978-1337094740 Chapter 7 Part 1

CHAPTER 7 –SIMPLE CONNECTIONS

7.3–1

(a) Minimum spacing 22

3d2. 6677/82. 33 in. 2.75 in. (OK)

Minimum edge distance from AISC Table J3.4 1.125 in. actual ℓe(OK)

2. 4dtFu2. 47/83/85845. 68 kips 26.91 kips

∴use Rn26. 91 kips Rn26. 9 kips

2. 4dtFu45. 68 kips 47.32 kips ∴use Rn45. 68 kips Rn45. 7 kips

7.3–2

(a) Minimum spacing 22

3d2. 6673/42. 0 in. 2.5 in. (OK)

Minimum edge distance from AISC Table J3.4 1.0 in. 1. 5 in. (OK)

2. 4dtFu2. 43/43/87047. 25 kips 34.46 kips

∴use Rn34. 46 kips

2. 4dtFu47. 25 kips 68.92 kips ∴use Rn47. 25 kips

7.4–1

(a) Minimum spacing 22

3d2. 6677/82. 33 in. 3 in. (OK)

Minimum edge distance from AISC Table J3.4 1.125 in. 2 in. (OK)

2. 4dtFu2. 47/83/85845. 68 kips 66.06 kips

∴use Rn45. 68 kips

2. 4dtFu45. 68 kips 53.84 kips ∴use Rn45. 68 kips

(The bearing strength of the channel does not need to be checked. Its thickness is

greater than that of the gusset plate, and its ultimate tensile stress is larger.)

For the connection, the strength for each bolt is the minimum of the shear and bearing

7.4–2

(a) Minimum spacing 22

3d2. 6677/82. 33 in. 2.75 in. (OK)

Minimum edge distance from AISC Table J3.4 1.125 in. actual ℓe(OK)

(b) Bolt shear: Abd2/4 7/82/4 0. 601 3 in.2

7.4–3

Abd2/4 3/42/4 0. 441 8 in.2

2. 4dtFu2. 43/43/85839. 15 kips 47.71 kips for shear

∴Bearing strength controls for each bolt.

29. 36 1. 93 Use 2 bolts

(b) ASD solution: Rn

7.4–4

Abd2/4 7/82/4 0. 601 3 in.2

2. 4dtFu2. 47/83/85845. 68 kips 57.72 kips for shear

∴Bearing strength controls.

34. 26 2. 63, use 4 bolts for symmetry on each side.

Use 8 bolts

19. 58 3. 06, use 4 bolts for symmetry on each side.

Use 8 bolts

7.4–5

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

Nominal shear capacity of one bolt is

RnFnvAb540. 441823. 86 kips

Check bearing on tension member (it has the smaller edge distance; everything else is

Let 1. 2D1. 62D107. 2, Solution is: D24. 36

7.4–6

Bolt shear (assume that the threads are in shear): Abd2/4 7/82/4 0. 601 3

in.2

Nominal shear capacity of one bolt is

RnFnvAb2 (for double shear) 540. 6013264. 94 kips/bolt

Bearing deformation: 2. 4dtFu2. 47/83/85845. 68 kips 39.96 kips

∴use Rn39. 96 kips

For the other bolts,

Tear-out: ℓcs−h3−15

16 2. 063 in.

7.6–1

Bolt shear (assume that the threads are in shear): Abd2/4 7/82/4 0. 601 3

in.2

Nominal shear capacity of one bolt is

RnFnvAb2 (for double shear) 540. 6013264. 94 kips/bolt

Rn1. 2ℓctFu1. 21. 5315/85866. 60 kips

Bearing deformation: 2. 4dtFu2. 47/85/85876. 13 kips 66.60 kips

∴use Rn66. 60 kips

For the other bolts,

Tear-out: ℓcs−h3−15

PnFyAg5014. 26713. 0 kips

Net section:

AnAg−tdh27. 13 −25

8

7

81

811. 76 in.2

̄

1. 4D1. 48. 5L389. 6 L32. 74 kips

PDL8. 532. 7432. 74 311. 0 kips P311 kips

(b) ASD solution. Compute the allowable strength:

7.6–2

Determine the nominal shear and bearing strengths per bolt. The shear strength is

RnFnvAb680. 6013281. 78 kips/bolt

Bearing: hd1

16 7

81

16 15

16 in.

16 2. 063 in.

Rn1. 2ℓctFu1. 22. 0630. 55871. 79 kips/bolt

Bearing deformation: 2. 4dtFu60. 9 kips/bolt 71.79 kips/bolt ∴use Rn60. 9

kips/bolt

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Total Rn0. 75335. 88360. 9217. 8 kips 179.6 kips (OK)

Use 6 bolts each side, 12 total

(b) ASD Solution

PaPDPL35 86 121. 0 kips

Select a trial number of bolts based on shear, then check bearing. The shear strength is

Rn

40. 89 2. 96; Try 4 on each side for symmetry.

