Longitudinal strain in a general body

Today, we will discuss longitudinal strain, which refers to the change in length per unit original length of an object. Can anyone tell me how we might measure this strain in a simple bar?

Do we need to apply force to the bar?

Exactly! When we apply a force to a bar, it stretches. The longitudinal strain can be calculated as the difference between the original length and the new length divided by the original length.

If I understand correctly, that's B5 = (l - L) / L, right?

That's right! Keep in mind that this is a simplified expression, and strain can vary throughout the body depending on how it is loaded.

So if the bar is long and we apply a force, the strain could be different along its length?

Exactly! That's an important point. The strain isn't uniform across all points in a more complex body, which takes us to our next example.

Now, let's discuss a vertical bar hanging under gravitational load. How do you think this will differ from the horizontal bar example?

I think the strain would be greater near the clamped end because it has more force acting on it.

Exactly! As you go down the bar, the mid-section experiences a different amount of strain compared to the clamped end. This variability leads us to define local strain at different locations.

So, would we have different strain values even at infinitesimally close points on the bar?

Yes, and that's a critical understanding. For simplicity, we can consider each point with its unique strain, denoting it mathematically as B5(x, n) at point x along direction n.

Could you remind us why we use this mathematical representation?

Of course! The strain representation helps us quantify how strain differs across various points and orientations, allowing for deeper analysis in complex bodies.

Let's break down how we can calculate strain at a specific point in a deformed body. What do we need?

We need to identify the original and deformed configurations, right?

Exactly! By comparing these two, we can derive the equations to define the longitudinal strain at any point. Recall we represent this as B5(x, n).

And as the body deforms, we can think of these as vectors, am I right?

Yes! The displacement vector helps establish how points move from their original to their deformed positions, which is crucial in calculating the strains.

So, if we understand the displacement of points, we can accurately calculate longitudinal strain at any point?

Exactly! This establishes a foundation for understanding deformation in various materials.