So tension is not a force....?

In summary: More generally, if we imagine a cross-section of infinitesimal length ##ds## then it will have an infinitesimal mass ##\rho\ ds## and one can write a differential equation relating the acceleration of the infinitesimal section to the rope density and the tension gradient through that section. [Think of a rope hanging under the weight of a person, for example]. However, the acceleration due to a tension force applied at a point along the rope is always zero.
  • #1
fog37
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Hello Forum,
I have been reading that tension, which is identified as a force in introductory physics books, is not really a force. For example, at a certain point ##P## along a rope under tension, the tension can point both left and right.

I understand that there are the stress and strain tensors. Is tension just a component of the stress tensor? If we multiply the stress tensor by an infinitesimal area vector, what we get out is vectors. These vectors are normal or shear forces. Is the force normal to the area vectors are the tension force?

I am confused. What is the right way to look at tension in slightly more detail than just assuming it is a vector force? The same ideas, I guess, transfer to pressure, which is not a vector but a scalar hence it must be a component of the stress tensor as well but pointing in the direction opposite to tension...

Thanks!
 
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  • #2
fog37 said:
Hello Forum,
I have been reading that tension, which is identified as a force in introductory physics books, is not really a force. For example, at a certain point ##P## along a rope under tension, the tension can point both left and right.

Tension is just a number with the dimensions of force. Its connection with force is that if you consider a small section of the rope, the force on each end is equal to the tension in the rope (and is in the direction along the rope).
 
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  • #3
fog37 said:
I understand that there are the stress and strain tensors. Is tension just a component of the stress tensor? If we multiply the stress tensor by an infinitesimal area vector, what we get out is vectors. These vectors are normal or shear forces. Is the force normal to the area vectors are the tension force?
Yes, tension is a component of the stress tensor, similar to pressure. [Not quite the same since pressure is isotropic and tension normally has a direction].

In introductory treatments, this is usually ignored and one simply treats tension as a scalar quantity that manifests as a non-negative force exerted by a rope at its endpoints. That force is, of course, parallel to the direction of the rope.

Since an area vector is already normal to the area, I think you mean to say that the force parallel to an area vector is the tension force exerted through that area.

Caveat: I am just an interested amateur, never exposed to a formal treatment of any of this.
 
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  • #4
Hello jbriggs444,

Yes, that is what I meant: in the case of a horizontal rope under tension, if we consider a cross-section of the rope, the area vector is perpendicular to it. The area vector can point either left of right. the force parallel to the area vectors is the tension force. So, as far as terminology goes, is it still correct to call tension a force since it eventually leads to a force on the the rope cross-section?

The rope cross-section is at rest but there are two equal magnitude forces applied to it (one pointing left and one pointing right) which means that the net force is zero. But there is surely a nonzero state of internal stress on the rope at that specific point regardless of the net force being zero. I guess it is important (for me) to be remember that a zero net force does imply zero net acceleration but physically, it is not exactly as if no forces at all were acting on the object since a state of stress exists which would not exist if no forces whatsoever were applied...
 
  • #5
fog37 said:
But there is surely a nonzero state of internal stress on the rope at that specific point regardless of the net force being zero. I guess it is important (for me) to be remember that a zero net force does imply zero net acceleration but physically, it is not exactly as if no forces at all were acting on the object since a state of stress exists which would not exist if no forces whatsoever were applied...
Right. Zero force and zero net force are very, very different things. That's a common problem people have.
 
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  • #6
Yes, it sounds like you have a pretty good handle on it.
fog37 said:
zero net force does imply zero net acceleration
Zero net force either implies zero net acceleration or zero mass. The mass of an imaginary cross-section through which tension acts is zero. The acceleration of the rope through a cross-section of zero mass is not constrained by the existence of the tension.

More generally, if we imagine a cross-section of infinitesimal length ##ds## then it will have an infinitesimal mass ##\rho\ ds## and one can write a differential equation relating the acceleration of the infinitesimal section to the rope density and the tension gradient through that section. [Think of a rope hanging under the effects of gravity on its own mass]

Edit: In first year physics, ropes are almost always assumed to be of negligible mass. Such ropes can accelerate under negligible net force.
 
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  • #7
fog37 said:
Hello Forum,
I have been reading that tension, which is identified as a force in introductory physics books, is not really a force. For example, at a certain point ##P## along a rope under tension, the tension can point both left and right.

I understand that there are the stress and strain tensors. Is tension just a component of the stress tensor? If we multiply the stress tensor by an infinitesimal area vector, what we get out is vectors. These vectors are normal or shear forces. Is the force normal to the area vectors are the tension force?

I am confused. What is the right way to look at tension in slightly more detail than just assuming it is a vector force? The same ideas, I guess, transfer to pressure, which is not a vector but a scalar hence it must be a component of the stress tensor as well but pointing in the direction opposite to tension...

