978-1337094740 Chapter 10 Part 3

Type Homework Help

Pages 13

Words 1225

Textbook Steel Design 6th Edition

Authors William T. Segui

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Maximum stiffener width bf−tw

214 −0. 5

26. 75 in.

Compute the compressive strength:

LcKL 0. 757757. 75 in.

r57. 75

3. 008 19. 20

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

RnFyAg509. 0450. 0 kips

cRn0. 90450. 0405. 0 kips 294. 8 kips (OK)

Because the stiffener was sized for dimensional constraints, it cannot be made any

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

Required spacing:

Rn

20. 88

sVuQ

13. 5 Vu824. 3

83730 ,Solutionis:Vu157. 1 kips

Shear at first interior load, left of load, 295. 8 −5. 56922173. 3 kips, so

20. 88

10 Vu824. 3

83730 ,Solutionis:Vu212. 1 kips

This occurs where 295. 8 −5. 569x212. 1, Solution is: x15. 03 ft.

Check middle third of span. Shear on right side of load 173. 3 −112. 0 61. 3 kips

sRnIx

2in.

Welds for bearing stiffeners at the supports (1/2 6):

Minimum weld size 3

16 in. (AISC Table J2.4, based on a thickness of 1/2 in.)

Minimum length 43

16 0. 75 in. 1.5 in., use 1.5 in.

25. 05

s3. 944

s6. 351 in.

For a center-to-center spacing of 6 1

4in., the clear spacing is 6. 25 −1. 5 4. 75 in

25. 05

16. 75 −1. 5 15. 25 in.9 in. max. (9 1. 5 10. 5 in. c-c)

10.7-8

(a) Assume a girder weight of 600 lb/ft.

wu1. 2wD1. 6wL1. 20. 3 0. 61. 60. 72. 2 kips/ft

Pu1. 2PD1. 6PL1. 2501. 6150300. 0 kips

137. 3 115

137. 3 0. 837 6 in.

[10-43]

232. 0 115

232. 0 0. 495 7 in.

Try a 3

2. 5 10. 46 in.

Try a 2 1

2in. 15 1

2in. flange, Af2. 515. 538. 75 in.2

Girder weight 0. 75115238. 750. 490

144

Tension flange: MnFySxt 5060433. 022 105in.-kips

Compression flange: Check FLB:

bf

2tf

15. 5

22. 53. 1, p0. 38 E

0. 38 29, 000

50 9. 152

Since p,Fcr Fy50 ksi

0. 7Fy

0. 750345. 7 in. 28.81 ft

Since LpLbLr,

Fcr CbFy−0. 3Fy

Lb−Lp

≤Fy

2. 5Mmax 3MA4MB3MC

12. 517750

28. 81 −8. 442 41. 70 ksi (controls)

Compute the bending strength reduction factor.

Rpg 1−aw

1200 300aw

−5. 7 E

≤1. 0

1. 10 kvE

1. 10 5. 73529000

50 63. 44

1. 10 kvE

153. 3 0. 413 8

Vn0. 6FyAwCv10. 650900. 41381117 kips

vVn0. 9011171005 kips 560 kips (OK)

No intermediate stiffeners (other than the bearing stiffeners) are required.

(b) Intermediate stiffeners: Intermediate stiffeners are not required.

Rn0. 75675506. 3 kips VL560 kips (N.G.)

Try two plates, 3

4in. 7 in., with 1-in. cutouts.

Apb 7−10. 7529. 0 in.2

Rn1. 8FyApb 1. 8509810. 0 kips

3. 413 25. 27

4. 71 E

4. 71 29000

50 113. 4

Rn0. 90FcrAg0. 9047. 7217. 25740. 9 kips 560 kips (OK)

Use 2 PL 3

47 with 1-in. cutouts for bearing stiffeners at the supports.

2. 083 41. 41

4. 71 E

4. 71 29000

50 113. 4

Rn0. 90FcrAg0. 9044. 1120. 06796. 4 kips 300 kips (OK)

3. 626 1053. 517 kips/in.

