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Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
BuyFindarrow_forward

Steel Design (Activate Learning wi...

6th Edition
Segui + 1 other
Publisher: Cengage Learning
ISBN: 9781337094740
Chapter 6, Problem 6.6.7P
Textbook Problem
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The member shown in Figure P6.6-7 has lateral support at points A, B, and C. Bending is about the strong axis. The loads are service loads, and the uniform load includes the weight of the member. A992 steel is used. Is the member adequate?

a. Use LRFD.

b. Use ASD.

Chapter 6, Problem 6.6.7P, The member shown in Figure P6.6-7 has lateral support at points A, B, and C. Bending is about the

To determine

(a)

If the member is adequate using LRFD method.

Explanation of Solution

Given data:

The dead load is 70k.

The live load is 170k.

Concept used:

Write the equation that satisfies AISC specification by using LRFD method.

PuϕbPn+89(MuxϕbMnx+MuyϕbMny)1.0

Here, the compressive design strength is ϕbPn, the factored axial load is Pu the flexural load about xaxis is Mux, the nominal flexural strength about xaxis is ϕbMnx, the flexural load about yaxis is Muy, and the nominal flexural strength about yaxis is ϕbMny.

Calculation:

Write the expression for axial factored load Pu on the member.

Pu=1.6PL+1.2PD (I)

Here, the axial dead load is PD, the axial live load is PL, and the factored axial load is Pu.

Substitute 170k for PL and 70k for PD in Equation (I).

Pu=(1.6×170k)+(1.2×70k)=272k+84k=356k

Write the expression for transverse concentrated load Tu on the member.

Fu=1.6TL+1.2TD (II)

Here, the transverse dead load is TD and the transverse live load is TL and the transverse load is Fu.

Substitute 18k for TL and 7k for TD in Equation (II).

Fu=(1.6×18k)+(1.2×7k)=28.8k+8.4k=37.2k

Write the expression for the uniformly factored load on the member.

Wu=1.6WL+1.2WD (III)

Here, the uniformly factored dead load is WD and the uniformly factored live load is WL and the uniformly factored load is Wu.

Substitute 1.5k/ft for WL and 3.5k/ft for WD in Equation (III).

Wu=(1.6×3.5k/ft)+(1.2×1.5k/ft)=5.6k/ft+1.8k/ft=7.4k/ft

Write the expression for the bending moment.

Mnt=WuL28+FuL4 (IV)

Here, the factored bending moment of the load is Fu, and the uniformly factored load is Wu, the length of the member is L, and the bending moment of the load is Mnt.

Substitute 7.4k/ft for Wu, 16ft for L and 37.2k for Fu in Equation (IV).

Mnt=7.4k/ft(16ft)28+37.2k(16ft)4=236.8k-ft+148.8k-ft=385.6k-ft

When the factor for non uniform bending within the unbraced length is Lb=8ft, then Cb=1.

If Lb=8ft and Cb=1 then,

ϕbMn=488k-ftϕbMp=488k-ft

Since the member is transversely loaded, then Cm=1.

Write the formula for calculating buckling stress.

Pe=π2EI(KL)2 (V)

Here, the steels elastic modulus is E, the sections moment of inertia about major principal axis is I, and the members effective length about is KL.

Substitute 29000ksi for E, 623in4 for I, 1.0 for K and 16ft in Equation (V).

Pe=π2(29000ksi)(623in4)(1.0×(16ft)(12inch1ft))2=178314142.7k-in236864in2=4837k

Write the expression for moment amplification factor

To determine

(b)

If the member is adequate using ASD method.

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