Thanks!
You and @jbriggs444 are definitely on the right track with respect to both tension and pressure. To understand all this, you need to know a little about the stress tensor (a 2nd order tensor) and how it is used to get the force per unit area vector on a surface of arbitrary orientation within a solid or fluid. To get you started, I need to know a bit more about your understanding of 2nd order tensors? Are you familiar with dyadic notation? Are you familiar with the Cauchy stress relation?
 
  • #8
At the risk of hijacking fog37's thread, I'll try to put what I think I know into my own words. Not having studied linear algebra, I may stumble over terminology.

Per wiki, a tensor is an object that maps from scalars, vectors and tensors to tensors. That makes sense and can fit with what I already know.

Although tensors are commonly portrayed as rectangular matrices of numbers, my understanding is that such is only a representation relative to a particular basis. More generally, a tensor (and the objects that they might map from and to) exist independent of such. So speaking of specific "component" of a stress tensor is only meaningful if one has imposed some sort of coordinate system. To my mind, the notions of a [Cartesian] coordinate system and of a linear basis are essentially identical.

That said, coordinate systems are a useful crutch for neophytes like myself. Expressed as a rectangular array of scalars, a tensor can be used as a matrix. For instance, you can multiply a 3 dimensional column-vector by a 3x3 tensor to get a 3 dimensional row-vector.

The stress tensor in particular relates to the physical situation in a small neighborhood in a continuous body. In a coordinate-free sense, it takes as input a vector (whose directed magnitude indicates the orientation and area of a small flat imaginary cut through the material at that point) and produces as output another vector (whose value indicates the force exerted by the material on the pointed-to side of the cut on the material on the pointed-from side).

Given a coordinate system, one could express this tensor as a 3x3 array of numbers where the first row is the vector force on a unit area in the x direction, the second row is the vector force on a unit area in the y direction and the third row is the vector force on a unit area in the y direction. One would use the tensor by multiplying the 3 dimensional column vector area by the 3x3 tensor to yield a 3 dimensional row-vector force.

I have no familiarity with either dyadic notation or the Cauchy stress relation.

For a rope oriented horizontally in the +x direction, the stress tensor for tension t per unit area would be $${\begin{pmatrix}t & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}$$
Multiplying it by a unit area vector in the +x direction ##\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}## would give you a force of ##(t, 0, 0)##. If we imagine a wall in the +x direction and a rope in the -x direction, this is the force of wall on rope. Inverting the direction of the unit area vector, we get the force of rope on wall: ##(-t, 0, 0)##. For a rope whose cross-section is equal to the unit of area anyway.
 
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  • #9
jbriggs444 said:
At the risk of hijacking fog37's thread, I'll try to put what I think I know into my own words. Not having studied linear algebra, I may stumble over terminology.

Per wiki, a tensor is an object that maps from scalars, vectors and tensors to tensors. That makes sense and can fit with what I already know.

Although tensors are commonly portrayed as rectangular matrices of numbers, my understanding is that such is only a representation relative to a particular basis. More generally, a tensor (and the objects that they might map from and to) exist independent of such. So speaking of specific "component" of a stress tensor is only meaningful if one has imposed some sort of coordinate system. To my mind, the notions of a [Cartesian] coordinate system and of a linear basis are essentially identical.

That said, coordinate systems are a useful crutch for neophytes like myself. Expressed as a rectangular array of scalars, a tensor can be used as a matrix. For instance, you can multiply a 3 dimensional column-vector by a 3x3 tensor to get a 3 dimensional row-vector.

The stress tensor in particular relates to the physical situation in a small neighborhood in a continuous body. In a coordinate-free sense, it takes as input a vector (whose directed magnitude indicates the orientation and area of a small flat imaginary cut through the material at that point) and produces as output another vector (whose value indicates the force exerted by the material on the pointed-to side of the cut on the material on the pointed-from side).

Given a coordinate system, one could express this tensor as a 3x3 array of numbers where the first column is the vector force on a unit area in the x direction, the second column is the vector force on a unit area in the y direction and the third column is the vector force on a unit area in the y direction. One would use the tensor by multiplying the 3-vector area by the 3x3 tensor to yield a 3-vector force.

I have no familiarity with either dyadic notation or the Cauchy stress relation.

For a rope oriented horizontally in the +x direction, the stress tensor for tension t per unit area would be $${\begin{pmatrix}t & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}}$$
Multiplying it by a unit area vector in the +x direction ##\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}## would give you a force of ##(t, 0, 0)##. If we imagine a wall in the +x direction and a rope in the -x direction, this is the force of wall on rope. Inverting the direction of the unit area vector, we get the force of rope on wall: ##(-t, 0, 0)##. For a rope whose cross-section is equal to the unit of area anyway.
This is very close to what I had in mind, and, even though you are unfamiliar with the Cauchy stress relation, you applied it flawlessly. Depending on how the OP responds (if at all), I may want to expand a little on this.

Chet
 
  • #10
Hello, thank you jbriggs444 and everyone else.

