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
    $\begingroup$ Deforming/displacing material requires molecular bonds to be altered, based on your material this could take a lot of energy to do mechanically. Look at the power needed to cold work steel. This is significantly greater than the kinetic energy change simply based on mass and velocity. $\endgroup$
    – jko
    Apr 1 at 19:17
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    $\begingroup$ The rate of change of kinetic energy is caused by external forces while the stress power is due to internal forces. Internal forces balance external forces but do not affect them, by definition. I prefer to think in terms of momentum/energy balance as expressed in, e.g., my.eng.utah.edu/~banerjee/Notes/ThermoElastic.pdf $\endgroup$ Apr 1 at 20:35
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    $\begingroup$ Think about a simple example like applying a transverse force to a stretched string. The work (or power) required to produce the same transverse velocity is higher if the tension (axial stress) is higher. So you can not describe the behavior of the system using only kinetic energy. You need the internal stress distribution as well. $\endgroup$
    – alephzero
    Apr 1 at 23:07

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