Curved Beam in Bending (Winkler — Crane Hook, C-Clamp, Press Frame)

stress

Verified build 16 relations · 2 identities proven · 3 modeling steps · 3 parity samples

A crane hook, a C-clamp, the throat of a press frame — these are beams, but they are already curved before you load them, and that changes the stress in a way the straight-beam formula Mc/I cannot see. When a curved bar bends, the fibers on the inside of the curve are shorter than those on the outside, so the strain is no longer linear across the depth and the neutral axis shifts off the centroid, toward the inner fiber. The inside runs hotter than Mc/I predicts — often 1.5 to 2 times hotter in a tight hook — and that is exactly where these parts crack. This page is the Winkler curved-beam theory for a rectangular section.

$\sigma(r) = \frac{M\,(r_n - r)}{A\,e\,r}, \qquad r_n = \frac{h}{\ln(r_o/r_i)}, \qquad e = r_c - r_n$

The whole effect lives in the small eccentricity $e = r_c - r_n$ between the centroidal radius and the neutral-axis radius. That neutral radius is a logarithmic average of the geometry — the curved-beam analogue of the exponential in the belt/capstan equation — and it is always slightly less than the centroid, which is why the neutral axis moves inward.

Why the inside runs hot

Set the moment $M$ and watch the stress at each fiber:

$\sigma_i = \frac{M\,c_i}{A\,e\,r_i}\ \text{(inner, tension)}, \qquad \sigma_o = -\frac{M\,c_o}{A\,e\,r_o}\ \text{(outer, compression)}$

with $c_i = r_n - r_i$ and $c_o = r_o - r_n$ the distances from the shifted neutral axis. Because $r_i < r_o$ sits in the denominator, the inner fiber always carries the larger stress. The widget draws the curved-beam distribution (a hyperbola in $r$ ) against the straight-beam prediction (a straight line through the centroid): the two cross zero at different radii — $r_n$ for the real curved beam, $r_c$ for the straight-beam fiction — and the gap between them is the physics.

The one number that says how much this matters is the curvature penalty at the inner fiber,

$K_i = \frac{\sigma_{b,i}}{\sigma_{str}} = \frac{c_i\,h}{6\,e\,r_i},$

the ratio of the true inner-fiber bending stress to the straight-beam $Mc/I$ . It is pure geometry — no load, no material — and it climbs as the beam gets tighter (small $r_c/h$ ). Widen the radii or thin the section in the widget and watch $K_i \to 1$ as the two distributions slide into coincidence: at $r_c/h \gtrsim 10$ the curve is so gentle that ordinary $Mc/I$ is within a few percent and the widget tells you so. That the Winkler stress collapses exactly to $Mc/I$ in the straight limit is proven by series expansion in the build’s physics test — it is not asserted, it is machine-checked.

The crane hook: bending plus a direct pull

A hook is the combined-loading case (the same superposition idea as the eccentric column). Its load $P$ runs through the center of curvature, so the critical throat section carries a direct tension $P/A$ and a bending moment $M = P\,r_c$ about the centroid at the same time. Both stress the inner fiber in tension, so they add:

$\sigma_i = \frac{P}{A} + \frac{M\,c_i}{A\,e\,r_i}, \qquad M = P\,r_c.$

That total inner-fiber stress is what the safety factor $\mathrm{SF} = \sigma_y/\sigma_i$ is taken against — the inside of the throat is the one spot in the hook that decides whether it holds.

A strength-only material axis

Only $\sigma_y$ enters, and only through the safety factor and the yield warning. Stiffness $E$ and density $\rho$ do not appear at all: the Winkler stress distribution is pure statics and section geometry, just like the transverse-shear distribution. (Curved-beam deflection would bring $E$ back in through Castigliano’s theorem — that is a different page.) Swapping steel for titanium here moves the margin and nothing else. This is the same “geometry concentrates stress on the inside” family as the thick-walled cylinder and the pressure vessel, where the bore, not the outer wall, is the hot surface — the credibility of every number is recorded on the /verification/ page.

