 475

2844 in.-kips 237.0 ft-kips Mp

(a) LRFD solution:

bMn0. 90237. 0213 ft-kips

5.5-14

75. 0 1. 830 in.2

rts 1. 830 1. 353 in.

cho

17. 38

17. 6

0. 736

75. 017. 382. 811. 115

75. 017. 38

29000

[5-26]

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

229. 3 in. 19. 11 ft

19. 11 −4. 246

3195 in.-kips 266.3 ft-kips Mp

(a) LRFD solution:

bMn0. 90266. 3240 ft-kips

5.5-15

(a) LRFD solution: Pu1. 2PD1. 6PL1. 2121. 636

72. 0 kips

10'

6.125 kips/ft

ABC

5' 5' 5' 5'

139.9 k 115.9 k

78.65 k

115.9 k

1.086'

Mu139. 911. 086−6. 12511. 0862/2 −72. 01. 086

1096 ft-kips

Verify that this shape is compact. For the flange,

p0. 38 29, 000

8. 50 p∴flange

2IyCw

Lr1. 95rts E

0. 7Fy

Sxho

Jc

Sxho

1. 953. 42129000

0. 750

4. 721. 0

25823. 354. 721. 0

25823. 35

6. 76 0. 750

29000

350. 4 in. 29. 20 ft

For Lb20 ft, LpLbLr,so

2. 51096310494852. 83502. 91. 268

29. 20 −10. 28

14, 790 in.-kips Mp

[5-29]

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

10'

4.104 kips/ft

ABC

5' 5' 5' 5'

93.56 k 77.56 k

52.52 k

-77.56 k

1.101'

From the LRFD solution, the shape is compact. Also from the LRFD

solution,

Lp10.28 ft., Lr29. 20 ft

2. 5732. 93701. 74570. 43336. 51. 267

(Note that this is essentially the same as the value found in Part (a) with

29. 20 −10. 28

14, 780 in.-kips Mp

5.5-16

(a) For segment ABC, when beam weight is included,

wD3. 5 0. 132 3. 632 kips/ft

wu1. 2wD1. 6wL1. 23. 6321. 61. 05. 958 kips/ft

For segment CD,wu1. 20. 1320. 158 4 kips/ft

0. 750

20913. 712. 31. 0

20913. 7

29000

[5-32]

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

670. 2 in. 55. 85 ft

Check unbraced segment BC.ForLb9ft,LbLpand

2. 5Mmax 3Ma4Mb3Mc

12. 5855. 4

56. 02 −13. 28

1. 854 104in.-kips 1545 ft-kips Mp

17.11'

-30.63 k

-28.25 k

From Part (a), Lp13. 28 ft, Lr55. 85 ft, Mp11, 700 in.-kips 

975.0 ft-kips.

Check unbraced segment BC.ForLb9ft,LbLpand

MnMp975 ft-kips

b

975

1. 67 584 ft-kips

2. 5530. 03401. 54270. 33136. 51. 648

Mn1. 648 11700 −11700 −0. 7 50 20918 −13. 28

56. 02 −13. 28

1. 848 104in.-kips 1540 ft-kips Mp

5.6-1

Check for compactness. For the flange,

24. 08 −9. 152

4749 in.-kips 395. 8 ft-kips

(a) Available strength design strength bMn

1. 2wD1. 6wL1. 2wD1. 63wD1. 140

wD0. 1900 kips/ft

wwDwL0. 19 30. 190. 760 kips/ft

5.6-2

Check for compactness. For the flange,

24. 08 −9. 152

[5-36]

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

5.6-3

Verify that the shape is noncompact: For the flange,

bf

18

0. 75018. 95

90. 55 ∴web is compact.

Compute properties of the cross section.

19. 5 13. 5 18. 47 in.

ZxA

2a19. 5 13. 5218. 471219 in.3

[5-37]

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

18. 95 −9. 152

5.6-4

Because the beam has continuous lateral support, lateral-torsional

buckling does not need to be checked. Verify that the shape is

0. 75018. 29

pr∴flange is noncompact.

For the web, h

40

12. 0 10. 0 15. 66 in.

ZxA

2a12. 0 10. 0215. 66689. 0 in.3

MpFyZx50689. 034, 450 in.-kips

18. 29 −9. 152

5.8-1

71. 0, 2. 24 E

2. 24 29000

53. 69, ∴Cv1

1. 10 kvE



1. 10 5. 3429000

5.8-2

For an M12 11.8 (tf0. 225 in.) of A572 Grade 50, Fy50 ksi

62. 5, 2. 24 E

2. 24 29000

61. 22, ∴Cv1

1. 10 kvE

61. 22

5.8-3

(a) Pu1. 6PL1. 66096 kips

1' 1'

96 k 96 k

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978-1337094740 Chapter 5 Part 2

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978-1337094740 Chapter 5 Part 2

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