978-1337094740 Chapter 8 Part 3

Type Homework Help

Pages 14

Words 3162

Textbook Steel Design 6th Edition

Authors William T. Segui

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f2xMy

J68. 991. 5

28. 67 3. 610 kips/in. 

8.4-8

2.50"

3.5"

Load when eccentricity is considered: Locate centroid with respect to right side.

̄632

19 1. 895 in., y

̄3. 5 in.

8.4-9

(a) LRFD Solution

Compute the tension member available strength. For the gross section:

tPn0. 90FyAg0. 90364. 38141. 9 kips

6. 96 82. 71

6. 96 11. 9 in. Use 12 in.

L2P2

6. 96 17. 43

6. 96 2. 50 in. Use 2½ in.

For the second option, the strength of the longitudinal welds is 0. 856. 965. 916

[8-42]

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

5. 916 6. 99

5. 916 1. 182 in. Use 1½ in.

(Min. length 4w45/161. 25 in.)

The total length of weld is nearly the same for both options. Try the first option and

check the value of U.

U1−x

2½

(not to scale)

(b) ASD Solution

Compute the tension member available strength. For the gross section:

FyAg

364. 38

1. 67 94. 42 kips

Net section: Assume that U0.80. Check once the connection is designed.

AeAgU4. 380. 803. 504 in.2

4. 640 55. 01

4. 640 11. 86 in. Use 12 in.

L2P2

4. 640 11. 59

4. 640 2. 498 in. Use 2½ in.

For the second option, the strength of the longitudinal welds is 0. 854. 6403. 944

kips/in.

3. 944 4. 61

3. 944 1. 169 in. Use 1½ in.

(Min. length 4w45/161. 25 in.)

The total length of weld is nearly the same for both options. Try the first option and

check the value of U.

8.4-10

Use Table 8-10, Angle 0°

0.8 0.875 0.9

C1.671 0.9361 1.0065 1.0299

1.8 0.871 0.957

C1. 006 by interpolation.

If the 9-kip load is a factored load,

8.4-11

Consider the weld on one angle. Use Table 8-8, Angle 0°

kℓ5 in., kkℓ

3. 00 −2. 80 3. 068 −3. 001. 096

(a) LRFD solution

Pu1. 2D1. 6L1. 20. 25 101. 60. 75 1015. 0 kips

For one angle, Pu15

0.884"

1.5"

Use Table 8-4, Angle 0°

0.7 0.75 0.8

C0.154 3.5488 3.5392 3.5296

0.2 3.42 3.41

[8-46]

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

C3. 540 by interpolation. If the 112-kip load is treated as a factored load,

8.4-13

(a) Pu1. 2D1. 6L1. 2181. 64390. 4 kips

Pux 90. 4 sin 15 ° 23. 40 kips Puy90. 4 cos 15 ° 87. 32 kips ↓

f1x23. 40

212 40. 731 3 kips/in. f1y87. 32

32 2. 729 kips/in. ↓

1. 392 3. 27 sixteenths. Use a ¼-in. fillet weld.

[8-47]

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

(b) Use Table 8-6, Angle 15°

0.3 0.3333 0.4

C0.241 4.6532 4.7982 5.0886

0.25 4.61 5.04

C4. 80 by interpolation

8.4-14

(a) PaDL18 43 61 kips

Pax 61sin15° 15. 79 kips Pay 61 cos 15 ° 58. 92 kips ↓

f1x15. 79

212 40. 493 4 kips/in. f1y58. 92

32 1. 841 kips/in. ↓

0. 9279 4. 42 sixteenths. Use a 5

(b) Use Table 8-6, Angle 15°

0.2 4.85 5.31

0.25 4.61 5.04

C4. 80 by interpolation

CC1ℓ261

4. 801122. 12 sixteenths Use a 3

8.4-15

(a) Pu1. 2D1. 6L1. 2311. 63186. 8 kips

For t5/16 in., min. w3/16 in. and max w5/16 −1/16 1/4 in.

4. 176 20. 79 in.

Length of longitudinal welds 20. 79 −6

3. 55026. 93 in.; use 7 in.

The second option requires shorter longitudinal welds:

86.8 k

1.60"

4. 176 10. 62

4. 176 2. 54 in. Use 3 in.

