Sticking Together. Steven Abbott
so the adhesion value in N m−2=0.001/0.000625=1.6 N m−2, a very small adhesion value. Readers may notice that including the overlap length is a cheat; it is the only way to get a value in the same units as the others. This is an early indication of the problems ahead. I am cheating deliberately and openly; many in the adhesion world are unaware that these numbers are meaningless.
If you measure the force required to fail the same joint by pulling in shear mode, you find it needs a force that depends not only on the surface energy but on the thickness and the modulus of the rubber, i.e. how much stretch you get divided by the force required to create that stretch. Using typical values, we find that F=1.25 N (not the mN of peel!) giving a N m−2=2000.
If you now try to pull it apart vertically, in the butt joint test mode, it requires F=27 N and a N m−2=55 000, 30 000 times more than the peel. This now explains why the two strong men could not pull two pieces of rubber apart (butt joint) while the little girl had no problem (peel joint). Although the men were stronger, they weren't 30 000 times stronger!
You can find all the formulae (which come from Prof. Kendall) and play with the key values in an app I wrote: https://www.stevenabbott.co.uk/practical-adhesion/weak-strong.php. More important is the message that Adhesion is a Property of the System. If your system is one where the forces are always equivalent to a butt joint, then the 55 000 N m−2 might be good enough. Yet if there is any risk of some peel forces, then the 1.6 N m−2 is likely to be very worrying.
Why do the values change by more than 30 000? Because joints fail whenever the local forces exceed a limit. In the peel test, the forces are focussed exactly at one point – the point where the joint will fail. With the lap and butt joints, the work put in to breaking the joint is also expended in stretching and distorting the material layers involved in the joint so that the force at the edge of the joint is much smaller.
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