At a point in a plate subjected to plane stress loading along its surface, the cartesian stress components acting on the x-face of an element oriented along the (x,y) frame are σx=40MPa and τxy=70MPa,
The maximum allowable shear stress, to prevent yielding of the material, is τo=74MPa.
What is the permissible range (minimum and maximum values in MPa) for the normal stress on the y-face, σy, to ensure that the material will not yield?
To find the permissible range for the normal stress on the y-face, σy, we need to consider the yield condition of the material. The yield condition states that the maximum shear stress should not exceed the maximum allowable shear stress, τo.
In this case, we are given σx = 40 MPa and τxy = 70 MPa. The maximum shear stress can be calculated using the formula:
τmax = (σx - σy) / 2 + sqrt((σx - σy)^2 / 4 + τxy^2)
We know that the permissible range for the normal stress on the y-face, σy, should ensure that τmax is less than or equal to τo.
So, we can rearrange the formula for τmax to find the range of σy:
σy = σx - 2 * (τmax - τxy)
To find the minimum value of σy, we substitute τmax = τo and solve:
σy(min) = σx - 2 * (τo - τxy)
Similarly, to find the maximum value of σy, we use the formula with τmax = -τo:
σy(max) = σx - 2 * (-τo - τxy)
Substituting the given values:
σy(min) = 40 MPa - 2 * (74 MPa - 70 MPa)
= 40 MPa - 2 * 4 MPa
= 32 MPa
σy(max) = 40 MPa - 2 * (-74 MPa - 70 MPa)
= 40 MPa - 2 * -144 MPa
= 376 MPa
Therefore, the permissible range for the normal stress on the y-face, σy, is 32 MPa to 376 MPa (minimum to maximum values).