Thanks a lot Aydin, Jorge and Simon,
This is the first of the discussion in this forum from which I really learned something.
Very interesting and worth while.
Thanks, Dirk Nieuwland
On Sep 27, 2011, at 5:50 PM, Simon Kattenhorn wrote:
>
> Dear Aydin
>
> To add to the response of Jorge - tractions are vectors that represent the limit as area tends to zero of the ratio of force over area. So it is essentially the vector representing force divided by area, acting on a point element. This traction vector represents the cumulative result of all forces acting on the body containing this point (whether these forces are external or internal). The orientation of this traction is also dependent on the orientation of any hypothetical (or real) surface containing this point. So traction is expressed as t(n), where n is the normal vector representing the orientation of the plane.
>
> If you were to consider all possible plane orientations passing through the point in 3D (i.e., an infinite number of planes), there would be a corresponding infinite number of traction vectors, even for the identical loading case of forces acting on the body containing the point. Furthermore, each traction acting on a particular plane would have a corresponding equal and opposite traction acting on the other side of the plane (Newton's 3rd law). These two tractions can be considered components of the stress tensor. In fact, the state of stress around the point can be represented by a family of tractions (an infinite number) that form the shape of an ellipsoid around that point (if the traction vectors are scaled to their magnitudes). This is the stress ellipsoid. Fortunately, being a simple mathematical construct, an ellipsoid can be fully defined by its 3 principal axes. These axes are defined by 6 traction vectors (or 3 pairs of equal but opposite tractions). Furthermore, these 6 tractions will always be acting on hypothetical planes passing through the point in such a way that the tractions are perpendicular to those planes. For that reason, they are planes of zero shear stress ("principal planes") and the 3 pairs of tractions acting upon them are normal stresses ("normal" here meaning "perpendicular to"). They are, in fact, the three principal stresses that we typically use to represent stress. This is why each stress component must be represented by 2 arrows, equal and opposite. They are the two traction vectors acting orthogonal to the principal planes.
>
> As Jorge mentioned, stress can also be represented as a tensor of 9 components (read an appropriate text to learn more about this). In essence, the stress tensor allows us to represent the state of stress acting on an infinitesimally small cube (a "volume element") in some pre-definined coordinate system (typically Cartesian). Each face of the cube will have one normal traction and two shear tractions acting upon it (in the 3 coordinate axis directions). This means the cube will have 18 traction components; however, again invoking Newton's 3rd law, opposing faces of the cube will have equal but opposite traction vectors (for the case of equilibrium), giving us 9 components of stress. There are thus 3 normal components and 6 shear components of stress in the stress tensor. This method of representing stress allows us to define the state of stress in terms of our pre-definined coordinate system, without concern about which directions the principal stresses are acting relative to this coordinate system.
>
> I hope this was somewhat helpful.
>
> Best wishes,
> Simon Kattenhorn
> University of Idaho
>
> On Sep 27, 2011, at 7:17 AM, aydin CICEK wrote:
>
>> Dear GeoTectonics Community,
>>
>>
>> What is the difference between stress and traction in terms of structural geology ?
>> I found a bunch of information about them. But, I am not satisfied.
>>
>> Any help will be appreciated!
>>
>> Kind regards,
>>
>> Aydın ÇİÇEK
>> Mineral Research & Exploration of Turkey
>> Geothermal Energy & Hot Springs Surveys Unit
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