what is the difference between stress and pressure in a deformed rigid body.? How come stress is a point function?

regards, Yogesh

- posted
18 years ago

what is the difference between stress and pressure in a deformed rigid body.? How come stress is a point function?

regards, Yogesh

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- posted
18 years ago

Dear yogesh:

Pressure is an applied distributed external force. A "boundary condition", as it were. Stress is how a "dimensioned" physical body counteracts the applied force/pressure.

Stress is a mathematically modelled "force balance" equation on an infinitely divisible "chunk" of matter. So it is largely imaginary, but somehow very descriptive and useful.

Do I get your "A"?

David A. Smith

- posted
18 years ago

- posted
18 years ago

Ahhh. But... what material exist ( that we know of ) this is truly rigid.

If you take a rigid body ( sitting on the ground or otherwise supported in some way ) and you apply some type of pressure to it's surface... then, isn't there a stress buildup on the surface whether you can measure a deformation or not? If there isn't any stress... the pressure forces could not be transfered through the rigid object to the supporting ground.

Dan :-)

I tend to think of stress as being "internal" to an object. I think of pressure as being internal or external. When all is said and done, I think it's not "really" much more than a symantic difference between the two.

Dan :-)

- posted
18 years ago

You mean you don't have a supply of Rigidium handy? I keep min next to the Unobtanium... Too long with matrix algebra and FEA...We get that way, some daze... At least I thought it was funny... :D

Roger

Dan Tex1 wrote:

- posted
18 years ago

Simple answer. Pressure is external loading on a body. Stress is a body's internal reaction to external loads. Stress can vary with position in a general 3D body.

- posted
18 years ago

pressure is physically a force/area, given in units of force per unit area

- f/A = ma/A = m (d/t^2)/d^2 = force/area

stress is physically the energy/volume, given in units of force per unit area (the why of that transformation has to do with the way the basic force determination equations are derived from the basic energy equations and what the final units end up as being, especially for the most basic of the stress conditions - see an advanced engineering text such as Faupel for the derivation)

- E/V = mv^2/d^3 = m (d^2/t^2)/d^3 = m (d/t^2)/d^2 = force/area

The technical Pressure is from a force distributed across an area, in units of force per unit area - e..g., the pressure is 10 pounds per square inch A lay use of the term "pressure" is sometimes used by non-technical persons, e.g., How much pressure can the board take before it breaks? It has no meaning in technical fields

The technical Stress is energy density. i.e., how the part is distributing the input energy seen as force or the energy wave, Thus no load on a beam, no energy in and no stress - add a weight, energy in the beam increases as stress increases -(see the basic energy-to-force equations)

- add a weight and don't let it deflect back when you remove the weight, and you know there is energy stored when you let it go (In many advanced mechanical engineering areas, the term "energy density" replaces the term "stress" in prediciting failure points)

It works easiest as a point in calculus

In force levels of engineering, where the input of the energy is slower than the natural frequency of the part, (e.g, weight on beams) the energy distributes and stress is NOT a point function but rather a continuous distribution. It is distributed according to its geometry and the location of the force. Using the point will tell you what stress is at that given point in that geometry.

In energy density areas of engineering such as shock (explosives) or crack propagation, the input of energy is faster than the natural frequency of the part. That energy density (stress) is localized and a point is the only way to predict material response.

- posted
18 years ago

Are you sure?

Perfectly rigid bar in uniaxial tension with 5 sq in cross section.

100 kip 100 kip(1) sum forces X dir == 0 = +100 kip + (-100kip) :: True

Pass a section through the bar and both free bodies.

100 kip 100 kip 100 kip(2) for both bars the sum of forces X-dir == 0 = +100 kip + (-100kip) :: True

For static equilibrium the force across the cutting plane for the left member will be

(3) Stress == F/A = 100 kip/5 sq in = 20 ksi :: True

If (1), (2) and (3) are true then in what way is stress not present. Now you 'might' talk about stress concentrations.

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