Buckling, Torsion & Special Effects

6% of exam

Euler and inelastic column buckling, effective length, torsion of shafts and members, fatigue, thermal deformation, bearing, and progressive-collapse considerations.

5 concepts

Exam areas

G. Special topics (buckling, torsion)

Euler Column Buckling & Effective Length

The Euler critical load, the effective-length factor K for ideal end conditions, the radius of gyration and slenderness KL/r, and the critical stress.

A slender compression member does not fail by crushing — it fails by suddenly bowing sideways at a load far below its squash load

P_y = F_y A

. That instability is buckling, and predicting the load at which it happens is the most-tested idea in the structural special-topics area. The whole subject rides on one equation and one number: the Euler critical load

P_{cr} = \pi^2 EI/(KL)^2

and the effective-length factor

K

that translates real end restraint into an equivalent pinned length. The NCEES PE Civil Reference Handbook gives both in §1.6.8 Columns (Chapter 1, General Engineering), including the theoretical

K

values for the four ideal end conditions — so this is a handbook-grounded concept. Points are lost not on the formula but on which

I

to use (always the smaller, weak-axis value unless that axis is separately braced), on

K

, and on confusing the Euler elastic result with the inelastic AISC strength that actually governs stocky columns.

The Euler critical load

Picture a pinned-pinned column under a growing axial load

P

. Below a threshold the straight shape is stable; at the threshold a slightly bent shape becomes an equally valid equilibrium, and any disturbance pushes the column into that bent shape — it buckles. Solving the governing differential equation

EIy'' = -Py

for the lowest load that admits a non-trivial sine-shaped deflection gives the Euler buckling load. Here

E

is the modulus of elasticity (ksi),

I

is the moment of inertia about the axis of buckling (in

^4

), and

L

is the actual unbraced length; the effective-length factor

K

(introduced next) accounts for end conditions other than pinned-pinned.

P_{cr} = \dfrac{\pi^2 E I}{(KL)^2}

Effective length and the ideal K factors

The effective length

KL

is the distance between points of zero moment (inflection points) on the buckled shape — the length of an equivalent pinned-pinned column with the same critical load. Stiffer end restraint pushes those inflection points closer together, shortens

KL

, and raises

P_{cr}

. The handbook lists the four theoretical values: pinned-pinned

K = 1.0

, fixed-fixed

K = 0.5

, fixed-pinned

K = 0.7

, and fixed-free (the cantilevered flagpole)

K = 2.0

. Because

P_{cr}

scales with

1/K^2

, a fixed-fixed column (

K = 0.5

) carries four times the Euler load of the same pinned-pinned column (

K = 1.0

), while a flagpole (

K = 2.0

) carries one-quarter. Real connections never reach the ideal fixity, so AISC recommends design values larger than these theoretical ones for the restrained cases:

0.65

for fixed-fixed,

0.80

for fixed-pinned,

1.0

for pinned-pinned, and

2.10

for fixed-free.

K_{\text{p-p}} = 1.0, \quad K_{\text{f-f}} = 0.5, \quad K_{\text{f-p}} = 0.7, \quad K_{\text{f-free}} = 2.0

Radius of gyration and slenderness

Rewrite the Euler load in terms of stress and you discover that only one geometric property matters: the radius of gyration

r = \sqrt{I/A}

(in), which measures how far the area is spread from the buckling axis. The dimensionless effective slenderness ratio

KL/r

then sets everything. A column always buckles about the axis with the larger

KL/r

— usually the weak axis, with the smaller

I

and smaller

r

— so for a wide-flange you check

r_y

unless that axis is separately braced. Slender means high

KL/r

(Euler governs); stocky means low

KL/r

(yielding or inelastic buckling governs).

r = \sqrt{\dfrac{I}{A}}, \qquad \lambda = \dfrac{KL}{r}

Critical buckling stress

Dividing

P_{cr}

by the area gives the critical (Euler) stress, a function of the slenderness ratio alone — independent of the material's strength

F_y

. This is the key insight the exam probes: making a column out of higher-strength steel does NOT raise its elastic buckling capacity, because

\sigma_{cr}

depends only on

E

and

KL/r

. Two steel columns of identical geometry buckle at the same

P_{cr}

whether they are A36 or A992. Strength only matters at low slenderness, where the column yields before it can buckle.

