Question

PROBLEM 7.82 - 100 MPa Solve Prob. 7.81, using the maximum-shearing-stress criterion. PROBLEM 7.81 The state of plane stress shown occurs in a machine component made of a steel with oy = 325 MPa. Using the maximum-distortion-energy criterion, determine whether yield will occur when (a) o = 200 MPa, (b) o = 240 MPa, (C) 00 = 280 MPa. If yield does not occur, determine the corresponding factor of safety.

Answer 1

Sigma _yp = 1/squareroot 2 squareroot (sigma x - sigma _y)^2 + (sigma_y - sigma _z)^2 + (sigma_x - sigma_z)^2 + 6 (Z_xy^2 + Z^2_yz + Z^2_zx) where sigma _yp is the yield point of material a) sigma_x = sigma_y = - 200 (compression) Z_xy = Z_yx = 100 sigma _yp + 1/ squareroot 2 squareroot (- 200 + 200)^2 + (- 200)^2 + (- 200) + 6 (100)^2 = 264.57 mPa Since sigma _y = 325 m P a: yield doesn't occur therefore Factor of Safety = 325/264.57 = 1.23 sigma_x = sigma_y = - 240 (compressive) sigma__yp = 1/squareroot 2 squareroot (- 240)^2 + (- 240)^2 + 6 (100^2) = 295.97 m P a < 325 m P a: yield doesn't occur therefore F. O. S = 325/295.97 = 1.098 sigma_x = sigma_y = - 280 (comp.) sigma_yp = 1/ squareroot 2 squareroot (- 200)^2 + (- 200)^2 + 6 (100)^2 = 329.24 m P A 329.24 > 375 mPa: yield does occur in this case