Eccentric Column (Secant Formula)

stabilitystressmass-cost

Verified build 12 relations · 1 identities proven · 3 modeling steps · 3 parity samples

The Euler column is a beautiful idealization: perfectly straight, loaded exactly through the centroid, perfectly content until — at one critical load — it bifurcates. Real columns never get that story. A bracket bolts on a few millimeters off-axis; a crane column carries its rail on one flange; even “centered” loads ride manufacturing tolerances. Offset the load by an eccentricity ee and the column carries a bending moment M=P(e+v)M = P(e + v) from the first newton — it bows immediately, the bow lengthens the moment arm, the longer arm deepens the bow. The self-amplification has an exact elastic answer, the secant formula:

σmax=PA[1+ecr2sec ⁣(L2rPEA)]\sigma_{max} = \frac{P}{A}\left[1 + \frac{ec}{r^2}\,\sec\!\left(\frac{L}{2r}\sqrt{\frac{P}{EA}}\right)\right]

Two things to read out of it. First, the dimensionless eccentricity ratio ec/r2ec/r^2 is the only place the offset enters — it families every column design chart ever printed. Second, the secant’s argument reaches π/2\pi/2 exactly when P=PE=π2EI/L2P = P_E = \pi^2 EI/L^2: the deflection and the stress diverge as the load approaches the Euler load. The imperfect column never buckles in the clean eigenvalue sense — it just runs away asymptotically toward the limit its perfect twin would have reached. (Dial PP upward in the widget and watch the bow do exactly this.)

Now the point this page exists to make. The stress is nonlinear in the load — double PP and σmax\sigma_{max} more than doubles, because the secant grows too. So the familiar margin σy/σmax\sigma_y/\sigma_{max} — fine for every linear THING in this catalog — overstates your safety. The honest question is: what load first yields a fiber? That means solving

σy=PyA[1+ecr2sec ⁣(L2rPyEA)]\sigma_y = \frac{P_y}{A}\left[1 + \frac{ec}{r^2}\,\sec\!\left(\frac{L}{2r}\sqrt{\frac{P_y}{EA}}\right)\right]

for PyP_y — and no algebra will ever isolate it: PyP_y sits both outside and inside the secant. Shigley’s text says it plainly: the load “appears twice … this equation needs an iterative process.” This page does exactly that, live — the root is bracketed inside (0,PE)(0, P_E), where it provably exists and is unique (the residual falls monotonically from σy\sigma_y to -\infty), and a verified root-finder pins it on every knob drag. The build proves the bracket’s sign change at dozens of sampled states and cross-checks the browser’s roots against 60-digit bisection. Both margins are displayed side by side: SFσ\mathrm{SF}_\sigma (deceptive) and SFP\mathrm{SF}_P (honest). At the default state they disagree by about 17% — in the unsafe direction.

One more cascade worth clicking through the material picker: swap A36 for quenched 4340 at the same EE and the stress readouts do not move — elastic stress amplification is set by stiffness and geometry alone — but PyP_y jumps by multiples. Strength is back in the story precisely because failure here is yielding, not the geometric escape of the perfect column.

Try it

Material

T3, bare flat sheet 0.010-0.128 in. thick, AMS 4037 / AMS-QQ-A-250/4 (MIL-HDBK-5J Table 3.2.3.0(b1), p. 3-71)

Bound properties of 2024-T3 aluminum sheet (bare)
E10.5 Msitypicalmil-hdbk-5j
sigma_y47 ksidesign min.mil-hdbk-5j
rho0.1 lb/inch**3typicalmil-hdbk-5j
Inputs
Cross-section area
Second moment of area
m⁴
Extreme-fiber distance
Radius of gyration
Eccentricity ratio
Euler load (asymptote)
Max compressive stress (midspan)
Midspan deflection
Margin on stress (deceptive)
Load at first yield
Margin on load (honest)
Column mass
kg

3 materials in the database are not listed here: no published value in our cited sources for every property this THING needs.

