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A simply supported beam is subjected to the end couples (bending is about the strong axis) and the axial load shown in Figure P6.6-5. These moments and axial load are from service loads and consist of equal parts dead load and live load. Lateral support is provided only at the ends. Neglect the weight of the beam and investigate this member as a beam—column. Use F y = 50 ksi. a. Use LRFD. b. Use ASD.

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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 6.6.5P
Textbook Problem
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A simply supported beam is subjected to the end couples (bending is about the strong axis) and the axial load shown in Figure P6.6-5. These moments and axial load are from service loads and consist of equal parts dead load and live load. Lateral support is provided only at the ends. Neglect the weight of the beam and investigate this member as a beam—column. Use

F y = 50 ksi.

a. Use LRFD.

b. Use ASD.

Chapter 6, Problem 6.6.5P, A simply supported beam is subjected to the end couples (bending is about the strong axis) and the

To determine

(a)

If the member is a beam-column using LRFD method.

Explanation of Solution

Given data:

The live load is 16k.

The dead load is 16k.

The factored bending moment is 60ft-k.

Concept used:

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

Pu2ϕbPn+(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 and the axial live load is PL and the factored axial load is Pu.

Substitute 16k for PL and 16k for PD in Equation (I).

Pu=(1.6×16k)+(1.2×16k)=25.6k+19.2k=44.8k

Write the expression for factored bending moment.

Mnt=1.6ML+1.2MD (II)

Here, the factored bending moment of dead load is MD, the factored bending moment of live load is ML, and the factored bending moment of the load is Mnt.

Substitute 30ft-k for MD and 30ft-k for ML in Equation (II).

Mnt=1.6(30ft-k)+1.2(30ft-k)=84ft-k

Write the expression for the factor Cm.

Cm=0.60.4(M1M2) (III)

Here, the moment which has lesser value is M1 and the moment which has higher value is M2

Substitute 84ft-k for M1 and 84ft-k for M2 in Equation (III).

Cm=0.60.4(84ft-k84ft-k)=0.60.4(1)=0.6+0.4=1.0

Write the formula for calculating buckling stress.

Pe=π2EI(KL/r)2 (IV)

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

Substitute 29000ksi for E, 171in4 for I, 10ft for KL/r in Equation (IV).

Pe=π2(29000ksi)(171in4)((10ft)(12inch1ft))2=48943368.23k-in214400in2=3399k

Write the expression for moment amplification factor.

B1=Cm1(αPuPe) (V)

Here, the modification factor is Cm, Eulers buckling stress is Pe, and the factored axial load is Pu

To determine

(b)

If the member is a beam-column using ASD method.

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