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,

Question

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=

Answer 1

3....MPa

use this equation
(sigma)max=(sigma)avg+R

Answer 2

To find the maximum permissible value for the normal stress on the x-face, σmaxx, we need to determine the principal stresses acting on the element and ensure that none of them exceed the maximum allowable tensile stress, σo.

The principal stresses can be obtained by solving the following equation:

σ1,2 = (σx + σy)/2 ± √[(σx - σy)/2]^2 + τxy^2

Given σy = 8MPa and τxy = 16MPa, we can substitute these values into the equation:

σ1,2 = (σx + 8)/2 ± √[(σx - 8)/2]^2 + 16^2

To ensure that the material will not fracture, we need to ensure that both principal stresses, σ1 and σ2, are less than or equal to σo.

Let's start by finding σ1:

σ1 = (σx + 8)/2 + √[(σx - 8)/2]^2 + 16^2

Now, substitute σo = 40MPa and solve the equation:

40 = (σx + 8)/2 + √[(σx - 8)/2]^2 + 16^2

Simplify and solve for σx.

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,

3....MPa

32 MPa

To find the maximum permissible value for the normal stress on the x-face, σmaxx, we need to determine the principal stresses acting on the element and ensure that none of them exceed the maximum allowable tensile stress, σo.