Euler, Johnson and tangent-modulus column allowables with a crippling cutoff, following Section 2 of the Air Force Flight Dynamics Laboratory Stress Analysis Manual (AFFDL-TR-69-42), and a beam-column check for an eccentric or side-loaded strut. Pick a cross-section and a material and the page works the section properties, the local crippling stress and the column curve, and shows what governs.

Inputs
Cross-section
Column
Load
Material
Advanced options
Caps the column curve below the crippling stress, for a limit set elsewhere (a joint, inter-rivet buckling, a test value). Bruhn C2 treats such a cutoff the same way.
Margins of safety
Show detailed calculation steps
Loading condition & section
Elevation — end fixity and load
Cross-section — drawn to your dimensions
Column curve
Reference Guide — Columns in Compression
01

The Two Ways a Column Fails

A compression member has two failure modes and a real design has to clear both. Column buckling is the member bowing as a whole, its cross-section keeping its shape. Crippling is the thin elements of that cross-section buckling locally — a flange rolling over, a web dimpling — while the member as a whole stays straight.

Which one arrives first depends on how slender the member is and how thin its walls are. They are not independent: once a section has crippled it cannot carry more, so the crippling stress sets the ceiling that the column curve is allowed to reach. That coupling is the whole point of this page.

A stocky, thin-walled strut is limited by crippling. A long, chunky one is limited by Euler. Most real members sit between the two, which is where the short-column branch applies.
The two modes side by side
Column buckling versus crippling: a slender column bows as a whole with its cross-section unchanged, while a short stub stays straight along its axis and a thin flange buckles into waves
02

Section Properties and the Weak Axis

Buckling happens about whichever axis is weakest, so the section property that matters is the minimum second moment of area and the radius of gyration that follows from it.

Radius of gyration \[ \rho = \sqrt{I_{min}/A} \]

For a symmetric shape the weak axis is a drawing axis and there is nothing to think about. For an angle or a zee it is not: the product of inertia is non-zero and the true weak axis is rotated. This page finds the principal axes rather than assuming them.

Principal second moments \[ I_{max,min} = \frac{I_x + I_y}{2} \pm \sqrt{\left(\frac{I_x - I_y}{2}\right)^2 + I_{xy}^2} \]
An equal-leg angle has its weak axis at 45° — the diagonal perpendicular to the one through its heel. Taking Iy instead of Imin overstates the allowable, so the error is unconservative.
Where an angle’s principal axes actually lie
Equal-leg angle with both principal axes at 45 degrees: the strong axis runs through the heel and the opposite corner, the weak axis is perpendicular to it

Section properties here come from a rectangle decomposition of the shape, so area, centroid and second moments are exact for the thin-walled idealisation rather than approximated with a formula per shape.

03

Crippling of Thin Sections

The manual's method is to break a built-up section into angles, find the crippling stress of each, and area-weight them. Everything on this page except the round tube and the solid bars is handled this way.

Element crippling stress — Equation 2-26 \[ F_{cc} = \frac{C_e \sqrt{F_{cy} E}}{\left(b'/t\right)^{0.75}}, \qquad b' = \frac{b+h}{2} \]
Built-up section — Equation 2-29 \[ F_{cc,\;section} = \frac{\sum F_{cc,i}\,A_i}{\sum A_i}, \qquad A_i = (b + h - t)\,t \]

The coefficient Ce is set by how the element's edges are held. An unsupported edge is free to roll, so the fewer free edges the element has the more it carries.

Edge restraint and the coefficient
Three edge-restraint cases for a flange and the crippling coefficient each one takes
Edge conditionCeWhere it applies
Two edges free0.316Plain angle
One edge free0.342Flange outstand of an I-section, channel or zee
No edge free0.366Wall of a closed box or rectangular tube
At the same b′/t a box wall crips at 0.366/0.316 = 1.16× the stress of a plain angle leg.
How each shape is divided
An I-section, a channel and a box divided into the angle elements the crippling method uses
ShapeElementsLegs b, hCe
Angle1b, h0.316
Channel2bf, h/20.342
Zee2bf, h/20.342
I-section4bf/2, h/20.342
Rectangular tube4b/2, h/20.366
Solid bar or round——no local buckling; plateau is Fcy
Nothing above the compressive yield has any meaning, so every element is capped at Fcy. A stocky section simply reaches Fcy and stops.
Needham’s crippling curves (Eq 2-26, the curves of Figures 2-71 and 2-72)
Dimensionless crippling stress Fcc/√(FcyE) = Ce/(b′/t)0.75, drawn from the equation the page uses; the dot is the most slender element of the current section. Above the Fcy cap the curve no longer applies.
04

Round Tubes Are Their Own Case

A tube has no flanges to roll over, so the angle method does not apply. Its wall buckles as a cylinder instead.

