3 6 9 10 k at a load of 18 kn life l 1 is given by 9

Info icon This preview shows pages 2–4. Sign up to view the full content.

3 6 9 20.3 10 8.365 10 K At a load of 18 kN, life L 1 is given by: 9 6 1 3 1 8.365 10 1.434 10 rev 18 a K L F For a load of 30 kN, life L 2 is: 9 6 2 3 8.365 10 0.310 10 rev 30 L Therefore; 1 2 1 2 1 l l L L Substituting, 2 6 6 200 000 1 1.434 10 0.310 10 l Problem 3 Solution Total life in revolutions l = total turns f 1 = fraction of turns at F 1 f 2 = fraction of turns at F 2 From the solution of previous problem L 1 = 1.434(10 6 ) rev and L 2 = 0.310(10 6 ) rev. Therefore: 1 2 1 2 1 2 1 2 1 l l f l f l L L L L from which 1 1 2 2 1 / / l f L f L 6 6 1 0.40 / 1.434 10 0.60 / 0.310 10 451 585 rev . l Ans
Image of page 2

Info icon This preview has intentionally blurred sections. Sign up to view the full version.