978-1337094740 Chapter 3 Part 1

Type Homework Help

Pages 13

Words 1225

Textbook Steel Design 6th Edition

Authors William T. Segui

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CHAPTER 3 -TENSION MEMBERS

3.2-1

For yielding of the gross section,

For rupture of the net section,

a) The design strength based on yielding is

The design strength based on rupture is

b) The allowable strength based on yielding is

The allowable strength based on rupture is

The allowable service load is the smaller value: Pn/t56. 6 kips

Alternate solution using allowable stress: For yielding,

and the allowable load is FtAg21. 62. 62556. 7 kips

For rupture,

and the allowable load is FtAe29. 02. 18063. 22 kips

[3-1]

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

3.2-2

For yielding of the gross section,

For rupture of the net section,

a) The design strength based on yielding is

The design strength based on rupture is

b) The allowable strength based on yielding is

The allowable strength based on rupture is

Alternate solution using allowable stress: For yielding,

and the allowable load is

FtAg 30. 02. 25  67. 5 kips

For rupture,

The allowable service load is the smaller value 67.5

kips

[3-2]

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

3.2-3

For yielding of the gross section,

For rupture of the net section,

a) The design strength based on yielding is

The design strength based on rupture is

b) The allowable strength based on yielding is

The allowable strength based on rupture is

t

2. 00 80. 95 kips

Alternate computation of allowable load using allowable stress: For yielding,

and the allowable load is

For rupture,

and the allowable load is

[3-3]

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

3.2-4

For A242 steel and t½ in., Fy50 ksi and Fu70 ksi. For yielding of the gross

section,

For rupture of the net section,

a) The design strength based on yielding is

The design strength based on rupture is

b) The allowable strength based on yielding is

The allowable strength based on rupture is

Alternate solution using allowable stress: For yielding,

and the allowable load is

For rupture,

and the allowable load is

[3-4]

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

3.2-5

For a thickness of t3/8 in., Fy50 ksi and Fu70 ksi. First, compute the nominal

strengths. For the gross section,

Net section:

a) The design strength based on yielding is

The design strength based on rupture is

Factored load:

The member is unsatisfactory.

Since Pu  tPn, (102 kips 96.0 kips),

b) For the gross section, the allowable strength is

Alternately, the allowable stress is

For the net section, the allowable strength is

[3-5]

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

Alternately, the allowable stress is

The smaller value controls; the allowable strength is 64.0 kips. When the only loads

are dead load and live load, ASD load combination 2 will always control:

3.2-6

Compute the strength for one angle, then double it. For the gross section,

Net section:

a) The design strength based on yielding is

The design strength based on rupture is

The member has enough strength.

b) For the gross section, the allowable strength is

Alternately, the allowable stress is

[3-6]

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and the allowable strength is FtAg21. 621. 2051. 84 kips

For the net section, the allowable strength is

t

2. 00 48. 38 kips

Alternately, the allowable stress is

The net section strength controls; the allowable strength is 48.4 kips. When the

only loads are dead load and live load, ASD load combination 2 will always control:

3.3-1

(b) Plate with longitudinal welds only:

[3-7]

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3.3-2

3.3-3

U1−x

3.3-4

For yielding of the gross section,

For rupture of the net section,

From AISC Table D3.1, Case 8, U0.80

a) The design strength based on yielding is

The design strength based on rupture is

[3-8]

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The design strength is the smaller value: tPn182. 8 kips

b) The allowable strength based on yielding is

The allowable strength based on rupture is

Alternate computation of allowable load using allowable stress: For yielding,

and the allowable load is

For rupture,

and the allowable load is

3.3-5

(a) The design strength based on yielding is

[3-9]

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

The design strength based on rupture is

Load combination 2 controls:

Since PutPn, (220 kips 169 kips), The member is not adequate.

The smaller value controls; the allowable strength is 112 kips.

Load combination 6 controls:

Since 145 kips 112 kips, The member is not adequate.

Alternate ASD solution using allowable stress:

3.3-6

For yielding of the gross section,

For rupture of the net section, from AISC Table D3.1, case 4,

[3-10]

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

a) The design strength based on yielding is

The design strength based on rupture is

b) The allowable strength based on yielding is

The allowable strength based on rupture is

3.3-7

Connection is through the flanges with four bolts per line.

(a) The design strength based on yielding is

The design strength based on rupture is

(b) For the gross section, The allowable strength is Pn

t

515. 0

1. 67 308 kips

[3-11]

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

t

3.3-8

Gross section: PnFyAg505. 17258. 5 kips

Net section:

(a) The design strength based on yielding is

The design strength based on rupture is

The design strength is the smaller value: tPn233 kips

Load combination 3:

Load combination 4:

Load combination 3 controls. Since PutPn, (205 kips 233 kips),

The member is adequate.

(b) For the gross section, The allowable strength is Pn

t

258. 5

1. 67 155 kips

[3-12]

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

3.4-1

Gross section: Ag101/25in.

Possibilities for net area:

3.4-2

Compute the strength of one plate, then double it.

Possibilities for net area:

[3-13]

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

3.4-3

Gross section: Ag83/83. 0 in.2,PnFyAg363. 0108 kips

3.4-4

Gross section: Ag5. 87 in.2,PnFyAg505. 87293. 5 kips

[3-14]

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

(a) Gross section: tPn0. 90293. 5264 kips

t

3.4-5

For two angles, Pn2200. 0400. 0 kips

[3-15]

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

(a) LRFD Solution

(b) ASD Solution

t

3.4-6

4.563"

Hole diameter 3

41

87

8in.

AnAg−∑tdor d′

3.5-1

Shear areas:

with an upper limit of

3.5-2

Shear areas:

with an upper limit of

[3-17]

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

3.5-3

Tension member:

For this type of connection, Ubs 1. 0, and from AISC Equation J4-5,

with an upper limit of

The nominal block shear strength of the tension member is therefore 148.4 kips.

Gusset Plate:

From AISC Equation J4-5,

with an upper limit of

The nominal block shear strength of the gusset plate is therefore 143.4 kips

The gusset plate controls, and the nominal block shear strength of the connection is

143.4 kips

3.5-4

Gross section nominal strength:

Net section nominal strength:

Block shear strength of tension member:

The tension area is

For this type of connection, Ubs 1. 0, and from AISC Equation J4-5,

with an upper limit of

The nominal block shear strength of the tension member is therefore 173.5 kips.

Block shear strength of gusset plate:

83−0. 5 0. 57/8 0. 796 9 in.2

From AISC Equation J4-5,

[3-19]

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978-1337094740 Chapter 3 Part 1

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978-1337094740 Chapter 3 Part 1

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