Plant-fibre strength and fair material comparisons

What you’ll learn

• Convert a breaking load to force and interpret tensile stress.
• Evaluate diameter, hydration and gripping as sources of variation.

Understand the idea

Core practical 8 compares the tensile strength of plant fibres. A breaking load gives the force at failure. A thicker fibre can support a larger force simply because it has more material across its cross-section, so breaking force alone is not always a fair comparison of the material. When cross-sectional area is supplied or measured appropriately, breaking stress is force divided by area. A value in $\mathrm{N\,mm^{-2}}$ is numerically equal to a value in MPa.

Control the length between grips, loading arrangement, hydration and preparation of the fibres. Biological fibres are variable, so sample multiple independent fibres and describe the spread as well as the mean. An assumed circular cross-section may be only an approximation; measuring a diameter poorly can strongly affect the calculated area. Record whether failure occurs in the free fibre or at a damaged grip.

A good evaluation distinguishes a material property from an apparatus artefact. Slipping in a clamp does not establish a fibre's breaking force, and damage caused by the clamp may create an artificially weak point. A supervised setup protects against falling masses and snapping fibres, while consistent loading reduces uncontrolled shocks. Comparisons of sustainability need additional evidence about production and disposal, rather than being inferred from strength alone.

Key formulae

$$F=mg$$

$$\sigma=\frac{F}{A}$$

$$A=\pi r^2\quad\text{for an assumed circular cross-section}$$

Worked example

A fibre fails under a supported mass of $0.800\ \mathrm{kg}$. Take $g=9.80\ \mathrm{N\,kg^{-1}}$. Its cross-sectional area is $0.0400\ \mathrm{mm^2}$. Calculate the breaking force and stress.

1. Convert the mass loading to force: $F=0.800\times9.80=7.84\ \mathrm{N}$.
2. Divide force by the stated cross-sectional area: $\sigma=7.84/0.0400$.
3. The stress is $196\ \mathrm{N\,mm^{-2}}$, equivalent to $196\ \mathrm{MPa}$.

Answer: Breaking force $7.84\ \mathrm{N}$; breaking stress $196\ \mathrm{MPa}$.