Question

+ 25 MPa Determine the equivalent state of stress on an element at the same point which represents (a) the principal stress, and (b) the maximum in-plane shear stress and the associated average normal stress Also, for each case, determine the corresponding orientation of the element with respect to the element shown and sketch the results on the element. - 100 MPa

Answer 1

Given that Equivalent state of stress on element at same point which represent sigma_x = -100 Mpa sigma_y = 0 sigma_xy = 25 Mpa = major principal stress = sigma_x + sigma_y/2 (+) squareroot (sigma_x - sigm,a_y/2)^2 + tau ay^2 = -100/2 + squareroot [- 100/2]^2 + 25^2 = -50 + 55.90 = 5.90 Mpa sigma_3 = Minor principal stress = sigma_x + sigma_y/2 (-) squareroot (sigma_x - sigma_y/2)^2 + tau^2_xy = -50 - 55.90 = -105.9 Mpa Orientation of principal stress \r\n Rightarrow Op = 1/2 times tan^- 1 [2 tau xy/sigma_y - sigma_y] = 1/2 times tan^- 1 [2 times 25/- 100] = (-) 13.2 (-) indicates anti clockwise direction Average normal stress (- avg) = sigma_x + sigma_y/2 sigma_avg = -50 Mpa Maximum M - plane shear stress tau_max = squareroot (sigma_x - sigma_y/2)^2 + tan xy^2 = 55.90 Mpa theta = Orientation of shear stress plane with respect to tau_max plane = 1/2 times tan^- 1 [sigma_x - sigma_y/ - 2 tau xy] = 1/2 times tan^- 1 [- 100/ - 2 times 25] = (+) 31.71 (+) indicates clockwise \r\n rightarrow principal stresses representing we draw the metric circle Op = 13.20 makes angle with sigma_3