I Distributed Forces and force density....

AI Thread Summary
The discussion centers on the distinction between force density and pressure density, specifically regarding the variable σ=f(x). It clarifies that σ represents stress, which is a force per unit area and can include both normal and shear components, while pressure is strictly normal to a surface. The conversation also highlights that stress is a rank-2 tensor, contrasting it with force density, which is defined as force per unit volume. Additionally, the participants debate the implications of integrating stress over an area versus integrating force density over a volume, concluding that stress is not a force density but rather a measure of internal forces within materials. The nuances of these definitions are essential for understanding mechanics in engineering and physics contexts.
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force density or pressure
Hello,
Forces can be concentrated (when acting at a single point) or distributed (when acting over a surface or line).
In the case of distributed forces, we can find the resultant concentrated force by calculating a surface or line integral of the force density ##f(x)## w.r.t. an area or length differential.

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This is equivalent to adding together small products of force times infinitesimal areas. Statics books show these types of calculations. For example, $$\int \sigma dA$$
Is the force density ##\sigma =f(x)## to be considered a pressure density or a force density? Pressure is fundamentally normal force per unit area. Is ##\sigma## a pressure density only when the force direction is exactly perpendicular to the surface? The force density may be at an angle at different points on the surface with both a parallel and normal components relative to the surface.

Thanks!
 
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fog37 said:
Summary: force density or pressure

Is the force density σ=f(x) to be considered a pressure density or a force density?
##\sigma## is stress, which is related to pressure but not the same thing. Stress is force density.

Pressure is the isotropic part of stress. Pressure is always normal to a surface, but in general stress can have shear components also.
 
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fog37 said:
... Statics books show these types of calculations. For example, $$\int \sigma dA$$
Is the force density ##\sigma =f(x)## to be considered a pressure density or a force density? Pressure is fundamentally normal force per unit area. Is ##\sigma## a pressure density only when the force direction is exactly perpendicular to the surface? The force density may be at an angle at different points on the surface with both a parallel and normal components relative to the surface.

Thanks!
Sigma refers only to a normal internal force in the cross-section, which is an indirect effect of the bending load on those beams.
Please, see:
https://en.m.wikipedia.org/wiki/Bending
 
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Stress is a symmetric 2nd-rank tensor. Let the components be ##\sigma_{jk}=\sigma_{jk}##. It's the force per unit area acting on a surface ##A## of a continuous medium. The total force thus is
$$F_k=\int_A \mathrm{d}^2 A_j \sigma_{jk}.$$
Pressure is a special case. E.g., in an ideal fluid the stress tensor is given (in the local rest frame of the fluid) by
$$\sigma_{jk}=-P \delta_{jk},$$
where ##P## is the pressure.

It's not a force density. This would be a quantity "force per unit volume", e.g., ##\vec{f}=\rho \vec{g}## for the gravitational force close to Earth acting on a fluid.
 
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vanhees71 said:
Stress is a symmetric 2nd-rank tensor. Let the components be ##\sigma_{jk}=\sigma_{jk}##. It's the force per unit area acting on a surface ##A## of a continuous medium. The total force thus is
$$F_k=\int_A \mathrm{d}^2 A_j \sigma_{jk}.$$
Pressure is a special case. E.g., in an ideal fluid the stress tensor is given (in the local rest frame of the fluid) by
$$\sigma_{jk}=-P \delta_{jk},$$
where ##P## is the pressure.

It's not a force density. This would be a quantity "force per unit volume", e.g., ##\vec{f}=\rho \vec{g}## for the gravitational force close to Earth acting on a fluid.
Thank you. So what I am calling ##\sigma## is the stress tensor (rank 2) whose integral with respect to area is the total force. So far so good. And ##\sigma## is NOT a force density. Why not? Because it is a tensor and not a vector field? The components of the stress tensor are forces either perpendicular or parallel to the surface...

In your last comment about the gravitational force you mention that ##\vec{f}## is instead a force density, force per unit volume, which I get...So the force density is a vector field, a different thing than a rank-2 tensor...Is my understanding correct? To find the overall gravitational force we calculate the volume integral of the force density ##\int f dV##...

To find the overall force on a surface with calculate the surface integral of the stress tensor...

Could we have a volume integral of stress tensor? If so, what does it represent?
 
vanhees71 said:
It's not a force density. This would be a quantity "force per unit volume"
I disagree. Current density is current per unit area. Not all densities are per unit volume. Stress is force density: force per unit area.
 
fog37 said:
Could we have a volume integral of stress tensor? If so, what does it represent?
Yes, energy.
 
Dale said:
I disagree. Current density is current per unit area. Not all densities are per unit volume. Stress is force density: force per unit area.
It's again a matter of definitions. The standard nomenclature in the field-theory literature is that a density of a quantity is per volume and current density of a quantity is per unit area and unit time. In natural units with ##c=1## the dimensions are the same ;-)).
 
This is an engineering question, not a field-theory question.
 
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