Curved Beam Stress Calculator

Q: Why is the neutral axis not at the centroid in a curved beam?

Because the fibres are not the same length to start with. A fibre at radius r has length rθ, so for a given rotation of the section the short inner fibres are strained more than the long outer ones and the strain varies as 1/r rather than linearly. Balancing the forces then requires the zero-stress line to sit where r_n = A ÷ ∫dA/r, which is a harmonic mean of radii and is always inside the centroid. The shift is small — a few per cent of the section depth — but it appears in the denominator of the stress, so it matters far more than its size suggests.

Q: When can I use the ordinary beam formula on a curved part?

When the centroidal radius is more than about five times the section depth, the straight formula is within about 7 per cent at the inner fibre and it is fair to use it. At ten times it is within 3.3 per cent, which is inside material scatter. Below five it goes wrong quickly and always in the unsafe direction at the inner fibre: 17 per cent low at R/h = 2 and a quarter low at R/h = 1.25. Watch the definition, though — many books write the rule as R/c with c the half-depth, so their “10” is this page's 5.

Q: What is the formula for stress in a curved beam?

σ = M(r_n − r) / (A·e·r), with r_n the neutral axis radius, e = R − r_n the eccentricity, r the radius of the point you want and M the moment about the CENTROIDAL axis. Some sources write it as σ = (M/Ae)(r_n/r − 1), which is identical. Add P/A by superposition if the section also carries a direct load. The distribution is hyperbolic in r, and the two things it must satisfy — no net force, and a moment of exactly M about the centroid — are what the formula is derived from.

Q: Why is a crane hook's section a trapezoid with the wide face inside?

Because that is where the stress is. The inner fibre carries the bending tension and the direct tension together, and it carries the bending stress amplified by the curvature; the outer fibre carries much less. Putting the wide face inside does two things at once: it adds material where the stress is highest, and it moves the centroid outwards, which lengthens the moment arm slightly but improves the distribution far more. Matched on weight rather than on depth, a trapezoid hook section carries about 55 per cent more load than a round bar: a 67 by 27 mm trapezoid 60 mm deep off a 50 mm inner radius takes 25 kN at 90 MPa where a 60 mm round bar of exactly the same area takes it at 140.

Q: Does this page design crane hooks?

No, and it should not be used to. A crane hook is a lifting component and lifting equipment is regulated: in Great Britain LOLER 1998 requires it to be of adequate strength, marked with its safe working load, and thoroughly examined at set intervals by a competent person, and ASME B30.10 covers hooks in the United States. A hook must be a rated, marked, certified part bought to a standard. This page is for understanding why a hook is the shape it is, and for checking unregulated curved parts — C-frames, clamps, press and punch frames.

Q: How do I find the neutral axis radius for a section that is not listed?

Integrate. r_n = A ÷ ∫dA/r, and for any section that can be sliced into strips of constant width between two radii, ∫dA/r for each strip is simply the width times ln(r_outer/r_inner). Add them up. That is exactly how the T section on this page is done and it extends to an I, a channel, a stepped section or anything you can draw. For a curved profile, integrate numerically — twenty strips is plenty. The section properties page will give you A, the centroid and I for eight shapes to go with it.

Q: Does a curved beam deflect differently too?

Yes, and this page does not compute it. A curved member's deflection has contributions from bending, from the axial force and from the shear, and on a sharply curved part the axial and shear terms are not negligible the way they are in a slender straight beam — Castigliano's theorem on the strain energy of the curved member is the usual route. The throat opening of a C-frame press under load, which is often the controlling design criterion, is a deflection problem and not a stress one.

Winkler’s theory for hooks, C-frames and chain links: the neutral axis shifted off the centroid, the hyperbolic stress it produces, and how far wrong the straight-beam formula would have been.

Last updated 24 Sep 2026

Curved beam stress

Section, radius and load → inner and outer fibre stress

Section shapeAll four neutral-axis radii come from the same definition, r_n = A ÷ ∫dA/r, and each was checked here against a numerical integration of the real section. The trapezoid with its wide face inside is the classic crane hook section, and it is that shape for a reason: it puts material where the stress is.

INNER radius r_i (mm) — to the surface nearest the centre of curvatureMeasured from the centre of curvature to the inside face of the section, not to the centroid and not to the outside. This is the radius that does the damage: the stress carries a 1/r in it, so the smaller this is the worse the inner fibre gets.

