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Homework Help: Determine the strain rate for a material fiber

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine the Strain Rate for a Material Fiber in the direction of the surface normal.

    The Velocity Field is
    V=((4y-3x)i+(5x+3y)j) ft/s

    http://puu.sh/9hQ7Q/2bda80620f.jpg [Broken] is the picture

    which describes a steady, planar flow

    where i and j are unit vectors.

    2. Relevant equations
    n * (n * ∇) V

    where n is the unit normal, and * are dot products.

    3. The attempt at a solution
    I know that the solution to this itself is very simple, it is just math, my biggest issue though is how do I find the unit normal vector? I have no idea where to begin, any hints would be greatly appreciated!
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 6, 2014 #2
    In your figure, the z axis lies within the plane, and the point 1,1,1 lies within the plane. Do you know how to determine a unit normal to this plane?

    Also, the rate of deformation tensor is 1/2 the velocity gradient tensor and its transpose, not just the velocity gradient tensor.

  4. Jun 7, 2014 #3
    The only thing I can think of is a cross product of two vectors.
    If I defined two vectors from the origin to the coordinates (0,0,1) and (1,1,0) and crossed them, this would give me -i+j
    however the solution is sqrt(2)/2(-i+j) and I'm not sure where that common factor is coming from
  5. Jun 7, 2014 #4
    (1,1,0) is not a unit vector. Divide it by its magnitude, and you will see where the sqrt(2)/2 came from. Another way to get the unit vector normal to the plane is just to draw a diagram of the intersection of the plane with the x-y plane.

  6. Jun 8, 2014 #5
    In general, when given something like, this how would I know which vectors to use for my cross product?
  7. Jun 8, 2014 #6
    Any two convenient in-plane unit vectors will do the trick. But often, it's easier to draw a diagram with a unit normal to the plane, and resolve it into components in the coordinate directions.

  8. Jun 8, 2014 #7
    Ah I see, thank you.
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