\sigma_{cr} = \dfrac{P_{cr}}{A} = \dfrac{\pi^2 E}{(KL/r)^2}

Where Euler stops: the inelastic transition

Euler's elastic curve is only valid while

\sigma_{cr}

stays below the proportional limit. At low slenderness the predicted Euler stress would exceed

F_y

, which is physically impossible — residual stresses and yielding take over first. AISC 360-16 codifies the dividing line at

KL/r = 4.71\sqrt{E/F_y}

(equivalently where the elastic stress

F_e = \pi^2 E/(KL/r)^2

falls to

0.44 F_y

): above it use the elastic formula

F_{cr} = 0.877 F_e

, below it the inelastic formula

F_{cr} = (0.658^{F_y/F_e}) F_y

. For

F_y = 50\,\text{ksi}

the transition sits at

KL/r \approx 113

. On the exam, after computing the Euler load always ask whether

\sigma_{cr} < F_y

; if not, the elastic Euler value overstates the real strength and the AISC inelastic equation governs.

\dfrac{KL}{r}\bigg|_{\text{trans}} = 4.71\sqrt{\dfrac{E}{F_y}}

Exam strategy

Lead with the weak axis: pick the smaller

I

(and

r

) unless the problem braces that axis separately, in which case compare

K_xL_x/r_x

against

K_yL_y/r_y

and use the larger. Convert

L

to inches before squaring — a foot-versus-inch slip is a factor-of-144 error in

P_{cr}

. Read

K

from the end conditions (use the theoretical values when the problem says 'ideal' and the AISC recommended values when it says 'design'). Compute

\sigma_{cr} = \pi^2 E/(KL/r)^2

and immediately compare to

F_y

: if

\sigma_{cr} \ge F_y

the column is too stocky for Euler and you must use the AISC inelastic strength. Carry

E = 29{,}000\,\text{ksi}

for steel throughout.

Key equations

Euler critical load

P_{cr} = \dfrac{\pi^2 E I}{(KL)^2}

Elastic buckling load (kips). $E$ = modulus (ksi), $I$ = moment of inertia about the buckling axis (in $^4$ ), $K$ = effective-length factor, $L$ = unbraced length (in).

Effective length

L_e = KL

Distance between inflection points of the buckled shape; the equivalent pinned-pinned length.

Theoretical K factors

K = 1.0,\ 0.5,\ 0.7,\ 2.0

Pinned-pinned, fixed-fixed, fixed-pinned, fixed-free respectively (NCEES handbook §1.6.8 Columns). AISC recommended design values are larger: 0.65, 0.80, 1.0, 2.10 for fixed-fixed, fixed-pinned, pinned-pinned, fixed-free.

Radius of gyration

r = \sqrt{\dfrac{I}{A}}

Spread of area about the buckling axis (in). Use the smaller (weak-axis) value unless that axis is braced.

Effective slenderness ratio

\lambda = \dfrac{KL}{r}

Dimensionless governing parameter; the column buckles about the axis with the larger $KL/r$ .

Critical (Euler) stress

\sigma_{cr} = \dfrac{\pi^2 E}{(KL/r)^2}

Buckling stress (ksi); depends only on $E$ and $KL/r$ , NOT on $F_y$ . Valid only while $\sigma_{cr} < F_y$ .

Squash (yield) load

P_y = F_y A

Upper bound on capacity (kips); buckling governs when $P_{cr} < P_y$ .

Inelastic-elastic transition (AISC 360-16)

\dfrac{KL}{r}\Big|_{\text{trans}} = 4.71\sqrt{\dfrac{E}{F_y}}

Above this slenderness elastic buckling governs ( $F_{cr}=0.877F_e$ ); below it inelastic buckling/yielding governs. About 113 for $F_y=50$ ksi.

Worked examples

Critical load of a pinned wide-flange column

Problem. A W10 $\times$ 33 column ( $A = 9.71\,\text{in}^2$ , $I_y = 36.6\,\text{in}^4$ , $r_y = 1.94\,\text{in}$ , $F_y = 50\,\text{ksi}$ , $E = 29{,}000\,\text{ksi}$ ) is $14\,\text{ft}$ tall and pinned at both ends about both axes. Find the Euler critical load, the slenderness ratio, and the critical stress, and state whether Euler is valid.

