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.

(a)

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

Given data:

The live load is 16 k.

The dead load is 16 k.

The factored bending moment is 60 ft-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 x−axis is Mux, the nominal flexural strength about x−axis is ϕbMnx, the flexural load about y−axis is Muy, and the nominal flexural strength about y−axis 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 16 k for PL and 16 k for PD in Equation (I).

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

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

Mnt=1.6(30 ft-k)+1.2(30 ft-k)=84 ft-k

Write the expression for the factor Cm.

Cm=0.6−0.4(−M1M2) (III)

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

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

Cm=0.6−0.4(−84 ft-k84 ft-k)=0.6−0.4(−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 29000 ksi for E, 171 in4 for I, 10 ft for KL/r in Equation (IV).

Pe=π2(29000 ksi)(171 in4)((10 ft)(12 inch1 ft))2=48943368.23 k-in214400 in2=3399 k

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

(b)

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