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 $\sigma = My/I$.

Data

$b = 30$ mm, $h = 50$ mm, $r_i = 40$ mm, $r_o = 40 + 50 = 90$ mm, $P = 12$ kN.

$$\begin{aligned} A &= bh = 1500\ \text{mm}^2 \\ r_c &= r_i + h/2 = 65\ \text{mm} \end{aligned}$$

Neutral axis radius

For a rectangle:

$$r_n = \frac{h}{\ln(r_o/r_i)} = \frac{50}{\ln(90/40)} = \frac{50}{0.8109} = 61.658\ \text{mm}$$ $$e = r_c - r_n = 65 - 61.658 = 3.342\ \text{mm}$$ $$y_i = r_n - r_i = 21.658\ \text{mm},\qquad y_o = r_o - r_n = 28.342\ \text{mm}$$

Loads on the section

The load acts along the line through the centre of curvature, so the section is subjected to a direct tension and a bending moment about the centroid:

$$M = P\,r_c = 12\times10^3\times65 = 780\times10^3\ \text{N mm} = 780\ \text{N m}$$

The moment tends to straighten (open) the hook, so the inner fibres are in tension and the outer ones in compression.

Stresses

$$\sigma_{direct} = \frac{P}{A} = \frac{12000}{1500} = 8.00\ \text{MPa (tension)}$$ $$\begin{aligned} \sigma_{b,i} &= \frac{M\,y_i}{A\,e\,r_i} = \frac{780\times10^3\times21.658}{1500\times3.342\times40} = 84.24\ \text{MPa (tension)} \\ \sigma_{b,o} &= \frac{M\,y_o}{A\,e\,r_o} = \frac{780\times10^3\times28.342}{1500\times3.342\times90} = -48.99\ \text{MPa (compression)} \end{aligned}$$

Total:

$$\begin{aligned} \sigma_{inner} &= 8 + 84.24 = 92.24\ \text{MPa (tension)} \\ \sigma_{outer} &= 8 - 48.99 = -40.99\ \text{MPa (compression)} \end{aligned}$$

Comparison with straight-beam formula

$$\sigma_{b,straight} = \frac{6M}{bh^2} = \frac{6\times780\times10^3}{30\times50^2} = 62.4\ \text{MPa}$$

The inner fibre bending stress of the curved beam is $84.2/62.4 \approx 1.35$ times the straight-beam value, so the straight beam formula is unsafe for such a sharply curved section.

Answer: Maximum tensile stress $= 92.2$ MPa (inner fibre); maximum compressive stress $= -41.0$ MPa magnitude (outer fibre). The straight-beam formula gives only $62.4$ MPa bending stress.

Data

$R = 150$ mm, $P = 1000$ N, $b = 25$ mm, $t = 12$ mm, $E = 200\times10^3$ MPa.

$$I = \frac{bt^3}{12} = \frac{25\times12^3}{12} = 3600\ \text{mm}^4$$

Statically indeterminate moment

The ring is symmetric about both diameters. Cut it at the horizontal section: by symmetry the shear is zero, the normal force is $P/2$ (compression) and the moment $M_0$ is unknown. The slope of this section does not change, so $\partial U/\partial M_0 = 0$.

          P
          |
        .-v-.
      /       \
  M0 |-- ring --| M0     (cut sections with P/2 each)
      \       /
        '-^-'
          |
          P

For a section at angle $\theta$ from the horizontal cut (towards the load):

$$M_\theta = M_0 - \frac{P}{2}R(1 - \cos\theta),\qquad \frac{\partial M_\theta}{\partial M_0} = 1$$ $$\begin{aligned} \frac{\partial U}{\partial M_0} &= \frac{4R}{EI}\int_0^{\pi/2}\left[M_0 - \frac{PR}{2}(1-\cos\theta)\right]d\theta = 0 \\ M_0\,\frac{\pi}{2} &= \frac{PR}{2}\left(\frac{\pi}{2} - 1\right) \\ M_0 &= PR\left(\frac12 - \frac1\pi\right) = 0.1817\,PR \end{aligned}$$

Bending moments

• At the horizontal section ($\theta = 0$, $90^\circ$ from the load): $M_0 = 0.1817\times1000\times150 = 27254$ N mm, tension on the outer surface.
• At the load point ($\theta = \pi/2$): $M = M_0 - PR/2 = -PR/\pi$, magnitude $0.3183\times1000\times150 = 47746$ N mm, tension on the inner surface (opposite sign to $M_0$).

More simply, $M_\theta = PR\left(\dfrac{\cos\theta}{2} - \dfrac1\pi\right)$.

Maximum bending stress

The maximum moment is at the load points:

$$\sigma_{max} = \frac{M c}{I} = \frac{47746\times6}{3600} = 79.6\ \text{MPa}$$

At the horizontal section: $\sigma = 27253\times6/3600 = 45.4$ MPa. (The direct stress $P/2A = 500/300 = 1.7$ MPa is small and is neglected, as stated.)

Change in loaded diameter

$$\begin{aligned} \delta_V &= \frac{\partial U}{\partial P} = \frac{4R}{EI}\int_0^{\pi/2} M_\theta\,\frac{\partial M_\theta}{\partial P}\,d\theta = \frac{4PR^3}{EI}\int_0^{\pi/2}\left(\frac{\cos\theta}{2} - \frac1\pi\right)^2 d\theta \\ &= \frac{4PR^3}{EI}\left(\frac{\pi}{16} - \frac{1}{2\pi}\right) = \frac{PR^3}{EI}\left(\frac\pi4 - \frac2\pi\right) = 0.1488\,\frac{PR^3}{EI} \end{aligned}$$ $$\delta_V = 0.1488\times\frac{1000\times150^3}{200\times10^3\times3600} = 0.697\ \text{mm}$$

The loaded diameter decreases by this amount, and the horizontal diameter increases by about $0.137\,PR^3/EI$.

Answer: $M_{load} = 47746$ N mm (0.318 $PR$), $M_{90^\circ} = 27254$ N mm (0.182 $PR$, opposite sign), $\sigma_{max} = 79.6$ MPa, decrease in loaded diameter $= 0.70$ mm.