Section depth h (mm), radially — the DIAMETER for a round barThe dimension along the radius, from the inner face to the outer one. Everything on this page is governed by the ratio of the centroidal radius to this number: below about 5 the straight-beam formula is unsafe, above about 10 it is fine.

Width b (mm) — the INNER width of a trapezoid, the flange width of a TFor a rectangle this is simply the width. For a trapezoid it is the width at the inner radius, which on a hook is the wide face. For a T it is the flange width, and the flange is taken to be on the inside.

Outer width of a trapezoid (mm)Trapezoid only. Making this smaller than the inner width moves the centroid outwards and area towards the inner fibre, which is exactly what a hook section is for.

Flange thickness of a T (mm)T section only, and measured radially. The flange sits at the inner radius, where the stress is.

Web thickness of a T (mm)T section only. It is clamped to the flange width, because a web wider than its flange is a rectangle.

What is loading itA crane hook, a C-frame press, a punch frame and a G-clamp all share one feature: the load line passes through or near the centre of curvature, so the bending moment at the critical section is the load times the centroidal radius, and the section carries P/A on top of it. Choose that mode and the moment field below locks and shows P·R.

Direct load P through the section (N)Positive pulls the section apart. On a hook this is the lifted load, and it does two things at once: it makes the moment and it adds a uniform P/A tension. Leave it at zero for pure bending.

Bending moment M (N·m)Positive when it opens the curve — when it puts the inner fibre into tension, which is what a hook’s load does. About the CENTROIDAL axis, not the neutral axis: that is the convention the formula σ = M(r_n−r)/(Aer) is written in, and getting it wrong shifts every stress on the page.

Not a circuit: the section on the left with its centre of curvature below the picture, and the stress across it on the right. The curve is the real, hyperbolic distribution; the straight line is what an ordinary beam calculation would have given for the same section and the same moment, drawn to the same scale. Two things to see. There are two horizontal lines across the plot, and they are not in the same place: the upper one is the centroid, halfway up the section, and the lower one is the NEUTRAL AXIS, which has moved towards the inner fibre and towards the centre of curvature. The gap between them is the eccentricity e printed below, and it is what the stress divides by. And the hyperbola runs outside the straight line at the inner fibre and inside it at the outer one: a curved beam does not simply carry more stress, it has moved its stress inwards. The distribution is drawn for a RECTANGLE at your radius-to-depth ratio, so that the shape shows the effect of curvature alone; the figures underneath are for the section you actually chose, and for an asymmetric one they are further apart still.

151.78MPaExample

A crane hook section: a trapezoid 60 mm deep with a 40 mm face inside and a 16 mm face outside, on a 50 mm inner radius, lifting 25 kN through the centre of curvature

Plane sections stay plane, but the inner fibres are shorter — so the stress is hyperbolic

r_n = A ÷ ∫dA/r · e = R − r_n · σ(r) = M(r_n − r) / (A·e·r) · σ_total = σ(r) + P/A

why it is not linear: a curved beam’s fibres are not the same length to begin with. The fibre at radius r has length r·θ, so a rotation of the section through Δθ strains it by Δθ/θ times (r_n−r)/r — and that 1/r is the whole difference from a straight beam. Plane sections still stay plane; it is the fibres, not the assumption, that have changed
r_n < R: the neutral axis moves INWARDS, towards the centre of curvature, because the inner fibres are stiffer per unit of rotation and the force balance has to shift towards them. It is a harmonic mean of radii rather than an arithmetic one, which is always the smaller of the two
e: the eccentricity, and the number the whole page hangs on, because it sits in the DENOMINATOR of the stress. It is small — a few per cent of the section depth even on a sharply curved part — and it is the difference of two much larger numbers, so it must be computed to full precision and never from rounded values of R and r_n. Rounding R and r_n to three figures each can change e by ten per cent and the stress with it
M about the CENTROID: the moment in this formula is taken about the centroidal axis, not the neutral axis. The distribution satisfies ∫σdA = 0 and ∫σ(R−r)dA = M, and both integrals are checked numerically here for every section shape. Take the moment about the wrong axis and every stress on the page moves
P/A: the direct stress, added by superposition. On a hook it is what makes the inner fibre unambiguously the critical point: the bending stress there is tensile and the direct stress is tensile too, so they add, while at the outer fibre they subtract

