Winkler-Bach formula for curved beams

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, $h = r_o - r_i$, $e = r_c - r_n$.

        ___ 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 $\sigma = My/I$.