Displacement Calculations – Energy Methods (2)

Welcome to the second part of the article on displacement calculations in beams. In this section, I will discuss the energy approach, which makes very friendly use of information about internal forces.

In this part of the article, I will cover the following topics:

• Changes in mechanical energy
• The energy approach – Maxwell – Mohr equation

Changes in mechanical energy – sometimes high, sometimes fast…

If we try to recall the basics of physics related to changes in energy, we might come across terms like work, potential energy, or kinetic energy.

Usually, these terms are associated with something like throwing a ball (e.g., a rubber ball) from a certain height. In this explanation, it’s often said: “Since the ball is at some height, it has potential energy (as it is in a gravitational field). If we let the ball go, its potential energy starts to decrease while its kinetic energy, related to the ball’s speed, increases (according to the law of conservation of mechanical energy). Thus, just before hitting the ground (our reference point), all the potential energy converts to kinetic energy, and the ball reaches its maximum speed.”

At this point, high school level discussions usually end, and few people think about what happens when the ball touches the ground. We all understand that if we assume that the collision didn’t disperse energy (which isn’t very accurate, since we usually hear the impact), the ball will shortly bounce up with the same initial speed it had when it hit the ground, and with some luck, it might return (almost) to the same height.

But how does its speed suddenly change direction? What magic causes the body to change its velocity direction in a split second, almost violating Newton’s second law?

Well… no magic at all. Nothing unusual happens here—it just happens quickly. So, what’s going on? – Let me explain. There’s a conversion of kinetic energy into potential energy and then back into kinetic energy. However, the potential energy in this transformation is not related to gravity but to the material’s elasticity.

When the ball hits the ground, due to the forces of compression and inertia (yes—the second law…), the material compresses, changing its shape—this way, it starts storing energy—yes, all the kinetic energy (at a certain point in time, the compressed ball is basically motionless—though it might vibrate a little…). This energy isn’t needed for ball happiness, so the material quickly releases it—applying force to the ground and accelerating upwards—increasing kinetic energy.

To sum up the above story, we can say that

various objects, when deforming, store energy, which we call elastic potential energy.

The energy approach – look how elastic I am…

This also applies to beams, frames, or trusses, where it should be noted that for static systems, this energy is the result of the interaction of loads on the system. Let me remind you here—if a force acts on an object, causing it to move, that force does work, and just like in the story of the ball—for the universe to remain in balance (hello, Thanos ????)

the work of external forces on the displacements of a static system is equal to the elastic energy U generated in the structure.

For the previously analyzed example of a tensile bar, we can quickly derive a formula for the elastic energy stored under the influence of a normal force N (we use Clapeyron’s theorem):

dU= {1\over2} N⋅du={1\over2} N⋅{N\over{EA}} dx={1\over2}  {N^2\over{EA}} dx→U={1\over2} ∫_L{N^2\over{EA}} dx

The normal forces appearing in the above formula can be constant or functions of the length of the bar, and they also depend on all the loads acting on the analyzed bar. If we already know the form of the system’s elastic energy, we can calculate the displacement of the bar at any point by using Castigliano’s theorem, which states that

the displacement at a given point is equal to the partial derivative of the elastic energy with respect to the force (or moment, in the case of rotation angles) applied at that point:

u={∂U\over∂F}

This theorem can be derived in several ways (I encourage the daredevils to read further ????), but now let’s focus on the trick needed to obtain the final formula for displacement determined by the energy method…

To calculate displacement according to the above formula, we need to introduce a virtual unit force F at the point of interest—after all, we need something to differentiate with respect to…

This virtual force changes the internal forces, so (also virtually) it affects the internal energy:

U={1\over2} ∫_L{(N+ \overline1⋅N_F)^2\over{EA}} dx

Now we simply need to take the partial derivative with respect to F according to Castigliano’s theorem, and then recall that in reality, the force F does not exist—it equals 0…

u={∂U\over∂F}={1\over2} ∫_L 2⋅{N+ \overline1⋅N_F\over{EA}}⋅N_Fdx →u=  ∫_L {N+ \overline0⋅N_F\over{EA}} N_Fdx

u=∫_L{N⋅N_F\over{EA}}⋅dx

The above formula is known as the Maxwell-Mohr equation, and it leads to obtaining the displacement of any point in the system, assuming we know the distribution of the normal forces caused by the actual loads and, additionally, the distribution of the normal forces caused by the unit force applied at the point where we are seeking the displacement. By transforming these formulas, we can also analyze statically indeterminate systems (e.g., using the force method).

What can be said in conclusion about energy methods?

First of all, that there are many ways to derive energy equations, which means there are many different methods that ultimately lead to the same displacement.

These methods affect the steps of the process, which I will describe in further articles on displacement analysis. In the next part of the article, we will focus on how the analytical and energy approaches change for different strength cases—check out the second part of the article.