How does a constant load affect the elongation of a steel cable?

Question

I 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

Answer 1

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.

Answer 2

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!

Answer 3

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.

Answer 4

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.

Answer 5

I got a -1 for my answer. I don't know what it means but it probably isn't good. So I will try again. There have been what I thought were some very good answers but I’m not sure if the question has been totally resolved. The analogy to a spring seems to be a good one. For stresses below the proportional limit, a tensile force applied to a steel rod behaves just like a spring.

In the figure below, the linear elastic portion of the stress-strain curve is shown. The data for these curves is collected in the laboratory using a pull tester linked to a computer. The computer uses the data to plot the curves. Loading and unloading is applied to the specimen. When an applied force produces stress in a specimen below the proportional limit, no yielding will occur. If the force is removed from the laboratory specimen, the unloading curve will follow the linear-elastic portion all the way back to the origin. This behavior is consistent with the definition of elasticity. If the elastic limit is exceeded by a small amount prior to removal of the load, a permanent strain or "offset" will result. The unloading curve will follow the same slope as the elastic curve. However, the strain will not return to zero even though no loading is applied. In general, structures are designed to ensure that they remain elastic.