# Why doesn't the Stress Power in the theorem of the expended mechanical power contribute to the kinetic energy?

The following is the formula of the expended mechanical power:

$$P(t)$$ $$=$$ $$\frac{d}{dt}$$ $$\int \frac{1}{2} \rho\ v^2 dV$$ $$+$$ $$\int \sigma \colon d\ dV$$

where $$P(t)$$ is the mechanical power entering a continuum medium, $$\rho$$ is the density, $$v$$ is the velocity, $$dV$$ is a differential volume, $$\sigma$$ is the stress tensor, and $$d$$ is the strain rate tensor.

The first part of the right hand term is the rate change in kinetic energy. It is logical that the forces (body forces and surface forces) exerted on the continuum body will do some work, and this work per unit of time is a power and will contribute to the total power of the system.

But as for the second part, it is referred to as the Stress Power. By other words, it is the work per unit of time done by the stress in the deformation process of the medium.

What I don't understand is that the Stress Power is said to be the mechanical power that is not spent in changing the kinetic energy. How is that?

During the deformation process, a displacement of particles will take place. So, the particles will be changing their positions in the medium. And such a movement should as well create a velocity for them, hence a contribution to the kinetic energy. So stating that the stress power is not spent in changing the kinetic energy seems not logical for me.

By the way, I got the information from this video

## 1 Answer

I can think of an example to understand the difference between the 2 parts.

A hot bar comes from a roller table to a pair of rolls of a stand in a rolling mill. During the rolling process its thickness is reduced and its width is a little increased. But the increase of the width doesn't compensate for the decrease of thickness, so that its cross sectional area is decreased by 20%.

During rolling, the portion of the bar not still rolled have a speed of 1 m/s. Considering the same volume before and after rolling, the speed of the portion already rolled is 1.25 m/s. So that $$A_0V_0 = A_1V_1$$.

The speed increases due to plastic deformation, but the power to deform the bar is much greater than the necessary to increase its kinetic energy.

Suppose the total rolling time is 4s. A billet of 1000 kg at 1 m/s becomes a rolled stock of 1000 kg at 1.25 m/s. The required power to increase the kinetic energy is just:$$P = \frac{1000}{2}\frac{1.25^2 - 1^2}{4} = 70.3W$$ The typical required power for this type of rolling pass is greater than 100kW!

Practically all the power used for plastic deformation is transformed in heat, not in increasing kinetic energy.

• The heat generated in the material due to plastic deformations is regarded as thermal power, and also part of the internal energy. Power is work/time. So, what kind of work is there in the case of thermal power? – user134613 Apr 2 at 10:25
• It is mechanical work done by the rolls on the bar. It resuls in plastic deformation that is a dissipative process, releasing energy in the form of heat. I don't think it is appropriate to call it thermal power. Thermal power can however be the original source of plastic deformation if the electricity driving the motors of the rolling mill comes from a thermoeletric power plant. – Claudio Saspinski Apr 2 at 14:25