Materials modeled here: 2024-T3 aluminum sheet (bare) 304 stainless steel 6061-T6 aluminum 7075-T6 aluminum AISI 1045 medium-carbon steel AISI 4340 low-alloy steel (Ni-Cr-Mo) ASTM A36 structural steel (hot-rolled) C26000 Cartridge Brass (70/30) Nylon 6/6 (PA66), unfilled Ti-6Al-4V

Governing relations

A=πd24A = \frac{\pi d^2}{4}

Assumes: solid circular cross-section

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

I=πd464I = \frac{\pi d^4}{64}

Assumes: solid circular cross-section

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

c=d2c = \frac{d}{2}

Assumes: the bending stress peaks at the surface, on the concave side

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

r=I/Ar = \sqrt{I/A}

Assumes: definition — the section's bending stiffness expressed as a length

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

ecr2\frac{ec}{r^2}

Assumes: the dimensionless offset that families all secant-formula design charts

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

PE=π2EIL2P_E = \frac{\pi^2 E I}{L^2}

Assumes: pinned-pinned reference; for other end conditions read L as the effective length KL; the perfect-column limit this imperfect column approaches but never reaches

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

σmax=PA[1+ecr2sec ⁣(L2rPEA)]\sigma_{max} = \frac{P}{A}\left[1 + \frac{ec}{r^2}\,\sec\!\left(\frac{L}{2r}\sqrt{\frac{P}{EA}}\right)\right]

Assumes: linear-elastic material, plane sections (Euler-Bernoulli); pinned ends, load offset e at BOTH ends (single-curvature bowing); the exact elastic solution — no small-deflection linearization of the moment · Valid while: P ≥ P_E: at or beyond the Euler load no bent equilibrium exists for the elastic column — the deflection grows without bound and the elastic readouts are refused. The honest margin SF_P < 1 told the story long before this point. σ_max > σ_y: the extreme fiber at midspan has yielded, and the elastic secant solution is fiction past first yield. The load at first yield P_y and the load margin SF_P remain exact — read those.

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

δ=e[sec ⁣(L2rPEA)1]\delta = e\left[\sec\!\left(\frac{L}{2r}\sqrt{\frac{P}{EA}}\right) - 1\right]

Assumes: midspan bow of the elastic beam-column; diverges as P approaches P_E

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

SFσ=σyσmax\mathrm{SF}_\sigma = \frac{\sigma_y}{\sigma_{max}}

Assumes: the linear-thinking margin — WRONG to design by here, shown for contrast; stress grows faster than load, so this overstates the true margin

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

σy=PyA[1+ecr2sec ⁣(L2rPyEA)]\sigma_y = \frac{P_y}{A}\left[1 + \frac{ec}{r^2}\,\sec\!\left(\frac{L}{2r}\sqrt{\frac{P_y}{EA}}\right)\right]

Assumes: first yield of the extreme midspan fiber defines the usable load ceiling; transcendental in P_y — no closed form exists; solved by bracketed root-finding inside (0, P_E), where a root always exists and is unique

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §4-13 (Columns with Eccentric Loading): the secant equation for the critical load, and the design rule that the factor of safety must be applied to the load because the stress is a nonlinear function of it.

SFP=PyP\mathrm{SF}_P = \frac{P_y}{P}

Assumes: the margin a nonlinear member must quote on LOAD, not stress (Shigley's design rule)

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §4-13 (Columns with Eccentric Loading): the secant equation for the critical load, and the design rule that the factor of safety must be applied to the load because the stress is a nonlinear function of it.

m=ρALm = \rho A L

Assumes: prismatic column, uniform density

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §11.5 (Columns with Eccentric Axial Loads) and §11.6 (The Secant Formula for Columns).

