I want to do some experiments on a neoprene rubber sheet (ex. supplier). When I went to order the sheet I was confused because the listed elongation and tensile strength for all sheet sizes (with different thicknesses) are the same.
The tensile strength is defined by force per cross sectional unit area, so if the thickness changes shouldn't the tensile strength change? Or am I missing something here?
Best Answer
They specify the ultimate tensile strength and % elongation of Nitrile Rubber (properties defined by the material's stress-strain curve). The ultimate tensile strength defines the stress at failure. Similarly, % elongation defines strain at failure.
All sizes have the same material properties (ie. tensile strength and elongation), but each size will fail at a different applied force (because each size has a different cross-sectional area).
The tensile strength is a property of the material (in this case, Nitrile Rubber) - it is constant and does not depend on the shape of the material body. This concept is often confused because tensile testing is used to determine the ultimate tensile strength: An increasing force ($F$) is applied to a material specimen (with known cross-sectional area, $A_{o}$) until failure. The change in length is also recorded at failure, where % elongation defines the ratio of elongation at failure ($L_{failure}$) to the original length ($L_{o}$).
$$\sigma_{ut} = \frac{F_{failure}}{A_{o}} \qquad \text{and} \qquad \text{% elongation} = \frac{L_{failure}}{L_{o}}$$
In contrast, the applied stress is different for each size (thickness and width) because their cross-sectional areas ($A_{size} = t \times w$) are different. If the applied stress is greater than the ultimate tensile strength ($\text{if:} \; \sigma_{applied} \gt \sigma_{ut}$), the material will fail **. Larger sizes result in lower applied stresses, and are able to transmit more force before failure.
$$\sigma_{applied} = \frac{F_{applied}}{A_{size}}$$
The factor of safety ($FOS$) compares the applied stress to the failure criteria. Note that when $FOS = 1$, the material fails because the applied stress equals the failure criteria. Engineers use large safety factors to ensure that design stresses stay below failure criteria. In this example: $$FOS = \frac{\sigma_{ut}}{\sigma_{size}}$$
** For completeness, materials fail according to different failure criteria, but the concepts discussed are equally applicable.