Solution. Pinned-pinned about the governing weak axis:

K = 1.0

L = 14\,\text{ft} = 168\,\text{in}

. Euler load:

P_{cr} = \dfrac{\pi^2 (29{,}000)(36.6)}{(1.0\times168)^2} = \dfrac{1.0476\times10^{7}}{28{,}224} = 371\,\text{kips}

. Slenderness:

\lambda = KL/r_y = 168/1.94 = 86.6

. Critical (Euler) stress:

\sigma_{cr} = \pi^2(29{,}000)/86.6^2 = 286{,}219/7{,}500 = 38.2\,\text{ksi}

. Cross-check:

\sigma_{cr} A = 38.2(9.71) = 371\,\text{kips}

, matching

P_{cr}

. Validity check: the transition is

4.71\sqrt{E/F_y} = 4.71\sqrt{29{,}000/50} = 113

. Here

\lambda = 86.6 < 113

, so the column is in the INELASTIC range — Euler is NOT valid as a strength, and the elastic value

38.2\,\text{ksi}

over-states the real capacity. The AISC inelastic equation governs: with

F_e = \sigma_{cr} = 38.2\,\text{ksi}

F_{cr} = 0.658^{F_y/F_e}F_y = 0.658^{50/38.2}(50) = 28.9\,\text{ksi}

, giving a usable nominal strength

P_n = F_{cr}A = 28.9(9.71) = 281\,\text{kips}

. **Answer: Euler

P_{cr} = 371\,\text{kips}

\lambda = 86.6

\sigma_{cr} = 38.2\,\text{ksi}

— but because

\lambda < 113

the column is inelastic, so the AISC strength

F_{cr} = 28.9\,\text{ksi}

(

P_n = 281\,\text{kips}

) governs, NOT the Euler value.**

P_{cr} = \dfrac{\pi^2 (29{,}000)(36.6)}{(168)^2} = 371\ \text{kips}

Effect of end fixity on a flagpole column

Problem. A solid round steel bar $3\,\text{in}$ in diameter ( $E = 29{,}000\,\text{ksi}$ ) projects $12\,\text{ft}$ as a cantilever (fixed-base, free-top). Find its Euler buckling load using the theoretical $K$ , and compare with the same bar pinned at both ends.

Solution. Section properties:

I = \pi d^4/64 = \pi(3)^4/64 = 3.976\,\text{in}^4

;

A = \pi(3)^2/4 = 7.069\,\text{in}^2

;

r = d/4 = 0.750\,\text{in}

. Length

L = 12\,\text{ft} = 144\,\text{in}

. Fixed-free:

K = 2.0

, so

KL = 288\,\text{in}

P_{cr} = \dfrac{\pi^2 (29{,}000)(3.976)}{288^2} = \dfrac{1.1380\times10^{6}}{82{,}944} = 13.7\,\text{kips}

. Pinned-pinned (

K = 1.0

P_{cr} = \dfrac{\pi^2 (29{,}000)(3.976)}{144^2} = 54.9\,\text{kips}

— exactly four times larger, as expected from the

1/K^2

scaling. Validity:

\sigma_{cr} = 13.7/7.069 = 1.94\,\text{ksi}

(fixed-free) is far below

F_y

, so Euler governs comfortably for this very slender bar (

KL/r = 288/0.75 = 384

). **Answer:

P_{cr} = 13.7\,\text{kips}

fixed-free vs.

54.9\,\text{kips}

pinned-pinned (4:1 ratio).** Sanity check: the flagpole is the weakest of the four ideal cases, consistent with its largest

K

\dfrac{P_{cr,\text{f-f}}}{P_{cr,\text{p-p}}} = \left(\dfrac{K_{\text{p-p}}}{K_{\text{f-free}}}\right)^2 = \left(\dfrac{1.0}{2.0}\right)^2 = \dfrac{1}{4}

Stocky column: when the AISC inelastic strength governs

Problem. A W8 $\times$ 31 column ( $A = 9.13\,\text{in}^2$ , $r_y = 2.02\,\text{in}$ , $F_y = 50\,\text{ksi}$ , $E = 29{,}000\,\text{ksi}$ ) is $12\,\text{ft}$ tall, pinned-pinned ( $K = 1.0$ ). Show that the Euler value over-states the strength and find the governing AISC critical stress and nominal compressive strength.

