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Who's up for a mechanical engineering brainteaser?

Here's the brainteaser:

Does there exist a purely mechanical system which, when load P is applied to it, experiences a stress with a magnitude of S, but when a load 2P is applied to it, experiences a stress with a magnitude less than S? If so, describe that system.

Note: The load is doubled, and its direction is unchanged. The mechanical system must of course have a nonlinear response. If its response were linear, the stress would be exactly 2S. Any type of mechanism may be employed, but the system must be in static equilibrium at load P, and also at load 2P. A dynamical system is considered cheating.

To restate the problem:

load P --> stress S in static equilibrium

load 2P --> stress less than S, but greater than 0, in static equilibrium

Describe the system and how it achieves such non-linearity.

Update:

Gintable - that is indeed a valid solution. I could add the stipulation that we only care about the peak stress in the system, but then you'd just make two springs with wildly differently elastic moduli, such that they have equal force but very different stresses. So, your system would still work.

I'll wait and see if anyone else comes up with a different solution.

Update 2:

On second thought, I will make the stipulation that the system must have zero stress when the load is zero.

Update 3:

RidingTheLight - Hydraulics are OK, but the system must be under static equilibrium. A fluid in static equilibrium has zero shear by definition (only the diagonal terms in the stress tensor are nonzero). Therefore, shear-dependent properties of the fluid cannot come into play.

2 Answers

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  • 1 decade ago
    Favorite Answer

    How about we have an object that is suspended in place by opposing springs (of equal k-values, but not of equal breaking strength), on either side of it...both initially pretensioned to a tension of 2*P.

    Note: we will not be breaking or yielding either spring

    Suppose we are only concerned about the stress in the spring on the right. Because, suppose the right spring is weaker than the left spring. Suppose the left spring is infinitely strong, so stress in it is unconcerned.

    Apply force P to the suspended object in a rightward direction. This will relieve the load of the right spring, so that the right spring carries tension of 1.5 P. This will of course increase the tenison in the left spring, so it carries tension of 2.5 P.

    Now increase force to 2*P.

    What will this do to the tensions?

    Well, the tension on the right spring will now be equal to value P.

    The tension on the left spring will now be equal to 3*P (but we didn't care).

    There you have it:

    Case of load P, the spring on the right carries stress S.

    Case of load 2*P, the spring on the right carries stress 2/3*S.

  • ?
    Lv 5
    1 decade ago

    The first answer would work but you asked for another so I'll just give another. Look into a shear-thinning non-Newtonian fluid, i.e. paint, ice cream. Shear-thinning non-Newtonian fluids are fluids that will become easier to manipulate with a higher shear rate. And since shear and stress are related you would be able to design a system that could implement the principles behind non-Newtonian fluids. The design would be complicated but it would work but instituting hydraulics.

    However, after rereading your question, I realize it says 'purely mechanical system' so that idea would have to be thrown out (unless you could say hydraulics are mechanical systems) haha. Either way, it would be neat a neat way to solve that brainteaser of a reduce stress with a higher load.

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