3.1. Traction

We want to know the state of stress at any point in the water body. So, we first think of a small infinitesimal plane at that point. As fluids cannot sustain shear when they are in static equilibrium, there will be no shear component of traction on any plane. The traction would be the same as pressure (p) and would act along the plane normal but pointing into the plane due to the compressive nature of pressure. Thus, at an arbitrary point x in the fluid and on an arbitrary plane at that point (with plane normal given by n), traction t will be given by:

$t(x;n) = -p(x)n$

In this portion, we explore how traction within a fluid behaves. Since fluids are incapable of withstanding shear stress when at rest, they only exert normal stress, which is equivalent to pressure. The traction force acts perpendicular to any surface within the fluid and is directed inward, highlighting that as pressure increases, it effectively compresses the fluid. The formula shows this relationship mathematically — the traction force (t) at a point x acts in the direction of the normal vector (n) and is proportional to the negative of the pressure. The negative sign indicates the direction of the force is inward, consistent with the behavior of pressure acting on a surface.

Imagine pushing your hand down into a bucket of water. As you do so, the water pushes back against your hand; this push is the pressure in the water, which we can think of as the water's traction. Just as you feel that force pushing back at you, the traction force acts on any surface within the fluid, demonstrating that fluids will always exert pressure directed toward the center of any object submerged in them.

Stress matrix in (e1, e2, e3) coordinate system can be found by writing the tractions as columns. Traction in one plane will be equal to −pe.

This chunk elaborates on formulating the stress matrix in a fluid. The stress matrix is constructed using the tractions acting on various planes of the fluid, resulting in the stress matrix being composed of pressure values on the diagonal, marking the normal stresses due to pressure. The shear components all equal zero, which is expected in static fluids. Therefore, the stress tensor, represented by a scalar multiple of the identity matrix (-pI), emphasizes this uniform response of the fluid due to pressure, showing that no shear stress is present.

Consider a soft balloon filled with water. When you squeeze the balloon, the water inside transmits the pressure uniformly throughout the material of the balloon. Each point of the balloon experiences a traction that pushes outward against the squeezing force, but within the water, that pressure is equal in all directions, effectively resembling how a stress tensor is represented with uniform pressure acting evenly at all points.

We observe that all the shear components in the stress matrix are zero. We can verify as shown below that σn will give us the traction given by equation (21):

$t = σn = -pI n = -p n$

Thus, the stress tensor for a fluid body in statics is −pI.

In this segment, we prove that the normal stress (σ) can directly represent the traction (t) acting at a point within the fluid. The relationship demonstrates that the total stress acting on the fluid at any point is a result of the negative pressure multiplied by the normal vector, reinforcing that pressure emanates uniformly from every direction. This confirms that the stress tensor for quiescent fluids is isotropic (the same in all directions) and depends solely on pressure.

Think of a filled soda can. When you apply pressure, the pressure inside pushes outwards evenly against the walls of the can. The relationship between pressure and traction inside the fluid reflects how every bit of liquid responds uniformly to pressure, similar to how the fluid's stress matrix behaves. If you consider the can's surface as a local point, the of fluid at that point can be viewed as being directly tied to the internal pressure, following the stress tensor dynamics discussed.

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