Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transformation relations tensors

  1. Aug 24, 2013 #1
    I'm trying to understand the transformation relations for 2d stress and the book doesn't show the derivation of the 2d stress transformation relations from the directional cosines. The 2d stress transformation relations are found by using the transformation equation and the 2d directional cosines matrix. I'm really confused as to how they go about performing the math.

    2d stress transformation relations:

    [itex]\sigma_{xx}^{'}= \sigma_{xx} \cos^2 \theta + \sigma_{yy} \sin^2 \theta + 2\sigma_{xy} \cos \theta \sin \theta[/itex]

    [itex]\sigma_{yy}^{'}= \sigma_{xx} \sin^2 \theta + \sigma_{yy} \cos^2 \theta - 2\sigma_{xy} \cos \theta \sin \theta[/itex]

    [itex]\sigma_{xy}^{'}= \sigma_{xx}(\cos^2 \theta - sin^2 \theta) + (\sigma_{yy} - \sigma_{xx}) \sin \theta \cos \theta[/itex]

    transformation equation:

    [itex]\sigma_{ij}^{'} = m_{ip} m_{jp} \sigma_{pq}[/itex]

    2d directional cosine matrix:

    [itex]m_{ij} = \left[\stackrel{\cos \theta}{ -\sin \theta}\ \stackrel{\sin \theta}{\cos \theta} \right][/itex]

    I guess the thing that I'm confused about is [itex]\sigma_{pq}[/itex]. What does that matrix look like?
     
  2. jcsd
  3. Aug 24, 2013 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I suspect σpq is simply the original matrix elements.
    σ11 = σxx etc.

    [itex]\sigma_{ij}^{'} = m_{ip} m_{jp} \sigma_{pq}[/itex]
    should be
    [itex]\sigma_{ij}^{'} = m_{ip} m_{jq} \sigma_{pq}[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Transformation relations tensors
  1. Tensor transformations (Replies: 8)

  2. What is a tensor? (Replies: 34)

  3. Modulus of a tensor (Replies: 1)

Loading...