# Ratio of shear stress to normal stress along principle place

In summary: This equation for ##\tau_{x'y'}## can be rewritten as:$$\tau_{x'y'}=\frac{(\tau_x+\tau_y)+(\sigma_{x'}-\sigma_{y'})}{2}$$where$$\tau_x=\frac{\tau_{x'x}}{\tau_{x'+1}}$$and$$\tau_y=\frac{\tau_{y'y}}{\tau_{y'+1}}$$

## 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

σx' = (σxy)/2 + ((σxy)/2)*cos(2θ) + τxy*sin(2θ)

σy' = (σxy)/2 - ((σxy)/2)*cos(2θ) - τxy*sin(2θ)

τmax,in plane = √[((σxy)/2)2 + τxy2]

σ1,2 = (σxy)/2) ± [((σxy)/2)2 + τxy2]

τmax,absolute = σ1/2

## The Attempt at a Solution

Ratio of shear stress to normal stress?? ->
τmax,in planex' ->
max,in plane = √[((σxy)/2)2 + τxy2]] / [σx' = (σxy)/2 + ((σxy)/2)*cos(2θ) + τxy*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

What is the definition of a principal direction of stress?

Chestermiller said:
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

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.

Chestermiller said:
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

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.

Chestermiller said:
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 τx'y' = - ((σxy)/2)*sin(2θ) + τxy*cos(2θ)

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 τx'y' = - ((σxy)/2)*sin(2θ) + τxy*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?

## 1. What is the ratio of shear stress to normal stress along the principle plane?

The ratio of shear stress to normal stress along the principle plane is known as the shear stress ratio. It is defined as the ratio of the maximum shear stress to the maximum normal stress along the principle plane.

## 2. How is the shear stress ratio calculated?

The shear stress ratio can be calculated by dividing the maximum shear stress by the maximum normal stress along the principle plane. This can be represented by the equation: Shear Stress Ratio = Maximum Shear Stress / Maximum Normal Stress.

## 3. What is the significance of the shear stress ratio in material behavior?

The shear stress ratio is an important factor in determining the strength and ductility of a material. It can indicate how a material will behave under different types of loading, such as tension, compression, or shear.

## 4. How does the shear stress ratio affect the failure of a material?

The shear stress ratio can have a significant impact on the failure of a material. A higher shear stress ratio can lead to a higher likelihood of failure due to shear, while a lower shear stress ratio can result in a material that is more ductile and less likely to fail.

## 5. Can the shear stress ratio change under different loading conditions?

Yes, the shear stress ratio can change depending on the type of loading that a material is subjected to. For example, a material may have a different shear stress ratio under tension than it would under compression. It is important for engineers and scientists to consider the different loading conditions when analyzing the shear stress ratio of a material.

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