4
$\begingroup$

enter image description hereI am sorry if this question has been asked before or if the answer is obvious, but I cannot seem to find the answer. A lot of the structural engineering problems we do involve calculating the elongation of certain types of beams/cables etc while under tensile stress. In particular there was a relatively straightforward question that asked for the total elongation of a structure comprised of two connected steel cables which have different cross sectional areas. Each of the two cables had consistent cross sectional area, and the structure was placed under an axial load of 800N.

Solving for the total elongation was simple with delta = (FL)/(EA) for each cable and adding the two deltas. My confusion however is with why this is considered the total elongation.

After the structure has undergone this elongation, it will become a 'new' structure, with a different length and different cross=sectional area for each cable. My question is, if the load continues to be applied after this 'initial' elongation (delta1+ delta1), then wouldn't the new structure still be under direct stress, and thus based on the new E of each of the now deformed cables, would each deformed cable undergo deformation again? And would this process not repeat until the object yields?

I understand that in tensile testing a specimen's stress-strain graph will be obtained by deforming the specimen until it has yielded. This is why I am confused because under load, would these cables not just deform until they yielded? And if so, wouldn't total elongation then be based on the final length just before the yield point?

Edit: The two steel cables in the questions are connected together, with the cross-sectional areas of the two cables being distinct, i.e. a smaller circle stuck onto a larger one

|improve this question
$\endgroup$
  • $\begingroup$ Hi @masiewpao, welcome to Engineering Stack Exchange. I am not quite clear about your two-cable structure - are they connected one after the other, or one next to each other? $\endgroup$ – AndyT Jul 15 '16 at 8:22
  • $\begingroup$ Hi Andy, sorry, they are connected together! $\endgroup$ – masiewpao Jul 15 '16 at 8:29
3
$\begingroup$

This answer will assume the applied load is not enough to damage your structure (the stresses generated are lower than the elastic yield stress). I am also going to ignore the distinction between engineering and true stress. Both are common basic assumptions in any structural analysis class.

The problem with your reasoning is that you're pretending that a bar under load is the same as when it is unloaded. It isn't.

Indeed, the bar deforms so that the system can reach a state of equilibrium. The instant you apply the load to the bar, the system is unbalanced. There is a force applied by the rope on the bar, but the bar is unable to react according to Newton's Third Law (for every action, there is an equal and opposite reaction). The system is therefore momentarily dynamic, and the bar is stretched.

As the bar is stretched, this generates an internal stress state, which is felt by the rope as a reaction to the force it is applying on the bar. At some point in time, this internal stress state (and therefore the reaction to the rope's applied force) will be equal to the force applied by the rope and the system will reach equilibrium.

One of the fundamental laws of structural analysis is Hooke's Law, which states that

$$\sigma = E\epsilon$$

This can be reworked into

$$\begin{align} F_I &= EA\epsilon \\ F_I &= \frac{EA}{L}\delta \end{align}$$

where $F_I$ is the internal force in the beam (equal to the product of the stress and cross-sectional area) and $\delta$ is the beam's total displacement, in relation to it's original length $L$.

If you were to then get the new length $\overline{L} = L+\delta$ and apply this equation again, you'd get a value of $F_1$ which is greater than the applied external force, which would mean the system is unbalanced, at which point the bar would shorten so as to reach a state of equilibrium once again.

|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Ahh thank you very much! So basically the system is momentarily out of equilibrium, and becomes a dynamic one, but during the dynamic phase, the internal stress resultants increase until it equals the load and it is now in equilibrium again and under this load will no longer undergo elongation! $\endgroup$ – masiewpao Jul 15 '16 at 16:35
1
$\begingroup$

The entire structure acts like a spring; it stretches/expands until the point when it is strained enough to carry the load. It will not stretch any more if you keep applying the force because it is at equilibrium. Like the scale at home we weigh ourselves with. By stepping on the scale you apply force to the spring, it will contract or expand till its loaded sufficiently to support your weight. It won't keep expanding if you don't get off it.

You can test the equilibrium by jumping on the scale or adding a short impact load to your example by adding a 20 kg load and removing it fast. the system start vibrating reacting to this perturbation and depending on the mechanical damping this vibration will die out over a few periods. The beauty of using elastic materials for structural members is just that: the resist large loads by expanding a bit and go back to their original position after removal of the force/load. No member will start to carry any load before it has been deformed in one way or the other.

A similar configuration of what you show here is used as suspension spring for cars: a soft but long spring in series with a short but hard spring. the soft spring takes most of road's bumps with soft reaction and comfort for passengers but the hard spring comes into action when the is big pot holes on the road!

|improve this answer
$\endgroup$
0
$\begingroup$

For your specific case:

Total elongation = elongation of cable 1 plus elongation of cable 2. The word "total" is used to make sure it's obvious that it doesn't just want elongation of cable 1.