Equation 2-23 \[ F_{cc} = C\,\frac{E\,t}{r} \]

The coefficient C falls as the wall gets thinner relative to the radius, along the curve the manual prints as Figure 2-67. It is digitized here and interpolated, and the same cap at Fcy applies.

The theoretically correct value, Fcc = 0.605 E t/r, runs far above test. The manual's own example gives 12,100 psi from theory against 4,400 psi from this curve for the same tube — which is why the empirical coefficient is used.
Figure 2-67 — coefficient C against r/t
Digitized from the printed chart; hover for a value.
05

The Column Curve

Above a transition slenderness the member fails elastically and Euler governs. Below it the material has already yielded locally, and a parabola tangent to the Euler curve is used instead.

Euler — long columns \[ F_c = \frac{\pi^2 E}{\left(L'/\rho\right)^2} \]
Johnson parabola — short columns \[ F_c = F_{cc}\left[1 - \frac{F_{cc}\left(L'/\rho\right)^2}{4\pi^2 E}\right] \]
Transition \[ \left(\frac{L'}{\rho}\right)_{cr} = \sqrt{\frac{2\pi^2 E}{F_{cc}}} \;\Rightarrow\; F_c = \frac{F_{cc}}{2} \]

Both branches give Fcc/2 at the transition and share a slope there, so the allowable is continuous across the changeover.

Note the parabola starts from Fcc, not Fcy. That one substitution is what carries the crippling result into the column answer, and it is why a thin-walled section is worth less as a column than its yield strength suggests.
How the branches fit together
The column curve: a crippling plateau, a Johnson parabola and the Euler hyperbola meeting tangentially
06

End Fixity

The coefficient c describes how much the ends are held, and enters through the effective length L′ = L/√c. Fixing both ends halves the effective length; a cantilever doubles it.

The four end-fixity cases with their buckled shapes and coefficients
End conditioncL′/LEffect
Pinned – pinned1.01.00The reference case
Fixed – pinned2.050.70About twice the load of pinned–pinned
Fixed – fixed4.00.50Four times, in theory
Fixed – free0.252.00A quarter — the worst case
These are the theoretical values. Real joints are never perfectly fixed and most structures departments require a reduced figure — commonly c = 3 or less in place of 4. Check what your programme allows before taking credit for full fixity.
Elastic end restraint (Figure 2-38)

A real end is held by whatever it is attached to, which acts as a rotational spring of stiffness μ (in-lb per radian). With the same spring at both ends, or one spring and a pin, the manual gives c in closed form through the restraint parameter μL/EI:

\[ \text{both ends:}\quad \frac{\mu L}{EI} = -\pi\sqrt{c}\,\cot\frac{\pi\sqrt{c}}{2} \qquad\qquad \text{one end:}\quad \frac{\mu L}{EI} = \frac{\pi^2 c}{\pi\sqrt{c}\,\cot\pi\sqrt{c} - 1} \]

The manual prints the first without its minus sign. Between c = 1 and 4 the cotangent is negative, so the sign is needed for a positive restraint. The page solves each equation for c. c rises fast at first and then levels off: μL/EI = 20 already gives 3.3 of the possible 4.0 for two restrained ends. It is the EI of the buckling axis (Imin) that counts, so a stiff member gains less from the same joint than a slender one.

Lateral supports along the span (Figures 2-40 and 2-41)

A pinned column braced sideways by a member of spring stiffness K (lb/in) buckles at a higher load than its full length suggests. Figure 2-40 covers one support at x from an end, with q = KL³/EI. Figure 2-41 covers two supports a distance b apart, centred, with q = KL³/(8EI). Note that the second chart uses 8EI.

