Question

answer E,F,G and H please

2. A bar of steel undergoes cyclic loading such that max = 60 kpsi and Omin = -20 kpsi. For the material, Sut = 80 kpsi, 5y = 65 kpsi,ka = 1.107. ky = 0.858, ke = ka = k, = k, = 1, and f = 0.9. (a) Calculate the alternating stress. (2 points) (b) Calculate the meani median) stress. (2 points) (c) Calculate the fully corrected endurance limit. (6 points) (d) Calculate the fatigue factor of safety using the Soderberg criterion. (6 points) (e) Does the fatigue factor of safety indicate a finite or infinite life? What do you base this on? (2 points) O Is it necessary to calculate the factor of safety for yielding? Why or why not? (2 points) (8) Calculate an equivalent completely reversed stress using the Soderberg criterion that gives the same life as the given stresses. (6 points) (h) Assuming a finite life, estimate the number of cycles to a fatigue failure using the Soderberg criterion. (6 points)

Answer 1

+421 2 - 0.9 - 20 + 40 - 40 = 20 k Psi 65 77 Чо Gaz 60- (-20) 130 = 40 KPSi. Sez 40X130 - 67.53 KPsi 77.

so se = 67.53: Set - Sul iob No. of cyclo (ye y) = y 2 - Y, (2-2). (se - 0:49 Sut ) - [Set - 0.9954 - n, 6-3 103 - X (2-3) (67.53-0.99 880) x 3 (80 – 0.99 X80) +3 = a vi +3 = 3.89 = x -35:0| - 39.2 - 10 = no. of cycles - - 103989 - 7762-47 cycles.

Safety actor = Strength Man stress. Factor of safety does not only indicate safety before fracture in componen Even Yielding to can lead bea of preraing sa of component component lead to failure as thtendene intended function of the component Is affected. Therefore Factor for Factor yielding of safety is important
E) factor of safety indicates finite life. But i didn't find exact reason.

Rest all three parts are answered order g, h, f.