Worked example

A crane hook section: a trapezoid 60 mm deep with a 40 mm face inside and a 16 mm face outside, on a 50 mm inner radius, lifting 25 kN through the centre of curvature
THE SECTION FIRST. Area = (40 + 16) × 60 / 2 = 1,680 mm². The centroid of a trapezoid sits nearer its wide face, so R = 75.714 mm, which is 25.71 mm out from the inner face rather than the 30 mm it would be for a rectangle
THE NEUTRAL AXIS. r_n = A ÷ ∫dA/r, and for a trapezoid ∫dA/r works out to ((b_i r_o − b_o r_i)/h)·ln(r_o/r_i) − (b_i − b_o). That gives r_n = 72.080 mm — 3.634 mm inside the centroid. That small number is the eccentricity e, and it is the number everything else divides by
THE MOMENT. The load line passes through the centre of curvature, so the moment at this section is P·R = 25,000 × 75.714 = 1,892.9 N·m. This is the step people skip: a hook is not loaded by a moment somebody hands them, it makes its own moment out of the load and the radius, and a deeper section makes a bigger one
THE INNER FIBRE. σ = M(r_n − r_i)/(A e r_i) = 1,892,857 × 22.080 / (1,680 × 3.634 × 50) = 136.9 MPa in tension. Add the direct stress P/A = 25,000/1,680 = 14.88 MPa, which is tensile too, and the inner fibre carries 151.8 MPa
THE OUTER FIBRE, for comparison. σ = -106.9 MPa in compression from bending, plus the same 14.88 MPa of tension, giving -92.0 MPa. The inner fibre is the critical one, and on a hook it always is, because that is the one place where the bending and the direct stress have the same sign
AND NOW THE NUMBER THAT MATTERS. A straight-beam calculation on this section would have given Mc/I = 1,892,857 × 25.71 / 1,337,143 = 36.4 MPa. That is 3.76 times too low — it misses 73 per cent of the real stress. Part of that is the curvature, which at R/h = 1.26 accounts for a factor of about 1.37 on its own; the rest is the trapezoid's own asymmetry, which pushes the centroid away from the inner fibre and so makes the straight formula's c small. Both effects are real and they multiply
WHAT TO DO WITH IT. 152 MPa is comfortable in any alloy steel and tight in a mild one — but a crane hook is not sized by this calculation. It is a rated lifting component: it must be bought to a standard, marked with its working load limit, certified, and thoroughly examined on a schedule the law sets. Use this page to understand a hook, to check a C-frame or a clamp, or to see how far a section is from adequate. Do not use it to design one

The four neutral-axis radii, and where they come from

Not copied from a table. Every one of these follows from the single definition r_n = A ÷ ∫dA/r, and ∫dA/r for a strip of width b between two radii is b·ln(r_o/r_i) — which is why a logarithm appears in three of the four and why a composite section is just a sum of logarithms. The round bar is the exception and the prettiest: ∫dA/r = 2π(R − √(R²−c²)), and dividing πc² by it collapses to (R + √(R²−c²))/2 with the c² cancelling. Each of the four was checked here against a Simpson integration of the actual section and they agree to better than two parts in a million. Shigley’s Table 3-4 prints the same four, and is cited rather than reproduced. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Section	Area A	r_n	Centroidal radius R
Rectangle, width b	b·h	h / ln(r_o/r_i)	r_i + h/2
Round bar, radius c	πc²	(R + √(R² − c²)) / 2	r_i + c
Trapezoid, b_i inside, b_o outside	(b_i + b_o)h/2	A ÷ [ ((b_i r_o − b_o r_i)/h)·ln(r_o/r_i) − (b_i − b_o) ]	h(b_i(2r_i+r_o) + b_o(r_i+2r_o)) / 6A
T, flange b_f × t_f inside, web t_w	b_f t_f + t_w(h − t_f)	A ÷ [ b_f·ln(r_2/r_i) + t_w·ln(r_o/r_2) ]	ΣA_j r_j / A

