---
title: "Winkler-Bach formula for curved beams"
subject: "Mechanics of Solids"
chapter: 6
key: "semester_5/mechanics_of_solids:winkler-bach-curved-beam-derivation"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/winkler-bach-curved-beam-derivation
site: BCT Engineering
---

# Winkler-Bach formula for curved beams

Mechanics of Solids · Chapter 6: Curved Beams

- Asked in: year not given, 8 marks
- Video: [Curved Beams](https://www.youtube.com/watch?v=eaaveRCA_3Q) — Youngstone Videos
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/winkler-bach-curved-beam-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-6

## Question

Derive the Winkler-Bach expression for the circumferential (bending) stress in a curved beam of rectangular cross-section subjected to a pure bending moment. Show the location of the neutral axis and explain why it shifts from the centroidal axis.

## Short answer

The Winkler-Bach formula gives circumferential stress σ = M(r_n - r)/(A e r), with neutral axis radius r_n = A / ∫(dA/r). The neutral axis shifts toward the centre of curvature because inner fibres are shorter, so they experience larger strain for the same rotation, requiring the axis to move inward to balance forces.

## Answer

#### Assumptions

1. The material is linear elastic, homogeneous and isotropic.
2. Plane sections remain plane (as in straight beams) and rotate about the neutral axis.
3. The section is symmetrical about the plane of bending; radial stress is neglected in the main derivation.
4. The beam is thick: the radius of curvature is comparable to the depth, so the fibres are not of equal length.

#### Geometry

Let $r_i$, $r_o$ be the inner and outer radii, $r_c$ the radius of the centroidal axis, $r_n$ the radius of the neutral axis, $h = r_o - r_i$, $e = r_c - r_n$.

```text
        ___ fibre at radius r
      /     \
     /  r_n  \
    /  r_c    \      O = centre of curvature
   |    .      |
    \         /
     \_______/
         O
```

Two sections separated by an angle $d\phi$ rotate relatively by $\Delta d\phi$ about the neutral axis. A fibre at radius $r$ has length $r\,d\phi$ and elongates by $(r_n - r)\Delta d\phi$ (for a moment that tends to reduce curvature, the inner fibres are in tension):

$$
\varepsilon = \frac{(r_n - r)\,\Delta d\phi}{r\,d\phi},\qquad \sigma = E\varepsilon = Ek\,\frac{r_n - r}{r},\quad k = \frac{\Delta d\phi}{d\phi}
$$

The stress is therefore **hyperbolic**, not linear.

#### Neutral axis

Net axial force is zero:

$$
\begin{aligned}
\int \sigma\,dA &= Ek\int \frac{r_n - r}{r}\,dA = 0 \\
r_n \int \frac{dA}{r} &= \int dA = A \quad\Rightarrow\quad r_n = \frac{A}{\displaystyle\int \dfrac{dA}{r}}
\end{aligned}
$$

For a rectangle ($dA = b\,dr$): $r_n = \dfrac{h}{\ln(r_o/r_i)}$. For a circle of radius $c$: $r_n = \dfrac{r_c + \sqrt{r_c^2 - c^2}}{2}$.

#### Moment relation

The moment of the stress about the neutral axis:

$$
\begin{aligned}
M &= \int \sigma\,(r_n - r)\,dA = Ek\int\frac{(r_n - r)^2}{r}\,dA \\
&= Ek\left[r_n^2\int\frac{dA}{r} - 2r_nA + \int r\,dA\right] \\
&= Ek\left[r_nA - 2r_nA + r_cA\right] = Ek\,A\,(r_c - r_n) = Ek\,A\,e
\end{aligned}
$$

so $Ek = M/(Ae)$. Hence

$$
\boxed{\sigma = \frac{M\,(r_n - r)}{A\,e\,r}}
$$

- At the inner fibre $(r = r_i)$ the stress is $\sigma_i = \dfrac{M\,y_i}{A\,e\,r_i}$, with $y_i = r_n - r_i$ (largest in magnitude).
- At the outer fibre $\sigma_o = -\dfrac{M\,y_o}{A\,e\,r_o}$, with $y_o = r_o - r_n$.

With a direct force $N$ at the centroid, add $N/A$ (superposition).

#### Why the neutral axis shifts

Fibres near the centre of curvature are shorter, so the same angular rotation produces larger strain in them than in the longer outer fibres. For zero net force the neutral axis must move **toward the centre of curvature** from the centroid, giving $r_n < r_c$ (i.e. $e>0$). As $r_c/h$ increases the beam becomes straight, $e \to 0$ and the formula reduces to $\sigma = My/I$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the Winkler-Bach theory, why does the bending stress vary hyperbolically across the depth of a curved beam?
   A. Because the material is non-linear
   B. Because a fibre at radius r has length r dphi, so the same rotation gives strain proportional to (rn - r)/r
   C. Because radial stress is large and cannot be neglected
   D. Because plane sections do not remain plane
   Answer: B. Strain is the elongation (rn - r)Δdφ divided by the fibre length r dφ, giving sigma = Ek(rn - r)/r, which is hyperbolic in r. (Key point: Hyperbolic stress distribution)

2. Which condition gives the radius of the neutral axis rn = A / ∫(dA/r)?
   A. Net bending moment about the centroid is zero
   B. Net axial force on the section is zero
   C. Net shear force is zero
   D. Strain at the centroid is zero
   Answer: B. Setting ∫sigma dA = 0 gives rn ∫dA/r = A, so rn = A/∫(dA/r). (Key point: Neutral axis location)

3. For a rectangular section of depth h with inner and outer radii ri and ro, the neutral axis radius is
   A. rn = (ri + ro)/2
   B. rn = h ln(ro/ri)
   C. rn = h / ln(ro/ri)
   D. rn = h^2 / (ro - ri)
   Answer: C. With dA = b dr, ∫dA/r = b ln(ro/ri), so rn = bh/(b ln(ro/ri)) = h/ln(ro/ri). (Key point: Rectangular section rn)

4. Why does the neutral axis move from the centroid towards the centre of curvature (rn < rc)?
   A. Inner fibres are shorter, so they strain more for the same rotation and the axis must shift inward for zero net force
   B. Outer fibres are stronger than inner fibres
   C. The load is applied off the centroid
   D. Radial stress pushes the axis inward
   Answer: A. Shorter inner fibres have larger strain for equal rotation, so to keep zero net axial force the axis moves toward the centre of curvature, giving e = rc - rn > 0. (Key point: Why the axis shifts)

## Related in Mechanics of Solids

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- [Stress in a crane hook section](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Acrane-hook-curved-beam-stress) — Mechanics of Solids, chapter 6
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