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)
sigma_y	47 ksi	design min.	mil-hdbk-5j

Inputs

Hook load (through the center of curvature) (P)

Inner radius (r_i)

Outer radius (r_o)

Section width (into the page) (w)

Section depth (radial): —
Centroidal radius: —
Cross-sectional area: —
Second moment of area (straight-beam): — m⁴
Neutral-axis radius: —
Eccentricity (r_c − r_n): —
Inner-fiber distance from NA: —
Outer-fiber distance from NA: —
Bending moment about the centroid: — N·m
Bending stress, inner fiber: —
Total stress, inner fiber (hot): —
Total stress, outer fiber: —
Straight-beam stress (Mc/I): —
Curvature penalty, inner fiber: —
Curvature ratio r_c/h: —
Safety factor (inner-fiber yield): —

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

M = P\,r_c

Assumes: crane-hook idealization: the load line passes through the center of curvature, so the moment about the section centroid is the load times the centroidal radius

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §3-18 (Curved Beams in Bending): the Winkler formula σ = M y/(A e (r_n − y)), the rectangular-section neutral radius r_n = h/ln(r_o/r_i) (Table 3-4), the eccentricity e = r_c − r_n, and the crane-hook combination of direct and bending stress.

h = r_o - r_i

Assumes: radial depth of the rectangular section

r_c = \dfrac{r_i + r_o}{2}

Assumes: centroid of a rectangle sits at the mean radius

A = b\,h

Assumes: solid rectangular cross-section, width b into the page

I = \dfrac{b h^3}{12}

Assumes: second moment about the centroidal axis, used only for the straight-beam comparison

r_n = \dfrac{h}{\ln(r_o/r_i)}

Assumes: Winkler assumption: plane sections remain plane and rotate about the neutral axis; the neutral axis sits where the net axial force vanishes, r_n = A / ∫(dA/r); for a rectangle ∫(dA/r) = b·ln(r_o/r_i), giving the log formula — always LESS than r_c; requires r_i > 0 (enforced by the variable's positivity + bounds): as r_i → 0 the log diverges and there is no material at the center of curvature; the exact closed form; the log is the curved-beam analogue of belt-drive's exponential · Valid while: The outer radius must exceed the inner radius — with r_o ≤ r_i there is no section and the neutral-axis logarithm is undefined. This is a geometry error, not a beam.

e = r_c - r_n

Assumes: the small positive gap between centroid and neutral axis — the whole curved-beam effect lives in this difference of nearly equal radii (watch the conditioning at large r_c/h)

c_i = r_n - r_i

Assumes: distance from the neutral axis out to the inner fiber

c_o = r_o - r_n

Assumes: distance from the neutral axis out to the outer fiber

\sigma_{b,i} = \dfrac{M\,c_i}{A\,e\,r_i}

Assumes: curved-beam bending stress at the inner fiber; σ(r) = M(r_n − r)/(A e r) evaluated at r = r_i — the eccentricity e in the denominator is why the inside runs hot

\sigma_i = \dfrac{P}{A} + \dfrac{M\,c_i}{A\,e\,r_i}

Assumes: total inner-fiber stress for the crane hook: the direct tension P/A superposed on the curved bending stress; both are tensile at the inner fiber, so they add — the worst spot; linear elastic, plane sections; the inner fiber governs against first yield · Valid while: Inner-fiber stress has reached the yield strength — past first yield the elastic Winkler distribution stops being the truth and the hook is into plastic behavior.

\sigma_o = \dfrac{P}{A} - \dfrac{M\,c_o}{A\,e\,r_o}

Assumes: outer-fiber stress: the curved bending puts the outside in compression (negative), and the direct tension P/A offsets part of it — the outer fiber is never the governing one here

\sigma_{str} = \dfrac{M\,(h/2)}{I} = \dfrac{6M}{b h^2}

Assumes: what a STRAIGHT beam of the same section and moment would predict at the extreme fiber (Mc/I, c = h/2): symmetric tension/compression about the centroid — the wrong answer here, shown for contrast

K_i = \dfrac{\sigma_{b,i}}{\sigma_{str}} = \dfrac{c_i\,h}{6\,e\,r_i}

Assumes: curvature stress-concentration factor at the inner fiber: how many times the straight-beam Mc/I the curved bending stress actually reaches — pure geometry, independent of load and material; > 1, and it grows as the beam gets tighter (small r_c/h)

\dfrac{r_c}{h}

Assumes: how curved the beam is; the one number that says whether you need this page at all · Valid while: r_c/h ≥ 10: the beam is barely curved, the neutral axis has all but returned to the centroid, and the straight-beam Mc/I result (shown alongside) is within a few percent — you do not need the curved-beam formula here.

