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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 3, Problem 3.3.6P
Textbook Problem
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A PL 1 4  × 5 is used as a tension member and is connected by a pair of longitudinal welds along its edges. The welds are each 7 inches long. A36 steel is used.

a. What is the design strength for LRFD?

b. What is the allowable strength for ASD?

To determine

(a)

The design strength using load and resistance factor design (LRFD) method.

Explanation of Solution

Section is PL14×5 tension member.

Grade of steel is A36.

Write the expression to calculate the gross area.

Ag=t×w …… (I)

Here, gross area is Ag, thickness is t and width of the section is w.

Substitute 0.25in for t and 5in for w in Equation (I).

Ag=0.25in×5in=1.25in2

Write the expression to calculate the nominal strength on the basis of yielding.

Pny=fyAg …… (II)

Here, nominal strength in yielding is Pny, yield strength is fy and gross area is Ag.

Substitute 36ksi for fy and 1.25in2 for Ag in Equation (II).

Pny=36ksi×1.25in2=45kips

Write the expression to calculate the slenderness ratio.

λ=lw …… (III)

Here, the slenderness ratio is λ length of the section is l and width of the section is w.

Substitute 7in for l and 5in for w in Equation (III).

λ=7in5in=1.4

According to AISC steel manual,

If the slenderness ratio is near to 1.5, the reduction factor is 0.75.

Calculate the effective net area of section.

Ae=UAg …… (IV)

Here, effective net area is Ae, gross area is Ag and reduction factor is U.

Substitute 0.75 for U and 1.25in2 for Ag in Equation (IV)

To determine

(b)

The allowable strength using allowable strength design (ASD) method.

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Chapter 3 Solutions

Steel Design (Activate Learning with these NEW titles from Engineering!)
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