978-1337094740 Chapter 7 Part 3

For the hole nearest the edge, ℓcℓ

e−h

21. 5 −0. 9375

21. 031 in.

Rn

1

1. 2ℓctFu1

2. 00 1. 21. 0313/858

Summary:

Use 6 bolts for tension member-to-gusset plate connection;

use 6 bolts for gusset-plate-to connection angles connection;

use 12 bolts for connection angles to column flange connection;

use edge distances of 1 1

2in. and bolt spacings of 2 1

2in. throughout.

7.9–5

Let angle that load makes with the horizontal arctan12. 5/1343. 88 °

Fnt

0. 7554frv ≤90

117 −2. 222frv ≤90

′0. 75117 −2. 222frv ≤0. 7590

214. 8

87. 75 −1. 667 206. 6

214. 8 87. 75Ab−344. 3, Solution is: Ab6. 372 in.2

For 12 bolts, required AbAb

12 6. 372

12 0. 531 in.2/bolt

d2

40. 531 , d0. 822 in.

For the holes nearest the edge, ℓcℓ

e−h

22. 5 −0. 9375

22. 031 in.

Rn1. 2ℓctFu0. 751. 22. 0311. 1165132 kips 17.2 kips (OK)

2. 4dtFu0. 752. 47/81. 1165114 kips/bolt 17.2 kips (OK)

Ta200 cos43. 88 °144. 2 kips, Va200 sin43. 88 °138. 6 kips

Assume that tension controls:

Fnt

′1. 390−2. 0090

54 frv ≤90

117 −3. 333frv ≤90

Fnt

′

144. 2

58. 5 −1. 667 138. 6

144. 2 58. 5Ab−231, Solution is: Ab6. 414 in.2

Required AbAb

12 6. 414

12 0. 534 5 in.2

Required diameter db4Ab

40. 5345

0. 824 9 in.

2. 4dtFu

2. 47/81. 1165

2. 00 75. 8 kips/bolt 11.6 kips (OK)

For the other holes, ℓcs−h4−0. 9375 3. 063 in. Since this is larger than ℓc

7.11–1

(a) Tension member gross section strength

PnFyAg502. 50125. 0 kips

tPn0. 90125112. 5 kips

Net section strength: U1−x

̄

1. 2D1. 62. 5D108. 6, Solution is: 20. 88

PDL20. 88 320. 8883. 52 kips P83. 5 kips

(b) ASD solution

Tension member gross section strength

PnFyAg502. 50125. 0 kips

Pn

125

Rn

5. 568

2. 00 2. 784 kips/in.

Base metal shear strength of the plate:

Yielding: Rn

0. 6Fyt

1. 50 0. 6363/8

1. 50 5. 4 kips/in 2.784 kips/in.

Rupture: Rn

7.11–2

(a) LRFD solution. Gross section: Ag5. 12 in.2

tPn0. 90FyAg0. 90365. 12165. 9 kips

̄

1. Use the basic weld strength for both the longitudinal and transverse welds.

Rn5. 56855583. 52 kips (for one angle)

2. Use 0.85 times the basic weld strength for the longitudinal welds and 1.5 times the

89. 09 2178. 2 kips

Check block shear on the gusset plate in lieu of base metal shear strength.

Agv Anv 3

0. 6FyAgv UbsFuAnt 0. 6363. 7501. 0581. 875189. 8 kips

1. 2D1. 63D142. 4, Solution is: 23. 73 kips

[7-46]

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

PDL23. 73 323. 7394. 92 kips P94. 9 kips

(b) ASD solution

2. Use 0.85 times the basic weld strength for the longitudinal welds and 1.5 times the

59. 39 2118. 8 kips

Check block shear on the gusset plate in lieu of base metal shear strength. From the

7.11–3

1. 2D1. 62. 0D156. 0, Solution is: 35. 45

PDL35. 45 2. 035. 45106. 4 kips P106 kips

(b) ASD solution

7.11–4

(a) LRFD solution. Gross section: For the outer member,

Ag5/16321. 875 in.2

2. 0.85 times the basic weld strength for the longitudinal welds and 1.5 times the basic

weld strength for the transverse weld:

Rn0. 855. 568331. 55. 5683299. 39 kips

Use the larger value of 100.2 kips.

0. 6FyAgv UbsFuAnt 0. 6363. 01. 0581. 5151. 8 kips

1. 2D1. 6D60. 75, Solution is: 21. 70

PDL21. 7 21. 7 43. 4 kips P43. 4 kips

(b) ASD solution

1. Basic weld strength for both the longitudinal and transverse welds:

Rn

2. 0.85 times the basic weld strength for the longitudinal welds and 1.5 times the basic

weld strength for the transverse weld:

Rn

0. 853. 712331. 53. 7123271. 27 kips

7.11–5

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

0. 6Fyt0. 6363

88. 1 kips/in.

0. 6Fyt0. 6505

16 218. 75 kips/in.

and the shear rupture strength is 0. 45Fut0. 45653

5. 568 28. 02 in.

0. 855. 5684. 733 kips/in.

and the strength of the transverse weld is

1. 55. 5688. 352 kips/in.

114. 2

24. 73312. 06 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)

0. 4Fyt0. 4363

0. 3Fut0. 3583