How do we define the durability as per density?

Question

We define the durability with the equation as seen below.

$$x\propto \frac{\text{Cross-Sectional Area}}{\text{Cross-Sectional Area}\times \text{Length} }$$

Hence

$$\therefore x\propto \text{Length}^{-1}$$

This equation was only for matters which has same density. Let us imagine that there is a tree with $3\text h$ and an iron pole with $3\text h$. Now we see that our equation doesn't work in this condition. The iron pole will be more durable than the tree even if they have same lenght. Then, how do we define the durability as per density?

According to our equation,

$$\text {Tree} = \frac {1}{3h}$$

$$\text {Iron Pole} = \frac {1}{3h}$$

It seems like their durabilities are same. However, Iron pole has greater density than tree. By the way, Iron pole is more durable. How do you explain this? or Is there any equation that we can use when there are two matter which doesn't have same density?

Answer 1

This is clearly building on this question, and specifically its accepted answer.

Not to toot my own horn, but I recommend you also read my answer to that question. More specifically, the first paragraph.

What's important to notice here is that the equation given above isn't describing an equation, but a relationship of proportionality. When you say that "durability" (an unclear term) $x \propto 1/L$, you are stating that durability is inversely proportional to length. As I stated in my answer to your other question, that is correct for most common failure states.

However, this statement makes no attempt at being a complete description of the variables that determine an element's "durability". It merely states that one of the variables that goes into "durability" is the inverse of the element's length. There may be (and indeed are) other variables that also influence "durability".

Abandoning that useless term and taking the example of a column under uniaxial load, we can calculate that the buckling load is equal to

$$P_E = \frac{\pi^2 EI}{(KL)^2}$$

I won't bother describing what the variables are because that's immaterial to this question. What's important to note here is that this is an equation, as described by the use of the equality symbol ($=$) instead of the proportionality symbol ($\propto$). An equation attempts to be a complete description of the dependent variable (in this case $P_E$).

So long as you accept the assumptions that go into deriving this equation, then $P_E$ is exactly and exclusively equal to $\dfrac{\pi^2 EI}{(KL)^2}$. That is a true, undeniable fact.

However, here are a few other true, undeniable facts:

• $P_E \propto E$
• $P_E \propto I$
• $P_E \propto 1/K^2$
• $P_E \propto 1/L^2$

That is, $P_E$ is directly proportional to $E$. If you double $E$, you double $P_E$. There are other variables that are relevant to actually determine the value of $P_E$, sure, but there's no denying that a greater $E$ implies a greater $P_E$. The same applies to all of the other statements of proportionality above.

That's what's wrong with your case. You are looking at a statement of proportionality describing that "durability" is proportional to the inverse of the element's length and assuming this states that there are no other relevant variables. That is not true. All that statement is saying is that an increase in length reduces "durability". It makes no assertions as to the existence (or lack thereof) of other variables which may also influence the "durability".