Question

Required information The cold-drawn AISI 1040 steel bar shown in the figure is subjected to a completely reversed axial load fluctuating between 16 kN in compression to 16 kN in tension Estimate the fatigue factor of safety based on achieving infinite life and the yielding factor of safety. If infinite life is not predicted, estimate the number of cycles to failure. 6-D 25 man 10m What is the number of cycles to failure for this part? The value of the number of cycles is

Answer 1

AISI 1040 steel # paoperties OF - 590mpa Tensile strengih (Sut) yield storenglas (sy) = 490 mpa Ofsteel burs 10 (25-6) sectional area = 190mm2 ER cross Forsin -16kN # given 216kcal Frax 16x103 max 84.2105 MPa 190 the factor of safety by considesing sy omax storess freed ny 490 = =5081875 84.2105 # 20.5 X 590 endurance limit =0.5X Sut 295 mpa Se=kakokcid ke sel Id=1 kb=1 Kc=0185 Ka= a (sur 35 ke=l a=4151 b=0:265

this for this plate in tension orsimple 0.265 S00831 ka fr.51) (590) Then Ser 0831 *8185 *1x1x295 = 208.4 mpa # for stoess ceskestration Falter CkCf) ☆ from fecom 6 25 20:24 w hole. central with daunsverse compression d this Kf=2045 Ô 124 W chorst !Kloten sensitive chant for of Form steel amel aluminium alloys Sudjelsey reversed bending or axial or axial Load. rad/ = 3mm = 0.12 m sute 85 lepsi ( 590mpa 2 Thes q=0.8

# Slness ancentration Kf = q (k-1). Faltor =lf 0.8 (2.454) 01.16 forest finin It kg Gas altemating stress fonat = -Fomin 16x103 cas 2-16 x 2x 30 tú 1811895 MPa 2 X 190 fonas + trosios If oma meus stress 2 A taxi6x10 = 246 2440 de G68181 = Elch 20

L Ozen Reversed storess Ga Oriev - 1- om sut brev 181.845 Miſa (sinu omso mpa) # SE = an ano SE + Ferrigeee sporength fousins - 103 cylk Sur= 590 10² cycle se=sef = 0.5sut ft fatigue storength factor F-0183 (2 sur 1² Thes los 7x5903 = Se 4083 1265 b= -1 log thur I log 0.87X590 2o8.3 se b= -0.1306 at then Grew N= a 0.1306 1818845 N= ( 1265 N = 2819335.22