The page does not read the charts. It solves the buckling problem behind them exactly: y = a sin kx + b cos kx + cx + d in each bay, deflection, slope and moment continuous through every support, the shear jumping by the spring force, and pinned ends. The lowest load at which that system has a solution gives c. The results match the printed curves wherever they can be read (q = 40 at mid-span: 1.81 here, 1.80 on the chart) and recover the limiting cases:

  • a mid-span support is rigid once q ≥ 16π² = 158, when the column buckles in two half-waves (c = 4);
  • a rigid support right beside a pin clamps that end (c = 2.05);
  • two rigid supports at the third points give three equal bays (c = 9).
Figure 2-40, computed: one lateral spring at x/L, q = KL³/EI. The red dot is the current column when that case is selected.
Figure 2-41, computed: two lateral springs, centred and b apart, q = KL³/(8EI). The printed chart draws some curves as straight segments to a sharp peak; the exact curves are rounded.
Load spread along the length (Figure 2-44)

A column loaded along its length, such as a stringer picking up shear flow or a heavy vertical member under its own weight, is less critical than one carrying the whole load at its end. Each section only carries the load from the part beyond it. Enter the total load as P; the larger c accounts for the spreading. The case is reacted at the top, as drawn in the manual:

Top (reacting) endBottom endcEnd-loaded c
PinnedPinned1.871.0
FixedPinned3.552.05
FixedFixed7.54.0
FixedFree0.7940.25

The fixed–free value is Greenhill’s heavy column, Pcr = 7.84 EI/L². The manual also notes that welded steel-tube truss members are usually taken as c = 2.0 (1.0 for engine mounts, or where every member at a joint is in compression). Stringers between flexible ribs are taken as 1.0, and about 1.5 where clipped to stiff bulkheads. Use Custom coefficient for those.

07

What This Page Computes

Three checks, reported side by side, with the lowest governing:

  • Column allowable — the governing answer. Fc·A using whichever branch applies at this slenderness.
  • Crippling — Fcc·A, the local capacity with no column effect at all. This is the ceiling; a very short member is limited by it.
  • Euler — π²EI/L′² on its own. Above the transition it equals the column allowable; below it, it reads high and is shown only for comparison.

With Beam-column ticked, a fourth check adds the bending from end eccentricity and side load, amplified by the axial load (card 10). The Column curve selector sets which curve the column allowable is read from (card 09).

Material properties are read from the same MIL-HDBK-5 dataset behind Material Property Lookup and Multi-Material Comparison. Only conditions publishing both Fcy and E are offered, since the method cannot run without them. Tick Full MIL-HDBK-5 database for the whole set, with grain direction and a service temperature that derates both properties off the effect-of-temperature curves.

08

Limitations

Torsion is checked elastically, for the four open shapes. Angles, channels, zees and I-sections get a flexural-torsional check (card 14) built on thin-wall section constants with uniform thickness, and lips, bulbs and fillets are ignored. The same effective length is used for bending and twisting, so an end that is pinned in bending but prevented from warping is treated conservatively. Closed sections have enough torsional stiffness that the bending mode always governs.

Beam-column assumptions. The amplification formulas are the pinned-end solutions, applied on the effective length L′. Eccentricity is equal at both ends and on the same side (single curvature, the worst case); the side load is uniform over the span or a single load at midspan. An initial bow of amplitude a0 can be entered as a side load w = 8 P a0/L² for a first estimate.

Uniform thickness. The crippling decomposition assumes one thickness through the section, which is what formed sheet and most extrusions have. A section with a heavy flange and a light web needs its elements treated with their own b/t.

Which column curve. The Johnson parabola is the conservative, widely used approximation, but it is not reliably conservative — it can sit either side of the measured curve depending on how rounded the alloy’s knee is. The tangent-modulus options follow the material instead; the Ramberg–Osgood one is only as good as the nc entered. Where no MIL-HDBK-5 curve exists for a condition, the MIL-HDBK-5 option falls back to Johnson and says so.

No joint or end-detail check. Fastener bearing, tension clips and local crushing of a tube end are outside the scope of this page, and any of them can govern before the column does.

09

Tangent Modulus: Two Sources

Above the proportional limit a column buckles on the slope of the stress-strain curve, not on E. Engesser’s tangent-modulus formula puts that slope into Euler’s:

\[ F_c = \frac{\pi^2 E_t(F_c)}{(L'/\rho)^2} \]

It is implicit, because Et depends on the stress being solved for, so the page solves it numerically. The two options differ only in where Et comes from.

