Ratio of shear stress to normal stress along principle place

[AnotherParadox]

Homework Statement

"In a component under multi-axial state of stress, the ratio of shear stress to normal stress along principle places is _____.

A) 0.0 B) 0.5 C) 1.0 D) 1.5 E)2.0"

Homework Equations

σ_x' = (σ_x+σ_y)/2 + ((σ_x-σ_y)/2)*cos(2θ) + τ_xy*sin(2θ)

σ_y' = (σ_x+σ_y)/2 - ((σ_x-σ_y)/2)*cos(2θ) - τ_xy*sin(2θ)

τ_{max,in plane} = √[((σ_x-σ_y)/2)² + τ_xy²]

σ_1,2 = (σ_x+σ_y)/2) ± [((σ_x-σ_y)/2)² + τ_xy²]

τ_max,absolute = σ₁/2

The Attempt at a Solution

Ratio of shear stress to normal stress?? ->
τ_{max,in plane}/σ_x' ->
[τ_{max,in plane} = √[((σ_x-σ_y)/2)² + τ_xy²]] / [σ_x' = (σ_x+σ_y)/2 + ((σ_x-σ_y)/2)*cos(2θ) + τ_xy*sin(2θ)]

Impossible to simplify? Wrong ratio equations?

I don't know what to do. Please help. I have heard from different sources that it is either 0 or 0.5 neither with explanations or work shown.

Thank you

 

[Chestermiller]

What is the definition of a principal direction of stress?

 

[AnotherParadox]

What is the definition of a principal direction of stress?

If I understand correctly the definition of a principal direction of stress is the stress vector at which the transformed normal stress vector is maximized.

According to the book If I substitute the relations for ThetaP1 or ThetaP2, which are 90deg apart the shear stress will equal 0 and thus the ratio will equal zero. However I'm missing how it is that the shear stress would always equal 0 given the substitution

 

[Chestermiller]

My understanding is that the principal directions are those for which the shear component of the traction vector on a plane normal to the principal direction is equal to zero.

 

[AnotherParadox]

My understanding is that the principal directions are those for which the shear component of the traction vector on a plane normal to the principal direction is equal to zero.

If a plane is at some angle and the stresses are transformed such that one is normal and one is parallel with the plane then wouldn't the vector that is parallel be the shear component and thus not be 0 ? I'm definitively missing something here

 

[Chestermiller]

One equation you omitted from your original post is the shear stress in the transformed coordinate system. Please include that too. The principal directions are those for which this shear stress is zero. Let's determine the angles for which that is zero, and then see what we get for the normal stress components.

 

[AnotherParadox]

One equation you omitted from your original post is the shear stress in the transformed coordinate system. Please include that too. The principal directions are those for which this shear stress is zero. Let's determine the angles for which that is zero, and then see what we get for the normal stress components.

Ah yea I used the wrong equation in my original post. However even with the other one I couldn't come up with any results.

Here's the other one τ_x'y' = - ((σ_x-σ_y)/2)*sin(2θ) + τ_xy*cos(2θ)

 

[Chestermiller]

Ah yea I used the wrong equation in my original post. However even with the other one I couldn't come up with any results.

Here's the other one τ_x'y' = - ((σ_x-σ_y)/2)*sin(2θ) + τ_xy*cos(2θ)

OK. So, from this equation, in terms theta, what is the equation for this shear stress being zero?

The equation for $\sigma_{x'}$ can be rewritten as:
$$\sigma_{x'}=\frac{(\sigma_x+\sigma_{y})}{2}+\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\ \left(\cos{\phi}\cos{2\theta}+\sin{\phi}\sin{2\theta}\right)$$where$$\cos{\phi}=\frac{\frac{\sigma_x-\sigma_y}{2}}{\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}}$$and $$\sin{\phi}=\frac{\tau_{xy}}{\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}}$$
OK so far? For $\tau_{x'y'}$, can you figure out how to express this in a similar way?