Solved Example: Design of A Steel Moment Frame by Direct Analysis Method Per AISC (ASD & LRFD)

Articles > Solved Example: Design of A Steel Moment Frame by Direct Analysis Method Per AISC (ASD & LRFD)


Determine the required strengths and effective length factors for the columns in the rigid frame shown below for the maximum gravity load combination, using LRFD and ASD. Use the direct analysis method. All members are ASTM A992 material.

Columns are unbraced between the footings and roof in the x- and y-axes and are assumed to have pinned bases.



The W12x65 has A = 19.1 in2.

The beams from grid lines A to B, and C to E and the columns at A, D and E are pinned at both ends and do not contribute to the lateral stability of the frame. There are no P-Δ effects to consider in these members and they may be designed using K=1.0.

The moment frame between grid lines B and C is the source of lateral stability and therefore must be designed using the provisions of Chapter C of the AISC Specification. Although the columns at grid lines A, D and E do not contribute to lateral stability, the forces required to stabilize them must be considered in the analysis. For the analysis, the entire frame could be modeled or the model can be simplified as shown in the figure below, in which the stability loads from the three “leaning” columns are combined into a single column.

From Chapter 2 of ASCE/SEI 7, the maximum gravity load combinations are:

wu = 1.2D + 1.6L
= 1.2(0.400 kip/ft) + 1.6(1.20 kip/ft)
= 2.40 kip/ft
wa = D + L
= 0.400 kip/ft + 1.20 kip/ft
= 1.60 kip/ft

Per AISC Specification Section C2.1, for LRFD perform a second-order analysis and member strength checks using the LRFD load combinations. For ASD, perform a second-order analysis using 1.6 times the ASD load combinations and divide the analysis results by 1.6 for the ASD member strength checks.


Frame Analysis Gravity Loads

The uniform gravity loads to be considered in a second-order analysis on the beam from B to C are:

w'u = 2.40 kip/ft wa' = 1.6(1.60 kip/ft)
= 2.56 kip/ft

Concentrated gravity loads to be considered in a second-order analysis on the columns at B and C contributed by adjacent beams are:

Pu' = (15.0 ft)(2.40 kip/ft)
= 63.0 kips
Pa' = 1.6(15.0 ft)(1.60 kip/ft)
= 8.34 kips


Concentrated Gravity Loads on the Pseudo “Leaning” Column

The load in this column accounts for all gravity loading that is stabilized by the moment frame, but is not directly applied to it.

PuL' = (60.0 ft)(2.40 kip/ft)
= 144 kips
PaL' = 1.6(60.0 ft)(1.60 kip/ft)
= 154 kips


Frame Analysis Notional Loads

Per AISC Specification Section C2.2, frame out-of-plumbness must be accounted for either by explicit modeling of the assumed out-of-plumbness or by the application of notional loads. Use notional loads.

From AISC Specification Equation C2-1, the notional loads are:

α = 1.0
Yi = (120 ft)(2.40 kip/ft)
= 288 kips
Ni = 0.002αYi (Spec. Eq. C2-1)
= 0.002(1.0)(288 kips)
= 0.576 kips
α = 1.6
Yi = (120 ft)(1.60 kip/ft)
= 192 kips
Ni = 0.002αYi (Spec. Eq. C2-1)
= 0.002(1.6)(192 kips)
= 0.614 kips


Summary of Applied Frame Loads


Per AISC Specification Section C2.3, conduct the analysis using 80% of the nominal stiffnesses to account for the effects of inelasticity. Assume, subject to verification, that αPr /Py is no greater than 0.5; therefore, no additional stiffness reduction is required.

50% of the gravity load is carried by the columns of the moment resisting frame. Because the gravity load supported by the moment resisting frame columns exceeds one third of the total gravity load tributary to the frame, per AISC Specification Section C2.1, the effects of P-δ upon P-Δ must be included in the frame analysis. If the software used does not account for P-δ effects in the frame analysis, this may be accomplished by adding joints to the columns between the footing and beam.

Using analysis software that accounts for both P-Δ and P-δ effects, the following results are obtained:

First-order results


Δ1st = 0.149 in.

Δ1st = 0.159 in. (prior to dividing by 1.6)

Second-order results


Δ2nd = 0.217 in.


Δ2nd = 0.239 in. (prior to dividing by 1.6)


Check the assumption that αPr/Py <= 0.5 and therefore, Tb = 1.0:

Py = FyAg = 50 ksi(19.1 in.2) = 955 kips


frac{alpha P_r}{P_y}=frac{1.0(72.6kips)}{955kips}=0.760leq0.5 


frac{alpha P_r}{P_y}=frac{1.6(48.4kips)}{955kips}=0.0811leq0.5


The stiffness assumption used in the analysis, Tb = 1.0, is verified.

Although the second-order sway multiplier is approximately 1.5, the change in bending moment is small because the only sway moments are those produced by the small notional loads. For load combinations with significant gravity and lateral loadings, the increase in bending moments is larger.

Verify the column strengths using the second-order forces shown above, using the following effective lengths (calculations not shown):

Use KLx = 20.0 ft
Use KLy = 20.0 ft

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