Solution. Slenderness:

\lambda = KL/r_y = (1.0)(144)/2.02 = 71.3

. Transition:

4.71\sqrt{E/F_y} = 4.71\sqrt{29{,}000/50} = 113

. Since

\lambda = 71.3 < 113

, the column is INELASTIC — Euler does not give the strength. Elastic (Euler) reference stress:

F_e = \pi^2 E/\lambda^2 = \pi^2(29{,}000)/71.3^2 = 56.3\,\text{ksi}

. Note this exceeds

F_y = 50\,\text{ksi}

, which is physically impossible as a real stress — a clear flag that the inelastic branch must be used. AISC inelastic strength:

F_{cr} = 0.658^{F_y/F_e}F_y = 0.658^{50/56.3}(50) = 0.658^{0.888}(50) = 34.5\,\text{ksi}

. Nominal strength:

P_n = F_{cr}A = 34.5(9.13) = 315\,\text{kips}

. **Answer:

F_{cr} = 34.5\,\text{ksi}

P_n = 315\,\text{kips}

— far below the Euler-implied

F_e A = 56.3(9.13) = 514\,\text{kips}

.** Sanity check: the inelastic

F_{cr}

correctly lands below both

F_e

and

F_y

, as it must for a stocky column where residual stress and yielding cut the elastic prediction.

F_{cr} = 0.658^{F_y/F_e}F_y = 0.658^{50/56.3}(50) = 34.5\ \text{ksi}, \quad P_n = 315\ \text{kips}

Common pitfalls

•Using the strong-axis $I$ when both axes are unbraced. A column buckles about the axis with the larger $KL/r$ — usually the weak axis with the smaller $I$ and $r$ . Always check the governing (smaller-capacity) axis.
•Leaving $L$ in feet inside $(KL)^2$ . The formula needs consistent units; with $E$ in ksi and $I$ in in $^4$ , $L$ must be in inches — a foot-versus-inch slip is a factor-of-144 error in $P_{cr}$ .
•Believing higher-strength steel raises the elastic buckling load. $\sigma_{cr} = \pi^2 E/(KL/r)^2$ depends only on $E$ and slenderness; A36 and A992 buckle at the same $P_{cr}$ for identical geometry.
•Reporting the Euler load for a stocky column. If $\sigma_{cr} \ge F_y$ (or $KL/r < 4.71\sqrt{E/F_y}$ ), elastic Euler overstates capacity — the AISC inelastic equation $F_{cr} = 0.658^{F_y/F_e}F_y$ governs.
•Confusing the theoretical $K$ (handbook §1.6.8 Columns: 0.5, 0.7, 1.0, 2.0) with the AISC recommended design $K$ (0.65, 0.80, 1.0, 2.10). Real connections are never perfectly fixed, so design values are larger; read which the problem wants.
•Computing $r$ as $I/A$ instead of $\sqrt{I/A}$ . The radius of gyration is the square root of $I/A$ ; dropping the root mangles the slenderness ratio.

References

NCEES PE Civil Reference Handbook — §1.6.8 Columns (Chapter 1 General Engineering; Euler's Formula and theoretical K)
AISC 360-16, Chapter E — Design of Members for Compression (Steel Construction Manual, 15th ed.) — inelastic/elastic transition and recommended design K values
Hibbeler, Mechanics of Materials, 10th ed. — column buckling background

Torsion of Shafts & Thin-Walled Sections

Circular-shaft torsion (τ = Tr/J, φ = TL/JG), closed thin-wall torsion and shear flow (q = T/2Aₘ), open-section torsion, and combined torsion-plus-shear.

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Effective Length & K-Factors (Alignment Charts)

Sidesway-inhibited vs uninhibited frames, the joint stiffness ratio G, the AISC alignment-chart K, and the leaning-column effect on stability.

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Bearing, Thermal & Support Movements

Bearing stress and contact at supports, thermal expansion and the forces it induces in restrained members, and the moments caused by support settlement.

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Fatigue & Progressive Collapse

The S-N fatigue method and stress-range categories, the role of stress concentrations, and progressive-collapse robustness through alternate load paths and tie forces.

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