978-1337094740 Chapter 8 Part 2

Let 0. 062 5Ru29. 82, Ru477 kips

The reaction capacity is therefore based on the slip-critical strength. Ru111 kips

(b) ASD solution

0. 1Ra7. 380, Ra73. 8 kips

Tension:

Let 0. 062 5Ra19. 88, Ra318 kips

The reaction capacity is therefore based on the slip-critical strength. Ra73. 8 kips

8.3–8

Assume that tension controls. Select bolt size based on tension, then check the other

Required diameter db4Ab

40. 5744

0. 855 2 in.

Try 7

8-in. diameter bolts, with Ab7/82/4 0. 601 3 in.2

Slip-critical strength will control over shear. In this type of connection, the tensile

22−15/16

21. 531 in.

Rn1. 2ℓctFu0. 751. 21. 5310. 59065

52. 8 kips 8.88 kips (OK)

2. 4dtFu0. 752. 47/80. 5906560. 4 kips/bolt 52.8 kips

From 42rt903, rt21. 5 kips

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Total tensile load per bolt 4. 7 21. 5 26. 2 kips. Let

Fnt 26. 2

Ab

Ab26. 2

DuTbnb

1. 1339100. 940 5

kscRn/0. 940511. 5310. 84 kips/bolt 60/10 6 kips/bolt (OK)

Check bearing (flange of WT controls). h11

89

8in.

8.3–9

(a) Factored load: Neglect the beam weight initially, and account for it later.

wu1. 2wD1. 6wL1. 646. 4 kips/ft

Mu1

8wuL21

86. 4302720 ft-kips

Let 35. 79NbVu: 35. 79Nb97. 4 Nb2. 72

Try 4 bolts in beam-to-angle connection and 8 bolts in angle-to-column connection.

Use a minimum spacing of 3d33/42. 25 in.

Minimum edge distance from AISC Table J3.4 1in.

Let 584. 0tVu97. 4 t0. 167 in.

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Check beam web. tw0. 440 in. There are no edge bolts. For the other bolts,

Rn1. 2ℓctFu0. 751. 22. 1880. 4406556. 32 kips/bolt

Beam connection Column connection

Shear yield strength of angles (AISC J4.2):

Rn0. 60FyAg21. 00. 603612 1

42

0. 6FyAgv UbsFuAnt 0. 6365. 251. 0580. 5313144. 2 kips

Rn0. 75144. 2108 kips 97. 4 kips (OK)

Use 2L4 41

4as shown, with four 3

4-in. diameter A325

bearing-type bolts in the beam web, and eight 3

Upper right bolt is critical:

pmx My

∑x2y2243. 54. 5

45 24. 35 kips

p∑px2∑py224. 35224. 35234. 4 kips

Upper right bolt is critical:

pmx My

∑x2y2243. 56

90 16. 23 kips

p∑px2∑py216. 23219. 482

0. 4418 22. 05 ksi

′1. 3Fnt −Fnt

0. 755422. 0568. 0 ksi 90 ksi

RnFnt

′Ab0. 7568. 00. 4418

8.3–10

(a) Neglect the beam weight initially, and account for it later.

wa4kips/ft

Ma1

8waL21

84302450 ft-kips

Let 23. 86NbVa: 23. 86Nb61. 14 Nb2. 56

Try 4 bolts in beam-to-angle connection and 8 bolts in angle-to-column connection.

Use a minimum spacing of 3d33/42. 25 in.

Minimum edge distance from AISC Table J3.4 1in.

Try s3 in. and ℓe11

Let 388.4t Va 61. 14 t 0.157 in.

Check beam web. tw 0.440 in. There are no edge bolts. For the interior bolts,

Rn

1

1. 2LctFu1

2. 00 1. 22. 1880. 4406537. 55 kips/bolt

1

2. 4dtFu1

2. 00 2. 43/40. 4406525. 74 kips/bolt 37. 55 kips/bolt

3 @ 3″

2.5″

1.5″1.5″

2.5″ 2.5″

Beam connection Column connection

Shear yield strength of angles (AISC J4.2):

Rn

1

0. 60FyAg21

1. 5 0. 603612 1

42

0. 6FyAgv UbsFuAnt 0. 6365. 251. 0580. 5313144. 2 kips

Rn

1

2. 00 144. 272. 1 kips 61. 14 kips (OK)

Use 2L4 41

4as shown, with four 3

4-in. diameter A325 bearing-type bolts in the

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pmx My

∑x2y2152. 94. 5

45 15. 29 kips

p∑px2∑py215. 29215. 29221. 6 kips

Rn

1

FnvAb1

2. 00 540. 44182 shear planes

Upper right bolt is critical:

0. 4418 13. 84 ksi

Fnt

′1. 3Fnt −Fnt

Fnv

frv ≤Fnt

1. 390−2. 0090

(d)

8.4–1

Direct shearing stress: f1y9

780. 6 kips/in. ↓

Shearing stress caused by moment: Locate centroid with respect to upper left corner.

x

̄73. 5

̄84

8.4–2

Direct shearing stress: f1y15

5351. 154 kips/in. ↓

Shearing stress caused by moment: Locate centroid with respect to lower left corner.

8.4–3

Px75 cos 75 ° 19. 41 kips Py75 sin 75 ° 72. 44 kips ↓

f1x19. 41

18 1. 078 kips/in. f1y72. 44

18 4. 024 kips/in. ↓

Shearing stress caused by moment: Locate centroid with respect to lower left corner.

8.4–4

(a) LRFD solution:

Pu1. 2D1. 6L1. 22. 51. 67. 515. 0 kips

Load on one angle 15

27. 5 kips, fy7. 5

350. 5 kips/in. ↓

Since 4.59 kips/in. 4.18 kips/in., weld is not adequate.

(b) ASD solution:

Pa10 kips

Load on one angle 10

25kips, fy5

350. 333 3 kips/in. ↓

8.4–5

Locate centroid with respect to left side.

x

̄6432

18 4. 667 in.

Ix63

1. 392 5. 34 sixteenths.

For t1/2 in., min. w3/16 in. and max w1/2 −1/16 7/16 in.

Use a 3

8-in. fillet weld.

(b) ASD solution

0. 9279 5. 72 sixteenths.

8.4–6

(a) LRFD solution

DLD2D20 D6. 667 kips, L13. 33 kips

Pu1. 2D1. 6L1. 26. 6671. 613. 3329. 33 kips

f1y29. 33

f2xMy

J96. 793. 6

46. 42 7. 506 kips/in. 

f2yMx

J96. 791. 3

46. 42 2. 711 kips/in. ↓

fv∑fx2∑fy27. 50625. 866 2. 711211. 4 kips/in.

M2021. 366. 0 in.-kips 

12 31. 5 −1. 3223

12 20. 323. 217 in. 4

JIxIy43. 2 3. 217 46. 42 in. 4

8.4–7



0.884″

1.5″

4″

Load when eccentricity is considered:

Ix241. 5218. 0 in. 4,Iy243