For the second option, the strength of the longitudinal welds is

0. 854. 1763. 550 kips/in.

and the strength of the transverse weld is 1. 54. 1766. 264 kips/in.

3. 550 4. 36

3. 550 1. 23 in. Use 1½ in. Use the weld shown.

[8-50]

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

8.4-16

(a) PaDL31 31 62 kips

For t5/16 in., min. w3/16 in. and max w5/16 −1/16 1/4 in.

2. 784 22. 27 in.

0. 852. 7842. 366 kips/in.

2. 36627. 81 in.; use 8 in.

The second option requires shorter longitudinal welds:

1.60"

2. 784 37. 12

2. 784 13. 3 in. Use 13½ in.

L2P2

0. 852. 7842. 366 kips/in.

and the strength of the transverse weld is 1. 52. 7844. 176 kips/in.

2. 366 32. 94

2. 366 13. 9 in. Use 14 in.

L2P2

2. 366 4. 00

2. 366 1. 69 in. Use 2 in. Use the weld shown.

8.4-17

(a) Pu1. 2D1. 6L1. 2401. 680176. 0 kips

For t3/8 in., min. w3/16 in. and max w5/8 −1/16 9/16 in.

4. 176 42. 15 in.

0. 854. 1763. 550 kips/in.

3. 550220. 4 in.; use 20½ in.

1.47"

176 k

(Not to scale)

(b) f1x176

21954. 093 kips/in. 

Locate centroid with respect to left side.

4. 176 113. 8

4. 176 27. 3 in. Use 27½ in.

L2P2

4. 176 41. 3

4. 176 9. 89 in. Use 10 in.

Option 2: The strength of the longitudinal welds is 0. 854. 1763. 550 kips/in.

3. 550 108. 6

3. 550 30. 6 in. Use 31 in.

3. 550 36. 08

3. 550 10. 2 in. Use 10½ in.

Option 1 results in a shorter connection. Use the connection shown

10½

(not to scale)

6. 96 25. 29 in.

0. 856. 965. 916 kips/in.

5. 916210. 5 in.; use 10½ in.

1.47"

176 k

[8-55]

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

(b) f1x176

210. 556. 769 kips/in. 

Locate centroid with respect to left side.

6. 96 106. 9

6. 96 15. 4 in. Use 15½ in.

L2P2

6. 96 34. 34

6. 96 4. 93 in. Use 5 in.

Option 2: The strength of the longitudinal welds is 0. 856. 965. 916 kips/in.

[8-56]

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

5. 916 25. 64

5. 916 4. 33 in. Use 4½ in.

Option 1 results in a shorter connection. Use the connection shown

(not to scale)

15½

8.4-18

(a) Pa40 80 120 kips

For t3/8 in., min. w3/16 in. and max w5/8 −1/16 9/16 in.

2. 784 43. 1 in.

Length of longitudinal welds 43. 1 −5

0. 852. 7842. 366 kips/in.

2. 366220. 95 in.; use 21 in.

Use 3/16-in. E70 fillet welds, across the end and 21 inches on each side

(b) f1x120

21952. 791 kips/in. 

Locate centroid with respect to left side.

2. 784 77. 76

2. 784 27. 93 in. Use 28 in.

L2P2

4. 176 28. 32

2. 784 10. 17 in. Use 10½ in.

Option 2: The strength of the longitudinal welds is 0. 852. 7842. 366 kips/in.

[8-58]

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

2. 366 24. 84

2. 366 10. 50 in. Use 10½ in.

Option 1 results in a shorter connection.

Use 3/16-in. welds: 5-in. transverse, 28-in. longitudinal bottom and 10½-in. longitudinal top.

4. 640 25. 86 in.

0. 854. 6403. 944 kips/in.

3. 944210. 8 in.; use 11 in.

Use 5/16-in. E70 fillet welds, across the end and 10½ inches on each side

(b) f1x120

210. 554. 615 kips/in. 

Locate centroid with respect to left side.

4. 64 73. 12

4. 64 15. 76 in. Use 16 in.

L2P2

4. 64 23. 68

4. 64 5. 103 in. Use 5½ in.

Option 2: The strength of the longitudinal welds is 0. 854. 643. 944 kips/in.

3. 944 17. 88

3. 944 4. 533 in. Use 4½ in.

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978-1337094740 Chapter 8 Part 3

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978-1337094740 Chapter 8 Part 3

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