978-1337094740 Chapter 8 Part 5

Type Homework Help

Pages 9

Words 1223

Textbook Steel Design 6th Edition

Authors William T. Segui

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Shear areas: Agv 1

41. 5 32. 51

492. 25 in.2

Anv 1

0. 6FyAgv UbsFuAnt 0. 6362. 251. 0580. 281364. 92 kips

The nominal block shear strength is therefore 64.92 kips. The design block shear

strength is

Rn0. 7564. 9248. 7 kips 43.2 kips (OK)

UseaPL 1

147. 0

24. 35 6. 037 try 8 bolts (4 pair)

Determine plate thickness required for bearing: h7

81

16 15

16 in

Minimum ℓe11

53. 82t147. 0

8,Solutionis:t0. 341 4 in. Try 3/8 in.

Brg. strength of edge bolts 53. 823/820. 18 kips/bolt

0. 90Fy

0. 9364. 537 in.2

0. 75Fu

0. 75583. 379 in.2

Try a plate width of wg7in.

For gross area requirement, tAg

4. 537

70. 648 in.

For net area requirement, hole diameter 7

0. 6FyAgv UbsFuAnt 0. 63613. 51. 0581. 875400. 4 kips

The nominal block shear strength is therefore 395.9 kips. The design block shear

strength is

Rn0. 75395. 9297 kips 147.0 kips (OK)

0. 6FyAgv UbsFuAnt 0. 6367. 651. 0580. 6375202. 2 kips

The nominal block shear strength is therefore 199.7 kips. The design block shear

strength is

Rn0. 75199. 7149. 8 kips 147.0 kips (OK)

Afg

2. 55 57. 62496 in.-kips 208 ft-kips

bMn0. 90208187 ft-kips 216. 8 ft-kips (N.G.)

7. 95 18. 49 try 22 bolts (11 pair)

[8-84]

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

Bearing and block shear will be satisfactory.

Afg

2. 55 57. 62964 in.-kips 247 ft-kips

bMn0. 90247222. 3 ft-kips 216. 8 ft-kips required (OK)

Plate length 102. 51. 5 1. 5 0. 5 28. 5 in.

UseaPL 3

′′ with 22 bolts

8.7-1

W16 45 beam, W10 45 column, A992 steel. Flange plates are 5/8 7.

16. 1 130. 1 kips Pbf

Local flange bending: from AISC Eq. J10-1,

20. 906. 25500. 6202108. 1 kips 130.1 kips (N.G.)



0. 75 0. 800. 3502130. 625

10. 1

0. 620

1.5 29000500. 620

0. 350 1. 0

127. 1 kips 130.1 kips (N.G.)

Local flange bending is the worst case. Required stiffener area is

Ast Pbf −Rnmin

stFyst

130. 1 −108. 1

0. 90360. 679 in.2

16. 1 87. 95 kips

Check local flange bending with AISC Equation J10-1:

1. 67 71. 93 kips 87. 95 kips (N.G.)

Check local web yielding with AISC Equation J10-2:



0. 800. 3502130. 625

0. 350

1.5 29000500. 620

2. 00 1. 0

84. 71 kips 87. 95 kips (N.G.)

Stiffeners are required. The smallest strength is 71.93 kips, for the limit state of local

flange bending. The required stiffener area is

Ast Pbf −Rn/min

8.7-2

LRFD solution

Force developed in the beam flange plate is

HFyA36 11



0. 75 0. 800. 68021311/16

0. 680

1.5 29, 000501. 09

8.7-3

(a) Web plate: neglect eccentricity.

Try 3

17. 89 2. 515 try 4 bolts

Determine plate thickness required for bearing. Assume that

Rn2. 4dtFu0. 752. 43/4t5878. 3t

Load resisted by each bolt 45

411. 25 kips. Let

ℓcℓ

e−h

21. 5 −13/16

21. 094 in.

Tearout: Rn1. 2ℓctFu0. 751. 21. 0943/165810. 71 kips

Bearing deformation: 2. 4dtFu0. 752. 43/43/165814. 68 kips

4-in. diameter A325 bolts .

Determine plate thickness required for shear. Shear yielding strength is

Rn1. 000. 60FyAgv 

Let 45 1. 000. 603610. 5tt0. 198 in. Try t1

4in.

49−3. 57/8 1. 484 in.2

Tension area: Ant 1

41. 5 −0. 57/8 0. 265 6 in.2

Rn0. 6FuAnv UbsFuAnt

0. 6581. 4841. 0580. 265667. 1 kips

0. 6FyAgv UbsFuAnt 0. 6362. 251. 0580. 265664. 0 kips

The nominal block shear strength is therefore 64.0 kips. The design block shear

strength is

Rn0. 7564. 048. 0 kips 45 kips (OK)

Use a plate 1

Weld strength controls: Total length required 45/2. 784 16. 16 in.

Use a continuous 1

8-in. fillet weld on each side of plate.

Check for column stiffener requirements. The force developed in the beam flange is

HMu

22012

20. 906. 25500. 6602122. 5 kips 152.8 kips (N.G.)

13. 9

0. 660

0. 370 1. 0

137. 2 kips 152.8 kips (N.G.)

Local flange bending is the worst case. Required stiffener area is

Ast Pbf −Rnmin

stFyst

152. 8 −122. 5

0. 90360. 935 in.2

Ast 1

4321. 5 in.20.935 in.2(OK)

Check for tst b/16 :

16 3

16 in. 1

4in. (OK)

Full-depth stiffeners are not required for this case. Use a depth of approximately

0. 6 (Given)

Pr1. 00. 6Py0. 6Pr0. 4Py, so use AISC Equation J10-10:

Rn0. 60Fydctw1. 4 −Pr

0. 605013. 90. 3701. 4 −0. 6123. 4 kips

Rn0. 90123. 4111. 1 kips 152.8 kips (N.G.)

0. 900. 603613. 91. 4 −0. 60. 192 9 in.

where 152. 8 −111. 1 is the extra strength, in kips, to be furnished by the doubler plate.

Use a 1/4-inch doubler plate.

(The welds will not be designed. For an explanation, see Example 8.12 in the

textbook.)

13. 9 51. 86 °

P41. 7

cos51. 86 °67. 52 kips

0. 9036Ast 67. 52 Ast 2. 08 in.2

Try2stiffeners,33

8, one on each side of the web.

Ast provided 233

1. 392L67. 52

1. 39290. 040. 539 sixteenths

J2.2b,

Minimum length 4w43

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978-1337094740 Chapter 8 Part 5

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978-1337094740 Chapter 8 Part 5

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