978-1337094740 Chapter 8 Part 1

CHAPTER 8 –ECCENTRIC CONNECTIONS

8.2–1

Direct shear components:

Px3

5159kips, Py4

51512 kips

8.2–2

Eccentricity: ex33410 in., ey5. 5

22. 75 in.

∑x2y27. 382Pu6

255. 6 0. 173 3Pukips →

[8-1]

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8.2–3

Direct shear component: pcy 40

58kips↓

Determine location of centroid with respect to lower left bolt:

8.2–4

Direct shear component: pcx 70

3″

1.42″

1.58″

70 k

∑x2y2110. 64. 5

50. 63 9. 83 kips ↑

8.2–5

Direct shear components:

pcx 20

∑x2y21881. 2

46. 8 4. 821 kips ↓

8.2–6

Determine location of centroid with respect to lower left bolt:

x

̄3161

61. 5 in., y

̄316193

66. 0 in.

∑x2y23444. 5

103. 5 14. 96 kips ↓

8.2–7

Determine location of centroid with respect to lower right bolt:

x

̄2216

61. 667 in., y

̄2418

62. 667 in.

∑x2y297. 211. 667

80. 67 2. 009 kips ↓

∑x2y297. 211. 667

80. 67 2. 009 kips ↓

∑px11. 41 3. 214 14. 62 kips ←∑py8. 56 2. 009 10. 57 kips ↓

p∑px2∑py214. 62210. 57218. 04 kips (controls)

∑x2y265. 861. 667

80. 67 1. 361 kips ↓

∑px7. 733 2. 177 9. 91 kips ←∑py5. 800 1. 361 7. 161 kips ↓

8.2–8

LRFD solution

Determine location of centroid with respect to lower left bolt:

x

̄35. 5

̄23. 517210. 5

∑x2y2212. 45. 5 −2. 357

170. 9 3. 906 kips ↓

∑px2. 571 6. 214 8. 785 kips ←

For 0. 339Tb12. 13, Solution is: 35. 78 kips.

From AISC Table J3.1, Tb39 kips for a 7

8.2–9

Location of centroid with respect to lower row of bolts: y

̄3631

63. 5 in.

∑x2y2202. 53

79. 5 7. 642 kips ↓

∑px2. 5 8. 915 11. 42 kips →∑py7. 642 kips ↓

p∑px2∑py211. 4227. 642213. 74 kips

Assuming that the threads are in shear (N bolts),

∑x2y21353

79. 5 5. 094 kips ↓

∑px1. 667 5. 943 7. 610 kips →∑py5. 094 kips ↓

p∑px2∑py27. 61025. 09429. 158 kips

2. 4dtFu

2. 43/43/858

2. 00 19. 6 kips 20.8 kips

8.2–10

(a) Direct shear components: pcx 43

67. 167 kips ←

M438344 in.-kips 

8.2–11

Use Table 7-8, Angle 30°. n5ands3in.

3″ 3″ 3″

30º

4″

e

x

3″

8.2–12

Use Table 7-13, Angle 30°. n3ands3in.

8.2–13

From the solution to problem 8.2-12, Use Table 7-13, Angle 30°, s3in.

ex442−3 tan 30 ° 8. 268 in.

23. 86 10. 90

8.3–1

Nominal bearing strength: h13

16 1. 188 in.

2. 4dtFu02. 41. 00. 5606587. 36 kips

∴use Rn70. 47 kips/bolt

Shear strength: Abd2/4 1. 02/4 0. 785 4 in. 2

Assume that threads are in the plane of shear.

0. 7854 32. 34 ksi

0. 755432. 3445. 13 ksi 90 ksi

RnFnt

′Ab0. 7545. 130. 785426. 6 kips/bolt 10.9 kips/bolt (OK)

Bolts are adequate

0. 7854 21. 65 ksi

Fnt

′1. 3Fnt −Fnt

Fnv

frv ≤Fnt

8.3–2

(a) LRFD solution

Pu1. 2D1. 6L1. 2241. 658121. 6 kips

Shear/bearing load per bolt 121. 6/8 15. 2 kips

Bearing strength: h7

0. 755423. 9163. 87 ksi 90 ksi

RnFnt

′Ab0. 7563. 870. 601328. 8 kips/bolt 15.2 kips/bolt (OK)

Bolts are adequate

2. 4dtFu

2. 47/80. 57565/2. 00 39. 24 kips 51.5 kips

∴use Rn

34. 33 kips/bolt 8.5 kips/bolt (OK)

Fnv

0. 6013 14. 14 ksi

1. 390−2. 0090

8.3–3

Nominal bearing strength: h7

2. 4dtFu2. 47/80. 6706591. 46 kips

2. 4dtFu2. 47/80. 6056582. 58 kips

∴use Rn82. 58 kips/bolt for these bolts

Shear strength: Abd2/4 7/82/4 0. 601 3 in. 2

Assume that threads are in the plane of shear.

0. 6013 24. 95 ksi

′1. 3Fnt −Fnt

0. 755424. 9561. 56 ksi 90 ksi

RnFnt

′Ab0. 7561. 560. 601327. 76 kips/bolt 25 kips/bolt (OK)

Bolts are adequate

0. 6013 16. 63 ksi

Fnt

′1. 3Fnt −Fnt

Fnv

frv ≤Fnt

8.3–4

(a) LRFD solution

Factored load Pu1. 2D1. 6L1. 2101. 62856. 8 kips

Shear/bearing load per bolt 56. 8/6 9. 467 kips

[8-15]

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Rn1. 2ℓctFu0. 751. 22. 1880. 5606571. 7 kips

2. 4dtFu0. 752. 43/40. 5606549. 1 kips 71.7 kips

∴use Rn49. 1 kips/bolt 9.467 kips/bolt (OK)

Shear strength: Abd2/4 3/42/4 0. 441 8 in. 2

Assume that threads are in the plane of shear.

0. 755421. 4369. 38 ksi 90 ksi

RnFnt

′Ab0. 7569. 380. 4418

22. 99 kips/bolt 8.52 kips/bolt (OK)

Bolts are adequate

0. 4418 12. 9 ksi

Fnt

′1. 3Fnt −Fnt

Fnv

frv ≤Fnt

8.3–5

(a) LRFD solution:

Pu1. 2D1. 6L1. 20. 33 1101. 60. 67 110161. 5 kips

Vertical component of load 3

5161. 596. 90 kips

0. 755416. 1281. 18 ksi 90 ksi

RnFnt

′Ab0. 7581. 180. 6013

36. 6 kips/bolt 29.1 kips/bolt (OK)

Bolts are adequate

Fnv

0. 6013 10. 98 ksi

1. 390−2. 0090

8.3–6

Pu1. 2D1. 6L1. 2151. 64590. 0 kips

Vertical component of load 2

59080. 50 kips

Horizontal component of load 1

0. 755417. 0979. 02 ksi 90 ksi

′Ab0. 7579. 020. 7854

8.3–7

Nominal bearing strength: h3

41

16 13

16 in.

2. 4dtFu2. 43/45/165832. 63 kips

∴use Rn23. 79 kips/bolt

Shear strength: Abd2/4 3/42/4 0. 441 8 in. 2

Assume that threads are in the plane of shear.

0. 1Ru11. 07, Ru111 kips

Tension:

∑MNA 2rt36236rt