A chain is only as strong as its weakest link?

If all links were the exact same strength and the force enacted on them was also the same then when the chain’s tensil strength is exceeded they would all fail at the same time.

This is a quote I usually apply to workers’ unions. A union is a good idea until it is protecting someone who is NOT doing their job to their fullest ability, and the union protects them from reprimand. If you cannot pull your weight, you sholdn’t have a job that demands that you do. That, and many unions are on their way out, thanks to their greed.

A union is only as strong as its weakest link.

Hmm… wouldn’t the end (if the chain is horizontal) or uppermost (if the chain is vertical) link(s) break first, since they’re bearing more of the weight of the chain itself (unless you assume zero-G)?

@Harp – I agree with you that if the chain were vertical, you’d have to factor in the additional weight on the uppermost link and it would be the first to go.

Basically, the effect of outside forces, things other than the strength of the links, would necessarily determine the breaking point.

With the horizontal example, I’m not sure. On one hand it seems like the one in the middle would be the first to break, because of the inherent “stretching” attributes of all materials; if all the links from each end going toward the middle have enough pressure applied to cause them each to be stretched to their maximum tolerable point, would that amount of combined tolerance be more than the tolerable point of the single link in the middle, meaning it would break first?

On the other hand, it seems maybe the combined effect of the absorption of the total pressure by all the links up to the middle one might leave it with the least pressure being put on it, suggesting the end ones may go first? (Plus the minuscule amount of additional weight borne by them from the rest of the chain.)

Wow, I need a drink and I don’t drink. Someone email Stephen Hawking.

All the links would break at the same time in this impossible example, unless more strain was placed on one link due to being pulled around a post or similar.

I have never liked this as an analogy for human behavior. We are all our brother’s and neighbor’s and friend’s next links. It is up to us to lift the entire tensile strength of the chain.

@SecondGlance, I’m seconding Harp here. In a static situation, forces don’t add up in series; if the chain were being stretched across a table top, each link would be stretched with the same force as the next one, and when the table is removed, the situation remains the same, exept that the inevitable sagging of the chain means that each link must also exert a force to cancel the gravitation pull on all links below it (directly below, meaning down to the middle), including itself. Naturally, this means that the middle link need only support an additional force to account for it’s own weight, while the links at the end must support the weights of half of the links.

The solution is stop yanking my chain.

Oh, and in the absence of gravity, The end at which the force originates will probably break first. Whenever there is a force on a chain, the bonds between the atoms of the chain are stretched, somewhat like little springs, with the collective force of all of the bonds in any given cross-section of the chain being equal to the total force on the spring. When you increase the tension of the chain, you do so by gripping one end (say the right end) and pulling on it, moving the rightmost atoms slightly away from the rest of the chain. This stretches the bonds between those atoms and the adjoining atoms, increasing the pull between them, which in turn moves the adjoining atoms to the right, which then stretches the bonds to their left, moving the next set of toms, and so on in a chain (heh!) reaction until the bonds at the other end (presumed to be affixed) are stretched, and all forces are again at equilibrium. What this means is that when the force is first exerted, before the longitudinal wave of stretching has had a chance to travel along the chain, the atomic bonds at the right of the chain are under more stress than those further along; so if the force is strong enough to break any of the bonds, it will start by breaking the bonds on the end at which it originated, and having broken those bonds, will not be transmitted to the rest of the chain. In fact, even if the force, applied slowly, would not stretch the chain enough to break it, that same degree of stretching applied quickly may do so, because as the wave travels it must first overextend the bonds before allowing them to retract and then oscillate back and forth down to their final length; the bonds may break when overextended even if they can support their final length at equilibrium. This effect is greatest when the chain is stretched quickly. Please excuse my failure to prune this monster of a paragraph.

Just a thought. Assuming that the force is applied at one end of the chain, isn’t there a delay before the force is felt by the rest of the chain? That would seem to imply that the end link, or perhaps a few of the end links, would fail, relieving the stress from the rest of the chain.

@SecondGlance I picture it like this: If the chain is suspended by its ends so that the two ends are at the same height but the chain is very slack, then obviously the end links will be under much more strain than the middle link. As the two ends are moved farther apart, the proportional difference in strain between the ends and the middle would decrease, but the difference would never go away entirely, even when the chain is under great tension.

it’s just an expression…not for all of the possibilities having to do with chains…in the job world, where there are hierarchies and different abilities, there is always a weaker link

@Simone_De_Beauvoir it is actually true for other then just the job world. When climbing for example the rope is only as strong as the weakest point. (the rope wares down from use). And I am SURE that a real chain will really break at the weakest link.

@Harp , We can get around the problem of the weight of the chain by simply imagining that the strengths of the links are adjusted so that they their breaking points are reached by the combination of load plus the weight of the part of the chain beneath them. I still think there is a delay factor. Imagine a 100 foot long chain. There must be a delay between the time that the bottom link feels the strain of the weight and the time that the top link does.

if it was a loop then I don’t think the chain would break..would it?

@LostInParadise; yes, there is a delay; see my second post for more detail. But adjusting the strength of the links to compensate for weight would rather contradict the terms of the question, so in the presence of gravity and in the absence of a support like a table under a horizontal chain, then in order to find the point at which the chain will break, the tendency of a chain to break at the ends due to weight (which is strongest for a longer chain) must be considered alongside the tendency to break at the point of the force’s origin (which is strongest if the force is applied quickly).

If the strengths are identical, the link experiencing the greatest stress will break first. This would be the link with the smallest cross sectional area (neglecting weight in the case of a vertical chain). Perhaps you also meant the chain links would be of identical cross-sectional area?

To consider the weight you would simply have to consider each link indivually given it’s individual cross sectional area (based on a measurement of the diameter of the material the link is made from) and how much additional load it supports due to the weight of the chain below it.

On another point: If the chain is a loop, it can still be broken, you just need to find a way to apply tension to it. Imagine an infinitely strong pressure vessel with a chain wrapped around it. Inflate the pressure vessel enough & surely the chain will break. This is a similar concept to hoop stress.

On yet another point: The time delay argument has validity in this discussion. However, in reality, chains are not intended for use in applications where they are subjected to instantaneous loads. In the rare event a chain was intentionally used in such an application (such as an emergency chain for a trailer hitch or a fall-protection safety device), it would be designed with a ridiculously high safety factor. What I’m getting at is that in a typical tension chain applications, the load is applied so gradually time can be neglected.

don’t have an answer, just a musing: In most of what I have read about the meaning of this saying it is assumed that the weakest link is faulty. One definition even refers to the weak link as the lazy person on the team. I really liked what marinelife had to say, that we are each other’s next link and it is up to us to bring up the strength of the chain as a whole.