Shear strain

[Bempel]

Hello,

I have problems with the "Engineering Shear Strain" formulas from the following website :

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/strain.cfm

I agree with the formulas and the graph on the left, but I don't with it on the right. On the left graph du/dy and dv/dx are separated. On the right graph (total shear strain) dv/dx and du/dy are put together on the x-axis. One can't simply add the deformation u along the y-axis to the deformation v along the x-axis. One can't change d(something)/dx by d(something)/dy or say its the same. Could anyone give me some explanation on this?

Note that there are some errors on the graphs with x's and y's exchanged.

Could anyone please help me on this? I am working on the subject and I want to understand. If you know some references of literature would be appreciated to

Best regards

Stephane

 

[dg]

Well I agree that adding "strains" along different directions might not sound so physical, nevertheless I do not see why it is ok for you adding up du/dy and dv/dx in the graph on the left (to define the strain tensor components) while it is not ok to add them on the one on the right (to define total strain).

The only difference in the two graphs is a rotation of axis.

Remember here we are simply talking geometry, there is no dynamical concepts involved, we are simply describing how a certain distribution of points in space has changed; no forces!

Here we are simply adding infinitesimal "angles" of rotation pretty much as we do when we sum different component of angular velocity in mechanics.

I used "angles" because I am not sure how exactly angles and strains are related to each other...

I will bve happy to work it out together if you like...

Dario

 

[teddy]

please read it through,it will help:

both dv/dx and du/dy are along the same direcdtion viz the z-direction.in other words,what we are doing is measuring how much the body's vertex angles have changed from initial 90 degrees.in many books i have seen a subscript 'z' instead of 'xy' with the shear strain which makes things clear.(the angles in a plane in any prob whatsoever are measued from an axis perpendicular to the plane.)

the shear strain along a direction (here z-direction)is a SCALAR because it is a componenet of a tensor(dont get bogged down by terms,this is true for more familiar vecors also,for example the horizontal componenet of velocity is scalar) .So we can add them up.

may be you have noticed it that on the right hand side the figure has just been rotated and you see that the shear starin suffered by the particle remains same.this provides proof for the shear strain along a direction being a scalar because scalars are invariant under transformation of axes.

cheers
bye