978-1337094740 Chapter 3 Part 2

Type Homework Help

Pages 12

Words 1176

Textbook Steel Design 6th Edition

Authors William T. Segui

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with an upper limit of

The nominal block shear strength of the gusset plate is therefore 167.7 kips. The gusset

plate controls, and the nominal block shear strength of the connection is 167.7 kips

(a) Design strength for LRFD:

Block shear controls. Maximum factored load design strength 126 kips

(b) Allowable strength for ASD:

Block shear controls. Maximum service load allowable strength 83.9 kips

3.6-1

(a) Pu1. 2D1. 6L1. 2281. 684168 kips

Try L5 3½ ¾

[3-20]

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

Use an L5 3½ ¾

(b) PaDL28 84 112 kips

Try L5 3½ ¾

3.6-2

(a) Pu1. 2D1. 6L1. 21001. 650200. 0 kips

Try C12 25

(Although this value for the radius of gyration does not quite satisfy the AISC

recommendation for maximum slenderness, tensile strength is not affected by

slenderness, so some leeway is permitted. Therefore, we will consider this value

acceptable.)

[3-21]

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

(b) PaDL100 50 150 kips

Try C12 25

(Although this value for the radius of gyration does not quite satisfy the AISC

recommendation for maximum slenderness, tensile strength is not affected by

slenderness, so some leeway is permitted. Therefore, we will consider this value

acceptable.)

3.6-3

(a) Pu1. 2D1. 6L1. 2301. 690180. 0 kips

The angle leg must be at least 5 in. long to accommodate two lines of bolts (See

workable gages for angles, Manual Table 1-7A).

From AISC Table D4.1, for 4 or more bolts per line, U0. 80

(b) PaDL30 90 120 kips

The angle leg must be at least 5 in. long to accommodate two lines of bolts (See

workable gages for angles, Manual Table 1-7A).

From AISC Table D4.1, for 4 or more bolts per line, U0. 80

3.6-4

(a) Load combination 2 controls:

Try C10 20:

[3-23]

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

(b) Load combination 6 controls:

Try C12 25:

3.6-5

Try C10 20

3.6-6

[3-24]

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

Try W10 49

Ag14. 4 in.210. 9 in.2(OK)

3.7-1

(a) LRFD: Load combination 1 controls: Pu1. 44563. 00 kips

(b) ASD: Load combination 2 controls: PaDL45 550 kips

[3-25]

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

3.7-2

(a) Dead load beam weight 0.048 kips/ft

Because of symmetry, the tension is the same in both rods.

(b) Maximum force in rod occurs when live load is an A or D. Entire live load is

taken by one rod.

3.7-3

(a) Dead load beam weight 0.048 kips/ft

Because of symmetry, the tension is the same in both rods.

(b) Maximum force in rod occurs when live load is an A or D. Entire live load is

taken by one rod.

[3-26]

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

3.7-4

All members are pin-connected, and all loads are applied at the joints; therefore, all

members are two-force members (either tension members or compression members).

Load combination 4 controls.

At joint C,

3.7-5

(a) LRFD: Pu1. 2D1. 6L1. 63556 kips

15'

56 k







tan ( / ) .

1915 3096

∑MC5615−Tsin30. 96°150, T108.9 kips

(b) ASD: PaDL35 kips

[3-27]

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

35 k







tan ( / ) .

1915 3096

3.7-6

From Part 1 of the Manual, the inside diameter is d10. 0 in.

Volume of water per foot of length d2

412 10. 02

412 942. 5 in.3

[3-28]

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

(b) The load at each support is 74.53 lb/ft 10 ft 745.3 lb.

lb7.372

03.7452





TFy

3.8-1

Interior joint load:

30.15

(The assumption that the truss weight is distributed equally to the joints is approximate

but is consistent with the approximate nature of the estimate of total truss weight.)

(a) Load combination 3 controls:

[3-29]

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

Joint B:

Try WT5 11

Ag3. 24 in.22. 79 in.2(OK) rmin 1. 33 in. 0.402 in. (OK)

(b) Load combination 3 controls:

[3-30]

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

For a free-body diagram of the entire truss,

For a free body of joint B:

3.8-2

The diagonal web members are the tension members, and member AL has the largest

force.

[3-31]

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

Try L3 1

231

Ag1. 58 in.21. 13 in.2(OK) rmin 0. 628 in. 0.509 in. (OK)

An1. 58 −3

Use L3 1

231

4for member AL

This shape can be used for all of the web tension members. Although each member

could be a different size, this would not usually be practical. The following table

shows the relatively small difference in requirements for all the web tension members.

Force Req’d Ag

Member (kips) (in.2

3.8-3

Use load combination 3: 1. 2D1. 6S.

[3-33]

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

3.8-4

(a) Load combination 3 controls:

10' 10' 10' 10'

Bottom chord: Member FE has the largest tension force.

Use a free body of joint E:

(This controls the gross area requirement.)

[3-34]

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

Web members: Design for the maximum tensile force, which occurs in member AH,

and use one shape for all tension web members (the diagonal web members). Using the

method of sections (see figure), consider the force in member AH to act at H.

(b) Load combination 3 controls:

10' 10' 10' 10'

Bottom chord: Member FE has the largest tension force.

[3-35]

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

Use a free body of joint E:

∑Fy39. 76 −8

40. 79 −FFE 0, FFE 198. 8 kips

Required AgFFE

0. 6Fy

198. 8

0. 6506. 627 in.2

(This controls the gross area requirement.)

12. 81 FAH30−11. 3610 20 300, FAH 36. 38 kips

Required AgFAH

36. 38

0. 5Fu

0. 5651. 119 in.2

From AeAgU, Required AgRequired Ae

U1. 119

0. 85 1. 32 in.2(controls)

3.8-5

Use sag rods at midspan of purlins.

40. 79 600. 0 lb

Since the design is for dead load only, use load combination 1:

Pu1. 4D1. 4600840 lb

21. 75 0. 02759 in.2

Let d2

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978-1337094740 Chapter 3 Part 2

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978-1337094740 Chapter 3 Part 2

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