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Derivation of normal strain

  1. Nov 24, 2012 #1

    What are the mathematical steps and assumptions to reach the conclusion that length(ab) ≈ dx + ∂u/∂x*dx ?

    If you consider the the squares of the gradients to be negligible, you still have a square root and multiplication by the constant "2". What other assumptions do we make to derive the final equation?

    *Edit, I should have posted this in calculus, I apologize.
    Last edited: Nov 24, 2012
  2. jcsd
  3. Nov 25, 2012 #2
    You were right to post this in engineering : it is an engineering issue.

    The statement

    [tex]{\rm{length(ab)}} = \sqrt {{{\left( {dx + \frac{{\partial {u_x}}}{{\partial x}}dx} \right)}^2} + {{\left( {\frac{{\partial {u_y}}}{{\partial x}}dx} \right)}^2}} [/tex]

    is nothing more than pythagoras theorem for the horizontal and vertical components of ab.
    I assume you are comfortable with this.

    Now AB was originally horizontal and it is stated that the strain is very small. Thus the angles alpha and beta in the diagram are very small.

    If beta is very small then the hypotenuse (ab) is very nearly the same as the horizontal component, which is

    [tex]dx + \frac{{\partial {u_x}}}{{\partial x}}dx[/tex]


    [tex]{\rm{length(ab)}} \approx dx + \frac{{\partial {u_x}}}{{\partial x}}dx[/tex]
  4. Nov 26, 2012 #3
    OP, I think your question is simply: how did they go from:

    [itex]length(ab)= \sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}[/itex]


    [itex]length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx[/itex] ?

    The math doesn't work, I agree.

    From a geometric point-of-view, as Studiot suggested, they assume that [itex]length(ab)[/itex] is equal to its horizontal projection, for small deformations.

    However, I'm not sure that I buy that, to be honest.

    In terms of the actual physics, and in looking at the provided diagram, I can tell you that if it were only a simple shear, you could get the angle change [itex]\alpha + \beta[/itex] but there has to be some sort of homogeneous (axial) deformation in order for BOTH [itex]dx[/itex] and [itex]dy[/itex] to change lengths.

    For example, one way to arrive at the apparent deformed shape would be:

    1) apply a homogenous deformation (e.x. to the right, of magnitude ([itex]length(ab)-dx[/itex]) -- i.e. [itex]\sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx\right)^2+\left(\frac{\partial u_y}{\partial x}dx\right)^2}-dx[/itex])

    2) apply a simple shear (e.x. to the right, of amount [itex]\alpha + \beta[/itex])

    3) apply a rigid body rotation (e.x. counter-clockwise, of amount [itex]\alpha[/itex])

    Does that make sense?

    You can play with this though.

    Take 1) to be zero. No deformation to the right means [itex]length(ab)=dx[/itex].

    2) and 3) still apply - and so we have a simple shear and a rigid body rotation.

    We should still get [itex]length(ab)=dx[/itex] in this case under either a small shear or a large shear. However, due to the rigid body rotation, [itex]\frac{\partial u_x}{\partial x}dx[/itex] in their diagram would be nonzero and so their expression [itex]length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx[/itex] is not equal to [itex]dx[/itex]. This doesn't mean that they are wrong, but I cannot immediately justify approximating [itex]length(ab)[/itex] as its horizontal projection, for the general case that they are showing.

    In other words, I don't like their expression [itex]length(ab)\approx dx+\frac{\partial u_x}{\partial x}dx[/itex] unless someone can prove to me that it agrees with more advanced solid mechanics.
  5. Nov 26, 2012 #4
    I thought this at first but came to the conclusion that the source material authors were simply replacing or substituting a simpler calculation, not simplyfying a more complicated one.

    This is not unusual, for instance the substitution of the chord for the arc or the other way in circular calculations of small angle.

    We will not know more without more information from the source.
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