Correct understanding of the tension force

[fog37]

Hello,
I am reviewing the concept of tension, a force that expresses the internal tensional state in a body (rope, chain, string, solid body). My understanding is that the force of tension "derives" from the stress tensor and relates (it is the product) the diagonal tensor components to an infinitesimal area $dA$ perpendicular to the tension force direction pointing away from the area.

• The stress tensor diagonal components are scalars (they are essentially what we call pressure, I guess, and when positive, they are used to derive tension). If the rope lies along the x-axis, then there is only one of nonzero diagonal components which multiplied by the rope cross-section, results in the tension force, correct?
• In general, a "true" vector is one that is uniquely defined in both magnitude and direction at a point in space. Inside a rope under tension, the tension force is always directed axially and at an arbitrary point $P$ inside the rope seems to point in two opposite directions at the same time...Is that because, considering an imaginary cut at point $P$, adjacent pieces of the rope have zero relative motion and experience an action-reaction force pair? The piece of rope to the left of the imaginary cut experiences a force pointing to the right and the piece of rope on the right of the cut experience a left directed force...Is that correct?

Thank you!

 

[vanhees71]

I think it's pretty well explained in the Wikipedia:

https://en.wikipedia.org/wiki/Cauchy_stress_tensor

 

[fog37]

Thank you!

Imaginary Cut
As far as the imaginary surface, I guess there could be two ways to look locally at the rope and its state of tension:
a) The infinitesimal segment $M$ of the rope located exactly at the imaginary transverse cut is pulled by the two neighboring rope segments $A$ and $B$ with equal and opposite forces, equal to the tension, stressing the middle segment $M$.
b) There is not rope segment at the cut, just a rope segment to the left and a rope segment the right of the imaginary cut. Each segment is pulled by the other segment with a force, the tension (force). The two forces are equal in magnitude and opposite in direction.

Stress tensor
The Cauchy symmetric stress tensor pertains to internal surface forces (not body forces). Surface forces are shear forces and tension/compression forces. In a real (not massless) rope with its entire length along the x-axis under tension, the tension varies along the rope and the stress tensor is a field tensor with components that vary with x.
The traction vector t(n) for a given current position x acts on an arbitrarily oriented surface element characterized by an outward unit normal area vector n. In our simple example, the normal vector n points along the x-axis. The traction vector t(n), which is a linear combination of the stress tensor components, comprises both shear and perpendicular effects. The traction is a vector has units N/m^2.

I am still confused on how to get the tension force T, which is a force and has units of N, from the traction vector t(n). Are the almost the same thing? Do I just multiply the traction vector t(n) by the infinitesimal area $dA$: $T= t(n) dA$?

 

[Chestermiller]

The traction vector acting on an area is equal to the stress tensor dotted (contracted) with the unit outwardly directed normal to the area. This is the force exerted by the material on the side of the area toward which the unit vector is pointing and on the material on the side of the area from which the unit normal vector is pointing. So, if you have a rope aligned along the x axis, the traction vector on the right side of a cross section is equal to the stress tensor dotted with $\mathbf{i_x}$; the traction vector on the left side of the cross section is equal to the stress tensor dotted with $(-\mathbf{i_x})$. So the two traction vectors have opposite signs, as required by Newton's 3rd law. If the area is not oriented normal to one of the principal stress directions, there is also a shear (tangential) component to the traction vector.

 

[fog37]

Thank you Chestermiller.
So once we find the traction vector, the tension force is simply the traction vector multiplied by the area...

 

[Chestermiller]

Thank you Chestermiller.
So once we find the traction vector, the tension force is simply the traction vector multiplied by the area...

Yes.