Shear stress

The clue is in the words 'longer' and 'shear'

Surely 'shear' implies 'across' rather than 'along' the axis of a beam or whatever.

Maybe you should post a picture/diagram of what is troubling you...it may ellicit more intelligent replies than mine?

Del

Are you familiar with Poisson's Ratio? If so, it seems to apply. If not, you must learn that.

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http://img201.imageshack.us/my.php?image=bilde004gg3.jpg¨

As you can see from the picture, the shear stress does not make the sides of the square longer(the sides only rotate), why?

Guest: I am familier with Poisson's ratio, but this subject has nothing to do with that. Normal strains are affected by Poisson's ratio, shear strains are not.

Looking at it from a pure mathematical point of view:

For an isotropic material subjected to a general state of stress in which all stress components are present, the strain and stress are related by:

ε_x=(1/E)[σ_x-ν(σ_y+σ_z)]

ε_y=(1/E)[σ_y-ν(σ_x+σ_z)]

ε_z=(1/E)[σ_z-ν(σ_x+σ_y)]

τ_xy=Gγ_xy

τ_yz=Gγ_yz

τ_xz=Gγ_xz

Then in the special case of plane stress where the normal stresses in the x and y-direction are zero as well, the above equations reduce to:

ε_x=ε_y=ε_z=0

γ_xy=τ_xy/G

Therefore - you don't have a change in length of the sides.

(in defence of Guest above: Poisson's ratio is indirectly relevant through the relation G=E/[2(1+ν)]

Guest(1),

I apologize; I misunderstood your question.

In the case of pure shear stress, as seen in theory books, Poisson's Ratio is not relevant (except as noted by Guest(3) above). I thought you were asking an applied question, in which case it is relevant.

Guest(2)

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