Minimum weld size 1

4in. (AISC Table J2.4)

Minimum length 41

For a 1.5-in. length, Rn1. 511. 1416. 71 kips

Required spacing:

Rn

16. 71

sVuQ

13. 5 Vu2277

3. 626 105,Solutionis:Vu197. 1 kips

Shear at first interior load, left of load, 560 −2. 225505. 0 kips, so maximum

spacing will not be used in the first third of the span.

Spacing required at left side of first interior load:

RnIx

4in. 11

2in. intermittent E70 fillet welds.

67 spaces at 4 1

2in. each end.

Spacing between interior loads 12 1

2in.

33. 41

s4. 956, Solution is: 6. 74 in.

For a center-to-center spacing of 6 1

2in., the clear spacing is 6. 5 −1. 5 5. 0 in.

25. 05

s2. 655, Solution is: s9. 435 kips/in.

For a center-to-center spacing of 9 in., the clear spacing is

10.7-9

(a) Assume a girder weight of 875 lb/ft.

wu1. 2wD1. 6wL1. 21. 9 0. 8751. 62. 87. 81 kips/ft

137. 3 114

137. 3 0. 830 3 in.

From h

≤11. 7 E

12. 0 29000

Try a 0. 75 in. 114 in. web.

114

0. 75 152. 0, Aw0. 75114 2390. 0 in.2

0. 9hFy

63823012

0. 911450−90

674. 43 in. 2

bf≥74. 43

324. 81 in.

Trya3in.28 in. flange, Af32884. 0 in.2

[10-51]

cIx

h/2 tf667500

114/2 31. 113 104in.3

Tension flange: MnFySxt 501. 113 1045. 565 105 in.-kips

Compression flange: Check FLB:

bf

28

0. 38 29, 000

Compute the bending strength reduction factor.

Rpg 1−aw

1200 300aw

−5. 7 E

≤1. 0

awhctw

bfctfc

1140. 75

2831. 018 10

152 61. 22 ∴Cv1

1. 10 kvE

61. 22

152 0. 402 8

Vn0. 6FyAwCv10. 650900. 40281088 kips

h4st

1.3

Fyw

1.5

Ist h4st

1.3

Fyw

1.5 11441

29000

1.5 302. 3 in.4

Try two 5/16 12 plates.

16 in. 12 in. 9 ft-2 in. for intermediate stiffeners.

Maximum stiffener width bf−tw

228 −3/4

213. 63 in.

Try b7 in. and t≥1. 79bFy

A18. 753/4279/1621. 94 in.2

rI

A151. 1

21. 94 2. 624 in.

[10-54]

Compute the compressive strength:

2. 624 32. 58

4. 71 E

4. 71 29000

50 113. 4

6. 675 1057. 899 kips/in.

Minimum weld size 1

4in. (AISC Table J2.4)

Minimum length 41

41. 0 in. 1.5 in., use 1.5 in.

Shear at first interior load, left of load, 1073 −8. 3730821. 9 kips, so the

maximum spacing will not be used in the first quarter of the span.

Spacing required at left side of first interior load:

sVuQ

13. 5 Vu4914

6. 675 105Vu169. 1 kips

Let Vu1073 −7. 81x−403. 2 169. 1 x64. 11 ft

Distance past first interior load 64. 11 −30 34. 11 ft (slighty past midspan).

Try an intermediate spacing of 10 inches. This will be required where

VuRnIx

and from C to D.

Welds for intermediate stiffeners 5

16 in. 12 in. :

Minimum weld size 3

16 in. (AISC Table J2.4)

25. 05

s10. 65 kips/in. s2. 35 in.

For a center-to-center spacing of 2 1

4in., the clear spacing is 3 −2. 25 0. 75 in.

Use continuous 3

22. 27

s3. 6 kips/in. s6. 186 in.

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978-1337094740 Chapter 10 Part 3

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978-1337094740 Chapter 10 Part 3

Unlock document.

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Trusted by Thousands of
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