When can you ignore all of this? A rectangular section against R/h

The second column is the whole mechanism in one number: the neutral axis moves inwards by e, and at R/h = 20 that shift is a four-hundredth of the section depth while at R/h = 1 it is a twelfth. Everything else follows from it, because the stress is M·y/(A·e·r) and e is in the denominator. The conventional rule — curved-beam theory below R/h = 5, straight-beam above — puts the switch where the error is 6.6 per cent, which matches Hibbeler’s “about 7 per cent” exactly. Be careful with the rule as it is usually QUOTED, though: some sources write it as R/c with c the half-depth, which doubles every threshold. R/c = 10 and R/h = 5 are the same statement. This page uses R/h throughout and prints it above. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
R/h	Eccentricity e ÷ h	Inner fibre: curved ÷ straight	Outer fibre: curved ÷ straight	Inner stress the straight formula misses (%)
1.00	0.08976	1.5235	0.7300	34.4
1.25	0.06978	1.3701	0.7777	27.0
1.50	0.05730	1.2875	0.8104	22.3
2.00	0.04238	1.1996	0.8531	16.6
3.00	0.02799	1.1244	0.8984	11.1
4.00	0.02092	1.0905	0.9222	8.3
5.00	0.01671	1.0711	0.9370	6.6
6.00	0.01391	1.0586	0.9470	5.5
8.00	0.01043	1.0433	0.9598	4.2
10.00	0.00834	1.0344	0.9676	3.3
15.00	0.00556	1.0227	0.9782	2.2
20.00	0.00417	1.0169	0.9836	1.7

The same 25 kN hook in four sections, all 60 mm deep off a 50 mm inner radius

Four sections at the same inner radius, the same depth and the same load. Read the last column rather than the stress column, because the stresses are deceptively similar while the AREAS differ by a factor of 2.7. At a fixed allowable stress the load a section carries per unit of material goes as 1/(stress × area), and on that measure the round bar is the worst of the four: the trapezoid with its wide face inside carries 55 per cent more load for the same weight of steel, and the T section 2.16 times as much. Check it the other way and you get the same answer: make a trapezoid of exactly the round bar’s area — 67 mm inside, 27.25 outside, 60 deep — and it carries the 25 kN at 90.3 MPa against the round bar’s 140.1. Two different comparisons, the same factor of 1.55. That is why hooks are forged as trapezoids. The round bar survives in small hooks and in chain links because it is what a forging or a bent link wants to be, has no corners to start a crack from, and is easy to inspect — and the T, which wins on paper, loses on all three. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Section	Area (mm²)	R (mm)	r_n (mm)	e (mm)	Total inner stress (MPa)	Total outer (MPa)	Load per kg of steel, against the round bar
Rectangle 28 × 60	1,680	80.00	76.10	3.902	174.1	-79.1	1.35
Round bar, 60 dia	2,827	80.00	77.08	2.919	140.1	-63.7	1.00
Trapezoid 40 inside, 16 outside, 60 deep	1,680	75.71	72.08	3.634	151.8	-92.0	1.55
T, 40 flange × 12, 12 web, 60 deep	1,056	72.36	68.20	4.167	173.3	-132.6	2.16

The neutral axis moves inwards, the stress goes hyperbolic, and the inside of the curve pays for it

In a curved beam the neutral axis does not pass through the centroid, and the stress distribution is hyperbolic rather than linear. The reason is not subtle: the fibres are different lengths before anything is loaded. A fibre at radius r has length rθ, so rotating the section by a given angle strains the short inner fibres more than the long outer ones, and the strain carries a 1/r in it. Force balance then pulls the neutral axis inwards, towards the centre of curvature, to a radius r_n = A ÷ ∫dA/r — which is a harmonic mean of radii and therefore always smaller than the arithmetic mean that gives the centroid. That one sentence is why a crane hook is not a beam.

The eccentricity is small, and it is in the denominator. e = R − r_n is a few per cent of the section depth even on a sharply curved part, and it divides the stress. It is also the difference of two much larger numbers, which makes it the one quantity on this page that must be carried to full precision: round R and r_n to three significant figures each before subtracting and you can change e by ten per cent and every stress with it. This is the commonest arithmetic failure in hand curved-beam calculations and it is completely silent.

