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Find in-plane shearing stress parallel and perpendicular to the grain

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data
    The grains of the wooden member forms an angle of 15° with the vertical. For the state of the stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the normal stress perpendicular to the plane.

    I have attached an image of the question.


    2. Relevant equations

    τx'y' = (-σxy)/2*sin(2θ) + τxycos(2θ)

    3. The attempt at a solution

    I'm having some trouble understanding what is going on here. I can see that the is a 600 psi shearing stress applied at the sides and that there are no normal stresses, hence the first portion of the equation I gave above is zero, leaving me with:

    τx'y' = τxycos(2θ)

    where τxy is 600 psi.

    Hence I get:

    τx'y' = 600cos(2*15)
    = 520 psi which i know is correct

    Can someone describe what is physically happening?

    For part b I know the answer is - 300 psi which I can get from:

    τx'y' = τxysin(2θ)

    = 300 psi but what makes it negative? and why does the cos change to a sin?
     

    Attached Files:

  2. jcsd
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