Buckling of Columns 57theory and agrees better than the double-modulus theory with test results. These inelastic buckling analyses using effective modulus are just academic history today. The present-day finite element codes capable of conducting incremental analyses of the geometric and material nonlinearities, as refined in their final form in the 1980s, can correctly evaluate the inelastic column strengths, including the effects of initial imperfections, inelastic material properties including strain hardening, and residual stresses. 1.13.1. Double-Modulus (Reduced Modulus) Theory Assumptions 1) Small displacement theory holds. 2) Plane sections remain plane. This assumption is called Bernoulli, or Euler, or Navier hypothesis. 3) The relationship between the stress and strain in any longitudinal fiber is given by the stress-strain diagram of the material (compression and tension, the same relationship). 4) The column section is at least singly symmetric, and the plane of bending is the plane of symmetry. 5) The axial load remains constant as the member moves from the straight to the deformed position. P Loading Et governs P = Pcr cr p Unloading E governs NA h2 h1 dx e R dx cg d z2 z1 2 s2 1 = E 1 s1 = Et Inside 1 2 2 concave Compressive Outside cr Et: the slope of stress-strain stress convex curve at = cr Figure 1-28 Reduced modulus model 13. s1 ðtensionÞ ¼ Ez1 y00 and s2 ðcompressionÞ ¼ Et z2 y00 .13.13.3) and the corresponding stresses are s1 ¼ Eh1 y00 (1.13.5) where Et ¼ tangent modulus.8) 0 0 Z h1 Z h2 Let Q1 ¼ z1 dA and Q2 ¼ z2 dA0EQ1 þ Et Q2 ¼ 0 0 0 (1.4) s2 ¼ Et h2 y00 (1.7) is expanded to ! Z h1 Z h2 " 00 y E z21 dA þ Et z22 dA ¼ Py (1. the curvature of the bent column is 1 d2y df ^ 2 ¼ (1.13.1) R dx dx From a similar triangle relationship.6) 0 0 Equating the internal moment to the external moment yields Z h1 Z h2 s1 z1 dA þ s2 z2 dA ¼ Py (1. The pure bending portion (no net axial force) requires Z h1 Z h2 s1 dA þ s2 dA ¼ 0 (1.13.13.13.6) is expanded to Z h1 Z h2 00 00 Ey z1 dA þ Et y z2 dA ¼ 0 (1.2) 32 ¼ z2 y00 (1.58 Chai Yoo In small displacement theory.7) 0 0 Equation (1. the flexural strains are computed 31 ¼ z1 y00 (1.13.13.13.13.10) 0 0 .9) Equation (1. 13.3).3) except that E has been replaced by E.10) takes the form EIy00 þ Py ¼ 0 (1.12) 0 0 Equation (1.13.cr ¼ 2 (1.13.13.18) I I A ‘ r .13. Corresponding critical load and critical stress based on the reduced modulus are p2 EI Pr. I1 is the moment of inertia of the tension side cross section about the neutral axis and I2 is the moment of inertia of the compression side cross section such that Z h1 Z h2 2 I1 ¼ z1 dA and I2 ¼ z22 dA (1.14) ‘ and p2 E sr.17) and I2 I1 Pr.15) ‘ r Introducing E Et I 2 I 1 Et sr ¼ ¼ þ < 1:0 and s ¼ < 1:0 (1.13) is the differential equation of a column stressed into the inelastic range identical to Eq. the reduced modulus. except that E is replaced by E.13. (1. then Eq. and its solution is identical to that of Eq. Buckling of Columns 59 EI1 þ Et I2 Let E ¼ (1.13) Equation (1.3.11) I which is called the reduced modulus that depends on the stress-strain relationship of the material and the shape of the cross section.cr p2 Esr sr ¼ s þ and sr.13. (1.cr ¼ ! "2 (1. If it can be assumed that E is constant.13.3.13.13) is a linear differential equation with constant coefficients.13.16) E E I I E the differential equation based on the reduced modulus becomes EIsr y00 þ Py ¼ 0 (1. (1.13.cr ¼ ¼ ! "2 (1. 13.2. prepare sr ð‘=rÞ curve. 1.21) ‘ 2 ! A E r Hence.13.cr ¼ ¼ with s ¼ (1. prepare sr s curve.13. prepare st ð‘=rÞ.60 Chai Yoo The procedure for determining sr. Tangent-Modulus Theory Assumptions The assumptions are the same as those used in the double-modulus theory. P+ P Et P s y cr M p r E 1 P+ P P M = – EtIy Figure 1-29 Tangent-modulus model .cr may be summarized as follows: 1) For s 3 diagram. The compressive stress increases at all points. If the load increment is assumed to be negligibly small such that DP <<< P (1. st vs ‘=r curve is not affected by the shape of the cross section. The procedure for determining the st ð‘=rÞ curve may be summarized as follows: 1) From s 3 diagram. prepare s s diagram. except assumption 5.20) and the corresponding critical stress is Pt. 3) From the result of step 2. 2) From the result of step 1.19) then Et Iy00 þ Py ¼ 0 (1. the tangent modulus governs the entire cross section. 2) From the result of step 1.cr p2 :Es Et st. such that the increase in average stress in compression is greater than the decrease in stress due to bending at the extreme fiber on the convex side.13. The axial load increases during the transition from the straight to slightly bent position. establish s s curve.