At a point in a strained material the principal stresses are 60MPa and 40MPa. Find the position of the plane across which the resultant stress is most inclined to the normal and determine the value
To find the position of the plane across which the resultant stress is most inclined to the normal, we need to calculate the principal stress angles. The principal stress angles can be determined using the following formula:
θ = 0.5 * atan(2τxy / (σx - σy))
Where:
θ = Principal stress angle
τxy = Shear stress
σx, σy = Principal stresses
But since only the principal stresses are given, we need to assume that there is no shear stress present. Therefore, τxy = 0.
Now, let's calculate the principal stress angles:
θ1 = 0.5 * atan(2 * 0 / (60 - 40))
= 0 (since 2 * 0 / (60 - 40) = 0)
θ2 = 0.5 * atan(2 * 0 / (40 - 60))
= 0 (since 2 * 0 / (40 - 60) = 0)
Since both angles are 0, it means that the principal stresses are equal and have the same orientation. In this case, any plane can be considered. Let's assume the horizontal plane (θ = 0) for simplicity.
Therefore, the position of the plane across which the resultant stress is most inclined to the normal is any plane (such as a horizontal plane).
As for determining the value of the resultant stress, we need to calculate the magnitude of the resultant stress. The magnitude of the resultant stress can be determined using the following formula:
σr = sqrt((σx - σy)^2 + 4τxy^2)
But since τxy = 0, the formula simplifies to:
σr = |σx - σy|
σr = |60 - 40|
= 20 MPa
So, the value of the resultant stress is 20 MPa.