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?

Q: Expert Answer

To determine (a) The design strength using load and resistance factor design (LRFD) method. Explanation Section is PL 1 4 × 5 tension member. Grade of steel is A 36 . Write the expression to calculate the gross area. A g = t × w …… (I) Here, gross area is A g , thickness is t and width of the section is w . Substitute 0.25 in for t and 5 in for w in Equation (I). A g = 0.25 in × 5 in = 1.25 in 2 Write the expression to calculate the nominal strength on the basis of yielding. P ny = f y A g …… (II) Here, nominal strength in yielding is P ny , yield strength is f y and gross area is A g . Substitute 36 ksi for f y and 1.25 in 2 for A g in Equation (II). P ny = 36 ksi × 1.25 in 2 = 45 kips Write the expression to calculate the slenderness ratio. λ = l w …… (III) Here, the slenderness ratio is λ length of the section is l and width of the section is w . Substitute 7 in for l and 5 in for w in Equation (III). λ = 7 in 5 in = 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. A e = U A g …… (IV) Here, effective net area is A e , gross area is A g and reduction factor is U . Substitute 0.75 for U and 1.25 in 2 for A g in Equation (IV) To determine (b) The allowable strength using allowable strength design (ASD) method.

Question

Still sussing out bartleby?

Answer 1

BuyFind

Steel Design (Activate Learning wi...

6th Edition

Segui + 1 other

Publisher: Cengage Learning

ISBN: 9781337094740

Solutions

Chapter

Section

Answer 2

BuyFind

Steel Design (Activate Learning wi...

6th Edition

Segui + 1 other

Publisher: Cengage Learning

ISBN: 9781337094740

Solutions

Chapter

Section

Answer 3

Solutions

Chapter

Section

Answer 4

Chapter

Section

Answer 5

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

Ag=0.25 in×5 in=1.25 in2

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

Pny=36 ksi×1.25 in2=45 kips

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

λ=7 in5 in=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 1.25 in2 for Ag in Equation (IV)

Answer 6

To determine

(b)

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

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?

Steel Design (Activate Learning wi...

Solutions

Steel Design (Activate Learning wi...