Question

A solid round bar, 25 mm in diameter, has a groove 2.5-mm deep with a 2.5-mm radius machined into it. The bar is made of AISI 1020 normalized steel and is subjected to a purely reversing torque of 130 N. m. (a) Estimate the number of cycles to failure. (b) If the bar is also placed in an environment with a temperature of 500 degrees C, estimate the number of cycles to failure. 2.5 mm 20 mm 25 mm

Answer 1

In order to solve this problem please use the proper design data hand book values .
25мм 20 MM 20mm The gun rokum 2 Aist 102.0 - Normalized steel. The values of various strengths of this material can be taken from the Denijn hand data book. from the given figure % = 0.5 mm = 0.125 aneda na casa alias And the stress concentration factor (kts) = 14 We know the relation b/w fatigue factur notch factor and stress concentration fachur, => 145 = 1 +9 (541) We can obtain q=0.94 5) ks = 1+0.94 (104-1) = 1,376 Design Hand data book for Alsi 1020 Normalized steel (Normalized at 870°c generally) the values are given by Ultimate Tennile stregth (Sut) = 440 MPa Tensile strength, yield (Sy) = 345 MPa we know, Ultivo ate shear strey th (sou) = 0.67 x Sut =) Ssu = 0.67(440) = 295 MPa In the given problem the steel is subjected to a purely reverning torque of 1300-m

16T max TT 3 16130 : 16x sex to N-mm x 11376 TT X (20 mm) -0.265 -0.265 Imene +(1.376x8 2. 46) MPa i 113.878 MPa We know, endurance streyth (s) = 0.504 (440) 222 MPa How we have to calculate Marin factors, - KQ = 4.51(sut) on ne 4.51(440) 20.899 Kb = 620.) 10 10.7 0.902 We can take, k = 0.59 kg =1, ke=1 : S, = 0.899 x 0.902 X 0.59 X 222 = 106.2 MPa calculation of a, b, N :- a- (0.9 x 295] . 664 lo 6.2 bi -Yo log 0.9 X295) = -0.132648 120.13265 106.2 -X Sut 3265 N = 113.878 664 N = 5,92,306 cycles Approximate estimated number of cycles to failure is 5,92, 306 cycled. Anowed)

b) Now, the bar in placed in an environment with a temperature of 500°c, from Design data table, kd= 0.843 = 0.899 X 0.902 X0:59 X0.843 X 222 - 89.5 MPa calculation of a, b, N :- az (0.9x295) = 788 89.5 b = -1/3 log o.qxd -0.15741 0.9 X 295 89.5 N- ( 93.878 ) %-0.15241 788 399 NE 2, 17,235 cycles 2, 17,200 cycles No. of cycles to failure aa Answer