Calculating Principal Stress, σx and τxy

[mm391]

Homework Statement

The principal stresses at a point in a material are 50 MN/m^2 and 30 MN/m^2 , both in tension. If the angle between the Ox axis and the direction in which the greater stress acts is 40º measured anticlockwise from the direction of the greater principal stress, find σx and τxy.
τx’y’ = 0.

Homework Equations

σn =1/2(σx +σy) + or - 1/2(σx −σy)cos2φ+τxysin2φ
τn =+ or - 1/2(σx −σy)sin2φ+τxycos2φ

tan2φ=(2τxy)/(σx −σy)

σ1 or σ2 = 1/2(σx +σy) + or - ((1/2(σx -σy))2+τxy^2)^(1/2)

The Attempt at a Solution

I have rearranged this equation to give

x=σ1 or σ2. I am not sure if this is right and not sure how to start this or what equations to?use.

 

[KataKoniK]

Thank you for posting your question. I would like to help you find the solution to your problem. First, let's review the given information. We have two principal stresses, 50 MN/m^2 and 30 MN/m^2, both in tension. The angle between the Ox axis and the direction of the greater stress is 40º measured anticlockwise. We are asked to find σx and τxy.

To find the values of σx and τxy, we can use the equations you have provided. Let's start with the equation for tan2φ. We can rearrange this equation to solve for τxy:

τxy = (σx - σy)tan2φ/2

Next, we can use the equation for σ1 or σ2 to find σx and σy:

σ1 or σ2 = 1/2(σx + σy) ± ((1/2(σx - σy))^2 + τxy^2)^(1/2)

Since we know that σ1 = 50 MN/m^2 and σ2 = 30 MN/m^2, we can plug these values into the equation and solve for σx and σy:

50 MN/m^2 = 1/2(σx + σy) ± ((1/2(σx - σy))^2 + τxy^2)^(1/2)
30 MN/m^2 = 1/2(σx + σy) ± ((1/2(σx - σy))^2 + τxy^2)^(1/2)

We now have two equations with two unknowns (σx and σy) and can solve for them. Once we have these values, we can plug them into the equation for τxy to find its value.

I hope this helps you solve the problem. If you have any further questions or need clarification, please don't hesitate to ask. Best of luck!