53:134 Structural Design IICombined Bending and Axial Loads Related Material AISC LRFD Manual: Part 6 AISC LRFD Specifications: Chapter C, Chapter H Two New Concepts Superposition of stresses due to bending and axial loads Secondary moment due to axial loads; moment amplification Interaction Equations • Strength interaction equations relating axial compression Pu to bending moment M u have been recognized as the practical procedure for design. • Axial compression strength requirement Required axial strength ≤ Design axial strength of the section Pu ≤1 Pu ≤ φc Pn , or φc Pn • Bending moment strength requirement Required bending strength ≤ Design bending strength of the section Mu ≤1 M u ≤ φb M n , or φb M n • Combined bending and axial compression: Interaction equation Pu Mu + ≤1 φc Pn φb M n J.S. Arora/Q. Wang 1 CombinedLoading.doc Problem is nonlinear.S.2 . axial load term is reduced. φ c Pn ⎛ Mu ⎞ ⎟ ≤1 +⎜ 2φc Pn ⎜ φb M n ⎟ ⎝ ⎠ Pu φ c = 0.doc . φ c Pn 8 ⎛ Mu ⎞ ⎟ ≤1 + ⎜ φc Pn 9 ⎜ φb M n ⎟ ⎝ ⎠ Pu • For Pu < 0. called secondary moment that must be accounted for in design. φ c Pn φc Pn Pu < 0. AISC permits use of moment amplification method or second order analysis.2. φ b = 0. Arora/Q.85. Wang 2 CombinedLoading. bending term is slightly reduced. large axial load.53:134 Structural Design II LRFD Criteria • For Pu ≥ 0. J. Axial load induces additional moment. small axial load.2.2. Pn = Nominal axial strength of the section M n = Nominal bending strength of the section Moment Amplification • Beam-column: the member subjected to axial compression and bending.90 Pu = Factored axial compression load M u = Factored bending moment (moment magnification used) Pu ≥ 0. requiring second order analysis. doc . Arora/Q.S.(2) kL . M z max wL2 ⎛ 8 ⎞⎛ kL ⎟ sec − 1⎞ ⎜ = ⎟ ⎜ 2 ⎟⎜ 8 ⎝ (kL ) ⎠⎝ 2 ⎠ It can be shown that M max J.53:134 Structural Design II • Maximum moment: Note that the coordinate system here is different from the one used earlier.(1) If w( z ) = 0 and M 1 = M 2 = M . d2y M z = M i + Py = − EI 2 dz Solution: -. M z max = M sec -. where k = 2 P / EI If w( z ) = w and M 1 = M 2 = 0 . α = Pu = M0⎜ ⎟ Pe ⎝1− α ⎠ 3 CombinedLoading. Wang ⎛ 1 ⎞. The factored moment is given as M u = B1M nt + B2 M lt (AISC Equation C1-1) M nt = maximum moment assuming that no side sway occurs. side sway is prevented. Wang 4 CombinedLoading. ⎝1− α ⎠ Pe = π 2 EAg ( KL / r ) 2 • Two Types of Amplification Factors – Braced vs Unbraced Frames 1.S. Arora/Q. whether the frame is actually braced or not (subscript nt is for “no translation”) M lt = maximum moment caused by side sway (subscript lt is for “lateral translation”). The second one accounts for the effect of sway when the member is part of an unbraced frame. This moment can be caused by lateral loads or by unbalanced gravity loads. The first one accounts for only member deflection.doc . B1 = amplification factor for the moment occurring in the member when it is braced against side sway. 2. α = Pu ⎜ ⎟ Pe . J. B2 = amplification factor for the moment resulting from side sway.53:134 Structural Design II Moment amplification factor ⎛ 1 ⎞. and the primary and secondary moments are additive. α = Pu Pe1 . For transversely loaded members.4⎜ ⎜M ⎟ ⎝ 2⎠ (AISC Equation C1-2) where M1/M2 is the ratio of the bending moments at the ends.6 − 0. Cm can be taken as 0. This factor applies only for braced conditions. Arora/Q.53:134 Structural Design II • Members in Braced Frames The amplification factor 1 / (1 − α ) is for members braced against side sway. the maximum moment in a member depends on distribution of the moment. J. The ratio is positive for members bent in reverse curvature and negative for single curvature. This distribution is accounted for by a factor Cm applied to the amplification factor. Cm = 1. the amplification factor was derived for this condition) • Members in Unbraced Frames Maximum secondary moment is always at the end. Wang 5 CombinedLoading. there is no need for Cm. Pe1 = π 2 EI ( KL ) 2 = π 2 EAg ( KL / r )2 Evaluation of Cm 1. i.e.0 if they are unrestrained (pinned. When no transverse loads are acting on the member ⎛ M1 ⎞ ⎟ Cm = 0.85 if the ends are restrained against rotation and 1.. Therefore. The members can bend in single curvature or reverse-curvature. Thus. 2.doc .S. M1 is smaller in absolute value. B1 = 1−α Cm ≥ 1. doc . use KL/r for the axis of bending and a value of K corresponding to the unbraced conditions) • Aminmansour (2000) developed a simplified design procedure. ⎧bPu + mM u ≤ 1.0. AISC 37.2 6 CombinedLoading. Arora/Q. u ⎩ u J. ⎪ ⎨ ⎪bP / 2 + 9mM / 8 ≤ 1. 41-72. • Refer to Pages 6-10 and 6-11 of the AISC Manual for more details. Wang Pu /(φc Pn ) ≥ 0. which was later adopted by LRFD.S. 2.2 Pu /(φc Pn ) < 0. Ref: Engineering Journal.0. No.53:134 Structural Design II Amplification Factor for Side sway Moments B2 = Or 1 (∑ Pu ) ⎛ Δoh ⎞ (AISC Equation C1-4) ⎜ 1− ⎜ L ⎟ ⎟ (∑ H ) ⎝ ⎠ 1 ⎛ ∑ Pu ⎞ ⎟ 1− ⎜ ⎜∑P ⎟ ⎝ e2 ⎠ B2 = (AISC Equation C1-5) where ∑ Pu = sum of factored loads on all columns in the Δoh = story under consideration drift (side sway displacement) of the story under consideration ∑ H = sum of all horizontal forces causing Δoh L= story height ∑ Pe 2 = sum of the Euler loads for all columns in the story (when computing Pe2. Arora/Q. 4.1 2. Use the values of b and m for the shape selected to check the AISC interaction equation. Select a shape from Table 6.S. J.2 that has values of b and m close to those needed.1 and Table 6. Procedure: 1.2 for all W-shape steels. Solve for m from the AISC interaction equation 3. φc Pn m= 8 which can be obtained from 9(φb M n ) AISC-LRFD Manual Table 6. Select an average b value from Table 6.doc .53:134 Structural Design II where b = 1 . Wang 7 CombinedLoading.