  • MIL-HDBK-5 Et curve — the compressive tangent-modulus curve published for the alloy, digitized from the handbook figures and matched to the selected condition by its compressive yield.
  • Ramberg–Osgood — the three-parameter fit \( \varepsilon = \sigma/E + 0.002\,(\sigma/F_{cy})^{n} \), whose slope gives \( E_t = \sigma \big/ \left[\sigma/E + 0.002\,n\,(\sigma/F_{cy})^{n}\right] \). It needs only E, Fcy and the shape factor nc, so it works for any condition. The stress is held at Ftu at the short end.

Both are capped by the crippling stress, and by the optional upper stress limit, exactly as the Johnson curve is.

10

Beam-Columns

A strut that also bends — from a load applied off its axis, or a side load along its length — deflects, and the axial load acting through that deflection adds more moment. The exact pinned-end solutions, with \( k = \sqrt{P/EI} \) and \( u = kL'/2 \), are:

\[ M_e = P\,e\,\sec u, \qquad M_w = \frac{w}{k^2}\left(\sec u - 1\right), \qquad M_Q = \frac{Q}{2k}\tan u \]

for equal end eccentricity e, a uniform side load w and a midspan load Q. At the same axial load they add, and the peak stress is \( f = P/A + M_{max}\,c/I \) about the weak axis. As P approaches the Euler load, u approaches π/2 and the moment runs away.

The check holds that peak stress to the column allowable stress from the selected column curve. Because the moment is not proportional to the load, the margin is found as the load factor that brings the peak stress to the allowable with every load scaled together, less one. With eccentricity alone this is the secant formula (Timoshenko & Gere §1.11; Bruhn C4).

11

Verification

The column math (js/column-math.js) is run in automated tests against closed-form solutions and the AFFDL charts.

CheckReferenceThis page
Ramberg–Osgood tangent modulus at Fcy (E 10.5 Msi, Fcy 65 ksi, n 15)Fcy/(Fcy/E + 0.002n), NACA TN 902exact
Tangent-modulus column (Engesser)Fc = π²Et(Fc)/(L′/ρ)²solved to 0.001 psi
Eccentric strutSecant formula (Timoshenko & Gere)exact
Singly symmetric flexural-torsional bucklingAISC 360 eq. E4-5exact
Both ends restrained, μL/EI = 203.3 (Fig. 2-38)3.32
One lateral spring at mid-span, q = 401.80 (Fig. 2-40)1.81
Two rigid supports at the third points9.0 (Fig. 2-41)9.000
Heavy column, fixed–free (Fig. 2-44)q L³ = 7.837 EI (Greenhill)c = 0.794

The worked example below carries one strut through every step by hand.

12

Worked Example: A Square Steel Strut

A 2″ × 2″ × 0.0625″ square tube of normalized 4130 steel (Fcy = 75 ksi, E = 29.0 Msi, the MIL-HDBK-5 S-basis values) is 30″ long, pinned at both ends, and carries 5,000 lb. These are the page’s opening values; load the example to follow along, with the beam-column check switched on.

Step 1 — section \[ A = 0.4844\ \text{in}^2, \qquad I = 0.3034\ \text{in}^4, \qquad \rho = \sqrt{I/A} = 0.7914\ \text{in} \]
Step 2 — crippling (Needham) \[ \text{four corners, } \frac{b'}{t} = \frac{(1.0 + 1.0)/2}{0.0625} = 16.0,\quad C_e = 0.366 \text{ (no edge free)} \] \[ F_{cc} = \frac{0.366\sqrt{75{,}000 \times 29\times10^6}}{16.0^{0.75}} = 67.5\ \text{ksi} \]
Step 3 — column \[ \frac{L'}{\rho} = \frac{30}{0.7914} = 37.9 \;<\; \sqrt{\frac{2\pi^2 E}{F_{cc}}} = 92.1 \;\Rightarrow\; \text{Johnson} \] \[ F_c = 67.5\left[1 - \frac{67.5\,(37.9)^2}{4\pi^2(29{,}000)}\right] = 61.8\ \text{ksi}, \qquad P_{allow} = 61.8 \times 0.4844 = 29{,}900\ \text{lb}, \qquad MS = +4.98 \]

The tube is short: it sits well left of the transition slenderness, so its strength is set by local crippling of the walls with only a 9% column knockdown. The tangent-modulus option shows the same thing from the material side. The handbook Et curve for normalized 4130 would allow 70.7 ksi, above crippling, so crippling caps it at 67.5 ksi; Johnson is 9% conservative here.