\mathrm{SF} = \dfrac{\sigma_y}{\sigma_i}

Assumes: factor of safety against first yield, taken at the hot inner fiber

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.

r_{n} = \frac{h}{\log{\left(\frac{r_{o}}{r_{i}} \right)}}

1. Winkler's move: assume plane cross-sections stay plane and simply rotate about some neutral axis. A fiber at radius r then has strain proportional to (r_n − r)/r — not linear in r, because the fibers started at different arc lengths. Requiring zero NET axial force over a pure-bending section, ∫σ dA = 0, pins the neutral axis at r_n = A / ∫(dA/r); for the rectangle ∫(dA/r) = b·ln(r_o/r_i), so r_n = h/ln(r_o/r_i). Because the harmonic-style average undershoots the arithmetic one, r_n is always a hair LESS than the centroid r_c. — modeling: plane sections rotate about the NA; zero net axial force locates it modeling step

\sigma_{bi} = \frac{M c_{i}}{A e r_{i}}

2. The stress is σ(r) = M(r_n − r)/(A·e·r). Taking the moment of that distribution about the centroid returns the applied moment M and fixes the constant, with e = r_c − r_n the small eccentricity. Evaluate it at the inner fiber (r = r_i, distance c_i = r_n − r_i) and it peaks: σ_{b,i} = M·c_i/(A·e·r_i). That e sitting in the denominator — a difference of nearly equal radii — is exactly why the inner fiber runs so much hotter than a straight beam's. — modeling: moment balance about the centroid fixes the stress constant modeling step

\sigma_{i} = \sigma_{bi} + \frac{P}{A}

3. A crane hook whose load P runs through the center of curvature loads the throat section two ways at once: a direct tension P/A spread over the area, AND a bending moment M = P·r_c about the centroid. Both put the inner fiber in tension, so superpose them — σ_i = P/A + σ_{b,i} — and the inside of the throat is the single worst-stressed point in the whole hook. (Set the hook aside and the P/A term is simply absent: pure curved bending.) — superposition: direct tension + curved bending (the crane hook)

K_{i} = \frac{\sigma_{bi}}{\sigma_{str}}

4. Put the SAME moment on a straight bar of depth h and you would predict σ_{str} = M(h/2)/I, equal and opposite about the centroid. The curved beam's inner fiber overshoots that by the factor K_i = σ_{b,i}/σ_{str}, which collapses to the pure-geometry ratio c_i·h/(6·e·r_i) — no load, no material in it. As r_c/h → ∞ the neutral axis slides back to the centroid (e → I/(A r_c)) and K_i → 1: the Winkler formula becomes ordinary Mc/I. The build's physics test proves that limit by series expansion. — definition: curvature penalty factor; the straight-beam limit

SF = \frac{\sigma_{y}}{\sigma_{i}}

5. The inner fiber is the hot one, so the margin against first yield is taken there: SF = σ_y/σ_i. Swap the material and only this number moves — the Winkler stresses are pure statics and geometry (see the material-axis note in the overview). — definition: safety factor on the governing inner fiber modeling step

How it fails

The widget answers a narrow elastic question — given this curved section and this load, what is the peak inner-fiber stress and the margin against first yield? Real hooks, clamps, and frames get into trouble in ways that mostly live in what the Winkler model deliberately leaves out.

The inner fiber yields first, and the elastic formula stops being true there. The whole point of the page is that the inside runs hottest, so that is where a hook first yields — and once it does, the elastic Winkler distribution over-predicts the stress (the material sheds load by flowing). The widget warns at $\sigma_i = \sigma_y$ rather than pretending the elastic number still holds. Real hooks are often deliberately designed to yield a little at the inner fiber before they fracture, so a bent-open hook is a warning, not yet a failure.
Rectangular is the easy case; real hooks are trapezoidal or round. A crane hook is thicker on the inside than the outside — a trapezoidal or teardrop section — precisely to put more material where the stress is highest. That shifts $r_n$ and changes $\int(dA/r)$ : Shigley Table 3-4 lists the neutral-axis formula for trapezoid, T, and round sections, and each is its own future THING. This page is the rectangle, where $r_n = h/\ln(r_o/r_i)$ is clean.
The eccentricity is a difference of nearly equal numbers. $e = r_c - r_n$ is small, and at large $r_c/h$ it is tiny — a difference of two radii that agree to several digits. Compute it carelessly (subtracting two rounded radii) and you can lose all your significant figures and get a wildly wrong stress. The build computes $r_n$ and $e$ at high precision and the widget warns you once $r_c/h \gtrsim 10$ that you are in the regime where the straight-beam formula is both simpler and better conditioned. This is a real numerical trap, not a theoretical footnote.
Stress concentrations sit on top of this. The Winkler stress is the nominal curved-beam stress. A hook with a cross-hole for a latch, a machining groove, or a sharp fillet multiplies it further by a stress-concentration factor the page does not include — and that local peak, on the already-hot inner fiber, is a classic crack initiation site.
Static, elastic, in-plane only. No fatigue (hooks fail in service by fatigue at the inner fiber under repeated lifts, not by a single overload), no curved-beam deflection or wind-up, no out-of-plane or twisting loads, no residual stresses from forging or heat treatment. Read the page as where the stress concentrates and how curvature makes it worse, not as a finished hook design.

