978-1337094740 Chapter 7 Part 4

will be used. To determine the required length of the longitudinal welds, investigate

3. 712 28. 29 in.

1. 53. 7125. 568 kips/in.

77. 16

23. 15512. 23 in.

The first option requires slightly shorter longitudinal welds. Try a 5-inch transverse

0. 6FyAgv UbsFuAnt 0. 63691. 0581. 875

303. 2 kips (controls)

7.11–6

From AISC Table J2.4, the minimum weld size is 1/8 inch (based on the angle

thickness). Maximum size 1/4 −1/16 3/16 in.

0. 6Fyt0. 6363

16 4. 05 kips/in.

2. 784 10. 2 in.

length of each longitudinal weld 10. 2

25. 1 in.

Use 1/8-in. fillet welds as shown.

0. 4Fyt0. 4363

0. 3Fut0. 3583

1. 856 10. 51 in.

length of longitudinal welds 10. 51

7.11–7

From AISC Table J2.4, the minimum weld size is 3/16 inch (based on the gusset plate

0. 6Fyt0. 6363

88. 1 kips/in.

Shear rupture strength is 0. 45Fut0. 45583

4. 176 59. 77 in.

0. 854. 1763. 550 kips/in.

and the strength of the transverse weld is

249. 6 −96. 264193. 2 kips

so the required length of the longitudinal welds is

193. 2

23. 55027. 21 in.

6. 96 35. 86 in.

1. 56. 9610. 44 kips/in.

The load to be carried by the longitudinal welds is

155. 6

25. 91613. 15 in.

Each option requires about the same length longitudinal welds. Try a 9-inch transverse

0. 6FyAgv UbsFuAnt 0. 63610. 131. 0583. 375414. 6 kips (controls)

The design strength is

Rn0. 75414. 6311. 0 kips 249.6 kips (OK)

0. 4Fyt0. 4363

0. 3Fut0. 3583

86. 525 kips/in.

4. 640 36. 21 in.

0. 854. 6403. 944 kips/in.

and the strength of the transverse weld is

1. 54. 6406. 96 kips/in.

The load to be carried by the longitudinal welds is

105. 4

23. 94413. 36 in.

Each option requires about the same length longitudinal welds. Try a 9-inch transverse

7.11–8

Tension member gross section: tPn0. 90FyAg0. 90361. 9362. 53 kips

For the net section, assume U0. 80: AeAgU1. 930. 801. 544 in.2

0. 6Fyt0. 6361

45. 4 kips/in.

4. 176 14. 97 in.

Check assumed value of U:

̄

7.11–9

0. 5Fu

0. 5580. 828 in.2

Min. rL

300 1212

300 0. 48 in.

0. 4Fyt0. 4361

43. 6 kips/in.

1. 856 12. 93 in., try two 6½-in. longitudinal welds.

Check assumed value of U:

̄

7.11–10

(a) LRFD solution

0. 9Fy

0. 9504. 28 in.2

Req’d AePu

0. 75Fu

192. 8

0. 75653. 96 in.2

Min. rL

300 17. 512

300 0. 7 in.

Try C10 15.3, Ag4. 48 in.24.28 in.2(OK)

0. 6Fyt0. 6363

88. 1 kips/in.

Shear rupture strength is 0. 45Fut0. 45583

2. 784 69. 25 in.

1. 52. 7844. 176 kips/in.

The load to be carried by the longitudinal welds is

151. 0

22. 36631. 91 in

The first option results in a slightly shorter connection. Use a transverse weld and two

30-inch longitudinal welds. Check assumed value of U:

̄

0. 6FyAgv UbsFuAnt 0. 63622. 51. 0583. 75703. 5 kips

Use Rn703. 5 kips, Rn0. 75703. 5527. 6 kips 192. 8 kips (OK)

[7-60]

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

0. 6Fy

0. 6504. 93 in.2

Req’d AePa

0. 5Fu

147. 8

0. 5654. 55 in.2

Min. rL

300 17. 512

300 0. 7 in.

Try C12 20.7, Ag6. 08 in.24.93 in.2(OK)

0. 4Fyt0. 4363

85. 4 kips/in.

Shear rupture strength is 0. 3Fut0. 3583

1. 856 79. 63 in.

length of longitudinal welds 79. 63 −12. 0

1. 51. 8562. 784 kips/in.

The load to be carried by the longitudinal welds is

114. 4

21. 57836. 25 in

The first option results in a shorter connection. Use a transverse weld and two 34-inch

longitudinal welds. Check assumed value of U:

̄

0. 6FyAgv UbsFuAnt 0. 63625. 51. 0584. 5811. 8 kips

Use Rn811. 8 kips

Rn