The inner fibre is much higher than a straight-beam calculation predicts, and the outer fibre is lower. At R/h = 5 the straight formula misses 6.6 per cent of the true inner fibre stress and overstates the outer one by 6.7 per cent; at R/h = 2 it is 16.6 per cent low inside and 17.2 per cent high outside; at R/h = 1.25 it is 27 per cent low inside. The conventional rule — use curved-beam theory below R/h = 5 and not above it — puts the switch at about 7 per cent, and that is a reasonable place for it. Two warnings about the rule as it is usually quoted. It is often written as R/c with c the half-depth, which doubles every threshold and is the same statement; this page uses R/h and prints it. And it is stated for SYMMETRIC sections. An asymmetric one — a hook’s trapezoid, a T — adds its own error on top, because the straight-beam formula measures c from the centroid and the centroid has moved away from the fibre that matters.

On a hook, the direct stress is what settles the argument. A hook’s load line passes through the centre of curvature, so the section carries a moment P·R and a tension P/A at the same time. At the inner fibre both are tensile and they add; at the outer fibre the bending is compressive and they subtract. That is why the inside of the throat is unambiguously the critical point on a hook and why hooks are inspected by measuring the throat opening: an overloaded hook yields at the inside first and stretches, and the stretch is the warning that arrives before the break. A hook is also a piece of regulated lifting equipment, and this page does not design one. Lifting equipment is governed by regulation everywhere — LOLER 1998 in Great Britain, ASME B30.10 in the United States — and a hook must be a rated, marked, certified component, thoroughly examined at the intervals the law sets by a competent person. Use this page to understand a hook, or to check an unregulated curved part such as a C-frame, a clamp or a press frame. Never to design, modify, repair or weld to a lifting hook.

Where this applies. Crane hooks, C-frames, chain links, G-clamps, press frames, punch frames, the throat of a bench vice, spring clips, snap rings and circlips, and the curved portion of any bent-up bracket. What they have in common is a small radius of curvature compared with the depth of the section and a load line that misses the centroid. The section properties page computes the area, the centroid and the second moment for eight shapes if you need them for a section this page does not carry; bring A, R and I here and use the rectangle or the trapezoid as an approximation, or integrate ∫dA/r yourself.

What the theory leaves out. Winkler assumes plane sections remain plane, that the section does not distort, that the material is linear elastic, and that the beam is loaded in its plane of curvature. The first two fail for thin-walled open sections: the bending stresses in an outstanding flange have a radial component that curls the flange towards the neutral axis and sheds load, an effect named after Bleich which needs a reduced effective flange width. Nothing here accounts for stress concentrations at fillets or holes, which on a real frame are often larger than the curvature effect — the stress concentration page has those. And nothing here is a fatigue calculation: a press frame or a hook is a cyclically loaded part, and the endurance limit page is where that belongs.

Frequently asked questions

Why is the neutral axis not at the centroid in a curved beam?

Because the fibres are not the same length to start with. A fibre at radius r has length rθ, so for a given rotation of the section the short inner fibres are strained more than the long outer ones and the strain varies as 1/r rather than linearly. Balancing the forces then requires the zero-stress line to sit where r_n = A ÷ ∫dA/r, which is a harmonic mean of radii and is always inside the centroid. The shift is small — a few per cent of the section depth — but it appears in the denominator of the stress, so it matters far more than its size suggests.

When can I use the ordinary beam formula on a curved part?

When the centroidal radius is more than about five times the section depth, the straight formula is within about 7 per cent at the inner fibre and it is fair to use it. At ten times it is within 3.3 per cent, which is inside material scatter. Below five it goes wrong quickly and always in the unsafe direction at the inner fibre: 17 per cent low at R/h = 2 and a quarter low at R/h = 1.25. Watch the definition, though — many books write the rule as R/c with c the half-depth, so their “10” is this page’s 5.

What is the formula for stress in a curved beam?

σ = M(r_n − r) / (A·e·r), with r_n the neutral axis radius, e = R − r_n the eccentricity, r the radius of the point you want and M the moment about the CENTROIDAL axis. Some sources write it as σ = (M/Ae)(r_n/r − 1), which is identical. Add P/A by superposition if the section also carries a direct load. The distribution is hyperbolic in r, and the two things it must satisfy — no net force, and a moment of exactly M about the centroid — are what the formula is derived from.

Why is a crane hook’s section a trapezoid with the wide face inside?