I get that a tensor is a generalization of the concept of vector. In 2D, a tensor is a 2x2 matrix whose components, as jbriggs444 mentions, depend on the choice of the basis of independent vectors within a certain coordinate system. For example, once we pick a certain coordinate system (Cartesian, cylindrical, spherical, etc.) and fix its origin O, we can choose among an infinite amount of bases each one having two independent vectors and the matrix representation (its components) will change.

For example, a 2x2 matrix multiplied by a 2D vector produces a 2D vector with two components as output. In the case of the stress tensor, the input vector is a infinitesimal area vector. The output is the force on that infinitesimal area either perpendicular or parallel to the area. So tension is a component of the stress tensor and represents force per unit area perpendicular to the area and parallel to the area vector, correct? Pressure is also part of the stress tensor but it produces a force with direction opposite to the infinitesimal area vector...

So, in physics problems, is it correct to talk about the tension force or is it better to talk about the force produced by the tension (which is a component of the stress tensor)?

As far as the Cauchy's theorem, I think it relates the stress tensor to the strain tensor... Does stress cause strain or vice versa? I don't think there is a cause-effect relation...
 
  • #11
I must say, I am puzzled by this whole thing.

Why are there issues with "tension" being considered as a "force", while there's no issue with a rod pushing on something? Why is there problems in dealing with something being pulled, but no problems in dealing with a compression force?

Zz.
 
  • #12
I'm going to present the approach that I have found simple and easy to understand and apply, and am sure that you will too. But, before I do, I'd just like to answer a couple of your questions.

fog37 said:
So, in physics problems, is it correct to talk about the tension force or is it better to talk about the force produced by the tension (which is a component of the stress tensor)?

In my expert judgment (based on my PhD training in material science and many years of practical experience), it is unquestionably better fundamentally to talk about it in relation to the stress tensor.

As far as the Cauchy's theorem, I think it relates the stress tensor to the strain tensor... Does stress cause strain or vice versa? I don't think there is a cause-effect relation...
No. Cauchy's relationship is not related to the strain tensor at all. It establishes the relationship between the stress tensor and the traction vector (also sometimes referred to as the stress vector) on a plane of specified arbitrary orientation within a material.

VECTORS (FIRST ORDER TENSORS) AND SECOND ORDER TENSORS

Using 3D Cartesian coordinates, we can express a vector (AKA a first order tensor) as a summation of three terms, involving 3 components (and three unit vectors) as follows: $$\mathbf{v}=v_x\mathbf{i_x}+v_y\mathbf{i_y}+v_z\mathbf{i_z}$$
By analogy, we can express a 2nd order tensor as a summation of 9 terms, involving 9 components as follows:
$$\mathbf{T}=T_{xx}\mathbf{i_x}\mathbf{i_x}+T_{xy}\mathbf{i_x}\mathbf{i_y}+T_{xz}\mathbf{i_x}\mathbf{i_z}+T_{yx}\mathbf{i_y}\mathbf{i_x}+T_{yy}\mathbf{i_y}\mathbf{i_y}+T_{yz}\mathbf{i_y}\mathbf{i_z}+T_{zx}\mathbf{i_z}\mathbf{i_x}+T_{zy}\mathbf{i_z}\mathbf{i_y}+T_{zz}\mathbf{i_z}\mathbf{i_z}$$
You will notice in this summation that there are pairs of unit vectors (e.g. ##\mathbf{i_x}\mathbf{i_y}##) placed right next to one another (i.e., in juxtaposition) with no mathematical operation (such as the dot product or the cross product) implied between them.

Such a juxtaposition of vectors is called a dyadic, and, more generally, a dyadic can be formed in terms of any two arbitrary vectors ##\mathbf{v_1}## and ##\mathbf{v_2}## as ##\mathbf{v_1}\mathbf{v_2}##. A dyadic of two vectors has no special physical significance until it fulfills its main mission in life, which is getting dotted with another vector according to the following rule:$$(\mathbf{v_1}\mathbf{v_2})\centerdot \mathbf{v_3}=\mathbf{v_1}(\mathbf{v_2}\centerdot \mathbf{v_3})$$This equation says that dotting the vector ##\mathbf{v_3}## with the dyadic ##\mathbf{v_1}\mathbf{v_2}## maps ##\mathbf{v_3}## into a new vector what has the same direction as ##\mathbf{v_1}##, and with a new magnitude.

It is also possible to dot a dyadic by a vector in front of it by writing:$$\mathbf{v_3} \centerdot(\mathbf{v_1}\mathbf{v_2})=(\mathbf{v_3} \centerdot \mathbf{v_1})\mathbf{v_2}$$In this case, the equation says that dotting the vector ##\mathbf{v_3}## in front with the dyadic ##\mathbf{v_1}\mathbf{v_2}## maps ##\mathbf{v_3}## into a new vector what has the same direction as ##\mathbf{v_2}##, and with a new magnitude.

STRESSES, TENSIONS, AND PRESSURES
Now the big question is how all this dyadic stuff relates to the stresses and tensions within materials. Well, first of all, we can represent the stress tensor in a material in terms of this notation simply by writing:
$$\boldsymbol{\sigma}=\sigma_{xx}\mathbf{i_x}\mathbf{i_x}+\sigma_{xy}\mathbf{i_x}\mathbf{i_y}+\sigma_{xz}\mathbf{i_x}\mathbf{i_z}+\sigma_{yx}\mathbf{i_y}\mathbf{i_x}+\sigma_{yy}\mathbf{i_y}\mathbf{i_y}+\sigma_{yz}\mathbf{i_y}\mathbf{i_z}+\sigma_{zx}\mathbf{i_z}\mathbf{i_x}+\sigma_{zy}\mathbf{i_z}\mathbf{i_y}+\sigma_{zz}\mathbf{i_z}\mathbf{i_z}$$
Secondly, Cauchy showed that we can calculate the "traction" vector ##\boldsymbol {\tau}## (force vector per unit area aka stress vector) on a differential surface of arbitrary orientation within a material simply by dotting the stress tensor ##\boldsymbol{\sigma}## with a unit normal vector ##\mathbf{n}## to the surface:
$$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}$$with ##\mathbf{n}=n_x\mathbf{i_x}+n_y\mathbf{i_y}+n_z\mathbf{i_z}##
Now, if we apply our dyadic multiplication rule with this equation, we obtain:
$$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}=(\sigma_{xx}n_x+\sigma_{xy}n_y+\sigma_{xz}n_z)\mathbf{i_x}+(\sigma_{yx}n_x+\sigma_{yy}n_y+\sigma_{yz}n_z)\mathbf{i_y}+(\sigma_{zx}n_x+\sigma_{zy}n_y+\sigma_{zz}n_z)\mathbf{i_z}$$

Note that the traction vector on the surface can be resolved into components both normal and tangent to the surface. Therefore, in general, there will be both a normal stress on any surface and a tangential (shear) stress on the surface. Only for special orientation of the surface is the shear stress zero.

APPLICATION TO TENSION IN ROPES AND PRESSURES IN FLUIDS

For a rope or a rod under uniaxial tension load, the stress tensor reduces simply to: $$\boldsymbol{\sigma}=\frac{T}{A}\mathbf{i_x}\mathbf{i_x}$$where T is the tension and A is the cross sectional area. If we dot this with a unit vector perpendicular to the rope cross section, in the positive x direction ##\mathbf{n}=\mathbf{i_x}##, we have $$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}=\frac{T}{A}\mathbf{i_x}$$This is the stress vector on the surface exerted by the material to the right of the surface on the material to the left of the surface. The tension force exerted by the material to the right on the material to the left is obtained simply by multiplying by the cross sectional area A.

If we want to determine stress vector on the cross section exerted by the material to the left on the material to the right, we simply dot the stress tensor by the normal unit vector in the negative x direction: $$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}=\frac{T}{A}\mathbf{i_x}\mathbf{i_x}\centerdot (-\mathbf{i_x})=-\frac{T}{A}\mathbf{i_x}$$

In the case of hydrostatic pressure in a fluid, the (tensile) stress tensor is "isotropic," and can be expressed as:
$$\boldsymbol{\sigma}=-p\mathbf{i_x}\mathbf{i_x}-p\mathbf{i_y}\mathbf{i_y}-p\mathbf{i_z}\mathbf{i_z}$$where p is the pressure. If we dot this stress tensor with a unit normal vector in any arbitrary direction, we obtain
$$\boldsymbol{\tau}=\boldsymbol{\sigma}\centerdot \mathbf{n}=-p(n_x\mathbf{i_x}+n_y\mathbf{i_y}+n_z\mathbf{i_z})=-p\mathbf{n}$$Note from this equation that the magnitude of the stress vector is equal to the fluid pressure and that its direction is normal to our arbitrarily oriented surface. Thus, under hydrostatic conditions, there is no shear stress component on any surface.
 
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  • #13
Thank you Chestermiller! That is an awesome educational piece. I am slowly digesting it. Some comments in relation to the two initial sections:
  • The dot product between a dyadic and an arbitrary vector is not commutative like it is between two vectors.
  • The stress tensor ##\sigma## is simply a linear combination of dyadics (dyadic products). Each dyadic product has a magnitude, like ##\sigma_{xy}##, and is calculated by using two different (or the same) unit vectors. What does the magnitude ##\sigma_{xy}## indicate? Is ##\sigma_{xy}## a force per unit area in the x-direction on a surface pointing in the y-direction.
  • The ##n## vector is essentially a unit vector perpendicular to the elemental surface area.
  • The traction vector ##\tau## is calculated through the dot product between the stress tensor ##\sigma## and the unit area vector ##n##.
  • The resulting traction vector ##\tau## has 3 components: the ##i_{x}## component, the ##i_{y}## component, the ##i_{z}## component. The traction vector on the surface can be resolved into two components, one normal (normal stress) to the surface and calculated via ##\tau \cdot n##, and one tangent (tangential (shear) stress) to the surface calculated via ##\tau \cross n##.
  • You mention that only for a special orientation of the elemental surface, the shear stress can be zero. That happens when the physical elemental surface is exactly perpendicular to the traction vector which happens when ##n## and ##\tau## are exactly parallel.
Hope this is all correct. I will move to the next section soon...
 
  • #14
fog37 said:
Thank you Chestermiller! That is an awesome educational piece. I am slowly digesting it. Some comments in relation to the two initial sections:
  • The dot product between a dyadic and an arbitrary vector is not commutative like it is between two vectors.
Yes.
  • The stress tensor ##\sigma## is simply a linear combination of dyadics (dyadic products). Each dyadic product has a magnitude, like ##\sigma_{xy}##, and is calculated by using two different (or the same) unit vectors. What does the magnitude ##\sigma_{xy}## indicate? Is ##\sigma_{xy}## a force per unit area in the x-direction on a surface pointing in the y-direction.
Yes, the force per unit area in the x-direction on a surface of constant y.
  • The ##n## vector is essentially a unit vector perpendicular to the elemental surface area.
Yes.
  • The traction vector ##\tau## is calculated through the dot product between the stress tensor ##\sigma## and the unit area vector ##n##.
Yes.
  • The resulting traction vector ##\tau## has 3 components: the ##i_{x}## component, the ##i_{y}## component, the ##i_{z}## component. The traction vector on the surface can be resolved into two components, one normal (normal stress) to the surface and calculated via ##\tau \cdot n##, and one tangent (tangential (shear) stress) to the surface calculated via ##\tau \cross n##.
Excellent. The shear stress can also be determined by subtracting the norma stress ##(\tau \cdot n)n## from ##\tau##
  • You mention that only for a special orientation of the elemental surface, the shear stress can be zero. That happens when the physical elemental surface is exactly perpendicular to the traction vector which happens when ##n## and ##\tau## are exactly parallel.
Yes. This is called a "principal direction" of the stress tensor.
You really did an amazing job of digesting all this. I commend you.

Chet
 
  • #15
Uhhh, than you!
This is a math question: we can perform the dyadic product between two 3D vectors (which are tensors of rank 1 with 3 components) and the result is a linear combination of nine dyadic products. But that same linear combination also symbolizes a tensor, since the nine tensor elements are the magnitudes, ##T_{xx##, ##T_{xy}##, etc. , of the various dyadic products in the linear combination.

We can also perform the dyadic between two tensors. (rank 2) This dyadic product is essentially what people call tensor product. In this case we get a combination of ##3^3=27## dyadic products, i.e. a tensor with 27 components. Is there a physical example of a dyadic product between two tensors?

Principal direction of the stress tensor: we are free to orient the ##x,y,z## Cartesian axes any way we prefer. Once possible choice is to orient the triad of unit vectors ##i_x##, ##i_y##, ,##i_z## so that all the cross terms of the stress tensor become zero and only the terms ##\sigma_{xx}##, ##\sigma_{yy}##, ##\sigma_{zz}## are nonzero, i.e. the tensor representation is a diagonal matrix. The infinitesimal area unit vector, which is always perpendicular to the infinitesimal area, points where it needs to point: the choice of orientation of the Cartesian axes should not affect the physical orientation of the vector ##n## but will affect the value of its 3 components. To make the the stress tensor diagonal, do we need to orient the ##i_x##, ##i_y##, ,##i_z## so that the ##n## vector points exactly and only along one of the Cartesian unit vectors ##i_x##, ##i_y##, ,##i_z## ?

When there is a real, physical shear stress effect on the infinitesimal area the shear stress component should be nonzero, and it seem strange that we can convert the shear stress to zero just by a judicious choice of the orientation of the Cartesian axes...What am I missing?
 
  • #16
Also, I think the stress tensor turns out to be symmetric. So among the 9 terms, there are only 6 meaningful terms...
 
  • #17
fog37 said:
Also, I think the stress tensor turns out to be symmetric. So among the 9 terms, there are only 6 meaningful terms...
Yes, but I didn't want to complicate things by mentioning that yet.
 
  • #18
fog37 said:
Uhhh, than you!
This is a math question: we can perform the dyadic product between two 3D vectors (which are tensors of rank 1 with 3 components) and the result is a linear combination of nine dyadic products. But that same linear combination also symbolizes a tensor, since the nine tensor elements are the magnitudes, ##T_{xx##, ##T_{xy}##, etc. , of the various dyadic products in the linear combination.
Yes. Sometimes we represent the momentum flux tensor as ##\rho \mathbf{v}\mathbf{v}##, where ##\mathbf{v}## is the fluid velocity and ##\rho## is its density.
We can also perform the dyadic between two tensors. (rank 2) This dyadic product is essentially what people call tensor product. In this case we get a combination of ##3^3=27## dyadic products, i.e. a tensor with 27 components. Is there a physical example of a dyadic product between two tensors?
I don't quite follow this. Can you provide a specific example please?
Principal direction of the stress tensor: we are free to orient the ##x,y,z## Cartesian axes any way we prefer. Once possible choice is to orient the triad of unit vectors ##i_x##, ##i_y##, ,##i_z## so that all the cross terms of the stress tensor become zero and only the terms ##\sigma_{xx}##, ##\sigma_{yy}##, ##\sigma_{zz}## are nonzero, i.e. the tensor representation is a diagonal matrix. The infinitesimal area unit vector, which is always perpendicular to the infinitesimal area, points where it needs to point: the choice of orientation of the Cartesian axes should not affect the physical orientation of the vector ##n## but will affect the value of its 3 components. To make the the stress tensor diagonal, do we need to orient the ##i_x##, ##i_y##, ,##i_z## so that the ##n## vector points exactly and only along one of the Cartesian unit vectors ##i_x##, ##i_y##, ,##i_z## ?
I think you are asking how we find the principal directions of the stress tensor, given its cartesian components for an arbitrary orientation of the material. What we would do would be to solve the following eigenvalue problem: $$\boldsymbol{\sigma}\centerdot \mathbf{n}=\lambda \mathbf{n}$$where ##\lambda## is a value of one of the principle stresses.
When there is a real, physical shear stress effect on the infinitesimal area the shear stress component should be nonzero, and it seem strange that we can convert the shear stress to zero just by a judicious choice of the orientation of the Cartesian axes...What am I missing?
You're not missing anything. A mathematical characteristic of a symmetric 2nd order tensor is that its components can always be diagonalized by judicious choice of the orientation of the Cartesian axes.
 
  • #19
I think it's interesting how much of learning physics (or maybe other technical subjects as well) has the pattern:
  1. You are told something when you're introduced to a topic.
  2. Later, you're told: Actually, that was a lie, but the truth is an advanced topic that you weren't ready for.
So first tension is a force, then later you find out it is a tensor. Centrifugal force is a force, then later you find out it is a connection coefficient. Etc.
 
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  • #20
stevendaryl said:
I think it's interesting how much of learning physics (or maybe other technical subjects as well) has the pattern:
  1. You are told something when you're introduced to a topic.
  2. Later, you're told: Actually, that was a lie, but the truth is an advanced topic that you weren't ready for.
So first tension is a force, then later you find out it is a tensor. Centrifugal force is a force, then later you find out it is a connection coefficient. Etc.
I think we should start a thread with other examples of this kind of thing.
 
  • #21
Thank you Chet!

Maybe I did not explain myself clearly on my last diagonalization question.

For example, let's say the elemental area ##dA## is perfectly horizontal, which means that the unit area vector ##n## points straight up and is fixed that way .
Let's assume that this small area is subjected to a physical force ##dF## that causes both tension (pull on the area) and shear (acts tangentially on the area) on the area. This implies that both the effects of tension and shear exist and the stress tensor with its 9 components should capture them. However, diagonalizing the stress tensor (achieved by suitable orientation of the Cartesian triad of axes) turns the stress tensor's shear terms (which correlate with the shear effect on the area) equal to zero. So there is the presence of a nonzero shear effect (which physically exists) but the stress tensor cross terms are zero. I can see how different Cartesian axes orientations change the values of the stress tensor components but it is strange that the cross terms, being an indication of shear effect acting on the area, become zero when there is shear existing on the unit area.
 
  • #22
fog37 said:
Thank you Chet!

Maybe I did not explain myself clearly on my last diagonalization question.

For example, let's say the elemental area ##dA## is perfectly horizontal, which means that the unit area vector ##n## points straight up and is fixed that way .
Let's assume that this small area is subjected to a physical force ##dF## that causes both tension (pull on the area) and shear (acts tangentially on the area) on the area. This implies that both the effects of tension and shear exist and the stress tensor with its 9 components should capture them. However, diagonalizing the stress tensor (achieved by suitable orientation of the Cartesian triad of axes) turns the stress tensor's shear terms (which correlate with the shear effect on the area) equal to zero. So there is the presence of a nonzero shear effect (which physically exists) but the stress tensor cross terms are zero. I can see how different Cartesian axes orientations change the values of the stress tensor components but it is strange that the cross terms, being an indication of shear effect acting on the area, become zero when there is shear existing on the unit area.
Screen Shot 2018-11-18 at 10.28.18 AM.png
 

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  • #23
Ok, so the red arrow pointing upward is the acting physical force which can be decomposed into:
a) tangential shear force component along the sloped surface.
b) tensile force component perpendicular to the sloped surface.

No problem with all of that. If we diagonalize the stress tensor, its shear terms are zero and disappear but that does not obviously make the force component a) physically disappear.
 
  • #24
fog37 said:
Ok, so the red arrow pointing upward is the acting physical force which can be decomposed into:
a) tangential shear force component along the sloped surface.
b) tensile force component perpendicular to the sloped surface.

No problem with all of that. If we diagonalize the stress tensor, its shear terms are zero and disappear but that does not obviously make the force component a) physically disappear.
What is the shear stress on the base?
 
  • #25
Let's try something else. Suppose that the stress tensor expressed in terms of the x-y coordinate system is simply $$\boldsymbol {\sigma}=\sigma_{xx}\mathbf{i_x}\mathbf{i_x}\tag{1}$$ Now we want to express this exact same stress tensor in terms of its components in the x'-y' coordinate system where the unit vectors are $$\mathbf{i_{x'}}=\frac{(\mathbf{i_x}+\mathbf{i_y})}{\sqrt{2}}$$ and $$\mathbf{i_{y'}}=\frac{(-\mathbf{i_x}+\mathbf{i_y})}{\sqrt{2}}$$In terms of ##\mathbf{i_{x'}}## and ##\mathbf{i_{y'}}##, what are ##\mathbf{i_x}## and ##\mathbf{i_y}##?If we substitute these equations for ##\mathbf{i_x}## and ##\mathbf{i_y}## into Eqn. 1, what do we obtain?
 
  • #26
Thank you Chet.

1) What you wrote is a 45 degree rotation of the Cartesian axes x and y where the rotated unit vectors are primed. The old unit vectors are expressed in terms of the primed vectors as follows: $$i_{x}= 0.707i_{x^{'}} - 0.707 i_{y^{'}}$$ $$i_{y}= 0.707i_{x^{'}} +0.707 i_{y^{'}}$$
2) In terms of the primed unit vectors, the stress tensor becomes $$\sigma= 0.5 i_{x^{'}} i_{x^{'}} - 0.5 i_{x^{'}}i_{y^{'}} - 0.5i_{y^{'}} i_{x^{'}} + 0.5 i_{y^{'}}i_{y^{'}}$$. In this case, we have two nonzero shear cross terms.

3) On your previous post, you asked "What is the shear stress on the base?".
In that case the red force exerts a purely tensile action on the base and no shear. The normal vector points straight down: $$n=- i_{y}$$ or $$n=- (0.707i_{x^{'}} +0.707 i_{y^{'}})$$. The force on the area is given by $$\tau= \sigma \cdot n$$. To get a force pointing down that only causes tension and no shear, the stress tensor should be $$\sigma= \sigma_{yy} i_{y} i_{y}$$ or, in the primed coordinate system, $$\sigma= \sigma_{yy} i_{x^{'}i_{x^{'}}} + \sigma_{yy} i_{x^{'}} i_{y^{'}} $$

 
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  • #27
fog37 said:
Thank you Chet.

1) What you wrote is a 45 degree rotation of the Cartesian axes x and y where the rotated unit vectors are primed. The old unit vectors are expressed in terms of the primed vectors as follows: $$i_{x}= 0.707i_{x^{'}} - 0.707 i_{y^{'}}$$ $$i_{y}= 0.707i_{x^{'}} +0.707 i_{y^{'}}$$
2) In terms of the primed unit vectors, the stress tensor becomes $$\sigma= 0.5 i_{x^{'}} i_{x^{'}} - 0.5 i_{x^{'}}i_{y^{'}} - 0.5i_{y^{'}} i_{x^{'}} + 0.5 i_{y^{'}}i_{y^{'}}$$. In this case, we have two nonzero shear cross terms.
This is basically done correctly, but you are missing a factor of ##\sigma_{xx}## on the right hand side. So you can see how, in one set of coordinates you can have shear stresses while, in another set of coordinates (for the same tensor), there are no shear stresses.
3) On your previous post, you asked "What is the shear stress on the base?".
In that case the red force exerts a purely tensile action on the base and no shear. The normal vector points straight down: $$n=- i_{y}$$
In this case, I was just referring to the figure in the previous example. I didn't expect you to consider the other coordinate system.
 
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  • #28
Ok, thanks.

1) Corrected, the stress tensor in the primed coordinate system becomes $$\sigma= \sigma_{xx} [0.5 i_{x^{'}} i_{x^{'}} - 0.5 i_{x^{'}}i_{y^{'}} - 0.5i_{y^{'}} i_{x^{'}} + 0.5 i_{y^{'}}i_{y^{'}}]$$ Because the stress tensor is always symmetric, the terms ##- 0.5 i_{x^{'}}i_{y^{'}}## and ##- 0.5 i_{x^{'}}i_{y^{'}}## are identical so we can write $$\sigma= \sigma_{xx} [0.5 i_{x^{'}} i_{x^{'}} - i_{x^{'}}i_{y^{'}} + 0.5 i_{y^{'}}i_{y^{'}}]$$ , correct?

2) The object in the figure is a wedge with three different areas (base, sloped and vertical). Each area has a different total area and orientation. Each different surface can be subjected to a different stress in terms of tension, pressure and shear. Shouldn't the stress tensor be position-dependent (i,e, the tensor components be a function of ##x## and ##y##)? Or are we considering this wedge object to be infinitesimally small that the stress tensor components are constant so we can use ##\sigma= \sigma_{xx} i_{x}i_{x}## for all three surfaces?
 
  • #29
fog37 said:
Ok, thanks.

1) Corrected, the stress tensor in the primed coordinate system becomes $$\sigma= \sigma_{xx} [0.5 i_{x^{'}} i_{x^{'}} - 0.5 i_{x^{'}}i_{y^{'}} - 0.5i_{y^{'}} i_{x^{'}} + 0.5 i_{y^{'}}i_{y^{'}}]$$ Because the stress tensor is always symmetric, the terms ##- 0.5 i_{x^{'}}i_{y^{'}}## and ##- 0.5 i_{x^{'}}i_{y^{'}}## are identical so we can write $$\sigma= \sigma_{xx} [0.5 i_{x^{'}} i_{x^{'}} - i_{x^{'}}i_{y^{'}} + 0.5 i_{y^{'}}i_{y^{'}}]$$ , correct?
No, but you can write $$\sigma= \sigma_{xx} [0.5 i_{x^{'}} i_{x^{'}} - 0.5( i_{x^{'}}i_{y^{'}} +i_{y^{'}} i_{x^{'}}) + 0.5 i_{y^{'}}i_{y^{'}}]$$
2) The object in the figure is a wedge with three different areas (base, sloped and vertical). Each area has a different total area and orientation. Each different surface can be subjected to a different stress in terms of tension, pressure and shear. Shouldn't the stress tensor be position-dependent (i,e, the tensor components be a function of ##x## and ##y##)?
Not if the loadings along the edges are uniformly distributed. We are assuming that the state of stress in the material is homogeneous.
Or are we considering this wedge object to be infinitesimally small that the stress tensor components are constant so we can use ##\sigma= \sigma_{xx} i_{x}i_{x}## for all three surfaces?
This works too, and is more of what I had in mind.
 
  • #30
Ok. All clear. Thanks again. Looking forward to your explanation about the stress tensor being symmetric (or not). Throughout this discussion, the stress tensor was homogeneous but not symmetric. Is the stress tensor always symmetric in the case of fluids and can be symmetric or asymmetric in the case of solid materials?

When the stress tensor is homogenous (not a function of position), it means that the internal stress is the same at every point in the material. What are a couple of simple examples of a material described by a homogeneous stress tensor and a couple of examples of a material whose stress tensor is instead inhomogeneous?

For the stress tensor to be homogeneous, does the material need to have a constant density, be homogeneous and isotropic, and does the external force need to be uniformly applied to material?
 
  • #31
fog37 said:
Ok. All clear. Thanks again. Looking forward to your explanation about the stress tensor being symmetric (or not).
To be clear, the stress tensor is symmetric (meaning that the components of the dyadics with corresponding different indices are equal). So the way you represented it, it was not symmetric.

Throughout this discussion, the stress tensor was homogeneous but not symmetric. Is the stress tensor always symmetric in the case of fluids and can be symmetric or asymmetric in the case of solid materials?
For the vast majority of practical materials, the stress tensor is symmetric for both solids and liquids.
When the stress tensor is homogenous (not a function of position), it means that the internal stress is the same at every point in the material. What are a couple of simple examples of a material described by a homogeneous stress tensor and a couple of examples of a material whose stress tensor is instead inhomogeneous?
Homogeneous: A rod under tension with the load distributed uniformly at its ends. A fluid under hydrostatic pressure.
Inhomogeneous: A rod under tension with the load distributed non-uniformly at its ends. A beam being bent. Most objects encountered in practice.
For the stress tensor to be homogeneous, does the material need to have a constant density, be homogeneous and isotropic, and does the external force need to be uniformly applied to material?
Pretty much yes. When you solve some problems, you will get the idea.
 
  • #32
A 2nd order tensor is said to be symmetric if it is equal to its "transpose." The transpose of a tensor is obtained by switching the order of the unit vectors in each of its dyadics.
 
  • #33
Thanks. I see my mistake in post #29: I assumed that being symmetric meant that the terms ##i_{x^{'}}i_{y^{'}}## and ##i_{y^{'}}i_{x^{'}}## were the same. Symmetry instead means that the components of these two dyadics are the same...
 
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Related to So tension is not a force....?

1. What is tension?

Tension is a pulling force that occurs when an object is stretched or pulled apart.

2. Is tension a force?

No, tension is not considered a force. It is a type of force that results from the interaction between two objects, but it is not a force in and of itself.

3. How is tension different from other forces?

Tension differs from other forces such as gravity or friction because it only occurs when an object is being stretched or pulled apart. It does not act on objects that are at rest or in motion.

4. Can tension be measured?

Yes, tension can be measured using a variety of instruments such as a spring scale or a force meter. It is typically measured in units of newtons (N).

5. What are some examples of tension?

Some common examples of tension include a rubber band being stretched, a rope being pulled taut, or a person hanging from a pull-up bar. Tension can also be seen in structures such as bridges and suspension cables.

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