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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 8, Problem 8.4.16P
Textbook Problem
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Same as Problem 8.4-15, but use ASD.

To determine

(a)

The design of the welded connection by using elastic method and ASD.

Explanation of Solution

Given:

L6×6×516

38in. -thick gusset plate

A36 steel

Dead load = 31 kips

Live load = 31 kips

Calculation:

Determine the factored loads:

Pa=D+L

where, D is dead load and L is live load

Pa=31+31=62kips

For t=5/16in.,

Minimum w=3/16in.

Maximum = w=5/161/16=4/16=1/4in.

Use E70 fillet welds.

Try w=3/16in., one transverse weld and two longitudinal welds.

Investigate the two options:

First assuming the same strength for both the longitudinal and transverse welds,

Capacity per inch of length is

RnΩ=0.9279D=0.9279(3)=2.784kips/in.

Base metal shear strength of the angle:

Yielding:

RnΩ=0.4Fyt=0.4(36)(5/16)=4

To determine

(b)

The design of a welded connection by using the elastic method and LRFD and balancing the welds.

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Chapter 8 Solutions

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