Pure Bending

Assumptions

1. The beam is initially straight, with constant cross-section, and the material is homogeneous and isotropic.
2. The material obeys Hooke's law and $E$ is the same in tension and compression.
3. Plane sections remain plane after bending (Bernoulli's assumption).
4. The beam bends in a plane of symmetry and the loads act in that plane.
5. The radius of curvature is large compared with the depth of the beam.
6. Each layer is free to expand or contract, with no stress between layers.

Derivation

       ___ M                 M ___
   ---/    \_____N_____A_____/    \---
        top: compression      O = centre of curvature
        NA : unchanged length
        bottom: tension

Consider a length $dx$ of beam bent to a radius $R$ (to the neutral layer) subtending angle $d\theta$. A layer at distance $y$ from the neutral layer has original length $dx = R\,d\theta$ and new length $(R+y)d\theta$.

$$\varepsilon = \frac{(R+y)d\theta - R\,d\theta}{R\,d\theta} = \frac{y}{R}$$

By Hooke's law:

$$\sigma = E\varepsilon = \frac{E\,y}{R} \qquad (1)$$

Pure bending means no net axial force, so over the section

$$\int \sigma\,dA = \frac{E}{R}\int y\,dA = 0 \;\Rightarrow\; \int y\,dA = 0$$

This says the first moment of area about the neutral axis is zero, so the neutral axis passes through the centroid.

Moment of resistance equals the applied moment $M$:

$$M = \int \sigma\,y\,dA = \frac{E}{R}\int y^2\,dA = \frac{E\,I}{R} \qquad (2)$$

where $I = \int y^2\,dA$ is the second moment of area about the neutral axis. From (1) and (2):

$$\boxed{\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}}$$

Use

The maximum stress is at the extreme fibre $y_{max}$:

$$\sigma_{max} = \frac{M\,y_{max}}{I} = \frac{M}{Z}, \qquad Z = \frac{I}{y_{max}}$$

$Z$ is the section modulus. For a rectangle $Z = bh^2/6$; for a circle $Z = \pi d^3/32$.

(a) Neutral axis and I

Measure $\bar y$ from the bottom of the web.

$$\bar y = \frac{931500}{8100} = 115\ \text{mm from bottom}, \quad 65\ \text{mm from top}$$

Using the parallel axis theorem:

$$I = \left[\frac{120\times30^3}{12} + 3600(165-115)^2\right] + \left[\frac{30\times150^3}{12} + 4500(115-75)^2\right]$$ $$I = (0.27 + 9.0)\times10^6 + (8.4375 + 7.2)\times10^6 = 24.91\times10^6\ \text{mm}^4$$

(b) Stresses at mid-span

$$M_{max} = \frac{wL^2}{8} = \frac{12\times4^2}{8} = 24\ \text{kN m} = 24\times10^6\ \text{N mm}$$

The load sags the beam, so the top (flange) is in compression and the bottom is in tension.

$$\sigma_c = \frac{M\,y_{top}}{I} = \frac{24\times10^6\times65}{24.91\times10^6} = 62.6\ \text{MPa}$$ $$\sigma_t = \frac{M\,y_{bot}}{I} = \frac{24\times10^6\times115}{24.91\times10^6} = 110.8\ \text{MPa}$$

(c) Safe UDL

Limit moment from tension: $M_t = \dfrac{\sigma_{t,all} I}{y_{bot}} = \dfrac{40\times24.91\times10^6}{115} = 8.66\times10^6$ N mm.

Limit moment from compression: $M_c = \dfrac{150\times24.91\times10^6}{65} = 57.5\times10^6$ N mm.

Tension governs (the bottom fibre is farther from the neutral axis and cast iron is weak in tension): $M = 8.66$ kN m.

$$w = \frac{8M}{L^2} = \frac{8\times8.66}{16} = 4.33\ \text{kN/m}$$

Answer: NA 115 mm above the bottom; $I = 24.91\times10^6\ \text{mm}^4$; $\sigma_t = 110.8$ MPa and $\sigma_c = 62.6$ MPa (the 12 kN/m load is unsafe in tension); safe load $w = 4.33$ kN/m. Placing the flange at the bottom (tension side) would be more suitable for cast iron.

Pure and simple bending

Pure bending is bending under a constant bending moment with zero shear force, for example the central portion of a beam loaded by two equal loads symmetrically placed. Simple bending is general bending in which the bending moment varies along the beam and shear force is also present.

 P              P      pure bending between the loads
 |              |
 v______________v
 A  a   |   L   | a  B    SF = 0, BM = Pa between loads

Neutral layer and neutral axis

• Neutral layer (surface): the layer of fibres that is neither stretched nor shortened during bending, so its stress and strain are zero. The fibres above it are in compression and those below it in tension for a sagging beam.
• Neutral axis (NA): the line of intersection of the neutral layer with a cross-section of the beam. Bending stress varies linearly from zero at the NA to the maximum at the extreme fibres.

Position of the neutral axis

For linearly elastic material the net axial force over the section is zero, $\int \sigma\,dA = 0$, hence $\int y\,dA = 0$. The neutral axis therefore passes through the centroid of the cross-section. For unsymmetrical sections (such as a T-section) it is not at mid-depth, so the maximum tensile and compressive stresses differ.

Section modulus

The section modulus is $Z = I/y_{max}$, where $I$ is the second moment of area about the neutral axis and $y_{max}$ the distance to the extreme fibre. From the flexure formula the moment of resistance is

$$M_r = \sigma_{all}\,Z$$

so for a given stress, the larger $Z$, the larger the moment the section can carry. Unit: mm$^3$ or m$^3$.

Comparison for equal area

Rectangle of width $b$ and depth $h = 2b$:

$$A = 2b^2, \qquad Z_r = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = 0.667\,b^3$$

Circle with the same area:

$$\frac{\pi d^2}{4} = 2b^2 \;\Rightarrow\; d = 1.596\,b$$ $$Z_c = \frac{\pi d^3}{32} = \frac{\pi (1.596b)^3}{32} = 0.399\,b^3$$ $$\frac{M_{r,\text{rect}}}{M_{r,\text{circle}}} = \frac{Z_r}{Z_c} = \frac{0.667}{0.399} = 1.67$$

For example, with $b = 100$ mm: $Z_r = 666.7\times10^3$ mm$^3$ and $Z_c = 398.9\times10^3$ mm$^3$. With $\sigma_{all} = 10$ MPa they carry 6.67 kN m and 3.99 kN m respectively. The rectangle (on edge) is about 67 % stronger in bending for the same weight of material.

Why I-sections

In bending the material near the neutral axis carries little stress. An I-section puts most of the area in the flanges, far from the NA, so $I$ and $Z$ are large for a small area. Its web resists shear. It therefore gives the highest strength-to-weight ratio, which is why rolled I-sections are used for beams.