Given sigmaxx 20 MPa sigmayy 10 MPa and sigmaxy 20 MPa fi

Given sigma_xx = 20 MPa, sigma_yy =-10 MPa, and sigma_xy = - 20 MPa, find the principal stresses and principal strains with LEHI behavior and E= 16 GPa and v = 0.325.

Solution

Principal stress is when sigma_xy is equal to zero and can be found using the following equations:

alpha_p = 0.5 tan ^-1 (2 * _xy / (_xx - _yy)) = -.46

Principle stresses are following:

\'_xx = ((_xx + _yy) / 2) + ((_xx - _yy) / 2)*cos(2 * alpha_p) + _xy * sin(2 * alpha_p) = 20MPa

\'_yy = ((_xx + _yy) / 2) + ((_yy - _xx) / 2)*cos(2 * alpha_p) - _xy * sin(2 * alpha_p) = -2MPa

\'_xy = ((_yy - _xx) / 2)*sin(2 * alpha_p) + _xy * cos(2 * alpha_p) = 0


Same deal for strain. Principal strain is when _xx and _yy are maximum.

_xx = 1/E(_xx - v*_yy) = 1.64mPa
_yy = 1/E(_yy - v*_xx) =1.03mPa
_xy = 1/2G * _xy where G = shear modulus = E/2(1+v) = -1.6mPa

Find alpha_p for principle strain

alpha_p = 0.5 tan^-1 (_xy / ((_xx - _yy)/2)) = .55

Then substitute values into following equations:

\'_xx = _xx*cos^2(alpha_p) + 2*_xy*sin(alpha_p)*cos(alpha_p) + _yy*sin^2(alpha_p) = -0.05mPa

\'_yy = _xx*sin^2(alpha_p) - 2*_xy*sin(alpha_p)*cos(alpha_p) + _yy*cos^2(alpha_p) = 2.61mPa

\'_xy = 2*sin(alpha_p)*cos(alpha_p)*((_yy - _xx) / 2) + (cos^2(alpha_p) - sin^2(alpha_p)*_xy = -0.974mPa