3. Traction and Stress inside a fluid body at rest

Think of astaticfluid, say bucketfilled with wateras shownin Figure5. Weknowthat the pressure(p) inside thewateris given byρghwhereh is thedepthfrom thetopsurface,g is theacceleration dueto gravity andρis thedensityofwater.

In this section, we start by discussing the pressure inside a static fluid. When you fill a bucket with water, the weight of the water creates pressure at any point under the surface. The formula for this pressure is given by the equation p = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth from the top of the fluid. This relationship tells us that the deeper you go in the fluid, the greater the pressure due to the weight of the water above.

Imagine you are diving into a swimming pool. When you dive down, you feel a stronger push against your ears the deeper you go. This feeling is because of the increased water pressure as you descend, which is essentially the weight of the water above you pressing down.

We want to know the state of stress at any point in the water body. So, we first think of a small infinitesimalplaneatthat point.Asfluids cannotsustainshearwhentheyarein staticequilibrium, there willbenoshearcomponentoftraction onanyplane. The traction wouldbethesameaspressure(p)and would act along the plane normal but pointing into the plane due to compressive nature of pressure. Thus, at anarbitrary pointxin thefluidandonanarbitrary planeatthatpoint (withplanenormal givenbyn),traction twill begivenby t(x;n)=−p(x)n.

Next, we delve into the concept of traction in a fluid. Since fluids cannot sustain shear stress when in static equilibrium, they only experience normal stress which is equal to the pressure at that point. Therefore, the traction, which is the force per unit area acting on a surface, can be expressed as the pressure (p) acting inwards, along the normal to the surface. The formula t(x;n) = -p(x)n represents this relationship where 'n' is a vector normal to the arbitrary plane within the fluid.

Think of a balloon filled with water. If you press on the wall of the balloon with your hand, the water inside will push back against that pressure equally in all directions, but only against the walls of the balloon. Since the water cannot slide sideways (because it is a liquid), it creates equal pressure on all sides – this is the traction described in the section, where the fluid pressure acts inward on the walls of its container.

Stressmatrix in(e ,e , e )coordinate systemcan befoundbywriting thetractions ascolumns. Tractionone planewill beequalto−pe.o We observe that all theshearcomponentsin thestressmatrix are zero. We can verify as shown below that σnwillgive usthetraction givenbyequation(21): t=σn=−pIn=−pn. Thus, thestresstensorforafluid bodyin statics is −pI.

In this part, we look at how the stress matrix for a fluid at rest can be constructed. The normal tractions on a plane are represented as columns in a stress matrix. In this case, shear components are zero, indicating that the only forces at play are due to pressure acting normally to the plane. We confirm that the traction is created by the stress tensor, which ultimately turns out to be -pI, where I is the identity matrix. This indicates that all the diagonal components of stress are equal to negative pressure, while there are no shear stresses since they remain at zero.

You can think of a still cup of water on a table. The water exerts pressure on the bottom of the cup due to gravity. However, the water doesn't slide around to create shear—it's just pressing downwards uniformly, which can be described mathematically by the stress matrix, where pressure is represented in all direct forces without any shear.

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