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## Homework Statement

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.

## Homework Equations

τ

_{x'y'}= (-σ

_{x}-σ

_{y})/2*sin(2θ) + τ

_{xy}cos(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:

τ

_{x'y'}= τ

_{xy}cos(2θ)

where τ

_{xy}is 600 psi.

Hence I get:

τ

_{x'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:

τ

_{x'y'}= τ

_{xy}sin(2θ)

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