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
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
Push uniform pressure on a flat circular plate — a tank head, a porthole, a valve cover — and how hard it deflects and where it cracks depend entirely on the RIM. Bolt it down (clamped) and it is stiff and hottest at the edge; rest it on a ring (simply supported) and it sags four times as far and is hottest at the center. This is the page where Poisson's ratio moves a STRESS: the simply-supported stress carries ν, the clamped-edge stress carries no material property at all.
- stress
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
Propped Cantilever (UDL)
A cantilever with a prop under its free end: one redundant support turns a determinate beam into a statically indeterminate one. Equilibrium alone cannot find the three reactions — compatibility (the prop deflects to zero) supplies the missing equation, and the build solves the coupled 3×3 system exactly.
- stress
- mass-cost
Simply Supported Beam (Center Load + UDL)
The floor joist under you right now: pinned at both ends, carrying a point load and a distributed load at once. Because the governing equation is linear, the two answers simply add — superposition, the single most-used trick in structural analysis, made visible.
- stress
- mass-cost
Transverse Shear in Beams (τ = VQ/Ib, Shear Flow, Fastener Spacing)
A beam does not only bend — the shear force V drags its layers past one another, and that longitudinal shear is what a built-up beam's nails or bolts actually carry. The stress is a parabola (peak 3V/2A at the neutral axis, zero at the surfaces), and the shear flow q = VQ/I sets the fastener spacing. Statics and geometry only: no stiffness enters at all.
- stress

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.

Composite Bar (Core + Sleeve)
- A A_1
- A A_2
- c_i L
- c_o L
- e L
- h L
- r_c L
- r_n L
Impact Loading (Falling Mass, Energy Method)
- c_i b
- c_i d
- c_i h
- c_i L
- c_o b
- c_o d
- c_o h
- c_o L
- e b
- e d
- e h
- e L
- h b
- h d
- h h
- h L
- r_c b
- r_c d
- r_c h
- r_c L
- r_n b
- r_n d
- r_n h
- r_n L
Symmetric Two-Bar Truss
- c_i d
- c_i L
- c_o d
- c_o L
- e d
- e L
- h d
- h L
- r_c d
- r_c L
- r_n d
- r_n L
Thermal Assembly (Two-Segment Bar Between Rigid Walls)
- A A_1
- A A_2
- c_i L_1
- c_i L_2
- c_o L_1
- c_o L_2
- e L_1
- e L_2
- h L_1
- h L_2
- r_c L_1
- r_c L_2
- r_n L_1
- r_n L_2
Cantilever Beam (End Load)
- c_i b
- c_i h
- c_i L
- c_o b
- c_o h
- c_o L
- e b
- e h
- e L
- h b
- h h
- h L
- r_c b
- r_c h
- r_c L
- r_n b
- r_n h
- r_n L
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
- c_i a
- c_i t
- c_o a
- c_o t
- e a
- e t
- h a
- h t
- r_c a
- r_c t
- r_n a
- r_n t
- sigma_bi q
- sigma_bi sigma_allow
- sigma_i q
- sigma_i sigma_allow
- sigma_o q
- sigma_o sigma_allow
- sigma_str q
- sigma_str sigma_allow
Fixed-Fixed Beam (UDL)
- c_i b
- c_i h
- c_i L
- c_o b
- c_o h
- c_o L
- e b
- e h
- e L
- h b
- h h
- h L
- r_c b
- r_c h
- r_c L
- r_n b
- r_n h
- r_n L
Propped Cantilever (UDL)
- c_i b
- c_i h
- c_i L
- c_o b
- c_o h
- c_o L
- e b
- e h
- e L
- h b
- h h
- h L
- r_c b
- r_c h
- r_c L
- r_n b
- r_n h
- r_n L

+ 24 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources

Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §3-18 (Curved Beams in Bending): the Winkler formula σ = M y/(A e (r_n − y)), the rectangular-section neutral radius r_n = h/ln(r_o/r_i) (Table 3-4), the eccentricity e = r_c − r_n, and the crane-hook combination of direct and bending stress.
Young, W. C., Budynas, R. G., & Sadegh, A. M., Roark's Formulas for Stress and Strain, 8th ed., McGraw-Hill, 2012 — Chapter 9 (Curved Beams): the same Winkler/Bleich curved-beam bending formulas and the rectangular-section neutral-axis radius, used here as an independent oracle.