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.
τx'y' = (-σx-σy)/2*sin(2θ) + τxycos(2θ)
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?
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