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

ke2=PEIk_{e}^{2} = \frac{P}{E I}

1. Offset the load by e and the column carries a bending moment from the very first newton: M(x) = −P·[e + v(x)]. The elastic curve obeys EI·v″ = M, i.e. v″ + k²v = −k²e with k² = P/EI — the same harmonic operator as Euler's perfect column, but now FORCED by the offset instead of waiting for an eigenvalue. There is no bifurcation to wait for. — modeling: beam-column with end eccentricity modeling step

δmid=e(1+1cos(Lke2))\delta_{mid} = e \left(-1 + \frac{1}{\cos{\left(\frac{L k_{e}}{2} \right)}}\right)

2. Solve with pinned ends v(0) = v(L) = 0: the bow peaks at midspan as δ = e[sec(kL/2) − 1]. At kL/2 = π/2 — which is exactly P = P_E — the secant blows up: the imperfect column does not buckle AT the Euler load, it deflects without bound APPROACHING it. (The test suite re-solves this boundary-value problem independently with dsolve.) — modeling: solve the beam-column BVP (re-derived in pipeline tests) modeling step

σmax=Pc(δmid+e)I+PA\sigma_{max} = \frac{P c \left(\delta_{mid} + e\right)}{I} + \frac{P}{A}

3. The midspan section carries the axial force AND the amplified moment M_max = P(e + δ). Superpose the two stresses at the concave extreme fiber — substituting δ and writing Ar² = I collapses this to the secant formula above. From here on the algebra is machine-verified. — superposition of axial and bending stress at midspan

PSFy=PyP SF_{y} = P_{y}

4. Because σ_max is nonlinear in P, the familiar σ_y/σ_max margin lies — doubling the load far more than doubles the stress. The honest question is "what load first yields a fiber?", and answering it means solving the secant equation for P_y, a transcendental with no closed form. This page brackets the root inside (0, P_E) — where one always exists, uniquely — and lets a verified root-finder answer live. SF_σ is displayed anyway, struck through in spirit: compare it to SF_P and never trust it again. — definition: the margin is taken on load (Shigley) — solved numerically, bracketed modeling step

How it fails

The secant formula is exact elastic theory for an idealized member — and each idealization marks a way the real column quits earlier:

  • Euler Column (Buckling)

    Push down on a slender strut and it fails sideways — at a load set entirely by stiffness and geometry. Yield strength is nowhere in Euler's formula: the "strong" steel column buckles at exactly the same load as the mild one. Strength only decides where the formula stops being true.

    • stability
    • stress
    • mass-cost
  • Symmetric Two-Bar Truss

    Two identical pin-jointed bars share a load at a common joint — the member force is P/(2cos α), which blows up as the truss flattens toward horizontal. Statically determinate by construction: equilibrium alone fixes the forces, no compatibility needed. Flatten it and watch the forces (and the joint deflection) diverge; in compression each bar must also clear Euler buckling.

    • statics
    • stress
    • stability
    • mass-cost
  • Cantilever Beam (End Load)

    A beam fixed at one end, loaded at the other — the fruit-fly of structures. One widget shows why stiffness (E) and strength (σ_y) are independent axes: swap steel for titanium and deflection goes UP while the safety factor also goes up.

    • stress
    • mass-cost
  • Composite Bar (Core + Sleeve)

    A solid core inside a concentric sleeve, bonded between rigid end plates and pushed by a centric axial load. The two materials must stretch together, so the load splits in proportion to each member's axial stiffness A·E — and the build solves that coupled 2×2 share exactly. Swap the sleeve's metal and watch the load migrate to the stiffer member.

    • stress
    • mass-cost
  • Compound Cylinder (Shrink Fit)

    Where the monobloc wall gave up: shrink a jacket over a liner and the interference squeezes the bore into hoop compression before the pressure ever arrives. Service tension must spend that compression first — and at the balanced fit with the interface at √(r_i·r_o), the elastic pressure ceiling approaches DOUBLE the one no solid wall could pass.

    • stress
    • mass-cost
  • Fixed-Fixed Beam (UDL)

    A beam built rigidly into a wall at BOTH ends under a uniform load. Two equilibrium equations, four unknown reactions — indeterminate to the second degree — so two compatibility conditions (zero slope and zero deflection at a released end) close the system, and the build solves the coupled 4×4 group exactly. The fixing moment at each wall governs, and none of the reactions cares about the material.

    • stress
    • mass-cost

Chains with

Outputs whose SI dimension and quantity kind match another THING's input — the only wires the planner's connectionLegal accepts (invariant 2, computed at build time, not hand-listed). Wire these on the chaining demo.

+ 26 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources