Look at the image, extracted from the pdf-file attached here.
Why 1/2 coefficient is much important and determines whether being tensor or not.
Dear Mr. Chomachar,
your question forgets some principles of mathematical definition of what is a tensor. There are special properties that comes from that.
But lets talk a little different. You must understand in the basic field of Elasticity (before going to more complicated matters) what means this tensor. So it is not a question of a number 'half' but what did it represents to the point of the body where such stress are present (measured or calculated, doesn't matter). You will always represent this situation like a line subjected to a pair of forces of opposite sense. The deformation called elongation comes from of this action.
See that when some axial stress happens, the portion of the body increases (or reduces) its length in this direction -elongation- not other direction is involved. So, the deformation is linear (i.e. in one dimension, it does not mean necessarily 'proporcional').
However, when you work with shearing stress, the deformation involves two axis (e.g. x and y) and four forces which are themselves in opposite sense at each pair (two for x axis and two for y axis). If you put this on paper, you will see a square becomes a parallelogram. Also you can imagine that one of its diagonal will raise while the other will reduce, like before in the axial elongation. You have two diagonals, you can see that now you have two effects not one, mean two deformations, an angular deformation. As you decompose this deformations to the corresponding axis, it will be clear for you that you will find two deformations, one for x caused by y, another for y by x. Half of each deformation will come to each side, that is the meaning of divide by two. The real deformation is the sum of the deformation in y caused by x, with deformation in x caused by y divided by two. This is not only a coefficient but a relationship about what is happening to the body, this is my point.
Complementing, the tensor is a special mathematical matrix which have properties that let us do some ordered operations and that provide us some complementary information, when we don't have ways to determine precisely some of its components. And we can consider some of its components negligible for studying a particular situation. Or also, can provide us a guide to what will happening in more complex situations and how to analyse the state of stress and strain in a point of the body.
But depending on the formulation adopted to some problem or study, you can put the deformations in a vector for simplicity and to make more confortable some matrix operations. This is just a way to work, which is different from what is the deformation state (or stress state) of a point of the body.
I suggest you to search and read the old but very good books from Schaum's outline series "Theory and problems of continuous mechanics" from PhD George E. Mase 1970 published by McGraw Hill.
For the theory there are several books including the classic and famous Timoshenko and Gere. But let me recommend you the "Elasticity in Engineering Mechanics" from (another Arthur, lgs) Arthur P. Boresi and Ken P. Chong 1987 published by PTR Prentice Hall. ISBN 0-13-500471-3.
One last word: the "continuum mechanic" science is not the same thing as "Mechanical Engineering", but we (of this engineering) had much complex situations to analyse generally harder than that of usual Civil engineering ones, in the past. Today, both of us has the same complex problems, for this reason, we both need the same study, capacitation and understanding in this field of Science. You are wellcome. Good luck.
PS. I made some English mistakes which I tried to repair now. (22/3).
@Mohamed Larbi Kadri. What should I verify to be linear? the strain?! how?
Dear Mr. Chomachar,
your question forgets some principles of mathematical definition of what is a tensor. There are special properties that comes from that.
But lets talk a little different. You must understand in the basic field of Elasticity (before going to more complicated matters) what means this tensor. So it is not a question of a number 'half' but what did it represents to the point of the body where such stress are present (measured or calculated, doesn't matter). You will always represent this situation like a line subjected to a pair of forces of opposite sense. The deformation called elongation comes from of this action.
See that when some axial stress happens, the portion of the body increases (or reduces) its length in this direction -elongation- not other direction is involved. So, the deformation is linear (i.e. in one dimension, it does not mean necessarily 'proporcional').
However, when you work with shearing stress, the deformation involves two axis (e.g. x and y) and four forces which are themselves in opposite sense at each pair (two for x axis and two for y axis). If you put this on paper, you will see a square becomes a parallelogram. Also you can imagine that one of its diagonal will raise while the other will reduce, like before in the axial elongation. You have two diagonals, you can see that now you have two effects not one, mean two deformations, an angular deformation. As you decompose this deformations to the corresponding axis, it will be clear for you that you will find two deformations, one for x caused by y, another for y by x. Half of each deformation will come to each side, that is the meaning of divide by two. The real deformation is the sum of the deformation in y caused by x, with deformation in x caused by y divided by two. This is not only a coefficient but a relationship about what is happening to the body, this is my point.
Complementing, the tensor is a special mathematical matrix which have properties that let us do some ordered operations and that provide us some complementary information, when we don't have ways to determine precisely some of its components. And we can consider some of its components negligible for studying a particular situation. Or also, can provide us a guide to what will happening in more complex situations and how to analyse the state of stress and strain in a point of the body.
But depending on the formulation adopted to some problem or study, you can put the deformations in a vector for simplicity and to make more confortable some matrix operations. This is just a way to work, which is different from what is the deformation state (or stress state) of a point of the body.
I suggest you to search and read the old but very good books from Schaum's outline series "Theory and problems of continuous mechanics" from PhD George E. Mase 1970 published by McGraw Hill.
For the theory there are several books including the classic and famous Timoshenko and Gere. But let me recommend you the "Elasticity in Engineering Mechanics" from (another Arthur, lgs) Arthur P. Boresi and Ken P. Chong 1987 published by PTR Prentice Hall. ISBN 0-13-500471-3.
One last word: the "continuum mechanic" science is not the same thing as "Mechanical Engineering", but we (of this engineering) had much complex situations to analyse generally harder than that of usual Civil engineering ones, in the past. Today, both of us has the same complex problems, for this reason, we both need the same study, capacitation and understanding in this field of Science. You are wellcome. Good luck.
PS. I made some English mistakes which I tried to repair now. (22/3).
Being transformed to another coordinate system the engineering strain does not define the same strain state, so it is not a tensor.
Maybe the next reasoning will be more convincing. Suppose that in the current coordinate system
[e] not =[eeng]. (*)
Suppose that [eeng] is a tensor, then [t]=[e]-[eeng] is a tensor too. In the coordinate system where [e] has a diagonal form [e'] =[eeng'], and [t']=0. Being a tensor, [t] must have zero components in any coordinate system, but this contradicts to (*) and shows that [eeng] is not a tensor.
@Victor J. Аdlucky. Good analysis. Could you please possibly give a definition for [e] not. What does "not" denote?
" not = " is " not equal ". I mean that [e] and [eeng] have different non-diagonal terms.
@Victor J. Аdlucky. Marvelous analysis. That's it. Thank you.
Both of them are not tensors.
Tensor is unitary invariant thing, but not a set of nine numbers written as 3 rows x 3 columns, whichever are these numbers somewhere halved or are not. Matrix is not tensor. Components of vector in some basis are not vector.
Infinitesimal (linear) strain is defined as symmetrized gradient of field of small displacements, and symmetrization means 0.5 x (the tensor + the same tensor transposed). Wikipedia looks pretty nice to begin with https://en.wikipedia.org/wiki/Infinitesimal_strain_theory (especially in contrast to your chap_04.pdf)
- Badges
- Science topic
- Similar topics
- Engineering
- Mechanical Engineering