Structural Mechanics, Column Stability & Non-Linear P-Delta Bending
Eccentric Column Secant Formula Max Compressive Stress
Calculate Eccentric Column Direct Axial Stress P/A (MPa), Stability Secant Angle (rad), Exact Secant Maximum Fiber Compressive Stress (MPa), and Naive Linear Stress Without P-Delta (MPa).
Superposition P/A + M*c/I fails on slender eccentric columns—use 1/COS(angle) for P-Delta amplification!
When an axial compressive load P acts at eccentricity e from a slender column’s centroid, lateral mid-height deflection delta creates an extra secondary bending moment P * (e + delta) that invalidates linear superposition! Solving the exact differential beam-column equation yields the Secant Formula for peak outer-fiber compressive stress: sigma_max = (P / A) * (1 + (e * c / r^2) / COS(theta)), where the stability half-angle in radians is theta = (L_e / (2 * r)) * SQRT(P / (E * A))! As theta approaches PI() / 2 = 1.5708 rad (Euler’s concentric buckling load!), COS(theta) -> 0 and sigma_max blows up non-linearly!
Verify on a Steel Strut with Eccentric Bracket Load—Axial_Load P = 200.0 kN (200,000 N, B2), Area A = 2,500 mm² (B3 -> Direct P/A = 80.00 MPa in E2), Eccentricity_Ratio e*c/r^2 = 0.30 (B4), Effective_Length L_e = 3,000 mm (B5), Radius_of_Gyration r = 25.0 mm (L_e/r = 120, B6), and Young_Modulus E = 200 GPa (200,000 MPa, B7): Stability Angle theta equals (3,000 / 50) * SQRT(200,000 / 500,000,000) = 1.2000 rad (E3), Exact Secant Peak Stress hits 80 * (1 + 0.30 / COS(1.20)) = 146.23 MPa (E4), whereas Naive Linear Stress P/A*(1 + ec/r^2) predicts only 104.00 MPa (E5—missing +42.23 MPa of P-Delta bending)!
| Beam-Column Secant KPI (P=200 kN, A=2,500 mm², ec/r²=0.30, Le/r=120, E=200 GPa) | Formula | Verified Output |
|---|---|---|
| Direct Axial Stress P/A (MPa) | (Axial_Load_kN * 1000) / Area_mm2 | 80.00 MPa (125.00 @ 312.5 kN) |
| Stability Angle theta (rad) | (Le / (2*r)) * SQRT(P_N / (E_MPa * A)) | 1.2000 rad (1.5000 @ 312.5 kN) |
| Exact Secant Max Stress (MPa) | (P / A) * (1 + (ec_r2) / COS(theta)) | 146.23 MPa (655.13 @ 312.5 kN!) |
| Naive No-P-Delta Stress (MPa) | (P / A) * (1 + ec_r2) | 104.00 MPa (162.50 False Safe!) |
Exact Beam-Column Secant Stress & Euler-Limit Guard Formula
Validate Load > 0, Area > 0, Eccentricity_Ratio >= 0, Length > 0, Radius_r > 0, Modulus > 0, and guard theta >= PI()/2 ("EULER BUCKLED: Angle >= PI/2")!
=LET(ld_k, B2, ar_m, B3, ec_r, B4, le_m, B5, rg_m, B6, md_g, B7, IF(OR(ld_k <= 0, ar_m <= 0, ec_r < 0, le_m <= 0, rg_m <= 0, md_g <= 0), "Invalid Column Input", LET(p_nw, ld_k * 1000, th_r, (le_m / (2 * rg_m)) * SQRT(p_nw / (md_g * 1000 * ar_m)), IF(th_r >= PI() / 2, "EULER BUCKLED: Angle >= PI/2", ROUND((p_nw / ar_m) * (1 + ec_r / COS(th_r)), 2)))))Near-Euler Non-Linear Blow-Up Check (Set Axial Load P = 312.5 kN vs 360 kN)
Raise Axial_Load B2 by +56% from 200.0 kN to 312.5 kN (theta E3 = 1.5000 rad): while Naive Linear Stress E5 rises modestly to 162.50 MPa, Exact Secant Stress E4 explodes 4.5x to 655.13 MPa (yielding S355 steel)! Raising B2 further to 360.0 kN pushes theta past PI()/2, triggering "EULER BUCKLED: Angle >= PI/2"!
=IF(OR(B2 <= 0, B3 <= 0, B4 < 0, B5 <= 0, B6 <= 0, B7 <= 0), "Invalid Column Input", ROUND(((B2 * 1000) / B3) * (1 + B4), 2))Try the example
Use the original values or change them to test your understanding.