Step 4 — the same strut with a 0.05″ load offset \[ u = \frac{L'}{2}\sqrt{\frac{P}{EI}} = 0.358, \qquad M = Pe\sec u = 5{,}000 \times 0.05 \times 1.068 = 267\ \text{in-lb} \] \[ f_{max} = \frac{P}{A} + \frac{Mc}{I} = 10.32 + \frac{267 \times 1.0}{0.3034} = 11.20\ \text{ksi} \]

Scaling the load until fmax reaches the 61.8 ksi column allowable gives a load factor of 5.35, MS = +4.35. A modest offset of 1/40 of the tube width costs about 13% of the margin. At this low P/PE = 0.05 the secant amplification is only 7%; it grows without limit as the load nears the Euler load of 96,500 lb.

13

References

  1. Air Force Flight Dynamics Laboratory (1986). Stress Analysis Manual, AFFDL-TR-69-42, Section 2, Columns: end fixity, Johnson-Euler curves, crippling (Needham), round tubes (Figure 2-67). Open Section 2.
  2. Needham, R. A. (1954). “The ultimate strength of aluminum-alloy formed structural shapes in compression.” Journal of the Aeronautical Sciences, 21(4), 217–229. The crippling method.
  3. Bruhn, E. F. (1973). Analysis and Design of Flight Vehicle Structures. Jacobs Publishing. Chapters C1 and C2 (columns, beam-columns) and C7 (crippling).
  4. Shanley, F. R. (1947). “Inelastic column theory.” Journal of the Aeronautical Sciences, 14(5), 261–268. Why the tangent modulus is the right one.
  5. Ramberg, W. and Osgood, W. R. (1943). Description of stress-strain curves by three parameters. NACA TN 902. Download PDF (2.3 MB).
  6. Timoshenko, S. P. and Gere, J. M. (1961). Theory of Elastic Stability, 2nd ed. McGraw-Hill. Chapter 1, beam-columns; chapter 2, elastic buckling.
  7. Young, W. C., Budynas, R. G. and Sadegh, A. M. (2020). Roark’s Formulas for Stress and Strain, 9th ed. Table 8.8, beam-columns.
  8. Gerard, G. and Becker, H. (1957). Handbook of Structural Stability, NACA TN 3781–3785. Crippling and local buckling. Part I, buckling of flat plates (PDF, 39 MB); Part IV, failure of plates and crippling (PDF, 3.6 MB).
  9. Department of Defense (2003). MIL-HDBK-5J. Design allowables and the tangent-modulus curves.
14

Flexural-Torsional Buckling

An open section twists easily. When its shear centre does not coincide with its centroid, as in an angle or a channel, the compressive load acting through the centroid twists the member as it bends, and the two modes combine into one that fails below either alone (AFFDL-TR-69-42 §2.3.3, Figures 2-56 to 2-63; Timoshenko & Gere §5.3). With principal axes x, y and the shear centre at (x₀, y₀) from the centroid, the elastic critical load is the smallest root of

\[ r_0^2(P-P_x)(P-P_y)(P-P_z) - P^2x_0^2(P-P_y) - P^2y_0^2(P-P_x) = 0 \] \[ P_x = \frac{\pi^2EI_x}{L'^2},\quad P_y = \frac{\pi^2EI_y}{L'^2},\quad P_z = \frac{1}{r_0^2}\left(GJ + \frac{\pi^2EC_w}{L'^2}\right),\quad r_0^2 = x_0^2 + y_0^2 + \frac{I_x+I_y}{A} \]

J = Σbt³/3 is the St Venant constant of the thin walls and Cw the warping constant: about zero for an angle, whose legs meet at the shear centre, and large for an I-section, whose flanges must bend to let it twist. A doubly symmetric section (I) and a point-symmetric one (zee) have the shear centre at the centroid, so the modes stay separate and only pure torsion can undercut bending. A channel couples torsion with bending about its axis of symmetry, and an angle couples all three.

Thin equal-leg angles are the classic victims: for an angle the torsional mode is really the local buckling of the legs about their common corner, so it barely depends on length. The page turns the elastic stress Fe = Pcr/A into an equivalent slenderness π√(E/Fe) and reads the selected column curve there, so the torsional allowable gets the same crippling cap and plasticity as the bending one, and the lower of the two is the column allowable.