Question

4 Stress Tensor 1. Given: T=14 1 01 MPa. 0 0 1 (a) find the stress vector on the plane whose normal is in the direction of +ý (b) Find the normal stress on the same plane. (c) Find the magnitude of the tangential stress on the same plane. (d) Find the three principal directions of stress. Hint: the determinant of a 3 x3 11 a12 13 det a a2 a23a11 (a22033-a3223) a12 a1a33-a31a23)+a13(a2132-a31022) 031 a32 33 Given the following stress distribution: 6 find Try(x, y) so that the stress distribution is in equilibrium with zero body force and so that the stress vector on the plane r- 1 is given by デー(1+y)2 + (5-y)

Answer 1

Given stress tensor T = [1 4 0 4 1 0 0 0 1] MP_a stress vector on the plane whose normal is x + y, we have to find. Representing the plane in matrix From, we get u = (1 1 0) Norma using it u = 1/squareroot 2(1 1 0) stress vector 5n = Tu = 1/squareroot 291 4 0 4 1 0 0 0 1)(1 1 0) = 1/squareroot 2(5 5 0) = 5/squareroot 2(1 1 0) Normal stress Normal stress, 5n u = 5/2 times 2 = 5 MP_a = T_n Tangential stress/Tangential stress vector on u is given by 5n - T_n u = 5/squareroot 2(1 1 0) - 5/squareroot 2(1 1 0) = 0 principal directions of stress principal stresses and their direction are simple the Eigen values and Eigen vectors OF the stress matrix T = (1 4 0 4 1 0 0 0 1) Eigen values are 5, -3, 1 i.e Lambda-1 = 5, Lambda_2 = -3,