Why Roof Beams Are Laid on Their Edge, Not Flat

Pull a steel wire and a copper wire of the same size with the same force, and the copper stretches more. To compare materials fairly, physics uses moduli of elasticity — stress divided by strain — which depend only on the material.

Those same numbers decide how thick a crane rope must be and why a beam is shaped the way it is.

This part covers the three moduli, Poisson's ratio, elastic energy, and how elasticity guides the design of structures. Take $g=9.8$ m/s${}^2$.

**Each modulus is stress divided by the matching strain: Young's modulus $Y=\frac{F/A}{\Delta L/L}$ for stretching, bulk modulus $B=-\frac{p}{\Delta V/V}$ for squeezing from all sides, and shear modulus $G=\frac{F/A}{\theta }$ for sliding layers.

Worked example 1 — Young's modulus.** A $2$ m steel wire of radius $0.5$ mm stretches $2$ mm under $157$ N.

$$A=\pi (0.5\times 10^{-3}{)}^2\approx 7.85\times 10^{-7}{\text{ m}}^2\text{stress}\approx 2.0\times 10^8\text{ Pa}\text{strain}=\frac{0.002}{2}=10^{-3}$$

$$Y=\frac{2.0\times 10^8}{10^{-3}}=2.0\times 10^{11}\text{ Pa}$$

Worked example 2 — bulk modulus. Water has $B\approx 2.2\times 10^9$ Pa. Under an extra pressure of $10^7$ Pa, $1$ litre shrinks by

$$\frac{\Delta V}{V}=\frac{10^7}{2.2\times 10^9}\approx 4.5\times 10^{-3}\Rightarrow \Delta V\approx 4.5{\text{ cm}}^3$$

Worked example 3 — shear modulus. Suppose a $0.1$ m metal cube's top face shifts $10^{-5}$ m under a $2\times 10^4$ N sideways force.

$$\text{stress}=\frac{2\times 10^4}{0.01}=2\times 10^6\text{ Pa}\theta =\frac{10^{-5}}{0.1}=10^{-4}G=2\times 10^{10}\text{ Pa}$$

An everyday example. Steel bars in a concrete roof are chosen because steel's large Young's modulus keeps the roof from stretching and cracking.

The substance. Liquids and gases have a bulk modulus but no Young's or shear modulus, because they cannot hold a fixed shape.

**Poisson's ratio is the lateral strain divided by the longitudinal strain, $\sigma =\frac{\Delta d/d}{\Delta L/L}$, and it tells you how much thinner a body gets as it is stretched.**

It has no unit. For most metals it lies roughly between $0.25$ and $0.35$, and its theoretical upper limit is $0.5$.

Worked example. A $2$ m wire of diameter $2$ mm is stretched by $2$ mm, and its diameter falls by $0.0006$ mm.

$$\text{longitudinal strain}=\frac{2}{2000}=10^{-3}\text{lateral strain}=\frac{0.0006}{2}=3\times 10^{-4}$$

$$\sigma =\frac{3\times 10^{-4}}{10^{-3}}=0.3$$

A check on volume. For a small stretch, the fractional change in volume is about $(1-2\sigma )$ times the longitudinal strain — here $0.4\times 10^{-3}$, so the wire's volume increases slightly.

An everyday example. Stretch a rubber band and watch it become visibly narrower — rubber has a Poisson's ratio close to $0.5$.

The substance. **When $\sigma =0.5$, the volume stays constant** as the body stretches, which is nearly true for rubber.

**The work done in stretching a wire is stored as elastic potential energy $U=\frac{1}{2}F\Delta L$, so the energy per unit volume is $u=\frac{1}{2}\times \text{stress}\times \text{strain}=\frac{1}{2}Y(\text{strain}{)}^2$.

Derivation.** When the extension is $l$, the tension is $F=\frac{YAl}{L}$. The work to stretch from $0$ to $\Delta L$ is

$$W={\int }_0^{\Delta L}\frac{YA}{L}ldl=\frac{1}{2}\frac{YA}{L}(\Delta L{)}^2=\frac{1}{2}F\Delta L$$

Dividing by the volume $AL$:

$$u=\frac{1}{2}\cdot \frac{F}{A}\cdot \frac{\Delta L}{L}$$

Worked example. For the steel wire above (stress $2.0\times 10^8$ Pa, strain $10^{-3}$):

$$u=\frac{1}{2}(2.0\times 10^8)(10^{-3})=1.0\times 10^5{\text{ J/m}}^3$$

$$V=7.85\times 10^{-7}\times 2\approx 1.57\times 10^{-6}{\text{ m}}^{3}U=uV\approx 0.157\text{ J}$$

Check: $\frac{1}{2}F\Delta L=\frac{1}{2}(157)(0.002)=0.157$ J.

An everyday example. Pulling back the rubber of a catapult (gulel) stores elastic energy that is handed to the stone on release.

The substance. **The stored energy is $\frac{1}{2}F\Delta L$, not $F\Delta L$**, because the force grows from zero as the wire stretches.

Structures are designed so that stresses stay well below the elastic limit: rope thickness is set by the safe stress, and beams are shaped to resist bending, which depends strongly on their depth and on Young's modulus.

Beams. A beam of length $l$, breadth $b$ and depth $d$, supported at its ends and loaded at the centre with $W$, sags by

$$\delta =\frac{W{l}^3}{4b{d}^{3}Y}$$

**Doubling the depth cuts the sag by $2^3=8$ times, while doubling the breadth only halves it. That is why beams are laid on their edge, and why steel girders have an I-shaped cross-section — material far from the middle resists bending while saving weight.

Worked example — a crane rope.** A crane must lift $10^4$ kg with steel of yield strength $3\times 10^8$ Pa. The minimum radius is

$$r=\sqrt{\frac{mg}{\pi {\sigma }_y}}=\sqrt{\frac{10^4\times 9.8}{\pi \times 3\times 10^8}}\approx 0.010\text{ m}=1\text{ cm}$$

In practice a safety margin makes it much thicker, and it is made of many thin wires twisted together for flexibility.

Columns and bridges. Pillars carrying loads must avoid buckling, so they are made thicker, and bridges must not bend too much under traffic and wind.

An everyday example. A wooden plank used as a bridge over a drain sags far less when turned on its edge than when laid flat.

The substance. Elastic limit sets safety; Young's modulus sets stiffness — a design must satisfy both.