Short term effects:

The L in the formula is "undeformed length without F". Therefore you don't need to recalculate L and apply the formula again. The cross-sectional area A is similarly the "area without F". E does not change. Therefore delta = (FL) / (EA) only needs to be done once.

Long term effects:

Over the long term, steel can relax. (This is similar in concept to creep in concrete, if you have heard of that.) The micro-structure rearranges, making the steel element longer. This additional deformation is plastic (non-recoverable when the load is removed) and is much smaller than the initial elastic deformation.

|improve this answer
$\endgroup$
  • $\begingroup$ Hi there, sorry I think I've done an awful job at explaining the situation, because the extensions are different for the different sections; I'll include an image of the question! With regards to the short term effects, I'm not sure I understand. The moment you apply F, the structure should begin to deform, so could you not redo the calculation for every slightly different deformed structure, and thus see continuous deformation? Cheers! $\endgroup$ – masiewpao Jul 15 '16 at 9:47
  • $\begingroup$ @masiewpao - No, you can't redo the calculation for every slightly deformed structure, because L in the equation is the undeformed length. If you put on 0.1F, then the structure will deform delta1 = 0.1FL/(EA). If you then add another 0.1F, making the total 0.2F, the total elongation is delta1+delta2 = 0.2FL/(EA). It is not correct to say delta2 = 0.1F(L+delta1)/(EA), because the delta=FL/(EA) formula is based on undeformed length L. $\endgroup$ – AndyT Jul 15 '16 at 11:24
  • $\begingroup$ That way lies madness. Obviously, delta = (FL) / (AE) has been verified for single cables being pulled, and there was no need to iterate a solution as long as the stress F/A remains in the elastic range of the material. If you go beyond the elastic limit, then you no longer have a linear problem to analyze, and an iterative approach may be required. $\endgroup$ – user16622 Jul 15 '16 at 11:27
  • $\begingroup$ "That way lies madness." That is not true. There is nothing "mad" about formulating continuum mechanics using different stress and strain measures from the "engineering" small stress and strain used in most beginning courses. For example it is a big advantage (not "madness") to use stress and strain measures that are zero for large rigid body rotations - consider what happens to the cable if you pull it so that it rotates 90 degrees from its original direction. Of course if the displacements of the structure are "small", the differences between the various formulations are also small. $\endgroup$ – alephzero Jul 16 '16 at 15:58
  • $\begingroup$ @AndyT You don't need to "redo the calculations for every slightly deformed structure". You just formulate the problem measuring strain in a different way. See en.wikipedia.org/wiki/Deformation_(mechanics) for a summary of several different strain measures that are all used in real-world engineering. The fact that beginning courses in mechanics of materials only use one measure of stress and strain may give students the wrong idea that this is the only way to measure them. $\endgroup$ – alephzero Jul 16 '16 at 16:06
0
$\begingroup$

This is a good question, and shows that the OP is thinking about what he/she is being taught!

It is perfectly possible to model the situation the way the OP proposes. The definition of "engineering strain" that is taught in most first courses on "strength of materials" is not the only way (and often is not the most useful way) to measure strain, but if students are only told about one way to measure strain, they may end up in a situation where "if the only tool you have is a hammer, you have to solve every problem as if it was a nail".

The OP's idea leads to the formulating mechanics using logarithmic strain and true stress. Some industry-standard computer software (for example Abaqus) uses this formulation for every problem, because

  • it is no more complicated than using engineering stresses and strains
  • it is applicable to a wider class of problems where the displacements and strains are not small, and
  • in the limit where the displacements and strains are small, the results are identical for all practical purposes.

The third bullet point explains one reason why this formulation is not taught in a first course - the other reason being that it requires more mathematics (including calculus) to understand it.

https://en.wikipedia.org/wiki/Deformation_(mechanics) gives a summary of several different ways to measure strains (some of which are even more general than logarithmic strain). Googling for "logarithmic strain" and "true stress" will find plenty of other relevant material.

|improve this answer
$\endgroup$
  • $\begingroup$ Good answer, but how can it require more mathematics and yet be no more complicated than using engineering stress? The general gist of structural analysis is relatively simple to understand. Its entirely complexity rests in the mathematics of it (in my opinion). $\endgroup$ – Wasabi♦ Jul 17 '16 at 3:13
  • $\begingroup$ This answer is missing an actual conclusion which answers OP's question(s). OP suggests that the incremental strain will continue to be added until the cable yields - this is incorrect, but is not covered at all in this answer. $\endgroup$ – AndyT Jul 18 '16 at 9:54

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.