Question

Problem 1: A component shown in the figure is subjected to cyclic force P. Find the state of the stress at point A in the component. Show stress in an element Pmax-10000 lb 2" 1.75 Pmin-1000 lb Problem 1 If the component is machined from AISI 1040 HR steel. What is the safety factor against fatigue failure for infinite life. How much load can be increased (Pma10000K and Pin-1000K) to have the component survived at least 104 cycles? Find value of K

Answer 1

Pmax- lo000lb P Pmin--1000lb d175" D-2". Given material: AlsI 1040 HR steel Assuming the steel to be quenched and tempered From Design- Databook , for "Notch sensivity () .93 For上='/g" T-75" 0.0714旱ー=1-143 → Stress concentration 뎌 Kt-1 . → kf- Fatigue stress concentration factor 2069 093-kf 2-15-1 : Stresses at A Ormax = Kr. fnax = 2.069x loooo = 8601.902 Rsǐ 4Xd2T452 Gmin= kf.fmin -20698-100--860-BoPsi TT/4x2/4 1-152 State of stress at A Ơmax=S60l.go2 Asi For AISI 1040-HR Steel : Sut = 76000fs1 520MPa For steels : se =05 Sut = 05 x 76000 = 38000 Ksi Calculation of corrected endurance limit For machine surface finish b-0265 a=4.51 Surface finish factora (Sut)b 4.51x (520) Ka 0.859

Size factor Kp 08 for d 175 Assuming 99% reliability Ke 0 814 0 859x 0.85x 0 814x 38000 22584.99 Psi For infinite life Ơamp ":"Se FOS 0amp max-0min8601 902 +860 190 471-046 Psi 2 2 4731 04622584.99 Fos Fos 4-773 For life oF atleast 10t cycles From similar AS 4-3 0-9 6-3 ogto(sut)-logi,se logidse 3 1%;0(09x 76000)-len(22584 99) 47276, 3604 PSİ cycles iF oamps Sp 1/4 42 2069x Pamp 47276-3604 Pamp S 54960 354 lb T/4X 1752 Amplitude oF load can be Increased from 5500 lb to 54960 354 Ib for survilp at least 104cydes Once amplitude tses than 54360-354 lb component car'b Survive upto l04 cycles and win fail earlier-