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

[Northbysouth]

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_xycos(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_xycos(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_xysin(2θ)

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

 

[KataKoniK]

Thank you for posting your question. It seems like you are on the right track in solving this problem. Let me try to explain what is happening physically and why the answer for part b is negative.

First, let's consider what is happening to the wooden member. It is being subjected to a shearing stress of 600 psi at the sides, as you correctly pointed out. This shearing stress is causing the grains of the wood to deform and rotate, forming an angle of 15° with the vertical.

Now, let's focus on the point where we are trying to determine the normal stress perpendicular to the plane. This point is located on the surface of the wooden member, at a distance from the sides of the member. At this point, the shearing stress is acting in the direction of the grain, which is at an angle of 15° with the vertical. This means that the shearing stress is not acting perpendicular to the surface at this point. In order to determine the normal stress perpendicular to the surface, we need to find the component of the shearing stress that is acting perpendicular to the surface.

This is where the trigonometric functions come into play. The normal stress perpendicular to the surface is equal to the component of the shearing stress that is acting perpendicular to the surface, which can be found by using the sine function. This is why in your equation for part b, the cosine function changes to a sine function.

Now, let's consider the direction of the normal stress. Since the shearing stress is acting in the direction of the grain, the normal stress will be acting perpendicular to the grain. In this case, the normal stress will be compressive, meaning it will be pushing the grains of the wood together. This is why the answer for part b is negative.

I hope this explanation helps you understand the physical concept behind this problem. Please let me know if you have any further questions.