Because that is where the stress is. The inner fibre carries the bending tension and the direct tension together, and it carries the bending stress amplified by the curvature; the outer fibre carries much less. Putting the wide face inside does two things at once: it adds material where the stress is highest, and it moves the centroid outwards, which lengthens the moment arm slightly but improves the distribution far more. Matched on weight rather than on depth, a trapezoid hook section carries about 55 per cent more load than a round bar: a 67 by 27 mm trapezoid 60 mm deep off a 50 mm inner radius takes 25 kN at 90 MPa where a 60 mm round bar of exactly the same area takes it at 140.

Does this page design crane hooks?

No, and it should not be used to. A crane hook is a lifting component and lifting equipment is regulated: in Great Britain LOLER 1998 requires it to be of adequate strength, marked with its safe working load, and thoroughly examined at set intervals by a competent person, and ASME B30.10 covers hooks in the United States. A hook must be a rated, marked, certified part bought to a standard. This page is for understanding why a hook is the shape it is, and for checking unregulated curved parts — C-frames, clamps, press and punch frames.

How do I find the neutral axis radius for a section that is not listed?

Integrate. r_n = A ÷ ∫dA/r, and for any section that can be sliced into strips of constant width between two radii, ∫dA/r for each strip is simply the width times ln(r_outer/r_inner). Add them up. That is exactly how the T section on this page is done and it extends to an I, a channel, a stepped section or anything you can draw. For a curved profile, integrate numerically — twenty strips is plenty. The section properties page will give you A, the centroid and I for eight shapes to go with it.

Does a curved beam deflect differently too?

Yes, and this page does not compute it. A curved member’s deflection has contributions from bending, from the axial force and from the shear, and on a sharply curved part the axial and shear terms are not negligible the way they are in a slender straight beam — Castigliano’s theorem on the strain energy of the curved member is the usual route. The throat opening of a C-frame press under load, which is often the controlling design criterion, is a deflection problem and not a stress one.

Related calculators

References

Emil Winkler, Formänderung und Festigkeit gekrümmter Körper, insbesondere der Ringe, Der Civilingenieur 4 (1858), 232–246. The origin of the theory this page implements, and of the one idea that makes it different from beam bending: plane sections stay plane, but the fibres nearer the centre of curvature are SHORTER, so the same rotation strains them more and the stress varies as 1/r rather than linearly. Cited for the theory, not reproduced — every formula on the page is re-derived from ∫σdA = 0 and ∫σ(R−r)dA = M, and both integrals are checked numerically.
Richard Budynas and Keith Nisbett, Shigley’s Mechanical Engineering Design, section 3-18 and Table 3-4 (“Formulas for sections of curved beams”). Cited by section and table number as the standard published collection of neutral-axis formulas. The four used here were not copied from it: each was derived from r_n = A / ∫dA/r and then checked against a Simpson integration of the real section, which agrees to better than two parts in a million.
RoyMech, Curved Beams (roymech.org, read 29 September 2026). The source consulted for the stress formula in the form σ = (M/Ae)(r_n/r − 1), which is identical to the M(r_n−r)/(Aer) written here. Its worked crane hook carries a 25 kN load and reaches 126.76 N/mm² tension at the inner surface against −42.25 at the outer — a three-to-one split that is the whole reason a hook is drawn the shape it is. The hook’s dimensions are given only as a drawing on that page, so the example could not be reproduced here and is quoted, not checked.
R. C. Hibbeler, Mechanics of Materials, the curved beam section, as quoted by EngineerExcel, Bending Stress in Curved Beams (engineerexcel.com, read 29 September 2026): for a rectangular section at a radius-of-curvature-to-depth ratio of 5, the straight-beam flexure formula gives a maximum normal stress about 7 per cent below the curved-beam value, and curved beam analysis is called for when the radius of curvature is less than five times the depth. Reproduced here rather than taken on trust: the calculation in this page’s own chart gives 6.64 per cent at R/h = 5.
The Lifting Operations and Lifting Equipment Regulations 1998 (LOLER, SI 1998/2307) in Great Britain, and ASME B30.10 Hooks in the United States. Cited by name because a crane hook is the archetypal curved beam and is also a piece of regulated lifting equipment: it must be a rated, marked, certified component, thoroughly examined at the intervals the regulations set, and it may not be designed from a first-pass stress calculation on a general-purpose page. Neither document is reproduced here and neither was fetched; they are named so that the boundary is unmistakable.