2
$\begingroup$

I'm reading the book The New Science of Strong Materials by J.E. Gordon. He writes

when we plotted (...) breaking strain, against thickness, we found it did not matter what the whiskers were made of, for they all plotted on the same curve. (p87, Penguin)

He proves this point with a graph where the y-axis is "Fracture stress $\%E$". However, he previously stated that stress is very different to strain—stress is the load per area, while strain is the extension of the material under said load. Thus, the graph doesn't seem to correlate with the argument. I'm thinking that the $\%E$ is therefore relevant. Can someone clarify this for me?

Here is the graph: enter image description here

$\endgroup$
3
$\begingroup$

The $E$ he is refering to is the Young's Modulus, which is the elastic modulus that tells you how the tensile stress for a wire relates to its extensional strain, i.e.

$$ \text{stress} = E \text{ strain} $$ in a thin wire. In the experiment he is summarizing in the graph, you pull a thin wire until it breaks, while measuring the stress on each end. The measured stresses at the point of breaking will differ for different materials, but by reporting the breaking stress here as a percentage of the young's modulus, he is really plotting the breaking strains.

$$ \text{ breaking stress } = E \text{ breaking strain } $$ or $$ \text{ breaking strain } = \frac{ \text{ breaking stress }}{E} $$

the strain is dimensionless, and in his experiments was at about $0.01 = 1\%$ for wires with a diameter of about 10 microns.

The actual stresses that registered on his apparatus would have been different for the different materials, as they have a different Young's modulus, but, taken as a percentage of their Young's modulus (or equivalently, computing the breaking strain), they come out as equivalent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.