Question

ln (Figure 1), -47.3 MPa. Part A Determine the principal stresses. Specify the orientation of the element Express your answers, separated by commas, to three significant figures. MPa, MPa, Submit Request Answer Part B Determine the maximum in-plane shear stress and average normal stress. Specify the orientation of the element Express your answers, separated by commas, to three significant figures ΑΣΦ vec Figure 1 of 1 MPa, MPa, Submit Request Answer Provide Feedback 30 MPa

Answer 1

sigma x = 0 sigma y = 47.3 mpa T_ny = 30 mpa tan 20 p = 2 Txy/sigma x - sigma y T, sigma 2 = sigma x + sigma y/2 plusminus squareroot (sigma x - sigma y/2)^2 + T_ny^2 = 47.3/2 plusminus squareroot (-47.3/2)^2 + (80)^2 = 23.65 plusminus 38.20 sigma 1 = 691.85 mpa sigma 2 = -14.55 mpa tan 20 p = 2 T xy/sigma x sigma y = 2 times 30/0 - 47.3 = -1.268 2 theta p = -51.75 degree theta p = -25.87 degree T max = sigma 1 - sigma 2/2 = > 61.85 + 14.55/2 = > 38.2 T max = 38.2 mpa sigma avg = sigma 1 + sigma 2/2 = > 61.85 - 14.55/2 T avg = 23.65 mpa Absolute mass shear stress Tabs max = sigma 4/2 = 47.3/2 = 23.65 mpa