For the connection, the strength for each bolt is the minimum of the shear and bearing

7.6–3

LRFD Solution

Pu1. 2D1. 6L1. 2451. 690198. 0 kips

The bearing strength of the gusset plate is smaller than the bearing strength of the

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Number required 198/30. 67 6. 46 Use 8 bolts

ASD Solution

PaDL45 90 135 kips

The bearing strength of the gusset plate is smaller than the bearing strength of the

member. For one bolt,

Rn

7.6–4

(a)

Bolt Single-shear Slip-critical design Single-shear Slip-critical allowable

1 31.809 17.289 21.206 11.526

1 1/8 40.258 18.984 26.839 12.656

7.6–5

Determine the nominal strengths for all limit states. For bolt shear,

Abd2/4 1. 1252/4 0. 994 in.2

2. 4dtFu2. 41. 1251/258

1. 2ℓctFu1. 21. 6871/25858. 71 kips/bolt

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2. 4dtFu78. 3 kips/bolt 58.71 kips/bolt ∴use Rn58. 71 kips/bolt

Tension on the gross section: Ag0. 56. 53. 25 in.2

21. 313 1. 937 in.2

PnFuAe581. 937112. 3 kips

Check block shear (tension member controls).

Agv 21

0. 6FyAgv UbsFuAnt 0. 6365. 01. 0580. 8435156. 9 kips

The nominal block shear strength is therefore 154.4 kips.

(a) LRFD solution. Compute design strengths:

For bolt shear, Rn0. 7553. 6840. 26 kips/bolt

1. 2D1. 6L1. 2D1. 63D84. 23, D14.04 kips

PDL14. 04 314. 0456. 16 kips P56. 2 kips

(b) ASD solution. Compute allowable strengths:

For bolt shear, Rn

7.6–6

Before proceeding to the LRFD and ASD solutions, compute the nominal bolt shear,

rmin

rz

0. 756 143 300 (OK)

For shear,

Abd2/4 1. 1252/4 0. 994 in.2

RnFnvAb540. 99453. 68 kips/bolt

2. 4dtFu58. 73 kips/bolt ∴use Rn58. 73 kips/bolt

Tension on the gross section: PnFyAg364. 50162. 0 kips

Tension on the net section: Use a hole diameter of 1 1

81

16 1. 313 in.

AnAg−tdh4. 50 −1

0. 6FyAgv UbsFuAnt 0. 6369. 751. 0580. 9219264. 1 kips

The nominal block shear strength is therefore 264.1 kips.

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(a) LRFD solution. Compute design strengths:

Bolt strength:

For load combination 2, Pu1. 2D1. 6L1. 2201. 660120. 0 kips

For load combination 4, Pu1. 2D1. 6W0. 5L1. 2201. 6200. 560

86. 0 kips

∴combination 2 controls; Pu120 kips130.2 kips (OK)

member and connection are satisfactory.

For block shear, Rn/264. 1/2. 00 132. 1 kips

Slip-critical strength controls. Rn/86. 82 kips

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For load combination 2, PaDL20 60 80 kips

For load combination 6, PaD0. 75W0. 75L20 0. 75200. 756080

7.7–1

32. 9 5. 350 try 6 bolts for symmetry.

Minimum spacing 22

3d2. 667 1 3

83. 667 in., try 4 in.

32. 65 5. 39; Use 6 bolts

21. 9 5. 479 try 6 bolts.

Minimum spacing 22

3d2. 667 1 3

83. 667 in., try 4 in.

21. 75 5. 517; Use 6 bolts

7.7–2

(a) LRFD Solution:

Load Combination 2: 1. 2D1. 6L1. 2501. 6100220. 0 kips

Load Combination 4: 1. 2D1. 0W0. 5L1. 2501. 0450. 5100155. 0

kips

0. 9Fy

0. 9366. 79 in.2

Required AePu

0. 75Fu

220

0. 75585. 06 in.2

Required rmin L

300 20 12

300 0. 80 in.

Try L8 65

8with the 8-inch leg connected to the gusset plate.

25. 69 8. 56 bolts; for symmetry, use two rows of 5 10 bolts.

Check block shear on the gusset plate. For hole diameters, use 1 1

43

16 1. 438 in.

Agv 3

0. 6FyAgv UbsFuAnt 0. 6369. 3751. 0580. 5858236. 5 kips

The nominal block shear strength of the gusset plate is therefore 228.9 kips

The design block shear strength is