At a point in a plate subjected to plane stress loading along its surface, the cartesian stress components acting on the y-face of an element oriented along the (x,y) frame are σy=8MPa and τxy=16MPa,
The maximum allowable tensile (normal) stress, to prevent fracture of the material, is σo=40MPa.
What is the maximum permissible value (in MPa) for the normal stress on the x-face, σmaxx, to ensure that the material will not fracture?
σmaxx=
To find the maximum permissible value for the normal stress on the x-face, σmaxx, we need to consider the failure criteria for the material.
In this case, we are given the maximum allowable tensile (normal) stress, σo, which is 40 MPa. This means that the material will fracture if the normal stress exceeds this value.
To determine the normal stress on the x-face, we can use the stress transformation equations. The stress transformation equations relate the normal stress components (σx, σy) and the shear stress component (τxy) to the normal and shear stress components in another coordinate system.
The stress transformation equation for the normal stress on the x-face is:
σx = (σy + σy) / 2 + √[(σy - σy) / 2]^2 + τxy^2
where σx is the normal stress on the x-face, σy is the normal stress on the y-face (given as 8 MPa), and τxy is the shear stress on the xy-plane (given as 16 MPa).
Plugging in the values, we have:
σx = (8 + 8) / 2 + √[(8 - 8) / 2]^2 + (16)^2
= 16 + √(0)^2 + 256
= 16 + 0 + 256
= 272 MPa
Therefore, the maximum permissible value for the normal stress on the x-face, σmaxx, is 272 MPa to ensure that the material will not fracture.
To find the maximum permissible value for the normal stress on the x-face, σmaxx, we can use the formula for the maximum normal stress criterion. According to this criterion, the material will not fracture if the maximum principal stress is less than or equal to the allowable tensile stress.
The maximum principal stress can be calculated using the following formula:
σmax = (σx + σy)/2 + √((σx - σy)/2)^2 + τxy^2)
Given that σy = 8 MPa and τxy = 16 MPa, we can substitute these values into the formula:
σmax = (σx + 8)/2 + √((σx - 8)/2)^2 + 16^2)
To ensure that the material will not fracture, σmax must be less than or equal to the allowable tensile stress, σo = 40 MPa. So, we can set up the following inequality:
(σx + 8)/2 + √((σx - 8)/2)^2 + 16^2) ≤ 40
Simplifying this inequality will allow us to find the maximum permissible value for σx.